Measure Theory Fremlin Vol4

MEASURE THEORY Volume 4

D.H.Fremlin

By the same author: Topological Riesz Spaces and Measure Theory, Cambridge University Press, 1974. Consequences of Martin’s Axiom, Cambridge University Press, 1982. Companions to the present volume: Measure Theory, vol. 1, Torres Fremlin, 2000. Measure Theory, vol. 2, Torres Fremlin, 2001. Measure Theory, vol. 3, Torres Fremlin, 2002.

First printing November 2003

MEASURE THEORY Volume 4 Topological Measure Spaces

D.H.Fremlin Research Professor in Mathematics, University of Essex

Dedicated by the Author to the Publisher

This book may be ordered from the publisher at the address below. For price and means of payment see the author’s Web page http://www.essex.ac.uk/maths/staff/fremlin/mtsales.htm, or enquire from [email protected].

First published in 2003 by Torres Fremlin, 25 Ireton Road, Colchester CO3 3AT, England c D.H.Fremlin 2003 ° The right of D.H.Fremlin to be identified as author of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988. This work is issued under the terms of the Design Science License as published in http://dsl.org/copyleft/dsl.txt. For the source files see http://www.essex. ac.uk/maths/staff/fremlin/mt4.2003/index.htm. Library of Congress classification QA312.F72 AMS 2000 classification 28A99 ISBN 0-9538129-4-4 Typeset by AMS-TEX Printed in England by Biddles Short Run Books, King’s Lynn

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Contents General Introduction

10

Introduction to Volume 4

11

Chapter 41: Topologies and measures I Introduction 411 Definitions

13 13

Topological, inner regular, τ -additive, outer regular, locally finite, effectively locally finite, quasi-Radon, Radon, completion regular, Baire, Borel and strictly positive measures; measurable and almost continuous functions; self-supporting sets and supports of measures; Stone spaces; Dieudonn´ e’s measure.

412 Inner regularity

19

Exhaustion; Baire measures; Borel measures on metrizable spaces; completions and c.l.d. versions; complete locally determined spaces; inverse-measure-preserving functions; subspaces; indefinite-integral measures; products; outer regularity.

413 Inner measure constructions

31

Inner measures; constructing a measure from an inner measure; the inner measure defined by a measure; complete locally determined spaces; extension of functionals to measures; countably compact classes; constructing measures dominating given functionals.

414 τ -additivity

50

Semi-continuous functions; supports; strict localizability; subspace measures; regular topologies; density topologies; lifting topologies.

415 Quasi-Radon measure spaces

58

Strict localizability; subspaces; regular topologies; hereditarily Lindel¨ of spaces; products of separable metrizable spaces; comparison and specification of quasi-Radon measures; construction of quasi-Radon measures extending given functionals; indefinite-integral measures; Lp spaces; Stone spaces.

416 Radon measure spaces

73

Radon and quasi-Radon measures; specification of Radon measures; c.l.d. versions of Borel measures; locally compact topologies; constructions of Radon measures extending or dominating given functionals; additive functionals on Boolean algebras and Radon measures on Stone spaces; subspaces; products; Stone spaces of measure algebras; compact and perfect measures; representation of homomorphisms of measure algebras; the split interval.

417 τ -additive product measures

88

The product of two effectively locally finite τ -additive measures; the product of many τ -additive probability measures; Fubini’s theorem; generalized associative law; measures on subproducts as image measures; products of strictly positive measures; quasi-Radon and Radon product measures; when ‘ordinary’ product measures are τ -additive; continuous functions and Baire σ-algebras in product spaces.

418 Measurable functions and almost continuous functions

110

Measurable functions; into (separable) metrizable spaces; and image measures; almost continuous functions; continuity, measurability, image measures; expressing Radon measures as images of Radon measures; Prokhorov’s theorem on projective limits of Radon measures; representing measurable functions into L0 spaces.

419 Examples

126

A nearly quasi-Radon measure; a Radon measure space in which the Borel sets are inadequate; a nearly Radon measure; the Stone space of the Lebesgue measure algebra; measures with domain Pω1 ; notes on Lebesgue measure.

Chapter 42: Descriptive set theory Introduction 421 Souslin’s operation

138 138

Souslin’s operation; is idempotent; as a projection operator; Souslin-F sets; *constituents.

422 K-analytic spaces

148

Usco-compact relations; K-analytic sets; and Souslin-F sets; *First Separation Theorem.

423 Analytic spaces

155

Analytic spaces; are K-analytic spaces with countable networks; Souslin-F sets; Borel measurable functions; injective images of Polish spaces; non-Borel analytic sets; von Neumann-Jankow selection theorem; *constituents of coanalytic sets.

424 Standard Borel spaces Elementary properties; isomorphism types; subspaces; Borel measurable actions of Polish groups.

165

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Chapter 43: Topologies and measures II Introduction 431 Souslin’s operation

172 172

The domain of a complete locally determined measure is closed under Souslin’s operation; the kernel of a Souslin scheme is approximable from within.

432 K-analytic spaces

176

Topological measures on K-analytic spaces; extensions to Radon measures; expressing Radon measures as images of Radon measures.

433 Analytic spaces

180

Measures on spaces with countable networks; inner regularity of Borel measures; expressing Radon measures as images of Radon measures; measurable and almost continuous functions; the von NeumannJankow selection theorem; products; extension of measures on σ-subalgebras; standard Borel spaces.

434 Borel measures

184

Classification of Borel measures; Radon spaces; universally measurable sets and functions; Borel-measurecompact, Borel-measure-complete and pre-Radon spaces; countable compactness and countable tightness; quasi-dyadic spaces and completion regular measures; first-countable spaces and Borel product measures.

435 Baire measures

203

Classification of Baire measures; extension of Baire measures to Borel measures (Maˇrik’s theorem); measure-compact spaces; sequential spaces and Baire product measures.

436 Representation of linear functionals

209

Smooth and sequentially smooth linear functionals; measures and sequentially smooth functionals; Baire measures; products of Baire measures; quasi-Radon measures and smooth functionals; locally compact spaces and Radon measures.

437 Spaces of measures

219

Smooth and sequentially smooth duals; signed measures; embedding spaces of measurable functions in the bidual of Cb (X); vague and narrow topologies; product measures; extreme points; uniform tightness; Prokhorov spaces.

438 Measure-free cardinals

240

Measure-free cardinals; point-finite families of sets with measurable unions; measurable functions into metrizable spaces; Radon and measure-compact metric spaces; metacompact spaces; hereditarily weakly θ-refinable spaces; when c is measure-free.

439 Examples

254

Measures with no extensions to Borel measures; universally negligible sets; Hausdorff measures are rarely semi-finite; a smooth linear functional not expressible as an integral; a first-countable non-Radon space; Baire measures not extending to Borel measures; N c is not Borel-measure-compact; the Sorgenfrey line; Q is not a Prokhorov space.

Chapter 44: Topological groups Introduction 441 Invariant measures on locally compact spaces

271 271

Measures invariant under homeomorphisms; Haar measures; measures invariant under isometries.

442 Uniqueness of Haar measure

280

Two (left) Haar measures are multiples of each other; left and right Haar R measures; R Haar measurable and Haar negligible sets; the modular function of a group; formulae for f (x−1 )dx, f (xy)dx.

443 Further properties of Haar measure

287

The Haar measure algebra of a group carrying Haar measures; actions of the group on the Haar measure algebra; locally compact groups; actions of the group on L0 and Lp ; the bilateral uniformity; Borel sets are adequate; completing the group; expressing an arbitrary Haar measure in terms of a Haar measure on a locally compact group; completion regularity of Haar measure; invariant measures on the set of left cosets of a closed subgroup of a locally compact group; modular functions of subgroups and quotient groups; transitive actions of compact groups on compact spaces.

444 Convolutions Convolutions of quasi-Radon measures; the Banach algebra of signed τ -additive measures; continuous actions and corresponding actions on L0 (ν) for an arbitrary quasi-Radon measure ν; convolutions of measures and functions; indefinite-integral measures over a Haar measure µ; convolutions of functions; Lp (µ); approximate identities.

311

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445 The duality theorem

334

Dual groups; Fourier-Stieltjes transforms; Fourier transforms; identifying the dual group with the maximal ideal space of L1 ; the topology of the dual group; positive definite functions; Bochner’s theorem; the Inversion Theorem; the Plancherel Theorem; the Duality Theorem.

446 The structure of locally compact groups

357

Finite-dimensional representations separate the points of a compact group; groups with no small subgroups have B-sequences; chains of subgroups.

447 Translation-invariant liftings

372

Translation-invariant liftings and lower densities; Vitali’s theorem and a density theorem for groups with B-sequences; Haar measures have translation-invariant liftings.

448 Invariant measures on Polish spaces

383

Countably full local semigroups of Aut A; σ-equidecomposability; countably non-paradoxical groups; G-invariant additive functions from A to L∞ (C); measures invariant under Polish group actions (the Nadkarni-Becker-Kechris theorem).

449 Amenable groups

393

Amenable groups; permanence properties; locally compact amenable groups; Tarski’s theorem; discrete amenable groups.

Chapter 45: Perfect measures, disintegrations and processes Introduction 451 Perfect, compact and countably compact measures

413 414

Basic properties of the three classes; subspaces, completions, c.l.d. versions, products; measurable functions from compact measure spaces to metrizable spaces; *weakly α-favourable spaces.

452 Integration and disintegration of measures

427

Integrating families of probability measures; τ -additive and Radon measures; disintegrations and regular conditional probabilities; disintegrating countably compact measures; disintegrating Radon measures; *images of countably compact measures.

453 Strong liftings

441

Strong and almost strong liftings; existence; on product spaces; disintegrations of Radon measures over spaces with almost strong liftings; Stone spaces; Losert’s example.

454 Measures on product spaces

454

Perfect, compact and countably compact measures on product spaces; extension of finitely additive functions with perfect countably additive marginals; Kolmogorov’s extension theorem; measures defined from conditional distributions; distributions of random processes; measures on C(T ) for Polish T .

455 Markov process and Brownian motion

464

Definition of Markov process from conditional distributions; existence of a measure representing Brownian motion; continuous sample paths.

456 Gaussian distributions

472

Gaussian distributions; supports; universal Gaussian distributions; cluster sets of n-dimensional processes; τ -additivity; Gaussian processes.

457 Simultaneous extension of measures

488

Extending families of finitely additive functionals; Strassen’s theorem; extending families of measures; examples.

458 Relative independence and relative products

497

Relatively independent families of σ-algebras and random variables; relative distributions; relatively independent families of closed subalgebras of a probability algebra; relative free products of probability algebras; relative products of probability spaces; existence of relative products.

459 Symmetric measures and exchangeable random variables

510

Exchangeable families of inverse-measure-preserving functions; de Finetti’s theorem; countably compact symmetric measures on product spaces disintegrate into product measures; symmetric quasi-Radon measures.

Chapter 46: Pointwise compact sets of measurable functions Introduction 461 Barycenters and Choquet’s theorem

522 522

Barycenters; elementary properties; sufficient conditions for existence; closed convex hulls of compact sets; Kreˇın’s theorem; measures on sets of extreme points.

462 Pointwise compact sets of continuous functions Angelic spaces; the topology of pointwise convergence on C(X); weak convergence and weakly compact sets in C0 (X); Radon measures on C(X); separately continuous functions; convex hulls.

532

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463 Tp and Tm

538

Pointwise convergence and convergence in measure on spaces of measurable functions; compact and sequentially compact sets; perfect measures and Fremlin’s Alternative; separately continuous functions.

464 Talagrand’s measure

550

The usual measure on PI; the intersection of a sequence of non-measurable filters; Talagrand’s measure; the L-space of additive functionals on PI; measurable and purely non-measurable functionals.

465 Stable sets

563

Stable sets of functions; elementary properties; pointwise compactness; pointwise convergence and convergence in measure; a law of large numbers; stable sets and uniform convergence in the strong law of large numbers; stable sets in L0 and L1 ; *R-stable sets.

466 Measures on linear topological spaces

590

Quasi-Radon measures for weak and strong topologies; Kadec norms; constructing weak-Borel measures; characteristic functions of measures on locally convex spaces; universally measurable linear operators.

*467 Locally uniformly rotund norms

598

Locally uniformly rotund norms; separable normed spaces; long sequences of projections; K-countably determined spaces; weakly compactly generated spaces; Banach lattices with order-continuous norms; Eberlein compacta.

Chapter 47: Geometric measure theory Introduction 471 Hausdorff measures

610 610

Metric outer measures; Increasing Sets Lemma; analytic spaces; inner regularity; Vitali’s theorem and a density theorem; Howroyd’s theorem.

472 Besicovitch’s Density Theorem

626

Besicovitch’s Covering Lemma; Besicovitch’s Density Theorem; *a maximal theorem.

473 Poincaré’s inequality

633

Differentiable and Lipschitz functions; smoothing by convolution; the Gagliardo-Nirenberg-Sobolev inequality; Poincar´ e’s inequality for balls.

474 The distributional perimeter

647

The divergence of a vector field; sets with locally finite perimeter, perimeter measures and outwardnormal functions; the reduced boundary; invariance under isometries; isoperimetric inequalities; Federer exterior normals; the Compactness Theorem.

475 The essential boundary

670

Essential interior, closure and boundary; the reduced boundary; perimeter measures; characterizing sets with locally finite perimeter; the Divergence Theorem; calculating perimeters from cross-sectional counts; Cauchy’s Perimeter Theorem; the Isoperimetric Theorem for convex sets.

476 Concentration of measure

690

Hausdorff metrics; Vietoris topologies; concentration by partial reflection; concentration of measure in R r ; the Isoperimetric Theorem; concentration of measure on spheres.

Chapter 48: Gauge integrals Introduction 481 Tagged partitions

704 704

Tagged partitions and Riemann sums; gauge integrals; gauges; residual sets; subdivisions; examples (the Riemann integral, the Henstock integral, the symmetric Riemann-complete integral, the McShane integral, box products, the approximately continuous Henstock integral).

482 General theory

714

Saks-Henstock lemma; when gauge-integrable functions are measurable; when integrable functions are gauge-integrable; Iν (f × χH); integrating derivatives; B.Levi’s theorem; Fubini’s theorem.

483 The Henstock integral

729

The Henstock and Lebesgue integrals; indefinite Henstock integrals; Saks-Henstock lemma; fundamental theorem of calculus; the Perron integral; ACG∗ functions.

484 The Pfeffer integral The Tamanini-Giacomelli theorem; a family of tagged-partition structures; the Pfeffer integral; the SaksHenstock indefinite integral of a Pfeffer integrable function; Pfeffer’s Divergence Theorem; differentiating the indefinite integral; invariance under lipeomorphisms.

746

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Chapter 49: Further topics Introduction 491 Equidistributed sequences

765 765

The asymptotic density ideal Z; equidistributed sequences; when equidistributed sequences exist; Z = PN/Z; effectively regular measures; equidistributed sequences and induced embeddings of measure algebras in Z.

492 Combinatorial concentration of measure

782

Concentration of measure in product spaces; concentration of measure in permutation groups.

493 Extremely amenable groups

789

Extremely amenable groups; concentrating additive functionals; measure algebras under 4 ; L0 ; automorphism groups of measure algebras; isometry groups of spheres in inner product spaces; locally compact groups.

494 Cubes in product spaces

800

Subsets of measure algebras with non-zero infima; product sets included in given sets of positive measure.

495 Poisson point processes

802

Poisson distributions; Poisson point processes; disintegrations; transforming disjointness into stochastic independence; representing Poisson point processes by Radon measures; exponential distributions and Poisson point processes on [0, ∞[.

Appendix to Volume 4 Introduction 4A1 Set theory

823 823

Cardinals; closed cofinal sets and stationary sets; ∆-system lemma; free sets; Ramsey’s theorem; the Marriage Lemma; filters; normal ultrafilters; Ostaszewski’s ♣; cardinals of σ-algebras.

4A2 General topology

827

Glossary; general constructions; Fσ , Gδ , zero and cozero sets; countable chain condition; separation ˇ axioms; compact and locally compact spaces; Lindel¨ of spaces; Stone-Cech compactifications; uniform spaces; first-countable, sequential, countably tight, metrizable spaces; countable networks; secondcountable spaces; separable metrizable spaces; Polish spaces; order topologies.

4A3 Topological σ-algebras

848

Borel σ-algebras; measurable functions; hereditarily Lindel¨ of spaces; second-countable spaces; Polish spaces; ω1 ; Baire σ-algebras; product spaces; compact spaces; Baire property algebras; cylindrical σalgebras.

4A4 Locally convex spaces

857

Linear topological spaces; locally convex spaces; Hahn-Banach theorem; normed spaces; inner product spaces; max-flow min-cut theorem.

4A5 Topological groups

863

Group actions; topological groups; uniformities; quotient groups; metrizable groups.

4A6 Banach algebras

869

Stone-Weierstrass theorem (fourth form); multiplicative linear functionals; spectral radius; invertible elements; exponentiation.

Concordance

873

References for Volume 4

874

Index to Volumes 1-4 Principal topics and results General index

881 893

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General introduction In this treatise I aim to give a comprehensive description of modern abstract measure theory, with some indication of its principal applications. The first two volumes are set at an introductory level; they are intended for students with a solid grounding in the concepts of real analysis, but possibly with rather limited detailed knowledge. As the book proceeds, the level of sophistication and expertise demanded will increase; thus for the volume on topological measure spaces, familiarity with general topology will be assumed. The emphasis throughout is on the mathematical ideas involved, which in this subject are mostly to be found in the details of the proofs. My intention is that the book should be usable both as a first introduction to the subject and as a reference work. For the sake of the first aim, I try to limit the ideas of the early volumes to those which are really essential to the development of the basic theorems. For the sake of the second aim, I try to express these ideas in their full natural generality, and in particular I take care to avoid suggesting any unnecessary restrictions in their applicability. Of course these principles are to to some extent contradictory. Nevertheless, I find that most of the time they are very nearly reconcilable, provided that I indulge in a certain degree of repetition. For instance, right at the beginning, the puzzle arises: should one develop Lebesgue measure first on the real line, and then in spaces of higher dimension, or should one go straight to the multidimensional case? I believe that there is no single correct answer to this question. Most students will find the one-dimensional case easier, and it therefore seems more appropriate for a first introduction, since even in that case the technical problems can be daunting. But certainly every student of measure theory must at a fairly early stage come to terms with Lebesgue area and volume as well as length; and with the correct formulations, the multidimensional case differs from the one-dimensional case only in a definition and a (substantial) lemma. So what I have done is to write them both out (§§114-115). In the same spirit, I have been uninhibited, when setting out exercises, by the fact that many of the results I invite students to look for will appear in later chapters; I believe that throughout mathematics one has a better chance of understanding a theorem if one has previously attempted something similar alone. As I write this Introduction (September 2003), the plan of the work is as follows: Volume Volume Volume Volume Volume

1: 2: 3: 4: 5:

The Irreducible Minimum Broad Foundations Measure Algebras Topological Measure Spaces Set-theoretic Measure Theory.

Volume 1 is intended for those with no prior knowledge of measure theory, but competent in the elementary techniques of real analysis. I hope that it will be found useful by undergraduates meeting Lebesgue measure for the first time. Volume 2 aims to lay out some of the fundamental results of pure measure theory (the Radon-Nikod´ ym theorem, Fubini’s theorem), but also gives short introductions to some of the most important applications of measure theory (probability theory, Fourier analysis). While I should like to believe that most of it is written at a level accessible to anyone who has mastered the contents of Volume 1, I should not myself have the courage to try to cover it in an undergraduate course, though I would certainly attempt to include some parts of it. Volumes 3 and 4 are set at a rather higher level, suitable to postgraduate courses; while Volume 5 will assume a wide-ranging competence over large parts of analysis and set theory. There is a disclaimer which I ought to make in a place where you might see it in time to avoid paying for this book. I make no attempt to describe the history of the subject. This is not because I think the history uninteresting or unimportant; rather, it is because I have no confidence of saying anything which would not be seriously misleading. Indeed I have very little confidence in anything I have ever read concerning the history of ideas. So while I am happy to honour the names of Lebesgue and Kolmogorov and Maharam in more or less appropriate places, and I try to include in the bibliographies the works which I have myself consulted, I leave any consideration of the details to those bolder and better qualified than myself. The work as a whole is not yet complete; and when it is finished, it will undoubtedly be too long to be printed as a single volume in any reasonable format. I am therefore publishing it one part at a time. However, drafts of most of the rest are available on the Internet; see http://www.essex.ac.uk/ maths/staff/fremlin/mt.htm for detailed instructions. For the time being, at least, printing will be in short runs. I hope that readers will be energetic in commenting on errors and omissions, since it should be possible to correct these relatively promptly. An inevitable consequence of this is that paragraph references may go out of date rather quickly. I shall be most flattered if anyone chooses to rely on this book as a source


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for basic material; and I am willing to attempt to maintain a concordance to such references, indicating where migratory results have come to rest for the moment, if authors will supply me with copies of papers which use them. I mention some minor points concerning the layout of the material. Most sections conclude with lists of ‘basic exercises’ and ‘further exercises’, which I hope will be generally instructive and occasionally entertaining. How many of these you should attempt must be for you and your teacher, if any, to decide, as no two students will have quite the same needs. I mark with a > those which seem to me to be particularly important. But while you may not need to write out solutions to all the ‘basic exercises’, if you are in any doubt as to your capacity to do so you should take this as a warning to slow down a bit. The ‘further exercises’ are unbounded in difficulty, and are unified only by a presumption that each has at least one solution based on ideas already introduced. Occasionally I add a final ‘problem’, a question to which I do not know the answer and which seems to arise naturally in the course of the work. The impulse to write this book is in large part a desire to present a unified account of the subject. Cross-references are correspondingly abundant and wide-ranging. In order to be able to refer freely across the whole text, I have chosen a reference system which gives the same code name to a paragraph wherever it is being called from. Thus 132E is the fifth paragraph in the second section of the third chapter of Volume 1, and is referred to by that name throughout. Let me emphasize that cross-references are supposed to help the reader, not distract her. Do not take the interpolation ‘(121A)’ as an instruction, or even a recommendation, to lift Volume 1 off the shelf and hunt for §121. If you are happy with an argument as it stands, independently of the reference, then carry on. If, however, I seem to have made rather a large jump, or the notation has suddenly become opaque, local cross-references may help you to fill in the gaps. Each volume will have an appendix of ‘useful facts’, in which I set out material which is called on somewhere in that volume, and which I do not feel I can take for granted. Typically the arrangement of material in these appendices is directed very narrowly at the particular applications I have in mind, and is unlikely to be a satisfactory substitute for conventional treatments of the topics touched on. Moreover, the ideas may well be needed only on rare and isolated occasions. So as a rule I recommend you to ignore the appendices until you have some direct reason to suppose that a fragment may be useful to you. During the extended gestation of this project I have been helped by many people, and I hope that my friends and colleagues will be pleased when they recognise their ideas scattered through the pages below. But I am especially grateful to those who have taken the trouble to read through earlier drafts and comment on obscurities and errors. Introduction to Volume 4 I return in this volume to the study of measure spaces rather than measure algebras. For fifty years now measure theory has been intimately connected with general topology. Not only do a very large proportion of the measure spaces arising in applications carry topologies related in interesting ways to their measures, but many questions in abstract measure theory can be effectively studied by introducing suitable topologies. Consequently any course in measure theory at this level must be frankly dependent on a substantial knowledge of topology. With this proviso, I hope that the present volume will be accessible to graduate students, and will lead them to the most important ideas of modern abstract measure theory. The first and third chapters of the volume seek to provide a thorough introduction into the ways in which topologies and measures can interact. They are divided by a short chapter on descriptive set theory, on the borderline between set theory, logic, real analysis and general topology, which I single out for detailed exposition because I believe that it forms an indispensable part of the background of any measure theorist. Chapter 41 is dominated by the concepts of inner regularity and τ -additivity, coming together in Radon measures (§416). Chapter 43 concentrates rather on questions concerning properties of a topological space which force particular relationships with measures on that space. But plenty of side-issues are treated in both, such as Lusin measurability (§418), the definition of measures from linear functionals (§436) and measure-free cardinals (§438). Chapters 45 and 46 continue some of the same themes, with particular investigations into ‘disintegrations’ or regular conditional probabilities (§§452-453), the abstract theory of stochastic processes (§§454-455), Talagrand’s theory of Glivenko-Cantelli classes (§465) and the theory of measures on normed spaces (§§466-467). In contrast with the relatively amorphous structure of Chapters 41, 43, 45 and 46, four chapters of this volume have definite topics. I have already said that Chapter 42 is an introduction to descriptive set theory;

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like Chapters 31 and 35 in the last volume, it is a kind of appendix brought into the main stream of the argument. Chapter 44 deals with topological groups. Most of it is of course devoted to Haar measure, giving the Pontryagin-van Kampen duality theorem (§445) and the Ionescu Tulcea theorem on the existence of translation-invariant liftings (§447). But there are also sections on Polish groups (§448) and amenable groups (§449), and some of the general theory of measures on measurable groups (§444). Chapter 47 is a second excursion, after Chapter 26, into geometric measure theory. It starts with Hausdorff measures (§471), gives a proof of the Di Giorgio-Federer Divergence Theorem (§475), and then examines a number of examples of ‘concentration of measure’ (§476). In Chapter 48, I set out the elementary theory of gauge integrals, with sections on the Henstock and Pfeffer integrals (§§483-484). Finally, in Chapter 49, I give notes on five special topics: equidistributed sequences (§491), combinatorial forms of concentration of measure (§492), extremely amenable groups (§493), subproducts in product spaces (§494) and Poisson point processes (§495). I had better mention prerequisites, as usual. To embark on this material you will certainly need a solid foundation in measure theory. Since I do of course use my own exposition as my principal source of references to the elementary ideas, I advise readers to ensure that they have easy access to all three previous volumes before starting serious work on this one. But you may not need to read very much of them. It might be prudent to glance through the detailed contents of Volume 1 and the first five chapters of Volume 2 to check that most of the material there is more or less familiar. But Volume 3, and the last three chapters of Volume 2, can probably be left on one side for the moment. Of course you will need the Lifting Theorem (Chapter 34) for §§447, 452 and 453, and Chapter 26 is essential background for Chapter 47, while Chapter 28 (on Fourier analysis) may help to make sense of Chapter 44, and parts of Chapter 27 (on probability theory) are necessary for §§455-456. And measure algebras are mentioned in every chapter except (I think) Chapter 48; but I hope that the cross-references are precise enough to lead you to just what you need to know at any particular point. Even Maharam’s theorem is hardly used in this volume. What you will need, apart from any knowledge of measure theory, is a sound background in general topology. This volume calls on a great many miscellaneous facts from general topology, and the list in §4A2 is not a good place to start if continuity and compactness and the separation axioms are unfamiliar. My primary reference for topology is Engelking 89. I do not insist that you should have read this book (though of course I hope you will do so sometime); but I do think you should make sure that you can use it. In the general introduction to this treatise, I wrote ‘I make no attempt to describe the history of the subject’, and I have generally been casual – some would say negligent – in my attributions of results to their discoverers. Through much of the first three volumes I did at least have the excuse that the history exists in print in far more detail than I am qualified to describe. In the present volume I find my position more uncomfortable, in that I have been watching the evolution of the subject relatively closely over the last thirty years, and ought to be able to say something about it. Nevertheless I remain reluctant to make definite statements crediting one person rather than another with originating an idea. My more intimate knowledge of the topic makes me even more conscious than elsewhere of the danger of error and of the breadth of reading that would be necessary to produce a balanced account. In some cases I do attach a result to a specific published paper, but these attributions should never be regarded as an assertion that any particular author has priority; at most, they declare that a historian should examine the source cited before coming to any decision. I assure my friends and colleagues that my omissions are not intended to slight either them or those we all honour. What I have tried to do is to include in the bibliography to this volume all the published work which (as far as I am consciously aware) has influenced me while writing it, so that those who wish to go into the matter will have somewhere to start their investigations.

411B

Definitions

13

Chapter 41 Topologies and Measures I I begin this volume with an introduction to some of the most important ways in which topologies and measures can interact, and with a description of the forms which such constructions as subspaces and product spaces take in such contexts. By far the most important concept is that of Radon measure (411H, §416). In Radon measure spaces we find both the richest combinations of ideas and the most important applications. But, as usual, we are led both by analysis of these ideas and by other interesting examples to consider wider classes of topological measure space, and the greater part of the chapter, by volume, is taken up by a description of the many properties of Radon measures individually and in partial combinations. I begin the chapter with a short section of definitions (§411), including a handful of more or less elementary examples. The two central properties of a Radon measure are ‘inner regularity’ (411B) and ‘τ -additivity’ (411C). The former is an idea of great versatility which I look at in an abstract setting in §412. I take a section (§413) to describe some methods of constructing measure spaces, extending the rather limited range of constructions offered in earlier volumes. There are two sections on τ -additive measures, §§414 and 417; the former covers the elementary ideas, and the latter looks at product measures, where it turns out that we need a new technique to supplement the purely measure-theoretic constructions of Chapter 25. On the way to Radon measures in §416, I pause over ‘quasi-Radon’ measures (411H, §415), where inner regularity and τ -additivity first come effectively together. The possible interactions of a topology and a measure on the same space are so varied that even a brief account makes a long chapter; and this is with hardly any mention of results associated with particular types of topological space, most of which must wait for later chapters. But I include one section on the two most important classes of functions acting between topological measure spaces (§418), and another describing some examples to demonstrate special phenomena (§419).

411 Definitions In something of the spirit of §211, but this time without apologising, I start this volume with a list of definitions. The rest of Chapter 41 will be devoted to discussing these definitions and relationships between them, and integrating the new ideas into the concepts and constructions of earlier volumes; I hope that by presenting the terminology now I can give you a sense of the directions the following sections will take. I ought to remark immediately that there are many cases in which the exact phrasing of the definitions is important in ways which may not be immediately apparent. 411A I begin with a phrase which will be a useful shorthand for the context in which most, but not all, of the theory here will be developed. Definition A topological measure space is a quadruple (X, T, Σ, µ) where (X, Σ, µ) is a measure space and T is a topology on X such that T ⊆ Σ, that is, every open set (and therefore every Borel set) is measurable. 411B Now I come to what are in my view the two most important concepts to master; jointly they will dominate the chapter. Definition Let (X, Σ, µ) be a measure space and K a family of sets. I say that µ is inner regular with respect to K if µE = sup{µK : K ∈ Σ ∩ K, K ⊆ E} for every E ∈ Σ. (Cf. 256Ac, 342Aa.) Remark Note that in this definition I do not assume that K ⊆ Σ, nor even that K ⊆ PX. But of course µ will be inner regular with respect to K iff it is inner regular with respect to K ∩ Σ. It is convenient in this context to interpret sup ∅ as 0, so that we have to check the definition only when µE > 0, and need not insist that ∅ ∈ K.

14

Topologies and measures I

411C

411C Definition Let (X, Σ, µ) be a measure space and T a topology on X. I say that µ is τ -additive (the phrase τ -regular hasSalso been used)Sif whenever G is a non-empty upwards-directed family of open sets such that G ⊆ Σ and G ∈ Σ then µ( G) = supG∈G µG. Remark Note that in this definition I do not assume that every open set is measurable. Consequently we cannot take it for granted that an extension of a τ -additive measure will be τ -additive; on the other hand, the restriction of a τ -additive measure to any σ-subalgebra will be τ -additive. 411D Complementary to 411B we have the following. Definition Let (X, Σ, µ) be a measure space and H a family of subsets of X. Then µ is outer regular with respect to H if µE = inf{µH : H ∈ Σ ∩ H, H ⊇ E} for every E ∈ Σ. 411E I delay discussion of most of the relationships between the concepts here to later in the chapter. But it will be useful to have a basic fact set out immediately. Proposition Let (X, Σ, µ) be a measure space and T a topology on X. If µ is inner regular with respect to the compact sets, it is τ -additive. S proof Let G be a non-empty upwards-directed family of measurable open sets such that H = G ∈ Σ. If γ < µH, there is a compact set K ⊆ H such that µK ≥ γ; now there must be a G ∈ G which includes K, so that µG ≥ γ. As γ is arbitrary, supG∈G µG = µH. 411F In order to deal efficiently with measures which are not totally finite, I think we need the following ideas. Definitions Let (X, Σ, µ) be a measure space and T a topology on X. (a) I say that µ is locally finite if every point of X has a neighbourhood of finite measure, that is, if the open sets of finite outer measure cover X. (b) I say that µ is effectively locally finite if for every non-negligible measurable set E ⊆ X there is a measurable open set G ⊆ X such that µG < ∞ and E ∩ G is not negligible. Note that an effectively locally finite measure must measure many open sets, while a locally finite measure need not. (c) This seems a convenient moment at which to introduce the following term. A real-valued function f defined R on a subset of X is locally integrable if for every x ∈ X there is an open set G containing x such that G f is defined (in the sense of 214D) and finite. 411G Elementary facts (a) If µ is a locally finite measure on a topological space X, then µ∗ K < ∞ for every compact set K ⊆ X. P P The family G of open sets of finite outer measure is upwards-directed and covers X, so there must be some G ∈ G including K, in which case µ∗ K ≤ µ∗ G is finite. Q Q (b) A measure µ on R r is locally finite iff every bounded set has finite outer measure (cf. 256Ab). P P (i) If every bounded set has finite outer measure then, in particular, every open ball has finite outer measure, so that µ is locally finite. (ii) If µ is locally finite and A ⊆ Rr is bounded, then its closure A is compact Q (2A2F), so that µ∗ A ≤ µ∗ A is finite, by (a) above. Q (c) I should perhaps remark immediately that a locally finite topological measure need not be effectively locally finite (419A), and an effectively locally finite measure need not be locally finite (411P). (d) An effectively locally finite measure must be semi-finite.

411K

Definitions

15

(e) A locally finite measure on a Lindelöf space X (definition: 4A2A) is σ-finite. P P Let G be the family of open sets of finite outer measure. Because µ is locally finite, G is a cover of X. Because X is Lindelöf, there is a sequence hGn in∈N in G covering X. For each n ∈ N, there is a measurable set En ⊇ Gn of finite measure, and now hEn in∈N is a sequence of sets of finite measure covering X. Q Q (f ) Let (X, T, Σ, µ) be a topological measure space such that µ is locally finite and inner regular with respect to the compact sets. Then µ is effectively locally finite. P P Suppose that µE > 0. Then there is a measurable compact set K ⊆ E such that µK > 0. As in the argument for (a) above, there is an open set G of finite measure including K, so that µ(E ∩ G) > 0. Q Q (g) Corresponding to (a) above, we R have the following fact. If µ is a measure on a topological space and f ∈ L0 (µ) is locally integrable, then K f dµR is finite for every compact K ⊆ X, because K can be covered by a finite family of open sets G such that G |f |dµ < ∞. (h) If µ is a locally finite measure space X, and f ∈ Lp (µ) for some p ∈ [1, ∞], then f is R on a topological R locally integrable; this is because G |f | ≤ E f ≤ kf kp kχEkq is finite whenever G ⊆ E and µE < ∞, where 1 1 older’s inequality (244Eb). p + q = 1, by H¨ (i) If (X, T) is a completely regular space and µ is a locally finite topological measure on X, then the set of open sets with negligible boundaries is a base for T. P P If x ∈ G ∈ T, let H ⊆ G be an open set of finite measure containing x, and f : X → [0, 1] a continuous function such that f (x) = 1 and f (y) = 0 for y ∈ X \ H. Then {f −1 [{α}] : 0 < α < 1} is an uncountable disjoint family of measurable subsets of H, so there must be some α ∈ ]0, 1[ such that f −1 [{α}] is negligible. Set U = {y : f (y) > α}; then U is an open neighbourhood of x included in G and ∂U ⊆ f −1 [{α}] is negligible. Q Q 411H Two particularly important combinations of the properties above are the following. Definitions (a) A quasi-Radon measure space is a topological measure space (X, T, Σ, µ) such that (i) (X, Σ, µ) is complete and locally determined (ii) µ is τ -additive, inner regular with respect to the closed sets and effectively locally finite. (b) A Radon measure space is a topological measure space (X, T, Σ, µ) such that (i) (X, Σ, µ) is complete and locally determined (ii) T is Hausdorff (iii) µ is locally finite and inner regular with respect to the compact sets. 411I Remarks(a) You may like to seek your own proof that a Radon measure space is always quasiRadon, before looking it up in §416 below. (b) Note that a measure on Euclidean space R r is a Radon measure on the definition above iff it is a Radon measure as described in 256Ad. P P In 256Ad, I said that a measure µ on R r is ‘Radon’ if it is a locally finite complete topological measure, inner regular with respect to the compact sets. (The definition of ‘locally finite’ in 256A was not the same as the one above, but I have already covered this point in 411Gb.) So the only thing to add is that µ is necessarily locally determined, because it is σ-finite (256Ba). Q Q 411J The following special types of inner regularity are of sufficient importance to have earned separate names. Definitions (a) If (X, T) is a topological space, I will say that a measure µ on X is tight if it is inner regular with respect to the closed compact sets. (b) If (X, T, Σ, µ) is a topological measure space, I will say that µ is completion regular if it is inner regular with respect to the zero sets (definition: 3A3Pa). 411K Borel and Baire measures If (X, T) is a topological space, I will call a measure with domain (exactly) the Borel σ-algebra of X (4A3A) a Borel measure on X, and a measure with domain (exactly) the Baire σ-algebra of X (4A3K) a Baire measure on X. Of course a Borel measure is a topological measure in the sense of 411A. On a metric space, the Borel and Baire measures coincide (4A3Kb). The most important measures in this chapter will be c.l.d. versions of Borel measures.

16


411L

411L When we come to look at functions defined on a topological measure space, we shall have to relate ideas of continuity and measurability. Two basic concepts are the following. Definition Let X be a set, Σ a σ-algebra of subsets of X and (Y, S) a topological space. I will say that a function f : X → Y is measurable if f −1 [G] ∈ Σ for every open set G ⊆ Y . Remarks (a) Note that a function f : X → R is measurable on this definition (when R is given its usual topology) iff it is measurable according to the familiar definition in 121C, which asks only that sets of the form {x : f (x) < α} should be measurable (121Ef). (b) For any topological space (Y, S), a function f : X → Y is measurable iff f is (Σ, B(Y ))-measurable, where B(Y ) is the Borel σ-algebra of Y (4A3Cb). 411M Definition Let (X, Σ, µ) be a measure space, T a topology on X, and (Y, S) another topological space. I will say that a function f : X → Y is almost continuous or Lusin measurable if µ is inner regular with respect to the family of subsets A of X such that f ¹A is continuous. 411N Finally, I introduce some terminology to describe ways in which (sometimes) measures can be located in one part of a topological space rather than another. Definitions Let (X, Σ, µ) be a measure space and T a topology on X. (a) I will call a set A ⊆ X self-supporting if µ∗ (A ∩ G) > 0 for every open set G such that A ∩ G is non-empty. (Such sets are sometimes called of positive measure everywhere.) (b) A support of µ is a closed self-supporting set F such that X \ F is negligible. (c) Note that µ can have at most one support. P P If F1 , F2 are supports then µ∗ (F1 \F2 ) ≤ µ∗ (X \F2 ) = 0 so F1 \ F2 must be empty. Similarly, F2 \ F1 = ∅, so F1 = F2 . Q Q (d) If µ is a τ -additive topological measure it has a support. P P Let G be the familySof negligible open S sets, and F the closed set X \ G. Then G is an upwards-directed family in T ∩ Σ and G ∈ T ∩ Σ, so S µ(X \ F ) = µ( G) = supG∈G µG = 0. If G is open and G ∩ F 6= ∅ then G ∈ / G so µ∗ (G ∩ F ) = µ(G ∩ F ) = µG > 0; thus F is self-supporting and is the support of µ. Q Q (e) Let X and Y be topological spaces with topological measures µ, ν respectively and a continuous inverse-measure-preserving function f : X → Y . Suppose that µ has a support E. Then f [E] is the support of ν. P P We have only to observe that for an open set H ⊆ Y νH > 0 ⇐⇒ µf −1 [H] > 0 ⇐⇒ f −1 [H] ∩ E 6= ∅ ⇐⇒ H ∩ f [E] 6= ∅ ⇐⇒ H ∩ f [E] 6= ∅. Q Q (f ) µ is strictly positive (with respect to T) if µ∗ G > 0 for every non-empty open set G ⊆ X, that is, X itself is the support of µ. *(g) If (X, T) is a topological space, and µ is a strictly positive σ-finite measure on X such that the domain Σ of µ includes a π-base U for T, then X is ccc. P P Let hEn in∈N be a sequence of sets of finite measure covering X. Let G be a disjoint family of non-empty open sets. For each G ∈ G, take P UG ∈ U \ {∅} such that UG ⊆ G; then µUG > 0, so there is an n(G) such that µ(En(G) ∩ UG ) > 0. Now G∈G,n(G)=k µ(Ek ∩ UG ) ≤ µEk is finite for every k, so {G : n(G) = k} must be countable and G is countable. Q Q 411O Example Lebesgue measure on R r is a Radon measure (256Ha); in particular, it is locally finite and tight. It is therefore τ -additive and effectively locally finite (411E, 411Gf). It is completion regular (because every compact set is a zero set, see 4A2Lc), outer regular with respect to the open sets (134F) and strictly positive.

411Q

Definitions

17

411P Example: Stone spaces (a) Let (Z, T, Σ, µ) be the Stone space of a semi-finite measure algebra (A, µ ¯), so that (Z, T) is a zero-dimensional compact Hausdorff space, (Z, Σ, µ) is complete and semi-finite, the open-and-closed sets are measurable, the negligible sets are the nowhere dense sets, and every measurable set differs by a nowhere dense set from an open-and-closed set (311I, 321K, 322Bd, 322Qa). (b) µ is inner regular with respect to the open-and-closed sets (322Qa); in particular, it is completion regular and tight. Consequently it is τ -additive (411E). (c) µ is strictly positive, because the open-and-closed sets form a base for T (311I) and a non-empty openand-closed set has non-zero measure. µ is effectively locally finite. P P Suppose that E ∈ Σ is not negligible. There is a measurable set F ⊆ E such that 0 < µF < ∞; now there is a non-empty open-and-closed set G included in F , in which case µG < ∞ and µ(E ∩ G) > 0. Q Q (d) The following are equiveridical, that is, if one is true so are the others: (i) (A, µ ¯) is localizable; (ii) µ is strictly localizable; (iii) µ is locally determined; (iv) µ is a quasi-Radon measure. P P The equivalence of (i)-(iii) is Theorem 322N. (iv)⇒(iii) is trivial. If one, therefore all, of (i)-(iii) are true, then µ is a topological measure, because if G ⊆ Z is open, then G is open-and-closed, by 314S, therefore measurable, and G \ G is nowhere dense, therefore also measurable. We know already that µ is complete, effectively locally finite and τ -additive, so that if it is also locally determined it is a quasi-Radon measure. Q Q (e) The following are equiveridical: (i) µ is a Radon measure; (ii) µ is totally finite; (iii) µ is locally finite; (iv) µ is outer regular with respect to the open sets. P P (ii)⇒(iv) If µ is totally finite and E ∈ Σ, then for any ² > 0 there is a closed set F ⊆ Z \ E such that µF ≥ µ(Z \ E) − ², and now G = Z \ F is an open set including E with µG ≤ µE + ². (iv)⇒(iii) Suppose that µ is outer regular with respect to the open sets, and z ∈ Z. Because Z is Hausdorff, {z} is closed. If it is open it is measurable, and because µ is semi-finite it must have finite measure. Otherwise it is nowhere dense, therefore negligible, and must be included in open sets of arbitrarily small measure. Thus in both cases z belongs to an open set of finite measure; as z is arbitrary, µ is locally finite. (iii)⇒(ii) Becasue Z is compact, this is a consequence of 411Ga. (i)⇒(iii) is part of the definition of ‘Radon measure’. Finally, (ii)+(iii)⇒(i), again directly from the definition and the facts set out in (a)-(b) above. Q Q 411Q Example: Dieudonn´ e’s measure Recall that a set E ⊆ ω1 is a Borel set iff either E or its complement includes a cofinal closed set (4A3J). So we may define a Borel measure µ on ω1 by saying that µE = 1 if E includes a cofinal closed set and µE = 0 if E is disjoint from a cofinal closed set. If E is disjoint from some cofinal closed set, so is any subset of E, so µ is complete. Since µ takes only the values 0 and 1, it is a purely atomic probability measure. µ is a topological measure; being totally finite, it is surely locally finite and effectively locally finite. It is inner regular with respect to the closed sets (because if µE > 0, there is a cofinal closed set F ⊆ E, and now F is a closed set with µF = µE), therefore outer regular with respect to the open sets. It is not τ -additive (because ξ = [0, ξ[ is an open set of zero measure for every ξ < ω1 , and the union of these sets is a measurable open set of measure 1). µ is not completion regular, because the set of countable limit ordinals is a closed set (4A1Bb) which does not include any uncountable zero set (see 411Ra below). The only self-supporting subset of ω1 is the empty set (because there is a cover of ω1 by negligible open sets). In particular, µ does not have a support. Remark There is a measure of this type on any ordinal of uncountable cofinality; see 411Xj.

18


411R

411R Example: The Baire σ-algebra of ω1 The Baire σ-algebra Ba(ω1 ) of ω1 is the countablecocountable algebra (4A3P). The countable-cocountable measure µ on ω1 is therefore a Baire measure on the definition of 411K. Since all sets of the form ]ξ, ω1 [ are zero sets, µ is inner regular with respect to the zero sets and outer regular with respect to the cozero sets. Since sets of the form [0, ξ[ (= ξ) form a cover of ω1 by measurable open sets of zero measure, µ is not τ -additive. 411X Basic exercises (a) Let (X, Σ, µ) be a totally finite measure space and T a topology on X. Show that µ is inner regular with respect to the closed sets iff it is outer regular with respect to the open sets, and is inner regular with respect to the zero sets iff it is outer regular with respect to the cozero sets. (b) Let µ be a Radon measure on R r , where r ≥ 1, and f ∈ L0 (µ). Show that f is locally integrable in R the sense of 411Fc iff it is locally integrable in the sense of 256E, that is, E f dν < ∞ for every bounded set E ⊆ Rr. (c) Let µ be a measure on a topological space, µ ˆ its completion and µ ˜ its c.l.d. version. Show that µ is locally finite iff µ ˆ is locally finite, and in this case µ ˜ is locally finite. > (d) Let µ be an effectively locally finite measure on a topological space X. (i) Show that the completion and c.l.d. version of µ are effectively locally finite. (ii) Show that if µ is complete and locally determined, then the union of the measurable open sets of finite measure is conegligible. (iii) Show that if X is hereditarily Lindelöf then µ must be σ-finite. (e) Let X be a topological space and µ a measure on X. Let U ⊆ L0 (µ) be the set of equivalence classes of locally integrable functions in L0 (µ). Show that U is a solid linear subspace of L0 (µ). Show that if µ is locally finite then U includes Lp (µ) for every p ∈ [0, ∞]. (f ) Let X be a topological space. (i) Let µ, ν be two totally finite Borel measures which agree on the closed sets. Show that they are equal. (Hint: 136C.) (ii) Let µ, ν be two totally finite Baire measures which agree on the zero sets. Show that they are equal. (g) Let (X, T) be a topological space, µ a measure on X, and Y a subset of X; let TY , µY be the subspace topology and measure. Show that if µ is a topological measure, or locally finite, or a Borel measure, so is µY . (h) Let h(Xi , Σi , µi )ii∈I be a family of measure spaces, with direct sum (X, Σ, µ); suppose that we are given a topology Ti on each Xi , and let T be the disjoint union topology on X. Show that µ is a topological measure, or locally finite, or effectively locally finite, or a Borel measure, or a Baire measure, or strictly positive, iff every µi is. (i) Let (X, Σ, µ) and (Y, T, ν) be two measure spaces, with c.l.d. product measure λ on X × Y . Suppose we are given topologies T, S on X, Y respectively, and give X × Y the product topology. Show that λ is locally finite, or effectively locally finite, if µ and ν are. (j) Let κ be any cardinal of uncountable cofinality (definition: 3A1Fb). Show that there is a complete topological probability measure µ on κ defined by saying that µE = 1 if E includes a cofinal closed set in κ, 0 if E is disjoint from some cofinal closed set. Show that µ is inner regular with respect to the closed sets but is not completion regular. 411Y Further exercises (a) Show that a function f : R r → R s is measurable iff it is almost continuous (where R r is endowed with Lebesgue measure and its usual topology, of course). (Hint: 256F.) (b) Let (X, ρ) be a metric space, r ≥ 0, and write µHr for r-dimensional Hausdorff measure on X (264K, §471). (i) Show that µHr is a topological measure, outer regular with respect to the Borel sets. (ii) Show that if X is complete then the c.l.d. version of µHr is tight, therefore completion regular.

412A

Inner regularity

19

(c) Let (X, T, Σ, µ) be a topological measure space. Set E = {E : E ⊆ X, µ(∂E) = 0}, where ∂E is the boundary of A. (i) Show that E is a subalgebra of PX, and that every member of E is measured by the completion of µ. (E is sometimes called the Jordan algebra of (X, T, Σ, µ). Do not confuse with the ‘Jordan algebras’ of abstract algebra.) (ii) Suppose that µ is complete and totally finite and inner regular with respect to the closed sets, and that T is normal. Show that {E • : E ∈ E} is dense in the measure algebra of µ endowed with its usual topology. (Hint: if f : X → R is continuous, then {x : f (x) ≤ α} ∈ E for all but countably many α.) (iii) Suppose that µ is a quasi-Radon measure and T is completely regular. Show that {E • : E ∈ E} is dense in the measure algebra of µ. (Hint: 414Aa.) 411 Notes and comments Of course the list above can give only a rough idea of the ways in which topologies and measures can interact. In particular I have rather arbitrarily given a sort of priority to three particular relationships between the domain Σ of a measure and the topology: ‘topological measure space’ (in which Σ includes the Borel σ-algebra), ‘Borel measure’ (in which Σ is precisely the Borel σ-algebra) and ‘Baire measure’ (in which Σ is the Baire σ-algebra). Abstract topological measure theory is a relatively new subject, and there are many technical questions on which different authors take different views. For instance, the phrase ‘Radon measure’ is commonly used to mean what I would call a ‘tight locally finite Borel measure’ (cf. 416F); and some writers enlarge the definition of ‘topological measure’ to include Baire measures as defined above. I give very few examples at this stage, two drawn from the constructions of Volumes 1-3 (Lebesgue measure and Stone spaces, 411O-411P) and one new one (‘Dieudonné’s measure’, 411Q), with a glance at the countable-cocountable measure of ω1 (411R). The most glaring omission is that of the product measures on {0, 1}I and [0, 1]I . I pass these by at the moment because a proper study of them requires rather more preparation than can be slipped into a parenthesis. (I return to them in 416U.) I have also omitted any discussion of ‘measurable’ and ‘almost continuous’ functions, except for a reference to a theorem in Volume 2 (411Ya), which will have to be repeated and amplified later on (§418). There is an obvious complementarity between the notions of ‘inner’ and ‘outer’ regularity (411B, 411D), but it works well only for totally finite spaces (411Xa); in other cases it may not be obvious what will happen (411O, 411Pe, 412W).

412 Inner regularity As will become apparent as the chapter progresses, the concepts introduced in §411 are synergic; their most interesting manifestations are in combinations of various kinds. Any linear account of their properties will be more than usually like a space-filling curve. But I have to start somewhere, and enough results can be expressed in terms of inner regularity, more or less by itself, to be a useful beginning. After a handful of elementary basic facts (412A) and a list of standard applications (412B), I give some useful sufficient conditions for inner regularity of topological and Baire measures (412D, 412E, 412G), based on an important general construction (412C). The rest of the section amounts to a review of ideas from Volume 2 and Chapter 32 in the light of the new concept here. I touch on completions (412H), c.l.d. versions and complete locally determined spaces (412H, 412J, 412L), strictly localizable spaces (412I), inverse-measurepreserving functions (412K, 412M), measure algebras (412N), subspaces (412O, 412P), indefinite-integral measures (412Q) and product measures (412R-412V), with a brief mention of outer regularity (412W); most of the hard work has already been done in Chapters 21 and 25. 412A I begin by repeating a lemma from Chapter 34, with some further straightforward facts. Lemma (a) Let (X, Σ, µ) be a measure space and K a family of sets such that whenever E ∈ Σ and µE > 0 there is a K ∈ K ∩ Σ such that K ⊆ E and µK > 0. ThenPwhenever E ∈ Σ there is a countable disjoint family hKi ii∈I in K ∩ Σ such that Ki ⊆ E for every i and i∈I µKi = µE. If moreover (†) K ∪ K 0 ∈ K whenever K, K 0 are disjoint members of K, S then µ is inner regular with respect to K. If i∈I Ki ∈ K for every countable disjoint family hKi ii∈I in K, then for every E ∈ Σ there is a K ∈ K ∩ Σ such that K ⊆ E and µK = µE.

20

Topologies and measures

412A

(b) Let (X, Σ, µ) be a measure space, T a σ-subalgebra of Σ, and K a family of sets. If µ is inner regular with respect to T and µ¹ T is inner regular with respect to K, then µ is inner regular with respect to K. (c) Let (X, Σ, µ) be a semi-finite measure space and hKn in∈N a sequence of families of sets such that µ is inner regular with respect to Kn and T (‡) if hKi ii∈N is a non-increasing sequence in Kn , then i∈N Ki ∈ Kn T for every n ∈ N. Then µ is inner regular with respect to n∈N Kn . proof (a) This is 342B-342C. (b) If E ∈ Σ and γ < µE, there are an F ∈ T such that F ⊆ E and µF > γ, and a K ∈ K ∩ T such that K ⊆ F and µK ≥ γ. (c) Suppose that E ∈ Σ and that 0 ≤ γ < µE. Because µ is semi-finite, there is an F ∈ Σ such that F ⊆ E and γ < µF < ∞ (213A). Choose hKi ii∈N inductively, as follows. Start with K0 = F . Given that Ki ∈ Σ and γ < µKi , then let ni ∈ N be such that 2−ni (i + 1) is an odd integer, and choose Ki+1 ∈ Kni such that Ki+1T⊆ Ki and µKi+1 > γ; this will be possible because µ is inner regular with respect to Kni . Consider K = i∈N Ki . Then K ⊆ E and µK = limi→∞ µKi ≥ γ. But also T K = j∈N K2n (2j+1) ∈ Kn T because hK2n (2j+1) ij∈N is a non-increasing sequence in Kn , for each n. So K ∈ n∈N Kn . As E and γ are T arbitrary, µ is inner regular with respect to n∈N Kn . 412B Corollary Let (X, Σ, µ) be a measure space and T a topology on X. Suppose that K is either the family of Borel subsets of X or the family of closed subsets of X or the family of compact subsets of X or the family of zero sets in X, and suppose that whenever E ∈ Σ and µE > 0 there is a K ∈ K ∩ Σ such that K ⊆ E and µK > 0. Then µ is inner regular with respect to K. proof In every case, K satisfies the condition (†) of 412Aa. 412C The next lemma provides a particularly useful method of proving that measures are inner regular with respect to ‘well-behaved’ families of sets. Lemma Let (X, Σ, µ) be a semi-finite measure space, and suppose that A ⊆ Σ and K are such that ∅ ∈ A ⊆ Σ, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K, T (‡) n∈N Kn ∈ K for every sequence hKn in∈N in K, X \ A ∈ A for every A ∈ A, whenever A ∈ A, F ∈ Σ and µ(A ∩ F ) > 0, there is a K ∈ K ∩ A such that K ⊆ A and µ(K ∩ F ) > 0. Let T be the σ-subalgebra of Σ generated by A. Then µ¹ T is inner regular with respect to K. proof (a) Write A for the measure algebra of (X, Σ, µ), and L = K ∩ T, so that L is also closed under finite unions and countable intersections. Set H = {E : E ∈ Σ, supL∈L,L⊆E L• = E • } in A, T0 = {E : E ∈ H, X \ E ∈ H}, so that the last two conditions tell us that A ⊆ T0 . (b) The intersection of any sequence in H belongs to H. P P Let hHn in∈N be a sequence in H with intersection H. Write An for {L• : L ∈ L, L ⊆ Hn } ⊆ A for each n ∈ N. Since A is weakly (σ, ∞)distributive (322F), An is upwards-directed, and sup An = Hn• for each n ∈ N,

412F

Inner regularity

21

H • = inf Hn• n∈N

(because F 7→ F : Σ → A is sequentially order-continuous, by 321H) = inf sup An = sup{ inf an : an ∈ An for every n ∈ N} •

n∈N

(316J) = sup{(

n∈N

\

Ln )• : Ln ∈ L, Ln ⊆ Hn for every n ∈ N}

n∈N

⊆

{L• : L ∈ L, L ⊆ H}

⊆

H •,

(by (‡))

and H ∈ H. Q Q (c) The union of any sequence in H belongs to H. P P If hHn in∈N is a sequence in H with union H then supL∈L,L⊆H L• ⊇ supn∈N supL∈L,L⊆En L• = supn∈N Hn• = H • , so H ∈ H. Q Q (d) T0 is a σ-subalgebra of Σ. P P (i) ∅ and X belong to A ⊆ H, so ∅ ∈ T0 . (ii) Obviously X \ E ∈ T0 0 wheneverTE ∈ T . (iii) If hEn in∈N is a sequence in T0 with union E then E ∈ H, by (c); but also Q X \ E = n∈N (X \ En ) belongs to H, by (b). So E ∈ T0 . Q (e) Accordingly T ⊆ T0 , and E • = supL∈L,L⊆E L• for every E ∈ T. It follows at once that if E ∈ T and µE > 0, there must be an L ∈ L such that L ⊆ E and µL > 0; since (†) is true, and L ⊆ T, we can apply 412Aa to see that µ¹ T is inner regular with respect to L, therefore with respect to K. 412D As corollaries of the last lemma I give two-and-a-half basic theorems. Theorem Let (X, T) be a topological space and µ a semi-finite Baire measure on X. Then µ is inner regular with respect to the zero sets. proof Write Σ for the Baire σ-algebra of X, the domain of µ, K for the family of zero sets, and A for K ∪ {X \ K : K ∈ K}. Since the union of two zero sets is a zero set (4A2C(b-ii)), the intersection of a sequence of zero sets is a zero set (4A2C(b-iii)), and the complement of a zero set is the union of a sequence of zero sets (4A2C(b-vi)), the conditions of 412C are satisfied; and as the σ-algebra generated by A is just Σ, µ is inner regular with respect to K. 412E Theorem Let (X, T) be a perfectly normal topological space (e.g., any metrizable space). Then any semi-finite Borel measure on X is inner regular with respect to the closed sets. proof Because the Baire and Borel σ-algebras are the same (4A3Kb), this is a special case of 412D. 412F Lemma Let (X, Σ, µ) be a measure space and T a topology on X such that µ is effectively locally finite with respect to T. Then µE = sup{µ(E ∩ G) : G is a measurable open set of finite measure} for every E ∈ Σ. proof Apply 412Aa with K the family of subsets of measurable open sets of finite measure.

22


412G

412G Theorem Let (X, Σ, µ) be a measure space with a topology T such that µ is effectively locally finite with respect to T and Σ is the σ-algebra generated by T ∩ Σ. If µG = sup{µF : F ∈ Σ is closed, F ⊆ G} for every measurable open set G of finite measure, then µ is inner regular with respect to the closed sets. proof In 412C, take K to be the family of measurable closed subsets of X, and A to be the family of measurable sets which are either open or closed. If G ∈ Σ ∩ T, F ∈ Σ and µ(G ∩ F ) > 0, then there is an open set H of finite measure such that µ(H ∩ G ∩ F ) > 0, because µ is effectively locally finite; now there is a K ∈ K such that K ⊆ H ∩ G and µK > µ(H ∩ G) − µ(H ∩ G ∩ F ), so that µ(K ∩ F ) > 0. This is the only non-trivial item in the list of hypotheses in 412C, so we can conclude that µ¹ T is inner regular with respect to K, where T is the σ-algebra generated by A; but of course this is just Σ. Remark There is a similar result in 416F(iii) below. 412H Proposition Let (X, Σ, µ) be a measure space and K a family of sets. (a) If µ is inner regular with respect to K, so are its completion µ ˆ (212C) and c.l.d. version µ ˜ (213E). (b) Now suppose that T (‡) n∈N Kn ∈ K whenever hKn in∈N is a non-increasing sequence in K. If either µ ˆ is inner regular with respect to K or µ is semi-finite and µ ˜ is inner regular with respect to K, then µ is inner regular with respect to K. proof (a) If F belongs to the domain of µ ˆ, then there is an E ∈ Σ such that E ⊆ F and µ ˆ(F \ E) = 0. So if 0 ≤ γ < µ ˆF = µE, there is a K ∈ K ∩ Σ such that K ⊆ E ⊆ F and µ ˆK = µK ≥ γ. If H belongs to the domain of µ ˜ and 0 ≤ γ < µ ˜H, there is an E ∈ Σ such that µE < ∞ and µ ˆ(E ∩ H) > γ (213D). Now there is a K ∈ K ∩ Σ such that K ⊆ E ∩ H and µK ≥ γ. As µK < ∞, µ ˜K = µK ≥ γ. (b) Write µ ˇ for whichever of µ ˆ, µ ˜ is supposed to be inner regular with respect to K. Then µ ˇ is inner regular with respect to Σ (212Ca, 213Fc), so is inner regular with respect to K ∩ Σ (412Ac). Also µ ˇ extends µ (212D, 213Hc). Take E ∈ Σ and γ < µE = µ ˇE. Then there is a K ∈ K ∩ Σ such that K ⊆ E and γ<µ ˇK = µK. As E and γ are arbitrary, µ is inner regular with respect to K. 412I Lemma Let (X, Σ, µ) be a strictly localizable measure space and K a family of sets such that whenever E ∈ Σ and µE > 0 there is a K ∈ K ∩ Σ such that K ⊆ E and µK > 0. (a) There is a decomposition hXi ii∈I of X such that at most one Xi does not belong to K, and that exceptional one, if any, is negligible. P (b) There is a disjoint family L ⊆ K ∩ Σ such that µ∗ A = L∈L µ∗ (A ∩ L) for every A ⊆ X. (c) If µ is σ-finite then the family hXi ii∈I of (a) and the set L of (b) can be taken to be countable. proof (a) Let hEj ij∈J be any decomposition of X. For each j ∈ J, let Kj be a maximal disjoint subset of {K : K ∈ K ∩ Σ, K ⊆ Ej , µK > 0}. S Because µEj < ∞, Kj must be countable. Set Ej0 = Ej \ Kj . By the maximality of Kj , Ej0 cannot include S any non-negligible set in K ∩ Σ; but this means that µEj0 = 0. Set X 0 = j∈J Ej0 . Then P P µX 0 = j∈J µ(X 0 ∩ Ej ) = j∈J µEj0 = 0. S Note that if j, j 0 ∈ J are distinct, and K ∈ Kj , K 0 ∈ Kj 0 , then K ∩ K 0 = ∅; thus L = j∈J Kj is disjoint. Let hXi ii∈I be any indexing of {X 0 } ∪ L. This is a partition (that is, disjoint cover) of X into sets of finite measure. If E ⊆ X and E ∩ Xi ∈ Σ for every i ∈ I, then for every j ∈ J S E ∩ Ej = (E ∩ X 0 ∩ Ej ) ∪ K∈Kj E ∩ K belongs to Σ, so that E ∈ Σ and µE =

P

P j∈J

K∈Kj

µ(E ∩ K) =

P i∈I

µ(E ∩ Xi ).

Thus hXi ii∈I is a decomposition of X, and it is of the right type because every Xi but one belongs to L ⊆ K.

412K

Inner regularity

(b) If now A ⊆ X is any set, µ∗ A = µA A =

P i∈I

µA (A ∩ Xi ) =

23

P i∈I

µ∗ (A ∩ Xi )

by 214Ia, writing µA for the subspace measure on A. So we have P P µ∗ A = µ∗ (A ∩ X 0 ) + L∈L µ∗ (A ∩ L) = L∈L µ∗ (A ∩ L), while L ⊆ K is disjoint. (c) If µ is σ-finite we can take J to be countable, so that I and L will also be countable. 412J Proposition Let (X, Σ, µ) be a complete locally determined measure space, and K a family of sets such that µ is inner regular with respect to K. (a) If E ⊆ X is such that E ∩ K ∈ Σ for every K ∈ K ∩ Σ, then E ∈ Σ. (b) If E ⊆ X is such that E ∩ K is negligible for every K ∈ K ∩ Σ, then E is negligible. (c) For any A ⊆ X, µ∗ A = supK∈K∩Σ µ∗ (A ∩ K). R (d) Let f be a non-negative [0, ∞]-valued function defined R R on a subset of X. If K f is defined in [0, ∞] for every K ∈ K, then f is defined and equal to supK∈K K f . R (e) If f is a µ-integrable function and ² > 0, there is a K ∈ K such that X\K |f | ≤ ². Remark In (c), we must interpret sup ∅ as 0 if K ∩ Σ = ∅. proof (a) If F ∈ Σ and µF < ∞, then E ∩ F ∈ Σ. P P If µF = 0, this is trivial, because µ is complete and E ∩ F is negligible. Otherwise, there is a sequence hKn in∈N in K ∩ Σ such that Kn ⊆ F for each n and S supn∈N µKn = µF . Now E ∩ F \ n∈N Kn is negligible, therefore measurable, while E ∩ Kn is measurable for every n ∈ N, by hypothesis; so E ∩ F is measurable. Q Q As µ is locally determined, E ∈ Σ, as claimed. (b) By (a), E ∈ Σ; and because µ is inner regular with respect to K, µE must be 0. (c) Let µA be the subspace measure on A. Because µ is complete and locally determined, µA is semi-finite (214Ic). So if 0 ≤ γ < µ∗ A = µA A, there is an H ⊆ A such that µA H is defined, finite and greater than γ. Let E ∈ Σ be a measurable envelope of H (132Ee), so that µE = µ∗ H > γ. Then there is a K ∈ K ∩ Σ such that K ⊆ E and µK ≥ γ. In this case µ∗ (A ∩ K) ≥ µ∗ (H ∩ K) = µ(E ∩ K) = µK ≥ γ. As γ is arbitrary, µ∗ A ≤ supK∈K∩Σ µ∗ (A ∩ K); but the reverse inequality is trivial, so we have the result. (d) Applying (b) with E = X \ dom f , we see that f is defined almost everywhere R in X. Applying (a) with E = {x : Rx ∈ dom f, f (x) ≥ α} for each α ∈ R, we see that f is measurable. So f is defined in [0, ∞], R R and of course f ≥ sup f . If γ < f , there is a non-negative simple function g such that g ≤a.e. K∈K K R R f and g R> γ; taking E = {x : g(x) > 0}, there R R is a K ∈ K Rsuch that K ⊆ E and µ(E \ K)kgk∞ ≤ γ − g, so that K f ≥ K g ≥ γ. As γ is arbitrary, f = supK∈K K f . R R (e) By (d), there is a K ∈ K such that K |f | ≥ |f | − ². Remark See also 413F below. 412K Proposition Let (X, Σ, µ) be a complete locally determined measure space, (Y, T, ν) a measure space and f : X → Y a function. Suppose that K ⊆ T is such that (i) ν is inner regular with respect to K; (ii) f −1 [K] ∈ Σ and µf −1 [K] = νK for every K ∈ K; (iii) whenever E ∈ Σ and µE > 0 there is a K ∈ K such that νK < ∞ and µ(E ∩ f −1 [K]) > 0. Then f is inverse-measure-preserving for µ and ν. proof (a) If F ∈ T, E ∈ Σ and µE < ∞, then E ∩ f −1 [F ] ∈ Σ. P P Let H1 , H2 ∈ Σ be measurable envelopes for E ∩ f −1 [F ] and E \ f −1 [F ] respectively. ?? If µ(H1 ∩ H2 ) > 0, there is a K ∈ K such that νK is finite

24


412K

and µ(H1 ∩ H2 ∩ f −1 [K]) > 0. Because ν is inner regular with respect to K, there are K1 , K2 ∈ K such that K1 ⊆ K ∩ F , K2 ⊆ K \ F and νK1 + νK2 > ν(K ∩ F ) + ν(K \ F ) − µ(H1 ∩ H2 ∩ f −1 [K]) = νK − µ(H1 ∩ H2 ∩ f −1 [K]). Now µ(H1 ∩ f −1 [K2 ]) = µ∗ (E ∩ f −1 [F ] ∩ f −1 [K2 ]) = 0, µ(H2 ∩ f −1 [K1 ]) = µ∗ (E ∩ f −1 [K1 ] \ f −1 [F ]) = 0, so µ(H1 ∩ H2 ∩ f −1 [K1 ∪ K2 ]) = 0 and µ(H1 ∩ H2 ∩ f −1 [K]) ≤ µ(f −1 [K] \ f −1 [K1 ∪ K2 ]) = µf −1 [K] − µf −1 [K1 ] − µf −1 [K2 ] = νK − νK1 − νK2 < µ(H1 ∩ H2 ∩ f −1 [K]), which is absurd. X X Now (E ∩ H1 ) \ (E ∩ f −1 [F ]) ⊆ H1 ∩ H2 is negligible, therefore measurable (because µ is complete), and E ∩ f −1 [F ] ∈ Σ, as claimed. Q Q (b) It follows (because µ is locally determined) that f −1 [F ] ∈ Σ for every F ∈ T. (c) If F ∈ T and νF = 0 then µf −1 [F ] = 0. P P?? Otherwise, there is a K ∈ K such that νK < ∞ and 0 < µ(f −1 [F ] ∩ f −1 [K]) = µf −1 [F ∩ K]. Let K 0 ∈ K be such that K 0 ⊆ K \ F and νK 0 > νK − µf −1 [F ∩ K]. Then f −1 [K 0 ] ∩ f −1 [F ∩ K] = ∅, so νK = µf −1 [K] ≥ µf −1 [K 0 ] + µf −1 [F ∩ K] > νK 0 + νK − νK 0 = νK, which is absurd. X XQ Q (d) Finally, µf −1 [F ] P = νF for every F ∈ T. P P Let hKi ii∈I be S a countable disjoint family in K such that Ki ⊆ F for every i and i∈I νKi = νF (412Aa). Set F 0 = F \ i∈I Ki . Then P P µf −1 [F ] = µf −1 [F 0 ] + i∈I µf −1 [Ki ] = µf −1 [F 0 ] + i∈I νKi = µf −1 [F 0 ] + νF . If νF = ∞ then surely µf −1 [F ] = ∞ = νF . Otherwise, νF 0 = 0 so µf −1 [F 0 ] = 0 (by (c)) and again µf −1 [F ] = νF . Q Q Thus f is inverse-measure-preserving. 412L Corollary Let X be a set and K a family of subsets of X. Suppose that µ, ν are two complete locally determined measures on X, with domains including K, agreeing on K, and both inner regular with respect to K. Then they are identical (and, in particular, have the same domain). proof Apply 412K with X = Y and f the identity function to see that µ extends ν; similarly, ν extends µ and the two measures are the same. 412M Corollary Let (X, Σ, µ) be a complete probability space, (Y, T, ν) a probability space and f : X → Y a function. Suppose that whenever F ∈ T and νF > 0 there is a K ∈ T such that K ⊆ F , νK > 0, f −1 [K] ∈ Σ and µf −1 [K] ≥ νK. Then f is inverse-measure-preserving. proof Set K∗ = {K : K ∈ T, f −1 [K] ∈ Σ, µf −1 [K] ≥ νK}. Then K∗ is closed under countable disjoint unions and includes K, so for every F ∈ T there is a K ∈ K∗ such that K ⊆ F and νK = νF , by 412Aa. But this means that µf −1 [K] = νK for every K ∈ K∗ . P P There is a K 0 ∈ K∗ such that K 0 ⊆ Y \ K and 0 νK = 1 − νK; but in this case µf −1 [K 0 ] + µf −1 [K] ≤ 1 = νK 0 + νK, so µf −1 [K] must be equal to νK. Q Q Moreover, there is a K ∗ ∈ K∗ such that νK ∗ = νY = 1, so −1 ∗ −1 µf [K ] = µX = 1 and µ(E ∩ f [K ∗ ]) > 0 whenever µE > 0. Applying 412K to K∗ we have the result.

412Q

Inner regularity

25

412N Lemma Let (X, Σ, µ) be a measure space and K a family of subsets of X such that µ is inner regular with respect to K. Then E • = sup{K • : K ∈ K ∩ Σ, K ⊆ E} in the measure algebra A of µ, for every E ∈ Σ. In particular, {K • : K ∈ K ∩ Σ} is order-dense in A; and if K is closed under finite unions, then {K • : K ∈ K ∩ Σ} is topologically dense in A for the measure-algebra topology. proof ?? If E • 6= sup{K • : K ∈ K ∩ Σ, K ⊆ E}, there is a non-zero a ∈ A such that a ⊆ E • \ K • whenever K ∈ K ∩ Σ and K ⊆ E. Express a as F • where F ⊆ E. Then µF > 0, so there is a K ∈ K ∩ Σ such that K ⊆ F and µK > 0. But in this case 0 6= K • ⊆ a, while K ⊆ E. X X It follows at once that D = {K • : K ∈ K ∩ Σ} is order-dense. If K is closed under finite unions, and a ∈ A, then Da = {d : d ∈ D, d ⊆ a} is upwards-directed and has supremum a, so a ∈ Da ⊆ D (323D(a-ii)). 412O Lemma Let (X, Σ, µ) be a measure space and K a family of subsets of X such that µ is inner regular with respect to K. (a) If E ∈ Σ, then the subspace measure µE (131B) is inner regular with respect to K. (b) Let Y ⊆ X be any set such that the subspace measure µY (214A-214B) is semi-finite. Then µY is inner regular with respect to KY = {K ∩ Y : K ∈ K}. proof (a) This is elementary. (b) Suppose that F belongs to the domain ΣY of µY and 0 ≤ γ < µY F . Because µY is semi-finite there is an F 0 ∈ ΣY such that F 0 ⊆ F and γ < µY F 0 < ∞. Let E be a measurable envelope for F 0 with respect to µ, so that µE = µ∗ F 0 = µY F 0 > γ. There is a K ∈ K ∩ Σ such that K ⊆ E and µK ≥ γ, in which case K ∩ Y ∈ KY ∩ ΣY and µY (K ∩ Y ) = µ∗ (K ∩ Y ) = µ∗ (K ∩ F 0 ) = µ(K ∩ E) = µK ≥ γ. As F and γ are arbitrary, µY is inner regular with respect to KY . Remark Recall from 214I that if (X, Σ, µ) has locally determined negligible sets (in particular, is either strictly localizable or complete and locally determined), then all its subspaces are semi-finite. 412P Proposition Let (X, Σ, µ) be a measure space, T a topology on X and Y a subset of X; write TY for the subspace topology of Y and µY for the subspace measure on Y . Suppose that either Y ∈ Σ or µY is semi-finite. (a) If µ is a topological measure, so is µY . (b) If µ is inner regular with respect to the Borel sets, so is µY . (c) If µ is inner regular with respect to the closed sets, so is µY . (d) If µ is inner regular with respect to the zero sets, so is µY . (e) If µ is effectively locally finite, so is µY . proof (a) is an immediate consequence of the definitions of ‘subspace measure’, ‘subspace topology’ and ‘topological measure’. The other parts follow directly from 412O if we recall that (i) a subset of Y is Borel for TY whenever it is expressible as Y ∩ E for some Borel set E ⊆ X (4A3Ca); (ii) a subset of Y is closed in Y whenever it is expressible as Y ∩ F for some closed set F ⊆ X; (iii) a subset of Y is a zero set in Y whenever it is expressible as Y ∩ F for some zero set F ⊆ X (4A2C(b-v)); (iv) µ is effectively locally finite iff it is inner regular with respect to subsets of open sets of finite measure. 412Q Proposition Let (X, Σ, µ) be a measure space, and ν an indefinite-integral measure over µ (definition: 234B). If µ is inner regular with respect to a family K of sets, so is ν. proof Because µ and its completion µ ˆ give the same integrals, ν is an indefinite-integral measure over µ ˆ; and as µ ˆ is still inner regular with respect to K (412H), we may suppose that µ itself is complete. Let f

26


412Q

be a Radon-Nikod´ ym derivative of ν with respect to µ; by 234Ca, we may suppose that f : X → [0, ∞[ is Σ-measurable. Suppose that F ∈ dom(ν) and that γ < νF . Set G = {x : f (x) > 0}, so that F ∩ G ∈ Σ (234D). For n ∈ N, set Hn = {x : x ∈ F, 2−n ≤ f (x) ≤ 2n }, so that Hn ∈ Σ and R R νF = f × χF dµ = limn→∞ f × χHn dµ. R Let n ∈ N be such that f × χHn dµ > γ. n If µHn = ∞, there is a K ∈ K such that R K ⊆ Hn and µK ≥ 2 Rγ, so that νK ≥ γ. If µHnR is finite, there is a K R∈ K such that 2n (µHn − µK) ≤ f × χHn dµ − γ, so that f × χ(Hn \ K)dµ + γ ≤ f × χHn and νK = f × χK dµ ≥ γ. Thus in either case we have a K ∈ K such that K ⊆ F and νK ≥ γ; as F and γ are arbitrary, ν is inner regular with respect to K. 412R Lemma Let (X, Σ, µ) and (Y, T, ν) be measure spaces, with c.l.d. product space (X × Y, Λ, λ) (251F). Suppose that K ⊆ PX, L ⊆ PY , M ⊆ P(X × Y ) are such that (i) µ is inner regular with respect to K; (ii) ν is inner regular with respect to L; (iii) K × L ∈ M for all K ∈ K, L ∈ L; (iv) T M ∪ M 0 ∈ M whenever M , M 0 ∈ M; (v) n∈N Mn ∈ M for every sequence hMn in∈N in M. Then λ is inner regular with respect to M. proof Write A = {E × Y : E ∈ Σ} ∪ {X × F : F ∈ T}. Then if V ∈ A, W ∈ Λ and λ(W ∩ V ) > 0, there is an M ∈ M ∩ A such that M ⊆ W and λ(M ∩ V ) > 0. P P Suppose that V = E × Y where E ∈ Σ. There must be E0 ∈ Σ and F0 ∈ T, both of finite measure, such that λ(W ∩ V ∩ (E0 × F0 )) > 0 (251F). Now there is a K ∈ K such that K ⊆ E ∩ E0 and µ((E ∩ E0 ) \ K) · νF0 < λ(W ∩ V ∩ (E0 × F0 )); but this means that M = K × Y is included in V and µ(W ∩ M ) > 0. Reversing the roles of the coordinates, the same argument deals with the case in which V = X × F for some F ∈ T. Q Q b is inner regular with respect to K. But λ is inner regular with respect to Σ⊗T b By 412C, λ0 = λ¹Σ⊗T (251Ib) so is also inner regular with respect to M (412Ab). 412S Proposition Let (X, Σ, µ) and (Y, T, ν) be measure spaces, with c.l.d. product space (X × Y, Λ, λ). Let T, S be topologies on X and Y respectively, and give X × Y the product topology. (a) If µ and ν are inner regular with respect to the closed sets, so is λ. (b) If µ and ν are tight (that is, inner regular with respect to the closed compact sets), so is λ. (c) If µ and ν are inner regular with respect to the zero sets, so is λ. (d) If µ and ν are inner regular with respect to the Borel sets, so is λ. (e) If µ and ν are effectively locally finite, so is λ. proof We have only to read the conditions (i)-(v) of 412R carefully and check that they apply in each case. (In part (e), recall that ‘effectively locally finite’ is the same thing as ‘inner regular with respect to the subsets of open sets of finite measure’.) 412T Lemma Let h(Xi , Σi , µi )ii∈I be a family of probability spaces, with product probability space (X, Λ, λ) (§254). Suppose that Ki ⊆ PXi , M ⊆ PX are such that (i) µi is inner regular with respect to Ki for each i ∈ I; (ii) πi−1 [K] ∈ M for every i ∈ I, K ∈ Ki , writing πi (x) = x(i) for x ∈ X; (iii) T M ∪ M 0 ∈ M whenever M , M 0 ∈ M; (iv) n∈N Mn ∈ M for every sequence hMn in∈N in M. Then λ is inner regular with respect to M. proof The argument is nearly identical to that of 412R. Write A = {πi−1 [E] : i ∈ I, E ∈ Σi }. Then if V ∈ A, W ∈ Λ and λ(W ∩ V ) > 0, express V as πi−1 [E], where i ∈ I and E ∈ Σi , and take K ∈ Ki such that K ⊆ E and µi (E \ K) < λ(W ∩ V ); then M = πi−1 [K] belongs to M ∩ A, is included in W , and meets V in a non-negligible set. So, just as in 412R, the conditions of 412C are met. N It follows that λ0 = λ¹ c i∈I Σi is inner regular with respect to K. But λ is the completion of λ0 (254Fd, 254Ff), so is also inner regular with respect to M (412Ha).

*412W

Inner regularity

27

412U Proposition Let h(Xi , Σi , µi )ii∈I be a family of probability spaces, with product probability space (X, Λ, λ). Suppose that we are given a topology Ti on each Xi , and let T be the product topology on X. (a) If every µi is inner regular with respect to the closed sets, so is λ. (b) If every µi is inner regular with respect to the zero sets, so is λ. (c) If every µi is inner regular with respect to the Borel sets, so is λ. proof This follows from 412T just as 412S follows from 412R. 412V Corollary Let h(Xi , Σi , µi )ii∈I be a family of probability spaces, with product probability space (X, Λ, λ). Suppose that we are given a Hausdorff topology Ti on each Xi , and let T be the product topology on X. Suppose that every µi is tight, and that Xi is compact for all but countably many i ∈ I. Then λ is tight. proof By 412Ua, λ is inner regular with respect to the closed sets. If W ∈ Λ and γ < λW , let V ⊆ W be a measurable closed set such that λV > γ. Let J be the set of those i ∈ I such that Xi is not compact; we P are supposing that J is countable. Let h²i ii∈J be a family of strictly positive real numbers such that i∈J ²j ≤ λV − γ (4A1P). For each i ∈ J, let KiQ⊆ Xi be a compact measurable set such that µi (Xi \ Ki ) ≤ ²i ; and for i ∈ I \ J, set Ki = Xi . Then K = i∈I Ki is a compact measurable subset of X, and P λ(X \ K) ≤ i∈J µ(Xi \ Ki ) ≤ λV − γ, so λ(K ∩ V ) ≥ γ; while K ∩ V is a compact measurable subset of W . As W and γ are arbitrary, λ is tight. *412W Outer regularity I have already mentioned the complementary notion of ‘outer regularity’ (411D). In this book it will not be given much prominence. It is however a useful tool when dealing with Lebesgue measure (see, for instance, the proof of 225K), for reasons which the next proposition will make clear. Proposition Let (X, Σ, µ) be a measure space and T a topology on X. (a) Suppose that µ is outer regular with respect to the open sets. Then for any integrable function f : X → [0,R ∞] and ²R > 0, there is a lower semi-continuous measurable function g : X → [0, ∞] such that f ≤ g and g ≤ ² + f . (b) Now suppose that there is a sequence of measurable open sets of finite measure covering X. Then the following are equiveridical: (i) µ is inner regular with respect to the closed sets; (ii) µ is outer regular with respect to the open sets; (iii) for any measurable set E ⊆ X and ² > 0, there are a measurable closed set F ⊆ E and a measurable open set H ⊇ E such that µ(H \ F ) ≤ ²; (iv) for every measurable function f : X → R [0, ∞[ and ² > 0, there is a lower semi-continuous measurable function g : X → [0, ∞] such that f ≤ g and g − f ≤ ²; (v) for every measurable function f : X → R and ² > 0, there is a lower semi-continuous measurable function g : X → ]−∞, ∞] such that f ≤ g and µ{x : g(x) ≥ f (x) + ²} ≤ ². R proof (a) Let η ∈ ]0, 1] be such that η(7 + f dµ) ≤ ². For n ∈ Z, set En = {x : (1 + η)n ≤ f (x) < (1 + η)n+1 }, and let En0 ∈ Σ be a measurable cover of En ; let Gn ⊇ En0 be a measurable open set such P∞ −|n| 0 that µGn ≤ 3 η + µEn . Set g = n=−∞ (1 + η)n+1 χGn . Then g is lower semi-continuous (4A2B(d-iii), 4A2B(d-v)), f ≤ g and Z gdµ =

∞ X

(1 + η)n+1 µGn

n=−∞

≤ (1 + η)

∞ X

(1 + η)n µEn0 +

n=−∞

Z

≤ (1 + η)

f dµ + 7η ≤

Z

∞ X n=−∞

f dµ + ²,

(1 + η)n+1 3−|n| η

28


*412W

as required. (b) Let hGn in∈N be a sequence of open sets of finite measure covering X; replacing it by h if necessary, we may suppose that hGn in∈N is non-decreasing and that G0 = ∅.

S i
Gi in∈N

(i)⇒(iii) Suppose that µ is inner regular with respect to the closed sets, and that E ∈ Σ, ² > 0. −n−2 For each ². Then S n ∈ N let Fn ⊆ Gn \ E be a measurable closed set such that µF1n ≥ µ(Gn \ E) − 2 H = n∈N (Gn \ Fn ) is a measurable open set including E and µ(H \ E) ≤ 2 ². Applying the same argument to X \ E, we get a closed set F ⊆ E such that µ(E \ F ) ≤ 12 ², so that µ(H \ F ) ≤ ². (ii)⇒(iii) The same idea works. Suppose that µ is outer regular with respect to the open sets, and that E ∈ Σ, that µ(Hn \ En ) ≤ 2−n−2 ²; S ² > 0. For each n ∈ N, let Hn ⊇ Gn ∩ En be an open set such 1 then H = n∈N Hn is a measurable open set including E, and µ(H \ E) ≤ 2 ². Now repeat the argument on X \ E to find a measurable closed set F ⊆ E such that µ(E \ F ) ≤ 12 ². (iii)⇒(iv) Assume (iii), and let f : X S → [0, ∞[ be a measurable function, ² > 0. Set ηn = 2−n ²/(16 + 4µGn ) for each n ∈ N. For k ∈ N set Ek = n∈N {x : x ∈ Gn , kηn ≤ f (x) < (k + 1)ηn }, and choose an open set Hk ⊇ Ek such that µ(Hk \ Ek ) ≤ 2−k . Set g = supk,n∈N (k + 1)ηn χ(Gn ∩ Hk ). Then g : X → [0, ∞] is lower semi-continuous (4A2B(d-v) again). Since supk,n∈N kηn χ(Gn ∩ Ek ) ≤ f ≤ supk,n∈N (k + 1)ηn χ(Gn ∩ Ek ), f ≤ g and g − f ≤ supk,n∈N (k + 1)ηn χ(Gn ∩ Hk \ Ek ) + supk,n∈N ηn χ(Gn ∩ Ek ) has integral at most

P∞ P∞ k=0

n=0 (k

+ 1)ηn 2−k +

P∞ n=0

ηn µGn ≤ ².

(i)⇒(v) Assume (i), and suppose that f : X → R is measurable and ² > 0. For each n ∈ N, let αn ≥ 0 be such that µEn < 2−n−1 ², where En = {x : x ∈ Gn+1 \ Gn , f (x) ≤ −αn }. Let Fn ⊆ (Gn+1 \ G ) \ En be a Pn∞ measurable closed set such that µ((Gn+1 \Gn )\Fn ) ≤ 2−n−2 ². Because hFn in∈N is disjoint, h =S n=0 αn χFn is defined as a function from X to R. {Fn : n ∈ N} is locally finite, so {x : h(x) ≥ α} = n∈N,αn ≥α Fn is closed for every α > 0 (4A2B(h-ii)), and h is upper semi-continuous. Now f1 = f + h is a measurable function. Since (i)⇒(iii)⇒(iv), there is a measurable lower semi-continuous function g1 : X → [0, ∞] such R that f1+ ≤ g1 and g1 − f1+ ≤ 21 ²2 , where f1+ = max(0, f1 ). But if we now set g = g1 − h, g is lower semi-continuous, f ≤ g and {x : f (x) + ² ≤ g(x)} ⊆ {x : f1+ (x) + ² ≤ g1 (x)} ∪ {x : f1 (x) < 0} [ ⊆ {x : f1+ (x) + ² ≤ g1 (x)} ∪ (Gn+1 \ Gn ) \ Fn n∈N

has measure at most ², as required. (iv)⇒(ii) and (v)⇒(ii) Suppose that either (iv) or (v) is true, and that E ∈ Σ, ² > 0. Then there is a measurable lower semi-continuous ≤ g and µ{x : χE(x) + 21 ≤ g(x)} ≤ R function g1 : X → ]0, ∞] such that χE 1 ², since this is certainly true if g − χE ≤ 2 ². Set G = {x : g(x) > 2 }; then E ⊆ G and µ(G \ E) ≤ ². (iii)⇒(i) is trivial. Assembling these fragments, the proof is complete. 412X Basic exercises (a) Let (X, Σ, µ) be a measure space and T a topology on X such that µ is inner regular with respect to the closed sets and with respect to the compact sets. Show that it is tight. (b) Explain how 213A is a special case of 412Aa. >(c) Let (X, Σ, µ) be a measure space, and Σ0 a σ-subalgebra of Σ such that µ is inner regular with respect to Σ0 . Show that if 1 ≤ p < ∞ then every member of Lp (µ) is of the form f • for some Σ0 -measurable f : X → R.

412Xp

Inner regularity

29

> (d) Let (X, Σ, µ) be a semi-finite T measure space and A ⊆ Σ an algebra of sets such that the σ-algebra generated by A is Σ. Write K for { n∈N En : En ∈ A for every n ∈ N}. Show that µ is inner regular with respect to K. (e) Let (X, T, Σ, µ) be an effectively locally finite Hausdorff topological measure space such that µ is inner regular with respect to the Borel sets. Suppose that µG = sup{µK : K ⊆ G is compact} for every open set G ⊆ X. Show that µ is tight. (f ) Let (X, T) be a topological space such that every open set is an Fσ set. Show that any effectively locally finite Borel measure on X is inner regular with respect to the closed sets. (g) Let (X, T) be a normal topological space and µ a topological measure on X which is inner regular with respect to the closed sets. Show that µG = sup{µH : H ⊆ G is a cozero set} for every open set G ⊆ X. Show that if µ is totally finite, then µF = inf{µH : H ⊇ F is a zero set} for every closed set F ⊆ X. (h) Let (X, Σ, µ) be a complete locally determined measure space, and suppose that µ is inner regular with respect to a family K of sets. Let Σ0 be the σ-algebra of subsets of X generated by K ∩ Σ. (i) Show that µ is the c.l.d. version of µ¹Σ0 . (Hint: 412J-412L.) (ii) Show that if µ is σ-finite, it is the completion of µ¹Σ0 . > (i) (i) Let (X, Σ, µ) be a σ-finite measure space and T a σ-subalgebra of Σ. Show that if µ is inner regular with respect to T then the completion of µ¹T extends µ, so that µ and µ¹T have the same negligible sets. (ii) Show that if µ is a σ-finite topological measure which is inner regular with respect to the Borel sets, then every µ-negligible set is included in a µ-negligible Borel set. (j) Devise a direct proof of 412L, not using 412K, by (i) showing that µ∗ (A ∩ K) = ν ∗ (A ∩ K) whenever A ⊆ X, K ∈ K (ii) showing that µ∗ = ν ∗ (iii) quoting 213C. (k) Let (X, Σ, µ) be a complete locally determined measure space, Y a set and f : X → Y a function. Show that the following are equiveridical: (i) µ is inner regular with respect to {f −1 [B] : B ⊆ Y } (ii) f −1 [f [E]] \ E is negligible for every E ∈ Σ. (l) Let h(Xi , Σi , µi )ii∈I be a family of measure spaces, with direct sum (X, Σ, µ). Suppose that for each i ∈ I we are given a topology Ti on Xi , and let T be the corresponding disjoint union topology on X. Show that (i) µ is inner regular with respect to the closed sets iff every µi is (ii) µ is inner regular with respect to the compact sets iff every µi is (iii) µ is inner regular with respect to the zero sets iff every µi is (iv) µ is inner regular with respect to the Borel sets iff every µi is. (m) Use 412L and 412Q to shorten the proof of 253I. (n) Let hXi ii∈I be a family of sets, and suppose that we are given, for each i ∈ I, a σ-algebra Σi of subsets Q N of Xi and a topology Ti on Xi . Let T be the product topology on X = i∈I Xi , and Σ = c i∈I Σi . Let µ be a totally finite measure with domain Σ, and set µi = µπi−1 for each i ∈ I, where πi (x) =Sx(i) for Q i ∈ I, x ∈ X. (i) Show that µ is inner regular with respect to the family K of sets expressible as X \ n∈N i∈I Eni where Eni ∈ Σi for every n, i and {i : Eni 6= Xi } is finite for each n. (ii) Show that if every µi is inner regular with respect to the closed sets, so is µ. (iii) Show that if every µi is inner regular with respect to the zero sets, so is µ. (iv) Show that if every µi is inner regular with respect to the Borel sets, so is µ. (v) Show that if every µi is tight, and all but countably many of the Xi are compact, then µ is tight. (o) Let (X, Σ, µ) be a measure space and T a Lindelöf topology on X such that µ is locally finite. (i) Show that µ is σ-finite. (ii) Show that µ is inner regular with respect to the closed sets iff it is outer regular with respect to the open sets. (p) Let X be a topological space and µ a measure on X which is outer regular with respect to the open sets. Show that for any Y ⊆ X the subspace measure on Y is outer regular with respect to the open sets.

30


412Xq

(q) Let X be a topological space and µ a measure on X which is outer regular with respect to the open sets. Show that if f : RX → R is integrable and ² > 0 then there is a lower semi-continuous g : X → ]−∞, ∞] such that f ≤ g and g − f ≤ ². (r) Let (X, Σ, µ) be a semi-finite measure space and hfn in∈N a sequence in L0 (µ) which converges almost everywhere to f ∈ L0 (µ). Show that µ is inner regular with respect to {E : hfn ¹Ein∈N is uniformly convergent}. (Cf. 215Yb.) (s) In 216E, give {0, 1}I its usual compact Hausdorff topology. Show that the measure µ described there is inner regular with respect to the zero sets. 412Y Further exercises (a) Let K be the family of subsets of R which are homeomorphic to the Cantor set. Show that Lebesgue measure is inner regular with respect to K. (Hint: show that if F ⊆ R \ Q is an uncountable compact set, then {x : [x − δ, x + δ] ∩ F is uncountable for every δ > 0} belongs to K.) (b) (i) Show that if X is a perfectly normal space then any semi-finite topological measure on X which is inner regular with respect to the Borel sets is inner regular with respect to the closed sets. (ii) Show that any subspace of a perfectly normal space is perfectly normal. (iii) Show that the split interval I k (343J, 343Yc, 419L) is perfectly normal. (iv) Show that ω1 , with its order topology, is completely regular, normal and Hausdorff, but not perfectly normal. (v) Show that I k × I k is not perfectly normal. (vi) Show that [0, 1]I is perfectly normal iff I is countable. (c) Let (X, Σ, µ) be a measure space, and suppose that µ is inner regular with respect to a family K ⊆ Σ such that K ∪ K 0 ∈ K for all K, K 0 ∈ K. Write Σf for {E : E ∈ Σ, µE < ∞}. Show that {E • : E ∈ K ∩ Σf } is dense in {E • : E ∈ Σf } for the strong measure-algebra topology. (d) Let X be a normal topological space and Y a closed subset of X. Show that every Baire subset of Y is the intersection of Y with a Baire subset of X. (Hint: use Tietze’s theorem.) (e) Let (X, Σ, µ) be [0, 1] with Lebesgue measure, and Y = [0, 1] with counting measure ν; give X its usual topology and Y the discrete topology, and let λ be the c.l.d. product measure on X × Y . (i) Show that µ, ν and λ are all tight (for the appropriate topologies) and therefore completion regular. (ii) Let λ0 be the primitive product measure on X × Y (definition: 251C). Show that λ0 is not tight. (Hint: 252Yf.) Remark : it is undecidable in ZFC whether λ0 is inner regular with respect to the closed sets. (f ) Give an example of a Hausdorff topological measure space (X, T, Σ, µ) such that µ is complete, strictly localizable and outer regular with respect to the open sets, but not inner regular with respect to the closed sets. 412 Notes and comments In this volume we are returning to considerations which have been left on one side for almost the whole of Volume 3 – the exceptions being in Chapter 34, where I looked at realization of homomorphisms of measure algebras by functions between measure spaces, and was necessarily dragged into an investigation of measure spaces which had enough points to be adequate codomains (343B). The idea of ‘inner regularity’ is to distinguish families K of sets which will be large enough to describe the measure entirely, but whose members will be of recognisable types. For an example of this principle see 412Ya. Of course we cannot always find a single type of set adequate to fill a suitable family K, though this happens oftener than one might expect, but it is surely easier to think about an arbitrary zero set (for instance) than an arbitrary measurable set, and whenever a measure is inner regular with respect to a recognisable class it is worth knowing about it. I have tried to use the symbols † and ‡ (412A, 412C) consistently enough for them to act as a guide to some of the ideas which will be used repeatedly in this chapter. Note the emphasis on disjoint unions and countable intersections; I mentioned similar conditions in 136Xi-136Xj. You will recognise 412Aa as an exhaustion principle; note that it is enough to use disjoint unions, as in 313K. In the examples of this section this disjointness is not important. Of course inner regularity has implications for the measure algebra (412N), but it is important to recognise that ‘µ is inner regular with respect to K’ is saying much more than

§413 intro.

Inner measure constructions

31

‘{K • : K ∈ K} is order-dense in the measure algebra’; the latter formulation tells us only that whenever µE > 0 there is a K ∈ K such that K \ E is negligible and µK > 0, while the former tells us that we can take K to be actually a subset of E. 412D, 412E and 412G are all of great importance. 412D looks striking, but of course the reason it works is just that the Baire σ-algebra is very small. In 412E the Baire and Borel σ-algebras coincide, so it is nothing but a special case of 412D; but as metric spaces are particularly important it is worth having it spelt out explicitly. In 412D and 412E the hypothesis ‘semi-finite’ is sufficient, while in 412G we need ‘effectively locally finite’; this is because in both 412D and 412E the open sets we are looking at are countable unions of measurable closed sets. There are interesting non-metrizable spaces in which the same thing happens (412Yb). As you know, I am strongly prejudiced in favour of complete and locally determined measures, and the Baire and Borel measures dealt with in these three results are rarely complete; but they can still be applied to completions and c.l.d. versions of these measures, using 412Ab or 412H. 412O-412V are essentially routine. For subspace measures, the only problem we need to come to terms with is the fact that subspaces of semi-finite measure spaces need not be semi-finite (216Xa). For product measures the point is that the c.l.d. product of two measure spaces, and the product of any family of probability spaces, as I defined them in Chapter 25, are inner regular with respect to the σ-algebra of sets generated by the cylinder sets. This is not in general true of the ‘primitive’ product measure (412Ye), which is one of my reasons for being prejudiced against it. I should perhaps warn you of a trap in the language I use here. I say that if the factor measures are inner regular with respect to the closed sets, so is the c.l.d. product measure. But I do not say that all closed sets in the product are measurable for the product measure, even if closed sets in the factors are measurable for the factor measures. So the path is open for a different product measure to exist, still inner regular with respect to the closed sets; and indeed I shall be going down that path in §417. The uniqueness result in 412L specifically refers to complete locally determined measures defined on all sets of the family K. ThereQis one special difficulty in 412V: in order to ensure that there are enough compact measurable sets in X = i∈I Xi , we need to know that all but countably many of the Xi are actually compact. When we come to look more closely at products of Radon probability spaces we shall need to consider this point again (417Q, 417Xq). In fact some of the ideas of 412U-412V are not restricted to the product measures considered there. Other measures on the product space will have inner regularity properties if their images on the factors, their ‘marginals’ in the language of probability theory, are inner regular; see 412Xn. I will return to this in §454. This section is almost exclusively concerned with inner regularity. The complementary notion of outer regularity is not much use except in σ-finite spaces (415Xh), and not always then (416Yd). In totally finite spaces, of course, and some others, any version of inner regularity corresponds to a version of outer regularity, as in 412Wb(i)-(ii); and when we have something as strong as 412Wb(iii) available it is worth knowing about it.

413 Inner measure constructions I now turn in a different direction, giving some basic results on the construction of inner regular measures. The first step is to describe ‘inner measures’ (413A) and a construction corresponding to the Carathéodory construction of measures from outer measures (413C). Just as every measure gives rise to an outer measure, it gives rise to an inner measure (413D). Inner measures form an effective tool for studying complete locally determined measures (413F). The most substantial results of the section concern the construction of measures as extensions of functionals defined on various classes K of sets. Typically, K is closed under finite unions and countable intersections, though it is sometimes possible to relax the hypotheses to some extent. The methods here make it possible to distinguish arguments which produce finitely additive functionals (413H, 413N, 413P, 413Q) from the succeeding steps to countably additive measures (413I, 413O, 413S). 413H-413M investigate conditions on a functional φ : K → [0, ∞[ sufficient to produce a measure extending φ, necessarily unique, which is inner regular with respect to K or Kδ , the set of intersections of sequences in K. 413N-413O look instead at functionals defined on sublattices of the class K of interest, and at sufficient conditions to ensure the existence of a measure, not normally unique, defined on the whole of K, inner regular with respect to K and extending

32


§413 intro.

the given functional. Finally, 413P-413S are concerned with majorizations rather than extensions; we seek a measure µ such that µK ≥ λK for K ∈ K, while µX is as small as possible. 413A I begin with some material from the exercises of earlier volumes. Definition Let X be a set. An inner measure on X is a function φ : PX → [0, ∞] such that φ∅ = 0; (α) φ(A ∪ B) ≥ φA + φB for all disjoint A, B ⊆ X; T (β) if hAn in∈N is a non-increasing sequence of subsets of X and φA0 < ∞ then φ( n∈N An ) = inf n∈N φAn ; (∗) φA = sup{φB : B ⊆ A, φB < ∞} for every A ⊆ X. 413B The following fact will be recognised as an element of Carathéodory’s method. There will be an application later in which it will be useful to know that it is not confined to proving countable additivity. Lemma Let X be a set and φ : X → [0, ∞] any function such that φ∅ = 0. Then Σ = {E : E ⊆ X, φA = φ(A ∩ E) + φ(A \ E) for every A ⊆ X} is an algebra of subsets of X, and φ(E ∪ F ) = φE + φF for all disjoint E, F ∈ Σ. proof The symmetry of the definition of Σ ensures that X \ E ∈ Σ whenever E ∈ Σ. If E, F ∈ Σ and A ⊆ X, then

φ(A ∩ (E∪F )) + φ(A \ (E ∪ F )) = φ(A ∩ (E ∪ F ) ∩ E) + φ(A ∩ (E ∪ F ) \ E) + φ(A \ (E ∪ F )) = φ(A ∩ E) + φ((A \ E) ∩ F ) + φ((A \ E) \ F ) = φ(A ∩ E) + φ(A \ E) = φA. As A is arbitrary, E ∪ F ∈ Σ. Finally, if A ⊆ X, φ(A ∩ ∅) + φ(A \ ∅) = φ∅ + φA = φA because φ∅ = 0; so ∅ ∈ Σ. Thus Σ is an algebra of sets. If E, F ∈ Σ and E ∩ F = ∅, then φ(E ∪ F ) = φ((E ∪ F ) ∩ E) + φ((E ∪ F ) \ E) = φE + φF . 413C Measures from inner measures I come now to a construction corresponding to Carathéodory’s method of defining measures from outer measures. Theorem Let X be a set and φ : X → [0, ∞] an inner measure. Set Σ = {E : E ⊆ X, φ(A ∩ E) + φ(A \ E) = φA for every A ⊆ X}. Then (X, Σ, φ¹Σ) is a complete measure space. proof (Compare 113C.) (a) The first step is to note that if A ⊆ B ⊆ X then φB ≥ φA + φ(B \ A) ≥ φA. Next, a subset E of X belongs to Σ iff φA ≤ φ(A ∩ E) + φ(A \ E) whenever A ⊆ X and µA < ∞. P P Of course any element of Σ satisfies the condition. If E satisfies the condition and A ⊆ X, then φA = sup{φB : B ⊆ A, φB < ∞} ≤ sup{φ(B ∩ E) + φ(B \ E) : B ⊆ A} = φ(A ∩ E) + φ(A \ E) ≤ φA,

413D


33

so E ∈ Σ. Q Q (b) By 413B, Σ is an algebra of subsets of X. Now suppose that hEn in∈N is a non-decreasing sequence in Σ, with union E. If A ⊆ X and φA < ∞, then φ(A \ E) = inf n∈N φ(A \ En ) = limn→∞ φ(A \ En ) because hA \ En in∈N is non-increasing and φ(A \ E0 ) is finite; so φ(A ∩ E) + φ(A \ E) ≥ limn→∞ φ(A ∩ En ) + φ(A \ En ) = φA. By (a), E ∈ Σ. So Σ is a σ-algebra. (c) If E, F ∈ Σ and E ∩ F = ∅ then φ(E ∪ F ) = φE + φF , by 413B. If hEn in∈N is a disjoint sequence in Σ with union E, then S Pn µE ≥ µ( i≤n Ei ) = i=0 µEi P∞ P∞ P∞ for every n, so µE ≥ Si=0 µEi . ?? If µE > i=0 µEi , there is an A ⊆ E such that i=0 µEi < φA < ∞. But now, setting Fn = i≤n Ei for each n, we have limn→∞ φ(A \ Fn ) = 0, so that P∞ φA = limn→∞ φ(A ∩ Fn ) + φ(A \ Fn ) = i=0 φ(A ∩ Ei ) < φA, P∞ which is absurd. X X Thus µE = i=0 µEi . As hEn in∈N is arbitrary, µ is a measure. (d) Finally, suppose that B ⊆ E ∈ Σ and µE = 0. Then for any A ⊆ X we must have φ(A ∩ B) + φ(A \ B) ≥ φ(A \ E) = φ(A ∩ E) + φ(A \ E) = φA, so B ∈ Σ. Thus µ is complete. Remark For a simple example see 213Yc. 413D The inner measure defined by a measure Let (X, Σ, µ) be any measure space. has an associated outer measure µ∗ defined by the formula

Just as µ

µ∗ A = inf{µE : A ⊆ E ∈ Σ} (132A-132B), it gives rise to an inner measure µ∗ defined by the formula µ∗ A = sup{µE : E ∈ Σf , E ⊆ A}, where I write Σf for {E : E ∈ Σ, µE < ∞}. P P µ∗ ∅ = µ∅ = 0. (†) If A ∩ B = ∅, and E ⊆ A, F ⊆ B belong to Σf , then E ∪ F ⊆ A ∪ B also has finite measure, so µ∗ (A ∪ B) ≥ µ(E ∪ F ) = µE + µF ; taking the supremum over E and F , µ∗ (A ∪ B) ≥ µ∗ A + µ∗ B. (‡) If hAn in∈N is a non-increasing sequence of sets with intersection A and µ∗ A0 < ∞, then for each n ∈ N we can find an En ⊆ An such that µEn ≥ µ∗ An − 2−n . In this case, S S µ( m∈N Em ) = supn∈N µ( m≤n Em ) ≤ µ∗ A0 < ∞. Set E=

T

S n∈N

m≥n

Em ⊆ A.

Then E ∈ Σf , so µ∗ A ≥ µE ≥ lim supn→∞ µEn = limn→∞ µ∗ An ≥ µ∗ A. (*) If A ⊆ X and µ∗ A = ∞ then sup{µ∗ B : B ⊆ A, µ∗ B < ∞} ≥ sup{µE : E ∈ Σf , E ⊆ A} = ∞. Q Q Warning Many authors use the formula µ∗ A = sup{µE : A ⊇ E ∈ Σ}. In ‘ordinary’ cases, when (X, Σ, µ) is semi-finite, this agrees with my usage (413Ed); but for non-semi-finite spaces there is a difference.

34


413D

For elementary properties of the construction here see 213Xe. 413E I note the following elementary facts concerning inner measures defined from measures. Proposition Let (X, Σ, µ) be a measure space. Write Σf for {E : E ∈ Σ, µE < ∞}. (a) For every A ⊆ X there is an E ∈ Σ such that E ⊆ A and µE = µ∗ A. (b) µ∗ A ≤ µ∗ A for every A ⊆ X. (c) If E ∈ Σ and A ⊆ X, then µ∗ (E ∩ A) + µ∗ (E \ A) ≤ µE, with equality if either (i) µE < ∞ or (ii) µ is semi-finite. (d) In particular, µ∗ E ≤ µE for every E ∈ Σ, with equality if either µE < ∞ or µ is semi-finite. (e) If µ is inner regular with respect to K, then µ∗ A = sup{µK : K ∈ K ∩ Σf , K ⊆ A} for every A ⊆ X. (f) If A ⊆ X is such that µ∗ A = µ∗ A < ∞, then A is measured by the completion of µ. (g) If µ ˆ, µ ˜ are the completion and c.l.d. version of µ, then µ ˆ∗ = µ ˜ ∗ = µ∗ . (h) If (Y, T, ν) is another measure space, and f : X → Y is an inverse-measure-preserving function, then µ∗ (f −1 [B]) ≤ ν ∗ B,

µ∗ (f −1 [B]) ≥ ν∗ B

for every B ⊆ Y , and ν ∗ (f [A]) ≥ µ∗ A for every A ⊆ X. (i) Suppose that µ is semi-finite. If A ⊆ E ∈ Σ, then E is a measurable envelope of A iff µ∗ (E \ A) = 0. f proof (a) S There is a sequence hEn in∈N in Σ such that En ⊆ A for each n and limn→∞ µEn = µ∗ A; now set E = n∈N En .

(b) If E ⊆ A ⊆ F we must have µE ≤ µF . (c) If F ⊆ E ∩ A and F ∈ Σf , then µF + µ∗ (E \ A) ≤ µF + µ(E \ F ) = µE; taking the supremum over F , µ∗ (E ∩ A) + µ∗ (E \ A) ≤ µE. If µE < ∞, then µ∗ (E ∩ A) = sup{µF : F ∈ Σ, F ⊆ E ∩ A} = µE − inf{µ(E \ F ) : F ∈ Σ, F ⊆ E ∩ A} = µE − inf{µF : F ∈ Σ, E \ A ⊆ F ⊆ E} = µE − µ∗ (E \ A). If µ is semi-finite, then µ∗ (E ∩ A) + µ∗ (E \ A) ≥ sup{µ∗ (F ∩ A) + µ∗ (F \ A) : F ∈ Σf , F ⊆ E} = sup{µF : F ∈ Σf , F ⊆ E} = µE. (d) Take A = E in (c). (e) µ∗ A = sup{µE : E ∈ Σf , E ⊆ A} = sup{µK : K ∈ K ∩ Σ, ∃ E ∈ Σf , K ⊆ E ⊆ A} = sup{µK : K ∈ K ∩ Σf , K ⊆ A}. (f ) By (a) above and 132Aa, there are E, F ∈ Σ such that E ⊆ A ⊆ F and µE = µ∗ A = µ∗ A = µF < ∞; now µ(F \ E) = 0, so F \ A and A are measured by the completion of µ. ˇ for its domain, and let A ⊆ X. (i) If γ < µ∗ A, there is an E ∈ Σ such (g) Write µ ˇ for either µ ˆ or µ ˜, and Σ that E ⊆ A and γ ≤ µE < ∞; now µ ˇE = µE (212D, 213Fa), so µ ˇ∗ A ≥ γ. As γ is arbitrary, µ∗ A ≤ µ ˇ∗ A.

413F


35

ˇ such that E ⊆ A and γ ≤ µ (ii) If γ < µ ˇ∗ A, there is an E ∈ Σ ˇE < ∞. Now there is an F ∈ Σ such that F ⊆ E and µF = µ ˇE (212C, 213Fc), so that µ∗ A ≥ γ. As γ is arbitrary, µ∗ A ≥ µ ˇ∗ A. (h) This is elementary; all we have to note is that if F , F 0 ∈ T and F ⊆ B ⊆ F 0 , then f −1 [F ] ⊆ f −1 [B] ⊆ f [F 0 ], so that −1

νF = µf −1 [F ] ≤ µ∗ f −1 [B] ≤ µ∗ f −1 [B] ≤ µf −1 [F 0 ] = νF 0 . Now, for A ⊆ X, µ∗ A ≤ µ∗ (f −1 [f [A]]) ≤ ν ∗ (f [A]). (i)(i) If E is a measurable envelope of A and F ∈ Σ is included in E \ A, then µF = µ(F ∩ E) = µ∗ (F ∩ A) = 0; as F is arbitrary, µ∗ (E \ A) = 0. (ii) If E is not a measurable envelope of A, there is an F ∈ Σ such that µ∗ (F ∩ A) < µ(F ∩ E). Let G ∈ Σ be such that F ∩ A ⊆ G and µG = µ∗ (F ∩ A). Then µ(F ∩ E \ G) > 0; because µ is semi-finite, µ∗ (E \ A) ≥ µ∗ (F ∩ E \ G) > 0. 413F The language of 413D makes it easy to express some useful facts about complete locally determined measure spaces, complementing 412J. Lemma Let (X, Σ, µ) be a complete locally determined measure space and K a family of subsets of X such that µ is inner regular with respect to K. Then for E ⊆ X the following are equiveridical: (i) E ∈ Σ; (ii) E ∩ K ∈ Σ whenever K ∈ Σ ∩ K; (iii) µ∗ (K ∩ E) + µ∗ (K \ E) = µ∗ K for every K ∈ K; (iv) µ∗ (K ∩ E) + µ∗ (K \ E) = µ∗ K for every K ∈ K; (v) µ∗ (E ∩ K) = µ∗ (E ∩ K) for every K ∈ K ∩ Σ; (vi) min(µ∗ (K ∩ E), µ∗ (K \ E)) < µK whenever K ∈ K ∩ Σ and 0 < µK < ∞; (vii) max(µ∗ (K ∩ E), µ∗ (K \ E)) > 0 whenever K ∈ K ∩ Σ and µK > 0. proof (a) Assume (i). Then of course E ∩ K ∈ Σ for every K ∈ Σ ∩ K, and (ii) is true. For any K ∈ K there is an F ∈ Σ such that F ⊇ K and µF = µ∗ K (132Aa); now µ∗ K ≤ µ∗ (K ∩ E) + µ∗ (K \ E) ≤ µ(F ∩ E) + µ(F \ E) = µF = µ∗ K, so (iii) is true. Next, for any K ∈ K, µ∗ (K ∩ E) + µ∗ (K \ E) ≤ µ∗ K = sup{µF : F ∈ Σf , F ⊆ K} (writing Σf for {F : F ∈ Σ, µF < ∞}) = sup{µ(F ∩ E) + µ(F \ E) : F ∈ Σf , F ⊆ K} ≤ µ∗ (K ∩ E) + µ∗ (K \ E). So (iv) is true. If K ∈ K ∩ Σ, then µ∗ (E ∩ K) = sup{µF : F ∈ Σf , F ⊆ E ∩ K} = µ(E ∩ K) = µ∗ (E ∩ K) because µ is semi-finite. So (v) is true. Since (iii)⇒(vi) and (iv)⇒(vii), we see that all the conditions are satisfied. (b) Now suppose that E ∈ / Σ; I have to show that (ii)-(vii) are all false. Because µ is locally determined, there is an F ∈ Σf such that E ∩ F ∈ / Σ. Take measurable envelopes H, H 0 of F ∩ E, F \ E respectively 0 (132Ee). Then F \ H ⊆ F ∩ E ⊆ F ∩ H, so G = (F ∩ H) \ (F \ H 0 ) = F ∩ H ∩ H 0 cannot be negligible. Take K ∈ K ∩ Σ such that K ⊆ G and µK > 0. As G ⊆ F , µK < ∞. Now µ∗ (K ∩ E) = µ∗ (K ∩ F ∩ E) = µ(K ∩ H) = µK,

36


413F

µ∗ (K \ E) = µ∗ (K ∩ F \ E) = µ(K ∩ H 0 ) = µK. But this means that µ∗ (K ∩ E) = µK − µ∗ (K \ E) = 0,

µ∗ (K \ E) = µK − µ∗ (K ∩ E) = 0,

by 413Eb. Now we see that this K witnesses that (ii)-(vii) are all false. 413G The ideas of 413F can be used to give criteria for measurability of real-valued functions. I spell out one which is particularly useful. Lemma Let (X, Σ, µ) be a complete locally determined measure space and suppose that µ is inner regular with respect to K ⊆ Σ. Suppose that f : X → R is a function, and for α ∈ R set Eα = {x : f (x) ≤ α}, Fα = {x : f (x) ≥ β}. Then f is Σ-measurable iff min(µ∗ (Eα ∩ K), µ∗ (Fβ ∩ K)) < µK whenever K ∈ K, 0 < µK < ∞ and α < β. proof (a) If f is measurable, then µ∗ (Eα ∩ K) + µ∗ (Fβ ∩ K) = µ(Eα ∩ K) + µ(Fβ ∩ K) ≤ µK whenever K ∈ Σ and α < β, so if 0 < µK < ∞ then we must have min(µ∗ (Eα ∩ K), µ∗ (Fβ ∩ K)) < µK. (b) If f is not measurable, then there is some α ∈ R such that Eα is not measurable. 413F(iii) tells us that there is a K ∈ K such that 0 < µK < ∞ and µ∗ (Eα ∩ K) = µ∗ (K \ Eα ) = µK. Note that K is a measurable envelope of K ∩ Eα (132Eb). Now hK ∩ Fα+2−n in∈N is a non-decreasing sequence with union K \ Eα , so there is some β > α such that K ∩ Fβ is not negligible. Let H ⊆ K be a measurable envelope of K ∩ Fβ , and K 0 ∈ K such that K 0 ⊆ H and µK 0 > 0; then µ∗ (K 0 ∩ Eα ) = µ∗ (K 0 ∩ K ∩ Eα ) = µ(K 0 ∩ K) = µK 0 , µ∗ (K 0 ∩ Fβ ) = µ∗ (K 0 ∩ H ∩ Fβ ) = µ(K 0 ∩ H) = µK 0 , so K 0 , α and β witness that the condition is not satisfied. 413H Inner measure constructions based on 413C are important because they offer an efficient way of setting up measures which are inner regular with respect to given families of sets. Two of the fundamental results are 413I and 413J. I proceed by means of a lemma on finitely additive functionals. Lemma Let X be a set and K a family of subsets of X such that ∅ ∈ K, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, (‡) K ∩ K 0 ∈ K for all K, K 0 ∈ K. Let φ0 : K → [0, ∞[ be a functional such that (α) φ0 K = φ0 L + sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K. Set φA = sup{φ0 K : K ∈ K, K ⊆ A} for A ⊆ X, Σ = {E : E ⊆ X, φA = φ(A ∩ E) + φ(A \ E) for every A ⊆ X}. Then Σ is an algebra of subsets of X, including K, and φ¹Σ : Σ → [0, ∞] is an additive functional extending φ0 . proof (a) To see that Σ is an algebra of subsets and φ¹Σ is additive, all we need to know is that φ∅ = 0 (413B); and this is because, applying hypothesis (α) with K = L = ∅, φ0 ∅ = φ0 ∅ + φ0 ∅, so φ0 ∅ = 0. (α) also assures us that φ0 L ≤ φ0 K whenever K, L ∈ K and L ⊆ K, so φK = φ0 K for every K ∈ K. (b) To check that K ⊆ Σ, we have a little more work to do. First, observe that (†) and (α) together tell us that φ0 (K ∪ K 0 ) = φ0 K + φ0 K 0 for all disjoint K, K 0 ∈ K. So if A, B ⊆ X and A ∩ B = ∅ then

413I


φA + φB =

sup K∈K,K⊆A

=

φ0 K +

sup K,L∈K,K⊆A,L⊆B

sup L∈K,L⊆A

37

φ0 L

φ0 (K ∪ L) ≤ φ(A ∪ B).

(c) K ⊆ Σ. P P Take K ∈ K, A ⊆ X. If L ∈ K and L ⊆ A, then φ0 L = φ0 (K ∩ L) + sup{φ0 L0 : L0 ∈ K, L0 ⊆ L \ K} ≤ φ(A ∩ K) + φ(A \ K). (Note the use of the hypothesis (‡).) As L is arbitrary, φA ≤ φ(A ∩ K) + φ(A \ K). We already know that φ(A ∩ K) + φ(A \ K) ≤ φA; as A is arbitrary, K ∈ Σ. Q Q This completes the proof. 413I Theorem Let X be a set and K a family of subsets of X such that ∅ ∈ K, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, T (‡) n∈N Kn ∈ K whenever hKn in∈N is a sequence in K. Let φ0 : K → [0, ∞[ be a functional such that (α) φ0 K = φ0 L + sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K, (β) inf n∈N φ0 Kn = 0 whenever hKn in∈N is a non-increasing sequence in K with empty intersection. Then there is a unique complete locally determined measure µ on X extending φ0 and inner regular with respect to K. proof (a) Set φA = sup{φ0 K : K ∈ K, K ⊆ A} for A ⊆ X, Σ = {E : E ⊆ X, φA = φ(A ∩ E) + φ(A \ E) for every A ⊆ X}. Then 413H tells us that Σ is an algebra of subsets of X, including K, and µ = φ¹Σ is an additive functional extending φ0 . T P Set L = T (b) Now µ( n∈N Kn ) = inf n∈N µKn whenever hKn in∈N is a non-increasing sequence in K. P K . Of course µL ≤ inf µK . For the reverse inequality, take ² > 0. Then (α) tells us that there n n∈N n n∈N is a K 0 ∈ K such that K 0 ⊆ K0 \ L and µK0 ≤ µL + µK 0 + ². Since hKn ∩ K 0 in∈N is a non-increasing sequence in K with empty intersection, (β) tells us that there is an n ∈ N such that µ(Kn ∩ K 0 ) ≤ ². Now µK0 − µL = µ(K0 \ L) = µ(K0 \ (K 0 ∪ L)) + µK 0 ≤ ² + µ(Kn ∩ K 0 ) + µ(K 0 \ Kn ) ≤ 2² + µ(K0 \ Kn ) = 2² + µK0 − µKn . (These calculations depend, of course, on the additivity of µ and the finiteness of µK0 .) So µL ≥ µKn − 2². As ² is arbitrary, µL = inf n∈N µKn . Q Q (c) If hAn in∈N is a non-increasing sequence of subsets of X, with intersection A, and φA0 < ∞, then φA = inf n∈N φAn . P P Of course φA ≤ φAn for every n. Given ² > 0, then for each n ∈ N choose Kn ∈ K such that K ⊆ A and φ0 KnT≥ φAn − 2−n ² (this is where I use the hypothesis that φA0 is finite); set n n T Ln = i≤n Ki for each n, L = n∈N Ln . Then we have φAn+1 − µLn+1 = φAn+1 − µ(Kn+1 ∩ Ln ) = φAn+1 − µKn+1 − µLn + µ(Kn+1 ∪ Ln ) ≤ 2−n−1 ² − µLn + φAn because Kn+1 ⊆ An+1 ⊆ An and Ln ⊆ Kn ⊆ An . Inducing on n, we see that µLn ≥ φAn − 2² + 2−n ² for every n. So φA ≥ µL = inf n∈N µLn ≥ inf n∈N φAn − 2²,

38


413I

using (b) above for the middle equality. As ² is arbitrary, φA = inf n∈N φAn . Q Q (d) It follows that φ is an inner measure. P P The arguments of parts (a) and (b) of the proof of 413H tellTus that φ∅ = 0 and φ(A ∪ B) ≤ φA + φB whenever A, B ⊆ X are disjoint. We have just seen that φ( n∈N An ) = inf n∈N φAn whenever hAn in∈N is a non-increasing sequence of sets and φA0 < ∞. Finally, φK = φ0 K is finite for every K ∈ K, so φA = sup{φB : B ⊆ A, φB < ∞} for every A ⊆ X. Putting these together, φ is an inner measure. Q Q (e) So 413C tells us that µ is a complete measure, and of course it is inner regular with respect to K, by the definition of φ. It is semi-finite because µK = φ0 K is finite for every K ∈ K. Now suppose that E ⊆ X and that E ∩ F ∈ Σ whenever µF < ∞. Take any A ⊆ X. If L ∈ K and L ⊆ A, we have L ∈ Σ and µL < ∞, so φ0 L = µL = µ(L ∩ E) + µ(L \ E) = φ(L ∩ E) + φ(L \ E) ≤ φ(A ∩ E) + φ(A \ E); taking the supremum over L, φA ≤ φ(A ∩ E) + φ(A \ E). As A is arbitrary, E ∈ Σ; as E is arbitrary, µ is locally determined. (f ) Finally, µ is unique by 412L. 413J Theorem Let X be a set and K a family of subsets of X such that ∅ ∈ K, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, (‡) K ∩ K 0 ∈ K whenever K, K 0 ∈ K. Let φ0 : K → [0, ∞[ be a functional such that (α) φ0 K = φ0 L + sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K, (β) inf n∈N φ0 Kn = 0 whenever hKn in∈N is a non-increasing sequence in K with empty intersection. Then there is a unique complete locally determined measure µ on X extending φ0 and inner regular with respect to Kδ , the family of sets expressible as intersections of sequences in K. proof (a) Set ψA = sup{φ0 K : K ∈ K, K ⊆ A} for A ⊆ X, T = {E : E ⊆ X, ψA = ψ(A ∩ E) + ψ(A \ E) for every A ⊆ X}. Then 413H tells us that T is an algebra of subsets of X, including K, and ν = ψ¹ T is an additive functional extending φ0 . (b) Write Tf for {E : E ∈ T, νE < ∞}. If hEn in∈N is a non-increasing sequence in Tf with empty intersection, limn→∞ νEn = 0. P P Given ² > 0, we can choose a sequence hKn in∈N in K such that Kn ⊆ En and for each n. Set Ln =

νKn = φ0 Kn ≥ νEn − 2−n ²

T i≤n

Ki for each n; then limn→∞ νLn = limn→∞ φ0 Ln = 0

by hypothesis (β). But also, for each n, νEn ≤ νLn +

Pn

ν(Ei \ Ki ) ≤ νLn + 2², S because ν is additive and non-negative and En ⊆ Ln ∪ i≤n (Ei \ Ki ). So lim supn→∞ νEn ≤ 2²; as ² is arbitrary, limn→∞ νEn = 0. Q Q i=0

(c) Write Tfδ for the family of sets expressible as intersections of sequences in Tf , and for H ∈ Tfδ set φ1 H = inf{νE : H ⊆ E ∈ T}. Note that because E ∩ F ∈ Tf for every E, F ∈ Tf , every member of Tfδ can be expressed as the intersection of a non-increasing sequence in Tf . (i) If hEn in∈N is a non-increasing sequence in Tf with intersection H ∈ Tfδ , φ1 H = limn→∞ νEn . P P Of course

413J


39

φ1 H ≤ inf n∈N νEn = limn→∞ νEn . On the other hand, if H ⊆ E ∈ T, then hEn \ Ein∈N is a non-increasing sequence in Tf with empty intersection, and νE ≥ limn→∞ ν(En ∩ E) = limn→∞ νEn − limn→∞ ν(En \ E) = limn→∞ νEn T by (b) above. As E is arbitrary, φ1 ( n∈N En ) = limn→∞ νEn . Q Q (ii) Because K ⊆ Tf , Kδ ⊆ Tfδ . Now for any H ∈ Tfδ , φ1 H = sup{φ1 L : L ∈ Kδ , L ⊆ H}. P P Express T H as n∈N En where hEn in∈N is a non-increasing sequence in Tf . Given ² > 0, weTcan choose a sequence hKn in∈N in K such that Kn ⊆ En and νKn ≥ νEn − 2−n ² for each n. Setting Ln = i≤n Ki for each n and T L = n∈N Ln , we have L ∈ Kδ , L ⊆ H and Pn φ1 H = limn→∞ νEn ≤ limn→∞ (νLn + i=0 ν(Ei \ Ki )) ≤ φ1 L + 2². As ² is arbitrary, this gives the result. Q Q (d) We find that Tfδ , φ1 satisfy the conditions of 413I. P P Of course ∅ ∈ Tfδ . If G, H ∈ Tfδ and G ∩ H = ∅, T T express them as n∈N En , n∈N Fn where hEn in∈N , hFn in∈N are non-increasing sequences in Tf . Then T G ∪ H = n∈N En ∪ Fn belongs to Tfδ , and φ1 (G ∪ H) = lim ν(En ∪ Fn ) = lim νEn + νFn − ν(En ∩ Fn ) n→∞

n→∞

= lim νEn + νFn n→∞

(by (b)) = φ1 G + φ1 H. The definition of Tfδ as the set of intersections of sequences in Tf ensures that the intersection of any sequence in Tfδ will belong to Tfδ . T T Now suppose that G, H ∈ Tfδ and that G ⊆ H. Express them as intersections n∈N En , n∈N Fn of non-increasing sequences in Tf , so that φ1 G = limn→∞ νEn , φ1 H = limn→∞ νFn . For each n, set T Hn = m∈N Fm \ En , so that Hn ∈ Tfδ , Hn ⊆ H \ G, and φ1 Hn = lim ν(Fm \ En ) = lim νFm − ν(Fm ∩ En ) m→∞

m→∞

≥ lim νFm − νEn = φ1 H − νEn . m→∞

Accordingly sup{φ1 G0 : G0 ∈ Tfδ , G0 ⊆ H \ G} ≥ supn∈N φ1 H − νEn = φ1 H − φ1 G. On the other hand, if G0 ∈ Tfδ and G0 ⊆ H \ G, then φ1 G + φ1 G0 = φ1 (G ∪ G0 ) ≤ φ1 H because of course φ1 is non-decreasing, as well as being additive on disjoint sets. So sup{φ1 G0 : G0 ∈ Tfδ , G0 ⊆ H \ G} = φ1 H − φ1 G as required by condition (α) of 413I. Finally, suppose that hHn in∈N is a non-increasing sequence in Tfδ with empty intersection. For each n ∈ N, let hEni ii∈N be a non-increasing sequence in Tf with intersection T Hn , and set Fm = n≤m Enn for each m. Then hFm im∈N is a non-increasing sequence in Tf with empty intersection, while Hm ⊆ Fm for each m, so limm→∞ φ1 Hm ≤ limm→∞ νFm = 0. Thus condition 413I(β) is satisfied, and we have the full list. Q Q

40


413J

(e) By 413I, we have a complete locally determined measure µ, extending φ1 , and inner regular with respect to Tfδ . Since φ1 K = νK = φ0 K for K ∈ K, µ extends φ0 . If G belongs to the domain of µ, and γ < µG, there is an H ∈ Tfδ such that H ⊆ G and γ < µH = φ1 H; by (c-ii), there is an L ∈ Kδ such that L ⊆ H and γ ≤ φ1 L = µL. Thus µ is inner regular with respect to Kδ . To see that µ is T unique, observe that if µ0 is any other measure with these properties, and L ∈ Kδ , then L is expressible as n∈N Kn where hKn in∈N is a sequence in K. Now T T µL = limn→∞ µ( i≤n Ki ) = limn→∞ φ0 ( i≤n Ki ) = µ0 L. So µ and µ0 must agree on Kδ , and by 412L they are identical. 413K Corollary (a) Let X be a set, Σ a subring of PX, and ν : Σ → [0, ∞[ a non-negative finitely additive functional such that limn→∞ νEn = 0 whenever hEn in∈N is a non-increasing sequence in Σ with empty intersection. Then ν has a unique extension to a complete locally determined measure on X which is inner regular with respect to the family Σδ of intersections of sequences in Σ. (b) Let X be a set, Σ a subalgebra of PX, and ν : Σ → [0, ∞[ a non-negative finitely additive functional such that limn→∞ νEn = 0 whenever hEn in∈N is a non-increasing sequence in Σ with empty intersection. Then ν has a unique extension to a measure defined on the σ-algebra of subsets of X generated by Σ. proof (a) Take Σ, ν in place of K, φ0 in 413J. (b) Let ν1 be the complete extension as in (a), and let ν10 be the restriction of ν1 to the σ-algebra Σ0 generated by Σ; this is the extension required here. To see that ν10 is unique, use the Monotone Class Theorem (136C). Remark You will sometimes see (b) above stated as ‘an additive functional on an algebra of sets extends to a measure iff it is countably additive’. But this formulation depends on a different interpretation of the phrase ‘countably additive’ from the one used in this book; see the note after the definition in 326E. 413L It will be useful to have a definition extending an idea in §342. Definition A countably compact class (or semicompact T paving) is a family K of sets such that T i≤n Ki 6= ∅ for every n ∈ N. n∈N Kn 6= ∅ whenever hKn in∈N is a sequence in K such that 413M Corollary Let X be a set and K a countably compact class of subsets of X such that ∅ ∈ K, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, (‡) K ∩ K 0 ∈ K whenever K, K 0 ∈ K. Let φ0 : K → [0, ∞[ be a functional such that (α) φ0 K = φ0 L + sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K. Then there is a unique complete locally determined measure µ on X extending φ0 and inner regular with respect to Kδ , the family of sets expressible as intersections of sequences in K. proof The point is that the hypothesis (β) of 413J is necessarily satisfied: if hKn in∈N is a non-increasing sequence in K with empty intersection, then, because K is countably compact, there must be some n such that Kn = ∅. Since hypothesis (α) here is already enough to ensure that φ0 ∅ = 0, φ0 K ≥ 0 for every K ∈ K, we must have inf n∈N φ0 Kn = 0. So we apply 413J to get the result. 413N I now turn to constructions of a different kind, being extension theorems in which the extension is not uniquely defined. Again I start with a theorem on finitely additive functionals. Theorem Let X be a set, T0 a subring of PX, and ν0 : T0 → [0, ∞[ a finitely additive functional. Suppose that K ⊆ PX is a family of sets such that (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, (‡) K ∩ K 0 ∈ K for all K, K 0 ∈ K, every member of K is included in some member of T0 ,

413N


41

and ν0 is inner regular with respect to K in the sense that (α) ν0 E = sup{ν0 K : K ∈ K ∩ T0 , K ⊆ E} for every E ∈ T0 . Then ν0 has an extension to a non-negative finitely additive functional ν1 , defined on a subring T1 of PX including T0 ∪ K, inner regular with respect to K, and such that whenever E ∈ T1 , ² > 0 there is an E0 ∈ T0 such that ν1 (E4E0 ) ≤ ². proof (a) Let P be the set of all non-negative additive real-valued functionals ν, defined on subrings of PX, inner regular with respect to K, and such that (∗) whenever E ∈ dom ν, ² > 0 there is an E0 ∈ T0 such that ν(E4E0 ) ≤ ². Order P by extension of functions, so that P is a partially ordered set. (b) It will be convenient to borrow some notation from the theory of countably additive functionals. If T is a subring of PX and ν : T → [0, ∞[ is a non-negative additive functional, set ν ∗ A = inf{νE : A ⊆ E ∈ T},

ν∗ A = sup{νE : A ⊇ E ∈ T}

for every A ⊆ X (interpreting inf ∅ as ∞ if necessary). Now if A ⊆ X and E, F ∈ T are disjoint, ν ∗ (A ∩ (E ∪ F )) = ν ∗ (A ∩ E) + ν ∗ (A ∩ F ), ν∗ (A ∩ (E ∪ F )) = ν∗ (A ∩ E) + ν∗ (A ∩ F ).

P P ν ∗ (A ∩ (E ∪ F )) = inf{νG : G ∈ T, A ∩ (E ∪ F ) ⊆ G} = inf{νG : G ∈ T, A ∩ (E ∪ F ) ⊆ G ⊆ E ∪ F } = inf{ν(G ∩ E) + ν(G ∩ F ) : G ∈ T, A ∩ (E ∪ F ) ⊆ G ⊆ E ∪ F } = inf{νG1 + νG2 : G1 , G2 ∈ T, A ∩ E ⊆ G1 ⊆ E, A ∩ F ⊆ G2 ⊆ F } = inf{νG1 : G1 ∈ T, A ∩ E ⊆ G1 ⊆ E} + inf{νG2 : G2 ∈ T, A ∩ F ⊆ G2 ⊆ F } = ν ∗ (E ∩ A) + ν ∗ (F ∩ A), ν∗ (A ∩ (E ∪ F )) = sup{νG : G ∈ T, A ∩ (E ∪ F ) ⊇ G} = sup{ν(G ∩ E) + ν(G ∩ F ) : G ∈ T, A ∩ (E ∪ F ) ⊇ G} = sup{νG1 + νG2 : G1 , G2 ∈ T, A ∩ E ⊇ G1 , A ∩ F ⊇ G2 } = sup{νG1 : G1 ∈ T, A ∩ E ⊇ G1 } + sup{νG2 : G2 ∈ T, A ∩ F ⊇ G2 } = ν∗ (E ∩ A) + ν∗ (F ∩ A). Q Q (c) The key to the proof is the following fact: if ν ∈ P and M ∈ K, there is a ν 0 ∈ P such that ν 0 extends ν and M ∈ dom ν 0 . P P Set T = dom ν, T0 = {(E ∩ M ) ∪ (F \ M ) : E, F ∈ T}. For H ∈ T0 , set ν 0 H = ν ∗ (H ∩ M ) + ν∗ (H \ M ). Now we have to check the following. (i) T0 is a subring of PX, because if E, F , E 0 , F 0 ∈ T then ((E ∩ M ) ∪ (F \ M )) ∗ ((E 0 ∩ M ) ∪ (F 0 \ M )) = ((E ∗ E 0 ) ∩ M ) ∪ ((F ∗ F 0 ) \ M ) for both the Boolean operations ∗ = 4 and ∗ = ∩. T0 ⊇ T because E = (E ∩ M ) ∪ (E \ M ) for every E ∈ T. (Cf. 312M.) M ∈ T0 because there is some E ∈ T0 such that M ⊆ E, so that M = (E ∩ M ) ∪ (∅ \ M ) ∈ T0 . (ii) ν 0 is finite-valued because if H = (E ∩ M ) ∪ (F \ M ), where E, F ∈ T, then ν 0 H ≤ νE + νF . If H, H ∈ T are disjoint, they can be expressed as (E ∩ M ) ∪ (F \ M ), (E 0 ∩ M ) ∪ (F 0 \ M ) where E, F , E 0 , F 0 belong to T; replacing E 0 , F 0 by E 0 \ E and F 0 \ F if necessary, we may suppose that E ∩ E 0 = F ∩ F 0 = ∅. Now 0

42


413N

ν 0 (H ∪ H 0 ) = ν ∗ ((E ∪ E 0 ) ∩ M ) + ν∗ ((F ∪ F 0 ) ∩ (X \ M )) = ν ∗ (E ∩ M ) + ν ∗ (E 0 ∩ M ) + ν∗ (F ∩ (X \ M )) + ν∗ (F 0 ∩ (X \ M )) (by (b) above) = ν 0H + ν 0H 0. Thus ν 0 is additive. (iii) If E ∈ T, then ν∗ (E \ M ) = sup{νF : F ∈ T, F ⊆ E \ M } = sup{νE − ν(E \ F ) : F ∈ T, F ⊆ E \ M } = sup{νE − νF : F ∈ T, E ∩ M ⊆ F ⊆ E} = νE − inf{νF : F ∈ T, E ∩ M ⊆ F ⊆ E} = νE − ν ∗ (E ∩ M ). So ν 0 E = ν ∗ (E ∩ M ) + ν∗ (E \ M ) = νE. Thus ν 0 extends ν. (iv) If H ∈ T0 and ² > 0, express H as (E ∩ M ) ∪ (F \ M ), where E, F ∈ T. Then we can find (α) a K ∈ K ∩ T such that K ⊆ E and ν(E \ K) ≤ ² (β) an F 0 ∈ T such that F 0 ⊆ F \ M and νF 0 ≥ ν∗ (F \ M ) − ² (γ) a K 0 ∈ K ∩ T such that K 0 ⊆ F 0 and νK 0 ≥ νF 0 − ². Set L = (K ∩ M ) ∪ K 0 ∈ T0 ; by the hypotheses (†) and (‡), L ∈ K. Now L ⊆ H and ν 0 L = ν 0 (K ∩ M ) + ν 0 K 0 = ν 0 (E ∩ M ) − ν 0 ((E \ K) ∩ M ) + νK 0 = ν ∗ (H ∩ M ) − ν ∗ ((E \ K) ∩ M ) + νK 0 ≥ ν ∗ (H ∩ M ) − ν(E \ K) + νF 0 − ² ≥ ν ∗ (H ∩ M ) + ν∗ (F \ M ) − 3² = ν 0 H − 3². As H and ² are arbitrary, ν is inner regular with respect to K. (v) Finally, given H ∈ T0 and ² > 0, take E, F ∈ T such that H ∩ M ⊆ E, F ⊆ H \ M , νE ≤ ν (H ∩ M ) + ² and νF ≥ ν∗ (H \ M ) − ². In this case, ∗

ν 0 (E \ (H ∩ M )) = ν 0 E − ν 0 (H ∩ M ) = νE − ν ∗ (H ∩ M ) ≤ ², ν 0 ((H \ M ) \ F ) = ν 0 (H \ M ) − ν 0 F = ν∗ (H \ M ) − νF ≤ ². But as H4(E ∪ F ) ⊆ (E \ (H ∩ M )) ∪ ((H \ M ) \ F ), ν 0 (H4(E ∪F )) ≤ 2². Now ν satisfies the condition (∗), so there is an E0 ∈ T0 such that ν((E ∪F )4E0 ) ≤ ², and ν 0 (H4E0 ) ≤ 3². As H and ² are arbitrary, ν 0 satisfies (∗). This completes the proof that ν 0 is a member of P extending ν. Q Q (d) It is easy to check that if Q ⊆ P is a non-empty totally ordered subset, the smallest common extension ν 0 of the functions in Q belongs to P . (To see that ν 0 is inner regular with respect to K, observe that if E ∈ dom ν 0 and γ < ν 0 E, there is some ν ∈ Q such that E ∈ dom ν; now there is a K ∈ K ∩ dom ν such that K ⊆ E and νK ≥ γ, so that K ∈ K ∩ dom ν 0 and ν 0 K ≥ γ.) And of course P is not empty, because ν0 ∈ P . So by Zorn’s Lemma P has a maximal element ν1 say; write T1 for the domain of ν1 . If M ∈ K there is an element of P , with a domain containing M , extending ν1 ; as ν1 is maximal, this must be ν1 itself, so M ∈ T1 . Thus K ⊆ T1 , and ν1 has all the required properties.

413P


43

413O Corollary Let (X, Σ0 , µ0 ) be a measure space and K a countably compact class of subsets of X such that (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, T (‡) n∈N Kn ∈ K for every sequence hKn in∈N in K, µ∗0 K < ∞ for every K ∈ K, µ0 is inner regular with respect to K. Then µ0 has an extension to a complete locally determined measure µ, defined on every member of K, inner regular with respect to K, and such that whenever E ∈ dom µ and µE < ∞ there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0. proof (a) Set T0 = {E : E ∈ Σ0 , µ0 E < ∞}, ν0 = µ0 ¹ T0 . Then ν0 , T0 satisfy the conditions of 413N; take ν1 , T1 as in 413N. If K, L ∈ K and L ⊆ K, then ν1 L + sup{ν1 K 0 : K 0 ∈ K, K 0 ⊆ K \ L} = ν1 L + ν1 (K \ L) = ν1 K. So ν1 ¹K satisfies the conditions of 413M and there is a complete locally determined measure µ, extending ν1 ¹K, and inner regular with respect to K. (b) Write Σ for the domain of µ. Then T1 ⊆ Σ. P P If E ∈ T1 and K ∈ K, µ∗ (K ∩ E) + µ∗ (K \ E) ≥ sup{µK 0 : K 0 ∈ K, K 0 ⊆ K ∩ E} + sup{µK 0 : K 0 ∈ K, K 0 ⊆ K \ E} = sup{ν1 K 0 : K 0 ∈ K, K 0 ⊆ K ∩ E} + sup{ν1 K 0 : K 0 ∈ K, K 0 ⊆ K \ E} = ν1 (K ∩ E) + ν1 (K \ E) = ν1 K = µK. By 413F(iv), E ∈ Σ. Q Q It follows at once that µ extends ν1 , since if E ∈ T1 ν1 E = sup{ν1 K : K ∈ K, K ⊆ E} = sup{µK : K ∈ K, K ⊆ E} = µE. (c) In particular, µ agrees with µ0 on T0 . Now in fact µ extends µ0 . P P Take E ∈ Σ0 . If K ∈ K, there is an E0 ∈ Σ0 such that K ⊆ E0 and µ0 E0 < ∞. Since E ∩ E0 ∈ T0 ⊆ Σ, E ∩ K = E ∩ E0 ∩ K ∈ Σ. As K is arbitrary, E ∈ Σ, by 413F(ii). Next, because every member of K is included in a member of T0 , µ0 E = sup{µ0 K : K ∈ K ∩ Σ0 , K ⊆ E} = sup{µ0 (E ∩ E0 ) : E0 ∈ T0 } = sup{µ(E ∩ E0 ) : E0 ∈ T0 } = sup{µK : K ∈ K, K ⊆ E} = µE. Q Q (d) Finally, suppose that E ∈ Σ and µE < ∞. For each n ∈ N P we can find Kn ∈ K, En ∈ Σ0 such that ∞ 0 −n −n Kn ⊆ E, µ(E \ K ) ≤ 2 and ν (K 4E ) ≤ 2 . In this case n 1 n n n=0 µ(En 4E) < ∞, so µ(E4E ) = 0, S T 0 where E = n∈N m≥n Em ∈ Σ0 . Thus µ has all the required properties. 413P I now describe an alternative route to some of the applications of 413N. As before, I do as much as possible in the context of finitely additive functionals. Lemma Let X be a set and K a sublattice of PX containing ∅. Let λ : K → [0, ∞[ be a bounded functional such that λ∅ = 0,

λK ≤ λK 0 whenever K, K 0 ∈ K and K ⊆ K 0 ,

λ(K ∪ K 0 ) + λ(K ∩ K 0 ) ≥ λK + λK 0 for all K, K 0 ∈ K. Then there is a finitely additive functional ν : PX → [0, ∞[ such that νX = supK∈K λK,

νK ≥ λK for every K ∈ K.

0 0 proof (a) The key to the proof isPthe followingPfact: if K0 , . . P . , Kn ∈ K, then Pn there are K0 , . . . , Kn ∈ K n n n 0 0 0 0 0 such that K0 ⊆ K1 ⊆ . . . ⊆ Kn , i=0 χKi = i=0 χKi and i=0 λKi ≥ i=0 λKi . P P Induce on n. If

44


413P

n = 0 the result is trivial. For the inductive step to n + 1, given K0 , . . . , Kn+1 ∈ K, set Ln+1 = Kn+1 and use the inductive hypothesis to find L0 , . . . , Ln ∈ K such that Pn Pn Pn Pn L0 ⊆ . . . ⊆ Ln , i=0 χLi = i=0 χKi and i=0 λLi ≥ i=0 λKi . Now set L0n = Ln+1 ∩ Ln , L0i = Li for i < n, and use the inductive hypothesis again to find K00 , . . . , Kn0 ∈ K such that Pn Pn Pn Pn K00 ⊆ . . . ⊆ Kn0 , i=0 χKi0 = i=0 χL0i and i=0 λKi0 ≥ i=0 λL0i . Set 0 Kn+1 = L0n+1 = Ln ∪ Ln+1 . Pn Then all the Li , L0i and Ki0 belong to K. We know that χKn0 ≤ i=0 χL0i , so S 0 Kn0 ⊆ i≤n L0i ⊆ Ln ⊆ Kn+1 ; 0 accordingly we have K00 ⊆ K10 ⊆ . . . ⊆ Kn0 ⊆ Kn+1 . Next, n+1 X

χKi0 =

i=0

n X

χL0i + χL0n+1 =

i=0

=

n−1 X

n−1 X

χLi + χ(Ln ∩ Ln+1 ) + χ(Ln ∪ Ln+1 )

i=0

χLi + χLn + χLn+1 =

i=0

n+1 X

λKi0 ≥

i=0

n X

≥

χLi + χKn+1 =

i=0

λL0i + λL0n+1 =

i=0 n−1 X

n X

n−1 X

n+1 X

χKi ,

i=0

λLi + λ(Ln ∩ Ln+1 ) + λ(Ln ∪ Ln+1 )

i=0

λLi + λLn + λLn+1

i=0

(using the hypothesis on λ) =

n X

λLi + λKn+1 ≥

n+1 X

i=0

λKi ,

i=0

0 , and the induction continues. Q Q so we have an appropriate family K00 , . . . , Kn+1

(b) For the moment, suppose that supK∈K P λK = 1. In T this case, if hKi ii∈I is a finite indexed family in P If I = ∅ this is trivial, so we may K, there is a set J ⊆ I such that #(J) ≥ i∈I λKi and i∈J Ki 6= ∅. P 0 0 , . . . , K as in (a). If Kn0 = ∅ then every Ki is empty so suppose that I = {0, . . . , n} for some n. Take K n 0 Pn (because λ∅ = 0) i=0 λKi = 0 and we may take J = ∅. TOtherwise, let m be the first number such that 0 0 Km 6= ∅, and take x ∈ Km , J = {i : i ≤ n, x ∈ Ki }. Then i∈J Ki 6= ∅ and #(J) = ≥

n X i=0 n X i=m

χKi (x) =

n X

χKi0 (x) = n − m + 1

i=0

λKi0 =

n X i=0

λKi0 ≥

n X

λKi ,

i=0

as claimed. Q Q By 391F, there is an additive functional ν : PX → [0, 1] such that νX = 1 and νK ≥ λK for every K ∈ K, as required. (c) For the general case, set γ = supK∈K λK. If γ = 0, take ν to be the zero functional; if γ > 0, apply (a)-(b) to the functional γ −1 λ. Remark If P is a lattice, a function f : P → R such that f (p ∨ q) + f (p ∧ q) ≥ f (p) + f (q) for all p, q ∈ P is called supermodular.

413Q


45

413Q Theorem Let X be a set and K a sublattice of PX containing ∅. Let Σ be the algebra of subsets of X generated by K, and ν0 : Σ → [0, ∞[ a finitely additive functional. Then there is a finitely additive functional ν : Σ → [0, ∞[ such that (i) νX = supK∈K ν0 K (ii) νK ≥ ν0 K for every K ∈ K (iii) ν is inner regular with respect to K in the sense that νE = sup{νK : K ∈ K, K ⊆ E} for every E ∈ Σ. proof (a) Set γ = supK∈K ν0 K. Let P be the set of all functionals λ : K → [0, γ] such that λK + λK 0 ≤ λ(K ∪ K 0 ) + λ(K ∩ K 0 ) for every K, K 0 ∈ K. Give P the natural partial ordering inherited from R K . Note that ν0 ¹K belongs to P . If Q ⊆ P is non-empty and upwards-directed, then sup Q, taken in R K , belongs to P ; so there is a maximal λ ∈ P such that ν0 ¹K ≤ λ. By 413P, there is a non-negative additive functional ν on PX such that νK ≥ λK for every K ∈ K and νX = γ. Since ν¹K also belongs to P , we must have νK = λK for every K ∈ K. (b) Now for any K0 ∈ K, νK0 + sup{νL : L ∈ K, L ⊆ X \ K0 } = γ. P P (i) Set L = {L : L ∈ K, L ⊆ X \ K0 }. For A ⊆ X, set θ0 A = supL∈L ν(A ∩ L). Because L is upwards-directed, θ0 : PX → R is additive, and of course 0 ≤ θ0 ≤ ν. Set θ1 = ν − θ0 , so that θ1 is another additive functional, and write λ0 K = θ0 K + sup{θ1 M : M ∈ K, M ∩ K0 ⊆ K} for K ∈ K. (ii) If K, K 0 ∈ K and ² > 0, there are M , M 0 ∈ K such that M ∩ K0 ⊆ K, M 0 ∩ K0 ⊆ K 0 and θ0 K + θ1 M ≥ λ0 K − ²,

θ0 K 0 + θ1 M 0 ≥ λ0 K 0 − ².

Now M ∪ M 0 , M ∩ M 0 ∈ K, (M ∪ M 0 ) ∩ K0 ⊆ K ∪ K 0 ,

(M ∩ M 0 ) ∩ K0 ⊆ K ∩ K 0 ,

so λ0 (K ∪ K 0 ) + λ0 (K ∩ K 0 ) ≥ θ0 (K ∪ K 0 ) + θ1 (M ∪ M 0 ) + θ0 (K ∩ K 0 ) + θ1 (M ∩ M 0 ) = θ0 K + θ1 M + θ0 K 0 + θ1 M 0 ≥ λ0 K + λ0 K 0 − 2². As ² is arbitrary, λ0 (K ∪ K 0 ) + λ0 (K ∩ K 0 ) ≥ λK + λK 0 . (iii) Suppose that K, M ∈ K are such that M ∩ K0 ⊆ K. If L ∈ L, then ν(K ∩ L) + θ1 M = ν(K ∩ L) + νM − θ0 M = ν(M ∩ K ∩ L) + ν(M ∪ (K ∩ L)) − θ0 M ≤ γ because K ∩ L ∈ L; taking the supremum over L and M , λ0 K ≤ γ. As K is arbitrary, λ0 ∈ P . (iv) If K ∈ K, then of course K ∩ K0 ⊆ K, so λ0 K ≥ θ0 K + θ1 K = νK = λK. Thus λ0 ≥ λ. Because λ is maximal, λ0 = λ. But this means that λK0 = λ0 K0 = θ0 K0 + sup{θ1 M : M ∈ K, M ∩ K0 ⊆ K0 } = supM ∈K θ1 M . Now given ² > 0 there is an M ∈ K such that γ − ² ≤ ν0 M ≤ λM = νM , so that νK0 = λK0 ≥ θ1 M = νM − θ0 M ≥ γ − ² − θ0 M ≥ γ − ² − supL∈L νL,

46


413Q

and νK0 + supL∈L νL ≥ γ − ². As ² is arbitrary, νK0 + supL∈L νL ≥ γ. But of course νK0 + νL ≤ νX = γ for every L ∈ L, so νK0 + supL∈L νL = γ, as claimed. Q Q (c) It follows that if K, L ∈ K and L ⊆ K, νK = νL + sup{νK 0 : K 0 ∈ K, K 0 ⊆ K \ L}. P P Because ν is additive and non-negative, we surely have νK ≥ νL + sup{νK 0 : K 0 ∈ K, K 0 ⊆ K \ L}. On the other hand, given ² > 0, there is an M ∈ K such that M ⊆ X \ L and νL + νM ≥ γ − ², so that M ∩ K ∈ K, M ∩ K ⊆ K \ L and νL + ν(M ∩ K) = νL + νK + νM − ν(M ∪ K) ≥ νK + γ − ² − γ = νK − ². As ² is arbitrary, νK ≤ νL + sup{νK 0 : K 0 ∈ K, K 0 ⊆ K \ L} and we have equality. Q Q (d) By 413H, we have an additive functional ν 0 : Σ → [0, ∞[ such that ν 0 E = sup{νK : K ∈ K, K ⊆ E} for every E ∈ Σ. It is easy to show that ν 0 and ν must agree on Σ, but even without doing so we can see that ν 0 has the properties (i)-(iii) required in the theorem. 413R The following lemma on countably compact classes, corresponding to 342Db, will be useful. Lemma (Marczewski 53) Let K be a countably compact class of sets. Then there is a countably compact T class K∗ ⊇ K such that K ∪ L ∈ K∗ and n∈N Kn ∈ K∗ whenever K, L ∈ K∗ and hKn in∈N is a sequence in K∗ . proof (a) Write Ks for {K0 ∪ . .T . ∪ Kn : K0 , . . . , Kn ∈ K}. Then Ks is countably compact. P P Let hLn iS n∈N 6 ∅ for each n ∈ N. Then there is an ultrafilter F on X = K be a sequence in Ks such that i≤n Li = containing every Ln . For each n, T Ln is a finite union of members T of K, so thereTmust be a Kn ∈ K such that Kn ⊆ Ln and Kn ∈ F. Now i≤n Ki 6= ∅ for every n, so n∈N Kn 6= ∅ and n∈N Ln 6= ∅. As hLn in∈N is arbitrary, Ks is countably compact. Q Q Note that L ∪ L0 ∈ Ks for all L, L0 ∈ Ks . (b) Write K∗ for

T

{

L0 : L0 ⊆ Ks is non-empty and countable}.

T Then K∗ is countably compact. P P If hMn in∈N is any sequence in K∗ such that i≤n Mi 6= ∅ for every n ∈ N, T then for each n ∈ N letSLn ⊆ Ks be aTcountable non-empty set such = Ln . Let hLn in∈N be a T that Mn T sequence running over n∈N Ln ; then i≤n Li 6= ∅ for every n, so n∈N Ln = n∈N Mn is non-empty. As hMn in∈N is arbitrary, K∗ is countably compact. Q Q (c) Of course K ⊆ Ks ⊆ K∗ . It is immediate from the definition of K∗ that it is closed under T countable ∗ intersections. Finally, if M , M ∈ K , let L , L ⊆ K be countable sets such that M = 1 2 1 2 s 1 T T L1 and M2 = L2 ; then L = {L1 ∪ L2 : L1 ∈ L1 , L2 ∈ L2 } is a countable subset of Ks , so M1 ∪ M2 = L belongs to K∗ . 413S Corollary Let X be a set and K a countably compact class of subsets of X. Let T be a subalgebra of PX and ν : T → R a non-negative finitely additive functional. (a) There is a complete measure µ on X such that µX ≤ νX, K ⊆ dom µ and µK ≥ νK for every K ∈ K ∩ T. (b) If (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K, T (‡) n∈N Kn ∈ K for every sequence hKn in∈N in K, we may arrange that µ is inner regular with respect to K.

413Xj


47

proof By 413R, there is always a countably compact class K∗ ⊇ K satisfying (†) and (‡); for case (b), take K∗ = K. By 391G, there is an extension of ν to a finitely additive functional ν 0 : PX → R. Let T1 be the subalgebra of PX generated by K∗ . By 413Q, there is a non-negative additive functional ν1 : T1 → R such that ν1 X ≤ ν 0 X = νX, ν1 K ≥ ν 0 K = νK for every K ∈ K∗ ∩ T and ν1 E = sup{ν1 K : K ∈ K∗ , K ⊆ E} for every E ∈ T1 . In particular, if K, L ∈ K∗ , ν1 L + sup{ν1 K 0 : K 0 ∈ K∗ , K 0 ⊆ K \ L} = ν1 L + ν1 (K \ L) = ν1 K. So K∗ and ν1 ¹K∗ satisfy the hypotheses of 413M. Accordingly we have a complete measure µ extending ν1 ¹K∗ and inner regular with respect to K∗ = Kδ∗ ; in which case µK = ν1 K ≥ νK for every K ∈ K ∩ T, µX = supK∈K∗ µK = supK∈K∗ ν1 K ≤ ν1 X ≤ νX, as required. 413X Basic exercises (a) Define φ : PN → [0, ∞[ by setting φA = 0 if A is finite, ∞ otherwise. Check that φ satisfies conditions (α) and (β) of 413A, but that if we attempt to reproduce the construction of 413C then we obtain Σ = PN, µ = φ, so that µ is not countably additive. (b) Let φ1 , φ2 be two inner measures on a set X, inducing measures µ1 , µ2 by the method of 413C. (i) Show that φ = φ1 + φ2 is an inner measure. (ii) Show that the measure µ induced by φ extends the measure µ1 + µ2 defined on dom µ1 ∩ dom µ2 . > (c) Let X be a set, φ an inner measure on X, and µ the measure constructed from it by the method of 413C. (i) Let Y be a subset of X. Show that φ¹ PY is an inner measure on Y , and that the measure on Y defined from it extends the subspace measure µY induced on Y by µ. (ii) Let Y be a set and f : X → Y a function. Show that B 7→ φf −1 [B] is an inner measure on Y , and that it defines a measure on Y which extends the image measure µf −1 . (d) Let (X, Σ, µ) be any measure space. Set θA = 21 (µ∗ A + µ∗ A) for every A ⊆ X. Show that θ is an outer measure on X, and that the measure defined from θ by Carathéodory’s method extends µ. S > (e) Show that there is a partition hAn in∈N of [0, 1] such that µ∗ ( i≤n Ai ) = 0 for every n, where µ∗ is Lebesgue inner measure. (Hint: set An = (A + qn ) ∩ [0, 1] where hqn in∈N is an enumeration of Q and A is a suitable set; cf. 134B.) (f ) Let (X, Σ, µ) be a measure space. (i) Show that µ∗ ¹Σ is the semi-finite version µsf of µ as constructed in 213Xc. (ii) Show that if A is any subset of X, and ΣA the subspace σ-algebra, then µ∗ ¹ΣA is a semi-finite measure on A. > (g) Let (X, Σ, µ) and (Y, T, ν) be two measure spaces, and λ the c.l.d. product measure on X × Y . Show that λ∗ (A × B) = µ∗ A · ν∗ B for all A ⊆ X and B ⊆ Y . (Hint: use Fubini’s theorem to show that λ∗ (A × B) ≤ µ∗ A · ν∗ B.) (h) Let X be a set and µ, ν two complete locally determined measures on X with domains Σ, T respectively, both inner regular with respect to K ⊆ Σ ∩ T. Suppose that, for K ∈ K, µK = 0 iff νK = 0. Show that Σ = T and that µ and ν give rise to the same negligible sets. > (i) Let (X, T, ν) be a measure space. (i) Show that the measure constructed by the method of 413C from the inner measure ν∗ is the c.l.d. version of ν. (ii) Set K = {E : E ∈ T, νE < ∞}, φ0 = ν¹K. Show that K, φ0 satisfy the conditions of 413I, and that the measure constructed by the method there is again the c.l.d. version of ν. (j) Let (X, Σ, µ) be a complete locally determined measure space and L a family of subsets of X such that µ is inner regular with respect to L. Set K = {K : K ∈ L ∩ Σ, µK < ∞} and φ0 = µ¹K. Show that K, φ0 satisfy the conditions of 413I and that the measure constructed from K, φ0 by the method there is just µ.

48


413Xk

> (k) Let K be the family of subsets of R expressible as finite unions of bounded closed intervals. (i) Show from first principles that there is a unique functional φ0 : K → [0, ∞[ such that φ0 [α, β] = β − α whenever α ≤ β and φ0 satisfies the conditions of 413J. (ii) Show that the measure on R constructed from φ0 by the method of 413J is Lebesgue measure. (l) Let X be a set, Σ a subring of PX, and ν : Σ → [0, ∞[ a non-negative additive functional such that limn→∞ νEn = 0 whenever hEn in∈N is a non-increasing sequence in Σ with empty intersection, as in 413K. Define θ : PX → [0, ∞] by setting P∞ θA = inf{ n=0 νEn : hEn in∈N is a sequence in Σ covering A} for A ⊆ X, interpreting inf ∅ as ∞ if necessary. Show that θ is an outer measure. Let µθ be the measure defined from θ by Carathéodory’s method. Show that the measure defined from ν by the process of 413K is the c.l.d. version of µθ . (Hint: the c.l.d. version of µθ is inner regular with respect to Σδ .) > (m) Let X be a set, Σ a subring of PX, and ν : Σ → [0, ∞[ a non-negative additive functional. Show that the following are equiveridical: (i) ν has an extension to a measure on X; n = 0 whenever S (ii) limn→∞ PνE ∞ hEn in∈N is a non-increasing sequence in Σ withSempty intersection; (iii) ν( n∈N En ) = n=0 νEn whenever hEn in∈N is a disjoint sequence in Σ such that n∈N En ∈ Σ. (n) Let h(X Q n , Σn , µn )in∈N be a sequence of probability spaces, and F a non-principal ultrafilter on N. equivalence relation; For x, y ∈ n∈N Xn , write x ∼ y if {n : x(n) = y(n)} ∈ F. (i) Show that ∼ is an Q (Compare write X for the set of equivalence classes, and x• ∈ X for the equivalence class of x ∈ n∈N Xn .Q 351M.) (ii) Let Σ be the set of subsets of X expressible in the form Q(hEn in∈N ) = {x• : x ∈ n∈N En }, where En ∈ Σn for each n ∈ N. Show that Σ is an algebra of subsets of X, and that there is a well-defined additive functional ν : Σ → [0, 1] defined by setting ν(Q(hEn in∈N )) = limn→F T µn En . (iii) Show that for any non-decreasing sequence hHi ii∈N in Σ there is an H ∈ Σ such that H ⊆ i∈N Hi and νH = limn→∞ νHn . (Hint: express each Hi as Q(hEin in∈N ). Do this in such a way that Ei+1,n ⊆ Ein for all i, n. Take a decreasing sequence hJi ii∈N in F, with empty intersection, such that νHi ≤ µEin + 2−i for n ∈ Ji . Set En = Ein for n ∈ Ji \ Ji+1 .) (iv) Show that there is a unique extension of ν to a complete probability measure µ on X which is inner regular with respect to Σ. (This is a kind of Loeb measure.) (o) Let A be a Boolean algebra and K ⊆ A a sublattice containing 0. Suppose that λ : K → [0, ∞[ is a bounded functional such that λ0 = 0, λa ≤ λa0 whenever a, a0 ∈ K and a ⊆ a0 , and λ(a ∪ a0 ) + λ(a ∩ a0 ) ≥ λa + λa0 for all a, a0 ∈ K. Show that there is a non-negative additive functional ν : A → R such that νa ≥ λa for every a ∈ K and ν1 = supa∈K λa. (p) Let X be a set and K a sublattice of PX containing ∅. Let λ : K → R be a bounded order-preserving function such that λ∅ = 0 and λ(K ∪ K 0 ) + λ(K ∩ K 0 ) = λK + λK 0 for all K, K 0 ∈ K. Show that there is a non-negative additive functional ν : PX → R extending λ. (Hint: for finite K, induce on #(K). For the general case, recall that if γ = supK∈K λK, the additive functionals form a compact subset of [0, γ]PX .) (If P is a lattice, a functional f : P → R such that f (p ∨ q) + f (p ∧ q) = f (p) + f (q) for all p, q ∈ P is called modular.) (q) Let X be a set and K a sublattice of PX containing K. Let λ : K → [0, 1] be a functional such that λK ≤ λK 0 whenever K, K 0 ∈ K and K ⊆ K 0 ,

inf K∈K λK = 0,

λ(K ∪ K 0 ) + λ(K ∩ K 0 ) ≤ λK + λK 0 for all K, K 0 ∈ K (λ is submodular). Show that there is a finitely additive functional ν : PX → [0, 1] such that νX = supK∈K λK,

νK ≤ λK for every K ∈ K.

413Y Further exercises (a) Give an example of two inner measures φ1 , φ2 on a set X such that the measure defined by φ1 + φ2 strictly extends the sum of the measures defined by φ1 , φ2 .

413 Notes


49

Q (b) Let h(XQ be any family of probability spaces, and λ the product measure on X = i∈I Xi . i , Σi , µi )ii∈IQ Show that λ∗ ( i∈I Ai ) ≤ i∈I (µi )∗ Ai whenever Ai ⊆ Xi for every i, with equality if I is countable. (c) Let (X, Σ, µ) be a totally finite measure space, and Z the Stone space of the Boolean algebra Σ. For b for the corresponding open-and-closed subset of Z. Show that there is a unique function E ∈ Σ write E b = E for every E ∈ Σ. Show that there is a measure ν on Z, inner regular with f : X → Z such that f −1 [E] respect to the open-and-closed sets, such that f is inverse-measure-preserving with respect to µ and ν, and that f represents an isomorphism between the measure algebras of µ and ν. Use this construction to prove (vi)⇒(i) in Theorem 343B without appealing to the Lifting Theorem. (d) Let X be a set, T a subalgebra of PX, and ν : T → [0, ∞[ a finitely additive functional. Suppose that there is a set K ⊆ T, containing ∅, such that (i) µF = sup{µK : K ∈ K, K ⊆ F } for every F ∈ T (ii) T K is monocompact, that is, n∈N Kn 6= ∅ for every non-decreasing sequence in K. Show that ν extends to a measure on X. (e) (i) Let X be a topological space. Show that the family of closed countably compact subsets of X is a countably compact class. (ii) Let X be a Hausdorff space. Show that the family of sequentially compact subsets of X is a countably compact class. 413 Notes and comments I gave rather few methods of constructing measures in the first three volumes of this treatise; in the present volume I shall have to make up for lost time. In particular I used Carathéodory’s construction for Lebesgue measure (Chapter 11), product measures (Chapter 25) and Hausdorff measures (Chapter 26). The first two, at least, can be tackled in quite different ways if we choose. The first alternative approach I offer is the ‘inner measure’ method of 413C. Note the exact definition in 413A; I do not think it is an obvious one. In particular, while (α) seems to have something to do with subadditivity, and (β) is a kind of sequential order-continuity, there is no straightforward way in which to associate an outer measure with an inner measure, unless they both happen to be derived from measures (132B, 413D), even when they are finite-valued; and for an inner measure which is allowed to take the value ∞ we have to add the semi-finiteness condition (∗) of 413A (see 413Xa). Once we have got these points right, however, we have a method which rivals Carathéodory’s in scope, and in particular is especially well adapted to the construction of inner regular measures. As an almost trivial example, we have a route to the c.l.d. version of a measure µ (413Xi(i)), which can be derived from the inner measure µ∗ defined from µ (413D). Henceforth µ∗ will be a companion to the familiar outer measure µ∗ , and many calculations will be a little easier with both available, as in 413D-413F. The intention behind 413I-413J is to find a minimal set of properties of a functional φ0 which will ensure that it has an extension to a measure. Indeed it is easy to see that, in the context of 413I, given a family K with the properties (†) and (‡) there, a functional φ0 on K can have an extension to an inner regular measure iff it satisfies the conditions (α) and (β), so in this sense 413I is the best possible result. Note that while Carathéodory’s construction is liable to produce wildly infinite measures (like Hausdorff measures, or primitive product measures), the construction here always gives us locally determined measures, provided only that φ0 is finite-valued. We have to work rather hard for the step from 413I to 413J. Of course 413I is a special case of 413J, and I could have saved a little space by giving a direct proof of the latter result. But I do not think that this would have made it easier; 413J really does require an extra step, because somehow we have to extend the functional φ0 from K to Kδ . The method I have chosen uses 413B and 413H to cast as much of the argument as possible into the context of algebras of sets with additive functionals, where I hope the required manipulations will seem natural. (But perhaps I should insist that you must not take them too much for granted, as some of the time we have a finitely additive functional taking infinite values, and must take care not to subtract illegally, as well as not to take limits in the wrong places.) Note that the progression φ0 → φ1 → µ in the proof of 413J involves first an approximation from outside (if K ∈ Kδ , then φ1 K will be inf{φ0 K 0 : K ⊆ K 0 ∈ K}) and then an approximation from inside (if E ∈ Σ, then µE = sup{φ1 K : K ∈ Kδ , K ⊆ E}). The essential difficulty in the proof is just that we have to take successive non-exchangeable limits. I have slipped 413K in as a corollary of 413J; but it can be regarded as one of the fundamental results of measure theory. A non-negative finitely additive functional ν on an algebra Σ of sets can be extended to a countably additive

50


413 Notes

S P∞ measure iff it is ‘relatively countably additive’ in the sense that ν( n∈N En ) = n=0 µEn whenever hEn in∈N S is a disjoint sequence in Σ such that n∈N En ∈ Σ (413Xm). Of course the same result can easily be got from an outer measure construction (413Xl). Note that the outer measure construction also has repeated limits, albeit simpler ones: in the formula P∞ θA = inf{ n=0 νEn : hEn in∈N is a sequence in Σ covering A} P∞ Pn the sum n=0 νEn = supn∈N i=0 νEi can be regarded as a crude approximation from inside, while the infimum is an approximation from outside. To get the result as stated in 413K, of course, the outer measure construction needs a third limiting process, to obtain the c.l.d. version automatically provided by the inner measure method, and the inner regularity with respect to Σδ , while easily checked, also demands a few words of argument. Many applications of the method of 413I-413J pass through 413M; if the family K is a countably compact class then the sequential order-continuity hypothesis (β) of 413I or 413J becomes a consequence of the other hypotheses. The essence of the method is the inner regularity hypothesis (α). I have tried to use the labels †, ‡, α and β consistently enough to suggest the currents which I think are flowing in this material. In 413N we strike out in a new direction. The object here is to build an extension which is not going to be unique, and for which choices will have to be made. As with any such argument, the trick is to specify the allowable intermediate stages, that is, the partially ordered set P to which we shall apply Zorn’s Lemma. But here the form of the theorem makes it easy to guess what P should be: it is the set of functionals satisfying the hypotheses of the theorem which have not wandered outside the boundary set by the conclusion, that is, which satisfy the condition (∗) of part (a) of the proof of 413N. The finitistic nature of the hypotheses makes it easy to check that totally ordered subsets of P have upper bounds (that is to say, if we did this by transfinite induction there would be no problem at limit stages), and all we have to prove is that maximal elements of P are defined on adequately large domains; which amounts to showing that a member of P not defined on every element of K has a proper extension, that is, setting up a construction for the step to a successor ordinal in the parallel transfinite induction (part (c) of the proof). Of course the principal applications of 413N in this book will be in the context of countably additive functionals, as in 413O. It is clear that 413N and 413Q overlap to some extent. I include both because they have different virtues. 413N can be applied to infinite measures in a way that 413Q, as given, cannot; but its chief advantage, from the point of view of the work to come, is the approximation of members of T1 , in measure, by members of T0 . This will eventually enable us to retain control of the Maharam types of measures constructed by the method of 413O. In 413S we have a different kind of control; we can specify a lower bound for the measure of each member of our basic class K, provided only that our specifications are consistent with some finitely additive functional.

414 τ -additivity The second topic I wish to treat is that of ‘τ -additivity’. Here I collect results which do not depend on any strong kind of inner regularity. I begin with what I think of as the most characteristic feature of τ -additivity, its effect on the properties of semi-continuous functions (414A), with a variety of corollaries, up to the behaviour of subspace measures (414K). A very important property of τ -additive topological measures is that they are often strictly localizable (414J). The theory of inner regular τ -additive measures belongs to the next section, but here I give two introductory results: conditions under which a τ -additive measure will be inner regular with respect to closed sets (414M) and conditions under which a measure which is inner regular with respect to closed sets will be τ -additive (414N). I end the section with notes on ‘density’ and ‘lifting’ topologies (414P-414R). 414A Theorem Let (X, T) be a topological space and µ an effectively locally finite τ -additive measure on X with domain Σ. S (a) Suppose that G is a non-empty family in Σ ∩ T such that H = G also belongs to Σ. Then supG∈G G• = H • in the measure algebra A of µ.

414D

τ -additivity

51

(b) Write L for the family of Σ-measurable lower semi-continuous functions from X to R. Suppose that ∅ 6= A ⊆ L and set g(x) = supf ∈A f (x) for every x ∈ X. If g is Σ-measurable and finite almost everywhere, then g˜• = supf ∈A f • in L0 (µ), where g˜(x) = g(x) whenever g(x) is finite. T • T(c) •Suppose that F is a non-empty family of measurable closed sets such that F ∈ Σ. Then inf F ∈F F = ( F) in A. (d) Write U for the family of Σ-measurable upper semi-continuous functions from X to R. Suppose that A ⊆ U is non-empty and set g(x) = inf f ∈A f (x) for every x ∈ X. If g is Σ-measurable and finite almost everywhere, then g˜• = inf f ∈A f • in L0 (µ), where g˜(x) = g(x) whenever g(x) is finite. proof (a) ?? If H • 6= supG∈G G• , there is a non-zero a ∈ A such that a ⊆ H • but a ∩ G• = 0 for every G ∈ G. Express a as E • where E ∈ Σ and E ⊆ H. Because µ is effectively locally finite, there is a measurable open set H0 of finite measure such that µ(H0 ∩ E) > 0. Now {H0 ∩ G : G ∈ G} is an upwards-directed family of measurable open sets with union H0 ∩ H ⊇ H0 ∩ E; as µ is τ -additive, there is a G ∈ G such that µ(H0 ∩ G) > µH0 − µ(H0 ∩ E). But in this case µ(G ∩ E) > 0, which is impossible, because G• ∩ E • = 0. X X (b) For any α ∈ R, {x : g(x) > α} =

S

f ∈A {x

: f (x) > α},

and these are all measurable open sets. Identifying {x : g(x) > α}• ∈ A with [[˜ g • > α]] (364Jb), we see from • • • (a) that [[˜ g > α]] = supf ∈A [[f > α]] for every α. But this means that g˜ = supf ∈A f • , by 364Mb. (c) Apply (a) to G = {X \ F : F ∈ F}. (d) Apply (b) to {−f : f ∈ A}. 414B Corollary Let X be a topological space and µ an effectively locally finite τ -additive topological measure on X. (a) Suppose that A is a non-empty upwards-directed family of lower semi-continuous functions from X R R to [0, ∞]. Set g(x) = supf ∈A f (x) in [0, ∞] for every x ∈ X. Then g = supf ∈A f in [0, ∞]. (b) Suppose that A is a non-empty downwards-directed family of non-negative continuous real-valued functions on RX, and that g(x) = inf x∈A f (x) for every x ∈ X. If any member of A is integrable, then R g = inf f ∈A f . proof (a) Of course all the f ∈ A, and also g, are measurable functions. Set gn = g ∧ nχX for every n ∈ N. Then gn (x) = supf ∈A (f ∧ nχX)(x) for every x ∈ X, so gn = supf ∈A (f ∧ nχX)• , by 414Ab, and •

R

gn =

R

gn• = supf ∈A

by 365Dh. But now, of course,

R

g = supn∈N

R

R

(f ∧ nχX)• = supf ∈A

gn = supn∈N,f ∈A

R

R

f ∧ nχX

f ∧ nχX = supf ∈A

R

f,

as claimed. (b) Take an integrable f0 ∈ A, and apply (a) to {(f0 − f )+ : f ∈ A}. 414C Corollary Let (X, T, Σ, µ) be an effectively locally finite τ -additive topological measure spaceTand F a non-empty downwards-directed family of closed sets. If inf F ∈F µF is finite, this is the measure of F. T proof Setting F0 = F, then F0• = inf F ∈F F • , by 414Ac; now µF0 = µ ¯F0• = inf F ∈F µ ¯F • = inf F ∈F µF by 321F. 414D Corollary Let µ be an effectively locally finite τ -additive measure on a topological space X. If ν is a totally finite measure with the same domain as µ, truly continuous with respect to µ, then ν is τ -additive. In particular, if µ is σ-finite and ν is absolutely continuous with respect to µ, then ν is τ -additive.

52


414D

proof We have a functional ν¯ : A → [0, ∞[, where A is the measure algebra of µ, such that ν¯E • = νE for every E in the common domain Σ of µ and ν. Now ν¯ is continuous for the measure-algebra topology of A (327Cd), therefore completely additive (327Ba), therefore order-continuous (326Kc). So if G is an upwards-directed family of open sets belonging to Σ with union G0 ∈ Σ, supG∈G νG = supG∈G ν¯G• = ν¯G•0 = νG0 because G•0 = supG∈G G• . The last sentence follows at once, because on a σ-finite space an absolutely continuous countably additive functional is truly continuous (232Bc). 414E Corollary Let (X, T, Σ, µ) be an effectively locally finiteS τ -additive topological measure space. Suppose that G ⊆ T is non-empty and upwards-directed, and H = G. Then (a) µ(E ∩ H) = supG∈G µ(E ∩ G) for every E ∈ Σ; virtually measurable real-valued function defined almost everywhere on X, then R (b) if f is a non-negative R f = sup f in [0, ∞]. G∈G H proof (a) In the measure algebra (A, µ ¯) of µ, (E ∩ H)• = E • ∩ H • = E • ∩ sup G• G∈G

= sup E ∩ G = sup (E ∩ G)• , •

•

G∈G

G∈G

using 414Aa and the distributive law 313Ba. So µ(E ∩ H) = µ ¯(E ∩ H)• = supG∈G µ ¯(E ∩ G)• = supG∈G µ(E ∩ G) by 321D, because G and {(E ∩ G)• : G ∈ G} are upwards-directed. (b) For each G ∈ G,

R G

f=

R

f × χG =

R

(f × χG)• =

R

f • × χG• ,

where χG• can be interpreted either as (χG)• (in L0 (µ)) or as χ(G• ) (in L0 (A), where A is the measure algebra of µ); see 364K. Now H • = supG∈G G• (414Aa); since χ and × are order-continuous (364Kc, 364P), f • × χH • = supG∈G f • × χG• ; so

R

H

f=

R

f • × χH • = supG∈G

R

f • × χG• = supG∈G

R

G

f

by 365Dh. 414F Corollary Let (X, T, Σ, µ) be an effectively locally finite τ -additive topological measure space. Then for every E ∈ Σ there is a unique relatively closed self-supporting set F ⊆ E such that µ(E \ F ) = 0. proof Let G be the set {G : G S∈ T, µ(G ∩ E) = 0}. Then G is upwards-directed, so µ(E ∩ G∗ ) = supG∈G µ(E ∩ G) = 0, where G∗ = G. Set F = E \ G∗ . Then F ⊆ E is relatively closed, and µ(E \ F ) = 0. If H ∈ T and H ∩ F 6= ∅, then H ∈ / G so µ(F ∩ H) = µ(E ∩ H) > 0; thus F is self-supporting. If F 0 ⊆ E is another self-supporting relatively closed set such that µ(E \ F 0 ) = 0, then µ(F \ F 0 ) = µ(F 0 \ F ) = 0; but as F \ F 0 is relatively open in F , and F 0 \ F is relatively open in F 0 , these must both be empty, and F = F 0 . 414G Corollary If (X, T, Σ, µ) is a Hausdorff effectively locally finite τ -additive topological measure space and E ∈ Σ is an atom for µ (definition: 211I), then there is an x ∈ E such that E \ {x} is negligible. proof Let F ⊆ E be a self-supporting set such that µ(E \ F ) = 0. Since µF = µE > 0, F is not empty; take x ∈ F . ?? If F 6= {x}, let y ∈ F \ {x}. Because T is Hausdorff, there are disjoint open sets G, H containing x, y respectively; and in this case µ(E ∩ G) = µ(F ∩ G) and µ(E ∩ H) = µ(F ∩ H) are both non-zero, which is impossible, since E is an atom. X X So F = {x} and E \ {x} is negligible. 414H Corollary If (X, T, Σ, µ) is an effectively locally finite τ -additive topological measure space and ν is an indefinite-integral measure over µ (definition: 234B), then ν is a τ -additive topological measure.

414K

τ -additivity

53

proof Because ν measures every set in Σ (234Da), it is a topological measure. To see that it is τ -additive, apply 414Eb to a Radon-Nikod´ ym derivative of ν. 414I Proposition Let (X, T, Σ, µ) be a complete locally determined effectively locally finite τ -additive S topological measure space. If E ⊆ X, G ⊆ T are such that E ⊆ G and E ∩ G ∈ Σ for every G ∈ G, then E ∈ Σ. proof Set K = {K : K ∈ Σ, E ∩ K S ∈ Σ}. Then whenever F ∈ Σ and µF > 0 there is a K ∈ K included in F with µK > 0. P P Set K1 = F \ G. Then K1 is a member of K included in F . If µK1 > 0 then we can stop. Otherwise, G ∗ = {G0 ∪ . . . ∪ Gn : G0 , . . . , Gn ∈ G} is an upwards-directed family of open sets, and S supG∈G ∗ µ(F ∩ G) = µ(F ∩ G ∗ ) = µF > 0, by 414E. So there is a G ∈ G ∗ such that µ(F ∩ G) > 0; but now E ∩ G ∈ Σ so F ∩ G ∈ K. Q Q By 412Aa, µ is inner regular with respect to K; by 412Ja, E ∈ Σ. 414J Theorem Let (X, T, Σ, µ) be a complete locally determined effectively locally finite τ -additive topological measure space. Then µ is strictly localizable. proof Let F be a maximal disjoint family of self-supporting measurable sets of finite measure. Then whenever E ∈ Σ and µE > 0, there is an F ∈ F such that µ(E ∩ F ) > 0. P P?? Otherwise, let G be an open set of finite measure such that µ(G ∩ E) > 0, and set F0 = {F : F ∈ F, F ∩SG 6= ∅}. Then µ(F ∩ G) S> 0 0 for every F ∈ F0 , while µG < ∞ and F is disjoint, so F is countable and F ∈ Σ. Set E = E \ F0 ; 0 0 0 S then E \ E 0 = E ∩ F0 is negligible, so µ(G ∩ E 0 ) > 0. By 414F, there is a self-supporting set F 0 ⊆ G ∩ E 0 such that µF 0 > 0. But in this case F 0 ∩ F = ∅ for every F ∈ F, so we ought to have added F 0 to F. X XQ Q This means that F satisfies the criterion of 213O. Because (X, Σ, µ) is complete and locally determined, it is strictly localizable. 414K Proposition Let (X, Σ, µ) be a measure space and T a topology on X, and Y ⊆ X a subset such that the subspace measure µY is semi-finite (see the remark following 412O). If µ is an effectively locally finite τ -additive topological measure, so is the subspace measure µY . proof By 412P, µY is an effectively locally finite topological measure. Now suppose that H is a non-empty upwards-directed family in TY with union H ∗ . Set S G = {G : G ∈ T, G ∩ Y ∈ H}, G∗ = G, so that G is upwards-directed and H ∗ = Y ∩G∗ . Let K be the family of sets K ⊆ X such that K ∩G∗ \G = ∅ for some G ∈ G. If E ∈ Σ, µE = µ(E \ G∗ ) + µ(E ∩ G∗ ) = µ(E \ G∗ ) + sup µ(E ∩ G) G∈G

(414Ea) = sup µ(E \ (G∗ \ G)), G∈G

so µ is inner regular with respect to K. By 412Ob, µY is inner regular with respect to {K ∩ Y : K ∈ K}. So if γ < µY H ∗ , there is a K ∈ K such that K ∩ Y ⊆ H ∗ and µY (K ∩ Y ) ≥ γ. But now there is a G ∈ G such that K ∩ G∗ \ G = ∅, so that K ∩ Y ⊆ G ∩ Y ∈ H and supH∈H µH ≥ γ. As H and γ are arbitrary, µ is τ -additive. Remarks Recall from 214I that if (X, Σ, µ) has locally determined negligible sets (in particular, is either strictly localizable or complete and locally determined), then all its subspaces are semi-finite. In 419C below I describe a tight locally finite Borel measure with a subset on which the subspace measure is not semi-finite, therefore not effectively locally finite or τ -additive. In 419A I describe a σ-finite locally finite τ -additive topological measure, inner regular with respect to the closed sets, with a closed subset on which the subspace measure is totally finite but not τ -additive.

54


414L

414L Lemma Let (X, T) be a topological space, and µ, ν two effectively locally finite Borel measures on X which agree on the open sets. Then they are equal. proof Write Tf for the family of open sets of finite measure. (I do not need to specify which measure I am using here.) For G ∈ Tf , set µG E = µ(G ∩ E), νG E = ν(G ∩ E) for every Borel set E. Then µG and νG are totally finite Borel measures which agree on T. By the Monotone Class Theorem (136C), µG and νG agree on the σ-algebra generated by T, that is, the Borel σ-algebra B. Now, for any E ∈ B, µE = supG∈Tf µG E = supG∈Tf νG E = νE, by 412F. So µ = ν. 414M Proposition Let (X, Σ, µ) be a measure space with a regular topology T such that µ is effectively locally finite and τ -additive and Σ includes a base for T. (a) µG = sup{µF : F ∈ Σ is closed, F ⊆ G} for every open set G ∈ Σ. (b) If µ is inner regular with respect to the σ-algebra generated by T ∩ Σ, it is inner regular with respect to the closed sets. proof (a) For U ∈ Σ ∩ T, the set HU = {H : H ∈ Σ ∩ T, H ⊆ U } S is an upwards-directed family of open sets, and HU = U because T is regular and Σ includes a base for T. Because µ is τ -additive, µU = sup{µH : H ∈ HU }. Now, given γ < µG, we can choose hUn in∈N in Σ ∩ T inductively, as follows. Start by taking U0 ⊆ G such that γ < µU0 < ∞ (using the hypothesis that µ is effectively locally finite). Given Un ∈ Σ ∩ T and µUn > γ, take Un+1 ∈ Σ ∩ T such that U n+1 ⊆ Un and µUn+1 > γ. On completing the induction, set T T F = n∈N Un = n∈N U n ; then F is a closed set belonging to Σ, F ⊆ G and µF ≥ γ. As γ is arbitrary, we have the result. (b) Let Σ0 be the σ-algebra generated by Σ ∩ T and set µ0 = µ¹Σ0 . Then Σ0 ∩ T = Σ ∩ T is still a base for T and µ0 is still τ -additive and effectively locally finite, so by (a) and 412G it is inner regular with respect to the closed sets. Now we are supposing that µ is inner regular with respect to Σ0 , so µ is inner regular with respect to the closed sets, by 412Ab. 414N Proposition Let (X, Σ, µ) be a measure space and T a topology on X. Suppose that (i) µ is semifinite and inner regular with respect to the closed sets (ii) whenever F is a non-empty downwards-directed family of measurable closed sets with empty intersection and inf F ∈F µF < ∞, then inf F ∈F µF = 0. Then µ is τ -additive. proof Let G be a non-empty upwards-directed family of measurable open sets with measurable union H. Take any γ < µH. Because µ is semi-finite, there is a measurable set E ⊆ H such that γ < µE < ∞. Now there is a measurable closed set F ⊆ E such that µF ≥ γ. Consider F = {F \ G : G ∈ G}. This is a downwards-directed family of closed sets of finite measure with empty intersection. So inf G∈G µ(F \ G) = 0, that is, γ ≤ µF = supG∈G µ(F ∩ G) ≤ supG∈G µG. As γ is arbitrary, µH = supG∈G µG; as G is arbitrary, µ is τ -additive. 414O The following elementary result is worth noting. Proposition If X is a hereditarily Lindelöf space (e.g., if it is separable and metrizable) then every measure on X is τ -additive. proof If µ is a measure on X, with domain Σ, and G ⊆ Σ is a non-empty S Supwards-directed family of measurable open sets, then there is a sequence hGn in∈N in G such that G = n∈N Gn . Now S S µ( G) = limn→∞ µ( i≤n Gi ) ≤ supG∈G µG. As G is arbitrary, µ is τ -additive.

414Xb

τ -additivity

55

414P Density topologies Recall that a lower density of a measure space (X, Σ, µ) is a function φ : Σ → Σ such that φE = φF whenever E, F ∈ Σ and µ(E4F ) = 0, µ(E4φE) = 0 for every E ∈ Σ, φ∅ = ∅ and φ(E ∩ F ) = φE ∩ φF for all E, F ∈ Σ (341C). Proposition Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lower density such that φX = X. Set T = {E : E ∈ Σ, E ⊆ φE}. Then T is a topology on X, the density topology associated with φ, and (X, T, Σ, µ) is an effectively locally finite τ -additive topological measure space; µ is strictly positive and inner regular with respect to the open sets. proof (a)(i) For any E ∈ Σ, φ(E ∩ φE) = φE because E \ φE is negligible; consequently E ∩ φE ∈ T. In particular, ∅ = ∅ ∩ φ∅ and X = X ∩ φX belong to T. If E, F ∈ T then φ(E ∩ F ) = φE ∩ φF ⊇ E ∩ F , so E ∩ F ∈ T.

S (ii) Suppose that G ⊆ T and H = G. By 341M, µ is (strictly) localizable, so G has an essential supremum F ∈ Σ such that F • = supG∈G G• in the measure algebra A of µ; that is, for E ∈ Σ, µ(G \ E) = 0 for every G ∈ G iff µ(F \ E) = 0. Now F \ H is negligible, by 213K. On the other hand, G ⊆ φG = φ(G ∩ F ) ⊆ φF for every G ∈ G, so H ⊆ φF , and H \ F ⊆ φF \ F is negligible. But as µ is complete, this means that H ∈ Σ. Also φH = φF ⊇ H, so H ∈ T. Thus T is closed under arbitrary unions and is a topology. (b) By its definition, T is included in Σ, so µ is a topological measure. If E ∈ Σ then E ∩ φE belongs to T, is included in E and has the same measure as E; so µ is inner regular with respect to the open sets. If E ∈ T is non-empty, then φE ⊇ E is non-empty, so µE > 0; thus µ is strictly positive. Finally, if G is a S non-emptySupwards-directed family in T, then the argument of (a-ii) shows that ( G)• = supG∈G G• in A, so that µ( G) = supG∈G µG. Thus µ is τ -additive. If E ∈ Σ and µE > 0 then there is an F ⊆ E such that 0 < µF < ∞, and now E ∩ φF is an open set of non-zero finite measure included in E; so µ is effectively locally finite. 414Q Lifting topologies Let (X, Σ, µ) be a measure space and φ : Σ → Σ a lifting, that is, a Boolean homomorphism such that φE = ∅ whenever µE = 0 and µ(E4φE) = 0 for every E ∈ Σ (341A). The lifting topology associated with φ is the topology generated by {φE : E ∈ Σ}. Note that {φE : E ∈ Σ} is a topology base, so is a base for the lifting topology. 414R Proposition Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lifting with lifting topology S and density topology T. Then S ⊆ T ⊆ Σ, and µ is τ -additive, effectively locally finite and strictly positive with respect to S. Moreover, S is zero-dimensional. proof Of course φ is a lower density, so we can talk of its density topology, and since φ2 E = φE, φE ∈ T for every E ∈ Σ, so S ⊆ T. Because µ is τ -additive and strictly positive with respect to T, it must also be τ -additive and strictly positive with respect to S. If E ∈ Σ and µE > 0 there is an F ⊆ E such that 0 < µF < ∞, and now φF is an S-open set of finite measure meeting E in a non-negligible set; so µ is effectively locally finite with respect to S. Of course S is zero-dimensional because φ[Σ] is a base for S consisting of open-and-closed sets. 414X Basic exercises (a) Let (X, Σ, µ) and (Y, T, ν) be measure spaces with topologies T and S, and f : X → Y a continuous inverse-measure-preserving function. Show that if µ is τ -additive with respect to T then ν is τ -additive with respect to S. Show that if ν is locally finite, so is µ. (b) Let h(Xi , Σi , µi )ii∈I be a family of measure spaces, with direct sum (X, Σ, µ); suppose that we are given a topology Ti on Xi for each i, and let T be the disjoint union topology on X. Show that µ is τ -additive iff every µi is.

56


414Xc

> (c) Let (X, T) be a topological space and µ a totally finite measure on X which is inner regular with respect to the closed sets. Suppose that µX = supG∈G µG whenever G is an upwards-directed family of measurable open sets covering X. Show that µ is τ -additive. (d) Let µ be an effectively locally finite τ -additive σ-finite measure on a topological space X, and ν : Σ → [0, ∞[ a countably additive functional which is absolutely continuous with respect to µ. Show from first principles that ν is τ -additive. (e) Give an example of an indefinite-integral measure over Lebesgue measure on R which is not effectively locally finite. (Hint: arrange for every non-trivial interval to have infinite measure.) (f ) Let (X, T) be a topological space and µ a complete locally determined effectively locally finite τ additive topological measure on X. Show that if f is a real-valued function, defined on a subset of X, which is locally integrable in the sense of 411Fc, then f is measurable. (g) Let (X, T) be a topological space and µ an effectively locally finite τ -additive measure on X. Let G be a cover of X consisting of measurable open sets, and K the ideal of subsets of X generated by G. Show that µ is inner regular with respect to K. (h) Let (X, T, Σ, µ) be a complete locally determined effectively locally finite τ -additive topological measure space, and A a subset of X. Suppose that for every x ∈ X there is an open set G containing x such that A ∩ G is negligible. Show that A is negligible. (i) Give an alternative proof of 414K based on the fact that the canonical map from the measure algebra of µ to the measure algebra of µY is order-continuous (322Yd). > (j) (i) If µ is an effectively locally finite τ -additive Borel measure on a regular topological space, show that the c.l.d. version of µ is a quasi-Radon measure. (ii) If µ is a locally finite, effectively locally finite τ -additive Borel measure on a locally compact Hausdorff space, show that µ is tight, so that the c.l.d. version of µ is a Radon measure. > (k) Let (X, Σ, µ) be a complete locally determined measure space and φ a lower density on X such that φX = X; let T be the corresponding density topology. (i) Show that a dense open subset of X must be conegligible. (ii) Show that a subset of X is nowhere dense for T iff it is negligible iff it is meager for T. (iii) Show that a function f : X → R is Σ-measurable iff it is T-continuous at almost every point of X. (Hint: if f is measurable, set Eq = {x : f (x) > q}, Fq = {x : f (x) < q}; show that f is continuous at every point S of X \ q∈Q ((Eq \ φEq ) ∪ (Fq \ φFq )).) (l) Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lower density such that φX = X, with density topology T. Show that if A ⊆ X and E is a measurable envelope of A then the T-closure of A is just A ∪ (X \ φ(X \ E)). (m) Let µ be Lebesgue measure on R r , Σ its domain, φ : Σ → Σ lower Lebesgue density (341E) and T the corresponding density topology. Show that for any A ⊆ R, the closure of A for T is just A ∪ {x : lim supδ↓0

µ∗ (A∩B(x,δ)) µB(x,δ)

> 0}, and the interior is A ∩ {x : limδ↓0


= 1}.

(n) Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lower density such that φX = X; let T be the associated density topology. Let A be a subset of X and E a measurable envelope of A; let ΣA be the subspace σ-algebra and µA the subspace measure on A. (i) Show that we have a lower density φA : ΣA → ΣA defined by setting φA (F ∩ A) = A ∩ φ(E ∩ F ) for every F ∈ Σ. (ii) Show that φA A = A iff A ⊆ φE, and that in this case the density topology on A derived from φA is just the subspace topology. (o) Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lifting, with density topology T and lifting topology S. (i) Show that T = {H ∩ G : G ∈ S, H is conegligible} = {H ∩ φE : E ∈ Σ, H is conegligible}. (ii) Show that if A ⊆ X and E is a measurable envelope of A then the T-closure of A is A ∪ φE.

414Yf

τ -additivity

57

(p) Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lifting; let S be its lifting topology. Let A be a subset of X such that A ⊆ φE for some (therefore any) measurable envelope E of A. Let ΣA be the subspace σ-algebra and µA the subspace measure on A. (i) Show that we have a lifting φA : ΣA → ΣA defined by setting φA (F ∩ A) = A ∩ φF for every F ∈ Σ. (ii) Show that the lifting topology on A derived from φA is just the subspace topology. (q) Let (X, Σ, µ) and (Y, T, ν) be complete locally determined measure spaces and f : X → Y an inversemeasure-preserving function. (i) Suppose that we have lower densities φ : Σ → Σ and ψ : T → T such that φX = X, ψY = Y and φf −1 [F ] = f −1 [ψF ] for every F ∈ T. Show that f is continuous for the density topologies of φ and ψ. (ii) Show that if φ and ψ are liftings then f is continuous for the lifting topologies. (r) Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lifting, with associated lifting topology S. Show that a function f : X → R is Σ-measurable iff there is a conegligible set H such that f ¹H is S-continuous. (Compare 414Xk.) (s) Let (X, Σ, µ) be a complete locally determined measure space and φ : Σ → Σ a lifting. Let (Z, T, ν) be the Stone space of the measure algebra of µ, and f : X → Z the inverse-measure-preserving function associated with φ (341P). Show that the lifting topology on X is just {f −1 [G] : G ⊆ Z is open}. (t) Let (X, Σ, µ) be a strictly localizable measure space and φ : Σ → Σ a lifting. Write L∞ for the Banach lattice of bounded Σ-measurable real-valued functions on X, identified with L∞ (Σ) (363H); let T : L∞ → L∞ be the Riesz homomorphism associated with φ (363F). (i) Show that T 2 = T . (ii) Show that if X is given the lifting topology S defined by φ, then T [L∞ ] is precisely the space of bounded continuous real-valued functions on X. (iii) Show that if f ∈ L∞ , x ∈ X and ² > 0 there is an S-open set U containing 1 R x such that |(T f )(x) − f dµ| ≤ ² for every non-negligible measurable set V included in U . V µV

414Y Further exercises (a) Let (X, T, Σ, µ) be a totally finite topological measure space. For E ∈ Σ set S µτ E = inf{µ(E \ G) + supG∈G µ(E ∩ G) : G ⊆ T is an upwards-directed set}. Suppose either that µ is inner regular with respect to the closed sets or that T is regular. Show that µτ is a τ -additive measure, the largest τ -additive measure with domain Σ which is dominated by µ. (b) Let X be a set, Σ an algebra of subsets of X, and T a topology on X. Let M be the L-space of bounded finitely additive real-valued functionals on Σ (362B). Let N ⊆ M be the set of those functionals ν such that inf G∈G |ν|(H \ G) = 0 whenever G ⊆ T ∩ Σ is a non-empty upwards-directed family with union H ∈ Σ. Show that N is a band in M . (Cf. 362Xi.) (c) Find a probability space (X, Σ, µ) and a topology T on X such that Σ includes a base for T and µ is τ -additive, but there is a set E ∈ Σ such that the subspace measure µE is not τ -additive. (d) Let (X, Σ, µ) be a complete locally determined measure space and φ a lower density on X such that φX = X; let T be the density topology. Show that Σ is precisely the Baire property algebra for T, so that (X, T) is a Baire space. (e) Let φ be lower Lebesgue density on R r , and T the associated density topology. Show that every T-Borel set is an Fσ set for T. (f ) Let (X, ρ) be a metric space and µ a strictly positive locally finite quasi-Radon measure on X; write T for the topology of X and Σ for the domain of µ. For E ∈ Σ set φ(E) = {x : x ∈ X, limδ↓0

µ(E∩B(x,δ)) µB(x,δ)

= 1}.

Suppose that E \ φ(E) is negligible for every E ∈ Σ (cf. 261D, 472D). (i) Show that φ is a lower density for µ, with φ(X) = X. Let Td be the associated density topology. (ii) Suppose that H ∈ Td and that K ⊆ H is T-closed and ρ-totally bounded. Show that there is a T-closed, ρ-totally bounded K 0 ⊆ H such that K is ´ & Zaj´ıc ˇek included in the Td -interior of K 0 . (iii) Show that Td is completely regular. (Hint: Lukeˇ s Maly 86.)

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414Yg

(g) Show that the density topology on R associated with lower Lebesgue density is not normal. (h) Let µ be Lebesgue measure on R r , Σ its domain, φ : Σ → Σ lower Lebesgue density and T the corresponding density topology. (i) Show that if f : Rr → Rr is a bijection such that f and f −1 are both differentiable everywhere, with continuous derivatives, then f is a homeomorphism for T. (Hint: 263D.) (ii) Show that if φ : Σ → Σ is a lifting and S the corresponding lifting topology, then x 7→ −x is not a homeomorphism for S. (Hint: 345Xc.) 414 Notes and comments I have remarked before that it is one of the abiding frustrations of measure theory, at least for anyone ambitious to apply the power of modern general topology to measure-theoretic problems, that the basic convergence theorems are irredeemably confined to sequences. In Volume 3 I showed that if we move to measure algebras and function spaces, we can hope that the countable chain condition or the countable sup property will enable us to replace arbitrary directed sets with monotonic sequences, thereby giving theorems which apply to apparently more general types of convergence. In 414A and its corollaries we come to a quite different context in which a measure, or integral, behaves like an order-continuous functional. Of course the theorems here depend directly on the hypothesis of τ -additivity, which rather begs the question; but we shall see in the rest of the chapter that this property does indeed often appear. For the moment, I remark only that as Lebesgue measure is τ -additive we certainly have a non-trivial example to work with. The hypotheses of the results above move a touch awkwardly between those with the magic phrase ‘topological measure’ and those without. The point is that (as in 412G, for instance) it is sometimes useful to be able to apply these ideas to Baire measures on completely regular spaces, which are defined on a base for the topology but may not be defined on every open set. I hope that no confusion will arise between the two topologies associated with a lifting on a complete locally determined space. I have called them the ‘density topology’ and the ‘lifting topology’ because the former can be defined directly from a lower density; but it would be equally reasonable to call them the ‘fine’ and ‘coarse’ lifting topologies. The density topology has the apparent advantage of giving us a measure which is inner regular with respect to the Borel sets, but at the cost of being wildly irregular regarded as a topological space. It has the important advantage that there are densities (like the Lebesgue lower density) which have some claim to be called canonical, and others with useful special properties, as in §346, while liftings are always arbitrary and invariance properties for them sometimes unachievable. So, for instance, the Lebesgue density topology on R r is invariant under diffeomorphisms, which no lifting topology can be (414Yh). The lifting topology is well-behaved as a topology, but only in special circumstances (as in 453Xd) is the measure inner regular with respect to its Borel sets, and even the closure of a set can be difficult to determine. As with inner regularity, τ -additivity can be associated with the band structure of the space of bounded additive functionals on an algebra (414Yb); there will therefore be corresponding decompositions of measures into τ -additive and ‘purely non-τ -additive’ parts (cf. 414Ya).

415 Quasi-Radon measure spaces We are now I think ready to draw together the properties of inner regularity and τ -additivity. Indeed this section will unite several of the themes which have been running through the treatise so far: (strict) localizability, subspaces and products as well as the new concepts of this chapter. In these terms, the principal results are that a quasi-Radon space is strictly localizable (415A), any subspace of a quasi-Radon space is quasi-Radon (415B), and the product of a family of strictly positive quasi-Radon probability measures on separable metrizable spaces is quasi-Radon (415E). I describe a basic method of constructing quasi-Radon measures (415K), with details of one of the standard ways of applying it (415L, 415N) and some notes on how to specify a quasi-Radon measure uniquely (415H-415I). I spell out useful results on indefinite-integral measures (415O) and Lp spaces (415P), and end the section with a discussion of the Stone space Z of a localizable measure algebra A and an important relation in Z × X when A is the measure algebra of a quasi-Radon measure space X (415Q-415R). It would be fair to say that the study of quasi-Radon spaces for their own sake is a minority interest. If you are not already well acquainted with Radon measure spaces, it would make good sense to read this

415D

Quasi-Radon measure spaces

59

section in parallel with the next. In particular, the constructions of 415K and 415L derive much of their importance from the corresponding constructions in §416. 415A Theorem A quasi-Radon measure space is strictly localizable. proof This is a special case of 414J. 415B Theorem Any subspace of a quasi-Radon measure space is quasi-Radon. proof Let (X, T, Σ, µ) be a quasi-Radon measure space and (Y, TY , ΣY , µY ) a subspace with the induced topology and measure. Because µ is complete, locally determined and localizable (by 415A), so is µY (214Id). Because µY is semi-finite and µ is an effectively locally finite τ -additive topological measure, so is µY (414K). Because µ is inner regular with respect to the closed sets and µY is semi-finite, µY is inner regular with respect to the relatively closed subsets of Y (412Pc). So µY is a quasi-Radon measure. 415C In regular topological spaces, the condition ‘inner regular with respect to the closed sets’ in the definition of ‘quasi-Radon measure’ can be weakened or omitted. Proposition Let (X, T) be a regular topological space. (a) If µ is a complete locally determined effectively locally finite τ -additive topological measure on X, inner regular with respect to the Borel sets, then it is a quasi-Radon measure. (b) If µ is an effectively locally finite τ -additive Borel measure on X, its c.l.d. version is a quasi-Radon measure. proof (a) By 414Mb, µ is inner regular with respect to the closed sets, which is the only feature missing from the given hypotheses. (b) The c.l.d. version of µ satisfies the hypotheses of (a). 415D In separable metric spaces, among others, we can even omit τ -additivity. Proposition Let (X, T) be a hereditarily Lindelöf topological space; e.g., a separable metrizable space (4A2P(a-iii)). (i) If µ is a complete effectively locally finite measure on X, inner regular with respect to the Borel sets, and its domain includes a base for T, then it is a quasi-Radon measure. (ii) If µ is an effectively locally finite Borel measure on X, then its completion is a quasi-Radon measure. (iii) Any quasi-Radon measure on X is σ-finite. (iv) If X is regular, any quasi-Radon measure on X is completion regular. proof (a) The basic S fact we S need is that if G is any family of open sets in X, then there is a countable G0 ⊆ G such that G0 = G (4A2H(c-i)). Consequently any effectively locally finite measure µ on X is σ-finite. P P Let G be the family of measurable S open sets of finite measure. Let G0 ⊆ G be a countable set with the same union as G. Then E = X \ G0 is measurable, and E ∩ G = ∅ for every G ∈ G, so µE = 0; accordingly G0 ∪ {E} is a countable cover of X by sets of finite measure, and µ is σ-finite. Q Q Moreover, any measure on X is τ -additive. P P If G is a non-empty upwards-directed family of open S measurable sets, there is a sequence hG i in G with union G. If n ∈ N there is a G ∈ G such that n n∈N S G ⊆ G, so i i≤n S S S µ( G) = µ( n∈N Gn ) = supn∈N µ( i≤n Gi ) ≤ supG∈G µG. As G is arbitrary, µ is τ -additive. Q Q (b)(i) Now let µ be a complete effectively locally finite measure on X, inner regular with respect to the Borel sets, and with domain Σ including a base for the topology of X. If H ∈ T, then G = {G : G ∈ Σ ∩ T, G ⊆ H} S has union H, because Σ ∩ T is a base for T; but in this case there is a countable G0 ⊆ G such that H = G0 , so that H ∈ Σ. Thus µ is a topological measure. We know also from (a) that it is τ -additive and σ-finite, therefore locally determined. By 415Ca, it is a quasi-Radon measure. (ii) If µ is an effectively locally finite Borel measure on X, then its completion µ ˆ satisfies the conditions of (i), so is a quasi-Radon measure.

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(iii) If µ is a quasi-Radon measure on X, it is surely effectively locally finite, therefore σ-finite. (iv) If X is regular, then every closed set is a zero set (4A2H(c-iii)), so any measure which is inner regular with respect to the closed sets is completion regular. 415E I am delaying most of the theory of products of (quasi-)Radon measures to §417. However, there is one result which is so important that I should like to present it here, even though some of the ideas will have to be repeated later. Theorem Let h(Xi , Ti , Σi , µi )ii∈I be a family of separable metrizableQquasi-Radon probability spaces such that every µi is strictly positive, and λ the product measure on X = i∈I Xi . Then (i) λ is a completion regular quasi-Radon measure; (ii) if F ⊆ X is a closed self-supporting set, there is a countable set J ⊆ I such that F is determined by coordinates in J, so F is a zero set. Q proof (a) Write Λ for the domain of λ, and U for the family of subsets of X of the form i∈I Gi where Gi ∈ Ti for every i ∈ I and {i : Gi 6= Xi } is finite. Then U is a base for the topology of X, included in Λ. Q (J) For J ⊆ I let λJ be the product measure on X = X for the family J i and ΛJ its domain. Write U i∈J Q of subsets of XJ of the form i∈J Gi where Gi ∈ Ti for each i ∈ J and {i : Gi 6= Xi } is finite. (b) Consider first the case in which I is countable. In this case X also is separable and metrizable (4A2P(a-v)), while Λ includes a base for its topology. Also λ is a complete probability measure and inner regular with respect to the closed sets (412Ua), so must be a quasi-Radon measure, by 415D(i). (c) Now consider uncountable I. The key to the proof is the following fact: if V ⊆ U has union W , then W ∈ Λ and λW = supV ∈V ∗ λV , where V ∗ is the set of unions of finite subsets of V. P P (i) S By 215B(iv), there is a countable set V1 ⊆ V such that λ(U \ W1 ) = 0 for every U ∈ V, where W1 = V1 . Every member of U is determined by coordinates in some finite set (see 254M for this concept), so there is a countable set J ⊆ I such that every member of V1 is determined by coordinates in J, and W1 also is determined by coordinates in J. Let πJ : X → XJ be the canonical map. Because it is an open map (4A2B(f-i)), πJ [W ] and πJ [W1 ] are open in XJ , and belong to ΛJ , by (b). S (ii) ?? Suppose, if possible, that λJ πJ [W ] > λJ πJ [W1 ]. Since πJ [W ] = {πJ [U ] : U ∈ V}, while λJ is quasi-Radon and all the sets πJ [U ] are open, there must be some U ∈ V such that λJ (πJ [U ] \ πJ [W1 ]) > 0 (414Ea). Now πJ is inverse-measure-preserving (254Oa), so 0 < λπJ−1 [πJ [U ] \ πJ [W1 ]] = λ(πJ−1 [πJ [U ]] \ πJ−1 [πJ [W1 ]]) = λ(πJ−1 [πJ [U ]] \ W1 ), because W1 is determined by coordinates in J. Q At this point note that U is of the form i∈I Gi , where Gi ∈ Ti for each I, so we can express U as −1 U 0 ∩ U 00 , where U 0 = πJ−1 [πJ [U ]] and U 00 = πI\J [πI\J [U ]]. U 0 depends on coordinates in J and U 00 depends on coordinates in I \ J. In this case λ(U \ W1 ) = λ(U 00 ∩ U 0 \ W1 ) = λU 00 · λ(U 0 \ W1 ), because U 00 depends on coordinates in I \ J and U 0 \ W1 depends on coordinates in J, and we can identify λ with the product λI\J × λJ (254N). But now recall that every µQ i is strictly positive. Since U is surely not empty, no Gi can be empty and no µi Gi can be 0. Consequently i∈I µi Gi > 0 (because only finitely many terms in the product are less than 1) and λU > 0; more to the point, λU 00 > 0. Since we already know that λ(U 0 \ W1 ) > 0, we have λ(U \ W1 ) > 0. But this contradicts the first sentence of (i) just above. X X (iii) Thus λJ πJ [W ] = λJ πJ [W1 ]. But this means that λπJ−1 [πJ [W ]] = λW1 . Since λ is complete and W1 ⊆ W ⊆ πJ−1 [πJ [W ]], λW is defined and equal to λW1 . Taking hVn in∈N to be a sequence running over V1 ∪ {∅}, we have S S λW = λW1 = λ( n∈N Vn ) = supn∈N λ( i≤n Vi ) ≤ supV ∈V ∗ λV ≤ λW, so λW = supV ∈V ∗ λV , as required. Q Q (d) Thus we see that λ is a topological measure. But it is also τ -additive. P P If W is an upwards-directed family of open sets in X with union W ∗ , set

415G


∗

S

61

V = {U : U ∈ U , ∃ W ∈ W, U ⊆ W }. ∗

Then W = V, so λW = supV ∈V ∗ λV , where V ∗ is the set of finite unions of members of V. But because W is upwards-directed, every member of V ∗ is included in some member of W, so λW ∗ = supV ∈V ∗ λV ≤ supW ∈W λW ≤ λW ∗ . As W is arbitrary, λ is τ -additive. Q Q (e) As in (b) above, we know that λ is a complete probability measure and is inner regular with respect to the closed sets, so it is a quasi-Radon measure. Because λ is inner regular with respect to the zero sets (412Ub), it is completion regular. (f ) Now suppose that F ⊆ X is a closed self-supporting set. By 254Oc, there is a set W ⊆ X, determined by coordinates in some countable set J ⊆ I, such that W 4F is negligible. ?? Suppose, if possible, that x ∈ F and y ∈ X \ F are such that x¹J = y¹J. Then there is a U ∈ U such that y ∈ U ⊆ X \ F . As in (b-ii) above, we can express U as U 0 ∩ U 00 where U 0 , U 00 ∈ U are determined by coordinates in J and I \ J respectively. In this case, λ(F ∩ U ) = λ(W ∩ U ) = λ(W ∩ U 0 ) · λU 00 = λ(F ∩ U 0 ) · λU 00 > 0, because x ∈ F ∩ U 0 and F is self-supporting, while U 00 6= ∅ and λ is strictly positive. But F ∩ U = ∅, so this is impossible. X X Thus F is determined by coordinates in the countable set J. Consequently it is of the form πJ−1 [πJ [F ]]. But πJ [X \ F ] is open (4A2B(f-i)), so its complement πJ [F ] is closed. Now XJ is metrizable (4A2P(a-v)), so πJ [F ] is a zero set (4A2Lc) and F is a zero set (4A2C(b-iv)). 415F Corollary(a) If Y is either [0, 1[ or ]0, 1[, endowed with Lebesgue measure, and I is any set, then Y I , with the product topology and measure, is a quasi-Radon measure space. (b) If hνi ii∈I is a family of probability distributions on R, in the sense of §271 (that is, Radon probability measures), and every νi is strictly positive, then the product measure on R I is a quasi-Radon measure. Remark See also 416U below, and 453I, where there is an alternative proof of the main step in 415E, I applicable to some further cases. Yet another approach, most immediately applicable to [0, 1[ , is in 443Xq. For further facts about these product measures, see §417, particularly 417G and 417M. 415G Comparing quasi-Radon measures: Proposition Let X be a topological space, and µ, ν two quasi-Radon measures on X. Then the following are equiveridical: (i) µF ≤ νF for every closed set F ⊆ X; (ii) dom ν ⊆ dom µ and µE ≤ νE for every E ∈ dom ν. If ν is locally finite, we can add (iii) µG ≤ νG for every open set G ⊆ X; (iv) there is a base U for the topology of X such that G ∪ H ∈ U for all G, H ∈ U and µG ≤ νG for G ∈ U. proof (a) Of course (ii)⇒(i). Suppose that (i) is true. Observe that if E ∈ dom µ ∩ dom ν (for instance, if E ⊆ X is Borel), then µE = sup{µF : F ⊆ E is closed} ≤ sup{νF : F ⊆ E is closed} = νE. S Set H = {H : H ⊆ X is open, νH < ∞}, and W = H. Then ν(X \ W ) = 0, because ν is effectively locally finite, so µ(X \ W ) = 0. Set F = {F : F ⊆ X is closed, µF < ∞}. Take any E ∈ dom ν. If F ∈ F and ² > 0, then µ(F ∩ W ) = µF , so there is an H ∈ H such that µ(F ∩ H) ≥ µF − ². Now there are closed sets F1 ⊆ F ∩ H ∩ E, F2 ⊆ F ∩ H \ E such that νF1 + νF2 ≥ ν(F ∩ H) − ², that is, ν((F ∩ H) \ (F1 ∪ F2 )) ≤ ², so that µ((F ∩ H) \ (F1 ∪ F2 )) ≤ ² and µF1 + µF2 ≥ µ(F ∩ H) − ². This means that µ∗ (F ∩ E) + µ∗ (F \ E) ≥ µ(F ∩ H) − ² ≥ µF − ².

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As ² is arbitrary, µ∗ (F ∩ E) + µ∗ (F \ E) ≥ µF ; as µ is inner regular with respect to F, µ measures E, by 413F(vii). Thus dom ν ⊆ dom µ; and we have already observed that µE ≤ νE whenever E is measured by both. (b) The first sentence in the proof of (a) shows that (i)⇒(iii), and (iii)⇒(iv) is trivial. If (iv) is true and G ⊆ X is open, then V = {V : V ∈ U , V ⊆ G} is upwards-directed and has union G, so µG = supV ∈V µV ≤ supV ∈V νV = νG. Thus (iv)⇒(iii). Now assume that ν is locally finite and that (iii) is true. ?? Suppose, if possible, that F ⊆ X is a closed set such that νF < µF . Then H, as defined in part (a) of the proof, is upwards-directed and has union X, so there is an H ∈ H such that νF < µ(F ∩ H). Now there is a closed set F 0 ⊆ H \ F such that νF 0 > ν(H \ F ) − µ(F ∩ H) + νF ≥ νH − µ(F ∩ H). Set G = H \ F 0 , so that F ∩ H ⊆ G and νG = νH − νF 0 < µ(F ∩ H) ≤ µG, which is impossible. X X This shows that (provided that ν is locally finite) (iii)⇒(i). 415H Uniqueness of quasi-Radon measures: Proposition Let (X, T) be a topological space and µ, ν two quasi-Radon measures on X. Then the following are equiveridical: (i) µ = ν; (ii) µF = νF for every closed set F ⊆ X; (iii) µG = νG for every open set G ⊆ X; (iv) there is a base U for the topology of X such that G ∪ H ∈ U for every G, H ∈ U and µ¹U = ν¹U ; (v) there is a base U for the topology of X such that G ∩ H ∈ U for every G, H ∈ U and µ¹U = ν¹U. proof Of course (i) implies all the others. (ii)⇒(i) is immediate from 415G (see also 412L). If (iii) is true, then, for any closed set F ⊆ X, µF = sup{µ(G ∩ F ) : G ∈ T, µG < ∞} = sup{µG − µ(G \ F ) : G ∈ T, µG < ∞} = sup{νG − ν(G \ F ) : G ∈ T, νG < ∞} = νF ; so (iii)⇒(ii). (iv)⇒(iii) by the argument of (iv)⇒(iii) in the proof of 415G. Finally, suppose that (v) is true. Then µ(G0 ∪ . . . ∪ Gn ) = ν(G0 ∪ . . . ∪ Gn ) for all G0 , . . . , Gn ∈ U. P P Induce on n. For the inductive step to n ≥ 1, if any Gi has infinite measure (for either measure) the result is trivial. Otherwise, [ [ µ(G0 ∪ . . . ∪ Gn ) = µ( Gi ) + µGn − µ( (Gn ∩ Gi )) i
= ν(

[

i
i
Gi ) + νGn − ν(

[

(Gn ∩ Gi )) = ν(G0 ∪ . . . ∪ Gn ). Q Q

i
So µ and ν agree on the base {G0 ∪ . . . ∪ Gn : G0 , . . . , Gn ∈ U}, and (iv) is true. 415I Proposition Let RX be a completely regular topological space and µ, ν two quasi-Radon measures R on X such that f dµ = f dν whenever f : X → R is a bounded continuous function integrable with respect to both measures. Then µ = ν. proof ?? Otherwise, there is an open set G ⊆ X such that µG 6= νG; suppose µG < νG. Because ν is effectively locally finite, there is an open set G0 ⊆ G such that µG < νG0 < ∞. Now the cozero sets form a base for the topology of X, so H = {H : H ⊆ G0 is a cozero set} has union G0 ; as ν is τ -additive, there is an H ∈ H such that νH > µG. Express H as {x : g(x) > 0} where g : X → [0, ∞[ is continuous. For each n ∈ N, setRfn = ng ∧ χX; then hfn in∈N is a non-decreasing sequence with limit χH, so there is an n ∈ N R such that fn dν > µG ≥ fn dµ. But fn is both µ-integrable and ν-integrable because µG and νH are both finite. X X

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415J Proposition Let X be a regular topological space, Y a subspace of X, and ν a quasi-Radon measure on Y . Then there is a quasi-Radon measure µ on X such that µE = ν(E ∩ Y ) whenever µ measures E, that is, Y has full outer measure in X and ν is the subspace measure on Y . proof Write B for the Borel σ-algebra of X, and set µ0 E = ν(E ∩ Y ) for every E ∈ B. Then it is easy to see that µ0 is a τ -additive Borel measure on X. Moreover, µ0 is effectively locally finite. P P If E ∈ B and µ0 E > 0, there is a relatively open set H ⊆ Y such that νH < ∞ and ν(H ∩ E ∩ Y ) > 0. Now H is of the form G ∩ Y where G ⊆ X is open, and we have µ0 G = νH < ∞, µ0 (E ∩ G) = ν(H ∩ E ∩ Y ) > 0. Q Q By 415Cb, the c.l.d. version µ of µ0 is a quasi-Radon measure on X. If E ∈ dom µ, then E ∩ Y ∈ dom ν. P P Let FY be the set of relatively closed subsets of Y of finite measure for ν. If F ∈ FY , it is expressible as F 0 ∩ Y where F 0 is a closed subset of X, and µF 0 = µ0 F 0 = νF is finite. So there are E1 , E2 ∈ B such that E1 ⊆ E ∩ F 0 ⊆ E2 and µE1 = µ(E ∩ F 0 ) = µE2 . Accordingly E1 ∩ Y ⊆ E ∩ Y ∩ F ⊆ E2 ∩ Y and ν(E1 ∩ Y ) = ν(E2 ∩ Y ) = µ(E ∩ F 0 ) is finite. This means that E ∩ Y ∩ F ∈ dom ν; because ν is complete and locally determined and inner regular with respect to FY , E ∩ Y ∈ dom ν, by 412Ja. Q Q If E ∈ dom µ, then µE = sup{µF : F ⊆ E is closed} = sup{ν(F ∩ Y ) : F ⊆ E is closed} ≤ ν(E ∩ Y ). On the other hand, if γ < ν(E ∩ Y ), there is a relatively open set H ⊆ Y such that νH < ∞ and ν(E ∩ Y ∩ H) ≥ γ (412F). Let G ⊆ X be an open set such that G ∩ Y = H. Then µE ≥ µG − µ(G \ E) = νH − ν(H \ E) = ν(E ∩ Y ∩ H) ≥ γ. As γ is arbitrary, µE = ν(E ∩ Y ). Thus µE = ν(E ∩ Y ) whenever µ measures E. So if E, F ∈ dom µ and E ∩ Y ⊆ F , µE = ν(E ∩ Y ) ≤ ν(F ∩ Y ) = µF ; ∗

as F is arbitrary, µ (E ∩ Y ) = µE; as E is arbitrary, Y has full outer measure in X. Moreover, if µY is the subspace measure on Y , µY H = µ∗ H = νH whenever H ∈ dom µY , that is, H = E ∩Y for some E ∈ dom µ. Now µY , like ν, is a quasi-Radon measure on Y (415B), and they agree on the (relatively) closed subsets of Y , so are equal, by 415H. 415K I come now to a couple of basic results on the construction of quasi-Radon measures. The first follows 413J. Theorem Let X be a topological space and K a family of closed subsets of X such that ∅ ∈ K, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, (‡) F ∈ K whenever K ∈ K and F ⊆ K is closed. Let φ0 : K → [0, ∞[ be a functional such that (α) φ0 K = φ0 L + sup{φ0 K 0 : K 0 ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K, (β) inf K∈K0 φ0 K = 0 whenever K0 is a non-empty downwards-directed subset of K with empty intersection, (γ) whenever K ∈ K and φ0 K > 0, there is an open set G such that the supremum supK 0 ∈K,K 0 ⊆G φ0 K 0 is finite, while φ0 K 0 > 0 for some K 0 ∈ K such that K 0 ⊆ K ∩ G. Then there is a unique quasi-Radon measure on X extending φ0 and inner regular with respect to K. proof By 413J, there is a complete locally determined measure µ on X, inner regular with respect to K, and extending φ0 ; write Σ for the domain of µ. If F ⊆ X is closed, then K ∩ F ∈ K ⊆ Σ for every K ∈ K, so F ∈ Σ, by 413F(ii); accordingly every open set is measurable. Because µ is inner regular with respect to

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K it is surely inner regular with respect to the closed sets. If E ∈ Σ and µE > 0, there is a K ∈ K such that K ⊆ E and µK > 0; now (γ) tells us that there is an open set G such that µG < ∞ and µ(G ∩ K) > 0, so that µ(G ∩ E) > 0. As E is arbitrary, µ is effectively locally finite. Now suppose that G is a non-empty upwards-directed family of open sets with union H, and that γ < µH. Then there is a K ∈ K such that K ⊆ H and µK > γ. Applying the hypothesis (β) to K0 = {K \G : G ∈ G}, we see that inf G∈G µ(K \G) = 0, so that supG∈G µG ≥ supG∈G µ(K ∩ G) = µK ≥ γ. As G and γ are arbitrary, µ is τ -additive. So µ is a quasi-Radon measure. 415L Proposition Let (X, Σ0 , µ0 ) be a measure space and T a topology on X such that µ0 is τ -additive, effectively locally finite and inner regular with respect to the closed sets, and Σ0 includes a base for T. Then µ0 has a unique extension to a quasi-Radon measure µ on X such that (i) µF = µ∗0 F whenever F ⊆ X is closed and µ∗0 F < ∞, (ii) µG = (µ0 )∗ G whenever G ⊆ X is open, (iii) the embedding Σ0 ⊆ Σ identifies the measure algebra (A0 , µ ¯0 ) of µ0 with an order-dense subalgebra of the measure algebra (A, µ ¯) of µ, so that the subrings Af0 , Af of elements of finite measure coincide, and Lp (µ0 ) may be identified with Lp (µ) for 1 ≤ p < ∞, (iv) whenever E ∈ Σ and µE < ∞, there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0, (v) for every µ-integrable real-valued function f there is a µ0 -integrable function g such that f = g µ-a.e. If µ0 is complete and locally determined, then we have (i)0 µF = µ∗0 F for every closed F ⊆ X. If µ0 is localizable, then we have (iii)0 A0 = A, so that L0 (µ) ∼ = L0 (µ0 ), L∞ (µ) ∼ = L∞ (µ0 ), (iv)0 for every E ∈ Σ there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0, (v)0 for every Σ-measurable real-valued function f there is a Σ0 -measurable real-valued function g such that f = g µ-a.e. proof (a) Let K be the set of closed subsets of X of finite outer measure for µ0 . Note that µ0 is inner regular with respect to K, because it is inner regular with respect to the closed sets and also with respect to the sets of finite measure. It is obvious from its definition that K satisfies (†) and (‡) of 415K. For K ∈ K, set φ0 K = µ∗0 K. Then φ0 satisfies (α)-(γ) of 415K. α) If K, L ∈ K and L ⊆ K, take measurable envelopes E0 , E1 ∈ Σ0 of K, L respectively. (i) Let P P (α ² > 0. Because µ0 is inner regular with respect to the closed sets, there is a closed set F ∈ Σ0 such that F ⊆ E0 \ E1 and µF ≥ µ0 (E0 \ E1 ) − ². Set K 0 = F ∩ K. Then K 0 ∈ K and φ0 K 0 = µ∗0 (F ∩ K) = µ0 (F ∩ E0 ) = µ0 F ≥ µ0 E0 − µ0 E1 − ² = φ0 K − φ0 L − ². As ² is arbitrary, we have φ0 K ≤ φ0 L + sup{φ0 K 0 : K 0 ∈ K, K 0 ⊆ K \ L}. (ii) On the other hand, ?? suppose, if possible, that there is a closed set K 0 ⊆ K \ L such that µ∗0 L + µ∗0 K 0 > µ∗ K. Let E2 be a measurable envelope of K 0 , so that µ0 E1 + µ0 E2 > µ0 E0 ; since µ0 (E1 \ E0 ) = µ∗0 (L \ E0 ) = µ∗0 ∅ = 0,

µ0 (E2 \ E0 ) = µ∗0 (K 0 \ E0 ) = 0,

µ0 (E1 ∩ E2 ) > 0. Because µ0 is effectively locally finite, there is a measurable open set G0 , of finite measure, such that µ0 (G0 ∩ E1 ∩ E2 ) > 0. Set G = {G ∪ G0 : G, G0 ∈ Σ0 ∩ T, G ⊆ G0 \ L, G0 ⊆ G0 \ K 0 }. Then G is an upwards-directed family of measurable open sets, and because Σ0 includes a base for the topology of X, its union is (G0 \L)∪(G0 \K 0 ) = G0 . So there is an H ∈ G such that µ0 H > µ0 G0 −µ0 (E1 ∩E2 ),

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65

that is, there are open sets G, G0 ∈ Σ0 such that G ⊆ G0 \ L, G0 ⊆ G0 \ K 0 and µ0 ((G ∪ G0 ) ∩ E1 ∩ E2 )) > 0. But we must have µ0 (G ∩ E1 ) = µ∗0 (G ∩ L) = 0,

µ0 (G0 ∩ E2 ) = µ∗0 (G0 ∩ K 0 ) = 0,

so this is impossible. X X Accordingly φ0 K ≥ φ0 L + sup{φ0 K 0 : K 0 ∈ K, K 0 ⊆ K \ L}, so that φ0 satisfies condition (α) of 415K. β ) Let K0 ⊆ K be a non-empty downwards-directed family with empty intersection. Fix K0 ∈ K0 and (β ² > 0. Let E0 be a measurable envelope of K0 and G0 a measurable open set of finite measure such that µ0 (G0 ∩ E0 ) ≥ µ0 E0 − ². Then G = {G : G ∈ Σ0 ∩ T, G ⊆ G0 \ K for some K ∈ K0 such that K ⊆ K0 } T is an upwards-directed family of measurable open sets, and its union is G0 \ K0 = G0 , again because Σ0 includes a base for the topology T. So there is a G ∈ G such that µ0 G ≥ µ0 G0 − ². Let K ∈ K0 be such that K ⊆ K0 and G ∩ K = ∅; then φ0 K = µ∗0 K ≤ µ0 (E0 \ G) ≤ µ0 (E0 \ G0 ) + µ0 (G0 \ G) ≤ 2². As ² is arbitrary, inf K∈K0 φ0 K = 0. (γγ ) If K ∈ K and φ0 K > 0, let E0 be a measurable envelope of K. Then there is a measurable open set G of finite measure such that µ0 (G ∩ E0 ) > 0. Of course supK 0 ∈K,K 0 ⊆G φ0 K 0 ≤ µ0 G < ∞; but also there is a measurable closed set K 0 ⊆ G ∩ E0 such that µ0 K 0 > 0, in which case φ0 (K ∩ K 0 ) = µ0 (E0 ∩ K 0 ) > 0. So φ0 satisfies condition (γ). Q Q (b) By 415K, φ0 has an extension to a quasi-Radon measure µ on X which is inner regular with respect to K. Write Σ for the domain of µ. Note that, for K ∈ K, µK = φ0 K = µ∗0 K, so we can already be sure that the conclusion (i) of the proposition is satisfied. Now µ extends µ0 . P P(i) Take any K ∈ K. Let E0 ∈ Σ0 be a measurable envelope of K for the measure µ0 . If E ∈ Σ0 , then surely µ∗ (K ∩ E) = sup{µK 0 : K 0 ∈ K, K 0 ⊆ K ∩ E} = sup{µ∗0 K 0 : K 0 ∈ K, K 0 ⊆ K ∩ E} ≤ µ∗0 (K ∩ E). On the other hand, given γ < µ∗0 (K ∩ E) = µ0 (E0 ∩ E), there is a closed set F ∈ Σ0 such that F ⊆ E0 ∩ E and µ0 F ≥ γ, so that µ∗ (K ∩ E) ≥ µ(K ∩ F ) = µ∗0 (K ∩ F ) = µ0 (E0 ∩ F ) ≥ γ. Thus µ∗ (K ∩ E) = µ∗0 (K ∩ E) for every K ∈ K, E ∈ Σ0 . (ii) If K ∈ K, E ∈ Σ0 then µ∗ (K ∩ E) + µ∗ (K \ E) = µ∗0 (K ∩ E) + µ∗0 (K \ E) = µ∗0 K = µK. Because µ is complete and locally determined and inner regular with respect to K, E ∈ Σ (413F(iv)). Thus Σ0 ⊆ Σ. (iii) For any E ∈ Σ0 , we now have µE = sup{µK : K ∈ K, K ⊆ E} = sup{µ∗0 K : K ∈ K, K ⊆ E} ≤ µ0 E = sup{µ0 K : K ∈ K ∩ Σ0 , K ⊆ E} ≤ µE. As E is arbitrary, µ extends µ0 . Q Q

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(c) Because Σ0 ∩ T is a base for T, closed under finite unions, µ is unique, by 415Hc. (d) Now for the conditions (i)-(v). I have already noted that (i) is guaranteed by the construction. Concerning (ii), if G ⊆ X is open, we surely have (µ0 )∗ G ≤ µ∗ G = µG because µ extends µ0 . On the other hand, writing G = {G0 : G0 ∈ Σ0 ∩ T, G0 ⊆ G}, G is upwards-directed and has union G, so µG = supG0 ∈G µG0 = supG0 ∈G µ0 G0 ≤ (µ0 )∗ G. So (ii) is true. Because µ extends µ0 , the embedding Σ0 ⊆ Σ corresponds to a measure-preserving embedding of A0 as a σ-subalgebra of A. To see that A0 is order-dense in A, take any non-zero a ∈ A. This is expressible as E • for some E ∈ Σ with µE > 0. Now there is a K ∈ K such that K ⊆ E and µK > 0. There is an E0 ∈ Σ0 which is a measurable envelope for K with respect to µ0 , so that µE0 = µ0 E0 = µ∗0 K = µK. But this means that 0 6= E0• = K • ⊆ E • = a in A, while E0• ∈ A0 . As a is arbitrary, A0 is order-dense in A. If a ∈ Af , then B = {b : b ∈ A0 , b ⊆ a} is upwards-directed and supb∈B µ ¯0 b ≤ µ ¯a is finite; accordingly B has a supremum in A0 (321C), which must also be its supremum in A, which is a (313O, 313K). So a ∈ A0 . Thus Af can be identified with Af0 . But this means that, for any p ∈ [1, ∞[, Lp (µ) ∼ ¯) is identified = Lp (A, µ p p ∼ with L (A0 , µ ¯0 ) = L (µ) (366H). This proves (iii). Of course (iv) and (v) are just translations of this. If E ∈ Σ and µE < ∞, then E • ∈ Af ⊆ A0 , that is, there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0. If f is µ-integrable, then f • ∈ L1 (µ) = L1 (µ0 ), that is, there is a µ0 -integrable function f0 such that f = f0 µ-a.e. (e) If µ0 is complete and locally determined and F ⊆ X is an arbitrary closed set, then µ∗0 F = supK∈K µ∗0 (F ∩ K) = supK∈K µ(F ∩ K) = supK∈K,K⊆F µK = µF by 412Jc, because µ and µ0 are both inner regular with respect to K. (f ) If µ0 is localizable, A0 is Dedekind complete; as it is order-dense in A, the two must coincide (314Ia). Consequently L0 (µ) ∼ = L0 (A) = L0 (A0 ) ∼ = L0 (µ0 ), L∞ (µ) ∼ = L∞ (A) = L∞ (A0 ) ∼ = L∞ (µ0 ). So (iii)0 is true; now (iv)0 and (v)0 follow at once. 415M Corollary Let (X, T) be a regular topological space and µ0 an effectively locally finite τ -additive measure on X, defined on the σ-algebra Σ0 generated by a base for T. Then µ0 has a unique extension to a quasi-Radon measure on X. proof By 414Mb, µ0 is inner regular with respect to the closed sets. So 415L gives the result. 415N Corollary Let (X, T) be a completely regular topological space, and µ0 a τ -additive effectively locally finite Baire measure on X. Then µ0 has a unique extension to a quasi-Radon measure on X. proof This is a special case of 415M, because the domain Σ0 of µ0 , the Baire σ-algebra, is generated by the family of cozero sets, which form a base for T (4A2Fc). 415O Proposition (a) Let (X, T) be a topological space, and µ, ν two quasi-Radon measures on X. Then ν is an indefinite-integral measure over µ iff νF = 0 whenever F ⊆ X is closed and µF = 0. (b) Let (X, T, Σ, µ) be a quasi-Radon measure space, and ν an indefinite-integral measure over µ (definition: 234B). If ν is effectively locally finite it is a quasi-Radon measure. proof (a) If ν is an indefinite-integral measure over µ, then of course it is zero on all µ-negligible closed sets. So let us suppose that the condition is satisfied. Write Σ = dom µ and T = dom ν. (i) If E ⊆ X is a µ-negligible Borel set it is ν-negligible, because every closed subset of E must be µ-negligible, therefore ν-negligible, and ν is inner regular with respect to the closed sets. In particular,

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taking U ∗ to be the union of the family U = {U : U ∈ T, µU < ∞}, ν(X \ U ∗ ) = µ(X \ U ∗ ) = 0 because µ is effectively locally finite. Also, of course, taking V ∗ to be the union of the family V = S {V : V ∈ T, νV < ∞}, ν(X \ V ∗ ) = 0 because ν is effectively locally finite. Setting G = U ∩ V and G∗ = G, we have G∗ = U ∗ ∩ V ∗ , so G∗ is ν-conegligible. (ii) In fact, every µ-negligible set E is ν-negligible. P P?? Otherwise, ν ∗ (E ∩ G∗ ) > 0. Because the subspace measure νE is quasi-Radon (415B), there is a G ∈ G such that ν ∗ (E ∩ G) > 0. But there is an Fσ set H ⊆ G \ E such that µH = µ(G \ E), and now E ∩ G is included in the µ-negligible Borel set G \ H, so that ν(E ∩ G) = ν(G \ H) = 0. X XQ Q (iii) Let K be the family of closed subsets F of X such that either F is included in some member of G or F ∩ G∗ = ∅. If E ∈ dom µ and µE > 0, then there is an F ∈ K such that F ⊆ E and µF > 0. P P If µ(E \ G∗ ) > 0 take any closed set F ⊆ E \ G∗ with µF > 0. Otherwise, µ(E ∩ G∗ ) > 0. Because the subspace measure µE is quasi-Radon, there is a G ∈ G such that µ(E ∩ G) > 0; and now we can find a closed set F ⊆ E ∩ G with µF > 0, and F ∈ K. Q Q (iv) By 412I, there is a decomposition hXi ii∈I for µ such that every Xi except perhaps one belongs to K and that exceptional one, if any, is µ-negligible. Now hXi ii∈I is a decomposition for ν. P P Every Xi is measured by ν because it is either closed or µ-negligible, and of finite measure for ν because it is included in either a member of G or the ν-negligible set X \ G∗ . If E ⊆ X and νE > 0, then ν(E ∩ G∗ ) > 0, so there must be some G ∈SG such that ν(E ∩ G) > 0. Now J = {i : i ∈ I, µ(Xi ∩ G) > 0} is countable, and S ν(G \ i∈J Xi ) = µ(G \ i∈J Xi ) = 0, so there is an i ∈ J such that ν(Xi ∩ E) > 0. By 213O, hXi ii∈I is a decomposition for ν. Q Q (v) It follows that Σ ⊆ T. P P If E ∈ Σ, then for every i ∈ I there is an Fσ set H ⊆ E ∩ Xi such that E ∩ Xi \ H is µ-negligible, therefore ν-negligible, and E ∩ Xi ∈ T. As i is arbitrary, E ∈ T. Q Q In fact, ν is the completion of ν¹Σ. P P If F S ∈ T, then for every i ∈ I there is an Fσ set Hi ⊆ F ∩ Xi such that F ∩ Xi \ Hi isPν-negligible. Set H = i∈I Hi ; because H ∩ Xi = Hi belongs to Σ for every i, H ∈ Σ; 0 0 and ν(F \ H) = i∈I ν(F ∩ Xi \ H) = 0. Similarly, there is an H ∈ Σ such that H ⊆ X \ F and 0 0 0 ν((X \ F ) \ H ) = 0, so that H ⊆ F ⊆ X \ H and ν((X \ H ) \ H) = 0. So F is measured by the completion of ν¹Σ. Since ν itself is complete, it must be the completion of ν¹Σ. Q Q (vi) By (iv), ν is inner regular with respect to {E : E ∈ Σ, µE < ∞}. By 234G, ν is an indefiniteintegral measure over µ. R (b) Let f ∈ L0 (µ) be a non-negative function such that νF = f × χF dµ whenever this is defined. Because µ is complete and locally determined, so is ν (234Fb). Because µ is an effectively locally finite τ -additive topological measure, ν is a τ -additive topological measure (414H). Because µ is inner regular with respect to the closed sets, so is ν (412Q). Since we are assuming in the hypotheses that ν is effectively locally finite, it is a quasi-Radon measure. 415P Proposition Let (X, T, Σ, µ) be a quasi-Radon measure space. (a) Suppose that (X, T) is completely regular. If 1 ≤ p < ∞ and f ∈ Lp (µ), then for any ² > 0 there is a bounded continuous function g : X → R such that µ{x : g(x) 6= 0} < ∞ and kf − gkp ≤ ². (b) Suppose that (X, T) is regular and Lindelöf. Let f ∈ L0 (µ) be locally integrable. Then for any ² > 0 there is a continuous function g : X → R such that kf − gk1 ≤ ². proof (a) Write C for the set of bounded continuous functions g : X → R such that {x : g(x) 6= 0} has finite measure. Then C is a linear subspace of RX included in Lp = Lp (µ). Let U be the closure of C in Lp , that is, the set of h ∈ Lp such that for every ² > 0 there is a g ∈ C such that kh − gkp ≤ ². Then U is closed under addition and scalar multiplication. Also χE ∈ U whenever µE < ∞. P P Let ² > 0. Set δ = 41 ²1/p . Write G for the family of open sets of finite measure. Because µ is effectively locally finite, there is a G ∈ G such that µ(E \ G) ≤ δ. Let F ⊆ G \ E be a closed set such that µF ≥ µ(G \ E) − δ; then µ(E4(G \ F )) ≤ 2δ. Write H for the family of cozero sets. Because T is completely regular, H is a base for T; because H is closed under finite unions (4A2C(b-iii)) and µ is τ -additive, there is an H ∈ H such that H ⊆ G \ F and µH ≥ µ(G \ F ) − δ, so that µ(E4H) ≤ 3δ. Express H as {x : g(x) > 0} where g : X → R is a continuous function. For each n ∈ N, set gn = ng ∧ χX ∈ C; then

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|χE − gn |p ≤ χ(E4H) + (χH − gn )p for every n, so

R

|χE − gn |p ≤ µ(E4H) +

R

(χH − gn )p → µ(E4H) R as n → ∞, because gn → χH. So there is an n ∈ N such that |χE − gn |p ≤ 4δ, that is, kχE − gn kp ≤ ². As ² is arbitrary, χE ∈ U. Q Q Accordingly every simple function belongs to U. But if f ∈ Lp and ² > 0, there is a simple function h such that kf − hkp ≤ 21 ² (244Ha); now there is a g ∈ C such that kh − gkp ≤ 21 ² and kf − gkp ≤ ², as claimed. R (b) This time, write G for the family of open subsets of X such that G f is finite, so that G is an open cover of X. As X is paracompact (4A2H(b-i)), there is a locally finite family G0 ⊆ G covering X, which must be countable (4A2H(b-ii)). P Let h²G iG∈G0 be a family of strictly positive real numbers such that G∈G0 ²G ≤ ² (4A1P). Since X is completely regular (4A2H(b-i)),R we can apply (a) to see that, for each G ∈ G0 , there is a continuous function gG : X → R such that |gG − f × χG| ≤ ²G . Next, because X is normal (4A2H(b-i)), there is P a family hhG iG∈G0 of continuous functions from X to [0, 1] such that hG ≤ χG for every G ∈ G0 and G∈G0 hG (x) = P1 for every x ∈ X (4A2F(d-ix)). Set g(x) = G∈G0 gG (x)hG (x) for every x ∈ X. Because G0 is locally finite, g : X → R is continuous (4A2Bh). Now Z Z X X Z |f − g| = | (f − gG ) × hG | ≤ |(f − gG ) × hG | G∈G0

≤

X Z G∈G0

|f − gG | ≤ G

X

G∈G0

²G ≤ ²,

G∈G0

as required. 415Q Recall (411P) that if (A, µ ¯) is a localizable measure algebra, with Stone space (Z, S, T, ν), then ν is a strictly positive completion regular quasi-Radon measure, inner regular with respect to the open-and-closed sets (which are all compact). The following construction is primarily important for Radon measure spaces (see 416V), but is also of interest for general quasi-Radon spaces. Proposition Let (X, T, Σ, µ) be a quasi-Radon measure space and (A, µ ¯) its measure algebra. Let (Z, S, T, ν) be the Stone space of (A, µ ¯). For E ∈ Σ let E ∗ ⊆ Z be the open-and-closed set corresponding to the image E • of E in A. Define R ⊆ Z × X by saying that (z, x) ∈ R iff x ∈ F whenever F ⊆ X is closed and z ∈ F ∗ . Set Q = R−1 [X]. (a) R is a closed subset of Z × X. (b) For any E ∈ Σ, R[E ∗ ] is the smallest closed set such that µ(E \ R[E ∗ ]) = 0. In particular, if F ⊆ X is closed then R[F ∗ ] is the self-supporting closed set included in F such that µ(F \ R[F ∗ ]) = 0; and R[Z] is the support of µ. (c) Q is of full outer measure in Z. (d) For any E ∈ Σ, R−1 [E]4(Q∩E ∗ ) is negligible; consequently ν ∗ R−1 [E] = µE and R−1 [E]∩R−1 [X \E] is negligible. (e) For any A ⊆ X, ν ∗ R−1 [A] = µ∗ A. (f) If (X, T) is regular, then R−1 [G] is relatively open in Q for every open set G ⊆ X, R−1 [F ] is relatively closed in Q for every closed set F ⊆ X and R−1 [X \ E] = Q \ R−1 [E] for every Borel set E ⊆ X. proof (a) R=

\

((Z \ F ∗ ) × X) ∪ (Z × F )

F ⊆X is closed

is an intersection of closed sets, therefore closed.

S (b) Let G be the family of open sets G ⊆ X such that µ(E ∩ G) = 0, and G0 = G; then G0 ∈ G (414Ea). Set F0 = X \ G0 , so that F0 is the smallest closed set such that E \ F0 is negligible, and F0∗ ⊇ E ∗ .

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If (z, x) ∈ R and z ∈ E ∗ we must have x ∈ F0 . Thus R[E ∗ ] ⊆ F0 . On the other hand, if x ∈ F0 , and G is an open set containing x, then G ∈ / G so µ(G ∩ E) > 0 and (E ∩ G)∗ 6= ∅. Accordingly {(G ∩ E)∗ : x ∈ G ∈ T} is a downwards-directed family of non-empty open-and-closed sets in the compact space Z and has non-empty intersection, containing a point z say. If H ⊆ X is closed and z ∈ H ∗ , then X \ H is open and z ∈ / (X \ H)∗ , ∗ so x cannot belong to X \ H, that is, x ∈ H; as H is arbitrary, (z, x) ∈ R and x ∈ R[E ]; as x is arbitrary, R[E ∗ ] = F0 , as claimed. Of course, when E is actually closed, R[E ∗ ] = F0 ⊆ E. Taking E = X we see that R[Z] = R[X ∗ ] is the support of µ. (c) If W ∈ T and νW > 0, there is a non-empty open-and-closed set U ⊆ W , by 322Qa, which must be of the form E ∗ for some E ∈ Σ. By (b), R[E ∗ ] cannot be empty; but E ∗ ⊆ W , so R[W ] 6= ∅, that is, W ∩ Q 6= ∅. As W is arbitrary, ν∗ (Z \ Q) = 0, that is, Z is a measurable envelope of Q (413Ei). • • (d)(i) S Let F be the set of closed subsets of X included in E. Then supF ∈F F = E in A (412N), so E ∗ \ F ∈F F ∗ is nowhere dense and negligible. Now for each F ∈ F, R[F ∗ ] ⊆ F , so Q ∩ F ∗ ⊆ R−1 [F ] ⊆ R−1 [E]. Accordingly S Q ∩ E ∗ \ R−1 [E] ⊆ E ∗ \ F ∈F F ∗

is nowhere dense and negligible. (ii) ?? Suppose, if possible, that ν ∗ (R−1 [E] \ E ∗ ) > 0. Then there is an open-and-closed set U of finite measure such that ν ∗ (R−1 [E] ∩ U \ E ∗ ) > 0 (use 412Jc). Express U as H ∗ , where µH < ∞, and let F ⊆ H \ E be a closed set such that µ((H \ E) \ F ) < ν ∗ (R−1 [E] ∩ H ∗ \ E ∗ ). Then we must have ν ∗ (R−1 [E] ∩ F ∗ ) > 0. But R[F ∗ ] ⊆ F ⊆ X \ E so F ∗ ∩ R−1 [E] = ∅, which is impossible. X X (iii) Putting these together, (Q ∩ E ∗ )4R−1 [E] is negligible. (iv) It follows at once that (because Z is a measurable envelope for Q) ν ∗ R−1 [E] = ν ∗ (Q ∩ E ∗ ) = νE ∗ = µE. Moreover, applying the result to X \ E, R−1 [X \ E] ∩ R−1 [E] ⊆ (R−1 [X \ E]4(Q ∩ (X \ E)∗ )) ∪ (R−1 [E]4(Q ∩ E ∗ )) is negligible. (e)(i) Take E ∈ Σ such that A ⊆ E and µE = µ∗ A; then R−1 [A] ⊆ R−1 [E], so ν ∗ R−1 [A] ≤ ν ∗ R−1 [E] ≤ µE = µ∗ A. (ii) ?? Suppose, if possible, that ν ∗ R−1 [A] < µ∗ A. Let W ∈ T be such that R−1 [A] ⊆ W and νW = ν ∗ R−1 [A]. Then there is an F ∈ Σ such that ν(W 4F ∗ ) = 0. Since µF = νF ∗ = νW < µ∗ A, µ∗ (A \ F ) > 0; let G be a measurable envelope of A \ F disjoint from F . Then G∗ ∩ F ∗ = ∅ so ν(G∗ \ W ) = νG∗ = µG > 0 and there is a non-empty open-and-closed V ⊆ G∗ \ W ; let H ∈ Σ be such that H ⊆ G and V = H ∗ . In this case, R[V ] is closed and µ(H \ R[V ]) = 0, by (a), so that H ∩ R[V ] is measurable, not negligible, and included in G. But H ∩ R[V ] ∩ A is empty, because V ∩ R−1 [A] is empty, so µ∗ (H ∩ R[V ] ∩ A) < µ(H ∩ R[V ]), and G cannot be a measurable envelope of A \ F . X X Thus ν ∗ R−1 [A] = µ∗ A, as claimed. (f ) Suppose now that (X, T) is regular. (i) If G ⊆ X is open, R−1 [G] ∩ R−1 [X \ G] = ∅. P P If z ∈ R−1 [G], then there is an x ∈ G such that (z, x) ∈ R. Let H be an open set containing x such that H ⊆ G. Then x ∈ / X \ H so z ∈ / (X \ H)∗ , that is, ∗ z ∈ H . But ∗

R[H ∗ ] ⊆ R[H ] ⊆ H ⊆ G, ∗

so H ∩ R−1 [X \ G] = ∅ and z ∈ / R−1 [X \ G]. Q Q (ii) It is easy to check that

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E = {E : E ⊆ X, R−1 [E] ∩ R−1 [X \ E] = ∅} = {E : E ⊆ X, R−1 [X \ E] = Q \ R−1 [E]} is a σ-algebra of subsets of X (indeed, an algebra closed under arbitrary unions), just because R ⊆ Z × X and R−1 [X] = Q. Because it contains all open sets, E must contain all Borel sets. (iii) Now suppose once again that G ⊆ X is open and that z ∈ R−1 [G]. As in (i) above, there is an open set H ⊆ G such that z ∈ H ∗ ⊆ Z \ R−1 [X \ G], so that z ∈ H ∗ ∩ Q ⊆ R−1 [G]. As z is arbitrary, R−1 [G] is relatively open in Q. (iv) Finally, if F ⊆ X is closed, R−1 [F ] = Q \ R−1 [X \ F ] is relatively closed in Q. 415R Proposition Let (X, T, Σ, µ) be a Hausdorff quasi-Radon measure space and (Z, S, T, ν) the Stone space of its measure algebra. Let R ⊆ Z × X be the relation described in 415Q. Then (a) R is (the graph of) a function f ; (b) f is inverse-measure-preserving for the subspace measure νQ on Q = dom f , and in fact µ is the image measure νQ f −1 ; (c) if (X, T) is regular, then f is continuous. proof (a) If z ∈ Z and x, y ∈ X are distinct, let G, H be disjoint open sets containing x, y respectively. Then (X \ G)∗ ∪ (X \ H)∗ = ((X \ G) ∪ (X \ H))∗ = Z, defining ∗ as in 415Q, so z must belong to at least one of (X \ G)∗ , (X \ H)∗ . In the former case (z, x) ∈ /R and in the latter case (z, y) ∈ / R. This shows that R is a function; to remind us of its new status I will henceforth call it f . The domain of f is just Q = R−1 [X]. (b) By 415Qd, f is inverse-measure-preserving for νQ and µ. Suppose that A ⊆ X and f −1 [A] is in the domain TQ of νQ , that is, is of the form Q ∩ U for some U ∈ T. Take any E ∈ Σ such that µE > 0; then either ν(E ∗ ∩ U ) > 0 or ν(E ∗ \ U ) > 0. (α) Suppose that ν(E ∗ ∩ U ) > 0. Because ν is inner regular with respect to the open-and-closed sets, there is an H ∈ Σ such that H ∗ ⊆ E ∗ ∩ U and µH = νH ∗ > 0. Now there is a closed set F ⊆ E ∩H with µF > 0. In this case, f [F ∗ ] ⊆ F ⊆ E, by 415Qb, while F ∗ ∩U ⊆ f −1 [A], so f [F ∗ ] ⊆ E ∩ A. But this means that µ∗ (E ∩ A) ≥ µf [F ∗ ] = µF > 0. (β) If ν(E ∗ \ U ) > 0, then the same arguments show that µ∗ (E \ A) > 0. (γ) Thus µ∗ (E ∩ A) + µ∗ (E \ A) > 0 whenever µE > 0. Because µ is complete and locally determined, A ∈ Σ (413F(vii)). Thus we see that {A : A ⊆ X, f −1 [A] ∈ TQ } is included in Σ, and µ is the image measure νQ f −1 . (c) If T is regular, then 415Qf tells us that f is continuous. 415X Basic exercises >(a) Let (X, T, Σ, µ) be a quasi-Radon measure space and E ∈ Σ an atom for the measure. Show that there is a closed set F ⊆ E such that µF > 0 and F is an atom of Σ, in the sense that the only measurable subsets of F are ∅ and F . (Hint: 414G.) Show that µ is atomless iff all countable subsets of X are negligible. (b) Let h(Xi , Ti , Σi , µi )ii∈I be any family of quasi-Radon measure spaces. Show that the direct sum measure µ on X = {(x, i) : i ∈ I, x ∈ Xi } is a quasi-Radon measure when X is given its disjoint union topology. (c) Let S be the right-facing Sorgenfrey topology or lower limit topology on R, that is, the topology generated by the half-open intervals of the form [α, β[. Show that Lebesgue measure is completion regular and quasi-Radon for S. (Hint: 114Yj or 221Yb, or 419L.) (d) Let X be a topological space and µ a complete measure on X, and suppose that there is a conegligible closed measurable set Y ⊆ X such that the subspace measure on Y is quasi-Radon. Show that µ is quasiRadon.

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71

(e) Let (X, T, Σ, µ) be a quasi-Radon measure space. Show that µ is inner regular with respect to the family of self-supporting closed sets included in open sets of finite measure. (f ) Let (X, T, Σ, µ) be a quasi-Radon measure space. Show that for any E ∈ Σ, ² > 0 there is an open set G such that µG ≤ µE + ² and E \ G is negligible. (g) Find a compact Hausdorff quasi-Radon measure space which is not σ-finite. (h) Let (X, T, Σ, µ) be an atomless quasi-Radon measure space which is outer regular with respect to the open sets. Show that it is σ-finite. (Hint: if not, take a decomposition hXi ii∈I in which every Xi except one is self-supporting, and a set A meeting every Xi in just one point.) (i) Let (X, Σ, µ) be a σ-finite measure space in which Σ is countably generated as a σ-algebra. Show that, for a suitable topology on X, the completion of µ is a quasi-Radon measure. (Hint: take the topology generated by a countable subalgebra of Σ, and use the arguments of 415D.) (j) Let h(Xi , Ti , Σi , µi )ii∈I be a family of Q quasi-Radon probability spaces such that every µi is strictly positive, and λ the product measure on X = i∈I Xi . Show that if every Ti has a countable network, λ is a quasi-Radon measure. Q (k) Let hXi ii∈I be a family of separable metrizable spaces, and µ a quasi-Radon measure on X = i∈I Xi . Show that µ is completion regular iff every self-supporting closed set in X is determined by coordinates in a countable set. (Hint: 4A2Eb.) (l) Find two quasi-Radon measures µ, ν on the unit interval such that µG ≤ νG for every open set G but there is a closed set F such that νF < µF . (m) Let X be a topological space and µ, ν two quasi-Radon measures on X. (i) Suppose that µF = νF whenever F ⊆ X is closed and both µF and νF are finite. Show that µ = ν. (ii) Suppose that µG = νG whenever G ⊆ X is open and both µG and νG are finite. Show that µ = ν. (n) In 415L, write µ ˜0 for the c.l.d. version of µ0 (213E). Show that µ extends µ ˜0 . Show that µ ˜0 is τ -additive and inner regular with respect to the closed sets. (o) Let (X, T, Σ, µ) be a σ-finite paracompact Hausdorff quasi-Radon measure space, and f ∈ L0 (µ) a Rlocally integrable function. Show that for any ² > 0 there is a continuous function g : X → R such that |f − g| ≤ ². > (p) Let (X, T, Σ, µ) be a σ-finite completely regular quasi-Radon measure space. (i) Show that for every E ∈ Σ there is an F in the Baire σ-algebra Ba(X) of X such that µ(E4F ) = 0. (Hint: start with an open set E of finite measure.) (ii) Show that for every Σ-measurable function f : X → R there is a Ba(X)-measurable function equal almost everywhere to f . (q) Let (X, Σ, µ) be a measure space and f a µ-integrable real-valued function. Show that there is a unique quasi-Radon measure λ on R such that λ{0} = 0 and λ [α, ∞[ = µ∗ {x R : x ∈ dom R f , f (x) ≥ α}, λ ]−∞, −α] = µ∗ {x : x ∈ dom f , f (x) ≤ −α} whenever α > 0; and that h dλ = hf dµ whenever h ∈ L0 (λ) and h(0) = 0 and either integral is defined in [−∞, ∞]. (Hint: set λE = µ∗ f −1 [E \ {0}] for Borel sets E ⊆ R, and use 414Mb, 414O and 235Ib.) 415Y Further exercises (a) Give an example of two quasi-Radon measures µ, ν on R such that their sum, as defined in 112Xe, is not effectively locally finite, therefore not a quasi-Radon measure. (b) Show that any quasi-Radon measure space is isomorphic, as topological measure space, to a subspace ˆ be its of a compact quasi-Radon measure space. (Hint: if X is a T1 quasi-Radon measure space, let X Wallman compactification (Engelking 89, 3.6.21).)

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415Yc

(c) Let (X, T, Σ, µ) be a quasi-Radon measure space. Show that the following are equiveridical: (i) µ is outer regular with respect to the open sets; (ii) every negligible subset of X is included in an open set of finite measure; (iii) {x : µ{x} = 0} can be covered by a sequence of open sets of finite measure. (d) Show that + : R × R → R is continuous for the right-facing Sorgenfrey topology. (e) Let r ≥ 1. On R r let S be the topology generated by the half-open intervals [a, b[ where a, b ∈ R r (as defined in §115). (i) Show that S is the product topology if each factor is given the right-facing Sorgenfrey topology (415Xc). (ii) Show that Lebesgue measure is quasi-Radon for S. (Hint: induce on r. See also 417Yi.) (f ) Let Y ⊆ [0, 1] be a set of full outer measure and zero inner measure for Lebesgue measure µ. Give [0, 1] the topology T generated by S∪{Y } where S is the usual topology. Show that the subspace measure ν = µY is quasi-Radon for the subspace topology TY , but that there is no measure λ on X which is quasi-Radon for T and such that the subspace measure λY is equal to ν. (g) Find a base U for the topology of X = {0, 1}N and two totally finite (quasi-)Radon measures µ, ν on X such that G ∩ H ∈ U for all G, H ∈ U, µG ≤ νG for every G ∈ U , but νX < µX. (h) Let X be a topological space and G an open cover of X. Suppose that for each G ∈ G we are given a quasi-Radon measure µG on G such that µG (U ) = µH (U ) whenever G, H ∈ G and U ⊆ G ∩ H is open. Show that there is a unique quasi-Radon measure on X such that each µG is the subspace measure on G. (Hint: if hµG iG∈G is a maximal family with the given properties, then G is upwards-directed.) (i) Let (X, Σ, µ) be a measure space and T a topology on X, and suppose that there is a family U ⊆ Σ ∩ T such that (i) µU < ∞ for every U ∈ U (ii) for every U ∈ U, T S∩ Σ ∩ PU is a base S for the subspace topology of U (iii) if G is an upwards-directed family in T ∩ Σ and G ∈ U , then µ( G) = supG∈G µG (iv) µ is inner regular with respect to the closed sets (v) if E ∈ Σ and µE > 0 then there is a U ∈ U such that µ(E ∩ U ) > 0. Show that µ has an extension to a quasi-Radon measure on X. (j) Let (X, T, Σ, µ) be a quasi-Radon measure space such that T is normal (but not necessarily Hausdorff or regular). Show that if 1 ≤ p < ∞, f ∈ Lp (µ) and ² > 0, there is a bounded continuous function g : X → R such that kf − gkp ≤ ² and {x : g(x) 6= 0} has finite measure. (k) Let (X, T, Σ, µ) be a completely regular quasi-Radon measure space and suppose that we are given a uniformity defining the topology T. Show that if 1 ≤ p < ∞, f ∈ Lp (µ) and ² > 0, there is a bounded uniformly continuous function g : X → R such that kf − gkp ≤ ² and {x : g(x) 6= 0} has finite measure. (l) Let (X, T, Σ, µ) be a completely regular quasi-Radon measure space and τ an extended Fatou norm on L0 (µ) such that (i) τ ¹Lτ is an order-continuous norm (ii) whenever E ∈ Σ and µE > 0 there is an open set G such that µ(E ∩ G) > 0 and τ (χG• ) < ∞. Show that Lτ ∩ {f • : f : X → R is continuous} is norm-dense in Lτ . (m) Let (X, T, Σ, µ) be a quasi-Radon measure space. Show that µ is a compact measure in the sense of §342 iff there is a locally compact topology S on X such that (X, S, Σ, µ) is quasi-Radon. 415 Notes and comments 415B is particularly important because a very high proportion of the quasiRadon measure spaces we study are actually subspaces of Radon measure spaces. I would in fact go so far as to say that when you have occasion to wonder whether all quasi-Radon measure spaces have a property, you should as a matter of habit look first at subspaces of Radon measure spaces; if the answer is affirmative for them, you will have most of what you want, even if the generalization to arbitrary quasi-Radon spaces gives difficulties. Of course the reverse phenomenon can also occur. Stone spaces (411P) can be thought of as quasi-Radon compactifications of Radon measure spaces (416V). But this is relatively rare. Indeed the reason why I give so few examples of quasi-Radon spaces at this point is just that the natural ones arise from Radon measure spaces. Note however that the quasi-Radon product of an uncountable family of Radon

§416 intro.

Radon measure spaces

73

probability spaces need not be Radon (see 417Xq), so that 415E here and 417O below are sources of nonRadon quasi-Radon measure spaces. Density and lifting topologies can also provide us with quasi-Radon measure spaces (453Xd, 453Xg). 415K is the second in a series of inner-regular-extension theorems; there will be a third in 416J. I have been saying since Volume 1 that the business of measure theory, since Lebesgue’s time, has been to measure as many sets and integrate as many functions as possible. I therefore take seriously any theorem offering a canonical extension of a measure. 415L and its corollaries can all be regarded as improvement theorems, showing that a good measure can be made even better. We have already had such improvement theorems in Chapter 21: the completion and c.l.d. version of a measure (212C, 213E). In all such theorems we need to know exactly what effect our improvement is having on the other constructions we are interested in; primarily, the measure algebra and the function spaces. The machinery of Chapter 36 shows that if we understand the measure algebra(s) involved then the function spaces will give us no further surprises. Completion of a measure does not affect the measure algebra at all (322Da). Taking the c.l.d. version does not change Af = {a : µ ¯a < ∞} or L1 (213Fc, 213G, 322Db, 366H), but can affect the rest of the measure algebra and therefore L0 and L∞ . In this respect, what we might call the ‘quasi-Radon version’ behaves like the c.l.d. version (as could be expected, since the quasi-Radon version must itself be complete and locally determined; cf. 415Xn). The archetypal application of 415L is 415N. We shall see later how Baire measures arise naturally when studying Banach spaces of continuous functions (436E). 415N will be one of the keys to applying the general theory of topological measure spaces in such contexts. A virtue of Baire measures is that inner regularity with respect to closed sets comes almost free (412D); but there can be unsurmountable difficulties if we wish to extend them to Borel measures (439M), and it is important to know that τ -additivity, even in the relatively weak form allowed by the definition I use here (411C), is often enough to give a canonical extension to a well-behaved measure defined on every Borel set. In 415C we have inner regularity for a different reason, and the measure is already known to be defined on every Borel set, so in fact the quasi-Radon version of the measure is just the c.l.d. version (415Xd). For a volume and a half I have neglected indefinite-integral measures, though they are mentioned in the exercises; but we shall need them later, and in 415O I spell out a result which it will be useful to be able to quote. The exact hypotheses are not perhaps instantly predictable; see 414Xe. One interpretation of the Lifting Theorem is that for a complete strictly localizable measure space (X, Σ, µ) there is a function g : X → Z, where Z is the Stone space of the measure algebra of µ, such that E4g −1 [E ∗ ] is negligible for every E ∈ Σ, where E ∗ ⊆ Z is the open-and-closed set corresponding to the image of E in the measure algebra (341Q). For a Hausdorff quasi-Radon measure space we have a function f : Q → X, where Q is a dense subset of Z, such that (Q ∩ E ∗ )4f −1 [E] is negligible for every E ∈ Σ (415Qd, 415R); moreover, there is a canonical construction for this function. For completeness’ sake, I have given the result for general, not necessarily Hausdorff, spaces X (415Q); but evidently it will be of greatest interest for regular Hausdorff spaces (415Rc), especially if they happen to be ‘compact’ in the sense of §342. Perhaps I should remark that in the most important applications, Q is the whole of Z (416Xw). Of course the question arises, whether f g can be the identity. (Z typically has larger cardinal than X, so asking for gf to be the identity is a bit optimistic.) This is in fact an important question; I will return to it in 453M.

416 Radon measure spaces We come now to the results for which the chapter so far has been preparing. The centre of topological measure theory is the theory of ‘Radon’ measures (411Hb), measures inner regular with respect to compact sets. Most of the section is devoted to pulling the earlier work together, and in particular to re-stating theorems on quasi-Radon measures in the new context. Of course this has to begin with a check that Radon measures are quasi-Radon (416A). It follows immediately that Radon measures are (strictly) localizable (416B). After presenting a miscellany of elementary facts, I turn to the constructions of §413, which take on simpler and more dramatic forms in this context (416J-416P). I proceed to investigate subspace measures (416R-416T) and some special product measures (416U). I end the section with further notes on the forms which earlier theorems on Stone spaces (416V) and compact measure spaces (416W) take when applied to Radon measure spaces.

74


416A

416A Proposition A Radon measure space is quasi-Radon. proof Let (X, T, Σ, µ) be a Radon measure space. Because T is Hausdorff, every compact set is closed, so µ is inner regular with respect to the closed sets. By 411E, µ is τ -additive; by 411Gf, it is effectively locally finite. Thus all parts of condition (ii) of 411Ha are satisfied, and µ is a quasi-Radon measure. 416B Corollary A Radon measure space is strictly localizable. proof Put 416A and 415A together. 416C In order to use the results of §415 effectively, it will be helpful to spell out elementary conditions ensuring that a quasi-Radon measure is Radon. Proposition Let (X, T, Σ, µ) be a locally finite Hausdorff quasi-Radon measure space. Then the following are equiveridical: (i) µ is a Radon measure; (ii) whenever E ∈ Σ and µE > 0 there is a compact set K such that µ(E ∩ K) > 0; (iii) sup{K • : K ⊆ X is compact} = 1 in the measure algebra of µ. If µ is totally finite we can add (iv) sup{µK : K ⊆ X is compact} = µX. proof (i)⇒(ii) and (ii) ⇐⇒ (iii) are trivial. For (ii)⇒(i), observe that if E ∈ Σ and µE > 0 there is a compact set K ⊆ E such that µK > 0. P P There is a compact set K 0 such that µ(E ∩ K 0 ) > 0, by hypothesis; now there is a closed set K ⊆ E ∩ K 0 such that µK > 0, because µ is inner regular with respect to the closed sets, and K is compact. Q Q By 412B, µ is tight. Being a complete, locally determined, locally finite topological measure, it is a Radon measure. When µX < ∞, of course, we also have (ii) ⇐⇒ (iv). 416D Some further elementary facts are worth writing out plainly. Lemma (a) In a Radon measure space, every compact set has finite measure. (b) Let (X, T, Σ, µ) be a Radon measure space, and E ⊆ X a set such that E ∩ K ∈ Σ for every compact K ⊆ X. Then E ∈ Σ. (c) A Radon measure is inner regular with respect to the self-supporting compact sets. (d) Let X be a Hausdorff space and µ a locally finite complete locally determined measure on X which is tight (that is, inner regular with respect to the compact sets). If every compact set belongs to the domain of µ, µ is a Radon measure. proof (a) 411Ga. (b) We have only to remember that µ is complete, locally determined and tight, and apply 413F(ii). (c) If (X, T, Σ, µ) is a Radon measure space, E ∈ Σ and γ < µE, there is a compact set K ⊆ E such that µK ≥ γ. By 414F, there is a self-supporting relatively closed set L ⊆ K such that µL = µK; but now of course L is compact, while L ⊆ E and µL ≥ γ. (d) Let K be the family of compact subsets of X; write Σ for the domain of µ. If F ⊆ X is closed, then F ∩ K ∈ K ⊆ Σ for every K ∈ K; accordingly F ∈ Σ. But this means that every closed set, therefore every open set, belongs to Σ, and µ is a Radon measure. 416E Specification of Radon measures In 415H I described some conditions which enable us to be sure that two quasi-Radon measures on a given topological space are the same. In the case of Radon measures we have a similar list. This time I include a note on the natural ordering of Radon measures. Proposition Let X be a Hausdorff space and µ, ν two Radon measures on X. (a) The following are equiveridical: (i) dom ν ⊆ dom µ and µE ≤ νE for every E ∈ dom ν; (ii) µK ≤ νK for every compact set K ⊆ X.

416E


75

(iii) µG ≤ νG for every open set G ⊆ X; (iv) µF ≤ νF for every closed set F ⊆ X. If X is locally R compact, R we can add (v) f dµ ≤ f dν for every non-negative continuous function f : X → R with compact support. (b) The following are equiveridical: (i) µ = ν; (ii) µK = νK for every compact set K ⊆ X. (iii) µG = νG for every open set G ⊆ X; (iv) µF = νF for every closed set F ⊆ X. If X is locally R compact, R we can add (v) f dµ = f dν for every continuous function f : X → R with compact support. proof (a)(i)⇒(iv)⇒(ii) and (i)⇒(iii) are trivial. (ii)⇒(i) If (ii) is true, then µE = supK⊆E

µK ≤ supK⊆E

is compact

is compact

νK = νE

for every set E measured by both µ and ν. Also dom ν ⊆ dom µ. P P Suppose that E ∈ dom ν and that K ⊆ X is a compact set such that µK > 0. Then there are compact sets K1 ⊆ K ∩ E, K2 ⊆ K \ E such that νK1 + νK2 ≥ ν(K ∩ E) + ν(K \ E) − µK = νK − µK. So µ(K \ (K1 ∪ K2 )) ≤ ν(K \ (K1 ∪ K2 )) < µK and µK1 + µK2 > 0. This shows that µ∗ (K ∩ E) + µ∗ (K \ E) > 0. As K is arbitrary, E ∈ dom µ (413F(vii)). Q Q So (i) is true. (iii)⇒(ii)SThe point is that if K ⊆ X is compact, then µK = inf{µG : G ⊆ X is open, K ⊆ G}. P P Because X = {µG : G ⊆ X is open, µG < ∞}, there is an open set G0 of finite measure including K. Now, for any γ > µK, there is a compact set L ⊆ G0 \ K such that µL ≥ µG0 − γ, so that µG ≤ γ, where G = G0 \ L is an open set including K. Q Q The same is true for ν. So, if (iii) is true, µK = inf G⊇K

is open

µG ≤ inf G⊇K

is open

νG = νK

for every compact K ⊆ X, and (ii) is true. (iii)⇒(v) If (iii) is true and f : X → [0, ∞[ is a non-negative continuous function, then Z

Z

∞

f dµ =

µ{x : f (x) > t}dt 0

(252O)

Z

Z

∞

≤

ν{x : f (x) > t}dt =

f dν.

0

(v)⇒(iii) If X is locally compact and (v) is true, take any open set G ⊆ X, and consider A = {f : f is a continuous function with compact support from X to [0, 1] and f ≤ χG}. Then A is upwards-directed and supf ∈A f (x) = χG(x) for every x ∈ X, by 4A2G(e-i). So µG = supf ∈A by 414Ba. As G is arbitrary, (iii) is true. (b) now follows at once, or from 415H.

R

f dµ ≤ supf ∈A

R

f dν = νG

76


416F

416F Proposition Let X be a Hausdorff space and µ a Borel measure on X. Then the following are equiveridical: (i) µ has an extension to a Radon measure on X; (ii) µ is locally finite and tight; (iii) µ is locally finite and effectively locally finite, and µG = sup{µK : K ⊆ G is compact} for every open set G ⊆ X; (iv) µ is locally finite, effectively locally finite and τ -additive, and µG = sup{µ(G ∩ K) : K ⊆ X is compact} for every open set G ⊆ X. In this case the extension is unique; it is the c.l.d. version of µ. proof (a)(i)⇒(iv) If µ = µ ˜¹B where µ ˜ is a Radon measure and B is the Borel σ-algebra of X, then of course µ is locally finite and effectively locally finite and τ -additive because µ ˜ is (see 416A) and every open set belongs to B. Also µG = sup{µK : K ⊆ G is compact} ≤ sup{µ(G ∩ K) : K ⊆ X is compact} ≤ µG for every open set G ⊆ X, because µ ˜ is tight and compact sets belong to B. (b)(iv)⇒(iii) Suppose that (iv) is true. Of course µ is locally finite and effectively locally finite. Suppose that G ⊆ X is open and that γ < µG. Then there is a compact K ⊆ X such that µ(G ∩ K) > γ. By 414K, the subspace measure µK is τ -additive. Now K is a compact Hausdorff space, therefore regular. By 414Ma there is a closed set F ⊆ G ∩ K such that µK F ≥ γ. Now F is compact, F ⊆ G and µF ≥ γ. As G and γ are arbitrary, (iii) is true. (c)(iii)⇒(ii) I have to show that if µ satisfies the conditions of (iii) it is tight. Let K be the family of compact subsets of X and A the family of subsets of X which are either open or closed. Then whenever A ∈ A, F ∈ Σ and µ(A ∩ F ) > 0, there is a K ∈ K such that K ⊆ A and µ(K ∩ F ) > 0. P P Because µ is effectively locally finite, there is an open set G of finite measure such that µ(G ∩ A ∩ F ) > 0. (α) If A is open, then there will be a compact set K ⊆ G ∩ A such that µK > µ(G ∩ A) − µ(G ∩ A ∩ F ), so that µ(K ∩ F > 0. (β) If A is closed, then let L ⊆ G be a compact set such that µL > µG − µ(G ∩ A ∩ F ); then K = L ∩ A is compact and µ(K ∩ F ) > 0. Q Q By 412C, µ is inner regular with respect to K, as required. (d)(ii)⇒(i) If µ is locally finite and tight, let µ ˜ be the c.l.d. version of µ. Then µ ˜ is complete, locally determined, locally finite (because µ is), a topological measure (because µ is) and tight (because µ is, using 412Ha); so is a Radon measure. Every compact set has finite measure for µ, so µ is semi-finite and µ ˜ extends µ (213Hc). (e) By 416Eb there can be at most one Radon measure extending µ, and we have observed in (c) above that in the present case it is the c.l.d. version of µ. 416G One of the themes of §434 will be the question: on which Hausdorff spaces is every locally finite quasi-Radon measure a Radon measure? I do not think we are ready for a general investigation of this, but I can give one easy special result. Proposition Let (X, T) be a locally compact Hausdorff space and µ a locally finite quasi-Radon measure on X. Then µ is a Radon measure. proof µ satisfies condition (ii) of 416C. P P Take E ∈ dom µ such that µE > 0. Let G be the family of relatively compact open subsets of X; then G is upwards-directed and has union X. By 414Ea, there is a G ∈ G such that µ(E ∩ G) > 0. But now G is compact and µ(E ∩ G) > 0. Q Q By 416C, µ is a Radon measure. 416H Corollary Let (X, T) be a locally compact Hausdorff space, and µ a locally finite, effectively locally finite, τ -additive Borel measure on X. Then µ is tight and its c.l.d. version is a Radon measure, the unique Radon measure on X extending µ. proof By 415Cb, the c.l.d. version µ ˜ of µ is a quasi-Radon measure extending µ. Because µ is locally finite, so is µ ˜; by 416G, µ ˜ is a Radon measure. By 416Eb, the extension is unique. Now

416K


µE = µ ˜E = supK⊆E

is compact

µ ˜K = supK⊆E

77 is compact

µK

for every Borel set E ⊆ X, so µ itself is tight. 416I While on the subject of locally compact spaces, I mention an important generalization of a result from Chapter 24. Proposition Let (X, T, Σ, µ) be a locally compact Radon measure space. Write Ck for the space of continuous real-valued functions on X with compact supports. If 1 ≤ p < ∞, f ∈ Lp (µ) and ² > 0, there is a g ∈ Ck such that kf − gkp ≤ ². proof By 415Pa, there is a bounded continuous function h1 : X → R such that G = {x : h1 (x) 6= 0} has finite measure and kf − h1 kp ≤ 21 ². Let K ⊆ G be a compact set such that kh1 k∞ (µ(G \ K))1/p ≤ 21 ², and let h2 ∈ Ck be such that χK ≤ h2 ≤ χG (4A2G(e-i)). Set g = h1 × h2 . Then g ∈ Ck and

R

|h1 − g|p ≤

R

G\K

|h1 |p ≤ µ(G \ K)kh1 kp∞ ,

so kh1 − gkp ≤ 21 ² and kf − gkp ≤ ², as required. 416J I turn now to constructions of Radon measures based on ideas in §413. Theorem Let X be a Hausdorff space. Let K be the family of compact subsets of X and φ0 : K → [0, ∞[ a functional such that (α) φ0 K = φ0 L + sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K, (γ) for every x ∈ X there is an open set G containing x such that sup{φ0 K : K ∈ K, K ⊆ G} is finite. Then there is a unique Radon measure on X extending φ0 . proof By 413M, there is a unique complete locally determined measure µ on X, extending φ0 , which is inner regular with respect to K. By (γ), µ is locally finite; by 416Dd, it is a Radon measure. 416K Proposition Let X be a regular Hausdorff space. Let K be the family of compact subsets of X, and φ0 : K → [0, ∞[ a functional such that (α1 ) φ0 K ≤ φ0 (K ∪ L) ≤ φ0 K + φ0 L for all K, L ∈ K, (α2 ) φ0 (K ∪ L) = φ0 K + φ0 L whenever K, L ∈ K and K ∩ L = ∅, (γ) for every x ∈ X there is an open set G containing X such that sup{φ0 K : K ∈ K, K ⊆ G} < ∞. Then there is a unique Radon measure µ on X such that µK = inf G⊆X

is open,K⊆G

supL⊆G is compact φ0 L

for every K ∈ K. proof (a) For open sets G ⊆ X set ψG = supL∈K,L⊆G φ0 L, and for compact sets K ⊆ X set φ1 K = inf{ψG : G ⊆ X is open, K ⊆ G}. Evidently ψG ≤ ψH whenever G ⊆ H. We need to know that ψ(G ∪ H) ≤ ψG + ψH for all open sets G, H ⊆ X. P P If L ⊆ G ∪ H is compact, then the disjoint compact sets L \ G, L \ H can be separated by disjoint open sets H 0 , G0 (4A2F(h-i)); now L \ G0 ⊆ H, L \ H 0 ⊆ G are compact and cover L, so φ0 L ≤ φ0 (L \ G0 ) + φ0 (L \ H 0 ) ≤ ψH + ψG. As L is arbitrary, ψ(G ∪ H) ≤ ψG + ψH. Q Q Moreover, ψ(G ∪ H) = ψG + ψH if G ∩ H = ∅. P P If K ⊆ G, L ⊆ H are compact, then φ0 K + φ0 L = φ0 (K ∪ L) ≤ ψ(G ∪ H). As K and L are arbitrary, ψG + ψH ≤ ψ(G ∪ H). Q Q

78


416K

(b) It follows that φ1 K is finite for every compact K ⊆ X. P P Set G = {G : G ⊆ X is open, ψG < ∞}. Then (a) tells us that G is upwards-directed. But also we are supposing that G covers X, by (γ). So if K ⊆ X is compact there is a member of G including K and φ1 K < ∞. Q Q (c) Now φ1 satisfies the conditions of 416J. α) Suppose that K, L ∈ K and L ⊆ K. Set γ = sup{φ1 M : M ∈ K, M ⊆ K \ L}. Take any ² > 0. P P(α Let G be an open set such that K ⊆ G and ψG ≤ φ1 K + ². If M ∈ K and M ⊆ K \ L, there are disjoint open sets U , V such that L ⊆ U and M ⊆ V (4A2F(h-i) again); we may suppose that U ∪ V ⊆ G. In this case,

φ1 L + φ1 M ≤ ψU + ψV = ψ(U ∪ V ) (by the second part of (a) above) ≤ ψG ≤ φ1 K + ². As M is arbitrary, γ ≤ φ1 K − φ1 L + ². On the other hand, there is an open set H such that L ⊆ H and ψH ≤ φ1 L + ². Set F = K \ H, so that F is a compact subset of K \ L. Then there is an open set V such that F ⊆ V and ψV ≤ φ1 F + ². In this case K ⊆ H ∪ V , so φ1 K ≤ ψ(H ∪ V ) ≤ ψH + ψV ≤ φ1 L + φ1 F + 2² ≤ φ1 L + γ + 2², so γ ≥ φ1 K − φ1 L − 2². As ² is arbitrary, γ = φ1 K − φ1 L; as K and L are arbitrary, φ1 satisfies condition (α) of 416J. (γγ ) Any x ∈ X is contained in an open set G such that ψG < ∞; but now sup{φ1 K : K ∈ K, K ⊆ G} ≤ ψG is finite. So φ1 satisfies condition (γ) of 416J. Q Q (d) By 416J, there is a unique Radon measure on X extending φ1 , as claimed. 416L Corollary Let X be a locally compact Hausdorff space. Let K be the family of compact subsets of X, and φ0 : K → [0, ∞[ a functional such that φ0 K ≤ φ0 (K ∪ L) ≤ φ0 K + φ0 L for all K, L ∈ K, φ0 (K ∪ L) = φ0 K + φ0 L whenever K, L ∈ K and K ∩ L = ∅. Then there is a unique Radon measure µ on X such that µK = inf{φ0 K 0 : K 0 ∈ K, K ⊆ int K 0 } for every K ∈ K. proof Observe that φ0 satisfies the conditions of 416K; 416K(γ) is true because X is locally compact. Define ψ, φ1 as in the proof of 416K, and set φ01 K = inf{φ0 K 0 : K 0 ∈ K, K ⊆ int K 0 } for every K ∈ K. Then φ01 = φ1 . P P Let K ∈ K, ² > 0. (i) There is an open set G ⊆ X such that K ⊆ G and ψG ≤ φ1 K + ². Now the relatively compact open subsets with closures included in G form an upwards-directed cover of K, so there is a K 0 ∈ K such that K ⊆ int K 0 and K 0 ⊆ G. Accordingly φ01 K ≤ φ0 K 0 ≤ ψG ≤ φK + ². (ii) There is an L ∈ K such that K ⊆ int L and φ0 L ≤ φ01 K + ², so that φ1 K ≤ ψ(int L) ≤ φ0 L ≤ φ01 K + ². (iii) As K, ² are arbitrary, φ01 = φ1 . Q Q Now 416K tells us that there is a unique Radon measure extending φ1 , and this is the measure we seek.

416O


79

416M The extension theorems in the second half of §413 also have important applications to Radon measures. Henry’s Theorem (Henry 69) Let X be a Hausdorff space and µ0 a measure on X which is locally finite and tight. Then µ0 has an extension to a Radon measure µ on X; and the extension may be made in such a way that whenever µE < ∞ there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0. proof All the work has been done in §413; we need to check here only that the family K of compact subsets of X and the measure µ0 satisfy the hypotheses of 413O. But (†) and (‡) there are elementary, and µ∗0 K < ∞ for every K ∈ K by 411Ga. Now take the measure µ from 413O. It is complete, locally determined and inner regular with respect to K; also K ⊆ dom µ. Because µ0 is locally finite and µ extends µ0 , µ is locally finite. By 416Dd, µ is a Radon measure. And the construction of 413O ensures that every set of finite measure for µ differs from a member of Σ0 by a µ-negligible set. 416N Proposition Let X be a Hausdorff space and T a subalgebra of PX. Let ν : T → R be a non-negative finitely additive functional. Then there is a Radon measure µ on X such that µX ≤ νX and µK ≥ νK for every compact set K ∈ T. proof Use 413S, with K the family of compact subsets of X, and 416Dd. 416O Theorem Let X be a Hausdorff space and T a subalgebra of PX. Let ν : T → R be a non-negative finitely additive functional such that νE = sup{νF : F ∈ T, F ⊆ E, F is closed} for every E ∈ T, νX = supK⊆X is compact inf F ∈T,F ⊇K νF . Then there is a Radon measure µ on X extending ν. proof (a) For A ⊆ X, write ν ∗ A = inf F ∈T,F ⊇A νF .

S Let hKn in∈N be a sequence of compact subsets of X such that limn→∞ ν ∗ Kn = νX; replacing Kn by i
νn0 E

+

νn0 F.

As E and F are arbitrary, νn0 is additive. Q Q 0 (b) For each n ∈ N, set νn E = νn+1 E − νn0 E for every E ∈ T; then νn is additive. Because Kn+1 ⊇ Kn , νn is non-negative. 0 If E ∈ T and E ∩ Kn+1 = ∅, then νn E = νn+1 E = 0. So if we set Tn = {E ∩ Kn+1 : E ∈ T}, we have an additive functional νñ : Tn → [0, ∞[ defined by setting νñ (E ∩ Kn+1 ) = νn E for every E ∈ T. Also νñ H = sup{˜ νn K : K ∈ Tn , K ⊆ H, K is compact} for every H ∈ Tn . P P Express H as E ∩ Kn+1 where E ∈ T. Given ² > 0, there is a closed set F ∈ T such that F ⊆ E and νF ≥ νE − ²; but now K = F ∩ Kn+1 ∈ Tn is a compact subset of H, and 0 νñ (H \ K) = νn (E \ F ) ≤ νn+1 (E \ F ) ≤ ν(E \ F ) ≤ ²,

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so νñ K ≥ νñ H − ². Q Q (c) For each n ∈ N, we have a Radon measure µn on Kn+1 , with domain Σn say, such that µn Kn+1 ≤ νñ Kn+1 and µn K ≥ νñ K for every compact set K ⊆ Kn+1 (416N). Since Kn+1 is itself compact, we must have µn Kn+1 = νñ Kn+1 . But this means that µn extends νñ . P P If H ∈ Tn , ² > 0 there is a compact set K ∈ Tn such that νñ K ≥ νñ H − ², so that (µn )∗ H ≥ µn K ≥ νñ H − ²; as ² is arbitrary, (µn )∗ H ≥ νñ H. So there is an F1 ∈ Σn such that F1 ⊆ H and µn F1 ≥ νñ H. Similarly, there is an F2 ∈ Σn such that F2 ⊆ Kn+1 \ H and µn F2 ≥ νñ (Kn+1 \ H). But in this case H \ F1 ⊆ Kn+1 \ (F1 ∪ F2 ) is µn -negligible, because µn F1 + µn F2 ≥ νñ H + νñ (Kn+1 \ H) = νñ Kn+1 = µn Kn+1 . So H \ F1 and H belong to Σn and µn H = µn F1 = νñ H. Q Q (d) Set Σ = {E : E ⊆ X, E ∩ Kn+1 ∈ Σn for every n ∈ N}, P∞ µE = n=0 µn (E ∩ Kn+1 ) for every E ∈ Σ. Then µ is a Radon measure on X extending ν. P P (i) It is easy to check that Σ is a σ-algebra of subsets of X including T, just because each Σn is a σ-algebra of subsets of Kn+1 including Tn ; and that µ is a complete measure because every µn is. (ii) If E ∈ T, then µE =

∞ X

µn (E ∩ Kn+1 ) =

n=0

= lim

∞ X

νñ (E ∩ Kn+1 ) =

n=0 n X

n→∞

∞ X

νn E = lim

n→∞

n=0

n X

νi E

i=0

ν ∗ (E ∩ Ki+1 ) − ν ∗ (E ∩ Ki ) = lim ν ∗ (E ∩ Kn ) ≤ νE. n→∞

i=0

On the other hand, µX = limn→∞ ν ∗ Kn+1 = νX, so in fact µE = νE for every E ∈ T, that is, µ extends ν. In particular, µ is totally finite, therefore locally determined and locally finite. (iii) If G ⊆ X is open, then G ∩ Kn+1 ∈ Σn for every n, so G ∈ Σ; thus µ is a topological measure. If µE > 0, there is some n ∈ N such that µn (E ∩ Kn+1 ) > 0; now there is a compact set K ⊆ E ∩ Kn+1 such that µn K > 0, so that µK > 0. This shows that µ is tight, so is a Radon measure, as required. Q Q Remark Observe that in this construction µKn+1 = =

∞ X i=0 ∞ X

µi (Kn+1 ∩ Ki+1 ) =

∞ X

ν˜i (Kn+1 ∩ Ki+1 ) =

i=0

∞ X

νi (Kn+1 ∩ Ki+1 )

i=0

0 νi+1 (Kn+1 ∩ Ki+1 ) − νi0 (Kn+1 ∩ Ki+1 )

i=0

=

∞ X

ν ∗ (Kn+1 ∩ Ki+1 ) − ν ∗ (Kn+1 ∩ Ki )

i=0

=

n X

ν ∗ (Kn+1 ∩ Ki+1 ) − ν ∗ (Kn+1 ∩ Ki ) = ν ∗ Kn+1

i=0

for every n ∈ N. What this means is that if instead of the hypothesis νX = supK⊆X

is compact

inf F ∈T,F ⊇K νF

we are presented with a specified non-decreasing sequence hLn in∈N of compact subsets of X such that νX = supn∈N ν ∗ Ln , then we can take Kn+1 = Ln in the argument above and we shall have µLn = ν ∗ Ln for every n.

416P


81

416P Theorem Let X be a Hausdorff space and µ a locally finite measure on X which is inner regular with respect to the closed sets. Then the following are equiveridical: (i) µ has an extension to a Radon measure on X; (ii) for every non-negligible measurable set E ⊆ X there is a compact set K ⊆ E such that µ∗ K > 0. If µ is totally finite, we can add (iii) sup{µ∗ K : K ⊆ X is compact} = µX. proof Write Σ for the domain of µ. (a)(i)⇒(ii) If λ is a Radon measure extending µ, and µE > 0, then λE > 0, so there is a compact set K ⊆ E such that λK > 0; but now, because λ is an extension of µ, µ∗ K ≥ λ∗ K = λK > 0. α) Let E be the family of measurable envelopes of compact sets. Then µE < ∞ (b)(ii)⇒(i) & (iii)(α for every E ∈ E. P P If E ∈ E, there is a compact set K such that E is a measurable envelope of K. Now µE = µ∗ K is finite by 411G. Q Q Next, E is closed under finite unions, by 132Ed. The hypothesis (ii) tells us that if µE > 0 then there is some F ∈ E such that F ⊆ E and µF > 0; for there is a compact set K ⊆ E such that µ∗ K > 0, K has a measurable envelope F0 , and F = E ∩ F0 is still a measurable envelope of K. So in fact µ is inner regular with respect to E (412Aa). In particular, µ is semi-finite. If γ < µX there is an F ∈ E such that µF ≥ ², and now there is a compact set K such that F is a measurable envelope of K, so that µ∗ K = µF ≥ γ. As γ is arbitrary, (iii) is true. β ) Because µ is inner regular with respect to E, D = {E • : E ∈ E} is order-dense in the measure (β algebra (A, µ ¯) of µ (412N), so there is a family hdi ii∈I in D which is a partition of unity in A (313K). For each i ∈ I, take Ei ∈ E such that Ei• = di . Then X

µ(E ∩ Ei ) =

i∈I

X

µ ¯(E • ∩ di ) = µ ¯E •

i∈I

(321E) = µE for every E ∈ Σ. (γγ ) For each i ∈ I, let µi be the subspace measure on Ei . Then there is a Radon measure λi on Ei extending µi . P P Because µ is inner regular with respect to the closed sets, µi is inner regular with respect to the relatively closed subsets of Ei (412O). Also there is a compact subset K ⊆ Ei such that µi Ei = µEi = µ∗ K = µ∗i K, so µi satisfies the conditions of 416O and has an extension to a Radon measure. Q Q (δδ ) Define λE =

P i∈I

λi (E ∩ Ei )

whenever E ⊆ X is such that λi measures E ∩ Ei for every i ∈ I. Then λ is a Radon measure on X extending µ. P P It is easy to check that it is a measure, just because every λi is a measure, and it extends µ by (β) above. If G ⊆ X is open, then G ∩ Ei is relatively open for every i ∈ I, so λ measures G; thus λ is a topological measure. If λE = 0 and A ⊆ E, then λi (A ∩ Ei ) ≤ λ(E ∩ Ei ) = 0 for every i, so λA = 0; thus λ is complete. For all distinct i, j ∈ I, λi (Ei ∩ Ej ) = µi (Ei ∩ Ej ) = µ(Ei ∩ Ej ) = µ ¯(di ∩ dj ) = 0, so λEi = λi Ei = µi Ei is finite. This means that if E ⊆ X is such that λ measures E ∩ F whenever λF < ∞, then λ must measure E ∩ Ei for every i, and λ measures E; thus λ is locally determined. If λE > 0 there are an i ∈ I such that λi (E ∩ Ei ) > 0 and a compact K ⊆ E ∩ Ei such that 0 < λi K = λK; consequently λ

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is tight. Finally, if x ∈ X, there is an E ∈ Σ such that x ∈ int E and λE = µE < ∞, so λ is locally finite. Thus λ is a Radon measure. Q Q So (i) is true. (c) Finally, suppose that µ is totally finite and (iii) is true. Then we can appeal directly to 416O to see that (i) is true. 416Q Proposition (a) Let X be a compact Hausdorff space and E the algebra of open-and-closed subsets of X. Then any non-negative finitely additive functional from E to R has an extension to a Radon measure on X. If X is zero-dimensional then the extension is unique. (b) Let A be a Boolean algebra, and Z its Stone space. Then there is a one-to-one correspondence between non-negative additive functionals ν on A and Radon measures µ on Z given by the formula νa = µb a for every a ∈ A, where for a ∈ A I write b a for the corresponding open-and-closed subset of Z. proof (a) Let ν : E → [0, ∞[ be a non-negative additive functional. Then ν satisfies the conditions of 416O (because every member of E is closed, while X is compact), so has an extension to a Radon measure µ. If X is zero-dimensional, E is a base for the topology of X closed under finite unions and intersections, so µ is unique, by 415H(iv) or 415H(v). (b) The map a 7→ b a is a Boolean isomorphism between A and the algebra E of open-and-closed subsets of Z, so we have a one-to-one correspondence between non-negative additive functionals ν on A and nonnegative additive functionals ν 0 on E defined by the formula ν 0 b a = νa. Now Z is compact, Hausdorff and zero-dimensional, so ν 0 has a unique extension to a Radon measure on Z, by part (a). And of course every Radon measure µ on Z gives us a non-negative additive functional µ¹E on E, corresponding to a non-negative additive functional on A. 416R Theorem (a) Any subspace of a Radon measure space is a quasi-Radon measure space. (b) A measurable subspace of a Radon measure space is a Radon measure space. (c) If (X, T, Σ, µ) is a Hausdorff complete locally determined topological measure space, and Y ⊆ X is such that the subspace measure µY on Y is a Radon measure, then Y ∈ Σ. proof (a) Put 416A and 415B together. (b) Let (X, T, Σ, µ) be a Radon measure space, and (E, TE , ΣE , µE ) a member of Σ with the induced topology and measure. Because µ is complete and locally determined, so is µE (214Ja). Because T is Hausdorff, so is TE (4A2F(a-i)). Because µ is locally finite, so is µE . Because µ is tight (and a subset of E is compact for TE whenever it is compact for T), µE is tight (412Oa). (c) ?? If Y ∈ / Σ, then there is a set F ∈ Σ such that µ∗ (Y ∩ F ) < µ∗ (Y ∩ F ) (413F(v)). But now µ (Y ∩ F ) = µY (Y ∩ F ), so there is a compact set K ⊆ Y ∩ F such that µY K > µ∗ (Y ∩ F ). When regarded as a subset of X, K is still compact; because T is Hausdorff, K is closed, so belongs to Σ, and ∗

µ∗ (Y ∩ F ) ≥ µK = µY K > µ∗ (Y ∩ F ), which is absurd. X X 416S Corresponding to 415O, we have the following. Proposition Let (X, T, Σ, µ) be a Radon measure space, and ν an indefinite-integral measure over µ (definition: 234B). If ν is locally finite, it is a Radon measure. proof Because µ is complete and locally determined, so is ν (234Fb). Because µ is tight (412Q). So if ν is also locally finite, it is a Radon measure. 416T I said in the notes to §415 that the most important quasi-Radon measure spaces are subspaces of Radon measure spaces. I do not know of a useful necessary and sufficient condition, but the following deals with completely regular spaces.

416V


83

Proposition Let (X, T, Σ, µ) be a locally finite completely regular Hausdorff quasi-Radon measure space. Then it is isomorphic, as topological measure space, to a subspace of a locally compact Radon measure space. ˇ proof (a) Write βX for the Stone-Cech compactification of X (4A2I); I will take it that X is actually a subspace of βX. Let U be the set of those open subsets U of βX such S that µ(U ∩ X) < ∞; then U is upwards-directed and covers X, because µ is locally finite. Set W = U ⊇ X. Then W is an open subset of βX, so is locally compact. (b) Let B be the Borel σ-algebra of W . Then V ∩ X is a Borel subset of X for every V ∈ B (4A3Ca), so we have a measure ν : B → [0, ∞] defined by setting νV = µ(X ∩ V ) for every V ∈ B. Now ν satisfies the conditions of 415Cb. P P (α) If νV > 0, then, because µ is effectively locally finite, there is an open set G ⊆ X such that µ(G ∩ V ) > 0 and µG < ∞. There is an open set U ⊆ βX such that U ∩ X = G, in which case U ⊆ W , νU < ∞ and ν(U ∩ V ) > 0. Thus ν is effectively locally finite. (β) If U is an upwards-directed family of open subsets of W , then {U ∩ X : U ∈ U} is an upwards-directed family of open subsets of X, so [ [ [ ν( U) = µ(X ∩ U ) = µ( {U ∩ X : U ∈ U}) = sup µ(X ∩ U ) = sup νU. Q Q U ∈U

U ∈U

So the c.l.d. version ν˜ of ν is a quasi-Radon measure on W (415Cb). (c) The construction of W ensures that ν and ν˜ are locally finite. By 416G, ν˜ is a Radon measure. So the subspace measure ν˜X is a quasi-Radon measure on X (416Ra). But ν˜X G = µG for every open set G ⊆ X. P P Note first that as ν effectively locally finite, therefore semi-finite, ν˜ extends ν (213Hc). If K ⊆ W is a compact set not meeting X, then ν˜K = νK = µ(K ∩ X) = 0; accordingly ν˜∗ (W \ X) = 0, by 413Ee. Now there is an open set U ⊆ W such that G = X ∩ U , and ν˜X G = ν˜∗ G ≤ ν˜U = νU = µ(U ∩ X) = µG = ν˜∗ (U ∩ X) + ν˜∗ (U \ X) (by 413E(c-ii), because ν˜ is semi-finite) ≤ ν˜∗ (U ∩ X) + ν˜∗ (W \ X) = ν˜∗ G. Q Q So 415H(iii) tells us that µ = ν˜X is the subspace measure induced by ν. 416U Theorem If h(Xi , Ti , Σi , µi )ii∈I is a family of compact metrizable Radon probability spaces such Q that every µi is strictly positive, the product measure on X = i∈I Xi is a completion regular Radon measure. In particular, the usual measures on {0, 1}I and [0, 1]I and PI are completion regular Radon measures, for any set I. proof By 415E, it is a completion regular quasi-Radon probability measure; but X is a compact Hausdorff space, so it is a Radon measure, by 416G or otherwise. Remark I suppose it is obvious that by the ‘usual measure on [0, 1]I ’ I mean the product measure when each copy of [0, 1] is given Lebesgue measure. Recall also that the ‘usual measure on PI’ is just the copy of the usual measure on {0, 1}I induced by the standard bijection A ↔ χA (254Jb), which is a homeomorphism (4A2Ud). 416V Stone spaces The results of 415Q-415R become simpler and more striking in the present context. Theorem Let (X, T, Σ, µ) be a Radon measure space, and (Z, S, T, ν) the Stone space of its measure algebra (A, µ ¯). For E ∈ Σ let E ∗ be the open-and-closed set in Z corresponding to the image E • of E in A. Define R ⊆ Z × X by saying that (z, x) ∈ R iff x ∈ F whenever F ⊆ X is closed and z ∈ F ∗ .

84


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S (a) R is the graph of a function f : Q → X, where Q = R−1 [X]. If we set W = {K ∗ : K ⊆ X is compact}, then W ⊆ Q is a ν-conegligible open set, and the subspace measure νW on W is a Radon measure. (b) Setting g = f ¹ W , g is continuous and µ is the image measure νW g −1 . (c) If X is compact, W = Q = Z and µ = νg −1 . proof (a) By 415Ra, R is the graph of a function. If K ⊆ X is compact and z ∈ K ∗ , then F = {F : F ⊆ X is closed, z ∈ F ∗ } is a family of non-empty closed subsets of X, closed under finite intersections, and containing the compact set K; so it has non-empty intersection, and there is an x ∈ K such that (z, x) ∈ R, that is, z ∈ Q and f (z) ∈ K. Thus W ⊆ Q. Of course W is an open set, being the union of a family of open-and-closed sets; but it is also conegligible, because sup{K • : K ⊆ X is compact} = 1 in A (412N), so Z \ W must be nowhere dense, therefore negligible. Now the subspace measure νW is quasi-Radon because ν is (411P(d-iv), 415B); but W is a union of compact open sets of finite measure, so νW is locally finite and W is locally compact; by 416G, νW is a Radon measure. (b) g is continuous. P P Let G ⊆ X be an open set and z ∈ g −1 [G]. Let K ⊆ X be a compact set such ∗ that z ∈ K . As remarked above, g(z) = f (z) belongs to K. K, being a compact Hausdorff space, is regular (3A3Bb), so there is an open set H containing g(z) such that L = H ∩ K ⊆ G. Note that L is compact, so L∗ ⊆ W . Now g(z) does not belong to the closed set X \ H, so z ∈ / (X \ H)∗ and z ∈ H ∗ ; accordingly ∗ ∗ ∗ ∗ −1 z ∈ (H ∩ K) ⊆ L . If w ∈ L , g(w) ∈ L ⊆ G; so L ⊆ g [G], and z ∈ int g −1 [G]. As z is arbitrary, g −1 [G] is open; as G is arbitrary, g is continuous. Q Q By 415Rb, we know that µ = νQ f −1 , where νQ is the subspace measure on Q. But as ν is complete and both Q and W are conegligible, we have νQ f −1 [A] = νf −1 [A] = νg −1 [A] = νW g −1 [A] whenever A ⊆ X and any of the four terms is defined, so that µ = νQ f −1 = νW g −1 . (c) If X is compact, then Z = X ∗ ⊆ W , so W = Q = Z and νg −1 = νW g −1 = µ. 416W Compact measure spaces Recall that a semi-finite measure space (X, Σ, µ) is ‘compact’ (as T a measure space) if there is a family K ⊆ Σ such that µ is inner regular with respect to K and K0 6= ∅ whenever K0 ⊆ K has the finite intersection property (342A); while (X, Σ, µ) is ‘perfect’ if whenever f : X → R is measurable and µE > 0, there is a compact set K ⊆ f [E] such that µf −1 [K] > 0 (342K). In §342 I introduced these concepts in order to study the realization of homomorphisms between measure algebras. The following result is now very easy. Proposition (a) Any Radon measure space is a compact measure space, therefore perfect. (b) Let (X, T, Σ, µ) be a Radon measure space, with measure algebra (A, µ ¯), and (Y, T, ν) a complete strictly localizable measure space, with measure algebra (B, ν¯). If π : A → B is an order-continuous Boolean homomorphism, there is a function f : Y → X such that f −1 [E] ∈ T and f −1 [E]• = πE • for every E ∈ Σ. If π is measure-preserving, f is inverse-measure-preserving. proof (a) If (X, T, Σ, µ) is a Radon measure space, µ is inner regular with respect to the compact class consisting of the compact subsets of X, so (X, Σ, µ) is a compact measure space. By 342L, it is perfect. (b) Use (i)⇒(v) of Theorem 343B. (Of course f is inverse-measure-preserving iff π is measure-preserving.) 416X Basic exercises > (a) Let (X, T, Σ, µ) be a Radon measure space, and E ∈ Σ an atom for the measure. Show that there is a point x ∈ E such that µ{x} = µE. (b) Let X be a topological space and µ a point-supported measure on X, as described in 112Bd. (i) Show that µ is tight, so is a Radon measure iff it is locally finite. In particular, show that if X has the discrete topology then counting measure on X is a Radon measure. (ii) Show that every purely atomic Radon measure is of this type. (c) Let h(Xi , Ti , Σi , µi )ii∈I be a family of Radon measure spaces, with direct sum (X, Σ, µ) (214K). Give X its disjoint union topology. Show that µ is a Radon measure.

416Xr


85

(d) Let (X, T) be a Hausdorff space, and µ1 , µ2 two Radon measures on X, with domains Σ1 , Σ2 . (i) Show that µ1 + µ2 , defined on Σ1 ∩ Σ2 (112Xe), is a Radon measure. (ii) Show that αµ1 , defined on Σ1 , is a Radon measure for any α > 0. (e) Let (X, T, Σ, µ) be a Radon measure space. (i) Show that µ has a decomposition hXi ii∈I in which every Xi except at most one is a self-supporting compact set, and the exceptional one, if any, is negligible. (ii) Show that µ has a decomposition hXi ii∈I in which every Xi is expressible as the intersection of a S closed set with an open set. (Hint: enumerate the open sets of finite measure as hGξ iξ<κ , and set Xξ = Gξ \ η<ξ Gη .) (f ) Let X be a Hausdorff space and µ, ν two Radon measures on X such that νG = µG whenever G ⊆ X is open and min(µG, νG) < ∞. Show that µ = ν. (g) Explain how to prove 416H from 416F, without appealing to §415. (h) Give a direct proof of 416I not relying on 415O. (i) Let (X, T) be a completely regular topological space and µ a locally finite topological measure on X which is inner regular with respect to the closed sets. Show that µK = inf{µG : G ⊇ K is a cozero set}

Z

= inf{µF : F ⊇ K is a zero set} = inf{

f dµ : χK ≤ f ∈ C(X)}

for every compact set K ⊆ X. (j) Let (X, T, Σ, µ) be a locally compact Radon measure space, and Ck the space of continuous realvalued functions on X with compact supports. Show that {f • : f ∈ Ck } is dense in L0 (µ) for the topology of convergence in measure. (k) Let X be a completely regular Hausdorff space and ν a locally finite Baire measure on X. (i) Show that ν ∗ K = inf{νG : G ⊆ X is a cozero set, K ⊆ G} for every compact set K ⊆ X. (ii) Show that there is a Radon measure µ on X such that µK = ν ∗ K for every compact set K ⊆ X. (Hint: in the language of the proof of 416K, φ1 = ν ∗ ¹K.) (l) Let X be a Hausdorff space and ν a non-negative finitely additive functional defined on some algebra of subsets of X. Show that there is a Radon measure µ on X such that µX ≤ νX and µK ≥ νE whenever E ∈ T, K ⊆ X is compact and E ⊆ K. (Hint: start by extending ν to PX.) (m) Let (X, T) be a Hausdorff space, Σ ⊇ T a σ-algebra of subsets of X, and ν : Σ → [0, ∞[ a finitely additive functional such that νE = sup{νK : K ⊆ E is compact} for every E ∈ Σ. Show that ν is countably additive and that its completion is a Radon measure on X. (n) Explain how to prove 416Rb from 416C and 415B. (o) Let X be a Hausdorff space, µ a complete locally finite measure on X, and Y a conegligible subset of X. Show that µ is a Radon measure iff the subspace measure on Y is a Radon measure. (p) Let X be a Hausdorff space, Y a subset of X, and ν a Radon measure on Y . Define a measure µ on X by setting µE = ν(E ∩ Y ) whenever ν measures E ∩ Y . Show that if either Y is closed or ν is totally finite, µ is a Radon measure on X. (Cf. 418I.) (q) Let (X, T, Σ, µ) be a Radon measure space and E ⊆ Σ a non-empty upwards-directed family. Set νF = supE∈E µ(E ∩ F ) whenever F ⊆ X is such that µ measures E ∩ F for every E ∈ E. Show that ν is a Radon measure on X. (r) Let hXn in∈N be a sequence of Hausdorff spaces with product X; write B(Xn ) for the Borel σ-algebra N of Xn . Let T be the σ-algebra c n∈N B(Xn ) (definition: 254E). Let ν : T → [0, ∞[ be a finitely additive functional such that E 7→ νπn−1 [E] : B(Xn ) → [0, ∞[ is countably additive and tight for each n ∈ N, writing πn (x) = x(n) for x ∈ X, n ∈ N. Show that there is a unique Radon measure on X extending ν.

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416Xs

S (s) Set S2∗ = n∈N {0, 1}n , and let φ : S2∗ → [0, ∞[ be a functional such that φ(σ) = φ(σ a 0) + φ(σ a 1) for every σ ∈ S2∗ , writing σ a 0 and σ a 1 for the two members of {0, 1}n+1 extending any σ ∈ {0, 1}n . Show that there is a unique Radon measure µ on {0, 1}N such that µ{x : x¹{0, . . . , n − 1} = σ} = φ(σ) whenever n ∈ N, σ ∈ {0, 1}N . (Hint: use 416Xr or 416Q.) (t) Let (X, T, Σ, µ) be a Radon measure space. Show that a measure ν on X is an indefinite-integral measure over µ iff (α) ν is a complete, locally determined topological measure (β) ν is tight (γ) νK = 0 whenever K ⊆ X is compact and µK = 0. (u) Let (X, T, Σ, µ) be a compact Hausdorff quasi-Radon measure space. Let W ⊆ X be the union of the open subsets of X of finite measure. Show that the subspace measure on W is a Radon measure. (v) Let (X, T, Σ, µ) be a completely regular Radon measure space. Show that it is isomorphic, as topological measure space, to a measurable subspace of a locally compact Radon measure space. (w) Let (X, T, Σ, µ) be a compact Radon measure space and (Z, S, T, ν) the Stone space of its measure algebra. For E ∈ Σ let E ∗ be the corresponding open-and-closed subset of Z, as in 416V. Show that the function described in 416V is the unique continuous function h : Z → X such that ν(E ∗ 4h−1 [E]) = 0 for every E ∈ Σ. (x) Show that the right-facing Sorgenfrey line (415Xc), with Lebesgue measure, is a quasi-Radon measure space which, regarded as a measure space, is compact, but, regarded as a topological measure space, is not a Radon measure space. 416Y Further exercises (a) Let X be a Hausdorff space and ν a countably additive real-valued functional defined on a σ-algebra Σ of subsets of X. Show that the following are equiveridical: (i) |ν| : Σ → [0, ∞[, defined as in 362B, is a Radon measure on X; (ii) ν is expressible as µ1 − µ2 , where µ1 , µ2 are Radon measures on X and Σ = dom µ1 ∩ dom µ2 . (b) Let X be a topological space and µ0 a semi-finite measure on X which is inner regular with respect to the family Kccc of closed countably compact sets. Show that µ0 has an extension to a complete locally determined topological measure µ on X, still inner regular with respect to Kccc ; and that the extension may be done in such a way that whenever µE < ∞ there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0. (Hint: use the argument of 416M, but with K = {K : K ∈ Kccc , µ∗0 K < ∞}.) (c) Let X be a topological space and µ0 a semi-finite measure on X which is inner regular with respect to the family Ksc of sequentially compact sets. Show that µ0 has an extension to a complete locally determined topological measure µ on X, still inner regular with respect to Ksc ; and that the extension may be done in such a way that whenever µE < ∞ there is an E0 ∈ Σ0 such that µ(E4E0 ) = 0. P 1 (d) Let X ⊆ βN be the union of all those open sets G ⊆ βN such that n∈G∩N n+1 is finite. For E ⊆ X P 1 set µE = n∈E∩N n+1 . Show that µ is a σ-finite Radon measure on the locally compact Hausdorff space X. Show that µ is not outer regular with respect to the open sets. S P∞ (e) Set S ∗ = n∈N N n , and let φ : S ∗ → [0, ∞[ be a functional such that φ(σ) = i=0 φ(σ a i) for every σ ∈ S ∗ , writing σ a i for the members of N n+1 extending any σ ∈ N n . Show that there is a unique Radon measure µ on N N such that µIσ = φ(σ)P for every σ ∈ S ∗ , where Iσ S = {x : x¹{0, . . . , n − 1} = σ} for any N n ∈ N, σ ∈ N . (Hint: set θA = inf{ σ∈R φ(σ) : R ⊆ S ∗ , A ⊆ σ∈R Iσ } for every A ⊆ N N , and use Carathéodory’s method.) (f ) Let (X, T, Σ, µ) be a Radon measure space. Show that a measure ν on X is an indefinite-integral measure over µ iff (i) there is a topology S on X, including T, such that ν is a Radon measure with respect to S (ii) νK = 0 whenever K is a T-compact set and µK = 0. (g) Let hxn in∈N enumerate subset of X = {0, 1}c (4A2B(e-ii)). Let λ be the usual measure on P∞ a dense 1 −n−2 χE(xn ) for E ∈ dom λ. (i) Show that µ is a strictly positive Radon X, and set µE = 2 λE + n=0 2

416 Notes


87

probability measure on X of Maharam type c. (ii) Let I ∈ [c]≤ω be such that xm ¹I 6= xn ¹I whenever I m that R 6= n. Set Z = {0, 1} and let π : X → Z be the canonical map. Show that if f ∈ C(X) is such f × gπ dµ = 0 for every g ∈ C(Z), then f R= 0. (Hint: otherwise, take n ∈ N Rsuch that |f (xn )| ≥ 12 kf k∞ , and let g ≥ 0 be such that g(πxn ) = 1 and g d(µπ −1 ) < 3 · 2−n−3 ; show that f × gπ dµ > 0.) (iii) Show that there is no orthonormal basis for L2 (µ) in {f • : f ∈ C(X)}. (See Hart & Kunen 99.) 416 Notes and comments The original measures studied by Radon (Radon 13) were, in effect, what I call differences of Radon measures on R r , as introduced in §256. Successive generalizations moved first to Radon measures on general compact Hausdorff spaces, then to locally compact Hausdorff spaces, and finally to arbitrary Hausdorff spaces, as presented in this section. I ought perhaps to remark that, following Bourbaki 65, many authors use the term ‘Radon measure’ to describe a linear functional on a space of continuous functions; I will discuss the relationship between such functionals and the measures of this chapter in §436. For the moment, observe that by 415I a Radon measure on a completely regular space can be determined from the integrals it assigns to continuous functions. It is also common for the phrase ‘Radon measure’ to be used for what I would call a tight Borel measure; you have to check each author to see whether local finiteness is also assumed. In my usage, a Radon measure is necessarily the c.l.d. version of a Borel measure. The Borel measures which correspond to Radon measures are described in 416F. In §256, I discussed Radon measures on R r as a preparation for a discussion of convolutions of measures. It should now be becoming clear why I felt that it was impossible, in that context, to give you a proper idea of what a Radon measure, in the modern form, ‘really’ is. In Euclidean space, too many concepts coincide. As a trivial example, the simplest definition of ‘local finiteness’ (256Ab, 411Fa) is not the right formulation in other spaces. Next, because every closed set is a countable union of compact sets, there is no distinction between ‘inner regular with respect to closed sets’ and ‘inner regular with respect to compact sets’, so one cannot get any intuition for which is important in which arguments. (When we come to subspace measures on non-measurable subsets, of course, this changes; quasi-Radon measures on subsets of Euclidean space are important and interesting.) Third, the fact that the c.l.d. product of two Radon measures on Euclidean space is already a Radon measure (256K) leaves us with no idea of what to do with a general product of Radon measures. (There are real difficulties at this point, which I will attack in the next section. For the moment I offer just 416U.) And finally, we simply cannot represent a product of uncountably many Radon probability measures on Euclidean spaces as a measure on Euclidean space. As you would expect, a very large proportion of the results of this chapter, and many theorems from earlier volumes, were originally proved for compact Radon measure spaces. The theory of general totally finite Radon measures is, in effect, the theory of measurable subspaces of compact Radon measure spaces, while the theory of quasi-Radon measures is pretty much the theory of non-measurable subspaces of Radon measure spaces. Thus the theorem that every quasi-Radon measure space is strictly localizable is almost a consequence of the facts that every Radon measure space is strictly localizable and any subspace of a strictly localizable space is strictly localizable. The cluster of results between 416J and 416P form only a sample, I hope a reasonably representative sample, of the many theorems on construction of Radon measures from functionals on algebras or lattices of sets. (See also 416Ye.) The essential simplification, compared with the theorems in §413 and §415, is that we do not need to mention any σ- or τ -additivity condition of the type 413I(β) or 415K(β), because we are dealing with a ‘compact class’, the family of compact subsets of a Hausdorff space. We can use this even at some distance, as in 416O (where the hypotheses do not require any non-empty compact set to belong to the domain of the original functional). The particular feature of 416O which makes it difficult to prove from such results as 413J and 413O above is that we have to retain control of the outer measures of a sequence hKn in∈N of non-measurable sets. In general this is hard to do, and is possible here principally because the sequence is non-decreasing, so that we can make sense of the functionals νn E = ν ∗ (E ∩ Kn+1 ) − ν ∗ (E ∩ Kn ).

88

Topologies and Measures

§417 intro.

417 τ -additive product measures The ‘ordinary’ product measures introduced in Chapter 25 have served us well for a volume and a half. But we come now to a fundamental obstacle. If we start with two Radon measure spaces, their product measure, as defined in §251, need not be a Radon measure (419E). Furthermore, the counterexample is one of the basic compact measure spaces of the theory; and while it is dramatically non-metrizable, there is no other reason to set it aside. Consequently, if we wish (as we surely do) to create Radon measure spaces as products of Radon measure spaces, we need a new construction. This is the object of the present section. It turns out that the construction can be adapted to work well beyond the special context of Radon measure spaces; the methods here apply to general effectively locally finite τ -additive topological measures (for the product of finitely many factors) and to τ -additive topological probability measures (for the product of infinitely many factors). Elsewhere in this volume I will show how these constructions are related to some others which have been described for special spaces. The fundamental theorems are 417C and 417E, listing the essential properties of what I call ‘τ -additive product measures’, which are extensions of the c.l.d. product measures and product probability measures of Chapter 25. They depend on a straightforward lemma on the extension of a measure to make every element of a given class of sets negligible (417A). It is relatively easy to prove that the extensions are more or less canonical (417D, 417F). We still have Fubini’s theorem for the new product measures (417H), and the basic operations from §254 still apply (417J, 417K, 417M). It is easy to check that if we start with quasi-Radon measures, then the τ -additive product measure is again quasi-Radon (417N, 417O). The τ -additive product of two Radon measures is Radon (417P), and the τ -additive product of Radon probability measures with compact supports is Radon (417Q). In the last part of the section I look at continuous real-valued functions and Baire σ-algebras; it turns out that for these the ordinary product measures are adequate (417U, 417V). 417A Lemma Let (X, Σ, µ) be a semi-finite measure space, and A ⊆ PX a family of sets such that S µ∗ ( n∈N An ) = 0 for every sequence hAn in∈N in A. Then there is a measure µ0 on X, extending µ, such that (i) µ0 A is defined and zero for every A ∈ A, (ii) µ0 is complete if µ is, (iii) for every F in the domain Σ0 of µ0 there is an E ∈ Σ such that µ0 (F 4E) = 0. In particular, µ and µ0 have isomorphic measure algebras, so that µ0 is localizable if µ is. proof Let A∗ be the collection of subsets of X which can be covered by a countable subfamily of A. Then A∗ is a σ-ideal of subsets of X and µ∗ A = 0 for every A ∈ A∗ . Set Σ0 = {E4A : E ∈ Σ, A ∈ A∗ }. Then Σ0 is a σ-algebra of subsets of X. P P (i) ∅ = ∅4∅ ∈ Σ0 . (ii) If E ∈ Σ, A ∈ A∗ then X \ (E4A) = 0 (X \ E)4A ∈ Σ . (iii) If hEn in∈N , hAn in∈N are sequences in Σ, A∗ respectively, then S S S E = n∈N En ∈ Σ, A = E4 n∈N (En 4An ) ⊆ n∈N An ∈ A∗ , S Q so n∈N (En 4An ) = E4A ∈ Σ0 . Q If E, E 0 ∈ Σ, A, A0 ∈ A∗ and E4A = E 0 4A0 , then E4E 0 = A4A0 ∈ A∗ and µ∗ (E4E 0 ) = 0; because µ is semi-finite, µ(E4E 0 ) = 0 and µE = µE 0 . Accordingly we can define µ0 : Σ0 → [0, ∞] by setting µ0 (E4A) = µE whenever E ∈ Σ, A ∈ A∗ . Evidently µ0 extends µ and µ0 A = 0 for every A ∈ A. Also µ0 is a measure. P P (i) µ0 ∅ = µ∅ = 0. (ii) If 0 ∗ hFn in∈N S is a disjoint sequence in ∗Σ , with0 union F , express each Fn as En 4An where En ∈ Σ, An ∈ A ; set E = n∈N En , so that F 4E ∈ A and µ F = µE. If m 6= n, then Em ∩ En ⊆ Am ∪ An , so µ(Em ∩ En ) = 0; accordingly P∞ P∞ µ0 F = µE = n=0 µEn = n=0 µ0 Fn . Q Q A subset of X is µ0 -negligible iff it can be included in a set of the form E4A where µE = 0 and A ∈ A∗ , so µ0 is complete if µ is. The embedding Σ ⊆ Σ0 induces a measure-preserving homomorphism from the measure algebra of µ to the measure algebra of µ0 which is an isomorphism just because every member of Σ0 is the symmetric difference of a member of Σ and a µ0 -negligible set.

417C

τ -additive product measures

89

417B Lemma Let X and Y be topological spaces, and ν a τ -additive topological measure on Y . (a) If W ⊆ X × Y is open, then x 7→ νW [{x}] : X → [0, ∞] is lower semi-continuous. (b) If ν is effectively locally finite and σ-finite and W ⊆ X × Y is a Borel set, then x 7→ νW [{x}] is Borel measurable. R (c) If f : X × Y → [0, ∞] is a lower semi-continuous function, then x 7→ f (x, y)ν(dy) : X → [0, ∞] is lower semi-continuous. R (d) If ν is totally finite and f : X × Y → R is a bounded continuous function, then x 7→ f (x, y)ν(dy) is continuous. (e) If ν is totally finite and W ⊆ X × Y is a Baire set, then x 7→ νW [{x}] is Baire measurable. proof (a) If x ∈ X and νW [{x}] > α, then H = {H : H ⊆ Y is open, there is an open set G containing x such that G × H ⊆ W } is an upwards-directed family of open sets with union W [{x}], so there is an H ∈ H such that νH ≥ α. Now there is an open set G containing x such that G × H ⊆ W , so that νW [{x0 }] ≥ α for every x0 ∈ G. (b)(i) Suppose to begin with that ν is totally finite. In this case, the set {W : W ⊆ X × Y, x 7→ νW [{x}] is a Borel measurable function defined everywhere on X} is a Dynkin class containing every open set, so contains every Borel set, by the Monotone Class Theorem (136B). (ii) For the general case, let hYn in∈N be a disjoint sequence of sets of finite measure covering Y , and for n ∈ N let νn be the subspace measure on Yn . Then νn is effectively locally finite and τ -additive (414K). If W ⊆ X × Y is a Borel set, then Wn = W ∩ (X × Yn ) isPa relatively Borel set for each n, so that ∞ x 7→ νn Wn [{x}] is Borel measurable, by (i). Since νW [{x}] = n=0 νn Wn [{x}] for every x, x 7→ νW [{x}] is Borel measurable. (c) For i, n ∈ N set Wni = {(x, y) : f (x, y) > 2−n i}, so that Wni ⊆ X × Y is open. Set fn = P 4n sequence with supremum f . For n ∈ N and x ∈ X, 2 i=1 χWni ; then hfnnin∈N is a non-decreasing R R P4 −n fn (x, y)ν(dy) is lower semi-continuous, by (a) and 4A2B(dfn (x, y)ν(dy) =R 2 i=1 νWni [{x}], soRx 7→ R iii). By 414Ba, f (x, y)ν(dy) = supn∈N fn (x, y)ν(dy) for every x, so x 7→ f (x, y)ν(dy) is lower semicontinuous (4A2B(d-v)). R (d) Applying (c) to f +kf k∞ χ(X ×Y ), we see that x 7→ f (x, y)ν(dy) is lower semi-continuous. Similarly, R R x 7→ − f (x, y)ν(dy) is lower semi-continuous, so x 7→ f (x, y)ν(dy) is continuous (4A2B(d-vi)). −n

(e) Suppose first that W is a cozero set; let f : X × Y → [0, 1] be a continuous function such that W = {(x, y) : f (x, y) > 0}. For n ∈ N set fn = nf ∧ χ(X × Y ). Then hfn in∈N R is a non-decreasing sequence of continuous functions with supremum χW . By (d), all the functions x 7→ fn (x, y)ν(dy) are continuous, so their limit x 7→ νW [{x}] is Baire measurable. Now {W : W ⊆ X × Y, x 7→ νW [{x}] is a Baire measurable function defined everywhere on X} is a Dynkin class containing every cozero set, so contains every Baire set, by the Monotone Class Theorem again. 417C Theorem Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure spaces, with c.l.d. product (X × Y, Λ, λ). Then λ has an extension to a τ -additive topological measure ˜ on X × Y . Moreover, we can arrange that: λ ˜ is complete, locally determined and effectively locally finite, therefore strictly localizable; (i) λ ˜ there is a Q1 ∈ Λ such that λ(Q4Q ˜ ˜ of λ, (ii) if Q belongs to the domain Λ 1 ) = 0; that is to say, the ˜ ˜ embedding Λ ⊆ Λ induces an isomorphism between the measure algebras of λ and λ;

90


417C

˜ then (iii) if Q ∈ Λ, ˜ = sup{λ(Q ˜ ∩ (G × H)) : G ∈ T, µG < ∞, H ∈ S, νH < ∞}; λQ ˜ , and (iv) if W ⊆ X × Y is open, then there is an open set W0 ∈ Λ such that W0 ⊆ W and λW0 = λW ˜ = λ∗ W = λW

R

νW [{x}]µ(dx) =

R

µW −1 [{y}]ν(dy);

˜ × H) = µG · νH for every G ∈ T, H ∈ S; in particular, λ(G (v) the support of λ is the product of the supports of µ and ν; ˜ (vi) if µ and ν are both inner regular with respect to the Borel sets, so is λ; ˜ (vii) if µ and ν are both inner regular with respect to the closed sets, so is λ; ˜ (viii) if µ and ν are both tight (that is, inner regular with respect to the closed compact sets), so is λ. proof Write Σf = {E : E ∈ Σ, µE < ∞}, Tf = T ∩ Σf ,

Tf = {F : F ∈ T, νF < ∞}, Sf = S ∩ Tf .

(a) Let U be {G×H : G ∈ Tf , H ∈ Sf }. Because T ⊆ Σ and S ⊆ T, U ⊆ Λ. U need not be a base for the topology of X × Y , unless µ and ν are locally finite, but if an open subset of X × Y is included in a member of U it is the union of the members of U it includes. Moreover, if Q ∈ Λ, then λQ = supU ∈U λ(Q ∩ U ). P P S Q By 412R, λ is inner regular with respect to U ∈U PU . Q Write Us for the set of finite unions of members of U, and V for the set of non-empty upwards-directed families V ⊆ Us such that supV ∈V λV < ∞. For each V ∈ V, fix on a countable V 0 ⊆ V such that supV ∈V 0 λV = supV ∈V λV ; because V is upwards-directed, we may suppose that V 0 = {Vn : n ∈ N} for some S S non-decreasing sequence hVn in∈N in V. Set A(V) = V \ V 0 . (b)(i) For V ∈ Us , set fV (x) = νV [{x}] for every x ∈ X. This is always defined because V is open; moreover,RfV is lower semi-continuous, by 417Ba. Because V is a finite union of products of sets of finite measure, fV dµ = λV . (ii) The key to the proof is the following fact: for any V ∈ V, almost every vertical section of A(V) is negligible. P P hfV iV ∈V is a non-empty upwards-directed set of lower semi-continuous functions. Set S S g(x) = ν( V ∈V V [{x}]), h(x) = ν( V ∈V 0 V [{x}]) for every x ∈ X. Because V is upwards-directed and ν is τ -additive, g(x) = supV ∈V νV [{x}] = supV ∈V fV (x) in [0, ∞] for each x, so, by 414Ba,

R

g dµ = supV ∈V

R

fV dµ = supV ∈V λV = supV ∈V 0 λV =

R

h dµ.

Since h ≤ g and supV ∈V λV is finite, g(x) = h(x) < ∞ for µ-almost every x. But for any such x, we must have S S ν( V)[{x}] = ν( V 0 )[{x}] < ∞, so that A(V)[{x}] = (

S

V)[{x}] \ (

S

V 0 )[{x}]

is negligible. Q Q S (c) ?? Suppose, if possible, that there is a sequence hVn in∈N in V such that λ∗ ( n∈N A(Vn )) > 0. Take S W ∈ Λ such that W ⊆ n∈N A(Vn ) and λW > 0. Because almost every vertical section of every A(Vn ) is negligible, almost every vertical section of W is negligible. But this contradicts Fubini’s theorem (252F). X X (d) We may therefore apply the construction of 417A to form a measure λ0 on the σ-algebra Λ0 = {W 4A : ˜ be the c.l.d. version of λ0 W ∈ Λ, A ∈ A∗ }, where A∗ is the σ-ideal generated by {A(V) : V ∈ V}. Let λ ˜ its domain. (213E), and Λ

417C


91

˜ Also, because λ is semi-finite, (i) If W ∈ Λ, then W ∈ Λ0 ⊆ Λ. λ0 W = λW = sup{λW 0 : W 0 ⊆ W, W ∈ Λ, λW 0 < ∞} ˜ ≤ λ0 W. ≤ sup{λ0 W 0 : W 0 ⊆ W, W ∈ Λ0 , λ0 W 0 < ∞} = λW ˜ ; as W is arbitrary, λ ˜ extends λ. Thus λW = λW ˜ there is a U ∈ U such ˜ and γ < λQ, (ii) It will be useful if we go directly to one of the targets: if Q ∈ Λ 0 0 ˜ that λ(Q ∩ U ) ≥ γ. P P There is a Q1 ∈ Λ such that Q1 ⊆ Q and γ < λ Q1 < ∞. There is a Q2 ∈ Λ such that λ0 (Q1 4Q2 ) = 0, so that λQ2 = λ0 Q2 = λ0 Q1 > γ. There is a U ∈ U such that λ(Q2 ∩ U ) ≥ γ, by (a). Now ˜ ∩ U ) ≥ λ0 (Q1 ∩ U ) = λ0 (Q2 ∩ U ) = λ(Q2 ∩ U ) ≥ γ. Q λ(Q Q ˜ is a topological measure. P ˜ > 0. ˜ and λQ (iii) λ P Let W ⊆ X × Y be an open set. Suppose that Q ∈ Λ ˜ By (ii), there is a U ∈ U such that λ(Q ∩ U ) > 0. Let V be {V : V ∈ Us , V ⊆ W ∩ U }. Then V ∈ V, so S ˜ λA(V) = λ0 A(V) = 0; since V 0 ∈ Λ, S ˜ W ∩ U = V ∈ Λ0 ⊆ Λ. But this means that ˜ ∗ (Q ∩ W ) + λ ˜ ∗ (Q \ W ) ≥ λ ˜ ∗ (Q ∩ U ∩ W ) + λ ˜ ∗ (Q ∩ U \ W ) = λ(Q ˜ ∩ U ) > 0. λ ˜ is complete and locally determined, and Q is arbitrary, this is enough to ensure that W ∈ Λ ˜ Because λ (413F(vii)). Q Q ˜ is τ -additive. P (iv) λ P?? Suppose, if possible, otherwise; that there is a non-empty upwards-directed S ∗ ˜ ∗ > γ = sup ˜ family W of open sets in X × Y such that λW W. In this case, we W ∈W λW , where W = can find a Q0 ∈ Λ0 such that Q0 ⊆ W ∗ and λ0 Q0 > γ, a Q1 ∈ Λ such that λ0 (Q0 4Q1 ) = 0, and a U ∈ U such that λ(Q1 ∩ U ) > S γ (using (a) again). Let V ∈ V be the set of those V ∈ Us such that V ⊆ W ∩ U for some W ∈ W. Then V = W ∗ ∩ U , so

γ < λ(Q1 ∩ U ) = λ0 (Q1 ∩ U ) = λ0 (Q0 ∩ U ) [ [ ∗ ˜ ˜ ˜ ≤ λ(W ∩ U ) = λ( V) = λ( V 0) ˜ (because λA(V) = 0) ˜ = sup λV V ∈V 0 0

(because V is countable and upwards-directed) ˜ ≤ γ, ≤ sup λW W ∈W

which is absurd. X XQ Q (e) Now for the supplementary properties (i)-(viii), in order. ˜ was constructed to be complete and locally determined. If λQ ˜ > 0, then by (d-ii) there is a U ∈ U (i) λ ˜ ˜ ˜ is effectively locally finite. such that λ(Q ∩ U ) > 0; since U is open and λU = λU is finite, this shows that λ By 414J, it is strictly localizable. (ii) The point is that λ is also (strictly) localizable. P P Let µ ˜, ν˜ be the c.l.d. versions of µ and ν. These are τ -additive topological measures (because µ and ν are), complete and locally determined (by construction), and are still effectively locally finite (cf. 412Ha), so are strictly localizable (414J again). Now λ is the c.l.d. product of µ ˜ and ν˜ (251S), therefore strictly localizable (251N). Q Q

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417C

˜ ˜ differs by a λ-negligible As remarked in 417A, it follows that λ0 is localizable, so that every member Q of Λ 0 0 set from a member Q0 of Λ (213Hb). Now there is a Q1 ∈ Λ such that λ (Q1 4Q0 ) = 0, in which case ˜ 1 4Q) = 0. λ(Q (iii) This is just (d-ii) above. (iv) Let W ⊆ X × Y be an open set. Set V = {V : V ∈ Us , V ⊆ W }. Then V is upwards-directed and has union W ∩ (X ∗ × Y ∗ ), where X ∗ and Y ∗ are the unions of the open sets of finite measure in X and Y respectively. Because µ and ν are effectively locally finite, X ∗ and Y ∗ are both conegligible. Now, as before, ˜ = λ(W ˜ ˜ = sup ˜ λW ∩ (X ∗ × Y ∗ )) = supV ∈V λV V ∈V λV ≤ λ∗ W ≤ λW . S If we take a countable set V0 ⊆ V such that supV ∈V0 λV = supV ∈V λV , and set W0 = V0 , then W0 is an ˜ . open set, belonging to λ and included in W , and λW0 = λW R Next, defining the functions fV as in part (a) of this proof, we have fV dµ = λV for every V ∈ V; and setting g(x) = νW [{x}], we have g(x) = supV ∈V fV (x) for every x ∈ X ∗ . So 414Ba tells us that

R

νW [{x}]µ(dx) =

R

R

g dµ = supV ∈V

R

˜ . fV dµ = supV ∈V λV = λW

˜ . µW −1 [{y}]ν(dy) = λW

(The point here is that while the arguments of part (b) of this proof give different roles to µ and ν, the actual construction performed in part (d) is symmetric between them.) Now, given G ∈ T and H ∈ S, set W = G × H; then R R ˜ × H) = µW −1 [{y}]ν(dy) = λ(G µGν(dy) = µG · νH. H

(v) Let E and F be the supports of µ, ν respectively. (By 411Nd these are defined.) Then E × F is a closed subset of X × Y . Because X \ E and Y \ F are negligible, E × F is conegligible. If W ⊆ X × Y is an open set and (x, y) ∈ W ∩ (E × F ), there are open sets G ⊆ X, H ⊆ Y such that (x, y) ∈ G × H ⊆ W . Now λ(W ∩ (E × F )) ≥ λ((G × H) ∩ (E × F )) = µ(G ∩ E) · ν(H ∩ F ) > 0. This shows that E × F is self-supporting, so is the support of λ. (vi), (vii), (viii) The same method works for all of these. Take K to be either the family of Borel subsets of X × Y (for (vi)) or the family of closed subsets of X × Y (for (vii)) or the family of closed compact subsets of X × Y (for (viii)); the essential features of K, valid for all three cases, are that (α) K \ W ∈ K whenever K ∈ K and W ⊆ X × Y is open, (β) K ∪ K 0 ∈ K for all K, K 0 ∈ K, T (γ) n∈N Kn ∈ K for every sequence hKn in∈N in K. Now the hypotheses of each part are just what we need in order to be sure that λ is inner regular with ˜ also is, without assuming that λ ˜ is respect to K (412Sd, 412Sa, 412Sb), and I am trying to show that λ inner regular with respect to Λ, because this need not be true. 0 ˜ ˜ and γ < λQ; Take Q ∈ Λ then there is a Q0 ∈ Λ0 such that S Q0 ⊆ Q and γ < λ Q0 < ∞.0 We have a Q1 ∈ Λ and a sequence hVn in∈N in V such that Q0 4Q1 ⊆ n∈N A(Vn ). Of course λQ1 = λ Q0 > γ; set ² = 41 (λQ1 − γ) > 0. For each n ∈ N, take Kn , Ln ∈ K ∩ Λ such that S S Kn ⊆ Q1 ∩ Vn0 , λ((Q1 ∩ Vn0 ) \ Kn ) ≤ 2−n ²,

Set Kn0 = Kn ∪ (Ln \

S

λ(Q1 \ Ln ) ≤ 2−n ².

Ln ⊆ Q1 , Vn ), so that Kn0 ∈ K and

[ ˜ 1 \ K 0 ) ≤ λ(Q ˜ 1 \ Ln ) + λA(V ˜ ˜ λ(Q Vn0 \ Kn ) n ) + λ(Q1 ∩ n [ = λ(Q1 \ Ln ) + λ(Q1 ∩ Vn0 \ Kn ) ≤ 2−n+1 ². Now K =

T n∈N

Kn0 belongs to K and K ⊆ Q1 \

S n∈N

A(Vn ) ⊆ Q0 ⊆ Q

417D


93

because Kn0 ⊆ Q1 \ A(Vn ) for every n. And λ(Q1 \ K) ≤

∞ X

˜ 1 \ K 0 ) ≤ 4², λ(Q n

n=0

˜ = λK ≥ γ. As Q and γ are arbitrary, λ ˜ is inner regular with respect to K. so λK 417D Multiple products Just as with the c.l.d. product measure (see 251W), we can apply the construction of 417C repeatedly to obtain measures on the products of finite families of τ -additive measure spaces. Proposition (a) Let h(Xi , Ti , Σi , µi )ii∈I be a finite family of effectively locally finite τ -additive topological measure spaces. Then there is a unique complete locally determined effectively locally finiteQτ -additive ˜ on X = Q Xi , inner regular with respect to the Borel sets, such that λ( ˜ topological measure λ i∈I i∈I Gi ) = Q µ G whenever G ∈ T for every i ∈ I. i i i∈I i i ˜ k is the product measure defined by the construction of (a) (b) If now hIk ik∈K is a partition of I, and λ Q Q ˜ with the on Zk = i∈Ik Xi for each k ∈ K, then the natural bijection between X and k∈K Zk identifies λ ˜ k defined by the construction of (a). product of the λ proof (a)(i) Suppose first that every µi is inner regular with respect to the Borel sets. Then a direct ˜ with the required induction on #(I), using 417C for the inductive step, tells us that there is a measure λ properties. Note that, in 417C, (vi) ensures that (in the present context) all our product measures will be inner regular with respect to the Borel sets. (ii) For the general case, apply (a) to µi ¹B(Xi ), where B(Xi ) is the Borel σ-algebra of Xi for each i. ˜ is unique, suppose that λ0 is another measure with the same properties. Let U be (iii) To see that λ Q ˜ the set { i∈I Gi : Gi ∈ Ti for every i ∈ SI}, and Q Us the set of finite unions of members of U. Then λ and λ0 agree on Us . P P Suppose that U = j≤n i∈I Gji , where Gji ∈ Ti for j ≤ n, i ∈ I. If there is any j Q Q 0 ˜ such that i∈I µi Gji > 0} and i∈I µi Gji = ∞, then λU = λ U = ∞. Otherwise, set L = {j : j ≤ n, S G∗i = j∈L Gji for i ∈ I. Then ˜ \ (Q G∗ )) = 0 = λ0 (U \ (Q G∗ )). λ(U i i i∈I i∈I On the other hand, µi Gji must be finite whenever j ∈ L and i ∈ I, so µi G∗i is finite for every i. Consider Q ˜ λ0 agree on I = { i∈I Gi : Gi ∈ Ti , Gi ⊆ G∗i for every i ∈ I}. Then V ∩ V 0 ∈ I for all V , V 0 ∈ I, and λ, 0 ˜ I. It follows from the Monotone Class Theorem (136C), or otherwise, that λ and λ agree on the algebra of Q ˜ ∩ (G∗ × H ∗ )) = λ0 (U ∩ (G∗ × H ∗ )), so that λU ˜ = λ0 U . subsets of i∈I G∗i generated by I. In particular, λ(U Q Q ˜ and λ0 are τ -additive, But now, because λ ˜ = sup{λU ˜ : U ∈ Us , U ⊆ W } = sup{λ0 U : U ∈ Us , U ⊆ W } = λ0 W λW ˜ for every open set W ⊆ X. Writing B(X) for the Borel σ-algebra of X, λ¹B(X) and λ0 ¹B(X) are effectively ˜ and λ0 locally finite Borel measures which agree on the open sets, so must be equal, by 414L. Since both λ are complete locally determined measures defined on B and inner regular with respect to B, they also are equal, by 412L. 0 ˜ k on Q (b) Let λ0 be the measure on X corresponding to the τ -additive product of the λ k∈K Zk . Then λ is an effectively locally finite complete locally determined τ -additive topological measure inner regular with respect to the zero sets, and if Gi ∈ Ti for every i ∈ I then Q Q Q Q Q ˜ k (Q λ0 ( Gi ) = λ Gi ) = µi Gi = µi Gi , i∈I

˜ so λ0 = λ.

k∈K

i∈Ik

k∈K

i∈Ik

i∈I

94


417E

417E Theorem Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces, with product probability space (X, Λ, λ). Then λ is τ -additive, and has an extension to a τ -additive topological ˜ on X. Moreover, we can arrange that: measure λ ˜ is complete; (i) λ ˜ there is a Q1 ∈ Λ such that λ(Q4Q ˜ ˜ (ii) if Q is measured by λ, 1 ) = 0; that is to say, the embedding Λ ⊆ Λ ˜ induces an isomorphism between the measure algebras of λ and λ; ˜ = λ∗ W for every open set W ⊆ X, and λF ˜ = λ∗ F for every closed set F ⊆ X; (iii) λW (iv) the support of λ is the product of the supports of the µi ; ˜ (v) if λ is inner regular with respect to the Borel sets, so is λ; ˜ (vi) if λ is inner regular with respect to the closed sets, so is λ; ˜ (vii) if λ is tight, so is λ. proof The strategy of the proof is the same S as in 417C, subject to some obviously necessary modifications. The key step, showing that every union n∈N A(Vn ) has zero inner measure, is harder, but we do save a little work because we no longer have to worry about sets of infinite measure. Q (a) I begin by setting up some machinery. Let C be the family of subsets of X expressible in the form i∈I Ei , where Ei ∈ Σi for every i and {i : Ei 6=QXi } is finite. Let U ⊆ C be the standard basis for the topology T of X, consisting of sets expressible as i∈I Gi where Gi ∈ Ti for every i ∈ I and {i : Gi 6= Xi } is finite. Write Us for the set of finite unions of members of U , and V for the set of non-empty upwardsdirected families in Us . Note that every member of Us is determined by coordinates in some finite subset of I (definition: 254M). Q If J ⊆ I, write λJ for the product measure on i∈J Xi ; we shall Q need λ∅ , which is the unique probability Q measure on the single-point set {∅} = i∈∅ Xi . For J ⊆ I, v ∈ i∈J Xi , W ⊆ X set if this is defined, identifying

fW (v) = λI\J {w : (v, w) ∈ W } Q i∈I\J Xi with X. i∈J Xi ×

Q

(b) We need two easy facts. R Q N (i) fW (v) = Q fW (v a t)µj (dt) whenever W ∈ c i∈I Σi , J ⊆ I, v ∈ i∈J Xi and j ∈ I \ J, writing v a t P Let A be the for the member of i∈J∪{j} Xi extending v and taking the value t at the coordinate j. P family of sets W satisfying the property. Then A is a Dynkin class including C, so includes the σ-algebra N Q generated by C, which is c i∈I Σi . Q Q a (ii) If J ⊆ I, v ∈ i∈J Xi , j ∈ I \ J and S V ∈ QUs , and we set g(t) = fV (v t) for t ∈ Xj , then g is lower semi-continuous. P P We can express V as n≤m i∈I Gni , where Gni ⊆ Xi is open for every n ≤ m, i ∈ I. Now if t ∈ Xj , we shall have {w : (v a t, w) ∈ V } ⊆ {w : (v a t0 , w) ∈ V } whenever t0 ∈ H = Xj ∩

T

{Gnj : n ≤ m, t ∈ Gnj }.

So g(t0 ) ≥ g(t) for every t0 ∈ H, which is an open neighbourhood of t. As t is arbitrary, g is lower semi-continuous. Q Q (c) For each V ∈ V, fix, for the remainder of this proof, a countable V 0 ⊆ V such that supV ∈V 0 λV = 0 supV ∈V λV ; because V is upwards-directed, S Swe0may suppose that V = {Vn : n ∈ N} for some non-decreasing sequence hVn in∈N in V. Set A(V) = V \ V . S ?? Suppose, if possible, that there is a sequence hVn in∈N in V such that λ∗ ( n∈N A(Vn )) > 0. S (i) We have λ∗ (X \ n∈N A(Vn )) < 1; let hCn in∈N be a sequence in C such that S S P∞ X \ n∈N A(Vn ) ⊆ n∈N Cn , n=0 λCn = γ0 < 1 0 (see 254A-254C). S 0 S For each n ∈ N, express Vn as {Vnr : r ∈ N} where hVnr ir∈N is non-decreasing, and set Wn = Vn = r∈N Vnr . Let J S ⊆ I be a countable set such that every Cn and every Vnr is dependent on coordinates in J. Express J as k∈N Jk where J0 = ∅ and, for each k, Jk+1 is equal either to Jk or to Jk

417E


95

with one point added. (As in the proof of 254Fa, I am using a formulation which will apply equally to finite and infinite I, though of course the case of finite I is elementary once we have 417C.) (ii) For each n ∈ N, set Wn0 =

S

k∈N {x

: x ∈ X, fWn (x¹Jk ) = 1}.

λ(Wn0

Then \ Wn ) = 0. P P and of fWn as a k Q For any k ∈ N, if we think of λ as the product of λJk and λI\JQ measurable function on i∈Jk Xi , we see that {x : fWn (x¹Jk ) = 1} is of the form Fk × i∈I\Jk Xi , where Q Fk ⊆ i∈Jk Xi is measurable; and λ((Fk ×

Q

i∈I\Jk

Xi ) \ Wn ) =

R

Fk

(1 − fWn (v))λJk (dv) = 0.

Summing over k, we see that Wn0 \ Wn is negligible. Q Q S 0 , like W , is determined by coordinates in J. So n∈N Wn0 \ Wn is of the form Observe that every W n n Q E × i∈I\J Xi where λJ E = 0 (254Ob). There is therefore a sequence hDn in∈N of measurable cylinders in Q S P∞ such that E ⊆ n∈N Dn and n=0 λJ Dn < 1 − γ0 . Set Cn0 = {x : x ∈ X, x¹J ∈ Dn } ∈ C for each i∈J Xi S S n. Then n∈N Wn0 \ Wn ⊆ n∈N Cn0 , so S S S S (X \ n∈N A(Vn )) ∪ n∈N Wn0 \ Wn ⊆ n∈N Cn ∪ n∈N Cn0 , γ=

P∞ n=0

λCn +

P∞ n=0

λCn0 < 1,

while each Cn and each Cn0 is dependent on coordinates in a finite subset of J. Q (iii) For k ∈ N, let Pk be the set of those v ∈ i∈Jk Xi such that P∞ 0 (v) ≤ γ, fV (v) ≤ fWn (v) for every n ∈ N, V ∈ Vn . n=0 fCn (v) + fCn Our hypothesis is that

P∞ n=0

fCn (∅) + fCn0 (∅) =

P∞ n=0

λCn + λCn0 ≤ γ,

and the Vn0 were chosen such that fV (∅) = λV ≤ λWn = fWn (∅) for every n ∈ N, V ∈ Vn ; that is, ∅ ∈ P0 . (iv) Now if k ∈ N, v ∈ Pk there is a v 0 ∈ Pk+1 extending v. P P If Jk+1 = Jk we can take v 0 = v. Otherwise, Jk+1 = Jk ∪ {j} for some j ∈ I \ Jk . Now

γ≥

∞ X

fCn (v) + fCn0 (v) =

n=0

((b-i) above) =

Z X ∞

∞ Z X

fCn (v a t) + fCn0 (v a t) µj (dt)

n=0

fCn (v a t) + fCn0 (v a t) µj (dt),

n=0

so H = {t : t ∈ Xj ,

P∞ n=0

fCn (v a t) + fCn0 (v a t)µj (dt) ≤ γ}

has positive measure. Next, for V ∈ Us , set gV (t) = fV (v a t) for each t ∈ Xj . Then gV is lower semi-continuous, by (b-ii) above. For each n ∈ N, {gV : V ∈ Vn } is an upwards-directed family of lower semi-continuous functions, so its supremum gn∗ is also lower semi-continuous, and because µj is τ -additive,

R

gn∗ dµj = supV ∈Vn

R

gV dµj = supV ∈Vn fV (v) ≤ fWn (v) =

R

fWn (v a t)µj (dt)

(using 414B and (b-i) again). But also, because hVnr ir∈N is non-decreasing and has union Wn , fWn (v a t) = supr∈N fVnr (v a t) ≤ gn∗ (t) for every t ∈ Xj . So we must have

96


417E

fWn (v a t) = gn∗ (t) a.e.(t). And this is true for every n ∈ N. There is therefore a t ∈ H such that fWn (v a t) = gn∗ (v a t) for every n ∈ N. Q Fix on such a t and set v 0 = v a t ∈ i∈Jk+1 Xi ; then v 0 ∈ Pk+1 , as required. Q Q (v) We can therefore choose a sequence hvk ik∈N such that vk ∈ Pk and vk+1 extends vk for each k. Choose x ∈ X such that x(i) = vk (i) whenever k ∈ N, i ∈ Jk and x(i) belongs to the support of µi whenever i ∈ I \ J. (Recall from 411Nd that µi does have a support.) We need to know that if k, n ∈ N and V ∈ Vn then fV \Wn (vk ) = 0. P P For any r ∈ N there is a V 0 ∈ Vn 0 such that V ∪ Vnr ⊆ V , so fV ∪Vnr (vk ) ≤ fV 0 (vk ) ≤ fWn (vk ), and fV \Wn (vk ) ≤ fV \Vnr (vk ) = fV ∪Vnr (vk ) − fVnr (vk ) ≤ fWn (vk ) − fVnr (vk ) → 0 as r → ∞. Q Q P Cn and Cn0 are determined by coordinates in a finite subset (vi) If n ∈ N, then x ∈ / Cn ∪ Cn0 . P of J, so must be determined by coordinates in Jk for some k ∈ N. Now fCn (vk ) + fCn0 (vk ) ≤ γ < 1, so Q / Cn ∪ Cn0 . Q {y : y¹Jk = vk } cannot be included in Cn ∪ Cn0 , and must be disjoint from it; accordingly x ∈ (vii) Because (X \

S n∈N

there is some n ∈ N such that

A(Vn )) ∪

S n∈N

Wn0 \ Wn ⊆

S n∈N

Cn ∪

S n∈N

Cn0 ,

S x ∈ A(Vn ) \ (Wn0 \ Wn ) ⊆ ( Vn ) \ Wn0 ,

that is, there is some V ∈ Vn such that x ∈ V \ Wn0 . Let U ∈ U be such that x ∈ U ⊆ V . Express U as U 0 ∩ U 00 where U 0 ∈ U is determined by coordinates in a finite subset of J, and U 00 ∈ U is determined by coordinates in a finite subset of I \ J. Let k ∈ N be such that U 0 is determined by coordinates in Jk . Then fU \Wn (vk ) ≤ fV \Wn (vk ) = 0 by (v) above. Now {w : w ∈

Q i∈I\Jk

Xi , (vk , w) ∈ U \ Wn } = {w : (vk , w) ∈ U 00 \ Wn }

(because (vk , w) = (x¹Jk , w) ∈ U 0 for every w), while {w : (vk , w) ∈ U 00 },

{w : (vk , w) ∈ Wn }

are stochastically independent because the former depends on coordinates in I \ J, while the latter depends on coordinates in J \ Jk . So we must have 0 = fU \Wn (vk ) = λI\Jk {w : (vk , w) ∈ U \ Wn } = λI\Jk {w : (vk , w) ∈ U 00 \ Wn } = λI\Jk {w : (vk , w) ∈ U 00 }(1 − λI\Jk {w : (vk , w) ∈ Wn }). At this point, recall that x(i) belongs to the support of µi for every i ∈ I \ J, while x ∈ U 00 . So if U 00 = {y : y(i) ∈ Hi for i ∈ K}, where K ⊆ I \ J is finite and Hi ⊆ Xi is open for every i, we must have µi Hi > 0 for every i, and Q λI\Jk {w : (vk , w) ∈ U 00 } = i∈K µi Hi > 0. On the other hand, we are also supposing that x ∈ / Wn0 , so λI\Jk {w : (vk , w) ∈ Wn } = fWn (vk ) < 1. But this means that we have expressed 0 as the product of two non-zero numbers, which is absurd. X X

417F


97

S (d) Thus λ∗ ( n∈N A(Vn )) = 0 for every sequence hVn in∈N in V. Accordingly there is an extension of λ ˜ on X such that λA(V) ˜ ˜ being to a measure λ = 0 for every V ∈ V, the domain of λ ˜ = {W 4A : W ∈ Λ, A ∈ A∗ }, Λ where A∗ is the σ-ideal generated by {A(V) : V ∈ V} (417A). ˜ Now measure. P P If W ⊆ X is open, then V = {V : V ∈ Us , V ⊆ W } belongs to V, S λ is a topological S and V = W . Now V 0 ∈ Λ (because V 0 is countable), so S W = V 0 ∪ A(V) ˜ Q is measured by λ. Q ˜ Also, λ is τ -additive. P P Let W be a non-empty upwards-directed family of open subsets of X with union W ∗ . Set Then V ∈ V and

S

V = {V : V ∈ Us , ∃ W ∈ W, V ⊆ W }. ˜ V = W ∗ , so λA(V) = 0 and S 0 ∗ ˜ ˜ ˜ ≤ sup ˜ ˜ ∗ λW = λ( V ) = supV ∈V 0 λV W ∈W λW ≤ λW

˜ is τ -additive. Q (using the fact that V 0 is upwards-directed). As W is arbitrary, λ Q Of course it follows at once that λ is also τ -additive. (e) Now for the supplementary properties (i)-(vi) listed in the theorem. ˜ = 0 and Q0 ⊆ Q, then Q is expressible as W 4A where λW = 0 and A ∈ A∗ . But in ˜ λQ (i) If Q ∈ Λ, ˜ is complete. ˜ Thus λ this case (because λ is complete) λ(Q0 ∩ W ) = 0 while Q0 \ W ⊆ A ∈ A∗ , so Q0 ∈ Λ. ˜ ˜ differs by a λ-negligible (ii) As always, the construction ensures that every member of Λ set from some member of Λ. (iii) Let W ⊆ X be an open set. Set V = {V : V ∈ Us , V ⊆ W }. Then ˜ = sup ˜ ˜ λW V ∈V λV = supV ∈V λV ≤ λ∗ W ≤ λW ˜ is a τ -additive extension of λ. Now if F ⊆ X is closed, just because λ ˜ = 1 − λ(X ˜ λF \ F ) = 1 − λ∗ (X \ F ) = λ∗ F . Q (iv) For each i ∈ I write Fi for the support of µi , and set F = i∈I Fi . This is closed because every Fi is. Its complement is covered by the negligible open sets {x : x ∈ X, x(i) ∈ Xi \ Fi } as i runs over I; as λ is τ -additive, the union of the negligible open sets is negligible, and Q F is conegligible. If W ⊆ X is open and x ∈ F ∩ W , let U ∈ U be such that x ∈ U ⊆ W . Express U as i∈I Gi where Gi ∈ Ti for every i ∈ I and J = {i : Gi 6= Xi } is finite. Then x(i) ∈ Gi ∩ Fi , so µi Gi > 0, for every i. Accordingly Q λ(W ∩ F ) = λW = λU = i∈J µi Ui > 0. Thus F is self-supporting and is the support of λ. (v), (vi), (vi) Use the same arguments as in the corresponding parts of 417C, this time using 412U to confirm that λ is inner regular with respect to the given family of sets. 417F Corollary Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that µi is inner regular with respect to the Borel sets for each i. Then there is a unique complete τ -additive ˜ on X = Q Xi which extends the ordinary product measure and is inner regular topological measure λ i∈I with respect to the Borel sets. ˜ with the right properties. If λ0 is any other complete τ -additive proof By 417E(v) we have a measure λ ˜ topological measure, extending λ and inner regular with respect to the family B of Borel sets, then λ0 W = λW ˜ .Q for every open set W ⊆ X. P P By the argument of (e-iii) of the proof of 417E, λ0 W = λ∗ W = λW Q By ˜ ˜ for every Borel set W . Now λ0 and λ ˜ are 414L, applied to the Borel measures λ0 ¹B and λ¹B, λ0 W = λW supposed to be complete topological probability measures inner regular with respect to B, so they must be identical, by 412L or otherwise.

98


417G

˜ Q 417G Notation In the context of 417D or 417F, I will call λ the τ -additive product measure on i∈I Xi . Note that the uniqueness assertions in 417D and 417F mean that for the products of finitely many probability spaces we do not need to distinguish between the two constructions. The latter also shows that we can relate 415E to the new method: if every Ti is separable and metrizable and every µi is strictly positive, then the ‘ordinary’ product measure λ is a complete topological measure. Since it is also inner regular with respect to the Borel sets (412Uc), it must be exactly the τ -additive product measure as described here. 417H Fubini’s theorem for τ -additive product measures Let (X, T, Σ, µ) and (Y, S, T, ν) be two complete locally determined effectively locally finite τ -additive topological measure spaces such that both µ ˜ be the τ -additive product measure on X × Y , and ν are inner regular with respect to the Borel sets. Let λ ˜ and Λ its domain. R ˜ is defined in [−∞, ∞] and (X × Y ) \ {(x, y) : (a) Let f be a [−∞, ∞]-valued function such that f dλ S (x, y) ∈ dom f, f (x, y) = 0} can be covered by a set of the form X × n∈N Yn where νYn < ∞ for every RR R ˜ n ∈ N. Then the repeated integral f (x, y)ν(dy)µ(dx) is defined and equal to f dλ. (b) Let f : X × Y → [0, ∞] be lower semi-continuous. Then

RR

f (x, y)ν(dy)µ(dx) =

RR

f (x, y)µ(dx)ν(dy) =

R

˜ f dλ

in [0, ∞]. RR ˜ ˜ (c) Let f be a Λ-measurable real-valued function defined λ-a.e. on X × Y . If either |f (x, y)|ν(dy)µ(dx) RR ˜ or |f (x, y)|µ(dx)ν(dy) is defined and finite, then f is λ-integrable. proof (a) I use 252B. R ˜ such that νW [{x}]µ(dx) is defined in [0, ∞] and equal to (i) Write W for the set of those W ∈ Λ ˜ . Then open sets belong to W, by 417C(iv). Next, any Borel subset of an open set of finite measure λW belongs to W. P P If W0 is an open set of finite measure, then {W : W ⊆ X × Y, W ∩ W0 ∈ W} is a Dynkin class containing every open set, so contains all Borel subsets of X × Y . Q Q S ˜ Now suppose that W ⊆ X × Y is λ-negligible and included in X × n∈N Yn , where νYn < ∞ for every n. Then W ∈ W. P P Set A = {x : x ∈ X, ν ∗ W [{x}] > 0}. For each n, let Hn ⊆ Y be an openSset of finite −n measure such that ν(Y S n \ Hn ) ≤ 2 ; we may arrange that Hn+1 ⊇ Hn for each n. Set H = n∈N Hn , so that W [{x}] \ H ⊆ n∈N Yn \ H is negligible for every x ∈ X. ˜ is inner regular with Fix an open set G ⊆ X of finite measure and n ∈ N for the moment. Because λ ˜ ˜ respect to the Borel sets, there is a Borel set V ⊆ (G × Hn ) \ W such that λV = λ((G × Hn ) \ W ), that is, ˜ 0 = 0, where V 0 = (G × Hn ) \ V ⊇ (G × Hn ) ∩ W . We know that V 0 ∈ W, so λV

R

˜ 0 = 0, νV 0 [{x}]dx = λV

and νV 0 [{x}] = 0 for almost every x ∈ X; but this means that Hn ∩ W [{x}] is negligible for almost every x ∈ G. At this point, recall that n was arbitrary, so H ∩ W [{x}] and W [{x}] are negligible for almost every x ∈ G, that is, A ∩ G is negligible. This is true for every open set G ⊆ X of finite measure. Because µ is inner regular with respect to subsets of open sets R of finite measure, and is complete and locally determined, A is negligible (412Jb). But this means that νW [{x}]µ(dx) is defined and equal to zero, so that W ∈ W. Q Q R ˜ is defined in [−∞, ∞] and that there is a sequence hYn in∈N of sets of (ii) Now suppose that f dλ S finite Smeasure in Y such that f (x, y) is defined and zero whenever x ∈ X and y ∈ Y \ n∈N Yn . Set Z = n∈N Yn . Write λ for the c.l.d. product measure on X × Y and Λ for its domain. Then there is a ˜ Λ-measurable function g : X × Y → [−∞, ∞] which is equal λ-almost everywhere to f . P P For q ∈ Q set ˜ q 4Vq ) = 0 (417C(ii)); ˜ and choose Vq ∈ Λ such that λ(W Wq = {(x, y) : (x, y) ∈ dom f, f (x, y) ≥ q} ∈ Λ, set g(x, y) = sup{q : q ∈ Q, (x, y) ∈ Vq } for x ∈ X, y ∈ Y , interpreting sup ∅ as −∞. Q Q Adjusting g if necessary, we may suppose that it is zero on X × (Y \ Z). Set A = (X × Y ) \ {(x, y) : f (x, y) = g(x, y)},

417I


99

˜ so that A is λ-negligible and included in X × Z. By (i), νA[{x}] = 0, that is, y 7→ f (x, y) and y 7→ g(x, y) are equal ν-a.e., for µ-almost every x. Write λX×Z for the subspace measure induced by λ on X × Z; note that this is the c.l.d. product of µ with the subspace measure νZ on Z, by 251P(ii-α). Now we have Z

Z

Z

˜= f dλ

˜= g dλ

g dλ

˜ to (X × Y, λ) is inverse-measure-preserving) (by 235Ib, because the identity map from (X × Y, λ) Z Z ZZ = g dλ = g dλX×Z = g(x, y)νZ (dy)µ(dx) X×Z

X×Z

Z

(by 252B, because νZ is σ-finite) ZZ = g(x, y)ν(dy)µ(dx) (because g(x, y) = 0 if y ∈ Y \ Z) ZZ = f (x, y)ν(dy)µ(dx). (b) If f is non-negative and lower semi-continuous, set Wni = {(x, y) : f (x, y) > 2−n i} for n, i ∈ N, and fn = 2−n

P 4n i=1

χWni

for n ∈ N. Applying 417C(iv) we see that

R

˜= fn dλ

RR

fn (x, y)dydx =

in [0, ∞] for every n; taking the limit as n → ∞,

R

˜= f dλ

RR

f (x, y)dydx =

RR

RR

fn (x, y)dxdy f (x, y)dxdy,

because hfn in∈N is a non-decreasing sequence with limit f . RR ˜ is (c) ?? Suppose, if possible, that γ = |f (x, y)|dydx is finite, but that f is not integrable. Because λ R ˜ ˜ > γ (213B). semi-finite, there must be a non-negative λ-simple function g such that g ≤a.e. |f | and g dλ ˜ For each n ∈ N, there are open sets Gn ⊆ X, Hn ⊆ Y of finite measure such that λ({(x, y) : g(x, y) ≥ −n −n 2R } \ (Gn × Hn )) ≤ 2 , by 417C(iii); now g × χ(Gn × Hn ) → g a.e., so there is some n such that ˜ > γ. In this case, setting g 0 (x, y) = min(g(x, y), |f (x, y)|) for (x, y) ∈ (Gn × Hn ) ∩ dom f , 0 g dλ Gn ×Hn R ˜ > γ. But we can apply (a) to g 0 to see that otherwise, we have g = g 0 a.e. on Gn × Hn , so that g 0 dλ γ<

R

˜= g 0 dλ

RR

g 0 (x, y)dydx ≤

RR

|f (x, y)|dydx ≤ γ,

which isRR absurd. X X ˜ So if |f (x, y)|dydx is finite, f must be λ-integrable. Of course the same arguments, reversing the roles RR ˜ of X and Y , show that f is λ-integrable if |f (x, y)|dxdy is defined and finite. 417I The constructions here have most of the properties one would hope for. I give several in the exercises (417Xd-417Xf, 417Xj). One fact which is particularly useful, and also has a trap in it, is the following. Proposition Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure ˜ the τ -additive product spaces in which the measures are inner regular with respect to the Borel sets, and λ measure on X × Y . Suppose that A ⊆ X and B ⊆ Y , and write µA , νB for the corresponding subspace measures; assume that both µA and νB are semi-finite. Then these are also effectively locally finite, τ ˜ A×B induced by λ ˜ on additive and inner regular with respect to the Borel sets, and the subspace measure λ A × B is just the τ -additive product measure of µA and νB .

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proof (a) To check that µA and νB are effectively locally finite, τ -additive and inner regular with respect to ˜ A×B inherits the same properties from λ, ˜ and is also complete the Borel sets, see 414K and 412P. Of course λ and locally determined, by 214Id. ˜ ∗ (C × D) = µA C · νB D. P (b) Now if C ∈ dom µA and D ∈ dom νB , then λ P (α) There are E ∈ Σ, F ∈ T ∗ ∗ such that C ⊆ E, D ⊆ F , µE = µ C and νF = ν D; in which case ˜ ∗ (C × D) ≤ λ(E ˜ × F ) = λ(E × F ) = µE · νF λ (251J) = µ∗ C · ν ∗ D = µA C · νB D. (β) If γ < µA C ·νB D then, because µA and νB are semi-finite, there are C 0 ⊆ C, D0 ⊆ D such that both have finite outer measure and µ∗ C 0 · ν ∗ D0 ≥ γ. In this case, take E 0 ∈ Σ, F 0 ∈ T such that C 0 ⊆ E 0 , D0 ⊆ F 0 and ˜ and C × D ⊆ W , we have C 0 × D0 ⊆ W ∩ (E × F ), both E 0 and F 0 have finite measure. Now if W ∈ dom λ ∗ 0 0 so that ν(W ∩ (E × F ))[{x}] ≥ ν D for every x ∈ C , and ˜ ≥ λW

R

E

ν(W ∩ (E × F ))[{x}]µ(dx) ≥ µ∗ C 0 · ν ∗ D0 ≥ γ,

˜ ∗ (C × D) ≥ γ; as γ is arbitrary, λ ˜ ∗ (C × D) ≥ µA C · νB D. Q by 417Ha. As W is arbitrary, λ Q (c) In particular, if U ⊆ A and V ⊆ B are relatively open, ˜ A×B (U × V ) = λ ˜ ∗ (U × V ) = µA U · νB V . λ ˜ A×B must be exactly the τ -additive product measure of µA and νB . But now 417D tells us that λ 417J In order to use 417H effectively in the theory of infinite products, we need an ‘associative law’ corresponding to 254N. Theorem Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that every ˜j µi is inner regular with respect to the Borel sets, and hKj ij∈J a partition of I. For each j ∈ J let λ Q ˜ be the τ -additive product measure on Zj = i∈Kj Xi , and write λ for the τ -additive product measure on Q X = i∈I Xi . Then the natural bijection Q x 7→ φ(x) = hx¹Kj ij∈J : X → j∈J Zj ˜ with the τ -additive product of the family hλ ˜ j ij∈J . identifies λ ˜ In particular, if K Q ⊆ I is any set,Qthen λ can be identified with the τ -additive product of the τ -additive product measures on i∈K Xi and i∈I\K Xi . proof We have a lot of measures to keep track of; I hope that the following notation will not be too ˜ j for the ordinary and opaque. Write λ for the ordinary product measure on X, and for j ∈ J write λj , λ Q τ -additive product measures on Zj . Write θ for the ordinary product measure on Z = j∈J Zj of the ˜ j , and θ˜ for the τ -additive product of the λ ˜ j . Write λ ˜ # for the measure on X τ -additive product measures λ # −1 −1 ˜ ˜ ˜ corresponding to θ on Z. (If you like, λ is the image measure θ(φ ) defined from θ˜ and the function ˜ # , like θ, ˜ is a complete τ -additive topological measure, inner regular with respect φ−1 : Z → X.) Then λ to theQBorel sets, because φ : X → Z is a homeomorphism. If C ⊆ X is a measurable cylinder, itQ is of the form i∈I Ei where Ei ∈ Σi for each i and {i : i ∈ I, Ei 6= Σi } is finite. So φ[C] is of the form j∈J Cj , Q where Cj = i∈Kj Ei , and Y Y Y ˜ # C = θ( ˜ ˜ j Cj λ Cj ) = θ( Cj ) = λ j∈J

=

Y

j∈J

λj C j =

j∈J

Y Y j∈J i∈Kj

j∈J

Ei =

Y

Ei = λC.

i∈I

˜ # ) to (X, λ), that λ ˜ # extends λ. So it But this means, applying 254G to the identity map from (X, λ is a complete τ -additive topological measure, inner regular with respect to the Borel sets, extending the ordinary product measure, and by the uniqueness declared in 417F, must be identical to the τ -additive product measure on X, as claimed.

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417K Proposition Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such ˜ be their τ -additive product. ˜ λ) that every µi is inner regular with respect to the Borel sets, and let (X, Λ, ˜ J be the τ -additive product measure on XJ = Q ˜ For J ⊆ I let λ X , and Λ i J its domain; let πJ : X → XJ i∈J −1 ˜ ˜ ˜ is determined by be the canonical map. Then λJ is the image measure λπJ . In particular, if W ∈ Λ ˜ ˜ ˜ J and λJ πJ [W ] = λW . coordinates in J ⊆ I, then πJ [W ] ∈ Λ ˜ is an extension of the ordinary product measure λ on X, λπ ˜ −1 is an extension of λπ −1 , proof Because λ J J ˜ is a τ -additive topological measure and which is the ordinary product measure on XJ (254Oa). Because λ ˜ −1 is a τ -additive topological measure; because λ ˜ is a complete probability measure, πJ is continuous, λπ J −1 −1 ˜ ˜ ˜ so is λπJ . Finally, λπJ is inner regular with respect to the Borel sets. P P Recall that we may identify λ −1 −1 ˜ J and λ ˜ I\J (417J). If V ∈ dom λπ ˜ ˜ we can think of with the τ -additive product of λ , that is, π [V ] ∈ Λ, πJ−1 [V ] ⊆ X as V × XI\J ⊆ XI × XJ . In this case, we must have ˜ −1 [V ] = λπ J

R

J

J

˜ J V dλI\J , λ

˜ J V must be defined and equal to λπ ˜ −1 [V ]. by Fubini’s theorem for τ -additive products (417Ha); that is, λ J ˜ −1 [V ], there must be a Borel set V 0 ⊆ V such that λ ˜ J V 0 ≥ γ. In this case, because πJ is Now if γ < λπ J ˜ −1 [V 0 ] is defined. As with V , this measure must be λ ˜ J V 0 ≥ γ. continuous, πJ−1 [V 0 ] is also Borel, and λπ J −1 ˜ Since V and γ are arbitrary, λπ Q J is inner regular with respect to the Borel sets, as claimed. Q −1 ˜ ˜ By the uniqueness assertion in 417F, λπJ must be λJ exactly. ˜ is determined by coordinates in J, then If now W ∈ Λ ˜ J πJ [W ] = λπ ˜ −1 [πJ [W ]] = λW ˜ . λ J

417L Corollary Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that ˜ their τ -additive product. Let hKj ij∈J ˜ λ) every µi is inner regular with respect to the Borel sets, and (X, Λ, ˜ j for the σ-algebra of members of Λ determined by be a disjoint family of subsets of I, and for j ∈ J write Λ ˜ j ij∈J is a stochastically independent family of σ-algebras (definition: 272Ab). coordinates in Kj . Then hΛ proof It is enough to consider the case in which J is finite (272Bb), no Kj is empty (since if Kj = ∅ ˜ j for ˜ j = {∅, X}) and S an extra term if necessary). In this case, if Wj ∈ Λ then Λ j∈J Kj = I (adding T Q Q each j, then the identification between X and j∈J i∈Kj Xi , as described in 417J, matches j∈J Wj with Q Q ˜ i∈Kj Xi , we j∈J πKj [Wj ], writing πKj (x) for x¹Kj . Now if λj is the τ -additive product measure on Zj = ˜ ˜ ˜ ˜ have λj πKj [Wj ] = λWj , by 417K. Since λ can be identified with the τ -additive product of hλj ij∈J (417J), Q Q ˜ ˜ T ˜ λ( j∈J λWj . j∈J Wj ) = j∈J λj πKj [Wj ] = ˜ j ij∈J is independent. As hWj ij∈J is arbitrary, hΛ 417M Proposition Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that every µi is inner regular withQ respect to the Borel sets and strictly positive. For J ⊆ I let πJ be the ˜ J for the ordinary and τ -additive product measures canonical map from X onto XJ = i∈J Xi ; write λJ , λ ˜ ˜ ˜ ˜ ˜ I , λ = λI , Λ = Λ I . on XJ , and ΛJ , ΛJ for their domains. Set λ = λI , Λ = Λ (a) Let F ⊆ X be a closed self-supporting set, and J the smallest subset of I such that F is determined by coordinates in J (4A2B(g-ii)). Then ˜ ˜ is such that W 4F is λ-negligible (i) if W ∈ Λ and determined by coordinates in K ⊆ I, then K ⊇ J; (ii) J is countable; ˜ (iii) there is a W ∈ Λ, determined by coordinates in J, such that W 4F is λ-negligible. T ˜ ˜ is (b) λ is inner regular with respect to the family of sets of the form n∈N Vn where each Vn ∈ Λ determined by finitely many coordinates. ˜ there are a countable J ⊆ I and sets W 0 , W 00 ∈ Λ, ˜ determined by coordinates in J, such (c) If W ∈ Λ, −1 0 00 00 0 ˜ ˜ ˜ . that W ⊆ W ⊆ W and λ(W \ W ) = 0. Consequently λπJ [πJ [W ]] = λW

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417M

proof (a)(i) ?? Suppose, if possible, otherwise. Then F is not determined by coordinates in K, so there are x Q ∈ F , y ∈ X \ F such that x¹K = y¹K. Let U be an open set containing y, disjoint from F , and of the form i∈I Gi , where Gi ∈ Ti for every i and L = {i : Gi 6= Xi } is finite. Set U 0 = {z : z ∈ X, z(i) ∈ Gi for every i ∈ L ∩ K}, U 00 = {z : z(i) ∈ Gi for every i ∈ L \ K}. Then U 0 ∩ W is determined by coordinates in K, while U 00 is determined by coordinates in I \ K, so ˜ ∩ U ) = λ(W ˜ ˜ ˜ ˜ 00 0 = λ(F ∩ U ) = λ(W ∩ U 0 ∩ U 00 ) = λ(W ∩ U 0 ) · λU (by 417L) ˜ ∩ U 0 ) · λU ˜ 00 = λ(F ˜ ∩ U 0) · = λ(F

Y

µi Gi .

i∈L\K

˜ ∩ U 0 ) > 0. Because every µi is But y ∈ U 0 , and x¹K = y¹K, so x ∈Q F ∩ U 0 ; as F is self-supporting, λ(F strictly positive, and no Gi is empty, i∈L\K µi Gi > 0; and this is impossible. X X ˜ 4W0 ) = 0. By 254Oc there is a W1 ∈ Λ, determined (ii) By 417E(ii), there is a W0 ∈ Λ such that λ(F ˜ 4W1 ) = 0, so (i) tells us by coordinates in a countable subset K of I, such that λ(W0 4W1 ) = 0. Now λ(F that J ⊆ K is countable. ˜ J -negligible. ˜ J . By 417E(ii), there is a V ∈ ΛJ such that V 4πJ [F ] is λ (iii) By 417K, πJ [F ] ∈ Λ −1 −1 Set W = πJ [V ]. Then W ∈ Λ, W is determined by coordinates in J, and W 4F = πJ [V 4πJ [F ]] is ˜ λ-negligible. ˜ which are determined by finitely many coordinates, (b)(i) Write V for the set of those members of Λ and Vδ for the set of intersections of sequences in V. Because V is closed under finite unions, so is Vδ ; Vδ is surely closed under countable intersections, and ∅, X belong to Vδ . (ii) We need to know that every self-supporting closed set F ⊆ X belongs to Vδ . P P By (a), F is determined by a countable set J of coordinates. Express J as the union of a non-decreasing sequence T F ∈ V . Q Q hJn in∈N of finite sets. Then Fn = πJ−1 [π [F ]] ∈ V for each n, and F = δ Jn n∈N n n ˜ and (iii) Let A be the family of subsets of X which are either open or closed. Then if A ∈ A, V ∈ Λ ˜ ∩ V ) > 0, there is a K ∈ Vδ ∩ A such that K ⊆ A and λ(K ˜ λ(A ∩ V ) > 0. P P (α) If A is open, set U = {U : U ∈ V is open, U ⊆ A}. S ˜ is τ -additive and V is closed under Because V includes a base for the topology of X, U = A; because λ ˜ ˜ ˜ ˜ ∩ V ) > 0. (β) If A finite unions, there is a U ∈ U such that U ⊆ A and λU > λA − λ(A ∩ V ), so that λ(U is closed, then it includes a self-supporting closed set V of the same measure (414F), which belongs to Vδ , by (ii) just above. Q Q ˜ ˜ to the Borel σ-algebra of X is inner regular with respect to Vδ . (iv) By 412C, the restriction λ¹B of λ ˜ ˜ But λ is just the completion of λ¹B, so is also inner regular with respect to Vδ (412Ha). ˜ n ≥ λW ˜ − 2−n (c) By (b), we have sequences hVn in∈N , hVn0 in∈N in Vδ such that Vn ⊆ W , Vn0 ⊆ X \ W , λV 0 −n 0 ˜ ˜ and λVn ≥ λ(X \ W ) − 2 for every n ∈ N. Each Vn , Vn is determined by a countable set of coordinates, so there isSa single countable set every Vn and every Vn0 is determined by coordinates in J. Set S J ⊆ 0I such that 0 00 0 W = n∈N Vn , W = X \ n∈N Vn ; then W , W 00 are both determined by coordinates in J, W 0 ⊆ W ⊆ W 00 00 ˜ and λ(W \ W 0 ) = 0, as required. 417N Theorem Let (X, T, Σ, µ) and (Y, S, T, ν) be two quasi-Radon measure spaces. Then the τ ˜ on X × Y is a quasi-Radon measure, the unique quasi-Radon measure on X × Y additive product measure λ ˜ × F ) = µE · νF for every E ∈ Σ, F ∈ T. such that λ(E ˜ is a complete, locally determined, effectively locally finite, τ -additive topological measure, inner proof λ regular with respect to the closed sets (417C(vii)). But this says just that it is a quasi-Radon measure. By 417D, it is the unique quasi-Radon measure with the right values on measurable rectangles.

417S


103

417O Theorem Let h(Xi , Ti , Σi , µi )ii∈I be a family of quasi-Radon probability spaces. Then the τ ˜ on X = Q Xi is a quasi-Radon measure, the unique quasi-Radon measure additive product measure λ i∈I on X extending the ordinary product measure. ˜ is inner regular with respect to the closed sets, so is a quasi-Radon measure, which proof By 417E(vi), λ is unique by 417F. 417P Theorem Let (X, T, Σ, µ) and (Y, S, T, ν) be Radon measure spaces. Then the τ -additive product ˜ on X ×Y is a Radon measure, the unique Radon measure on X ×Y such that λ(E ˜ ×F ) = µE · νF measure λ whenever E ∈ Σ, F ∈ T. ˜ is locally finite (because λ(G ˜ × H) = µG · νH is finite whenever proof Of course X × Y is Hausdorff, and λ ˜ is tight, so is a Radon measure. As in 417N, it is uniquely defined µG and νH are finite). By 417C(viii), λ by its values on measurable rectangles. ˜ 417Q Theorem Let h(XQ i , Ti , Σi , µi )ii∈I be a family of Radon probability spaces, and λ the τ -additive product measure on X = i∈I Xi . For each i ∈ I, let Zi ⊆ Xi be the support of µi . Suppose that ˜ is a Radon measure, the unique Radon measure on J = {i : i ∈ I, Zi is not compact} is countable. Then λ X extending the ordinary product measure. ˜ being totally finite, is locally proof Of course X, being a product of Hausdorff spaces, is Hausdorff, and λ, P finite. Now, given ² ∈ ]0, 1], let h²j ij∈J be a family of strictly positive numbers such that j∈J ²j ≤ ², and for j ∈ J choose a compact set Kj ⊆ XQ j such that µj Kj ≥ 1 − ²j ; for i ∈ I \ J, set Ki = Zi , so that Ki is compact and µi Ki = 1. Consider K = i∈I Ki . Then, using 417E(iii) and 254Lb for the two equalities, ˜ = λ∗ K = Q µi Ki ≥ Q λK j∈J 1 − ²j ≥ 1 − ², i∈I ˜ satisfies the condition (iv) of 416C, and where λ is the ordinary product measure on X. As ² is arbitrary, λ is a Radon measure. As in 417F, it is the unique Radon measure on X extending λ. 417R Notation I will use the phrase quasi-Radon product measure for a τ -additive product measure which is in fact a quasi-Radon measure; similarly, a Radon product measure is a τ -additive product measure which is a Radon measure. 417S Later I will give an example in which a τ -additive product measure is different from the corresponding c.l.d. product measure (419E). In 415E-415F, 415Ye and 416U I have described cases in which c.l.d. measures are τ -additive product measures. It remains very unclear when to expect this to happen. I can however give a couple of results which show that sometimes, at least, we can be sure that the two measures coincide. Proposition (a) Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure spaces such that both µ and ν are inner regular with respect to the Borel sets, and λ the c.l.d. product measure on X ×Y . If every open subset of X ×Y is measured by λ, then λ is the τ -additive product measure on X × Y . (b) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces suchQthat every µi is inner regular with respect to the Borel sets, and λ the ordinary product measure on X = i∈I Xi . If every open subset of X is measured by λ, then λ is the τ -additive product measure on X. Q ˜ J the τ -additive (c) In (b), let λJ be the ordinary product measure on XJ = i∈J Xi for each J ⊆ I, and λ ˜ ˜ I is the product measure. If λJ = λJ for every finite J ⊆ I, and every µi is strictly positive, then λ = λ τ -additive product measure on X. proof (a), (b) In both cases, λ is a complete locally determined effectively locally finite τ -additive measure which is inner regular with respect to the Borel sets (assembling facts from 251I, 254F, 412S, 412U, 417C and 417E). The extra hypothesis added here is that λ is a topological measure, so itself satisfies the conditions of 417D or 417F, and is the τ -additive product measure.

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S ˜ J for every countable J ⊆ I. P (c)(i) The first step is to note that λJ = λ P Express J as n∈N Jn where T hJn in∈N is a non-decreasing sequence of finite sets. If F ⊆ XJ is closed, then it is n∈N πn−1 [πn [F ]], where πn : XJ → XJn is the canonical map for each n. But every πn [F ] is a closed subset of XJn , therefore measured by λJn ; because πn is inverse-measure-preserving (417K), πn−1 [πn [F ]] ∈ dom λJ for each n, and F ∈ dom λJ . Thus every closed set, therefore every open set is measured by λJ , and λJ is a topological ˜J . Q measure; by (b), λJ = λ Q ˜ such that W 0 ⊆ W ⊆ W 00 , (ii) Suppose that W ⊆ X is open. By 417M, there are W 0 , W 00 measured by λ 00 0 00 ˜ I (W \ W 0 ) = 0. Let J ⊆ I be a both W and W are determined by coordinates in a countable set, and λ ˜ J measures πJ [W 0 ], by 417K, countable set such that W 0 and W 00 depend on coordinates in J. Then λJ = λ −1 0 00 ˜ I (W 00 \W 0 ) = 0, so λ measures W = πJ [πJ [W ]], by 254Oa. Similarly, λ measures W . Now λ(W 00 \W 0 ) = λ so λ measures W . As W is arbitrary, λ is a topological measure and must be the τ -additive product measure, by (a). 417T Proposition Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure spaces such that both µ and ν are inner regular with respect to the Borel sets, and λ the c.l.d. product measure on X × Y . If X has a conegligible subset with a countable network (e.g., if X is separable and metrizable), then λ is the τ -additive product measure on X × Y . proof (a) Suppose to begin with that µ and ν are totally finite, and that X has a countable network; let ˜ be the τ -additive ˆ its domain. Let λ hAn in∈N run over a network for X. Let µ ˆ be the completion of µ and Σ ˜ I will use a product measure on X × Y . (We are going to need Fubini’s theorem both for λ and for λ. sprinkling of references to §§251-252 to indicate which parts of the argument below depend on the properties of λ.) Let W ⊆ X × Y be an open set. For each n ∈ N, set S Hn = {H : H ∈ S, An × H ⊆ W }, S P Of course An × Hn ⊆ W for every n ∈ N. If (x, y) ∈ W , so that Hn is open. Then W = n∈N An × Hn . P there are open sets G ⊆ X, H ⊆ Y such that (x, y) ∈ G × H ⊆ W ; now there is an n ∈ N such that x ∈ An ⊆ G, so that H ⊆ Hn and (x, y) ∈ An × Hn . Q Q ˜ By 417C(iv), there is an open set W0 in the domain Λ of λ such that W0 ⊆ W and λ(W \ W0 ) = 0. By 417Ha, applied to χ(W \ W0 ), A = {x : ν(W [{x}] \ W0R[{x}]) > 0} is µ-negligible. For each n ∈ N, x ∈ X set fn (x) = ν(Hn ∩ W0 [{x}]); then 252B tells us that fn dµ is defined and equal to λ(W0 ∩ (X × Hn )). ˆ ˆ If x ∈ An , then Hn ⊆ W [{x}], so In particular, fn is Σ-measurable. Set En = {x : fn (x) = νHn } ∈ Σ. An \ En ⊆ A. Now, by 252B again, Z λ((En × Hn ) \ W0 ) = ν(Hn \ W0 [{x}])µ(dx) En Z = νHn − ν(Hn ∩ W0 [{x}])µ(dx) = 0. So if we set W1 =

En

S n∈N

En × Hn , W1 \ W ⊆ W1 \ W0 is λ-negligible. On the other hand, S W \ W1 ⊆ n∈N (An \ En ) × Hn ⊆ A × Y

is also λ-negligible. Because λ is complete, W ∈ Λ. As W is arbitrary, λ is a topological measure and is ˜ by 417Sa. equal to λ, (b) Now consider the general case. Let Z be a conegligible subset of X with a countable network; since any subset of a space with a countable network again has a countable network (4A2Na), we may suppose that Z ∈ Σ. Again let W be an open set in X ×Y . This time, take arbitrary E ∈ Σ, F ∈ T of finite measure, and consider the subspace measures µE∩Z and νF . These are still effectively locally finite and τ -additive (414K), and are now totally finite. Also E ∩ Z has a countable network. So (a) tells us that the relatively open set W ∩ ((E ∩ Z) × F ) is measured by the c.l.d. product of µE∩Z and νF , which is the subspace measure on (E ∩ Z) × F induced by λ (251P). Since λ surely measures E × F , it measures W ∩ (Z × Y ) ∩ (E × F ). As

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E and F are arbitrary, λ measures W ∩ (Z × Y ) (251H). But λ((X \ Z) × Y ) = µ(X × Z) · νY = 0 (251Ia), so λ also measures W . As W is arbitrary, λ is the τ -additive product measure. 417U Proposition Let h(Xi , Ti , Σi , µi )ii∈I be a family Q of τ -additive topological probability spaces. Let λ be the ordinary product probability measure on X = i∈I Xi and Λ its domain. Then every continuous function f : X → R is Λ-measurable, so Λ includes the Baire σ-algebra of X. ˜ be a τ -additive topological measure extending λ (417E), and Λ ˜ its domain; then f is proof (a) Let λ ˜ is a topological measure. For α ∈ R, set ˜ Λ-measurable, just because λ Gα = {x : x ∈ X, f (x) < α},

Hα = {x : x ∈ X, f (x) > α},

Fα = {x : x ∈ X, f (x) = α}. ˜ α > 0} is countable, and A0 = R \ A is dense in R; let Then hFα iα∈R is disjoint, so A = {α : α ∈ R, λF 0 Q ⊆ A be a countable dense set. For each q ∈ Q, let Vq ⊆ Gq , Wq ⊆ Hq be such that ˜ q, λVq = λ∗ Gq = λG

˜ q λWq = λ∗ Hq = λH

(413Ea, 417E(iii)). Then ˜ λ∗ (Gq \ Vq ) ≤ λ(X \ (Vq ∪ Wq )) = 1 − λVq − λWq = λ(X \ (Gq ∪ Hq )) = 0. Because λ is complete, Gq \ Vq and Gq belong to Λ. But now, if α ∈ R, S {x : f (x) < α} = q∈Q,q<α Gq ∈ Λ, so f is Λ-measurable. (b) It follows that every zero set belongs to Λ, so that Λ must include the Baire σ-algebra of X. 417V Proposition Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure spaces, and (X × Y, Λ, λ) their c.l.d. product. Then every continuous function f : X × Y → R is Λ-measurable, and the Baire σ-algebra of X × Y is included in Λ. proof Let Z ⊆ X × Y be a zero set. If E ∈ Σ, F ∈ T are sets of finite measure, then Z ∩ (E × F ) is a zero set for the relative topology of E × F . Now the subspace measures µE and νF are τ -additive topological measures (414K), so Z ∩ (E × F ) is measured by the c.l.d. product µE × νF of µE and νF . P P If either µE or νF is zero, this is trivial. Otherwise, they have scalar multiples µ0E , νF0 which are probability measures, and of course are still τ -additive topological measures. By 417U, Z ∩ (E × F ) is measured by µ0E × νF0 . Since µE × νF is just a scalar multiple of µ0E × νF0 , Z ∩ (E × F ) is measured by µE × νF . Q Q But µE × νF is the subspace measure λE×F (251P), so Z ∩ (E × F ) ∈ Λ. As E and F are arbitrary, Z ∈ Λ (251H). Thus every zero set belongs to Λ; accordingly Λ must include the Baire σ-algebra, and every continuous function must be Λ-measurable. 417X Basic exercises (a) Let (X, Σ, µ) be a semi-finite measure space and A a family of subsets of X. Show that the following S are equiveridical: (i) there is a measure µ0 on X, extending µ, such that µ0 A = 0 for every A ∈ A; (ii) µ∗ ( n∈N An ) = 0 for every sequence hAn in∈N in A. (b) Let (X, Σ, µ) and (Y, T, ν) be measure spaces with topologies with respect to which µ and ν are locally finite. Show that the c.l.d. product measure on X × Y is locally finite. > (c) Let (X, T, Σ, µ) and (Y, S, T, ν) be topological measure spaces such that µ and ν are both effectively locally finite τ -additive Borel measures. Show that there is a unique effectively locally finite τ -additive Borel measure λ0 on X × Y such that λ0 (G × H) = µG · νH for all open sets G ⊆ X, H ⊆ Y . >(d) Let h(Xi , Ti , Σi , µi )ii∈I be a family of topological probability spaces in whichQevery µi is a τ additive Borel Q measure. Show that there is a unique τ -additive Borel measure λ0 on X = i∈I Xi such that Q 0 λ ( i∈I Fi ) = i∈I µi Fi whenever Fi ⊆ Xi is closed for every i ∈ I.

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(e) Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure spaces in ˜ the τ -additive product measure which the measures are inner regular with respect to the Borel sets, and λ on X × Y . Let hXi ii∈I , hYj ij∈J be decompositions for µ, ν respectively (definition: 211E). Show that ˜ (Cf. 251N.) hXi × Yj ii∈I,j∈J is a decomposition for λ. > (f ) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that µi is inner ˜ the τ -additive product measure on X = Q Xi . regular with respect to the Borel sets for every i, and λ i∈I ˜ Take Ai ⊆ Xi for each i ∈ I. (i) Show that if µ∗i Ai = 1 for every i, then the subspace measure induced by λ Q # ˜ on A = i∈I Ai is just the τ -additive product λ of the subspace measures on the Ai . (Hint: show that if ˜ # (W ∩ A) for Borel sets W ⊆ X, then λ0 satisfies the conditions of 417Xd.) (ii) Show that we set λ0 W = λ ∗ ˜ A = Q µ∗ Ai . (Cf. 254L.) in any case λ i∈I i (g) Let h(Xi , Ti , Σi , µi )ii∈I and h(Yi , Si , Ti , νi )ii∈I be two families of τ -additive topological probability ˜ λ ˜ 0 be the spaces in which every µi and every Q νi is inner regularQwith respect to the Borel sets. Let λ, τ -additive product measures on X = i∈I Xi and Y = i∈I Yi respectively. Suppose that for each i ∈ I we are given a continuous inverse-measure-preserving function φi : Xi → Yi . Show that the function φ : X → Y defined by setting φ(x)(i) = φi (x(i)) for x ∈ X, i ∈ I is inverse-measure-preserving. (h) Let (X, T, Σ, µ) and (Y, S, T, ν) be two complete locally determined effectively locally finite τ -additive ˜ topological measure spaces such that both µ and ν are inner regular with respect to the Borel sets. Let λ ˜ its domain. Suppose that ν is σ-finite. Show that for be the τ -additive product measure on X × Y , and Λ ˜ W [{x}] ∈ T for almost every x ∈ X, and x 7→ νW [{x}] is measurable. any W ∈ Λ, > (i) Let (X, T, Σ, µ) be [0, 1] with its usual topology and Lebesgue measure, and let (Y, S, T, ν) be [0, 1] with the discrete topology and counting measure. (i) Show that both are Radon measure spaces. (ii) Show that the c.l.d. product measure on X × Y is a Radon measure. (Hint: 252Kc, or use 417T and 417P.) (iii) Show that 417Ha can fail if we omit the hypothesis on {(x, y) : f (x, y) 6= 0}. (j) Let (X, T, Σ, µ) and (Y, S, T, ν) be two effectively locally finite τ -additive topological measure spaces. ˜ the τ -additive product measure on X ×Y . Show that λ∗ (A×B) = Let λ be the c.l.d. product measure and λ ∗ ˜ λ (A × B) for all sets A ⊆ X, B ⊆ Y . (Hint: start with A, B of finite outer measure, so that 417I applies.) (k) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces with strictly positive ˜ their τ -additive product. For ˜ λ) measures all inner regular with respect to the Borel sets, and (X, T, Λ, ˜ J be the τ -additive product measure on XJ = Q ˜ X , and Λ J ⊆ I let λ i J its domain. (i) Show that if f is i∈J ˜ ˜ a real-valued Λ-measurable function defined λ-almost everywhere on X, we can find a countable set J ⊆ I ˜ J -almost everywhere on XJ , such that f extends gπJ . (ii) In (i), ˜ J -measurable function g, defined λ and a Λ R R ˜ ˜ show that f dλ = g dλJ if either is defined in [−∞, ∞]. (l) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that every µi is ˜ their τ -additive product. Show that for any ˜ λ) inner regular with respect to the Borel sets, and (X, T, Λ, ˜ ˜ determined by coordinates in J, with W ∈ Λ there is a smallest set J ⊆ I for which there is a W 0 ∈ Λ, 0 ˜ λ(W 4W ) = 0. (Hint: 254R.) (m) What needs to be added to 417M and 415Xk to complete a proof of 415E? (n) Let (X, T, Σ, µ) be an atomless τ -additive topological probability space such that µ is inner regular with respect to the Borel sets, and I a set of cardinal at most that of the support of µ. Show that the set of injective functions from I to X has full outer measure for the τ -additive product measure on X I . > (o) Let (X, T, Σ, µ) and (Y, S, T, ν) be Radon measure spaces. Show that the Radon product measure ˜ such that λ(K ˜ × L) = µK · νL for all compact sets K ⊆ X, L ⊆ Y . on X × Y is the unique Radon measure λ >(p) Let I be an uncountable set, and for each i ∈ I let Xi be {0, 1}, Ti = P{0, 1} the usual topology, and µi the measure on Σi = P{0, 1} defined by saying that µi A = 1 if 0 ∈ A, 0 otherwise. Check that

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107

˜ be the ordinary and τ -additive product measures on (Xi , Ti , Σi , µi ) is a Radon probability space. Let λ, λ Q I ˜ is not determined by any X = i∈I Xi = {0, 1} . Show that they are different. Show that the support of λ ˜ countable set of coordinates. Find a λ-negligible open set W ⊆ X such that its projection onto {0, 1}J is conegligible for every countable J ⊆ I. ˜ the quasi-Radon product (q) Let h(Xi , Ti , Σi , µi )ii∈I be a family of Radon probability spaces, and λ Q ˜ is a Radon measure measure on X = i∈I Xi . For each i ∈ I, let Zi ⊆ Xi be the support of µi . Show that λ iff {i : i ∈ I, Zi is not compact} is countable. In particular, show that the ordinary product measure on I [0, 1[ , where I is uncountable and each copy of [0, 1[ is given Lebesgue measure, is a quasi-Radon measure, but not a Radon measure. (r) Let h(Xn , Tn , Σn , µn )in∈N be a sequence of Radon probability spaces. Show that the Radon product Q Q∞ ˜ on X such that λ( ˜ Q measure on X = n∈N Xn is the unique Radon measure λ n∈N Kn ) = n=0 µn Kn whenever Kn ⊆ Xn is compact for every n. (s) Let (X, T, Σ, ν) and (Y, S, T, ν) be two topological measure spaces, of which (X, T) has a countable network. (i) Show that the c.l.d. product measure λ on X × Y is a topological measure. (ii) Show directly, without relying on ideas from 417D, that λ is τ -additive if ν is. (t) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that every µi ˜ the ordinary and τ -additive product measures on is inner sets, and λ, λ Q regular with respect to the Borel ˜ boundary, then A is measured by λ. X = i∈I Xi . Show that if A ⊆ X has λ-negligible (u) Let us say that a topological space X is chargeable if there is an additive functional ν : PX → [0, ∞[ such that νG > 0 for every non-empty open set G ⊆ X. (i) Show that if there is a σ-finite measure µ on X such that µ∗ G > 0 for every non-empty open set G, then X is chargeable. (Hint: 215B(vii), 391G.) (ii) Show that any separable space is chargeable. (iii) Show that X is chargeable iff its regular open algebra is chargeable in the sense of 391X. (Hint: see the proof of 314P.) (iv) Show that any open subspace of a chargeable space is chargeable. (v) Show that if Y ⊆ X is dense, then X is chargeable iff Y is chargeable. (vi) Show that if X is expressible as the union of countably many chargeable subspaces, then it is chargeable. (vii) Show that any product of chargeable spaces is chargeable. (Cf. 391Xb(iii).) (viii) Show that if hXi ii∈I is a family of chargeable spaces with product X, then every regular open subset of X and every Baire subset of X is determined by coordinates in a countable set. (Hint: 4A2Eb, 4A3Mb.) (v) Let (X, T, Σ, µ) and (Y, S, T, ν) be quasi-Radon measure spaces such that µX · νY > 0. Show that the quasi-Radon product measure on X × Y is completion regular iff it is equal to the c.l.d. product measure and µ and ν are both completion regular. (Hint: 412Sc; if µE, νF are finite and Z ⊆ E × F is a zero set of positive measure, use Fubini’s theorem to show that Z has sections of positive measure.) (w) Let h(Xi , Ti , Σ Qi , µi )ii∈I be a family of quasi-Radon probability spaces. Show that the quasi-Radon product measure on i∈I Xi is completion regular iff it is equal to the ordinary product measure and every µi is completion regular. (x) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that Q every µi is inner regular with respect to the Borel sets, and λ the τ -additive product measure on X = i∈I Xi ; write Λ for its domain. (i) Show that if W ∈ Λ, λW > 0 and ² > 0 then there are a finite J ⊆ I and a W 0 ∈ Λ such that λW 0 ≥ 1 − ² and for every x ∈ W 0 there is a y ∈ W such that x¹I \ J = y¹I \ J. (Cf. 254Sb.) (ii) Show that if A ⊆ X is determined by coordinates in I \ {i} for every i ∈ I then λ∗ A ∈ {0, 1}. (Cf. 254Sa.) 417Y Further exercises (a) Give an example to show that, in 417A, λ can be strictly localizable while λ0 is not. (b) Let (X, T, Σ, µ) and (Y, S, T, ν) be effectively locally finite τ -additive topological measure spaces such that µ and ν are both inner regular with respect to the Borel sets. (i) Fix open Rsets G ⊆ X, H ⊆ Y of finite measure. Let WGH be the set of those W ⊆ X × Y such that θGH (W ) = G νˆ(W [{x}] ∩ H)dx is defined, where νˆ is the completion of ν. (α) Show that every

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S open set belongs to WGH . (β) Show that θGH is countably additive in the sense that θGH ( n∈N Wn ) = P∞ S n=0 θGH (Wn ) for every disjoint sequence hWn in∈N in WGH , and τ -additive in the sense that θGH ( V) = supV ∈V θGH (V ) for every non-empty upwards-directed family V of open sets in X × Y . (γ) Show that every Borel set belongs to WGH . (Hint: Monotone Class Theorem.) (δ) Writing B for the Borel σ-algebra of X × Y , show that θGH ¹B is a τ -additive Borel measure; let λGH be its completion. (²) Show that λGH = θGH ¹ΛGH , where ΛGH = dom λGH . (ζ) Show that λGH (E × F ) is defined and equal to µE · νF whenever E ∈ Σ, F ∈ T, E ⊆ G and F ⊆ H. (Hint: start with open E and F , move to Borel E and F with the Monotone Class Theorem.) (η) Writing λ for the c.l.d. product measure on X × Y , show that λGH (W ) is defined and equal to λ(W ∩ (G × H)) whenever W ∈ dom λ. ˜ = sup ˜ to be T{ΛGH : G ∈ T, H ∈ S, µG < ∞, νH < ∞} and λW (ii) Now take Λ G,H λGH (W ) for ˜ ˜ W ∈ Λ. Show that λ is an extension of λ to a complete locally determined effectively locally finite τ -additive topological measure on X × Y which is inner regular with respect to the Borel sets, so is the τ -additive product measure as defined in 417G. (c) Let (X, Σ, µ) and (Y, T, ν) be complete measure spaces with topologies T, S. Suppose that µ and ν are effectively locally finite and τ -additive and moreover that their domains include bases for the two topologies. Show that the c.l.d. product measure on X × Y has the same properties. (Hint: start by assuming that µX and νY are both finite. If V is an upwards-directed family of measurable open sets with measurable open union W , look at gV (x) = νV [{x}] for V ∈ V.) (d) Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces such that every µi ˜ their τ -additive product. (i) Show that the ˜ λ) is inner regular with respect to the Borel sets, and (X, T, Λ, ˜ following are equiveridical: (α) µi is strictly positive for all but countably many i ∈ I; (β) whenever W ∈ Λ ˜ there are a countable J ⊆ I and W1 , W2 ∈ Λ, determined by coordinates in J, such that W1 ⊆ W ⊆ W2 and ˜ 2 \ W1 ) = 0. (ii) Show that when these are false, λ ˜ cannot be equal to the ordinary product measure λ(W on X. (e) Let (X, Σ, µ) and (Y, T, ν) be measure spaces with topologies T, S such that both µ and ν are inner regular with respect to the families of sequentially compact sets in each space. Show that the c.l.d. product measure λ on X × Y is also inner regular with respect to the sequentially compact sets, so has an extension to a topological measure which is inner regular with respect to the sequentially compact sets. (Hint: 412R, 416Yc.) (f ) Let h(Xi , Σi , µi )ii∈I be a family of probability spaces with topologies Ti such that every µi is inner regular with respect to the family of closedQcountably compact sets in Xi and every Xi is compact. Show that the ordinary product measure λ on X = i∈I Xi is also inner regular with respect to the closed countably ˜ which is inner regular with respect to the compact sets, so has an extension to a topological measure λ ˜ closed countably compact sets in X. Show that this can be done in such a way that for every W ∈ dom λ ˜ there is a V ∈ dom λ such that λ(W 4V ) = 0. (Hint: 412T, 416Yb.) (g) Let h(Xn , Σn , µn )in∈N be a sequence of probability spaces with topologies Tn such that every µn is inner regular with respect to the family of sequentially compact sets in Xn . Show that the ordinary product Q measure λ on X = n∈N Xn is also inner regular with respect to the sequentially compact sets, so has an ˜ which is inner regular with respect to the sequentially compact sets in extension to a topological measure λ ˜ there is a V ∈ dom λ such that X. Show that this can be done in such a way that for every W ∈ dom λ ˜ λ(W 4V ) = 0. ˜ (h) Let h(Xi , Ti , Σi , µi )ii∈I be a family Q of quasi-Radon probability spaces, and λ, λ the ordinary and quasi-Radon product measures on X = i∈I Xi . Suppose that all but one of the Ti have countable networks ˜ and all but countably many of the µi are strictly positive. Show that λ = λ. (i) Let us say that a quasi-Radon measure space (X, T, Σ, µ) has the simple product property if the c.l.d. product measure on X × Y is equal to the quasi-Radon product measure for every quasi-Radon measure space (Y, S, T, ν). (i) Show that if (X, T) has a countable network then (X, T, Σ, µ) has the simple

417 Notes


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product property. (ii) Show that if a quasi-Radon measure space has the simple product property so do all its subspaces. (iii) Show that the quasi-Radon product of two quasi-Radon measure spaces with the simple product property has the simple product property. (iv) Show that the quasi-Radon product of any family of quasi-Radon probability spaces with the simple product property has the simple product property. (v) Show that the real line with the right-facing Sorgenfrey topology (415Xc) and Lebesgue measure has the simple product property. 417 Notes and comments The general problem of determining just when a measure can be extended to a measure with given properties is one which will recur throughout this volume. I have more than once mentioned the Banach-Ulam problem; if you like, this is the question of whether there can ever be an extension of the countable-cocountable measure on a set X to a measure defined on the whole algebra PX. This particular question appears to be undecidable from the ordinary axioms of set theory; but for many sets (for instance, if X = ω1 ) it is known that the answer is ‘no’. (See 419G, 438C.) This being so, we have to take each manifestation of the general question on its own merits. In 417C and 417E the challenge is to take a product measure λ defined in terms of the factor measures alone, disregarding their topological properties, and extend it to a topological measure, preferably τ -additive. Of course there are important cases in which λ is itself already a topological measure; for instance, we know that the c.l.d. product of Lebesgue measure on R with itself is Lebesgue measure on R 2 (251M), and other examples are in 415E, 415Ye, 416U, 417S-417T and 417Yk. But in general not every open set in the product belongs to the domain of λ, even when we have the product of two Radon measures on compact Hausdorff spaces (419E). Once we have resolved to grasp the nettle, however, there is a natural strategy for the proof. It is easy ˜ then we to see that if λ, in 417C or 417E, is to have an extension to a τ -additive topological measure λ, ˜ must have λA(V) = 0 for every V belonging to the class V. Now 417A descibes a sufficient (and obviously necessary) condition for there to be an extension of λ with this property. So all we have to do is check. The check is not perfectly straightforward; in 417E it uses all the resources of the original proof that there is a product measure on an arbitrary product of probability spaces (which I suppose is to be expected), with 414B (of course) to apply the hypothesis that the factor measures are τ -additive, and a couple of extra wrinkles (the Wn0 and Cn0 of part (c-ii) of the proof of 417E, and the use of supports in part (c-v)). ˜ are It is worth noting that (both for finite and for infinite products) the measure algebras of λ and λ ˜ and the identical (417C(ii), 417E(ii)), so there is no new work to do in identifying the measure algebra of λ associated function spaces. An obstacle we face in 417C-417E is the fact that not every τ -additive measure µ has an extension to a τ additive topological measure, even when µ is totally finite and its domain includes a base for the topology. (I give an example in 419H.) Consequently it is not enough, in 417C or 417E, to show that the ordinary product measure λ is τ -additive. But perhaps I should remark that if λ is inner regular with respect to the closed sets, this obstacle evaporates (415L). Accordingly, for the principal applications (to quasi-Radon and Radon product measures, and in particular whenever the topological spaces involved are regular) we have rather easier proofs available, based on the constructions of §415. For completely regular spaces, there is yet another approach, because the product measures can be described in terms of the integrals of continuous functions (415I), which by 417U and 417V can be calculated from the ordinary product measures. Of course the proof that λ itself is τ -additive is by no means trivial, especially in the case of infinite products, corresponding to 417E; but for finite products there are relatively direct arguments, applying indeed to slightly more general situations (417Yc). If we have measures which are inner regular with respect to countably compact classes of sets, then there may be other ways of approaching the extension, using theorems from §413 (see 417Ye-417Yg), and for compact Radon measure spaces, λ becomes tight (412Sb, 412V), so its τ -additivity is elementary. As always, it is important to recognise which constructions are in some sense canonical. The arguments of 417C and 417E allow for the possibility that the factor measures are defined on σ-algebras going well beyond the Borel sets. For all the principal applications, however, the measures will be c.l.d. versions of Borel measures, and in particular will be inner regular with respect to the Borel sets. In such a context it is natural to ask for product measures with the same property, and in this case we can identify a canonical τ -additive topological product measure, as in 417D and 417F. (If you prefer to restrict your measures to Borel σ-algebras, you again get canonical product Borel measures (417Xc-417Xd).) Having done so, we

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can reasonably expect ‘commutative’ and ‘associative’ and ‘distributive’ laws, as in 417D, 417J and 417Xe. Subspaces mostly behave themselves (417I, 417Xf). Of course extending the product measure means that we get new integrable functions on the product, so that Fubini’s theorem has to be renegotiated. Happily, it remains valid, at least in the contexts in which it was effective before (417Ha); we still need, in effect, one of the measures to be σ-finite. The theorem still fails for arbitrary integrable functions on products of Radon measure spaces, and the same example works as before (417Xi). In fact this means that we have an alternative route to the construction of the τ -additive product of two measures (417Yb). But note that on this route ‘commutativity’, the identification of the product measure on X × Y with thatR on Y × X, becomes something which can no longer be taken ˜ to be νW [{x}]dx we have to worry about when, and why, this will for granted, Rbecause if we define λW −1 be equal to µW [{y}]dy. A version of Tonelli’s theorem follows from Fubini’s theorem, as before (417Hc). We also have results corresponding to most of the theorems of §254. But note that there are two traps. In the theorem that a measurable set can be described in terms of a projection onto a countable subproduct (254O, 417M) we need to suppose that the factor measures are strictly positive, and in the theorem that a product of Radon measures is a Radon measure (417Q) we need to suppose that the factor measures have compact supports. The basic examples to note in this context are 417Xp and 417Xq. It is not well understood when we can expect c.l.d. product measures to be topological measures, even in the case of compact Radon probability spaces. Example 419E remains a rather special case, but of course much more effort has gone into seeking positive results. Note that the ordinary product measures of this section are always effectively locally finite and τ -additive (417C, 417E), so that they will be equal to the τ additive products iff they measure every open set (417S). Regarding infinite products, the τ -additive product measure can fail to be the ordinary product measure in just two ways: if one of the finite product measures is not a topological measure, or if uncountably many of the factor measures are not strictly positive (417Sc, 417Xp, 417Yd). So it is finite products which need to be studied. Q Whenever we have a subset F of an infinite product X = i∈I Xi , it is important to know when F is determined by coordinates in a proper subset of I; in measure theory, we are particularly interested in sets determined by coordinates in countable subsets of I (254Mb). It may happen that there is a smallest set J such that F is determined by coordinates in J; for instance, when we have a topological product and F is closed (4A2Bg). When we have a product of probability spaces, we sometimes wish to identify sets J such that F is ‘essentially’ determined by coordinates in J, in the sense that there is an F 0 , determined by coordinates in J, such that F 4F 0 is negligible. In this context, again, there is a smallest such set (254Rd), which can be identified in terms of the probability algebra free product of the measure algebras (325Mb). In 417Ma the two ideas come together: under the conditions there, we get the same smallest J by either route. In 417Ma, we have a product of strictly positive τ -additive topological probability measures. If we keep the ‘strictly positive’ but abandon everything else, we still have very striking results just because the product topology is ccc, so that we can apply 4A2Eb. An abstract expression of this idea is in 417Xu.

418 Measurable functions and almost continuous functions In this section I work through the basic properties of measurable and almost continuous functions, as defined in 411L and 411M. I give the results in the full generality allowed by the terminology so far introduced, but most of the ideas are already required even if you are interested only in Radon measure spaces as the domains of the functions involved. Concerning the codomains, however, there is a great difference between metrizable spaces and others, and among metrizable spaces separability is of essential importance. I start with the elementary properties of measurable functions (418A-418C) and almost continuous functions (418D). Under mild conditions on the domain space, almost continuous functions are measurable (418E); for a separable metrizable codomain, we can expect that measurable functions should be almost continuous (418J). Before coming to this, I spend a couple of paragraphs on image measures: a locally finite image measure under a measurable function is Radon if the measure on the domain is Radon and the function is almost continuous (418I).

418B

Measurable functions and almost continuous functions

111

418L-418Q are important results on expressing given Radon measures as image measures associated with continuous functions, first dealing with ordinary functions f : X → Y (418L) and then coming to Prokhorov’s theorem on projective limits of probability spaces (418M). The machinery of the first part of the section can also be used to investigate representations of vectorvalued functions in terms of product spaces (418R-418T). 418A Proposition Let X be a set, Σ a σ-algebra of subsets of X, Y a topological space and f : X → Y a measurable function. (a) f −1 [F ] ∈ Σ for every Borel set F ⊆ Y . (b) If A ⊆ X is any set, endowed with the subspace σ-algebra, then f ¹A : A → Y is measurable. (c) Let (Z, T) be another topological space. Then gf : X → Z is measurable for every Borel measurable function g : Y → Z; in particular, for every continuous function g : Y → Z. proof (a) The set {F : F ⊆ Y, f −1 [F ] ∈ Σ} is a σ-algebra of subsets of Y containing every open set, so contains every Borel subset of Y . (b) is obvious from the definition of ‘subspace σ-algebra’ (121A). (c) If H ⊆ Z is open, then g −1 [H] is a Borel subset of Y so (gf )−1 [H] = f −1 [g −1 [H]] belongs to Σ. 418B Proposition Let X be a set and Σ a σ-algebra of subsets of X. (a) If Y is a metrizable space and hfn in∈N is a sequence of measurable functions from X to Y such that f (x) = limn→∞ fn (x) is defined in Y for every x ∈ X, then f : X → Y is measurable. (b) If Y is a topological space, Z is a separable metrizable space and f : X → Y , g : X → Z are functions, then x 7→ (f (x), g(x)) : X → Y × Z is measurable iff f and g are measurable. (c) If Y is a hereditarily Lindelöf space, U a family of open sets generating its topology, and f : X → Y a function such that f −1 [U ] ∈ Σ for every U ∈ U, then f is measurable. (d) If hYi ii∈I is a countable family of separable metrizable spaces, with product Y , then a function f : X → Y is measurable iff πi f : X → Yi is measurable for every i, writing πi (y) = y(i) for y ∈ Y , i ∈ N. proof (a) Let ρ be a metric defining the topology of Y . Let G ⊆ Y be any open set, and for each n ∈ N set Fn = {y : y ∈ Y, ρ(y, z) ≥ 2−n for every z ∈ Y \ G}. Then Fn is closed, so fi−1 [Fn ] ∈ Σ for every n, i ∈ N. But this means that S T f −1 [G] = n∈N i≥n fi−1 [Fi ] ∈ Σ. As G is arbitrary, f is measurable. (b)(i) The functions (y, z) 7→ y, (y, z) 7→ z are continuous, so if x 7→ (f (x), g(x)) is measurable, so are f and g, by 418Ac. (ii) Now suppose that f and g are measurable, and that W ⊆ Y × Z is open. By 4A2P(a-i), the topology of Z has a countable base H; let hHn in∈N be a sequence running over H ∪ {∅}. For each n, set S Gn = {G : G ⊆ Y is open, G × Hn ⊆ W }; S S then Gn is open and Gn × Hn ⊆ W . Accordingly W ⊇ n∈N Gn × Hn . But in fact W = n∈N Gn × Hn . P P If (y, z) ∈ W , there are open sets G ⊆ Y , H ⊆ Z such that (y, z) ∈ G × H ⊆ W . Now there is an n ∈ N such that z ∈ Hn ⊆ H, in which case G × Hn ⊆ W and G ⊆ Gn and (y, z) ∈ Gn × Hn . Q Q Accordingly S {x : (f (x), g(x)) ∈ W } = n∈N f −1 [Gn ] ∩ g −1 [Hn ] ∈ Σ. As W is arbitrary, x 7→ (f (x), g(x)) is measurable. (c) This is just 4A3Db. (d) If f is measurable, so is every πi f , by 418Ac. If every πi f is measurable, set U = {πi−1 [H] : i ∈ I, H ⊆ Yi is open}. Then U generates the topology of Y , and if U = πi−1 [H] then f −1 [U ] = (πi f )−1 [H], so f −1 [U ] ∈ Σ for every U . Also Y is hereditarily Lindelöf (4A2P(a-iii)), so f is measurable, by (c).

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418C Proposition Let (X, Σ, µ) be a measure space and Y a Polish space. Let hfn in∈N be a sequence of measurable functions from X to Y . Then {x : x ∈ X, limn→∞ fn (x) is defined in Y } belongs to Σ. proof (Compare 121H.) Let ρ be a complete metric on Y defining the topology of Y . (a) For m, n ∈ N and δ > 0, the set {x : ρ(fm (x), fn (x)) ≤ δ} belongs to Σ. P P The function x 7→ (fm (x), fn (x)) : X → Y 2 is measurable, by 418Bb, and the function ρ : Y 2 → R is continuous, so x → 7 ρ(fm (x), fn (x)) is measurable and {x : ρ(fm (x), fn (x)) ≤ δ} ∈ Σ. Q Q (b) Now hfn (x)in∈N is convergent iff it is Cauchy, because Y is complete. But {x : x ∈ X, hfn (x)in∈N is Cauchy} =

\ [ \

{x : ρ(fi (x), fm (x)) ≤ 2−n }

n∈N m∈N i≥m

belongs to Σ. 418D Proposition Let (X, Σ, µ) be a measure space and T a topology on X. (a) Suppose that Y is a topological space. Then any continuous function from X to Y is almost continuous. (b) Suppose that Y and Z are topological spaces, f : X → Y is almost continuous and g : Y → Z is continuous. Then gf : X → Z is almost continuous. (c) Suppose that (Y, S, T, ν) is a σ-finite topological measure space, Z is a topological space, g : Y → Z is almost continuous and f : X → Y is inverse-measure-preserving and almost continuous. Then gf : X → Z is almost continuous. (d) Suppose that µ is semi-finite, and that hYi ii∈I is a countable family of topological spaces with product Y . Then a function f : X → Y is almost continuous iff fi = πi f is almost continuous for every i ∈ I, writing πi (y) = y(i) for i ∈ I, y ∈ Y . proof (a) is trivial. (b) The set {A : A ⊆ X, gf ¹A is continuous} includes {A : A ⊆ X, f ¹A is continuous}; so if µ is inner regular with respect to the latter, it is inner regular with respect to the former. (c) Take E ∈ Σ and γ < µE; take ² > 0. We have a cover of Y by a non-decreasing sequence hYn in∈N of measurable sets of finite measure; now hf −1 [Yn ]in∈N is a non-decreasing sequence covering E, so there is an n ∈ N such that µ(E ∩ f −1 [Yn ]) ≥ γ. Because f is inverse-measure-preserving, E ∩ f −1 [Yn ] has finite measure. Now we can find measurable sets F ⊆ Yn , E1 ⊆ E ∩ f −1 [Yn ] such that f ¹E1 , g¹F are continuous and νF ≥ νYn − ², µE1 ≥ µ(E ∩ f −1 [Yn ] \ E1 ) − ². In this case E0 = E1 ∩ f −1 [F ] has measure at least γ − 2² and gf ¹E0 is continuous. As E, γ and ² are arbitrary, gf is almost continuous. (d)(i) If f is almost continuous, every fi must be almost continuous, by (b). (ii) Now suppose that every fi is almost continuous. Take E ∈ Σ and γ < µE. There is an E0 ⊆ E such P that E0 ∈ Σ and γ < µE0 < ∞. Let h²i ii∈I be a family of strictly positive real numbers such that i∈I ²i ≤ µE0 − γ. For each Ti ∈ I choose a measurable set Fi ⊆ E0 such that µFi ≥ µE0 − ²i and fi ¹Fi is continuous. Then F = E0 ∩ i∈I Fi is a subset of E with measure at least γ, and f ¹F is continuous because fi ¹F is continuous for every i (3A3Ib). 418E Theorem Let (X, T, Σ, µ) be a complete locally determined topological measure space, Y a topological space, and f : X → Y an almost continuous function. Then f is measurable. proof Set K = {K : K ∈ Σ, f ¹K is continuous}; then µ is inner regular with respect to K. If H ⊆ Y is open and K ∈ K, then K ∩ f −1 [H] is relatively open in K, that is, there is an open set G ⊆ X such that K ∩ f −1 [H] = K ∩ G. Because µ is a topological measure, G ∈ Σ so K ∩ f −1 [H] ∈ Σ. As K is arbitrary, and µ is complete and locally determined, f −1 [H] ∈ Σ (412Ja). As H is arbitrary, f is measurable.

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418F Proposition Let (X, T, Σ, µ) be a semi-finite topological measure space, Y a metrizable space, and f : X → Y a function. Suppose there is a sequence hfn in∈N of almost continuous functions from X to Y such that f (x) = limn→∞ fn (x) for almost every x ∈ X. Then f is almost continuous. proof Suppose that E ∈ Σ and that γ < µE, ² > 0. Then there is a measurable set F ⊆ E such that γ ≤ µF < ∞; discarding a negligible set if necessary, we may arrange that f (x) = limn→∞ fn (x) for every x ∈ F . Let ρ be a metric on Y defining its topology. For each T n ∈ N, let Fn ⊆ F be a measurable set such that fn ¹Fn is continuous and µ(Fn \ F ) ≤ 2−n ²; set G = n∈N Fn , so that µG ≥ γ − 2² and fn ¹G is continuous for every n ∈ N. For m, n ∈ N, the functions x 7→ (fm (x), fn (x)) : G → Y 2 and x 7→ ρ(fm (x), fn (x)) : G → R are continuous, therefore measurable, because µ is a topological measure. Also hfn (x)in∈N is a Cauchy sequence for every x ∈ G. So if we set Gkn = {x : x ∈ G, ρ(fi (x), fj (x)) ≤ 2−k for all i, j ≥ n}, hGkn in∈N is a nondecreasing sequence of measurable sets with union G for each k ∈ N, and T we can find a strictly increasing sequence hnk ik∈N such that µ(G\Gknk ) ≤ 2−k ² for every k. Setting H = k∈N Gknk , µH ≥ µG−2² ≥ γ −4² and ρ(fi (x), fnk (x)) ≤ 2−k whenever x ∈ H and i ≥ nk ; consequently ρ(f (x), fnk (x)) ≤ 2−k whenever x ∈ H and k ∈ N. But this means that hfnk ik∈N converges to f uniformly on H, while every fnk is continuous on H, so f ¹H is continuous (3A3Nb). And of course H ⊆ E. As E, γ and ² are arbitrary, f is almost continuous. 418G Proposition Let (X, T, Σ, µ) be a σ-finite quasi-Radon measure space, Y a metrizable space and f : X → Y an almost continuous function. Then there is a conegligible set X0 ⊆ X such that f [X0 ] is separable. proof (a) Let K be the family of self-supporting measurable sets K of finite measure such that f ¹K is continuous. Then µ is inner regular with respect to K. P P If E ∈ Σ and γ < µE, there is an F ∈ Σ such that F ⊆ E and γ < µF < ∞; there is an H ∈ Σ such that H ⊆ F , γ ≤ µH and f ¹H is continuous; and there is a measurable self-supporting K ⊆ H with the same measure as H (414F), in which case K ∈ K and K ⊆ E and µK ≥ γ. Q Q (b) Now f [K] is ccc for every K ∈ K. P P If G is a disjoint family of non-empty relatively open subsets of f [K], then hK ∩ P f −1 [G]iG∈G is a disjoint family of non-empty relatively open subsets of K, because f ¹K is continuous, and G∈G µ(K ∩ f −1 [G]) ≤ µK. Because K is self-supporting, µ(K ∩ f −1 [G]) > 0 for every G ∈ G; because µK is finite, G is countable. As G is arbitrary, f [K] is ccc. Q Q Because Y is metrizable, f [K] must be separable (4A2Pd). S (c) BecauseSµ is σ-finite, there is a countable family L ⊆ K such that X0 = L is conegligible (412Ic). Now f [X0 ] = L∈L f [L] is a countable union of separable spaces, so is separable (4A2B(e-i)). 418H Proposition (a) Let X and Y be topological spaces, µ an effectively locally finite τ -additive measure on X, and f : X → Y an almost continuous function. Then the image measure µf −1 is τ -additive. (b) Let (X, T, Σ, µ) be a totally finite quasi-Radon measure space, (Y, S) a regular topological space, and f : X → Y an almost continuous function. Then there is a unique quasi-Radon measure ν on Y such that f is inverse-measure-preserving for µ and ν. proof (a) Let of Y , all measured by µf −1 , and suppose S H be an upwards-directed family of open−1subsets ∗ ∗ that H = H is also measurable. Take any γ < (µf )(H ) = µf −1 [H ∗ ]. Then there is a measurable set E ⊆ f −1 [H ∗ ] such that µE ≥ γ and f ¹E is continuous. Consider {E ∩ f −1 [H] : H ∈ H}. This is an upwards-directed family of relatively open measurable subsets of E with measurable union E. By 414K, the subspace measure on E is τ -additive, so γ ≤ µE ≤ supH∈H µ(E ∩ f −1 [H]) ≤ supH∈H µf −1 [H]. As γ is arbitrary, µf −1 [H ∗ ] ≤ supH∈H µf −1 [H]; as H is arbitrary, µf −1 is τ -additive. (b) By 418E, f is measurable. Let ν0 be the restriction of µf −1 to the Borel σ-algebra of Y ; by (a), ν0 is τ -additive, and f is inverse-measure-preserving with respect to µ and ν0 . Because Y is regular, the completion ν of ν0 is a quasi-Radon measure (415Cb). Because µ is complete, f is still inverse-measurepreserving with respect to µ and ν (235Hc).

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To see that ν is unique, observe that its values on Borel sets are determined by the requirement that f be inverse-measure-preserving, so that 415H gives the result. 418I The next theorem is one of the central properties of Radon measures. I have already presented what amounts to a special case in 256G. Theorem Let (X, T, Σ, µ) be a Radon measure space, Y a Hausdorff space, and f : X → Y an almost continuous function. If the image measure ν = µf −1 is locally finite, it is a Radon measure. proof (a) By 418E, f is measurable, that is, f −1 [H] ∈ Σ for every open set H ⊆ Y ; but this means that the domain T of ν contains every open set, and ν is a topological measure. (b) ν is tight (that is, inner regular with respect to the compact sets). P P If F ∈ T and νF > 0, then µf −1 [F ] > 0, so there is an E ⊆ f −1 [F ] such that µE > 0 and f ¹E is continuous. Next, there is a compact set K ⊆ E such that µK > 0. In this case, L = f [K] is a compact subset of F , and νL = µf −1 [L] ≥ µK > 0. By 412B, this is enough to prove that ν is tight. Q Q Note that because ν is locally finite, νL < ∞ for every compact L ⊆ Y (411Ga). (c) Because µ is complete, so is ν (212Bd). Next, ν is locally determined. P P Suppose that H ⊆ Y is such that H ∩ F ∈ T whenever νF < ∞. Then, in particular, H ∩ f [K] ∈ T whenever K ⊆ X is compact and f ¹K is continuous. But setting K = {K : K ⊆ X is compact, f ¹K is continuous}, µ is inner regular with respect to K (412Ac). And if K ∈ K, K ∩ f −1 [H] = K ∩ f −1 [H ∩ f [K]] ∈ Σ. Because µ is complete and locally determined, this is enough to show that f −1 [H] ∈ Σ (412Ja), that is, H ∈ T. As H is arbitrary, ν is locally determined. Q Q (d) Thus ν is a complete locally determined tight locally finite topological measure; that is, it is a Radon measure. 418J Theorem Let (X, Σ, µ) be a semi-finite measure space and T a topology on X such that µ is inner regular with respect to the closed sets. Suppose that Y is a separable metrizable space and f : X → Y is measurable. Then f is almost continuous. proof Let H be a countable base for the topology of Y , and hHn in∈N a sequence running over H ∪ {∅}. Take E ∈ Σ and γ < µE. Choose hEn in∈N inductively, as follows. There is an E0 ∈ Σ such that E0 ⊆ E and γ < µE0 < ∞. Given En ∈ Σ with γ < µEn < ∞, En \ f −1 [Hn ] ∈ Σ, so there is a closed set Fn ∈ Σ such that Fn ⊆ En \ f −1 [Hn ],

µ((En \ f −1 [Hn ]) \ Fn ) < µEn − γ;

set En+1 = (En ∩ f −1 [Hn ]) ∪ Fn , so that En+1 ∈ Σ,

En+1 ⊆ En ,

µEn+1 > γ,

En+1 \ f −1 [Hn ] = Fn .

Continue. T At the end of the induction, set F = n∈N En . Then F ⊆ E, µF ≥ γ, and for every n ∈ N F ∩ f −1 [Hn ] = F ∩ En+1 ∩ f −1 [Hn ] = F \ Fn is relatively open in F . It follows that f ¹F is continuous (4A2B(a-ii)). As E, γ are arbitrary, f is almost continuous. Remark For variations on this idea, see 418Yg, 433E and 434Yb; also 418Yh. 418K Corollary Let (X, T, Σ, µ) be a quasi-Radon measure space and Y a separable metrizable space. Then a function f : X → Y is measurable iff it is almost continuous. proof Put 418E and 418J together. Remark This generalizes 256F.

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418L In all the results above, the measure starts on the left of the diagram f : X → Y ; in 418I-418K, it is transferred to an image measure on Y . If X has enough compact sets, a measure can move in the reverse direction, as follows. Theorem Let (X, T) be a Hausdorff space, (Y, S, T, ν) a Radon measure space and f : X → Y a continuous function such that whenever F ∈ T and νF > 0 there is a compact set K ⊆ X such that ν(F ∩ f [K]) > 0. Then there is a Radon measure µ on X such that ν is the image measure µf −1 and the inverse-measurepreserving function f induces an isomorphism between the measure algebras of ν and µ. proof (a) Note first that ν is inner regular with respect to L = {f [K] : K ∈ K}, where K is the family of compact subsets of X. P P If νF > 0, there is a K ∈ K such that ν(F ∩ f [K]) > 0; now there is a closed set F 0 ⊆ F ∩ f [K] such that νF 0 > 0, and K 0 = K ∩ f −1 [F 0 ] is compact, while f [K 0 ] ⊆ F has non-zero measure. As L is closed under finite unions, this is enough to show that ν is inner regular with respect to L (412Aa). Q Q (b) Consequently there is a disjoint set L0 ⊆ L such that every non-negligible F ∈ T meets some member of L0 S in a non-negligible set (412Ib). We can express L0 as {f [K] : K ∈ K0 } where K0 ⊆ K is disjoint. Set X0 = K0 . (c) Set Σ0 = {X0 ∩ f −1 [F ] : F ∈ T}. Then Σ0 is a σ-algebra of subsets of X0 . If F , F 0 ∈ T and νF 6= νF 0 , then there must be some K ∈ K0 such that f [K]∩(F 4F 0 ) 6= ∅, so that X0 ∩f −1 [F ] 6= X0 ∩f −1 [F 0 ]; we therefore have a functional µ0 : Σ0 → [0, ∞] defined by setting µ0 (X0 ∩ f −1 [F ]) = νF whenever F ∈ T. It is easy to check that µ0 is a measure on X0 . Now µ0 is inner regular with respect to K. P P If E ∈ Σ0 and µE > 0, there is an F ∈ T such that E = X0 ∩f −1 [F ] and νF > 0. There are a K ∈ K0 such that ν(F ∩f [K]) > 0, and a closed set F 0 ⊆ F ∩f [K] such that νF 0 > 0; now K ∩ f −1 [F 0 ] = X0 ∩ f −1 [F 0 ] belongs to Σ0 ∩ K, is included in E and has measure greater than 0. Because K is closed under finite unions, this is enough to show that µ0 is inner regular with respect to K. Q Q (c) Set Σ1 = {E : E ⊆ X, E ∩ X0 ∈ Σ0 },

µ1 E = µ0 (E ∩ X0 ) for every E ∈ Σ1 .

Then µ1 is a measure on X (being the image measure µ0 ι−1 , where ι : X0 → X is the identity map), and is inner regular with respect to K. If F ∈ T, then µ1 f −1 [F ] = µ0 (X0 ∩ f −1 [F ]) = νF , so f is inverse-measure-preserving for µ1 and ν. Consequently µ1 is locally finite. P P If x ∈ X, there is an open set H ⊆ Y such that f (x) ∈ H and νH < ∞; now f −1 [H] is an open subset of X of finite measure containing x. Q Q In particular, µ∗1 K < ∞ for every compact K ⊆ X (411Ga). (d) By 413O, there is an extension of µ1 to a complete locally determined measure µ on X which is inner regular with respect to K, defined on every member of K, and such that whenever E belongs to the domain Σ of µ and µE < ∞, there is an E1 ∈ Σ1 such that µ(E4E1 ) = 0. Now µ is locally finite because µ1 is, so µ is a Radon measure; and f is inverse-measure-preserving for µ and ν because it is inverse-measurepreserving for µ1 and ν. (e) The image measure µf −1 extends ν, so is locally finite, and is therefore a Radon measure (418I); since it agrees with ν on the compact subsets of Y , it must be identical with ν. (f ) I have still to check that the corresponding measure-preserving homomorphism π from the measure algebra B of ν to the measure algebra A of µ is actually an isomorphism, that is, is surjective. If a ∈ A and µ ¯a < ∞, we can find E ∈ Σ such that E • = a and E1 ∈ Σ1 such that µ(E4E1 ) = 0. Now E1 ∩ X0 = −1 f [F ] ∩ X0 for some F ∈ T; but in this case µ(E1 4f −1 [F ]) = µ1 (E1 4f −1 [F ]) = 0,

a = E1• = (f −1 [F ])• = πF • .

Accordingly π[B] includes {a : µ ¯a < ∞}, and is order-dense in A. But as π is injective and B is Dedekind complete (being the measure algebra of a Radon measure, which is strictly localizable), it follows that π[B] = A (314Ib). Thus π is an isomorphism, as required.

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Remarks Of course this result is most commonly applied when X and Y are both compact and f is a surjection, in which case the condition (*) whenever F ∈ T and νF > 0 there is a compact set K ⊆ X such that ν(F ∩ f [K]) > 0 is trivially satisfied. Evidently (*) is necessary if there is to be any Radon measure on X for which f is inverse-measurepreserving, so in this sense the result is best possible. In 433D, however, there is a version of the theorem in which f is not required to be continuous. 418M Prokhorov’s theorem Suppose that (I, ≤), h(Xi , Ti , Σi , µi )ii∈I , hfij ii≤j∈I , (X, T) and hgi ii∈I are such that (I, ≤) is a non-empty upwards-directed partially ordered set, every (Xi , Ti , Σi , µi ) is a Radon probability space, fij : Xj → Xi is an inverse-measure-preserving function whenever i ≤ j in I, (X, T) is a Hausdorff space, gi : X → Xi is a continuous function for every i ∈ I, gi = fij gj whenever i ≤ j in I. Suppose moreover that for every ² > 0 there is a compact set K ⊆ X such that µi gi [K] ≥ 1 − ² for every i ∈ I. Then there is a Radon probability measure µ on X such that every gi is inverse-measure-preserving for µ. proof (a) Set T = {gi−1 [E] : i ∈ I, E ∈ Σi } ⊆ PX. Then T is a subalgebra of PX. P P (i) There is an i ∈ I, so ∅ = gi−1 [∅] belongs to T. (ii) If H ∈ T there are −1 i ∈ I, E ∈ Σi such that H = gi [E]; now X \ H = gi−1 [Xi \ E] belongs to T. (iii) If G, H ∈ T, there are i, j ∈ I and E ∈ Σi , F ∈ Σj such that G = gi−1 [E] and H = gj−1 [F ]. Now I is upwards-directed, so there is −1 −1 a k ∈ I such that i ≤ k and j ≤ k. Because fik and fjk are inverse-measure-preserving, fik [E] and fjk [F ] belong to Σk , so that G ∩ H = gi−1 [E] ∩ gj−1 [F ] = (fik gk )−1 [E] ∩ (fjk gk )−1 [F ] −1 −1 = gk−1 [fik [E] ∩ fjk [F ]] ∈ T. Q Q

(b) There is an additive functional ν : T → [0, 1] defined by writing νgi−1 [E] = µi E whenever i ∈ I and E ∈ Σi . P P (i) Suppose that i, j ∈ I and E ∈ Σi , F ∈ Σj are such that gi−1 [E] = gj−1 [F ]. Let k ∈ I be such that i ≤ k and j ≤ k. Then −1 −1 gk−1 [fik [E]4fjk [F ]] = gi−1 [E]4gj−1 [F ] = ∅, −1 −1 so gk [X] ∩ (fik [E]4fjk [F ]) = ∅. But now remember that for every ² > 0 there is a set K ⊆ X such that −1 −1 µk gk [K] ≥ 1 − ². This means that µk gk [X] must be 1, so that fik [E]4fjk [F ] must be negligible, and −1 −1 µi E = µk fik [E] = µk fjk [F ] = µj F .

Thus the proposed formula for ν defines a function on T. (ii) Now suppose that G, H ∈ T are disjoint. Again, take i, j ∈ I and E ∈ Σi , F ∈ Σj such that G = gi−1 [E] and H = gj−1 [F ], and k ∈ I such that i ≤ k and j ≤ k. Then −1 −1 νG + νH = µi E + µj F = µk fik [E] + µk fjk [F ] −1 −1 −1 −1 = µk (fik [E] ∪ fjk [F ]) + µk (fik [E] ∩ fjk [F ]) −1 −1 −1 −1 = νgk−1 [fik [E] ∪ fjk [F ]] + νgk−1 [fik [E] ∩ fjk [F ]]

= ν(G ∪ H) + ν(G ∩ H).

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But as ν∅ is certainly 0, we get ν(G ∪ H) = νG + νH. As G, H are arbitrary, ν is additive. Q Q Note that νX = 1. (c) νG = sup{νH : H ∈ T, H ⊆ G, H is closed} for every G ∈ T. P P If γ < νG, there are an i ∈ I and an E ∈ Σi such that G = gi−1 [E]. In this case µi E = νG > γ; let L ⊆ E be a compact set such that µi L ≥ γ; then H = gi−1 [L] is a closed subset of G and νH = µi L ≥ γ. Q Q νX = supK⊆X is compact inf G∈T,G⊇K νG. P P If ² > 0, there is a compact K ⊆ X such that µi gi [K] ≥ 1 − ² for every i ∈ I, by the final hypothesis of the theorem. If G ∈ T and G ⊇ K, there are an i ∈ I and an E ∈ Σi such that G = gi−1 [E], in which case gi [K] ⊆ E, so that νG = µi E ≥ µi gi [K] ≥ 1 − ². Thus inf G∈T,G⊇K νG ≥ 1 − ²; as ² is arbitrary, we have the result. Q Q This means that the conditions of 416O are satisfied, and there is a Radon measure µ extending ν. Of course this means that every gi is inverse-measure-preserving. 418N Remarks (a) Taking I to be a singleton, we get a version of 418L in which Y is a probability space, and omitting the check that the function g induces an isomorphism of the Q measure algebras. Taking I to be the family of finite subsets of a set T , and every Xi to be a product t∈i Zt of Radon probability spaces with its product Radon measure, we obtain a method of constructing products of arbitrary families of compact probability spaces from finite products. (b) In the hypotheses of 418M, I asked only that the fij should be measurable, and omitted any check on the compositions fij fjk when i ≤ j ≤ k. But it is easy to see that the fij must in fact be almost continuous, and that fij fjk must be equal almost everywhere to fik (418Xu), just as in 418P below. (c) In the theorem as written out above, the space X and the functions gi : X → Xi are part of the data. Of course in many applications we start with a structure (h(Xi , Ti , Σi , µi )ii∈I , hfij ii≤j∈I ), and the first step is to find a suitable X and gi , as in 418O and 418P. (d) There are important questions concerning possible relaxations of the hypotheses in 418M, especially in Q Q the special case already mentioned, in which Xi = t∈i Zt , fij (x) = x¹i when i ⊆ j ∈ [T ]<ω , X = t∈T Zt , and gi (x) = x¹i for x ∈ X and i ∈ I, but there is no suggestion that the µi are product measures. For a case in which we can dispense with auxiliary topologies on the Xi , see 451Yb. (e) A typical class of applications of Prokhorov’s theorem is in the theory of stochastic processes, in which we have large families hXt it∈T of random variables; for definiteness, imagine that T = [0, ∞[, so that we are looking at a system evolving over time. Not infrequently our intuition leads us to a clear description of the joint distributions νJ of finite subfamilies hXt it∈J without providing any suggestion of a measure space on which the whole family hXt it∈T might be defined. (As I tried to explain in the introduction to Chapter 27, probability spaces themselves are often very shadowy things in true probability theory.) Each νJ can be thought of as a Radon measure on R J , and for I ⊆ J ∈ [T ]<ω we have a natural map fIJ : R J → R I , setting fIJ (y) = y¹I for y ∈ R J . If our distributions νJ mean anything at all, every fIJ will surely be inversemeasure-preserving; this is simply saying that νI is the joint distribution of a subfamily of hXt it∈J . If we can find a Hausdorff space Ω and a continuous function g : Ω → R T such that, for every finite J ⊆ T and ² > 0, there is a compact set K ⊆ Ω such that νJ gJ [K] ≥ 1 − ² (where gJ (x) = g(x)¹J), then Prokhorov’s theorem will give us a measure µ on Ω which will then provide us with a suitable realization of hXt it∈T as a family of random variables on a genuine probability space, writing Xt (ω) = g(ω)(t). That they become continuous functions on a Radon measure space is a valuable shield against irrelevant complications. Clearly, if this can be done at all it can be done with Ω = R T ; but some of the central results of probability theory are specifically concerned with the possibility of using other sets Ω (e.g., Ω = C(T ), as in 455D). (f ) In (e) above, we do always have the option of regarding each νJ as a measure on the compact space [−∞, ∞]J . In this case, by 418O or otherwise, we can be sure of finding a measure on [−∞, ∞]T to support functions Xt , at the cost of either allowing the values ±∞ or (as I should myself ordinarily do) accepting

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that each Xt would be undefined on a negligible set. The advantage of this is just that it gives us confidence in applying the Kolmogorov-Lebesgue theory to the whole family hXt it∈T at once, rather than to finite or countable subfamilies. For an example of what can happen if we try to do similar things with non-compact measures, see 419K. For an example of the problems which can arise with uncountable families, see 418Xv. 418O I mention two cases in which we can be sure that the projective limit (X, hgi ii∈I ) required in Prokhorov’s theorem will exist. Proposition Suppose that (I, ≤), h(Xi , Ti )ii∈I and hfij ii≤j∈I are such that (I, ≤) is a non-empty upwards-directed partially ordered set, every (Xi , Ti ) is a compact Hausdorff space, fij : Xj → Xi is a continuous surjection whenever i ≤ j in I, fij fjk = fik whenever i ≤ j ≤ k in I. Then there are a compact Hausdorff space (X, T) and a family hgi ii∈I such that gi : X → Xi is a continuous surjection and fij gj = fi whenever i ≤ j ∈ I; so that if we endow the Xi with Radon probability measures for which the fij are inverse-measure-preserving, (I, hXi ii∈I , hfij ii≤j∈I , X, hgi ii∈I ) will satisfy all the hypotheses of 418M. proof Set X = {x : x ∈

Q i∈I

Xi , fij x(j) = x(i) whenever i ≤ j ∈ I},

gi (x) = x(i) for x ∈ X, i ∈ I. Of course gi = fij gj whenever i ≤ j. To see that every gi is surjective, observe that if y ∈ Xi and J ⊆ I is finite, then Q FJ = {x : x ∈ i∈I Xi , x(i) = y, fjk x(k) = x(j) whenever j ≤ k ∈ J} is a closed set. FJ is always non-empty, because if k is an upper bound of J ∪ {i} there is a z ∈ Xk such that fik (z) = y, in which case x ∈ FJ whenever x(j) = fjk (z) for every j ∈ J ∪Q {i}. Now {FJ : J ∈ [I]<ω } is a downwards-directed family of non-empty closed sets in the compact space j∈I Xj , so has non-empty intersection, and if x is any point of the intersection then x ∈ X and gi (x) = y. 418P Proposition Let (I, ≤), h(Xi , Ti , Σi , µi )ii∈I and hfij ii≤j∈I be such that (I, ≤) is a countable non-empty upwards-directed partially ordered set, every (Xi , Ti , Σi , µi ) is a Radon probability space, fij : Xj → Xi is an inverse-measure-preserving almost continuous function whenever i ≤ j in I, fij fjk = fik µk -a.e. whenever i ≤ j ≤ k in I. Then there are a Radon probability space (X, T, Σ, µ) and continuous inverse-measure-preserving functions gi : X → Xi such that gi = fij gj whenever i ≤ j in I. proof (a) We can use the same formula as in 418O: Q X = {x : x ∈ i∈I Xi , fij x(j) = x(i) whenever i ≤ j ∈ I}, gi (x) = x(i) for x ∈ X, i ∈ I. As before, the consistency relation gi = fij gj is a trivial consequence of the definition of X. For the rest, we have to check that the final condition P of 418M is satisfied. Fix ² ∈ ]0, 1[. Start by taking a family h²ij ii≤j∈I of strictly positive numbers such that i≤j∈I ²ij ≤ 12 ². (This is where we need to know that I is countable.) P P Set ²j = i≤j ²ij for each j, so that j∈I ²j ≤ 12 ². For i ≤ j ≤ k in I, set Eijk = {x : x ∈ Xk , fik (x) = fij fjk (x)},

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T so that Eijk is µk -conegligible; set Ek = i≤j≤k Eijk , so that Ek is µk -conegligible. For i ≤ j ∈ I, choose compact sets Kij ⊆ Ej such that µj Kij ≥ 1 − ²ij and fij ¹Kij is continuous. Now we seem to need a three-stage construction, as follows: T for j ∈ I, set Kj = i≤j Kij ; T −1 for j ∈ I, set Kj∗ = Kj ∩ i≤j fij [Ki ]; Q ∗ finally, set K = X ∩ i∈I Ki . Let us trace the properties of these sets stage by stage. (b) For each j ∈ I, Kj ⊆ Kjj ⊆ Ej is compact and P P µj (Xj \ Kj ) ≤ i≤j µj (Xj \ Kij ) ≤ i≤j ²ij = ²j , so that µj Kj ≥ 1 − ²j . Note that fik agrees with fij fjk on Kk whenever i ≤ j ≤ k, and that fij ¹Kj is continuous whenever i ≤ j. (c) Every Kj∗ is compact, and if i ≤ j ≤ k then fik agrees with fij fjk on Kk∗ , while fij ¹Kj∗ is always continuous. Also X −1 µj (Xj \ Kj∗ ) ≤ µj (Xj \ Kj ) + µj (Xj \ fij [Ki ]) ≤ ²j +

X

i≤j

²i ≤ ²,

i≤j

so µj Kj∗ ≥ 1 − ², for every j ∈ I. −1 The point of moving from Kj to Kj∗ is that fjk [Kk∗ ] ⊆ Kj∗ whenever j ≤ k in I. P P Kk∗ ⊆ fjk [Kj ], so ∗ fjk [Kk ] ⊆ Kj . If i ≤ j, then −1 −1 −1 Kk∗ = Kk∗ ∩ fik [Ki ] = Kk∗ ∩ fjk [fij [Ki ]] −1 because fij fjk agrees with fik on Kk∗ . So fjk [Kk∗ ] ⊆ fij [Ki ]. As i is arbitrary, fjk [Kk∗ ] ⊆ Kj∗ . Q Q Again because fik agrees with fij fjk on Kk∗ , we have fik [Kk∗ ] = fij [fjk [Kk∗ ]] ⊆ fij [Kj∗ ] whenever i ≤ j ≤ k. And because fij ¹Kj∗ is always continuous, all the sets fij [Kj∗ ] are compact.

(d)(i) K is compact. P P Q

K = {x : x ∈

Q i∈I

Ki∗ , fij x(j) = x(i) whenever i ≤ j ∈ I}

is closed in i∈I Ki∗ because fij ¹Kj∗ is always continuous (and every Xi is Hausdorff). Since compact, so is K. Q Q

Q i∈I

Ki∗ is

(ii) µi gi [K] ≥ 1−² for every i ∈ I. P P By (c), fik [Kk∗ ] ⊆ fij [Kj∗ ] whenever i ≤ j ≤ k. So {fij [Kj∗ ] : j ≥ i} is a downwards-directed family of compact sets; write L for their intersection. Since −1 µi fij [Kj∗ ] = µj fij [fij [Kj∗ ]] ≥ µj Kj∗ ≥ 1 − ²

for every j ≥ i, µi L ≥ 1 − ² (414C). If z ∈ L, then for every k ≥ i the set Q Fk = {x : x ∈ j∈I Kj∗ , x(k) = z, fjk x(k) = x(j) whenever j ≤ k} Q is a non-empty closed set in j∈I Kj∗ , while Fk ⊆ Fj when j ≤ k; so that {Fk : k ≥ i} is a downwards-directed T family of non-empty closed sets in a compact space, and has non-empty intersection. But if x ∈ k≥i Fk , then x ∈ K and x(k) = z, so z ∈ gi [K]. Thus gi [K] ⊇ L and µi gi [K] ≥ 1 − ². Q Q (e) As ² is arbitrary, the final condition of 418M is satisfied. But now 418M tells us that (β) is true. 418Q Corollary Let h(Xn , Tn , Σn , µn )in∈N be a sequence of Radon probability spaces, and suppose we are given an inverse-measure-preserving almost continuous function fn : Xn+1 → Xn for each n. Set Q X = {x : x ∈ n∈N Xn , fn (x(n + 1)) = x(n) for every n ∈ N}. Then there is a unique Radon probability measure µ on X such that all the coordinate maps x 7→ x(n) : X → Xn are inverse-measure-preserving.

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proof (a) For i ≤ j ∈ N, define fij : Xj → Xi by writing fii (x) = x for every x ∈ Xi , fi,j+1 = fij fj for every j ≥ i. It is easy to check that fij fjk = fik whenever i ≤ j ≤ k, and that every fij is inverse-measure-preserving and almost continuous (using 418Dc). So we are exactly in the situation of 418P, and we know that there is a Radon probability measure on X for which every gi is inverse-measure-preserving. (b) To see that µ is unique, observe that if K ⊆ X is compact and x ∈ X \ K then there is some finite set I ⊆ N such that {y : y¹I = x¹I} is disjoint from K. But as gn = fn gn+1 for every n, this means that {y : gn (y) = gn (x)} is disjoint from K for all n large enough. As x is arbitrary, we see that hgn−1 [gn [K]]in∈N is a non-increasing sequence with intersection K, so that µK = limn→∞ µgn−1 [gn [K]] = limn→∞ µn gn [K]. Thus the values of µ on compact sets are determined by the construction. By 416E(b-ii), µ is uniquely defined. 418R I turn now to a special kind of measurable function, corresponding to a new view of product spaces. Theorem Let X be a set, Σ a σ-algebra of subsets of X, and (Y, T, ν) a σ-finite measure space. Give L0 (ν) b b the topology of convergence in measure (§245). Write L0 (Σ⊗T) for the space of Σ⊗T-measurable functions b is the σ-algebra of subsets of X × Y generated by {E × F : E ∈ Σ, F ∈ T}. h : X × Y → R, where Σ⊗T Then for a function f : X → L0 (ν) the following are equiveridical: (i) f [X] is separable and f is measurable; b (ii) there is an h ∈ L0 (Σ⊗T) such that f (x) = h•x for every x ∈ X, where hx (y) = h(x, y) for x ∈ X, y ∈Y. proof Let hYn in∈N be a non-decreasing sequence of subsets of Y of finite measure covering Y . (a)(i)⇒(ii) pseudometric on L0 (ν) defined by saying that R For each n ∈ N, let ρn be the continuous 0 • • ρn (g1 , g2 ) = Yn min(1, |g1 − g2 |)dν for g1 , g2 ∈ L (T), writing L0 (T) for the space of T-measurable realvalued functions on Y (245A). Then {ρn : n ∈ N} defines the topology of L0 (ν) (see the proof of 245Eb). Because f [X] is separable, there is a sequence hvk ik∈N in L0 (ν) such that f [X] ⊆ {vk : k ∈ N}. For each k, choose gk ∈ L0 (T) such that gk• = vk . For n, k ∈ N set Enk = {x : x ∈ X, ρn (f (x), vk ) < 2−n }, S Hnk = Enk \ i
Yn

≤ ρn (gk•m+1 , f (x)) + ρn (f (x), gk•m ) ≤ ρm+1 (gk•m+1 , f (x)) + ρm (f (x), gk•m ) ≤ 3 · 2−m−1 . But this means that

P∞

R

m=0 Yn

(m+1)

min(1, |hx

(m)

(m)

−hx |) is finite, so that hhx im∈N must be convergent almost (m)

everywhere on Yn . As this is true for every n, hhx im∈N is convergent a.e. on Y . Moreover, (m)

limm→∞ (hx )• = limm→∞ gk•m = f (x) in L0 (ν).

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Since this is true for every x, W = {(x, y) : hh(m) (x, y)im∈N converges in R} b because every h(m) is Σ⊗T-measurable b has conegligible vertical sections, while of course W ∈ Σ⊗T (418C). (m) b If we set h(x, y) = limm→∞ h (x, y) for (x, y) ∈ W , 0 for other (x, y) ∈ X × Y , then h ∈ L0 (Σ⊗T), while (by 245Ca) (m)

h•x = limm→∞ (hx )• = f (x) in L0 (ν) for every x ∈ X. So we have a suitable h. b (ii)⇒(i) Let Φ be the set of those h ∈ L0 (Σ⊗T) such that (i) is satisfied; that is, x 7→ h•x is measurable, and {h•x : x ∈ X} is separable. Then ˜ belong to Φ, set A = {h• : x ∈ X}, A˜ = {h ˜ • : x ∈ X}. Then α) Φ is closed under addition. P (α P If h, h x x ˜• ) : both A and A˜ are separable metrizable spaces, so A × A˜ is separable and metrizable and x 7→ (h•x , h x 0 ˜ X → A × A is measurable (418Bb). But addition on L (ν) is continuous (245D), so ˜ • = (h + h) ˜ • x 7→ h•x + h x x is measurable (418Ac), and ˜ • : x ∈ X} ⊆ {u + u ˜ ˜ : u ∈ A, u ˜ ∈ A} {(h + h) x ˜ ∈ Φ. Q is separable (4A2Be). Thus h + h Q β ) Φ is closed under scalar multiplication, just because u 7→ αu : L0 (ν) → L0 (ν) is always continuous. (β (γγ ) If hh(n) in∈N is a sequence in Φ and h(x, y) = limn→∞ h(n) (x, y) for all x ∈ X, y ∈ Y , then h ∈ Φ. S (n) P P Setting An = {(hx )• : x ∈ X} for each n, then A = {h•x : x ∈ X} is included in n∈N An , which is (n) separable (4A2B(e-i)), so A is separable (4A2P(a-iv)); moreover, h•x = limn→∞ (hx )• for every x ∈ X, so • Q x 7→ hx is measurable, by 418Ba. Q b χW ∈ Φ}, then W \ W 0 ∈ W whenever (δδ ) What this means is that if weSset W = {W : W ∈ Σ⊗T, 0 0 W , W ∈ W and W ⊆ W , and that n∈N Wn ∈ W whenever hWn in∈N is a non-decreasing sequence in W. Also, it is easy to see that E × F ∈ W whenever E ∈ Σ and F ∈ T. By the Monotone Class Theorem b (136B), W includes the σ-algebraP generated by {E × F : E ∈ Σ, F ∈ T}, that is, is equal to Σ⊗T. It follows n b at once, from (α) and (β), that i=0 αi χWi ∈ Φ whenever W0 , . . . , Wn ∈ Σ⊗T and α0 , . . . , αn ∈ R, and b hence (using (γ)) that L0 (Σ⊗T) ⊆ Φ, which is what we had to prove. 418S Corollary Let (X, Σ, µ) and (Y, T, ν) be σ-finite measure spaces with c.l.d. product (X × Y, Λ, λ). Give L0 (ν) the topology of convergence in measure. Write L0 (λ) for the space of Λ-measurable real-valued functions defined λ-a.e. on X × Y , as in §241. (a) If h ∈ L0 (λ), set hx (y) = h(x, y) whenever this is defined. Then {x : f (x) = h•x is defined in L0 (ν)} is µ-conegligible, and includes a conegligible set X0 such that f : X0 → L0 (ν) is measurable and f [X0 ] is separable. (b) If f : X → L0 (ν) is measurable and there is a conegligible set X0 ⊆ X such that f [X0 ] is separable, then there is an h ∈ L0 (λ) such that f (x) = h•x for almost every x ∈ X. b (251K). So there is a conegligible proof (a) The point is that λ is just the completion of its restriction to Σ⊗T ˜ b b set W ∈ Σ⊗T such that h¹ W is Σ⊗T-measurable (212Fa). Setting h(x, y) = h(x, y) for (x, y) ∈ W , 0 ˜ • for every x ∈ X, we see from 418R that f˜ is measurable and that f˜[X] is otherwise, and setting f˜(x) = h x separable. But 252D tells us that X0 = {x : ((X × Y ) \ W )[{x}] is negligible} ˜ x ν-a.e., so that f (x) is defined and equal to f˜(x). This proves is conegligible; and if x ∈ X0 then hx = h the result.

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b (b) f ¹X0 satisfies 418R(i). So, setting f1 (x) = f (x) for x ∈ X0 , 0 otherwise, there is some h ∈ L0 (Σ⊗T) such that f1 (x) = h•x for every x, so that f (x) = h•x for almost every x, and (ii) is true. 418T Corollary (Mauldin & Stone 81) Let (Y, T, ν) be a σ-finite measure space, and (B, ν¯) its measure algebra, with its measure-algebra topology (§323). (a) Let X be a set, Σ a σ-algebra of subsets of X, and f : X → B a function. Then the following are equiveridical: (i) f [X] is separable and f is measurable; b such that f (x) = W [{x}]• for every x ∈ X. (ii) there is a W ∈ Σ⊗T (b) Let (X, Σ, µ) be a σ-finite measure space and Λ the domain of the c.l.d. product measure λ on X × Y . (i) Suppose that ν is complete. If W ∈ Λ, then {x : f (x) = W [{x}]• is defined in B} is µ-conegligible, and includes a conegligible set X0 such that f : X0 → B is measurable and f [X0 ] is separable. (ii) If f : X → B is measurable and there is a conegligible set X0 ⊆ X such that f [X0 ] is separable, b such that f (x) = W [{x}]• for almost every x ∈ X. then there is a W ∈ Σ⊗T proof Everything follows directly from 418R and 418S if we observe that B is homeomorphically embedded in L0 (ν) by the function F • 7→ (χF )• for F ∈ T (323Xf, 367R). We do need to check, for (i)⇒(ii) of part b b such that (a), that if h ∈ L0 (Σ⊗T) and h•x is always of the form (χF )• , then there is some W ∈ Σ⊗T • • hx = (χW [{x}]) for every x; but of course this is true if we just take W = {(x, y) : h(x, y) = 1}. Now (b-ii) follows from (a) just as 418Sb followed from 418R. 418X Basic exercises > (a) Let (X, Σ, µ) be a measure space, Y a set and h : X → Y a function; give Y the image measure µh−1 . Show that for any function g from Y to a topological space Z, g is measurable iff gh : X → Z is measurable. > (b) Let X be a set, Σ a σ-algebra of subsets of X, hYn in∈N a sequence ofQ topological spaces with product Y , and f : X → Y a function. Show that f is measurable iff ψn f : X → i≤n Yi is measurable for every n ∈ N, where ψn (y) = (y(0), . . . , y(n)) for y ∈ Y , n ∈ N. (c) Let (X, Σ, µ) be a semi-finite measure space, (Y, S) a metrizable space, and hfn in∈N a sequence of measurable functions from X to Y such that hfn (x)in∈N is convergent for almost every x ∈ X. Show that µ is inner regular with respect to {E : hfn ¹Ein∈N is uniformly convergent}. (Cf. 412Xr.) > (d) Set Y = [0, 1][0,1] , with the product topology. For x ∈ [0, 1], n ∈ N define fn (x) ∈ Y by saying that fn (x)(t) = max(0, 1 − 2n |x − t|) for t ∈ [0, 1]. Check that (i) each fn is continuous, therefore measurable; (ii) f (x) = limn→∞ fn (x) is defined in Y for every x ∈ [0, 1]; (iii) for each t ∈ [0, 1], the coordinate functional x 7→ f (x)(t) is continuous except at t, and in particular is almost continuous and measurable; (iv) f ¹F is not continuous for any infinite closed set F ⊆ [0, 1], and in particular f is not almost continuous; (v) every subset of [0, 1] is of the form f −1 [H] for some open set H ⊆ Y ; (vi) f is not measurable; (vii) the image measure µf −1 , where µ is Lebesgue measure on [0, 1], is neither a topological measure nor tight. (e) Let (X, T, Σ, µ) be a quasi-Radon measure space, Y a topological space, and f : X → Y a function. Suppose that for every x ∈ X there is an open set G containing x such that f ¹G is almost continuous with respect to the subspace measure on G. Show that f is almost continuous. (f ) For i = 1, 2 let (Xi , Ti , Σi , µi ) and (Yi , Si , Ti , νi ) be quasi-Radon measure spaces, and fi : Xi → Yi an almost continuous inverse-measure-preserving function. Show that (x1 , x2 ) 7→ (f1 (x1 ), f2 (x2 )) is inversemeasure-preserving for the quasi-Radon product measures. (g) Let h(Xi , Ti , Σi , µi )ii∈I and h(Yi , Ti , Σi , νi )ii∈I be two families of topological Q spaces with τ -additive Q Borel probability measures, and let µ, ν be the τ -additive product measures on X = i∈I Xi , Y = i∈I Yi . Suppose that every νi is strictly positive. Show that if fi : Xi → Yi is almost continuous and inversemeasure-preserving for each i, then x 7→ hfi (x(i))ii∈I : X → Y is inverse-measure-preserving, but need not be almost continuous.

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(h) Let (X, T, Σ, µ) and (Y, S, T, ν) be quasi-Radon measure spaces, (Z, U) a topological space and f : X × Y → Z a function which is almost continuous with respect to the quasi-Radon product measure on X × Y . Suppose that ν is σ-finite. Show that y 7→ f (x, y) is almost continuous for almost every x ∈ X. (i) Let (X, T, Σ, µ) be an effectively locally finite τ -additive topological measure space, Y a topological space and f : X → Y an almost continuous function. (i) Show that the image measure µf −1 is τ -additive. (ii) Show that if µ is a totally finite quasi-Radon measure and the topology on Y is regular, then µf −1 is quasi-Radon. (j) Let (X, T, Σ, µ) be a topological measure space and U a linear topological space. Show that if f : X → U and g : X → U are almost continuous, then f + g : X → U is almost continuous. (k) Let (X, T, Σ, µ) and (Y, S, T, ν) be topological measure spaces, and (Z, U) a topological space; let f : X → Y be almost continuous and inverse-measure-preserving, and g : Y → Z almost continuous. Show that if either µ is a Radon measure and ν is locally finite or µ is τ -additive and effectively locally finite and ν is effectively locally finite, then gf : X → Z is almost continuous. (Hint: show that if µE > 0 there is a set F such that νF < ∞ and µ(E ∩ f −1 [F ]) > 0.) (l) Let (X, Σ, µ) be a complete strictly localizable measure space, φ : Σ → Σ a lower density such that φX = X, and T the associated density topology on X (414P). Let f : X → R be a function. Show that the following are equiveridical: (i) f is measurable; (ii) f is almost continuous; (iii) f is continuous at almost every point; (iv) there is a conegligible set H ⊆ X such that f ¹H is continuous. (Cf. 414Xk.) (m) Let (X, Σ, µ) be a complete strictly localizable measure space, φ : Σ → Σ a lifting, and S the lifting topology on X (414Q). Let f : X → R be a function. Show that the following are equiveridical: (i) f is measurable; (ii) f is almost continuous; (iii) there is a conegligible set H ⊆ X such that f ¹H is continuous. (Cf. 414Xr.) (n) Let (X, T, Σ, µ) be a quasi-Radon measure space, (Y, S) a regular topological space and f : X → Y an almost continuous function. Show S that there is a quasi-Radon measure ν on Y such that f is inversemeasure-preserving for µ and ν iff {f −1 [H] : H ⊆ Y is open, µf −1 [H] < ∞} is conegligible in X. (o) Let X and Y be Hausdorff spaces and f : X → Y a continuous injective function. Show that if µ1 and µ2 are distinct totally finite Radon measures on X then µ1 f −1 6= µ2 f −1 . (p) Let (X, T, Σ, µ) be a Radon measure space, (Y, S) and (Z, U) Hausdorff spaces, f : X → Y an almost continuous function such that ν = µf −1 is locally finite, and g : Y → Z a function. Show that g is almost continuous with respect to ν iff gf is almost continuous with respect to µ. (q) Let (X, T, Σ, µ) and (Y, S, T, ν) be topological probability spaces, and Rf : X → Y a measurable R function such that µf −1 [H] ≥ νH for every H ∈ S. Show that (i) gf dµ = g dν for every g ∈ Cb (Y ) (ii) µf −1 [F ] = νF for every Baire set F ⊆ Y (iii) if µ is a Radon measure and f is almost continuous, then µf −1 [F ] = νF for every Borel set F ⊆ Y , so that if in addition ν is complete and inner regular with respect to the Borel sets then it is a Radon measure. (r) Let (X, T, Σ, µ) be a totally finite topological measure space in which the topology T is normal and µ is inner regular with respect to the closed sets. Show that if f : X → R is a measurable function and ² > 0 there is a continuous g : X → R such that µ{x : g(x) 6= f (x)} ≤ ². (s) Let X and Y be Hausdorff spaces, ν a totally finite Radon measure on Y , and f : X → Y an injective continuous function. Show that the following are equiveridical: (i) there is a Radon measure µ on X such that f is inverse-measure-preserving; (ii) f [X] is conegligible and f −1 : f [X] → X is almost continuous. (t) Let (X, T, Σ, µ) and (Y, S, T, ν) be Radon measure spaces and f : X → Y an almost continuous inverse-measure-preserving function. Show that (i) µ∗ A ≤ ν∗ f [A] for every A ⊆ X (ii) ν is precisely the image measure µf −1 .

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(u) In 418M, show that all the fij must be almost continuous. Show that if i ≤ j ≤ k then fij fjk = fik almost everywhere on Xk . > (v) Let I be the family of finite subsets of [0, 1], and let (XI , TI , ΣI , µI ) be [0, 1] \ I with its subspace topology and measure for each I ∈ I. For I ⊆ J ∈ I, y ∈ XJ set fIJ (y) = y. Show that these XI , fIJ satisfy nearly all the hypotheses of 418O, but that there are no X, gI which satisfy the hypotheses of 418M. (w) Let T be any set, and X the set of total orders on T . (i) Regarding each member of X as a subset of T × T , show that X is a closed subset of P(T × T ). (ii) Show that there is a unique Radon measure µ on X such that Pr(t1 ≤ t2 ≤ . . . ≤ tn ) =

1 n!

for all distinct t1 , . . . , tn ∈ T . (Hint: for I ∈ [T ]<ω , let XI be the

set of total orderings on I with the uniform probability measure giving the same measure to each singleton; show that the natural map from XI to XJ is inverse-measure-preserving whenever J ⊆ I.) (x) In 418S, suppose that f1 : X → L0 (ν) and f2 : X → L0 (ν) correspond to h1 , h2 ∈ L0 (λ). Show that f1 (x) ≤ f2 (x) µ-a.e.(x) iff h1 ≤ h2 λ-a.e. Hence show that (if we assign appropriate algebraic operations to the space of functions from X to L0 (ν)) we have an f -algebra isomorphism between L0 (λ) and the space of equivalence classes of measurable functions from X to L0 (ν) with separable ranges. 418Y Further exercises (a) Let X be a set, Σ a σ-algebra of subsets of X, Y a topological space and f : X → Y a function. Set T = {F : F ⊆ Y, f −1 [F ] ∈ Σ}. Suppose that Y is hereditarily Lindelöf and its topology is generated by some subset of T. Show that f is measurable. (b) Let (X, Σ, µ) be a measure space, Y and Z topological spaces and f : X → Y , g : X → Z measurable functions. Show that if Z has a countable network consisting of Borel sets (e.g., Z is second-countable, or Z is regular and has a countable network), then x 7→ (f (x), g(x)) : X → Y × Z is measurable. (c) Let X be a set, Σ a σ-algebra of subsets of X, and hYi ii∈I a countable family of topological spaces with product Y . Suppose that every Yi has a countable network, and that f : X → Y is a function such that πi f is measurable for every i ∈ I, writing πi (y) = y(i). Show that f is measurable. (d) Find strictly localizable Hausdorff topological measure spaces (X, T, Σ, µ), (Y, S, T, ν) and (Z, U, Λ, λ) and almost continuous inverse-measure-preserving functions f : X → Y , g : Y → Z such that gf is not almost continuous. (e) Let (X, Σ, µ) be a σ-finite measure space and T a topology on X such that µ is effectively locally finite and τ -additive. Let Y be a topological space and f : X → Y an almost continuous function. Show that there is a conegligible subset X0 of X such that f [X0 ] is ccc. (f ) Show that if µ is Lebesgue measure on R, T is the usual topology on R and S is the right-facing Sorgenfrey topology, then the identity map from (R, T, µ) to (R, S) is measurable, but not almost continuous, and the image measure is not a Radon measure. (g) Let (X, Σ, µ) be a semi-finite measure space and T a topology on X such that µ is inner regular with respect to the closed sets. Suppose that Y is a topological space with a countable network consisting of Borel sets, and that f : X → Y is measurable. Show that f is almost continuous. (h) Find a topological probability space (X, T, Σ, µ) in which µ is inner regular with respect to the closed sets, a topological space Y with a countable network and a measurable function f : X → Y which is not almost continuous. (i) Let (X, Σ, µ) be a semi-finite measure space and T a topology on X such that µ is inner regular with respect to the closed sets. Let A ⊆ L∞ (µ) be a norm-compact set. Show that there is a set B of bounded real-valued measurable functions on X such that (i) A = {f • : f ∈ B} (ii) B is norm-compact in `∞ (X) (iii) µ is inner regular with respect to {E : f ¹E is continuous for every f ∈ B}.

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(j) Let µ be Lebesgue measure on [0, 1]. For t ∈ [0, 1] set ut = χ[0, t]• ∈ L0 (µ). Show that A = {ut : t ∈ [0, 1]} is norm-compact in Lp (µ) for every p ∈ [1, ∞[ and also compact for the topology of convergence in measure on L0 (µ). Show that if B is a set of measurable functions such that A = {f • : f ∈ B} then µ is not inner regular with respect to {E : f ¹E is continuous for every f ∈ B}. (k) Suppose that (I, ≤), h(Xi , Ti , Σi , µi )ii∈I and hfij ii≤j∈I are such that (α) (I, ≤) is a non-empty upwards-directed partially ordered set (β) every (Xi , Ti , Σi , µi ) is a completely regular Hausdorff quasiRadon probability space (γ) fij : Xj → Xi is a continuous inverse-measure-preserving function whenever i ≤ j in I (δ) fij fjk = fik whenever i ≤ j ≤ k in I. Let Xi0 be the support of µi for each i; show that ˇ fij [Xj0 ] is a dense subset of Xi0 whenever i ≤ j. Let Zi be the Stone-Cech compactification of Xi0 and let ˜i be the Radon probability measure on f˜ij : Zj → Zi be the continuous extension of fij ¹Xj0 for i ≤ j; let µ Zi corresponding to µi ¹ PXi0 (416V). Show that Zi , f˜ij satisfy the conditions of 418O, so that we have a projective limit Z, hgi ii∈I , µ as in 418M. (l) Suppose that (I, ≤), h(Xi , Σi , µi )ii∈I , X, hfi ii∈I and hfij ii≤j∈I are such that (α) (I, ≤) is a nonempty upwards-directed partially ordered set (β) every (Xi , Σi , µi ) is a probability space (γ) fij : Xj → Xi is inverse-measure-preserving whenever i ≤ j in I (δ) fij fjk = fik whenever i ≤ j ≤ k (²) fi : X → Xi is a function for every i ∈ I (ζ) fi = fij fj whenever i ≤ j (η) whenever hin in∈N , hxn in∈N are such that hin in∈N is a non-decreasing sequence in I, xn ∈ Xin for every n ∈ N and fin in+1 (xn+1 ) = xn for every n ∈ N, then there is an x ∈ X such that fin (x) = xn for every n. Show that there is a probability measure on X such that every fi is inverse-measure-preserving. 418 Notes and comments The message of this section is that measurable functions are dangerous, but that almost continuous functions behave themselves. There are two fundamental problems with measurable functions: a function x 7→ (f (x), g(x)) may not be measurable when the components f and g are measurable (419Xg), and an image measure under a measurable function can lose tightness, even when both domain and codomain are Radon measure spaces and the function is inverse-measure-preserving (419Xh). (This is the ‘image measure catastrophe’ mentioned in 235J and the notes to §343.) Consequently, as long as we are dealing with measurable functions, we often have to impose strong conditions on the range spaces – commonly, we have to restrict ourselves to separable metrizable spaces (418B, 418C), or something similar, which indeed often means that a measurable function is actually almost continuous (418J, 433E). Indeed, for functions taking values in metrizable spaces, ‘almost continuity’ is very close to ‘measurable with essentially separable range’ (418G, 418J). The condition ‘separable and metrizable’ is a little stronger than is strictly necessary (418Yb, 418Yc, 418Yg), but covers the principal applications other than 433E. If we keep the ‘metrizable’ we can very substantially relax the ‘separable’ (438E, 438F), and it is in fact the case that a measurable function from a Radon measure space to any metrizable space is almost continuous (451S). These extensions apply equally to the results in 418R-418T (438Xg-438Xh, 451Xn). But both take us deeper into set theory than seems appropriate at the moment. For almost continuous functions, the two problems mentioned above do not arise (418Dd, 418I, 418Xt). Indeed we rather expect almost continuous functions to behave as if they were continuous. But we still have to be careful. The limit of a sequence of almost continuous functions need not be almost continuous (418Xd), unless the codomain is metrizable (418F); and if we have a function f from a topological measure space to an uncountable product of topological spaces, it can happen that every coordinate of f is an almost continuous function while f is not (418Xd again). But for many purposes, intuitions gained from the study of measurable functions between Euclidean spaces can be transferred to general almost continuous functions. Theorems 418L and 418M are of a quite different kind, but seem to belong here as well as anywhere. Even in the simplest application of 418L (when Y = [0, 1] with Lebesgue measure, and X ⊆ [0, 1]2 is a closed set meeting every vertical line) it is not immediately obvious that there will be a measure with the right projection onto the horizontal axis, though there are at least two proofs which are easier than the general case treated in 413N-413O-418L. As I explain in 418N, the really interesting question concerning 418M is when, given the projective system h(Xi , Ti , Σi , µi )ii∈I , hfij ii≤j∈I , we can expect to find X and hgi ii∈I satisfying the rest of the hypotheses, and once past the elementary results 418O-418Q this can be hard to determine. I describe a method in

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418Yk which can sometimes be used, but (like the trick in 418Nf) it is too easy and too abstract to be often illuminating. See 454G below for something rather deeper. The results of 418R-418T stand somewhat aside from anything else considered in this chapter, but they form part of an important technique. A special case has already been mentioned in 253Yg. I do not discuss vector-valued measurable functions in this book, except incidentally, but 418R is one of the fundamental results on their representation; it means, for instance, that if V is any of the Banach function spaces of Chapter 36 we can expect to represent Bochner integrable V -valued functions (253Yf) in terms of functions on product spaces, because V will be continuously embedded in an L0 space (367P). The measure-algebra version in 418T will be very useful in Volume 5 when establishing relationships between properties of measure spaces and corresponding properties of measure algebras.

419 Examples In §216, I went much of the way to describing examples of spaces with all the possible combinations of the properties considered in Chapter 21. When we come to topological measure spaces, the number of properties involved makes it unreasonable to seek any such comprehensive list. I therefore content myself with seven examples to indicate some of the boundaries of the theory developed here. The first example (419A) is supposed to show that the hypothesis ‘effectively locally finite’ which appears in so many of the theorems of this chapter cannot as a rule be replaced by ‘locally finite’. The next two (419C-419D) address technical questions concerning the definition of ‘Radon measure’, and show how small variations in the definition can lead to very different kinds of measure space. The fourth example (419E) shows that the τ -additive product measures of §417 are indeed new constructions. 419H is there to show that extension theorems of the types proved in §415 and §417 cannot be taken for granted. The classical example 419K exhibits one of the obstacles to generalizations of Prokhorov’s theorem (418M, 418Q). Finally, I return to the split interval (419L) to describe its standard topology and its relation to the measure introduced in 343J. 419A Example There is a locally compact Hausdorff space X with a complete, σ-finite, locally finite, τ -additive topological measure µ, inner regular with respect to the closed sets, which has a closed subset Y , of measure 1, such that the subspace measure µY on Y is not τ -additive. In particular, µ is not effectively locally finite. proof (a) Let Q be a countably P 1 infinite set, not containing any ordinal. Fix an enumeration hqn in∈N of Q, and for A ⊆ Q set νA = { n+1 : qn ∈ A}. Let I be the ideal {A : A ⊆ Q, νA < ∞}. For any sets I, J, say that I ⊆∗ J if I \ J is finite; then ⊆∗ is a reflexive transitive relation. Let κ be the smallest cardinal of any family K ⊆ I for which there is no I ∈ I such that K ⊆∗ I for every K ∈ K. Then κ is uncountable. P P (i) Of course κ is infinite. (ii) If hIn in∈N is any S sequence in I, then for each n ∈ N we can find a finite In0 ⊆ In such that ν(In \ In0 ) ≤ 2−n ; setting I = n∈N In \ In0 , we have νI ≤ 2 < ∞, while In ⊆∗ I for every n. Thus κ > ω. Q Q (b) There is a family hIξ iξ<κ in I such that (i) Iη ⊆∗ Iξ whenever η ≤ ξ < κ (ii) there is no I ∈ I such that Iξ ⊆∗ I for every ξ < κ. P P Take a family hKξ iξ<κ in I such that there is no I ∈ I such that Kξ ⊆∗ I for every ξ < κ. Choose hIξ iξ<κ in I inductively in such a way that Kξ ⊆∗ Iξ ,

Iη ⊆∗ Iξ for every η < ξ.

(This can be done because {Kξ } ∪ {Iη : η < ξ} will always be a subset of I of cardinal less than κ.) If now Iξ ⊆∗ I for every ξ < κ, then Kξ ⊆∗ I for every ξ < κ, so I ∈ / I. Q Q Hence, or otherwise, we see that κ is regular. P P If A ⊆ κ and #(A) < κ, then there is an I ∈ I such that Iζ ⊆∗ I for every ζ ∈ A; now there must be a ξ < κ such that Iξ 6⊆∗ I, in which case ζ < ξ for every ζ ∈ A, and A is not cofinal with κ. Q Q (c) Set X = Q ∪ κ. (This is where it is helpful to have arranged at the start that no ordinal belongs to Q, so that Q ∩ κ = ∅.) Let T be the family of sets G ⊆ X such that G ∩ κ is open for the order topology of κ, for every ξ ∈ G ∩ κ \ {0} there is an η < κ such that Iξ \ Iη ⊆∗ G,

419A

Examples

127

if 0 ∈ G then I0 ⊆∗ G. S (i) This is a Hausdorff topology on X. P P (α) It is easy to check that X ∈ T, ∅ ∈ T and G ∈ T for every G ⊆ T. (β) Suppose that G, H ∈ T. Then (G ∩ H) ∩ κ = (G ∩ κ) ∩ (H ∩ κ) is open for the order topology of κ. If ξ ∈ G ∩ H ∩ κ \ {0} there are η, ζ < ξ such that Iξ \ Iη ⊆∗ G and Iξ \ Iζ ⊆∗ H, and now α = max(η, ζ) < ξ,

Iη ∪ Iζ ⊆∗ Iα ,

so Iξ \ Iα ⊆∗ (Iξ \ Iη ) ∩ (Iξ \ Iζ ) ⊆∗ G ∩ H. Finally, if 0 ∈ G ∩ H then I0 ⊆∗ G ∩ H. So G ∩ H ∈ T. Thus T is a topology on X. (γ) For any ξ < κ, the set Eξ = (ξ + 1) ∪ Iξ is open-and-closed for T; for any q ∈ Q, {q} is open-and-closed. Since these sets separate the points of X, T is Hausdorff. Q Q (ii) The sets Eξ of the last paragraph are all compact for T. P P Let F be an ultrafilter on X containing Eξ . (α) If a finite set K belongs to F, then F must contain {x} for some x ∈ K, and converges to x. So suppose henceforth that F contains no finite set. (β) If E0 ∈ F, then for any open set G containing 0, E0 \ G is finite, so does not belong to F, and G ∈ F; as G is arbitrary, F → 0. (γ) If E0 ∈ / F, let η ≤ ξ be the least ordinal such that Eη ∈ F. If G is an open set containing η, there are ζ 0 , ζ 00 < η such that Iη \ Iζ 0 ⊆∗ G, ]ζ 00 , η] ⊆ G; so that Eη \ Eζ ⊆∗ G, where ζ = max(ζ 0 , ζ 00 ) < η. Now Eη ∈ F, Eζ ∈ / F and (Eη \ Eζ ) \ G ∈ / F, so that G ∈ F . As G is arbitrary, F → η. (δ) As F is arbitrary, Eξ is compact. Q Q (iii) It follows that T is locally compact. P P For q ∈ Q, {q} is a compact open set containing q; for ξ < κ, Eξ is a compact open set containing ξ. Q Q (iv) The definition of T makes it clear that Q ∈ T, that is, that κ is a closed subset of X. We need also to check that the subspace topology Tκ on κ induced by T is just the order topology of κ. P P (α) By the definition of T, G ∩ κ is open for the order topology of κ for every G ∈ T. (β) For any ξ < κ, Eξ is open-and-closed for T so ξ + 1 = Eξ ∩ κ is open-and-closed for Tκ . But this means that all sets of the forms S [0, ξ[ = η<ξ η + 1 and ]ξ, κ[ = κ \ (ξ + 1) belong to Tκ ; as these generate the order topology, every open set for the order topology belongs to Tκ , and the two topologies are equal. Q Q (d) Now let F be the filter on X generated by the cofinal closed sets in κ. Because the intersection of any sequence of closed cofinal sets in κ is another (4A1Bc), the intersection of any sequence in F belongs to F. So Σ = F ∪ {X \ F : F ∈ F} is a σ-algebra of subsets of X, and we have a measure µ1 : Σ → {0, 1} defined by saying that µ1 F = 1, µ1 (X \ F ) = 0 if F ∈ F. (e) Set µE = ν(E ∩ Q) + µ1 E for E ∈ Σ. Then µ is a measure. Let us work through the properties called for. (i) If µE = 0 and A ⊆ E, then X \ A ⊇ X \ E ∈ F , so A ∈ Σ. Thus µ is complete. (ii) µ(κ) = 1 and µ{q} is finite for every q ∈ Q, so µ is σ-finite. (iii) If G ⊆ X is open, then κ \ G is closed, in the order topology of κ; if it is cofinal with κ, it belongs to F; otherwise, κ ∩ G ∈ F . Thus in either case G ∈ Σ, and µ is a topological measure. (iv) The next thing to note is that µG = ν(G ∩ Q) for every open set G ⊆ X. P P If G ∈ / F this is trivial. If G ∈ F, then κ \ G cannot be cofinal with κ, so there is a ξ < κ such that κ \ ξ ⊆ G. ?? If G ∩ Q ∈ I, then (G ∩ Q) ∪ Iξ ∈ I. There must be a least η < κ such that Iη 6⊆∗ (G ∩ Q) ∪ Iξ ; of course η > ξ, so η ∈ G. There is some ζ < η such that Iη \ Iζ ⊆∗ G; but as Iζ ⊆∗ G ∪ Iξ , by the choice of η, we must also have Iη ⊆∗ G ∪ Iξ , which is impossible. X X Thus G ∩ Q ∈ / I and µG = ν(G ∩ Q) = ∞. Q Q (v) It follows that µ is τ -additive. P P Suppose that G ⊆ T is a non-empty upwards-directed set with union H. Then µH = ν(H ∩ Q) = supG∈G ν(G ∩ Q) = supG∈G µG

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because ν is τ -additive (indeed, is a Radon measure) with respect to the discrete topology on Q. Q Q (vi) µ is inner regular with respect to the closed sets. P P Take E ∈ Σ, γ < µE. Then γ − µ1 (E ∩ κ) < ν(E ∩ Q), so there is a finite I ⊆ E ∩ Q such that νI > γ − µ(E ∩ κ). If µ1 (E ∩ κ) = 0, then I ⊆ E is already a closed set with µI > γ. Otherwise, E ∩ κ ∈ F , so there is a cofinal closed set F ⊆ κ such that F ⊆ E; now F is closed in X (because κ is closed in X and the subspace topology on κ is the order topology), so I ∪ F is closed, and µ(I ∪ F ) > γ. As E and γ are arbitrary, µ is inner regular with respect to the closed sets. Q Q (vii) µ is locally finite. P P For any ξ < κ, Eξ is an open set containing ξ, and µEξ = νIξ is finite. For any q ∈ Q, {q} is an open set containing q, and µ{q} = ν{q} is finite. Q Q (viii) Now consider Y = κ. This is surely a closed set, and µκ = 1. I noted in (b-iv) above that the subspace topology Tκ is just the order topology of κ. But this means that {ξ : ξ < κ} is an upwards-directed family of negligible relatively open sets with union κ, so that the subspace measure µκ = µ1 is not τ -additive. (ix) It follows from 414K that µ cannot be effectively locally finite; but it is also obvious from the work above that κ is a measurable set, of non-zero measure, such that µ(κ ∩ G) = 0 whenever G is an open set of finite measure. 419B Lemma For any non-empty set I, there is a dense Gδ set in [0, 1]I which is negligible for the usual measure on [0, 1]I . proof Fix on some i0 ∈ I, and set π(x) = x(i0 ) for each x ∈ [0, 1]I , so that π is continuous and inversemeasure-preserving for the usual topologies and measures on [0, 1]I and [0, 1]. For each n ∈ N let Gn ⊇ [0, 1] ∩ Q be an open subset [0, 1] with measure at most 2−n , so that π −1 [Gn ] is an open set of measure T of −1 I −n is any non-empty open set, at most 2 , and E = n∈N π [Gn ] is a Gδ set of measure 0. If H ⊆ [0, 1]T its image π[H] is open in [0, 1], so contains some rational number, and meets n∈N Gn ; but this means that H ∩ E 6= ∅, so E is dense. 419C Example (Fremlin 75b) There is a completion regular Radon measure space (X, T, Σ, µ) such that (i) there is an E ∈ Σ such that µ(F 4E) > 0 for every Borel set F ⊆ X, that is, not every element of the measure algebra of µ can be represented by a Borel set; (ii) µ is not outer regular with respect to the Borel sets; (iii) writing ν for the restriction of µ to the Borel σ-algebra of X, ν is a locally finite, effectively locally finite, tight (that is, inner regular with respect to the compact sets) τ -additive topological measure, and there is a set Y ⊆ X such that the subspace measure νY is not semi-finite. proof (a) For each ξ < ω1 set Xξ = [0, 1]ω1 \ξ , and take µξ to be the usual measure on Xξ ; write Σξ for its domain. S Note that µξ is a completion regular Radon measure for the usual topology Tξ of Xξ (416U). Set X = ξ<ω1 Xξ , and let µ be the direct sum measure on X (214K), that is, write Σ = {E : E ⊆ X, E ∩ Xξ ∈ Σξ for every ξ < ω1 }, P µE = ξ<ω1 µξ (E ∩ Xξ ) for every E ∈ Σ. Then µ is a complete locally determined (in fact, strictly localizable) measure on X. Write Σ for its domain. (b) For each η < ω1 let hβξη iξ≤η be a summable family of strictly positive real numbers with βηη = 1 (4A1P). Define gη : X → R by setting gη (x) =

1 x(η) βξη

if x ∈ Xξ where ξ ≤ η,

= 0 if x ∈ Xξ where ξ > η. Now define f : X → ω1 × R

ω1

by setting f (x) = (ξ, hgη (x)iη<ω1 )

if x ∈ Xξ . Note that f is injective. Let T be the topology on X defined by f , that is, the family {f −1 [W ] : W ⊆ ω1 × R ω1 is open}, where ω1 and R ω1 are given their usual topologies (4A2S, 3A3K), and their product

419C

Examples

129

is given its product topology. Because f is injective, T can be identified with the subspace topology on f [X]; it is Hausdorff and completely regular. (c) For ξ, η < ω1 , gη ¹Xξ is continuous for the compact topology Tξ . Consequently f ¹Xξ is continuous, and the subspace topology on Xξ induced by T must be Tξ exactly. It follows that µ is a Radon measure for T. P P (i) We know already that µ is complete and locally determined. (ii) If G ∈ T then G ∩ Xξ ∈ Tξ ⊆ Σξ for every ξ < ω1 , so G ∈ Σ; thus µ is a topological measure. (iii) If E ∈ Σ and µE > 0, there is a ξ < ω1 such that µξ (E ∩ Xξ ) > 0. Because µξ is a Radon measure, there is a Tξ -compact set F ⊆ E ∩ Xξ such that µξ F > 0. Now F is T-compact and µF > 0. As E is arbitrary, µ is tight (using 412B). (iv) If x ∈ X, take that ξ < ω1 such that x ∈ Xξ , and consider G = f −1 [(ξ + 1) × {w : w ∈ Rω1 , w(ξ) < 2}]. Because ξ + 1 is open in ω1 , G ∈ T. Because gξ (x) = x(ξ) ≤ 1, x ∈ G. Now for ζ ≤ ξ, µζ (G ∩ Xζ ) = µζ {x : x ∈ Xζ , gξ (x) < 2} −1 = µζ {x : x ∈ Xζ , βζξ x(ξ) < 2}

= µζ {x : x ∈ Xζ , x(ξ) < 2βζξ } ≤ 2βζξ , so µG =

P ζ≤ξ

µζ (G ∩ Xζ ) ≤ 2

P ζ≤ξ

βζξ < ∞.

As x is arbitrary, µ is locally finite, therefore a Radon measure. Q Q We also find that µ is completion regular. P P If E ⊆ X and µE > 0, then there is a ξ < ω1 such that µ(E ∩ Xξ ) > 0. Because µξ is completion regular, there is a set F ⊆ E ∩ XSξ , a zero set for Tξ , such that µF > 0. Now Xξ is a Gδ set in X (being the intersection of the open sets η<ζ<ξ+1 Xζ for η < ξ, unless ξ = 0, in which case Xξ itself is open), so F is a Gδ set in X (4A2C(a-iv)); being a compact Gδ set in a completely regular space, it is a zero set (4A2F(h-v)). Thus every set of positive measure includes a zero set of positive measure. So µ is inner regular with respect to the zero sets (412B). Q Q (d) The key to the example is the following fact: if G ⊆ X is open, then either there is a cofinal closed set V ⊆ ω1 such that G ∩ Xξ = ∅ for every ξ ∈ V or {ξ : µ(G ∩ Xξ ) 6= 1} is countable. P P Suppose that A = {ξ : G ∩ Xξ 6= ∅} meets every cofinal closed set, that is, is stationary (4A1C). Then B = A ∩ Ω is stationary, where Ω is the set of non-zero countable limit ordinals (4A1Bb, 4A1Cb). Let H ⊆ ω1 × R ω1 be an open set such that G = f −1 [H]. For each ξ ∈ B choose xξ ∈ G ∩ Xξ . Then f (xξ ) ∈ H, so there must be a ζξ < ξ, a finite set Iξ ⊆ ω1 , and a δξ > 0 such that z ∈ H whenever z = (γ, htη iη∈ω1 ) ∈ ω1 × R ω1 , ζξ < γ ≤ ξ and |tη − gη (x)| < δξ for every η ∈ Iξ . Because ξ is a non-zero limit ordinal, ζξ0 = sup({ζξ } ∪ (Iξ ∩ ξ)) < ξ. By the Pressing-Down Lemma (4A1Cc), there is a ζ < ω1 such that C = {ξ : ξ ∈ B, ζξ0 = ζ} is uncountable. ?? Suppose, if possible, that ζ < η < ω1 and µ(G ∩ Xη ) < 1. Then there is a measurable subset F of Xη \ G, determined by coordinates in a countable set J ⊆ ω1 \ η, such that µF = µη F > 0 (254Ob). Let ξ ∈ C be such that η < ξ and J ⊆ ξ, and take any y ∈ F . If we define y 0 ∈ Xη by setting y 0 (γ) = y(γ) for γ ∈ ξ \ η, = xξ (γ) for γ ∈ ω1 \ ξ, then y 0 ∈ F . But also ζξ < η < ξ and ξ \ η ⊆ ξ \ ζξ0 is disjoint from Iξ , so gγ (y 0 ) = gγ (xξ ) for every γ ∈ Iξ , since both are zero if γ ≤ η and otherwise y 0 (γ) = xξ (γ). By the choice of ζξ , Iξ we must have f (y 0 ) ∈ H and y 0 ∈ F ∩ G; which is impossible. X X Thus µ(G ∩ Xη ) = 1 for every η > ζ, as required by the second alternative. Q Q (e) For each ξ < ω1 , let Iξ be the family of negligible meager subsets of Xξ . Then Iξ is a σ-ideal; note that it contains every closed negligible set, because µξ is strictly positive. Set Tξ = Iξ ∪ {Xξ \ F : F ∈ Iξ }, so that Tξ is a σ-algebra of subsets of Xξ , containing every conegligible open set, and µξ F ∈ {0, 1} for every F ∈ Tξ . Set

130


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T = {E : E ∈ Σ, {ξ : E ∩ Xξ ∈ / Tξ } is non-stationary}. Then T is a σ-subalgebra of Σ (because the non-stationary sets form a σ-ideal of subsets of ω1 , 4A1Cb), and contains every open set, by (d); so includes the Borel σ-algebra B of X. If we set S Eξ = {x : x ∈ Xξ , x(ξ) ≤ 21 } for each ξ < ω1 , E = ξ<ω1 Eξ , then E ∈ Σ. But if F ⊆ X is a Borel set, F ∈ T so µ(E4F ) = ∞. This proves the property (i) claimed for the example. S (f ) Next, for each ξ < ω1 , take a negligible dense Gδ set Eξ0 ⊆ Xξ (419B). Set Y = ξ<ω1 Eξ0 , so that µY = 0. If F ⊇ Y is a Borel set, then F ∩ Xξ ⊇ Eξ ∈ / Iξ for every ξ < ω1 , while F ∈ T, so {ξ : µξ (F ∩ Xξ ) = 0} is non-stationary and µF = ∞. Thus µ is not outer regular with respect to the Borel sets. Taking ν = µ¹B, the subspace measure νY is not semi-finite. P P We have just seen thatSνY Y = ν ∗ Y is infinite. If F ∈ B and νF < ∞, then A = {ξ : µξ (F ∩ Xξ ) > 0} is countable, so F0 = ξ∈A Eξ0 and S F1 = F \ ξ∈A Xξ are negligible Borel sets; since F ∩ Y ⊆ F0 ∪ F1 , νY (F ∩ Y ) = 0. But this means that νY takes no values in ]0, ∞[ and is not semi-finite. Q Q Remark X here is not locally compact. But as it is Hausdorff and completely regular, it can be embedded as a subspace of a locally compact Radon measure space (X 0 , T0 , Σ0 , µ0 ) (416T). Now µ0 still has the properties (i)-(iii). 419D Example (Fremlin 75b) There is a complete locally determined τ -additive completion regular topological measure space (X, T, Σ, µ) in which µ is tight and compact sets have finite measure, but µ is not localizable. proof (a) Let I be a set of cardinal greater than c. Set X = [0, 1]I . For i ∈ I, t ∈ [0, 1] set Xit = {x : x ∈ X, x(i) = t}. Give Xit its natural topology Tit and measure µit , with domain Σit , defined from the expression of Xit as [0, 1]I\{i} × {t}, each factor [0, 1] being given its usual topology and Lebesgue measure, and the singleton factor {t} being given its unique (discrete) topology and (atomic) probability measure. By 416U, µit is a completion regular Radon measure. Set T = {G : G ⊆ X, G ∩ Xit ∈ Tit for all i ∈ I, t ∈ [0, 1]}, Σ = {E : E ⊆ X, E ∩ Xit ∈ Σit for all i ∈ I, t ∈ [0, 1]}, P µE = i∈I,t∈[0,1] µit (E ∩ Xit ) for every E ∈ Σ. (Compare 216D.) Then it is easy to check that T is a topology. T is Hausdorff because it is finer(= larger) than the usual topology S on X; because each Tit is the subspace topology induced by S, it is also the subspace topology induced by T. Next, the definition of µ makes it a locally determined measure; it is a tight complete topological measure because every µit is. (b) If K ⊆ X is compact, µK < ∞. P P?? Otherwise, M = {(i, t) : i ∈ I, t ∈ [0, 1], µit (K ∩ Xit ) > 0} must be infinite. Take any sequence h(in , tn )in∈N of distinct elements of M . Choose a sequence hxn in∈N in K inductively, as follows. Given hxm im
419F

Examples

131

S point recall that #(I) > c, so there is some j ∈ I \ ({k} ∪ t∈[0,1] Jt ). Since Xj0 ∩ Xkt is negligible for every R1 t ∈ [0, 1], Xj0 ∩ E must be negligible, and 0 νHt dt = 0, where Ht = {y : y ∈ [0, 1]I\{j,k} , (y, 0, t) ∈ E} and ν is the usual measure on [0, 1]I\{j,k} , identifying X with [0, 1]I\{j,k} × [0, 1] × [0, 1]. But because Ft is determined by coordinates in I \ {j}, we can identify it with Ft0 × [0, 1] × {t} where Ft0 is a ν-conegligible subset of [0, 1]I\{j,k} , and Ft0 ⊆ Ht , so νHt = 1 for every t, which is absurd. X X Thus E has no essential supremum in Σ, and µ cannot be localizable. Q Q (d) I have still to check that µ is completion regular. P P If E ∈ Σ and µE > 0, there are i ∈ I, t ∈ [0, 1] such that µit (E ∩ Xit ) > 0, and an F ⊆ E ∩ Xit , a zero set for the subspace topology of Xit , such that µit F > 0. But now observe that Xit is a zero set in X for the usual topology S, so that F is a zero set for S (4A2Gc) and therefore for the finer topology T. By 412B, this is enough to show that µ is inner regular with respect to the zero sets. Q Q Remark It may be worth noting that the topology T here is not regular. See Fremlin 75b, p. 106. 419E Example (Fremlin 76) Let (Z, S, T, ν) be the Stone space of the measure algebra of Lebesgue measure on [0, 1], so that ν is a strictly positive completion regular Radon probability measure (411P). Then the c.l.d. product measure λ on Z × Z is not a topological measure, so is not equal to the τ -additive product ˜ and λ ˜ is not completion regular. measure λ, ˜ described in 346K. We have W ∈ Λ = dom λ and W ˜ = S V, where proof Consider the sets W , W V = {G × H : G, H ⊆ Z are open-and-closed, (G × H) \ W is negligible}. ˜ is a union of open sets, therefore must be open in Z 2 . And λ∗ W ˜ ≤ λW . P W P?? Otherwise, there is a ˜ V ∈ Λ such that V ⊆ W and λV > λW . Now λ is tight, by 412Sb, so there is a compact set K ⊆ V such S that K ∈ Λ and λK > λW . There must be U0 , . . . , Un ∈ V such that K ⊆ i≤n Ui . But λ(Ui \ W ) = 0 for every i, so λ(K \ W ) = 0 and λK ≤ λW . X XQ Q ˜ should be 1 and λW strictly less than 1. So However, the construction of 346K arranged that λ∗ W ∗ ˜ ˜ ˜ λ∗ W < λ W and W ∈ / Λ. Accordingly λ is not a topological measure and cannot be equal to the Radon ˜ of 417P. measure λ We know that λ is inner regular with respect to the zero sets (412Sc) and is defined on every zero set ˜ properly extends λ. But this means that λ ˜ cannot be inner regular with respect to the zero (417V), while λ sets, by 412L, that is, cannot be completion regular. b 419F Theorem (Rao 69) P(ω1 × ω1 ) = Pω1 ⊗Pω 1 , the σ-algebra of subsets of ω1 generated by {E × F : E, F ⊆ ω1 }. proof (a) Because ω1 ≤ c, there is an injection h : ω1 → {0, 1}N ; set Ei = {ξ : h(ξ)(i) = 1} for each i ∈ N. b (b) Suppose that A ⊆ ω1 has countable vertical sections. Then A ∈ Pω1 ⊗Pω P Set B = A−1 [ω1 ] 1. P and for ξ ∈ B let fξ : N → A[{ξ}] be a surjection. Set gn (ξ) = fξ (n) for ξ ∈ B and n ∈ N, and An = {(ξ, fξ (n)) : ξ ∈ B} for n ∈ N. Then An = {(ξ, η) : ξ ∈ B, η = gn (ξ)} = {(ξ, η) : ξ ∈ B, η < ω1 , h(gn (ξ)) = h(η)} \¡ ¢ = {(ξ, η) : ξ ∈ gn−1 [Ei ], η ∈ Ei } ∪ {(ξ, η) : ξ ∈ B \ gn−1 [Ei ], η ∈ ω1 \ Ei } i∈N

b ∈ Pω1 ⊗Pω 1. So A=

S n∈N

b An ∈ Pω1 ⊗Pω Q 1. Q

b (c) Similarly, if a subset of ω1 × ω1 has countable horizontal sections, it belongs to Pω1 ⊗Pω 1 . But for 0 00 any A ⊆ ω1 × ω1 , A = A ∪ A where

132


419F

A0 = {(ξ, η) : (ξ, η) ∈ A, η ≤ ξ} has countable vertical sections, A00 = {(ξ, η) : (ξ, η) ∈ A, ξ ≤ η} has countable horizontal sections, b so both A0 and A00 belong to Pω1 ⊗Pω 1 and A also does. 419G Corollary (Ulam 30) Let Y be a set of cardinal at most ω1 and µ a σ-finite measure with domain PY . Then µ is point-supported: in particular, there is a countable conegligible set A ⊆ Y . P proof (a) Let µ0 be the point-supported part of µ, that is, µ0 A = y∈A µ{y} for every A ⊆ Y ; then µ0 is a measure (112Bd) and so is ν = µ − µ0 . If A ⊆ Y is countable, then of course µ0 A = µA, so νA = 0. Because µ is σ-finite, so is ν. (b) If Y is countable, we can stop. For the case in which #(Y ) = ω1 , it is enough to consider the case in which Y is actually equal to ω1 . Let λ = ν × ν be the product measure on ω1 × ω1 . By 419F, the domain of λ is the whole of P(ω1 × ω1 ); in particular, it contains the set V = {(ξ, η) : ξ ≤ η < ω1 }. Now by Fubini’s theorem λV =

R

and also λV =

νV [{ξ}]ν(dξ) =

R

R

ν(ω1 \ ξ)ν(dξ) = (νω1 )2 ,

νV −1 [{η}]ν(dη) =

R

ν(η + 1)ν(dη) = 0.

So νω1 = 0 and µ = µ0 . Now {y : µ{y} > 0} is a countable set (because µ is σ-finite) and is conegligible. Remark I ought to remark that this result, though not 419F, is valid for many other cardinals besides ω1 ; see, in particular, 438C below. There will be more on this topic in Volume 5. 419H Example There is a complete probability space (X, Σ, µ) with a topology T such that µ is τ additive and inner regular with respect to the Borel sets, T is generated by T ∩ Σ, but µ has no extension to a topological measure. proof (a) Set Y = ω1 + 1 = ω1 ∪ {ω1 }. Let T be the σ-algebra of subsets of Y generated by the σ-ideal of countable subsets of ω1 . Let ν be the probability measure with domain T defined by the formula 1 2

νF = #(F ∩ {0, ω1 }) for every F ∈ T. Set S = {∅, Y } ∪ {H : 0 ∈ H ⊆ ω1 }. This is a topology on Y , and every subset of Y is a Borel set for S; so ν is surely inner regular with respect to the Borel sets. Note that {{0, α} : α < ω1 } ∪ {Y } is a base for S included in T. (b) Let λ be the product probability measure on Y N , and Λ its domain; let Λ0 ⊆ Λ be the σ-algebra of subsets of Y N generated by sets of the form {y : y(i) ∈ F }, where i ∈ N and F ∈ T. Let S∗ be the product topology on Y N , so that S∗ ∩ Λ0 is a base for S∗ . Then λ is inner regular with respect to the Borel sets (412Uc, or otherwise). Define φ : {0, 1}N → Y by setting φ(u)(n) = 0 if u(n) = 0, = ω1 if u(n) = 1. Let νω be the usual measure on {0, 1}N ; then φ is inverse-measure-preserving for νω and λ, by 254H. If V ∈ Λ0 there is an α < ω1 such that whenever x ∈ V , y ∈ Y N and y(i) = x(i) whenever min(x(i), y(i)) < α, then y ∈ V .

419I

Examples

133

P P Let W be the family of sets V ∈ Λ with this property. Then W is a σ-algebra of subsets of Y N including the family C of measurable cylinders, so includes Λ0 . Q Q (c) Let X be Y N \ {0, ω1 }N . Then λ∗ X = 1. P P Take V ∈ Λ such that V ⊇ X and λV = λ∗ X. 0 0 Then there is a V ∈ Λ0 such that V ⊆ V and λV 0 = λV (254Ff). As remarked in (b), there is an α ∈ ]0, ω1 [ such that y ∈ V 0 whenever x ∈ V 0 , y ∈ Y N and y(i) = x(i) whenever min(x(i), y(i)) < α. But as {0, α}N \ {0} ⊆ X ⊆ V 0 , {0, ω1 }N \ {0} ⊆ V 0 . Accordingly λ∗ X = λV = λV 0 = νω φ−1 [V 0 ] = νω ({0, 1}N \ {0}) = 1. Q Q Give X the subspace measure µ induced by λ, with domain Σ = {X ∩ W : W ∈ Λ}, and the subspace topology T induced by S∗ . Then µX = λ∗ X = 1, and µ(X ∩ V ) = λV for every V ∈ Λ. Σ ∩ T is a base for T, just because Λ ∩ S∗ is a base for S∗ ; and µ is inner regular with respect to the Borel sets, by 412Pb. (d) ?? Suppose, if possible, that µ is not τ -additive. Let G ⊆ Σ ∩ T be a non-empty upwards-directed set with S union W ∈ Σ ∩ T such that µW > supG∈GSµG. Let G0 Sbe a countable subset of G such that µ(G \ G0 ) = 0 for every G ∈ G0 , and set W1 = W \ G0 ; then µ( G0 ) ≤ supG∈G µG, so µW1 > 0, while µ(G ∩ W1 ) = 0 for every G ∈ G. Express W1 as V1 ∩ X where V1 ∈ Λ. Then there is a V2 ∈ Λ0 such that V2 ⊆ V1 and λ(V1 \ V2 ) = 0, so that λV2 = λV1 = µW1 > 0. Let α < ω1 be such that y ∈ V2 whenever x ∈ V2 , y ∈ Y N and y(i) = x(i) whenever min(x(i), y(i)) < α. Let F ⊆ φ−1 [V2 ] be a non-negligible measurable self-supporting set for the Radon measure νω (416U, 414F). By Lemma 345E, there are u, u0 ∈ F which differ at exactly one coordinate; let i ∈ N be that coordinate, and suppose that u(i) = 1, u0 (i) = 0. We know that φ(u) ∈ V2 , and φ(u)(i) = ω1 . Define x ∈ X S by saying that x(j) = φ(u)(j) for j 6= i, x(i) = α. Then x ∈ V2 , by the choice of α. Now V2 ∩ X ⊆ W1 ⊆ G, soQthere is a G ∈ G containing x. Let V ⊆ Y N be a basic open set such that x ∈ X ∩ V ⊆ G; express V as j∈N Hj where Hj ∈ S for every j and J = {j : Hj 6= Y } is finite. Observe that 0 ∈ Hj and x(j) < ω1 for every j ∈ J, just because 0 belongs to every non-empty open subset of Y and the only open set containing ω1 is Y itself. But this means that u(j) = 0 for every j ∈ J \ {i}, so u0 (j) = 0 for every j ∈ J, and φ(u0 ) ∈ V ; thus the open set U = φ−1 [V ] meets F . Because F is self-supporting, 0 < νω (F ∩ U ) ≤ λ(V2 ∩ V ) = µ(V2 ∩ V ∩ X) ≤ µ(W1 ∩ G), which is impossible. X X Thus µ is τ -additive. (e) ?? But suppose, if possible, that there were a topological measure µ ˜ on X agreeing with µ on every open set in the domain of µ. For each i ∈ N, set πi (x) = x(i) for x ∈ X. Every subset of Y is a Borel set for S; because πi is continuous, the image measure µ ˜πi−1 is defined on PY . Now #(Y ) = ω1 , so there must be a countable conegligible set (419G), and there must be some αi < ω1 such that µ ˜πi−1 (ω1 \ αi ) = 0. On the other hand, µ ˜πi−1 (αi \ {0}) = µπi−1 (αi \ {0}) = λ{y : 0 < y(i) < αi } = ν(αi \ {0}) = 0, so µ ˜πi−1 (ω1 \ {0}) = 0. But the definition of X was exactly devised so that S X = i∈N πi−1 (ω1 \ {0}), so this is impossible. X X So we have the required example. Remark I note that the topology of X is not regular. Of course the phenomenon here cannot arise with regular spaces, by 415M. 419I For the next example it will be helpful to know some basic facts about Lebesgue measure which seemed a little advanced for Volume 1 and for which I have not found a suitable place since.

134


419I

Lemma (a) If (X, T, Σ, µ) is an atomless Radon measure space and E ∈ Σ has non-zero measure, then #(E) ≥ c. (b) The number of closed subsets of R is c. proof (a) There must be a compact set K ⊆ E such that µK > 0, and a self-supporting closed K 0 ⊆ K such that µK 0 = µK. Because µ{x} = 0 for every x ∈ X, K 0 can have no isolated points. So #(K 0 ) ≥ c (4A2G(i-ii)) and #(E) ≥ c. (b) Write E for the family of closed subsets of R. Let hUn in∈N enumerate a base for the topology of R (4A2Ua). For each I ⊆ N, set S FI = R \ n∈I Un . Because every open set is expressible as a union of some Ui , the map I 7→ FI : PN → E is surjective. So #(E) ≤ #(PN) = c. On the other hand, the map x → 7 [0, x] : [0, 1] → E is injective, so #(E) ≥ #([0, 1]) = c. Remark In fact, of course, the number of Borel subsets of R is c; see 4A1O. 419J The next result is a strengthening of 134D. Lemma Let µ be Lebesgue measure on R, and H any measurable subset of R. Then there is a disjoint family hAα iα 0. Let E be the family of closed subsets of H of non-zero measure. By 419Ib, #(E) ≤ c; enumerate E × c as h(Fξ , αξ )iξ 0 (413Ei), so there is a non-negligible measurable set E ⊆ H \ Aα . Now there is an F ∈ E such that F ⊆ E. Let ξ < c be such that F = Fξ and α = αξ ; then xξ ∈ Aα ∩ F , which is impossible. X X Thus H is always a measurable envelope of Aα . It follows from the definition of ‘measurable envelope’ that µ∗ Aα = µH. But also, if α < c, µ∗ Aα ≤ µ∗ (H \ Aα+1 ), which is 0, as we have just seen. So we have a suitable family. 419K Example (Blackwell 56) There are sequences hXn in∈N , hTn in∈N and hνn in∈N such that (i) for Q each n, (Xn , Tn ) is a separable metrizable space and νn is a quasi-Radon probability measure on Zn = map πmn : Zn → Zm is inverse-measure-preserving (iii) there is no i≤n Xi (ii) for m ≤ n the canonical Q probability measure on Z = i∈N Xi such that all the canonical maps from Z to Zn are inverse-measurepreserving. proof Let hAn in∈N be a disjoint sequence of subsets of [0, 1] such thatS µ∗ ([0, 1] \ An ) = 0, that is, µ∗ An = 1 for every n, where µ is Lebesgue measure (using 419J). Set Xn = i≥n Ai , so that hXn in∈N is a nonincreasing sequence of sets of outer measure 1 with empty intersection. For each n ≥ 1, we have a map fn : Xn → Zn defined by setting fn (x)(i) = x for every i ≤ n, x ∈ Xn . Let νn be the image measure µXn fn−1 , where µXn is the subspace measure on Xn induced by µ. Note that fn is a homeomorphism between Xn and the diagonal ∆n = {z : z ∈ Zn , z(i) = z(j) for all i, j ≤ n}, which is a closed subset of Zn ; so that νn , like µXn , is a quasi-Radon probability measure. If m ≤ n, then πmn is inverse-measure-preserving, where πmn (z)(i) = z(i) for z ∈ Zn , i ≤ m. P P If −1 W ⊆ Zm is measured by νm , then fm [W ] is measured by µXm , so is of the form Xm ∩ E where E is −1 −1 Lebesgue measurable. But in this case fm [πmn [W ]] = Xn ∩ E, so that −1 −1 νn (πmn [W ]) = µXn (fn−1 [πmn [W ]]) = µ∗ (Xn ∩ E) = µE = µ∗ (Xm ∩ E) = νm W . Q Q Q ?? But suppose, if possible, that there is a probability measure ν on Z = i∈N Xi such that πn : Z → Zn is inverse-measure-preserving for every n, where πn (z)(i) = z(i) for z ∈ Z, i ≤ n. Then

419Xc

Examples

135

νπn−1 [∆n ] = νn ∆n = µXn fn−1 [∆n ] = 1 for each n, so because

T n∈N

T 1 = ν( n∈N πn−1 [∆n ]) = ν{z : z ∈ Z, z(i) = z(j) for all i, j ∈ N} = ν∅, Xn = ∅; which is impossible. X X

419L The split interval again (a) For the sake of an example in §343, I have already introduced the ‘split interval’ or ‘double arrow space’. As this construction gives us a topological measure space of great interest, I repeat it here. Let I k be the set {a+ : a ∈ [0, 1]} ∪ {a− : a ∈ [0, 1]}. Order it by saying that a+ ≤ b+ ⇐⇒ a− ≤ b+ ⇐⇒ a− ≤ b− ⇐⇒ a ≤ b,

a+ ≤ b− ⇐⇒ a < b.

Then it is easy to check that I k is a totally ordered space, and that it is Dedekind complete. (If A ⊆ [0, 1] is a non-empty set, then supa∈A a− = (sup A)− , while supa∈A a+ is either (sup A)+ or (sup A)− , depending on whether sup A belongs to A or not.) Its greatest element is 1+ and its least element is 0− . Consequently the order topology on I k is a compact Hausdorff topology (4A2Rc, 4A2Ri). Note that Q = {q + : q ∈ [0, 1] ∩ Q} ∪ {q − : q ∈ [0, 1] ∩ Q} is dense, because it meets every non-trivial interval in I k . By 4A2E(a-ii) and 4A2Rn, I k is ccc and hereditarily Lindelöf. (b) If we define h : I k → [0, 1] by writing h(a+ ) = h(a− ) = a for every a ∈ [0, 1], then h is continuous, because {x : h(x) < a} = {x : x < a− }, {x : h(x) > a} = {x : x > a+ } for every a ∈ [0, 1]. Now we can describe the Borel sets of I k , as follows: a set E ⊆ I k is Borel iff there is a Borel set F ⊆ [0, 1] such that E4h−1 [F ] is countable. P P Write Σ0 for the family of subsets E of I k such that E4h−1 [F ] is countable for some Borel set F ⊆ [0, 1]. It is easy to check that Σ0 is a σ-algebra of subsets of I k . −1 (If is S countable, so is (I k \ E)4h−1 [[0, 1] \ F ]; if En 4h−1 [Fn ] is countable for every n, so is S E4h [F ]−1 ( n∈N En )4h [ n∈N Fn ].) Because the topology of I k is Hausdorff, every singleton set is closed, so every countable set is Borel. Also h−1 [F ] is Borel for every Borel set F ⊆ [0, 1], because h is continuous (4A3Cd). So if E4h−1 [F ] is countable for some Borel set F ⊆ [0, 1], E = h−1 [F ]4(E4h−1 [F ]) is a Borel set in I k . Thus Σ0 is included in the Borel σ-algebra B of I k . On the other hand, if J ⊆ I k is an interval, h[J] is also −1 k an interval, therefore a Borel S set, and h [h[J]] \ J can contain at most two points, so J ∈ Σ0 . If G ⊆ I is open, it is expressible as i∈I Ji , where hJi ii∈I is a disjoint family of non-empty open intervals (4A2Rj). As X is ccc, I must be countable. Thus G is expressed as a countable union of members of Σ0 and belongs to Σ0 . But this means that the Borel σ-algebra B must be included in Σ0 , by the definition of ‘Borel algebra’. So B = Σ0 , as claimed. Q Q (c) In 343J I described the standard measure µ on I k ; its domain is the set Σ = {h−1 [F ]4M : F ∈ ΣL , M ⊆ I k , µL h[M ] = 0}, where ΣL is the set of Lebesgue measurable subsets of [0, 1] and µL is Lebesgue measure, and µE = µL h[E] for E ∈ Σ. h is inverse-measure-preserving for µ and µL . The new fact I wish to mention is: µ is a completion regular Radon measure. P P I noted in 343Ja that it is a complete probability measure; a fortiori, it is locally determined and locally finite. If G ⊆ I k is open, then we can express it as h−1 [F ]4C for some Borel set F ⊆ [0, 1], countable set C ⊆ I k ((b) above), so it belongs to Σ; thus µ is a topological measure. If E ∈ Σ and µE > γ, then F = [0, 1] \ h[I k \ E] is Lebesgue measurable, and µE = µL F . So there is a compact set L ⊆ F such that µL L ≥ γ. But now K = h−1 [L] ⊆ E is closed, therefore compact, and µK ≥ γ. Moreover, L is a zero set, being a closed set in a metrizable space (4A2Lc), so K is a zero set (4A2C(b-iv)). As E and γ are arbitrary, µ is inner regular with respect to the compact zero sets, and is a completion regular Radon measure. Q Q 419X Basic exercises (a) Show that the topological space X of 419A is zero-dimensional. (b) Give an example of a compact Radon probability space in which every dense Gδ set is conegligible. (Hint: 411P.) (c) In 419E, show that we can start from any atomless probability measure in place of Lebesgue measure on [0, 1].

136


419Xd

> (d) (i) Show that if E ⊆ R 2 is Lebesgue measurable, with non-zero measure, then it cannot be covered by fewer than c lines. (Hint: if H = {t : µ1 E[{t}] > 0}, where µ1 is Lebesgue measure on R, then µ1 H > 0, so #(H) = c. So if we have a family L of lines, with #(L) < c, there must be a t ∈ H such that Lt = {t} × R does not belong to L. Now #(Lt ∩ E) = c and each member of L meets Lt ∩ E in at most one point.) (ii) Show that there is a subset A of R 2 , of full outer measure, which meets every vertical line and every horizontal line in exactly one point. (Hint: enumerate R as htξ iξ (h) (i) Again writing I k for the split interval, show that the function which exchanges x+ and x− for every x ∈ [0, 1] is a Borel automorphism and an automorphism for the usual Radon measure ν on I k , but is not almost continuous. (ii) Show that if we set f (x) = x+ for x ∈ [0, 1], then f is inverse-measure-preserving for Lebesgue measure µL on [0, 1], but the image measure µL f −1 is not ν (nor, indeed, a Radon measure). 419Y Further exercises (a) In the example of 419E, show that there is a Borel set V ⊆ Z 2 such that ˜ = 0 and λ∗ V = 1. λV (b) Show that if A ⊆ Pω1 is any family with #(A) ≤ ω1 , there is a countably generated σ-algebra Σ of subsets of ω1 such that A ⊆ Σ. (c) Show that the split interval with its usual topology and measure has the simple product property (417Yi). 419 Notes and comments The construction of the locally compact space X in 419A from the family hIξ iξ<κ is a standard device which has been used many times. The relation ⊆∗ also appears in many contexts. In effect, part of the argument is taking place in the quotient algebra A = PQ/[Q]<ω , since I ⊆∗ J iff I • ⊆ J • in A; setting I # = {I • : I ∈ I}, the cardinal κ is min{#(A) : A ⊆ I # has no upper bound in I # }, the ‘additivity’ of the partially ordered set I # . Additivities of partially ordered sets will be one of the important concerns of Volume 5. I remark that we do not need to know whether (for instance) κ = ω1 or κ = c. This is an early taste of the kind of manoeuvre which has become a staple of set-theoretic analysis. It happens that the cardinal κ here is one of the most important cardinals of set-theoretic measure

419 Notes

Examples

137

theory; it is ‘the additivity of Lebesgue measure’ (529Xa), and under that name will appear repeatedly in Chapter 52. Observe that the measure µ of 419A only just fails to be a quasi-Radon measure; it is locally finite instead of being effectively locally finite. And it would be a Radon measure if it were inner regular with respect to the compact sets, rather than just with respect to the closed sets. 419C and 419D are relevant to the question: have I given the ‘right’ definition of Radon measure space? 419C is perhaps more important. Here we have a Radon measure space (on my definition) for which the associated Borel measure is not localizable. (If A is the measure algebra of the measure µ, and B the measure algebra of µ¹B where B is the Borel σ-algebra of X, then the embedding B ⊆ Σ induces an embedding of B in A which represents B as an order-dense subalgebra of A, just because µ is inner regular with respect to B. Property (i) of 419C shows that B 6= A, so B cannot be Dedekind complete in itself, by 314Ia.) Since (I believe) localizable versions of measure spaces should almost always be preferred, I take this as strong support for my prejudice in favour of insisting that ‘Radon’ measure spaces should be locally determined as well as complete. Property (ii) of 419C is not I think of real significance, but is further evidence, to be added to 415Xh, that outer regularity is like an exoskeleton: it may inhibit growth above a certain size. In 419D I explore the consequences of omitting the condition ‘locally finite’ from the definition of Radon measure. Even if we insist instead that compact sets should have finite measure, we are in danger of getting a non-localizable measure. Of course this particular space is pathological in terms of most of the criteria of this chapter – for instance, every non-empty open set has infinite measure, and the topology is not regular. Perhaps the most important example in the section is 419E. The analysis of τ -additive product measures in §417 was long and difficult, and if these were actually equal to the familiar product measures in all important cases the structure of the theory would be very different. But we find that for one of the standard compact Radon probability spaces of the theory, the c.l.d. product measure on its square is not a Radon measure, and something has to be done about it. I present 419H here to indicate one of the obstacles to any simplification of the arguments in 417C and 417E. It is not significant in itself, but it offers a welcome excuse to describe some fundamental facts about ω1 (419F-419G). Similarly, 419K asks for some elementary facts about Lebesgue measure (419I-419J) which seem to have got left out. This example really is important in itself, as it touches on the general problem of representing stochastic processes, to which I will return in Chapter 45.

138

Descriptive set theory

Chapter 42 Descriptive set theory At this point, I interpolate an auxiliary chapter, in the same spirit as Chapters 31 and 35 in the last volume. As with Boolean algebras and Riesz spaces, it is not just that descriptive set theory provides essential tools for modern measure theory; it also offers deep intuitions, and for this reason demands study well beyond an occasional glance at an appendix. Several excellent accounts have been published; the closest to what we need here is probably Rogers 80; at a deeper level we have Moschovakis 80, and an admirable recent treatment is Kechris 95. Once again, however, I indulge myself by extracting those parts of the theory which I shall use directly, giving proofs and exercises adapted to the ideas I am trying to emphasize in this volume and the next. The first section describes Souslin’s operation and its basic set-theoretic properties up to the theory of ‘constituents’ (421N-421Q), mostly steering away from topological ideas, but with some remarks on σalgebras and Souslin-F sets. §422 deals with usco-compact relations and K-analytic spaces, working through the topological properties which will be useful later, and giving a version of the First Separation Theorem (422I-422J). §423 looks at ‘analytic’ or ‘Souslin’ spaces, treating them primarily as a special kind of Kanalytic space, with the von Neumann-Jankow selection theorem (423N). §424 is devoted to ‘standard Borel spaces’; it is largely a series of easy applications of results in §423, but there is one substantial theorem on Borel measurable actions of Polish groups (424H).

421 Souslin’s operation I introduce Souslin’s operation S (421B) and show that it is idempotent (421D). I describe alternative characterizations of members of S(E), where E ⊆ PX, as projections of sets in N N ×X (421G-421J). I briefly mention Souslin-F sets (421J-421L) and a special property of ‘inner Souslin kernels’ (421M). At the end of the section I set up an abstract theory of ‘constituents’ for kernels of Souslin schemes and their complements (421N-421Q). 421A Notation Throughout this chapter, and frequently in the next, I shall regard a member of N as the set of its predecessors, so that N k can be identified with the set of functions from k to N, and if φ ∈ N N and k ∈ N, we can speak of the restriction φ¹k ∈ N k . In the same spirit, identifying functions with their graphs, I can write ‘σ ⊆ φ’ when σ ∈ N k , φ ∈ N N and φ extends σ. On occasion I may write #(σ) for the ‘length’ of a finite function σ – again identifying σ with its graph – so that #(σ) = k if σ ∈ X k . And if k = 0, identified with ∅, then the only function from k to X is the empty function, so X 0 becomes {∅}. I shall sometimes refer to the ‘usual topology of N N ’; this is the product topology if each copy of N is S given its discrete topology. Writing Iσ = {φ : φ ∈ N N , φ ⊇ σ} for σ ∈ S ∗ = k∈N N k , {Iσ : σ ∈ S ∗ } is a base for the topology of N N consisting of open-and-closed sets. If σ ∈ N k , i ∈ N I write σ a i for the member τ of N k+1 such that τ (k) = i and τ (j) = σ(j) for j < k. 421B Definition Let S be the set expressible in the form

S k≥1

N k . If E is a family of sets, I write S(E) for the family of sets [ \

Eφ¹k

φ∈N N k≥1

for some family hEσ iσ∈S in E. S T A family hEσ iσ∈S is called a Souslin scheme; the corresponding set φ∈N N k≥1 Eφ¹k is its kernel; the operation [ \ hEσ iσ∈S 7→ Eφ¹k φ∈N N k≥1

is Souslin’s operation or operation A. Thus S(E) is the family of sets obtainable from sets in E by Souslin’s operation. If E = S(E), we say that E is closed under Souslin’s operation.

421Ce

Souslin’s operation

139

S ∗ k Remark I should perhaps warn S you that T some authors use S = k∈N N here in place of S; so that their Souslin kernels are of the form φ∈N N k≥0 Eφ¹k ⊆ E∅ . Consequently, for such authors, any member of S(E) is included in some member of E. If E has a greatest member (or, fractionally more generally, if any sequence in E is bounded above in E) this makes no difference; but if, for instance, E is the family of compact subsets of a topological space, the two definitions of S may not quite coincide. I believe that on this point, for once, I am following the majority. facts (a) It is worth noting straight away that if E is any family of sets, then S 421C Elementary T P Set n∈N En and n∈N En belong to S(E) for any sequence hEn in∈N in E. P Fσ = Eσ(0) for every σ ∈ S, Gσ = Ek whenever k ∈ N, σ ∈ Nk+1 ; then

S n∈N

T n∈N

En =

En =

S

T φ∈N N

S

k≥1

T φ∈N N

k≥1

Fφ¹k ∈ S(E),

Gφ¹k ∈ S(E). Q Q

In particular, E ⊆ S(E). But note that there is no reason why E \ F should belong to S(E) for E, F ∈ E. (b) Let X and Y be sets, and f : X → Y a function. Let hFσ iσ∈S be a Souslin scheme in PY , with kernel B. Then f −1 [B] is the kernel of the Souslin scheme hf −1 [Fσ ]iσ∈S . P P T S T S Q f −1 [B] = f −1 [ φ∈N N n≥1 Fφ¹n ] = φ∈N N n≥1 f −1 [Fφ¹n ]. Q (c) Let X and Y be sets, and f : X → Y a function. Let F be a family of subsets of Y . Then {f −1 [B] : B ∈ S(F)} = S({f −1 [F ] : F ∈ F}). P P For a set A ⊆ X, A ∈ S({f −1 [F ] : F ∈ F }) iff there is some Souslin scheme hEσ iσ∈S in {f −1 [F ] : F ∈ F} such that A is the kernel of hEσ iσ∈S , that is, iff there is some Souslin scheme hFσ iσ∈S in F such that A is the kernel of hf −1 [Fσ ]iσ∈S , that is, iff A = f −1 [B] where B is the kernel of some Souslin scheme in F. Q Q (d) Let X and Y be sets, and f : X → Y a surjective function. Let F be a family of subsets of Y . Then S(F) = {B : B ⊆ Y, f −1 [B] ∈ S({f −1 [F ] : F ∈ F})}. P P If B ∈ S(F), then f −1 [B] ∈ S({f −1 [F ] : F ∈ F }), by (c) above. If B ⊆ Y and f −1 [B] ∈ S({f −1 [F ] : F ∈ F }), then there is a Souslin scheme hFσ iσ∈S in F such that f −1 [B] is the kernel of hf −1 [Fσ ]iσ∈S , that is, f −1 [B] = f −1 [C] where C is the kernel of hFσ iσ∈S . Because f is surjective, B = C ∈ S(F). Q Q (e) Souslin’s operation can be thought of as a projection operator, as follows. Let hEσ iσ∈S be a Souslin scheme with kernel A. Set T S R = n≥1 σ∈N n Iσ × Eσ , writing Iσ = {φ : σ ⊆ φ ∈ N N } as usual. Then R[N N ] = A. P P For any x, and any φ ∈ N N , (φ, x) ∈ R ⇐⇒ for every n ≥ 1 there is a σ ∈ N n such that x ∈ Eσ , φ ∈ Iσ ⇐⇒ x ∈ Eφ¹n for every n ≥ 1. But this means that x ∈ R[N N ] ⇐⇒ there is a φ ∈ N N such that (φ, x) ∈ R \ ⇐⇒ there is a φ ∈ N N such that x ∈ Eφ¹n ⇐⇒ x ∈ A. Q Q n≥1

140


421D

421D The first fundamental theorem is that the operation S is idempotent. Theorem (Souslin 17) For any family E of sets, S(E) is closed under Souslin’s operation. S T proof (a) Let hAσ iσ∈S be a family in S(E), and set A = φ∈N N k≥1 Aφ¹k ; I have to show that A ∈ S(E). S T For each σ ∈ S, let hEστ iτ ∈S be a family in E such that Aσ = ψ∈N N m≥1 Eσ,ψ¹m . Then [ \ [ \ [ \ Eφ¹k,ψ¹m = Eφ¹k,ψk ¹m , A= φ∈N N k≥1 ψ∈N N m≥1

k,m≥1 φ∈N N Ψ∈(N N )N\{0}

writing Ψ = hψk ik≥1 for Ψ ∈ (N N )N\{0} . The idea of the proof is simply that N N × (N N )N\{0} is essentially identical to N N , so that all we have to do is to organize new names for the Eστ . But as it is by no means a trivial matter to devise a coding scheme which really works, I give the details at length. (b) The first step is to note that S and S 2 are countable, so there is a sequence hHn in∈N running over {Eστ : σ, τ ∈ S}. Next, choose any injective function q : N × N → N \ {0} such that q(0, 0) = 1 and q(0, 1) = 2. For k, m ≥ 1 set Jkm = {(i, 0) : i < k} ∪ {(i, k) : i < m}, so that J11 = {(0, 0), (0, 1)}, and choose a family h(kn , mn )in≥3 running over (N \ {0})2 such that q[Jkn ,mn ] ⊆ n for every n ≥ 3. (The pairs (kn , mn ) need not all be distinct, so this is easy to achieve.) Now, for υ ∈ N n , where n ≥ 3, set Fυ = Eστ where σ ∈ N kn , σ(i) = υ(q(i, 0)) for i < kn , τ ∈ N mn , τ (i) = υ(q(i, kn )) for i < mn ; these are well-defined because q[Jkn ,mn ] ⊆ n. For υ ∈ N 1 ∪ N 2 , set Fυ = Hυ(0) . (c) This defines a Souslin scheme hFυ iυ∈S in E. Let A0 be its kernel, so that A0 ∈ S(E). The point is that A0 = A. T P P (i) If x ∈ A, there must be φ ∈ N N , Ψ ∈ (N N )N\{0} such that x ∈ k,m≥1 Eφ¹k,ψk ¹m . Choose θ ∈ N N such that Hθ(0) = Eφ¹1,ψ1 ¹1 , θ(q(i, 0)) = φ(i) for every i ∈ N, θ(q(i, k)) = ψk (i) for every k ≥ 1, i ∈ N. (This is possible because q : N 2 → N \ {0} is injective.) Now Fθ¹1 = Fθ¹2 = Hθ(0) = Eφ¹1,ψ1 ¹1 certainly contains x. And for n ≥ 3, Fθ¹n = Eστ where σ(i) = θ(q(i, 0)) for i < kn , τ (i) = θ(q(i, kn )) for i < mn , that is, σ = φ¹kn and τ = ψkn ¹mn , so again x ∈ Fθ¹n . Thus T x ∈ n≥1 Fθ¹n ⊆ A0 . As x is arbitrary, A ⊆ A0 . (ii) Now take any x ∈ A0 . Let θ ∈ N N be such that x ∈ setting

T n≥1

Fθ¹n . Define φ ∈ NN , Ψ ∈ (N N )N\{0} by

φ(i) = θ(q(i, 0)) for i ∈ N, ψk (i) = θ(q(i, k)) for k ≥ 1, i ∈ N. If k, m ≥ 1, let n ≥ 3 be such that k = kn , m = mn . Then x ∈ Fθ¹n = Eστ , where σ(i) = θ(q(i, 0)) for i < kn ,

τ (i) = θ(q(i, kn )) for i < mn ,

that is, σ = φ¹kn = φ¹k and τ = ψkn ¹mn = ψk ¹m. As m and n are arbitrary, T x ∈ m,n≥1 Eφ¹k,ψk ¹m ⊆ A. As x is arbitrary, A0 ⊆ A. Q Q Accordingly we must have A ∈ S(E), and the proof is complete.

421G


141

421E Corollary For any family E of sets, S(E) is closed under countable unions and intersections. proof For 421Ca tells us that the union and intersection of any sequence in S(E) will belong to SS(E) = S(E). 421F Corollary Let X be a set and E a family of subsets of X. Suppose that X and ∅ belong to S(E) and that X \ E ∈ S(E) for every E ∈ E. Then S(E) includes the σ-algebra of subsets of X generated by E. proof The set Σ = {F : F ∈ S(E), X \ F ∈ S(E)} is closed under complements (necessarily), contains ∅ (because ∅ and X belong to S(E)), and is also closed under countable unions, by 421E. So it is a σ-algebra; but the hypotheses also ensure that E ⊆ Σ, so that the σ-algebra generated by E is included in Σ and in S(E). 421G Proposition Let E be a family of sets such that ∅ ∈ E. For σ ∈ S = N N , φ ⊇ σ}. Then

S k≥1

N k set Iσ = {φ : φ ∈

S(E) = {R[N N ] : R ∈ S({Iσ × E : σ ∈ S, E ∈ E})} = {R[N N ] : R ∈ S({Iσ × E : σ ∈ S, E ∈ E}), R−1 [{x}] is closed for every x}. proof Set F = {Iσ × E : σ ∈ S, E ∈ E}. (a) Suppose first that A ∈ S(E). Let hEσ iσ∈S be a Souslin scheme in E with kernel A. Set T S R = k≥1 σ∈N k Iσ × Eσ . Then R ∈ S(F), by 421E, and R[N N ] = A, by 421Ce. Also T S R−1 [{x}] = k≥1 {Iσ : σ ∈ N k , x ∈ Eσ } is closed, for every x. (b) Now suppose that A = R[N N ] for some R ∈ S(F). Let hIτ (σ) × Eσ iσ∈S be a Souslin scheme in F with kernel R. For k ≥ 1, σ ∈ Nk set Fσ = Eσ if

\

Iτ (σ¹n) 6= ∅,

1≤n≤k

= ∅ otherwise. Then hFσ iσ∈S is a Souslin scheme in E, so its kernel A0 belongs to S(E). The point T is that A0 = A. P P (i) If x ∈ A, there are a φ ∈ NN such that (φ, x) ∈ R and a ψ ∈ N N such that (φ, x) ∈ n≥1 Iτ (ψ¹n) × Eψ¹n . Now, for any k ≥ 1, we have T T φ ∈ 1≤n≤k Iτ (ψ¹n) = 1≤n≤k Iτ ((ψ¹k)¹n) , T so that Fψ¹k = Eψ¹k contains x; thus x ∈ k≥1 Fψ¹k ⊆ A0 . As x is arbitrary, A ⊆ A0 . (ii) If x ∈ A0 , take T T ψ ∈ N N such that x ∈ n≥1 Fψ¹n . In this case we must have Fψ¹k 6= ∅, so 1≤n≤k Iτ (ψ¹n) 6= ∅, for every k ≥ 1. But what this means is that, setting τn = τ (ψ¹n) for each n ≥ 1, τn (i) = τm (i) whenever i ∈ N is T such that both are defined. So {τn : n ≥ 1} must have a common extension φ ∈ N N , and φ ∈ n≥1 Iτ (ψ¹n) . Now T (φ, x) ∈ n≥1 Iτ (ψ¹n) × Eψ¹n ⊆ R, so x ∈ A. Thus A0 ⊆ A and the two are equal. Q Q This shows that {R[N N ] : R ∈ S(F)} ⊆ S(E), and the proof is complete.

142


421H

421H When the class E is a σ-algebra, the last proposition can be extended. Proposition Let S X be a set, and Σ a σ-algebra of subsets of X. Let B be the algebra of Borel subsets of N N . For σ ∈ S = k≥1 N k , set Iσ = {φ : φ ∈ N N , φ ⊇ σ}. Then b S(Σ) = {R[N N ] : R ∈ B ⊗Σ} = {R[N N ] : R ∈ S({Iσ × E : σ ∈ S, E ∈ Σ})} b = {R[N N ] : R ∈ S(B ⊗Σ)}. b is the σ-algebra of subsets of N N × X generated by {H × E : H ∈ B, E ∈ Σ}. Notation Recall that B ⊗Σ proof (a) Suppose first that A ∈ S(Σ). As in 421G, let hEσ iσ∈S be a Souslin scheme in Σ with kernel A, and set T S R = k≥1 σ∈N k Iσ × Eσ , b so that A = R[N N ]. Because every Iσ is an open-and-closed set in N N , R ∈ B ⊗Σ. Thus b S(Σ) ⊆ {R[N N ] : R ∈ B ⊗Σ}. b = S(F). P (b) Set F = {Iσ × E : σ ∈ S, E ∈ Σ}. Then S(B⊗Σ) P If E ∈ Σ and σ ∈ N k then S (N N × X) \ (Iσ × E) = (Iσ × (X \ E)) ∪ τ ∈Nk ,τ 6=σ Iτ × X ∈ S(F). Also NN × X =

S σ∈N1

Iσ × X,

∅ = Iτ × ∅

(where τ is any member of S) belong to S(F). S By 421F, S(F) includes the σ-algebra Λ of sets generated by F. Now if E ∈ Σ and H ⊆ N N is open, H = σ∈T Iσ for some T ⊆ S; as T is necessarily countable, S H × E = σ∈T Iσ × E ∈ Λ. Since {F : F ⊆ N N , F × E ∈ Λ} is a σ-algebra of subsets of N N , and we have just seen that it contains all b ⊆ Λ ⊆ S(F), and the open sets, it must include B; thus F × E ∈ Λ for every F ∈ B, E ∈ Σ. So B ⊗Σ b ⊆ SS(F) = S(F) S(F) ⊆ S(B⊗Σ) (421D). Q Q (c) Now we have

b S(Σ) ⊆ {R[N N ] : R ∈ B ⊗Σ} (by (a)) b ⊆ {R[N N ] : R ∈ S(B ⊗Σ)} = {R[N N ] : R ∈ S(F)} (by (b)) = S(Σ) by 421G. 421I There is a particularly simple description of sets obtainable by Souslin’s operation from closed sets in a topological space. Lemma Let X be a topological space and R ⊆ N N × X a closed set. Then S T R[A] = φ∈A n≥1 R[Iφ¹n ]. for any A ⊆ N N , writing Iσ = {φ : σ ⊆ φ ∈ N N } as usual. In particular, R[N N ] is the kernel of the Souslin scheme hR[Iσ ]iσ∈S .

421M


proof Set B=

S

T φ∈A

n≥1

143

R[Iφ¹n ].

(i) If x ∈ R[A], there is a φ ∈ A such that (φ, x) ∈ R. In this case, φ ∈ Iφ¹n so x ∈ R[Iφ¹n ] ⊆ R[Iφ¹n ] for every n, and x ∈ B. Thus R[A] ⊆ B. (ii) If x ∈ B, let φ ∈ A be such that x ∈ R[Iφ¹n ] for every n ∈ N. ?? If (φ, x) ∈ / R, then (because R is closed) there are a σ ∈ S and an open G ⊆ X such that φ ∈ Iσ , x ∈ G and (Iσ × G) ∩ R = ∅. But this means that G ∩ R[Iσ ] = ∅ so G ∩ R[Iσ ] = ∅ and x ∈ / R[Iσ ]; which is absurd, because σ = φ¹n for some n ≥ 1. X X Thus (φ, x) ∈ R and x ∈ R[A]. As x is arbitrary, B ⊆ R[A] and B = R[A], as required. 421J Proposition Let X be a topological space, and F the family of closed subsets of X. Then a set A ⊆ X belongs to S(F) iff there is a closed set R ⊆ N N × X such that A is the projection of R on X. proof (a) Suppose that A ∈ S(F). Let hFσ iσ∈S be a Souslin scheme in F with kernel A. Set T S R = n≥1 σ∈N n Iσ × Fσ . For each n ≥ 1,

S σ∈N n

N

Iσ × Fσ = (N N × X) \

S σ∈Nn

Iσ × (X \ Fσ )

N

is closed in N × X, so R is closed; and the projection R[N ] is A, by 421Ce. (b) Suppose that R ⊆ N N is a closed set with projection A. Then A is the kernel of the Souslin scheme hR[Iσ ]iσ∈S , by 421I, so belongs to S(F). 421K Definition Let X be a topological space. A subset of X is a Souslin-F set in X if it is obtainable from closed subsets of X by Souslin’s operation; that is, is the projection of a closed subset of N N × X. For a subset of R r , or, more generally, of any Polish space, it is common to say ‘Souslin set’ for ‘Souslin-F set’; see 421Xm. 421L Proposition Let X be any topological space. Then every Baire subset of X is Souslin-F. proof Let Z be the family of zero sets in X. If F ∈ Z then X \ F is a countable union of zero sets (4A2C(b-vi)), so belongs to S(Z). By 421F, the σ-algebra generated by Z is included in S(Z) ⊆ S(F), where F is the family of closed subsets of X; that is, every Baire set is Souslin-F. T 421M Proposition Let E be any family of sets such that ∅ ∈ E and E ∪ E 0 , n∈N En belong to E for every E, E 0 ∈ E and all sequences hEn in∈N in E. (For instance, E could be the family of closed subsets of a topological space, or a σ-algebra of sets.) Let hESσ iσ∈STbe a Souslin scheme in E, and K ⊆ N N a set which is compact for the usual topology on N N . Then φ∈K n≥1 Eφ¹n ∈ E. S T proof Set A = φ∈K n≥1 Eφ¹n . For k ∈ N, set Kk = {φ¹k : φ ∈ K}; note that Kk ⊆ N k is compact, because φ 7→ φ¹k is continuous, therefore finite, because the topology of N k is discrete. Set T S T H = k≥1 σ∈Kk 1≤n≤k Eσ¹n . Because E is closed under finite unions and countable intersections, H ∈ T E. Now A = H. P P (i) If x ∈ A, take φ ∈ K such that x ∈ Eφ¹n for every n ≥ 1; then φ¹k ∈ Kk and x ∈ 1≤n≤k E(φ¹k)¹n for every k ≥ 1, T so x ∈ H. Thus A ⊆ H. (ii) If x ∈ H, then for each k ∈ N we have a σk ∈ Kk such that x ∈ 1≤n≤k Eσk ¹n . Choose φk ∈ K such that φk ¹k = σk for each k. Now K is supposed to be compact, so the sequence hφk ik∈N has a cluster point φ in K. If n ≥ 1, then Iφ¹n is a neighbourhood of φ in N N , so must contain φk for infinitely many k; let k ≥ n be such that φk ¹n = φ¹n. In this case x ∈ Eσk ¹n = Eφk ¹n = Eφ¹n . As n is arbitrary,

144


x∈

T n≥1

421M

Eφ¹n ⊆ A.

Q As x is arbitrary, H ⊆ A and H = A, as claimed. Q So A ∈ E. *421N I now embark on preparations for the theory of ‘constituents’ of analytic and coanalytic sets. It turns out that much of the work can be done in the abstract context of this section. S Trees and derived trees (a) Let T be the family of subsets T of S = n≥1 N n such that σ¹k ∈ T whenever σ ∈ T and 1 ≤ k ≤ #(σ). Note that the intersection and union of any non-empty family of members of T again belong to T . Members of T are often called trees. (b) For T ∈ T , set ∂T = {σ : σ ∈ S, ∃ i ∈ N, σ a i ∈ T }, so that ∂T ∈ T and ∂T ⊆ T . Of course ∂T0 ⊆ ∂T1 whenever T0 , T1 ∈ T and T0 ⊆ T1 . T (c) For T ∈ T , define h∂ ξ T iξ<ω1 inductively by setting ∂ 0 T = T and, for ξ > 0, ∂ ξ T = ∂( η<ξ ∂ η T ). An easy induction shows that ∂ ξ T ∈ T , ∂ ξ T ⊆ ∂ η T and ∂ ξ+1 T = ∂(∂ ξ T ) whenever η ≤ ξ < ω1 . T (d) For any T ∈ T , there is a ξ < ω1 such that ∂ ξ T = ∂ η T whenever ξ ≤ η < ω1 . P P Set T1 = ξ<ω1 ∂ ξ T . For each σ ∈ S \ T1 , there is a ξσ < ω1 such that σ ∈ / ∂ ξσ T . Set ξ = sup{ξσ : σ ∈ S \ T1 }; because S is ξ countable, ξ < ω1 , and we must now have ∂ T = T1 , so that ∂ ξ T = ∂ η T whenever ξ ≤ η < ω1 . Q Q (e) For T ∈ T , its rank is the first ordinal r(T ) < ω1 such that ∂ r(T ) T = ∂ r(T )+1 T ; of course ∂ r(T ) T = ∂ T whenever r(T ) ≤ η < ω1 , and ∂(∂ r(T ) T ) = ∂ r(T ) T . η

(f ) For T ∈ T , the following are equiveridical: (α) ∂ r(T ) T 6= ∅; (β) there is a φ ∈ N N such that φ¹n ∈ T for every n ≥ 1. P P (i) If σ ∈ ∂ r(T ) T then σ ∈ ∂(∂ r(T ) T ) so there is an i ∈ N such that σ a i ∈ ∂ r(T ) T . We can therefore choose hσn in∈N inductively so that σn ∈ ∂ r(T ) T and σn+1 properly extends σn for every n. At S the end of the induction, φ = n∈N σn belongs to N N and φ¹n = σn ¹n ∈ ∂ r(T ) T ⊆ T for every n ≥ 1. (ii) If φ ∈ N N is such that φ¹n ∈ T for every n ≥ 1, then an easy induction shows that φ¹n ∈ ∂ ξ T for every ξ < ω1 and every n ≥ 1, so that ∂ r(T ) T is non-empty. Q Q (g) Now suppose that hAσ iσ∈S is a Souslin scheme. For any x we have a tree Tx ∈ T defined by saying that T Tx = {σ : σ ∈ S, , x ∈ 1≤i≤#(σ) Aσ¹i }. Now the kernel of hAσ iσ∈S is just A = {x : ∃ φ ∈ N N , x ∈

\

Aφ¹n }

n≥1

= {x : ∃ φ ∈ N N , φ¹n ∈ Tx ∀ n ≥ 1} = {x : ∂ r(T ) T 6= ∅} by (f). The sets {x : x ∈ X \ A, r(Tx ) = ξ} = {x : x ∈ X, r(Tx ) = ξ, ∂ ξ Tx = ∅}, for ξ < ω1 , are called constituents of X \ A. (Of course they should properly be called ‘the constituents of the Souslin scheme hAσ iσ∈S ’.)

*421Q


145

*421O Theorem Let X be a set and Σ a σ-algebra of subsets of X. Let hAσ iσ∈S be a Souslin scheme in Σ with kernel A, and for x ∈ X set T Tx = {σ : σ ∈ S, x ∈ 1≤i≤#(σ) Aσ¹i } ∈ T as in 421Ng. (a) For every ξ < ω1 and σ ∈ S, {x : x ∈ X, σ ∈ ∂ ξ Tx } ∈ Σ. (b) For every ξ < ω1 , {x : x ∈ A, r(Tx ) ≤ ξ} and {x : x ∈ X \ A, r(Tx ) ≤ ξ} belong to Σ. In particular, all the constituents of X \ A belong to Σ. proof (a) Induce on ξ. For ξ = 0, we have {x : x ∈ X, σ ∈ ∂ 0 Tx } = {x : x ∈ X, σ ∈ Tx } =

T 1≤i≤#(σ)

Aσ¹i ∈ Σ.

For the inductive step to ξ > 0, we have {x : σ ∈ ∂ ξ Tx } = {x : σ ∈ ∂( =

[ \

\

∂ η Tx )} =

[

{x : σ a i ∈

i∈N

η<ξ

\

∂ η Tx }

η<ξ

{x : σ a i ∈ ∂ η Tx } ∈ Σ

i∈N η<ξ

because ξ is countable and all the sets {x : σ a i ∈ ∂ η Tx } belong to Σ by the inductive hypothesis. (b) Now, given ξ < ω1 , we see that r(Tx ) ≤ ξ iff ∂ ξ+1 Tx ⊆ ∂ ξ Tx , so that if we set Eξ = {x : x ∈ X, r(Tx ) ≤ ξ} then T / ∂ ξ Tx } Eξ = σ∈S {x : x ∈ X, σ ∈ ∂ ξ+1 Tx or σ ∈ belongs to Σ. Now, given that x ∈ Eξ , so that ∂ r(Tx ) Tx = ∂ ξ Tx , 421Ng tells us that x ∈ A iff ∅ ∈ ∂ ξ Tx ; so that Eξ ∩ A and Eξ \ A both belong to Σ. S Now the constituents of X \ A are the sets (Eξ \ A) \ η<ξ Eη for ξ < ω1 , so all belong to Σ. *421P Corollary Let X be a set and Σ a σ-algebra of subsets of X. If A ∈ S(Σ) then both A and X \ A can be expressed as the union of at most ω1 members of Σ. proof In the language of 421O, we have S A = ξ<ω1 Eξ ∩ A,

X \A=

S ξ<ω1

Eξ \ A.

*421Q Lemma Let X be a set and hAσ iσ∈S and hB Tσ iσ∈S two Souslin schemes of subsets of X. Suppose that whenever φ, ψ ∈ N N there is an n ≥ 1 such that 1≤i≤n Aφ¹i ∩ Bψ¹i = ∅. For x ∈ X set S T Tx = n≥1 {σ : σ ∈ N n , x ∈ 1≤i≤n Aσ¹i } as in 421Ng, and let B be the kernel of hBσ iσ∈S . Then supx∈B r(Tx ) < ω1 . T T proof For σ ∈ S set A0σ = 1≤i≤#(σ) Aσ¹i , Bσ0 = 1≤i≤#(σ) Bσ¹i . Then Tx = {σ : σ ∈ S, x ∈ A0σ } for each 0 = ∅. x ∈ X, B is the kernel of hBσ0 iσ∈S , and for every φ, ψ ∈ N N there is an n ∈ N such that A0φ¹n ∩ Bψ¹n Define hQξ iξ<ω1 inductively by setting Q0 = {(σ, τ ) : σ, τ ∈ S, A0σ ∩ Bτ0 6= ∅}, and, for 0 < ξ < ω1 , Qξ = {(σ, τ ) : σ, τ ∈ S, ∃ i, j ∈ N, (σ a i, τ a j) ∈

T η<ξ

Qη }.

Then the same arguments as in 421Na-421Nd show that there is a ζ < ω1 such that Qζ+1 = Qζ . ?? If Qζ 6= ∅, then, just as in 421Nf, there must be φ, ψ ∈ N N such that (φ¹m, ψ¹n) ∈ Qζ ⊆ Q0 for every m, 0 6= ∅ for every n ≥ 1, which is supposed to be impossible. X X n ≥ 1; but this means that A0φ¹n ∩ Bψ¹n 0 for every n ≥ 1. But this Now suppose that x ∈ B. Then there is a ψ ∈ N N such that x ∈ Bψ¹n means that (σ, ψ¹n) ∈ Q0 for every σ ∈ Tx and every n ≥ 1. An easy induction shows that (σ, ψ¹n) ∈ Qξ whenever ξ < ω1 , σ ∈ ∂ ξ Tx and n ≥ 1. But as Qζ = ∅ we must have ∂ ζ Tx = ∅ and r(Tx ) ≤ ζ. Thus supx∈B r(Tx ) ≤ ζ < ω1 , and the proof is complete.

146


421X

421X Basic exercises (a) Let X be a set and E a family of subsets of X. (i) Show that ∅ ∈ S(E) iff there is a sequence in E with empty intersection. (ii) Show that X ∈ S(E) iff there is a sequence in E with union X. (b) Let E be a family of sets and F any set. Show that S({E ∩ F : E ∈ E}) = {A ∩ F : A ∈ S(E)}, S({E ∪ F : E ∈ E}) = {A ∪ F : A ∈ S(E)}. (c) Let E be a family of sets and F any set. Show that S(E ∪ {F }) = {F } ∪ {A ∩ F : A ∈ S(E)} ∪ {B ∪ F : B ∈ S(E)} ∪ {(A ∩ F ) ∪ B : A, B ∈ S(E)}. (d) Suppose that E is a family of sets with #(E) ≤ c. Show that #(S(E)) ≤ c. (Hint: #(E S ) ≤ #((PN)S ) = #(P(N × S)).) (e) Let E be the family of half-open intervals [2−n k, 2−n (k + 1)[, where n ∈ N, k ∈ Z; let G be the set of open subsets of R; let F be the set of closed subsets of R; let K be the set of compact subsets of R; let B be the Borel σ-algebra of R. Show that S(E) = S(F) = S(G) = S(K) = S(B). (Hint: 421F.) S (f ) Let I be the family {Iσ : σ ∈ k∈N N k } (421A); let G be the set of open subsets of N N ; let F be the set of closed subsets of N N ; let K be the set of compact subsets of N N ; let B be the Borel σ-algebra of N N . Show that S(I) = S(F) = S(G) = S(B), but that S(K) is strictly smaller thanSthese. (Hint: if hKn in∈N is any sequence in K, set φ(i) = 1 + supψ∈Ki ψ(i) for each i ∈ N, so that φ ∈ / n∈N Kn ; hence show that NN ∈ / S(K).) (g) Let X be a separable metrizable space with at least two points; let U be any base for its topology, and B its Borel σ-algebra. Show that S(U) = S(B). What can happen if #(X) ≤ 1? (h) Let X be a topological space; let Z be the set of zero sets in X, G the set of cozero sets, and Ba the Baire σ-algebra. Show that S(Z) = S(G) = S(Ba). (i) Let X be a set, E a family of subsets of X, and Σ the σ-algebra of subsets of X generated by E. Show that if #(E) ≤ c then #(Σ) ≤ c. (Hint: #(cS ) = #(P(N × N)) = c and Σ ⊆ S(E ∪ {X \ E : E ∈ E}).) (j) Let X be a topological space such that every open set is Souslin-F. Show that every Borel set is Souslin-F. (k) Let X be a topological space and B(X) its Borel σ-algebra. Show that S(B(X)) is just the set of projections on X of Borel subsets of N N × X. (Hint: 4A3G.) (l) Let X and Y be topological spaces, f : X → Y a continuous function and F ⊆ Y a Souslin-F set. Show that f −1 [F ] is a Souslin-F set in X. (m) Let X be any metrizable space; let G be the set of open subsets of X, F the set of closed subsets, and B the Borel σ-algebra. Show that S(G) = S(F) = S(B). (n) Let us say that a Souslin scheme hEσ iσ∈S is regular if Eσ ⊆ Eτ whenever σ, τ ∈ S, #(τ ) ≤ #(σ) and σ(i) ≤ τ (i) for every i < #(σ). Let E be a family of sets such that E ∪ F , E ∩ F ∈ E for all E, F ∈ E. Show that every member of S(E) can be expressed as the kernelTof a regular Souslin S scheme in E. (Hint: if hEσ iσ∈S is any Souslin scheme in E with kernel A, set Fσ = τ ⊆σ Eτ , Gσ = τ ≤σ Fτ , where τ ≤ σ if τ (i) ≤ σ(i) for i < #(τ ) = #(σ); show that A is the kernel of hFσ iσ∈S and of hGσ iσ∈S , using an idea from 421M for the latter.)

421 Notes


147

(o) Let X be a Hausdorff space and hKσ iσ∈S a Souslin scheme in which every Kσ is a compact S topological T subset in X. Show that φ∈K n≥1 Kφ¹n is compact for any compact K ⊆ N N . 421Y Further exercises (a) Let X be a topological space, Y a Hausdorff space and f : X → Y a continuous function. Let K be the family of closed countably compact subsets of X. Show that for any E ⊆ K, {f [A] : A ∈ S(E)} = S({f [E] : E ∈ E}). (b) Let X be a topological space, and Ba its Baire σ-algebra. Show that S(Ba) is just the family of sets expressible as f −1 [B] where f is a continuous function from X to some metrizable space Y and B ⊆ Y is Souslin-F. (c) Let X be a set, E a family of subsets of X, and Σ the smallest σ-algebra of subsets of X including E and closed under Souslin’s operation.S Show that if #(E) ≤ c then #(Σ) ≤ c. (Hint: define hEξ iξ<ω1 by setting Eξ = S({X \ E : E ∈ E ∪ η<ξ Eξ }) for each ξ. Show that #(Eξ ) ≤ c for every ξ and that S Σ = ξ<ω1 Eξ .) (d) Let X be a compact S space and A a Souslin-F set in X. Show that there is a family hFξ iξ<ω1 of Borel sets such that X \ A = ξ<ω Fξ and whenever B ⊆ X \ A is a Souslin-F set there is a ξ < ω1 such that B ⊆ Fξ . (Hint: take Fξ = {x : r(Tx ) ≤ ξ} \ A as in 421Ob, and apply 421Q.) 421 Notes and comments In 111G, I defined the Borel sets of R to be the members of the smallest σ-algebra containing every open set. In 114E, I defined a set to be Lebesgue measurable if it behaves in the right way with respect to Lebesgue outer measure. The latter formulation, at least, provides some sort of testing principle to determine whether a set is Lebesgue measurable. But the definition of ‘Borel set’ does not. The only tool so far available for proving that a set E ⊆ R is not Borel is to find a σ-algebra containing all open sets and not containing E; conversely, the only method we have for proving properties of Borel sets is to show that a property is possessed by every member of some σ-algebra containing every open set. The revolutionary insight of Souslin 17 was a construction which could build every Borel set from rational intervals. (See 421Xe.) For fundamental reasons, no construction of this kind can provide all Borel sets without also producing other sets, and to actually characterize the Borel σ-algebra a further idea is needed (423Fa); but the class of analytic sets, being those constructible by Souslin’s operation from rational intervals (or open sets, or closed sets, or Borel sets – the operation is robust under such variations), turns out to have remarkable properties which make it as important in modern real analysis as the Borel algebra itself. The guiding principle of ‘descriptive set theory’ is that the properties of a set may be analysed in the light of a construction for that set. Thus we can think of a closed set F ⊆ R as S R \ (q,q0 )∈I ]q, q 0 [ where I ⊆ Q × Q. The principle can be effective because we often have such descriptions in terms of objects fundamentally simpler than the set being described. In the formula above, for instance, Q × Q is simpler than the set F , being a countable set with a straightforward description from N. The set P(Q × Q) is relatively complex; but a single subset I of Q × Q can easily be coded as a single subset of N (taking some more or less natural enumeration of Q2 as a sequence h(qn , qn0 )in∈N , and matching I with {n : (qn , qn0 ) ∈ I}). So, subject to an appropriate coding, we have a description of closed subsets of R in terms of subsets of N. At the most elementary level, this shows that there are at most c closed subsets of R. But we can also set out to analyse such operations as intersection, union, closure in terms of these descriptions. The details are complex, and I shall go no farther along this path here; but investigations of this kind are at the heart of some of the most exciting developments of twentieth-century real analysis. The particular descriptive method which concerns us in the present section is Souslin’s operation. Starting from a relatively simple class E, we proceed to the larger class S(E). The most fundamental property of S is 421D: SS(E) = S(E). This means, for instance, that if E ⊆ S(F) and F ⊆ S(E), then S(E) will be equal to S(F); consequently, different classes of sets will often have the same Souslin closures, as in 421Xe-421Xh.

148


421 Notes

After a little practice you will find that it is often easy to see when two classes E and F are at the same level in this sense; but watch out for traps like the class of compact subsets of N N (421Xf) and odd technical questions (421Xg). Souslin’s operation, and variations on it, will be the basis of much of the next chapter; it has dramatic applications in general topology and functional analysis as well as in real analysis and measure theory. An important way of looking at the kernel of a Souslin T scheme S hEσ iσ∈S is to regard it as the projection on the second coordinate of the corresponding set R = k≥1 σ∈N k Iσ × Eσ (421Ce). We find that many other sets R ⊆ N N × X will also have projections in S(E) (421G, 421H). Let me remark that it is essential here that the first coordinate should be of the right type. In one sense, indeed, N N is the only thing that will do; but its virtue transfers to analytic spaces, as we shall see in 423M-423O below. We shall often want to deal with members of S(E) which are most naturally defined in terms of some such auxiliary space. I have moved into slightly higher gear for 421N-421Q because these are not essential for most of the work of the next chapter. From the point of view of this section 421P is very striking but the significance of 421Q is unlikely to be apparent. It becomes important in contexts in which the condition T ∀ φ, ψ ∈ N N ∃ n ≥ 1, 1≤i≤n Aφ¹i ∩ Bψ¹i = ∅ is satisfied for natural reasons. I will expand on these in the next two sections. In the meantime, I offer 421Yd as an example of what 421O and 421Q together can tell us.

422 K-analytic spaces I introduce K-analytic spaces, defined in terms of usco-compact relations. The first step is to define the latter (422A) and give their fundamental properties (422B-422E). I reach K-analytic spaces themselves in 422F, with an outline of the most important facts about them in 422G-422K. 422A Definition Let X and Y be Hausdorff spaces. A relation R ⊆ X × Y is usco-compact if (α) R[{x}] is a compact subset of Y for every x ∈ X, (β) R−1 [F ] is a closed subset of X for every closed set F ⊆ Y . (Relations satisfying condition (β) are sometimes called ‘upper semi-continuous’.) 422B The following elementary remark will be useful. Lemma Let X and Y be Hausdorff spaces and R ⊆ X × Y an usco-compact relation. If H is an open subset of Y including R[{x}] where x ∈ X, there is an open set G ⊆ X, containing x, such that R[G] ⊆ H. proof Set G = X \ R−1 [Y \ H]. Because Y \ H is closed, so is R−1 [Y \ H], and G is open. Of course R[G] ⊆ H, and x ∈ G because R[{x}] ⊆ H. 422C Proposition Let X and Y be Hausdorff spaces. Then a subset R of X × Y is an usco-compact relation iff whenever F is an ultrafilter on X × Y , containing R, such that the first-coordinate image π1 [[F]] of F has a limit in X, then F has a limit in R. proof Recall that, writing π1 (x, y) = x and π2 (x, y) = y for (x, y) ∈ X × Y , π1 [[F]] = {A : A ⊆ X, π1−1 [A] ∈ F} = {A : A ⊆ X, A × Y ∈ F} (2A1Ib), and that F → (x, y) iff π1 [[F]] → x and π2 [[F]] → y (3A3Ic). (a) Suppose that R is usco-compact and that F is an ultrafilter on X × Y , containing R, such that π1 [[F]] has a limit x ∈ X. ?? If F has no limit in R, then, in particular, it does not converge to (x, y) for any y ∈ R[{x}]; that is, π2 [[F]] does not converge to any point of R[{x}], that is, every point of R[{x}] belongs to an open set not belonging to π2 [[F]]. Because R[{x}] is compact, it is covered by a finite union of open sets not belonging to π2 [[F]]; but as π2 [[F]] is an ultrafilter (2A1N), there is an open set H ⊇ R[{x}] such that Y \ H ∈ π2 [[F]]. Now 422B tells us that there is an open set G containing x such that R[G] ⊆ H. In this case, G ∈ π1 [[F]] so G × Y ∈ F; at the same time, X × (Y \ H) ∈ F. So

422D

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R ∩ (G × Y ) ∩ (X × (Y \ H)) ∈ F. But this is empty, by the choice of G; which is intolerable. X X Thus F has a limit in R, as required. (b) Now suppose that R has the property described. (i) Let x ∈ X, and suppose that G is an ultrafilter on Y containing R[{x}]. Set h(y) = (x, y) for y ∈ Y ; then F = h[[G]] is an ultrafilter on X × Y containing R. The image π1 [[F]] is just the principal filter generated by {x}, so certainly converges to x; accordingly F must converge to some point (x, y) ∈ R, and G = π2 [[F]] converges to y ∈ R[{x}]. As G is arbitrary, R[{x}] is compact (2A3R). (ii) Let F ⊆ Y be closed, and x ∈ R−1 [F ] ⊆ X. Consider E = {R, X × F } ∪ {G × Y : G ⊆ X is open, x ∈ G}. Then E has the finite intersection property. P P If G0 , . . . , GTn are open sets containing x, then R−1 [F ] meets G0 ∩ . . . ∩ Gn in z say, and now (z, y) ∈ R ∩ (X × F ) ∩ i≤n (Gi × Y ) for some y ∈ F . Q Q Let F be an ultrafilter on X × Y including E (4A1Ia). Because G × Y ∈ E ⊆ F for every open set G containing x, π1 [[F]] → x, so F converges to some point (x, y) of R. Because X × F is a closed set belonging to E ⊆ F, y ∈ F and x ∈ R−1 [F ]. As x is arbitrary, R−1 [F ] is closed; as F is arbitrary, R satisfies condition (β) of 422A, and is usco-compact. 422D Lemma (a) Let X and Y be Hausdorff spaces. If R ⊆ X × Y is an usco-compact relation, then R is closed in X × Y . (b) Let X and Y be Hausdorff spaces. If R ⊆ X × Y is an usco-compact relation and R0 ⊆ R is a closed set, then R0 is usco-compact. (c) Let X and Y be Hausdorff spaces. If f : X → Y is a continuous function, then its graph is an usco-compact relation. (d) Let hXi ii∈I and of Hausdorff spaces, and Ri ⊆ Xi × Yi an usco-compact relation Q Q hYi ii∈I be families for each i. Set X = i∈I Xi , Y = i∈I Yi and R = {(x, y) : x ∈ X, y ∈ Y, (x(i), y(i)) ∈ Ri for every i ∈ I}. Then R is usco-compact in X × Y . (e) Let X and Y be Hausdorff spaces, and R ⊆ X × Y an usco-compact relation. Then (i) R[K] is a compact subset of Y for any compact subset K of X (ii) R[L] is a Lindelöf subset of Y for any Lindelöf subset L of X. (f) Let X, Y and Z be Hausdorff spaces, and R ⊆ X × Y , S ⊆ Y × Z usco-compact relations. Then the composition S ◦ R = {(x, z) : there is some y ∈ Y such that (x, y) ∈ R and (y, z) ∈ S} is usco-compact in X × Z. (g) Let X and Y be Hausdorff spaces and Y0 any subset of Y . Then a relation R ⊆ X ×Y0 is usco-compact when regarded as a relation between X and Y0 iff it is usco-compact when regarded as a relation between X and Y . proof (a) If (x, y) ∈ R, there is an ultrafilter F containing R and converging to (x, y) (4A2Bc). By 422C, F must have a limit in R; but as X × Y is Hausdorff, this limit must be (x, y), and (x, y) ∈ R. As (x, y) is arbitrary, R is closed. (b) It is obvious that R0 will satisfy the condition of 422C if R does. (c) f [{x}] = {f (x)} is surely compact for every x ∈ X, and f −1 [F ] is closed for every closed set F ⊆ Y because f is continuous. (d) For i ∈ I, x ∈ X, y ∈ Y set φi (x, y) = (x(i), y(i)). If F is an ultrafilter on X × Y containing R such that π1 [[F]] has a limit in X, then π1 φi [[F]] = ψi π1 [[F]]

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has a limit in Xi for every i ∈ I, writing ψi (x) = x(i) for i ∈ I, x ∈ X. But φi [[F]] contains Ri , so has a limit (x0 (i), y0 (i)) in Xi × Yi , for each i. Accordingly (x0 , y0 ) is a limit of F in X × Y (3A3Ic). As F is arbitrary, R is usco-compact. (e) For the moment, let L be any subset of X. Let H be a family of open sets in Y covering R[L]. Let G be the family of those open sets G ⊆ X such that R[G] can be covered S by finitely many members of H. Then G covers L. P P If x ∈ L, then R[{x}] is a compact subset of R[L] ⊆ H,Sso there is a finite set H0 ⊆ H covering R[{x}]. Now there is an open set G containing x such that R[G] ⊆ H0 , by 422B. Q Q S (i) If L is compact, then there must be a finite subfamily G 0 of G covering L; now R[L] ⊆ R[ G 0 ] is covered by finitely many members of H. As H is arbitrary, R[L] is compact. S (ii) If L is Lindelöf, then there must be a countable subfamily G 0 of G covering L; now R[L] ⊆ R[ G 0 ] is covered by countably many members of H. As H is arbitrary, R[L] is Lindelöf. (f ) If x ∈ X then R[{x}] ⊆ Y is compact, so (SR)[{x}] = S[R[{x}] is compact, by (e-i). If F ⊆ Z is closed then S −1 [F ] ⊆ Y is closed so (SR)−1 [F ] = R−1 [S −1 [F ]] is closed. (g)(i) Suppose that R is usco-compact when regarded as a subset of X × Y0 . Set S = {(y, y) : y ∈ Y0 }; by (c), S is usco-compact when regarded as a subset of Y0 × Y , so by (f) R = SR is usco-compact when regarded as a subset of X × Y . (ii) If R is usco-compact when regarded as a subset of X × Y , and x ∈ X, then R[{x}] is a subset of Y0 which is compact for the topology of Y , therefore for the subspace topology of Y0 . If F ⊆ Y0 is closed for the subspace topology, it is of the form F 0 ∩ Y0 for some closed F 0 ⊆ Y , so R−1 [F ] = R−1 [F 0 ] is closed in X. As x and F are arbitrary, R is usco-compact in X × Y0 . 422E The following lemma is actually very important in the structure theory of K-analytic spaces (see 422Yb). It will be useful to us in 423C below. Lemma Let X and Y be Hausdorff spaces, and R ⊆ X × Y an usco-compact relation. If X is regular, so is R (in its subspace topology). proof ?? Suppose, if possible, otherwise; that there are a closed set F ⊆ R and an (x, y) ∈ R \ F which cannot be separated from F by open sets (in R). If G, H are open sets containing x, y respectively, then R ∩ (G × H), R \ (G × H) are disjoint relatively open sets in R, so the latter cannot include F ; that is, F ∩ (G × H) 6= ∅ whenever G, H are open and x ∈ G, y ∈ H. Accordingly there is an ultrafilter F on X × Y such that F ∩ (G × H) ∈ F whenever G ⊆ X, H ⊆ Y are open sets containing x, y respectively. In this case R ∈ F, and G ∈ π1 [[F]] for every open set G containing x. Because the topology of X is regular, every open set containing x includes G for some smaller open set G containing x, and belongs to π1 [[F]]; thus π1 [[F]] → x in X. Because R is usco-compact, F has a limit in R, which must be of the form (x, y 0 ). Because F ∈ F is closed (in R), (x, y 0 ) ∈ F . But also y 0 ∈ H for every open set H containing y, since X × H is a closed set belonging to F; because the topology of Y is Hausdorff, y 0 must be equal to y, and (x, y) ∈ F , which is absurd. X X 422F Definition (Frol´ık 61) Let X be a Hausdorff space. Then X is K-analytic if there is an usco-compact relation R ⊆ N N × X such that R[N N ] = X. If X is a Hausdorff space, we call a subset of X K-analytic if it is a K-analytic space in its subspace topology. 422G Theorem (a) Let X be a Hausdorff space. Then a subset A of X is K-analytic iff there is an usco-compact relation R ⊆ N N × X such that R[N N ] = A. (b) N N is K-analytic. (c) Compact Hausdorff spaces are K-analytic. (d) If X and Y are Hausdorff spaces and R ⊆ X × Y is an usco-compact relation, then R[A] is K-analytic whenever A ⊆ X is K-analytic. In particular, a Hausdorff continuous image of a K-analytic Hausdorff space is K-analytic. (e) A product of countably many K-analytic Hausdorff spaces is K-analytic.

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(f) A closed subset of a K-analytic Hausdorff space is K-analytic. (g) A K-analytic Hausdorff space is Lindelöf, so a regular K-analytic Hausdorff space is completely regular. proof (a) A is K-analytic iff there is an usco-compact relation R ⊆ N N × A with projection A. But a subset of N N × X with projection A is usco-compact in N N × A iff it is usco-compact in N N × X, by 422Dg. (b) The identity function from N N to itself is an usco-compact relation, by 422Dc. (c) If X is compact, then R = N N × X is an usco-compact relation (because R[{φ}] = X is compact for every φ ∈ N N , while R−1 [F ] is either N N or ∅ for every closed F ⊆ X), so X = R[N N ] is K-analytic. (d) By (a), there is an usco-compact relation S ⊆ N N × X such that S[N N ] = A. Now R ◦ S ⊆ N N × Y is usco-compact, by 422Df, and RS[N N ] = R[A]. In particular, if X itself is K-analytic and f : X → Y is a continuous surjection, f is an usco-compact relation (422Dc), so Y = f [X] is K-analytic. (e) Let hXi ii∈I be a countable family of K-analytic Hausdorff spaces with product X. If I = ∅ then X = {∅} is compact, therefore K-analytic. Otherwise, choose for each i ∈ I an usco-compact relation Ri ⊆ N N × Xi such that Ri [N N ] = Xi . Set R = {(φ, x) : φ ∈ (N N )I , x ∈ X, (φ(i), x(i)) ∈ Ri for every i ∈ I}. By 422Dd, R is an usco-compact relation in (N N )I × X, and it is easy to see that R[(N N )I ] = X. But (N N )I ∼ = N N×I is homeomorphic to N N , because I is countable, so we can identify R with a relation in N N × X which is still usco-compact, and X is K-analytic. (f ) Let X be a K-analytic Hausdorff space and F a closed subset. Let R ⊆ N N × X be an usco-compact relation such that R[N N ] = X. Set R0 = R ∩ (N N × F ). Then R0 is a closed subset of R, so is usco-compact (422Da). By (a), F = R0 [N N ] is K-analytic. (g) Let X be a K-analytic Hausdorff space. N N is Lindelöf (4A2Ub), and there is an usco-compact relation R such that R[N N ] = X, so that X is Lindelöf, by 422D(e-ii). 4A2H(b-i) now tells us that if X is regular it is completely regular. 422H Theorem (a) If X is a Hausdorff space, then any K-analytic subset of X is Souslin-F in X. (b) If X is a K-analytic Hausdorff space, then a subset of X is K-analytic iff it is Souslin-F in X. (c) For any Hausdorff space X, the family of K-analytic subsets of X is closed under Souslin’s operation. proof (a) If A ⊆ X is K-analytic, there is an usco-compact relation R ⊆ N N × X such that R[N N ] = A, by 422Ga. By 422Da, R is a closed set; so A is Souslin-F by 421J. (b) Now suppose that X itself is K-analytic, and that A ⊆ X is Souslin-F in X. Then there is a closed set R ⊆ N N × X such that R[N N ] = A (421J, in the other direction). N N × X is K-analytic (422Gb, 422Ge), and R is closed, therefore itself K-analytic (422Gf); so its continuous image A is K-analytic, by 422Gd. (c)(i) The first step is to show that the union of a sequence of K-analytic subsets of X is K-analytic. P P Let hAn in∈N be a sequence of K-analytic sets, with union A. For each n ∈ N, let Rn ⊆ N N × X be an usco-compact relation such that Rn [N N ] = An . In (N × N N ) × X let R be the set {((n, φ), x) : n ∈ N, (φ, x) ∈ Rn }. N

If (n, φ) ∈ N × N , then R[{(n, φ)}] = Rn [{φ}] is compact; if F ⊆ X is closed, then R−1 [F ] = {(n, φ) : n ∈ N, φ ∈ Rn−1 [F ]} is closed in N × N N . So R is usco-compact, and of course S R[N × N N ] = n∈N Rn [N N ] = A. As N × N N is K-analytic (in fact, homeomorphic to N N ), A is K-analytic. Q Q (ii) S Now suppose that hAσ iσ∈S is a Souslin scheme consisting of K-analytic sets with kernel A. Then X 0 = σ∈S Aσ is K-analytic, by (i). By (a), every Aσ is Souslin-F when regarded as a subset of X 0 . But since the family of Souslin-F subsets of X 0 is closed under Souslin’s operation, by 421D, A is also Souslin-F in X 0 . By (b) of this theorem, A is K-analytic. As hAσ iσ∈S is arbitrary, we have the result.

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422I It seems that for the measure-theoretic results of §432, at least, the following result (the ‘First Separation Theorem’) is not essential. However I do not think it possible to get a firm grasp on K-analytic and analytic spaces without knowing some version of it, so I present it here. It is most often used through the forms in 422J and 422Xd below. S T Lemma Let X be a Hausdorff space. Let E be a family of subsets of X such that (i) n∈N En , n∈N En ∈ E whenever hEn in∈N is a sequence in ∈ E (ii) whenever x, y are distinct points of X, there are disjoint E, F ∈ E such that x ∈ int E and y ∈ int F . Then whenever A, B are disjoint non-empty K-analytic subsets of X, there are disjoint E, F ∈ E such that A ⊆ E and B ⊆ F . proof (a) We need to know that if K, L are disjoint non-empty compact subsets of X, there are disjoint E, F ∈ E such that K ⊆ int E and L ⊆ int F . P P For any point (x, y) ∈ K × L, we can find disjoint Exy , Fxy ∈ E such that x ∈ int E and y ∈ int F . Because L is compact and S T non-empty, there S is for each x ∈ K a non-empty finite set Ix ⊆ L such that L ⊆ y∈Ix int Fxy . Set Ex = y∈Ix Exy , Fx = y∈Ix Fxy ; then Ex , Fx are disjoint members of E, x ∈ int Ex and K is compact and S L ⊆ int Fx . Because S T not empty, there is a non-empty finite set J ⊆ K such that K ⊆ x∈J int Ex . Set E = x∈J Ex , F = x∈J Fx ; then E, F ∈ E, E ∩ F = ∅, K ⊆ int E and L ⊆ int F , as required. Q Q (b) LetSQ, R ⊆ N N × X be usco-compact relations such that Q[N N ] = A and R[N N ] = B. For each σ ∈ S ∗ = n∈N N n , set Iσ = {φ : σ ⊆ φ ∈ N N }, S so that A = A∅ and Aσ = i∈N Aσa i for every σ.

Aσ = Q[Iσ ],

Bσ = R[Iσ ],

(c) Write T for the set of pairs {(σ, τ ) : σ, τ ∈ S ∗ and there are disjoint E, F ∈ E such that Aσ ⊆ E, Bτ ⊆ F }. If σ, τ ∈ S ∗ are such that (σ a i, τ a j) ∈ T for every i, j ∈ N, then (σ, τ ) ∈ T . P P For each i, j ∈ N take disjoint Eij , Fij ∈ E such that S

Then E = i∈N (σ, τ ) ∈ T . Q Q

T j∈N

Eij , F =

T

Aσa i ⊆ Eij , Bτ a j ⊆ Fij . S i∈N j∈N Fij are disjoint and belong to E, and Aσ ⊆ E, Bτ ⊆ F . So

(d) ?? Now suppose, if possible, that there are no disjoint E, F ∈ E such that A ⊆ E and B ⊆ F ; that is, that (∅, ∅) ∈ / T . By (c), used repeatedly, we can find sequences hφ(i)ii∈N , hψ(i)ii∈N such that (φ¹n, ψ¹n) ∈ /T for every n ∈ N. Set K = Q[{φ}], L = R[{ψ}]. These are compact (because R is usco-compact) and disjoint (because K ⊆ A and L ⊆ B). By (a), there are disjoint E, F ∈ E such that K ⊆ int E and L ⊆ int F . By 422B, there are open sets U , V ⊆ N N such that φ ∈ U,

Q[U ] ⊆ int E,

ψ ∈V,

R[V ] ⊆ int F .

But now there is some n ∈ N such that Iφ¹n ⊆ U and Iψ¹n ⊆ V , in which case Aφ¹n ⊆ E,

Bψ¹n ⊆ F ,

and (φ¹n, ψ¹n) ∈ T , contrary to the choice of φ and ψ. X X This contradiction shows that the lemma is true. 422J Corollary Let X be a Hausdorff space and A, B disjoint K-analytic subsets of X. Then there is a Borel set which includes A and is disjoint from B. proof Apply 422I with E the Borel σ-algebra of X. *422K I give the next step in the theory of ‘constituents’ begun in 412O-412R. Theorem Let X be a Hausdorff space. (i) Suppose that X is regular. Let A ⊆ X be a K-analytic set. Then there is a non-decreasing family hEξ iξ<ω1 of Borel sets in X, with union X \ A, such that every Souslin-F subset of X disjoint from A is included in some Eξ .

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(ii) Suppose that X is regular. Let A ⊆ X be a Souslin-F set. Then there is a non-decreasing family hEξ iξ<ω1 of Borel sets in X, with union X \ A, such that every K-analytic subset of X \ A is included in some Eξ . (iii) Let A ⊆ X be a K-analytic set. Then there is a non-decreasing family hEξ iξ<ω1 of Borel sets in X, with union X \ A, such that every K-analytic subset of X \ A is included in some Eξ . proof (a) The first two parts depend on the following fact: if X is regular, R ⊆ N N × X is usco-compact, hFσ iσ∈S is a Souslin scheme consisting of closed sets with kernel B, and R[N N ] ∩ B = ∅, then for any φ, T ψ ∈ N N there is an n ≥ 1 such that R[Iφ¹n ] ∩ 1≤i≤n Fψ¹i is empty, where I write Iσ = {θ : σ ⊆ θ ∈ N N } for S T σ ∈ S = n≥1 Nn . P P We know that K = R[{φ}] is a compact set disjoint from the closed set n≥1 Fψ¹n . T So there is some m ≥ 1 such that K ∩ F = ∅ where F = 1≤i≤m Fψ¹i . Because X is regular, there are disjoint open sets G, H ⊆ X such that K ⊆ G and F ⊆ H (4A2F(h-ii)). Now R−1 [X \ G] is a closed set not containing φ, so there is some n such that R[Iφ¹n ] ⊆ G. Of course we can take n ≥ m, and in this case T R[Iφ¹n ] ∩ 1≤i≤n Fψ¹i ⊆ G ∩ F = ∅, as required. Q Q (b)(i) Suppose that A ⊆ X is K-analytic. Then there is an usco-compact set R ⊆ N N × X such that R[N N ] = A, and R is closed (422Da), so that A is the kernel of the Souslin scheme hR[Iσ ]iσ∈S (421I). For x ∈ X set Tx = {σ : σ ∈ S, x ∈ R[Iσ ]}, as in 421Ng, and let r(Tx ) < ω1 be the rank of the tree Tx . Then Eξ = {x : x ∈ X \ A, r(Tx ) ≤ ξ} is a Borel set for every ξ < ω1 , by 421Ob. Now suppose that B ⊆ X \ A is a Souslin-F set. Then it is the kernel of a Souslin scheme hFσ iσ∈S consisting of closed sets. If φ, ψ ∈ N N T then by (a) above there is an n ≥ 1 such that R[Iφ¹n ] ∩ 1≤i≤n Fψ¹i is empty. By 421Q, there must be some ξ < ω1 such that B ⊆ Eξ . So hEξ iξ<ω1 is a suitable family. (ii) The other part is almost the same. Suppose that A ⊆ X is Souslin-F. Then it is the kernel of a Souslin scheme hFσ iσ∈S consisting of closed sets. For x ∈ X set T S Tx = n≥1 {σ : σ ∈ Nn , x ∈ 1≤i≤n Fσ¹i }, and let r(Tx ) < ω1 be the rank of the tree Tx . Then Eξ = {x : x ∈ X \ A, r(Tx ) ≤ ξ} is a Borel set for every ξ < ω1 . Now let B ⊆ X \ A be a K-analytic set. There is an usco-compact set R ⊆ N N × X such that R[N N ] = B, and B is the kernel of the Souslin scheme hR[Iσ ]iσ∈S . If φ, ψ ∈ N N then by (a) above there is T an n ≥ 1 such that 1≤i≤n Fψ¹i ∩ R[Iφ¹n ] is empty. So there must be some ξ < ω1 such that B ⊆ Eξ . Thus here again hEξ iξ<ω1 is a suitable family. (c) If X is not regular, we still have a version of the result in (a), as follows: if R, S ⊆ N N × X and R[N N ] ∩ S[N N ] = ∅, then for any φ, ψ ∈ N N there is an n ≥ 1 such that R[Iφ¹n] ∩ S[Iψ¹n ] is empty. P P This time, R[{φ}] and S[{ψ}] are disjoint compact sets, so there are disjoint open sets G, H with R[{φ}] ⊆ G and S[{ψ}] ⊆ H (4A2F(h-i)). Now R[Iφ¹n ] ⊆ G and S[Iψ¹n ] ⊆ H for all n large enough. Q Q Now the argument of (b-i), with Fσ = S[Iσ ], gives part (iii). 422X Basic exercises (a) Let X and Y be Hausdorff spaces and X0 a closed subset of X. Show that a relation R ⊆ X0 × Y is usco-compact when regarded as a relation between X0 and Y iff it is usco-compact when regarded as a relation between X and Y . (b) Show that a locally compact Hausdorff space is K-analytic iff it is Lindelöf iff it is σ-compact. > (c) Prove 422Hc from first principles, without using 421D. (Hint: if hRσ iσ∈S is a Souslin scheme of usco-compact relations in N N × X, {((φ, hψσ iσ∈S ), x) : (ψφ¹n , x) ∈ Rφ¹n for every n} is usco-compact in (N N × (N N )S ) × X.) (d) Let X be a completely regular Hausdorff space and A, B disjoint K-analytic subsets of X. Show that there is a Baire set including A and disjoint from B.

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(e) Let X be a Hausdorff K-analytic space. (i) Show that every Baire subset of X is K-analytic. (Hint: 136G.) (ii) Show that if X is regular, it is perfectly normal iff it is hereditarily Lindelöf iff every open subset of X is K-analytic. (f ) Let X be a Hausdorff space in which every open set is K-analytic. Show that every Borel set is K-analytic. (Hint: apply 136Xi to the family of K-analytic subsets of X.) ˇ 422Y Further exercises (a) Let X be a completely regular Hausdorff space, and βX its Stone-Cech compactification. Show that X is K-analytic iff it is a Souslin-F set in βX. (b) Show that a Hausdorff space is K-analytic iff T S it is a continuous image of a Kσδ set in a compact Hausdorff space, that is, a set expressible as m∈N n∈N Kmn where every Kmn is compact. (Hint: Write K∗ for the class of Hausdorff continuous images of Kσδ subsets of compact Hausdorff spaces. (i) Show that N N is a Kσδ set in Y N , where Y is the one-point compactification of N. (ii) Show that if X is a compact Hausdorff space and R ⊆ N N × X is closed, then R ∈ K∗ . (iii) Show that if X is a compact Hausdorff space, then every Souslin-F subset of X belongs to K∗ . (iv) Show that if X is a regular K-analytic Hausdorff space, then X ∈ K∗ . (v) Show that if X is any Hausdorff space and R ⊆ N N × X is an usco-compact relation, then R ∈ K∗ . See Jayne 76.) (c) Let X be a normal space and C the family of countably compact closed subsets of X. Let A, B be disjoint sets obtainable from C by Souslin’s operation. (For instance, if X itself is countably compact, A and B could be disjoint Souslin-F sets.) Show that there is a Borel set including A and disjoint from B. (Hint: adapt the proof of 422I.) T (d) Let X be a Hausdorff space and hAn in∈N a sequence of K-analytic subsets of X such that Tn∈N An = ∅. Show that there is a sequence hEn in∈N of Borel sets such that An ⊆ En for every n ∈ N and n∈N En = ∅. N (Hint: for each n ∈ N choose an usco-compact T Rn ⊆ N × X with projection An . Consider the set T = {hσn in∈N : ∃ Borel En , Rn [Iσn ] ⊆ En ∀ n, n∈N En = ∅}.) (e) Explain how to prove 422J from 421Q, without using 422I. (f ) Let X be a set and S, T two Hausdorff topologies on X such that S ⊆ T and (X, T) is K-analytic. Show that S, T yield the same K-analytic subspaces of X. 422 Notes and comments In a sense, this section starts at the deep end of its topic. ‘Descriptive set theory’ originally developed in the context of the real line and associated spaces, and this remains the centre of the subject. But it turns out that some of the same arguments can be used in much more general contexts, and in particular greatly illuminate the theory of Radon measures on Hausdorff spaces. I find that a helpful way to look at K-analytic spaces is to regard them as a common generalization of compact Hausdorff spaces and Souslin-F subsets of R; if you like, any theorem which is true of both these classes has a fair chance of being true of all K-analytic spaces. In the next section we shall come to the special properties of the more restricted class of ‘analytic’ spaces, which are much closer to the separable metric spaces of the original theory. The phrase ‘usco-compact’ is neither elegant nor transparent, but is adequately established and (in view of the frequency with which it is needed) seems preferable to less concise alternatives. If we think of a relation R ⊆ X × Y as a function x 7→ R[{x}] from x to PY , then an usco-compact relation is one which takes compact values and is ‘upper semi-continuous’ in the sense that {x : R[{x}] ⊆ H} is open for every open set H ⊆ Y ; just as a real-valued function is upper semi-continuous if {x : f (x) < α} is open for every α. This is not supposed to be a book on general topology, and in my account of the topological properties of K-analytic spaces I have concentrated on facts which are useful when proving that spaces are K-analytic, on the assumption that these will be valuable when we seek to apply the results of §432 below. Other properties are mentioned only when they are relevant to the measure-theoretic results which are my real concern, and readers already acquainted with this area may be startled by some of my omissions. For a proper treatment of the subject, I refer you to Rogers 80. As usual, however, I take technical details

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seriously in the material I do cover. I hope you will not find that such results as 422Dg and 422Ga try your patience too far. I think a moment’s thought will persuade you that it is of the highest importance that K-analyticity (like compactness) is an intrinsic property. In contrast, the property of being ‘Souslin-F’, like the property of being closed, depends on the surrounding space. A completely regular Hausdorff space ˇ is compact iff it must be a closed set in any surrounding Hausdorff space iff it is closed in its Stone-Cech compactification; and it is K-analytic iff it must be a Souslin-F set in any surrounding Hausdorff space iff it ˇ is a Souslin-F set in its Stone-Cech compactification (422Ya). For regular spaces, 422K gives us another version of the First Separation Theorem. But this one is simultaneously more restricted in its scope (it does not seem to have applications to Baire σ-algebras, for instance) and very much more powerful in its application. When all Borel sets are Souslin-F, as in the next section, it tells us something very important about the cofinal structure of the Souslin-F subsets of the complement of a K-analytic set.

423 Analytic spaces We come now to the original class of K-analytic spaces, the ‘analytic’ spaces. I define these as continuous images of N N (423A), but move as quickly as possible to their characterization as K-analytic spaces with countable networks (423C), so that many other fundamental facts (423E-423G) can be regarded as simple corollaries of results in §422. I give two versions of Lusin’s theorem on injective images of Borel sets (423I), and a form of the von Neumann-Jankow measurable selection theorem (423N). I end with notes on constituents of coanalytic sets (423P-423Q). 423A Definition A Hausdorff space is analytic or Souslin if it is either empty or a continuous image of N N . 423B Proposition (a) A Polish space (definition: 4A2A) is analytic. (b) A Hausdorff continuous image of an analytic Hausdorff space is analytic. (c) A product of countably many analytic Hausdorff spaces is analytic. (d) A closed subset of an analytic Hausdorff space is analytic. (e) An analytic Hausdorff space has a countable network consisting of analytic sets. proof (a) Let X be a Polish space. If X = ∅, we canS stop. Otherwise, let ρ be a metric on X, inducing its topology, under which X is complete. For σ ∈ S ∗ = n∈N N n choose Xσ ⊆ X as follows. X∅ = X. Given that Xσ is a closed non-empty subset of X, where σ ∈ S ∗ , then Xσ is separable, because X is separable and metrizable (4A2P(a-iv)), and we can choose a sequence hxσi ii∈N in Xσ such that {xσi : i ∈ N} is dense in Xσ . Set Xσa i = Xσ ∩ B(xσi , 2−n ) for each i ∈SN, where B(x, δ) = {y : ρ(y, x) ≤ δ}, and continue. Note that because {xσi : i ∈ N} is dense in Xσ , Xσ = i∈N Xσa i , for every σ ∈ S ∗ . For each φ ∈ N N , hXφ¹n in∈N is a non-increasing sequence of non-empty closed sets, and diam(Xφ¹n+1 ) T ≤ 2−n+1 for every n. Because X is complete under ρ, n∈N Xφ¹n is a singleton {f (φ)} say. (f (φ) is the limit of the Cauchy sequence hxφ¹n,φ(n) in∈N .) Thus we have a function f : N N → X. f is continuous because ρ(f (ψ), f (φ)) ≤ 2−n+1 whenever φ¹n + 1 = ψ¹n + 1 (since in this case both f (ψ) and f (φ) belong to Xφ¹n+1 ). f is surjective because, given x ∈ X, we can S choose hφ(i)ii∈N inductively so that x ∈ Xφ¹n for every n; at the inductive step, we have x ∈ Xφ¹n = i∈N X(φ¹n)a i , so we can take φ(n) such that x ∈ X(φ¹n)a φ(n) = Xφ¹n+1 . Thus X is a continuous image of N N , as claimed. (b) If X is an analytic Hausdorff space and Y is a Hausdorff continuous image of X, then either X is a continuous image of N N and Y is a continuous image of N N , or X = ∅ and Y = ∅. (c) Let hXi ii∈I be a countable family of analytic Hausdorff spaces, with product X. Then X is Hausdorff (3A3Id). If I = ∅ then X = {∅} is a continuous image of N N , therefore analytic. If there is some i ∈ I such that Xi = ∅, then X = ∅ is analytic. Otherwise, we have for each i ∈ I a continuous surjection fi : N N → Xi . Setting f (φ) = hfi (φ(i))ii∈I for φ ∈ (N N )I , f : (N N )I → X is a continuous surjection. But (N N )I ∼ = N N×I is N homeomorphic to N , so X is analytic.

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(d) Let X be an analytic Hausdorff space and F a closed subset of X. Then F is Hausdorff in its subspace topology (4A2F(a-i)). If X = ∅ then F = ∅ is analytic. Otherwise, there is a continuous surjection f : N N → X. Now H = f −1 [F ] is a closed subset of the Polish space N N , therefore Polish in its induced topology (4A2Qd). By (a), H is analytic, so its continuous image F = f [H] is also analytic, by (b). (e) Let X be an analytic Hausdorff space. If it is empty then of course it has a countable network consisting of analytic sets. Otherwise, there is a continuous surjection f : N N → X. For σ ∈ S ∗ set Iσ = {φ : σ ⊆ φ ∈ N N }; then {Iσ : σ ∈ S ∗ } is a base for the topology of N N , so {f [Iσ ] : σ ∈ S ∗ } is a network for the topology of X (see the proof of 4A2Nd). But Iσ is homeomorphic to N N , so f [Iσ ] is analytic, for every σ ∈ S ∗ , and {f [Iσ ] : σ ∈ S ∗ } is a countable network consisting of analytic sets. 423C Theorem A Hausdorff space is analytic iff it is K-analytic and has a countable network. proof (a) Let X be an analytic Hausdorff space. By 423Be, it has a countable network. If X = ∅ then surely it is K-analytic. Otherwise, X is a continuous image of N N . But N N is K-analytic (422Gb), so X also is K-analytic, by 422Gd. (b) Now suppose that X is a K-analytic Hausdorff space and has a countable network. (i) If X ⊆ N N then X is analytic. P P Let R ⊆ N N ×X be an usco-compact relation such that R[N N ] = X. Then R is still usco-compact when regarded as a subset of N N ×N N (422Dg), so is closed in N N ×N N (422Da). But N N × N N ∼ = N N is analytic, so R is in itself an analytic space (423Bd), and its continuous image X is analytic, by 423Bb. Q Q (ii) Now suppose that X is regular. By 4A2Ng, X has a countable network E consisting of closed sets. Adding ∅ to E if need be, we may suppose that E = 6 ∅. Let hEn in∈N be a sequence running over E. For each n ∈ N, let hFni ii∈N be a sequence running over {En } ∪ {E : E ∈ E, E ∩ En = ∅}. Now consider the relation T R = {(φ, x) : φ ∈ N N , x ∈ n∈N Fn,φ(n) } ⊆ N N × X. α) R is closed in N N × X. P (α P Because every Fni is closed, S / Fni } (N N × X) \ R = i,n∈N {(φ, x): φ(n) = i and x ∈ is open. Q Q β ) R[N N ] = X. P (β P For every n ∈ N, X \ En = because E is a network and En is closed, so such that x ∈ Fn,φ(n) , and (φ, x) ∈ R. Q Q

S S

{E : E ∈ E, E ⊆ X \ En }

i∈N

Fni = X. So, given x ∈ X, we can find for each n a φ(n)

(γγ ) R is the graph of a function. P P?? Suppose that we have (φ, x) and (φ, y) in R where x 6= y. Because the topology of X is Hausdorff, there is an n ∈ N such that x ∈ En and y ∈ / En . But in this case x ∈ En ∩ Fn,φ(n) , so Fn,φ(n) = En , while y ∈ Fn,φ(n) \ En , so Fn,φ(n) 6= En ; which is absurd. X XQ Q (δδ ) Set A = R−1 [X], so that R is the graph of a function from A to X; in recognition of its new status, give it a new name f . Then f is continuous. P P Suppose that φ ∈ A and that x = f (φ) ∈ G, where G ⊆ X is open. Then there is an n ∈ N such that x ∈ En ⊆ G. In this case x ∈ Fn,φ(n) , because (φ, x) ∈ R, so Fn,φ(n) = En . Now if ψ ∈ A and ψ(n) = φ(n), we must have f (ψ) ∈ Fn,ψ(n) = En ⊆ G. Thus f −1 [G] includes a neighbourhood of φ in A. As φ and G are arbitrary, f is continuous. Q Q (²²) At this point recall that X is K-analytic. It follows that N N × X is K-analytic (422Ge), so that its closed subset R is K-analytic (422Gf) and A, which is a continuous image of R, is K-analytic (422Gd). But now A is a K-analytic subset of N N , so is analytic, by (i) just above. And, finally, X is a continuous image of A, so is analytic. (iii) Thus any regular K-analytic space with a countable network is analytic. Now suppose that X is an arbitrary K-analytic Hausdorff space with a countable network. Let R ⊆ N N × X be an usco-compact

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relation such that R[N N ] = X. Then R is a closed subset of N N × X, so is itself a K-analytic space with a countable network. But it is also regular, by 422E. So R is analytic and its continuous image X is analytic. This completes the proof. 423D Corollary (a) Any analytic Hausdorff space is hereditarily Lindelöf. (b) In a regular analytic Hausdorff space, closed subsets are zero sets and the Baire and Borel σ-algebras coincide. (c) A compact subset of an analytic Hausdorff space is metrizable. (d) A metrizable space is analytic iff it is K-analytic. proof (a)-(c) These are true just because there is a countable network (4A2Nb, 4A2Na, 4A2Nh, 4A3Kb). (d) Let X be a metrizable space. If X is analytic, of course it is K-analytic. If X is K-analytic, it is Lindelöf (422Gg) therefore separable (4A2Pd) and has a countable network (4A2P(a-iii)), so is analytic. 423E Theorem (a) For any Hausdorff space X, the family of subsets of X which are analytic in their subspace topologies is closed under Souslin’s operation. (b) Let (X, T) be an analytic Hausdorff space. For a subset A of X, the following are equiveridical: (i) A is analytic; (ii) A is K-analytic; (iii) A is Souslin-F; (iv) A can be obtained by Souslin’s operation from the family of Borel subsets of X. In particular, all Borel sets in X are analytic. proof (a) Let X be a Hausdorff space and A the family of analytic subsets of X. Let hAσ iσ∈S be a Souslin scheme in A with kernel A. Then every Aσ is K-analytic, so A is K-analytic, by 422Hc. Also every Aσ has a S countable network, so A0 = σ∈S Aσ has a countable network (4A2Nc); as A ⊆ A0 , A also has a countable network (4A2Na) and is analytic. (b) Because X has a countable network, so does A. So 423C tells us at once that (i) ⇐⇒ (ii). In particular, X is K-analytic, so 422Hb tells us that (ii) ⇐⇒ (iii). Of course (iii)⇒(iv). Now suppose that G ⊆ X is open. Then G ∈ A. P P If X = ∅ then G = ∅ is open. Otherwise, there is a continuous surjection f : N N → X. Set H = f −1 [G], so that H ⊆ N N is open and G = f [H]. Being an open set in a metric space, H is Fσ (4A2Lc), so, in particular, is Souslin-F; but N N is analytic, so H is analytic and its continuous image G is analytic. Q Q We have already seen that closed subsets of X belong to A (423Bd). Because A is closed under Souslin’s operation, it contains every Borel set, by 421F. It therefore contains every set obtainable by Souslin’s operation from Borel sets, and (i)⇒(i). Remark See also 423Yb below. 423F Proposition Let (X, T) be an analytic Hausdorff space. (a) A set E ⊆ X is Borel iff both E and X \ E are analytic. (b) If S is a coarser(= smaller) Hausdorff topology on X, then S and T have the same Borel sets. proof (a) If E is Borel, then E and X \ E are analytic, by 423Eb. If E and X \ E are analytic, they are K-analytic (423Eb) and disjoint, so there is a Borel set F ⊇ E which is disjoint from X \ E (422J); but now of course F = E, so E must be Borel. (b) Because the identity map from (X, T) to (X, S) is continuous, S is an analytic topology (423Bb) and every S-Borel set is T-Borel. If E ⊆ X is T-Borel, then it and its complement are T-analytic, therefore S-analytic (423Bb), and E is S-Borel by (a). 423G Lemma Let X and Y be analytic Hausdorff spaces and f : X → Y a Borel measurable function. (a) (The graph of) f is an analytic set. (b) f [A] is an analytic set in Y for any analytic set (in particular, any Borel set) A ⊆ X. (c) f −1 [B] is an analytic set in X for any analytic set (in particular, any Borel set) B ⊆ Y .

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proof (a) Let E be a countable network for the topology of Y . Set T R = E∈E (X × E) ∪ ((X \ f −1 [E]) × Y ). Then R is a Borel set in X × Y . But also R is the graph of f . P P If f (x) = y, then surely y ∈ E whenever x ∈ f −1 [E], so (x, y) ∈ R. On the other hand, if x ∈ X, y ∈ Y and f (x) 6= y, there are disjoint open sets G, H ⊆ Y such that f (x) ∈ G and y ∈ H; now there is an E ∈ E such that f (x) ∈ E ⊆ G, so that f (x) ∈ E but y ∈ / E, and (x, y) ∈ / R. Q Q Because X × Y is analytic (423Bc), R is analytic (423Eb). (b) If A ⊆ X is analytic, then A × Y and R ∩ (A × Y ) are analytic (423Ea), so f [A] = R[A], which is a continuous image of R ∩ (A × Y ), is analytic. (c) Similarly, if B ⊆ Y is analytic, then f −1 [B] is a continuous image of R ∩ (X × B), so is analytic. 423H Lemma Let (X, T) be an analytic Hausdorff space, and hEn in∈N any sequence of Borel sets in X. Then the topology T0 generated by T ∪ {En : n ∈ N} is analytic. proof If X = ∅ this is trivial. Otherwise, there is a continuous surjection f : N N → X. Set Fn = f −1 [En ] for each n; then Fn is a Borel subset of N N , so there is a Polish topology S0 on N N , finer than the usual topology, for which every Fn is open, by 4A3I. But now f is continuous for S0 and T0 , so T0 is analytic, by 423Ba and 423Bb. (Of course T0 is Hausdorff, because it is finer than T.) 423I Theorem Let X be a Polish space, E ⊆ X a Borel set, Y a Hausdorff space and f : E → Y an injective function. (a) If f is continuous, then f [E] is Borel. (b) If Y has a countable network (e.g., is an analytic space or a separable metrizable space), and f is Borel measurable, then f [E] is Borel. proof (a)(i) Since there is a finer Polish topology on X for which E is closed (4A3I), therefore Polish in the subspace topology (4A2Qd), and f will still be continuous for this topology, we may suppose that E = X. (ii) Let hUn in∈N run over a base for the topology of X (4A2P(a-i)). For each pair m, n ∈ N such that Um ∩ Un is empty, f [Um ] and f [Un ] are analytic sets in Y (423Eb, 423Bb) and are disjoint (because f is injective), so there is a Borel set Hmn including f [Um ] and disjoint from f [Un ] (422J). Set T En = f [Un ] ∩ {Hnm \ Hmn : m ∈ N, Um ∩ Un = ∅} for each n ∈ N; then En is a Borel set in Y including f [Un ]. Note that if Um ∩ Un is empty, then Em ∩ En ⊆ (Hmn \ Hnm ) ∩ (Hnm \ Hmn ) is also empty. Fix a metric ρ on X, inducing its topology, for which X is complete, and for k ∈ N set S Fk = {En : n ∈ N, diam Un ≤ 2−k }, T so that Fk is Borel. Let F = k∈N Fk ; then F is also a Borel subset of Y . The point is that F = f [X]. P P (i) If x ∈ X, then for every k ∈ N there is an n ∈ N such that x ∈ Un ⊆ {y : ρ(y, x) ≤ 2−k−1 }; now diam Un ≤ 2−k , so f (x) ∈ f [Un ] ⊆ En ⊆ Fk . As k is arbitrary, f (x) ∈ Fk ; as x is arbitrary, f [X] ⊆ F . (ii) If y ∈ F , then for each k ∈ N we can find an n(k) such that y ∈ En(k) and diam Un(k) ≤ 2−k . Since f [Un(k) ] ⊇ En(k) is not empty, nor is Un(k) , and we can choose xk ∈ Un(k) . Indeed, for any j, k ∈ N, En(j) ∩ En(k) contains y, so is not empty, and Un(j) ∩ Un(k) cannot be empty; but this means that there is some x in the intersection, and ρ(xj , xk ) ≤ ρ(xj , x) + ρ(x, xk ) ≤ diam Un(j) + diam Un(k) ≤ 2−j + 2−k . This means that hxk ik∈N is a Cauchy sequence. But X is supposed to be complete, so hxk ik∈N has a limit x say.

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159

?? If f (x) 6= y, then (because Y is Hausdorff) there is an open set H containing f (x) such that y ∈ / H. Now f is continuous, so there is a δ > 0 such that f (x0 ) ∈ H whenever ρ(x0 , x) ≤ δ. There is a k ∈ N such that 2−k + ρ(xk , x) ≤ δ. If x0 ∈ Un(k) , then ρ(x0 , x) ≤ ρ(x0 , xk ) + ρ(xk , x) ≤ δ; thus f [Un(k) ] ⊆ H, and En(k) ⊆ f [Un(k) ] ⊆ H. But y ∈ En(k) \ H. X X Thus y = f (x) belongs to f [X]; as y is arbitrary, F ⊆ f [X]. Q Q Accordingly f [X] = F is a Borel subset of Y , as claimed. (b) By 4A2Nf, there is a countable family V of open sets in Y such that whenever y, y 0 are distinct points of Y there are disjoint V , V 0 ∈ V such that y ∈ V and y 0 ∈ V 0 . Let S0 be the topology generated by V; then S0 is Hausdorff. For each V ∈ V, f −1 [V ] is a Borel set in X, so there is a Polish topology T0 on X, finer than the original topology, for which every f −1 [V ] is open (4A3I). Now f is continuous for T0 and S0 (4A2B(a-ii)), and E is T0 -Borel, so f [E] is a S0 -Borel set in Y , by (a). Since S0 is coarser than the original topology S on Y , f [X] is also S-Borel. 423J Lemma If X is an uncountable analytic Hausdorff space, it has subsets homeomorphic to {0, 1}N and N N . S proof (a) Let f : N N → X be a continuous surjection. Write S ∗ = n∈N N n , Iσ = {φ : σ ⊆ φ ∈ N N } for σ ∈ S∗, T = {σ : σ ∈ S ∗ , f [Iσ ] is uncountable}. Then if σ ∈ T there are τ , τ 0 ∈ T , both extending σ, such that f [Iτ ] ∩ f [Iτ 0 ] = ∅. P P Set S ∗ A = {f [Iτ ] : τ ∈ S \ T }. Then A is a countable union of countable sets, so is countable. There must therefore be distinct points x, y of f [Iσ ] \ A; express x as f (φ) and y as f (ψ) where φ and ψ belong to Iσ . Because X is Hausdorff, there are disjoint open sets G, H such that x ∈ G and y ∈ H. Because f is continuous, there are m, n ∈ N such that Iφ¹m ⊆ f −1 [G] and Iψ¹n ⊆ f −1 [H]. Of course both τ = φ¹m and τ 0 = ψ¹n must extend σ, and they belong to T because x ∈ f [Iτ ] \ A, y ∈ f [Iτ 0 ] \ A. Q Q S (b) We can therefore choose inductively a family hτ (υ)iυ∈S2∗ in T , where S2∗ = n∈N {0, 1}n , such that τ (∅) = ∅, τ (υ a i) ⊇ τ (υ) whenever υ ∈ S2∗ , i ∈ {0, 1}, f [Iτ (υa 0 )] ∩ f [Iτ (υa 1 ] = ∅ for every υ ∈ S2∗ . Note that #(τ (υ)) ≥ #(υ) for every υ ∈ S2 . For each z ∈ {0, 1}N , hτ (z¹n)in∈N is a sequence in S ∗ in which each term strictly extends its predecessor, so there is a unique g(z) ∈ N N such that τ (z¹n) ⊆ g(z) for every n. Now g(z 0 )¹n = g(z)¹n whenever z¹n = z 0 ¹n, so g and f g : {0, 1}N → X are continuous. If w, z are distinct points of {0, 1}N , there is a first n such that w(n) 6= z(n), in which case f g(w) ∈ f [Iτ (w¹n)a w(n) ] and f g(z) ∈ f [Iτ (w¹n)a z(n) ] are distinct. So f g : {0, 1}N → X is a continuous injection, therefore a homeomorphism between {0, 1}N and its image, because {0, 1}N is compact (3A3Dd). (c) Thus X has a subspace homeomorphic to {0, 1}N . Now {0, 1}N has a subspace homeomorphic to N. P P For instance, setting dn (n) = 1, dn (i) = 0 for i 6= n, D = {dn : n ∈ N} is homeomorphic to N. Q Q Now DN is homeomorphic to N N and is a subspace of ({0, 1}N )N ∼ = {0, 1}N×N ∼ = {0, 1}N , so {0, 1}N has a subspace homeomorphic to N N . Accordingly X also has a subspace homeomorphic to N N . 423K Corollary Any uncountable Borel set in any analytic space has cardinal c.

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423K

proof If X is an analytic space and E ⊆ X is an uncountable Borel set, then E is analytic (423E), so includes a copy of {0, 1}N and must have cardinal at least #({0, 1}N ) = c. On the other hand, E is also a continuous image of N N , so has cardinal at most #(N N ) = c. 423L Proposition Let X be an uncountable analytic Hausdorff space. Then it has a non-Borel analytic subset. proof (a) I show first that there is an analytic set A ⊆ N N × N N such that every analytic subset of N N is a vertical section of A. P P Let U be a countable base for the topology of N N × N N , containing ∅, and hUn in∈N an enumeration of U. Write S M = (N N × N N × N N ) \ m,n∈N ({x : x(m) = n} × Un ). Then M is a closed subset of (N N × NN ) × N N , therefore analytic (423Ba, 423Bd), so its continuous image A = {(x, z) : there is some y such that (x, y, z) ∈ M } is analytic (423Bb). Now let E be any analytic subset of N N . By 423E, E is Souslin-F; by 421J, there is a closed set F ⊆ N N × N N such that E = {z : ∃ y, (y, z) ∈ F }. Let hx(m)im∈N be a sequence running over {n : n ∈ N, Un ∩ F = ∅}, so that S F = (N N × N N ) \ m∈N Ux(m) = {(y, z) : (x, y, z) ∈ M }. Now {z : (x, z) ∈ A} = {z : there is some y such that (x, y, z) ∈ M } = {z : there is some y such that (y, z) ∈ F } = E, and E is a vertical section of A, as required. Q Q (b) It follows that there is a non-Borel analytic set B ⊆ N N . P P Take A from (a) above, and try B = {x : (x, x) ∈ A}. Because B is the inverse image of A under the continuous map x 7→ (x, x), it is analytic (423Gc). ?? If B were a Borel set, then B 0 = N N \ B would also be Borel, therefore analytic (423E), and there would be an x ∈ NN such that B 0 = {y : (x, y) ∈ A}. But in this case x ∈ B ⇐⇒ (x, x) ∈ A ⇐⇒ x ∈ B 0 , which is a difficulty you may have met before. X XQ Q (c) Now return to our arbitrary uncountable analytic Hausdorff space X. By 423J, X has a subset Z homeomorphic to N N . By (b), Z has an analytic subset A which is not Borel in Z, therefore cannot be a Borel subset of X. 423M I devote a few paragraphs to an important method of constructing selectors. Theorem Let X be an analytic Hausdorff space, Y a set, and C ⊆ PY . Write T for the σ-algebra of subsets of Y generated by S(C), where S is Souslin’s operation, and V for S({F × C : F ⊆ X is closed, C ∈ C}). If W ∈ V, then W [X] ∈ T and there is a T-measurable function f : W [X] → X such that (f (y), y) ∈ W for every y ∈ W [X]. proof Write F for {F × C : F ⊆ X is closed, C ∈ C}. (a) Consider first the case in which X = N N and all the horizontal sections W −1 [{y}] of W are closed. S ∗ Let E be the family of closed subsets of Y . For σ ∈ S = n∈N N n set Iσ = {φ : σ ⊆ φ ∈ N N }. Then W ∩ (Iσ × Y ) ∈ V. P P Because Souslin’s operation is idempotent (421D), S(V) = V. The set {V : V ∩ (Iσ × Y ) ∈ V} is therefore closed under Souslin’s operation (apply 421Cc to the identity map from Iσ × Y to X × Y , or otherwise); since it includes F, it is the whole of V, and contains W . Q Q By 421G, W [Iσ ] = (W ∩(Iσ ×Y ))[N N ] belongs to S(C) ⊆ T for every σ. In particular, W [N N ] = W [I∅ ] ∈ T. Choose hYσ iσ∈S ∗ in T inductively, as follows. Y∅ = W [N N ]. Given that Yσ ∈ T and that Yσ ⊆ W [Iσ ], set

423N

Analytic spaces

Yσa j = Yσ ∩ W [Iσa j ] \

161

S i
W [Iσa i ]

for every j ∈ N. Continue. At the end of the induction, we have S S j∈N Yσ a j = Yσ ∩ j∈N W [Iσ a j ] = Yσ ∩ W [Iσ ] = Yσ for every σ ∈ S ∗ , while hYσa j ij∈N is always disjoint. So for each y ∈ Y∅ = W [N N ] we have a unique f (y) ∈ N N such that y ∈ Yf (y)¹n for every n. Since f −1 [Iσ ] = Yσ ∈ T for every σ ∈ S ∗ , f is T-measurable (4A2H(c-ii)). Also (f (y), y) ∈ W for every y ∈ W [N N ]. P P For each n ∈ N, y ∈ Yf (y)¹n = W [If (y)¹n ], so there is an xn ∈ N N such that xn ¹n = f (y)¹n and (xn , y) ∈ W . But this means that f (y) = limn→∞ xn ; since we are supposing that the horizontal sections of W are closed, (f (y), y) ∈ W . Q Q Thus the theorem is true if X = N N and W has closed horizontal sections. (b) Now suppose that X = N N and that W ⊆ N N × Y is any set in V. Then there is a Souslin scheme hFσ × Cσ iσ∈S in F with kernel W ; of course I mean you to suppose that Fσ ⊆ N N is closed and Cσ ∈ C for every σ. Set S N ˜ =T W × NN × Y . k Iσ × Fσ × Cσ ⊆ N k≥1

σ∈N

˜ onto the last two coordinates, by 421Ce. If y ∈ Y , then Then W is the projection of W S ˜}=T {(φ, ψ) : (φ, ψ, y) ∈ W {Iσ × Fσ : σ ∈ N k , y ∈ Cσ } k≥1

N

N

k

is closed in N × N . (If J is any subset of N , then S S S (N N × N N ) \ σ∈J Iσ × Fσ = σ∈J Iσ × (N N \ Fσ ) ∪ σ∈N k \J Iσ × N N is open.) Also Iσ × Fσ is a closed subset of N N × N N for every σ, and N N × N N is homeomorphic to N N . We ˜ , regarded as a subset of (N N × N N ) × Y , to see that W [N N ] = W ˜ [N N × N N ] ∈ T can therefore apply (a) to W N N N ˜ for and that there is a T-measurable function h = (g, f ) : W [N ] → N × N such that (g(y), f (y), y) ∈ W every y ∈ W [N N ]. Now, of course, f : W [N N ] → N N is T-measurable and (f (y), y) ∈ W for every y ∈ W [N N ]. (c) Finally, suppose only that X is an analytic Hausdorff space and that W ∈ V. If X is empty, so is Y , ˜ y) = (h(φ), y) for and the result is trivial. Otherwise, there is a continuous surjection h : N N → X. Set h(φ, ˜ : N N × Y → X × Y is a continuous surjection, and W ˜ −1 [W ] is the kernel of ˜ =h φ ∈ N N and y ∈ Y ; then h a Souslin scheme in ˜ −1 [F × C] : F ⊆ N N is closed, C ∈ C} = {h−1 [F ] × C : F ⊆ N N is closed, C ∈ C} {h ˜ [N N ] ∈ T and there is a T-measurable g : W [X] → N N by 421Cb. So we can apply (b) to see that W [X] = W ˜ for every y ∈ Y . Finally f = hg : W [X] → X is T-measurable and (f (y), y) ∈ W such that (g(y), y) ∈ W for every y ∈ Y . This completes the proof. 423N The expression V = S({F × C : F ⊆ X is closed, C ∈ C}) in 423M is a new formulation, and I had better describe one of the basic cases in which we can use the result. Corollary Let X be an analytic Hausdorff space and Y any topological space. Let T be the σ-algebra of subsets of Y generated by S(B(Y )), where B(Y ) is the Borel σ-algebra of Y . If W ∈ S(B(X × Y )), then W [X] ∈ T and there is a T-measurable function f : W [X] → X such that (f (y), y) ∈ W for every y ∈ W [X]. proof (a) Suppose to begin with that X = N N . In 423M, set C = B(Y ). S Then every open subset and every closed subset of X ×Y belongs to V as defined in 423M. P P For σ ∈ S ∗ = k∈N N k , set Iσ = {φ : σ ⊆ φ ∈ N N }. If V ⊆ X × Y is open, set S Hσ = {H : H ⊆ Y is open, Iσ × Hσ ⊆ V } S for each σ ∈ S ∗ . Because {Iσ : σ ∈ S ∗ } is a base for the topology of N N , V = σ∈S ∗ Iσ × Hσ ∈ V. As for the complement of V , we have

162


(N N × Y ) \ V =

\

423N

(N N × Y ) \ (Iσ × Hσ )

k∈N σ∈N k

=

\

k∈N σ∈N k

=

\

((N N \ Iσ ) × Y ) ∪ (N N × (Y \ Hσ )) [

(Iτ × Y ) ∪ (N N × (Y \ Hσ ))

k∈N τ ∈N k σ∈N k τ = 6 σ

again belongs to V, because V is closed under countable unions and intersections and contains Iτ × Y and N N × (Y \ Hσ ) for all σ, τ ∈ S ∗ . Q Q By 421F, V contains every Borel subset of N N × Y , so includes S(B(N N × Y )). So in this case we can apply 423M directly to get the result. (b) Now suppose that X is any analytic space. If X is empty, the result is trivial. Otherwise, let ˜ y) = (h(φ), y) for φ ∈ N N , y ∈ Y , so that h ˜ : N N ×Y → X×Y h : N N → X be a continuous surjection. Set h(φ, −1 −1 N ˜ ˜ ˜ ˜ ∈ S(B(N N ×Y )) is continuous. Set W = h [W ]. If V ∈ B(X ×Y ) then h [V ] ∈ B(N ×Y ) (4A3Cd), so W N N N ˜ ˜ ˜ (421Cc). By (a), W [N ] ∈ T and there is a T-measurable function g : W [N ] → N such that (g(y), y) ∈ W N N ˜ ˜ for every y ∈ W [N ]. It is now easy to check that W [X] = W [N ] ∈ T (this is where we need to know that h is surjective), that f = hg : W [X] → X is T-measurable, and that (f (y), y) ∈ W for every y ∈ W [X], as required. Remark This is a version of the von Neumann-Jankow selection theorem. 423O Corollary Let X and Y be analytic Hausdorff spaces, and f : X → Y a Borel measurable surjection. Let T be the σ-algebra of subsets of Y generated by the Souslin-F sets in Y . Then there is a T-measurable function g : Y → X such that f g is the identity on Y . proof The point is that the graph Γ = {(x, f (x)) : x ∈ X} of f is analytic (423Ga), therefore a Souslin-F set in X × Y (423Eb). Also Γ[X] = f [X] = Y . By 423N, there is a T-measurable function g : Y → X such that (g(y), y) ∈ Γ, that is, f (g(y)) = y, for every y ∈ Y . *423P Constituents of coanalytic sets: Theorem Let X be a Hausdorff space, and A ⊆ X an analytic subset of X. Then there is a non-decreasing family hEξ iξ<ω1 of Borel subsets of X, with union X \ A, such that every analytic subset of X \ A is included in some Eξ . proof Put 422K(iii) and 423C together. *423Q Remarks (a) Let A be an analytic set in an analytic space X and hEξ iξ<ω1 a family of Borel sets as in 423P. There is nothing unique about the Eξ . But if hEξ0 iξ<ω1 is another such family, then every Eξ0 is an analytic subset of X \ A, by 423E, so is included in some Eη ; and, similarly, every Eξ is included in some Eη0 . We therefore have a function f : ω1 → ω1 such that Eξ0 ⊆ Ef (ξ) and Eξ ⊆ Ef0 (ξ) for every ξ < ω. If = {ξ : ξ < ω1 , f (η) < ξ for every η < ξ}, then C is a closed cofinal set in ω1 (4A1Bd), and S we set C S 0 η<ξ Eη = η<ξ Eη for every ξ ∈ C. If X \ A is itself analytic, that is, if A is a Borel set, then we shall have to have X \ A = Eξ = Eξ0 for some ξ < ω1 . Another way of expressing the result in 423P is to say that if we write I = {B : B ⊆ X \ A is analytic}, then {E : E ∈ I, E is Borel} is cofinal with I (this is the First Separation Theorem) and cf I ≤ ω1 . (b) It is a remarkable fact that, in some models of set theory, we can have non-Borel coanalytic sets in Polish spaces such that all their constituents are countable (Jech 78, p. 529, Cor. 2). (Note that, by (a), this is the same thing as saying that X \ A is uncountable but all its Borel subsets are countable.) But in ‘ordinary’ cases we shall have, for every Borel subset E of X \ A, an uncountable Borel subset of (X \ A) \ E; so that for any family hGξ iξ<ω1 of Borel constituents of X \ A, there must be uncountably many uncountable Gξ . To see that this happens at least sometimes, take any non-Borel analytic subset A0 of N N (423L), and consider A = A0 × N N ⊆ (N N )2 . Then A is analytic (423B). If E ⊆ (N N )2 \ A is Borel, then

423Yc

Analytic spaces

163

π1 [E] = {x : (x, y) ∈ E} is an analytic subset of N N \ A, so is not the whole of N N \ A0 (by 423Fa). Taking any x ∈ (N N \ A0 ) \ π1 [E], {x} × N N is an uncountable Borel subset of ((N N )2 \ A) \ E. For an alternative construction, see 423Ye. 423X Basic exercises > (a) For a Hausdorff space X, show that the following are equiveridical: (i) X is analytic; (ii) X is a continuous image of a Polish space; (iii) X is a continuous image of a closed subset of N N . (b) Write out a direct proof of 423Ea, not quoting 423C or 421D. > (c) Let X be a set and S a Hausdorff topology on X, T an analytic topology on X such that S ⊆ T. Show that S and T have the same analytic sets. (Hint: 423F.) > (d) Let X be an analytic Hausdorff space. (i) Show that its Borel σ-algebra B(X) is countably generated as σ-algebra. (Hint: use 4A2Nf and 423Fb.) (ii) Show that there is an analytic subset Y of R such that (X, AX ) is isomorphic to (Y, AY ), where AX , AY are the families of Souslin-F subsets of X, Y respectively. (Hint: show that there is an injective Borel measurable function from X to R (cf. 343E), and use 423G.) (iii) Show that (X, B(X)) is isomorphic to (Y, B(Y )), where B(Y ) is the Borel σ-algebra of Y . (Hint: 423Fa.) (iv) Let TX , TY be the σ-algebras generated by AX , AY respectively. Show that (X, TX ) and (Y, TY ) are isomorphic. (e) Let S be the right-facing Sorgenfrey topology on R (415Xc). Show that S has the same Borel sets as the usual topology T on R. (Hint: S is hereditarily Lindelöf (419Xf) and has a base consisting of T-Borel sets.) Show that S is not analytic. (f ) Let X be an analytic Hausdorff space and Y any topological space. Let T be the σ-algebra of subsets of Y generated by the Souslin-F sets. Show that if W ⊆ X × Y is Souslin-F, then W [X] ∈ T and there is a T-measurable function f : W [X] → Y such that (f (y), y) ∈ W for every y ∈ W [X]. (Hint: start with X = N N , as in 423N.) (g) Let X and Y be Hausdorff spaces, and R ⊆ X × Y an analytic set such that R−1 [Y ] = X. Show that there is a function g : X → Y , measurable with respect to the σ-algebra generated by the Souslin-F subsets of X, such that (x, g(x)) ∈ R for every x ∈ X. > (h) (i) Show that the family of analytic subsets of [0, 1] has cardinal c. (Hint: 421Xd.) (ii) Show that the σ-algebra T of subsets of [0, 1] generated by the analytic sets has cardinal c. (Hint: 421Xi.) (iii) Show that there is a set A ⊆ [0, 1] which does not belong to T. (i) Let X = Y = [0, 1]. Give Y the usual topology, and give X the topology corresponding to the one-point compactification of the discrete topology on [0, 1[, that is, a set G ⊆ X is open if either 1 ∈ / G or G is cofinite. Show that the identity map f : X → Y is a Borel measurable bijection, but that f −1 is not measurable for the σ-algebra of subsets of Y generated by the Souslin-F sets. 423Y Further exercises (a) Show that a space with a countable network is hereditarily separable (that is, every subset is separable), therefore countably tight. (b) Show that if X is a Hausdorff space with a countable network, then every analytic subset of X is obtainable by Souslin’s operation from the open subsets of X. (c) Let (X, T) and (Y, S) be analytic Hausdorff spaces and f : X → Y a Borel measurable function. (i) Show that there is a zero-dimensional separable metrizable topology S0 on Y with the same Borel sets and the same analytic sets as S. (Hint: 423Xd.) (ii) Show that there is a zero-dimensional separable metrizable topology T0 on X, with the same Borel sets and the same analytic sets as T, such that f is continuous for the topologies T0 , S0 . (iii) Explain how to elaborate these ideas to deal with any countable family of analytic spaces and Borel measurable functions between them.

164


423Yd

(d) Let X be an analytic Hausdorff space, Y a Hausdorff space with a countable network, and f : X → Y a Borel measurable surjection. Let T be the σ-algebra of subsets of Y generated by the Souslin-F sets in Y . Show that there is a T-measurable function g : Y → X such that f g is the identity on Y . S (e) Set S = n≥1 N n and consider PS with its usual topology. Let T ⊆ PS be the set of trees (412O); show that T is closed, therefore a compact metrizable space. Set Fσ = {T : σ ∈ T ∈ T } for σ ∈ S, and let A be the kernel of the Souslin scheme hFσ iσ∈s . Show that the constituents of T \ A for this scheme are just the sets Gξ = {T : r(T ) = ξ}, where r is the rank function of 421Ne. Show by induction on ξ that all the Gξ is non-empty, so that A is not a Borel set. Show that #(Gξ ) = c for 1 ≤ ξ < ω1 . Show that if X is any topological space and B ⊆ X is a Souslin-F set, there is a Borel measurable function f : X → T such that B = f −1 [A]. 423 Notes and comments We have been dealing, in this section and the last, with three classes of topological space: the class of analytic spaces, the class of K-analytic spaces and the class of spaces with countable networks. The first is more important than the other two put together, and I am sure many people would find it more comfortable, if more time-consuming, to learn the theory of analytic spaces thoroughly first, before proceeding to the others. This was indeed my own route into the subject. But I think that the theory of K-analytic spaces has now matured to the point that it can stand on its own, without constant reference to its origin as an extension of descriptive set theory on the real line; and that our understanding of analytic spaces is usefully advanced by seeing how easily their properties can be deduced from the fact that they are K-analytic spaces with countable networks. ‘Selection theorems’ appear everywhere in mathematics. The axiom of choice is a selection theorem: if hAi ii∈I is a family of non-empty sets, there is a selector for the relation {(i, x) : i ∈ I, x ∈ Ai }. The Lifting Theorem (§341) is a special kind of selection theorem, where we seek a selector which is a Boolean homomorphism. In general topology we look for continuous selectors; in measure theory, for measurable selectors. Any selection theorem will have expressions either as a theorem on right inverses of functions, as in 423O, or as a theorem on selectors for relations, as in 423M-423N. In the language of this section, however, the strongest results are most easily set out in terms of relations, because the essence of the method is that we can find selectors taking values in analytic spaces, and the relations of 423M-423N can be very far from being analytic spaces in themselves, even when we have a natural topology on the space Y . The value of these results in measure theory will become clearer in §431, where we shall see that there are important σ-algebras which are closed under Souslin’s operation, so that they can include the algebras T of 423M-423O. Typical applications are in As in §422, I have made no attempt to cover the general theory of analytic spaces, nor even to give a balanced introduction. I have tried instead to give a condensed account of the principal methods for showing that spaces are analytic, with some of the ideas which can be applied to make them more accessible to the imagination (423J, 423Xc-423Xd, 423Yb-423Yc). Lusin’s theorem 423I does not mention ‘analytic’ sets in its statement, but it depends essentially on the separation theorem 422J, so cannot really be put with the other results on Polish spaces in 4A2Q. You must of course know that not all analytic sets are Borel (423L) and that not all sets are analytic (423Xh). For further information about this fascinating subject, see Rogers 80, Kechris 95 and Moschovakis 80. ‘Selection theorems’ appear everywhere in mathematics. The axiom of choice is a selection theorem; it says that whenever R ⊆ X × Y is a relation and R[X] = Y , there is a function f : Y → X such that (f (y), y) ∈ R for every y ∈ Y . The Lifting Theorem (§341) asks for a selector which is a Borel homomorphism. In general topology we look for continuous selectors. In measure theory, naturally, we are interested in measurable selectors, as in 423M-423O. Any selection theorem will have expressions either as a theorem on right inverses of functions, as in 423O, or as a theorem on selectors for relations, as in 423M-423N. In the language here, however, we get better theorems by examining relations, because the essence of the method is that we can find measurable functions into analytic spaces, and the relations of 423M can be very far from being analytic, even when there is a natural topology on the space Y . The value of these results will become clearer in §431, when we shall see that the σ-algebras T of 423M-423O are often included in familiar σ-algebras. Typical applications are in 433F-433G below.

424C

Standard Borel spaces

165

424 Standard Borel spaces This volume is concerned with topological measure spaces, and it will come as no surprise that the topological properties of Polish spaces are central to the theory. But even from the point of view of unadorned measure theory, not looking for topological structures on the underlying spaces, it turns out that the Borel algebras of Polish spaces have a very special position. It will be useful later on to be able to refer to some fundamental facts concerning them. 424A Definition Let X be a set and Σ a σ-algebra of subsets of X. We say that (X, Σ) is a standard Borel space if there is a Polish topology on X for which Σ is the algebra of Borel sets. Warning! Many authors reserve the phrase ‘standard Borel space’ for the case in which X is uncountable. I have seen the phrase ‘Borel space’ used for what I call a ‘standard Borel space’. 424B Proposition (a) If (X, Σ) is a standard Borel space, then Σ is countably generated as σ-algebra of sets. Q N (b) If h(Xi , Σi )ii∈I is a countable family of standard Borel spaces, then ( i∈I Xi , c i∈I Σi ) (definition: 254E) is a standard Borel space. (c) Let (X, Σ) and (Y, T) be standard Borel spaces and f : X → Y a (Σ, T)-measurable surjection. Then (i) if E ∈ Σ is such that f [E] ∩ f [X \ E] = ∅, then f [E] ∈ T; (ii) T = {F : F ⊆ Y, f −1 [F ] ∈ Σ}; (iii) if f is a bijection it is an isomorphism. (d) Let (X, Σ) and (Y, T) be standard Borel spaces and f : X → Y a (Σ, T)-measurable injection. Then Z = f [X] ∈ T and f is an isomorphism between (X, Σ) and (Z, TZ ), where TZ is the subspace σ-algebra. proof (a) Let T be a Polish topology on X such that Σ is the algebra of Borel sets. Then T has a countable base U , which generates Σ (4A3Da/4A3E). (b) For each i ∈ I let Ti be a Polish topology on Xi such that Σi is the algebra of Ti -Borel sets. Then N Q X = i∈I Ti , with the product topology T, is Polish (4A2Qc). By 4A3Dc/4A3E, Σ = c i∈I Σi is just the Borel σ-algebra of X, so (X, Σ) is a standard Borel space. (c) Let T, S be Polish topologies on X, Y respectively for which Σ and T are the Borel σ-algebras. Then f is Borel measurable. (i) By 423Gb and 423Eb, f [E] and f [X \ E] are analytic subsets of Y . But they are complementary, so they are Borel sets, by 423Fa. (ii) f −1 [F ] ∈ Σ for every F ∈ T, just because f is measurable. On the other hand, if F ⊆ Y and E = f −1 [F ] ∈ Σ, then F = f [E] ∈ T by (i). (iii) follows at once. (d) Give X and Y Polish topologies for which Σ, T are the Borel σ-algebras. By 423Ib, f [E] ∈ T for every E ∈ Σ; in particular, Z = f [X] belongs to T. Also f −1 [F ] ∈ Σ for every F ∈ TZ , so f is an isomorphism between (X, Σ) and (Z, TZ ). 424C Theorem Let (X, Σ) be a standard Borel space. (a) If X is countable then Σ = PX. (b) If X is uncountable then (X, Σ) is isomorphic to (N N , B(N N )), where B(N N ) is the algebra of Borel subsets of N N . proof Let T be a Polish topology on X such that Σ is its Borel σ-algebra. (a) Every singleton subset of X is closed, so must belong to Σ. If X is countable, every subset of X is a countable union of singletons, so belongs to Σ. (b) (Rao & Srivastava 94) The strategy of the proof is to find Borel sets Z ⊆ X, W ⊆ N N such that (Z, ΣZ ) ∼ = (N N , B(N N )) and (W, B(W )) ∼ = (X, Σ) (writing ΣZ , B(W ) for the subspace σ-algebras), and use a form of the Schröder-Bernstein theorem.

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424C

(i) By 423J, X has a subset Z homeomorphic to N N ; let h : N N → Z be a homeomorphism. By 424Bd, h is an isomorphism between (N N , B(N N )) and (Z, ΣZ ). (ii) Let hUn in∈N run over a base for the topology of X. Define g : X → {0, 1}N ⊆ N N by setting g(x) = hχUn (x)in∈N for every x ∈ N. Then g is injective, because X is Hausdorff. Also g is Borel measurable, by 4A3D(c-ii). By 424Bd, g is an isomorphism between (X, Σ) and (W, B(W )), where W = g[X] belongs to B(N N ). (iii) We have Z ∈ Σ, W ∈ B(N N ) such that (Z, ΣZ ) ∼ = (N N , B(N N )) and (W, B(W )) ∼ = (X, Σ). By N N ∼ 344D, (X, Σ) = (N , B(N )), as claimed. 424D Corollary (a) If (X, Σ) and (Y, T) are standard Borel spaces and #(X) = #(Y ), then (X, Σ) and (Y, T) are isomorphic. (b) If (X, Σ) is an uncountable standard Borel space then #(X) = #(Σ) = c. proof These follow immediately from 424C, if we recall that if B(N N ) is the Borel σ-algebra of N N then #(B(N N )) = c (4A3Fb). 424E Proposition Let X be a set and Σ a σ-algebra of subsets of X; suppose that (X, Σ) is countably separated in the sense that there is a countable set E ⊆ Σ separating the points of X. If A ⊆ X is such that (A, ΣA ) is a standard Borel space, where ΣA is the subspace σ-algebra, then A ∈ Σ. proof Give A a Polish topology T such that ΣA is the Borel σ-algebra of A, and let S be the topology on X generated by E ∪ {X \ E : E ∈ E}. Then S is second-countable (4A2Oa), so has a countable network (4A2Oc), and is Hausdorff because E separates the points of X. The identity map from A to X is Borel measurable for T and S, so 423Ib tells us that A is S-Borel; but of course the S-Borel σ-algebra is just the σ-algebra generated by E (4A3Da), so is included in Σ. 424F Corollary Let X be a Polish space and A ⊆ X any set which is not Borel. Let B(A) be the Borel σ-algebra of A. Then (A, B(A)) is not a standard Borel space. 424G Proposition Let (X, Σ) be a standard Borel space. Then (E, ΣE ) is a standard Borel space for every E ∈ Σ, writing ΣE for the subspace σ-algebra. proof Let T be a Polish topology on X for which Σ is the Borel σ-algebra. Then there is a Polish topology T0 ⊇ T for which E is closed (4A3I), therefore itself Polish in the subspace topology T0E (4A2Qd). But T0 and T have the same Borel sets (423Fb), so ΣE is just the Borel σ-algebra of E for T0E , and (E, ΣE ) is a standard Borel space. *424H For the full strength of a theorem in §448 we need a remarkable result concerning group actions on Polish spaces. Theorem (Becker & Kechris 96) Let G be a Polish group, (X, T) a Polish space and • a Borel measurable action of G on X. Then there is a Polish topology T0 on X, yielding the same Borel sets as T, such that the action is continuous for T0 and the given topology of G. proof (a) Fix on a right-translation-invariant metric ρ on G defining the topology of G (4A5Q), and let D be a countable dense subset of G; write e for the identity of G. Let Z be the set of 1-Lipschitz functions from G to [0, 1], that is, functions f : G → [0, 1] such that |f (g) − f (h)| ≤ ρ(g, h) for all g, h ∈ G. Then Z, with the topology of pointwise convergence inherited from the product topology of [0, 1]G , is a compact metrizable space. P P It is a closed subset of [0, 1]G , so is a compact Hausdorff space. Writing q(f ) = f ¹D for f ∈ Z, q : Z → [0, 1]D is injective, because D is dense and every member of Z is continuous; but this means that Z is homeomorphic to q[Z], which is metrizable, by 4A2Pc. Q Q We see also that the Borel σ-algebra of Z is the σ-algebra generated by sets of the form Wgα = {f : f (g) < α} where g ∈ G and α ∈ [0, 1]. P P This σ-algebra contains every set of the form {f : f ∈ Z, α < f (g) < β}, where g ∈ G and α, β ∈ R; since these sets generate the topology of Z, the σ-algebra they generate is the Borel σ-algebra of Z, by 4A3Da. Q Q

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(b) There is a continuous action of G on Z defined by setting (g •f )(h) = f (hg) for f ∈ Z and g, h ∈ G. P P (i) If f ∈ Z and g, h1 , h2 ∈ G, then |(g •f )(h1 ) − (g •f )(h2 )| = |f (h1 g) − f (h2 g)| ≤ ρ(h1 g, h2 g) = ρ(h1 , h2 ) because ρ is right-translation-invariant. So g •f ∈ Z for every f ∈ Z, g ∈ G. (ii) (e•f )(h) = f (he) = f (h) for every h ∈ G, so e•f = f for every f ∈ Z. (iii) If f ∈ Z and g1 , g2 , h ∈ G, then (g1 •(g2 •f ))(h) = (g2 •f )(hg1 ) = f (hg1 g2 ) = ((g1 g2 )•f )(h); as h is arbitrary, (g1 g2 )•f = g1 •(g2 •f ). Thus h ∈ G and ² > 0. Set

•

is an action of G on Z. (iv) Suppose that g0 ∈ G, f0 ∈ Z,

V = {g : g ∈ G, ρ(hg, hg0 ) < 21 ²}, W = {f : f ∈ Z, |f (hg0 ) − f0 (hg0 )| < 21 ²}. Then V is an open set in G containing g0 (because g 7→ ρ(hg, hg0 ) is continuous) and W is an open set in Z containing f0 . If g ∈ V and f ∈ W , |(g •f )(h) − (g0 •f0 )(h)| = |f (hg) − f0 (hg0 )| ≤ |f (hg) − f (hg0 )| + |f (hg0 ) − f0 (hg0 )| 1 2

≤ ρ(hg, hg0 ) + ² ≤ ². As f0 , g0 , ² are arbitrary, the map (g, f ) 7→ (g •f )(h) is continuous; as h is arbitrary, the map (g, f ) 7→ g •f is continuous. Q Q (c) Let B(X) be the Borel σ-algebra of X. For x ∈ X and B ∈ B(X), set PB (x) = {g : g ∈ G, g •x ∈ B}, QB (x) =

S

{V : V ⊆ G is open, V \ PB (x) is meager},

fB (x)(g) = inf({1} ∪ {ρ(g, h) : h ∈ G \ QB (x)}) for g ∈ G. It is easy to check that fB (x)(g 0 ) ≤ ρ(g, g 0 ) + fB (x)(g 0 ) for all g, g 0 ∈ G, so that every fB (x) belongs to Z. Every PB (x) is a Borel set, because • is Borel measurable. It follows that QB (x)4PB (x) is meager. P P By 4A3Ra, there is an open set V0 ⊆ G such that V0 4(X \ PB (x)) is meager and V0 ∩ V is empty whenever V ⊆ G is open and V \ PB (x) is meager. Now QB (x) ∩ V0 is empty, so QB (x) \ PB (x) ⊆ (X \ PB (x)) \ V0 is meager. Also (X \ V 0 ) \ PB (x) ⊆ (X \ PB (x)) \ V0 is meager, so X \ V 0 ⊆ QB (x) and PB (x) \ QB (x) ⊆ (V 0 \ V0 ) ∪ (V0 \ (X \ PB (x)) is meager. Q Q Consequently QB (x) \ PB (x) is empty, because the only meager open set in G is empty, by Baire’s theorem for complete metric spaces (4A2Ma). Let V be a countable base for the topology of G containing G. (d) For each B ∈ B(X), the map fB : X → Z is Borel measurable. P P Because the Borel σ-algebra of Z is generated by the sets Wgα of (a) above, it is enough to show that {x : fB (x) ∈ Wgα } = {x : fB (x)(g) < α}

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always belongs to B(X), because {W : fB−1 [W ] ∈ B} is surely a σ-algebra of subsets of Z. But if α > 1 this set is X, while if α ≤ 1 it is

{x : there is some h ∈ X \ QB (x) such that ρ(g, h) < α} = {x : there is some V ∈ V such that V \ QB (x) 6= ∅ and ρ(g, h) < α for every h ∈ V } [ = {x : V 6⊆ QB (x)} V ∈V 0 0

(where V = {V : V ∈ V, ρ(g, h) < α for every h ∈ V }) [ {x : V \ PB (x) is not meager}. = V ∈V 0

But for any fixed V ∈ V, {x : V \ PB (x) is not meager} = X \ {x : W [{x}] is meager} where W = {(y, g) : y ∈ X, g ∈ V, g •y ∈ X \ B} b is a Borel subset of X × G, because • is supposed to be Borel measurable; and therefore W ∈ B(X)⊗B(G), b writing B(G) for the Borel σ-algebra of G (4A3G). Now B(G) ⊆ B(G), the Baire property algebra of G b b b B(G). (4A3Rb), so W ∈ B(X)⊗ By 4A3Rc, the quotient algebra B/M has a countable order-dense set, where M is the σ-ideal of meager sets, so 4A3Sa tells us that {x : W [{x}] is meager} ∈ B(X). Accordingly {x : V \ PB (x) is not meager} is Borel for every V , and {x : fB (x)(g) < α} is Borel. Q Q (e) If g ∈ G, B ∈ B(X) and x ∈ X, then g •fB (x) = fB (g •x). P P PB (g •x) = {h : h•(g •x) ∈ B} = {h : hg ∈ PB (x)} = PB (x)g −1 . Because the map h 7→ hg −1 : G → G is a homeomorphism, QB (g •x) = QB (x)g −1 ; because it is an isometry, fB (g •x)(h) = min(1, ρ(h, X \ QB (g •x))) = min(1, ρ(h, X \ QB (x)g −1 )) = min(1, ρ(hg, X \ QB (x))) = fB (x)(hg) = (g •fB (x))(h) for every h ∈ G. Q Q (f ) Let hB0m im∈N run over a base for T containing X. We can now find a countable set E ⊆ B(X) such that (i) the topology T∗ generated by E is a Polish topology finer than T (ii) fB is T∗ -continuous for every B ∈ E (iii) X ∈ E. P P Let W be a countable base for the topology of Z. Enumerate N × N × W as h(kn , mn , Wn )in∈N in such a way that kn ≤ n for every n. Having chosen Borel sets Bij ⊆ X for i ≤ n, j ∈ N in such a way that the topology Sn generated by {Bij : i ≤ n, j ∈ N} is a Polish topology finer than T, consider the set Cn = {x : fBkn ,mn (x) ∈ Wn }. This is T-Borel, therefore Sn -Borel, so by 4A3H there is a Polish topology Sn+1 ⊇ Sn such that Cn ∈ Sn+1 . Let hBn+1,m im∈N run over a base for Sn+1 ; by 423Fb, every Bn+1,m belongs to B(X). Continue. ∗ Let T∗ be the topology S generated by E = {Bij : i, j ∈ N}. By 4A2Qf, T is Polish, because it is the topology generated by n∈N Sn . If W ∈ W, B ∈ E there are i, j ∈ N such that B = Bij and an n ∈ N such that (i, j, W ) = (kn , mn , Wn ); now fB−1 [W ] = Cn ∈ Sn+1 ⊆ T∗ . As W is arbitrary, fB is T∗ -continuous. Also X ∈ {B0m : m ∈ N} ⊆ E, as required. Q Q

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169

(g) Define θ : X → Z E by setting θ(x)(E) = fE (x) for x ∈ X, E ∈ E. Then θ is injective. P P Suppose that x, y ∈ X and that x 6= y. For every g ∈ G, g −1 •(g •x) = x 6= y = g −1 •(g •y), so g •x 6= g •S y and there is some m ∈ N such that g •x ∈ B0m while g •y ∈ / B0m , that is, g ∈ PB0m (x) \ PB0m (y). Thus G = m∈N PB0m (x) \ PB0m (y); by Baire’s theorem, there is some m ∈ N such that PB0m (x) \ PB0m (y) is non-meager. Because QB0m (x)4PB0m (x) and QB0m (y)4PB0m (y) are both meager, {g : fB0m (x)(g) > 0} = QB0m (x) 6= QB0m (y) = {g : fB0m (y)(g) > 0}, and θ(x)(B0m ) = fB0m (x) 6= fB0m (y) = θ(y)(B0m ). So θ(x) 6= θ(y). Q Q Because fE is T∗ -continuous for every E ∈ E, θ is T∗ -continuous. (h) Let T0 be the topology on X induced by θ; that is, the topology which renders θ a homeomorphism between X and θ[X]. Because θ[X] ⊆ Z E is separable and metrizable, T0 is separable and metrizable. Because θ is T∗ -continuous, T0 ⊆ T∗ . (i) The action of G on X is continuous for the given topology S on G and T0 on X. P P For any E ∈ E, (g, x) 7→ θ(g •x)(E) = fE (g •x) = g •fE (x) ((e) above) is S × T0 -continuous because the action of G on Z is continuous ((b) above) and fE : X → Z is T0 -continuous (by the definition of T0 ). But this means that (g, x) 7→ θ(g •x) is S × T0 -continuous, so that (g, x) 7→ g •x is (S × T0 , T0 )-continuous. Q Q (j) Let σ be a complete metric on G defining the topology S, and τ a complete metric on X defining the topology T∗ . (We do not need to relate σ to ρ in any way beyond the fact that they both give rise to the same topology S.) For E ∈ E, V ∈ V and n ∈ N let SEV n be the set of those φ ∈ Z E such that either φ(E)(g) = 0 for every g ∈ V or there is an F ∈ E such that F ⊆ E, diamτ (F ) ≤ 2−n and φ(F )(g) > 0 for some g ∈ V . Then SEV n is the union of a closed set and an open set, so is a Gδ set in Z E (4A2C(a-i)). Consequently T Y = {φ : φ ∈ Z E , φ(X) = χG} ∩ θ[X] ∩ E∈E,V ∈V,n∈N SEV n is a Gδ subset of Z E , being the intersection of countably many Gδ sets. (k) θ[X] ⊆ Y . P P Let x ∈ X. (i) PX (x) = G so QX (x) = G and θ(x)(X) = fX (x) = χG. (ii) Of course θ(x) ∈ θ[X]. (iii) Suppose that E ∈ E, V ∈ V and n ∈ N. If QE (x) ∩ V = ∅ then θ(x)(E)(g) = fE (x)(g) = 0 for every g ∈ V , and θ(x)(E) ∈ SEV n . Otherwise, V ∩ PE (x) is non-meager. But S E = {F : F ∈ E, F ⊆ E, diamτ (F ) ≤ 2−n }, so PE (x) =

S

{PF (x) : F ∈ E, F ⊆ E, diamτ (F ) ≤ 2−n };

because E is countable, this is a countable union and there is an F ∈ E such that F ⊆ E, diamτ (F ) ≤ 2−n and PF (x) ∩ V is non-meager. In this case QF (x) ∩ V is non-empty and θ(x)(F ) = fF (x) is non-zero at some point of V ; thus again θ(x) ∈ SEV n . As E, V and n are arbitrary, we have the result. Q Q (l) (The magic bit.) Y ⊆ θ[X]. P P Take any φ ∈ Y . Choose hEn in∈N in E, hVn in∈N in V and h˜ gn in∈N in G as follows. E0 = X and V0 = G. Given that φ(En ) is non-zero at some point of Vn , then, because φ ∈ SEn Vn n , there is an En+1 ∈ E such that En+1 ⊆ En , diamτ (En+1 ) ≤ 2−n and φ(En+1 ) is non-zero at some point of Vn ; say gñ ∈ Vn is such that φ(En+1 )(˜ gn ) > 0. Now we can find a Vn+1 ∈ V such that gñ ∈ Vn+1 ⊆ Vn and diamσ (Vn+1 ) ≤ 2−n . Continue. We are supposing also that φ ∈ θ[X], so we have a sequence hxi ii∈N in X such that hθ(xi )ii∈N → φ, that is, hfE (xi )ii∈N → φ(E) for every E ∈ E. In particular, limi→∞ fEn+1 (xi )(˜ gn ) = φ(En+1 )(˜ gn ) > 0 for every n. Let hin in∈N be a strictly increasing sequence in N such that fEn+1 (xin )(˜ gn ) > 0 for every n. Then

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*424H

gñ ∈ Vn+1 ∩ QEn+1 (xin ) ⊆ Vn+1 ∩ PEn+1 (xin ); there is therefore some gn ∈ Vn+1 ∩ PEn+1 (xin ), so that gn •xin ∈ En+1 . hVn in∈N is a non-increasing sequence of sets with σ-diameters converging to 0, so hgn in∈N is a Cauchy sequence for the complete metric σ. Similarly, hgn •xin in∈N is a Cauchy sequence for the complete metric τ , because diamτ (En+1 ) ≤ 2−n . We therefore have g ∈ G, y ∈ X such that hgn in∈N → g for S and hgn •xin in∈N → y for T∗ . In this case, hgn •xin in∈N → y for the coarser topology T0 , while hgn−1 in∈N → g −1 for S. Because the action is (S × T0 , T0 )-continuous, hxin in∈N = hgn−1 •(gn •xin )in∈N → g −1 •y for T0 . But θ is continuous for T0 , so θ(g −1 •y) = limn→∞ θ(xin ) = φ, and φ ∈ θ[X]. As φ is arbitrary, Y ⊆ θ[X]. Q Q (m) Thus θ[X] = Y is a Gδ set in the compact metric space Z E , and is a Polish space in its induced topology (4A2Qd). But this means that (X, T0 ), which is homeomorphic to θ[X], is also Polish. (n) I have still to check that T0 has the same Borel sets as T. But T, T∗ and T0 are all Polish topologies and T∗ is finer than both the other two. By 423Fb, T∗ has the same Borel sets as either of the others. This completes the proof. 424X Basic exercises > (a) Let h(Xi , Σi )ii∈I be a countable family of standard Borel spaces, and (X, Σ) their direct sum, that is, X = {(x, i) : i ∈ I, x ∈ Xi }, Σ = {E : E ⊆ X, {x : (x, i) ∈ E} ∈ Σi for every i}. Show that (X, Σ) is a standard Borel space. > (b) Let (X, Σ) be a standard Borel space and T a countably generated σ-subalgebra of Σ. Show that there is an analytic Hausdorff space Z such that T is isomorphic to the Borel σ-algebra of Z. (Hint: by 4A3I, we can suppose that X is a Polish space and T is generated by a sequence of open-and-closed sets, corresponding to a continuous function from X to {0, 1}N .) > (c) Let (X, Σ) be a standard Borel space and T1 , T2 two countably generated σ-subalgebras of Σ which separate the same points, in the sense that if x, y ∈ X then there is an E ∈ T1 such that x ∈ E, y ∈ X \ E iff there is an E 0 ∈ T2 such that x ∈ E 0 , y ∈ X \ E 0 . Show that T1 = T2 . (Hint: 424Xb, 423Fb.) In particular, if T1 separates the points of X then T1 = Σ. (d) Let A be a Dedekind σ-complete Boolean algebra. Show that A is isomorphic to the Borel σ-algebra of an analytic Hausdorff space iff it is isomorphic to a countably generated σ-subalgebra of the Borel σ-algebra of [0, 1]. > (e) Let U be a separable Banach space. Show that its Borel σ-algebra is generated, as σ-algebra, by the sets of the form {u : h(u) ≤ α} as h runs over the dual U ∗ and α runs over R. > (f ) Let X be a compact metrizable space. Show that the Borel σ-algebra of C(X) (the Banach space of continuous real-valued functions on X) is generated, as σ-algebra, by the sets {u : u ∈ C(X), u(x) ≥ α} as x runs over X and α runs over R. (g) Let (X, Σ) be a standard Borel space, Y any set, and T a σ-algebra of subsets of Y . Write T∗ for the σ-algebra of subsets of Y generated by S(T), the family of sets obtainable from sets in T by Souslin’s b operation. Let W ∈ S(Σ⊗T). Show that W [X] ∈ T∗ and that there is a (T∗ , Σ)-measurable function f : W [X] → X such that (f (y), y) ∈ W for every y ∈ W [X]. (Hint: 423M.) (h) Let (X, Σ) be a standard Borel space, Y any set, and T a countably generated σ-algebra of subsets of Y . Let f : X → Y be a (Σ, T)-measurable function, and write T∗ for the σ-algebra of subsets of Y generated by S(T), the family of sets obtainable from sets in T by Souslin’s operation. Show that (i) f [X] ∈ T∗ (ii) there is a (T∗ , Σ)-measurable function g : f [X] → X such that gf is the identity on X. (Hint: start with b the case in which T separates the points of Y , so that the graph of f belongs to Σ⊗T.)

424 Notes


171

> (i) Show that 424Xc and 424Xh are both false if we omit the hypothesis ‘countably generated’. (Hint: consider (i) the countable-cocountable algebra of R (ii) the split interval (iii) 423Xi.) (j) Let (X, Σ) be a standard Borel space. Show that if X is uncountable, Σ has a countably generated σ-subalgebra not isomorphic either to Σ or to PI for any set I. (Hint: 423L.) (k) Let (X, Σ) be a standard Borel space. (i) Show that the Boolean algebra A = Σ/[X]≤ω is homogeneous. (ii) Show that for every sequentially order-continuous Boolean homomorphism π : A → A there is a (Σ, Σ)-measurable f : X → X representing π in the sense that π(E • ) = (f −1 [E])• for every E ∈ Σ. (Hint: reduce to the case X = {0, 1}N .) (iii) Show that for every π ∈ Aut A there is an automorphism f of (X, Σ) representing π. (Hint: 344Ya.) 424Y Further exercises (a) Let (X, T) be a Polish space and F the family of closed subsets of X. Let Σ be the σ-algebra of subsets of F generated by the sets EH = {F : F ∈ F, F ∩ H 6= ∅} as H runs over the open subsets of X. (i) Show that S (F, Σ) is a standard Borel space. (Hint: take a complete metric ρ defining the topology of X. Set S ∗ = n∈N N n and choose a family hUσ iσ∈S ∗ of open sets in X such that U∅ = X, S diam Uσ ≤ 2−n whenever #(σ) = n + 1, Uσ = i∈N Uσa i for every σ, U σa i ⊆ Uσ for every σ, i. Define ∗ f : F → {0, 1}S by setting f (F )(σ) = 1 if F ∩ Uσ 6= ∅, 0 otherwise. Show that Z = f [F] is a Borel set and that f is an isomorphism between Σ and the Borel σ-algebra of Z.) (This is the Effros Borel structure on F.) (ii) Show that [X]n ∈ Σ for every n ∈ N. (b) Let (X, Σ) be a standard Borel space. Let T be the family of Polish topologies on X for which Σ is the Borel σ-algebra. Show that any sequence in T has an upper bound in T, and that any sequence with a lower bound has a least upper bound. (c) Let (X, Σ) be a standard Borel space. Say that C ⊆ X is coanalytic if its complement belongs to S(Σ). Show that for any such C the partially ordered set Σ ∩ PC has cofinality 1 if C ∈ Σ and cofinality ω1 otherwise. (Hint: 423P.) (d) Let I k be the split interval. Show that there is a σ-algebra Σ of subsets of I k such that (I k , Σ) is a b standard Borel space and {(x, y) : x, y ∈ I k , x ≤ y} ∈ Σ⊗Σ. 424 Notes and comments In this treatise I have generally indulged my prejudice in favour of ‘complete’ measures. Consequently Borel σ-algebras, as such, have taken subordinate roles. But important parts of the theory of Lebesgue measure, and Radon measures on Polish spaces in general, are associated with the fact that these are completions of measures defined on standard Borel spaces. Moreover, such spaces provide a suitable framework for a large part of probability theory. Of course they become deficient in contexts where we need to look at uncountable independent families of random variables, and there are also difficulties with σ-subalgebras, even countably generated ones, since these can correspond to the Borel algebras of general analytic spaces, which will not always be standard Borel structures (424F, 423L). 424Xf and 424Ya suggest the ubiquity of standard Borel structures; the former shows that they are not always presented as countably generated algebras, while the latter is an example in which we have to make a special construction in order to associate a topology with the algebra. The theory is of course dominated by the results of §423, especially 423Fb and 423I. I include 424H in this section because there is no other convenient place for it, but I have an excuse: the idea of ‘Borel measurable action’ can, in this context, be described entirely in terms of σ-algebras, since the Borel algebra of G × X is just the σ-algebra product of the Borel algebras of the factors (as in 424Bb). Of course for the theorem as expressed here we do need to know that G has a Polish group structure; but X could be presented just as a standard Borel space. The result is a dramatic expression of the fact that, given a standard Borel space (X, Σ), we have a great deal of freedom in defining a corresponding Polish topology on X.

172

Topological measure spaces II

Chapter 43 Topologies and measures II The first chapter of this volume was ‘general’ theory of topological measure spaces; I attempted to distinguish the most important properties a topological measure can have – inner regularity, τ -additivity – and describe their interactions at an abstract level. I now turn to rather more specialized investigations, looking for features which offer explanations of the behaviour of the most important spaces, radiating outwards from Lebesgue measure. In effect, this chapter consists of three distinguishable parts and two appendices. The first three sections are based on ideas from descriptive set theory, in particular Souslin’s operation (§431); the properties of this operation are the foundation for the theory of two classes of topological space of particular importance in measure theory, the K-analytic spaces (§432) and the analytic spaces (§433). The second part of the chapter, §§434-435, collects miscellaneous results on Borel and Baire measures, looking at the ways in which topological properties of a space determine properties of the measures it carries. In §436 I present the most important theorems on the representation of linear functionals by integrals; if you like, this is the inverse operation to the construction of integrals from measures in §122. The ideas continue into §437, where I discuss spaces of signed measures representing the duals of spaces of continuous functions. The first appendix, §438, looks at a special topic: the way in which the patterns in §§434-435 are affected if we assume that our spaces are not unreasonably complex in a rather special sense defined in terms of measures on discrete spaces. Finally, I end the chapter with a further collection of examples, mostly to exhibit boundaries to the theorems of the chapter, but also to show some of the variety of the structures we are dealing with. 431 Souslin’s operation I begin the chapter with a short section on Souslin’s operation (§421). The basic facts we need to know are that (in a complete locally determined measure space) the family of measurable sets is closed under Souslin’s operation (431A), and that the kernel of a Souslin scheme can be approximated from within in measure (431D). 431A Theorem Let (X, Σ, µ) be a complete locally determined measure space. Then Σ is closed under Souslin’s operation. proof Let hEσ iσ∈S be a Souslin scheme in Σ with kernel A. Write S S ∗ = k∈N N k = S ∪ {∅}. If F ∈ Σ and µF < ∞, then A ∩ F ∈ Σ. P P For each σ ∈ S ∗ , set S T Aσ = φ∈N N ,φ⊇σ n≥1 Eφ¹n , and let Gσ be a measurable envelope of Aσ ∩ F . Because Aσ ⊆ Eσ (writing E∅ = X), we may suppose that Gσ ⊆ Eσ ∩ F . Now, for any σ ∈ S ∗ , S S Aσ ∩ F = i∈N Aσa i ∩ F ⊆ i∈N Gσa i , so Hσ = Gσ \

S i∈N

Gσa i

is negligible. S Set H = σ∈S ∗ Hσ , so that H is negligible. Take any x ∈ G∅ \ H. Choose hφ(i)ii∈N inductively, as follows. Given that σ = hφ(i)ii
431D


173

431B Corollary If (X, T, Σ, µ) is a complete locally determined topological measure space, every SouslinF set in X (definition: 421K) is measurable. 431C Corollary Let X be a set and θ an outer measure on X. Let µ be the measure defined by Carathéodory’s method, and Σ its domain. Then Σ is closed under Souslin’s operation. proof Let hEσ iσ∈S be a Souslin scheme in Σ with kernel A. Take any C ⊆ X such that θC < ∞. Then θC = θ¹ PC is an outer measure on C; let µC be the measure on C defined from θC by Carathéodory’s method, and ΣC its domain. If σ ∈ S, D ⊆ C then θC (D ∩ C ∩ Eσ ) + θC (D \ (C ∩ Eσ )) = θ(D ∩ Eσ ) + θ(D \ Eσ ) = θD = θC D; as D is arbitrary, C ∩ Eσ ∈ ΣC . µC is a complete totally finite measure, so 431A tells us that the kernel of the Souslin scheme hC ∩ Eσ iσ∈S belongs to ΣC . But this is just C ∩ A (applying 421Cb to the identity map from C to X.) So θ(C ∩ A) + θ(C \ A) = θC (C ∩ A) + θC (C \ A) = θC C = θC. As C is arbitrary, A ∈ Σ (113D). As hEσ iσ∈S is arbitrary, we have the result. 431D Theorem Let (X, Σ, µ) be a complete locally determined measure space, and hEσ iσ∈S a Souslin scheme in Σ with kernel A. Then [ \ Eφ¹n ) : K ⊆ N N is compact} µA = sup{µ( φ∈K n≥1

= sup{µ(

[ \

Eφ¹n ) : ψ ∈ N N },

φ≤ψ n≥1

writing φ ≤ ψ if φ(i) ≤ ψ(i) for every i ∈ N. S T proof (a) By 431A, A is measurable. For K ⊆ N N , set HK = φ∈K n≥1 Eφ¹n . Of course HK ⊆ A, and we know from 421M (or otherwise) that H QK ∈ Σ if K is compact. So surely µA ≥ µHK for every compact K ⊆ N N . If ψ ∈ N N , then {φ : φ ≤ ψ} = i∈N (ψ(i) + 1) is compact. We therefore have µA ≥ sup{µ(

[ \

Eφ¹n ) : K ⊆ N N is compact}

φ∈K n≥1

≥ sup{µ(

[ \

Eφ¹n ) : ψ ∈ N N }.

φ≤ψ n≥1

So what I need to prove is that

S T µA ≤ sup{µ( φ≤ψ n≥1 Eφ¹n ) : ψ ∈ N N }.

S (b) Fix on a set F ∈ Σ of finite measure. For σ ∈ S ∗ = k∈N N k set S T Aσ = φ∈N N ,φ⊇σ n≥1 Eφ¹n . We need to know that Aσ belongs to Σ; this follows from 431A, because writing Eτ0 = Eτ if τ ⊆ σ or σ ⊆ τ , ∅ otherwise, S T 0 ∈ S(Σ) = Σ, Aσ = φ∈N N n≥1 Eφ¹n writing S for Souslin’s operation, as in §421. P Let ² > 0, and take S a family h²σ iσ∈S ∗ of strictly positive real numbers such that σ∈S ∗ ²σ ≤ ². For each σ ∈ S ∗ we have Aσ = i∈N Aσa i , so there is an mσ ∈ N such that S µ(F ∩ Aσ \ i≤mσ Aσa i ) ≤ ²σ .

174

Topological measure spaces II

431D

Define ψ ∈ N N by saying that ψ(k) = max{mσ : σ ∈ Nk , σ(i) ≤ ψ(i) for every i < k} for each k ∈ N. Set H= (c) Set G=

S

T φ∈N N ,φ≤ψ

S σ∈S ∗

n≥1

F ∩ Aσ \

Eφ¹n .

S i≤mσ

Aσ a i ,

so that µG ≤ ², by the choice of the ²σ and the mσ . Then F ∩ A \ G ⊆ H. P P If x ∈ F ∩ A \ G, choose hφ(i)ii∈N inductively, as follows. Given that φ(i) ≤ ψ(i) for i < k and x ∈ Aσ , where σ = hφ(i)ii
T

µ∗ E = µ ˜E = supK⊆N N

is compact

µFK ,

where FK = φ∈K n≥1 Eφ¹n for K ⊆ N N . But every FK is closed, by 421M. So µ∗ E ≤ supF ⊆E as the reverse inequality is trivial, we have the result.

is closed

µF ;

*431F There is a topological version of 431A, as follows. Theorem Let X be any topological space, and Bb its Baire property algebra. (a) For any A ⊆ X, there is a Baire property envelope of A, that is, a set E ∈ Bb such that A ⊆ E and b E \ F is meager whenever A ⊆ F ∈ B. (b) Bb is closed under Souslin’s operation. proof (a) By 4A3Ra, there is an open set H ⊆ X such that A \ H is meager and H ∩ G is empty whenever b G ⊆ X is open and A ∩ G is meager. Set E = A ∪ H; then E ⊇ A and E4H = A \ H is meager, so E ∈ B. b let G be an open set such that G4(X \ F ) is meager. Then G ∩ A ⊆ G ∩ F is meager, so If A ⊆ F ∈ B, G ∩ H is empty and E \ F ⊆ (E4H) ∪ (G4(X \ F )) is meager. Thus E is a Baire property envelope of A. (b) Let hEσ iσ∈S be a Souslin scheme in Bb with kernel A. Write S S ∗ = k∈N N k = S ∪ {∅}. For each σ ∈ S ∗ , set Aσ =

S

T φ∈N N ,φ⊇σ

n≥1

Eφ¹n ,

and let Gσ be a Baire property envelope of Aσ as described in (a). Because Aσ ⊆ Eσ (writing E∅ = X), we may suppose that Gσ ⊆ Eσ . Now, for any σ ∈ S ∗ , S S Aσ = i∈N Aσa i ⊆ i∈N Gσa i , so

431 Notes


Hσ = Gσ \

S i∈N

175

Gσa i

is meager. S Set H = σ∈S ∗ Hσ , so that H is negligible. Take any x ∈ G∅ \ H. Choose hφ(i)ii∈N inductively, as follows. Given that σ = hφ(i)ii (d) Let (X, Σ, µ) be a measure space with locally determined negligible sets (definition: 213I), and hEσ iσ∈S a Souslin scheme in Σ with kernel A. Show that S T µ∗ A = supψ∈N N µ( φ≤ψ n≥1 Eφ¹n ). > (e) Let (X, Σ, µ) be a semi-finite measure space, and hEσ iσ∈S a Souslin scheme in Σ with kernel A. Show that S T µ∗ A = supψ∈N N µ( φ≤ψ n≥1 Eφ¹n ). 431Y Further exercises (a) Let (X, Σ, µ) be a complete measure space with the measurable envelope property (213Xl). Show that Σ is closed under Souslin’s operation. (b) Let X be a set, Σ a σ-algebra of subsets of X, and I a σ-ideal of subsets of X such that I ⊆ Σ. Suppose that for every A ⊆ X there is an F ∈ Σ such that A ⊆ F and F \ E ∈ I whenever A ⊆ E ∈ Σ. Show that Σ is closed under Souslin’s operation. (c) Let X be a set, Σ a σ-algebra of subsets of X, and I an ω1 -saturated σ-ideal of Σ; suppose that A ∈ Σ whenever A ⊆ B ∈ I. Show that Σ is closed under Souslin’s operation. 431 Notes and comments From the point of view of measure theory, the most important property of Souslin’s operation, after its idempotence, is the fact that (for many measure spaces) the family of measurable sets is closed under the operation (431A). The proof I give here is based on the concept of measurable envelope, which can be used in other cases of great interest (431Yb, 431Yc). But for some applications it is also very important to know that if A is the S kernel T of a Souslin scheme hEσ iσ∈S , then A can be approximated from inside by sets of the form H = φ≤ψ n≥1 Eφ¹n (431D, 431Xd), which belong to the σ-algebra generated by the Eσ (421M). A typical application of this idea is when every Eσ is a Borel subset of R; then we find not only that A is Lebesgue measurable (indeed, measurable for every Radon measure on R) but that (for any given Radon measure µ) the Souslin scheme itself provides a Borel subset H of A of measure approximating the measure of A. Let me repeat that the essence of descriptive set theory is that we are not satisfied merely to know that a set of a certain type exists. We want also to know how to build it, because we expect that an explicit

176

Topologies and measures II

431 Notes

construction will be valuable later on. For instance, the construction given in 431D shows that if the Souslin scheme consists of closed compact sets, the set H will be compact (421Xo). I mention 431B as a typical application of 431A, even though it is both obvious and obviously less than what can be said. The algebras Σ of this section are algebras closed under Souslin’s operation. In a complete locally determined topological measure space, the algebra Σ of measurable sets includes the open sets (by definition), therefore the Borel algebra B, therefore S(B); but now we can take the algebra A1 generated by S(B), and A1 and S(A1 ) will also be included in Σ, so that Σ will included the algebra A2 generated by S(A1 ), and so on. (Note that S(A1 ) includes the σ-algebra generated by A1 , by 421F, so I do not need to mention that separately.) We haveSto run through all the countable ordinals before we can be sure of getting to the smallest algebra Aω1 = ξ<ω1 Aξ which contains every open set and is closed under Souslin’s operation, and we shall then have Aω1 ⊆ Σ. The result in 431D is one of the special features of measures. (A similar result, based on rather different hypotheses, is in 432J.) But the argument of 431A can be applied in many other cases; see 431Yb-431Yc. A striking one is 431F, which will be useful in Volume 5.

432 K-analytic spaces I describe the basic measure-theoretic properties of K-analytic spaces (§422). I start with ‘elementary’ results (432A-432C), assembling ideas from §§421, 422 and 431. The main theorem of the section is 432D, one of the leading cases of the general extension theorem 416P. An important corollary (432G) gives a sufficient condition for the existence of pull-back measures. I briefly mention ‘capacities’ (432I-432J). 432A Proposition Let (X, T, Σ, µ) be a complete locally determined Hausdorff topological measure space. Then every K-analytic subset of X is measurable. proof If A ⊆ X is K-analytic, it is Souslin-F (422Ha), therefore measurable (431B). 432B Theorem Let X be a K-analytic Hausdorff space, and µ a semi-finite topological measure on X. Then µX = sup{µK : K ⊆ X is compact}. proof If γ < µX, there is an E ∈ dom µ such that γ < µE < ∞; set νF = µ(E ∩ F ) for every Borel set F ⊆ X, so that ν is a totally finite Borel measure on X, and νX > γ. Let νˆ be the completion S of ν. Let R ⊆ N N × X be an usco-compact relation such that R[N N ] = X. Set Fσ = R[Iσ ] for σ ∈ S = k≥1 N k , where Iσ = {φ : σ ⊆ φ ∈ N N }. Because R is closed in N N × X (422Da), S XTis the kernel of the Souslin scheme hFσ iσ∈S (421I). By 431D, there is a compact L ⊆ N N such that νˆ( φ∈L n∈N Fφ¹n ) ≥ γ. But, by 421I, this is just νˆ(R[L]); and R[L] is compact, by 422D(e-i). So µR[L] is defined, with µR[L] ≥ νR[L] = νˆR[L], and we have a compact subset of X of measure at least γ. As γ is arbitrary, the proposition is proved. 432C Proposition Let X be a Hausdorff space such that all its open sets are K-analytic, and µ a Borel measure on X. (a) If µ is semi-finite, it is tight (that is, inner regular with respect to the compact sets). (b) If µ is locally finite, its completion is a Radon measure on X. proof (a) By 422Hb, every open subset of X is Souslin-F. Applying 421F to the family E of closed subsets of X, we see that every Borel subset of X is Souslin-F, therefore K-analytic (by 422Hb in the opposite direction). Now suppose that E ⊆ X is a Borel set. Then the subspace measure µE is a semi-finite Borel measure on the K-analytic space E, so by 432B µE = supK⊆E is compact µK. As E is arbitrary, µ is tight. (b) Because X is Lindelöf (422Gg), µ is σ-finite (411Ge), therefore semi-finite. So (a) tells us that µ is tight. By 416F, its c.l.d. version is a Radon measure. But (because µ is σ-finite) this is just its completion (213Ha).

432G

K-analytic spaces

177

432D Theorem (Aldaz & Render 00) Let X be a K-analytic Hausdorff space and µ a locally finite measure on X which is inner regular with respect to the closed sets. Then µ has an extension to a Radon measure on X. In particular, µ is τ -additive. proof The point is that if µE > 0 then there is a compact K ⊆ E such that µ∗ K > 0. P P Write Σ for the domain of µ. Take γ < µE. Because X is Lindelöf (422Gg), µ is σ-finite (411Ge), therefore semi-finite. Let E 0 ⊆ E be such that γ < µE 0 < ∞. Because µ is inner regular with respect to the closed sets, there is a closed set F ⊆ E such that µF > γ. S F is K-analytic (422Gf); let R ⊆ N N × F be an usco-compact relation such that R[N N ] = F . For σ ∈ S ∗ = n∈N N n set Fσ = {x : (φ, x) ∈ R for some φ ∈ N N such that φ(i) ≤ σ(i) for every i < #(σ)}. Then hFσa i ii∈N is a non-decreasing sequence with union Fσ , so µ∗ Fσ = supi∈N µ∗ Fσa i for every σ ∈ S ∗ . We can therefore find a sequence ψ ∈ N N such that µ∗ Fψ¹n > γ for every n ∈ N. Set Q

K = {φ : φ ∈ N N , φ(i) ≤ ψ(i) for every i ∈ N};

then K = n∈N (ψ(n) + 1) is compact, so R[K] is compact (422D(e-i)). ?? Suppose, if possible, that µ∗ R[K] < γ. Then there is an H ∈ Σ such that R[K] ⊆ H ⊆ F and µ(F \ H) > µF − γ. Because µ is inner regular with respect to the closed sets, there is a closed set F 0 ∈ Σ such that F 0 ⊆ F \ H and µF 0 > µF − γ. Since R[K] ∩ F 0 = ∅, K ∩ R−1 [F 0 ] = ∅. R−1 [F 0 ] is closed, because R is usco-compact, so there is some n such that L = {φ : φ ∈ N N , φ¹n = φ0 ¹n for some φ0 ∈ K} does not meet R−1 [F 0 ] (4A2F(h-vi)), and R[L] ∩ F 0 = ∅. But L is just {φ : φ(i) ≤ ψ(i) for every i < n}, so R[L] = Fψ¹n , and γ < µ∗ Fψ¹n ≤ µ(F \ F 0 ) < γ, which is absurd. X X Thus µ∗ R[K] ≥ γ. As γ > 0, we have the result. Q Q Now the theorem follows at once from 416P(ii)⇒(i). 432E Corollary Let X be a K-analytic Hausdorff space, and µ a locally finite quasi-Radon measure on X. Then µ is a Radon measure. proof By 432D, µ has an extension to a Radon measure µ0 . But of course µ and µ0 must coincide, by 415H or otherwise. 432F Corollary Let X be a K-analytic Hausdorff space, and ν a locally finite Baire measure on X. Then ν has an extension to a Radon measure on X; in particular, it is τ -additive. If the topology of X is regular, the extension is unique. proof Because X is Lindelöf (422Gg), ν is σ-finite, therefore semi-finite; by 412D, it is inner regular with respect to the closed sets. So 432D tells us that it has an extension to a Radon measure on X. Since the extension is τ -additive, so is ν. If X is regular, then it must be completely regular (4A2H(b-i)), and the family G of cozero sets is a base for the topology closed under finite unions. If µ, µ0 are Radon measures extending ν, they agree on G, and must be equal, by 415H(iv). 432G Corollary Let X be a K-analytic Hausdorff space, Y a Hausdorff space and ν a locally finite measure on Y which is inner regular with respect to the closed sets. Let f : X → Y be a continuous function such that f [X] has full outer measure in Y . Then there is a Radon measure µ on X such that f is inverse-measure-preserving for µ and ν. If ν is Radon, it is precisely the image measure µf −1 .

178


432G

proof (a) Write T for the domain of ν, and set Σ0 = {f −1 [F ] : f ∈ T}, so that Σ0 is a σ-algebra of subsets of X, and we have a measure µ0 on X defined by setting µ0 f −1 [F ] = νF whenever F ∈ T (132Yd). (b) If E ∈ Σ0 and γ < µ0 E, there is an F ∈ T such that E = f −1 [F ]. Now there is a closed set F 0 ⊆ F such that νF 0 ≥ γ. Because f is continuous, f −1 [F 0 ] is closed, and we have f −1 [F 0 ] ⊆ E and µ0 f −1 [F 0 ] ≥ γ. As E and γ are arbitrary, µ0 is inner regular with respect to the closed sets. If x ∈ X, then (because ν is locally finite) there are an open set H ⊆ Y and an F ∈ T such that f (x) ∈ H ⊆ F ,

νF < ∞.

In this case f −1 [H] is an open set containing x and µ∗0 f −1 [H] ≤ µ0 f −1 [F ] = νF < ∞. As x is arbitrary, µ0 is locally finite. (c) By 432D, there is a Radon measure µ on X extending µ0 . Because f is inverse-measure-preserving for µ0 and ν, it is surely inverse-measure-preserving for µ and ν. The image measure µf −1 extends ν, so must be locally finite; it is therefore a Radon measure (418I). So if ν itself is a Radon measure, it must be identical with µf −1 , by 416Eb. 432H Corollary Suppose that X is a set and that S, T are Hausdorff topologies on X such that (X, T) is K-analytic and S ⊆ T. Then the totally finite Radon measures on X are the same for S and T. proof Write f for the identity function on X regarded as a continuous function from (X, T) to (X, S). If µ is a totally finite T-Radon measure on X, then µ = µf −1 is S-Radon, by 418I. If ν is a totally finite S-Radon measure on X, then 432G tells us that it is of the form µ = µf −1 for some T-Radon measure µ, that is, is itself T-Radon. 432I Capacitability The next result is not exactly measure theory as studied in most of this treatise; but it is clearly very close to the other ideas of this section, and it has important applications to measure theory in the narrow sense. Definition Let (X, T) be a topological space. A (Choquet) capacity on X is a function c : PX → [0, ∞] such that (i) c(A) ≤ c(B) whenever A ⊆ B ⊆ X; (ii) limn→∞ c(An ) = c(A) whenever hAn in∈N is a non-decreasing sequence of subsets of X with union A; (iii) c(K) = inf{c(G) : G ⊇ K is open} for every compact set K ⊆ X. 432J Theorem (Choquet 55) Let X be a Hausdorff space and c a capacity on X. If A ⊆ X is K-analytic, then c(A) = sup{c(K) : K ⊆ A is compact}. proof Take γ < c(A). Let R ⊆ N N × X be an usco-compact relation such that R[N N ] = A; for σ ∈ S ∗ = S n n∈N N set Aσ = {x : (φ, x) ∈ R for some φ ∈ N N such that φ(i) ≤ σ(i) for every i < #(σ)}. Then hAσa i ii∈N is a non-decreasing sequence with union Aσ , so c(Aσ ) = supi∈N c(Aσa i ) for every σ ∈ S ∗ . We can therefore find a sequence ψ ∈ N N such that c(Aψ¹n ) > γ for every n ∈ N. Set Q

K = {φ : φ ∈ N N , φ(i) ≤ ψ(i) for every i ∈ N};

then K = n∈N (ψ(n) + 1) is compact, so R[K] is compact (422D(e-i)). ?? Suppose, if possible, that c(R[K]) < γ. Then, by (iii) of 432I, there is an open set G ⊇ R[K] such that c(G) < γ. Set F = X \ G, so that F is closed and K ∩ R−1 [F ] = ∅. R−1 [F ] is closed, because R is usco-compact, so there is some n such that L = {φ : φ ∈ N N , φ¹n = φ0 ¹n for some φ0 ∈ K} does not meet R−1 [F ] (4A2F(h-vi)), and R[L] ∩ F = ∅, that is, R[L] ⊆ G. But L is just {φ : φ(i) ≤ ψ(i) for every i < n}, so R[L] = Aψ¹n , and

432 Notes

K-analytic spaces

179

γ < c(Aψ¹n ) ≤ c(G) < γ, which is absurd. X X Thus c(R[K]) ≥ γ. As γ is arbitrary and R[K] is compact, we have the result. 432X Basic exercises (a) Put 422Xf, 431Xb and 432D together to prove 432C. (b) Let X be a K-analytic Hausdorff space, and µ a measure on X which is outer regular with respect to the open sets. Show that µX = supK⊆X is compact µ∗ K. (Hint: use the argument of part (a) of the proof of 432D.) > (c) Let X be a K-analytic Hausdorff space, and µ a semi-finite topological measure on X. Show that if either µ is inner regular with respect to the closed sets or X is regular and µ is a τ -additive Borel measure, then µ is tight. (d) Use 422Gf, 432B and 416C to prove 432E. > (e) Suppose that X is a set and that S, T are Hausdorff topologies on X such that (X, T) is K-analytic and S ⊆ T. Let (Z, U, T, ν) be a Radon measure space and f : Z → X a function which is almost continuous for U and S. Show that f is almost continuous for U and T. (Hint: it is enough to consider totally finite ν; show that νf −1 is T-Radon, so is inner regular for {K : TK = SK }, writing TK for the subspace topology induced by T on K.) (f ) Let X be a topological space and µ a locally finite measure on X which is inner regular with respect to the closed sets. Show that µ∗ is a capacity. (g) Let X be a topological space and F a closed subset of X. Define c : PX → {0, 1} by setting c(A) = 1 if A meets F , 0 otherwise. Show that c is a capacity on X. (h) Let X and Y be Hausdorff spaces, and R ⊆ X × Y an usco-compact relation. Show that if c is a capacity on Y , then A 7→ c(R[A]) is a capacity on X. (i) Use 432J and 432Xf to shorten the proof of 432D. 432Y Further exercises (a) Show that there are a K-analytic Hausdorff space X and a probability measure µ on X such that (i) µ is inner regular with respect to the Borel sets (ii) the domain of µ includes a base for the topology of X (iii) every compact subset of X is negligible. Show that there is no extension of µ to a topological measure on X. 432 Notes and comments The measure-theoretic properties of K-analytic spaces can largely be summarised in the slogan ‘K-analytic spaces have lots of compact sets’. I said above that it is sometimes helpful to think of K-analytic spaces as an amalgam of compact Hausdorff spaces and Souslin-F subsets of R. For the former, it is obvious that they have many compact subsets; for the latter, it is not obvious, but is of course one of their fundamental properties, deducible from 422De. 432B and part (a) of the proof of 432D are typical manifestations of the phenomenon. The real point of these theorems is that we can extend a Borel or Baire measure to a Radon measure with no assumption of τ -additivity (432F). A Radon measure must be τ -additive just because it is tight. A (locally finite) Borel or Baire measure must be τ -additive whenever the measurable open sets are K-analytic. The condition ‘every open set is K-analytic’ in 432C is of course a very strong one in the context of compact Hausdorff spaces (422Xe). But for analytic spaces it is automatically satisfied (423E), and that is the side on which the principal applications of 432C appear. The results which I call corollaries of 432D can mostly be proved by more direct methods (see 432Xd), but the line I choose here seems to be the most powerful technique. Indeed it can be used to deal with 432C as well (432Xa). In §434 I will discuss ‘universally measurable’ sets in topological spaces. In fact K-analytic sets are universally measurable in a particularly strong sense (432A). The point here is that K-analyticity is intrinsic;

180


432 Notes

a K-analytic space is measurable whenever embedded as a subspace of a (complete locally determined) topological measure space. The theorems here touch on two phenomena of particular importance. First, in 432G we have an example of ‘pulling back’ a measure, that is, we have a measure ν on a set Y and a function f : X → Y and seek a Radon measure µ on X such that f is inverse-measure-preserving, or, even better, such that ν = µf −1 . There was a similar result in 418L. In both cases we have to suppose that f is continuous and (in effect) that ν is a Radon measure. (This is not part of the hypotheses of 432G, but of course it is an easy consequence of them, using 432B.) In 418L, we need a special hypothesis to ensure that there are enough compact subsets of X to carry an appropriate Radon measure; in 432G, this is an automatic result of assuming that X is K-analytic. Both 418L and 432G can be regarded as consequences of Henry’s theorem (416M). The difficulty arises from the requirement that µ should be a Radon measure; if we do not insist on this there is a much simpler solution, since we need suppose only that f [X] is of full outer measure (132Yd). The next theme I wish to mention is a related one, the investigation of comparable topologies. If S and T are (Hausdorff) topologies on a set X, and S is coarser than T (so that (X, S) is a continuous image of (X, T)), then 418I tells us that any totally finite T-Radon measure is S-Radon. We very much want to know when the reverse is true, so that the (totally finite) Radon measures for the two topologies are the same. 432H provides one of the important cases in which this occurs. The hypothesis ‘(X, T) is K-analytic’ generalizes the alternative ‘(X, T) is compact’; in the latter case, S = T, so that the result is, from our point of view here, trivial. (But from the point of view of elementary general topology, of course, it is one of the pivots of the theory of compact Hausdorff spaces.) In a similar vein we have a variety of important topological consequences of the same hypotheses (422Ye, 423Fb). The paragraphs 432I-432J may appear to be no more that a minor extension of ideas already set out. I ought therefore to say plainly that the topological and measure theory of K-analytic spaces have co-evolved with the notion of capacity, and that 432J (‘K-analytic spaces are capacitable’) is one of the cornerstones of a theory of which I am giving only a minuscule part. For a idea of the vitality and scope of this theory, see Dellacherie 80.

433 Analytic spaces We come now to the special properties of measures on ‘analytic’ spaces, that is, continuous images of N N , as described in §423. I start with a couple of facts about spaces with countable networks. 433A Proposition Let (X, T) be a topological space with a countable network, and µ a localizable topological measure on X which is inner regular with respect to the Borel sets. Then µ has countable Maharam type. proof Let µ ˜ be the c.l.d. version of µ (213E). Then the measure algebra A of µ ˜ can be identified with the measure algebra of µ (322D(b-iii)). Also µ ˜ is complete, locally determined and localizable, so every subset ˜ be the domain of µ of X has a measurable envelope with respect to µ ˜ (213L). Let Σ ˜, and E a countable ˜ network for T. For each E ∈ E, let FE ∈ Σ be a measurable envelope of E. ˜ F • ∈ B}. Let B be the order-closed subalgebra of A generated by {FE• : E ∈ E}, and set T = {F : F ∈ Σ, ˜ Because B is an order-closed subalgebra P If G ∈ T, set S of A, T is a σ-subalgebra of Σ. Now T ⊆ T. P E0 = {E : E ∈ E, E ⊆ G}. Set F = E∈E0 FE , so that F ∈ T and G ⊆ F . For each E ∈ E0 , FE \ G is negligible, so F \ G is negligible, and G• = F • ∈ B, so G ∈ T. Q Q It follows that T includes the Borel σ-algebra of X. Because µ is inner regular with respect to the Borel sets, B is order-dense in A, and B = A. Thus the countable set {FE• : E ∈ E} τ -generates A, and the Maharam type of A, which is the Maharam type of µ, is countable. 433B Lemma If (X, T) is a Hausdorff space with a countable network, then any topological measure on X is countably separated in the sense of 343D. proof By 4A2Nf, there is a countable family of open sets separating the points of X.

433G

Analytic spaces

181

433C Theorem Let X be an analytic Hausdorff space, and µ a Borel measure on X. (a) If µ is semi-finite, it is tight. (b) If µ is locally finite, its completion is a Radon measure on X. proof X is K-analytic (423C); moreover, every open subset of X is again analytic (423Eb). So 432C gives the result at once. Remark Compare 256C. 433D Theorem Let X and Y be analytic Hausdorff spaces, ν a totally finite Radon measure on Y and f : X → Y a Borel measurable function such that f [X] is of full outer measure in Y . Then there is a Radon measure µ on X such that ν = µf −1 . proof By 423Ga, the graph R of f is an analytic set in X × Y , therefore K-analytic. Set π1 (x, y) = x, π2 (x, y) = y for (x, y) ∈ R, so that π1 and π2 are continuous. Now π2 [R] = f [X] is of full outer measure, so by 432G there is a Radon measure λ on R such that ν = λπ2−1 . Next, because π1 is continuous, the image µ = λπ1−1 is a Radon measure on X, by 418I. But π2 = f π1 , so µf −1 = (λπ1−1 )f −1 = λ(f π1 )−1 = λπ2−1 = ν, as required. 433E Proposition Let (X, Σ, µ) be a semi-finite measure space and T a topology on X such that µ is inner regular with respect to the closed sets. Let (Y, S) be an analytic space and f : X → Y a measurable function. Then f is almost continuous. proof Take E ∈ Σ and γ < µE. Then there is an F ∈ Σ such that F ⊆ E and γ < µF < ∞. For Borel sets H ⊆ Y , set νH = µ(F ∩ f −1 [H]). Then ν is a totally finite Borel measure on Y , so is tight (that is, inner regular with respect to the compact sets) (433C); let K ⊆ Y be a compact set such that νK > γ, so that µ(F ∩ f −1 [K]) > γ. The subspace measure on L = F ∩ f −1 [K] is still inner regular with respect to the (relatively) closed sets (412Pc), and f ¹L is still measurable; but f ¹L is a function from L to K, and K is metrizable, by 423Dc. So f ¹L is almost continuous, by 418J, and there is a set F 0 ⊆ L, of measure at least γ, such that f ¹F 0 is continuous. As E and γ are arbitrary, f is almost continuous. Remark Compare 418Yg. 433F I give two simple corollaries of the von Neumann-Jankow selection theorem (423N-423O). Proposition Let (X, T) and (Y, S) be analytic Hausdorff spaces, and f : X → Y a Borel measurable surjection. Let ν be a complete locally determined topological measure on Y , and T its domain. Then there is a T-measurable function g : Y → X such that gf is the identity on X. proof By 423O we know that there is a function g : Y → X such that gf is the identity and g is T1 measurable, where T1 is the σ-algebra generated by the Souslin-F subsets of Y . But T contains every Souslin-F subset of Y , by 431B, therefore includes T1 , and g is actually T-measurable. 433G Proposition Let (X, T) be an analytic Hausdorff space, (Y, T, ν) a complete locally determined measure space, and f : X → Y a surjection. Suppose that there is some countable family F ⊆ T such that F separates the points of Y (that is, whenever y, y 0 are distinct points of Y there is a member of F containing one and not the other) and f −1 [F ] is a Borel subset of X for every F ∈ F . Then there is a T-measurable function g : Y → X such that f g is the identity on Y . proof Set A = F ∪ {Y \ F : F ∈ F}. The topology T1 on X generated by T ∪ {f −1 [A] : A ∈ A} is still analytic (423H). If we take S to be the topology on Y generated by A, then S is Hausdorff and f is (T1 , S)-continuous, so S is analytic (423Bb). Because S is generated by a countable subset A of T, it is second-countable, and S ⊆ T (4A3Da/4A3E). So ν is a topological measure with respect to S. By 433F, there is a function g : Y → X, measurable for T and the topology T1 , such that gf is the identity on X; and of course g is still measurable for T and the coarser original topology T on X.

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433H Because analytic spaces have countable networks (423C), and their compact subsets are therefore metrizable (4A2Nh), their measure theory is very close to that of R or [0, 1] or {0, 1}N . I give some simple manifestations of this principle. Proposition Let hXi ii∈I be a family of analytic spaces, and forQeach i ∈ I let µi be a Radon probability measure on Xi . Let λ be the ordinary product measure on X = i∈I Xi , as defined in §254. (a) If I is countable then λ is a Radon measure. (b) If every µi is strictly positive, then λ is a quasi-Radon measure. proof (a) In this case, X is analytic (423Bc), therefore hereditarily Lindelöf (423Da). Let Λ be the domain of λ and T the topology of X. Then Λ ∩ T is a base for T; by 4A3Da, T ⊆ Λ. By 417Sb, λ is the τ -additive product measure on X; by 417Q, this is a Radon measure. Q (b) By (a), the ordinary product measure on i∈J Xi is a Radon measure for every finite set J ⊆ I. So 417Sc tells us that λ is the τ -additive product measure on X; by 417O, this is a quasi-Radon measure. 433I Proposition Let X be an analytic Hausdorff space, and T a countably generated σ-subalgebra of the Borel σ-algebra B(X) of X. Then any locally finite measure with domain T has an extension to a Radon measure on X. proof Let µ0 be a locally finite measure with domain T. (a) Consider first the case in which µ0 is totally finite. Let hFn in∈N be a sequence in T generating T as σ-algebra. Define f : X → {0, 1}N by setting f (x)(n) = χFn (x) for n ∈ N, x ∈ X. Then f is T-measurable (use 418Bd), so we have a Borel measure ν0 on {0, 1}N defined by setting ν0 E = µ0 f −1 [E] for every Borel set E ⊆ {0, 1}N . Now the completion ν of ν0 is a Radon measure (433C). Also f [X] must be analytic, by 423Gb, because f is B(X)-measurable. So ν measures f [X] (432A), and νf [X] = ν0∗ f [X] = ν0 {0, 1}N , that is, f [X] is ν-conegligible. By 433D, there is a Radon measure µ on X such that ν = µf −1 . Because every Fn is expressible as f −1 [E] for some E ∈ B({0, 1}N ), so is every member of T. If F ∈ T, take H ∈ B({0, 1}N ) such that F = f −1 [H]; then µF = νH = ν0 H = µ0 F . Thus µ extends µ0 and µ¹Σ will serve. (b) In general, because X is Lindelöf and µ0 is locally finite, µ0 is σ-finite. Let hXn in∈N be a partition (n) of X into members of T such that µ0 Xn is finite for every n, and set µ0 F = µ0 (F ∩ Xn ) for every n and P∞ (n) every F ∈ T; then every µ0 has an extension to a Radon measure µ(n) , and we can set µF = n=0 µ(n) F for every F ∈ Σ. Because µ0 is locally finite, so is µ, and it is now easy to check that µ is complete, locally determined and tight, just because every µ(n) is. 433J I turn now to a brief mention of ‘standard Borel spaces’. From the point of view of this chapter, it is natural to regard the following results as simple corollaries of theorems about Polish spaces. But, as remarked in §424, there are cases in which a standard Borel space is presented without any specific topology being attached; and in any case it is interesting to look at the ways in which we can express these ideas as theorems about σ-algebras rather than about topological spaces. Proposition Let (X, Σ) be a standard Borel space and T a countably generated σ-subalgebra of Σ. Then any σ-finite measure with domain T has an extension to Σ. proof Let µ0 be a σ-finite measure with domain T. (a) If µ0 is totally finite, give X a Polish topology for which Σ is the Borel σ-algebra of X, and use 433I. (b) In general, let hXn in∈N be a partition of X into members of T such that µ0 Xn < ∞ for every n, and (n) (n) set µ0 F = µ0 (F ∩ Xn ) for every n and every F ∈ T; then every µ0 has an extension to a measure µ(n) P∞ with domain Σ, and we can set µF = n=0 µ(n) F for every F ∈ Σ.

433Yb

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433K Proposition Let h(Xn , Σn , µn )in∈N be a sequence of probability spaces such that (Xn , Σn ) is a standard Borel space for every n. Suppose that for each n ∈ N we are given an inverse-measure-preserving function fn : Xn+1 → Xn . Then we can find a standard Borel space (X, Σ), a probability measure µ with domain Σ, and inverse-measure-preserving functions gn : X → Xn such that fn gn+1 = gn for every n. proof For each n, choose a Polish topology Tn on Xn such that Σn is the algebra of Tn -Borel sets. Let µ ˆn be the completion of µn ; then µ ˆn is a Radon measure (433C). Every fn is inverse-measure-preserving for µ ˆn+1 and µ ˆn , by 235Hc, and almost continuous, by 418J. By 418Q, we have a Radon measure µ ˆ on Q X = {x : x ∈ n∈N Xn , fn (x(n + 1)) = x(n) for every n ∈ N} such that the continuous maps xQ7→ x(n) = gn (x) : X → Xn are inverse-measure-preserving for every n. Now X is a Borel subset of Z = n∈N Xn . P P For each n ∈ N, let Un be a countable base for Tn . Then S S Z \ X = n∈N U,V ∈Un ,U ∩V =∅ {z : z(n) ∈ U, fn (z(n + 1)) ∈ V } is a countable union of Borel sets because {z : z(n) ∈ U } is open and {z : z(n + 1) ∈ fn−1 [V ]} is Borel whenever n ∈ N and U , V ∈ Un . So Z \ X and X are Borel sets. Q Q Accordingly (X, Σ) is a standard Borel space, where Σ is the Borel σ-algebra of X (424G). So if we take µ=µ ˆ¹Σ, we shall have a suitable measure on X. 433X Basic exercises (a) Let (X, T) be a topological space with a countable network, and µ a topological measure on X which is inner regular with respect to the Borel sets and has the measurable envelope property (213Xl). Show that µ has countable Maharam type. (b) Show that an effectively locally finite measure on a hereditarily Lindelöf space (in particular, on any analytic space) is σ-finite. (c) Let X ⊆ [0, 1] be a set of Lebesgue outer measure 1 and inner measure 0. Show that the subspace measure on X is a totally finite Borel measure which is not tight. (d) Let X be a Hausdorff space and µ a locally finite measure on X, inner regular with respect to the Borel sets, such that dom µ includes a base for the topology of X. Suppose that Y ⊆ X is an analytic set of full outer measure. Show that µ has a unique extension to a Radon measure µ ˜ on X, and that Y is µ ˜-conegligible. (e) Let (X, Σ) be a standard Borel space. (i) Show that any semi-finite measure with domain Σ is a compact measure (definition: 342Ac), therefore perfect. (Hint: if X is given a suitable topology, the measure is tight.) (ii) Show that any measure with domain Σ is countably separated. (f ) Let (X, Σ, µ) and (Y, T, ν) be atomless probability spaces such that (X, Σ) and (Y, T) are standard Borel spaces. Show that (X, Σ, µ) and (Y, T, ν) are isomorphic. (Hint: by 344I, their completions are isomorphic; by 344H, they have negligible sets of cardinal c; show that any isomorphism between the completions is (Σ, T)-measurable on a conegligible set; use 424Da to match residual negligible sets.) (g) Let X be [0, 1] × {0, 1}, with its usual topology, and I k the split interval (419L); define f : X → I k by setting f (t, 0) = t− , f (t, 1) = t+ for t ∈ [0, 1]. (i) Give I k its usual Radon measure ν (343J, 419Lc). Show that there is no Radon measure λ on X such that ν = λf −1 . (ii) Let µ be the product Radon probability measure on X, starting from Lebesgue measure on [0, 1] and the usual fair-coin measure on {0, 1}. Show that f is inverse-measure-preserving for µ and ν. Show that f is not almost continuous. 433Y Further exercises (a) Find a Hausdorff topological space X with a countable network and a semi-finite Borel measure on X which does not have countable Maharam type. (b) Let X be an analytic Hausdorff space and µ an atomless Radon measure on X. Show that (X, µ) is isomorphic to Lebesgue measure on some measurable subset of R. (Hint: 344I.)

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(c) Let (X, T) be a Polish space without isolated points, and µ a σ-finite topological measure on X. Show that there is a conegligible meager set. (Hint: Show that every non-empty open set is uncountable. Find a countable dense negligible set and a negligible Gδ set including it.) (d) Let hXn in∈N be a sequence of analytic spaces and for each n ∈ N let µn be a Borel probability measure on Xn . Suppose that for each n ∈ N we are given an inverse-measure-preserving function fn : Xn+1 → Xn . Show that we can find a standard Borel space (X, Σ), a probability measure µ with domain Σ, and inversemeasure-preserving functions gn : X → Xn such that fn gn+1 = gn for every n. 433 Notes and comments The measure-theoretic results of 433C-433E are of much the same type as those in §432. A characteristic difference is that Borel measurable functions between analytic spaces behave in many ways like continuous functions. (Compare 433D and 432G.) You may feel that 423Yc offers some explanation of this. For any question which refers to the Borel algebra of an analytic space X, or to the class of its analytic subsets, we can expect to be able to suppose that X is separable and metrizable (see 423Xd), and that any single Borel measurable function appearing is continuous. (424H is a particularly remarkable instance of this principle.) 433H here amounts to spelling out a special case of ideas already treated in 417S. As this territory is relatively unfamiliar, I give detailed examples (423Xi, 433Xc, 433Xg, 439A, 439K) to show that the theorems of this section are not generally valid for compact Hausdorff spaces (the archetype of K-analytic spaces which need not be analytic), nor for separable metric spaces (the archetypical spaces with countable network). They really do depend on the particular combination of properties possessed by analytic spaces. For large parts of probability theory, standard Borel spaces provide an adequate framework, and have a number of advantages; some of the technical problems concerning measurability which loom rather large in this treatise disappear in such contexts. Many authors accordingly give them great prominence. I myself believe that the simplifications are an entrapment rather than a liberation, that sooner or later everyone has to leave the comfortable environment of Borel algebras on Polish spaces, and that it is better to be properly equipped with a suitable general theory when one does. But it is surely important to know what the simplifications are, and the results in 433J-433K will I hope show at least that there are wonderful ideas here, even if my own presentation tends to leave them on one side. 434 Borel measures What one might call the fundamental question of topological measure theory is the following. What kinds of measures can arise on what kinds of topological space? Of course this question has inexhaustible ramifications, corresponding to all imaginable properties of measures and topologies and connexions between them. The challenge I face here is that of identifying particular ideas as being more important than others, and the chief difficulty lies in the bewildering variety of topological properties which have been studied, any of which may have implications for the measure theory of the spaces involved. In this section and the next I give a sample of what is known, necessarily biased and incomplete. I try however to include the results which are most often applied and enough others for the proofs to contain, between them, most of the non-trivial arguments which have been found effective in this area. In 434A I set out a crude classification of Borel measures on topological spaces. For compact Hausdorff spaces, at least, the first question is whether they carry Borel measures which are not, in effect, Radon measures; this leads us to the definition of ‘Radon’ space (434C) which is also of interest in the context of general Hausdorff spaces. I give a brief account of the properties of Radon spaces (434F, 434N). I look also at two special topics: ‘quasi-dyadic’ spaces (434O-434Q) and a construction of Borel product measures by integration of sections (434R). In the study of Radon spaces we find ourselves looking at ‘universally measurable’ subsets of topological spaces (434D). These are interesting in themselves, and also interact with constructions from earlier parts of this treatise (434S-434T). Three further classes of topological space, defined in terms of the types of topological measure which they carry, are the ‘Borel-measure-compact’, ‘Borel-measure-complete’ and ‘preRadon’ spaces; I discuss them briefly in 434G-434J. They provide useful methods for deciding whether Hausdorff spaces are Radon (434K).

434Da

Borel measures

185

434A Types of Borel measures In §411 I introduced the following properties which a Borel measure may or may not have: (i) inner regularity with respect to closed sets; (ii) inner regularity with respect to zero sets; (iii) tightness (that is, inner regularity with respect to closed compact sets); (iv) τ -additivity. These are of course interrelated. (ii)⇒(i) just because zero sets are closed, and (iii)⇒(iv) by 411E; in a Hausdorff space, (iii)⇒(i); and for an effectively locally finite measure on a regular topological space, (iv)⇒(i) (414Mb). On a regular Hausdorff space, therefore, we can divide totally finite Borel measures into four classes: (A) measures which are not inner regular with respect to the closed sets; (B) measures which are inner regular with respect to the closed sets, but not τ -additive nor tight; (C) measures which are τ -additive and inner regular with respect to the closed sets, but not inner regular with respect to the compact sets; (D) tight measures; and each of the classes (B)-(D) can be further subdivided into those which are completion regular (B1 , C1 , D1 ) and those which are not (B0 , C0 , D0 ). Examples may be found in 434Xf (type A), 411Q and 439K (type B0 ), 439J (type B1 ), 415Xc and 434Xa (type C1 ) and 434Xb (type D0 ), while Lebesgue measure itself is of type D1 , and any direct sum of spaces of types D0 and C1 will have type C0 . (The space in 439J depends for its construction on supposing that there is a cardinal which is not measure-free. It seems that no convincing example of a space of class B1 , that is, a completion regular, non-τ -additive Borel probability measure on a completely regular Hausdorff space, is known which does not depend on some special axiom beyond ordinary ZFC. For one of the obstacles to finding such a space, see 434Q below.) Note that a totally finite Borel measure µ on a regular Hausdorff space can be extended to a quasi-Radon measure iff µ is of class C or D (415M), and that in this case the quasi-Radon measure must be just the completion µ ˆ of µ. µ ˆ will be of the same type, on the classification here, as µ; in particular, µ ˆ will be a Radon measure iff µ is of class D (416F). 434B Compact, analytic and K-analytic spaces For any class of topological spaces, we can enquire which of the seven types of measure described above can be realized by measures on spaces of that class. The enquiry is limited only by our enterprise and diligence in seeking out new classes of topological space. For the spaces studied in §§432-433, however, we have something worth repeating here. On a K-analytic Hausdorff space, a semi-finite Borel measure which is inner regular with respect to the closed sets is tight (432B, 432D); consequently classes B and C of 434A cannot appear, and we are left with only the types A, D0 and D1 , all of which appear on compact Hausdorff spaces (434Xb, 434Xf). On an analytic Hausdorff space we have further simplifications: every semi-finite Borel measure is tight (433C), and (if X is regular) every closed set is a zero set (423Db). Thus on an analytic regular Hausdorff space only type D1 , of the seven types in 434A, can appear. (If the topology is not regular, we may also get measures of type D0 ; see 434Ya.) 434C Radon spaces: Definition For K-analytic Hausdorff spaces, therefore, we have a large gap between the ‘bad’ measures of class A and the ‘good’ measures of class D; furthermore, we have an important class of spaces in which type A cannot appear. It is natural to enquire further into the spaces in which every (totally finite) Borel measure is of class D, and (given that no exact description can be found) we are led, as usual, to a definition. A Hausdorff space X is Radon if every totally finite Borel measure on X is tight. 434D Universally measurable sets Before going farther with the study of Radon spaces it will be useful to spend a couple of paragraphs on the following concept. Let X be a topological space. (a) I will say that a subset E of X is universally measurable (in X) if it is measured by the completion of every totally finite Borel measure on X; that is, for every totally finite Borel measure µ on X there is a Borel set F ⊆ X such that E4F is µ-negligible.

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434Db

(b) A subset of X is universally measurable iff it is measured by every complete locally determined topological measure on X. P P (i) Suppose that A ⊆ X is universally measurable and that µ is a complete locally determined topological measure on X. Let F ⊆ X be such that µF is defined and finite. Then we have a totally finite Borel measure ν on X defined by setting νE = µ(F ∩ E) for every Borel set E ⊆ X. Now there are Borel sets E, B ⊆ X such that A4E ⊆ B and νB = 0. In this case, (A∩F )4(E ∩F ) ⊆ B ∩F and µ(B ∩ F ) = 0, so that (because µ is complete) A ∩ F is measured by µ. Because F is arbitrary and µ is locally determined, A is measured by µ. (ii) Suppose that A ⊆ X is measured by every complete locally determined topological measure on X. Then, in particular, it is measured by the completion of any totally finite Borel measure, so is universally measurable. Q Q (c) If X is a topological space then the family Σum of universally measurable subsets of X is a σ-algebra closed under Souslin’s operation and including the Borel σ-algebra. (For it is the intersection of the domains of a family of complete totally finite measures, and all these are σ-algebras including the Borel σ-algebra and closed under Souslin’s operation, by 431A.) In particular, Souslin-F sets are universally measurable, so (if X is Hausdorff) K-analytic and analytic sets are (422Ha, 423C). (d) Note that a function f : X → R is Σum -measurable iff it is µ-virtually measurable for every totally finite Borel measure µ on X (122Q, 212F). Generally, if Y is another topological space, I will say that f : X → Y is universally measurable if f −1 [H] ∈ Σum for every open set H ⊆ Y ; that is, if f is (Σum , B(Y ))-measurable, where B(Y ) is the Borel σ-algebra of Y . (Y )

(Y )

(e) In fact, if f : X → Y is universally measurable, then it is (Σum , Σum )-measurable, where Σum is the (Y ) algebra of universally measurable subsets of Y . P P Take F ∈ Σum and a totally finite Borel measure µ on X. If µ ˆ is the completion of µ, then the image measure ν = µ ˆf −1 is a complete totally finite topological −1 measure on Y , so measures F , and f [F ] ∈ dom µ ˆ. As µ is arbitrary, f −1 [F ] ∈ Σum ; as F is arbitrary, f (Y ) Q is (Σum , Σum )-measurable. Q (f ) It follows that if Z is a third topological space and f : X → Y , g : Y → Z are universally measurable, then gf : X → Z is universally measurable. 434E Universally Radon-measurable sets A companion idea is the following. (a) Let X be a Hausdorff space. I will say that a subset E of X is universally Radon-measurable if it is measured by every Radon measure on X. (b) If X is a Hausdorff space, the family ΣuRm of universally Radon-measurable subsets of X is a σalgebra closed under Souslin’s operation and including the algebra of universally measurable subsets of X (and, a fortiori, including the Borel σ-algebra). (Use 434Db and the idea of 434Dc.) (c) If X is a Hausdorff space, I will say that a function f from X to a topological space Y is universally Radon-measurable if f −1 [H] ∈ ΣuRm for every open set H ⊆ Y . A function f : X → R is ΣuRm measurable iff it is µ-virtually measurable for every totally finite tight Borel measure µ on X. (Compare 434Dd.) 434F Elementary properties of Radon spaces: Proposition Let X be a Hausdorff space. (a) The following are equiveridical: (i) X is a Radon space; (ii) every semi-finite Borel measure on X is tight; (iii) if µ is a locally finite Borel measure on X, its c.l.d. version µ ˜ is a Radon measure; (iv) whenever µ is a totally finite Borel measure on X, and G ⊆ X is an open set with µG > 0, then there is a compact set K ⊆ G such that µK > 0; (v) whenever µ is a non-zero totally finite Borel measure on X, there is a Radon subspace Y of X such that µ∗ Y > 0. (b) If Y ⊆ X is a subspace which is a Radon space in its induced topology, then Y is universally measurable in X.

434F

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187

(c) If X is a Radon space and Y ⊆ X, then Y is Radon iff it is universally measurable in X iff it is universally Radon-measurable in X. In particular, all Borel subsets and all Souslin-F subsets of X are Radon spaces. (d) The family of Radon subspaces of X is closed under Souslin’s operation and set difference. proof (a)(i)⇒(ii) Let µ be a semi-finite Borel measure on X, E ⊆ X a Borel set and γ < µE. Because µ is semi-finite, there is a Borel set H of finite measure such that µ(E ∩ H) > γ. Set νF = µ(F ∩ H) for every Borel set F ⊆ X; then ν is a totally finite Borel measure on X, and νE > γ. Because X is a Radon space, there is a compact set K ⊆ E such that νK ≥ γ, and now µK ≥ γ. As µ, E and γ are arbitrary, (ii) is true. (ii)⇒(i) and (i)⇒(v) are trivial. (v)⇒(iv) Assume (v), and let µ be a totally finite Radon measure on X and G a non-negligible open set. Set νE = µ(E ∩ G) for every Borel set E ⊆ X. Then ν is a non-zero totally finite Borel measure on X, so there is a Radon subspace Y of X such that ν ∗ Y > 0. The subspace measure νY on Y is a Borel measure on Y , so is tight. Since νY (Y \ G) = ν(X \ G) = 0, νY (Y ∩ G) > 0 and there is a compact set K ⊆ Y ∩ G such that νY K > 0. Now µK > 0. As µ and G are arbitrary, (iv) is true. not-(i)⇒not-(iv) If X is not Radon, there is a totally finite Borel measure µ on X which is not tight. By 416F(iii), there is an open set G ⊆ X such that µG > supK⊆G is compact µK = γ say. Let K be the family of compact subsets of G. By 215B(v), S there is a non-decreasing sequence hKn in∈N in K such that µ(K \ F ) = 0 for every K ∈ K, where F = n∈N Kn . Observe that µF = limn→∞ µKn ≤ γ < µG. Now set νE = µ(E ∩ G \ F ) for every Borel set E ⊆ X. Then ν is a Borel measure on X, and νG > 0. If K ⊆ G is compact, then νK = µ(K \ F ) = 0. So ν and G witness that (iii) is false. (i)⇒(iii) The point is that µ ˜ is tight. P P If µ ˜E > γ, then, because µ ˜ is semi-finite, there is a set E 0 ⊆ E such that γ < µ ˜E 0 < ∞; now there is a Borel set H ⊆ E 0 such that µH = µ ˜E 0 (213Fc). Setting νF = µ(H ∩ F ) for every Borel set F , ν is a totally finite Borel measure on X and νH > γ, so there is a compact set K ⊆ H such that νK ≥ γ. Now γ ≤ µK < ∞, so µ ˜K = µK ≥ γ (213Fa), while K ⊆ E. As E and γ are arbitrary, µ ˜ is tight. Q Q On the other hand, every point of X belongs to an open set of finite measure for µ, which is still of finite measure for µ ˜ (213Fa again). So µ ˜ is locally finite; since it is surely complete and locally determined, it is a Radon measure. (iii)⇒(i) Assume (iii), and let µ be a totally finite Borel measure on X. Then its c.l.d. version µ ˜ is tight. But µ ˜ extends µ (213Hc), so µ also is tight. As µ is arbitrary, X is a Radon space. (b) Let µ be a totally finite Borel measure on X, and µ ˆ its completion; let ² > 0. Let µY be the subspace measure on Y , so that µY is a totally finite Borel measure on Y , and is tight. There is a compact set K ⊆ Y such that νK ≥ µY Y − ². But this means that µ∗ Y = µY Y ≤ µY K + ² = µ∗ (K ∩ Y ) + ² = µK + ² ≤ µ∗ Y + ². As ² is arbitrary, µ∗ Y = µ∗ Y , and Y is measured by µ ˆ (413Ef); as µ is arbitrary, Y is universally measurable. (c) (i) If Y is Radon, it is universally measurable, by (b). (ii) If Y is universally measurable, it is universally Radon-measurable, by 434Eb. (iii) Suppose that Y is universally Radon-measurable, and that ν is a totally finite Borel measure on Y . For Borel sets E ⊆ X, set µE = ν(E ∩ Y ). Then µ is a totally finite Borel measure on X, so its c.l.d. version µ ˜ is a Radon measure on X, by (a-iii). We are supposing that Y is universally Radon-measurable, so, in particular, it must be measured by µ ˜. We have

µ ˜(X \ Y ) =

sup K⊆X\Y is compact

(213Ha, because µ is totally finite)

µ ˜K =

sup K⊆X\Y is compact

µK

188


=

sup

434F

ν(K ∩ Y ) = 0,

K⊆X\Y is compact

and Y is µ ˜-conegligible. Now suppose that E ⊆ Y is a (relatively) Borel subset of Y . Then E is of the form F ∩ Y where F is a Borel subset of X, so that νE = µF = µ ˜F = µ ˜(Y ∩ F ) = µ ˜E =

sup K⊆E is compact

µK =

sup

νK.

K⊆E is compact

As E is arbitrary, ν is tight; as ν is arbitrary, Y is a Radon space. By 434Dc, it follows that all Borel subsets and all Souslin-F subsets of X are Radon spaces. (d) The first step is to note that if hEn in∈N is a sequence of Radon subspaces of X with union E, then E is Radon; this is immediate from (a-v) above. Now S let hEσ iσ∈S be a Souslin scheme, consisting of Radon subsets of X, with kernel A. We know that E = σ∈S Eσ is a Radon space. Every Eσ is universally measurable in E, by (b), so A also is (434Dc), and must be Radon, by (c). Thus the family of Radon subspaces of X is closed under Souslin’s operation. If E and F are Radon subsets of X, then E ∪ F is Radon, and, just as above, F is universally measurable in E ∪ F . But this means that E \ F = (E ∪ F ) \ F is universally measurable in E ∪ F , so that E \ F is Radon. 434G Just as we can address the question ‘when can we be sure that every Borel measure is of class D’ in terms of the definition of ‘Radon’ space (434C), we can form other classes of topological space by declaring that the Borel measures they support must be of certain kinds. Three definitions which lead to interesting patterns of ideas are the following. Definitions (a) A topological space X is Borel-measure-compact (Gardner & Pfeffer 84) if every totally finite Borel measure on X which is inner regular with respect to the closed sets is τ -additive, that is, X carries no measure of class B in the classification of 434A. (b) A topological space X is Borel-measure-complete (Gardner & Pfeffer 84) if every totally finite Borel measure on X is τ -additive. (If X is regular and Hausdorff, this amounts to saying that X carries no measures of classes A or B in the classification of 434A.) (c) A Hausdorff space X is pre-Radon (also called ‘hypo-radonian’, ‘semi-radonian’) if every τ additive totally finite Borel measure on X is tight. (If X is regular, this amounts to saying that X carries no measure of class C in the classification of 434A.) 434H Proposition Let X be a topological space and B its Borel σ-algebra. (a) The following are equiveridical: (i) X is Borel-measure-compact; (ii) every semi-finite Borel measure on X which is inner regular with respect to the closed sets is τ -additive; (iii) every effectively locally finite Borel measure on X which is inner regular with respect to the closed sets has an extension to a quasi-Radon measure; (iv) every totally finite Borel measure on X which is inner regular with respect to the closed sets has a support; (v) if µ is a non-zero totally finite Borel measure on X, inner regular with respect to the closed sets, and G is an open cover of X, then there is some G ∈ G such that µG > 0. (b) If X is Lindelöf (in particular, if X is a K-analytic Hausdorff space), it is Borel-measure-compact. (c) If X is Borel-measure-compact and A ⊆ X is a Souslin-F set, then A is Borel-measure-compact in its subspace topology. In particular, any Baire subset of X is Borel-measure-compact. proof (a)(i)⇒(ii) Assume (i), and let µ be a semi-finite Borel measure on X which is inner regular with respect to the closed sets. Let G be an upwards-directed family of open sets with union G0 , and γ < µG0 .

434I

Borel measures

189

Because µ is semi-finite, there is an H ∈ B such that µH < ∞ and µ(H ∩ G0 ) ≥ γ. Set νE = µ(E ∩ H) for every E ∈ B; then ν is a totally finite Borel measure on X. For any E ∈ B, νE = µ(E ∩ H) = sup{µF : F ⊆ E ∩ H is closed} ≤ sup{νF : F ⊆ E is closed}, so ν is inner regular with respect to the closed sets, and must be τ -additive. Now γ ≤ νG0 = supG∈G νG ≤ supG∈G µG. As γ and G is arbitrary, µ is τ -additive. (ii)⇒(iii) Assume (ii), and let µ be an effectively locally finite Borel measure on X which is inner regular with respect to the closed sets. Then it is semi-finite (411Gd), therefore τ -additive. By 415L, it has an extension to a quasi-Radon measure on X. (iii)⇒(i) If (i) is true and µ is a totally finite Borel measure on X which is inner regular with respect to the closed sets, then µ has an extension to a quasi-Radon measure, which is τ -additive, so µ is also τ -additive (411C). (i)⇒(iv) Use 411Nd. (iv)⇒(v) Suppose that (iv) is true, that µ is a non-zero totally finite Borel measure on X which is inner regular with respect to the closed sets, and that G is an open cover of X. If F is the support of µ, then µF > 0 so F 6= ∅; there must be some G ∈ G meeting F , and now µG > 0. not-(i)⇒not-(v) Suppose that there is a totally finite Borel measure µ on X, inner regular with respect to the closed sets, which is not τ -additive. Let G be an upwards-directed family of open sets such S that µG∗ > γ, where G∗ = G and γ = supG∈G S µG. Let hGn in∈N be a non-decreasing sequence in G such that µ(G \ G∗0 ) for every G ∈ G, where G∗0 = n∈N Gn (215B(v)). Then µG∗0 ≤ γ, so there is a closed set F ⊆ G∗ \ G∗0 such that µF > 0. Let ν be the Borel measure on X defined by setting µE = µ(E ∩ F ) for every E ∈ B. As in the argument for (i)⇒(ii), ν is inner regular with respect to the closed sets. Consider H = G ∪ {X \ F }; this is an open cover of X. If G ∈ G then νG ≤ µ(G \ G∗0 ) = 0, so νH = 0 for every H ∈ H; thus ν and H witness that (iv) is false. (b) Use (a-v) and 422Gg. (c) Let µ be a Borel measure on A which is inner regular with respect to the closed sets, that is to say, the relatively closed sets in A. Let ν be the corresponding Borel measure on X, defined by setting νE = µ(A ∩ E) for every E ∈ B. Let νˆ be the completion of ν. Putting 431D and 421M together, we see that νÂ = sup{ˆ ν F : F ⊆ A is closed in X}, that is, νX = sup{µF : F ⊆ A is closed in X}. But this means that if E ∈ B and γ < νE, there is a closed set F in X such that F ⊆ A such that µ(E ∩ F ) > γ; now there is a relatively closed set F 0 ⊆ A such that F 0 ⊆ E ∩ F and µF 0 ≥ γ, and as F 0 must be relatively closed in F it is closed in X, while νF 0 ≥ γ. Since E and γ are arbitrary, ν is inner regular with respect to the closed sets, and will be τ -additive. Now suppose that G is an upwards-directed family of relatively open subsets of A. Set H = {H : H ⊆ X is open, H ∩ A ∈ G}. Then H is upwards-directed, so S S µ( G) = ν( H) = supH∈H νH = supG∈G µG. As µ and G are arbitrary, A is Borel-measure-compact. By 421L, it follows that any Baire subset of X is Borel-measure-compact. 434I Proposition Let X be a topological space. (a) The following are equiveridical: (i) X is Borel-measure-complete; (ii) every semi-finite Borel measure on X is τ -additive; (iii) every totally finite Borel measure on X has a support; (iv) S whenever µ is a totally finite Borel measure on X there is a base U for the topology of X such that µ( {U : U ∈ U , µU = 0}) = 0.

190


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(b) If X is regular, it is Borel-measure-complete iff every effectively locally finite Borel measure on X has an extension to a quasi-Radon measure. (c) If X is Borel-measure-complete, it is Borel-measure-compact. (d) If X is Borel-measure-complete, so is every subspace of X. (e) If X is hereditarily Lindelöf (for instance, if X is separable and metrizable, see 4A2P(a-iii)), it is Borel-measure-complete, therefore Borel-measure-compact. proof (a)(i)⇒(ii) Use the argument of (i)⇒(ii) of 434Ha; this case is simpler, because we do not need to check that the auxiliary measure ν is inner regular. (ii)⇒(i) is trivial. (i)⇒(iv) If X is Borel-measure-complete and µ is a totally finite Borel measure on X, take U to be the family of all open subsets of X. S This is surely a base for the topology, and setting U0 = {U : U ∈ U , µU = 0}, U0 is upwards-directed so µ( U0 ) = supU ∈U0 µU = 0, as required. (iv)⇒(iii) Assume (iv), and let µ be a totally finite Borel measure on X. Take S a base U as in (iv), S so that µ( U0 ) = 0, where U0 is the family of negligible members of U . Set F = X \ U0 , so that F is a conegligible closed set. If G ⊆ X is an open set meeting F , there is a member U of U such that U ⊆ G and U ∩ F 6= ∅; now U ∈ / U so µ(G ∩ F ) = µG ≥ µU > 0. As G is arbitrary, F is self-supporting and is the support of µ. (iii)⇒(i) Assume (iii), and let µ be a totally finite Borel measure on X. Let G be an upwards-directed family of open sets with union G∗ . Set γ = supG∈G S µG. Let hGn in∈N be a non-decreasing sequence in G such that µ(G \ G∗0 ) for every G ∈ G, where G∗0 = n∈N Gn (215B(v)). Then µG∗0 ≤ γ. Let ν be the Borel measure on X defined by setting µE = µ(E ∩ G∗ \ G∗0 ) for every E ∈ B. Then ν has a support F say. Now νG = 0 for every G ∈ G, so F ∩ G = ∅ for every G ∈ G, and F ∩ G∗ = ∅; but this means that µ(G∗ \ G∗0 ) = νX = νF = 0. Accordingly µG∗ = γ. As µ and G are arbitrary, X is Borel-measure-complete. (b) If X is Borel-measure-complete and µ is an effectively locally finite Borel measure on X, then µ is τ -additive, by (a-ii), so extends to a quasi-Radon measure on X, by 415Cb. If effectively locally finite Borel measures on X extend to quasi-Radon measures, then any totally finite Borel measure is τ -additive, by 411C, and X is Borel-measure-complete. (c) Immediate from the definitions. (d) If Y ⊆ X and µ is a totally finite Borel measure on Y , let ν be the Borel measure on X defined by setting νE = µ(E ∩ Y ) for every Borel set E ⊆ X. It is easy to check that ν is τ -additive. So if G is an upwards-directed family of relatively open subsets of Y , and we set H = {H : H ⊆ X is open, H ∩ Y ∈ G}, we shall get S S µ( G) = ν( H) = supH∈H νH = supG∈G µG. As µ and G are arbitrary, Y is Borel-measure-complete. (e) If µ is a totally finite Borel measure on X and G is a non-empty upwards-directed family of open subsets of X with union G∗ , then there is a sequence hGn in∈N in G with union G∗ , by 4A2H(c-i). Because G is upwards-directed, there is a non-decreasing sequence hG0n in∈N in G such that G0n ⊇ Gn for every n ∈ N, so that µG∗ = limn→∞ µG0n ≤ supG∈G µG. As µ and G are arbitrary, X is Borel-measure-complete.

434J

Borel measures

191

434J Proposition Let X be a Hausdorff space. (a) The following are equiveridical: (i) X is pre-Radon; (ii) every effectively locally finite τ -additive Borel measure on X is tight; (iii) whenever µ is a non-zero totally finite τ -additive Borel measure on X, there is a compact set K ⊆ X such that µK > 0; (iv) whenever µ is a totally finite τ -additive Borel measure on X, µX = supK⊆X is compact µK; (v) whenever µ is a locally finite effectively locally finite τ -additive Borel measure on X, the c.l.d. version of µ is a Radon measure on X. (b) If X is pre-Radon, then every locally finite quasi-Radon measure on X is a Radon measure. (c) If X is regular and every totally finite quasi-Radon measure on X is a Radon measure, then X is pre-Radon. (d) If X is pre-Radon, then any universally Radon-measurable subspace (in particular, any Borel subset or Souslin-F subset) A of X is pre-Radon. (e) If A ⊆ X is pre-Radon in its subspace topology, it is universally Radon-measurable in X. (f) If X is K-analytic (for instance, if it is compact), it is pre-Radon. ˇ (g) If X is completely regular and Cech-complete (for instance, if it is locally compact (4A2Gj), or metrizable and complete under a metric inducing its topology (4A2Md)), it is pre-Radon. Q (h) If X = i∈I Xi where hXi ii∈I is a countable family of pre-Radon Hausdorff spaces, then X is pre-Radon. (i) If every point of X belongs to a pre-Radon open subset of X, then X is pre-Radon. proof (a)(i)⇒(ii) Suppose that X is pre-Radon, that µ is an effectively locally finite τ -additive Borel measure on X, that E ⊆ X is Borel, and that γ < µE. Because µ is semi-finite, there is a Borel set H ⊆ X of finite measure such that µ(H ∩ E) > γ. Set νF = µ(F ∩ H) for every Borel set F ⊆ X; then ν is a totally finite Borel measure on X, and is τ -additive, by 414Ea. Now νE > γ, so there is a compact set K ⊆ E such that γ ≤ νK ≤ µK. As E is arbitrary, µ is tight. (ii)⇒(iii) is trivial. (iii)⇒(iv) Assume (iii), and let µ be a totally finite τ -additive Borel measure on X. Let K be the family of compact subsets of X and set α = supK∈K µK. ?? Suppose, if possible, that µX > α. Let hKn in∈N S be a sequence in K such that supn∈N µKn = α, and set L = n∈N Kn ; then S µL = limn→∞ µ( i≤n Ki ) = α. Set νE = µ(E \ L) for every Borel set E ⊆ X. Then ν is a non-zero totally finite Borel measure on X, and is τ -additive, by 414E again. So there is a K ∈ K such that νK > 0. But now there is an n ∈ N such that νK + µKn > α, and in this case K ∪ Kn ∈ K and µ(K ∪ Kn ) = µK + µKn − µ(K ∩ Kn ) ≥ νK + µKn > α, which is impossible. X X So µX = α, as required. (iv)⇒(i) Assume (iv), and let µ be a totally finite τ -additive Borel measure on X. Suppose that E ⊆ X is Borel and that γ < µE. By (iii), there is a compact set K ⊆ X such that µK > µX − µE + γ, so that µ(E ∩ K) > γ. Consider the subspace measure µK on K. By 414K, this is τ -additive, so inner regular with respect to the closed subsets of K (414Mb). There is therefore a relatively closed subset F of K such that F ⊆ K ∩ E and µK F ≥ γ; but now F is a compact subset of E and µF ≥ γ. As E and γ are arbitrary, µ is tight. As µ is arbitrary, X is pre-Radon. (ii)⇒(v) Assume (ii), and let µ be a locally finite effectively locally finite τ -additive Borel measure on X. Then µ is tight, so by 416F(ii) its c.l.d. version is a Radon measure. (v)⇒(iv) Assume (v), and let µ be a totally finite τ -additive Borel measure on X. Then the c.l.d. version µ ˜ of µ is a Radon measure; but µ ˜ extends µ (213Hc), so supK⊆X

is compact

µK supK⊆X

is compact

µ ˜K = µ ˜X = µX.

(b) Let µ be a locally finite quasi-Radon measure on X. By (a-ii), µ is tight; by 416C, µ is a Radon measure.

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(c) Let µ be a totally finite τ -additive Borel measure on X. Because X is regular, its c.l.d. version µ ˜ is a quasi-Radon measure (415C), therefore a Radon measure; but µ ˜ extends µ (213Hc again), so µ, like µ ˜, must be tight. As µ is arbitrary, X is pre-Radon. (d) A be a universally Radon-measurable subset of X, and µ a totally finite τ -additive Borel measure on A. Set νE = µ(E ∩ A) for every Borel set E ⊆ X; then ν is a totally finite τ -additive Borel measure on X. So its c.l.d. version (that is, its completion νˆ, by 213Ha) is a Radon measure on X, by (a-v). Now νˆ measures A, so µA = ν ∗ A = νÂ = sup{ˆ ν K : K ⊆ A is compact} = sup{µK : K ⊆ A is compact}. By (b-iv), A is pre-Radon. (e) Let µ be a totally finite Radon measure on X. Then the subspace measure µA is τ -additive (414K), so its restriction ν to the Borel σ-algebra of A is still τ -additive. Because A is pre-Radon, µ∗ A = µA A = νA = sup{νK : K ⊆ A is compact} = sup{µK : K ⊆ A is compact} = µ∗ A, and µ measures A (413Ef). As µ is arbitrary, A is universally Radon-measurable. (f ) Put 432B and (a-iv) together. (g) If we identify X with a Gδ set in a compact Hausdorff space Z, then Z is pre-Radon, by (f), so X is pre-Radon, by (d). (h) Let µ be a totally finite τ -additive Borel measure on X, and ² > 0. Let h²i ii∈I be a family of strictly P positive real numbers such that i∈I ²i ≤ ² (4A1P). For each i ∈ I, Borel set F ⊆ Xi set µi F = µπi−1 [F ], where πi (x) = x(i) for x ∈ X; because πi : X → Xi is continuous, µi is a totally finite τ -additive Borel measure on Xi . Because S set Ki ⊆ Xi such that µi (Xi \ Ki ) ≤ ²i , Q Xi is pre-Radon, we can find a compact by (a-iv). Now K = i∈I Ki is compact (3A3J), and X \ K ⊆ i∈I πi−1 [Xi \ Ki ], so P P µ(X \ K) ≤ i∈I µi (Xi \ Ki ) ≤ i∈I ²i ≤ ². As ² and µ are arbitrary, X satisfies the condition of (b-iv), and is pre-Radon. (i) Let G be a cover of X by pre-Radon open sets. Let µ be a non-zero totally finite τ -additive Borel S measure on X. Then µX = sup{µ( G0 ) : G0 ⊆ G is finite}, so there is some G ∈ G such that µG > 0. Now the subspace measure µG is a non-zero totally finite τ -additive Borel measure on G, so there is a compact set K ⊆ G such that µG K > 0, in which case µK > 0. As µ is arbitrary, X is pre-Radon, by (a-iii). 434K I return to criteria for deciding whether Hausdorff spaces are Radon. Proposition (a) A Hausdorff space is Radon iff it is Borel-measure-complete and pre-Radon. (b) An analytic Hausdorff space is Radon. In particular, any compact metrizable space is Radon and any Polish space is Radon. (c) ω1 and ω1 + 1, with their order topologies, are not Radon. (d) For a set I, [0, 1]I is Radon iff {0, 1}I is Radon iff I is countable. (e) A hereditarily Lindelöf K-analytic Hausdorff space is Radon; in particular, the split interval (343J, 419L) is Radon. proof (a) Put the definitions 434C, 434Gb and 434Gc together, recalling that a tight measure is necessarily τ -additive (411E). (b) 433Cb. (c) Dieudonné’s measure (411Q) is a Borel measure on ω1 which is not tight, so ω1 is certainly not a Radon space; as it is an open set in ω1 + 1, and the subspace topology inherited from ω1 + 1 is the order topology of ω1 (4A2S(a-iii)), ω1 + 1 cannot be Radon (434Fc). (d) If I is countable, then {0, 1}I and [0, 1]I are compact metrizable spaces, so are Radon. If I is uncountable, then ω1 + 1, with its order topology, is homeomorphic to a closed subset of {0, 1}I . P P Set

434M

Borel measures

193

κ = #(I). For ξ ≤ ω1 , η < κ set xξ (η) = 1 if η < ξ, 0 if ξ ≤ η. The map ξ 7→ xξ : ω1 + 1 → {0, 1}κ is injective because κ ≥ ω1 , and is continuous because all the sets {ξ : xξ (η) = 0} = (ω1 + 1) ∩ (η + 1) are open-and-closed in ω1 + 1. Since ω1 + 1 is compact in its order topology (4A2S(a-i)), it is homeomorphic to its image in {0, 1}κ ∼ Q = {0, 1}I . Q I By 434Fc, {0, 1} cannot be a Radon space. Since {0, 1}I is a closed subset of [0, 1]I , [0, 1]I is also not a Radon space. (e) Suppose that X is a hereditarily Lindelöf K-analytic Hausdorff space. Then it is Borel-measurecomplete by 434Ie and pre-Radon by 434Jf, so by (a) here it is Radon. Since the split interval is compact and Hausdorff and hereditarily Lindelöf (419La), it is a Radon space. 434L It is worth noting an elementary special property of metric spaces. Proposition If (X, ρ) is a metric space, then any quasi-Radon measure on X is inner regular with respect to the totally bounded subsets of X. proof Let µ be a quasi-Radon measure on X and Σ its domain. Suppose that E ∈ Σ and γ < µE. Then there is an open S set G of finite measure such that µ(E ∩ G) > γ; set δ = µ(E ∩ G) − γ. For n ∈ N, I ⊆ X set H(n, I) = x∈I {y : ρ(y, x) < 2−n }. Then {H(n, I) : I ∈ [X]<ω } is an upwards-directed family of open −n−1 sets covering X. δ. T Because µ is τ -additive, there is a finite set In ⊆ X such that µ(G \ H(n, In )) ≤ 2 Consider F = n∈N H(n, In ). This is totally bounded and µ(G \ F ) ≤ δ, so E ∩ F is totally bounded and µ(E ∩ F ) ≥ γ. As E and γ are arbitrary, µ is inner regular with respect to the totally bounded sets. 434M I turn next to a couple of ideas depending on countable compactness. Lemma Let X be a countably compact topological space and E a non-empty family of closed subsets of X with the finite intersection property. Then there is a Borel probability measure µ on X, inner regular with respect to the closed sets, such that µF = 1 for every F ∈ E. proof (a) By Zorn’s lemma, E is included in a maximal family E ∗ of closed subsets of X with the finite intersection property. (i) If F ⊆ X is closed and F ∩ F0 ∩ . . . ∩ Fn 6= ∅ for every F0 , . . . , Fn ∈ E ∗ , then E ∗ ∪ {F } has the finite intersection property, so F ∈ E ∗ . (ii) If F , F 0 ∈ E ∗ , then F ∩ F 0 ∩ F0 ∩ . . . ∩ Fn 6= ∅ for all F0 , . . . , Fn ∈ E ∗ , so F ∩ F 0 ∈ E ∗ . (iii) If F ⊆ X is closed and F ∩ F 0 ∈ E ∗ for every F 0 ∈ E ∗ , then F ∩ F0 ∩ . . . ∩ Fn ∈ E ∗ for every F0 , . . . , Fn ∈ E ∗ (because F0 ∩ . . . ∩ Fn ∈ E ∗ , by (ii)), so F ∈ E ∗ . T (iv) If hFn in∈N is a sequence in E ∗ , with intersection F , and F 0 ∈ E ∗ , then F 0 ∩ i≤n Fi 6= ∅ for every n ∈ N. Because X is countably compact, F 0 ∩ F 6= ∅ (4A2G(f-ii)). As F 0 is arbitrary, F ∈ E ∗ , by (iii). Thus E ∗ is closed under countable intersections. (b) Set Σ = {E : E ⊆ X, there is an F ∈ E ∗ such that either F ⊆ E or F ∩ E = ∅}, and define µ ˆ : Σ → {0, 1} by saying that µ Ê = 1 if there is some F ∈ E ∗ such that F ⊆ E, 0 otherwise. Then µ ˆ is a probability measure on X. P P (i) ∅ ∈ Σ because E ∗ ⊇ E is not empty. (ii) X \ E ∈ Σ whenever E ∈ Σ because the definition of Σ is symmetric between E and X \ E. (iii) If hEn in∈N is any sequence in Σ with union E, then either there are n ∈ N, F ∈ E ∗ such that F ⊆T En ⊆ E and E ∈ Σ, or for every n ∈ N there is an Fn ∈ E ∗ such that Fn ∩ En = ∅. In this case F = n∈N Fn ∈ E ∗ , by (a-iv), and E ∩ F = ∅, so again E ∈ Σ. Thus Σ is a σ-algebra of subsets of X. (iv) µ ˆ∅ = 0 because ∅ cannot belong to E ∗ . (v) If hEn in∈N is any disjoint sequence in Σ with union E, then either there is some n such thatP µ Ên = 1, ∞ in which case µ Êi = 0 for every i 6= n (because any two membersP of E ∗ must meet) and µ Ê = 1 = i=0 µ ˆ Ei , ∞ or µ Êi = 0 for every i, in which case, just as in (iii), µ Ê = 0 = i=0 µ Êi . Thus µ ˆ is a measure. (vi) Because E ∗ 6= ∅, µ ˆX = 1. Thus µ ˆ is a probability measure. Q Q (c) If F ⊆ X is a closed set, then either F itself belongs to E ∗ , so F ∈ Σ, or there is some F 0 ∈ E ∗ such that F ∩ F 0 = ∅, in which case again F ∈ Σ. So Σ contains every closed set, therefore every Borel set, and µ ˆ

194


434M

is a topological measure. By construction, µ ˆ is inner regular with respect to E ∗ and therefore with respect to the closed sets. Finally, if F ∈ E then F ∈ E ∗ , so µ ˆF = 1. We may therefore take µ to be the restriction of µ ˆ to the Borel σ-algebra of X, and µ will be a Borel measure on X, inner regular with respect to the closed sets, such that µE = 1 for every E ∈ E. 434N Proposition (a) Let X be a Borel-measure-compact topological space. Then closed countably compact subsets of X are compact. (b) Let X be a Borel-measure-complete topological space. Then closed countably compact subsets of X are compact. (c) Let X be a Hausdorff Borel-measure-complete topological space. Then compact subsets of X are countably tight. (d) In particular, any Radon compact Hausdorff space is countably tight. proof (a) Let C be a closed countably compact subset of X. Let F be an ultrafilter on C. Let E be the family of closed subsets of C belonging to F. Then E has the finite intersection property, so by 434M there is a Borel probability measure µ on C, inner regular with respect to the closed sets, such that µE = 1 for every E ∈ E. Let ν be the Borel measure on X defined by setting νH = µ(C ∩ H) for every Borel set H ⊆ X. Then ν is also inner regular with respect to the closed sets (of either C or X); because X is Borel-measure-compact, ν has a support F (434H(a-iv)). Since νF = νX = 1, F ∩ C 6= ∅; take x ∈ F ∩ C. If G is any open set (in X) containing x, then ν(X \ G) < 1, so C \ G ∈ / F and C ∩ G ∈ F. As G is arbitrary, F → x; as F is arbitrary, C is compact. (b) Repeat the argument of (a). Let C be a countably compact subset of X and F be an ultrafilter on C. Let E be the family of relatively closed subsets of C belonging to F. Then there is a Borel probability measure µ on C such that µE = 1 for every E ∈ E. Let ν be the Borel measure on X defined by setting νH = µ(C ∩ H) for every Borel set H ⊆ X. Because X is Borel-measure-complete, ν has a support F (434I(a-iii)). Since νF = νX = 1, F ∩ C 6= ∅; take x ∈ F ∩ C. If G is any open set containing x, then ν(X \ G) < 1, so C \ G ∈ / F and C ∩ G ∈ F. As G is arbitrary, F → x; as F is arbitrary, C is compact. S (c) Again let C be a (countably) compact subset of X. Take A ⊆ C, and set C0 = {B : B ∈ [A]≤ω }. Then C0 is countably compact. P P If hyn in∈N is any sequence in C0 , itShas a cluster point y ∈ C. For each n ∈ N there is a countable set Bn ⊆ A such that yn ∈ Bn . Now B = n∈N Bn is a countable subset of A, and y ∈ B ⊆ C0 , so y is a cluster point of hyn in∈N in C0 . As hyn in∈N is arbitrary, C0 is countably compact. Q Q By (b), C0 is compact, therefore closed, and must include A. Thus every point of A is in the closure of some countable subset of A. As A is arbitrary, C is countably tight. (d) Finally, a compact Radon Hausdorff space is Borel-measure-complete (434Ka) and countably compact, therefore countably tight. 434O Quasi-dyadic spaces I wish now to present a result in an entirely different direction. Measures of type B1 in the classification of 434A (completion regular, but not τ -additive) seem to be hard to come by. The next theorem shows that on a substantial class of spaces they cannot appear. First, we need a definition. Definition A topological space X is quasi-dyadic if it is expressible as a continuous image of a product of separable metrizable spaces. I give some elementary results to indicate what kind of spaces we have here. 434P Proposition (a) A continuous image of a quasi-dyadic space is quasi-dyadic. (b) Any product of quasi-dyadic spaces is quasi-dyadic. (c) A space with a countable network is quasi-dyadic. (d) A Baire subset of a quasi-dyadic space is quasi-dyadic. (e) If X is any topological space, a countable union of quasi-dyadic subspaces of X is quasi-dyadic. proof (a) Immediate from the definition.

434Q

Borel measures

195

Q (b) Again immediate; if Xi is a continuous image of j∈Ji Yij , where Yij is a separable metric space for Q Q every i ∈ I, j ∈ Ji , then i∈I Xi is a continuous image of i∈I,j∈Ji Yij . (c) Let E be a countable network for the topology of X. On X let ∼ be the equivalence relation in which x ∼ y if they belong to just the same members of E; let Y be the space X/ ∼ of equivalence classes, and φ : X → Y the canonical map. Y has a separable metrizable topology with base {φ[E] : E ∈ E} ∪ {φ[X \ E] : E ∈ E}. Let I be any set such that #({0, 1}I ) ≥ #(X), and for each y ∈ Y let fy : {0, 1}I → y be a surjection. Then we have a continuous surjection f : Y × {0, 1}I → X given by saying that f (y, z) = fy (z) for y ∈ Y , z ∈ {0, 1}I . (d) Let hYi ii∈I be a family of separable metric spaces with product Y and f : Y → X a continuous surjection. If W ⊆ Y is a Baire subset Q set, it is determined by coordinates in a countable Q Qof I (4A3Nb), so can be regarded as W 0 × i∈I\J Yi , where J ⊆ I is countable and W 0 ⊆ i∈J Yi ; as i∈J Yi and W 0 are separable metrizable spaces (4A2Pa), W can be thought of as a product of separable metrizable spaces. Now the set {E : E ⊆ X, f −1 [E] is a Baire set in Y } is a σ-algebra containing every zero set in X, so contains every Baire set. Thus every Baire subset of X is a continuous image of a Baire subset of Y , and is therefore quasi-dyadic. Q S (e) If En ⊆ X is quasi-dyadic for each n ∈ N, then Z = N ×S n∈N En is quasi-dyadic, and f : Z → n∈N En is a continuous surjection, where f (n, hxi ii∈N ) = xn . So n∈N En is quasi-dyadic. 434Q Theorem(Fremlin & Grekas 95) A semi-finite completion regular topological measure on a quasi-dyadic space is τ -additive. proof ?? Suppose, if possible, otherwise. (a) The first step is the standard reduction to the case in which µX = 1 and X is covered by open sets of zero measure. In detail: suppose that X is a quasi-dyadic space and µ0 is a semi-finite completion regular topological measure on X which is not τ -additive. Let G be an upwards-directed family of open sets in X S such that µ0 ( G) is strictly greater than supG∈G µ S S0 G = γ say. Let hGn in∈N be a non-decreasing sequence in G such that limn→∞ µ0 Gn = γ, and set H0 = n∈N Gn , so that µ0 H0 = γ; take a closed set Z ⊆ G such that γ < µ0 Z < ∞. Set µ1 E = µ0 (E ∩ Z \ H0 ) for every Borel set E ⊆ X. Then µ1 is a non-zero totally finite Borel measure on X, and is completion regular. P P If E ⊆ X is a Borel set and ² > 0, there is a zero set F ⊆ E ∩ Z \ H0 such that µ0 F ≥ µ0 (E ∩ X \ H0 ) − ², and now µ1 F ≥ µ1 E − ². Q Q Note that µ1 (X \ Z) = µ1 G = 0 for every G ∈ G. For Borel sets E ⊆ X, set µE = µ1 E/µ1 X; then µ is a completion regular Borel probability measure on X, and G ∪ {X \ Z} is a cover of X by open negligible sets. (b) Now let hYi ii∈I Q be a family of separable metric spaces such that there is a continuous surjection f : Y → X, where Y = i∈I Yi . For each i ∈ I let Bi be a countable base for the topology of Yi . For J ⊆ I let C(J) be the family of all open cylinders in Y expressible in the form {s : s(i) ∈ Bi ∀ i ∈ K}, where K is a finite subset of J and Bi ∈ Bi for each i ∈ K; thus C(I) is a base for the topology of Y . Set C0 (J) = {U : U ∈ C(J), µ∗ f [U ] = 0} for each J ⊆ I. Note that (because every Bi is countable) C(J) and C0 (J) are countable for every countable subset J of I. It is easy to see that C(J) ∩ C(K) = C(J ∩ K) for all J, K ⊆ I, because if U ∈ C(I) is not empty it belongs to C(J) iff its projection onto Xi is the whole of Xi for every i ∈ / J. For each negligible set E ⊆ X, let hFn (E)in∈N be a family of zero sets, subsets of X \ E, such that supn∈N µFn = 1. Then each f −1 [Fn (E)] is a zero set in Y , so there is a countable set M (E) ⊆ I such that all the sets f −1 [Fn (E)] are determined by coordinates in M (E) (4A3Nc). Let J be the family of countable subsets J of I such that M (f [U ]) ⊆ J for every U ∈ C0 (J); then J is cofinal with [I]≤ω , that is, every countable subset of I is included in some member of J . P P If we start from any countable subset J0 of I and set S Jn+1 = Jn ∪ {M (f [U ]) : U ∈ C0 (Jn )} S for S each n ∈ N, S then every Jn is countable, and n∈N Jn ∈ J , because hJn in∈N is non-decreasing, so C0 ( n∈N Jn ) = n∈N C0 (Jn ). Q Q

196


(c) For each J ∈ J , set QJ =

434Q

T S { n∈N Fn (f [U ]) : U ∈ C0 (J)}.

Then µQJ = 1 and f −1 [QJ ] is determined by coordinates in J, while f −1 [QJ ] ∩ U = ∅ whenever U ∈ C0 (J). If G ⊆ X is an open set, then G ∩ QJ = ∅ whenever J ∈ J and there is a negligible Baire set Q Q⊇G such that f −1 [Q] is determined by coordinates in J. P P Set H = πJ−1 [πJ [f −1 [G]]], where πJ : Y → i∈J Yi is the canonical map; then H is a union of members of C(J), because f −1 [G] is open in Y and πJ [f −1 [G]] Q is open in i∈J Yi . Also, because f −1 [Q] is determined by coordinates in J, H ⊆ f −1 [Q], so f [H] ⊆ Q and µ∗ f [H] = 0; thus all the members of C(J) included in H actually belong to C0 (J), and H ∩ f −1 [QJ ] = ∅. But this means that f −1 [G] ∩ f −1 [QJ ] = ∅ and (because f is a surjection) G ∩ QJ = ∅, as claimed. Q Q In particular, if G is a negligible open set in X, then G ∩ QJ = ∅ whenever J ∈ J and J ⊇ M (G). (d) If J ∈ J , there are s, s0 ∈ f −1 [QJ ] such that s¹J = s0 ¹J and f (s), f (s0 ) can be separated by open sets in X. P P Start from any x ∈ QJ and take a negligible open set G including x (recall that our hypothesis is that X is covered by negligible open sets). For each n ∈ N let hn : X → S R be a continuous function such that Fn (G) = h−1 [{0}]. We know that G ∩ Q = 6 ∅, while G ⊆ X \ ( J n n∈N Fn (G) ∩ QJ ), which is a S negligible Baire set; by (c), f −1 [X \ ( n∈N Fn (G) ∩ QJ )] is not determined by coordinates in J, and there must be some n such that f −1 [Fn (G) ∩ QJ ] is not determined by coordinates in J. Accordingly there must be s, s0 ∈ Y such that s¹J = s0 ¹J, s ∈ f −1 [Fn (G) ∩ QJ ] and s0 ∈ / f −1 [Fn (G) ∩ QJ ]. Now s ∈ f −1 [QJ ], 0 0 −1 which is determined by coordinates in J; since s¹J = s ¹J, s ∈ f [QJ ] and s0 ∈ / f −1 [Fn (G)]. Accordingly 0 0 hn (f (s)) = 0 6= hn (f (s )) and f (s), f (s ) can be separated by open sets. Q Q (e) We are now ready to embark on the central construction of the argument. We may choose inductively, for ordinals ξ < ω1 , sets Jξ ∈ J , negligible open sets Gξ , G0ξ ⊆ X, points sξ , s0ξ ∈ Y and sets Uξ , Vξ , Vξ0 ∈ C(I) such that Jη ⊆ Jξ , Uη , Vη , Vη0 all belong to C(Jξ ) and Gη ∩ QJξ = ∅ whenever η < ξ < ω1 (using the results of (b) and (c) to choose Jξ ); sξ ¹Jξ = s0ξ ¹Jξ , sξ ∈ f −1 [QJξ ] and f (sξ ) and f (s0ξ ) can be separated by open sets in X (using (d) to choose sξ , s0ξ ); Gξ , G0ξ are disjoint negligible open sets containing f (sξ ), f (s0ξ ) respectively (choosing Gξ , G0ξ ); Uξ ∈ C(Jξ ), Vξ , Vξ0 ∈ C(I \ Jξ ), sξ ∈ Uξ ∩ Vξ ⊆ f −1 [Gξ ], s0ξ ∈ Uξ ∩ Vξ0 ⊆ f −1 [G0ξ ] (choosing Uξ , Vξ , Vξ0 , using the fact that sξ ¹Jξ = s0ξ ¹Jξ ). On completing this construction, take for each ξ < ω1 a finite set Kξ ⊆ Jξ+1 such that Uξ , Vξ and Vξ0 all belong to C(Kξ ). By the ∆-system Lemma (4A1Db), there is an uncountable A ⊆ ω1 such that hKξ iξ∈A is a ˜ξ ∩ U 0 where U ˜ξ ∈ C(K), U 0 ∈ C(Kξ \ K). Then there ∆-system with root K say. For ξ ∈ A, express Uξ as U ξ ξ ˜ξ , so there is an uncountable B ⊆ A such that U ˜ξ is constant for are only countably many possibilities for U ˜ for the constant value. Let C ⊆ B be an uncountable set, not containing min A, such that ξ ∈ B; write U Kξ \ K does not meet Jη whenever ξ, η ∈ C and η < ξ (4A1Eb). Let D ⊆ C be such that D and C \ D are both uncountable. Note that K ⊆ Kη ⊆ Jξ whenever η, ξ ∈ A and η < ξ, so that K ⊆ Jξ for every ξ ∈ C. Consequently Uξ0 , Vξ and Vξ0 all belong to C(Kξ \ K) for every ξ ∈ C. (f ) Consider the open set G=

S ξ∈D

Gξ ⊆ X.

At this point the argument divides. ˜ ]) > 0. Then there is a Baire set Q ⊆ G such that µ∗ (Q ∩ f [U ˜ ]) > 0. case 1 Suppose µ∗ (G ∩ f [U Let J ⊆ I be a countable set such that f −1 [Q] is determined by coordinates in J. Let γ ∈ C \ D be so ˜ ] is not empty; take large that Kξ \ K does not meet J for any ξ ∈ A with ξ ≥ γ. Then Q ∩ QJγ ∩ f [U −1 ˜ s ∈ U ∩ f [Q ∩ QJγ ]. Because the Kξ \ K are disjoint from each other and from J ∪ Jγ for ξ ∈ D, ξ > γ, we may modify s to form s0 such that s0 ¹J ∪ Jγ = s¹J ∪ Jγ and s0 ∈ Uξ0 ∩ Vξ0 for every ξ ∈ D, ξ > γ; now ˜ (because K ⊆ Jγ ), so s0 ∈ U ˜ ∩ U 0 ∩ V 0 ⊆ f −1 [G0 ] and f (s0 ) ∈ s0 ∈ U / Gξ whenever ξ ∈ D, ξ > γ. On the ξ ξ ξ

434R

Borel measures

197

other hand, if ξ ∈ D and ξ < γ, Gξ ∩ QJγ = ∅, while s0 ∈ f −1 [QJγ ] (because f −1 [QJγ ] is determined by coordinates in Jγ ), so again f (s0 ) ∈ / Gξ . Thus f (s0 ) ∈ / G. But s0 ¹J = s¹J so f (s) ∈ Q ⊆ G; which is impossible. ˜ ]) > 0. This contradiction disposes of the possibility that µ∗ (G ∩ f [U ˜ ]) = 0. In this case there is a negligible Baire set Q ⊇ G ∩ f [U ˜ ]. Let case 2 Suppose that µ∗ (G ∩ f [U −1 J ⊆ I be a countable set such that f [Q] is determined by coordinates in J. Let ξ ∈ D be such that Kξ \ K does not meet J. Then ˜ ∩ U 0 ∩ Vξ ⊆ f −1 [Gξ ] ∩ U ˜ ⊆ f −1 [G ∩ f [U ˜ ]] ⊆ f −1 [Q], U ξ

˜ ⊆ f −1 [Q], because U 0 ∩ Vξ is a non-empty member of C(I \ J). But this means that µ∗ f [U ˜ ] = 0 and so U ξ ∗ −1 ∗ µ f [Uξ ] = 0. On the other hand, we have sξ ∈ Uξ ∩ f [QJξ ], so Uξ ∈ / C0 (Jξ ) and µ f [Uξ ] > 0. X X Thus this route also is blocked and we must abandon the original hypothesis that there is a quasi-dyadic space with a semi-finite completion regular topological measure which is not τ -additive. 434R There is a useful construction of Borel product measures which can be fitted in here. Proposition Let X and Y be topological spaces with Borel measures µ and ν; write B(X), B(Y ) for the Borel σ-algebras of X and Y respectively. If either X is first-countable or ν is τ -additive and effectively locally finite, there is a Borel measure λB on X × Y defined by the formula λB W = supF ∈B(Y ),νF <∞

R

ν(W [{x}] ∩ F )µ(dx)

for every Borel set W ⊆ X × Y . Moreover b (i) if µ is semi-finite, then λB agrees with the c.l.d. product measure λ on B(X)⊗B(Y ), and the c.l.d. ˜ version λB of λB extends λ; R (ii) if ν is σ-finite, then λB W = νW [{x}]µ(dx) for every Borel set W ⊆ X × Y ; (iii) if both µ and ν are τ -additive and effectively locally finite, then λB is just the restriction of the ˜ ( 417D, 417G) to the Borel σ-algebra of X × Y ; in particular, λB is τ -additive. τ -additive product measure λ proof (a) The point is that x 7→ ν(W [{x}] ∩ F ) is lower semi-continuous whenever W ⊆ X × Y is open and νF < ∞. P P Of course W [{x}] is always open, so ν always measures W [{x}] ∩ F . Take any α ∈ R and set G = {x : x ∈ X, ν(W [{x}] ∩ F ) > α}; let x0 ∈ G. α) Suppose that X is first-countable. Let hUn in∈N be a non-increasing sequence running over a (α base of open neighbourhoods of x0 . For each n ∈ N, set S Vn = {V : V ⊆ Y is open, Un × V ⊆ W }. Then hVn in∈N is a non-decreasing sequence with union W [{x}], so there is an n ∈ N such that ν(Vn ∩F ) > α. Now Vn ⊆ W [{x}] for every x ∈ Un , so Un ⊆ G. β ) Suppose that ν is τ -additive and effectively locally finite. Set (β V = {V : V ⊆ Y is open, U × V ⊆ W for some open set U containing x0 }. Then V is an upwards-directed family of open sets with union W [{x0 }], so there is a V ∈ V such that ν(V ∩ F ) > α (414Ea). Let U be an open set containing x0 such that U × V ⊆ W ; then V ⊆ W [{x}] for every x ∈ U , so U ⊆ G. (γγ ) Thus in either case we have an open set containing x0 and included in G. As x0 is arbitrary, G is open; as α is arbitrary, x 7→ ν(W [{x}] ∩ F ) is lower semi-continuous. Q Q (b) It follows that x 7→ ν(W [{x}] ∩ F ) is Borel measurable whenever W ⊆ X × Y is a Borel set and νF < ∞. P P Let W be the family of sets W ⊆ X × Y such that W [{x}] is a Borel set for every x ∈ X and x 7→ ν(W [{x}] ∩ F ) is Borel measurable. Then every S open subset of X × Y belongs to W (by (a) above), W \W 0 ∈ W whenever W , W 0 ∈ W and W 0 ⊆ W , and n∈N Wn ∈ W whenever hWn in∈N is a non-decreasing sequence in W. By the Monotone Class Theorem (136B), W includes the σ-algebra generated by the open sets, that is, the Borel σ-algebra of X × Y . Q Q R (c) It is now easy to check that W 7→ ν(W [{x}] ∩ F )µ(dx) is a Borel measure on X × Y whenever νF < ∞, and therefore that λB , as defined here, is a Borel measure.

198


434R

b (d) Now suppose that µ is semi-finite, and that W ∈ B(X)⊗B(Y ). Then

λW =

sup

λ(W ∩ (E × F ))

µE<∞,νF <∞

(by the definition of ‘c.l.d. product measure’, 251F) Z = sup ν(W [{x}] ∩ F )µ(dx) µE<∞,νF <∞

E

(by Fubini’s theorem, 252C, applied to the product of the subspace measures µE and νF ) Z = sup ν(W [{x}] ∩ F )µ(dx) νF <∞

(by 213B, because µ is semi-finite) = λB W. (e) If, on the other hand, ν is σ-finite, let hFn in∈N be a non-decreasing sequence of sets of finite measure covering Y ; then Z Z λB W = sup ν(W [{x}] ∩ F )µ(dx) ≥ sup ν(W [{x}] ∩ Fn )µ(dx) νF <∞ n∈N Z Z = sup ν(W [{x}] ∩ Fn )µ(dx) = νW [{x}]µ(dx) ≥ λB W n∈N

for any Borel set W ⊆ X. (f ) If both µ and ν are τ -additive and effectively locally finite, so that we have a τ -additive product ˜ then Fubini’s theorem for such measures (417H) tells us that λB W = λW ˜ at least when W ⊆ measure λ, ˜ is finite. If W is any Borel subset of X × Y , then, as in (d), X × Y is a Borel set and λW Z λB W = sup ν(W [{x}] ∩ F )µ(dx) µE<∞,νF <∞

=

sup

E

˜ ˜ λ(W ∩ (E × F )) = λW

µE<∞,νF <∞

by 417C(iii). Remark The case in which X is first-countable is due to Johnson 82. *434S The concept of ‘universally measurable’ set enables us to extend a number of ideas from earlier sections. First, recall a problem from the very beginning of measure theory on the real line: the composition of Lebesgue measurable functions need not be Lebesgue measurable (134Ib), while the composition of a Borel measurable function with a Lebesgue measurable function is measurable (121Eg). In fact we can replace ‘Borel measurable’ by ‘universally measurable’, as follows. Proposition Let (X, Σ, µ) be a complete locally determined measure space, Y and Z topological spaces, f : X → Y a measurable function and g : Y → Z a universally measurable function. Then gf : X → Z is measurable. In particular, f −1 [F ] ∈ Σ for every universally measurable set F ⊆ Y . proof Let H ⊆ Z be an open set and E ∈ Σ a set of finite measure. Let µE be the subspace measure on E. Then the image measure ν = µE (f ¹E)−1 is a complete totally finite topological measure on Y , so its domain contains g −1 [H], and E ∩ (gf )−1 [H] = (f ¹E)−1 [g −1 [H]] ∈ dom µE ⊆ Σ. As E is arbitrary and µ is locally determined, (gf )−1 [H] ∈ Σ; as H is arbitrary, gf is measurable. Applying this to g = χF , we see that f −1 [F ] ∈ Σ for every universally measurable F ⊆ Y .

434Xe

Borel measures

199

*434T The next remark concerns the concept [[u ∈ E]] of §364. Proposition Let (A, µ ¯) be a localizable measure algebra. Write Σum for the algebra of universally measurable subsets of R. (a) For any u ∈ L0 = L0 (A), we have a sequentially order-continuous Boolean homomorphism E 7→ [[u ∈ E]] : Σum → A defined by saying that [[u ∈ E]] = sup{[[u ∈ F ]] : F ⊆ E is Borel} = inf{[[u ∈ F ]] : F ⊇ E is Borel} for every E ∈ Σum . ¯ (b) For any u ∈ L0 and universally measurable function h : R → R we have a corresponding element h(u) 0 of L defined by the formula ¯ [[h(u) ∈ E]] = [[u ∈ h−1 [E]]] for every E ∈ Σum , u ∈ L0 . proof We can regard (A, µ ¯) as the measure algebra of a complete strictly localizable measure space (X, Σ, µ) (322N), in which case L0 can be identified with L0 (µ) (364J). Write B for the Borel σ-algebra of R. (a) If u ∈ L0 , let f : X → R be a Σ-measurable function representing u. Then f −1 [E] ∈ Σ for every E ∈ Σum , by 434S. Setting φE = (f −1 [E])• , φ : Σum → A is a sequentially order-continuous Boolean homomorphism. Now φE = sup{[[u ∈ F ]] : F ∈ B, F ⊆ E} = inf{[[u ∈ F ]] : F ∈ B, F ⊇ E} for every E ∈ Σum . P P If H ∈ Σ and µH < ∞, then (writing µH for the subspace measure on H) the image measure µH (f ¹H)−1 is a complete topological measure, and its restriction ν to the Borel σ-algebra B of R is a totally finite Borel measure. Now E is measured by the completion of ν, so there are F1 , F2 ∈ B such that F1 ⊆ E ⊆ F2 and ν(F2 \ F1 ) = 0. In this case, [[u ∈ F1 ]] = (f −1 [F1 ])• ⊆ φE ⊆ (f −1 [F2 ])• = [[u ∈ F2 ]], using the formula of 364Jb, while H • ∩ [[u ∈ F1 ]] = H • ∩ [[u ∈ F2 ]]. As (A, µ ¯) is semi-finite, sup{[[u ∈ F ]] : F ∈ B, F ⊆ E} and inf{[[u ∈ F ]] : F ∈ B, F ⊇ E} must both be equal to φE. Q Q We can therefore identify the sequentially order-continuous Boolean homomorphism φ with E 7→ [[u ∈ E]], as described. (b) If u ∈ L0 and h : R → R is universally measurable, then, once again identifying u with f • where f is Σ-measurable, we see that hf is Σ-measurable (by 434S), so we have a corresponding element (hf )• of L0 . If E ∈ Σum , then [[(hf )• ∈ E]] = ((hf )−1 [E])• = (f −1 [h−1 [E]])• = [[u ∈ h−1 [E]]], ¯ using 434De to check that h−1 [E] ∈ Σum , so that we can identify (hf )• with h(u), as described. 434X Basic exercises > (a) Let A ⊆ [0, 1] be any non-measurable set. Show that the subspace measure on A is completion regular and τ -additive but not tight. >(b) Let X be any Hausdorff space with a point x such that {x} is not a Gδ set; for instance, X = ω1 + 1 and x = ω1 , or X = {0, 1}I for any uncountable set I and x any point of X. Show that setting µE = χE(x) we get a tight Borel measure on X which is not completion regular. > (c) Let X be a topological space. (i) Show that if A ⊆ X is universally measurable in X, then A ∩ Y is universally measurable in Y for any set Y ⊆ X. (ii) Show that if Y ⊆ X is universally measurable in X, and A ⊆ Y is universally measurable in Y , then A is universally measurable in X. (iii) Suppose that X is the product of a countableQ family hXi ii∈I of topological spaces, and Ei ⊆ Xi is a universally measurable set for each i ∈ I. Show that i∈I Ei is universally measurable in X. (d) Let X be a topological space, Y an analytic Hausdorff space, and W ⊆ X × Y a Borel set. Show that W −1 [Y ] is a universally measurable subset of X. (Hint: 423N.) (e) Let X be a Hausdorff space. (i) Show that, for A ⊆ X, the following are equiveridical: (α) A is universally Radon-measurable in X; (β) A is measured by every atomless Radon probability measure on X;

200


434Xe

(γ) A ∩ K is universally Radon-measurable in K for every compact K ⊆ X. (ii) Show that if A ⊆ X is universally Radon-measurable in X, then A ∩ Y is universally Radon-measurable in Y for any set Y ⊆ X. (iii) Show that if Y ⊆ X is universally Radon-measurable in X, and A ⊆ Y is universally Radon-measurable in Y , then A is universally Radon-measurable in X. (iv) Show that if G is an open cover of X, and A ⊆ X is such that A ∩ G is universally Radon-measurable (in G or in X) for every G ∈ G, then A is universally (X) (Y ) Radon-measurable in X. (v) Show that if Y is another Hausdorff space, and ΣuRm , ΣuRm are the algebras of universally Radon-measurable subsets of X, Y respectively, then every continuous function from X to Y is (X) (Y ) hXi ii∈I of topological (ΣuRm , ΣuRm )-measurable. (vi) Suppose that X is the product of a countable family Q spaces, and Ei ⊆ Xi is a universally Radon-measurable set for each i ∈ I. Show that i∈I Ei is universally Radon-measurable in X. > (f ) (i) Let µ0 be Dieudonné’s measure on ω1 . Give ω1 + 1 = ω1 ∪ {ω1 } its compact Hausdorff order topology, and define a Borel measure µ on ω1 + 1 by setting µE = µ0 (E ∩ ω1 ) for every Borel set E ⊆ ω1 + 1. Show that µ is a complete probability measure and is neither τ -additive nor inner regular with respect to the closed sets. (ii) Show that the universally measurable subsets of ω1 + 1 are just its Borel sets. (Hint: 4A3J, 411Q.) (iii) Show that every totally finite τ -additive topological measure on ω1 + 1 has a countable support. (iv) Show that every subset of ω1 + 1 is universally Radon-measurable. (g) (i) Show that there is a set X ⊆ [0, 1] such that K ∩ X and K \ X are both of cardinal c for every uncountable compact set K ⊆ [0, 1]. (Hint: 419Ib, 423K.) (ii) Show that if we give X its subspace topology, then every subset of X is universally Radon-measurable, but not every subset is universally measurable. (Hint: every compact subset of X is countable, so every Radon measure on X is purely atomic, but X has full outer Lebesgue measure in [0, 1].) (h) Show that a Hausdorff space X is Radon iff (α) every compact subset of X is Radon (β) for every non-zero totally finite Borel measure µ on X there is a compact subset K of X such that µK > 0. (Hint: 434F(a-v).) > (i) (i) Let X and Y be K-analytic Hausdorff spaces and f : X → Y a continuous surjection. Suppose that F ⊆ Y and that f −1 [F ] is universally Radon-measurable in X. Show that F is universally Radonmeasurable in Y . (Hint: 432G.) (ii) Let X and Y be analytic Hausdorff spaces and f : X → Y a Borel measurable surjection. Suppose that F ⊆ Y and that f −1 [F ] is universally Radon-measurable in X. Show that F is universally Radon-measurable in Y . (Hint: 433D.) (j) Show that if X is a perfectly normal space then it is Borel-measure-compact iff it is Borel-measurecomplete. (k) Show that if we give ω1 + 1 its order topology, it is Borel-measure-compact but not Borel-measurecomplete or pre-Radon, and that its open subset ω1 is not Borel-measure-compact. (l) Set X = ω1 + 1, with its order topology, and let Σ be the σ-algebra of subsets of X generated by the countable sets and the set Ω of limit ordinals in X. Show that there is a unique probability measure µ on X with domain Σ such that µξ = µΩ = 0 for every ξ < ω1 . Show that µ is inner regular with respect to the Borel sets, is defined on a base for the topology of the compact Hausdorff space X, but has no extension to a topological measure on X. (m) Show that R, with the right-facing Sorgenfrey topology, is Borel-measure-complete and Borelmeasure-compact, but not Radon or pre-Radon. (n) Let X be a topological space. (i) Show that the family of Borel-measure-compact subsets of X is closed under Souslin’s operation. (ii) Show that the union of a sequence of Borel-measure-complete subsets of X is Borel-measure-complete. (iii) Show that if X is Hausdorff then the family of pre-Radon subsets of X is closed under Souslin’s operation. (Hint: in (i) and (iii), start by showing that the family under consideration is closed under countable unions.)

434Yd

(o) Show that ]0, 1[

Borel measures ω1

201

is not pre-Radon.

> (p) Show that a K-analytic Hausdorff space is Radon iff all its compact subsets are Radon. (Hint: 432B, 434Xh.) (q) Suppose that X is a K-analytic Hausdorff space such that every Radon measure on X is completion regular. Show that X is a Radon space. (r) Let X and Y be topological spaces, and suppose that Y has a countable network. (i) Show that if X is Borel-measure-complete, then X × Y is Borel-measure-complete. (ii) Show that if X and Y are Radon Hausdorff spaces, then X × Y is Radon. Q Q (s) Let hXn in∈N be a sequence of topological spaces; write X = n∈N Xn and Zn = i≤n Xi for each n. (i) Show that if every Zn is Borel-measure-complete, so is X. (ii) Show that if every Zn is Hausdorff and pre-Radon, so is X. (iii) Show that if every Zn is Hausdorff and Radon, so is X. Q (t) (i) Let hXn in∈N be a sequence of Radon Hausdorff spaces such that i≤n Ki is Radon whenever Q n ∈ N and Ki ⊆ Xi is compact for every i ≤ n. Show that X = n∈N XnQis Radon. (ii) Let hXn in∈N be a sequence of Radon Hausdorff spaces with countable networks. Show that n∈N Xn is Radon. (u) (i) Let X be a Radon Hausdorff space and Y a Borel-measure-compact space. Show that X × Y is Borel-measure-compact. (ii) Let X be a Radon Hausdorff space and Y a Borel-measure-complete space. Show that X × Y is Borel-measure-complete. R RR (v) Show that if, in 434R, ν is σ-finite, then g dλB = g(x, y)ν(dy)µ(dx) for every λB -integrable function g : X × Y → R. (w) Show that the product measure construction of 434R is ‘associative’ and ‘distributive’ in the sense that (under the product measures on (X × Y ) × Z and X × (Y × Z) agree, and S S S appropriate hypotheses) those on i,j∈N (Xi × Yj ) and ( i∈N Xi ) × ( j∈N Yj ) agree. > (x) Show that the product measure construction of 434R is not ‘commutative’; indeed, taking µ = ν to be Dieudonné’s measure on ω1 , show that the Borel measures λ1 , λ2 on ω12 defined by setting λ1 W =

R

νW [{ξ}]µ(dξ),

λ2 W =

R

µW −1 [{η}]ν(dη)

are different. X ∈ E)’ for random variables X and (y) Read through §271, looking for ways to apply the concept ‘Pr(X universally measurable sets E. 434Y Further exercises (a) Set X = N \ {0, 1}. For m, p ∈ X set Ump = m + pN; show that {Ump : m, p ∈ X are coprime} is a base for a connected Hausdorff topology on X. (Hint: pq ∈ U mp for every q ≥ 1. See Steen & Seebach 78, ex. 60.) Show that X is a second-countable analytic Hausdorff space and carries a Radon measure which is not completion regular. (b) Let (X, Σ, µ) be a semi-finite measure space and T a topology on X such that µ is inner regular with respect to the closed sets. Suppose that Y is a topological space with a countable network consisting of universally measurable sets, and that f : X → Y is measurable. Show that f is almost continuous. (c) Let X be a completely regular Hausdorff space. Show that the following are equiveridical: (α) X is ˇ Radon; (β) X is a universally measurable subset of its Stone-Cech compactification; (γ) whenever Y is a Hausdorff space and X 0 is a subspace of Y which is homeomorphic to X, then X 0 is universally measurable in Y . (d) Let X be a completely regular Hausdorff space. Show that the following are equiveridical: (α) ˇ X is pre-Radon; (β) X is a universally Radon-measurable subset of its Stone-Cech compactification; (γ) 0 whenever Y is a Hausdorff space and X is a subspace of Y which is homeomorphic to X, then X 0 is universally Radon-measurable in Y .

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(e) Let X be a metrizable space without isolated points, and µ a σ-finite Borel measure on X. Show that there is a conegligible meager set. (Hint: there is a dense set D ⊆ X such that {{d} : d ∈ D} is σ-metrically-discrete.) (f ) (Aldaz 97) A topological space X is countably metacompact if whenever G is a countable open cover of X then there is a point-finite open cover H of X refining G. (i) Show that X is countably metacompact iff whenever hFn in∈N is a non-increasing sequence of closed sets with empty intersection in X then there is a sequence hGn in∈N of open sets, with empty intersection, such that Fn ⊆ Gn for every n. (ii) Let X be any topological space and ν : PX → [0, 1] a finitely additive functional such that νX = 1. Show that there is a finitely additive ν 0 : PX → [0, 1] such that νF ≤ ν 0 F = inf{ν 0 G : G ⊇ F is open} for every closed F ⊆ X. (Hint: 413Q.) (iii) Show that if X is countably metacompact and µ is any Borel probability measure on X, there is a Borel probability measure µ0 on X, inner regular with respect to the closed sets, such that µF ≤ µ0 F for every closed set F ⊆ X; so that µ and µ0 agree on the Baire σ-algebra of X. (g) Show that βN is not countably tight, therefore not Borel-measure-compact. (h) Let X be a quasi-dyadic space. Suppose that h(Gξ , Hξ )iξ<ω1 T is a family of T pairs of disjoint non-empty open sets. Show that there is an uncountable A ⊆ ω1 such that ξ∈B Gξ ∩ ξ∈A\B Hξ is non-empty for every B ⊆ A. (Hint: start by supposing that X is itself a product of separable metric spaces, and that every Gξ , Hξ is an open cylinder set; use the ∆-system lemma.) (i) (i) Show that the split interval is not quasi-dyadic. (ii) Show that R, with the Sorgenfrey right-facing topology, is not quasi-dyadic. (iii) Show that ω1 and ω1 +1, with their order topologies, are not quasi-dyadic. (Hint: 434Yh.) (j) Show that a perfectly normal quasi-dyadic space is Borel-measure-compact. (k) Let hXn in∈N Q be a sequence of first-countable spaces, and µn a Borel measure on Xn for each n. For n ∈ N set Zn = i≤n Xi , and let λn be the product Borel measure on Zn constructed by repeatedly using Q the method of 434R (cf. 434Xw). Show that there is a unique Borel measure λ on Z = n∈N Xn such that all the canonical maps from Z to Zn are inverse-measure-preserving. Show that, for any n, λ can be Q identified with the product of λn and a suitable product measure on i>n Xi . (l) Let X be a totally ordered set with its order topology. Show that any τ -additive Borel probability • measure on X has countable Maharam type. (Hint: {]−∞, x] : x ∈ X} generates the measure algebra.) 434Z Problems (a) Must every Radon compact Hausdorff space be sequentially compact? (b) Must a Hausdorff continuous image of a Radon compact Hausdorff space be Radon? 434 Notes and comments I said that the fundamental question of topological measure theory is ‘which measures can appear on which topological spaces’ ? In this section I have concentrated on Borel measures, classified according to the scheme laid out in §411. (Of course there are other kinds of classification. One of the most interesting is the Maharam classification of Chapter 33: we can ask what measure algebras can appear from topological measures on a given topological space. But I pass this by for the moment, with only 434Yl to give a taste.) We can ask this question from either of two directions. The obvious approach is to ask, for a given class of topological spaces, which types of measure can appear. But having discovered that (for instance) there are several types of topological space on which all (totally finite) Borel measures are tight, we can use this as a definition of a class of topological spaces, and ask the ordinary questions about this class. Thus we have ‘Radon’, ‘Borel-measure-complete’, ‘Borel-measure-compact’ and ‘pre-Radon’ spaces (434C, 434H, 434I, 434J). I have given precedence to the first partly to honour the influence of Schwartz 73 and partly because a compact Hausdorff space is always Borel-measure-compact and pre-Radon (434Hb, 434Jf) and is Borel-measure-complete iff it is Radon (434Ka). In effect, ‘Borel-measure-complete’ means ‘Borel measures are quasi-Radon’ (434Ib), ‘pre-Radon’ means ‘quasi-Radon measures are Radon’ (434Jb), and ‘Radon’ means ‘Borel measures are Radon’ (434F(a-iii)). These slogans have to be interpreted with

435A

Baire measures

203

care; but it is true that a Hausdorff space is Radon iff it is both Borel-measure-complete and pre-Radon (434Ka). The concept of ‘Radon’ space is in fact one of the important contributions of measure theory to general topology, offering a variety of challenging questions. One which has attracted some attention is the problem of determining when products of Radon spaces are Radon. Uncountable products hardly ever are (434Ke); for countable products it is enough to understand products of finitely many compact spaces (434Xt); but the product of two compact spaces already seems to lead us into undecidable questions (438Xn, Wage 80). Two more very natural questions are in 434Z. One of the obstacles to the investigation is the rather small number of Radon compact Hausdorff spaces which are known. I should remark that if the continuum hypothesis (for instance) is true, then every compact Hausdorff space in which countably compact sets are closed is sequentially compact (Ismail & Nyikos 80, or Fremlin 84, 24Nc), so that in this case we have a quick answer to 434Za. You will recognise the construction of 434M as a universal version of Dieudonné’s measure (411Q). ‘Tightness’ is of great interest for other reasons (Engelking 89), and here is very helpful in giving quick proofs that spaces are not Radon (434Yg). A large proportion of the definitions in general topology can be regarded as different abstractions from the concept of metrizability. Countable tightness is an obvious example; so is ‘first-countability’ (434R). In quite a different direction we have ‘metacompactness’ (438J, 434Yf). The construction of the product measure in 434R is an obvious idea, as soon as you have seen Fubini’s theorem, but it is not obvious just when it will work. ‘Quasi-dyadic’ spaces are a relatively recent invention; I introduce them here only as a vehicle for the argument of 434Q. Of course a dyadic space (4A2A) is quasi-dyadic; for basic facts on dyadic spaces, see 4A2Dd, 4A5T and Engelking 89, §3.12 and 4.5.9-4.5.11.

435 Baire measures Imitating the programme of §434, I apply a similar analysis to Baire measures, starting with a simpleminded classification (435A). This time the central section (435D-435H) is devoted to ‘measure-compact’ spaces, those on which all (totally finite) Baire measures are τ -additive. 435A Types of Baire measures In 434A I looked at a list of four properties which a Borel measure may or may not possess: inner regularity with respect to closed sets, inner regularity with respect to zero sets, tightness (that is, inner regularity with respect to closed compact sets), and τ -additivity. Since every (semi-finite) Baire measure is inner regular with respect to the zero sets (412D), only two of the four are important considerations for Baire measures: tightness and τ -additivity. On the other hand, there is a new question we can ask. Given a Baire measure on a topological space, when can it be extended to a Borel measure? And in the case of a positive answer, we can ask whether the extension is unique, and whether we can find extensions to Borel measures satisfying the properties considered in 434A. We already have some information on this. If X is a completely regular space, and µ is a τ -additive effectively locally finite Baire measure on X, then µ has a (unique) extension to a τ -additive Borel measure (415N). While if µ is tight, the extension will also be tight (cf. 416C). Perhaps I should remark immediately that while there can be only one τ -additive Borel measure extending µ, there might be another Borel measure, not τ -additive, also extending µ; see 435Xa. Of course if there is any completion regular Borel measure extending µ, there is only one; moreover, if µ is σ-finite, and there is a completion regular Borel measure extending µ, this is the only Borel measure extending µ. (For every Borel set will belong to the domain of the completion of µ.) A possible division of Baire measures is therefore into classes (E) measures which are not τ -additive, (F) measures which are τ -additive, but not tight, (G) tight measures, and within these classes we can distinguish measures with no extension to a Borel measure (type E0 ), measures with more than one extension to a Borel measure (types E1 , F1 and G1 ), measures with exactly one extension to a Borel measure which is not completion regular (types E2 , F2 and G2 ) and measures with

204


435A

an extension to a completion regular Borel measure (types E3 , F3 and G3 ). For examples, see 439M and 439O (E0 ), 439N (E2 ), 439J (E3 ), 435Xc (F1 ), 435Xd (F2 ), 415Xc and 434Xa (F3 ), 435Xa (G1 ), 435Xb (G2 ) and the restriction of Lebesgue measure to the Baire subsets of R (G3 ); other examples may be constructed as direct sums of these. A separate question we can ask of a Baire measure is whether it can be extended to a Radon measure. For this there is a straightforward criterion (435B), which shows that (at least for totally finite measures on completely regular spaces) only the types F1 and F2 are divided by this question. (If a Baire measure µ can be extended to a Radon measure, it is surely τ -additive. If µ is tight, it satisfies the criteria of 435B, so has an extension to a Radon measure. If µ has an extension to a completion regular Borel measure µ1 and has an extension to a Radon measure µ2 , then the completion µ ˆ of µ extends µ1 , while µ2 extends µ ˆ; so µ1 is the restriction of µ2 to the Borel sets and µ2 = µ ˆ1 = µ ˆ and µ, like µ2 , is tight, by 412Hb or otherwise. Thus no measure of type F3 can be extended to a Radon measure.) As with the classification of Borel measures that I offered in §434, any restriction on the topology of the underlying space may eliminate some of these possibilities. For instance, because a semi-finite Baire measure is inner regular with respect to the closed sets, we can have no (semi-finite) measure of classes E or F on a compact Hausdorff space. On a locally compact Hausdorff space we can have no effectively locally finite Baire measure of class F (435Xe), while on a K-analytic Hausdorff space we can have no locally finite Baire measure of class E (432F). In a metric space, or a regular space with a countable network (e.g., a regular analytic Hausdorff space), the Baire and Borel σ-algebras coincide (4A3Kb), so we can have no measures of type E0 , E1 , F1 or G1 . 435B Theorem Let X be a Hausdorff space and µ a locally finite Baire measure on X. Then the following are equiveridical: (i) µ has an extension to a Radon measure on X; (ii) for every non-negligible Baire set E ⊆ X there is a compact set K ⊆ E such that µ∗ K > 0. If µ is totally finite, we can add (iii) sup{µ∗ K : K ⊆ X is compact} = µX. proof Because µ is inner regular with respect to the closed sets (412D), this is just a special case of 416P. ˇ´ık 57) Let X be a normal countably paracompact space. Then any semi-finite 435C Theorem (Mar Baire measure on X has an extension to a semi-finite Borel measure which is inner regular with respect to the closed sets. proof (a) Let ν be a semi-finite Baire measure on X. Let K be the family of those closed subsets of X which are included in zero sets of finite measure, and set φ0 K = ν ∗ K for K ∈ K. Then K and φ0 satisfy the conditions of 413I, that is, ∅ ∈ K, (†) K ∪ K 0 ∈ K whenever K, K 0 ∈ K are disjoint, T (‡) n∈N Kn ∈ K whenever hKn in∈N is a sequence in K, (α) φ0 K = φ0 L + sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L} whenever K, L ∈ K and L ⊆ K, (β) inf n∈N φ0 Kn = 0 whenever hKn in∈N is a non-increasing sequence in K with empty intersection. P P The first three are trivial. α) Take K, L ∈ K with L ⊆ K, and set γ = sup{φ0 K 0 : K ∈ K, K 0 ⊆ K \ L}. (i) If K 0 ⊆ K \ L is (α closed, then (because X is normal) there is a zero set F including K 0 and disjoint from L (4A2F(d-iv)), so φ0 K 0 + φ0 L = ν ∗ ((K 0 ∪ L) ∩ F ) + ν ∗ ((K 0 ∪ L) \ F ) = ν ∗ (K 0 ∪ L) ≤ ν ∗ K. As K 0 is arbitrary, γ + φ0 L ≤ φ0 K. (ii) Let ² > 0. Let F0 be a zero set of finite measure including K. Because ν is inner regular with respect to the zero sets (412D), there is a zero set F ⊆ F0 \ L such that νF ≥ ν∗ (F0 \ L) − ² (413Ee), so that ν(F0 \ F ) ≤ ν ∗ L + ² (413Ec). Set K 0 = K ∩ F . Then ν ∗ K = ν ∗ (K \ F ) + ν ∗ (K ∩ F ) ≤ ν(F0 \ F ) + ν ∗ K 0 ≤ ν ∗ L + ² + γ. As ² is arbitrary, ν ∗ K ≤ ν ∗ L + γ.

435Fb

Baire measures

205

β ) If hKn in∈N is a non-increasing sequence in K with empty intersection, then (because XTis countably (β paracompact) there is a sequence hGn in∈N of open sets such that Kn ⊆ Gn for every n and n∈N Gn = ∅ (4A2Ff). T Because X is normal, there are zero sets Fn such that Kn ⊆ Fn ⊆ Gn for each n (4A2F(d-iv)), so that n∈N Fn = ∅. We may suppose that F0 has finite measure. In this case, T limn→∞ ν ∗ Kn ≤ limn→∞ ν( i≤n Fi ) = 0. Thus K and φ0 satisfy the conditions (α) and (β) as well. Q Q (b) By 413I, there is a complete locally determined measure µ on X, extending φ0 and inner regular with respect to K. If F ⊆ X is closed, then F ∩ K ∈ K for every K ∈ K, so F ∈ dom µ (413F(ii)); accordingly µ is a topological measure, and because ν also is inner regular with respect to K, µ must extend ν. So the restriction of µ to the Borel sets is a Borel extension of ν which is inner regular with respect to the closed sets. Remark If X is normal, but not countably paracompact, the result may fail; see 439O. I have stated the result in terms of ‘countable paracompactness’, but the formally distinct ‘countable metacompactness’ is also sufficient (435Ya). If we are told that the Baire measure is τ -additive and effectively locally finite, we have a much stronger result (415M). 435D Just as with the ‘Radon’ spaces of §434, we can look at classes of topological spaces defined by the behaviour of the Baire measures they carry. The class which has aroused most interest is the following. Definition A completely regular topological space X is measure-compact (sometimes called almost Lindel¨ of) if every totally finite Baire measure on X is τ -additive, that is, has an extension to a quasiRadon measure on X (415N). 435E The following lemma will make our path easier. Lemma Let X be a completely regular topological space and ν a totally finite Baire measure on X. Suppose that supG∈G νG = νX whenever G is an upwards-directed family of cozero sets with union X. Then ν is τ -additive. S proof Let G be an upwards-directed family of open Baire sets such that G∗ = G is also a Baire set, and ² > 0. Because ν is inner regular with respect to the zero sets, there is a zero set F ⊆ G∗ such that νF ≥ νG∗ − ². Let G 0 be the family of cozero sets included S in members of G; because X is completely regular, so that the cozero sets are a base for its topology, G 0 = G∗ , and of course G 0 is upwards-directed. Now H = {G ∪ (X \ F ) : G ∈ G 0 } is an upwards-directed family of cozero sets with union X, so there is a G0 ∈ G 0 such that ν(G0 ∪ (X \ F )) ≥ νX − ². In this case supG∈G νG ≥ νG0 ≥ νX − ² − ν(X \ F ) ≥ νG∗ − 2². As G and ² are arbitrary, ν is τ -additive. 435F Elementary facts (a) If X is a completely regular space which is not measure-compact, there are a Baire probability measure µ on X and a cover of X by µ-negligible cozero sets. P P There is a totally finite Baire measure ν on X which is not τ -additive. By 435E, there is an upwards-directed family G of cozero sets, covering X, such that sup SG∈G νG < νX. Let hGn in∈N be a sequence in G such that supn∈N νGn = supG∈G νG. Then γ = ν(X \ n∈N Gn ) > 0. Set 1 γ

µH = ν(H \

S n∈N

Gn )

for Baire sets H ⊆ X; then µ is a Baire probability measure and G is a cover of X by µ-negligible cozero sets. Q Q

206


435Fb

(b) Regular Lindelöf spaces are measure-compact. (For if a Lindelöf space can be covered by negligible open sets, it can be covered by countably many negligible open sets, so is itself negligible.) In particular, compact Hausdorff spaces, indeed all regular K-analytic Hausdorff spaces (422Gg), are measure-compact. Note that regular Lindelöf spaces are normal and paracompact (4A2H(b-i)), so their measure-compactness is also a consequence of 435C and 434Hb. (c) An open subset of a measure-compact space need not be measure-compact (435Xi(i)). A continuous image of a measure-compact space need not be measure-compact (435Xi(ii)). N c is not measure-compact (439P). The product of two measure-compact spaces need not be measure-compact (439Q). (d) If X is a measure-compact completely regular space it is Borel-measure-compact. P P Let µ be a non-zero totally finite Borel measure on X and G an open cover of X. Let ν be the restriction of µ to the Baire σ-algebra of X, so that ν is τ -additive. Let U be the set of cozero sets S U ⊆ X included in members of G; because the family of cozero sets is a base for the topology of X, U = X, and there is some U ∈ U such that νU > 0. This means that there is some G ∈ G such that µG > 0. By 434H(a-v), X is Borel-measure-compact. Q Q 435G Proposition A Souslin-F subset of a measure-compact completely regular space is measurecompact. proof (a) Let X be a measure-compact completely regular space, hFσ iσ∈S a Souslin scheme consisting of closed subsets of X with kernel A, ν a totally finite Baire measure on A, and G an upwards-directed family of (relatively) cozero subsets of A covering A. Let ν1 be the Baire measure on X defined by setting ν1 H = ν(A ∩ H) for every Baire subset H of X. Because X is measure-compact, ν1 has an extension to a quasi-Radon measure µ on X. Let µA be the subspace measure on A. (b) By 431A, A is measured by µ. In fact µA = νA. P P The construction of µ given in 415K-415N ensures that µF = ν1∗ F for every closed set F , and this is in any case a consequence of the facts that µ is τ -additive and dom ν1 includes a base for the topology. For each σ ∈ S, in particular, µFσ = ν1∗ Fσ ; let Fσ0 ⊇ Fσ be a Baire set such that ν1 Fσ0 = ν1∗ Fσ . Then µFσ0 = ν1 Fσ0 = ν1∗ Fσ = µFσ and µ(Fσ0 \ Fσ ) = 0 for every σ ∈ S. Let A0 be the kernel of the Souslin scheme hFσ0 iσ∈S . Then A ⊆ A0 and P µ(A0 \ A) ≤ σ∈S µ(Fσ0 \ Fσ ) = 0, so µA = µA0 . On the other hand, writing νˆ1 for the completion of ν1 , A0 is measured by νˆ1 , by 431A again, so that (because µ extends ν1 ) µA = µA0 = µ∗ A0 ≤ (ν1 )∗ A0 = ν1∗ A0 ≤ µ∗ A0 = µA. Thus µA = ν1∗ A0 . But of course νA = ν1 X = ν1∗ A = ν1∗ A0 , so that µA = νA. Q Q Since we surely have µX = ν1 X = νA, we see that µ(X \ A) = 0. (c) It follows that µF = νF for every (relatively) zero set F ⊆ A. P P There is a closed set F 0 ⊆ X such 0 0 that F = A ∩ F . Now if H ⊆ X is a Baire set including F , H ∩ A is a (relatively) Baire set including F , so νF ≤ ν(H ∩ A) = ν1 H; as H is arbitrary, νF ≤ ν1∗ F 0 . But ν1∗ F 0 = µF 0 , as remarked above, and µ(X \ A) = 0, so µF = µF 0 = ν1∗ F 0 ≥ νF . On the other hand, A \ F is (relatively) cozero, so there is a non-decreasing sequence hFn in∈N of (relatively) zero subsets of A with union A \ F , and µ(A \ F ) = limn→∞ µFn ≥ limn→∞ νFn = ν(A \ F ).

435Xi

Baire measures

207

Since we already know that µA = νA, it follows that µF = µA − µ(A \ F ) ≤ νA − ν(A \ F ) = νF , and µF = νF . Q Q (d) The set {E : E ⊆ A is a (relative) Baire set, µE = νE} therefore contains every (relatively) zero set, and by the Monotone Class Theorem (136C) contains every (relatively) Baire set. What this means is that µ actually extends ν; so the subspace measure µA = µ¹ PA also extends ν. But µA is a quasi-Radon measure (415B), therefore τ -additive, and ν must also be τ -additive. 435H Corollary A Baire subset of a measure-compact completely regular space is measure-compact. proof Put 435G and 421L together. 435X Basic exercises >(a) Give ω1 + 1 its order topology. (i) Show that its Baire σ-algebra Σ is just the family of sets E ⊆ ω1 + 1 such that either E or its complement is a countable subset of ω1 . (ii) Show that there is a unique Baire probability measure ν on ω1 + 1 such that ν{ξ} = 0 for every ξ < ω1 . (iii) Show that ν is τ -additive. (iv) Show that there is exactly one Radon measure on ω1 + 1 extending ν, but that the measure µ of 434Xf is another Borel measure also extending ν. > (b) Let I be a set of cardinal ω1 , endowed with the discrete topology, and X = I ∪ {∞} its one-point compactification (3A3O). Let µ be the Radon probability measure on X such that µ{∞} = 1. (i) Show that every subset of X is a Borel set. (ii) Show that {∞} is not a zero set. (iii) Let ν be the restriction of µ to the Baire σ-algebra of X. Show that ν is tight. Show that µ is the unique Borel measure extending ν (hint: you will need 419G), but is not completion regular. (iv) Show that the subspace measure νI on I is the countable-cocountable measure on I, and is not a Baire measure, nor has any extension to a Baire measure on I. (v) Show that X is measure-compact. (c) On R ω1 let µ be the Baire measure defined by saying that µE = 1 if χω1 ∈ E, 0 otherwise. (i) Show that µ is τ -additive, but not tight. (Hint: 4A3P.) (ii) Show that the map ξ 7→ χξ : ω1 + 1 → R ω1 is continuous, so that µ has more than one extension to a Borel measure. (iii) Show that µ has an extension to a Radon measure. (d) Set X = ω1 + 1 with the topology Pω1 ∪ {X \ A : A ⊆ ω1 is countable}. Let µ be the Baire measure on X defined by saying that, for Baire sets E ⊆ X, µE = 1 if ω1 ∈ E, 0 otherwise. (i) Show that a function f : X → R is continuous iff {ξ : ξ ∈ X, f (ξ) 6= f (ω1 )} is countable; show that X is completely regular and Hausdorff. (ii) Show that µ is τ -additive. (iii) Show that every subset of X is Borel. (iv) Show that the only Borel measure extending µ is that which gives measure 1 to {ω1 }, and that this is a Radon measure. (v) Show that all compact subsets of X are finite, so that µ is not tight. *(vi) Show that X is Lindelöf. (e) Let X be a locally compact Hausdorff space and µ an effectively locally finite τ -additive Baire measure on X. Show that µ is tight. (Hint: the relatively compact cozero sets cover X; use 414Ea and 412D.) > (f ) Let X be a completely regular space and µ a totally finite τ -additive Borel measure on X. Let µ0 be the restriction of µ to the Baire σ-algebra of X. Show that µF = µ∗0 F for every closed set F ⊆ X. (g) Show that if a semi-finite Baire measure ν on a normal countably paracompact space is extended to a Borel measure µ by the construction in 435C, then the measure algebra of ν becomes embedded as an order-dense subalgebra of the measure algebra of µ, so that L1 (µ) can be identified with L1 (ν). (h) Show that a Borel-measure-compact normal countably paracompact space is measure-compact. (i) (i) Show that ω1 + 1 is measure-compact, in its order topology, but that its open subset ω1 is not (cf. 434Xk). (ii) Show that a discrete space of cardinal ω1 is measure-compact, but that it has a continuous image which is not measure-compact.

208


435Xj

(j) Let X be a metacompact completely regular space and ν a totally finite strictly positive Baire measure on X. Show that X is Lindelöf, so that ν has an extension to a quasi-Radon measure on X. (Hint: if H is a point-finite open cover of X, not containing ∅, then for each H ∈ H choose a non-empty cozero set GH ⊆ H; show that {H : νGH ≥ δ} is finite for every δ > 0.) (k) A completely regular space X is strongly measure-compact (Moran 69) if µX = sup{µ∗ K : K ⊆ X is compact} for every totally finite Baire measure µ on X. (i) Show that a completely regular Hausdorff space X is strongly measure-compact iff every totally finite Baire measure on X has an extension to a Radon measure. (ii) Show that a strongly measure-compact completely regular space is measurecompact. (iii) Show that a Souslin-F subset of a strongly measure-compact completely regular space is strongly measure-compact. (iv) Show that a discrete space of cardinal ω1 is strongly measure-compact. (v) Show that a countable product of strongly measure-compact completely regular spaces is strongly measurecompact. (vi) Show that N ω1 is not strongly measure-compact. (Hint: take a non-trivial probability measure on N and consider its power on N ω1 .) (vii) Show that if X and Y are completely regular spaces, X is measure-compact and Y is strongly measure-compact then X × Y is measure-compact. (viii) Show that if X is a strongly measure-compact completely regular Hausdorff space, then it is Borel-measure-compact and pre-Radon. 435Y Further exercises (a) Show that a normal countably metacompact space (434Yf) is countably paracompact. ˇ (b) Let X be a completely regular Hausdorff space and βX its Stone-Cech compactification. Show that X is measure-compact iff whenever ν is a Radon measure on βX such that νX = 0, there is a ν-negligible Baire subset of βX including X. 435 Notes and comments The principal reason for studying Baire measures is actually outside the main line of this chapter. For a completely regular Hausdorff space X, write Cb (X) for the M -space of bounded continuous real-valued functions on X. Then Cb (X)∗ = Cb (X)∼ is an L-space (356N), and inside Cb (X)∗ we have the bands generated by the tight, smooth and sequentially smooth functionals (see 437Xq and 437A below), all identifiable, if we choose, with spaces of ‘signed Baire measures’. Wheeler 83 argues convincingly that for the questions a functional analyst naturally asks, these Baire measures are often an effective aid. From the point of view of the arguments in this section, the most fundamental difference between ‘Baire’ and ‘Borel’ measures lies in their action on subspaces. If X is a topological space and A is a subset of X, then any Borel or Baire measure µ on A provides us with a measure µ1 of the same type on X, setting µ1 E = µ(A ∩ E) for the appropriate sets E. In the other direction, if µ is a Borel measure on X, then the subspace measure µA is a Borel measure on A, because the Borel σ-algebra of A is just the subspace σ-algebra derived from the Borel algebra of X (4A3Ca). But if µ is a Baire measure on X, it does not follow that µA is a Baire measure on A; this is because (in general) not every continuous function f : A → [0, 1] has a continuous extension to X, so that not every zero set in A is the intersection of A with a zero set in X (see 435Xb). The analysis of those pairs (X, A) for which the Baire σ-algebra of A is just the subspace algebra derived from the Baire sets in X is a challenging problem in general topology which I pass by here. For the moment I note only that avoiding it is the principal technical problem in the proof of 435G. I do not know if I ought to apologise for ‘countably tight’ spaces (434N), ‘first-countable’ spaces (434R), ‘metacompact’ spaces (438J), ‘normal countably paracompact’ spaces (435C), ‘quasi-dyadic’ spaces (434O) and ‘sequential’ spaces (436F). General topology is notorious for invoking arcane terminology to stretch arguments to their utmost limit of generality, and even specialists may find their patience tried by definitions which seem to have only one theorem each. In 438J, for instance, it is obvious that the original result concerned metrizable spaces (438H), and you may well feel at first that the extension is a baroque overelaboration. On the other hand, there are (if you look for them) some very interesting metacompact spaces (Engelking 89, §5.3), and metacompactness has taken its place in the standard lists. In this book I try to follow a rule of introducing a class of topological spaces only when it is both genuinely interesting, from the point of view of general topology, and also supports an idea which is interesting from the point of view of measure theory.

436C

Representation of linear functionals

209

436 Representation of linear functionals I began this treatise with the three steps which make measure theory, as we know it, possible: a construction of Lebesgue measure, a definition of an integral from a measure, and a proof of the convergence theorems. I used what I am sure is the best route: Lebesgue measure from Lebesgue outer measure, and integrable functions from simple functions. But of course there are many other paths to the same ends, and some of them show us slightly different aspects of the subject. In this section I come – rather later than many authors would – to an account of a procedure for constructing measures from integrals. I start with three fundamental theorems, the first and third being the most important. A positive linear functional on a truncated Riesz space of functions is an integral iff it is sequentially smooth (436D); a smooth linear functional corresponds to a quasi-Radon measure (436H); and if X is a compact Hausdorff space, any positive linear functional on C(X) corresponds to a Radon measure (436J-436K). 436A Definition Let X be a set, U a Riesz subspace of RX , and f : U → R a positive linear functional. I say that f is sequentially smooth if whenever hun in∈N is a non-increasing sequence in U such that limn→∞ un (x) = 0 for every x ∈ X, then limn→∞ f (un ) = 0. If (X, Σ, µ) is a measure space Rand U is a Riesz subspace of the space of real-valued µ-integrable functions defined everywhere on X, then dµ : U → R is sequentially smooth, by Fatou’s Lemma or Lebesgue’s Dominated Convergence Theorem. Remark It is essential to distinguish between ‘sequentially smooth’, as defined here, and ‘sequentially ordercontinuous’, as in 313Hb or 355G. In the context here, a positive linear operator f : U → R is sequentially order-continuous if limn→∞ f (un ) = 0 whenever hun in∈N is a non-increasing sequence in U such that 0 is the greatest lower bound for {un : n ∈ N} in U ; while f is sequentially smooth if limn→∞ f (un ) = 0 whenever hun in∈N is a non-increasing sequence in U such that 0 is the greatest lower bound for {un : n ∈ N} in RX . So there can be sequentially smooth functionals which are not sequentially order-continuous, as in 436Xi. A sequentially order-continuous positive linear functional is of course sequentially smooth. 436B Definition Let X be a set. I will say that a Riesz subspace U of RX is truncated (or satisfies Stone’s condition) if u ∧ χX ∈ U for every u ∈ U . In this case, u ∧ γχX ∈ U for every γ ≥ 0 and u ∈ U (being −u− if γ = 0, γ(γ −1 u ∧ χX) otherwise). 436C Lemma Let X be a set and U a truncated Riesz subspace of RX . Write K for the family of sets of the form {x : x ∈ X, u(x) ≥ 1} as u runs over U . Let f : U → R be a sequentially smooth positive linear functional, andR µ a measure on X such that µK is defined and equal to inf{f (u) : χK ≤ u ∈ U } for every K ∈ K. Then u dµ exists and is equal to f (u) for every u ∈ U . R proof It is enough to deal with the case u ≥ 0, since U = U + − U + and both f and are linear. Note that if v ∈ U , K ∈ K and v ≤ χK, then v ≤ w whenever χK ≤ w ∈ U , so f (v) ≤ µK. For k, n ∈ N set Knk = {x : u(x) ≥ 2−n k},

unk = u ∧ 2−n kχX.

Then, for k ≥ 1, Knk = {x : 2n k −1 u ≥ 1} ∈ K, 2n (un,k+1 − unk ) ≤ χKnk ≤ 2n (unk − un,k−1 ). So 2n f (un,k+1 − unk ) ≤ µKnk ≤ 2n f (unk − un,k−1 ), and f (un,4n +1 − un1 ) ≤

P4n k=1

2−n µKnk ≤ f (u).

But setting wn = un,4n +1 − un1 , hwn in∈N is a non-decreasing sequence of functions in U and supn∈N wn (x) = P 4n u(x) for every x, so limn→∞ f (u − wRn ) = 0 and limn→∞ f (wn ) = f (u). Also, setting vn = k=1 2−n χKnk , we have wn ≤ vn ≤ u and f (wn ) ≤ vn ≤ f (u) for each n, so

R

u = limn→∞

R

vn = f (u)

210


436C

by B.Levi’s theorem. 436D Theorem Let X be a set and U a truncated Riesz subspace of RX . Let f : U → R be a positive linear functional. Then the following are equiveridical: (i) f is sequentially smooth; R (ii) there is a measure µ on X such that u dµ is defined and equal to f (u) for every u ∈ U . proof I remarked in 436A that (ii)⇒(i) is a consequence of Fatou’s Lemma. So the argument here is devoted to proving that (i)⇒(ii). (a) Let K be the family of sets K ⊆ X such that χK = inf n∈N un for some sequence hun in∈N in U , taking the infimum in RX , so that (inf n∈N un )(x) = inf n∈N un (x) for every x ∈ X. Then K is closed under finite unions and countable intersections. P P (i) If K, K 0 ∈ K take sequences hun in∈N , hu0n in∈N in U such that 0 0 χK = inf n∈N un and χK = inf n∈N un ; then χ(K ∪ K 0 ) = inf m,n∈N um ∨ u0n , so K ∪ K 0 ∈ K. (ii) If hKn in∈N is a sequence in K, then for each n ∈TN we can choose a sequence huni ii∈N in U such that χKn = inf i∈N uni ; T now χ( n∈N Kn ) = inf n,i∈N uni , so n∈N Kn ∈ K. Q Q Note that ∅ ∈ K because 0 ∈ U . (b) We need to know that if u ∈ U then K = {x : u(x) ≥ 1} belongs to K. P P Set un = 2n ((u ∧ χX) − (u ∧ (1 − 2−n )χX)). Because U is truncated, every un belongs to U , and it is easy to check that inf n∈N un = χK. Q Q It follows that 1 α

{x : u(x) ≥ α} = {x : u(x) ≥ 1} ∈ K whenever u ∈ U and α > 0. (c) For K ∈ K, set φ0 K = inf{f (u) : u ∈ U, u ≥ χK}. Then φ0 satisfies the conditions of 413I. P P I have already checked (†) and (‡) of 413I. α) Fix K, L ∈ K with L ⊆ K. Set γ = sup{φ0 K 0 : K 0 ∈ K, K 0 ⊆ K \ L}. (α (i) Suppose that K 0 ∈ K is included in K \ L, and ² > 0. Let hun in∈N , hu0n in∈N be sequences in U such that χL = inf n∈N un and χK 0 = inf n∈N u0n , and let u ∈ U be such that u ≥ χK and f (u) ≤ φ0 K + ². Set vn = u ∧ inf i≤n ui , vn0 = u ∧ inf i≤n u0i for each n. Then hvn ∧ vn0 in∈N is a non-increasing sequence in U with infimum χL ∧ χK 0 = 0, so there is an n such that f (vn ∧ vn0 ) ≤ ². In this case φ0 L + φ0 K 0 ≤ f (vn ) + f (vn0 ) = f (vn + vn0 ) = f (vn ∨ vn0 ) + f (vn ∧ vn0 ) ≤ f (u) + ² ≤ φ0 K + 2². As ² is arbitrary, φ0 L + φ0 K 0 ≤ φ0 K. As K 0 is arbitrary, φ0 L + γ ≤ φ0 K. (ii) Next, given ² ∈ ]0, 1[, there are u, v ∈ U such that u ≥ χK, v ≥ χL and f (v) ≤ φ0 L + ². Consider K 0 = {x : x ∈ K, min(1, u(x)) − v(x) ≥ ²} ⊆ K \ L. By (a) and (b), K 0 ∈ K. If w ∈ U and w ≥ χK 0 , then v(x) + w(x) ≥ 1 − ² for every x ∈ K, so φ0 K ≤

1 f (v 1−²

+ w) ≤

1 (φ0 L + ² + f (w)). 1−²

As w is arbitrary, (1 − ²)φ0 K ≤ φ0 L + ² + φ0 K 0 ≤ φ0 L + ² + γ. As ² is arbitrary, φ0 K ≤ φ0 L + γ and we have equality, as required by (α) in 413I. β ) Now suppose that hKn in∈N is a non-increasing sequence in K with empty intersection. For each (β n ∈ N let huni ii∈N be a sequence in U with infimum χKn in RX . Set vn = inf i,j≤n uji for each n; then hvn in∈N is a non-increasing sequence in U with infimum inf n∈N χKn = 0, so inf n∈N f (vn ) = 0. But vn ≥ inf j≤n χKj = χKn ,

φ0 Kn ≤ f (vn )

436F


211

for every n, so inf n∈N φ0 Kn = 0, as required by (β) of 413I. Q Q (d) By 413I, there is a complete locally determined measure µ on X, inner regular with respect to K, R extending φ0 . By 436C, f (u) = u dµ for every u ∈ U , as required. 436E Proposition Let X be any topological space, and Cb (X) the space of bounded continuous realvalued functions on X. Then there is a one-to-one correspondence between totally finite Baire measures µ on X and sequentially smooth positive linear functionals f : Cb (X) → R, given by the formulae f (u) =

R

u dµ for every u ∈ Cb (X),

µZ = inf{f (u) : χZ ≤ u ∈ Cb (X)} for every zero set Z ⊆ X. proof (a) If µ is a totally finite Baire measure on X, then every continuous bounded real-valued function R is integrable, and f = dµ is a sequentially smooth positive linear operator on Cb (X), by Fatou’s Lemma, as usual. (b) If f : Cb (X) → R is a Rsequentially smooth positive linear operator, then 436D tells us that there is a measure µ0 on X such that u dµ0 is defined and equal to f (u) for every u ∈ Cb (X). By the construction in 436D, or otherwise, we may suppose that µ0 is complete, so that every u ∈ Cb (X) is Σ-measurable, where Σ is the domain of µ0 . It follows by the definition of the Baire σ-algebra Ba of X (4A3K) that Ba ⊆ Σ, so R that µ = µ0 ¹Σ is a Baire measure; of course we still have f (u) = u dµ for every u ∈ Cb (X). Also, if Z ⊆ X is a zero set, µZ = inf{f (u) : χZ ≤ u ∈ Cb (X)}. P P Express Z as {x : v(x) = 0} where v : X → [0, 1] is continuous. Set un = (χX − 2n v)+ for n ∈ N; then hun in∈N is a non-decreasing sequence in Cb (X) and hun (x)in∈N → (χZ)(x) for every x ∈ X, so Z µZ ≤ inf{ u dµ : χZ ≤ u ∈ Cb (X)} = inf{f (u) : χZ ≤ u ∈ Cb (X)} Z ≤ inf f (un ) = lim un dµ = µZ. Q Q n→∞

n∈N

(c) The argument of (b) shows that if two totally finite Baire measures give the same integrals to every member of Cb (X), then they must agree on all zero sets. By the Monotone Class Theorem (136C) they agreeRon the σ-algebra generated by the zero sets, that is, Ba, and are therefore equal. Thus the operator µ 7→ dµ from the set of totally finite Baire measures on X to the set of sequentially smooth positive linear R operators on Cb (X) is a bijection, and if f = dµ then µZ = inf{f (u) : χZ ≤ u ∈ Cb (X)} for every zero set Z, as required. 436F Corresponding to 434R, we have the following construction for product Baire measures, applicable to a slightly larger class of spaces. Proposition Let X be a sequential space, Y a topological space, and µ, ν totally finite Baire measures on X, Y respectively. Then there is a Baire measure λ on X × Y such that λW =

R

νW [{x}]µ(dx),

R

f dλ =

RR

f (x, y)ν(dy)µ(dx)

for every Baire set W ⊆ X × Y and every bounded continuous function f : X × Y → R. RR proof (a) φ(f ) = f (x, P R y)dydx is defined in R for every bounded continuous function f : X × Y → R. P For each x ∈ X, g(x) = f (x, y)dy is defined because y 7→ f (x, y) is continuous. If hxn in∈N is any sequence in X converging to x ∈ X, then g(x) =

R

f (x, y)dy =

R

limn→∞ f (xn , y)dy = limn→∞

R

f (xn , y)dy = limn→∞ g(xn )

by Lebesgue’s Dominated Convergence Theorem. RR R So g is sequentially continuous; because X is sequential, g is continuous (4A2Kd). So f (x, y)dydx = g(x)dx is defined in R. Q Q

212


436F

(b) Of course φ is a positive linear functional on Cb (X × Y ), and B.Levi’s theorem shows that it is R sequentially smooth. By 436E, there is a Baire measure λ on X × Y such that f dλ = φ(f ) for every f ∈ Cb (X × Y ). (c) If W ⊆ X × Y is a zero set, there is a non-increasing sequence hfn in∈N in Cb (X × Y ) such that χW = inf n∈N fn . So B.Levi’s theorem tells us that

R

νW [{x}]dx = limn→∞

R

fn (x, y)dydx = limn→∞

R

fn dλ = λW .

Now the Monotone Class Theorem (136B) tells us that R {W : W ⊆ X × Y is Baire, νW [{x}]dx exists = λW } includes the σ-algebra generated by the zero sets, that is, contains every Baire set in X × Y . So λ has the required properties. 436G Definition Let X be a set, U a Riesz subspace of RX , and f : U → R a positive linear functional. I say that f is smooth if whenever A is a non-empty downwards-directed family in U such that inf u∈A u(x) = 0 for every x ∈ X, then inf u∈A f (u) = 0. Of course a smooth functional is sequentially smooth. If (X, T, Σ, µ) is an effectively locally finite τ additive topological measure space and U is a Riesz subspace of RX consisting of integrable continuous R functions, then dµ : U → R is smooth, by 414Bb. Corresponding to the remark in 436A, note that an order-continuous positive linear functional must be smooth, but that a smooth positive linear functional need not be order-continuous. 436H Theorem Let X be a set and U a truncated Riesz subspace of RX . Let f : U → R be a positive linear functional. Then the following are equiveridical: (i) f is smooth; (ii) there are a topology T and a measure µ on X such that µ is a quasi-Radon measure with respect to R T, U ⊆ C(X) and u dµ is defined and equal to f (u) for every u ∈ U ; (iii) writing S for the coarsest topology on X for which every member of U R is continuous, there is a measure µ on X such that µ is a quasi-Radon measure with respect to S, and u dµ is defined and equal to f (u) for every u ∈ U . proof As remarked in 436G, in a fractionally more general context, (ii)⇒(i) is a consequence of 414B. Of course (iii)⇒(ii). So the argument here is devoted to proving that (i)⇒(iii). Except for part (b) it is a simple adaptation of the method of 436D. (a) Let K be the family of sets K ⊆ X such that χK = inf A in RX for some non-empty set A ⊆ U . Then K is closed under finite unions and arbitrary intersections. P P (i) If K, K 0 ∈ K take A, A0 ⊆ U such 0 0 0 0 that χK = inf A, χK = A ; then χ(K ∪ K ) = inf{uS∨ u : u ∈ A, u0 ∈ A0 }, so K ∪ K 0 ∈ K. (ii) If L is a non-empty subset of K with intersection K, set A = L∈L {u : χL ≤ u ∈ U }; then χK = inf A, so K ∈ K. Q Q Note that ∅ ∈ K because 0 ∈ U . As in part (b) of the proof of 436D, {x : u(x) ≥ α} ∈ K whenever α > 0 and u ∈ U . (b) Every member of K is closed for S, being of the form {x : u(x) ≥ 1 for every u ∈ A} for some A ⊆ U . We need to know that if K ∈ K and G ∈ S, then K \ G ∈ K. P P Take a non-empty set B ⊆ U such that χK = inf B. Because K is closed under finite unions and arbitrary intersections, SK = {G : G ⊆ X, K \G ∈ K} is a topology on X. (i) If u ∈ U and G = {x : u(x) > 0}, then χ(K \ G) = inf{(v − ku)+ : v ∈ B, k ∈ N} so K \ G ∈ K and G ∈ SK . (ii) If u ∈ U and α > 0, then {x : u(x) > α} = {x : (u − u ∧ αχX)(x) > 0} belongs to SK , by (i). (iii) If u ∈ U and α > 0, set G = {x : u(x) < α}. Then K \ G = K ∩ {x : u(x) ≥ α} ∈ K, so G ∈ SK . (iv) Thus every member of U + is SK -continuous (2A3Bc), so every member of U is SK continuous (2A3Be), and S ⊆ SK , that is, K \ G ∈ K for every G ∈ S. Q Q (c) For K ∈ K, set φ0 K = inf{f (u) : u ∈ U, u ≥ χK}. Then φ0 satisfies the conditions of 415K. P PI have already checked (†) and (‡) of 415K.

436I


213

α) Fix K, L ∈ K with L ⊆ K. Set γ = sup{φ0 K 0 : K 0 ∈ K, K 0 ⊆ K \ L}. (α (i) Suppose that K 0 ∈ K is included in K \ L and ² > 0. Set A = {u : χL ≤ u ∈ U }, A0 = {u : χK 0 ≤ u ∈ U }, so that χL = inf A and χK 0 = inf A0 , and let v ∈ U be such that v ≥ χK and f (v) ≤ φ0 K + ². Then {u ∧ u0 : u ∈ A, u0 ∈ A0 } is a downwards-directed family with infimum 0 in RX , so (because f is smooth) there are u ∈ A, u0 ∈ A0 such that f (u ∧ u0 ) ≤ ². In this case φ0 L + φ0 K 0 ≤ f (v ∧ u) + f (v ∧ u0 ) = f (v ∧ (u ∨ u0 )) + f (v ∧ u ∧ u0 ) ≤ φ0 K + 2². As ² is arbitrary, φ0 L + φ0 K 0 ≤ φ0 K. As K 0 is arbitrary, φ0 L + γ ≤ φ0 K. (ii) Next, given ² ∈ ]0, 1[, there are u, v ∈ U such that u ≥ χK, v ≥ χL and f (v) ≤ φ0 L + ². Consider K 0 = {x : x ∈ K, min(1, u(x)) − v(x) ≥ ²}. By the last remark in (a), K 0 ∈ K. If w ∈ U and w ≥ χK 0 , then v(x) + w(x) ≥ 1 − ² for every x ∈ K, so φ0 K ≤

1 f (v 1−²

+ w) ≤

1 (φ0 L + ² + f (w)). 1−²

As w is arbitrary, (1 − ²)φ0 K ≤ φ0 L + ² + φ0 K 0 ≤ φ0 L + ² + γ. As ² is arbitrary, φ0 K ≤ φ0 L + γ and we have equality, as required by (α) in 415K. β ) Now suppose that K0 is a non-empty downwards-directed subset of K with empty intersection. Set (β S A = K∈K0 {u : χK ≤ u ∈ U }. Then A is a downwards-directed subset of U and inf A = 0 in RX . Because f is smooth, 0 = inf u∈A f (u) = inf K∈K0 φ0 K. Thus (β) of 415K is satisfied. (γγ ) If K ∈ K and φ0 K > 0, take u ∈ U such that u ≥ χK, and consider G = {x : u(x) > 21 }. Then K ⊆ G, while G ⊆ {x : 2u(x) ≥ 1}, so sup{φ0 K 0 : K 0 ∈ K, K 0 ⊆ G} ≤ 2f (u) < ∞. Thus φ0 satisfies (γ) of 415K. Q Q (d) By 415K, there is a quasi-Radon measure µ on X extending φ0 . By 436C, f (u) = u ∈ U.

R

u dµ for every

Remark It is worth noting explicitly that µ, as constructed here, is inner regular with respect to the family K of sets K ⊆ X such that χK = inf A for some set A ⊆ U . 436I Lemma Let X be a topological space. Let C0 (X) be the space of continuous functions u : X → R which ‘vanish at infinity’ in the sense that {x : |u(x)| ≥ ²} is compact for every ² > 0. (a) C0 (X) is a norm-closed solid linear subspace of Cb (X), so is a Banach lattice in its own right. (b) C0 (X)∗ = C0 (X)∼ is an L-space (definition: 354M). (c) If A ⊆ C0 (X) is a non-empty downwards-directed set such that inf u∈A u(x) = 0 for every x ∈ X, then inf u∈A kuk∞ = 0. proof (a)(i) If u ∈ C0 (X), then K = {x : |u(x)| ≥ 1} is compact, so kuk∞ ≤ sup({1} ∪ {|u(x)| : x ∈ K}) is finite, and u ∈ Cb (X). (ii) If u, v ∈ C0 (X) and α ∈ R and w ∈ Cb (X) and |w| ≤ |u|, then for any ² > 0 1 2

1 2

{x : |u(x) + v(x)| ≥ ²} ⊆ {x : |u(x)| ≥ ²} ∪ {x : |v(x)| ≥ ²}, {x : |αu(x)| ≥ ²} ⊆ {x : |u(x)| ≥

² }, 1+|α|

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{x : |w(x)| ≥ ²} ⊆ {x : |u(x)| ≥ ²} are closed relatively compact sets, so are compact, and u + v, αu, w belong to C0 (X). Thus C0 (X) is a solid linear subspace of Cb (X). (iii) If hun in∈N is a sequence in C0 (X) which k k∞ -converges to u ∈ Cb (X), then for any ² > 0 there is an n ∈ N such that ku − un k∞ ≤ 21 ², so that 1 2

{x : |u(x)| ≥ ²} ⊆ {x : |un (x)| ≥ ²} is compact, and u ∈ C0 (X). Thus C0 (X) is norm-closed in Cb (X). (iv) Being a norm-closed Riesz subspace of a Banach lattice, C0 (X) is itself a Banach lattice. (b) By 356Dc, C0 (X)∗ = C0 (X)∼ is a Banach lattice. Now kf + gk = kf k + kgk for all non-negative f , g ∈ C0 (X)∗ . P P Of course kf + gk ≤ kf k + kgk. On the other hand, for any ² > 0 there are u, v ∈ C0 (X) such that kuk∞ ≤ 1, kvk∞ ≤ 1 and |f (u)| ≥ kf k − ², |g(v)| ≥ kgk − ². Set w = |u| ∨ |v|; then w ∈ C0 (X) and kwk∞ = max(kuk∞ , kvk∞ ) ≤ 1. So kf + gk ≥ (f + g)(w) ≥ f (|u|) + g(|v|) ≥ |f (u)| + |g(v)| ≥ kf k + kgk − 2². As ² is arbitrary, kf + gk ≥ kf k + kgk. Q Q So C0 (X)∗ is an L-space. (c) Let ² > 0. For u ∈ A set Ku = {x : u(x) ≥ ²}. Then {Ku : u ∈ A} is a downwards-directed family of closed compact sets with empty intersection, so there must be some u ∈ A such that Ku = ∅, and kuk∞ ≤ ². As ² is arbitrary, we have the result. Remark (c) is a version of Dini’s theorem. 436J Riesz Representation Theorem (first form) Let (X, T) be a locally compact Hausdorff space, and Ck (X) the space of continuous real-valued functions on X with compact support. If f R: Ck (X) → R is any positive linear functional, there is a unique Radon measure µ on X such that f (u) = u dµ for every u ∈ Ck (X). proof (a) The point is that f is smooth. P P Suppose that A ⊆ Ck (X) is non-empty and downwards-directed and that inf A = 0 in RX . Fix u0 ∈ A and set K = {x : u0 (x) > 0}, so that K is compact. Because X is locally compact, there is an open relatively compact set G ⊇ K. Now there is a continuous function u1 : X → [0, 1] such that u1 (x) = 1 for x ∈ K and u1 (x) = 0 for x ∈ X \ G (4A2F(h-iii)). Because G is relatively compact, u1 ∈ Ck (X). Take any ² > 0. By 436Ic, there is a v ∈ A such that kvk∞ ≤ ². Now there is a v 0 ∈ A such that 0 v ≤ v ∧ u0 , so that v 0 (x) ≤ ² for every x ∈ K and v 0 (x) = 0 for x ∈ / K. In this case v 0 ≤ ²u1 , and inf u∈A f (u) ≤ f (v 0 ) ≤ ²f (u1 ). As ² is arbitrary, inf u∈A f (u) = 0; as A is arbitrary, f is smooth. Q Q (b) Note that because T is locally compact, it is the coarsest topology on X for which every function in Ck (X) is continuous (4A2G(e-ii)). Also Ck (X) is a truncated Riesz subspace of R X . So 436H tells us that R there is a quasi-Radon measure µ on X such that f (u) = u dµ for every u ∈ Ck (X). And µ is locally finite. P P If x0 ∈ X, then (as in (a) above) there is a u1 ∈ Ck (X)+ such that u1 (x0 ) = 1; now G = {x : u1 (x) > 12 } is an open set containing x0 , and µG ≤ 2f (u1 ) is finite. Q Q By 416G, or otherwise, µ is a Radon measure. (c) By 416E(b-v), µ is unique. 436K Riesz Representation Theorem (second form) Let (X, T) be a locally compact Hausdorff space. If f : C0 (X) → RR is any positive linear functional, there is a unique totally finite Radon measure µ on X such that f (u) = u dµ for every u ∈ C0 (X).

*436L


215

proof (a) As noted in 436Ib, C0 (X)∗ = C0 (X)∼ , so f is k k∞ -continuous. Ck (X) is a linear subspace of C0 (X), and f ¹CkR(X) is a positive linear functional; so by 436J there is a unique Radon measure µ on X such that f (u) = u dµ for every u ∈ Ck (X). Now µ is totally finite. P P By 414Ab, µX = sup{f (u) : u ∈ Ck (X), 0 ≤ u ≤ χX} ≤ kf k < ∞. Q Q R (b) Accordingly u dµ is defined for every u ∈ Cb (X), and in particular for every u ∈ C0 (X). Next, Ck (X) is norm-dense in C0 (X). P P If u ∈ C0 (X)+ , then un = (u − 2−n χX)+ belongs to Ck (X) and −n ku − un k∞ for every n ∈ N, so u ∈ C k ; accordingly C0 (X) = C0 (X)+ − C0 (X)+ is included in C k . R ≤2 Q Q Since dµ, regarded as a linear functional on C0R(X), is positive, therefore continuous, and agrees with f on Ck (X), it must be identical to f . Thus f (u) = u dµ for every u ∈ C0 (X). (c) Because there is only one Radon measure giving the right integrals to members of Ck (X) (436J), µ is unique. *436L The results here, by opening a path between measure theory and the study of linear functionals on spaces of continuous functions, provide an enormously powerful tool for the analysis of dual spaces C(X)∗ and their relatives. I will explore some of these ideas in the next section. Here I will give only a sample theorem to show how measure theory can tell us something about Banach lattices which seems difficult to reach by other methods. Proposition Let X be a topological space and U a norm-closed linear subspace of Cb (X)∗ such that the functional u 7→ f (u × v) : Cb (X) → R belongs to U whenever f ∈ U and v ∈ Cb (X). Then U is a band in the L-space Cb (X)∗ . proof (a) Let e = χX be the standard order unit of Cb (X), and if f ∈ Cb (X)∗ and u, v ∈ Cb (X) write fv (u) for f (u × v). By 356Na, Cb (X)∗ = Cb (X)∼ is an L-space. (b) I show first that U is a Riesz subspace of Cb (X)∼ . P P If f ∈ U and ² > 0, there is a v ∈ Cb (X) such that |v| ≤ e and f (v) ≥ |f |(e) − ² (356B). Now fv ≤ |f | and k|f | − fv k = (|f | − fv )(e) ≤ ² (356Nb), while fv ∈ U . As ² is arbitrary, |f | ∈ U = U ; as f is arbitrary, U is a Riesz subspace of Cb (X)∗ (352Ic). Q Q (c) Now suppose that X is a compact Hausdorff space. Then U is a solid linear subspace of Cb (X)∼ = C(X)∼ . P P Suppose that f ∈ U and that 0 ≤ R g ≤ f . Let ² >R0. By either 436J or 436K, there are Radon measures µ, ν on X such that f (u) = u dµ and g(u) = u dν for every u ∈ C(X). By 416Ea, dom µ ⊆ dom ν and νE ≤ µE for every E ∈ dom R µ, so ν is an indefinite-integral measure over µ (415Oa, or otherwise); let w : X → [0,R1] be such that E w dµ = νE for every E ∈ dom ν. There is a continuous function v : X → R such that |w − v|dµ ≤ ² (416I), and now Z |g(u) − fv (u)| = |

Z u dν −

(235M)

Z u × v dµ| = |

u × (w − v)dµ|

Z ≤ kuk∞

|w − v|dµ ≤ ²kuk∞

for every u ∈ C(X), so kg − fv k ≤ ², while fv ∈ U . As ² is arbitrary, g ∈ U ; as f and g are arbitrary (and U is a Riesz subspace of C(X)∗ ), U is a solid linear subspace of C(X)∗ . Q Q Since every norm-closed solid linear subspace of an L-space is a band (use 354Ea), it follows that (provided X is a compact Hausdorff space) U is actually a band. (c) For the general case, let Z be the set of all Riesz homomorphisms z : Cb (X) → R such that z(e) = 1. Then we have a Banach lattice isomorphism T : Cb (X) → C(Z) given by the formula (T u)(z) = z(u) for u ∈ Cb (X), z ∈ Z (see the proofs of 353M and 354K). But note also that T is multiplicative (353Pd), and

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*436L

T 0 : C(Z)∗ → Cb (X)∗ is a Banach lattice isomorphism. Let V = (T 0 )−1 [U ] ⊆ C(Z)∗ ; then V is a closed linear subspace of C(Z)∗ . If g ∈ V and v, w ∈ C(Z), then g(v × w) = (T 0 g)(T −1 v × T −1 w), so gw , defined in C(Z)∗ by the convention of (a) above, is just (T 0 )−1 ((T 0 g)T −1 w ), and belongs to V . By (b), V is a band in C(Z)∗ so U is a band in Cb (X)∗ , as required. 436X Basic exercises > (a) Let (X, Σ, µ0 ) be a measure space, and U the R set of µ0 -integrable Σmeasurable real-valued functions defined everywhere on X. For u ∈ U set f (u) = u dµ0 . Show that U and f satisfy the conditions of 436D, and that the measure µ constructed from f by the procedure there is just the c.l.d. version of µ0 . R R (b) Let µ and ν be two complete locally determined measures on a set X, and suppose that f dµ = f dν for every function f : X → R for which either integral is defined in R. Show that µ = ν. > (c) Let X be a set, U a truncated Riesz subspace of RX , and f : U → R a sequentially smooth positive linear functional. For A ⊆ X set θA = inf{sup f (un ) : hun in∈N is a non-decreasing sequence in U + , n∈N

lim un (x) = 1 for every x ∈ A},

n→∞

taking inf ∅ = ∞ if need be. Show that θ is an R outer measure on X. Let µ0 be the measure defined from θ by Carathéodory’s method. Show that f (u) = u dµ0 for every u ∈ U . Show that the measure µ constructed in 436D is the c.l.d. version of µ0 . (d) Let X be a set and U a truncated Riesz subspace of RX . Let τ : U → [0, ∞[ be a seminorm such that (i) τ (u) ≤ τ (v) whenever |u| ≤ |v| (ii) limn→∞ τ (un ) = 0 whenever hun in∈N is a non-increasing sequence in U Rand limn→∞ un (x) = 0 for every x ∈ X. Show that for any u0 ∈ U + Rthere is a measure µ on x such that u dµ is defined, and less than or equal to τ (u), for every u ∈ U , and u0 dµ = τ (u0 ). (Hint: put the Hahn-Banach theorem together with 436D.) (e) Let X be any topological space. Show that every positive linear functional on C(X) is sequentially smooth (compare 375A), so corresponds to a totally finite Baire measure on X. (f ) Let X be a completely regular space. Show that it is measure-compact iff every sequentially smooth positive linear functional on Cb (X) is smooth. (Hint: 436Xj.) (g) A completely regular space is called realcompact if every Riesz homomorphism from C(X) to R is of the form u 7→ αu(x) for some x ∈ X and α ≥ 0. (i) Show that, for any topological space X, any Riesz homomorphism from C(X) to R is representable by a Baire measure on X which takes at most two values. (ii) Show that a completely regular space X is realcompact iff every {0, 1}-valued Baire measure on X is of the form E 7→ χE(x). (iii) Show that a completely regular space X is realcompact iff every purely atomic totally finite Baire measure on X is τ -additive. (iii) Show that a measure-compact completely regular space is realcompact. (iv) Show that the discrete topology on [0, 1] is realcompact. (Hint: if ν is a Baire measure taking only the values 0 and 1, set x0 = sup{x : ν[x, 1] = 1}.) (v) Show that any product of realcompact completely regular spaces is realcompact. (vi) Show that a Souslin-F subset of a realcompact completely regular space is realcompact. (For realcompact spaces which are not measure-compact, see 439Xm.) (h) In 436F, suppose that µ and ν are τ -additive. Let µ ˜ and ν˜ be the corresponding quasi-Radon measures ˜ the quasi-Radon product of µ ˜ to the Baire (415N), and λ ˜ and ν˜ (417N). Show that λ is the restriction of λ σ-algebra of X × Y . >(i) For u ∈ C([0, 1]) let f (u) be the Lebesgue integral of u. Show that f is smooth (therefore sequentially smooth) but not sequentially order-continuous (therefore not order-continuous). (Hint: enumerate Q ∩ [0, 1] as hqn in∈N , and set un (x) = mini≤n 2i+2 |x − qi | for n ∈ N, x ∈ [0, 1]; show that inf n∈N un = 0 in C([0, 1]) but limn→∞ f (un ) > 0.)

436Y


217

(j) In 436E, show that µ is τ -additive iff f is smooth. (k) Suppose that X is a set, U is a truncated Riesz subspace of RX and f : U → R is a smooth positive linear functional. For A ⊆ X set θA = inf{sup f (u) : B is an upwards-directed family in U + u∈B

such that sup u(x) = 1 for every x ∈ A}, u∈B

taking inf ∅ = ∞ if need be. Show that θ is an R outer measure on X. Let µ0 be the measure defined from θ by Carathéodory’s method. Show that f (u) = u dµ0 for every u ∈ U . Show that the measure µ constructed in 436H is the c.l.d. version of µ0 . (l) Let X be a completely regular topological space and f a smooth positive linear functional on Cb (X). R Show that there is a unique totally finite quasi-Radon measure µ on X such that f (u) = u dµ for every u ∈ Cb (X). > (m) For u ∈ C([0, 1]) let f (u) be the Riemann integral of u (134K). Show that the Radon measure on [0, 1] constructed by the method of 436J is just Lebesgue measure on [0, 1]. Explain how to construct Lebesgue measure on R from an appropriate version of the Riemann integral on R. (n) Let X be a topological space. Let f : Cb (X) → R be a positive linear functional. Show that the following are equiveridical: (i) f is tight, that is, for every ² > 0 there is a compact K ⊆ X Rsuch that f (u) ≤ ² whenever 0 ≤ u ≤ χ(X \ K) (ii) there is a tight Borel measure µ on X such that f (u) = u dµ for every u ∈ Cb (X). (Hint: show that a tight functional is smooth.) > (o) Let (X, R T, Σ, µ) and (Y, S, T, ν) be locally compact Radon measure spaces. (i) Show that the function x 7→ w(x, y)ν(dy) belongs to Ck (X) for every w ∈ Ck (X × Y ), so we have a positive linear functional h : Ck (X × Y ) → R defined by setting h(w) =

RR

w(x, y)ν(dy)µ(dx)

for w ∈ Ck (X × Y ). (ii) Show that the corresponding Radon measure on X × Y is just the product Radon measure as defined in 417P/417R. > (p) Let A be a Boolean algebra and Z its Stone space; identify L∞ (A) with C(Z), as in 363A. Let ν be a non-negative finitely additive functional on A, f the corresponding positive linear R functional on L∞ (A) (363K), and µ the corresponding Radon measure on Z (416Qb). Show that f (u) = u dµ for every u ∈ L∞ (A). (q) Let X be a locally compact Hausdorff space. Show that a sequence hfn in∈N in C0 (X) converges to 0 for the weak topology on C0 (X) iff supn∈N kfn k∞ is finite and limn→∞ fn (x) = 0 for every x ∈ X. (r) Let A be a Boolean algebra, and M (A) the L-space of bounded finitely additive functionals on A. Let U ⊆ M (A) be a norm-closed linear subspace such that a 7→ ν(a ∩ b) belongs to U whenever ν ∈ U and b ∈ A. Show that U is a band in M (A). (Hint: identify L∞ (A) with Cb (X), where X is the Stone space of A, and M (A) with Cb (X)∗ .) (s) Let X be a non-empty compact Hausdorff space, and φ : X → X a continuous function. Show that there is a Radon probability measure µ on X such that φ is inverse-measure-preserving for µ.R (Hint: let F be a non-principal ultrafilter on X, x0 any point of X, and define µ by the formula f dµ = 1 Pn k −1 = µ.) limn→F k=0 f (φ (x0 )) for every f ∈ C(X). Use 416E(b-v) to show that µφ n+1

436Y Further exercises (a) Let X be a set, U a Riesz subspace of RX , and f : U → R a sequentially smooth positive linear functional. (i) Write Uσ for the set of functions from X to [0, ∞] expressible as the supremum of a non-decreasing sequence hun in∈N in U + such that supn∈N f (un ) < ∞. Show that there is a

218


436Y

functional fσ : Uσ → [0, ∞[ such that fσ (u) = supn∈N f (un ) whenever hun in∈N is a non-decreasing sequence in U + with supremum u ∈ Uσ . (Compare 122I.) (ii) Show that u + v ∈ Uσ and fσ (u + v) = fσ (u) + fσ (v) for all u, v ∈ Uσ . (iii) Suppose that u, v ∈ Uσ , u ≤ v and u(x) = v(x) whenever v(x) is finite. Show that fσ (u) = fσ (v). (Hint: take non-decreasing sequences hun in∈N , hvn in∈N with suprema u, v. Consider hf (vk −un −δvn )+ in∈N where k ∈ N, δ > 0.) (iv) Let V be the set of functions v : X → R such that there are u1 , u2 ∈ Uσ such that v(x) = u1 (x) − u2 (x) whenever u1 (x), u2 (x) are both finite. Show that V is a linear subspace of RX and that there is a linear functional g : V → R defined by setting g(v) = fσ (u1 ) − fσ (u2 ) whenever v = u1 − u2 in the sense of the last sentence. (v) Show that V is a Riesz subspace of RX . (vi) Show that if hvn in∈N is a non-decreasing sequence in V and γ = supn∈N g(vn ) is finite, then there is a v ∈ V such that g(v) = supn∈N g(v ∧ vn ) = γ. (This is a version of the Daniell integral.) (b) Develop further the theory of 436Ya, finding a version of Lebesgue’s Dominated Convergence Theorem, a concept of ‘negligible’ subset of X, and an L-space of equivalence classes of ‘integrable’ functions. (c) Let X be a countably compact topological space. Show that every positive linear functional on Cb (X) is sequentially smooth, so corresponds to a totally finite Baire measure on X. (d) Let X be a sequential space, Y a topological space, µ a semi-finite Baire measure on X and ν a σ-finite Baire measure on Y . Let µ ˜ be the c.l.d. version of µ. Show that there is a semi-finite Baire measure λ on X × Y such that λW =

R

νW [{x}]˜ µ(dx),

R

f dλ =

RR

f (x, y)ν(dy)˜ µ(dx)

for every Baire set W ⊆ X × Y and every non-negative continuous function f : X × Y → R. Show that the c.l.d. version of λ extends the c.l.d. product measure of µ and ν. (e) Let X0 , . . . , Xn be sequential spaces and µi a totally finite Baire measure on Xi for each i. (i) Show that if f : X0 × . . . × Xn → R is a bounded separately continuous function, then φ(f ) =

R

...

R

f (x0 , . . . , xn )µn (dxn ) . . . µ0 (dx0 )

is defined, so that we have a corresponding Baire product measure on X0 × . . . × Xn . (ii) Show that this product is associative. (f ) Let X and Y be compact Hausdorff spaces. (i) Show that there is a unique bilinear map φ : C(X)∗ × C(Y )∗ → C(X × Y )∗ which is separately continuous for the weak* topologies and such that φ(δx , δy ) = δ(x,y) for all x ∈ X and y ∈ Y , setting δx (f ) = f (x) for f ∈ C(X) and x ∈ X.R(ii) Show thatR if µ R and ν are Radon measures on X, Y respectively with Radon measure product λ, then φ( dµ, dν) = dλ. (iii) Show that kφk ≤ 1 (definition: 253Ab). 436 Notes and comments From the beginning, integration has been at the centre of measure theory. My own view – implicit in the arrangement of this treatise, from Chapter 11 onward – is that ‘measure’ and ‘integration’ are not quite the same thing. I freely acknowledge that my treatment of ‘integration’ is distorted by my presentation of it as part of measure theory; on the other side of the argument, I hold that regarding ‘measure’ as a concept subsidiary to ‘integral’, as many authors do, seriously interferes with the development of truly penetrating intuitions for the former. But it is undoubtedly true that every complete locally determined measure can be derived from its associated integral (436Xb). Moreover, it is clearly of the highest importance that we should be able to recognise integrals when we see them; I mean, given a linear functional on a linear space of functions, then if it can be expressed as an integral with respect to a measure this is something we need to know at once. And thirdly, investigation of linear functionals frequently leads us to measures of great importance and interest. Concerning the conditions in 436D, an integral must surely be ‘positive’ (because measures in this treatise are always non-negative) and ‘sequentially smooth’ (because measures are supposed to be countably additive). But it is not clear that we are forced to restrict our attention to Riesz subspaces of RX , and even less clear that they have to be ‘truncated’. In 439I below I give an example to show that this last condition is essential for 436D and 436H as stated. However it is not necessary for large parts of the theory. In many cases, if U ⊆ RX is a Riesz subspace which is not truncated, we can take an element e ∈ U + and look at

437A

Spaces of measures

219

Y = {x : e(x) > 0}, Ve = {u : u(x) = 0 for every x ∈ X \ Y } ∼ = We , where We = {u/e : u ∈ Ve } is a truncated Riesz subspace of R Y . But there are applications in which this approach is unsatisfactory and a more radical revision of the basic theory of integration, as in 436Ya, is useful. I have based the arguments of this section on the inner measure constructions of §413. Of course it is also possible to approach them by means of outer measures (436Xc, 436Xk). I emphasize the exercise 436Xo because it is prominent in ‘Bourbakist’ versions of the theory of Radon measures, in which (following Bourbaki 65 rather than Bourbaki 69) Radon measures are regarded as linear functionals on spaces of continuous functions. By 436J, this is a reasonably effective approach as long as we are interested only in locally compact spaces, and there are parts of the theory of topological groups (notably the duality theory of §445 below) in which it even has advantages. The construction of 436Xo shows that we can find a direct approach to the tensor product of linear functionals which does not require any attempt to measure sets rather than integrate functions. I trust that the prejudices I am expressing will not be taken as too sweeping a disparagement of such methods. Practically all correct arguments in mathematics (and not a few incorrect ones) are valuable in some way, suggesting new possibilities for investigation. In particular, one of the challenges of measure theory (not faced in this treatise) is that of devising effective theories of vector-valued measures. Typically this is much easier with Riemann-type integrals, and any techniques for working directly with these should be noted. 436L revisits ideas from Chapter 35, and the result would be easier to find if it were in §356. I include it here as an example of the way in which familiar material from measure theory (in particular, the RadonNikod´ ym theorem) can be drafted to serve functional analysis. I should perhaps remark that there are alternative routes which do not use measure theory explicitly, and while longer are (in my view) more illuminating.

437 Spaces of measures Once we have started to take the correspondence between measures and integrals as something which operates in both directions, we can go a very long way. While ‘measures’, as dealt with in this treatise, are essentially positive, an ‘integral’ can be thought of as a member of a linear space, dual in some sense to a space of functions. Since the principal spaces of functions are Riesz spaces, we find ourselves looking at dual Riesz spaces as discussed in §356; while the corresponding spaces of measures are close to those of §362. Here I try to draw these ideas together with an examination of the spaces Uσ∼ and Uτ∼ of sequentially smooth and smooth functionals, and the matching spaces Mσ and Mτ of countably additive and τ -additive measures (437A-437I). Because a (sequentially) smooth functional corresponds to a countably additive measure, which can be expected to integrate many more functions than those in the original Riesz space (typically, a space of continuous functions), we find that relatively large spaces of bounded measurable functions can be canonically embedded into the biduals (Uσ∼ )∗ and (Uτ∼ )∗ (437C, 437H, 437I). The guiding principles of functional analysis encourage us not only to form linear spaces, but also to examine linear space topologies, starting with norm and weak topologies. The theory of Banach lattices described in §354, particularly the theory of M - and L-spaces, is an important part of the structure here. In addition, our spaces Uσ∼ have natural weak* topologies which can be regarded as topologies on spaces of measures; these are the ‘vague’ topologies of 437J, which have already been considered, in a special case, in §285. It turns out that on the positive cone of Mτ , at least, the vague topology may have an alternative description directly in terms of the behaviour of the measures on open sets (437L). This leads us to a parallel idea, the ‘narrow’ topology on non-negative additive functionals (437Jd). The second half of the section is devoted to the elementary properties of narrow topologies (437K-437R), with especial reference to compact sets in these topologies (437O-437U). Seeking to identify narrowly compact sets, we come to the concept of ‘uniform tightness’ (437T). Bounded uniformly tight sets are narrowly relatively compact (437U); in ‘Prokhorov spaces’ (437V) the converse is true; I end the section with a list of the best-known Prokhorov spaces (437W). 437A Smooth and sequentially smooth duals Let X be a set, and U a Riesz subspace of RX . Recall that U ∼ is the Dedekind complete Riesz space of order-bounded linear functionals on U , that Uc∼ is the band of differences of sequentially order-continuous positive linear functionals, and that U × is the band of

220


437A

differences of order-continuous positive linear functionals (356A). A functional f ∈ (U ∼ )+ is ‘sequentially smooth’ if inf n∈N f (un ) = 0 whenever hun in∈N is a non-increasing sequence in U and limn→∞ un (x) = 0 for every x ∈ X, and ‘smooth’ if inf u∈A f (u) = 0 whenever A ⊆ U is a non-empty downwards-directed set and inf u∈A u(x) = 0 for every x ∈ X (436A, 436G). (a) Set Uσ∼ = {f : f ∈ U ∼ , |f | is sequentially smooth}, the sequentially smooth dual of U . Then Uσ∼ is a band in U ∼ . P P (i) If f ∈ Uσ∼ , g ∈ U ∼ and |g| ≤ |f |, then |g|(un ) ≤ |f |(un ) → 0 whenever hun in∈N is a non-increasing sequence in U and limn→∞ un (x) = 0 for every x, so |g| is sequentially smooth and g ∈ Uσ∼ . Thus Uσ∼ is a solid subset of U ∼ . (ii) If f , g ∈ Uσ∼ and α ∈ R, then |f + g|(un ) ≤ |f |(un ) + |g|(un ) → 0,

|αf |(un ) = |α||f |(un ) → 0

whenever hun in∈N is a non-increasing sequence in U and limn→∞ un (x) = 0 for every x, so |f + g| and |αf | are sequentially smooth. Thus Uσ∼ is a Riesz subspace of U ∼ . (iii) Now suppose that B ⊆ (Uσ∼ )+ is an upwards-directed set with supremum g ∈ U ∼ , and that hun in∈N is a non-increasing sequence in U such that limn→∞ un (x) = 0 for every x ∈ X. Then, given ² > 0, there is an f ∈ B such that (g − f )(u0 ) ≤ ² (355Ed), so that g(un ) ≤ f (un ) + ² for every n, and lim supn→∞ g(un ) ≤ ² + limn→∞ f (un ) ≤ ². As ² and hun in∈N are arbitrary, g ∈ Uσ∼ ; as B is arbitrary, Uσ∼ is a band (352Ob). Q Q As remarked in 436A, sequentially order-continuous positive linear functionals are sequentially smooth, so Uc∼ ⊆ Uσ∼ . P (i) (b) Set Uτ∼ = {f : f ∈ U ∼ , |f | is smooth}, the smooth dual of U . Then Uτ∼ is a band in U ∼ . P Suppose that f , g ∈ Uτ∼ , α ∈ R, h ∈ U ∼ and |h| ≤ |f |. If A ⊆ U is a non-empty downwards-directed set and inf u∈A u(x) = 0 for every x ∈ X, and ² > 0, then there are u0 , u1 ∈ A such that |f |(u0 ) ≤ ² and |g|(u1 ) ≤ ², and a u ∈ A such that u ≤ u0 ∧ u1 . In this case |h|(u) ≤ |f |(u) ≤ ², |f + g|(u) ≤ |f |(u) + |g|(u) ≤ 2², |αf |(u) = |α||f |(u) ≤ |α|². As A and ² are arbitrary, h, f + g and αf all belong to Uτ∼ ; so that Uτ∼ is a solid Riesz subspace of U ∼ . (ii) Now suppose that B ⊆ (Uτ∼ )+ is an upwards-directed set with supremum g ∈ U ∼ , and that A ⊆ U is a non-empty downwards-directed set such that inf u∈A u(x) = 0 for every x ∈ X. Fix any u0 ∈ A. Then, given ² > 0, there is an f ∈ B such that (g − f )(u0 ) ≤ ², so that g(u) ≤ f (u) + ² whenever u ∈ A and u ≤ u0 . But A0 = {u : u ∈ A, u ≤ u0 } is also a downwards-directed set, and inf u∈A0 u(x) = 0 for every x ∈ X, so inf u∈A g(u) ≤ ² + inf u∈A0 f (u) ≤ ². As ² and A are arbitrary, g ∈ Uτ∼ ; as B is arbitrary, Uτ∼ is a band. Q Q Just as Uc∼ ⊆ Uσ∼ , U × ⊆ Uτ∼ . 437B Signed measures Collecting these ideas together with those of §§362-363, we are ready to approach ‘signed measures’. Recall that if X is a set and Σ is a σ-algebra of subsets of X, we can identify L∞ = L∞ (Σ), as defined in §363, with the space L∞ of bounded Σ-measurable real-valued functions (363Ha). Now, because L∞ is sequentially order-closed in RX , sequentially smooth functionals on L∞ are actually sequentially order∞ ∼ ∞ ∼ continuous, so (L∞ )∼ σ = (L )c . Next, we can identify (L )c with the space Mσ of countably additive functionals, or ‘signed measures’, on Σ (363K); if ν ∈ Mσ , the corresponding member of (L∞ )∼ c is the unique order-bounded (or norm-continuous) linear functional f on L∞ such that f (χE) = νE for every E ∈ Σ. If ν ≥ 0, so that ν is a totally finite measure with domain Σ, then of course f , when interpreted as a functional on L∞ , must be just integration with respect to ν. The identification between (L∞ )∼ in 363K is an L-space isomorphism. So it tells us, c and Mσ described R for instance, that if we are willing to use the symbol for the duality between L∞ and the space of bounded finitely additive functionals on Σ, as in 363L, then we can write

437C

Spaces of measures

R ∞

for every u ∈ L

u d(µ + ν) =

R

u dµ +

221

R

u dν

and all µ, ν ∈ Mσ .

437C Theorem Let X be a set and U a Riesz subspace of `∞ (X), the M -space of bounded real-valued functions on X, containing the constant functions. (a) Let Σ be the smallest σ-algebra of subsets of X with respect to which every member of U is measurable. Let Mσ = Mσ (Σ) be the L-space of countably additive functionals on Σ (326E, 362B). Then we have a R Banach lattice isomorphism T : Mσ → Uσ∼ defined by saying that (T µ)(u) = u dµ whenever µ ∈ Mσ+ and u ∈ U. (b) We now have a sequentially order-continuous norm-preserving Riesz homomorphism S, embedding the M -space L∞ = L∞ (Σ) of bounded real-valued Σ-measurableRfunctions on X (363Ha) into the M -space (Uσ∼ )∼ = (Uσ∼ )∗ = (Uσ∼ )× , defined by saying that (Sv)(T µ) = v dµ whenever v ∈ L∞ and µ ∈ Mσ+ . If u ∈ U , then (Su)(f ) = f (u) for every f ∈ Uσ∼ . proof (a)(i) The norm k k∞ is an order-unit norm on U (354Ga), so U ∗ = U ∼ is an L-space (356N), and the band Uσ∼ (437Aa) is an L-space in its own right (354O). (ii) As noted Banach lattice isomorphism T0 : Mσ → (L∞ )∼ c defined by saying R in 437B, we have a ∞ that (T0 µ)(u) = u dµ whenever u ∈ L and µ ∈ Mσ+ . If we set T µ = T0 µ ¹ U , then T is a positive linear operator from Mσ to U ∼ , just because U is a linear subspace of L∞ ; and since T µ ∈ Uσ∼ for every µ ∈ Mσ+ , + ∼ + P By 436D, T is an operator from Mσ to Uσ∼ . Now R every f ∈ (Uσ ) is of the form T µ for some µ ∈ Mσ . P there is some measure λ such that u dλ = f (u) for every u ∈ U . Completing λ if necessary, we see that we may suppose that every member of U is (dom λ)-measurable, that is, that Σ ⊆ dom λ; take µ = λ¹Σ. Q Q So T is surjective. (iii) Write K for the family of sets K ⊆ X such that χK = inf n∈N un for some sequence hun in∈N in U . (See the proof of 436D.) We need to know the following. (α) K ⊆ Σ. (β) If K ∈ K, then there is a non-increasing sequence hun in∈N in U such that χK = inf n∈N un . (For if hu0n in∈N is any sequence in U such that χK = inf n∈N u0n , we can set un = inf i≤n u0i for each i.) (γ) The σ-algebra of subsets of X generated by K is Σ. P P Let T be the σ-algebra of subsets of X generated by K. T ⊆ Σ because K ⊆ Σ. If u ∈ U and α > 0 then {x : u(x) ≥ α} ∈ K (see part (b) of the proof of 436D). So every member of U + , therefore every member of U , is T-measurable, and Σ ⊆ T. Q Q − (iv) T is injective. P P If µ1 , µ2 ∈ Mσ and T µ1 = T µ2 , set νi = µi + µ− 1 + µ2 for each i, so that νi is non-negative and T ν1 = T ν2 . If K ∈ K then there is a non-increasing sequence hun in∈N in U such that χK = inf n∈N un in RX , so

ν1 K = inf n∈N

R

un dν1 = inf n∈N

R

un dν2 = ν2 K.

Now K contains X and is closed under finite intersections and ν1 and ν2 agree on K. By the Monotone Class Theorem (136C), ν1 and ν2 agree on the σ-algebra generated by K, which is Σ; so ν1 = ν2 and µ1 = µ2 . Q Q Thus T is a linear space isomorphism between Mσ and Uσ∼ . (v) As noted in (ii), T [Mσ+ ] = (Uσ∼ )+ ; so T is a Riesz space isomorphism. (vi) Now if µ ∈ Mσ ,

kT µk = |T µ|(χX) (356Nb) = (T |µ|)(χX) (because T is a Riesz homomorphism) = |µ|(X) = kµk (362Ba). So T is norm-preserving and is an L-space isomorphism, as claimed. (b)(i) By 356Pb, (Uσ∼ )∗ = (Uσ∼ )∼ = (Uσ∼ )× is an M -space.

222


437C

× (ii) We have a canonical map S0 : L∞ → ((L∞ )∼ c ) defined by saying that (S0 v)(h) = h(v) for every ∞ ∞ ∼ v ∈ L , h ∈ (L )c ; and by 356F, S0 is a Riesz homomorphism. If hvn in∈N is a non-increasing sequence in + L∞ with infimum 0, then inf n∈N (S0 vn )(h) = inf n∈N h(vn ) = 0 for every h ∈ ((L∞ )∼ c ) , so inf n∈N S0 vn = 0 (355Ee); as hvn in∈N is arbitrary, S0 is sequentially order-continuous (351Gb). ∞ Also S0 is norm-preserving. P P (α) If h ∈ (L∞ )∼ c and v ∈ L , then

|(S0 v)(h)| = |h(v)| ≤ khkkvk∞ , so kS0 vk ≤ kvk∞ . (β) If v ∈ L∞ and 0 ≤ γ < kvk∞ , take x ∈ X such that |v(x)| > γ, and define ∞ hx ∈ (L∞ )∼ c by setting hx (w) = w(x) for every w ∈ L ; then khx k = 1 and |(S0 v)(hx )| = |hx (v)| = |v(x)| ≥ γ, so kS0 vk ≥ γ. As γ is arbitrary, kS0 vk ≥ kvk∞ and kS0 vk = kvk∞ . Q Q ∼ (iii) Now T0 : Mσ → (L∞ )∼ c and T : Mσ → Uσ are both norm-preserving Riesz space isomorphisms, ∗ ∼ ∗ ∞ ∼ −1 ∼ so T0 T : Uσ → (L )c is another, and its adjoint S1 : ((L∞ )∼ c ) → (Uσ ) must also be a norm-preserving Riesz space isomorphism. So if we set S = S1 S0 , S will be a norm-preserving sequentially order-continuous Riesz homomorphism from L∞ to (Uσ∼ )× = (Uσ∼ )∗ .

(iv) Setting the construction out in this way tells us a lot about the properties of the operator S, but undeniably leaves it somewhat obscure. So let us start again from v ∈ L∞ and µ ∈ Mσ+ , and seek to calculate (Sv)(T µ). We have (Sv)(T µ) = (S1 S0 v)(T µ) = (S0 v)(T0 T −1 T µ) (because S1 is the adjoint of T0 T −1 )

Z = (S0 v)(T0 µ) = (T0 µ)(v) =

v dµ,

as claimed. If u ∈ U , then (T µ)(u) = (T0 µ)(u) for every µ ∈ Mσ , so if f ∈ Uσ∼ then (Su)(f ) = (S1 S0 u)(f ) = (S0 u)(T0 T −1 f ) = (T0 T −1 f )(u) = (T T −1 f )(u) = f (u). This completes the proof. 437D Remarks What is happening here is that the canonical Riesz homomorphism u 7→ u ˆ from U to (Uσ∼ )∗ (356F) has a natural extension to L∞ (Σ). The original homomorphism u 7→ u ˆ is not, as a rule, sequentially order-continuous, just because Uσ∼ is generally larger than Uc∼ ; but the extension to L∞ is sequentially order-continuous. If you like, it is sequential smoothness which is carried over to the extension, and because the embedding of L∞ in RX is sequentially order-continuous, a sequentially smooth operator on L∞ is sequentially order-continuous. R R In the statement of 437C, I have used the formulae (T µ)(u) = u dµ and (Sv)(T µ) = v dµ onR the assumption that µ ∈ Mσ+ , so that µ is actually a measure on the definition used in this treatise, and dµ is the ordinary integral as constructed in §122. Since the functions u and v are bounded, measurable and defined everywhere, we can choose toR extend the notion of integration to signed measures, as in 363L, in R which case the formulae (T µ)(u) = u dµ and (Sv)(T µ) = v dµ become meaningful, and true, for all µ ∈ Mσ , u ∈ U and v ∈ L∞ . In fact the ideas here can be pushed farther, as in 437Ib, 437Xe and 437Yd. 437E Corollary Let X be a completely regular Hausdorff space, and Ba = Ba(X) its Baire σ-algebra. Then we can identify Cb (X)∼ σ with the L-space Mσ (Ba) of countably additive functionals on Ba, and we ∗ have a norm-preserving sequentially order-continuous Riesz homomorphism from L∞ (Ba) to (Cb (X)∼ σ) R ∞ ∼ + defined by setting (Sv)(f ) = v dµf for every v ∈ L and f ∈ (Cb (X)σ ) , where µf is the Baire measure associated with f . proof Put 437C and 436E together.

437G

Spaces of measures

223

437F Proposition Let X be a topological space and B = B(X) its Borel σ-algebra. Let Mσ be the L-space of countably additive functionals on B. (a) Write Mτ ⊆ Mσ for the set of differences of τ -additive totally finite Borel measures. Then Mτ is a band in Mσ , so is an L-space in its own right. (b) Write Mt ⊆ Mτ for the set of differences of totally finite Borel measures which are tight (that is, inner regular with respect to the closed compact sets). Then Mt is a band in Mσ , so is an L-space in its own right. proof (a)(i) Let µ1 , µ2 be totally finite τ -additive Borel measures on X, α ≥ 0, and µ ∈ Mσ such that 0 ≤ µ ≤ µ1 . Then µ1 + µ2 , αµ1 and µ are totally finite τ -additive Borel measures. P P They all belong to Mσ , that is, are Borel measures. Now let G be a non-empty upwards-directed family of open sets in X with union H, and ² > 0. Then there are G1 , G2 ∈ G such that µ1 G1 ≥ µ1 H − ² and µ2 G2 ≥ µ2 H − ², and a G ∈ G such that G ⊇ G1 ∪ G2 . In this case, (µ1 + µ2 )(G) ≥ (µ1 + µ2 )(H) − 2², (αµ1 )(G) ≥ (αµ1 )(H) − α² and µG = µH − µ(H \ G) ≥ µH − µ1 (H \ G) ≥ µH − ². As G and ² are arbitrary, µ1 + µ2 , αµ1 and µ are all τ -additive. Q Q It follows that Mτ is a solid linear subspace of Mσ . (ii) Now suppose that B ⊆ Mτ+ is non-empty and upwards-directed and has a supremum ν in Mσ . Then ν ∈ Mτ . P P If G is a non-empty upwards-directed family of open sets with union H, then νH = sup µH µ∈B

(362Be) =

sup µ∈B,G∈G

µG = sup νG; G∈G

as G is arbitrary, ν is τ -additive and belongs to Mτ . Q Q As B is arbitrary, Mτ is a band in Mσ . By 354O, it is itself an L-space. (b) We can use the same arguments. Suppose that µ1 , µ2 ∈ Mσ are tight, α ≥ 0, and µ ∈ Mσ is such that 0 ≤ µ ≤ µ1 . If E ∈ B and ² > 0, there are closed compact sets K1 , K2 ⊆ E such that µ1 (E \ K1 ) ≤ ² and µ2 (E \ K2 ) ≤ ²; now K = K1 ∪ K2 is a closed compact subset of E, and µK = µE − µ(E \ K) ≥ µE − µ1 (E \ K) − µ2 (E \ K) ≥ µE − 2², (αµ1 )(K) ≥ (αµ1 )(E) − α². As E and ² are arbitrary, µ and αµ1 are tight; as µ1 , µ2 and α are arbitrary, Mt is a solid linear subspace of Mσ . Now suppose that B ⊆ Mt+ is non-empty and upwards-directed and has a supremum ν in Mσ . Take any E ∈ B and ² > 0. Then there is a µ ∈ B such that µE ≥ νE − ²; there is a closed compact set K ⊆ E such that µK ≥ µE − ²; and now νK ≥ νE − 2². As E and ² are arbitrary, ν is tight; as B is arbitrary, Mt is a band in Mσ , and is in itself an L-space. 437G Definitions Let X be a topological space. A signed Baire measure on X will be a countably additive functional on the Baire σ-algebra Ba(X), which by the Jordan decomposition theorem (231F) is expressible as the difference of two totally finite Baire measures; a signed Borel measure will be a countably additive functional on the Borel σ-algebra B(X), that is, the difference of two totally finite Borel measures; a signed τ -additive Borel measure will be a member of the L-space Mτ as described in 437F, that is, the difference of two τ -additive totally finite Borel measures; and if X is Hausdorff, a signed tight Borel measure will be a member of the L-space Mt as described in 437F, that is, the difference of two tight totally finite Borel measures.

224


437H

437H Theorem Let X be a set and U a Riesz subspace of `∞ (X) containing the constant functions. Let T be the coarsest topology on X rendering every member of U continuous, and B = B(X) the corresponding Borel σ-algebra. (a) Let Mτ be the L-space of signed τ -additive Borel measures on X. Then we have a Banach lattice R isomorphism T : Mτ → Uτ∼ defined by saying that (T µ)(u) = u dµ whenever µ ∈ Mτ+ and u ∈ U . (b) We now have a sequentially order-continuous norm-preserving Riesz homomorphism S, embedding the M -space L∞ = L∞ (B) of bounded on X into (Uτ∼ )∼ = (Uτ∼ )∗ = (Uτ∼ )× , R Borel measurable functions ∞ defined by saying that (Sv)(T µ) = v dµ whenever v ∈ L and µ ∈ Mτ+ . If u ∈ U , then (Su)(f ) = f (u) for every f ∈ Uτ∼ . proof The proof follows the same lines as that of 437C. (a)(i) As before, the norm k k∞ is an order-unit norm on U , so U ∗ = U ∼ is an L-space, and the band (437Ab) is an L-space in its own right, like Mτ (437F).

Uτ∼

(ii) Let Mσ be the L-space of all countably additive functionals on B, so that Mτ is a band in MσR . Let u dµ T0 : Mσ → (L∞ )∼ c be the canonical Banach lattice isomorphism defined by saying that (T0 µ)(u) = whenever u ∈ L∞ and µ ∈ Mσ+ . If we set T µ = T0 µ¹U for µ ∈ Mτ , then T is a positive linear operator from Mτ to U ∼ , just because U is a Riesz subspace of L∞ ; and since T µ ∈ Uτ∼ for every µ ∈ Mτ+ (436H), T is ∼ + + an operator from Mτ to Uτ∼ . Now every P By 436H, there R f ∈ (Uτ ) is of the form T µ for some µ ∈ Mτ . P is a quasi-Radon measure λ such that u dλ = f (u) for every u ∈ U ; set µ = λ¹B. Q Q So T is surjective. (iii) Let K be the family of subsets K of X such that χK = inf A in RX for some non-empty subset A of U . Then K is just the family of closed sets for T. P P As noted in part (b) of the proof of 436H, every member of K is closed, and K \ G ∈ K whenever K ∈ K and G ∈ T; but as, in the present case, X ∈ K, every closed set belongs to K. Q Q − (iv) T is injective. P P If µ1 , µ2 ∈ Mτ and T µ1 = T µ2 , set νi = µi + µ− 1 + µ2 for each i, so that νi is non-negative and T ν1 = T ν2 . If K ∈ K, set A = {u : u ∈ U , u ≥ χK}, so that χK = inf A in RX , and A is downwards-directed. By 414Bb,

ν1 K = inf u∈A

R

u dν1 = inf u∈A

R

u dν2 = ν2 K.

Now K contains X and is closed under finite intersections and ν1 and ν2 agree on K. By the Monotone Class Theorem, ν1 and ν2 agree on the σ-algebra generated by K, which is B; so ν1 = ν2 and µ1 = µ2 . Q Q Thus T is a linear space isomorphism between Mτ and Uτ∼ . (v) As noted in (ii), T [Mτ+ ] = (Uτ∼ )+ ; so T is a Riesz space isomorphism. (vi) Now if µ ∈ Mτ , kT µk = |T µ|(χX) = (T |µ|)(χX) = |µ|(X) = kµk. So T is norm-preserving and is an L-space isomorphism, as claimed. (b)(i) By 356Pb, (Uτ∼ )∗ = (Uτ∼ )∼ = (Uτ∼ )× is an M -space. (ii) Because T0 : Mσ → (L∞ )∼ c is a Banach lattice isomorphism, and Mτ is a band in Mσ , W = T0 [Mτ ] ∞ is a band in (L∞ )∼ → W × defined by writing c . We therefore have a Riesz homomorphism S0 : L ∞ (S0 v)(h) = h(v) for v ∈ L , h ∈ W (356F). Just as in (b-ii) of the proof of 437C, S0 is sequentially order-continuous and norm-preserving. (We need to observe that hx in the second half of the argument there always belongs to W ; this is because hx = T0 (δx ), where δx ∈ Mτ is defined by setting δx (E) = χE(x) for every Borel set E.) ∼ (iii) Now T0 : Mσ → (L∞ )∼ c and T : Mτ → Uτ are both norm-preserving Riesz space isomorphisms, −1 ∼ so T0 T : Uτ → W is another, and its adjoint S1 : W ∗ → (Uτ∼ )∗ must also be a norm-preserving Riesz space isomorphism. So if we set S = S1 S0 , S will be a norm-preserving sequentially order-continuous Riesz homomorphism from L∞ to (Uτ∼ )× = (Uτ∼ )∗ .

(iv) If v ∈ L∞ and µ ∈ Mτ+ ,

437I

Spaces of measures

225

(Sv)(T µ) = (S1 S0 v)(T µ) = (S0 v)(T0 T −1 T µ) Z = (S0 v)(T0 µ) = (T0 µ)(v) = v dµ; if u ∈ U and f ∈ Uτ∼ , then (T µ)(u) = (T0 µ)(u) for every µ ∈ Mτ , so (Su)(f ) = (S1 S0 u)(f ) = (S0 u)(T0 T −1 f ) = (T0 T −1 f )(u) = (T T −1 f )(u) = f (u). 437I Proposition Let X be a locally compact Hausdorff space, B(X) its Borel σ-algebra, and L∞ the M -space of bounded Borel measurable real-valued functions on X. (a) Let Mt be the L-space of signed tight Borel measuresR on X. Then we have a Banach lattice isomorphism T : Mt → C0 (X)∗ defined by saying that (T µ)(u) = u dµ whenever µ ∈ Mt+ and u ∈ C0 (X). (b) Let ΣuRm be the algebra of universally Radon-measurable subsets of X (definition: 434E), and L∞ (ΣuRm ) the M -space of bounded ΣuRm -measurable real-valued functions on X. Then we have a normpreserving sequentially order-continuous Riesz homomorphism S : L∞ (ΣuRm ) → C0 (X)∗∗ defined by saying R that (Sv)(T µ) = v dµ whenever v ∈ L∞ (ΣuRm ) and µ ∈ Mt+ ; and (Su)(f ) = f (u) for every u ∈ C0 (X), f ∈ C0 (X)∗ . proof (a) The point is just that in this context Mt is equal to Mτ , as defined in 437F-437H (416H), while C0 (X)∗ = C0 (X)∼ τ (see part (a) of the proof of 436J), and the topology of X is completely regular, so we just have a special case of 437Ha. (b)(i) Write B for the Borel σ-algebra of X, and L∞ (B) ⊆ L∞ (ΣuRm ) for the space of bounded Borel measurable functions on X. As in 437Hb, we have a sequentially order-continuous Riesz homomorphism R S0 : L∞ (B) → C0 (X)∗∗ defined by saying that (S0 v)(T ν) = v dν whenever v ∈ L∞ (B) and ν ∈ Mt+ . (ii) If v ∈ L∞ (ΣuRm ), then sup{S0 w : w ∈ L∞ (B), w ≤ v} = inf{S0 w : w ∈ L∞ (B), w ≥ v} in C0 (X)∗∗ . P P Set A = {w : w ∈ L∞ (B), w ≤ v},

B = {w : w ∈ L∞ (B), w ≥ v}.

Because the constant functions belong to L∞ (B), A and B are both non-empty; of course w ≤ w0 and S0 w ≤ S0 w0 for every w ∈ A and w0 ∈ B; because C0 (X)∗∗ is Dedekind complete, φ = sup S0 [A] and ψ = inf S0 [B] are both defined in C0 (X)∗∗ , and φ ≤ ψ. If f ≥ 0 in C0 (X)∗ , then there is a ν ∈ Mt+ such that T ν = f . Since v is ν-virtually measurable (see 434Ec), there are (bounded) Borel measurable functions w, w0 such that w ≤ v ≤ w0 and w = w0 ν-a.e., that is, w ∈ A, w0 ∈ B and (S0 w)(f ) =

R

w dν =

R

w0 dν = (S0 w0 )(f ).

But as (S0 w)(f ) ≤ φ(f ) ≤ ψ(f ) ≤ (S0 w0 )(f ), φ(f ) = ψ(f ); as f is arbitrary, φ = ψ. Q Q (iii) We can therefore define S : L∞ (ΣuRm ) → C0 (X)∗∗ by setting Sv = sup{S0 w : w ∈ L∞ (B), w ≤ v} = inf{S0 w : w ∈ L∞ (B), w ≥ v} R for every v ∈RL∞ . The argument in (ii) tells us also that (Sv)(T ν) = v dν for every ν ∈ Mt+ ; that is, that (Sv)(T µ) = v dµ for every Radon measure µ on X. (iv) Now S is a norm-preserving sequentially order-continuous Riesz homomorphism. P P (Compare 355F.) (α) The non-trivial part of this is actually the check that S is additive. But the formula Sv = sup{S0 w : w ∈ L∞ (B), w ≤ v} ensures that Sv1 + Sv2 ≤ S(v1 + v2 ) for all v1 , v2 ∈ L∞ , while the formula

226


437I

Sv = inf{S0 w : w ∈ L∞ (B), w ≥ v} ensures that Sv1 + Sv2 ≥ S(v1 + v2 ) for all v1 , v2 . (β) It is easy to check that S(αv) = αSv whenever v ∈ L∞ (ΣuRm ) and α > 0, so that S is linear. (γ) Because S0 is a Riesz homomorphism, S(v1 ∧v2 ) = 0 whenever v1 ∧ v2 = 0, so S is a Riesz homomorphism. (δ) Now suppose that hvn in∈N is a non-increasing R sequence in L∞ (ΣuRm ) with infimum 0 in L∞ (ΣuRm ). Then inf n∈N vn (x) = 0 for every x ∈ X, so inf n∈N vn dν = 0 for every ν ∈ Mt+ and inf n∈N (Svn )(f ) = 0 for every f ∈ (C0 (X)∗ )+ . So inf n∈N Svn = 0; as hvn in∈N is arbitrary, S is sequentially order-continuous. (iv) If v ∈ L∞ (ΣuRm ), then |v| ≤ kvk∞ χX, so kSvk ≤ kvk∞ kS(χX)k = kvk∞ . On the other hand, for any x ∈ X we have the corresponding point-supported probability measure δx , with the matching functional hx ∈ C0 (X)∗ , and kSvk = k|Sv|k = kS|v|k ≥ (S|v|)(hx ) =

R

|v|dδx = |v(x)|,

so kSvk ≥ kvk∞ ; thus S is norm-preserving. Q Q

R Remark Once again, if we interpret integrals in the way set out in 363L, we shall have (Sv)(T µ) = v dµ whenever v ∈ L∞ (B) and u ∈ U and µ ∈ Mτ (in 437H) or v ∈ L∞ (ΣuRm ) and u ∈ C0 (X) and µ ∈ Mt (in 437I). 437J Vague and narrow topologies We are ready for another look at ‘vague’ topologies on spaces of measures. Let X be a topological space. (a) Let Σ be an algebra of subsets of X. I will say that Σ separates zero sets if whenever F , F 0 ⊆ X are disjoint zero sets then there is an E ∈ Σ such that F ⊆ E and E ∩ F 0 = ∅. (b) If Σ is any algebra of subsets of X, we can identify the Banach algebra and Banach lattice L∞ (Σ), as defined in §363, with the k k∞ -closed linear subspace of `∞ (X) generated by {χE : E ∈ Σ} (363F, 363Ha). If we do this, then Cb (X) ⊆ L∞ (Σ) iff Σ separates zero sets. P P (i) Suppose that Cb (X) ⊆ L∞ (Σ) and that F1 , F2 ⊆ X are disjoint zero sets. Let u1 , u2 : X → R be continuous functions such that Fi = u−1 i [{0}] for both i; then |u1 (x)| + |u(2)(x)| > 0 for every x; set v =

|u1 | , |u1 |+|u2 |

so that v : X → [0, 1] is continuous,

∞

v(x) = 0 for x ∈ F1 and v(x) = 1 for x ∈ F2 . Now v ∈ Cb (X) ⊆ L (Σ), so there is a w ∈ S(Σ), the linear subspace of L∞ (Σ) generated by {χE : E ∈ Σ}, such that kv − wk∞ < 21 (363C). Set E = {x : w(x) ≤ 21 }; then E ∈ Σ and F1 ⊆ E ⊆ X \ F2 . As F1 and F2 are arbitrary, Σ separates zero sets. (ii) Now suppose that Σ separates zero sets, that u : X → [0, 1] is continuous, and that n ≥ 1 is an integer. For i ≤ n, set Fi = {x : x ∈ X, u(x) ≤ ni }, Fi0 = {x : x ∈ X, u(x) ≥ i+1 }. Then Fi and Fi0 are Pnn 1 0 disjoint zero sets so there is an Ei ∈ Σ such that Fi ⊆ Ei ⊆ X \ Fi . Set w = n i=1 χEi ∈ S(Σ). If x ∈ X, let j ≤ n be such that nj ≤ u(x) < j+1 n ; then for i ≤ n i < j ⇒ x ∈ Fi0 ⇒ x ∈ Ei ⇒ x ∈ / Fi ⇒ i ≤ j, 1 and w(x) = n1 #({i : i ≤ n, x ∈ Ei }) is either nj or j+1 n . Thus |w(x) − u(x)| ≤ n . As x is arbitrary, 1 ∞ ∞ ∞ ku − wk∞ ≤ n ; as n is arbitrary, u ∈ L (Σ). As L (Σ) is a linear subspace of ` (X), this is enough to show that Cb (X) ⊆ L∞ (Σ). Q Q

(c) It follows that if Σ is an algebra of subsets of X separating the zero sets, andR ν : Σ → R is a R bounded additive functional, we have an interpretation Rof u dν for any u ∈ Cb (X); dν is the unique norm-continuous linear functional on L∞ (Σ) such that χE dν = νE for every E ∈ Σ (363L). The map R ν 7→ dν is a Banach lattice isomorphism from the L-space M (Σ) of bounded additive functionals on Σ to ∗ L∞ (Σ)∗ = L∞ (Σ)∼ R (363K). We therefore have a positive linear operator T : M (Σ) → Cb (X) defined by setting (T ν)(u) = u dν for every ν ∈ M (Σ) and u ∈ Cb (X). Except in the trivial case X = ∅, kT k = 1 (if x ∈ X, we have δx ∈ M (Σ) defined by setting δx (E) = χE(x) for E ∈ Σ, and kT (δx )k = 1).R The vague topology on M (Σ) is now the topology generated by the functionals ν 7→ u dν as u runs ∗ over Cb (X); that is, the coarsest topology on M (Σ) such that the canonical map R T : M (Σ) → Cb (X) ∗ is continuous for the weak* topology of Cb (X) . Because the functionals ν 7→ | u dν| are seminorms on M (Σ), the vague topology is a locally convex linear space topology.

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(d) There is a variant of the vague topology which can be applied directly to spaces of (non-negative) ˜ + be the set of all non-negative real-valued additive functionals defined on totally finite measures. Let M ˜ + is that generated by algebras of subsets of X which contain every open set. The narrow topology on M sets of the form ˜ + , νG > α}, {ν : ν ∈ M ˜ + , νX < α} {ν : ν ∈ M for open sets G ⊆ X and real numbers α. ˜ + → [0, ∞[ is continuous for the narrow topology, and if G ⊆ X is open Observe that ν 7→ νX : M ˜ + for the set of totally finite then ν 7→ νG is lower semi-continuous for the narrow topology. Writing M σ + ˜ topological measures on X, then ν 7→ νE : Mσ → [0, ∞[ is Borel measurable, for the narrow topology on ˜ σ+ , for every Borel set E ⊆ X (because the set of E for which ν 7→ νE is Borel measurable is a Dynkin M class containing the open sets). Writing P˜ for the set of topological probability measures, then the narrow topology on P˜ is generated by sets of the form {µ : µ ∈ P˜ , µG > α} for open sets G ⊆ X. (e) Vague topologies, being linear space topologies, are necessarily completely regular (3A4Ad, 4A2Ja). In the very general context of (c) here, in which we have a space M (Σ) of all finitely additive functionals on an algebra Σ, we do not expect the vague topology to be Hausdorff. But if we look at particular subspaces, such as the space Mσ (Ba(X)) of signed Baire measures, or the space Mτ of signed τ -additive Borel measures on a completely regular space X, we may well have a Hausdorff vague topology (437Xf). ˜ + is rarely Hausdorff. But on important subspaces we can get Similarly, the narrow topology on M + Hausdorff topologies. In particular, the narrow topology on the space MqR of totally finite quasi-Radon + measures is Hausdorff. P P Take distinct µ0 , µ1 ∈ MqR . If µ0 X 6= µ1 X then they can be separated by open sets of the form {µ : µX < α}, {µ : µX > α}. Otherwise, set γ = µ0 X = µ1 X. There is certainly an open set G such that µ0 G 6= µ1 G (415H); suppose that µ0 G < µ1 G. Because µ1 is inner regular with respect to the closed sets, there is a closed set F ⊆ G such that µ0 G < µ1 F . Now µ1 G + µ0 (X \ F ) = µ1 G + γ − µ0 F ≥ µ1 F + γ − µ0 G > γ, so there are α < µ0 (X \ F ) and β < µ1 G such that α + β > γ. Now {µ : µ(X \ F ) > α, µX < α + β} and {µ : µG > β, µX < α + β} are disjoint open sets containing µ0 , µ1 respectively. Q Q (f ) For a variant of the narrow topology, adapted to the space of all Radon measures on a Hausdorff space, see 495O below. ˜ + the set of all non-negative real-valued additive 437K Proposition Let X be a topological space, and M functionals defined on algebras of subsets of X containing every open set. R ˜ + → Cb (X)∗ defined by the formula (T ν)(f ) = u dν for every ν ∈ M ˜ +, (a) We have a function T : M u ∈ Cb (X). ˜ + and the weak* topology on Cb (X)∗ . (b) T is continuous for the narrow topology on M ˜ + is a family of τ -additive totally finite (c) Suppose now that X is completely regular, and that W ⊆ M topological measures such that two members of W which agree on the Borel σ-algebra are equal. Then T ¹W is a homeomorphism between W , with the narrow topology, and T [W ], with the weak* topology. proof (a) We have only to assemble the operators of 437Jc, noting that if an algebra of subsets of X contains every open set then it certainly separates the zero sets (indeed, it actually contains every zero set). ˜ + . Suppose that u ∈ Cb (X)+ and that γ ∈ R. Then (b) Write S forR the narrow topology on M ˜ + , u dµ > γ} belongs to S. P V = {ν : ν ∈ M P If γ < 0 this is trivial. Otherwise, for n, i ∈ N, let Gni be P 4n −n the cozero set {x : x ∈ X, u(x) > 2 i}, so that Gni ∈ dom ν. For n ∈ N set un = 2−n i=1 χGni . Then R S T P 4n Vn = {ν : un dν > γ} = { 1≤i≤4n {ν : νGni > γi } : 2−n i=1 γi > γ} S belongs to S. Since hun in∈N is a non-decreasing sequence converging to u for k k∞ , V = n∈N Vn and V ∈ S. Q Q R Next, W = {ν : u dν < γ} belongs to S. P P If u = 0 this is trivial. Otherwise, set v = kuk∞ χX − u, so that v ∈ Cb (X)+ . Then

228


W =

S

β>0 {ν : νX <

β , kuk∞

R

437K

v dν > β − γ}

belongs to S. Q Q R (ii) This shows that ν 7→ (T ν)(u) = u dν is S-continuous for every u ∈ Cb (X)+ and therefore for every u ∈ Cb (X). Since the weak* topology on Cb (X)∗ is the coarsest topology on Cb (X)∗ for which all the functionals f 7→ f (u) are continuous, T is continuous. (c)(i) Write T for the topology on W induced by T , that is, the family of sets of the form W ∩ T −1 [V ] where V ⊆ Cb (X)∗ is weak*-open. If G ⊆ X is open, then A = {u : u ∈ Cb (X), 0 ≤ u ≤ χG} is R upwards-directed and has supremum χG, so µG = sup u dµ for every µ ∈ W (414Ba). Accordingly u∈A S {µ : µ ∈ W , µG > α} = u∈A {µ : (T µ)(u) > α} belongs to T for every α ∈ R. Also, of course, {µ : µX < α} = {µ : (T µ)(χX) < α} ∈ T for every α. So if S0 is the narrow topology on W , S0 ⊆ T. Putting this together with (b), we see that S0 = T. (ii) Now the same formulae show that T ¹W is injective. P P Suppose that µ1 , µ2 ∈ W and that T µ1 = T µ2 . Then µ1 G = µ2 G for every open set G ⊆ X. By the Monotone Class Theorem, µ1 and µ2 agree on all Borel sets; but our hypothesis is that this is enough to ensure that µ1 = µ2 . Q Q Since T : W → T [W ] is continuous and open, it is a homeomorphism. 437L Corollary Let X be a completely regular topological space, and Mτ the space of signed τ -additive Borel measures on X. Then the narrow and vague topologies on Mτ+ coincide. ˜ τ+ (X) for the space of totally finite τ -additive 437M Theorem For a topological space X, write M + topological measures on X, Mτ (X) for the space of totally finite τ -additive Borel measures on X, Mτ (X) for the L-space of signed τ -additive Borel measures on X, and P˜τ (X) for the space of τ -additive topological probability measures on X. ˜ τ+ (X) and ν ∈ M ˜ τ+ (Y ), write µ × ν ∈ M ˜ τ+ (X × Y ) for the (a) Let X and Y be topological spaces. If µ ∈ M τ -additive product measure on X × Y (417G). Then (µ, ν) 7→ µ × ν is continuous for the narrow topologies ˜ τ+ (X), M ˜ τ+ (Y ) and M ˜ τ+ (X × Y ). on M (b) Let hXi ii∈I be a family of topological spaces, with product X. If hµi ii∈I is a family of probability Q measures such that µi ∈ P˜τ (Xi ) for each i, write i∈I µi ∈ P˜τ (X) for their τ -additive product. Then Q hµi ii∈I 7→ i∈I µi is continuous for the narrow topology on P˜τ (X) and the product of the narrow topologies Q on i∈I P˜τ (Xi ). (c) Let X and Y be topological spaces. (i) We have a unique bilinear map ψ : Mτ (X)×Mτ (Y ) → Mτ (X ×Y ) such that ψ(µ, ν) is the restriction of µ × ν to the Borel σ-algebra of X × Y whenever µ ∈ Mτ+ (X) and ν ∈ Mτ+ (Y ). (ii) kψk ≤ 1 (definition: 253Ab). (iii) ψ is separately continuous for the vague topologies on Mτ (X), Mτ (Y ) and Mτ (X × Y ). (d) In (c), suppose that X and Y are compact. If B ⊆ Mτ (X) and B 0 ⊆ Mτ (Y ) are norm-bounded, then ψ¹B × B 0 is continuous for the vague topologies. ˜ τ+ (X) × proof (a)(i) If W ⊆ X × Y is open and α ∈ R, then Q = {(µ, ν) : (µ × ν)(W ) > α} is open in M + 0 ˜ Mτ (Y ). P PS Suppose that (µ0 , ν0 ) ∈ Q. Because µ0 × ν0 is τ -additive, there is a subset W ⊆ W , expressible in the form i≤n Gi × Hi where Gi ⊆ X and Hi ⊆ Y are open for every i, such that α < (µ0 × ν0 )(W 0 ) =

R

ν0 W 0 [{x}]µ0 (dx)

(417C(iv)). Set u(x) = ν0 W 0 [{x}] for x ∈ X, so that u is lower semi-continuousP(417Ba). LetR η > 0 be such R ∞ that u dµ0 > α+(1+2µ0 X)η, and set Ei = {x : u(x) > ηi} for i ∈ N, so that η i=1 µ0 Ei > u dµ0 −ηµ0 X. ˜ τ+ (X) such that Because every Ei is open, there is a neighbourhood U of µ0 in M R R P∞ u dµ0 − ηµ0 X ≤ η i=1 µEi ≤ u dµ = (µ × ν0 )(W 0 ) for every µ ∈ U ; shrinking U if necessary, we can arrangeSat the same time that µX < µ0 X + 1 for every µ ∈ U . Next, observe that H = {W 0 [{x}] : x ∈ X} ⊆ { i∈I Hi : I ⊆ {0, . . . , n}} is finite, so there is a

437M

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229

˜ τ+ (Y ) such that νH ≥ ν0 H − η for every H ∈ H and ν ∈ V . If µ ∈ U and neighbourhood V of ν0 in M ν ∈ V , we have Z Z 0 0 (µ × ν)(W ) ≥ (µ × ν)(W ) = νW [{x}]µ(dx) ≥ u(x) − η µ(dx) Z Z = u dµ − ηµX ≥ u dµ0 − ηµ0 X − η(1 + µ0 X) > α. As µ0 and ν0 are arbitrary, Q is open. Q Q (ii) Since (µ × ν)(X × Y ) = µX · νY , the sets {(µ, ν) : (µ × ν)(X × Y ) < α} are also open for every α ∈ R. So (µ, ν) 7→ µ × ν is continuous (4A2B(a-ii)). (b) For finite sets I, this is a simple induction on #(I), using 417Db. For infinite I, let W ⊆ X be an open set and α ∈ R, and consider Q Q = {hµi ii∈I : µi ∈ P˜τ (Xi ) for each i, ( µi )(W ) > α}. i∈I

IfQhµi ii∈I ∈ Q, then there is an open set W 0 ⊆ W , determined Q by coordinates in a finite set J ⊆ I, such that ( i∈I µi )(W 0 ) > α. Setting V = {x¹J : x ∈ W 0 }, we have ( i∈J µi )(V ) > α. Now we can find open sets Ui Q Q in P˜τ (Xi ), for i ∈ J, such that ( i∈J νi )(V ) > α whenever νi ∈ Ui for i ∈ J. If now hνi ii∈I ∈ i∈I P˜τ (Xi ) is such that νi ∈ Ui for every i ∈ J, Q Q Q ( i∈I νi )(W ) ≥ ( i∈I νi )(W 0 ) = ( i∈J νi )(V ) > α, Q so i∈I νi ∈ Q. As hµi ii∈I is arbitrary, Q Qis open. As W and α are arbitrary, hµi ii∈I 7→ i∈I µi is continuous. (c)(i) Start by writing ψ(µ, ν) = (µ × ν)¹B(X × Y ) for µ ∈ Mτ+ (X) and ν ∈ Mτ+ (Y ), where B(X × Y ) is the Borel σ-algebra of X × Y . If µ, µ1 , µ2 ∈ Mτ+ (X) and ν, ν1 , ν2 ∈ Mτ+ (Y ) and α ≥ 0, then ψ(µ1 + µ2 , ν) = ψ(µ1 , ν) + ψ(µ2 , ν). P P On each side of the equation we have a τ -additive Borel measure, and the two measures agree on the standard base W for the topology of X × Y consisting of products of open sets; since W is closed under finite intersections, they agree on the algebra generated by W and therefore on all open sets and therefore (using the Monotone Class Theorem yet again) on all Borel sets. Q Q Similarly, ψ(µ, ν1 + ν2 ) = ψ(µ, ν1 ) + ψ(µ, ν2 ),

ψ(αµ, ν) = ψ(µ, αν) = αψ(µ, ν).

Now if µ01 , µ02 ∈ Mτ+ (X) and ν10 , ν20 ∈ Mτ+ (Y ) are such that µ1 − µ2 = µ01 − µ02 and ν1 − ν2 = ν10 − ν20 , we shall have ψ(µ1 , ν1 ) − ψ(µ1 ,ν2 ) − ψ(µ2 , ν1 ) + ψ(µ2 , ν2 ) = ψ(µ1 , ν1 + ν20 ) − ψ(µ1 + µ02 , ν20 ) + ψ(µ02 , ν20 ) − ψ(µ1 , ν10 + ν2 ) + ψ(µ1 + µ02 , ν10 ) − ψ(µ02 , ν10 ) − ψ(µ2 , ν1 + ν20 ) + ψ(µ2 + µ01 , ν20 ) − ψ(µ01 , ν20 ) + ψ(µ2 , ν10 + ν2 ) − ψ(µ2 + µ01 , ν10 ) + ψ(µ01 , ν10 ) = ψ(µ02 , ν20 ) − ψ(µ02 , ν10 ) − ψ(µ01 , ν20 ) + ψ(µ01 , ν10 ). We can therefore extend ψ to an operator on Mτ (X) × Mτ (Y ) by setting ψ(µ1 − µ2 , ν1 − ν2 ) = ψ(µ1 , ν1 ) − ψ(µ1 , ν2 ) − ψ(µ2 , ν1 ) + ψ(µ2 , ν2 ) whenever µ1 , µ2 ∈ Mτ+ (X) and ν1 , ν2 ∈ Mτ+ (Y ), and it is straightforward to check that ψ is bilinear. (ii) If µ ∈ Mτ (X), then kµk = µ+ (X) + µ− (X), where µ+ and µ− are evaluated in the Riesz space Mτ (X). Now if ν ∈ Mτ (Y ), |ψ(µ, ν)| = |ψ(µ+ , ν + ) − ψ(µ+ , ν − ) − ψ(µ− , ν + ) + ψ(µ− , ν − )| ≤ ψ(µ+ , ν + ) + ψ(µ+ , ν − ) + ψ(µ− , ν + ) + ψ(µ− , ν − ),

230


437M

so kψ(µ, ν)k = |ψ(µ, ν)|(X × Y ) ≤ ψ(µ+ , ν + )(X × Y ) + ψ(µ+ , ν − )(X × Y ) + ψ(µ− , ν + )(X × Y ) + ψ(µ− , ν − )(X × Y ) = µ+ (X) · ν + (Y ) + µ+ (X) · ν − (Y ) + µ− (X) · ν + (Y ) + µ− (X) · ν − (Y ) = kµkkνk. As µ and ν are arbitrary, kψk ≤ 1. (iii) Fix ν ∈ Mτ+ (Y ) and w ∈ Cb (X × Y )+ , and consider the map µ 7→ Note first that if µ ∈ Mτ+ (X),

R

w dψ(µ, ν) =

R

w d(µ × ν) =

RR

R

w dψ(µ, ν) : Mτ (X) → R.

w(x, y)ν(dy)µ(dx)

(417H). Since both sides of this equation are linear in µ, we have

R

R

w dψ(µ, ν) =

R

w(x, y)ν(dy)µ(dx)

for every µ ∈ Mτ (X). Now x 7→ w(x, y)ν(dy) is continuous. P P By 417Bc, it is lower semi-continuous; but R if α ≥ kwk∞ and w0 = αχ(X × Y ) − w, then x 7→ w0 (x, y)ν(dy) is lower semi-continuous, so x 7→ ανY −

R

w0 (x, y)ν(dy) =

R

w(x, y)ν(dy)

R is also upper semi-continuous, therefore continuous. Q Q It follows at once that µ 7→ w(x, y)ν(dy)µ(dx) is continuous for the vague topology on Mτ (X). The argument has R supposed that w and ν are positive; but taking positive and negative parts as usual, we see that µ 7→ w dψ(µ, ν) is vaguely continuous for every w ∈ Cb (X × Y ) and ν ∈ Mτ (Y ). As w is arbitrary, µ 7→ ψ(µ, ν) is vaguely continuous, for every ν. Similarly, ν 7→ ψ(µ, ν) is vaguely continuous for every µ, and ψ is separately continuous. (d) Now suppose that X and Y are compact. Let W be the linear subspace of C(X × Y ) generated by {u ⊗ v : u ∈ C(X), v ∈ C(Y )}, writing (u ⊗ v)(x, y) = u(x)v(y) as in 253B. Then W is a subalgebra of C(X × Y ) separating the points of X × Y and containing the constant functions, so is k k∞ -dense in C(X × Y ) (281E). Now (µ, ν) 7→

R

u ⊗ v dψ(µ, ν) =

is continuous whenever u ∈ C(X) and v ∈ C(Y ), so (µ, ν) 7→

R

R

u dµ ·

R

v dν

w dψ(µ, ν)

is continuous whenever w ∈ W . Now suppose that B ⊆ Mτ (X) and B 0 ⊆ Mτ (Y ) are bounded. Let γ ≥ 0 be such that kµk ≤ γ for every µ ∈ B and kνk ≤ γ for every ν ∈ B. If w ∈ C(X × Y ) and ² > 0, there is a w0 ∈ W such that kw − w0 k∞ ≤ ². In this case Z Z | w dψ(µ, ν) − w0 dψ(µ, ν)| ≤ kw − w0 k∞ kψ(µ, ν)k ≤ ²kµkkνk ≤ γ 2 ² R whenever µ ∈ B and ν ∈ B 0 . As ² is arbitrary, the function (µ, ν) 7→ w dψ(µ, ν) is uniformly approximated on B × B 0 by vaguely continuous functions, and is therefore itself vaguely continuous on B × B 0 . 437N (Quasi)-Radon measures Throughout the above discussion, I have tried to maintain the formal distinctions between ‘quasi-Radon measure’ and ‘τ -additive effectively locally finite Borel measure inner regular with respect to the closed sets’, and between ‘Radon measure’ and ‘tight locally finite Borel measure’. There are obvious problems in interpreting the sum and difference of measures with different domains, which are readily soluble (see, for instance, 416Ea and 416Xd) but in the context of this section are unilluminating. + If, however, we take MqR to be the set of totally finite quasi-Radon measures on X, and X is completely + regular, we have a canonical embedding of MqR into a cone in the L-space Cb (X)∗ ; more generally, even

437Q

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231

+ if our space X is not completely regular, the map µ 7→ µ¹B(X) : MqR → Mσ (B(X)) is still injective, and + we can identify MqR with a cone in the L-space Mτ of signed τ -additive Borel measures (often the whole positive cone of Mτ , as in 415M). Similarly, when X is Hausdorff, we can identify totally finite Radon measures with tight totally finite Borel measures (416F). It is even possible to extend these ideas to measures which are not totally finite (437Yj), though there may be new difficulties (415Ya).

437O As usual, the case in which X is compact and Hausdorff is particularly important. 0 for the set of Radon measures Proposition Let X be a compact Hausdorff space, and γ ∈ [0, ∞[. Write PRγ 0 µ on X such that µX ≤ γ, and PRγ for the set of Radon measures µ on X such that µX = γ. Then PRγ and PRγ are compact in their narrow topologies.

proof By 437Kc the narrow topologies on these sets can be identified with the weak* topologies on their ˜ + → C(X)∗ be the map described in 437K. Then images in Cb (X)∗ = C(X)∗ . Let T : M 0 T [PRγ ] = {f : f ∈ C(X)∗ , f (u) ≥ 0 for every u ∈ C(X)+ , f (χX) ≤ γ}. R 0 P P If µ ∈ PRγ , then of course (T µ)(u) = u dµ ≥ 0 whenever u ≥ 0, and (T µ)(χX) = µX ≤ γ. If f ∈ (C(X)∗ )+ and f (χX) ≤ γ, then the Riesz representation theorem (436J/436K) tells us that f = T µ for 0 some Radon measure µ, and µX = f (χX) ≤ γ, so µ ∈ PRγ .Q Q 0 This description of T [PRγ ] makes it plain that it is weak*-closed in C(X)∗ . But also, of course, kT µk = 0 0 0 µX ≤ γ for every µ ∈ PRγ ; since the unit ball of C(X)∗ is weak*-compact (3A5F), so is T [PRγ ]. So PRγ is compact for the narrow topology. R 0 Since PRγ = {µ : µ ∈ PRγ , χX dµ = γ}, it too is compact. 0 437P The sets PRγ , PRγ can be identified with convex sets in C(X)∗ , and can therefore have extreme points. These extreme points are easy to identify; I look at PR1 = PR .

Proposition Let X be a Hausdorff space, and PR the set of Radon probability measures on X. The natural embedding of PR into the linear space Mσ (B(X)) of signed Borel measures on X gives PR a convex structure, for which the extreme points are just the point-supported measures δx for x ∈ X, where δx (E) = χE(x) for every E ⊆ X. proof (a) Suppose that x ∈ X. If µ1 , µ2 ∈ PR are such that δx = 12 (µ1 + µ2 ), then (performing the linear operations in Mσ ) we must have µ1 E ≤ 2µE for every Borel set E; in particular, µ1 (X \ {x}) = 0 and µ1 {x} = 1, that is, µ1 = δx . Similarly, µ2 = δx ; as µ1 and µ2 are arbitrary, δx is an extreme point of PR . (b) Suppose that µ is an extreme point of PR . Let K be the support of µ. ?? If K has more than one point, take distinct x, y ∈ K. As X is Hausdorff, there are disjoint open sets G, H such that x ∈ G and y ∈ H. Set E = G ∩ K, α = µE. Because K is the support of µ, α > 0. But similarly µ(H ∩ K) > 0 and α < 1. Let µ1 , µ2 be the indefinite-integral measures defined over µ by α1 χE and β1 χ(X \ E) respectively. Then both are Radon probability measures on X (416S), so belong to PR . Now µF = αµ1 F + βµ2 F for every Borel set F , so (computing αµ1 + βµ2 in Mσ ) µ = αµ1 + βµ2 ; as neither µ1 nor µ2 is equal to µ, µ is not extreme in PR . X X Thus K = {x} for some x ∈ X. But this means that µ{x} = 1 and µ(X \ {x}) = 0, so µ = δx is of the declared form. 437Q For metrizable spaces we can go farther. It will help to have a straightforward lemma. Lemma (a) Let X and Y be Hausdorff spaces, and φ : X → Y a continuous function. Let MR+ (X), MR+ (Y ) ˜ be the spaces of totally finite Radon measures on X and Y respectively. Writing φ(µ) for the image measure + ˜ + + −1 µφ for µ ∈ MR , φ : MR (X) → MR (Y ) is continuous for the narrow topologies on MR+ (X) and MR+ (Y ). (b) If Y is a Hausdorff space, X a subset of Y , and φ : X → Y the identity map, then φ˜ is a homeomorphism between MR+ (X) and {ν : ν ∈ MR+ (Y ), ν(Y \ X) = 0}. proof (a) All we have to do is to recall from 418I that µφ−1 ∈ MR+ (Y ) for every µ ∈ MR+ (X), and observe that

232


{µ : (µφ−1 )(H) > α} = {µ : µφ−1 [H] > α},

437Q

{µ : (µφ−1 )(Y ) < α} = {µ : µX < α}

are narrowly open in MR+ (X) for every open set H ⊆ Y and α ∈ R. ˜ (b) First note that if µ ∈ MR+ (X), then certainly φ(µ)(Y \ X) = 0; while if ν ∈ MR+ (Y ) and ν(Y \ X) = 0, ˜ then µ = ν¹PX is a Radon measure on X (416Rb) and ν = φ(µ). Thus φ˜ is a continuous bijection from MR+ (X) to {ν : ν ∈ MR+ (Y ), ν(Y \ X) = 0}. Now if G ⊆ X is relatively open and α ∈ R, there is an open set H ⊆ Y such that G = H ∩ X, so that ˜ {µ : µ ∈ M + (X), µG > α} = {µ : φ(µ)(H) > α} R

is the inverse image of a narrowly open set in MR+ (Y ); and of course ˜ {µ : µ ∈ MR+ (X), µX < α} = {µ : φ(µ)(Y ) < α} is also the inverse image of an open set. So φ˜ is a homeomorphism between MR+ (X) and {ν : ν ∈ MR+ (Y ), ν(Y \ X) = 0}. 437R Theorem For a Hausdorff space X, write PR (X) for the set of Radon probability measures on X, with its narrow topology. (a) If X is a compact metrizable space, PR (X) is compact and metrizable. (b) If X is a Polish space, PR (X) is Polish. proof (a) Let U be a countable base for the topology of X which contains ∅ and is closed under finite unions. For U ∈ U, q ∈ Q set HU q = {µ : µ ∈ PR (X), µU > q}. Then every HU q is open for the narrow topology T on PR (X); let S be the topology generated by {HU q : U ∈ U, q ∈ Q}, so that S is a second-countable topology coarser than T. If G ⊆ X is open and α ∈ R, then {U : U ∈ U, U ⊆ G} is an upwards-directed set with union G, so S S {µ : µ ∈ PR (X), µG > α} = U ∈U ,U ⊆G q∈Q,q≥α HU q ∈ S. Accordingly S must be actually equal to T, and T is second-countable; since we already know that T is compact (437O), it must be metrizable (4A2Nh). (b) By 4A2Qg, we can suppose that X is a Gδ subset of a compact metrizable space Z. Then we have a homeomorphism betweenTPR (X) and Q = {ν : ν ∈ PR (Z), ν(Z \ X) = 0} (437Qb). Now Q is a Gδ set in PR (Z). P P Express X as n∈N Hn where Hn ⊆ Z is open for each n. Then T T Q = n∈N {ν : νHn = 1} = m,n∈N {ν : νHn > 1 − 2−m } is a Gδ set in PR (Z). Q Q By (a), PR (Z) is a compact metrizable space; by 4A2Qd, its Gδ subset Q is Polish; so PR (X) also is Polish, as claimed. 437S We now have a language in which to express a fundamental result in the theory of dynamical systems. Theorem Let X be a non-empty compact Hausdorff space, and φ : X → X a continuous function. Write Qφ for the set of Radon probability measures on X for which φ is inverse-measure-preserving. Then Qφ is not empty, and is compact for the narrow topology; regarded as a subset of the linear space Mt of signed tight Borel measures on X, Qφ is convex. proof (a) Write PR for the set of Radon probability measures on X. For µ ∈ PR , write T µ for the correR sponding linear functional on C(X), so that (T µ)(u) = u dµ for u ∈ C(X). By the Riesz Representation Theorem (436J/436K), T is a bijection between PR and the set K of positive linear functionals f on C(X) such that f (χX) = 1. By 437H/437I, the linear structure of Mt matches that of C(X)∗ , so the convex structure of K, when transferred to PR by T , corresponds to the convex structure of {µ¹B(X) : µ ∈ PR } ⊆ Mt . By 437Kc, the narrow topology on PR corresponds to the weak* topology Ts (C(X)∗ , C(X)) on K. As noted in 437O, PR and K are compact. (b) If µ ∈ PR , then µ ∈ Qφ iff (T µ)(uφ) = (T µ)(u) for every u ∈ C(X). P P For µ ∈ PR , consider the image measure µφ−1 . This is a Radon measure (418I). Now

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µ ∈ Qφ ⇐⇒ µφ−1 [E] = µE for every E ∈ dom µ (because µφ−1

⇐⇒ µφ−1 = µ and µ are Radon measures, so if they agree on all Borel sets they must be identical) Z Z −1 ⇐⇒ u d(µφ ) = u dµ for every u ∈ C(X)

(416E(b-v))

Z ⇐⇒

Z uφ dµ =

u dµ for every u ∈ C(X)

(235A) ⇐⇒ (T µ)(uφ) = (T µ)(u) for every u ∈ C(X). Q Q (c) Since {f : f ∈ C(X)∗ , f (u) = f (uφ) for every u ∈ C(X)} is convex and closed in C(X)∗ for the weak* topology, Qφ is convex and closed in PR , and is therefore compact. (d) To see that Qφ is not empty, take anyPx0 ∈ X and a non-principal ultrafilter F on N. Define n 1 i f : C(X) → R by setting f (u) = limn→F n+1 i=0 u(φ (x0 )) for every u ∈ C(X) = Cb (X). Then f is a positive linear functional and f (χX) = 1. So there is a µ ∈ PR such that f = T µ. If u ∈ C(X), then f (u) = f (uφ). P P 1 n→F n+1

|f (uφ) − f (u)| = | lim =

n X

u(φi+1 (x0 )) − u(φi (x0 ))|

i=0

1 | lim u(φn+1 (x0 )) − u(x0 )| n→F n+1

≤ lim

1

n→F n+1

|u(φn+1 (x0 )) − u(x0 )| ≤ lim

n→F

2kuk∞ n+1

= 0. Q Q

By the characterization in (b), µ ∈ Qφ . 437T Prokhorov spaces For topological spaces X which are not compact, weak* compactness in Cb (X)∗ is still interesting, and we can look for measure-theoretic criteria which will help us to recognise sets of measures which are relatively compact in the vague topology. The most important of these seems to be the following. Definition Let X be a Hausdorff space. (a) If A is a norm-bounded subset of Cb (X)∗ , we say that A is uniformly tight if for every ² > 0 there is a compact set K ⊆ X such that |f (u)| ≤ ² whenever f ∈ A, u ∈ Cb (X) and |u| ≤ χ(X \ K). (Cf. 436Xn.) ˜ be the space of bounded additive functionals defined on subalgebras of PX containing every (b) Let M ˜ define |ν| ∈ M ˜ by saying that if Σ is the domain of ν, then |ν|E = sup{νF − ν(E \ F ) : open set. For ν ∈ M F ∈ Σ, F ⊆ E} for every E ∈ Σ; that is, we calculate |ν| in the L-space M (Σ) of bounded additive functionals on Σ (362Ba). (Warning! if T is a subalgebra of Σ, |ν¹T| may be different from |ν|¹T.) I ˜ is tight if |ν|(E) = sup{|ν|(K) : K ⊆ E is closed and compact} for every say that a functional ν ∈ M ˜ is uniformly tight if for every ² > 0 E ∈ dom ν, and that a norm-bounded set A of tight functionals in M there is a compact set K ⊆ X such that |ν|(X \ K) ≤ ² for every ν ∈ A. 437U Proposition Let X be a Hausdorff space, andR MR+ the set of totally finite Radon measures on X. For µ ∈ MR+ define T µ ∈ Cb (X)∗ by setting (T µ)(u) = u dµ for every u ∈ Cb (X). (a) Let A ⊆ Cb (X)∗ be a norm-bounded set. Then the following are equiveridical: (i) A is uniformly tight; (ii) there is a solid convex weak*-closed uniformly tight set A˜ ⊆ Cb (X)∗ including A;

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(iii) there is a uniformly tight set B ⊆ MR+ such that {|f | : f ∈ A} = T [B]. (b) If A ⊆ MR+ is bounded and uniformly tight, then A is relatively compact in MR+ for the narrow topology on MR+ . proof (a)(i)⇒(ii) If A is uniformly tight, then for each n ∈ N we can choose a compact set Kn ⊆ X such that |f (u)| ≤ 2−n whenever u ∈ Cb (X) and |u| ≤ χ(X \ Kn ). Set A˜ = {f : f ∈ Cb (X)∗ , |f (u)| ≤ 2−n whenever u ∈ Cb (X), n ∈ N and |u| ≤ χ(X \ Kn )}. ˜ g ∈ Cb (X)∗ and |g| ≤ |f |, then If f ∈ A, |g(v)| ≤ |g|(|v|) ≤ |f |(|v|) = sup{|f (u)| : |u| ≤ |v|} ˜ Thus A˜ is solid (352Ja). It is easy to check that A˜ is for every v ∈ Cb (X); it follows at once that g ∈ A. convex, weak*-closed and uniformly tight. So (ii) is true. (ii)⇒(i) is trivial. (i)⇒(iii) Suppose that A is uniformly tight. For every n ∈ N choose a compact set Kn ⊆ X such S that |f (u)| ≤ 2−n whenever u ∈ Cb (X) and |u| ≤ χ(X \ Kn ); replacing Kn by i≤n Ki if necessary, we can arrange that hKn in∈N is non-decreasing. Let B1 be the set of Radon measures µ on X such that µX ≤ supf ∈A kf k and µ(X \ Kn ) ≤ 2−n for every n ∈ N; of course B1 is uniformly tight. α) Set A0 = {|f | : f ∈ A}. Then every g ∈ A0 is smooth. P (α P Let f ∈ A be such that g = |f |. Suppose that D ⊆ Cb (X)∗ is non-empty and downwards-directed, and that inf u∈D u(x) = 0 for every x ∈ X. Fix v0 ∈ D. Let ² > 0. Let n ∈ N be such that 2−n (kgk + kv0 k∞ ) ≤ ². Then there is a v ∈ D such that v ≤ v0 and v(x) ≤ 2−n whenever x ∈ Kn (apply 436Ic to {v¹Kn : v ∈ D, v ≤ v0 } ⊆ C(Kn )). Set v 0 = v ∧ 2−n χX, v 00 = v − v 0 . Then g(v 0 ) ≤ 2−n kgk. Also, if u ∈ Cb (X) and |u| ≤ v 00 , we have |u| ≤ kv0 k∞ χ(X \ Kn ), so |f (u)| ≤ 2−n kv0 k∞ ; as u is arbitrary, g(v 00 ) ≤ 2−n kv0 k∞ . Putting these together, g(v) ≤ 2−n (kgk + kv0 k∞ ≤ ². As D and ² are arbitrary, g is smooth. Q Q β ) For every g ∈ A0 there is a µ ∈ B1 such that g = T µ. P (β P ByR 436H there is a measure ν on X, quasi-Radon for the topology S generated by Cb (X), such that g(u) = u dν for every u ∈ Cb (X); and of course νX ≤ supf ∈A kf k. As noted in the remark following 436H, we can arrange that ν should be inner regular for the family L of sets L such that χL = inf D for some non-empty D ⊆ Cb (X). If L ∈ L and n ∈ N and Kn ∩ L = ∅ and ² > 0, set D = {u : u ∈ Cb (X), χL ≤ u ≤ χX}. Then D is downwards-directed and inf u∈D u(x) = 0 for every x ∈ Kn , so there is a u ∈ D such that u(x) ≤ ² for every x ∈ Kn ; in this case, (1 − ²)νL ≤ g((u − ²χX)+ ) ≤ 2−n . As ² is arbitrary, νL ≤ 2−n ; as L is arbitrary, ν∗ (X \ Kn ) ≤ 2−n and ν ∗ Kn ≥ νX − 2−n . What this means is that νX = supn∈N ν ∗ Kn , so by 416O there is a Radon measure µ on X extending ν such that µKn = ν ∗ Kn Rfor every Rn. In this case µ(X \ Kn ) = ν∗ (X \ Kn ) ≤ 2−n for every n, and µ ∈ B1 . Of course we now have u dµ = u dν = g(u) for every u ∈ Cb (X) (because the identity map from X to itself is inverse-measure-preserving for µ and ν), so T µ = g. Q Q (γγ ) So if we set B = {µ : µ ∈ B1 , T µ ∈ A0 }, then B witnesses that (iii) is true. (iii)⇒(i) If B ⊆ MR+ is uniformly tight, then {T µ : µ ∈ B} is uniformly tight. P P For every ² > 0 there is a compact set K ⊆ X such that µ(X \ K) ≤ ² for every µ ∈ B; now |(T µ)(u)| ≤ ² whenever µ ∈ B, u ∈ Cb (X) and |u| ≤ χ(X \ K). Q Q So (iii) implies that A0 = {|f | : f ∈ A} is uniformly tight; because (i)⇒(ii), it follows at once that A is uniformly tight. (b) Set γ = supµ∈A µX (taking γ = 0 in the trivial case A = ∅), and for each n ∈ N let Kn ⊆ X be a compact set such that µ(X \ Kn ) ≤ 2−n for every µ ∈ A. Let A˜ ⊇ A be the set of Radon measures µ on X ˜ and such that µX ≤ γ and µ(X \ Kn ) ≤ 2−n for every n ∈ N. Let F be an ultrafilter on MR+ containing A, set νE = limµ→F µE for every Borel set E ⊆ X. Then ν is a non-negative additive functional on the Borel σ-algebra, and νX ≤ γ, while ν(X \ Kn ) ≤ 2−n for every n. By 416N, there is a Radon measure λ on X such that λX ≤ νX and λK ≥ νK for every compact set K. Now since inf K⊆X is compact ν(X \ K) = 0, supK⊆X is compact ν(E ∩ K) = νE for every E ∈ dom ν. In particular, λX ≤ νX = supK⊆X

is compact

νK ≤ supK⊆X

is compact

λK = λX

437W

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and λX = νX. Now λKn ≥ νKn ≥ νX − 2−n = λX − 2−n ˜ Next, if G ⊆ X is open, λG ≤ νG. P for every n, so λ ∈ A. P If ² > 0, let K ⊆ X be a compact set such that νK ≥ νX − ². Then λG ≤ λX − λ(K \ G) ≤ νX − ν(K \ G) ≤ ² + νK − ν(K \ G) = ² + ν(K ∩ G) ≤ ² + νG; as ² is arbitrary, λG ≤ νG. Q Q Now suppose that U is a subset of MR+ which contains λ and is open for the narrow topology. Then there must be open sets G0 , . . . , Gn ⊆ X and α0 , . . . , αn , β ∈ R such that V = {µ : µ ∈ MR+ , µGi > αi for every i ≤ n, µX < β} contains λ and is included in U . In this case, however, νX < β and νGi > αi for every i, so V ∈ F and U ∈ F. Thus λ is a limit of F; as F is arbitrary, A˜ is compact and A is relatively compact. 437V In important cases, the narrowly compact subsets of MR+ (X) are exactly the bounded uniformly tight sets. Once again, it is worth introducing a word to describe when this happens. Definition Let X be a Hausdorff space and PR (X) the set of Radon probability measures on X. X is a Prokhorov space if every subset of PR (X) which is compact for the narrow topology is uniformly tight. 437W Theorem (a) Compact Hausdorff spaces are Prokhorov spaces. (b) A closed subspace of a Prokhorov Hausdorff spaces is a Prokhorov space. (c) An open subspace of a Prokhorov Hausdorff space is a Prokhorov space. (d) The product of a countable family of Prokhorov Hausdorff spaces is a Prokhorov space. (e) Any Gδ subset of a Prokhorov Hausdorff space is a Prokhorov space. ˇ (f) Cech-complete spaces are Prokhorov spaces. (g) Polish spaces are Prokhorov spaces. proof (a) This is trivial; on a compact Hausdorff space the set of all Radon probability measures is uniformly tight. (b) Let X be a Prokhorov Hausdorff space, Y a closed subset of X, and A ⊆ PR (Y ) a narrowly compact ˜ set. Taking φ to be the identity map from Y to X, and defining φ˜ : MR+ (Y ) → MR+ (X) as in 437Q, φ[A] is narrowly compact in PR (X), so is uniformly tight. For any ² > 0, there is a compact set K ⊆ X such that ˜ φ(µ)(X \ K) ≤ ² for every µ ∈ A. Now K ∩ Y is a compact subset of Y and µ(Y \ (K ∩ Y )) ≤ ² for every µ ∈ A. As ² is arbitrary, A is uniformly tight in PR (Y ). (c) Let X be a Prokhorov Hausdorff space, Y an open subset of X, and A ⊆ PR (Y ) a narrowly compact ˜ set. Once again, take φ to be the identity map from Y to X, so that φ[A] ⊆ PR (X) is narrowly compact and uniformly tight in PR (X). ?? Suppose, if possible, that A is not uniformly tight in PR (Y ). Then there is an ² > 0 such that AK = {µ : µ ∈ A, µ(Y \ K) ≥ 5²} is non-empty for every compact set K ⊆ Y . Note that AK ⊆ AK 0 whenever K ⊇ K 0 , so {AK : K ⊆ Y is compact} has the finite intersection property, and there is an ultrafilter F on PR (Y ) containing every AK . Because A is narrowly compact, there is a λ ∈ PR (Y ) such that F → λ. Let K ∗ ⊆ Y be a compact set such that λ(Y \ K ∗ ) ≤ ². ˜ ˜ As φ[A] is uniformly tight, there is a compact set L ⊆ X such that µ(Y \ L) = φ(µ)(X \ L) ≤ ² for every ∗ µ ∈ A. Now K and L \ Y are disjoint compact sets in the Hausdorff space X, so there are disjoint open sets G, H ⊆ X such that K ∗ ⊆ G and L \ Y ⊆ H (4A2Fh). Set K = L \ H ⊇ L ∩ G; then K is a compact subset of Y . As AK ∈ F, there must be a µ ∈ AK such that µY ≤ λY + ² and µ(G ∩ Y ) ≥ λ(G ∩ Y ) − ². Accordingly

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µ(Y ∩ L) ≤ ², µ(Y \ G) = µY − µ(G ∩ Y ) ≤ λY + ² − λ(G ∩ Y ) + ² = λ(Y \ G) + 2² ≤ λ(Y \ K ∗ ) + 2² ≤ 3², µ((Y ∩ L) ∪ (Y \ G)) = µ(Y \ (L ∩ G)) ≥ µ(Y \ K) ≥ 5², which is impossible. X X Thus A is uniformly tight. As A is arbitrary, Y is a Prokhorov space. (d) Let hXn in∈N be a sequence of Prokhorov Hausdorff spaces with product X. Let A ⊆ PR (X) be a narrowly compact set. Let ² > 0. For each n ∈ N let πn : X → Xn be the canonical map and π ñ : MR+ (X) → MR+ (Xn ) the associated function. Then π ñ [A] is narrowly compact in PR (Xn ),Qtherefore uniformly tight, and there is a compact set Kn ⊆ Xn such that (˜ πn µ)(Xn \Kn ) ≤ 2−n−1 ². Set K = n∈N Kn , S −1 so that K is a compact subset of X and X \ K = n∈N πn [Xn \ Kn ]. If µ ∈ A, then P∞ P∞ µ(X \ K) ≤ n=0 µπn−1 [Xn \ Kn ] ≤ n=0 2−n−1 ² = ². As ² is arbitrary, A is uniformly tight; as A is arbitrary, X is a Prokhorov space.

T (e) Let X be a Prokhorov Hausdorff space and Y a Gδ subset of X. Express Y as n∈N Yn where every Q Yn ⊆ X is open. Set Q Z = {z : z ∈ n∈N Yn , z(m) = z(n) for all m, n ∈ N}. Because X is Hausdorff, Z is a closed subspace of n∈N Yn homeomorphic to Y . Putting (c), (d) and (b) together, Z and Y are Prokhorov spaces. ˇ (f ) Put (a), (e) and the definition of ‘Cech-complete’ together. (g) This is a special case of (f) (4A2Md). 437X Basic exercises (a) Let X be a set, U a Riesz subspace of RX and f ∈ U ∼ . (i) Show that f ∈ Uσ∼ iff limn→∞ f (un ) = 0 whenever hun in∈N is a non-increasing sequence in U such that limn→∞ un (x) = 0 for every x ∈ X. (Hint: show that in this case, if 0 ≤ vn ≤ un , we can find k(n) such that f (vn ∧ uk(n) ) ≤ f (vn )+2−n .) (ii) Show that f ∈ Uτ∼ iff inf u∈A |f (u)| = 0 whenever A ⊆ U is a non-empty downwards-directed set and inf u∈A u(x) = 0 for every x ∈ X. (Hint: given ² > 0, set B = {v : f (v) ≥ inf u∈A f + (u) − ², ∃ w ∈ A, v ≥ w} and show that B is a downwards-directed set with infimum 0 in RX .) (b) Let X be a set, U a Riesz subspace of `∞ (X) containing the constant functions, and Σ the smallest σ-algebra of subsets of X with respect to which every member of U is measurable. Let µ and ν be two totally finite measures on X with domain Σ, and f , g the corresponding linear functionals on U . Show that f ∧ g = 0 in U ∼ iff there is an E ∈ Σ such that µE = ν(X \ E) = 0. (Hint: 326I.) (c) (i) Show that S, in 437C, is the unique sequentially order-continuous positive linear operator from L∞ to (Uσ∼ )∗ which extends the canonical embedding of U in (Uσ∼ )∗ . (ii) Show that S, in 437H, is the unique sequentially order-continuous positive linear operator from L∞ to (Uτ∼ )∗ which extends the canonical embedding of U in (Uτ∼ )∗ and is ‘τ -additive’ in the sense that whenever G is a non-empty upwards-directed family of open sets with union H then S(χH) = supG∈G S(χG) in (Uτ∼ )∗ . (d) Let X and Y be completely regular topological spaces and φ : X → Y a continuous function. Define T : Cb (Y ) → Cb (X) by setting T (v) = vφ for every v ∈ Cb (Y ), and let T 0 : Cb (Y )∗ → Cb (X)∗ be its adjoint. ∼ (i) Show that T 0 is a norm-preserving Riesz homomorphism. (ii) Show that T 0 [Cb (Y )∼ σ ] ⊆ Cb (X)σ , and ∼ 0 that if f ∈ Cb (X)σ corresponds to a Baire measure µ on X, then T f corresponds to the Baire measure ∼ ∼ µφ−1 ¹Ba(Y ). (iii) Show that T 0 [Cb (Y )∼ τ ] ⊆ Cb (X)τ , and that if f ∈ Cb (X)τ corresponds to a Borel measure 0 −1 ∞ µ on X, then T f corresponds to the Borel measure µφ ¹B(Y ). (iv) Write L∞ X and LY for the M -spaces ∼ ∗ of bounded real-valued Borel measurable functions on X, Y respectively, and SX : L∞ X → (Cb (X)τ ) , ∞ ∼ ∗ SY : LY → (Cb (Y )τ ) for the canonical Riesz homomorphisms as constructed in 437Hb. Show that if ∗ ∼ ∗ 0 ∼ 00 ∞ T 00 : (Cb (Y )∼ τ ) → (Cb (X)τ ) is the adjoint of T ¹Cb (X)τ , then T SY (v) = SX (vφ) for every v ∈ L (Y ). (e) Let X be a topological space, L∞ (Σum ) the space of bounded universally measurable real-valued functions on X, and Mσ the space of countably additive functionals on the Borel σ-algebra of X. Show that we have a sequentially order-continuous Riesz homomorphism S : L∞ (Σum ) → Mσ∗ defined by the formula R (Sv)(µ) = v dµ whenever v ∈ L∞ (Σum ) and µ ∈ Mσ+ .

437Xr

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237

(f ) Let X be a completely regular topological space. Show that the vague topology on the space of differences of τ -additive totally finite Borel measures on X is Hausdorff. (g) Let X be a topological space, and for x ∈ X set δx (E) = χE(x) for every E ∈ B = B(X). (i) Show that the linear span of {δx : x ∈ X} is dense in M (B) for the vague topology on M (B). (ii) Show that the convex hull of {δx : x ∈ X} is dense in {µ : µ ∈ M (B)+ , µX = 1} for the narrow topology. (Hint: for (ii), do not use the Hahn-Banach theorem.) > (h) Let X and Y be topological spaces, and φ : X → Y a continuous function. Write M# (X) for any of M (Ba(X)), Mσ (Ba(X)), M (B(X)), Mσ (B(X)), Mτ (X) or Mt (X), where Mτ (X) ⊆ Mσ (B(X)) is the space of signed τ -additive Borel measures and Mt (X) ⊆ Mτ (X) is the space of signed tight Borel measures; and M# (Y ) for the corresponding space based on Y . Show that there is a positive linear operator ˜ φ˜ : M# (X) → M# (Y ) defined by saying that φ(µ)(E) = µφ−1 [E] whenever µ ∈ M# (X) and E belongs to Ba(Y ) or B(Y ), as appropriate, and that φ˜ is continuous for the vague topologies on M# (X) and M# (Y ). (i) Let X be any topological space. Show that the narrow topology on the set of totally finite Borel measures on X is T0 . ˜ + the set of non-negative additive functionals defined on (j) Let X be any topological space and M ˜ + define µ + ν ∈ M ˜ + by setting (µ + ν)(E) = subalgebras of PX containing every open set. For µ, ν ∈ M ˜ + is continuous for the narrow topology. (ii) µE + νE for E ∈ dom µ ∩ dom ν. (i) Show that addition on M + + ˜ ˜ ˜ +. Show that (α, µ) 7→ αµ : [0, ∞[ × M → M is continuous for the narrow topology on M (k) Give X = ω1 + 1 its order topology. On X let µ be the Borel probability measure corresponding to Dieudonné’s measure on ω1 (see 434Xf) and ν the Borel probability measure such that ν{ω1 } = 1. (i) Show that µG ≥ νG for every open set G ⊆ X. (ii) Show that the narrow topology on the set of Borel probability measures on X is not T1 . > (l) Let X be a zero-dimensional compact Hausdorff space and E the algebra of open-and-closed subsets of X. (i) Show that E separates zero sets. (ii) Show that the vague topology on M (E) is just the pointwise topology induced by the usual topology of R E . (iii) Writing Mt for the space of signed tight Borel measures on X, show that µ 7→ µ¹E : Mt → M (E) is a Banach lattice isomorphism between the L-spaces Mt and M (E), and is also a homeomorphism when Mt and M (E) are given their vague topologies. ˜ + the space of non-negative real-valued additive functionals (m) Let X be a topological space, and M defined on algebras of subsets of X which contain every open set, R with its+narrow topology. Show that if ˜ → R is lower semi-continuous. u : X → R is a bounded lower semi-continuous function then ν 7→ u dν : M (n) In 437Mc, show that |ψ(µ, ν)| = ψ(|µ|, |ν|) for every µ ∈ Mτ (X) and ν ∈ Mτ (Y ). (o) (i) In 437P, give PR its narrow topology. Show that the map x 7→ δx is a homeomorphism between X and its image in PR . (ii) Show that if X is completely regular then x 7→ δx is a homeomorphism between X and its image in PR when PR is given the vague topology corresponding to its embedding in Mσ (B(X)). (p) Let X be a compact Hausdorff space and PR0 the set of Radon measures µ on X such that µX ≤ 1. Show that the extreme points of PR0 are the point-masses δx , as in 437P, together with the zero measure. (q) Let X be a topological space. Show that the set of tight functionals f ∈ Cb (X)∼ (436Xn) is a band in Cb (X)∼ included in Cb (X)∼ τ . ˜ + the set of non-negative additive functionals defined on subalgebras (r) Let X be a Hausdorff space and M of PX containing every open set, with the algebraic operations defined in 437Xj. Let A and B be norm˜ + and γ ≥ 0. Show that A ∪ B, A + B = {µ + ν : µ ∈ A, ν ∈ B} and bounded uniformly tight subsets of M {αµ : µ ∈ A, 0 ≤ α ≤ γ} are uniformly tight.

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> (s) Let X be a Prokhorov Hausdorff space, and A a set of Radon measures on X which is compact for the narrow topology. Show that A is uniformly tight. (Hint: (i) γ = supµ∈A µX is finite; (ii) for any 1 µ : µ ∈ A, µX ≥ ²} is narrowly compact, therefore uniformly tight; for any ² > 0 the set ² > 0 the set { µX {µ : µ ∈ A, µX ≥ ²} is uniformly tight.) (t) Give ω1 its order topology, and let Mt be the L-space of signed tight Borel measures on ω1 . (i) Show that ω1 is a Prokhorov space. (ii) For ξ < ω1 , define µξ ∈ Mt by setting µξ (E) = χE(ξ) − χE(ξ + 1) for every Borel set E ⊆ ω1 . Show that {µξ : ξ < ω1 } is relatively compact in Mt for the vague topology, but is not uniformly tight. 437Y Further exercises (a) Let X be a set and U a Riesz subspace of RX . Give formulae for the components of a given element of U ∼ in the bands Uσ∼ , (Uσ∼ )⊥ , Uτ∼ and (Uτ∼ )⊥ . (Hint: 356Yb.) (b) Let X be a compact Hausdorff space. Show that the dual C(X; C)∗ of the complex linear space of continuous functions from X to C can be identified with the space of ‘complex tight Borel measures’ on X, that is, the space of functionals µ : B(X) → C expressible as a complex linear combination of tight totally finite Borel measures; explain how this may be identified, as Banach space, with the complexification of the L-space Mt of signed tight Borel measures as described in 354Yk. Show that the complex Banach space ∗∗ L∞ C (B(X)) is canonically embedded in C(X; C) . (c) Write µc for counting measure on [0, 1], and µL for Lebesgue measure; write µc × µL for the product measure on [0, 1]2 , and µ for the direct sum of µc and µc ×µL . Show that the L-space C([0, 1])∼ is isomorphic, as L-space, to L1 (µ). (Hint: every Radon measure on [0, 1] has countable Maharam type.) (d) Let X be a set, U a Riesz subspace of `∞ (X) containing the constant functions, and Σ the smallest σ˜ for the intersection algebra of subsets of X with respect to which every member of U is measurable. Write Σ of the domains of the completions of the totally finite measures with domain Σ. Show that there is a unique ˜ to (Uσ∼ )∗ ∼ sequentially order-continuous norm-preserving Riesz homomorphism from L∞ (Σ) = Mσ∗ such that ∼ (Su)(f ) = f (u) whenever u ∈ U and f ∈ Uσ . (e) Let X be a completely regular Hausdorff space and Pτ the space of τ -additive Borel probability measures on X. Let B ⊆ Pτ be a non-empty set. Show that the following are equiveridical: (i) B is relatively compact in Pτ for the vague topology; (ii) whenever A ⊆ CRb (X) is non-empty and downwardsdirected and inf u∈A u(x) = 0 for every x ∈ A, then inf u∈A supµ∈B u dµ = 0; (iii) whenever G is an upwards-directed family of open sets with union X, then supG∈G inf µ∈B µG = 1. (f ) Explain how to express the proof of 285L(iii)⇒(ii) as (α) a proof that if the characteristic functions of a sequence hνn in∈N of Radon probability measures on R r converge pointwise to a characteristic function, then {νn : n ∈ N} is uniformly tight (β) the observation that any subalgebra of Cb (R r ) which separates the points of R r and contains the constant functions will define the vague topology on any vaguely compact set of measures. (g) Let X be a topological space. (i) Let Mσ (Ba) be the space of signed Baire measures on RX, and u : X → R a bounded Baire measurable function. Show that we have a linear functional µ 7→ u dµ : Mσ (Ba) → R agreeing with ordinary integration with respect to non-negative measures. Show that this functional is Baire measurable with respect to the vague topology on Mσ (Ba). (ii) Let RMτ be the space of signed τ -additive Borel measures on X. Show that we have a linear functional µ 7→ u dµ : Mτ → R agreeing with ordinary integration with respect to non-negative measures. Show that this functional is Borel measurable with respect to the vague topology on Mτ . (h) For a topological space X let Mτ (X) be the L-space of signed τ -additive Borel measures on X, and ψ : Mτ (X) × Mτ (X) → Mτ (X × X) the canonical bilinear map (437M); give Mτ (X) and Mτ (X × X) their vague topologies. (i) Show that if X = [0, 1] then ψ is not continuous. (ii) Show that X = Z and B is the unit ball of Mτ (X) then ψ¹B × B is not continuous.

437 Notes

Spaces of measures

239

(i) Let (X, ρ) be a metric space. For f ∈ Cb (X)∼ σ , set kf kH = sup{|f (u)| : u ∈ Cb (X), kuk∞ ≤ 1, u is 1-Lipschitz} (definition: 471J). (i) Show that k kH is a norm on Cb (X)∼ σ . The metric it induces is sometimes called Huntingdon’s metric. (ii) Show that the topology induced by Huntingdon’s metric on the set Mτ of signed τ -additive Borel measures on X is the vague topology. (Hint: 434L.) (iii) Show that if (X, ρ) is complete then the set Mt of signed tight Borel measures is complete under Huntingdon’s metric. (Hint: show that any Cauchy sequence in Mt must be uniformly tight.) (iv) Show that if (X, ρ) is R with its usual metric, then Huntingdon’s metric, interpreted as a metric on the set of Radon probability measures on X, is not uniformly equivalent to Lévy’s metric as described in 274Ya. (j) Let X be any Hausdorff space, and MR∞+ the set of Radon measures on X. Define addition and scalar multiplication (by positive scalars) on MR∞+ as in 416Xd, and ≤ by the formulae of 416Ea. (i) Show that there is a Dedekind complete Riesz space V such that the positive cone of V is isomorphic to MR∞+ . (ii) Show that every principal band in V is an L-space. (iii) Show that if X is metrizable then V is perfect. (k) Let X be a compact Hausdorff space and PR the set of Radon probability measures on X, with its narrow topology. Show that the weight w(PR ) of PR is at most max(ω, w(X)). (Hint: the weight of C(X) in its norm topology is at most max(ω, w(X)).) ˇ (l) Let X be a Cech-complete completely regular Hausdorff space and PR the set of Radon probability ˇ measures on X, with its narrow topology. Show that PR is Cech-complete. (m) Let X and Y be topological spaces, and ψ : Mτ (X)×Mτ (Y ) → Mτ (X×Y ) the bilinear map of 437Mc. Write Mt (X), etc., for the spaces of signed tight Borel measures. (i) Show that ψ(µ, ν) ∈ Mt (X × Y ) for every µ ∈ Mt (X), ν ∈ Mt (Y ). (ii) Show that if B ⊆ Mt (X), B 0 ⊆ Mt (Y ) are norm-bounded and uniformly tight, then ψ¹B × B 0 is continuous for the vague topologies. (n) (i) X be a metrizable space, and A a narrowly compact subset of the set of Radon probability measures on X. Show that there is a separable subset Y of X which is conegligible for every measure in A. (ii) Show that a metrizable space is Prokhorov iff all its closed separable subspaces are Prokhorov. (o) I say that a completely regular Hausdorff space X is strongly Prokhorov if every vaguely compact subset of the space Mt (X) of signed tight Borel measures on X is uniformly tight. (i) Check that a strongly Prokhorov completely regular Hausdorff space is Prokhorov. (ii) Show that a closed subspace of a strongly Prokhorov completely regular Hausdorff space is strongly Prokhorov. (iii) Show that the product of a countable family of strongly Prokhorov completely regular Hausdorff spaces is strongly Prokhorov. (iv) Show that a Gδ subset of a strongly Prokhorov metrizable space is strongly Prokhorov. (v) Show that a Polish space is strongly Prokhorov. 437 Notes and comments The ramifications of the results here are enormous. For completely regular topological spaces X, the theorems of §436 give effective descriptions of the totally finite Baire, quasi-Radon and Radon measures on X as linear functionals on Cb (X) (436E, 436Xl, 436Xn). This makes it possible, and natural, to integrate the topological measure theory of X into functional analysis, through the theory of Cb (X)∗ . (See Wheeler 83 for an extensive discussion of this approach.) For the rest of this volume we shall never be far away from such considerations. In 437C-437I I give only a sample of the results, heavily slanted towards the abstract theory of Riesz spaces in Chapter 35 and the first part of Chapter 36. Note that while the constructions of the dual spaces U ∼ , Uc∼ and U × are ‘intrinsic’ to a Riesz space U , in that we can identify these functions as soon as we know the linear and order structure of U , the spaces Uσ∼ and Uτ∼ are definable only when U is presented as a Riesz subspace of RX . In the same way, while the space Mσ (Σ) of countably additive functionals on a σ-algebra Σ depends only on the Boolean algebra structure, the spaces Mτ here (not to be confused with the space of completely additive functionals considered in 362B) depend on the topology as well as the Borel algebra. (For an example in which radically different ´ sz Kunen & Rudin 76.) topologies give rise to the same Borel algebra, see Juha You may have been puzzled by the shift from ‘quasi-Radon’ measures in 436H to ‘τ -additive’ measures in 437H; somewhere the requirement of inner regularity has got lost. The point is that the topologies being

240


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considered here, being defined by declaring certain families of functions continuous, are (completely) regular; so that τ -additive measures are necessarily inner regular with respect to the closed sets (414Mb). The theory of ‘vague’ and ‘narrow’ topologies’ in 437J-437W here hardly impinges on the questions considered in §§274 and 285, where vague topologies first appeared. This is because the earlier investigation was dominated by the very special position of the functions x 7→ eiy . x (what we shall in §445 come to call the ‘characters’ of the additive group of R or R r ). One idea which does appear essentially in the proof of 285L, and has a natural interpretation in the general theory, is that of a ‘uniformly tight’ family of Radon measures; see 437T-437W. In 445Yh I set out a generalization of 285L to abelian locally compact groups. In §461 I will return to the general theory of extreme points in compact convex sets. Here I remark only that it is never surprising that extreme points should be special in some way, as in 437P; but the precise ways in which they are special are often unexpected. A good deal of work has been done on relationships between the topological properties of a topological space X and the space P of Radon probability measures on X with the vague topology. Here I show only that if X is compact, so is P (437O), that if X is compact and metrizable so is P (437Ra), and that if X is Polish so is P (437Rb), with a couple of hints at generalizations (437Yk-437Yl). The terms ‘vague’ and ‘narrow’ both appear in the literature on this topic, and I take the opportunity to use them both, meaning slightly different things. Vague topologies, in my usage, are linear space topologies on linear spaces of functionals; narrow topologies are topologies on spaces of (finitely additive) measures, which are not linear spaces, though we can make an attempt to define addition and multiplication by non-negative scalars (437Xj). I must warn you that this distinction is not standard. I see that the word ‘narrow’ appears a good deal oftener than the word ‘vague’, which is in part a reflection of a simple prejudice against signed measures; but from the point of view of this treatise as a whole, it is more natural to work with a concept well adapted to measures with variable domains, even if we are considering questions (like compactness of sets of measures) which originate in linear analysis. I should mention also that the definition in 437Jc includes a choice. The duality considered there uses the space Cb (X); for locally compact X, we have the rival spaces C0 (X) and Ck (X) (see 436J and 436K), and there are occasions when one of these gives a more suitable topology on a space of measures (as in 495Xl below). The elementary theory of uniform tightness and Prokhorov spaces (437T-437W) is both pretty and useful. The emphasis I give it here, however, is partly because it provides the background to a remarkable construction by D.Preiss (439S below), showing that Q is not a Prokhorov space.

438 Measure-free cardinals At several points in §418, and again in §434, we had theorems about separable metrizable spaces in which the proofs undoubtedly needed some special property of these spaces (e.g., the fact that they are Lindelöf), but left it unclear whether something more general could be said. When we come to investigate further, asking (for instance) whether complete metric spaces in general are Radon (438H), we find ourselves once again approaching the Banach-Ulam problem, already mentioned at several points in previous volumes, and in particular in 363S. It seems to be undecidable, in ordinary set theory with the axiom of choice, whether or not every discrete space is Radon in the sense of 434C. On the other hand it is known that discrete spaces of cardinal at most ω1 (for instance) are indeed always Radon. While as a rule I am deferring questions of this type to Volume 5, this particular phenomenon is so pervasive that I think it is worth taking a section now to clarify it. The central definition is that of ‘measure-free cardinal’ (438A), and the basic results are 438B-438D. In particular, ‘small’ infinite cardinals are measure-free (438C). From the point of view of measure theory, a metrizable space whose weight is measure-free is almost separable, and most of the results in §418 concerning separable metrizable spaces can be extended (438E-438G). In fact ‘measure-free weight’ exactly determines whether a metrizable space is measure-compact (438J, 438Xk) and whether a complete metric space is Radon (438H). If c is measure-free, some interesting spaces of functions are Radon (438R). I approach these last spaces through the concept of ‘hereditary weak θ-refinability’ (438K), which enables us to do most of the work without invoking any special axiom.

438B

Measure-free cardinals

241

438A Measure-free cardinals: Definition A cardinal κ is measure-free or of measure zero if whenever µ is a probability measure with domain Pκ then there is a ξ < κ such that µ{ξ} > 0. In 363S I discussed some statements equiveridical with the assertion ‘every cardinal is measure-free’. 438B It is worth getting some basic facts out into the open immediately. Lemma Let (X, Σ, µ) be a semi-finite measure space and hEi ii∈I a point-finite family of subsets of X such S S that #(I) is measure-free andS i∈J Ei ∈ Σ for every J ⊆ I. Set E = i∈I Ei . (a) µE = supJ⊆I is finite µ( i∈J Ei ). P (b) If hEi ii∈I is disjoint, then µE = i∈I µEi . In particular, if Σ = PX and A ⊆ X has measure-free P cardinal, then µA = x∈A µ{x}. S (c) If µ is σ-finite, then L = {i : i ∈ I, µEi > 0} is countable and i∈I\L Ei is negligible. proof (a)(i) The first step is to show, by induction on n, that the result is true if µX < ∞ and every Ei is negligible and #({i : i ∈ I, x ∈ Ei }) ≤ n for every x ∈ X. If n = 0 this is trivial, since S every Ei must be empty. For the inductive step to n ≥ 1, define ν : PI → [0, ∞[ by setting νJ = µ( i∈J Ei ) for every S J ⊆ I. Then ν is a measure on I. P P Write FJ = i∈J Ej for J ⊆ I. ( α) If J, K ⊆ I are disjoint, then for i ∈ I set Ei0 = Ei ∩ FK for i ∈ J, ∅ for i ∈ I \ J. In this case, hEi0 ii∈I is a family of negligible subsets of X, S 0 0 0 0 i∈J 0 Ei = FJ ∩J ∩ FK is measurable for every J ⊆ I, and #({i : x ∈ Ei }) ≤ n − 1 for every x ∈ X; so the inductive hypothesis tells us that S S µ( i∈I Ei0 ) = supJ 0 ⊆I is finite µ( i∈J 0 µEi0 ) = 0, that is, FJ ∩ FK is negligible. But this means that ν(J ∪ K) = µFJ∪K = µ(FJ ∪ FK ) = µFJ + µFK = νJ + νK. As J and K are arbitrary, ν is additive. (β) If hJn in∈N is a disjoint sequence in PI, then [ [ [ ν( Jn ) = µ( FJn ) = lim µ( FJm ) n∈N

n→∞

n∈N

= lim

n→∞

n X

νJm =

m=0

∞ X

m≤n

νJn ,

n=0

so ν is countably additive and is a measure. Q Q At the same time, ν{i} = µEi = 0 for every i. Because #(I) is measure-free, νI = 0. P P?? Otherwise, let f : I → κ = #(I) be any bijection and set λA =

1 νf −1 [A] νI

for every A ⊆ κ; then λ is a probability

measureSwith domain Pκ which is zero on singletons, and κ is not measure-free. X XQ Q But this means just that µ( i∈I Ei ) = 0. Thus the induction proceeds. S (ii) ?? Now suppose, if possible, that the general result is false. For finite sets J ⊆ I set FJ = i∈J Ei , as before, and consider E = {FJ : J ∈ [κ]<ω } (see 3A1J for this notation). Then E is upwards-directed and γ = supH∈E µH is finite, because it is less than µE; S let hHn in∈N be a non-decreasing sequence in E such that µ(H \ H ∗ ) = 0 for every H ∈ E, where H ∗ = n∈N Hn and µH ∗ = γ (215Ab). Because µ is semi-finite, there is an F ∈ Σ such that F ⊆ E and γ < µF < ∞. For each n ∈ N, set Yn = {x : x ∈ F \ H ∗ , #({i : x ∈ Ei }) ≤ n}. Then there is some n ∈ N such that µ∗ Yn > 0. Let ν be the subspace measure S on Yn , so that ν isSnon-zero and totally finite. Now hEi ∩ Yn ii∈I is a family of negligible subsets of Yn , i∈J Ei ∩ Yn = Yn ∩ i∈J Ei is measured by ν for every J ⊆ I, and #({i : x ∈ Ei ∩ Yn }) ≤ n for every x ∈ Yn . But this contradicts (i) above. X X This proves (a). (b) If hEi ii∈I is disjoint, then

S P P supJ∈[I]<ω µ( i∈J Ei ) = supJ∈[I]<ω i∈J µEi = i∈I µEi .

Setting I = A, Ex = {x} for x ∈ A, we get the special case.

242


438B

(c) Let hXn in∈N be a non-decreasing sequence of measurable sets of finite measure covering X. For each −n n, set Ln = {i : i ∈ I, µ(Ei ∩ distinct elements SXn ) ≥ 2 }. ?? If Ln is infinite, take a sequence hik ik∈N of −n in Ln , and consider Gm = k≥m Eik for m ∈ N; then every Gm has measure at least 2 , G0 has finite T measure, hGm im∈N is non-increasing, and m∈N Gm is empty, because hEik ik∈N is point-finite. But this is impossible. X X S Thus every Ln is finite and L = n∈N Ln is countable. S Now (a), applied to hEi0 ii∈I where Ei0 = Ei if i ∈ I \ L, ∅ if i ∈ L, tells us that i∈I\L Ei is negligible. 438C I do not think we are ready for the most interesting set-theoretic results concerning measure-free cardinals. But the following facts may help to make sense of the general pattern. Theorem (Ulam 30) (a) ω is measure-free. (b) If κ is a measure-free cardinal and κ0 ≤ κ is a smaller cardinal, then κ0 is measure-free. (c) If hκξ iξ<λ is a family of measure-free cardinals, and λ is also measure-free, then κ = supξ<λ κξ is measure-free. (d) If κ is a measure-free cardinal so is κ+ . (e) The following are equiveridical: (i) c is not measure-free; (ii) there is a semi-finite measure space (X, PX, µ) which is not purely atomic; (iii) there is a measure µ on [0, 1] extending Lebesgue measure and measuring every subset of [0, 1]. (f) If κ ≥ c is a measure-free cardinal then 2κ is measure-free. proof (a) This is trivial. (b) If µ is a probability measure with domain Pκ0 , set νA = µ(κ0 ∩ A) for every A ⊆ κ. Then ν is a probability measure with domain Pκ, so there is a ξ < κ such that ν{ξ} > 0; evidently ξ < κ0 and µ{ξ} > 0. (c) Let µ be a probability measure on κ with domain Pκ. Define f : κ → λ by setting f (α) = min{ξ : α < κξ } for α < κ. Then the image measure µf −1 is a probability measure on λ with domain Pλ, so there is a ξ < λ such that µf −1 [{ξ}] > 0. Now µκξ > 0. Applying 438Bb to A = κξ , we see that there is an α < κξ such that µ{α} > 0. As µ is arbitrary, κ is measure-free. (d) By (a) and (b), we need consider only the case κ ≥ ω. ?? Suppose, if possible, that µ is a probability measure with domain Pκ+ such that µ{α} = 0 for every α < κ+ . For each α < κ+ , S let fα : α → κ be an injection. For β < κ+ , ξ < κ set A(β, ξ) = {α : β < α < κ+ , fα (β) = ξ}. Then κ \ ξ<κ A(β, ξ) = β + 1 S has cardinal at most κ, which is measure-free, so µ(β + 1) = 0 and µ( ξ<κ A(β, ξ)) > 0. Also hA(β, ξ)iξ<κ is disjoint. There is therefore a ξβ < κ such that µA(β, ξβ ) > 0, by 438Bb. Now κ+ > max(ω, κ), so there must be an η < κ such that B = {β : ξβ = η} is uncountable. In this case, however, hA(β, η)iβ∈B is an uncountable family of sets of measure greater than zero, and cannot be disjoint, because µ is totally finite (215B(iii)); but if α ∈ A(β, η) ∩ A(β 0 , η), where β 6= β 0 , then fα (β) = fα (β 0 ) = η, which is impossible, because fα is supposed to be injective. X X So there is no such measure µ, and κ+ is measure-free. (e)(i)⇒(ii) Suppose that c is not measure-free; let µ be a probability measure with domain Pc such that µ{ξ} = 0 for every ξ < c. Then µ is atomless. P P?? Suppose, if possible, that A ⊆ c is an atom for µ. Let f : c → PN be a bijection. For each n ∈ N, set En = {ξ : n ∈ f (ξ)}. Set D = {n T : µ(A ∩ ESn ) = µA}. Because A is an atom, µ(A ∩ En ) = 0 for every n ∈ N \ D. This means that B = n∈D En \ n∈N\D En has measure µA > 0; but f (ξ) = D for every ξ ∈ B, so #(B) ≤ 1, and µ{ξ} > 0 for some ξ, contrary to hypothesis. X XQ Q So (ii) is true. (ii)⇒(iii) Suppose that there is a semi-finite measure space (X, PX, µ) which is not purely atomic. Then there is a non-negligible set E ⊆ X which does not include any atom; let F ⊆ E be a set of non-zero finite measure. If we take ν to be

1 µF , µF

where µF is the subspace measure on F , then ν is an atomless

probability measure with domain PF . Consequently there is a function g : F → [0, 1] which is inversemeasure-preserving for ν and Lebesgue measure (343Cb). But this means that the image measure νg −1 is a measure defined on every subset of [0, 1] which extends Lebesgue measure.

438E


243

not-(i)⇒not-(iii) Conversely, if c is measure-free, then any probability measure on [0, 1] measuring every subset must give positive measure to some singleton, and cannot extend Lebesgue measure. (f ) We are supposing that κ ≥ c is measure-free, so, in particular, c is measure-free. Let µ be a probability measure with domain P(2κ ). By (e), it cannot be atomless; let E ⊆ 2κ be an atom. Let f : 2κ → Pκ be a bijection, and for ξ < κ set Eξ = {α : α < 2κ , ξ ∈ f (α)}; set D = {ξ : ξ < κ, µ(E ∩ Eξ ) = µE}. Note that µ(E ∩ Eξ ) must be zero for every ξ ∈ κ \ D, so that E ∩ {α : ξ ∈ D4f (α)} is always negligible. Consider Aξ = {α : α ∈ E, ξ = min(D4f (α))} for ξ < κ. Then hAξ iξ<κ is a disjoint family of negligible sets, so its union A is negligible, by 438Bb, because κ is measure-free. But E \ A ⊆ f −1 [{D}] has at most one element, and is not negligible; so µ{α} > 0 for some α. As µ is arbitrary, 2κ is measure-free. Remark This extends the result of 419G, which used a different approach to show that ω1 is measure-free. We see from (d) above that ω2 , ω3 , . . . are all measure-free; so, by (c), ωω also is; generally, if κ is any measure-free cardinal, so is ωκ (438Xa). I ought to point out that there are even more powerful arguments showing that any cardinal which is not measure-free must be enormous (Solovay 71, Fremlin 93). In this context, however, c = 2ω can be ‘large’, at least in the absence of an axiom like the continuum hypothesis to locate it in the hierarchy hωξ iξ∈On ; it is generally believed that it is consistent to suppose that c is not measure-free. 438D I turn now to the contexts in which measure-free cardinals behave as if they were ‘small’. Proposition Let (X, Σ, µ) be a σ-finite measure space, Y a metrizable space with measure-free weight, and f : X → Y a measurable function. Then there is a closed separable set Y0 ⊆ Y such that f −1 [Y0 ] is conegligible; that is, there is a conegligible measurable set X0 ⊆ X such that f [X0 ] is separable. S proof Let U be a σ-disjoint base for the topology of Y (4A2L(h-ii)); express it as n∈N Un where each Un is a disjoint family of open sets. If n ∈ N, #(Un ) S ≤ w(Y ) (4A2Db) is S a measure-free cardinal (438Cb), and hf −1 [U ]iU ∈Un is a disjoint family in Σ such that u∈V f −1 [U ] = f −1 [ V] is measurable for every V ⊆ Un ; so 438Bc tells us that there is a countable set Vn ⊆ Un such that S S f −1 [ (Un \ Vn )] = U ∈Un \Vn f −1 [U ] is negligible. Set Then f

−1

[Y \ Y0 ] =

S n∈N

f

−1

S

Y0 = Y \

S

S n∈N

(Un \ Vn ).

[ (Un \ Vn )] is negligible. On the other hand, S {U ∩ Y0 : U ∈ U} ⊆ {∅} ∪ {V ∩ Y0 : V ∈ n∈N Vn }

is countable, and is a base for the subspace topology of Y0 (4A2B(a-iv)); so Y0 is second-countable and must be separable (4A2Oc). Thus we have an appropriate Y0 . Now X0 = f −1 [Y0 ] is conegligible and measurable and f [X0 ] ⊆ Y0 is separable (4A2P(a-iv)). 438E Proposition (cf. 418B) Let (X, Σ, µ) be a complete locally determined measure space. (a) If Y is a topological space, Z is a metrizable space, w(Z) is measure-free, and f : X → Y , g : X → Z are measurable functions, then x 7→ (f (x), g(x)) : X → Y × Z is measurable. (b) If hYn in∈N is a sequence of metrizable spaces, with product Y , and w(Yn ) is measure-free Q for every n ∈ N, and fn : X → Yn is measurable for every n ∈ N, then x 7→ f (x) = hfn (x)in∈N : X → n∈N Yn is measurable. proof (a)(i) Consider first the case in which µ is totally finite. Then there is a conegligible set X0 ⊆ X such that g[X0 ] is separable (438D). Applying 418Bb to f ¹X0 and g¹X0 , we see that x 7→ (f (x), g(x)) : X0 → Y × g[Z0 ] is measurable. As µ is complete, it follows that x 7→ (f (x), g(x)) : X → Y × Z is measurable. (ii) In the general case, take any open set W ⊆ Y × Z and any measurable set F ⊆ X of finite measure. Set Q = {x : (f (x), g(x)) ∈ W }. By (i), applied to f ¹F and g¹F , F ∩ Q ∈ Σ; as F is arbitrary and µ is locally determined, Q ∈ Σ; as W is arbitrary, x 7→ (f (x), g(x)) is measurable.

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(b) As in (a), it is enough to consider the case in which µ is totally finite. In this case, we have for each T n ∈ N a conegligible set Xn such that fn [Xn ] is separable. Set X 0 = n∈N Xn ; then 418Bd tells us that f ¹X 0 is measurable, so that f is measurable. 438F Proposition (cf. 418J) Let (X, Σ, µ) be a semi-finite measure space and T a topology on X such that µ is inner regular with respect to the closed sets. Suppose that Y is a metrizable space, w(Y ) is measure-free and f : X → Y is measurable. Then f is almost continuous. proof Take E ∈ Σ and γ < µE. Then there is a measurable set F ⊆ E such that γ < µF < ∞. Let F0 ⊆ F be a measurable set such that F \ F0 is negligible and f [F0 ] is separable (438D). By 412Pc, the subspace measure on F0 is still inner regular with respect to the closed sets, so f ¹F0 is almost continuous (418J), and there is a measurable set H ⊆ F0 , of measure at least γ, such that f ¹H is continuous. As E and γ are arbitrary, f is almost continuous. 438G Corollary (cf. 418K) Let (X, T, Σ, µ) be a quasi-Radon measure space and Y a metrizable space such that w(Y ) is measure-free. Then a function f : X → Y is measurable iff it is almost continuous. 438H Now let us turn to questions which arose in §434. Proposition A complete metric space is Radon iff its weight is measure-free. proof Let (X, ρ) be a complete metric space, and κ = w(X) its weight. (a) If κ is measure-free, let µ be any totally finite Borel measure on X. Applying 438D to the identity map from X to itself, we see that there is a closed separable conegligible subspace X0 . Now X0 is complete, so is a Polish space, and by 434Kb is a Radon space. The subspace measure µX0 is therefore tight (that is, inner regular with respect to the compact sets); as X0 is conegligible, it follows at once that µ also is. As µ is arbitrary, X is a Radon space. S (b) If κ is not measure-free, take any σ-disjoint base U for the topology of X. Express U as n∈N Un where every Un is disjoint. Then κ ≤ #(U) and there is a probability measure ν on U, with domain PU, such that ν{U } = 0 for every U ∈ U . Let n ∈ N be such that νUn > 0. For each U ∈ Un choose x SU ∈ U . For Borel sets E ⊆SX set µE = ν{U : U ∈ Un , xU ∈ E}; then µ is a Borel measure on X and µ( Un ) = νUn > 0, while µ( V) = νV = 0 for every finite V ⊆ Un . Thus µ is not τ -additive and cannot be tight, and X is not a Radon space. 438I Proposition Let X be a metrizable space and hFξ iξ<κ a non-decreasing family of closed subsets of X, where κ is a measure-free cardinal. Then P S S µ( ξ<κ Fξ ) = ξ<κ µ(Fξ \ η<ξ Fη ) for every semi-finite Borel measure µ on X. S proof (a) I had better begin by remarking that Hξ = η<ξ Fη is an Fσ set for every ordinal ξ ≤ κ, by P 4A2Ld and 4A2Ka. So, setting Eξ = Fξ \ Hξ , ξ<κ µEξ is defined. P (b) I show by induction on ζ that µHζ = ξ<ζ µEξ for every ζ ≤ κ. The induction starts trivially with µH0 = 0. The inductive step to a successor ordinal ζ + 1 is also immediate, as Hζ+1 = Hζ ∪ Eζ . For the inductive step to a limit ordinal ζ of countable cofinality, let hζn in∈N be a non-decreasing sequence in ζ with supremum ζ; then P P µHζ = supn∈N µHζn = supn∈N ξ<ζn µEξ = ξ<ζ µEξ , as required. (c) So we P are left with the case in which ζ is a limit ordinal of uncountable cofinality. In this case, µ(E ∩ Hζ ) ≤ ξ<ζ µEξ whenever µE is finite. P P Let U be a σ-disjoint base for the topology of X (4A2L(hS ii)), and express U as n∈N Un where each Un is disjoint. For n ∈ N, ξ ≤ ζ set S Vnξ = {U : U ∈ Un , U ∩ Hξ 6= ∅}, Vnξ = Vnξ .

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Define φn : Vnζ → ζ by saying that φn (x) = min({ξ : ξ < ζ, x ∈ Vnξ }). Then, for any D ⊆ ζ, S S S φ−1 (Vnξ \ η<ξ Vnη ) n [D] = ξ∈D is a union of members of Un , so is open. We therefore have a measure νn on Pζ defined by saying that νn D = µ(E ∩ φ−1 point, recall that we are supposing that κ is measure-free, n [D]) for every D ⊆ ζ. At this P so #(ζ) is also measure-free (438Cb) and νn ζ = ξ<ζ νn {ξ} = supξ<ζ νn ξ. Interpreting this in X, we have µ(E ∩ Vnζ ) = supξ<ζ µ(E ∩ Vnξ ). This is true for every n ∈ N. So there is a countable set C ⊆ ζ such that µ(E ∩ Vnζ ) = supξ∈C µ(E ∩ Vnξ ) for every n ∈ N. Because cf(ζ) > ω, there is an α < ζ such that C ⊆ α, and µ(E ∩ Vnζ ) = µ(E ∩ Vnα ), that is, E ∩ Vnζ \ Vnα is negligible, for every n ∈ N. Now note that Fα is closed. So Hζ \ F α ⊆

[

{U : U ∈ U , Hζ ∩ U 6= ∅, U ∩ Fα = ∅} [ = Vnζ \ Vn,α+1 , n∈N

and E ∩ Hζ \ Fα is negligible. Accordingly, using the inductive hypothesis, P P µ(E ∩ Hζ ) ≤ µFα = µHα+1 ≤ ξ≤α µEξ ≤ ξ<ζ µEξ , as claimed. Q Q P Because µ is semi-finite, and E is arbitrary, µHζ ≤ ξ<ζ Eξ ; but the reverse inequality is trivial, so we have equality, and the induction proceeds in this case also. P (d) At the end of the induction we have µHκ = ξ<κ µEξ , as stated. 438J So far we have been looking at metrizable spaces, the obvious first step. But it turns out that the concept of ‘metacompactness’ leads to generalizations of some of the results above. Proposition (Moran 70, Haydon 74) Let X be a metacompact space with measure-free weight. (a) X is Borel-measure-compact. (b) If X is normal, it is measure-compact. (c) If X is perfectly normal (for instance, if it is metrizable), it is Borel-measure-complete. proof (a) ?? If X is not Borel-measure-compact, there are a non-zero totally finite Borel measure µ on X and a cover G of X by negligible open sets (434H(a-v)). Let H be a point-finite open cover of X refining S G. By 4A2Dc, #(H) ≤ max(ω, w(X)), so is measure-free, by 438C.SBecause µ is a Borel measure, H0 is measurable for every H0 ⊆ H; µH = 0 for every H ∈ H; while µ( H) = µX > 0. But this contradicts 438Ba. X X (b) Now suppose that X is normal, and that µ is a totally finite Baire measure on X. Because a normal metacompact space is countably paracompact (4A2F(g-iii)), µ has an extension to a Borel measure µ1 which is inner regular with respect to the closed sets, by Maˇr´ık’s theorem (435C). Now µ1 is τ -additive, by (a) above, so µ also is (411C). As µ is arbitrary, X is measure-compact. (c) Since on a perfectly normal space the Baire and Borel measures are the same, X is Borel-measurecomplete iff it is measure-compact, and we can use (b). Remark The arguments here can be adapted in various ways, and in particular the hypotheses can be weakened; see 438Yd-438Yf. 438K Hereditarily weakly θ-refinable spaces A topological space X is hereditarily weakly θ-refinable (also called hereditarily σ-relatively metacompact, hereditarily weakly submetacompact) if forSevery family G of open subsets of X there is a σ-isolated family A of subsets of X, refining G, such S that A = G.

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438L Lemma (a) Any subspace of a hereditarily weakly θ-refinable topological space is hereditarily weakly θ-refinable. (b) A hereditarily metacompact space (e.g., any metrizable space, see 4A2Lb) is hereditarily weakly θrefinable. (c) A hereditarily Lindelöf space is hereditarily weakly θ-refinable. (d) A topological space with a σ-isolated network is hereditarily weakly θ-refinable. proof (a) If X is hereditarily weakly θ-refinable, Y is a subspace of X, and H is a family of open subsets of S Y , set G = {G : G ⊆ X is open, G ∩ Y ∈ H}. Then there is a σ-isolated familySA, refining G, with union G; and {A ∩ Y : A ∈ A} is σ-isolated (4A2B(a-viii)), refines H, and has union H. As H is arbitrary, Y is hereditarily weakly θ-refinable. (b) If X is hereditarily metacompact, and G is any family of open sets in X with union W , then G is an open cover of the metacompact space W , so has aT point-finite open refinement H with the same union. For each x ∈ W , set Hx = {H : x ∈ H ∈ H}, Vx = Hx , so that Hx is a non-empty finite set and Vx is an open set containing x. For n ≥ 1, set An = {x : x ∈ W, #(Hx ) = n}; then for any distinct x, y ∈ An , either Hx = Hy and Vx = Vy , or #(Hx ∪ Hy ) > n and Vx ∩ Vy ∩ An = ∅. This means that An = {Vx ∩ An : x ∈ An } is a partition of ASn into relatively open sets, and is an isolated family. Also, S An is a refinement of H and therefore of G; so n≥1 An is a σ-isolated refinement of G, and its union is n≥1 An = W . As G is arbitrary, X is hereditarily weakly θ-refinable. (c) If X is hereditarily Lindelöf and G is a family of open subsets of X, there is a countable G0 ⊆ G with the same union; now G0 , being countable, is σ-isolated. As G is arbitrary, X is hereditarily weakly θrefinable. (d) If X has a σ-isolated network A, and G is any family of open subsets of X, then E = {A : A ∈ A, A is included in some member of G} S is a σ-isolated family (4A2B(a-viii)), refining G, with union G. 438M Proposition (Gardner 75) If X is a hereditarily weakly θ-refinable topological space with measure-free weight, it is Borel-measure-complete. proof Let µ be a Borel probability measure on X, and G the S S family of µ-negligible open sets. Let A be a σ-isolated family refining S G with union G. Express A as n∈N An where each An is an isolated family; for each n ∈ N, set Xn = An and let µn be the subspace measure on Xn . Then An is a disjoint family of relatively open µn -negligible sets; as #(An ) ≤ w(Xn ) ≤ w(X) (4A2D) is measure-free, and µn is a totally finite Borel measure on Xn , S µ∗ Xn = µn Xn = µn ( An ) = 0, S S by 438Bb. Now µ( G) = µ∗ ( n∈N Xn ) = 0. As µ is arbitrary, X is Borel-measure-complete (434I(a-iv)). 438N For the next few paragraphs, I will use the following notation. Let X be a topological space S and G a family of subsets of X. Then J (G) will be the family of subsets of X expressible as A for some S σ-isolated family A refining G. Observe that X is hereditarily weakly θ-refinable iff G belongs to J (G) for every family G of open subsets of X. (a) J (G) is always a σ-ideal of subsets of X. P P If A is a σ-isolated family of subsets of X, refining G, and B is any set, then {B ∩ A : A ∈ A} is still σ-isolated and still refines G; so any subset of a member of S J (G) belongs to J (G). If hAn in∈N is a sequence of σ-isolated families refining G, then n∈N An is σ-isolated and refines G; so the union of any sequence in J (G) belongs to J (G). Q Q (b) If H refines G, then J (H) ⊆ J (G). P P All we need to remember is that any family refining H also refines G. Q Q (c) If X and Y are topological spaces, A ⊆ X, f : A → Y is continuous, and H is a family of subsets of Y , set G = {f −1 [H] : H ∈ H}. Then J (G) ⊇ {f −1 [B] : B ∈ J (H)}. P P If B ∈ J (H), there is a

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σ-isolated family D of subsets of Y , refining H, S and with union B. Now A = {f −1 [D] : D ∈ D} refines −1 G and has union f [B]. We can express D as n∈N Dn , where each Dn is a relatively open family; set S An = {f −1 [D] : D ∈ Dn }, so that A = n∈N An . For each n, An is disjoint, because Dn is. Moreover, S S if D ∈ DS Dn for some open set H ⊆ Y , so that f −1 [D] = f −1 [H] ∩ An is relatively n , then D = H ∩ open in An ; this shows that An is an isolated family. Accordingly A is σ-isolated and witnesses that f −1 [B] ∈ J (G). As B is arbitrary, we have the result. Q Q (d)SIf X is a topological space, G is a family of subsets of X, and hDi ii∈I is an isolated family in J (G), then i∈I D P For each iS∈ I, let hAni in∈N be a sequence of isolated families, Si ∈ JS(G). P S Sall refining S G, such that Di = n∈N Ain . Set An = i∈I Ain for each n. Then A refines G, and A = n n n∈N i∈I Di . It S S is easy to check that every An is isolated, so that n∈N An witnesses that i∈I Di belongs to J (G). Q Q 438O Lemma Give R the topology S generated by the closed intervals ]−∞, t] for t ∈ R, and let r ≥ 1. Then R r , with the product topology corresponding to S, is hereditarily weakly θ-refinable. proof Induce on r. Write ≤ for the usual partial order of R r , and ]−∞, x] for {y : y ≤ x}; set VA = S r r r x∈A ]−∞, x] for A ⊆ R . The sets ]−∞, x], as x runs over R , form a base for the topology of R . The induction starts easily because S itself is hereditarily Lindelöf. P P If G ⊆ S, set A = {x : x ∈ R, there is some G ∈ G such that ]−∞, x] ⊆ G}. Then A has a countable cofinal set D, so that there is a corresponding countable subset of G with the same union as G. Q Q By 438Lc, S is hereditarily weakly θ-refinable. For the inductive step to r + 1, where r ≥ 1, let G be a family of open subsets of R r+1 , and set A = {x : x ∈ R r , there is some G ∈ G such that ]−∞, x] ⊆ G}. For each k ≤ r, q ∈ Q set Kk = (r + 1) \ {k} and let Bkq ⊆ R Kk be the set {z : z a q ∈ VA }, writing z a q for that member x of R r+1 such that x¹Kk = z and x(k) = q. (I am thinking of members of R r+1 as functions from r + 1 = {0, . . . , r} to R.) Set Akq = {x : x ∈ VA , x(k) = q}, Gkq = {]−∞, x] : x ∈ Akq }. Then, in the notation of 438N, VAkq ∈ J (G). P P Set f (x) = x¹Kk for each x ∈ VAkq , so that f : VAkq → R Kk is continuous. For x ∈ Akq , ]−∞, x] = f −1 [ ]−∞, f (x)] ]. Now R KkSis hereditarily weakly θ-refinable, by the inductive hypothesis, so if we set Hkq = {]−∞, f (x)] : x ∈ Akq }, Hkq ∈ J (Hkq ) and (by 438Nc) S VAkq = f −1 [ Hkq ] ∈ J (Gkq ) ⊆ J (G). Q Q S Accordingly W ∈ J (G), where W = k≤r,q∈Q VAkq , by 438Na. Now consider VA \ W . If x, x0 ∈ VA \ W and x ≤ x0 then x = x0 . P P?? Otherwise, there are a k ≤ n and a q ∈ Q such that x(k) ≤ q ≤ x0 (k). In this case, setting y¹Kk = x¹Kk and y(k) = q, we have y ∈ Akq and x ∈ VAkq . X XQ Q But this means that the subspace topology of VA \ W is discrete, so Sthat {{x} : x ∈ VA \ W } is an isolated family covering VA \ W and refining G; thus VA \ W ∈ J (G) and G = VA belongs to J (G). As G is arbitrary, R r+1 is hereditarily weakly θ-refinable and the induction proceeds. 438P Proposition Let X ⊆ R R be the family of functions x : R → R such that lims↑t x(s) and lims↓t x(s) are defined in R for every t ∈ R. Then X, with its topology of pointwise convergence inherited from the product topology of R R , is a K-analytic hereditarily weakly θ-refinable space. proof (a) I start by showing that it is K-analytic. (i) The first step is to find an alternative characterization of X. If x ∈ X and U , V ⊆ R are disjoint, then I say that a finite set D ⊆ R is a back-and-forth set for x, U and V if whenever s, t are successive points of D (in the usual ordering of R) then either x(s) ∈ U and x(t) ∈ V , or x(s) ∈ V and x(t) ∈ U . Now, if x ∈ X and k ∈ N and U , V ⊆ R have disjoint closures, ωkU V (x) = sup{#(D) : D ⊆ [−k, k] is a back-and-forth set for x, U and V } is finite. P P For each t ∈ [−k, k] there is a δ > 0 such that {x(s) : t − δ < s < t} meets at most one of U and V , and {x(s) : t < s < t + δ} meets at most one of U and V . So if D is a back-and-forth set for x, U and V , D can meet each of the intervals ]t − δ, t[ and ]t, t + δ[ in at most one point, and #(D ∩ ]t − δ, t + δ[) ≤ 3.

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By the Heine-Borel theorem, there is a cover of [−k, k] by m intervals of this type for some m ∈ N, so that ωkU V (x) ≤ 3m. Q Q (ii) If x ∈ X and htn in∈N is any bounded sequence in R, then it has a monotonic subsequence ht0n in∈N and hx(t0n )in∈N must be convergent in R. It follows that {x(t) : −k ≤ t ≤ k} is bounded in R for every k ∈ N. (iii) Conversely, let I be the family of non-empty open intervals in R with rational endpoints, and suppose that x ∈ R R is such that whenever U , V ∈ I have disjoint closures, and k ∈ N, then {x(t) : |t| ≤ k} is bounded and ωkU V (x) is finite. Then x ∈ X. P P Let t ∈ R. ?? If lims↓t x(s) is not defined in R, then (because {x(s) : t ≤ s ≤ t + 1} is bounded) there must be decreasing sequences htn in∈N , ht0n in∈N , both convergent to t, such that z = limn→∞ x(tn ), z 0 = limn→∞ x(t0n ) are both defined and different. Let U , V ∈ I be such that z ∈ U , z 0 ∈ V and U ∩ V = ∅; then for any m ∈ N there is a back-and-forth set D ⊆ {tn : n ∈ N} ∪ {t0n : n ∈ N} for x, U and V with #(D) ≥ m, so ωkU V (x) is infinite for some k. X X Thus lims↓t x(s) is defined in R. Similarly, lims↑t x(s) is defined in R; as t is arbitrary, x ∈ X. Q Q (iv) Now this means that X is a Souslin-F set in the compact space [−∞, ∞]R . P P For k, m ∈ N and disjoint U , V ∈ I, set Fkm = {x : x ∈ [−∞, ∞]R , |x(t)| ≤ m for every t ∈ [−k, k]}, 0 R FkU V m = {x : x ∈ [−∞, ∞] , ωkU V (x) ≤ m}. 0 0 Of course every Fkm is closed. But so is every FkU / FkU V m , because if x ∈ V m and D ⊆ [−k, k] is a back0 and-forth set for x, U and V of size greater than m, then D will witness that y ∈ / FkU V m for any y close enough to x. Now the characterization of X above tells us just that T T S S 0 X = k∈N m∈N Fkm ∩ k∈N,U,V ∈I,U ∩V =∅ m∈N FkU V m,

which is a countable intersection of Fσ sets, and is Souslin-F. Q Q So X is K-analytic (422Hb). (b) Now I turn to the proof that X is hereditarily weakly θ-refinable. Let G be a family of open sets in X. S (i) In the notation of S 438N, I seek to show that G belongs to J (G). Of course it will be enough to consider the case S in which G is non-empty. The following elementary remarks will be useful. (α) If D ⊆ G, and E is a partition of D into relatively open sets such that E refines G, then D ∈ J (G). (β) If hDi ii∈I S is any family in J (G), and hHi ii∈I is a family of open sets, and D ⊆ {x : #({i : x ∈ Hi }) = P hD ∩ Hi ∩ Di ii∈I is an isolated family in J (G); use 438Nd. 1}, then D ∩ i∈I Hi ∩ Di belongs to J (G). P Q Q (ii) Write Q for the family of all finite sequences q = (I0 , U0 , V0 , W0 , I1 , U1 , V1 , W1 , . . . , In , Un , Vn , Wn ) where I0 , I1 , . . . , In are disjoint members of I, all the Ui , Vi , Wi belong to I, and, for each i ≤ n, any pair of Ui , Vi , Wi are either equal or disjoint. Fix q = (I0 , . . . , Wn ) ∈ Q for the moment. Q (iii) Set Tq = i≤n Ii . For τ ∈ Tq , write Fq τ for the set of those x ∈ X such that, for every i ≤ n, S x(s) ∈ Ui for s ∈ Ii ∩ ]−∞, τ (i)[, x(τ (i)) ∈ Vi and x(s) ∈ Wi for s ∈ Ii ∩ ]τ (i), ∞[. Set Xq = {Fq τ : τ ∈ Tq }, and for τ ∈ Tq set Hq τ = {x : x ∈ Xq , x(τ (i)) ∈ Vi for every i ≤ n}. Finally, set Sq = {τ : τ ∈ Tq and Hq τ is included in some member of G}. (iv) If T ⊆ Sq then H =

S τ ∈T

Hq τ belongs to J (G). P P Induce on #(L(T )), where

L(T ) = {i : i ≤ n, there are τ , τ 0 ∈ T such that τ (i) 6= τ 0 (i)}. If L(T ) = ∅, then #(T ) ≤ 1, so H is either empty or included in some member of G, and the induction starts. For the inductive step to #(L(T )) = k ≥ 1, consider three cases.

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case 1 Suppose there is a j ∈ L(T ) such that Uj = Vj = Wj . Then x(t) ∈ Vj for every x ∈ Xq , t ∈ Ij . Fix any t∗ ∈ Ij and for τ ∈ Tq define τ ∗ ∈ Tq by setting τ ∗ (j) = t∗ , τ ∗ (i) = τ (i) for i 6= j; then Hq τ ∗ = Hq τ . Set T ∗ = {τ ∗ : τ ∈ST }; then L(T ∗ ) = L(T ) \ {j}, so #(L(T ∗ )) < #(L(T )), while T ∗ ⊆ Sq . By the inductive hypothesis, H = τ ∈T ∗ Hq τ is negligible. case 2 Suppose there is a j ∈ L(T ) such that Uj 6= Vj and Vj 6= Wj . Then Vj ∩ (Uj ∪ Wj ) = ∅. For s ∈ Ij set Ts∗ = {τ : S τ ∈ T, τ (j) = s}. Then #(L(Ts∗ )) < #(L(T )) so, by the inductive hypothesis, ∗ ∗ Hs ∈ J (G), where Hs = τ ∈T ∗ Hq τ . But, for τ ∈ T and x ∈ Hq τ , x(s) ∈ Vj iff s = τ (j); so Hs∗ = {x : x ∈ s H, x(s) ∈ Vj } and hHs∗ is∈Ij is a partition of H into relatively open sets. By (i-β), H ∈ J (G). case 3 Otherwise, L(T ) = J ∪ J 0 where J = {i : i ∈ L(T ), Ui = Vi } and J 0 = {i : i ∈ L(T ), Vi = Wi } are disjoint. For x ∈ H, i ∈ J we see that there is a largest t ∈ Ii such that x(t) ∈ Vi ; set φi (x) = −t. Similarly, for x ∈ H, i ∈ J 0 there is a smallest t ∈ Ii such that x(t) ∈ Vi ; in this case, set φi (x) = t. Observe that, for i ∈ J, s ∈ R, {x : x ∈ H, φi (x) ≤ s} = ∅ if s < −t for every t ∈ Ii , = H if − t < s for every t ∈ Ii , = {x : x ∈ H, x(−s) ∈ Vi } if − s ∈ Ii , so is always relatively open in H, and φi is continuous if R is given the topology S of Lemma 438O. Similarly, for i ∈ J 0 , s ∈ R, {x : x ∈ H, φi (x) ≤ s} = ∅ if s < t for every t ∈ Ii , = H if t < s for every t ∈ Ii , = {x : x ∈ H, x(s) ∈ Vi } if s ∈ Ii . So in this case also φi is continuous. Accordingly, giving R L(T ) the product topology corresponding to S, we have a continuous map φ : H → L(T ) R defined by setting φ(x) = hφi (x)ii∈L(T ) for x ∈ H. For τ ∈ T , set τ˜(i) = −τ (i) if i ∈ J, τ (i) if i ∈ J 0 , ˜ and Hτ = ]−∞, τ˜] ⊆ R L(T ) . Then Hq τ = {x : x ∈ H, x(τ (i)) ∈ Vi for every i ≤ n} = {x : x ∈ H, x(τ (i)) ∈ Vi for every i ∈ L(T )} (because if x ∈ H, i ≤ n, i ∈ / L(T ) then there is some τ 0 ∈ T such that x ∈ Hq τ 0 , so that x(τ 0 (i)) ∈ Vi and therefore x(τ (i)) ∈ Vi ) ˜ τ ]. = {x : x ∈ H, φi (x) ≤ τ˜(i) for every i ∈ L(T )} = φ−1 [H ˜ τ : τ ∈ T }. Because R L(T ) is hereditarily weakly θ-refinable (438O), and G˜ is a family of open Set G˜ = {H S˜ S −1 L(T ) ˜ By 438Nc, H = S [ G] belongs to J ({Hq τ : τ ∈ T } and subsets of R , G˜ =∈ J (G). τ ∈T Hq τ = φ therefore to J (G), by 438Nb. Thus in all three cases the induction proceeds. Q Q S (v) This means that, S for any q ∈SQ, Yq = {Hq τ : τ ∈ Sq } belongs to J (G). Since Q is countable, Y ∈ J (G), where Y = q ∈Q Yq . But G ⊆ Y . P P If x ∈ G ∈ G, there are t0 < . . . < tn and Vi0 ∈ I, for i ≤ n, such that x ∈ {y : y ∈ X, y(ti ) ∈ Vi0 for every i ≤ n} ⊆ G. Set zi = x(ti ), zi− = lims↑ti x(s), zi+ = lims↓ti x(s) for i ≤ n; let Ui , Vi , Wi ∈ I be such that zi− ∈ Ui , zi ∈ Vi ⊆ Vi0 , zi+ ∈ Wi and any pair of Ui , Vi , Wi are either equal or disjoint; and let I0 , . . . , In ∈ I be disjoint and such that ti ∈ Ii , x(s) ∈ Ui for s ∈ Ii ∩ ]−∞, ti [ and x(s) ∈ Wi for s ∈ Ii ∩ ]ti , ∞[ for each i ≤ n. Then, setting q = (I0 , . . . , Wn ) and τ (i) = ti for i ≤ n, x ∈ Xq τ ⊆ Hq τ ⊆ G, so that τ ∈ Sq and x ∈ Yq ⊆ Y . Q Q As G is arbitrary, X is hereditarily weakly θ-refinable.

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438Q Corollary (a) Let I k be the split interval (419L). Then any countable power of I k is a K-analytic hereditarily weakly θ-refinable space. (b) Let Y be the ‘Helly space’, the space of non-decreasing functions from [0, 1] to itself with the topology of pointwise convergence inherited from the product topology on [0, 1][0,1] (Kelley 55, Ex. 5M). Then Y is a K-analytic hereditarily weakly θ-refinable space. proof These are both (homeomorphic to) subspaces of the space X of Proposition 438P. To see this, argue as follows. For (a), observe that we have a function f : I k → X defined by setting f (t− )(s) = f (t+ )(s) = 1 if s < t, f (t− )(s) = f (t+ )(s) = 0 if s > t, and f (t− )(t) = 0, f (t+ )(t) = 1, and that f is a homeomorphism between I k and its image. Next, for any L ⊆ N, we can define g : (I k )L → X by setting g(t)(s) = f (tn )(s − 2n) if n ∈ L and 2n ≤ s ≤ 2n + 1 [ [2n, 2n + 1] = 0 if s ∈ R \ n∈L k L

for t = htn in∈N ∈ (I ) ; it is easy to check that g is a homeomorphism between (I k )L and its image in X. As for (b), if we take g(y) to be the extension of the function y : [0, 1] → [0, 1] to the function which is constant on each of the intervals ]−∞, 0] and [1, ∞[, then g : Y → X is a homeomorphism between Y and its image g[Y ]. Since both (I k )N and Y are compact, they are homeomorphic to closed subsets of X, and are K-analytic (422Gc) and hereditarily weakly θ-refinable (438La). 438R Proposition Assume that c is measure-free. Then the space X of 438P, (I k )N and the Helly space (438Qb) are all Radon spaces. proof By 438P and 438Q, they are K-analytic and hereditarily weakly θ-refinable, therefore pre-Radon and Borel-measure-complete (434Jf, 438M), therefore Radon (434Ka). 438S In 434R I described a construction of product measures. In accordance with my general practice of examining the measure algebra of any new measure, I give the following result. Proposition Let X and Y be topological spaces with σ-finite Borel measures µ, ν respectively. Suppose that either X is first-countable or ν is τ -additive and effectively locally finite. Write λ for the Borel measure on X × Y defined by the formula λW =

R

νW [{x}]µ(dx) for every Borel set W ⊆ X × Y

as in 434R(ii). If either the weight of X or the Maharam type of ν is a measure-free cardinal, then for every b Borel set W ⊆ X × Y there is a set W 0 ∈ B(X)⊗B(Y ) such that λ(W 4W 0 ) = 0; consequently, the measure algebra of λ can be identified with the localizable measure algebra free product of the measure algebras of µ and ν. proof (a) Write (B, ν¯) for the measure algebra of ν. With its measure-algebra topology, this is metrizable (323Gb). Let hYn in∈N be a non-decreasing sequence of Borel sets of finite measure in Y with union Y . (b) For the moment (down to the end of (e) below) fix on an open set W ⊆ X × Y . For x ∈ X, set f (x) = W [{x}]• in B. Then f : X → B is Borel measurable. P P (i) Let H ⊆ B be an open set. For k, n ∈ N set Enk = {x : x ∈ X, 2−n k ≤ ν(Yn ∩ W [{x}]) < 2−n (k + 1)}. Just as in part (a) of the proof of 434R, the function x 7→ ν(Yn ∩ W [{x}]) is lower semi-continuous, so Enk is a Borel set. Set S Gnk = {G : G ⊆ X is open, G ∩ Enk ⊆ f −1 [H]}; S then E = n,k∈N (Gnk ∩ Enk ) is a Borel set included in f −1 [H]. (ii) The point is that E = f −1 [H]. To see this, take any x such that f (x) ∈ H. Then there are b ∈ B, ² > 0 such that ν¯b < ∞ and c ∈ H whenever c ∈ B and ν¯(b ∩ (c 4 f (x))) ≤ 5². Since b = supn∈N b ∩ Yn• ,

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there is an n ∈ N such that ν¯(b \ Yn• ) ≤ ² and 2−n ≤ ². In this case c ∈ H whenever c ∈ B and ν¯(Yn• ∩ (c 4 f (x))) ≤ 4²; thus {x0 : ν(Yn ∩ (W [{x0 }]4W [{x}])) ≤ 4²} ⊆ f −1 [H]. Let k ∈ N be such that 2−n k ≤ ν(Yn ∩ W [{x}]) < 2−n (k + 1), that is, x ∈ Enk . Again using the ideas of part (a) of the proof of 434R, there are an open set G containing x and an open set V ⊆ Y such that G × V ⊆ W and ν(Yn ∩ V ) ≥ 2−n (k − 1). Now if x0 ∈ G ∩ Enk , V ⊆ W [{x0 }] ∩ W [{x}], so ν(Yn ∩ (W [{x0 }]4W [{x}])) ≤ ν(Yn ∩ W [{x0 }]) + ν(Yn ∩ W [{x}]) − 2ν(Yn ∩ V ) ≤ 2−n ((k + 1) + (k + 1) − 2(k − 1)) = 4 · 2−n ≤ 4². But this means that G ∩ Enk ⊆ f −1 [H], so G ⊆ Gnk and x ∈ G ∩ Enk ⊆ E. As x is arbitrary, f −1 [H] ⊆ E and E = f −1 [H]. (iii) Thus f −1 [H] is a Borel set. As H is arbitrary, f is Borel measurable. Q Q (c) We need to know also that if H is a disjoint family of open subsets of B all meeting f [X], then #(H) ≤ max(ω, min(w(X), τ (B))) P P Repeat the ideas of (b) above, setting S (H) Gnk = {G : G ⊆ X is open, G ∩ Enk ⊆ f −1 [H]} S (H) (H) for H ∈ H and k, n ∈ N, so that f −1 [H] = n,k∈N Gnk ∩Enk . For fixed n and k the family hGnk ∩Enk iH∈H is disjoint, so can have at most w(Enk ) ≤ w(X) non-empty members (4A2D). But this means that S (H) H = n,k∈N {H : Gnk ∩ Enk 6= ∅} has cardinal at most max(ω, w(X)). On the other hand, there is a set B ⊆ B, of cardinal τ (B), which τ -generates B. The algebra B0 generated by B has cardinal at most max(ω, #(B)) (331Gc), and B0 is topologically dense in B (323J), so every member of H meets B0 , and #(H) ≤ #(B0 ) ≤ max(ω, τ (B)). Putting these together, we have the result. Q Q In particular, under the hypotheses above, #(H) is measure-free whenever H is a disjoint family of open subsets of B all meeting f [X]. (d) The next step is to observe that there is a conegligible BorelSset Z ⊆ X such that f [Z] is separable. P P Let H be a σ-disjoint base for the topology of B; express it as n∈N Hn where each Hn is disjoint. Let hXm im∈N be a cover of X by Borel sets of finite measure. For n ∈ N consider Hn0 = {H : H ∈ Hn , H ∩f [X] 6= ∅}. For m ∈ N, we have a totally finite measure νnm with domain PHn0 defined by saying S νnm E = µ(Xm ∩ f −1 ( E)) for every E ⊆ Hn0 . Since Hn0 has measure-free cardinal, by (c), there must be a countable set Enm ⊆ Hn0 such that νnm (Hn0 \ Enm ) = 0. Set S S Z = X \ m,n∈N (Xm ∩ f −1 [ (Hn0 \ Enm )]); then Z is conegligible. If x ∈ Z and f (x) ∈ H ∈ Hn , then there is some m ∈ N such that x ∈ Xm , while H ∈ Hn0 , so H must belong to Enm . ButS this means that {f [Z] ∩ H : H ∈ H}, which is a base for the topology of f [Z], is just {f [Z] ∩ H : H ∈ m,n∈N Emn }, and is countable. So f [Z] is separable (4A2Oc), as required. Q Q b (e) 418T(a-ii) now tells us that there is a set W 0 ∈ B(X)⊗B(Y ) such that f (x) = W 0 [{x}]• for every 0 x ∈ Z, so that ν(W [{x}]4W [{x}]) = 0 for almost every x, that is, λ(W 4W 0 ) = 0. And this is true for every open set W ⊆ X × Y . b (f ) Now let W be the family of those Borel sets W ⊆ X × Y for which there is a W 0 ∈ B(X)⊗B(Y ) 0 such that λ(W 4W ) = 0. This is a σ-algebra containing every open set, so is the whole Borel σ-algebra, as required.

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b Since the c.l.d. product measure λ0 on X × Y is just the completion of its restriction to B(X)⊗B(Y ) b b (251K), and λ0 and λ agree on B(X)⊗B(Y ) (by Fubini’s theorem), the embedding B(X)⊗B(Y ) ⊆ B(X × Y ) induces an isomorphism between the measure algebras of λ and λ0 . As remarked in 325Eb, because µ and ν are strictly localizable, the latter may be identified with ‘the’ localizable measure algebra free product of the measure algebras of µ and ν. Remark The hypothesis on the weight of X can be slightly weakened; see 438Yg. To see that some restriction on (X, µ) and (Y, ν) is necessary, see 439L. 438X Basic exercises (a) Show that a cardinal κ is measure-free iff Mσ = Mτ , where Mσ , Mτ are the spaces of countably additive and completely additive functionals on the algebra Pκ, as in §362. (b) Let (X, Σ, µ) be a localizable measure space with magnitude (definition: 332Ga) which is either finite or a measure-free cardinal. Show that any absolutely continuous countably additive functional ν : Σ → R is truly continuous. (Hint: 363S.) (c) Let A be a Dedekind complete Boolean algebra with measure-free cellularity. Show that any countably additive functional ν : A → R is completely additive. (d) Let U be a Dedekind complete Riesz space such that any disjoint order-bounded family in U + has measure-free cardinal. Show that Uc∼ = U × . > (e) Show that if κ is a measure-free cardinal, so is ωκ . (Hint: show by induction on ordinals ξ that if #(ξ) is measure-free, then so is ωξ .) > (f ) Let (X, Σ, µ) be a complete locally determined measure space, (Y, ρ) a complete metric space, and hfn in∈N a sequence of measurable functions from X to Y . Suppose that w(Y ) is measure-free. Show that {x : limn→∞ fn (x) is defined in Y } is measurable. (Cf. 418C.) (g) Let (X, Σ, µ) and (Y, T, ν) be σ-finite measure spaces with c.l.d. product (X ×Y, Λ, λ). Give L0 (ν) the topology of convergence in measure. Suppose that f : X → L0 (ν) is measurable and there is a conegligible set X0 ⊆ X such that w(f [X0 ]) is measure-free. Show that there is an h ∈ L0 (λ) such that f (x) = h•x for every x ∈ X. (Cf. 418S.) (h) Let (Y, T, ν) be a σ-finite measure space, and (B, ν¯) its measure algebra, with its usual topology; assume that the Maharam type of B is measure-free. Let (X, Σ, µ) be a σ-finite measure space and Λ the domain of the c.l.d. product measure λ on X × Y . Show that if f : X → B is measurable, then there is a W ∈ Λ such that f (x) = Wx• for every x ∈ X. (Cf. 418T(b-ii).) (i) Let (X, Σ, µ) is a complete locally determined measure space and V a normed space such that w(V ) is measure-free. (i) Show that the space L of measurable functions from X to V is a linear space, setting (f + g)(x) = f (x) + g(x), etc. (ii) Show that if V is a Riesz space with a Riesz norm then L is a Riesz space under the natural operations. > (j) Let X be a topological space and G a point-finite open cover of X such that #(G) is measure-free. Suppose that E ⊆ X is such that E ∩G is universally measurable for every G ∈ G. Show that E is universally measurable. (Compare 434Xe(iv).) > (k) Show that for a metrizable space X, the following are equiveridical: (i) X is Borel-measure-compact; (ii) X is Borel-measure-complete; (iii) X is measure-compact; (iv) w(X) is measure-free. (l) Let Y be the Helly space. (i) Show that Y is a compact convex subset of R [0,1] with its usual topology. (ii) Show that there is a natural one-to-one correspondence between the split interval I k and the set of extreme points of Y , matching t− ∈ I k with the function χ [0, t[ and t+ with χ[0, t]. (iii) Let P be the set of Radon probability measures on I k with its narrow topology (437J). Show that there is a natural homeomorphism φ : P → Y defined by setting φ(µ)(t) = µ[0− , t− ] for µ ∈ P , t ∈ [0, 1].

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(m) Give R the right-facing Sorgenfrey topology (415Xc). Show that any countable power of R is hereditarily weakly θ-refinable. > (n) Let I k be the split interval. Show that I k × I k is a Radon space iff c is measure-free. (Hint: {(α+ , (1 − α)+ ) : α ∈ [0, 1]} is a discrete Borel subset of cardinal c.) (o) Give R the right-facing Sorgenfrey topology. Show that the following are equiveridical: (i) c is measure-free; (ii) R N , with the corresponding product topology, is Borel-measure-complete; (iii) R 2 , with the product topology, is Borel-measure-compact. (Compare 439Q.) (p) Suppose that c is measure-free. Let X ⊆ R R be the set of functions of bounded variation on R, with the topology of pointwise convergence inherited from the product topology of R R . Show that X is a Radon space. (Hint: X is an Fσ subset of the space of 438P.) 438Y Further exercises (a) Let S (X, Σ, µ) be a probability space and hEi ii∈I a point-finite family of measurable sets such that νJ = µ( i∈J Ei ) is defined for every J ⊆ I. Show directly that ν is a uniformly exhaustive Maharam submeasure on PI (definition: 392B), and use the Kalton-Roberts theorem (392J) to prove 438Ba. (b) Suppose that c is measure-free, but that κ > c is not measure-free. Show that there is a non-principal ω1 -complete ultrafilter on κ. (Hint: part (b) of the proof of 451P.) (c) Show that if X is a metrizable space and either c or w(X) is measure-free, then every σ-finite Borel measure on X has countable Maharam type. (d) Let X be a metacompact T1 space. Show that X is Borel-measure-compact iff every closed discrete subspace has measure-free cardinal. (e) Let X be a topological space such that every subspace of X is metacompact and has measure-free cellularity. Show that X is Borel-measure-complete. (f ) Let X be a normal metacompact Hausdorff space. Show that it is measure-compact iff every closed discrete subspace has measure-free cardinal. (g) In 438S, show that it would be enough to suppose that every discrete subset of X has cardinal of measure 0. ˜ ⊆ Z R be the set of functions x : R → Z such that lims↑t x(s), (h) Let Z be any Polish space, and let X ˜ with the topology of pointwise convergence lims↓t x(s) are defined in Z for every t ∈ R. Show that X, R inherited from Z , is a K-analytic hereditarily weakly θ-refinable space. (i) Suppose that c is measure-free. Let D be any subset of R and X ⊆ R D the set of functions of bounded variation on D, with the topology of pointwise convergence inherited from the product topology of R D . Show that X is a Radon space. (j) Let X be a totally ordered set with its order topology. Show that if c(X) is measure-free then every σ-finite Borel measure on X has countable Maharam type. (Cf. 434Yl.) 438 Notes and comments Since the axiom ‘every cardinal is measure-free’ is admissible – that is, will not lead to a paradox unless one is already latent in the Zermelo-Fraenkel axioms for set theory – it is tempting, in the context of this section, to assume it; so that ‘every complete metric space is Radon’ becomes a theorem, along with ‘every measurable function from a quasi-Radon measure space to a metrizable space is almost continuous’ (438G), ‘Uc∼ = U × for every Dedekind complete Riesz space U ’ (438Xd), ‘metacompact spaces are Borel-measure-compact’ (438J), ‘the sum of two measurable functions from a complete probability space to a normed space is measurable’ (438Xi) and ‘the Helly space is Radon’ (438R). Undoubtedly the consequent mathematical universe is tidier. In my view, the tidiness is the tidiness of poverty. Apart from

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anything else, it leads us to neglect such questions as ‘is every measurable function from a Radon measure space to a metrizable space almost continuous?’, which have answers in ZFC (451S). From the point of view of measure theory, the really interesting question is whether c is measure-free. It is not quite clear from the results above why this should be so; 438R is a very small part of the story. There is a larger hint in 438Ce-438Cf: if c is measure-free, but κ > c is not measure-free, then the witnessing measures will be purely atomic. I mean to return to this point in Volume 5. For a general exploration of universes in which c is not measure-free, see Fremlin 93. For fragments of what happens if we suppose that we have an atom for a measure which witnesses that κ is not measure-free, see 438Yb and the notes on normal filters in 4A1I-4A1L. There are many further applications of 438P besides those in 438Q and 438Xm-438Xp. But the most obvious candidate, the space C(R) of continuous real-valued functions on R, although indeed it is a Borel subset of the potentially Radon space of 438P, is in fact Radon whether or not c is measure-free (454Pa). As soon as we start using any such special axiom as ‘c = ω1 ’ or ‘c is measure-free’, we must make a determined effort to check, through such examples as 438Xn, that our new theorems do indeed depend on something more than ZFC.

439 Examples As in Chapter 41, I end this chapter with a number of examples, exhibiting some of the boundaries around the results in the rest of the chapter, and filling in a gap with basic facts about Lebesgue measure (439E). The first three examples (439A) are measures defined on σ-subalgebras of the Borel σ-algebra of [0, 1] which have no extensions to the whole Borel algebra. The next part of the section (439B-439G) deals with ‘universally negligible’ sets; I use properties of these to show that Hausdorff measures are generally not semi-finite (439H), closing some unfinished business from §§264, and that smooth linear functionals may fail to be representable by integrals in the absence of Stone’s condition (439I). In 439J-439R I set out some examples relevant to §§434-435, filling out the classification schemes of 434A and 435A, with spaces which just miss being Radon (439K) or measure-compact (439N, 439P, 439Q). In 439S I present the canonical example of a non-Prokhorov topological space, answering an obvious question from §437. 439A Example Let B be the Borel σ-algebra of [0, 1]. There is a probability measure ν defined on a σ-subalgebra T of B which has no extension to a measure on B. first construction Let A ⊆ [0, 1] be an analytic set which is not Borel (423L). Let I be the family of sets of the form E ∪ F where E, F are Borel sets, E ⊆ A and F ⊆ [0, 1] \ A. Then I is a σ-ideal of B not containing [0, 1]. Set T = I ∪ {[0, 1] \ H : H ∈ I}, and define ν : T → {0, 1} by setting νH = 0, ν([0, 1] \ H) = 1 for every H ∈ I; then ν is a probability measure (cf. countable-cocountable measures (211R) or Dieudonné’s measure (411Q)). ?? If µ : B → [0, 1] is a measure extending ν, then its completion µ ˆ measures A (432A). Also µ ˆ is a Radon measure (433Cb). Now every compact subset of A belongs to I, so µ Â = sup{ˆ µK : K ⊆ A is compact} = sup{νK : K ⊆ A is compact} = 0. Similarly µ ˆ([0, 1] \ A) = 0, which is absurd. X X second construction This time, let I be the family of meager Borel sets in [0, 1]. As before, let T be I ∪ {[0, 1] \ E : E ∈ I}, and set νE = 0, ν([0, 1] \ E) = 1 for E ∈ I. ?? If µ is a Borel measure extending ν, then µ([0, 1] \ Q) = 1, and µ is tight (that is, inner regular with respect to the compact sets), so there is a closed subset F of [0, 1] \ Q such that µF > 0. But F is nowhere dense, so νF = 0. X X third construction1 There is a function f : [0, 1] → {0, 1}c which is (B, Ba)-measurable, where Ba is the Baire σ-algebra of {0, 1}c , and such that f [ [0, 1] ] meets every non-empty member of Ba. P P Set X = C([0, 1])N with the product of the norm topologies, so that X is an uncountable Polish space (4A2Pe, 4A2Qc), and ([0, 1], B) is isomorphic to (X, B(X)), where B(X) is the Borel σ-algebra of X (424Da). Define g : X → {0, 1}[0,1] by saying that g(hui ii∈N )(t) = 1 iff limi→∞ ui (t) = 1. For each t ∈ [0, 1], 1I

am grateful to M.Laczkovich and D.Preiss for showing this to me.

439C

Examples

{hui ii∈N : lim ui (t) = 1} = i→∞

\ [

255

{hui ii∈N : |ui (t) − 1| ≤ 2−m for every i ≥ n}

m∈N n∈N

is a Borel subset of X, so g is (B(X), Ba({0, 1}[0,1] ))-measurable, where Ba({0, 1}[0,1] ) is the Baire σ-algebra of {0, 1}[0,1] (4A3Ne). If E ∈ Ba({0, 1}[0,1] ) is non-empty, there is a countable set I ⊆ [0, 1] such that E is determined by coordinates in I (4A3Nb), so that E ⊇ {w : w¹I = z} for some z ∈ {0, 1}I . Now we can find a sequence hui ii∈N in C([0, 1]) such that limi→∞ ui (t) = z(t) for every t ∈ I (if I ⊆ {tj : j ∈ N}, take ui such that |ui (tj ) − z(tj )| ≤ 2−i whenever j ≤ i), and in this case g(hui ii∈N ) ∈ E. Because (X, B(X)) ∼ = ([0, 1], B) and ({0, 1}[0,1] , Ba({0, 1}[0,1] )) ∼ = ({0, 1}c , Ba), we can copy g to a function f with the required properties. Q Q Now set T = {f −1 [H] : H ∈ Ba}, and let λ be the usual measure on {0, 1}c . Then T is a σ-subalgebra of B and there is a probability measure ν with domain T defined by saying that νf −1 [H] = λH for every H ∈ Ba. P P We need to know that if H, H 0 ∈ Ba and f −1 [H] = f −1 [H 0 ] then λH = λH 0 . But as 0 (H4H ) ∩ f [ [0, 1] ] = ∅, H = H 0 , so this is trivial. Thus the given formula defines a function ν : T → [0, 1]. It is now elementary to check that ν is a probability measure. Q Q In fact, because λ is completion regular (415E), f [ [0, 1] ] has full outer measure for λ, so the map H • 7→ f −1 [H]• from the measure algebra of λ to the measure algebra of ν is measure-preserving; since it is surely surjective, the measure algebras are isomorphic, and ν has Maharam type c. However, any probability measure on the whole algebra B has countable Maharam type (433A), so cannot extend ν. Remark Compare 433I-433J. 439B Definition Let X be a Hausdorff space. I will call X universally negligible if there is no Borel probability measure µ defined on X such that µ{x} = 0 for every x ∈ X. A subset of X will be ‘universally negligible’ if it is universally negligible in its subspace topology. 439C Proposition Let X be a Hausdorff space. (a) If A is a subset of X, the following are equiveridical: (i) A is universally negligible; (ii) µ∗ A = 0 whenever µ is a Borel probability measure on X such that µ{x} = 0 for every x ∈ X; (iii) µ∗ A = 0 whenever µ is a σ-finite topological measure on X such that µ{x} = 0 for every x ∈ A; (iv) for every σ-finite topological measure µ on X there is a countable set B ⊆ A such that µ∗ A = µB; (v) A is a Radon space and every compact subset of A is scattered. In particular, countable subsets of X are universally negligible. (b) The family of universally negligible subsets of X is a σ-ideal. (c) Suppose that Y is a universally negligible Hausdorff space and f : X → Y a Borel measurable function such that f −1 [{y}] is universally negligible for every y ∈ Y . Then X is universally negligible. (d) If the topology on X is discrete, X is universally negligible iff #(X) is measure-free. proof (a)(i)⇒(iii) If A is universally negligible and µ is a σ-finite topological measure on X such that µ{x} = 0 for every x ∈ A, let µA be the subspace measure on A. ?? If µ∗ A = α > 0, then (because µ is σ-finite) there is a measurable set E ⊆ X such that γ = µ∗ (E ∩ A) is finite and non-zero. The subspace measure µE∩A is a topological measure on E ∩ A; set νF = γ −1 µE∩A (E ∩ F ) for relatively Borel sets F ⊆ A; then ν is a Borel probability measure on A which is zero on singletons. X X So µ∗ A = 0. (iii)⇒(iv) If (iii) is true and µ is a σ-finite topological measure on X, set B = {x : x ∈ X, µ{x} > 0}. Because µ is σ-finite, B must be countable, therefore measurable, and if we set νE = µ(E \ B) for every Borel set E ⊆ X, ν is a σ-finite Borel measure on X and ν{x} = 0 for every x ∈ X. By (iii), ν ∗ A = 0, that is, there is a Borel set E ⊇ A such that µ(E \ B) = 0; in which case µ∗ A ≤ µ(E \ (B \ A)) = µ(E \ B) + µ(A ∩ B) = µ(A ∩ B) ≤ µ∗ A,

256


439C

so µ∗ A = µ(A ∩ B). As µ is arbitrary, (iv) is true. (iv)⇒(ii) is trivial. not-(i)⇒not-(ii) If A is not universally negligible, let µ be a Borel probability measure on A which is zero on singletons. Set νE = µ(E ∩ A) for any Borel set E ⊆ X; then ν is a Borel probability measure on X which is zero on singletons, and ν ∗ A = 1. (i)⇒(v) Suppose that A is universally negligible. Let µ be a totally finite Borel measure on A. Applying (i)⇒(iv) with X = A, we see that there is a countable set B ⊆ A such that µB = µA; but this means that µ is inner regular with respect to the finite subsets of B, which of course are compact. As µ is arbitrary, A is a Radon space. ?? Suppose, if possible, that A has a compact set K which is not scattered. In this case there is a continuous surjection f : K → [0, 1] (4A2G(i-iv)). Now there is a Radon probability measure ν on K such that f is inverse-measure-preserving for ν and Lebesgue measure on [0, 1] and induces an isomorphism of the measure algebras, so that ν is atomless (418L). Accordingly we have a Borel probability measure µ on A defined by setting µE = ν(K ∩ E) for every relatively Borel set E ⊆ A, and µ{x} = 0 for every x ∈ A, so A is not universally negligible. X X Thus all compact subsets of A are scattered, and (v) is true. (v)⇒(i) Now suppose that (v) is true and that that µ is a Borel probability measure on A. Then µ has an extension to a Radon measure µ ˜ (434F(a-iii)). Let K ⊆ A be a non-empty compact set which is self-supporting for µ ˜ (416Dc). K is scattered, so has an isolated point {x}; because K is self-supporting, µ{x} = µ ˜{x} > 0. As µ is arbitrary, A is universally negligible. (b) This is immediate from (a-ii). (c) Let µ be a Borel probability measure on X. Then F 7→ νf −1 [F ] is a Borel probability measure on Y . Because Y is universally negligible, there must be a y ∈ Y such that µf −1 [{y}] > 0. Set E = f −1 [{y}] and let µE be the subspace measure on E. Then µE is a non-zero totally finite Borel measure on E. Since E is supposed to be universally negligible, there must be some x ∈ E such that 0 < µE {x} = µ{x}. (d) This is just a re-phrasing of the definition in 438A. 439D Remarks (a) The following will be useful when interpreting the definition in 439B. Let X be a hereditarily Lindelöf Hausdorff space and µ a topological probability measure on X such that µ{x} = 0 for every x ∈ X. Then µ is atomless. P P Suppose that µH > 0. Write G = {G : G ⊆ X is open, µ(G ∩ H) = 0}. S S Then there is a countable G0 ⊆ G such that G0 = G (4A2H(c-i)), so S S µ(H ∩ G) = µ(H ∩ G0 ) = 0, S S and µ(H \ G) > 0. Because µ is zero on singletons, H \ G has at least two points x, y say. Now there are disjoint open sets G0 , G1 containing x, y respectively, and neither belongs to G, so H ∩ G0 , H ∩ G1 are disjoint subsets of H of positive measure. Thus H is not an atom. As H is arbitrary, µ is atomless. Q Q (b) The obvious applications of (a) are when X is separable and metrizable; but, more generally, we can use it on any Hausdorff space with a countable network, e.g., on any analytic space. 439E Lemma (a) Let E, B ⊆ R be such that E is measurable and µL E, µ∗L B are both greater than 0, where µL is Lebesgue measure. Then E − B = {x − y : x ∈ E, y ∈ B} includes a non-trivial interval. (b) If A ⊆ R and µ∗L A > 0, then A + Q is of full outer measure in R. proof (a) By 261Da, there are a ∈ E, b ∈ B such that lim

1

δ↓0 2δ

Let γ > 0 be such that

µ(E ∩ [a − δ, a + δ]) = lim

1

δ↓0 2δ

µ∗ (B ∩ [b − δ, b + δ]) = 1.

439F

Examples 3 2

257 3 2

µ∗L (B ∩ [b − δ, b + δ]) > δ

µL (E ∩ [a − δ, a + δ]) > δ,

whenever 0 < δ ≤ γ. Now suppose that 0 < δ ≤ γ. Then µL ((E + b) ∩ [a + b, a + b + δ]) = µL (E ∩ [a, a + δ]) 1 2

≥ µL (E ∩ [a − δ, a + δ]) − δ > δ, and similarly µ∗L ((B + a + δ) ∩ [a + b, a + b + δ]) = µ∗L (B ∩ [b − δ, b]) 1 2

≥ µ∗L (B ∩ [b − δ, b + δ]) − δ > δ. But this means that (E + b) ∩ (B + a + δ) cannot be empty. If u ∈ (E + b) ∩ (B + a + δ), then u − b ∈ E, u − a − δ ∈ B so a − b + δ = (u − b) − (u − a − δ) ∈ E − B. As δ is arbitrary, E − B includes the interval ]a − b, a − b + γ]. (b) ?? Suppose, if possible, otherwise; that there is a measurable set E ⊆ R such that µL E > 0 and E ∩ (A + Q) = ∅. Then E − A does not meet Q and cannot include any non-trivial interval. X X Remark There will be a dramatic generalization of (a) in 443Db. 439F Lemma Let κ be the least cardinal of any set of non-zero Lebesgue outer measure in R. (a) There is a set X ⊆ [0, 1] of cardinal κ and full outer Lebesgue measure. (b) If (Z, T, ν) is any atomless complete locally determined measure space and A ⊆ Z has cardinal less than κ, then ν ∗ A = 0. (c) (Grzegorek 81) There is a universally negligible set Y ⊆ [0, 1] of cardinal κ. proof (a) Take any set A ⊆ R such that #(A) = κ and µ∗L A > 0, where µL is Lebesgue measure. Set B = A + Q. Then (µL )∗ (R \ B) = 0, by 439Eb. Set X = [0, 1] ∩ B; then µ∗L X = 1 while #(X) ≤ #(B) = κ. By the definition of κ, #(X) must be exactly κ. (b) ?? Otherwise, by 412Jc, there is a set F ⊆ Z such that νF < ∞ and ν ∗ (F ∩ A) > 0. By 343Cc, there is a function f : F → [0, νZn ] which is inverse-measure-preserving for the subspace measure νF and Lebesgue measure on [0, νF ]. But f [A ∩ F ] has cardinal less than κ, so µL f [A ∩ F ] = 0 and 0 < ν ∗ (A ∩ F ) ≤ νf −1 [f [A ∩ F ]] = 0, which is absurd. X X (c)(i) Enumerate X as hxξ iξ<κ . For each ξ < κ, Aξ = {xη : η ≤ P ξ} has cardinal less than κ, so is Lebesgue ∞ negligible; let hIξn in∈N be a sequence of intervals covering Aξ with n=0 µL Iξn < 12 . Enlarging the intervals slightly if necessary, we may suppose that every Iξn has rational endpoints; let hJm im∈N enumerate the family of intervals in R with rational endpoints. Set Cmn = {ξ : ξ < κ, Iξn = Jm } for each m, n ∈ N. (ii) If ν is an atomless totally finite measure on κ which measures every Cmn , then νκ = 0. P P Note first that (by (b), applied to the completion of ν) ν ∗ ξ = 0 for every ξ < κ. Let λ be the (c.l.d.) product of µX , the subspace measure on X, with ν. Set S B = m,n∈N ((X ∩ Jm ) × Cmn ) ⊆ X × κ. Then B is measured by λ, so, by Fubini’s theorem,

R

νB[{x}]µX (dx) =

R

µX B −1 [{ξ}]ν(dξ)

258


439F

(252D). Now look at the sectional measures νB[{x}], µX B −1 [{ξ}]. (Because B is actually a countable union of measurable rectangles, these are always defined.) For any x ∈ X, there is an η < κ such that x = xη , and now B[{x}] = {ξ : there are m, n ∈ N such that x ∈ Jm and ξ ∈ Cmn } = {ξ : there are m, n ∈ N such that x ∈ Jm and Iξn = Jm } = {ξ : there is an n ∈ N such that x ∈ Iξn } ⊇ κ \ η by the choice of the Iξn . But as ν ∗ η = 0, this means that νB[{x}] = νκ. On the other hand, if ξ < κ, then B −1 [{ξ}] = {x : there are m, n ∈ N such that x ∈ Jm and ξ ∈ Cmn } = {x : there are m, n ∈ N such that x ∈ Jm and Iξn = Jm } [ = {x : there is an n ∈ N such that x ∈ Iξn } = X ∩ Iξn , n∈N

so that µX B −1 [{ξ}] ≤ Returning to the integrals, we have νκ =

R

νB[{x}]µX (dx) =

P∞ n=0

R

1 2

µL Iξn ≤ .

1 2

µX B −1 [{ξ}]ν(dξ) ≤ νκ,

so that νκ must be 0, as claimed. Q Q (iii) Now there is an injective function g : κ → [0, 1] such that g[Cmn ] is relatively Borel in g[κ] for every m, n ∈ N. P P Define h : κ → {0, 1}N×N×N by setting h(ξ)(m, n, k) = 1 if ξ ∈ Cmn and xξ ∈ Jk = 0 otherwise. Then h is injective (because if ξ 6= η then xξ 6= xη , so there is some k such that xξ ∈ Jk and xη ∈ / Jk ), and h[Cmn ] = h[κ] ∩ {w : there is some k such that w(m, n, k) = 1} is relatively Borel in h[κ] for every m, n ∈ N. But now recall that {0, 1}N×N×N ∼ = {0, 1}N is homeomorphic to the Cantor set C ⊆ [0, 1] (4A2Uc). If φ : {0, 1}N×N×N → C is any homeomorphism, then φh has the required properties. Q Q (iv) Set Y = g[κ]. Because g is injective, #(Y ) = κ. Also Y is universally negligible. P P Suppose that ν˜ is a Borel measure on Y which is zero on singletons. Then it is atomless, because Y is separable and metrizable (439D). So its copy ν = ν˜(g −1 )−1 on κ is atomless. Because g[Cmn ] is a Borel subset of Y , ν measures Cmn for all m, n ∈ N, so ν˜Y = νκ = 0, by (ii) above. Q Q 439G Corollary A metrizable continuous image of a universally negligible metrizable space need not be universally negligible. proof Take X and Y from 439Fa and 439Fb above, and let f : X → Y be any bijection. Let Γ be the graph of f . The projection map (x, y) 7→ y : Γ → Y is continuous and injective, so Γ is universally negligible, by 439Cc. On the other hand, the projection map (x, y) 7→ x : Γ → X is continuous and surjective, and X is surely not universally negligible, since it is not Lebesgue negligible. 439H Corollary One-dimensional Hausdorff measure on R 2 is not semi-finite. proof Let µH1 be one-dimensional Hausdorff measure on R 2 . Let X, Γ be the sets described in 439F and the proof of 439G.

439K

Examples

259

(a) µ∗H1 Γ > 0. P P The first-coordinate map π1 : R 2 → R is 1-Lipschitz, so, writing µL for Lebesgue measure on R, 1 = µ∗L X = µ∗L π1 [Γ] ≤ µ∗H1 Γ by 264G and 264I. Q Q (b) If E ⊆ R 2 and µH1 E < ∞, then µH1 (E ∩ Γ) = 0, because E ∩ Γ is universally negligible (439Ca) and µH1 is a topological measure (264E) which is zero on singletons. (c) ?? Suppose, if possible, that Γ is not measured by µH1 . Then there is a set A ⊆ R 2 such that < µ∗H1 (A ∩ Γ) + µ∗H1 (A ∩ Γ) (264C, 264Fb). Let E be a Borel set including A such that µH1 E = µ∗H1 A (264Fa); then µH1 (E ∩ Γ) = 0, so µ∗H1 A

µ∗H1 (A ∩ Γ) + µ∗H1 (A ∩ Γ) ≤ µH1 (E ∩ Γ) + µ∗H1 A = µ∗H1 A. X X (d) Since Γ is measurable, not negligible, and meets every measurable set of finite measure in a negligible set, it is purely infinite, and µH1 is not semi-finite. 439I Example There are a set X, a Riesz subspace U of RX and a smooth positive linear functional h : U → R which is not expressible as an integral. proof Take X and Y from 439F. Replacing Y by Y \ {0} if need be, we may suppose that 0 ∈ / Y . Let f : X → Y be any bijection. Let U be the Riesz subspace {u × f : u ∈ Cb } ⊆ RX , where Cb is the space of bounded continuous functions from X to R. Because f is strictly positive, u 7→ u × f : Cb → U is a bijection, therefore a Riesz space isomorphism; moreover, for a non-empty set A ⊆ Cb , inf u∈A u(x) = 0 for every x ∈ X iff inf u∈A u(x)f (x) = R0 for every x ∈ X. We therefore have a smooth linear functional h : U → R defined by setting h(u × f ) = u dµX for every u ∈ Cb , where µX is the subspace measure on X induced by Lebesgue measure. (By 415B, µX is quasi-Radon, so the integral it defines on Cb is smooth, as noted in 436H.) ?? But suppose, if possible, that h is the integral with respect to some measure ν on X. Since f ∈ U , it must be T-measurable, where T is the domain of the completion νˆ of ν. Note that νˆ{x} = 0 for every x ∈ X. P P Set un (y) = max(0, 1 − 2n |y − x|) for y ∈ X. Then Z f (x)ˆ ν {x} = lim un × f dν = lim h(un × f ) n→∞ n→∞ Z = lim un dµX = µX {x} = 0, n→∞

so νˆ{x} = 0. Q Q ˆ of λ is a Radon measure on [0, 1] For Borel sets E ⊆ [0, 1] set λE = νˆf −1 [E]. Then the completion λ −1 ˆ (433Cb or 256C). If t ∈ [0, 1] then f [{t}] contains at most one point, so λ{t} = λ{t} = 0. But Y is ∗ ∗ ˆ Y = 0 (439Ca), that is, there is a Borel set E ⊇ Y with supposed to be universally negligible, so λ Y = λ λE = 0; in which case νX = νˆf −1 [E] = 0, which is impossible. X X Thus h is not an integral, despite being a smooth linear functional on a Riesz subspace of RX . Remark This example is adapted from Fremlin & Talagrand 78. 439J Example Assume that there is some cardinal κ which is not measure-free (definition: 438A). Give κ its discrete topology, and let µ be a probability measure with domain Pκ such that µ{ξ} = 0 for every ξ < κ. Now every subset of κ is open-and-closed, so µ is simultaneously a Baire probability measure and a completion regular Borel probability measure. Of course it is not τ -additive. In the classification schemes of 434A and 435A, we have a measure which is of type B1 as a Borel measure and type E3 as a Baire measure. 439K Example There is a first-countable compact Hausdorff space which is not Radon. proof The construction starts from a compact metrizable space (Z, S) with an atomless Radon probability measure µ. The obvious candidate is [0, 1] with Lebesgue measure; but for technical convenience in a later application I will instead use Z = {0, 1}N with its usual product topology and measure (254J).

260


439K

(a) There is a topology Tc on Z such that (α) S ⊆ Tc ; (β) every point of Z belongs to a countable set which is compact and open for Tc ; T S (γ) if hFn in∈N is a sequence of Tc -closed sets with empty intersection, then n∈N F n is countable, where I write F

S

for the S-closure of F .

P P(i) Tc will be the last in a family hTξ iξ≤c of topologies. We must begin by enumerating Z as hzξ iξ
Tξ ∪ {{yξ }} ∪ {Ln : n ∈ N},

S m≥n

Km for each n.

(δδ ) Because Tξ ⊆ Tξ+1 , Xξ will be open in Xξ+1 . Because the Ki are always Tξ -open, and xξ , yξ are distinct points of Z \ Xξ , the topology on Xξ induced by Tξ+1 is just Tξ . Consequently (by the inductive hypothesis) the topology on Xη induced by Tξ+1 is Tη for every η ≤ ξ. We have t¹n = xξ ¹n for every t ∈ Ln , so Tξ+1 is finer than the usual topology on Xξ+1 . If x ∈ Xξ , then there is a countable Tξ -open Tξ -compact set containing x, which is still Tξ+1 -open and Tξ+1 -compact. Of course {yξ } is a countable Tξ+1 -open Tξ+1 -compact set containing yξ . As for xξ , L0 is surely countable and Tξ+1 -open. To see that it is Tξ+1 -compact, observe that any ultrafilter containing L0 either contains every Ln , and converges to xξ , or contains some Ki and converges to a point of Ki . Thus the induction proceeds at successor stages. S (iii) Inductive step to a limit ordinal If ξ ≤ κ is a non-zero limit ordinal, then we have Xξ = η<ξ Xη , S and can take Tξ to be the topology generated by η<ξ Tη . It is easy to check that this works (because the topologies Tη are consistent). (iv) At the end of the induction, we have Xc = Z because zξ ∈ Xξ+1 ⊆ Xc for every ξ. The final topology Tc on Z will have the properties (α) and (β) required. ?? Now suppose, if possible, that hFn in∈N is T S a sequence of Tc -closed sets with empty intersection, and that n∈N F n is S uncountable. For each n ∈ N, let Jn ⊆ Fn be a countable S-dense set. Then there is some ζ < c such that n∈N Jn ⊆ Xζ (because cf c > ω, see 4A1A(c-ii)). Let ξ ≥ ζ be such that Jn = Iξn for every n ∈ N. Then in the construction of Tξ+1 we must be in case (β) of (ii) above. Taking htm im∈N as described there, we have htm im∈N → xξ for TTξ+1 , and therefore for Tc . But for any n ∈ N, tm ∈ In ⊆ Fn for infinitely many m, so xξ ∈ Fn . Thus xξ ∈ n∈N Fn ; but this is impossible. X X

439K

Examples

261

So we have a topology of the type required. Q Q (b) There is a probability measure ν on Z, extending the usual measure µ, such that with respect to Tc ν is a topological measure inner regular with respect to the closed sets, but is not τ -additive. P P Let K be the family of Tc -closed subsets of Z. For F ∈ K, set φF = µ∗ F . S

(i) If E, F are disjoint Tc -closed sets, then E ∩ F (a-γ)). So

S

must be countable (take F2n = E, F2n+1 = F in S

S

φ(E ∪ F ) = µ∗ (E ∪ F ) = µ∗ ((E ∪ F ) ∩ E ) + µ∗ ((E ∪ F ) \ E ) = µ∗ E + µ∗ F = φE + φF. (ii) If E, F ∈ K, E ⊆ F and ² > 0, there is an S-open set G ⊇ E such that µG ≤ µ∗ E + ². Now F \ G ∈ K and φF = µ∗ (F ∩ G) + µ∗ (F \ G) ≤ µG + φ(F \ G) ≤ φE + φ(F \ G) + ². Putting this together with (i), we see that φF = φE + sup{φE 0 : E 0 ∈ K, E 0 ⊆ F \ E}. (iii) If hFn in∈N is a non-increasing sequence in K with empty intersection, then T S S limn→∞ φFn ≤ limn→∞ µF n = µ( n∈N F n ) = 0. (iv) Thus K and φ satisfy all the conditions of 413I, and there is a measure ν, extending φ, which is defined on every member of K and inner regular with respect to K, and therefore is (for Tc ) a topological measure inner regular with respect to the closed sets. If we write V for the family of Tc -compact Tc -open countable subsets of Z, then for any K ∈ V νK = φK = µK = 0, while V is upwards-directed and has union Z; so that ν is not τ -additive. Q Q (c) So far we seem to have very little more than is provided by ω1 with the order topology and Dieudonné’s measure. The point of doing all this work is the next step. Set X = Z × {0, 1} and give X the topology T generated by {G × {0, 1} : G ∈ S} ∪ {H × {1} : H ∈ Tc } ∪ {X \ (K × {1}) : K is Tc -compact}. (i) T is Hausdorff. P P If w, z are distinct points of X, then either their first coordinates differ and they are separated by sets of the form G0 × {0, 1}, G1 × {0, 1} where G0 , G1 belong to S, or they are of the form (x, 1), (x, 0) and are separated by open sets of the form K × {1}, X \ (K × {1}) for some set K which is compact and open for Tc . Q Q (ii) T is compact. P P Let F be an ultrafilter on X. Writing π1 (x, 0) = π1 (x, 1) = x for x ∈ Z, π1 [[F]] is S-convergent, to x0 say. If K × {1} ∈ F for some Tc -compact set K, then F is T-convergent to (x, 1); otherwise, it is T-convergent to (x, 0) (using 4A2B(a-v)). Q Q (iii) T is first-countable. P P If x ∈ Z, then {(x, 0), (x, 1)} = π1−1 [{x}] is a Gδ set in X because {x} is a Gδ set in Z and π1 is continuous (4A2C(a-iii)). Now {(x, 0)} and {(x, 1)} are relatively open in {(x, 0), (x, 1)}, so are Gδ sets in X (4A2C(a-iv)). Thus singletons are Gδ sets. Because T is compact and Hausdorff, it is first-countable (4A2Kf). Q Q (iv) (X, T) is not a Radon space. P P Z × {1} is an open subset of X, homeomorphic to Z with the topology Tc . But the measure ν of (b) above (or, if you prefer, its restriction to the Tc -Borel algebra) witnesses that Tc is not a Radon topology, so T also cannot be a Radon topology, by 434Fc. Q Q ´ sz Remark Aficionados will recognise Tc as a kind of ‘JKR-space’, derived from the construction in Juha Kunen & Rudin 76.

262


439L

439L Example Suppose that κ is a cardinal which is not measure-free; let µ be a probability measure with domain Pκ which is zero on singletons. Give κ the discrete topology, so that µ is a Borel measure and κ is first-countable. Let ν be the restriction of the usual measure on Y = {0, 1}κ to the algebra B of Borel subsets of Y , so that ν is a τ -additive probability measure, and λ the product measure on κ × Y constructed by the method of 434R. Then S W = {(ξ, y) : ξ < κ, y(ξ) = 1} = ξ<κ {ξ} × {y : y(ξ) = 1} is open in κ × Y . b then λ(W 4W 0 ) = 21 . P If W 0 ∈ Pκ⊗B P There is a countable set E ⊆ B such that W 0 belongs to the σ-algebra generated by {A × E : A ⊆ κ, E ∈ E} (331Gd). For J ⊆ κ, write πJ (y) = y¹J for y ∈ Y , let νJ be the usual measure on {0, 1}J and ΛJ its domain, and let Λ0J be the family of sets E ⊆ Y such that there are H, H 0 ∈ ΛJ such that πJ−1 [H] ⊆ E ⊆ πJ−1 [H 0 ] and νJ (H 0 \ H) = 0. Then Λ0J ⊆ Λ0K whenever J ⊆ K ⊆ κ, and every set measured by ν belongs to Λ0J for some countable J (254Ob). There is therefore a countable set J ⊆ κ such that E ⊆ Λ0J . Also, of course, Λ0J is a σ-algebra of subsets of Y . The set {V : V ⊆ κ × Y, V [{ξ}] ∈ Λ0J for every ξ < κ} is a σ-algebra of subsets of κ × Y containing A × E whenever A ⊆ κ and E ∈ E, so contains W 0 . But this means that if ξ ∈ κ \ J, W [{ξ}] and W 0 [{ξ}] are stochastically independent, and ν(W [{ξ}]4W 0 [{ξ}]) = 21 . Since µ(κ \ J) = 1, λ(W 4W 0 ) =

R

1 2

ν(W [{ξ}]4W 0 [{ξ}])µ(dξ) = ,

as claimed. Q Q b In particular, W • in the the measure algebra of λ cannot be represented by a member of Pκ⊗B. 439M Example There is a first-countable locally compact Hausdorff space X with a Baire probability measure µ which is not τ -additive and has no extension to a Borel measure. In the classification of 435A, µ is of type E0 . proof Let Ω be the set of non-zero countable limit ordinals, and for each ξ ∈ Ω let hθξ (i)ii∈N be a strictly increasing sequence of ordinals with supremum ξ. Set X = ω1 × (ω + 1), and define a topology T on X by saying that G ⊆ X is open iff {ξ : (ξ, n) ∈ G} is open in the order topology of ω1 for every n ∈ N, whenever ξ ∈ Ω and (ξ, ω) ∈ G then there is some n < ω such that (η, i) ∈ G whenever n ≤ i < ω and θξ (i) < η ≤ ξ. This is finer than the product of the order topologies, so is Hausdorff. For every ξ < ω1 , n ∈ N, (ξ + 1) × {n} is a countable compact open set containing (ξ, n); for every ξ ∈ ω1 \ Ω, {(ξ, ω)} is a countable compact open set containg (ξ, ω); and for every ξ ∈ Ω, {(ξ, ω)} ∪ {(η, i) : i < ω, θξ (i) < η ≤ ξ} is a countable compact open subset of X containing (ξ, ω). Thus T is locally compact, and every singleton subset of X is Gδ , so T is first-countable (4A2Kf). If f : X → R is continuous, then for every n ∈ N there is a ζn < ω1 such that f is constant on {(ξ, n) : ζn ≤ ξ < ω1 } (4A2S(b-iii)). Setting ζ = supn∈N ζn , f must be constant on {(ξ, ω) : ξ ∈ Ω, ξ > ζ}. P P If ξ, η ∈ Ω\(ζ +1), then f (ξ, ω) = limi→∞ f (θξ (i)+1, i) and f (η, ω) = limi→∞ f (θη (i)+1, i). But there is some n such that both θξ (i) and θη (i) are greater than ζ for every i ≥ n, so that f (θξ (i)+1, i) = f (θη (i)+1, i) for every i ≥ n and f (ξ, ω) = f (η, ω). Q Q Writing Σ for the family of subsets E of X such that {ξ : ξ ∈ Ω, (ξ, ω) ∈ E} is either countable or cocountable in Ω, Σ is a σ-algebra of subsets of X such that every continuous function is Σ-measurable, so every Baire set belongs to Σ. We therefore have a Baire measure µ0 on X defined by saying that µ0 E = 0 if E ∩ (Ω × {ω}) is countable, 1 otherwise. {(ξ + 1) × (ω + 1) : ξ < ω1 } is a cover of X by negligible open-and-closed sets, so µ0 is not τ -additive. ?? Suppose, if possible, that µ were a Borel measure on X extending µ0 . Then we must have µ(ω1 ×{n}) = µ0 (ω1 × {n}) = 0 for every n ∈ N, so µ(ω1 × {ω}) = 1. Let λ be the subspace measure on ω1 × {ω} induced

439O

Examples

263

by µ. If A ⊆ ω1 , (A × {ω}) ∪ (ω1 × ω) is an open set, so λ is defined on every subset of ω1 × {ω}; and if ξ < ω1 , then µ0 ((ξ + 1) × (ω + 1)) = 0, so λ is zero on singletons. And this contradicts Ulam’s theorem (419G, 438Cd). X X 439N Example Give ω1 its order topology. (i) ω1 is a normal Hausdorff space which is not measure-compact. (ii) There is a Baire probability measure µ0 on ω1 which is not τ -additive and has a unique extension to a Borel measure, which is not completion regular; that is, µ0 is of type E2 in the classification of 435A. proof (a) As noted in 4A2Rc, order topologies are always normal and Hausdorff. (b) Let µ be Dieudonné’s measure on ω1 (411Q), and µ0 its restriction to the Baire σ-algebra. Then µ is the only Borel measure extending µ0 . P P Let ν be any Borel measure extending µ0 . Every set [0, ξ] = [0, ξ + 1[, where ξ < ω1 , is open-and-closed, so ν[0, ξ] = µ0 [0, ξ] = µ[0, ξ] = 0; also, of course, νX = 1. Let F ⊆ X be any closed set. If F is countable, then it is included in some initial segment [0, ξ], so νF = µF = 0. Now suppose that F is uncountable. Set G = ω1 \ F . For each ξ ∈ F , set ζξ = min{η : ξ < η ∈ F } and Gξ = ]ξ, ζξ [. Then hGξ iξ∈F is a disjoint family of open sets. By 438Bb and 419G/438Cd, P S ν( ξ∈F Gξ ) = ξ∈F νGξ = 0. But now

S 1 = νω1 = νF + ν [0, min F [ + ν( ξ∈F Gξ ) = νF = µF .

Thus µ and ν agree on the family E of closed sets. By the Monotone Class Theorem (136C), they agree on the σ-algebra generated by E, which is their common domain; so they are equal. Q Q (c) I have already remarked in 411Q-411R that µ and µ0 are not τ -additive and µ is not completion regular. So of course ω1 is not measure-compact. 439O In 439M I described a Baire measure with no extension to a Borel measure. In view of Maˇr´ık’s theorem (435C), it is natural to ask whether this can be done with a normal space. This leads us into relatively deep water, and the only examples known need special assumptions. Example Assume ♣. Then there is a normal Hausdorff space with a Baire probability measure µ which is not τ -additive and not extendable to a Borel measure. (In the classification of 435A, µ is of type E0 .) proof (a) ♣ implies that there is a family hCξ iξ<ω1 of sets such that (i) Cξ ⊆ ξ for every ξ < ω1 (ii) Cξ ∩ η is finite whenever η < ξ < ω1 (iii) for any uncountable sets A, B ⊆ ω1 there is a ξ < ω1 such that A ∩ Cξ and B ∩ Cξ are both infinite (4A1N). For A ⊆ ω1 , set A0 = {ξ : ξ < ω1 , A ∩ Cξ is infinite}; then A0 ∩ B 0 is non-empty whenever A, B ⊆ ω1 are uncountable. But this means that A0 ∩ B 0 is actually uncountable for uncountable A, B, since A0 ∩ B 0 \ γ ⊇ (A \ γ)0 ∩ (B \ γ)0 is non-empty for every γ < ω1 . Set X = ω1 × N. For x = (ξ, n) ∈ X, say that Ix = Cξ × {n − 1} if n ≥ 1, = ∅ otherwise . (b) Define a topology T on X by saying that a set G ⊆ X is open iff Ix \ G is finite for every x ∈ G. The form of the construction ensures that T is T1 . In fact, Ix ∩ Iy is finite whenever x 6= y in X. P P Express x as (ξ, m) and y as (η, n) where η ≤ ξ. If either m = 0 or n = 0 or m 6= n, Ix ∩ Iy = ∅. If n ≥ 1 and η < ξ, then Ix ∩ Iy ⊆ (Cξ ∩ η) × {m − 1} is finite. Similarly, Ix ∩ Iy is finite if m ≥ 1 and ξ < η. Q Q Consequently {x} ∪ J is closed for every x ∈ X, J ⊆ Ix .

264


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Observe that (ξ + 1) × N is open and closed for every ξ < ω1 , again because Cη ∩ (ξ + 1) is finite whenever η ∈ Ω and η > ξ, while Cξ ⊆ ξ for every ξ. (c) The next step is to understand the uncountable closed subsets of X. First, if F ⊆ X is closed and n ∈ N, then F −1 [{n}]0 , as defined in (a), is a subset of F −1 [{n+1}], since if ξ ∈ F −1 [{n}]0 then I(ξ,n+1) ∩F is infinite. If F is uncountable, there is some n ∈ N such that F −1 [{n}] is uncountable, so that (inducing on m) F −1 [{m}] is uncountable for every m ≥ n. Finally, this means that if E, F ⊆ X are uncountable closed sets, there is an m ∈ N such that E −1 [{m}] and F −1 [{m}] are both uncountable, so that E −1 [{m}]0 ∩ F −1 [{m}]0 is non-empty and E ∩ F is non-empty. (d) It follows that X is normal. P P Let E and F be disjoint closed sets in X. By (c), at least one of them is countable; let us take it that E ⊆ ζ × N where ζ < ω1 . Enumerate the open-and-closed set W = (ζ + 1) × N as hxn in∈N . Choose hUn in∈N , hVn in∈N inductively, as follows. U0 = E, V0 = F ∩ W . If xn ∈ Un , then Un+1 = Un ∪ (Ixn \ Vn ) and Vn+1 = Vn ; if xn ∈ / Un , then Un+1 = Un and Vn+1 = Vn ∪ {xn } ∪ (Ixn \ Un ). An easy induction shows that, for every n, (α) Un ∩ Vn = ∅ (β) Un ∪ Vn ⊆ W (γ) Ix ∩ (Un ∪ Vn ) is finite for every x ∈ X \ (Un ∪ Vn ) (δ) Ix ∩ Vn is finite for every x ∈ Un (²) Ix ∩ Un is finite for every x ∈ Vn . S S At the end of the induction, set G = n∈N Un , H = n∈N Vn ∪ (X \ W ). Then E ⊆ G, F ⊆ H and G ∩ H = ∅. If x ∈ G, it is of the form xn for some n, in which case xn ∈ Un (because xn ∈ / Vn+1 ) and Ix \ Un+1 = Ix ∩ Vn is finite; thus G is open. If x ∈ H ∩ W , again it is of the form xn where this time xn ∈ / Un , so that Ix \ Vn+1 = Ix ∩ Un is finite; so H is open. Thus E and F are separated by open sets; since E and F are arbitrary, X is normal. Q Q Being T1 (see (b)), X is also Hausdorff. (e) Because disjoint closed sets in X cannot both be uncountable ((c) above), any bounded continuous function on X must be constant on a cocountable set. (Compare 4A2S(b-iii).) The countable-cocountable measure µ0 is therefore a Baire measure on X (cf. 411R). But it has no extension to a Borel measure. P P The point is that if A is any subset of ω1 , and n ∈ N, then (A × {n}) ∪ (ω1 × n),

ω1 × n

are both open, so A × {n} is Borel; accordingly every subset of X is a Borel set. But ω1 is measure-free (419G, 438Cd), so there can be no Borel probability measure on X which is zero on singletons. Q Q Of course µ0 is not τ -additive, because {(ξ + 1) × N : ξ < ω1 } is a cover of X by open-and-closed negligible sets. Remark Thus in Maˇr´ık’s theorem we really do need ‘countably paracompact’ as well as ‘normal’, at least if we want a theorem valid in ZFC. Observe that any example of this phenomenon must involve a Dowker space, that is, a normal Hausdorff space which is not countably paracompact. The one here is based on de Caux 76. Such spaces are hard to come by in ZFC if we do not allow ourselves to use special principles like ♣. ‘Real’ Dowker spaces have been described by Rudin 71 and Balogh 96; for a survey, see Rudin 84. I do not know if either of these can be adapted to provide a ZFC example to replace the one above. 439P Example (cf. Moran 68) N c is not Borel-measure-compact, therefore not Borel-measure-complete, measure-compact or Radon. proof Consider the topology Tc on Z = {0, 1}N , as constructed in 439K. Then (Z, Tc ) is homeomorphic to a closed subset of N Z × {0, 1}N , where in this product the second factor {0, 1}N is given its usual topology S. P P For each x ∈ Z, let Lx be a Tc -open Tc -compact subset of Z. The first thing to observe is that if x ∈ Z, and we write Vxm = {y : y ∈ Z, y¹m = x¹m} for each m ∈ N, then Ux = {Lx ∩ Vxm : m ∈ N} is a downwards-directed family of compact open neighbourhoods of x with intersection {x}, so is a base of neighbourhoods of x (4A2Gd); thus U = {Lx : x ∈ Z} ∪ S generates Tc . Now, for x ∈ Z, define φx : Z → N by setting φx (y) = 0 if y ∈ Lx , = m + 1 if y ∈ Vxm \ (Lx ∪ Vx,m+1 ).

439Q

Examples

265

Then every φx is Tc -continuous, so we have a Tc -continuous function φ : Z → N Z × {0, 1}N defined by setting φ(y) = (hφx (y)ix∈C , y) for y ∈ Z. Because every element of U is of the form φ−1 [H] for some open set H ⊆ N Z × {0, 1}N , Z is homeomorphic to its image φ[Z]. Now suppose that (w, z) ∈ φ[Z]. In this case, there is an ultrafilter containing φ[Z] which converges to (w, z) (4A2Bc), and this must be of the form φ[[F]] for some ultrafilter on Z (2A1Ib). Of course z = lim F for S. ?? If z is not the Tc -limit of F, then F can have no Tc -limit, and can contain no Tc -compact set (2A3R). In particular, Lz ∈ / F; but in this case Vzm \Lz ∈ F for every m, so that {(v, y) : v(x) > m} ∈ φ[[F]] for every m, and w(x) > m for every m, which is impossible. X X Thus F → z, and (as φ is continuous) (w, z) = φ(z). This shows that φ[Z] is closed, so we have the required homeomorphism between Z and a closed subset of N Z × {0, 1}N . Q Q Of course N Z × {0, 1}N is a closed subset of N Z × N N ∼ = N c . So Z is homeomorphic to a closed subset c of N . But Z, with Tc , carries a Borel probability measure ν which is inner regular with respect to the closed sets and is not τ -additive (439Kb). So (Z, Tc ) is not Borel-measure-compact. By 434Hc, N c is not Borel-measure-compact. By 434Ic, N c is not Borel-measure-complete; by 434K, it is not Radon; by 435Fd, it is not measure-compact. 439Q Example Let X be the real line with the right-facing Sorgenfrey topology, generated by sets of the form [a, b[ where a < b in R. Then X is measure-compact but X 2 is not. proof (a) Note that every set [a, b[ is open-and-closed in X, so that the topology is zero-dimensional, therefore completely regular; and it is finer than the usual topology of R, so is Hausdorff. X is Lindelöf. P P Let G be an open cover of X. For each q ∈ Q, set S

Aq = {a : a ∈ ]−∞, q[ and there is some G ∈ G such that [a, q[ ⊆ G}.

Then q∈Q Aq = R. For each q ∈ Q, there is a countable set A0q ⊆ Aq such that inf A0q = inf Aq in [−∞, ∞] 0 and A0q contains min Aq if Aq has a least element. Now, Sfor each pair 0(a, q) where q ∈ Q and a ∈ Aq , choose Gaq ∈ G such that [a, q[ ⊆ Gaq . It is easy to see that {Gaq : a ∈ Aq } ⊇ Aq , so that the countable family {Gaq : q ∈ Q, a ∈ A0q } covers X. As G is arbitrary, X is Lindelöf. Q Q It follows that X is measure-compact (435Fb). (b) Let S be the usual topology on R 2 , and T the product topology on X 2 . (i) Whenever G, H are disjoint T-open sets, there is an S-Borel set E such that G ⊆ E ⊆ X 2 \ H. P P For n ∈ N, set An = {(a, b) : [a, a + 2−n [ × [b, b + 2−n [ ⊆ G}. S

S

?? Suppose, if possible, that there is a point (x, y) ∈ An ∩ H, where I write to denote closure for the topology S. Let δ > 0 be such that [x, x + 2δ[ × [y, y + 2δ[ ⊆ H and 2δ < 2−n . Then there must be (a, b) ∈ An such that |a − x| ≤ δ and |b − y| ≤ δ. In this case, a ≤ x + δ < a + 2−n and b ≤ y + δ < b + 2−n , so (x + δ, y + δ) ∈ G; while δ was chosen so that (x + δ, y + δ) would belong to H. X X S S S Accordingly E = n∈N An is an S-Borel set disjoint from H. But G = n∈N An , so G ⊆ E. Q Q (ii) Consequently every T-continuous real-valued function is S-Borel measurable. P P If f : X 2 → R is T-continuous and α ∈ R, then there is an S-Borel set Eα such that {(x, y) : f (x, y) < α} ⊆ Eα ⊆ {(x, y) : f (x, y) ≤ α}. S But this means that {(x, y) : f (x, y) < α} = n∈N Eα−2−n is S-Borel. Q Q (iii) It follows that every T-Baire set is S-Borel. We therefore have a T-Baire probability measure ν on X defined by setting νE = µL {t : t ∈ [0, 1], (t, 1 − t) ∈ E} for every T-Baire subset of X 2 , where µL is Lebesgue measure on R. In this case every point (x, y) of X 2 belongs to a T-open set of zero measure for ν. P P Set K = {(t, 1−t) : t ∈ [0, 1]}. Then K is S-closed, therefore T-closed, and ν(X 2 \ K) = 0, so if (x, y) ∈ / K then we can stop. If (x, y) ∈ K, then [x, x + 1[ × [y, y + 1[ is a T-open T-closed set meeting K in the single point (x, y), so is a negligible T-neighbourhood of (x, y). Q Q

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Thus ν is not τ -additive and X 2 is not measure-compact. Remark Contrast this with 438Xo. 439R Example There are first-countable completely regular Hausdorff spaces X, Y with Baire probability measures µ, ν such that the Baire measures λ, λ0 on X × Y defined by the formulae

R

f dλW =

RR

f (x, y)ν(dy)µ(dx),

R

f dλ0 =

RR

f (x, y)µ(dx)ν(dy)

(436F) are different. proof Let X, Y be disjoint stationary subsets of ω1 (4A1Cd). Give each the topology induced by the order topology of ω1 . Let µ ˜ be Dieudonné’s measure on ω1 (411Q), and let µ ˜X , µ ˜Y be the subspace measures induced on X and Y by µ ˜; let µ and ν be the restrictions of µ ˜X , ν˜Y to the Baire σ-algebras of X, Y respectively. Then µX = µ ˜X X = µ ˜∗ X = 1 because X meets every cofinal closed set in ω1 ; similarly, νY = 1. Set W = {(x, y) : x ∈ X, y ∈ Y, x < y} = {(x, y) : x ∈ X, y ∈ Y, x ≤ y}. Then W is open-and-closed in X × Y (use 4A2Rl), so that f = χW is continuous. But

RR

RR

f (x, y)ν(dy)µ(dx) = f (x, y)µ(dx)ν(dy) =

R

R

ν{y : y ∈ Y, x < y}µ(dx) = 1, µ{x : x ∈ X, x < y}ν(dy) = 0.

Remark Contrast this with 434Xx and 439Yh. ˇ 439S The results of 437W leave open the question of which familiar spaces, beyond Cech-complete spaces, can be Prokhorov. In fact rather few are. The basis of any further investigation must be the following result. Theorem (Preiss 73) Q is not a Prokhorov space. proof (a) There is a non-decreasing sequence hXk ik∈N of non-empty compact subsets of X = Q ∩ [0, 1], with union X, such that whenever k ∈ N, x ∈ Xk and δ > 0, then Xk+1 ∩ [x − δ, x + δ] is infinite. P P Start by enumerating X as hqk ik∈N . Set X0 = {q0 }. Given that Xk ⊆ X is compact, then for each m ∈ N let Em be a finite cover of Xk by open intervals of length at most 2−m all meeting S Xk , and let Ikm be a finite subset of X \ Xk meeting every member of Em ; set Xk+1 = Xk ∪ {qk+1 } ∪ m∈N Ikm . If H is any cover of Xk+1 by open sets inSR, then there is a finite H0 ⊆ H covering XkS. There must be an m ∈ N suchSthat [x − 2−m , x + 2−m ] ⊆ H0 for every x ∈ Xk (2A2Ed), so that Ikl ⊆ H0 for every l ≥ m, and Xk+1 \ H0 is finite; accordingly there is a finite H1 ⊆ H covering Xk+1 . As H is arbitrary, Xk+1 is compact, and the induction can proceed. If x ∈ Xk and δ > 0, then for every m ∈ N there is an x0 ∈ Xk+1 \ Xk such that |x0 − x| ≤ 2−m , so that [x − δ, x + δ] ∩ Xk+1 must be infinite. Q Q (b) If h²k ik∈N is any sequence in ]0, ∞[, and F ⊆ [0, 1] is a countable closed set, then there is an x∗ ∈ X \ F such that ρ(x∗ , Xk ) < ²k for every k ∈ N. P P We can suppose that limk→∞ ²k = 0. Define hHk ik∈N inductively, as follows. H0 = R. Given Hk , set Hk+1 = Hk ∩ {x : ρ(x, Xk ∩ Hk ) < ²k }, where ρ(x, A) = inf y∈A |x − y| for x ∈ R, A ⊆ R. Observe that every THk is an open subset of R and that Xk ∩ Hk ⊆ Hk+1 ⊆ Hk for every k; consequently, setting E = k∈N Hk , E is a Gδ subset of R and Xk ∩Hk ⊆ E for every k. In particular, E∩X contains q0 and is not empty. Next, for each k, ρ(x, E∩Xk ) < ²k for every x ∈ Hk+1 and therefore for every x ∈ E; accordingly E ∩ X is dense in E. Moreover, if x ∈ E ∩ X, there is a k ∈ N such that x ∈ Xk ; we must have x ∈ Hk , and in this case Hk+1 is a neighbourhood of x. So every neighbourhood of x contains infinitely many points of Hk+1 ∩ Xk+1 ⊆ E ∩ X. Thus E ∩ X has no isolated points; it follows that E has no isolated points. By 4A2Mc and 4A2Me, E is uncountable. There is therefore a point z ∈ E \ F . Let m ∈ N be such that ρ(z, F ) ≥ ²m . As z ∈ Hm+1 , there is an x∗ ∈ Hm ∩ Xm such that |z − x∗ | < ²m and x∗ ∈ / F . Let k ∈ N. If k ≥ m then certainly ρ(x∗ , Xk ) = 0 < ²k . ∗ ∗ ∗ If k < m then x ∈ Hk+1 so ρ(x , Xk ) ≤ ρ(x , Hk ∩ Xk ) < ²k . So we have a suitable x∗ . Q Q

439Xe

Examples

267

(c) For n, k ∈ N set Gkn = {x : x ∈ R \ Xk , ρ(x, Xn ) > 2−k }. Then Gkn is an open subset of R. Let A be the set of Radon probability measures µ on X such that µ(Gkn ∩ X) ≤ 2−n for all n, k ∈ N. (d) Write A˜ for the set of Radon probability measures µ on [0, 1] such that µ(Gkn ∩ [0, 1]) ≤ 2−n for all k, n ∈ N. Then A˜ is a narrowly closed subset of the set of Radon probability measures on [0, 1], which is ˜ P itself narrowly compact (437O). Also µ([0, 1] \ X) = 0 for every µ ∈ A. P Let set K ⊆ [0, 1] \ X be compact, and n ∈ N. Then K and Xn are disjoint compact sets, so there is some k ∈ N such that |x − y| > 2−k for every x ∈ Xn and y ∈ K. In this case K ⊆ Gkn so µK ≤ 2−n . As n is arbitrary, µK = 0; as K is arbitrary, µ([0, 1] \ X) = 0. Q Q A is compact in the narrow topology. P P The identity map φ : X → [0, 1] induces a map φ˜ : MP+ (X) → + MP ([0, 1]) which is a homeomorphism between MP+ (X) and {µ : µ ∈ MP+ ([0, 1]), µ([0, 1] \ X) = 0} (437Qb). ˜ ˜ since A˜ ⊆ {µ : µ ∈ M + ([0, 1]), µ([0, 1] \ X) = 0}, φ¹A The definition of A makes it plain that it is φ˜−1 [A]; P ˜ is a homeomorphism between A and A, and A is compact. Q Q (e) A, regarded as a subset of MP+ (X), is not uniformly tight. P Let K ⊆ X be P compact. Consider the PP set C of those w ∈ [0, 1]X such that w(x) = 0 for every x ∈ K, x∈X w(x) ≤ 1 and x∈Gkn ∩X w(x) ≤ 2−n for all k, n ∈ N. Then C is a compact subset of [0, 1]X . If D ⊆ C is any non-empty upwards-directed X set, then sup D, taken in [0, 1] P , belongs to C. By Zorn’s Lemma, C has a maximal member w say. ?? Suppose, if possible, that x∈X w(x) = γ < 1. For each n ∈ N, let Ln ⊆ X be a finite set such P −n−1 w(x) ≥ γ − 2 , and mn ∈ N such that Ln ⊆ Xmn . By (b), there is an x∗ ∈ X \ K that x∈Ln ∗ −mn such that ρ(x , Xn ) < 2 for every n ∈ N. Let r ∈ N be such that x∗ ∈ Xr and γ + P 2−r ≤ 1, and set 0 ∗ ∗ −r 0 ∗ 0 X w (x ) = w(x )+2 , w (x) = w(x) w ∈ [0, 1] and x∈X w0 (x) ≤ 1. P }. Then certainly −n P for every x0 ∈ X \{x ∗ If k, n ∈ N and x ∈ / Gkn , then x∈Gkn ∩X w (x) = x∈Gkn ∩X w(x) ≤ 2 . If x∗ ∈ Gkn , then n < r and 2−k < ρ(x∗ , Xn ) < 2−mn , so mn < k and Ln ⊆ Xk and P P P −n−1 , x∈X\Ln w(x) ≤ 2 x∈X\Xk w(x) ≤ x∈Gkn ∩X w(x) ≤ P x∈Gkn ∩X

w0 (x) ≤ 2−n−1 + 2−r ≤ 2−n .

Thus w0 ∈ C and X Pw was not maximal. X Accordingly x∈X w(x) = 1 and the point-supported measure µ defined by w is a probability measure on X. By the definition of C, µ ∈ A and µ(X \ K) = 1. As K is arbitrary, A cannot be uniformly tight. Q Q (f ) Thus A witnesses that X = Q ∩ [0, 1] is not a Prokhorov space. Since X is a closed subset of Q, 437Wb tells us that Q is not a Prokhorov space. 439X Basic exercises (a) (i) Show that there is a set A ⊆ [0, 1] such that µ∗L A = 1, where µL is Lebesgue measure, and every member of [0, 1] is uniquely expressible as a + q where a ∈ A, q ∈ Q. (Hint: use an idea from 419J.) (ii) Define f : [0, 1] → A by setting f (x) = a when x ∈ a + Q. Show that the image measure µL f −1 takes only the values 0 and 1. (Aldaz 95. Compare 342Xg.) (b) Let X be a Radon Hausdorff space and A a subset of X. Show that A is universally negligible iff µA = 0 for every atomless Radon measure on X. > (c) Let X be a Hausdorff space. Show that a set A ⊆ X is universally negligible iff µA = 0 whenever µ is a complete locally determined topological measure on X such that µ{x} = 0 for every x ∈ X. (d) Let X be a Hausdorff space. Show that any universally negligible subset of X is universally measurable in the sense of 434D. (e) (i) Show that there is an analytic set A ⊆ R such that for any Borel subset E of R \ A there is an uncountable Borel subset of R \ (A ∪ E). (Hint: 423Qb, part (c) of the proof of 423L.) (ii) Show that A is universally measurable, but there is no Borel set E such that A4E is universally negligible.

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(f ) Show that a first-countable compact Hausdorff space is universally negligible iff it is scattered iff it is countable. (g) Show that the product of two universally negligible Hausdorff spaces is universally negligible. (h) Let us say that a Hausdorff space X is universally τ -negligible if there is no τ -additive Borel probability measure on X which is zero on singletons. (i) Show that if X is a Hausdorff space and A ⊆ X, then A is universally τ -negligible iff µ∗ A = 0 for every τ -additive Borel probability measure on X such that µ{x} = 0 for every x ∈ X. (ii) Show that if X is a regular Hausdorff space, then a subset A of X is universally τ -negligible iff µA = 0 for every atomless quasi-Radon measure on X. (iii) Show that if X is a completely regular Hausdorff space, it is universally τ -negligible iff whenever µ is an atomless Radon measure on a space Z, and X 0 ⊆ Z is homeomorphic to X, then µX 0 = 0. (iv) Show that a Hausdorff space X is universally negligible iff it is Borel-measure-complete and universally τ -negligible. (v) Show that if X is a Hausdorff space, Y is a universally τ -negligible Hausdorff space, and f : X → Y is a continuous function such that f −1 [{y}] is universally τ -negligible for every y ∈ Y , then X is universally τ -negligible. (vi) Show that the product of two universally τ -negligible Hausdorff spaces is universally τ -negligible. (vi) Show that a scattered Hausdorff space (in particular, any discrete space) is universally τ -negligible. (vii) Show that a compact Hausdorff space is universally τ -negligible iff it is scattered. > (i) Let κ be the smallest cardinal of any subset of R which is not Lebesgue negligible. Show that if (Z, T, ν) is any complete locally determined atomless measure space and A ⊆ Z has cardinal less than κ, then νA = 0. (j) Let (X, ≤) be any well-ordered set and µ a non-zero σ-finite measure on X such that every singleton has measure 0. Show that {(x, y) : x ≤ y} is not measurable for the (c.l.d.) product measure on X × X. (Hint: Reduce to the case in which µ is complete and totally finite, X = ζ is an ordinal and µξ = 0 for every ξ < ζ. You will probably need 251P.) (k) Show that 1-dimensional Hausdorff measure on R 2 is not inner regular with respect to the closed sets. (Hint: 439H, 471R.) (l) Show that N I is not pre-Radon for any uncountable set I. (Hint: 417Xq.) (m) (i) Suppose that X is a completely regular Hausdorff space and there is a continuous function f from X to a separable metrizable space Z such that f −1 [{z}] is Lindelöf for every z ∈ Z. Show that X is realcompact (definition: 436Xg). (ii) Show that the spaces X of 439K and X 2 of 439Q are realcompact. (iii) Show that c with the discrete topology, and N c with the product topology, are realcompact. 439Y Further exercises (a) (i) Show that a subset A of R is universally negligible iff f [A] is Lebesgue negligible for every continuous injective function f : R → R. (Hint: if ν is an atomless Borel probability measure on R, set f (x) = x + ν[0, x] for x ≥ 0, and show that µL f [E] ≥ νE for every Borel set E ⊆ [0, ∞[.) (ii) Let X be a separable metrizable space. Show that X is universally negligible iff f [X] is Lebesgue negligible for every injective Borel measurable function f : X → R. (b) For this exercise only, let us say that a ‘universally negligible measurable space’ is a pair (X, Σ) where X is a set and Σ a σ-algebra of subsets of X containing every countable subset of X such that there is no probability measure µ with domain Σ such that µ{x} = 0 for every x ∈ X. (i) Let X be a set, Σ a σ-algebra of subsets of X containing all countable subsets of X, A ⊆ X and ΣA the subspace σ-algebra. Show that (A, ΣA ) is universally negligible iff µ∗ A = 0 whenever µ is a probability measure with domain Σ which is zero on singletons. Show that if (X, Σ) is universally negligible so is (A, ΣA ). (ii) Let X and Y be sets, Σ and T σ-algebras of subsets of X and Y containing all appropriate countable sets, and f : X → Y a (Σ, T)-measurable function. Suppose that (Y, T) and (f −1 [{y}], Σf −1 [{y}] ) are universally negligible for every y ∈ Y . Show that (X, Σ) is universally negligible. (iii) Let X be a set and Σ a σ-algebra of subsets of X containing all countable subsets of X. Show that the set of those A ⊆ X such that (A, ΣA ) is universally negligible is a σ-ideal of subsets of X.

439 Notes

Examples

269

(c) Let X be an analytic space and A an analytic subset of X. Show that X \ A is universally negligible iff all the constituents of X \ A (for any Souslin scheme defining A) are countable. (d) (i) Let X be a metrizable space such that f [X] is Lebesgue negligible for every continuous function f : X → R. Show that X is universally negligible. (ii) Let X be a completely regular Hausdorff space such that f [X] is Lebesgue negligible for every continuous function f : X → R. Show that X is universally τ -negligible. (e) Let Tc be the topology on {0, 1}N constructed in the proof of 439K. (i) Show that it is normal. (ii) Show that any Tc -zero set is an S-Borel set. (f ) Show that there is no atomless Borel probability measure on ω1 endowed with its order topology. (Hint: 411R, 439N.) (g) Show that the Sorgenfrey right-facing topology on R is hereditarily Lindelöf, but that its square is not Lindelöf. (h) Show that if ω1 is given its order topology, and f : ω12 → R is continuous, then there is a ζ < ω1 such that f is constant on (ω1 \ ζ)2 . (Hint: 4A2S(b-iii).) Show that if µ and ν are Baire probability measures on ω1 , then the Baire probability measures µ × ν, ν × µ on ω12 defined by the formulae of 436F coincide. (i) Let X ⊆ [0, 1] be a dense set with no uncountable compact subset. Show that X is not a Prokhorov space. 439 Notes and comments I give three separate constructions in 439A because the phenomenon here is particularly important. For two chapters I have, piecemeal, been offering theorems on the extension of measures. The principal ones to date seem to be 413O, 415L, 416M, 417C, 417E and 435C, and I have used methods reflecting my belief that the essential feature on which each such theorem depends is inner regularity of an appropriate kind. I think we should simultaneously seek to develop an intuition for measures which do not extend, and those in 439A are especially significant because they refer to the Borel algebra of the unit interval, which in so many other contexts is comfortably clear of the obstacles which beset more exotic structures. Outside the context of Polish spaces, the terms ‘universally measurable’ and ‘universally negligible’ are not properly settled. I have tried to select definitions which lead to a reasonable pattern. At least a universally negligible subset of a Hausdorff space is universally measurable (439Xd), and both concepts can be expressed in terms of sets with σ-algebras, as in 439Yb. It is important to notice, in 439B, that I write ‘µ{x} = 0 for every x ∈ X’, not ‘µ is atomless’. For instance, Dieudonné’s measure shows that ω1 , with its order topology, is not universally negligible on the definition here; but it is easy to show that there is no atomless Borel probability measure on ω1 (439Yf). In many cases, of course, we do not need to make this distinction (439D). The cardinal κ of 439F (the ‘uniformity’ of the Lebesgue null ideal) is one of a large family of cardinals which will be examined in Chapter 52 in the next volume. In some of the arguments above (439J, 439L, 439O) I appeal to (different) principles (‘there is a cardinal which is not measure-free’, ♣) which are not theorems according to the rules I follow in this book. Such examples would in some ways fit better into Volume 5, where I mean to investigate such principles properly. I include the examples here because they do at least exhibit bounds on what can be proved in ZFC. I should not want anyone to waste her time trying to show, for instance, that all completion regular Borel measures are τ -additive. Nevertheless, the absence of a ‘real’ counter-example (obviously we want a probability measure on a completely regular Hausdorff space) remains in my view a significant gap. It remains conceivable that there is a mathematical world in which no such space exists. Clearly the discovery of such a world is likely to require familiarity with the many worlds already known, and I am not going to embark on any such exploration in this volume. On the other hand, it is also very possible that all we need is a bit of extra ingenuity to construct a counter-example in ZFC. In this section we have two examples of successes of this kind. In 439F-439H, for instance, we have results which were long known as consequences of the continuum hypothesis; the particular insight of Grzegorek 81 was the observation that they depended on

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determinate properties of the cardinal κ of 439F, and that its indeterminate position between ω1 and c was ´sz Kunen & Rudin 76, where a similar unimportant. In 439K I show how a re-working of ideas in Juha example was constructed (for an entirely different purpose) assuming the continuum hypothesis, provides us with an interesting space (a first-countable non-Radon compact Hausdorff space) in ZFC. Let me emphasize that these ideas were originally set out in a framework supported by an extra axiom, where some technical details were easier and the prize aimed at (a non-Lindelöf hereditarily separable space) more important. The examples in 439K-439R are mostly based on constructions more or less familiar from general topology. I have already mentioned the origins of 439K. 439M is related to the Tychonoff and Dieudonné planks (Steen & Seebach 78, §§86-89). 439N and 439Q revisit yet again ω1 and the Sorgenfrey line. 439O is adapted from one of the standard constructions of Dowker spaces. Products of disjoint stationary sets (439R) have also been used elsewhere.

441A

Invariant measures on locally compact spaces

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Chapter 44 Topological groups Measure theory begins on the real line, which is of course a group; and one of the most fundamental properties of Lebesgue measure is its translation-invariance (134A). Later we come to the standard measure on the unit circle (255M), and counting measure on the integers is also translation-invariant, if we care to notice; moreover, Fourier series and transforms clearly depend utterly on the fact that shift operators don’t disturb the measure-theoretic structures we are building. Yet another example appears in the usual measure on {0, 1}I , which is translation-invariant if we identify {0, 1}I with the group ZI2 (345Ab). Each of these examples is special in many other ways. But it turns out that a particular combination of properties which they share, all being locally compact Hausdorff spaces with group operations for which multiplication and inversion are continuous, is the basis of an extraordinarily powerful theory of invariant measures. As usual, I have no choice but to move rather briskly through a wealth of ideas. The first step is to set out a suitably general existence theorem, assuring us that every locally compact Hausdorff topological group has non-trivial invariant Radon measures, that is, ‘Haar measures’ (441E). As remarkable as the existence of Haar measures is their (essential) uniqueness (442B); the algebra, topology and measure theory of a topological group are linked in so many ways that they form a peculiarly solid structure. I have already mentioned that Fourier analysis depends on the translation-invariance of Lebesgue measure. It turns out that substantial parts of the abstract theory of Fourier series and transforms can be generalized to arbitrary locally compact groups. In particular, convolutions (§255) appear again, even in non-commutative groups (§444). But for the main part of the theory, a transform relating functions on a group X to functions on its ‘dual’ group X , we do need the group to be abelian. Actually I give only the foundation of this theory: if X is an abelian locally compact Hausdorff group, it is the dual of its dual (445U). (In ‘ordinary’ Fourier theory, where we are dealing with the cases X = X = R and X = S 1 , X = Z, this duality is so straightforward that one hardly notices it.) But on the way to the duality theorem we necessarily see many of the themes of Chapter 28 in more abstract guises. A further remarkable fact is that any Haar measure has a translation-invariant lifting (447J). The proof demands a union between the ideas of the ordinary Lifting Theorem (§341) and some of the elaborate structure theory which has been developed for locally compact groups (§446). For the last two sections of the chapter, I look at groups which are not locally compact. For the most important such groups, the Polish groups, we have a natural necessary and sufficient condition for the existence of a translation-invariant measure (448P); as in the locally compact case, this is best expressed as a theorem about measures invariant under group actions. (In fact it is the case that any topological group carrying a non-trivial translation-invariant quasi-Radon measure is very nearly locally compact (443L).) In a slightly different direction, we can look at those groups, the ‘amenable’ groups, for which all actions (on compact Hausdorff spaces) have invariant measures. This again leads to some very remarkable ideas, which I sketch in §449.

441 Invariant measures on locally compact spaces I begin this chapter with the most important theorem on the existence of invariant measures: every locally compact Hausdorff group has left and right Haar measures (441E). I derive this as a corollary of a general result concerning invariant measures on locally compact spaces (441C), which has other interesting consequences (441H). 441A Group actions I recall a fundamental definition from group theory. (a) If G is a group and X is a set, an action of G on X is a function (a, x) 7→ a•x : G × X → X such that (ab)•x = a•(b•x) for all a, b ∈ G, x ∈ X, e•x = x for every x ∈ X where e is the identity of G (4A5B). In this context I write

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a•A = {a•x : x ∈ A} for a ∈ G, A ⊆ X. (b) If a group G acts on a set X, a measure µ on X is G-invariant if µ(a−1 •E) is defined and equal to µE whenever a ∈ G and µ measures E. (Of course this is the same thing as saying that µ(a•E) = µE for every a ∈ G and measurable set E; I use the formula with a−1 so as to match my standard practice when a is actually a function from X to X.) (c) It will be convenient if I immediately introduce the following notation. Let • be an action of a group G on a set X. If f is any function defined on a subset of X, and a ∈ G, write a•f for the function defined by saying that (a•f )(x) = f (a−1 •x) whenever x ∈ X and a−1 •x ∈ dom f . It is easy to check that this defines an action of G on the class of all functions with domains included in X. If X = G, we have left, right and conjgacy actions •l , •r and •c of G on itself (4A5Ca). So we have corresponding actions •l , •r and •c of G on the class of functions with domains included in G; (a•l f )(x) = f (a−1 x),

(a•r f )(x) = f (xa),

(a•c f )(x) = f (a−1 xa)

whenever these are defined. (d) If • is an action of a group G on a set X, and Y ⊆ X is G-invariant in the sense that a•y ∈ Y whenever a ∈ G and y ∈ Y , then •¹G × Y is an action of G on Y . In the context of (c) above, this means that if V is any set of functions with domains included in X such that a•f ∈ V whenever a ∈ G and f ∈ V , then we have an action of G on V . 441B It will be useful later to be able to quote the following elementary results. Lemma Let X be a topological space, G a group, and • an action of G on X such that x 7→ a•x is continuous for every a ∈ G. (a) If µ is a quasi-Radon measure on X such that µ(a•U ) ≤ µU for every open set U ⊆ X and every a ∈ G, then µ is G-invariant. (b) If µ is a Radon measure on X such that µ(a•K) ≤ µK for every compact set K ⊆ X and every a ∈ G, then µ is G-invariant. proof Note first that the maps x 7→ a•x are actually homeomorphisms (4A5Bd), so that a•U and a•K will be open, or compact, as U and K are. Next, the inequality ≤ in the hypotheses is an insignificant refinement; since we must also have µ(a•U ) ≤ µ(a−1 •a•U ) = µU in (a), µ(a•K) ≤ µ(a−1 •a•K) = µK in (b), we always have equality here. Now fix a ∈ G, and set Ta (x) = a•x for x ∈ X. Then Ta is a homeomorphism, so the image measure µTa−1 will be quasi-Radon, or Radon, if µ is. In (a), we are told that µTa−1 agrees with µ on the open sets, while in (b) we are told that they agree on the compact sets; so in both cases we have µ = µTa−1 , by 415H(iii) or 416E(b-ii). Consequently we have µTa−1 [E] = µE whenever µ measures E. As a is arbitrary, µ is G-invariant. 441C Theorem (Steinlage 75) Let X be a non-empty locally compact Hausdorff space and G a group acting on X. Suppose that (i) x 7→ a•x is continuous for every a ∈ G; (ii) every orbit {a•x : a ∈ G} is dense; (iii) whenever K and L are disjoint compact subsets of X there is a non-empty open subset U of X such that, for every a ∈ G, at most one of K, L meets a•U . Then there is a non-zero G-invariant Radon measure µ on X.

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S proof (a) a∈G a•U = X for every non-empty open U ⊆ X. P P If x ∈ X, then the orbit of x must meet U , so there is a a ∈ G such that a•x ∈ U ; but this means that x ∈ a−1 •U . Q Q Fix some point z0 of X and write V for the set of open sets containing z0 . Then if K, L are disjoint compact subsets of X there is a U ∈ V such that, for every a ∈ G, at most one of K, L meets a•U . P P By hypothesis, there is a non-empty open set V such that, for every a ∈ G, at most one of K, L meets a•V . Now there is an b ∈ G such that b•z0 ∈ V ; set U = b−1 •V . Because b−1 acts on X as a homeomorphism, U ∈ V; and if a ∈ G, then a•U = (ab−1 )•V can meet at most one of K and L. Q Q (b) If U ∈ V and A ⊆ X is any relatively compact set, then {a•U : a ∈ G} is an open cover S S of X, so there is a finite set I ⊆ G such that A ⊆ a∈I a•U . Write dA : U e for min{#(I) : I ⊆ G, A ⊆ a∈I a•U }. (c) The following facts are now elementary. (i) If U ∈ V and A, B ⊆ X are relatively compact, then 0 ≤ dA : U e ≤ dA ∪ B : U e ≤ dA : U e + dB : U e, and dA : U e = 0 iff A = ∅. (ii) If U , V ∈ V and V is relatively compact, and A ⊆ X also is relatively compact, then P P If A ⊆

S a∈I

a•V and V ⊆

S

dA : U e ≤ dA : V e dV : U e. S • • Q b∈J b U , then A ⊆ a∈I,b∈J (ab) U . Q

(iii) is relatively compact and b ∈ G, then db•A : U e = dA : U e. P P If I ⊆ G and S If U ∈ V, A ⊆ X S • • A ⊆ a∈I a U , then b A ⊆ a∈I (ba)•U , so db•A : U e ≤ #(I); as I is arbitrary, db•A : U e ≤ dA : U e. On the other hand, the same argument shows that dA : U e ≤ db−1 •b•A : U e = dA : U e, so we must have equality. Q Q (d) Fix a relatively compact V0 ∈ V. (This is the first place where we use the hypothesis that X is locally compact.) For every U ∈ V and every relatively compact set A ⊆ X write λU A =

dA : U e . dV0 : U e

Then (c) tells us immediately that (i) if A, B ⊆ X are relatively compact, 0 ≤ λU A ≤ λU (A ∪ B) ≤ λU A + λU B; (ii) λU A ≤ dA : V0 e for every relatively compact A ⊆ X; (iii) λU (b•A) = λU A for every relatively compact A ⊆ X and every b ∈ G; (iv) λU V0 = 1. (e) Now for the point of the hypothesis (ii) of the theorem. If K, L are disjoint compact subsets of X, there is a V ∈ V such that λU (K ∪ L) = λU K + λU L whenever U ∈ V and U ⊆ V . P P By (a), there is a V ∈ V such that anyStranslate a•V can meet at most one of K and L. Take any U ∈ V included in V . Let I ⊆ G be such that a∈I a•U ⊇ K ∪ L and #(I) = dK ∪ L : U e. Then are disjoint. K ⊆

S

I 0 = {a : a ∈ I, K ∩ a•U 6= ∅}, a∈I 0

I 00 = {a : a ∈ I, L ∩ a•U 6= ∅}

a•U , so dK : U e ≤ #(I 0 ), and similarly dL : U e ≤ #(I 00 ). But this means that dK : U e + dL : U e ≤ #(I 0 ) + #(I 00 ) ≤ #(I) = dK ∪ L : U e, λU K + λU L ≤ λU (K ∪ L).

As we already know that λU (K ∪ L) ≤ λU K + λU L, we must have equality, as claimed. Q Q (f ) Now let F be an ultrafilter on V containing all sets of the form {U : U ∈ V, U ⊆ V } for V ∈ V. If A ⊆ X is relatively compact, 0 ≤ λU A ≤ dA : V0 e for every U ∈ V, so λA = limU →F λU A is defined in [0, dA : U0 e]. From (d-i) and (d-iii) we see that

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0 ≤ λ(b•A) = λA ≤ λ(A ∪ B) ≤ λA + λB for all relatively compact A, B ⊆ X and b ∈ G. From (d-iv) we see that λV0 = 1. Moreover, from (e) we see that if K, L ⊆ X are disjoint compact sets, {U : U ∈ V, λU (K ∪ L) = λU K + λU L} ∈ F, so λ(K ∪ L) = λK + λL. (g) By 416L, there is a Radon measure µ on X such that µK = inf{λL : L ⊆ X is compact, K ⊆ int L} for every compact set K ⊆ X. Now µ is G-invariant. P P Take b ∈ G. If K, L ⊆ X are compact and K ⊆ int L, then b•K ⊆ int b•L, because x 7→ b•x is a homeomorphism; so µ(b•K) ≤ λ(b•L) = λL. As L is arbitrary, µ(b•K) ≤ µK. As K is arbitrary, µ is G-invariant, by 441Bb. Q Q (h) Finally, µV0 ≥ λV0 ≥ 1, so µ is non-zero. 441D The hypotheses of 441C are deliberately drawn as widely as possible. The principal application is the one for which the chapter is named. Definition If G is a topological group, a left Haar measure on G is a non-zero quasi-Radon measure µ on G which is invariant for the left action of G on itself, that is, µ(aE) = µE whenever µ measures E and a ∈ G. Similarly, a right Haar measure is a non-zero quasi-Radon measure µ such that µ(Ea) = µE for every E ∈ dom µ, a ∈ G. (My reasons for requiring ‘quasi-Radon’ here will appear in §§442 and 443.) 441E Theorem A locally compact Hausdorff topological group has left and right Haar measures, which are both Radon measures. proof Both the left and right actions of G on itself satisfy the conditions of 441C. P P In both cases, condition (i) is just the (separate) continuity of multiplication, and (ii) is trivial, as every orbit is the whole of G. As for (iii), let us take the left action first. Given disjoint compact subsets K, L of G, then M = {y −1 z : y ∈ K, z ∈ L} is a compact subset of G not containing the identity e. Because the topology is Hausdorff, M is closed and X \ M is a neighbourhood of e. Because multiplication and inversion are continuous, there are open neighbourhoods V , V 0 of e such that uv −1 ∈ G \ M whenever u ∈ V and v ∈ V 0 . Set U = V ∩ V 0 ; then U is a non-empty open set in G. ?? Suppose, if possible, that there is a a ∈ G such that aU meets both K and L. Take y ∈ K ∩ aU and z ∈ L ∩ aU . Then a−1 y ∈ U ⊆ V and a−1 z ∈ U ⊆ V 0 , so y −1 z = (a−1 y)−1 a−1 z ∈ G \ M ; but also y −1 z ∈ M . X X Thus aU meets at most one of K, L for any a ∈ G. As K and L are arbitrary, condition (ii) of 441C is satisfied. For the right action, we use the same ideas, but vary the formulae. Set M = {yz −1 : y ∈ K, z ∈ L}, and choose V and V 0 such that uv −1 ∈ X \ M for u ∈ V , v ∈ V 0 . Then if a ∈ G, y ∈ K and z ∈ L, za(ya)−1 ∈ M and one of za, ya does not belong to U , that is, one of z, y does not belong to U a−1 = a•U . Q Q Then 441C provides us with non-zero left and right Haar measures on G, and also tells us that they are Radon measures. 441F A different type of example is provided by locally compact metric spaces. Definition If (X, ρ) is any metric space, its isometry group is the set of bijections g : X → X which are isometries, that is, ρ(g(x), g(y)) = ρ(x, y) for all x, y ∈ X.

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441G The topology of an isometry group Let (X, ρ) be a metric space and G the isometry group of X. (a) Give G the topology of pointwise convergence inherited from the product topology of X X . Then G is a Hausdorff topological group and the action of G on X is continuous. P P If x ∈ X, g0 , h0 ∈ G and ² > 0, then V = {g : ρ(gh0 (x), g0 h0 (x)) ≤ 21 ²} is a neighbourhood of g0 and V 0 = {h : ρ(h(x), h0 (x)) ≤ 21 ²} is a neighbourhood of h0 . If g ∈ V and h ∈ V 0 then 1 2

ρ(gh(x), g0 h0 (x)) ≤ ρ(gh(x), gh0 (x)) + ρ(gh0 (x), g0 h0 (x)) ≤ ρ(h(x), h0 (x) + ² ≤ ². As g0 , h0 and ² are arbitrary, the function (g, h) 7→ gh(x) is continuous; as x is arbitrary, multiplication on G is continuous. As for inversion, suppose that g0 ∈ G, ² > 0 and x ∈ X. Then V = {g : ρ(gg0−1 (x), x) ≤ ²} is a neighbourhood of g0 , and if g ∈ V then ρ(g −1 (x), g0−1 (x)) = ρ(x, gg0−1 (x)) ≤ ². Because g0 , ² and x are arbitrary, inversion on G is continuous, and G is a topological group. Because X is Hausdorff, so is G. To see that the action is continuous, take g0 ∈ G, x0 ∈ X and ² > 0. Then V = {g : g ∈ G, ρ(g(x0 ), g0 (x0 )) < 21 ²} is a neighbourhood of g0 . If g ∈ V and x ∈ U (x0 , 12 ²), then 1 2

ρ(g(x), g0 (x0 )) ≤ ρ(g(x), g(x0 )) + ρ(g(x0 ), g0 (x0 )) ≤ ρ(x, x0 ) + ² ≤ ². As g0 , x0 and ² are arbitrary, (g, x) 7→ g(x) : G × X → X is continuous. Q Q (b) If X is compact, so is G. P P By Tychonoff’s theorem (3A3J), X X is compact. Suppose that g ∈ X X X belongs to the closure of G in X . For any x, y ∈ X, the set {f : f ∈ X X , ρ(f (x), f (y)) = ρ(x, y) is closed and includes G, so contains g; thus g is an isometry. ?? If g[X] 6= X, take x ∈ X \ g[X] and set xn = g n (x) for every n ∈ N. Because g is continuous and X is compact, g[X] is closed and there is some δ > 0 such that U (x, δ) ∩ g[X] = ∅. But this means that ρ(xm , xn ) = ρ(g m (x), g m (xn−m )) = ρ(x, xn−m ) ≥ δ whenever m < n, so that hxn in∈N can have no cluster point in X; which is impossible, because X is supposed to be compact. X X This shows that g is surjective and belongs to G. As g is arbitrary, G is closed in X X , therefore compact. Q Q 441H Theorem If (X, ρ) is a non-empty locally compact metric space with isometry group G, then there is a non-zero G-invariant Radon measure on X. proof (a) Fix any x0 ∈ X, and set Z = {g(x0 ) : g ∈ G}; then Z is a closed subset of X, so is in itself locally compact. Let H be the isometry group of Z. (b) We need to know that g¹Z ∈ H for every g ∈ G. P P Because g : X → X is a homeomorphism, g[Z] = {gg 0 (x0 ) : g 0 ∈ G} = Z, so g¹Z is a bijection from Z to itself, and of course it is an isometry, that is, belongs to H. Q Q (c) Now Z and H satisfy the conditions of 441C. P P(i) is true just because all isometries are continuous. (ii) Take z ∈ Z and let U be a non-empty relatively open subset of Z. Then U = Z ∩ V for some open set V ⊆ X; as Z ∩ V 6= ∅, there must be a g0 ∈ G such that g0 (x0 ) ∈ V . At the same time, there is a sequence hhn in∈N in G such that z = limn→∞ hn (x0 ). Now ρ(g0 (x0 ), g0 h−1 n (z)) = ρ(hn (x0 ), z) → 0, −1 −1 so there is some n such that g0 h−1 n (z) ∈ V ; of course g0 hn (z) also belongs to U , while g0 hn ¹Z belongs to H, by (b) above. As U is arbitrary, the H-orbit of z is dense in Z; as z is arbitrary, H satisfies condition (ii) of 441C.

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Topological groups

441H

(iii) Given that K and L are disjoint compact subsets of Z, there must be a δ > 0 such that ρ(y, z) ≥ δ for every y ∈ K, z ∈ L. Let U be the relatively open ball {z : z ∈ Z, ρ(z, x0 ) < 12 δ}. Then for any h ∈ H, ρ(y, z) < δ for any y, z ∈ b•U = h[U ], so h[U ] cannot meet both K and L. Q Q (d) 441C therefore provides us with a non-zero H-invariant Radon measure ν on Z. Setting µE = ν(E∩Z) whenever E ⊆ X and E ∩ Z ∈ dom ν, it is easy to check that µ is G-invariant (using (b) again) and is a Radon measure on Z. 441I Remarks (a) Of course there is a degree of overlap between the cases above. In an abelian group, for instance, the left and right group actions necessarily give rise to the same invariant measures. If we take X = R 2 , it has a group structure (addition) for which we have invariant measures (e.g., Lebesgue measure); these are just the translation-invariant measures. But 441H tells us that we also have measures which are invariant under all isometries (rotations and reflections as well as translations); from where we now stand, there is no surprise remaining in the fact that Lebesgue measure is invariant under this much larger group. (Though if you look back at Chapter 26, you will see that a bare-handed proof of this takes a certain amount of effort.) If we turn next to the unit sphere {x : kxk = 1} in R 3 , we find that there is no useful group structure, but it is a compact metric space, so carries invariant measures, e.g., two-dimensional Hausdorff measure. (b) The arguments of 441C leave open the question of how far the invariant measures constructed there are unique. Of course any scalar multiple of an invariant measure will again be invariant. It is natural to give a special place to invariant probability measures, and call them ‘normalized’; whenever we have a non-zero totally finite invariant measure we shall have an invariant probability measure. Counting measure on any set will be invariant under any action of any group, and it is natural to say that these measures also are ‘normalized’; when faced with a finite set with two or more elements, we have to choose which normalization seems most reasonable in the context. (c) We shall see in 442B that Haar measures (with a given handedness) are necessarily scalar multiples of each other. If, in 441C, there are non-trivial G-invariant subsets of X, we do not expect such a result. But there are interesting cases in which the question seems to be open. 441J Of course we shall be much concerned with integration with respect to invariant measures. The results we need are elementary corollaries of theorems already dealt with at length, but it will be useful to have them spelt out. Proposition Let X be a set, G a group acting on X,R and µ a G-invariant measure on X. If f is a real-valued R function defined on a subset of X, and g ∈ G, then f (x)µ(dx) = f (a•x)µ(dx) if either integral is defined in [−∞, ∞]. proof Apply 235I to the inverse-measure-preserving functions x 7→ a•x and x 7→ a−1 •x. 441K Theorem Let X be a set, G a group acting on X, and µ a G-invariant measure on X with measure algebra A. (a) We have an action of G on A defined by setting a•E • = (a•E)• whenever a ∈ G and µ measures E. (b) We have an action of G on L0 = L0 (µ) defined by setting a•f • = (a•f )• for every a ∈ G, f ∈ L0 (µ). (c) For 1 ≤ p ≤ ∞ the formula of (b) defines actions of G on Lp = Lp (µ), and ka•ukp = kukp for every u ∈ Lp , a ∈ G. proof (a) If E, F ∈ dom µ and E • = F • , then (because x 7→ a−1 •x is inverse-measure-preserving) (a•E)• = (a•F )• . So the given formula does define a function from G × A to A. It is now easy to check that it is an action. (b) Let f ∈ L0 = L0 (µ), a ∈ G. Set φa (x) = a−1 •x for x ∈ X, so that φa : X → X is inversemeasure-preserving. Then a•f = f φa belongs to L0 . If f , g ∈ L0 and f =a.e. g, then f φa =a.e. gφa , so (a•f )• = (a•g)• . This shows that the given formula defines a function from G × L0 to L0 , and again it is easy to see that it is an action.

441Xe


(c) If f ∈ Lp (µ) then

R

|a•f |p dµ =

R

|f (a−1 •x)|p dµ =

R

277

|f |p dµ

by 441J. So a•f ∈ Lp (µ) and ka•f • kp = kf • kp . Thus we have a function from G × Lp to Lp , and once more it must be an action. 441L Proposition Let X be a locally compact Hausdorff space and G a group acting on X in such a wayR that x 7→ a•xR is continuous for every a ∈ G. If µ is a Radon measure on X, then µ is G-invariant iff f (x)µ(dx) = f (a•x)µ(dx) for every a ∈ G and every continuous function f : X → R with compact support. proof For a ∈ G, set Ta (x) = a•x for every x ∈ X. Then νa = µTa−1 is a Radon measure on X. If f ∈ Ck (X), then

R

f dνa =

R

f Ta dµ

by 235I. Now

µ is G-invariant ⇐⇒ νa = µ for every a ∈ G Z Z ⇐⇒ f dνa = f dµ for every a ∈ G, f ∈ Ck (X) (416E(b-v))

Z ⇐⇒

Z f Ta dµ =

f dµ for every a ∈ G, f ∈ Ck (X)

as claimed. 441X Basic exercises > (a) Let X be a set. (i) Show that there is a one-to-one correspondence between actions • of the group Z on X and bijections f : X → X defined by the formula n•x = f n (x). (ii) Show that if f : X → X is a bijection, a measure µ on X is Z-invariant for the corresponding action iff f and f −1 are both inverse-measure-preserving. (iii) Show that if X is a compact Hausdorff space and • is a continuous action of Z on X, then there is a Z-invariant Radon probability measure on X. (Hint: 437S.) (b) Let (X, T, ν) be a measure space and G a group acting on X. Set Σ = {E : E ⊆ X, g •E ∈ T for every g ∈ G}, and for E ∈ Σ set n X

µE = sup{

ν(gi •Fi ) : n ∈ N, F0 , . . . , Fn are disjoint subsets of E

i=0

belonging to Σ, gi ∈ G for each i ≤ n} (cf. 112Ye). Show that µ is a G-invariant measure on X. (c) Let X be a topological space and G a group acting on X such that (α) all the maps x 7→ a•x are continuous (β) all the orbits of G are dense. Show that any non-zero G-invariant quasi-Radon measure on X is strictly positive. > (d) Let G be a compact Hausdorff topological group. (i) Show that its conjugacy classes are closed. (ii) Show that if K, L ⊆ G are disjoint compact sets then {ac−1 da−1 : a ∈ G, c ∈ K, d ∈ L} is a compact set not containing e, so that there is a neighbourhood U of e such that whenever c−1 d ∈ U and a ∈ G then either aca−1 ∈ / K or ada−1 ∈ / L. (iii) Show that every conjugacy class of G carries a Radon probability measure which is invariant under the conjugacy action of G (definition: 4A5Ca). (e) Let (G, ·) be a topological group. (i) On G define a binary operation ¦ by saying that x ¦ y = y · x for all x, y ∈ G. Show that (G, ¦) is a topological group isomorphic to (G, ·), and that any element of G has the same inverse for either group operation. (ii) Suppose that µ is a left Haar measure on (G, ·). Show

278

Topological groups

441Xe

that µ is a right Haar measure on (G, ¦). (iii) Set φ(a) = a−1 for a ∈ G. Show that µφ−1 is a right Haar measure on (G, ·). (iv) Show that (G, ·) has a left Haar measure iff it has a right Haar measure. (v) Show that (G, ·) has a left Haar probability measure iff it has a totally finite left Haar measure iff it has a right Haar probability measure. (iv) Show that (G, ·) has a σ-finite left Haar measure iff it has a σ-finite right Haar measure. R 1 > (f ) (i) For Lebesgue measurable E ⊆ R \ {0}, set νE = E |x| dx. Show that ν is a (two-sided) Haar measure if R \ {0} is given the group operation of multiplication. (ii) For Lebesgue measurable E ⊆ C \ {0}, R identified with R2 \ {0}, set νE = E |z|1 2 µ(dz), where µ is two-dimensional Lebesgue measure. Show that ν is a (two-sided) Haar measure on C \ {0} if we take complex multiplication for the group operation. (Hint: 263D.) > (g) (i) Show that Lebesgue measure is a (two-sided) Haar measure on R r , for any r ≥ 1, if we take addition for the group operation. (ii) Show that the usual measure on {0, 1}I is a two-sided Haar measure on {0, 1}I , for any set I, if we give {0, 1}I the group operation corresponding to its identification with ZI2 . (iii) Describe the corresponding Haar measure on PI when PI is given the group operation 4. (h) Let G be a locally compact Hausdorff topological group. (i) Show that any (left) Haar measure on G must be strictly positive. (ii) Show that G has a totally finite (left) Haar measure iff it is compact. (iii) Show that G has a σ-finite (left) Haar measure iff it is σ-compact. > (i) (i) Let G and H be topological groups with left Haar measures µ and ν. Show that the quasi-Radon product measure on G×H (417N) is a left Haar measure on G×H. (ii) Let hGi ii∈I be a family of topological groups, and suppose that each Gi has a left Haar probability measure (as happens, for instance, if each Gi Q G is compact). Show that the quasi-Radon product measure on i (417O) is a left Haar measure on i∈I Q i∈I Gi . (j) (i) Show that any (left) Haar measure on a topological group, as defined in 441D, must be locally finite. (ii) Show that any (left) Haar measure on a locally compact Hausdorff group must be a Radon measure. (k) Let r ≥ 1 be an integer, and set X = {x : x ∈ Rr , kxk = 1}. Let G be the group of orthogonal r × r real matrices, so that G acts transitively on X. Show that (when given its natural topology as a subset of 2 R r ) G is a compact Hausdorff topological group. Let µ be a left Haar measure on G, and x any point of X; set φx (T ) = T x for T ∈ G. Show that µφ−1 x is a G-invariant measure on X, independent of the choice of x. (l) Let X be a compact metric space, and g : X → X any isometry. Show that g is surjective. (Hint: if x ∈ X, then ρ(g m x, g n x) ≥ ρ(x, g[X]) for any m < n.) > (m) Let (X, ρ) be a metric space, and G its isometry group with the topology T of pointwise convergence. (i) Show that if X is compact, T can be defined by the metric (g, h) 7→ maxx∈X ρ(g(x), h(x)). (ii) Show that if {y : ρ(y, x) ≤ γ} is compact for every x ∈ X and γ > 0, then G is locally compact. (iii) Show that if X is separable then G is metrizable. (iv) Show that if X is separable and (X, ρ) is complete then G is Polish. (Hint: if D ⊆ X is a countable dense set, then f 7→ (f ¹D, f −1 ¹D) is a homeomorphism between G and a closed subset of DX × DX .) (n) Give N the metric ρ which takes only the values 0 and 1. Let G be the isometry group of N (that is, the group of all permutations of N) with its pointwise topology. (i) Show that G is a Gδ subset of N N , so is a Polish group. (ii) Show that if we set ∆(g, h) = min{n : n ∈ N, g(n) 6= h(n)} and σ(g, h) = 1/(1 + ∆(g −1 , h−1 )) for distinct g, h ∈ G, then σ is a right-translation-invariant metric on G inducing its topology. (iii) Show that there is no complete right-translation-invariant metric on G inducing its topology. (Hint: any such metric must have the same Cauchy sequences as σ.) (iv) Show that G is not locally compact. > (o) Let r ≥ 1 be an integer, and Sr−1 the sphere {x : x ∈ R r , kxk = 1}. (i) Show that every isometry φ from Sr−1 to itself corresponds to an orthogonal r × r matrix T . (Hint: T = hφ(ei ) . ej ii,j
441Yj


279

(p) Let X be a locally compact metric space and G a subgroup of the isometry group of X. Show that for every x ∈ X there is a non-zero G-invariant Radon measure on {g(x) : g ∈ G}. 441Y Further exercises (a) Let (X, ρ) be a metric space, and C the family of closed subsets of X. For E, F ∈ C \ {∅}, set ρ˜(E, F ) = max(supx∈E ρ(x, F ), supy∈F ρ(y, E)), where ρ(x, F ) = inf y∈F ρ(x, y). (i) Show that ρ˜ is a metric on C \ {∅} (the Hausdorff metric). (ii) Show that if X is complete under ρ, C \ {∅} is complete under ρ˜. (iii) Show that if X is totally bounded under ρ, C \ {∅} is totally bounded under ρ˜. (iv) Show that if X is compact under ρ, C \ {∅} is compact under ρ˜. (v) Show that if G is the isometry group of X, (a, F ) 7→ g[F ] is an action of G on C. (b) Take 1 ≤ s ≤ r ∈ N, and in R r let X be {x : kxk = 1}. Let C be the set of closed subsets of X, and ρ˜ the Hausdorff metric on C \ {∅} defined from the usual metric of X, so that C \ {∅} is compact (441Ya). Let Cs ⊆ C be {X ∩ V : V ⊆ R r is an s-dimensional linear subspace}. Show that Cs is a closed subset of C \ {∅}, therefore a compact metric space in its own right, and that the isometry group G of X acts transitively on Cs , so that there is a G-invariant Radon probability measure on Cs . (c) Let (X, W) be a locally compact Hausdorff uniform space, and suppose that G is a group acting on X ‘uniformly equicontinuously’; that is, for every W ∈ W there is a V ∈ W such that (a•x, a•y) ∈ W whenever (x, y) ∈ V and a ∈ G. Show that there is a non-zero G-invariant Radon measure on X. (d) In 441Xk, show that µφ−1 x is a scalar multiple of Hausdorff (r − 1)-dimensional measure on X. (e) For any topological spaces X and Y , and any set G of functions from X to Y , the compact-open topology on G is the topology generated by sets of the form {g : g ∈ G, g[K] ⊆ H}, where K ⊆ X is compact and H ⊆ Y is open. Show that if (X, ρ) is a metric space and G is the isometry group of X, then the topology of pointwise convergence on G is its compact-open topology. (f ) Let X be a compact Hausdorff space and G the group of all homeomorphisms from X to itself. (i) Let P be the family of all continuous pseudometrics on X (see 4A2Jf). For ρ ∈ P and g, h ∈ G, set τρ (g, h) = maxx∈X ρ(g(x), h(x)). Show that every τρ is a right-translation-invariant pseudometric on G, and that G with the topology generated by {τρ : ρ ∈ P} is a topological group. (ii) Show that this is the compact-open topology as defined in 441Ye. (iii) Show that if X is metrizable then G is Polish. (g) Let X be a compact metric space and G the isometry group of X. Show that every orbit in X is closed. (h) For w, z ∈ C \ {0} set ρ(w, z) = | Ln( wz )|, where Ln v = ln |v| + i arg v for non-zero complex numbers (ρ) v. (i) Show that ρ is a metric on C \ {0}. (ii) Show that the 2-dimensional Hausdorff measure µH2 derived (ρ) from ρ (471A) is a Haar measure for the multiplicative group C \ {0}. (iii) Show that µH2 = π4 ν, where ν is the measure of 441Xf(ii). (Hint: 264I.) (i) Let X be the group of real r × r orthogonal matrices, where r ≥ 2 is an integer. Give X the 2 Euclidean metric, regarding it as a subset of R r . (i) Show that the left and right actions of X on itself -dimensional Hausdorff measure on X is a two-sided Haar are distance-preserving. (ii) Show that r(r−1) 2 measure. (j) Let T be any set, and ρ the {0, 1}-valued metric on T . Let X be the set of partial orders on T , regarded as a subset of P(T × T ). Show that X is compact (cf. 418Xw). Let G be the group of isometries of T with its topology of pointwise convergence. Set π •x = {(t, u) : (π −1 (t), π −1 (u)) ∈ x} for π ∈ G and x ∈ X. Show that • is a continuous action of G on X. Show that there is a strictly positive G-invariant Radon probability measure µ on X.

280

Topological groups

441Yk

(k) Let X be a set, G a group acting on X, and µ a G-invariant measure on X with measure algebra A. Show that if τ is any rearrangement-invariant extended Fatou norm on L0 (A) then the formula of 441Kb defines a norm-preserving action of G on the Banach space Lτ . 441 Notes and comments The proof I give of 441C is essentially the same as the proof of 441E in Halmos 50, §58. In part (f) of the proof of 441C I use an ultrafilter, relying on a fairly strong consequence of the Axiom of Choice. In this volume, as in the last, I generally employ the axiom of choice without stopping to consider whether it is really needed. But Haar measure, at least, is so important that I point out now that it can be built with much weaker principles. For a construction of Haar measure not dependent on anything more than countable choice, see Naimark 70, §27.4. I should remark that the argument there, while closely paralleling that of 441C in its ideas, relies for its conclusion on the results of §436 above, so there is a lot of checking to do to see that these also can be established with countable choice alone. Moreover, in the absence of the full axiom of choice, we may find that we have fewer locally compact topological groups than we expect. While Haar measure is surely the pre-eminent application of the theory here, I think that some of the other consequences of 441C (441H, 441Xd, 441Xk, 441Yb, 441Yc) are sufficiently striking to justify the trouble involved in the extra generality. I ought to remark that there are important examples of invariant measures which have nothing to do with 441C. Some of these will appear in §449; for the moment I note only 441Xa. Federer 69, §2.7, develops a general theory of ‘covariant’ measures µ (‘relatively invariant’ in Halmos 50) such that µ(a•E) = ψ(a)µE for appropriate sets E ⊆ X and a ∈ G, where ψ : G → ]0, ∞[ is a homomorphism; for instance, taking µ to be Lebesgue measure on R r , we have µT [E] = | det T |µE for every linear space isomorphism T : R r → R r and every measurable set E (263A). The theory I have described here can deal only with the subgroup of isometric linear isomorphisms (that is, the orthogonal group). Covariant measures arise naturally in many other contexts, such as 443S below. Hausdorff measures, being defined in terms of metrics, are necessarily invariant under isometries, so appear naturally in this context, starting with 264I. There are interesting challenges both in finding suitable metrics and in establishing exact constants, as in 441Yh-441Yi. It is worth pausing over the topology of an isometry group, as described in 441G. It is quite surprising that such an elementary idea should give us a topological group at all. I offer some exercises (441Xm-441Xo, 441Ye) to help you relate the construction to material which may be more familiar. It is of course a ‘weak’ topology, except when the underlying space is compact (441Xm(i)). See also 441Yf. These groups are rarely locally compact, and you may find them pushed out of your mind by the extraordinary theory which you will see developed in the next hundred pages; but in the last two sections of this chapter, and in §493, they will become leading examples.

442 Uniqueness of Haar measures Haar measure has an extraordinary wealth of special properties, and it will be impossible for me to cover them all properly in this chapter. But surely the second thing to take on board, after the existence of Haar measures on locally compact Hausdorff groups (441E), is the fact that they are, up to scalar multiples, unique. This is the content of 442B. We find also that while left and right Haar measures can be different (442Xf), they are not only direct mirror images of each other (442C) – as is, I suppose, to be expected – but even more closely related (442F, 442H, 442L). Investigating this relation, we are led naturally to the ‘modular function’ of a group (442I). 442A Lemma Let X be a topological group and µ a left Haar measure on X. (a) µ is strictly positive and locally finite. (b) If G ⊆ X is open and γ < µG, there are an open set H and an open neighbourhood U of the identity such that HU ⊆ G (writing HU for {xy : x ∈ H, y ∈ U }) and µH ≥ γ. (c) If X is locally compact and Hausdorff, µ is a Radon measure.

442B

Uniqueness of Haar measures

281

proof (a)(i) ?? If G ⊆ X were a non-empty open set such that µG = 0, then we should have µ(xG) = 0 for every x ∈ X, so that X would be covered by negligible open sets; but as µ is supposed to be τ -additive, µX = 0. X X (ii) Because µ is effectively locally finite, there is some non-empty open set G such that µG < ∞; but now {xG : x ∈ X} is a cover of X by open sets of finite measure. (b) Let U be the family of open sets containing the identity e, and H the family of open sets H such that HU ⊆ G S for some U ∈ U. Because U is downwards-directed, H is upwards-directed; because e ∈ U for every U ∈ U , H ⊆ G. If x S ∈ G, then x−1 G ∈ U , and there is a U ∈ U such that U U ⊆ x−1 G; but now xU U ⊆ G, so xU ∈ H. Thus H = G. Because µ is τ -additive, there is an H ∈ H such that µH ≥ γ. (c) Use (a) and 416G. 442B Theorem Let X be a topological group. If µ and ν are left Haar measures on X, they are multiples of each other. proof (a) Let G be the family of non-empty open sets G such that µG and νG are both finite; because µ and ν are locally finite (442Aa), G is a base for the topology of X. Note that G ∪ H ∈ G for all G, H ∈ G. Set U = {U : U ∈ G, U = U −1 , e ∈ U }, where e is the identity of G; then U is a base of neighbourhoods of e (4A5Ec). Let F be the filter on U generated by the sets {U : U ∈ U , U ⊆ V } as V runs over U. (b) (The key.) If G ∈ G and 0 < ² < 1, there is a V1 ∈ U such that (1 − ²)

µG νG

≤

µU νU

whenever U ∈ U and U ⊆ V1 . P P By 442Aa, µG and νG are both non-zero. By 442Ab, there are an open set H and a neighbourhood V1 of e such that HV1 ⊆ G and µH ≥ (1 − ²)µG; shrinking V1 if need be, we may suppose that V1 ∈ U . Take any U ∈ U such that U ⊆ V1 , so that HU ⊆ G. Consider the product quasi-Radon measure λ of µ and ν on X × X (417R), and the set W = {(x, y) : x, y ∈ G, x−1 y ∈ U }. Because the function (x, y) 7→ x−1 y is continuous (4A5Eb), W is open. Consequently

R

µ{x : (x, y) ∈ W }ν(dy) = λW =

R

ν{y : (x, y) ∈ W }µ(dx)

(417C(iv)). But we see that if x ∈ H then xU ⊆ G, so (x, y) ∈ W whenever y ∈ xU , and

R

ν{y : (x, y) ∈ W }µ(dx) ≥

R

H

ν(xU )µ(dx) = µH · νU .

On the other hand, if y ∈ X, then Z Z Z µ{x : (x, y) ∈ W }ν(dy) ≤ µ{x : x−1 y ∈ U }ν(dy) = µ{x : y −1 x ∈ U }ν(dy) G G Z = µ(yU )ν(dy) = µU · νG, G

so that (1 − ²)µG · νU ≤ µH · νU ≤ λW ≤ µU · νG. Dividing both sides by νU · νG, we have the result. Q Q (c) In the same way, there is a V2 ∈ V such that (1 − ²)

νG µG

≤

νU µU

whenever U ∈ U and U ⊆ V2 . So if U ∈ U and U ⊆ V1 ∩ V2 , we have ≤

µU νU

≤

1 µG . 1−² νG

limU →F

µU νU

=

µG . νG

(1 − ²)

µG νG

As ² is arbitrary,

282

Topological groups

442B

And this is true for every G ∈ G. (d) So if we set α = limU →F µU νU , we shall have µG = ανG for every G ∈ G. Now µ and αν are quasi-Radon measures agreeing on the base G, which is closed under finite unions, so are identical, by 415H(iv). 442C Proposition Let X be a topological group and µ a left Haar measure on X. Setting νE = µ(E −1 ) whenever E ⊆ X is such that E −1 = {x−1 : x ∈ E} is measured by µ, ν is a right Haar measure on X. proof Set φ(x) = x−1 for x ∈ X. Then φ is a homeomorphism, so the image measure ν = µφ−1 is a quasi-Radon measure. It is non-zero because νX = µX. If E ∈ dom ν and x ∈ X, then ν(Ex) = µ(x−1 E −1 ) = µE −1 = νE. So ν is a right Haar measure. 442D Remark Clearly all the arguments of 442A-442C must be applicable to right Haar measures; that is, any right Haar measure must be locally finite and strictly positive; two right Haar measures on the same group must be multiples of each other; and if X carries a right Haar measure ν then E 7→ νE −1 will be a left Haar measure on X. (If you are unhappy with such a bold appeal to the symmetry between ‘left’ and ‘right’ in topological groups, write the reflected version of 442C out in full, and use it to reflect 442A-442B.) Thus we may say that a topological group carries Haar measures if it has either a left or a right Haar measure. These can, of course, be the same; in fact it takes a certain amount of exploration to find a group in which they are different (e.g., 442Xf). 442E Lemma Let X be a topological group, µ a left Haar measure on X and ν a right Haar measure on X. If G, H ⊆ X are open, then µG · νH =

R

H

ν(xG−1 )µ(dx).

proof Let λ be the quasi-Radon product measure of µ and ν on X × X. The sets W1 = {(x, y) : y −1 x ∈ G, x ∈ H} and W2 = {(x, y) : x ∈ G, yx ∈ H} are both open, so 417C(iv) tells us that Z Z −1 ν(xG )µ(dx) = ν{y : x ∈ H, y −1 x ∈ G}µ(dx) = λW1 H Z Z = µ{x : x ∈ H, y −1 x ∈ G}ν(dy) = µ(H ∩ yG)ν(dy) Z Z −1 = µ(y H ∩ G)ν(dy) = µ{x : x ∈ G, yx ∈ H}ν(dy) Z = λW2 = ν{y : yx ∈ H}µ(dx) G Z = ν(Hx−1 )µ(dx) = µG · νH. G

442F Domains of Haar measures 442B tells us, in part, that any two left Haar measures on a topological group must have the same domain and the same negligible sets; similarly, any two right Haar measures have the same domain and the same negligible sets. In fact left and right Haar measures agree on both. Proposition Let X be a topological group which carries Haar measures. If µ is a left Haar measure and ν is a right Haar measure on X, then they have the same domains and the same null ideals. proof (a) Suppose that F ⊆ X is a closed set such that µF = 0. Then νF = 0. P P?? Otherwise, there is an open set H such that νH < ∞ and ν(F ∩ H) > 0. Let G be any open set such that 0 < µG < ∞. By 442E, µG · νH = But also

R

H

ν(xG−1 )µ(dx).

442Ia


µG · ν(H \ F ) =

R H\F

ν(xG−1 )µ(dx) =

R H

283

ν(xG−1 )µ(dx) = µG · νH,

so µG · ν(H ∩ F ) = 0. X XQ Q (b) It follows that νE = supF ⊆E

is closed

νF = 0

whenever E is a Borel set such that µE = 0. Now take any E ∈ dom µ. Set G = {G : G ⊆ X is open, µG < ∞, νG < ∞}. Because both µ and ν are locally finite, G covers X. If G ∈ G, there are Borel sets E 0 , E 00 such that E 0 ⊆ E ∩ G ⊆ E 00 and µ(E 00 \ E 0 ) = 0. In this case ν(E 00 \ E 0 ) = 0 so E ∩ G ∈ dom ν. Because ν is complete, locally determined and τ -additive, E ∈ dom ν (414I). If µE = 0, it follows that νE = supF ⊆E

is closed

νF = 0

just as above. (c) Thus ν measures E whenever µ measures E, and E is ν-negligible whenever it is µ-negligible. I am sure you will have no difficulty in believing that all the arguments above, in particular that of 442E, can be re-cast to show that dom ν ⊆ dom µ; alternatively, apply the result in the form just demonstrated to the left Haar measure ν 0 and the right Haar measure µ0 , where ν 0 E = νE −1 ,

µ0 E = µE −1

as in 442Cb. 442G Corollary Let X be a topological group and µ a left Haar measure on X with domain Σ. Then, for E ⊆ X and a ∈ X, E ∈ Σ ⇐⇒ E −1 ∈ Σ ⇐⇒ Ea ∈ Σ, µE = 0 ⇐⇒ µE −1 = 0 ⇐⇒ µ(Ea) = 0. proof Apply 442F with νE = µE −1 . 442H Remark From 442F-442G we see that if X is any topological group which carries Haar measures, there is a distinguished σ-algebra Σ of subsets of X, which we may call the algebra of Haar measurable sets, which is the domain of any Haar measure on X. Similarly, there is a σ-ideal N of PX, the ideal of Haar negligible sets, which is the null ideal for any Haar measure on X. Both Σ and N are translationinvariant and also invariant under the inversion operation x 7→ x−1 . Note that if A ⊆ X is such that A ∩ K ∈ Σ for every compact set K ⊆ X, then A ∈ Σ (412Ja); similarly, if A ⊆ X and A ∩ K ∈ N for every compact K ⊆ X, then A ∈ N (412Jb). If we form the quotient A = Σ/N , then we have a fixed Dedekind complete Boolean algebra which is the Haar measure algebra of the group X in the sense that any Haar measure on X, whether left or right, has measure algebra based on A. If a ∈ X, the maps x 7→ ax, x 7→ xa, x 7→ x−1 give rise to Boolean automorphisms of A. For a member of Σ, we have a notion of ‘σ-finite’ which is symmetric between left and right (442Xd). We do not in general have a corresponding two-sided notion of ‘finite measure’ (442Xg(i)); but of course we can if we wish speak of a set as having ‘finite left Haar measure’ or ‘finite right Haar measure’ without declaring which Haar measure we are thinking of. It is the case, however, that if the group X itself has finite left Haar measure, it also has finite right Haar measure; see 442Ic-d below. 442I The modular function Let X be a topological group which carries Haar measures. (a) There is a group homomorphism ∆ : X → ]0, ∞[ defined by the formula µ(Ex) = ∆(x)µE whenever µ is a left Haar measure on X and E ∈ dom µ. P P Fix on a left Haar measure µ ˜ on X. For x ∈ X, let µx be the function defined by saying

284

Topological groups

442Ia

µx E = µ ˜(Ex) whenever E ⊆ X, Ex ∈ dom µ, that is, for every Haar measurable set E ⊆ X. Because the function φx : X → X defined by setting φx (y) = yx−1 is a homeomorphism, µx = µ ˜φ−1 x is a quasi-Radon measure on X; and µx (yE) = µ ˜(yEx) = µ ˜(Ex) = µx E whenever µx measures E, so µx is a left Haar measure on X. By 442B, there is a ∆(x) ∈ ]0, ∞[ such that µx = ∆(x)˜ µ; because µ ˜ surely takes at least one value in ]0, ∞[, ∆(x) is uniquely defined. If µ is any other left Haar measure on X, then µ = α˜ µ for some α > 0, so that µ(Ex) = α˜ µ(Ex) = α∆(x)˜ µE = ∆(x)µE. Thus ∆ : X → ]0, ∞[ has the property asserted in the formula offered. To see that ∆ is a group homomorphism, take any x, y ∈ X and a Haar measurable set E such that 0<µ ˜E < ∞, and observe that ∆(xy)˜ µE = µ ˜(Exy) = ∆(y)˜ µ(Ex) = ∆(y)∆(x)˜ µE, so that ∆(xy) = ∆(y)∆(x) = ∆(x)∆(y). Q Q ∆ is called the left modular function of X. (b) We find now that ν(xE) = ∆(x−1 )νE whenever ν is a right Haar measure on X, x ∈ X and E ⊆ X is Haar measurable. P P Let µ be the left Haar measure derived from ν, so that µE = νE −1 whenever E is Haar measurable. If x ∈ X and E ∈ dom ν, then ν(xE) = µ(E −1 x−1 ) = ∆(x−1 )µE −1 = ∆(x−1 )νE. Q Q Thus we may call x 7→ ∆(x−1 ) =

1 ∆(x)

the right modular function of X.

(c) If X is abelian, then obviously ∆(x) = 1 for every x ∈ X, because µ(Ex) = µ(xE) = µE whenever x ∈ X, µ is a left Haar measure on X and E is Haar measurable. Equally, if any (therefore every) left (or right) Haar measure µ on X is totally finite, then µ(Xx) = µ(xX) = µX, so again ∆(x) = 1 for every x ∈ X. This will be the case, in particular, for any compact Hausdorff topological group (recall that by 441E any such group carries Haar measures), because its Haar measures are locally finite, therefore totally finite. Groups in which ∆(x) = 1 for every x are called unimodular. (d) From the definition of ∆, we see that a topological group carrying Haar measures is unimodular iff every left Haar measure is a right Haar measure. (e) In particular, if a group has any totally finite (left or right) Haar measure, its left and right Haar measures are the same, and it has a unique Haar probability measure, which we may call its normalized Haar measure. In the other direction, any group with the discrete topology is unimodular, since counting measure is a two-sided Haar measure. 442J Proposition For any topological group carrying Haar measures, its left modular function is continuous. proof Let X be a topological group carrying Haar measures, with left modular function ∆. (a) If ² > 0, there is an open set U² containing the identity e of X such that ∆(x) ≤ 1 + ² for every x ∈ U . P P Take any left Haar measure µ on X, and an open set G such that 0 < µG < ∞. By 442Ab, there are an open set H and a neighbourhood U² of the identity such that HU² ⊆ G and µG ≤ (1 + ²)µH. If x ∈ U² , then Hx ⊆ G, so ∆(x) =

µ(Hx) µH

≤ 1 + ². Q Q

(b) Now, given x0 ∈ X and ² > 0, V = {x : x−1 x0 ∈ U² , x−1 0 x ∈ U² } is an open set containing x0 . If x ∈ V , then ∆(x) = ∆(x0 )∆(x−1 0 x) ≤ (1 + ²)∆(x0 ),

442L


285

∆(x0 ) = ∆(x)∆(x−1 x0 ) ≤ (1 + ²)∆(x), so 1 ∆(x0 ) 1+²

≤ ∆(x) ≤ (1 + ²)∆(x).

As ² is arbitrary, ∆ is continuous at x0 ; as x0 is arbitrary, ∆ is continuous. 442K Theorem Let X be a topological group and µ a left Haar measure on X. Let ∆ be the left modular functionRof X. −1 (a) µ(E = E ∆(x−1R)µ(dx) for every E ∈ dom µ. R ) −1 (b)(i) f (x )µ(dx) = ∆(x−1 )f (x)µ(dx) whenever f is a real-valued function such that either integral is definedR in [−∞, ∞]; R (ii) f (x)µ(dx) = ∆(x−1 )f (x−1 )µ(dx) whenever f is a real-valued function such that either integral is defined R in [−∞, ∞]. R (c) f (xy)µ(dx) = ∆(y −1 ) f (x)µ(dx) whenever y ∈ X and f is a real-valued function such that either integral is defined in [−∞, ∞]. proof (a)(i) Setting ν1 E = µE −1 for Haar measurable sets E ⊆ X, we know that ν1 is a right Haar measure, so 442E tells us that

R

H

ν1 (xG−1 )µ(dx) =

R

H

µ(Gx−1 )µ(dx) = µG

R

∆(x−1 )µ(dx) R for all open sets G, H ⊆ X. Since there is an open set G such that 0 < µG < ∞, µH −1 = H ∆(x−1 )µ(dx) for every open set H ⊆ X. R (ii) Now let ν2 be the indefinite-integral measure defined by setting ν2 E = ∆(x−1 )χE(x)µ(dx) whenever this is defined in [0, ∞] (§234). Then ν2 is effectively locally finite. P P If ν2 E > 0, then µE > 0, so there is an n ∈ N such that µ(E ∩ H) > 0, where H is the open set {x : ∆(x−1 ) < n}. Now there is an open set G ⊆ H such that µG < ∞ and µ(E ∩ G) > 0, in which case ν2 G ≤ nµG < ∞ and ν2 (E ∩ G) > 0. Q Q Accordingly ν2 is a quasi-Radon measure (415O).R Since it agrees with the quasi-Radon measure ν1 on open sets, by (i), the two are equal; that is, µE −1 = E ∆(x−1 )µ(dx) for every E ∈ dom µ. µG · ν1 H =

H

(b)(i) Apply 235E with X = Y , Σ = T = dom µ, µ = ν and φ(x) = x−1 , J(x) = ∆(x−1 ), g(x) = ∆(x−1 )f (x) for x ∈ X. From (a) we have

R

J × χ(φ−1 [F ])dµ =

R

F −1

∆(x−1 )µ(dx) = µF = νF

for every F ∈ T (using 442G to see that F −1 ∈ Σ). So we get Z Z Z f (x−1 )µ(dx) = ∆(x−1 )g(x−1 )µ(dx) = J × gφ dµ Z Z = g dν = ∆(x−1 )f (x)µ(dx) if any of the integrals is defined in [−∞, ∞]. (ii) Set f˜(x) = f (x−1 ) whenever this is defined, and apply (i) to f˜. (c) Similarly, apply 235E with µ = ν, φ(x) = xy, J(x) = ∆(y) for every x ∈ X; then

R

J × χ(φ−1 [F ])dµ = ∆(y)µ(F y −1 ) = µF

for every F ∈ dom µ, so

R

f (xy)µ(dx) = ∆(y −1 )

R

J × f φ dµ = ∆(y −1 )

R

f (x)µ(dx).

442L Corollary Let X be a group carrying Haar measures. If µ is a left Haar measure on X and ν is a right Haar measure, then each is an indefinite-integral measure over the other. proof Let µ ˜Rbe the right Haar measure defined by setting µ ˜E = µE −1 for every Haar measurable E ⊆ X. −1 Then µ ˜E = E ∆(x )µ(dx) for every E ∈ dom µ = dom µ ˜, so µ ˜ is an indefinite-integral measure over µ; because ν is a multiple of µ ˜, it also is an indefinite-integral measure over µ. Similarly, or because ∆ is strictly positive, µ is an indefinite-integral measure over ν.

286

Topological groups

442X

442X Basic exercises > (a) Let X and Y be topological groups with (left) Haar probability measures µ and ν, and φ : X → Y a continuous surjective group homomorphism. Show that φ is inverse-measurepreserving for µ and ν. (b) (i) Let X and Y be two topological groups carrying Haar measures. Show that the product topological group X × Y (4A5G) carries Haar measures. (ii) Let hXi ii∈IQbe any family of topological groups carrying totally finite Haar measures. Show that the product group i∈I Xi carries a totally finite Haar measure. (Hint: 417O.) (c) Let X be a subgroup of the group (R, +). Show that X carries Haar measures iff it is either discrete (so that counting measure is a Haar measure on X) or of full outer Lebesgue measure (so that the subspace measure on X is a Haar measure). (Hint: if G has a Haar measure ν and is not discrete, then ν(G ∩ [α, β]) = (β − α)ν(G ∩ [0, 1]) whenever α ≤ β.) In particular, Q does not carry Haar measures. (d) Let X be a topological group carrying Haar measures; let Σ be the algebra of Haar measurable subsets of X. Let µ and ν be any Haar measures on X (either left or right). Show that a set E ∈ Σ can be covered by a sequence of sets of finite measure for µ iff it can be covered by a sequence of sets of finite measure for ν. (e) Let X be a topological group carrying Haar measures and A its Haar measure algebra (in the sense of 442H). Show that we have left, right and conjugacy actions of X on A given by the formulae z •E • = (zE)• , z •E • = (Ez −1 )• and z •E • = (zEz −1 )• for every Haar measurable E ⊆ X and every z ∈ X. > (f ) On R 2 define a binary operation ∗ by setting (ξ1 , ξ2 ) ∗ (η1 , η2 ) = (ξ1 + η1 , ξ2 + eξ1 η2 ). (i) Show that ∗ is a group operation under which R 2 is a locally compact topological group. (ii) Show that Lebesgue measure µ is aR right Haar measure for ∗. (iii) Let ν be the indefinite-integral measure on R 2 defined by setting νE = E e−ξ1 dξ1 dξ2 for Lebesgue measurable sets E ⊆ R 2 . Show that ν is a left Haar measure for ∗. (Hint: 263D.) (iv) Thus (R 2 , ∗) is not unimodular. (v) Show that the left modular function of (R 2 , ∗) is (ξ1 , ξ2 ) 7→ e−ξ1 . > (g) Let X be any topological group which is not unimodular. (i) Show that there is an open subset of X which is of finite measure for all left Haar measures on X and of infinite measure for all right Haar measures. (Hint: the modular function is unbounded.) (ii) Let µ be a left Haar measure on X and ν a right Haar measure. Show that L0 (µ) = L0 (ν) and L∞ (µ) = L∞ (µ), but that Lp (µ) 6= Lp (ν) for any p ∈ [1, ∞[. (h) Let X and Y be topological groups carrying Haar measures, with left modular functions ∆X and ∆Y respectively. Show that the left modular function of X × Y is (x, y) 7→ ∆X (x)∆Y (y). (i) Let X be any topological group and ∆ : X → ]0, ∞[ a group homomorphism such that {x : ∆(x) ≤ 1 + ²} is a neighbourhood of the identity in X for every ² > 0. Show that ∆ is continuous. (j) LetR X be a topological group with a right Haar measure ν and left modular function ∆. Show that νE −1 = E ∆(x)ν(dx) for every Haar measurable set E ⊆ X. 442Y Further exercises (a) In 441Yb, show that the only G-invariant Radon measures on Cs are multiples of Hausdorff s(r − s)-dimensional measure on Cs . (Hint: G itself is r(r−1) -dimensional (cf. 441Yi), 2 (r−s)(r−s−1) and for any C ∈ Cs the stabilizer of C is s(s−1) + -dimensional. See Federer 69, 3.2.28.) 2 2 (b) Let r ≥ 1, and let X be the group of non-singular r × r real matrices. Regarding X as an open subset R 2 1 of R r , show that a two-sided Haar measure µ can be defined on X by setting µE = E µL (dA), r | det A|

r2

where µL is Lebesgue measure on R ; so that X is unimodular. (c) Show that there is a set A ⊆ [0, 1], of Lebesgue outer measure 1, such that no countable set of translates of A covers any set of Lebesgue measure greater than 0. (Hint: let hFξ iξ
443A

Further properties of Haar measure

287

442Z Problem Let X be a compact Hausdorff space, and G the group of autohomeomorphisms of Z. Suppose that G acts transitively on X. Does it follow that there is at most one G-invariant Radon probability measure on X? 442 Notes and comments Haar measure dominates the theory of locally compact topological groups for two reasons: it is ubiquitous (the existence theorem, 441E) and essentially defined by the group structure (the uniqueness theorem, 442B). I have tried to show that these are rather different results by setting the theorems out with different hypotheses. I presented the existence theorem as a special case of 441C, which demands a locally compact space and a group, but allows them to be different. In the uniqueness theorem (roughly following Halmos 50) I demand a group with an invariant quasi-Radon measure, but do not (at this point) ask for any hypothesis of compactness. In fact it will become apparent in the next section that this is a somewhat spurious generality; 442B and 442I here can be deduced from the traditional forms in which the group is assumed to be locally compact and Hausdorff. From the point of view of the topological measure theory to which this volume is devoted, however, I think the small extra labour involved in tracing through the arguments without relying on the Riesz Representation Theorem is instructive. For instance, it emphasizes interesting features of the domains and null ideals of Haar measures (442H). There is however a more serious question concerning the uniqueness theorem. I do not know whether it really belongs to the theory of topological groups, as described here, or to the theory of group actions along with 441C. The trouble is that I know of no example of a Hausdorff space X and a transitive group G of homeomorphisms of X such that X carries G-invariant Radon measures which are not multiples of each other (see 442Z). 443T and 443Xz below eliminate the simplest possibilities. We do need to put some restriction on the measures; for instance, counting measure on R is translation-invariant, but has nothing to do with Lebesgue measure. There are also proper translation-invariant extensions of Lebesgue measure (442Yc); for far-reaching elaborations of this idea see Hewitt & Ross 63, §16.

443 Further properties of Haar measure I devote a section to filling in some details of the general theory of Haar measures before turning to the special topics dealt with in the rest of the chapter. The first question concerns the left and right shift operators acting on sets, on elements of the measure algebra, on measurable functions and on function spaces. All these operations can be regarded as group actions, and, if appropriate topologies are assigned, they are continuous actions (443C, 443G). As an immediate consequence of this I give an important result about product sets {ab : a ∈ A, b ∈ B} in a topological group carrying Haar measures (443D). The second part of the section revolves around a basic structure theorem: all the Haar measures considered here can be reduced to Haar measures on locally compact Hausdorff groups (443L). The argument involves two steps: the reduction to the Hausdorff case, which is elementary, and the completion of a Hausdorff topological group. Since a group carries more than one natural uniform structure we must take care to use the correct one, which in this context is the ‘bilateral’ uniformity (443H-443I, 443K). On the way I pick up an essential fact about the approximation of Haar measurable sets by Borel sets (443J). Finally, I give Halmos’ theorem that Haar measures are completion regular (443M). In the third part of the section I turn to the special properties of quotient groups of locally compact groups and the corresponding actions, following A.Weil. If X is a locally compact Hausdorff group and Y is a closed subgroup of X, then Y is again a locally compact Hausdorff group, so has Haar measures and a modular function; at the same time, we have a natural action of X on the set of left cosets of Y . It turns out that there is an invariant Radon measure for this action if and only if the modular function of Y matches that of X (443Q). In this case we can express a left Haar measure of X as an integral of measures supported by the cosets of Y (443P). When Y is a normal subgroup, so that X/Y is itself a locally compact Hausdorff group, we can relate the modular functions of X and X/Y (443S). We can apply these results whenever we have a continuous transitive action of a compact group on a compact space (443T). 443A Haar measurability I recall and expand on some facts already noted in 442H. Let X be a topological group carrying Haar measures.

288

Topological groups

443Aa

(a) All Haar measures on X, whether left or right, have the same domain Σ, which I call the algebra of ‘Haar measurable’ sets, and the same null ideal N , which I call the ideal of ‘Haar negligible’ sets. The corresponding quotient algebra A = Σ/N , the ‘Haar measure algebra’, is the Boolean algebra underlying the measure algebra of any Haar measure. Because Haar measures are (by the definition in 441D) quasi-Radon, therefore strictly localizable (415A), A is Dedekind complete (322Be). Recall that any semi-finite measure on A, and in particular any Haar measure on X, gives rise to the same topology and uniformity on A (324H), so we may speak of ‘the’ topology and uniformity of A. Because Σ is the domain of a left Haar measure, xE ∈ Σ for every E ∈ Σ, x ∈ X; because Σ is the domain of a right Haar measure, Ex ∈ Σ for every E ∈ Σ, x ∈ X. Similarly, xE and Ex are Haar negligible whenever E is Haar negligible and x ∈ X. Moreover, E −1 = {x−1 : x ∈ E} is Haar measurable or Haar negligible whenever E is. (b) We even have a symmetric notion of ‘measurable envelope’ in X: for any A ⊆ X, there is a Haar measurable set E ⊇ A such that µ(E ∩ F ) = µ∗ (A ∩ F ) for any Haar measurable F ⊆ X and any Haar measure µ on X. P P Start with a fixed Haar measure µ0 . Then A has a measurable envelope E for µ0 , by 213L. Now to say that ‘E is a measurable envelope for A’ means just that (i) A ⊆ E ∈ Σ (ii) if F ∈ Σ and F ⊆ E \ A then F ∈ N , so E is also a measurable envelope for A for any other Haar measure on X. Q Q In this context I will call E a Haar measurable envelope of A. (c) Similarly, we have a notion of full outer Haar measure: a subset A of X is of full outer Haar measure if X is a Haar measurable envelope of A, that is, A ∩ E 6= ∅ whenever E is a Haar measurable set which is not Haar negligible, that is, A is of full outer measure for any (left or right) Haar measure on X. (d) For any Haar measure µ on X, we can identify L∞ (µ) with L∞ (A) (363I) and L0 (µ) with L0 (A) (364Jc). Thus these constructions are independent of µ. The topology of convergence in measure of L0 is determined by its Riesz space structure (367T); so that this also is independent of the particular Haar measure we may select. Of course the same is true of the norm of L∞ . Note however that the spaces Lp , for 1 ≤ p < ∞, are different for left and right Haar measures on any group which is not unimodular (442Xg), and that even for left Haar measures µ the norm on Lp (µ) changes if µ is re-normalized. (e) When it seems appropriate, I will use the phrases Haar measurable function, meaning a function measurable with respect to the σ-algebra of Haar measurable sets, and Haar almost everywhere, meaning ‘on the complement of a Haar negligible set’. Note that we can identify L0 (A) with the set of equivalence classes in the space L0 , where L0 is the space of Haar measurable real-valued functions defined Haar-a.e. in X, and f ∼ g if f = g Haar-a.e. In the language of §241, L0 = L0 (µ) for any Haar measure µ on X. (f ) Because E −1 ∈ Σ for every E ∈ Σ, and E −1 ∈ N whenever E ∈ N , we have a canonical automorphism a 7→ a ˜ : A → A defined by writing (E • )∼ = (E −1 )• for every E ∈ Σ. Being an automorphism, this must be a homeomorphism for the measure-algebra topology of A (324G). In the same way, if f ∈ L0 then f˜ ∈ L0 , where f˜(x) = f (x−1 ) whenever this is defined, since {x : x ∈ dom f˜, f˜(x) > α} = {x : x ∈ dom f, f (x) > α}−1 ∈ Σ for every α; and we can define an f -algebra automorphism u 7→ u ˜ : L0 → L0 by saying that (f • )∼ = (f˜)• for 0 0 0 f ∈ L . If we identify L with L (A) rather than with a set of equivalence classes in L0 , then we can define the map u 7→ u ˜ as the Riesz homomorphism associated with the Boolean homomorphism a 7→ a ˜ : A → A, as in 364R. Note that k˜ uk∞ = kuk∞ for every u ∈ L∞ , but that (unless X is unimodular) other Lp spaces are not invariant under the involution ∼ (442Xg). (g) If X carries any totally finite (left or right) Haar measure, it is unimodular, and has a unique, twosided, Haar probability measure (442Ie). (In particular, this is the case whenever X is compact.) For such groups we have Lp -spaces, for 1 ≤ p ≤ ∞, defined by the group structure, with canonical norms. 443B Lemma Let X be a topological group and µ a left Haar measure on X. If E ⊆ X is measurable and µE < ∞, then for any ² > 0 there is a neighbourhood U of the identity e such that µ(E4xEy) ≤ ² whenever x, y ∈ U . proof Set δ = min(1, ²)/(10 + 3µE) > 0. Write U for the family of open neighbourhoods of e. Because µ is effectively locally finite, there is an open set G0 of finite measure such that µ(E \ G0 ) ≤ δ. Let F ⊆ G0 \ E

443C


289

be a closed set such that µF ≥ µ(G0 \ E) − δ, and set G = G0 \ F , so that µ(G \ E) ≤ δ. For U ∈ U set HU = int{x : U xU ⊆ G}. Then H = {HU : U ∈ U } is upwards-directed, and has union G, because if x ∈ G there is a U ∈ U such that U xU U ⊆ G, so that x ∈ HU . So there is a V ∈ U such that µ(G \ HV ) ≤ δ. Recall that the left modular function ∆ of X is continuous (442J). So there is a U ∈ U such that U ⊆ V and |∆(y) − 1| ≤ δ for every y ∈ U . Now suppose that x, y ∈ U . Set E1 = E ∩ HV . Then xE1 y ⊆ G, so µ(E1 ∪ xE1 y) ≤ µG ≤ µE + δ. On the other hand, µE1 ≥ µE − µ(E \ G) − µ(G \ HV ) ≥ µE − 2δ, µ(xE1 y) = ∆(y)µE1 ≥ (1 − δ)(µE − 2δ) ≥ µE − (2 + µE)δ. So µ(E ∩ xEy) ≥ µ(E1 ∩ xE1 y) = µE1 + µ(xE1 y) − µ(E1 ∪ xE1 y) ≥ µE − (5 + µE)δ. At the same time, µ(xEy) = ∆(y)µE ≤ (1 + δ)µE. So µ(E4xEy) = µE + µ(xEy) − 2µ(E ∩ xEy) ≤ (10 + 3µE)δ ≤ ², as required.

443C Theorem Let X be a topological group carrying Haar measures, and A its Haar measure algebra. Then we have continuous actions of X on A defined by writing x•l E • = (xE)• , x•r E • = (Ex−1 )• , x•c E • = (xEx−1 )• for Haar measurable sets E ⊆ X and x ∈ X. proof (a) The functions •l , •r and •c are all well defined because the maps E 7→ xE, E 7→ Ex and E 7→ xEx−1 are all Boolean automorphisms of the algebra Σ of Haar measurable sets preserving the ideal of Haar negligible sets (442G). It is elementary to check that they are actions of X on A. Fix a left Haar measure µ on X and let µ ¯ be the corresponding measure on A. Then the topology of A is defined by the pseudometrics ρa , for µ ¯a < ∞, where ρa (b, c) = µ ¯(a ∩ (b 4 c)). (b) Now suppose that x0 ∈ X, b0 ∈ A, µ ¯a < ∞ and ² > 0. Let E, F0 ∈ Σ be such that E • = a and 1 F0 = b0 ; set δ = 4 ² > 0. Note that •

−1 µ(x−1 0 E ∩ F0 ) ≤ µ(x0 E) = µE < ∞.

Let U be a neighbourhood of the identity e such that µ(E4yE) ≤ δ,

−1 µ((x−1 0 E ∩ F0 )4y(x0 E ∩ F0 )) ≤ δ

whenever y ∈ U (443B). Set −1 • • a0 = x−1 0 l a = (x0 E) .

Now suppose that x ∈ U x0 ∩ x0 U −1 and that ρa0 (b, b0 ) ≤ δ. Then ρa (x•l b, x0 •l b0 ) ≤ ². P P Let F ∈ Σ be −1 • −1 such that F = b. Then xx0 and x x0 both belong to U , so

290

Topological groups

443C

ρa (x•l b, x0 •l b0 ) = µ(E ∩ (xF 4x0 F0 )) = µ(E ∩ xF ) + µ(E ∩ x0 F0 ) − 2µ(E ∩ xF ∩ x0 F0 ) = µ(x−1 E ∩ F ) + µ(E ∩ x0 F0 ) − 2µ(x−1 x0 (x−1 0 E ∩ F0 ) ∩ F ) −1 −1 ≤ µ(x−1 E4x−1 0 E ∩ F ) + µ(x 0 E) + µ(x0 E ∩ F0 ) −1 −1 x0 (x−1 − 2µ(x−1 0 E ∩ F0 ∩ F ) + 2µ(x 0 E ∩ F0 )4(x0 E ∩ F0 )) −1 ≤ µ(x−1 0 E ∩ (F 4F0 )) + µ(E4xx0 E) + 2δ ≤ δ + δ + 2δ = 4δ = ². Q Q

As x0 , b0 , a and ² are arbitrary,

•

l

is continuous.

(c) The same arguments, using a right Haar measure to provide pseudometrics defining the topology of A, show that •r is continuous. (Or use the method of 443Xe.) (d) Accordingly the map (x, y, a) 7→ x•l (y •r a) is continuous. So (x, a) 7→ x•l (x•r a) = x•c a is continuous. 443D Proposition Let X be a topological group carrying Haar measures. If E ⊆ X is Haar measurable but not Haar negligible, and A ⊆ X is not Haar negligible, then (a) there are x, y ∈ X such that A ∩ xE, A ∩ Ey are not Haar negligible, (b) EA and AE both have non-empty interior. proof (a)(i) Let µ be any left Haar measure on X, and for Borel sets F ⊆ X set νF = sup{µ(F ∩ IE) : I ⊆ X is finite}. It is easy to check that ν is an effectively locally finite τ -additive Borel measure, inner regular with respect to the closed sets, because {IE : I ∈ [X]<ω } is upwards-directed and µ is quasi-Radon. Moreover, ν(xF ) = supI∈[X]<ω µ(xF ∩ IE) = supI∈[X]<ω µ(F ∩ x−1 IE) = νF for every Borel set F ⊆ X, x ∈ X. Accordingly the c.l.d. version ν˜ of ν is a left-translation-invariant quasiRadon measure on X (415Cb); since νE > 0, ν˜ is non-zero and is itself a left Haar measure. Consequently A is not ν˜-negligible. Let H be a measurable envelope of A for Haar measure (443Ab). Then H is not Haar negligible, so there is a closed set F ⊆ H which is not Haar negligible, and νF = ν˜F > 0. Thus there is an x ∈ X such that 0 < µ(F ∩ xE) ≤ µ(H ∩ xE) = µ∗ (A ∩ xE), and A ∩ xE is not Haar negligible. (ii) Applying the same arguments, but starting with a right Haar measure µ, we see that there is a y ∈ X such that A ∩ Ey is not Haar negligible. (b) Let µ be a left Haar measure on X, and F a Haar measurable envelope of A. The function x 7→ (xE −1 )• : X → A is continuous, where A is the Haar measure algebra of X (443C), so H = {x : µ∗ (A ∩ xE −1 ) > 0} = {x : F • ∩ (xE −1 )• 6= 0} is open. Now H ⊆ {x : A ∩ xE −1 6= ∅} = AE, so H ⊆ int AE; and E −1 is Haar measurable and not Haar negligible, so H 6= ∅, by (a). Thus int AE 6= ∅. Similarly, using a right Haar measure (or observing that EA = (A−1 E −1 )−1 ), we see that EA has non-empty interior.

443G


291

443E Corollary Let X be a Hausdorff topological group carrying Haar measures. Then the following are equiveridical: (i) X is locally compact; (ii) every Haar measure on X is a Radon measure; (iii) there is some compact subset of X which is not Haar negligible. proof (i)⇒(ii) Haar measures are locally finite quasi-Radon measures (441D, 442Aa), so on locally compact Hausdorff spaces must be Radon measures (416G). (ii)⇒(iii) is obvious, just because any Radon measure is tight (that is, inner regular with respect to the compact sets). (iii)⇒(i) If K ⊆ X is a compact set which is not Haar negligible, then KK is a compact set with non-empty interior, so X is locally compact (4A5Eg). 443F Later in the chapter we shall need the following straightforward fact. Lemma Let X be a topological group carrying Haar measures, and Y an open subgroup of X. If µ is a left Haar measure on X, then the subspace measure µY is a left Haar measure on Y . Consequently a subset of Y is Haar measurable or Haar negligible, when regarded as a subset of the topological group Y , iff it is Haar measurable or Haar negligible when regarded as a subset of the topological group X. proof By 415B, µY is a quasi-Radon measure; because µ is strictly positive, µY is non-zero, and of course it is left-translation-invariant. So it is a Haar measure on Y . The rest follows at once from 442H/443A. 443G We can repeat the ideas of 443C in terms of function spaces, as follows. Theorem Let X be a topological group with a left Haar measure µ. Let Σ be the domain of µ, L0 = L0 (µ) the space of Σ-measurable real-valued functions defined almost everywhere on X, and L0 = L0 (µ) the corresponding space of equivalence classes (§241). (a) a•l f , a•r f and a•c f (definitions: 441Ac) belong to L0 for every f ∈ L0 and a ∈ X. (b) If a ∈ X, then ess sup |a•l f | = ess sup |a•r f | = ess sup |f | for every f ∈ L∞ = L∞ (µ), where ess sup |f | is the essential supremum of |f | (243Da). For 1 ≤ p < ∞, ka•l f kp = kf kp and ka•r f kp = ∆(a)−1/p kf kp for every f ∈ Lp = Lp (µ), where ∆ is the left modular function of X. (c) We have actions of X on L0 defined by setting a•l f • = (a•l f )• , a•r f • = (a•r f )• , a•c f • = (a•c f )• for a ∈ X and f ∈ L0 . If we give L0 its topology of convergence in measure these actions are continuous. (d) For 1 ≤ p ≤ ∞ the formulae of (c) define actions of X on Lp = Lp (µ), and ka•l ukp = kukp for every u ∈ Lp , a ∈ X. (e) For 1 ≤ p < ∞ these actions are continuous. proof (a) Let f ∈ L0 . Then F = dom f is conegligible, so aF = dom a•l f and F a−1 = dom a•r f are conegligible (442G). For any α ∈ R, set Eα = {x : x ∈ F, f (x) < α}; then {x : (a•l f )(x) < α} = aEα and {x : (a•r f )(x) < α} = Eα a−1 are measurable, so a•l f and a•r f are measurable. Thus a•l f and a•r f belong to L0 . It follows at once that a•c f = a•l (a•r f ) belongs to L0 . (b)(i) For α ≥ 0, ess sup |f | ≤ α ⇐⇒ |f (x)| ≤ α for almost all x ⇐⇒ |(a•l f )(x)| ≤ α for almost all x ⇐⇒ |(a•r f )(x)| ≤ α for almost all x because the null ideal is invariant under both left and right translations. So ess sup |f | = ess sup |a•l f | = ess sup |a•r f |.

292

Topological groups

443G

(ii) For 1 ≤ p < ∞, Z ka•l f kpp =

Z |(a•l f )(x)|p µ(dx) =

Z |f (a−1 x)|p µ(dx) =

|f (x)|p µ(dx)

(441J) = kf kpp , Z Z Z p p p −1 ka•r f kp = |(a•r f )(x)| µ(dx) = |f (xa)| µ(dx) = ∆(a ) |f (x)|p µ(dx) (442Kc) = (∆(a)−1/p kf kp )p .

(c)(i) I have already checked that a•l f , a•r f and a•c f belong to L0 whenever f ∈ L0 and a ∈ X. If f , g ∈ L0 and f =a.e. g, let E be the conegligible set {x : x ∈ dom f ∩ dom g, f (x) = g(x)}; then aE and Ea−1 and aEa−1 are conegligible and (a•l f )(x) = (a•l g)(x) for every x ∈ aE,

(a•r f )(x) = (a•r g)(x) for every x ∈ Ea−1 ,

(a•c f )(x) = (a•c g)(x) for every x ∈ aEa−1 , so a•l f =a.e. a•l g, a•r f =a.e. a•r g and a•c f =a.e. a•c g. Accordingly the formulae given define functions •l , •r and •c from X × L0 to L0 . They are actions just because the original •l , •r and •c are actions of X on L0 (441Ac-441Ad). (ii) For the next step, it will be more convenient to work with the space L0strict of Haar measurable real-valued functions defined on the whole of X. I will use a characterization of convergence in measure from 245F: a subset W of L0 is open iff whenever f0• ∈ W there are a set E of finite measure and an ² > 0 such that f • ∈ W whenever µ{x : x ∈ E, |f (x) − f0 (x)| > ²} ≤ ². Now if E is a measurable set of finite measure, f ∈ L0strict and ² > 0, there is a neighbourhood U of the identity e of X such that µ{x : x ∈ E, |f (axb) − f (x)| > ²} ≤ ² for all a, b ∈ U . P P Let m ≥ 1 be such that µ{x : x ∈ E, |f (x)| ≥ m²} ≤ 12 ². For −m ≤ k < m, set Ek = {x : x ∈ E, k² ≤ f (x) < (k + 1)²}. By 443B, there is a neighbourhood U of e such that µ(Ek 4a−1 Ek b−1 ) ≤

² 4m

whenever a, b ∈ U and −m ≤ k < m. Now, for a, b ∈ U , {x : x ∈ E, |f (axb) − f (x)| > ²} ⊆ {x : x ∈ E, |f (x)| ≥ m²} ∪

Sm−1

k=−m (Ek 4a

−1

Ek b−1 )

has measure at most ² 2

+ 2m

² 4m

= ². Q Q

(iii) Let E be a measurable set of finite measure, a0 , b0 ∈ X, f0 ∈ L0strict and ² > 0. Set δ = ²/(2 + −1 3∆(b−1 0 )) > 0. Note that µ(a0 Eb0 ) = ∆(b0 )µE is finite. Let U be a neighbourhood of e such that µ{x : x ∈ a−1 0 Eb0 , |f0 (yxz) − f0 (x)| ≥ δ} ≤ δ, µ(yEz4E) ≤ δ,

∆(y) ≤ 2

whenever y, z ∈ U . Now suppose that a ∈ U a0 ∩ a0 U −1 and b ∈ U −1 b0 ∩ b0 U and that f ∈ L0strict is such that µ{x : x ∈ −1 a0 Eb0 , |f (x) − f0 (x)| ≥ δ} ≤ δ. In this case,

443G


293

{x : x ∈ E, |f (a−1 xb) − f0 (a−1 0 xb0 )| ≥ ²} ⊆ {x : x ∈ E, |f (a−1 xb) − f0 (a−1 xb)| ≥ δ} ∪ {x : x ∈ E, |f0 (a−1 xb) − f0 (a−1 0 xb0 )| ≥ δ} −1 −1 ⊆ (E4aa−1 ) ∪ {x : x ∈ aa−1 , |f (a−1 xb) − f0 (a−1 xb)| ≥ δ} 0 Eb0 b 0 Eb0 b −1 −1 ∪ a0 {w : w ∈ a−1 a0 wb−1 0 Eb0 , |f0 (a 0 b) − f0 (w)| ≥ δ}b0 −1 ⊆ (E4aa0−1 Eb0 b−1 ) ∪ a{w : w ∈ a−1 0 Eb0 , |f (w) − f0 (w)| ≥ δ}b −1 −1 ∪ a0 {w : w ∈ a−1 a0 wb−1 0 Eb0 , |f0 (a 0 b) − f0 (w)| ≥ δ}b0

has measure at most −1 −1 δ + δ∆(b−1 ) + δ∆(b−1 )∆(b−1 0 ) ≤ δ(1 + ∆(b0 b 0 ) + ∆(b0 ))

≤ δ(1 + 3∆(b−1 0 )) ≤ ² −1 because aa−1 a0 , b0 b−1 and b−1 0 , a 0 b all belong to U . Because E and ² are arbitrary, the function (a, b, u) 7→ a•l (b•r u) is continuous at (a0 , b0 , f0• ); as a0 , b0 and f0 are arbitrary, •l and •r are continuous. It follows at once that •c is continuous, since a•c u = a•l (a•r u).

(d) follows at once from (b) and (c). (e) Fix p ∈ [1, ∞[. (i) If u ∈ Lp and ² > 0, there is a neighbourhood U of e such that ku − y •l (z •r u)kp ≤ ² whenever y, z ∈ U. P P When u is of the form (χE)• , where µE < ∞, we have y •l (z •r u) = χ(yEz −1 )• ,

ku − y •l (z •r u)kp = µ(E4yEz −1 )1/p , Pn • so the result is immediate from 443B. If u = i=0 αi (χEi ) , where every Ei is of finite measure, then, • setting ui = (χEi ) for each i, Pn ku − y •l (z •r u)kp ≤ i=0 |αi |kui − y •l (z •r ui )kp ≤ ² whenever y and z are close enough to e. In general, there is a v of this form such that ku − vkp ≤ 41 ². If we take a neighbourhood U of e such that kv − y •l (z •r v)kp ≤ 14 ² and ∆(z)−1/p ≤ 2 whenever y, z ∈ U , then 1 2

ky •l (z •r u) − y •l (z •r v)kp = ∆(z)−1/p ku − vkp ≤ ² whenever z ∈ U , so ku − y •l (z •r u)kp ≤ ku − vkp + kv − y •l (z •r v)kp + ky •l (z •r v) − y •l (z •r u)kp ≤² whenever y, z ∈ U . Q Q (ii) Now suppose that u0 ∈ Lp , a0 , b0 ∈ X and ² > 0. Set v0 = a0 •l (b0 •r u0 ) and δ = ²/(1 + 2∆(b0 )−1/p ) > 0. Let U be a neighbourhood of e such that ∆(y)−1/p ≤ 2 and kv0 − y •l (z •r v0 )kp ≤ δ whenever y, z ∈ U . If a ∈ U a0 , b ∈ U b0 and ku − u0 kp ≤ δ, then ka•l (b•r u) − v0 kp ≤ ka•l (b•r u) − a•l (b•r u0 )kp + ka•l (b•r u0 ) − v0 kp −1 • • = ∆(b)−1/p ku − u0 kp + kaa−1 0 l (bb0 r v0 ) − v0 kp −1/p ≤ ∆(bb−1 ∆(b0 )−1/p δ + δ ≤ δ(1 + 2∆(b0 )−1/p ) = ². 0 )

As ² is arbitrary, (a, b, u) 7→ a•l (b•r u) is continuous at (a0 , b0 , u0 ). As in (c), this is enough to show that •l , •r and •c are all continuous actions. Remark I have written this out for a left Haar measure µ, since the spaces Lp (µ) depend on this; if ν is a right Haar measure, and X is not unimodular, then Lp (µ) 6= Lp (ν) for 1 ≤ p < ∞. But recall that the topology of convergence in measure on L0 depends only on its Riesz space structure (367T), while L0 (µ) can be identified, as Riesz space, with L0 (ν) (443Ad), so (c) above, and the case p = ∞ of (b) and (d), are two-sided; they belong to the theory of the Haar measure algebra.

294

Topological groups

443H

443H Theorem Let X be a topological group carrying Haar measures. Then there is a neighbourhood of the identity which is totally bounded for the bilateral uniformity on X. proof Let µ be a left Haar measure on X. Let V0 be a neighbourhood of the identity e such that µV0 < ∞ (442Aa). Let V be a neighbourhood of e such that V V ⊆ V0 and V −1 = V . ?? Suppose, if possible, that V is not totally bounded for the bilateral uniformity on X. By 4A5Oa, one of the following must occur: case 1 There is an open neighbourhood U case, we may choose a sequence hxn in∈N in V be an open neighbourhood of e such that U1 µ(xn U1 ) = µU1 > 0 for every n (by the other every n, so this is impossible.

of e such that V 6⊆ IU for any finite set I ⊆ X. In this inductively such that xn ∈ / xi U whenever i < n. Let U1 ⊆ V and U1 U1−1 ⊆ S U ; then hxn U1 in∈N is disjoint. Since clause in 442Aa), µ( n∈N xn U1 ) = ∞; but xn U1 ⊆ V0 for

case 2 There is an open neighbourhood U of e such that V 6⊆ U I for any finite set I ⊆ X. So we may choose a sequence hxn in∈N in V inductively such that xn ∈ / U xi whenever i < n. Let U1 be an open neighbourhood of e such that U1 ⊆ V and U1 U1−1 ⊆SU ; then hU1−1 xn in∈N is disjoint, so hx−1 n U1 in∈N is also −1 −1 disjoint. Since µ(x−1 n U1 ) = µU1 > 0 for every n, µ( n∈N xn U1 ) = ∞; but xn U1 ⊆ V0 for every n, so this also is impossible. X X Thus V is totally bounded for the bilateral uniformity on X, and we have the required totally bounded neighbourhood of e. 443I Lemma Let X be a topological group. If A ⊆ X is totally bounded for the bilateral uniformity of X, it has finite outer measure for any (left or right) Haar measure on X. proof If µ is a Haar measure on X, let U be an open neighbourhood of the identity e of finite measure. P There is a finite set I such that A ⊆ U I, so that µ∗ A ≤ µU x∈I ∆(x) (if µ is a left Haar measure) and µ∗ A ≤ #(I)µU (if µ is a right Haar measure); in either case, µ∗ A < ∞. 443J Proposition Let X be a topological group carrying Haar measures. (a) There is an open-and-closed subgroup Y of X such that, for any Haar measure µ on X, Y can be covered by countably many open sets of finite measure. (b)(i) If E ⊆ X is any Haar measurable set, there are an Fσ set E 0 ⊆ E and a Gδ set E 00 ⊇ E such that 00 E \ E 0 is Haar negligible. (ii) Every Haar negligible set is included in a Haar negligible Borel set, and for every Haar measurable set E there is a Borel set F such that E4F is Haar negligible. (iii) The Haar measure algebra A of X may be identified with B/I, where B is the Borel σ-algebra of X and I is the ideal of Haar negligible Borel sets. (iv) Every member of L0 (A) is the equivalence class of some Borel measurable function from X to R. Every member of L∞ (A) is the equivalence class of a bounded Borel measurable function from X to R. proof (a) Let V be an open neighbourhood of the identity which is totally bounded for the bilateral uniformity of X (443H); we may suppose that V −1 = V . Set Y = V ∪ V V ∪ V V V ∪ V V V V ∪ . . . . Then Y is an open subgroup of X, therefore also closed (4A5Ek). By 4A5Ob, every power of V is totally bounded, so Y is a countable union of totally bounded sets. If µ is any left Haar measure on X, then any totally bounded set has finite outer measure for µ (443I). Thus Y is a countable union of sets of finite measure for µ. The same argument applies to right Haar measures, so Y is a subgroup of the required form. (b)(i) Let E ⊆ X be a Haar measurable set, and fix a left Haar measure µ on X. Take the open subgroup Y of (a), and index the set of its left cosets as hYi ii∈I ; because any S translate of a totally bounded set is totally bounded (4A5Ob), each Yi is an open set expressible as n∈N Hin , where every Hin is a totally bounded open set, so that µHin is finite. −n For iS ∈ I and m, n ∈ N there is a closed set Fimn ⊆ E ∩ Him such that µFimn S ≥ µ(E ∩ Him ) − 2 . Set 0 Fmn = i∈I Fimn for each m, n ∈ N; then Fmn is closed (4A2Bb). So E = m,n∈N Fmn is Fσ . For each i ∈ I, ¡ ¢ S S (E \ E 0 ) ∩ Yi ⊆ m∈N E ∩ Hin \ n∈N Fimn

443L


295

is negligible; thus {G : G ⊆ X is open, µ(G ∩ E \ E 0 ) = 0} covers X and E \ E 0 must be negligible (414Ea). In the same way, there is an Fσ set F ∗ ⊆ X \ E such that (X \ E) \ F ∗ is negligible; now E 00 = X \ F ∗ is Gδ and E 00 \ E is negligible, so E 00 \ E 0 also is. (I am speaking here as if ‘negligible’ meant ‘µ-negligible’. But of course this is the same thing as the ‘Haar negligible’ of the statement of the proposition.) (ii), (iii), (iv) follow at once. 443K Theorem Let X be a Hausdorff topological group carrying Haar measures. Then the completion b of X under its bilateral uniformity is a locally compact Hausdorff group, and X is of full outer Haar X b Any (left or right) Haar measure on X is the subspace measure corresponding to a Haar measure in X. b measure (of the same type) on X. b is a locally compact Hausdorff group in which X is embedded as a dense subgroup. proof (a) By 4A5N, X b such that µ is the (b) Let µ be a left Haar measure on X. Then there is a Radon measure µ ˜ on X b subspace measure µ ˜X . P P For Borel sets E ⊆ X, set νE = µ(X ∩ E). Then ν is a Borel measure, and it is τ -additive because µ is. Because X is locally compact, it follows that νG = sup{νK : K ⊆ G is compact} b it has a compact neighbourhood V in X; b now V must be totally for every open set G ⊆ X. If z ∈ X, b bounded for the bilateral uniformity of X (4A2Jd), so V ∩ X is totally bounded for the bilateral uniformity of X (4A5Ma), and νV = µ(V ∩ X) is finite (443I). Thus ν is locally finite. If νE > 0, there is an open set H ⊆ X such that µH < ∞ and µ(H ∩ X ∩ E) > 0, because µ is effectively locally finite; now there is an b such that H = X ∩ G, so that νG < ∞ and ν(G ∩ E) > 0. Thus ν is effectively locally open set G ⊆ X finite. b Since ν˜K = νK = 0 whenever K ⊆ X b \X By 416F(iii), ν has an extension to a Radon measure µ ˜ on X. is compact, X is of full outer measure for µ ˜. Accordingly µ ˜X (G ∩ X) = µ ˜G = νG = µ(G ∩ X) b and µ for every open set G ⊆ X, ˜X = µ, because they are quasi-Radon measures agreeing on the open sets (415B, 415H(iii)). Q Q b P b be open, and z ∈ X. b (c) Continuing the argument of (b), µ ˜ is a left Haar measure on X. P Let G ⊆ X −1 b If K ⊆ zG is compact, then z K ⊆ G, and {w : w ∈ X, wK ⊆ G} is a non-empty open set (4A5Ei), so meets X. Take x ∈ X such that xK ⊆ G; then µ ˜G = µ(X ∩ G) ≥ µ(X ∩ xK) = µ(x(X ∩ K)) = µ(X ∩ K) = µ ˜K. b on itself, that is, is a As K is arbitrary, µ ˜(zG) ≤ µ ˜G. By 441Ba, µ ˜ is invariant under the left action of X left Haar measure. Q Q b We know that X is of full outer measure for µ ˜, so this shows that it is of full outer Haar measure in X. (d) The same arguments, looking at Gz and Kz −1 in (c), show that if µ is a right Haar measure on X b it is the subspace measure µ ˜X for a right Haar measure µ ˜ on X. 443L Corollary Let X be any topological group with a Haar measure µ. Then we can find a locally compact Hausdorff topological group Z, with a Haar measure ν˜, and a continuous homomorphism φ : X → Z, inverse-measure-preserving for µ and ν˜, such that µ is inner regular with respect to {φ−1 [K] : K ⊆ Z is compact}. If E ⊆ X is Haar measurable, we can find a Haar measurable set F ⊆ Z such that φ−1 [F ] ⊆ E and E \ φ−1 [F ] is Haar negligible. proof (a) Let Y ⊆ X be the closure of {e}, where e is the identity of X. Then Y is a closed normal subgroup of X, and if φ : X → X/Y is the quotient map, every open (or closed) subset of X is of the form φ−1 [H] for some open (or closed) set H ⊆ Y (4A5Kb). Consider the image measure ν = µφ−1 on X/Y . This is quasi-Radon. P P Because µ is a complete τ additive topological measure, so is ν. If F ∈ dom ν and νF > 0, there is an open set G ⊆ X such that µG < ∞ and µ(G ∩ φ−1 [F ]) > 0; now G = φ−1 [H] for some open set H ⊆ X/Y , and νH = µG is finite,

296

Topological groups

443L

while ν(H ∩ F ) = µ(G ∩ φ−1 [F ]) > 0. Thus ν is effectively locally finite (therefore semi-finite). Again, if F ∈ dom ν and νF > γ, there is a closed set E ⊆ φ−1 [F ] such that µE ≥ γ; now E is expressible as φ−1 [H] for some closed set H ⊆ X/Y ; because φ is surjective, H ⊆ F , and νH = µF ≥ γ. Thus ν is inner regular with respect to the closed sets. Finally, suppose that F ⊆ X/Y is such that F ∩ F 0 ∈ dom ν whenever νF 0 < ∞. If E ⊆ X is a closed set of finite measure, it is of the form φ−1 [F 0 ] where νF 0 = µE < ∞, so F 0 ∩ F ∈ dom ν and E ∩ f −1 [F ] ∈ dom µ; by 412Ja, we can conclude that φ−1 [F ] ∈ dom µ and F ∈ dom ν. Thus ν is locally determined and is a quasi-Radon measure. Q Q We find also that ν is a left Haar measure. P P If z ∈ X/Y and F ∈ dom ν, express z as φ(x) where x ∈ X; then φ−1 [zF ] = xφ−1 [F ], so ν(zF ) = µ(xφ−1 [F ]) = µφ−1 [F ] = νF . Q Q (b) Thus X/Y is a topological group with a left Haar measure ν. Because Y is closed, X/Y is Hausdorff [ , a locally compact Hausdorff group, and find a (4A5Jb). We can therefore form its completion Z = X/Y left Haar measure ν˜ on Z such that ν is the corresponding subspace measure on X/Y , which is of full outer measure in Z (443K). The embedding X/Y ⊆ Z is therefore inverse-measure-preserving for ν and ν˜, so that φ, regarded as a map from X to Z, is inverse-measure-preserving for µ and ν˜. Also, of course, φ : X → Z is a continuous homomorphism. If E ⊆ X and µE > γ, there is a closed set E 0 ⊆ E such that µE 0 > γ. Now E 0 is of the form φ−1 [F ] where F ⊆ X/Y is closed and νF = µE 0 > γ. Next, F is of the form (X/Y ) ∩ F 0 where F 0 ⊆ Z is closed and ν˜F 0 = νF > γ. So there is a compact set K ⊆ F 0 such that ν˜K ≥ γ, and we have φ−1 [K] ⊆ φ−1 [F 0 ] = φ−1 [F ] ⊆ E,

µφ−1 [K] = ν(K ∩ (X/Y )) = ν˜K ≥ γ.

As E and γ are arbitrary, µ is inner regular with respect to {φ−1 [K] : K ⊆ Z is compact}. (c) If E ⊆ X is Haar measurable, then by 443J(b-i) there is an Fσ set E 0 ⊆ E such that E \ E 0 is Haar negligible. Now there are an Fσ set H ⊆ X/Y such that E 0 = φ−1 [H] and an Fσ set F ⊆ Z such that H = (X/Y ) ∩ F , in which case E 0 = φ−1 [F ], and E \ φ−1 [F ] is Haar negligible. 443M Theorem (Halmos 50) Let X be a topological group and µ a Haar measure on X. Then µ is completion regular. proof (a) Suppose first that µ is a left Haar measure and that X is locally compact and Hausdorff. In this case any self-supporting compact set K ⊆ X is a zero set. P P For each n ∈ N, there is an open neighbourhood Un of the identity such that µ(K4xK) ≤ 2−n for every x ∈ Un (443B); we may suppose that U n+1 ⊆ Un T T U U K is a G set. ? ? If K = 6 for each n. Each set U K is open (4A5Ed), so δ n n∈N n K, there is an n∈N n T x ∈ n∈N Un K \ K. For each n ∈ N, there are yn ∈ Un , zn ∈ K such that x = yn zn . Let F be any non-principal ultrafilter on N. Then z = limn→F zn is defined in K, so xz −1 = limn→F xzn−1 = limn→F yn is defined in X; because yn ∈ U i for every i ≤ n, T T xz −1 ∈ i∈N U i = i∈N Ui . Consequently µ(xz −1 K4K) = 0; because µ is left-translation-invariant, µ(K \ zx−1 K) = 0. But as x ∈ / K, z ∈ K \ zx−1 K and K ∩ (X \ zx−1 K) is non-empty. And zx−1 K is closed, so X \ zx−1 K is open and K is not self-supporting, contrary to hypothesis. X X T Thus K = n∈N Un K is a Gδ set. Being a compact Gδ set in a completely regular Hausdorff space, it is a zero set (4A2F(h-v)). Q Q Since µ is surely inner regular with respect to the compact self-supporting sets (414F), it is inner regular with respect to the zero sets, and is completion regular. (b) Now suppose that µ is a left Haar measure on an arbitrary topological group X. By 443L, we can find a locally compact Hausdorff topological group Z, a continuous homomorphism φ : X → Z and a left Haar measure ν˜ on Z such that φ is inverse-measure-preserving for µ and ν˜ and µ is inner regular with respect to {φ−1 [K] : K ⊆ Z is compact}. Now if E ∈ dom µ and γ < µE, there is a compact set K ⊆ Z such that φ−1 [K] ⊆ E and νK > γ. Next, there is a zero set L ⊆ K such that νL ≥ γ; in which case φ−1 [L] ⊆ E is a zero set and µφ−1 [L] ≥ γ. Thus µ is inner regular with respect to the zero sets and is completion regular.

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297

(c) Finally, if µ is a right Haar measure on a topological group X, let µ0 be the corresponding left Haar measure, setting µ0 E = µE −1 for Haar measurable sets E. Then µ0 is inner regular with respect to the zero sets; because x 7→ x−1 : X → X is a homeomorphism, so is µ. 443N I give a simple result showing how the measure-theoretic properties of groups carrying Haar measures have consequences of unexpected kinds. Proposition Let be X is a topological group carrying Haar measures (for instance, X might be any locally compact Hausdorff group). (i) Let G be a regular open subset of X. Then G is a cozero set. (ii) Let F be a nowhere dense subset of X. Then F is included in a nowhere dense zero set. proof (a) Suppose to begin with that X is locally compact, σ-compact and Hausdorff. Let µ be a left Haar measure on X; then µ is σ-finite, because X is covered by a sequence of compact sets, which must all have finite measure. (i) Write U for the family of open neighbourhoods of the identity e of X. For each U ∈ U, set HU = int{x : xU ⊆ G}; then {HU : U ∈ U} is an upwards-directed family of open sets with union G, as in the proofs of 442Ab and 442B, so G• = supU ∈U HU• in the measure algebra A of µ. Because µ is σ-finite, A isSccc (322G) and there is a sequence hUn in∈N in U such that G• = supn∈N HU• n (316E). In this case, G \ n∈N HUn is negligible, so must have empty interior. T By 4A5S, there is a closed normal subgroup Y of X, included in n∈N Un , such that X/Y is metrizable. Let π : X → X/Y be the canonical map. For each n ∈ N, Qn = π[HUn ] is open (4A5Ja), and HUn ⊆ π −1 [Qn ] = HUn Y ⊆ HUn Un ⊆ G. So G= Setting Q = int

S n∈N

S n∈N

HUn =

S n∈N

π −1 [Qn ].

Qn , and using 4A5Ja and 4A2B(f-ii), we see that S π −1 [Q] = int n∈N π −1 [Qn ] = int G = G

(this is where I use the hypothesis that G is a regular open set). But Q, being an open set in a metrizable space, is a cozero set (4A2Lc), so G = π −1 [Q] is a cozero set (4A2C(b-iv)), as required by (i). (ii) Now consider the nowhere dense set F ⊆ X. This time, let G be a maximal disjoint family of S cozero subsets of X \ F . Then G is countable, again because µ is σ-finite, and G is dense, because the S topology of X is completely regular. So X \ G is a nowhere dense zero set including F . (b) Next, suppose that X is any locally compact Hausdorff topological group. Then X has a σ-compact open subgroup X0 (4A5El). By (a), any regular open set in X0 is a cozero set in X0 . The same applies to all the (left) cosets of X0 , because these are homeomorphic to X0 . If C is any coset of X0 , then G ∩ C is a regular open set in C, so is a cozero set in C. But as the left cosets of X0 form a partition of X into open sets, G is also a cozero set in X (4A2C(b-vii)). Similarly, F ∩ C is nowhere dense in C for every left coset C of X0 , so is included in a nowhere dense zero set in C, and the union of these will be a nowhere dense zero set in X including F . b under the (c) Now suppose that X is a Hausdorff group carrying Haar measures. Then its completion X c X b therefore a cozero bilateral uniformity is locally compact (443K). H = intX is a regular open set in X, cG b set in X, by (b). But it is easy to check that G = X ∩ H, so that G is a cozero set in X (4A2C(b-v)). c

X b so is included in a nowhere dense zero set F 0 ⊆ X, b and F 0 ∩ X Similarly, F is a nowhere dense set in X, is a zero set in X including F .

(d) Finally, let X be any topological group carrying Haar measures. Set Y = {e}, where e is the identity of X, and let π : X → X/Y be the canonical map. G = π −1 [π[G]] and π[G] is a regular open set in X/Y (4A5Kb). X/Y is a Hausdorff topological group carrying Haar measures (see the proof of 443L), so π[G] is a cozero set, by (c), and G = π −1 [π[G]] is a cozero set in X, as claimed.

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Similarly, π[F ] is nowhere dense (4A5K(b-iv)), so is included in a nowhere dense zero set V ⊆ X/Y (by (c) here), and π −1 [V ] is a nowhere dense zero set (4A5K(b-v), 4A2C(b-iv)) including F . This completes the proof. 443O Quotient spaces I come now to the relationship between the modular functions of §442, normal subgroups and quotient spaces. Lemma Let X be a locally compact Hausdorff topological group and Y a closed subgroup of X. Let Z = X/Y be the set of left cosets of Y in X with the quotient topology and π : X 7→ Z the canonical map, so that Z is a locally compact Hausdorff space and we have a continuous action of X on Z defined by writing a•πx = π(ax) for a, x ∈ X (4A5Jc). Let ν be a left Haar measure on Y and write Ck (X), Ck (Z) for the spaces of continuous real-valued functions with compact supports on X, Z respectively. (a) We have a positive linear operator T : Ck (X) → Ck (Z) defined by writing (T f )(πx) =

R

Y

f (xy)ν(dy)

for every f ∈ Ck (X) and x ∈ X. If f > 0 in Ck (X) then T f > 0 in Ck (Z). If h ≥ 0 in Ck (Z) then there is an f ≥ 0 in Ck (X) such that T f = h. (b) If a ∈ X and f ∈ Ck (X), then T (a•l f )(z) = (T f )(a−1 •z) for every z ∈ Z. (c) Now suppose that a belongs to the normalizer of Y (that is, aY a−1 = Y ). In this case, we can define ψ(a) ∈ ]0, ∞[ by the formula ν(aF a−1 ) = ψ(a)νF for every F ∈ dom ν, and (T (a•r f )(πx) = ψ(a) · (T f )(π(xa)) for every x ∈ X and f ∈ Ck (X). proof I should begin by remarking that because Y is a closed subgroup of a locally compact Hausdorff group, it is itself a locally compact Hausdorff group, so does have a left Haar measure, which is a Radon measure. (a)(i) The first thing to check is that if f ∈ Ck (X) then T f is well-defined as a member of R Z . P P (α) If R x ∈ X, then y 7→ f (xy) : Y → R is a continuous function with compact support, so f (xy)ν(dy) is defined in R. (β) If x1 , x2 ∈ X and πx1 = πx2 , then x−1 1 x2 ∈ Y , and

R

f (x2 y)ν(dy) =

R

f (x1 (x−1 1 x2 y))ν(dy) =

R

f (x1 y)ν(dy),

applying 441J R to the function y 7→ f (x1 y) and the left action of Y on itself. Thus we can safely write (T f )(πx) = f (xy)ν(dy) for every x ∈ X, and T f will be a real-valued function on Z. Q Q (ii) Now T f is continuous for every f ∈ Ck (X). P P Given z0 ∈ Z, take x0 ∈ X such that z = πx0 . We have an h ∈ Ck (X)+ such that for every ² > 0 there is an open set U² containing x0 such that |f (x0 y) − f (xy)| ≤ ²h(y) whenever x ∈ U² and y ∈ X (4A5Pb). In this case, |(T f )(πx) − (T f )(x0 )| = |

R

R

f (xy) − f (x0 y)ν(dy)| ≤ ² Y h(y)ν(dy) R for every x ∈ U² , so that |(T f )(z) − (T f )(z0 )| ≤ ² Y h dν for every z ∈ π[U² ]. Since each π[U² ] is an open neighbourhood of z0 (4A5Ja), T f is continuous at z0 ; as z0 is arbitrary, T f is continuous. Q Q (iii) Since Z {z : (T f )(z) 6= 0} = {πx :

f (xy)ν(dy) 6= 0} ⊆ {πx : f (xy) 6= 0 for some y ∈ Y }

= {πx : f (x) 6= 0} ⊆ π[{x : f (x) 6= 0}] is relatively compact, T f ∈ Ck (Z) for every f ∈ Ck (X). (iv) The formula for T f makes it plain that T : Ck (X) → Ck (Z) is a positive linear operator. (v) If f ∈ Ck (X)+ , x ∈ X are such that f (x) > 0, then {y : y ∈ Y, f (xy) > 0} is a non-empty open set in Y ; because ν is strictly positive,

443P


(T f )(πx) =

R

299

f (xy)ν(dy) > 0.

In particular, T f > 0 if f > 0. Moreover, if z ∈ Z there is an f ∈ Ck (X)+ such that (T f )(z) > 0. Now {{z : (T f )(z) > 0} : f ∈ Ck (X)+ } is an upwards-directed family of open subsets of Z, so if L ⊆ Z is any compact set there is an f ∈ Ck (X)+ such that (T f )(z) > 0 for every z ∈ L. (vi) Now suppose that h ∈ Ck (Z)+ . By (v), there is an f0 ∈ Ck (X)+ such that (T f0 )(z) > 0 whenever z ∈ {w : h(w) 6= 0}. Setting h0 (z) = h(z)/T f0 (z) when h(z) 6= 0, 0 for other z ∈ Z, we have h0 (z) ∈ Ck (Z) and h = h0 × T f0 . Set f (x) = f0 (x)h0 (πx) ≥ 0 for every x ∈ X. Because h0 and π are continuous, f ∈ Ck (X). For any x ∈ X, Z Z 0 0 (T f )(πx) = f0 (xy)h (π(xy))ν(dy) = h (πx) f0 (xy)ν(dy) = h0 (πx)T f0 (πx) = h(πx). Thus T f = h. (b) If z = πx, then a−1 •z = π(a−1 x), so (T f )(a−1 •z) =

R

f (a−1 xy)ν(dy) =

R

(a•l f )(xy)ν(dy) = T (a•l f )(z).

(c) Define φ : Y → Y by writing φ(y) = a−1 ya for y ∈ Y . Because φ is a homeomorphism, the image measure νφ−1 is a Radon measure on Y ; because φ is a group automorphism, νφ−1 is a left Haar measure. (If F ∈ dom νφ−1 and y ∈ Y , then νφ−1 (yF ) = ν(ayF a−1 ) = ν(F a−1 ) = ν(aF a−1 ) = νφ−1 (F ).) It must therefore be a multiple of ν; say νφ−1 = ψ(a)ν. If g ∈ Ck (Y ), then

R

g(a−1 ya)ν(dy) =

R

gφ dν =

R

g d(νφ−1 ) = ψ(a)

R

g dν.

Now take f ∈ Ck (X). Then Z T (a r f )(πx) =

f (xya)ν(dy) Z f (xa(a−1 ya))ν(dy) = ψ(a) f (xay)ν(dy) •

Z =

Z (a r f )(xy)ν(dy) =

•

(using the remark above with g(y) = f (xay)) = ψ(a)(T f )(π(xa)) for every x ∈ X, as claimed. 443P Theorem Let X be a locally compact Hausdorff topological group and Y a closed subgroup of X. Let Z = X/Y be the set of left cosets of Y in X with the quotient topology, and π : X → Z the canonical map, so that Z is a locally compact Hausdorff space and we have a continuous action of X on Z defined by writing a•πx = π(ax) for a, x ∈ X. Let ν be a left Haar measure on Y . Suppose that λ is a non-zero X-invariant Radon measure on Z. (a) For each z ∈ Z, we have a Radon measure νz on X defined by the formula νz E = ν(Y ∩ x−1 E) whenever πx = z and the right-hand side is defined. In this case, for a real-valued function f defined on a subset of X,

R

f dνz =

R

f (xy)ν(dy)

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Topological groups

443P

whenever either side is defined in [−∞, ∞]. (b) We have a left Haar measure µ on X defined by the formulae

R

f dµ =

for every f ∈ Ck (X), and

RR

R

µG =

f dνz λ(dz)

νz (G)λ(dz)

for every open set G ⊆ X. (c) If D ⊆ Z, then D ∈ dom λ iff π −1 [D] ⊆ X is Haar measurable, and λD = 0 iff π −1 [D] is Haar negligible. (d) If νY = 1, then λ is the image measure µπ −1 . R (e) Suppose nowR that X is Then µE = νz (E)λ(dz) for every Haar measurable set E ⊆ X. R σ-compact. R If f ∈ L1 (µ), then f dµ = ( f dνz )λ(dz). (f) Still supposing that X is σ-compact, take f ∈ L1 (µ), and for a ∈ X set fa (y) = f (ay) whenever y ∈ Y and ay ∈ dom f . Then Qf = {a : a ∈ X, fa ∈ L1 (ν)} is µ-conegligible, and the function a 7→ fa• : Qf → L1 (ν) is almost continuous. proof (a) First, we do have a function νz depending only on z, because if z = πx1 = πx2 then x−1 2 x1 ∈ Y , so −1 −1 −1 ν(Y ∩ x−1 1 E) = ν(x2 x1 (Y ∩ x1 E)) = ν(Y ∩ x2 E)

whenever either side is defined. Of course νz , being the image of the Radon measure ν under the continuous map y 7→ xy : Y → X whenever πx = z, is always a Radon measure on X (418I). We also have

R

Y

f (xy)ν(dy) =

R

X

f dνπx

whenever x ∈ X and f is a real-valued function such that either side is defined in [−∞, ∞], by 235L. I remark here that if z ∈ Z then the coset C = π −1 [{z}] is νz -conegligible, because if πx = z then Y = Y ∩ x−1 C. (b)(i) Let T : Ck (X) → Ck (Z) be the positive linear operator of 443O; that is, (T f )(z) =

R

f (xy)ν(dy) =

R

f dνz

whenever f ∈ Ck (X), R x ∈ X and z = πx. Then we have a positive linear functional θ : Ck (X) → R defined by setting θ(f ) = T f dλ for every f ∈ Ck (X). R By the Riesz Representation Theorem (436J), there is a Radon measure µ on X defined by saying that f dµ = θ(f ) for Revery f ∈ Ck (X). Note that µ is non-zero. P P Because λ is non-zero, there is Rsome h ∈ Ck (Z)+ such that h dλ 6= 0; now there is some f ∈ Ck (X) such that T f = h, by 443Oa, and f dµ 6= 0. Q Q (ii) µ is a left Haar measure. P P If f ∈ Ck (X) and a ∈ X, then we have T (a•l f )(z) = (T f )(a−1 •z) for every z ∈ Z, by 443Ob. So Z

Z a•l f dµ =

Z

Z T (a•l f )dλ =

T f (a−1 •z)λ(dz) =

T f (z)λ(dz)

(by 441J or 441L, because λ is X-invariant) Z = f dµ. By 441L in the other direction, µ is invariant under the left action of X on itself, that is, is a left Haar measure. Q Q R (iii) If GR ⊆ X is open then µG = νz (G)λ(dz). P P Set A = {f : f ∈ Ck (X), 0 ≤ f ≤ χG}. Then µG = supf ∈A f dµ and νz G = supf ∈A

R

f dνz = supf ∈A (T f )(z)

for every z ∈ Z, by 414Ba, because νz is τ -additive. But as T f is continuous for every f ∈ A, and λ is τ -additive, we also have

443P


R

νz (G)λ(dz) = supf ∈A

R

T f dλ = supf ∈A

R

301

f dµ = µG. Q Q

R (c)(i) Let S A be the family of those sets A ⊆ X such that µA and νz (A)λ(dz) are defined in [0, ∞] and equal. Then n∈N An belongs to A whenever hAn in∈N is a non-decreasing sequence in A, and A \ B ∈ A whenever A, B ∈ A, B ⊆ A and µA < ∞. Moreover, if A ∈ A and µA = 0, then every subset of A belongs to A, since A must be νz -negligible for λ-almost every z. We also know from (b) that every open set belongs to A. Applying the Monotone Class Theorem (136B) to {A : A ∈ A, A ⊆ G}, we see that if E ⊆ X is a Borel set included in an open set G of finite measure, then E ∈ A. So if E is a relatively compact Haar measurable set, E ∈ A (using 443J(b-i), or otherwise). (ii) If D ∈ dom λ then π −1 [D] ∈ dom µ. P P Let K ⊆ X be compact. Then π[K] ⊆ Z is compact, so there are Borel sets F1 , F2 ⊆ Z such that F1 ⊆ D ∩ π[K] ⊆ F2 and F2 \ F1 is λ-negligible. Now νz (K ∩ π −1 [F2 \ F1 ]) = 0 whenever z ∈ / F2 \ F1 , by the remark added to the proof of (a) above, so µ(K ∩ π −1 [F2 \ F1 ]) =

R

νz (K ∩ π −1 [F2 \ F1 ])λ(dz) = 0.

Since K ∩ π −1 [F1 ] ⊆ K ∩ π −1 [D] ⊆ K ∩ π −1 [F2 ], K ∩ π −1 [D] ∈ dom µ. As K is arbitrary, π −1 [D] is measured by µ, so is Haar measurable. Q Q If λD = 0 then the same arguments show that µ(K ∩ π −1 [D]) = 0 for every compact K ⊆ X, so that µπ −1 [D] = 0 and π −1 [D] is Haar negligible. (iii) Now suppose that D ⊆ Z is such that π −1 [D] ∈ dom µ. Let L ⊆ Z be compact. Then there is a relatively compact open set G ⊆ X such that π[G] ⊇ L (because {π[G] : G ⊆ X is open and relatively compact} is an upwards-directed family of open sets covering Z). In this case,

R

νz (G ∩ π −1 [D ∩ L])λ(dz) = µ(G ∩ π −1 [D] ∩ π −1 [L])

is well-defined, by (i). But if z = πx then νz (G ∩ π −1 [D ∩ L]) = 0 if z ∈ / D∩L = νz G = ν(Y ∩ x−1 G) > 0 if z ∈ D ∩ L, because if z ∈ L then πx ∈ π[G] and Y ∩ x−1 G 6= ∅. So D ∩ L = {z : νz (G ∩ π −1 [D ∩ L]) > 0} is measured by λ. As L is arbitrary, and λ is a Radon measure, D ∈ dom λ. (d) If νY = 1, then, for any open set H ⊆ Z, νz π −1 [H] = 1 if z ∈ H, 0 otherwise. So µπ −1 [H] =

R

νz (π −1 [H])λ(dz) = λH.

Thus λ and the image measure µπ −1 agree on the open sets and, being Radon measures (418I), must be equal (416E(b-iii)). R (e) If X is actually σ-compact, then (c)(i) of this proof tells us that µE = νz (E)λ(dz) for every Haar measurable setRE ⊆ X, since RR E is the union of an increasing sequence of relatively compact measurable sets. Consequently f dµ = f dνz λ(dz) for every µ-simple function f . Now suppose that f is a non-negative µ-integrable function. Then there is a non-decreasing sequence hfn in∈N of non-negative µ-simple functions converging to f everywhere on dom f . If we set A = {z : νz∗ (X \ dom f ) > 0}, then

R

so λA = 0. Since

R

νz (X \ dom f )λ(dz) = µ(X \ dom f ) = 0, R f dνz = limn→∞ fn dνz for every z ∈ Z \ A, Z Z f dµ = lim fn dµ n→∞ ZZ ZZ = lim fn dνz λ(dz) = f dνz λ(dz). n→∞

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Topological groups

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Applying this to the positive and negative parts of f , we see that the same formula is valid for any µintegrable function f . (f )(i) If f ∈ L1 (µ) and a ∈ X, then

R

fa dν =

R

R

f (ay)ν(dy) = f dνπa R 1 if any of these are defined. So if f , g ∈ L (µ), kfa − ga k1 = |f − g|dνπa if either is defined. (ii) Let Φ be the set of all almost continuous functions from µ-conegligible subsets of X to L1 (ν), where L1 (ν) is given its norm topology. (In terms of the definition in 411M, a member φ of Φ is to be almost continuous with respect to the subspace measure on dom φ.) If φ ∈ Φ and ψ is a function from a conegligible subset of X to L1 (ν) which is equal almost everywhere to φ, then ψ ∈ Φ. If hφn in∈N is a sequence in Φ converging µ-almost everywhere to ψ, then ψ ∈ Φ (418F). (iii) For f ∈ L1 (µ), set φf (a) = fa• whenever this is defined in L1 (ν). Set M = {f : f ∈ L1 (µ), φf ∈ Φ}. If hfP in∈N is a sequence in M , f ∈ L1 (µ) and kf (n) − f k1 ≤ 4−n for every n, then f ∈ M . P P Set ∞ g = n=0 2n |f (n) − f |, defined on T P∞ {x : x ∈ dom f ∩ n∈N dom f (n) , n=0 2n |f (n) (x) − f (x)| < ∞}; (n)

then g ∈ L1 (µ). Now D = {z : z ∈ Z, g is νz -integrable} is λ-conegligible, by (e), and E = {a : a ∈ X, πa ∈ D} is µ-conegligible, by (c). If a ∈ E, then (n)

|fa

− fa | ≤ 2−n g νπa -a.e.

for every n ≥ 1. So Z

Z

|fa − fa(n) |dν = Y Z −n ≤2 g dνπa → 0

kφf (a) − φf (n) (a)k1 =

|f − f (n) |dνπa X

as n → ∞. Thus φf = limn→∞ φf (n) almost everywhere, and φ ∈ Φ, by (ii). Q Q (iv) Ck (X) ⊆ M . P P If f ∈ Ck (X) and a0 ∈ X, then there is an h ∈ Ck (X)+ such that for every ² > 0 there is an open set G² containing a0 such that |f (a0 y) − f (ay)| ≤ ²h(y) whenever a ∈ G² and y ∈ X (4A5Pb). In this case, kφf (a0 ) − φf (a)k1 =

R

|f (a0 y) − f (ay)|ν(dy) ≤ ²

R

Y

h dν

whenever a ∈ G² . As ² is arbitrary, φf is continuous at a0 ; as a0 is arbitrary, φf is continuous, and f ∈ M . Q Q (v) Now take any f ∈ L1 (µ). Then for each n ∈ N we can find f (n) ∈ Ck (X) such that kf (n) −f k1 ≤ 4−n (416I), so f ∈ M , by (iii). This completes the proof. 443Q Theorem Let X be a locally compact Hausdorff topological group and Y a closed subgroup of X. Let Z = X/Y be the set of left cosets of Y in X with the quotient topology, so that Z is a locally compact Hausdorff space and we have a continuous action of X on Z defined by writing a•(xY ) = axY for a, x ∈ X. Let ∆X be the left modular function of X and ∆Y the left modular function of Y . Then the following are equiveridical: (i) there is a non-zero X-invariant Radon measure λ on Z; (ii) ∆Y is the restriction of ∆X to Y . proof Fix a left Haar measure ν on Y , and let T : Ck (X) → Ck (Z) be the corresponding linear operator as defined in 443Oa. (a)(i)⇒(ii) Suppose that λ is a non-zero X-invariant Radon measure on Z. Construct a left Haar measure µ on X as in 443P.

443Q


303

Suppose that a ∈ Y . In this case, a surely belongs to the normalizer of Y , and, in the language of 443Oc, we have ν(aF a−1 ) = ψ(a)νF for every F ∈ dom ν. But as ν(aF a−1 ) = ν(F a−1 ) = ∆Y (a−1 )νF , we must have ψ(a) = ∆Y (a−1 ). Fix some f > 0 in Ck (X). We have T (a•r f )(πx) = ψ(a) · (T f )(π(xa)) = ψ(a) · (T f )(πx) for every x, so that (using 442Kc) we have Z Z Z −1 • ∆X (a ) f dµ = a r f dµ = T (a•r f )dλ Z Z Z = ψ(a) T f dλ = ψ(a) f dµ = ∆Y (a−1 ) f dµ. As

R

f dµ > 0, we must have ∆X (a−1 ) = ∆Y (a−1 ); as a is arbitrary, ∆Y = ∆X ¹ Y , as required by (ii).

(b)(ii)⇒(i) Now suppose that ∆Y = ∆X ¹ Y . This time, start with a left Haar measure µ on X. R α) (The key.) If f ∈ Ck (X) is such that T f ≥ 0 in Ck (Z), then f dµ ≥ 0. P (α P There is an h ∈ Ck (Z)+ such that h(z) = 1 whenever z ∈ Z and (T f )(z) 6= 0; by 443Oa, we can find a g ∈ Ck (X)+ such that T g = h. Now observe that x 7→ (T f )(πx) is a non-negative continuous real-valued function on X, so Z 0≤ ZX = X

g(x)(T f )(πx)µ(dx) Z Z Z g(x) f (xy)ν(dy)µ(dx) = g(x)f (xy)µ(dx)ν(dy) Y

Y

X

(by 417Ha or 417Hb, because (x, y) 7→ g(x)f (xy) : X × Y → R is a continuous function with compact support)

Z

Z ∆X (y −1 )

= Y

g(xy −1 )f (x)µ(dx)ν(dy) X

(applying 442Kc to the function x 7→ g(xy −1 )f (x)) Z Z = f (x) ∆X (y −1 )g(xy −1 )ν(dy)µ(dx) X

Y

(because (x, y) 7→ ∆X (y −1 )g(xy −1 )f (x) is continuous and has compact support) Z Z = f (x) ∆Y (y −1 )g(xy −1 )ν(dy)µ(dx) X

Y

(because ∆X ¹ Y = ∆Y , by hypothesis) Z Z = f (x) g(xy)ν(dy)µ(dx) X

Y

(applying 442K(b-ii) to the function y 7→ g(xy)) Z Z = f (x)(T g)(πx)µ(dx) = f (x)µ(dx) X

X

because (T g)(πx) = 1 whenever f (x) 6= 0. Q Q

R R R β ) Applying this to f and −f , we see that f dµ = 0 whenever T f = 0, so that f dµ = g dµ (β whenever f , g ∈ Ck (X) and T f = T g. Accordingly (because T is surjective) we have a functional θ : R Ck (Z) → R defined by saying that θ(T f ) = f dµ whenever f ∈ Ck (X), and θ is positive Rand linear. By the Riesz Representation Theorem again, there is a Radon measure λ on Z such that θ(h) = h dλ for every h ∈ Ck (Z). If a ∈ X and h ∈ Ck (Z) take f ∈ Ck (X) such that T f = h. Then, for any x ∈ X, (T f )(a•πx) = (T f )(π(ax)) =

R

f (axy)ν(dy) =

R

(a−1 •l f )(xy)ν(dy) = T (a−1 •l f )(πx).

304

Topological groups

443Q

So Z

Z

Z

(T f )(a z)λ(dz) = T (a−1 •l f )(z)λ(dz) Z Z Z = a−1 •l f dµ = f dµ = h(z)λ(dz).

h(a z)λ(dz) = •

•

By 441L, λ is X-invariant. Also λ is non-zero because there is surely some f such that have the required non-zero X-invariant Radon measure on Z.

R

f dµ 6= 0. So we

443R Applications This theorem applies in a variety of cases. Let X be a locally compact Hausdorff topological group and Y a closed subgroup of X. (a) If Y is a normal subgroup of X, then ∆Y = ∆X ¹ Y . P P X/Y has a group structure under which it is a locally compact Hausdorff group (4A5Jb). It therefore has a left Haar measure, which is surely X-invariant in the sense of 443Q. Q Q (b) If Y is compact, then ∆Y = ∆X ¹ Y . P P Both ∆Y and ∆X ¹ Y are continuous homomorphisms from Y to ]0, ∞[; since the only compact subgroup of ]0, ∞[ is {1}, they are both constant with value 1. Q Q So 443Q tells us that we have an X-invariant Radon measure λ on X/Y . Since Y has a Haar probability measure, we can take λ to be the image of a left Haar measure under the canonical map (443Pd). (c) If Y is a compact normal subgroup, then we find that, for W ∈ dom λ and x ∈ X, λ(W · πx) = µ(π −1 [W ] · x) = ∆X (x)λW , so that ∆X/Y (πx) = ∆X (x), writing π : X → X/Y for the canonical map. This is a special case of 443S below, because (in the terminology there) ψ(a) = νY /νY = 1 for every a ∈ X. (d) If Y is open, ∆Y = ∆X ¹ Y . P P If µ is a left Haar measure on X, then the subspace measure µY is a left Haar measure on Y (443F). There is an open set G ⊆ Y such that 0 < µG < ∞, and now µ(Gy) = ∆X (y)µG = ∆Y (y)µG for every y ∈ Y . Q Q This time, X/Y is discrete, so counting measure is an X-invariant Radon measure on X/Y . 443S Theorem Let X be a locally compact Hausdorff topological group and Y a closed normal subgroup of X; let Z = X/Y be the quotient group, and π : X → Z the canonical map. Write ∆X , ∆Z for the left modular functions of X, Z respectively. Define ψ : X → ]0, ∞[ by the formula ν(aF a−1 ) = ψ(a)νF whenever F ∈ dom ν and a ∈ X, where ν is a left Haar measure on Y (cf. 443Oc). Then ∆Z (πa) = ψ(a)∆X (a) for every a ∈ X. proof Let T : Ck (X) → Ck (Z) be the map defined in 443O, and λRa left Haar R measure on Z; then, as in 443Pb, we have a left Haar measure µ on X defined by the formula f dµ = T f dλ for every f ∈ Ck (X). Fix on some f > 0 in Ck (X) and a ∈ X, and set w = πa. By 443Oc, we have T (a•r f )(z) = ψ(a)(T f )(π(xa)) = ψ(a)(T f )(zw) whenever πx = z. So Z −1

∆X (a

)

Z f dµ =

(442Kc)

a•r f dµ Z

=

Z

T (a•r f )dλ = ψ(a) (T f )(zw)λ(dz) Z Z = ψ(a)∆Z (w−1 ) T f (z)λ(dz) = ψ(a)∆Z (w−1 ) f dµ.

443T


305

Thus ∆X (a−1 ) = ψ(a)∆Z (w−1 ); because both ∆X and ∆Z are multiplicative, ∆Z (πa) = ∆Z (w) = ψ(a)∆X (a).

443T Transitive actions All the results from 443O onwards have been expressed in terms of groups acting on quotient groups. But the same structures can appear if we start from a group action. To simplify the hypotheses, I give the following result for compact groups only. Theorem Let X be a compact Hausdorff topological group, Z a non-empty compact Hausdorff space, and a transitive continuous action of X on Z. For z ∈ Z, write πz (x) = x•z for each x ∈ X. (a) For every z ∈ Z, Yz = {x : x ∈ X, x•z = z} is a compact subgroup of X. If we give the set X/Yz of left cosets of Yz in X its quotient topology, we have a homeomorphism φz : X/Yz → Z defined by the formula φz (xYz ) = x•z for every x ∈ X. (b) Let µ be a Haar probability measure on X. Then the image measure µπz−1 is an X-invariant Radon −1 probability measure on Z, and µπw = µπz−1 for all w, z ∈ Z. (c) Every non-zero X-invariant Radon measure on Z is of the form µπz−1 for a Haar measure µ on X and some (therefore any) z ∈ Z. (d) There is a strictly positive X-invariant Radon probability measure on Z, and any two non-zero X-invariant Radon measures on Z are scalar multiples of each other. (e) Take any z ∈ Z, and let ν be the Haar probability measure of Yz . If µ is a Haar measure on X, then •

µE =

R

ν(Yz ∩ x−1 E)µ(dx)

whenever E ⊆ X is Haar measurable. proof (a) Given z ∈ Z, then for x, y ∈ X we have x•z = y •z ⇐⇒ x−1 y ∈ Yz ⇐⇒ xYz = yYz . So the formula given for φz defines a bijection between X/Yz and Z. To see that φz is continuous, take any open set H ⊆ Z. Then −1 • {x : xYz ∈ φ−1 z [H]} = {x : x z ∈ H} = πz [H]

is open in X (because • is continuous), so φ−1 z [H] is open in X/Yz . Because X/Yz is compact and φz is a bijection, it is a homeomorphism (3A3Dd). (b) Because X is compact, therefore unimodular (442Ic), we can speak of ‘Haar measures’ on X without specifying ‘left’ or ‘right’. If µ is a Haar probability measure on X, then the image measure µφ−1 z is a Radon probability measure on Z (418I). To see that the measures µφ−1 are X-invariant, take any Borel set H ⊆ Z z and y ∈ X, and consider −1 • (µφ−1 H) = µ{x : x•z ∈ y −1 •H} = µ{x : yx•z ∈ H} z )(y −1 −1 = µ(y −1 φ−1 z [H]) = µ(φz [H]) = (µφz )(H).

By 441B, this is enough to ensure that µφ−1 z is invariant. If w, z ∈ Z and H ⊆ Z is a Borel set, then there is a y ∈ X such that y •w = z (because the action is transitive), and now −1 • • φ−1 ) z ∈ H} = (φ−1 w [H] = {x : x w ∈ H} = {x : (xy z [H])y.

But µ is a two-sided Haar measure, so −1 −1 −1 −1 (µφ−1 w )(H) = µ(φw [H]) = µ((φz [H])y) = µ(φz [H]) = (µφz )(H). −1 Thus µφ−1 w and µφz agree on the Borel sets and must be equal (416Eb).

(c) Now we come to the interesting bit. Suppose that λ is a non-zero X-invariant Radon measure on Z. Take any z ∈ Z and consider the Radon measure λ0 on X/Yz got by setting λ0 H = λφz [H] whenever H ⊆ X/Yz and φz [H] is measured by λ. In this case, if x ∈ X and H ⊆ X/Yz is measured by λ,

306

Topological groups

443T

λ0 (x•H) = λ{φz (x•w) : w ∈ H} = λ{φz (x•yYz ) : y ∈ X, yYz ∈ H} = λ{φz (xyYz ) : y ∈ X, yYz ∈ H} = λ{xy •z : y ∈ X, yYz ∈ H} = λ{x•(y •z) : y ∈ X, yYz ∈ H} = λ(x•φz [H]) = λφz [H] (because λ is X-invariant) = λ0 H. So λ0 is X-invariant. Now let ν be the Haar probability measure on the compact R Hausdorff group Yz . By 443Pb, we have a left Haar measure µ on X defined by the formula µG = νw (G)λ0 (dw) for every open G ⊆ X, where νxYz (G) = ν(Y ∩ x−1 G) for every y ∈ X and every open G ⊆ X. Let H ⊆ Z be an open set. Then for any x ∈ X, νxYz (πz−1 [H]) = ν{y : y ∈ Yz , xy ∈ πz−1 [H]} = ν{y : y ∈ Yz , xy •z ∈ H} = ν{y : y ∈ Yz , x•(y •z) ∈ H} = νYz = 1 if x•z ∈ H, = ν∅ = 0 otherwise. So µ(πz−1 [H]) = λ0 {xYz : x ∈ X, x•z ∈ H} = λ0 φ−1 z [H] = λH. As H is arbitrary, the image measure µπz−1 agrees with λ on the open subsets of Z; as they are both Radon measures, µπz−1 = λ, as required. (d) This is now easy. X carries a non-zero Haar measure, so by (b) there is an X-invariant Radon probability measure on Z. If λ1 and λ2 are non-zero X-invariant Radon measures on Z, then they are −1 and µ2 πz−1 where µ1 and µ2 are Haar measures on X and w, z ∈ Z. By (b) again, of the form µ1 πw −1 −1 µ1 πw = µ1 πz , and since µ1 and µ2 are multiples of each other (442B), so are λ1 and λ2 . To see that the invariant probability measure λ on X is strictly positive, take any non-empty open set H ⊆ Z. Then Z is covered by the open sets x•H, as x runs through X. Because Z is compact, it is covered by finitely many of these, so at least one of them has non-zero measure. But they all have the same measure as H, so λH > 0. (e) Let θ : X → X/Yz be the canonical map. For w ∈ X/Yz we have a Radon measure νw on X defined by setting νw E = ν(Yz ∩ x−1 E) whenever θ(x) = w and the right-hand side is defined (443Pa). By 443Q/443Rb, or otherwise, there is a non-zero X-invariant Radon measure λ on X/Yz ; re-scaling if necessary, we may suppose that λ(X/Yz ) = µX. By 443Pe, we have a Haar measure µ0 on X defined by R 0 setting µ E = νw E λ(dw) for every Haar measurable E; since µ0 X = 0

0 −1

µ = µ. Moreover, λ = µ θ

R

νw X λ(dw) =

R

νYz λ(dw) = λ(X/Yz ) = µX,

(443Pd). So Z µE = µ0 E = νw E λ(dw) Z Z = νθ(x) E µ(dx) = ν(Yz ∩ x−1 E)µ(dx)

for every Haar measurable E ⊆ X. 443X Basic exercises (a) Let X be a topological group carrying Haar measures. (i) Show that a set A ⊆ X is self-supporting (definition: 411Na) for one Haar measure on X iff it is self-supporting for every Haar measure on X. (ii) Show that a set A ⊆ X has non-zero inner measure for one Haar measure on X iff it has non-zero inner measure for every Haar measure on X.

443Xm


307

> (b) Let X be a topological group and µ, ν two left Haar measures on X. (i) Show that the quasi-Radon product measure λ on X ×X is a left Haar measure for the product group operation on X ×X. (ii) Show that the maps (x, y) 7→ (x, xy), (x, y) 7→ (y −1 x, y) are automorphisms of the measure space (X ×X, λ). (Hint: use 417C(iv) to show that they preserve the measures of open sets.) (iii) Show that the maps (x, y) 7→ (y −1 , xy), (x, y) 7→ (yx, x−1 ) are automorphisms of (X × X, λ). (Hint: express them as compositions of maps of the forms in (ii).) (iv) Show that the maps (x, y) 7→ (y, x) and (x, y) 7→ (y, y −1 x) are automorphisms of (X × X, λ). (c) Let X be a topological group, µ a Haar measure on X, and E, F measurable subsets of X. Show that (x, y, w, z) 7→ µ(xEy ∩ wF z) : X 4 → [0, ∞] is lower semi-continuous. (d) Let X be a topological group carrying Haar measures and A its Haar measure algebra. Show that we have a continuous action of X × X on A defined by the formula (x, y)•E • = (xEy −1 )• for x, y ∈ X and Haar measurable sets E ⊆ X. (e) Let X be a topological group carrying Haar measures and A its Haar measure algebra. Show that if x ∈ X, a ∈ A then x•r a = (x•l a ˜)∼ , where a ˜ is as defined in 443Af. (f ) Let X be a topological group carrying Haar measures, and F ⊆ X a Haar measurable set such that F ∩ U is not Haar negligible for any neighbourhood U of the identity. Show that for any A ⊆ X the set A0 = {x : x ∈ A, A ∩ xF is Haar negligible} is Haar negligible. (Hint: otherwise, take a closed non-negligible set H of finite measure, a Haar measurable envelope of H ∩ A0 , and show that H ∩ xF is negligible for every x ∈ H. Consider the set {(x, y) : x ∈ H, y ∈ F, xy ∈ H} = {(x, y) : y ∈ F, x ∈ H ∩ Hy −1 }; show that its vertical sections are negligible but some of its horizontal sections are not.) (g) Let X be a topological group with a left Haar measure µ. (i) Show that if Y is a subgroup of X such that µ∗ Y > 0, then Y is open. In particular, any non-negligible closed subgroup of X is open. (ii) Let Y be any subgroup of X which is not Haar negligible. Show that the subspace measure µY is a left Haar measure on Y . Show that Y is a Haar measurable envelope of Y . (Hint: apply 443Db inside the topological group Y .) (h) Write out a version of 443G for right Haar measures. (i) Let X be a topological group carrying Haar measures, and L0 the space of equivalence classes of Haar measurable functions, as in 443A; let u 7→ u ˜ : L0 → L0 be the operator of 443Af. (i) Show that if µ is a left Haar measure on X and ν is a right Haar measure, p ∈ [1, ∞] and u ∈ L0 , then u ˜ ∈ Lp (ν) iff u ∈ Lp (µ). (ii) Show that x•r u = (x•l u ˜)∼ for any x ∈ X, u ∈ L0 . (j) Let X be a topological group carrying Haar measures, and A its Haar measure algebra. Show that, in the language of 443C and 443G, χ(x•l a) = x•l χa and χ(x•r a) = x•r χa for every x ∈ X and a ∈ A. (k) Let X be a topological group carrying Haar measures. Show that X is totally bounded for its bilateral uniformity iff its Haar measures are totally finite. (l) Let X be a topological group, µ a left Haar measure on X, and A ⊆ X a set which is self-supporting for µ. Show that the following are equiveridical: (i) for every neighbourhood U of the identity e, there is a countable set I ⊆ X such that A ⊆ U I; (ii) for every neighbourhood U of e, there is a countable set I ⊆ X such that A ⊆ IU ; (iii) for every neighbourhood U of e, there is a countable set I ⊆ X such that A ⊆ IU I; (iv) A can be covered by countably many sets of finite measure for µ; (v) A can be covered by countably many open sets of finite measure for µ; (vi) A can be covered by countably many sets which are totally bounded for the bilateral uniformity on X. >(m) Let X be a topological group carrying Haar measures. (i) Show that the following are equiveridical: (α) X is ccc; (β) X has a σ-finite Haar measure; (γ) every Haar measure on X is σ-finite. (ii) Show that if X is locally compact and Hausdorff, we can add (δ) X is σ-compact.

308

Topological groups

443Xn

(n) Let X be a topological group carrying Haar measures. Show that every subset of X has a Haar measurable envelope which is a Borel set. (o) In 443L, show that ∆X = ∆Z φ, where ∆X , ∆Z are the left modular functions of X, Z respectively. > (p) Let X and Y be topological groups with left Haar measures µ and ν. Show that the c.l.d. and quasi-Radon product measures of µ and ν on X × Y coincide. (Hint: start with locally compact Hausdorff b spaces, and show that a compact Gδ set in X × Y belongs to B(X)⊗B(Y ), where B(X) and B(Y ) are the Borel σ-algebras of X and Y ; now use 443L.) Q (q) Let hXi ii∈I be a family of topological groups and X = i∈I Xi their product. Suppose that each Xi has a Haar probability measure µi . Show that the ordinary and quasi-Radon product measures on X coincide. (r) Let X be a topological group and µ a left Haar measureTon X. Show that µ is inner regular with respect to the family of closed sets F ⊆ X such that F = n∈N F Un , for some sequence hUn in∈N of neighbourhoods of the identity. (Hint: use 443M for the locally compact case, and 443L to extend the result to general Haar measures.) (s) Let X be a locally compact Hausdorff group and Y a closed subgroup of X; write X/Y for the space of left cosets of Y in X, with its quotient topology. Show that if λ1 and λ2 are non-zero X-invariant Radon measures on X/Y , then each is a multiple of the other. (Hint: look at the Haar measures they define on X.) > (t) Write S 1 for the circle group {s : s ∈ C, |s| = 1}, and set X = S 1 × S 1 , where the first copy of S is given its usual topology and the second copy is given the discrete topology, so that X is an abelian locally compact Hausdorff group. Set E = {(s, s) : s ∈ S 1 }. Show that E is a closed Haar negligible subset of X. Set Y = {(1, s) : s ∈ S 1 }; check that Y is a closed normal subgroup of X, and that the quotient group X/Y can be identified with S 1 with its usual topology; let λ be the Haar probability measure of X/Y . Let ν beR counting measure on Y . Show that, in the language of 443P, νz E = 1 for every z ∈ X/Y , so that µE 6= νz (E)λ(dz). Setting f = χE, show that the map a 7→ fa• described in 443Pf is not almost continuous. 1

(u) (i) In 443O, suppose that G ⊆ X is an open set such that GY = X. Show that for every h ∈ Ck (Z)+ there is an f ∈ Ck (X)+ such that T f = h and {x : f (x) > 0} ⊆ G. (ii) In 443Q, suppose that there is an open set G ⊆ X such that GY = X and G has finite measure for the left Haar measures of X. Show that Z has an X-invariant Radon probability measure. > (v) Let X = R 2 be the example of 442Xf. (i) Let Y1 be the subgroup {(ξ, 0) : ξ ∈ R}. Describe the left cosets of Y1 in X. Show that there is no non-trivial X-invariant Radon measure on the set X/Y of these left cosets. Find a base U for the topology of X/Y such that you can identify the sets x•U , where x ∈ X and U ∈ U , with sufficient precision to explain why the hypothesis (iii) of 441C is not satisfied. (ii) Let Y2 be the normal subgroup {(0, ξ) : ξ ∈ R}. Find the associated function ψ : X → ]0, ∞[ as described in 443Oc and 443S. (w) Let X be a locally compact Hausdorff group and Y a compact normal subgroup of X. Show that X is unimodular iff the quotient group X/Y is unimodular. (Hint: the function ψ of 443S must be constant.) (x) Take any integer r ≥ 1, and let G be the isometry group of R r with its topology of pointwise convergence (441G). (i) Show that G is metrizable and locally compact. (Hint: 441Xm.) (ii) Let H ⊆ G be the set of translations. Show that H is an abelian closed normal subgroup of G, and that Lebesgue measure on R r can be regarded as a Haar measure on H. (iii) Show that the quotient group G/H is compact. (iv) Show that G is unimodular. (Hint: the function ψ of 443S is constant.) > (y) Set X = R 3 with the operation (ξ1 , ξ2 , ξ3 ) ∗ (η1 , η2 , η3 ) = (ξ1 + η1 , ξ2 + eξ1 η2 , ξ3 + e−ξ1 η3 ).

443Yk


309

(i) Show that (with the usual topology of R 3 ) X is a topological group. (ii) Show that it is unimodular. (Hint: set Y = {(ξ1 , ξ2 , ξ3 ) : ξ1 = 0} C X, show that Lebesgue 2-dimensional measure on Y is a Haar measure, and that the conjugacy action of X on Y is measure-preserving, while X/Y ∼ = R is unimodular.) (iii) Show that X has both a closed subgroup and a Hausdorff quotient group which are not unimodular. > (z) Let (X, ρ) be a non-empty compact metric space such that the group G of isometries of X is transitive. Show that any two non-zero G-invariant Radon measures on X must be multiples of each other. (Hint: 441Gb, 443T.) 443Y Further exercises (a) Let X be a topological group carrying Haar measures and A its Haar measure algebra. Show that two principal ideals of A are isomorphic (as Boolean algebras) iff they have the same cellularity. (b) Let X be a locally compact Hausdorff group. Show that we have continuous actions X on the Banach space C0 (X) defined by formulae corresponding to those of 443G.

•

l,

•

r

and

•

c

of

(c) Give an example of a locally compact Hausdorff group, with left Haar measure µ, such that no open normal subgroup can be covered by a sequence of sets of finite measure for µ. (d) Let X be a topological group with a left Haar measure µ, and A ⊆ X. Show that the following are equiveridical: (i) A is totally bounded for the bilateral uniformity of X (ii) there are non-empty open sets G, H ⊆ X such that µ(AG), µ(A−1 H) are both finite. (e) Let X be a topological group carrying Haar measures. Let E ⊆ X be a Haar measurable set such that E ∩ U is not Haar negligible for any neighbourhood U of the identity. Show that there is a sequence hxn in∈N in E such that xi0 xi1 . . . xin ∈ E whenever n ∈ N and i0 < i1 < . . . < in in N. (See Plewik & Voigt 91.) (f ) Let X be a topological group and µ a Haar measure on X. Show that any closed self-supporting subset of X is a zero set. (Hint: start with the case in which X is locally compact and σ-compact.) (g) Let X be a topological group. Let Σ be the family of subsets of X expressible in the form φ−1 [F ] for some Borel subset F of a separable metrizable topological group Y and some continuous homomorphism φ : X → Y . Show that Σ is a σ-algebra of subsets of X and that multiplication, regarded as a function b Σ)-measurable. Show that any compact Gδ set belongs to Σ. Show that if X from X × X to X, is (Σ⊗Σ, is σ-compact, then Σ is the Baire σ-algebra of X. (h) Let X be any Hausdorff topological group of cardinal greater than c. Let B be the Borel σ-algebra b of X. Show that (x, y) 7→ xy is not (PX ⊗PX, B)-measurable. (Hint: show that for any sequence hAn in∈N in PX there are distinct x, y ∈ X such that {n : x ∈ An } = {n : y ∈ An }; consequently {(x, y) : xy −1 = b e} ∈ / PX ⊗PX.) (i) Find a compact Hausdorff space X with a strictly positive Radon measure such that there is a regular open set G ⊆ X which is not a cozero set. (Hint: 419A.) (j) Show that there is a subgroup X of the additive group R 2 such that X has full outer Lebesgue measure but {ξ : (ξ, 0) ∈ X} = Q. (Hint: enumerate the family K of compact subsets of R 2 of positive Lebesgue measure as hKξ iξ
310

Topological groups

443Yl

(l) Show that if X is a locally compact Hausdorff group, Y a compact subgroup of X, Z = X/Y the set of left cosets of Y and π : X → Z the canonical map, and Z is given its quotient topology, then R = {(πx, x) : x ∈ X} is an usco-compact relation in Z × X and R[L] = π −1 [L] is compact for every compact L ⊆ Z. (m) Let G be the isometry group of 443Xx. (i) Show that if we set ρ(g, h) = supkxk≤1 kg(x) − h(x)k, then ρ is a metric on G defining its topology. (ii) Describe Haar measures on X (α) in terms of Lebesgue measure on a suitable Euclidean space (β) in terms of Hausdorff measure of an appropriate dimension for the metric ρ. (n) Let r ≥ 1 be an integer, and X the group of invertible affine transformations of R r , with the topology r of pointwise convergence inherited from (R r )R . (i) Show that X is a topological group. (ii) Show that X is not unimodular, and find its modular functions. (Hint: use the method of 443Xx, supported by 442Yb.) (o) Let X be a topological group with a left Haar measure µ and left modular function ∆; suppose that X is not unimodular. Show that µ{x : α < ∆(x) < β} = ∞ whenever α < β. (Hint: begin by supposing that X is locally compact and Hausdorff. Set Y = {x : ∆(x) = 1}. By 443S, νY = ∞, where ν is any left Haar measure on Y , so Y cannot be totally bounded for its own bilateral uniformity nor for the bilateral uniformity of X. Consequently µ(Y G) = ∞ for any non-empty open set G ⊆ X, by 443Yd. For general X use 443L.) 443 Notes and comments Most of us, by the time we come to study measures on general topological groups, have come to trust our intuition concerning the behaviour of Lebesgue measure on R, and the principal discipline imposed by the subject is the search for the true path between hopelessness and overconfidence when extending this intuition to the general setting. The biggest step is the loss of commutativity, especially as the non-abelian groups of elementary courses in group theory are mostly finite, and are therefore untrustworthy guides to the concerns of this chapter. Accordingly we find ourselves going rather slowly and carefully through the calculations in such results as 443C. Note, for instance, that in an abelian group the actions I call •l and •r are still different; x•l E corresponds to x + E = E + x, but x•r E corresponds to E − x. (The action •c becomes trivial, of course.) When we come to translate the formulae of Fourier analysis in the next section, manoeuvres of this kind will often be necessary. In the present section, I have done my best to give results in ‘symmetric’ forms; you may therefore take it that when the words ‘left’ or ‘right’ appear in the statement of a proposition, there is some real need to break the symmetry. Subject to these remarks, such results as 443C and 443G are just a matter of careful conventional analysis. I see that I have used slightly different techniques in the two cases. Of course 443C can be thought of as a special case of 443Gc, if we remember that χ : A → L0 (A) embeds A topologically as a subspace of L0 (367R). 443B, 443D and 443E belong to a different family; they deal with actual sets rather than with members of a measure algebra or a function space. I suppose it is 443D which will most often be quoted. Its corollary 443E deals with the obvious question of when Haar measures are tight. Readers who have previously encountered the theory of Haar measures on locally compact groups will have been struck by how closely the more general theory here is able to follow it. The explanation lies in 443K-443L; my general Haar measures are really just subspace measures on subgroups of full Haar measure in locally compact groups. Knowing this, it is easy to derive all the results above from the locally compact theory. I use this method only in 443M, because (following Halmos 50) I feel that from the point of view of pure measure theory the methods show themselves more clearly if we do not use ideas depending on compactness, but instead rely directly on τ -additivity. But I note that 443Xp and 443Yd also go faster with the aid of 443L. Halmos 50 goes a little farther, with a theory of groups carrying translation-invariant measures for which b Σ)-measurable, where Σ is the domain of the measure. The essence of the the operation (x, y) 7→ xy is (Σ⊗Σ, theory here, and my reason for insisting on τ -additive measures, is that for these we have a suitable theory of product measures. If we start with quasi-Radon measures and use the quasi-Radon product measure, then multiplication is measurable just because it is continuous. In order to achieve similar results without either assuming metrizability or using the theory of τ -additive product measures, we have to restrict the measure to something smaller than the Borel σ-algebra, as in 443Yd. (See 443Yh.) 443J and 443Xp show

444B

Convolutions

311

that certain disconcerting features of general Radon measures (419C, 419E) cannot arise in the case of Haar measures. When I come to look at subgroups and quotient groups, I do specialize to the locally compact case. One obstacle to generalization is the fact that a closed subgroup of a group carrying Haar measures need not itself carry Haar measures (443Yj). 443O and 443Q are taken from Federer 69, who goes farther, with many other applications. I give a fairly detailed analysis of the relationship between a Haar measure on a group X and a corresponding X-invariant measure on a family of left cosets (443P) for the sake of applications in §447. One of the challenges here is to distinguish clearly those results which apply to all locally compact groups from those which rely on σ-compactness or some such limitation. There is a standard example (443Xt) which provides a useful test in such questions. You may recognise this as a version of a fundamental example related to Fubini’s theorem, given in 252K. Most of the results of this section begin with a topological group X. But starting from §441 it is equally natural to start with a group action. If we have a topological group X acting continuously on a topological space Z, and X carries Haar measures, then we have a good chance of finding an invariant measure on Z. In the simplest case, in which X and Z are both compact and the action is transitive, there is a unique invariant Radon probability measure on Z (443T).

444 Convolutions In this section, I look again at the ideas of §§255 and 257, seeking the appropriate generalizations to topological groups other than R. Following Hewitt & Ross 63, I begin with convolutions of measures (444A-444E) before proceeding to convolutions of functions (444O-444V); in between, I mention the convolution of a function and a measure (444G-444M) and a general result concerning continuous group actions on quasi-Radon measure spaces (444F). 444A Convolution of measures: Proposition If X is a topological group and λ and ν are two totally finite quasi-Radon measures on X, we have a quasi-Radon measure λ ∗ ν on X defined by saying that (λ ∗ ν)(E) = (λ × ν){(x, y) : xy ∈ E} Z Z = ν(x−1 E)λ(dx) = λ(Ey −1 )ν(dy) for every E ∈ dom(λ ∗ ν), where λ × ν is the quasi-Radon product measure on X × X. proof Because multiplication is continuous, Q(E) = {(x, y) : xy ∈ E} is a Borel set in X × X for every Borel set E ⊆ X, so the formula θE = (λ × ν)Q(E) defines a functional on the Borel σ-algebra of X. Of course θ is non-negative and countably additive, and it is τ -additive because λ × ν is. Because X, being a topological group, is regular (4A5Ha), the completion of θ is a quasi-Radon measure (415Cb); call this measure λ ∗ ν. If E ∈ dom(λ ∗ ν), we have Borel sets E 0 , E 00 such that E 0 ⊆ E ⊆ E 00 and (λ ∗ ν)(E 0 ) = (λ ∗ ν)(E 00 ) = (λ ∗ ν)(E), so that (λ × ν)Q(E 0 ) = (λ × ν)Q(E 00 ) = (λ ∗ ν)(E), and (λ × ν)Q(E) must be defined and equal to (λ ∗ ν)(E). As for the other formulae, Fubini’s theorem (417Ha) tells us that (λ ∗ ν)(E) = (λ × ν)Q(E) Z Z = νQ(E)[{x}]λ(dx) = ν(x−1 E)λ(dx) Z Z = λQ(E)−1 [{y}]ν(dy) = λ(Ey −1 )ν(dy) for any E ∈ dom(λ ∗ ν).

312

Topological groups

444B

444B Proposition If X is a topological group, λ1 ∗ (λ2 ∗ λ3 ) = (λ1 ∗ λ2 ) ∗ λ3 for all totally finite quasi-Radon measures λ1 , λ2 and λ3 on X. proof If E ⊆ X is Borel, then Z (λ2 ∗ λ3 )(x−1 E)λ1 (dx)

(λ1 ∗ (λ2 ∗ λ3 ))(E) = Z =

(λ2 × λ3 ){(y, z) : xyz ∈ E}λ1 (dx)

= (λ1 × (λ2 × λ3 )){(x, (y, z)) : xyz ∈ E} = ((λ1 × λ2 ) × λ3 ){((x, y), z) : xyz ∈ E} Z = (λ1 × λ2 ){(x, y) : xyz ∈ E}λ3 (dz) = ((λ1 ∗ λ2 ) ∗ λ3 )(E). (For the central identification between λ1 ×(λ2 ×λ3 ) and (λ1 ×λ2 )×λ3 , observe that as both are quasi-Radon measures it is enough to check that they agree on sets of the type G1 × (G2 × G3 ) ∼ = (G1 × G2 ) × G3 where G1 , G2 and G3 are open, as in 417J.) So λ1 ∗ (λ2 ∗ λ3 ) and (λ1 ∗ λ2 ) ∗ λ3 agree on the Borel sets and must be identical. 444C Proposition Let X be a topological group and λ, ν two totally finite quasi-Radon measures on X. Then

R

f d(λ ∗ ν) =

R

f (xy)(λ × ν)d(x, y) =

RR

f (xy)λ(dx)ν(dy) =

RR

f (xy)ν(dy)λ(dx)

for any (λ ∗ ν)-integrable real-valued function f . In particular, (λ ∗ ν)(X) = λX · νX. proof If f is of the form χE, so that

R

RR

f d(λ ∗ ν) = (λ ∗ ν)(E),

f (xy)λ(dx)ν(dy) =

R

R

f (xy)(λ × ν)d(x, y) = (λ × ν){(x, y) : xy ∈ E},

λ(Ey −1 )ν(dy),

RR

f (xy)ν(dy)λ(dx) =

R

ν(x−1 E)λ(dx)

this is covered by the result in 444A. Now it is easy to run through the standard progression to the cases of (i) simple functions (ii) non-negative Borel measurable functions defined everywhere (iii) functions defined, and zero, almost everywhere (iv) non-negative integrable functions and (v) arbitrary integrable functions. 444D Proposition Let X be an abelian topological group. Then λ ∗ ν = ν ∗ λ for all totally finite quasi-Radon measures λ, µ on X. proof For any Borel set E ⊆ X, (λ ∗ ν)(E) = (λ × ν){(x, y) : xy ∈ E} = (ν × λ){(y, x) : xy ∈ E} = (ν × λ){(y, x) : yx ∈ E} = (ν ∗ λ)(E). 444E The Banach algebra of τ -additive measures (a) Let X be a topological group. Recall ∼ from 437Ab that we have a band Cb (X)∼ τ in the L-space Cb (X) consisting of those order-bounded linear functionals f : Cb (X) → R such that |f | is smooth (equivalently, f + and f − are both smooth); that is, such that |f |, f + and f − can be represented by totally finite quasi-Radon measures on X. Because X is completely regular (4A5Ha), Cb (X)∼ τ can be identified with the band Mτ of signed τ -additive Borel measures on X, that is, the set of those countably additive functionals ν defined on the Borel σ-algebra B of X such that |ν| is τ -additive (437Ha). (b) For any τ -additive totally finite Borel measures λ, ν on X we can define their convolution λ ∗ ν by the formulae of 444A, that is, (λ ∗ ν)(E) =

R

ν(x−1 E)λ(dx) =

R

λ(Ey −1 )ν(dy)

444F

Convolutions

313

ˆ νˆ of λ and ν are quasi-Radon measures (415Cb), for any Borel set E ⊆ X, if we note that the completions λ, ˆ ∗ νˆ, as defined in 444A, to the Borel so that λ ∗ ν, as defined by these formulae, is just the restriction of λ σ-algebra. Now the formulae make it obvious that the map ∗ is bilinear in the sense that (λ1 + λ2 ) ∗ ν = λ1 ∗ ν + λ2 ∗ ν, λ ∗ (ν1 + ν2 ) = λ ∗ ν1 + λ ∗ ν2 , (αλ) ∗ ν = λ ∗ (αν) = α(λ ∗ ν) for all totally finite τ -additive Borel measures λ, λ1 , λ2 , ν, ν1 , ν2 and all α ≥ 0. Consequently, regarding elements of Mτ as functionals on the Borel σ-algebra, we have a bilinear map ∗ : Mτ × Mτ → Mτ defined by saying that (λ1 − λ2 ) ∗ (ν1 − ν2 ) = λ1 ∗ ν1 − λ1 ∗ ν2 − λ2 ∗ ν1 + λ2 ∗ ν2 for all λ1 , λ2 , ν1 , ν2 ∈ Mτ+ . (c) We see from 444B that ∗ is associative on Mτ+ , so it will be associative on the whole of Mτ . Observe that ∗ is positive in the sense that λ ∗ ν ≥ 0 if λ, ν ≥ 0; so that |λ ∗ ν| = |λ+ ∗ ν + − λ+ ∗ ν − − λ− ∗ ν + + λ− ∗ ν − | ≤ λ+ ∗ ν + + λ+ ∗ ν − + λ− ∗ ν + + λ− ∗ ν − = |λ| ∗ |ν| for any λ, ν ∈ Mτ . (d) If λ, ν ∈ Mτ+ then kλk = λX and kνk = νX (362Ba), so kλ ∗ νk = (λ ∗ ν)(X) = (λ × ν)(X × X) = λX · νX = kλkkνk. Generally, for any λ, ν ∈ Mτ , kλ ∗ νk = k|λ ∗ ν|k ≤ k|λ| ∗ |ν|k = k|λ|kk|ν|k =kλkkνk. Thus Mτ is a Banach algebra under the operation ∗, as well as being an L-space. If X is abelian then Mτ will be a commutative algebra, by 444D. 444F In preparation for the next construction I give a general result extending ideas already touched on in 443C and 443G. Theorem Let X be a topological space, G a topological group and • a continuous action of G on X. For A ⊆ X, a ∈ G write a•A = {a•x : x ∈ A}. Let ν be a quasi-Radon measure on X. R (a) If f : X → [0, ∞] is lower semi-continuous, then a 7→ a•f dν : G → [0, ∞] is lower semi-continuous. (See 441Ac for the definition of a•f .) In particular, if V ⊆ X is open, then a 7→ ν(a•V ) : G → [0, ∞] is lower semi-continuous. (b) If f : X → R is continuous, then a 7→ (a•f )• : G → L0 is continuous, if L0 = L0 (ν) is given the topology of convergence in measure. (c) If ν is σ-finite and E ⊆ X is a Borel set, then a 7→ (a•E)• : G → A is Borel measurable, if the measure algebra A of ν is given its measure-algebra topology. (d) If ν is σ-finite and f : X → R is Borel measurable, then a 7→ (a•f )• : G → L0 is Borel measurable. (e) If ν is σ-finite, then (i) a 7→ ν(a•E) : G → [0, R∞] is Borel measurable for any Borel set E ⊆ X (ii) ifR f : X → R is Borel measurable, then Q = {a : a•f dν is defined in [−∞, ∞]} is a Borel set, and a 7→ a•f dν : Q → [−∞, ∞] is Borel measurable. proof (a)(i) Note first that if f : X → [0, ∞] is Borel measurable, then, for each a ∈ G, a•f is the composition of f with the continuous function x 7→ a−1 •x, so is Borel measurable, and if f is finite-valued then (a•f )• is defined in L0 = L0 (ν). R (ii) Suppose that f : X → [0, ∞] is lower semi-continuous, that a ∈ G and that a•f > γ. Let U be the set of open neighbourhoods of a in G. For U ∈ U, x ∈ X set

314

Topological groups

444F

φU (x) = sup{inf b∈U,y∈V (b•f )(y) : V is an open neighbourhood of x in X}. Then φU is lower semi-continuous. P P If φU (x) > α, there is an open neighbourhood V of x such that inf b∈U,y∈V (b•f )(y) > α; now φU (y) > α for every y ∈ V . Q Q If U 0 ⊆ U then φU 0 ≥ φU , so {φU : U ∈ U} is X P Of course φU (x) ≤ (a•f )(x) for every x. If x ∈ X upwards-directed. Also supU ∈U φU = a•f in [0, ∞] . P and (a•f )(x) > α, then {y : f (y) > α} is an open set containing a−1 •x, so (because • is continuous) there are a U ∈ U and an open neighbourhood V of x such that f (b−1 •y) > α whenever b ∈ U , y ∈ V ; in which • case φU (x) ≥ α. Q R R As α is arbitrary,R supU ∈U φU (x) = (a f )(x). Q By 414Ba, a•f dν = supU ∈U φU dν, andR there is a U ∈ U such that φURdν > γ. Now suppose that b ∈ U ; then φU (x) R≤ (b•f )(x) for every x, so b•f dν > γ. This shows that {a : a•f dν > γ} is an open set in G, so that a 7→ a•f dν is lower semi-continuous. (iii) If V R⊆ X is open, then χV is lower semi-continuous, and χ(a•V ) = a•(χV ) for every a ∈ G. So a 7→ ν(a•V ) = a•(χV )dν is lower semi-continuous. Thus (a) is true. (b) Take any a ∈ G, E ∈ dom ν such that νE < ∞ and ² > 0. Let U be the family of open neighbourhoods of a in G, and for U ∈ U set HU = int{x : |(b•f )(x) − (a•f )(x)| ≤ ² whenever b ∈ U }. Then H = {HU : U ∈ U } is upwards-directed. Also, it has union X. P P If x ∈ X then, because (b, y) 7→ f (b−1 •y) is continuous, there are a U ∈ U and an open neighbourhood V of x such that |(b•f )(y)−(a•f )(x)| ≤ 1 • • 2 ² whenever b ∈ U and y ∈ V . But now |(b f )(y) − (a f )(y)| ≤ ² whenever b ∈ U and y ∈ V , so that HU includes V , which contains x. Q Q So there is a U ∈ U such that ν(E \ HU ) ≤ ² (414Ea). In this case, for any b ∈ U , we must have

R

E

min(1, |b•f − a•f |)dν ≤ ²(1 + νE).

As E and ² are arbitrary, b 7→ (b•f )• is continuous at a; as a is arbitrary, it is continuous everywhere. Thus (ii) is true. (c)(i) Let us start by supposing that E is an open set and that ν is totally finite. In this case the function a 7→ ν(a•E) is lower semi-continuous, by (a) above, therefore Borel measurable. Now let W ⊆ A be an open set, and write H = {a : a ∈ G, (a•E)• ∈ W }. For m, k ∈ N set Hmk = {a : 2−m k ≤ ν(a•E) < 2−m (k + 1)}, S so that Hmk is Borel, and Umk = G\Hmk \ H, so that Umk is open. Let H 0 be m,k∈N Hmk ∩Umk ; then H 0 is Borel and H 0 ⊆ H. In fact H 0 = H. P P If a ∈ H, then W is an open set containing (a•E)• . Let δ > 0 be such that a ∈ W whenever ν¯(a 4 (a•E)• ) ≤ δ, where ν¯ is the measure on A; let m, k ∈ N be such that 2−m ≤ 14 δ and 2−m k ≤ ν(a•E) < 2−m (k + 1). If we take ν 0 to be the indefinite-integral measure over ν defined by χ(a•E), then ν 0 is a quasi-Radon measure (415O), so (by (a) again) U = {b : ν 0 (b•E) > 2−m (k − 1)} is an open set, and of course it contains a. If b ∈ U ∩ Hmk , then ν((b•E)4(a•E)) = ν(b•E) + ν(a•E) − 2ν((b•E) ∩ (a•E)) ≤ 2−m (k + 1) + 2−m (k + 1) − 2 · 2−m (k − 1) ≤ 4 · 2−m ≤ δ, so (b•E)• ∈ W and b ∈ H. This shows that U ∩ (Hmk \ H) = ∅ and U ⊆ Umk and a ∈ U ∩ Hmk ⊆ H 0 . As a is arbitrary, H = H 0 . Q Q Thus H is a Borel subset of G. As W is arbitrary, the map a 7→ (a•E)• is Borel measurable. (ii) To extend this to a general σ-finite quasi-Radon measure ν, still supposing that E is open, let h : X → R be a strictly positive integrable function (215B(viii)) and ν 0 the corresponding indefinite-integral measure. As in (ii), this ν 0 also is a quasi-Radon measure. Since ν and ν 0 have the same domains and the same null ideals, the Boolean algebra A is still the underlying algebra of the measure algebra of ν 0 ; by 324H, the topologies on A induced by the measures ν¯, ν¯ 0 are the same. So we can apply (i) to the measure ν 0 to see that a 7→ (a•E)• : G → A is still Borel measurable. S (iii) Next, suppose that E is expressible as i≤n V2i \ V2i+1 where each Vi is open. Then a 7→ (a•E)• is Borel measurable. P P Set X 0 = X × {0, . . . , 2n + 1}, with the product topology (giving {0, . . . , 2n + 1} the discrete topology) and define a measure ν 0 on X and an action of G on X 0 by setting

444F

Convolutions

ν 0F =

P2n+1 i=0

315

ν{x : (x, i) ∈ F }

0

whenever F ⊆ X is such that {x : (x, i) ∈ F } ∈ dom ν for every i ≤ 2n + 1, a•(x, i) = (a•x, i) whenever a ∈ G, x ∈ X and i ≤ 2n + 1. Then V = {(x, i) : i ≤ 2n + 1, x ∈ Vi } is an open set in X 0 , while ν 0 is a σ-finite quasi-Radon measure, as is easily checked; so, by (ii), the map a 7→ (a•V )• : G → A0 is Borel measurable, where A0 is the measure algebra of ν 0 . On the other hand, we can identify A0 with the simple product A2n+2 (322Kb), and the map hci ii≤2n+1 7→ supi≤n c2i \ c2i+1 : A2n+1 → A is continuous, by 323B. So the map a 7→ (a•E)• = supi≤n (a•V2i )• \ (a•V2i+1 )• is the composition of a Borel measurable function with a continuous function, and is Borel measurable. Q Q (iv) Now the family E of all those Borel sets E ⊆ X such that a 7→ (a•E)• is Borel measurable is closed under unions and intersections of monotonic sequences. P P (α) If hEn in∈N is a non-decreasing sequence in E with union E, then (a•E)• = supn∈N (a•En )• = limn→∞ (a•E)• (323Ea) for every a ∈ G. So a 7→ (a•E)• is the pointwise limit of a sequence of Borel measurable functions into a metrizable space (323Gb, because (A, ν¯) is σ-finite), and is Borel measurable, by 418Ba. Thus E ∈ E. (β) If hEn in∈N is a non-increasing sequence in E the same argument applies, since this time (a•E)• = inf n∈N (a•En )• = limn→∞ (a•E)• (323Eb) for every a ∈ G. Q Q S Since E contains all sets of the form i≤n Vi ∩ Fi where every Vi is open and every Fi is closed, by (iii), E must be the whole Borel σ-algebra, by 4A3Cg. This completes the proof of (c). (d)(i) We need the following extension of (c): if hEn in∈N is any sequence of Borel sets in X, then a 7→ h(a•En )• in∈N : G 7→ AN is Borel measurable. P P I repeat the idea of (c-iii) above. On X 0 = X × N define a measure ν 0 by setting P∞ ν 0 F = n=0 ν{x : (x, n) ∈ F } whenever F ⊆ X 0 is such that {x : (x, n) ∈ F } ∈ dom ν for every n ∈ N. As before, it is easy to check that ν 0 is a σ-finite quasi-Radon measure, if we give N the discrete topology and X 0 the product topology. As before, set a•(x, n) = (a•x, n) for a ∈ G, x ∈ X and n ∈ N, to obtain a continuous action of G on X 0 . Applying (c) to this action, the map a 7→ (a•E)• : G → A0 is Borel measurable, where A0 is the measure algebra of ν 0 and E = {(x, n) : n ∈ N, x ∈ En }. But we can identify A0 (as Boolean algebra) with AN , by 322K, as before; so that if we re-interpret a 7→ (a•E)• : G → A0 as a 7→ h(a•En )• in∈N : G → AN it is still Borel measurable. (As in (c-ii), this time using 323L, the measure-algebra topology of A0 matches the product topology on AN .) Q Q (ii) Now suppose that f : X → [0, 1] is Borel measurable. Define hEn in∈N inductively by the formula P En = {x : x ∈ X, (f − i
316

Topological groups

444F

P = {u : u ∈ L0 , [[u ∈ ]0, 1[ ]] = 1} (367S). But now a•f = p ◦ q ◦ a•f = p ◦ a•(q ◦ f ) for every a, so that a 7→ (a•f )• is the composition of the Borel map a 7→ (a•(q ◦ f ))• with the continuous map p¯, and is Borel measurable. (e)(i) We need only recall that ν¯ : A → R is lower semi-continuous (323Cb), so that (applying (c) above) a 7→ ν(a•E) = ν¯(a•E)• is a composition of Borel measurable functions and is Borel measurable. (Of course there are much more direct arguments, using fragments of the proof above.) + − 0 0 + (ii) R The point is that the maps u 7→ u , u 7→ u : L → (L ) are continuous (245Db, 367S), while u 7→R u : (L0 )R+ → [0, ∞] is lower Rsemi-continuous (369Mb), therefore Borel measurable. Accordingly a 7→ (a•f )+ = ((a•f )• )+ and a 7→ (a•f )− are Borel measurable functions from G to [0, ∞], so that R R Q = {a : min( (a•f )+ , (a•f )− ) < ∞}

is a Borel set, and a 7→

R

a•f =

R

(a•f )+ −

R

(a•f )− : Q → [−∞, ∞]

is Borel measurable. 444G Corollary Let X be a topological group and ν a σ-finite R quasi-Radon measure on X. (a) If f : X → R is aRBorel measurable function, then {x : f (y −1 x)ν(dy) is defined in [−∞, ∞]} is a Borel set in X and x 7→ f (y −1 x)ν(dy) is Borel measurable. R (b) If f , g : X → R are measurable functions, then {x : f (xy −1 )g(y)ν(dy) is defined in [−∞, ∞]} R Borel −1 is a Borel set and x 7→ f (xy )g(y)ν(dy) is Borel measurable. R (c) If ν is totally finite and f : X → R is a bounded continuous function, then x 7→ f (y −1 x)ν(dy) : X → R is continuous. ˜ • proof (a) Set f˜(x) = f (x−1 ); then R f is Borel measurable. Let l be the left action of X on R itself. Then, in the language of 444Fe, Q = {x : x•l f˜dν is defined in [−∞, ∞]} is a Borel set, and x 7→ x•l f˜dν is Borel measurable. But (x•l f˜)(y) = f˜(x−1 y) = f (y −1 x) R R for all x, y, so (x•l f˜)dν = f (y −1 x)ν(dy) if either integral is defined. (b)(i) Set ν˜E = νE −1 when this is defined, writing E −1 = {x−1 : x ∈ E} for E ⊆ X; that is, ν˜ is the image measure νφ−1 = x−1 for x ∈ X. Because φ is a homeomorphism, ν˜ is a quasi-Radon R , where Rφ(x) −1 measure. By 235L, h d˜ ν = h(x )ν(dx) for any real-valued function h for which either integral is defined in [−∞, ∞]. (ii) Now suppose that f and g are non-negative Borel measurable functions from X to R. Set g˜(y) = g(y −1 ) for every y ∈ X, so that g˜ is also a non-negative Borel measurable function. We know from 444Fd that x 7→ (x−1 •l f )• : X → L0 (˜ ν ) is a Borel measurable function; now multiplication in L0 is continuous −1 • (245Dc), so the map x 7→ ((x l f ) × g˜)• is Borel measurable; since integration is lower semi-continuous on R −1 0 + (L ) , x 7→ (x •l f ) × g˜ d˜ ν : X → [0, ∞] is Borel measurable. But

R

(x−1 •l f ) × g˜ d˜ ν=

R

f (xy)g(y −1 )˜ ν (dy) =

R

f (xy −1 )g(y)ν(dy)

whenever any of the integrals is defined in [−∞, ∞], so this is the function we needed to know about. (iii) For general Borel measurable functions f and g, we have Z Z Z ¡ ¢ −1 + −1 + f (xy )g(y)ν(dy) = f (xy )g (y)ν(dy) + f − (xy −1 )g − (y)ν(dy) Z Z ¡ ¢ − f + (xy −1 )g − (y)ν(dy) + f − (xy −1 )g + (y)ν(dy)

444K

Convolutions

317

R exactly when the subtraction can be done in [−∞, ∞], so that x 7→ f (xy −1 )g(y)ν(dy) is a difference of Borel measurable functions and is Borel measurable, with Borel measurable domain. (c) Continuing the argument of (a), if f is bounded and continuous and ν is totally finite then all the R ˜ ˜ • • functions x l f ) are bounded and continuous, so x l f dν is defined and finite for every x. Next, the function x 7→ (x•l f˜)• : X → L0 (ν) is continuous for the topology of convergence in measure (444Fb), which agrees with the norm topology of L1 (ν) on k k∞ -bounded sets (246Jb). It follows that x 7→

R

(x•l f˜)• =

R

f (y −1 x)ν(dy)

is continuous. 444H Convolutions of measures and functions I introduce some notation which I shall use for the rest of the section. Let X be a topological group. If f is a real-valued function defined on a subset of X, and ν is a measure on X, set (ν ∗ f )(x) =

R

f (y −1 x)ν(dy)

whenever the integral is defined in R. 444I Proposition Let X be a topological group and λ, ν two totally finite quasi-Radon measures on X. (a) For any Borel measurable function f : X → R, ν ∗ f is a Borel measurable function with a Borel domain. (b) ν ∗ f ∈ Cb (X) for every f ∈ Cb (X). (c) For any real-valued function f defined on a subset of X, (λ ∗ (ν ∗ f ))(x) = ((λ ∗ ν) ∗ f )(x) whenever the right-hand side is defined. proof (a) This follows at once from 444Ga. (b) This is just a restatement of 444Gc. (c) If ((λ ∗ ν) ∗ f )(x) is defined, then Z (444C)

ZZ f (t−1 x)(λ ∗ ν)(dt) =

((λ ∗ ν) ∗ f )(x) =

f ((yz)−1 x)ν(dz)λ(dy)

ZZ f (z −1 y −1 x)ν(dz)λ(dy)

= Z =

(ν ∗ f )(y −1 x)λ(dy) = (λ ∗ (ν ∗ f ))(x).

444J Convolutions of functions and measures Let X be a topological group carrying Haar measures; let ∆ be its left modular function (442I). If f is a real-valued function defined on a subset of X, and ν is a measure on X, set (f ∗ ν)(x) =

R

f (xy −1 )∆(y −1 )ν(dy)

whenever the integral is defined in R. From 444Gb we see that if ν is a σ-finite quasi-Radon measure and f is Borel measurable, then f ∗ ν is a Borel measurable function with a Borel domain. If f is non-negative and ν-integrable, write f ν for the corresponding indefinite-integral measure over ν (§234). 444K Proposition Let X be a topological group with a left Haar measure µ. Let ν be a totally finite quasi-Radon measure on X. Then for any non-negative µ-integrable real-valued function f , f µ is a quasi-Radon measure; moreover, ν ∗ f and f ∗ ν are µ-integrable, and we have In particular,

R

(ν ∗ f )µ = ν ∗ f µ, (f ∗ ν)µ = f µ ∗ ν. R (ν ∗ f )dµ = (f ∗ ν)dµ = νX · f dµ. R

318

Topological groups

444K

proof (a) f µ is a quasi-Radon measure by 415O. (b) Suppose first that f is bounded, Borel measurable and defined everywhere on X, as well as being non-negative and µ-integrable. (i) In this case ν ∗ f is defined everywhere, and is Borel measurable, by 444Ga, so we can be sure that (ν ∗ f )µ is well-defined. Let G ⊆ X be any open set such that µG < ∞. The function (x, y) 7→ f (x−1 y)χG(y) : X × X → [0, ∞[ is Borel measurable, so Z (f µ)(x−1 G)ν(dx)

(ν ∗ f µ)(G) = (444A)

ZZ f (y)χ(x−1 G)(y)µ(dy)ν(dx)

= (by the definition of f µ, 234A)

ZZ f (x−1 y)χ(x−1 G)(x−1 y)µ(dy)ν(dx)

= (441J)

ZZ f (x−1 y)χG(y)µ(dy)ν(dx)

= ZZ

f (x−1 y)χG(y)ν(dx)µ(dy)

=

(by Fubini’s theorem, 417Ha, because (x, y) 7→ f (x−1 y)χG(y) is non-negative and Borel measurable and zero outside X × G) Z = (ν ∗ f )(y)µ(dy) = ((ν ∗ f )µ)(G). G

Thus the quasi-Radon measures (ν ∗ f )µ and ν ∗ f µ agree at least on open sets of finite measure for µ. By 415H(iv) (because µ is locally finite, by 442Aa), they are identical. (ii) Similarly, (f ∗ ν)(x) is defined in R for every x, and f ∗ ν is Borel measurable, by 444Gb. This time, if G ⊆ X is open and µG < ∞, Z

ZZ (f µ)(Gy −1 )ν(dy) = f (x)µ(dx)ν(dy) Gy −1 Z Z = ∆(y −1 ) f (xy −1 )µ(dx)ν(dy)

(f µ ∗ ν)(G) =

G

(by 442Kc)

Z Z ∆(y −1 )f (xy −1 )ν(dy)µ(dx)

= ZG =

(f ∗ ν)(x)µ(dx) = ((f ∗ ν)µ)(G). G

So once again the quasi-Radon measures f µ ∗ ν and (f ∗ ν)µ must be equal. (c) Now suppose that f is Borel measurable and non-negative and defined everywhere, but not necessarily bounded. In this case set fn (x) = min(n, f (x)) for every n ∈ N, x ∈ X, and observe that (ν ∗ f )(x) = limn→∞ (ν ∗ fn )(x), (f ∗ ν)(x) = limn→∞ (fn ∗ ν)(x) whenever either side is defined in R. So whenever E ⊆ X is a Borel set, we must (using B.Levi’s theorem twice) have

444L

Convolutions

Z

319

Z

((ν ∗ f )µ)(E) =

(ν ∗ f )(x)µ(dx) = lim

n→∞

E

(ν ∗ fn )µ(dx) E

= lim ((ν ∗ fn )µ)(E) = lim (ν ∗ fn µ)(E) n→∞ n→∞ Z Z −1 = lim ν(x E)fn (x)µ(dx) = ν(x−1 E)f (x)µ(dx) = (ν ∗ f µ)(E) n→∞

if any of these is finite, and (ν ∗ f )µ = ν ∗ f µ. Similarly, Z Z (fn ∗ ν)(x)µ(dx) ((f ∗ ν)µ)(E) = (f ∗ ν)(x)µ(dx) = lim n→∞

E

E

= lim ((fn ∗ ν)µ)(E) = lim (fn µ ∗ ν)(E) n→∞ n→∞ Z Z −1 = lim ν(Ey )fn (y)µ(dy) = ν(Ey −1 )f (y)µ(dy) = (f µ ∗ ν)(E) n→∞

for any Borel set E such that either ((f ∗ ν)µ)(E) or (f µ ∗ ν)(E) is finite, so that f µ ∗ ν = (f ∗ ν)µ. (d) Next, suppose that f is defined and zero µ-a.e. In this case there is a Borel set E such that µE = 0 and f (x) is defined and equal to zero for every x ∈ X \ E (443J(b-ii)). Set g = χE. Then gµ is the zero measure, so (ν ∗ g)µ = ν ∗ gµ, (g ∗ ν)µ = gµ ∗ ν are all zero; that is, there is some µ-conegligible set F such that 0 = (ν ∗ g)(x) = 0 = (g ∗ ν)(x) =

R

R

χE(y −1 x)ν(dy) = ν(xE −1 ),

χE(xy −1 )∆(y −1 )ν(dy) =

R E −1 x

∆(y −1 )ν(dy)

for every x ∈ F . But now, for x ∈ F , we must have ν(xE −1 ) = ν(E −1 x) = 0 (since ∆ is strictly positive), so that (ν ∗ f )(x) =

R

f (y −1 x)ν(dy) =

R

xE −1

f (y −1 x)ν(dy) = 0,

because if y ∈ / xE −1 then y −1 x ∈ / E and f (y −1 x) = 0. Similarly, (f ∗ ν)(x) =

R

f (xy −1 )∆(y −1 )ν(dy) =

R

E −1 x

f (xy −1 )∆(y −1 )ν(dy) = 0.

Thus ν ∗ f and f ∗ ν are defined, and zero, µ-almost everywhere. (e) For an arbitrary non-negative µ-integrable function f , we can express it in the form g + h where g is a non-negative µ-integrable Borel measurable function defined everywhere, and h is zero almost everywhere. In this case, ν ∗h+ , ν ∗h− , h+ ∗ν and h− ∗ν are defined, and zero, µ-a.e., so ν ∗f = ν ∗g and f ∗ν = g∗ν µ-a.e., and (ν ∗ f )µ = (ν ∗ g)µ, (f ∗ ν)µ = (g ∗ ν)µ. We therefore have (ν ∗ f )µ = (ν ∗ g)µ = ν ∗ gµ = ν ∗ f µ,

(f ∗ ν)µ = (g ∗ ν)µ = gµ ∗ ν = f µ ∗ ν,

as required. (f ) Finally, we have

R

R

ν ∗ f dµ = ((ν ∗ f )µ)(X) = (ν ∗ f µ)(X) = νX · (f µ)(X) = νX · f ∗ ν dµ = ((f ∗ ν)µ)(X) = (f µ ∗ ν)(X) = (f µ)(X) · νX = νX ·

R R

f dµ, f dµ.

444L Corollary Let X be a topological group carrying Haar measures. Suppose that ν is a non-zero quasi-Radon measure on X and E ⊆ X is a Borel set such that ν(xE) = 0 for every x ∈ X. Then E is Haar negligible. proof Let µ be a left Haar measure on X. There is a non-zero totally finite quasi-Radon measure ν 0 on X such that ν 0 (xE) = 0 for every x ∈ X. P P Take any F such that 0 < νF < ∞, and set ν 0 H = ν(H ∩ F ) whenever this is defined. Q Q Let G be any Borel set such that µG < ∞, and set f = χ(G ∩ E −1 ). Then (ν 0 ∗ f )(x) =

R

χ(G ∩ E −1 )(y −1 x)ν 0 (dy) = ν 0 (xG−1 ∩ xE) = 0

320

Topological groups

444L

for every x ∈ X, so ν 0 ∗ f µ = (ν 0 ∗ f )µ is the zero measure, and (f µ)(X) = 0, that is, µ(G ∩ E −1 ) = 0. As G is arbitrary, µE −1 = 0 and E is Haar negligible (442H). 444M Proposition Let X be a topological group and µ a left Haar measure on X. Let ν be a totally finite quasi-Radon measure on X and p ∈ [1, ∞]. (a) We have a bounded positive linear operator u 7→ ν ∗ u : Lp (µ) → Lp (µ), of norm at most νX, defined by saying that Rν ∗ f • = (ν ∗ f )• for every f ∈R Lp (µ). (b) Set γ = ∆(y)(1−p)/p ν(dy) if p < ∞, ∆(y)−1 ν(dy) if p = ∞, where ∆ is the left modular function of X. Suppose that γ < ∞. Then we have a bounded positive linear operator u 7→ u ∗ ν : Lp (µ) → Lp (µ), of norm at most γ, defined by saying that f • ∗ ν = (f ∗ ν)• for every f ∈ Lp (µ). proof I begin with some straightforward remarks. If νX = 0 the whole result is trivial, so we may suppose that νX > 0. Multiplying ν by a positive scalar does not affect the inequalities we need, so we may suppose that νX = 1. Note also that 444K tells us that changing a function f on a negligible set will change ν ∗ f and f ∗ ν on negligible sets. I will write Lp , Lp for Lp (µ), Lp (µ). (a) If f ≥ 0 is µ-integrable, then 444K tells us that ν ∗ f is µ-integrable and that Z kν ∗ f k1 = (ν ∗ f )(x)µ(dx) = ((ν ∗ f )µ)(X) X

= (ν ∗ f µ)(X) = νX · (f µ)(X) = kf k1 , using 444C or 444A for the penultimate equality. Since evidently ν ∗ (f + g) = ν ∗ f + ν ∗ g, ν ∗ (αf ) = αν ∗ f whenever the right-hand sides of the equations are defined, we have a positive linear operator T1 : L1 → L1 defined by saying that T1 g • = (ν ∗ g)• for every µ-integrable Borel measurable function g, with kT1 k ≤ 1. Similarly, if h : X → R is a bounded Borel measurable function, then ν ∗ h is also a Borel measurable function, by 444Ga. Of course it is bounded, since |(ν ∗ h)(x)| = |

R

h(y −1 x)ν(dy)| ≤ supy∈X |h(y)|

for every x. So we have a positive linear operator T∞ : L∞ → L∞ defined by saying that T∞ h• = (ν ∗ h)• for every bounded Borel measurable function h. Moreover, if u ∈ L∞ , there is a Borel measurable h such that h• = u and supy∈X |h(y)| = kuk∞ , so that kT∞ uk∞ ≤ supx∈X |(ν ∗ h)(x)| ≤ kuk∞ ; thus kT∞ k ≤ 1. Since T1 and T∞ agree on L1 ∩ L∞ , they have a common extension to a linear operator T : L1 + L∞ → L1 + L∞ . By 371Gd, kT ukp ≤ kukp for every p ∈ [1, ∞], u ∈ Lp . (Strictly speaking, I am relying on the standard identifications of L1 , L∞ and Lp with the corresponding subspaces of L0 (A), where (A, µ ¯) is the measure algebra of µ. Of course the argument for 371Gd applies equally well in L0 (µ).) Now suppose that f ∈ Lp . Then it is expressible as g + h where g ∈ L1 and h : X → R is a bounded Borel measurable function, so we shall have ν ∗ f = ν ∗ g + ν ∗ h wherever the right-hand side is defined; accordingly ν ∗ f is defined a.e. and is measurable, and kν ∗ f kp = k(ν ∗ g)• + (ν ∗ h)• kp = kT1 g • + T∞ h• kp = kT f • kp ≤ kf • kp = kf kp , as required. (b)(i) If p = 1, then γ = νX = 1. If f ∈ L1 is non-negative, then, by 444K, as in (a) above, kf ∗ νk1 = ((f ∗ ν)µ)(X) = (f µ ∗ ν)(X) = (f µ)(X) · νX = kf k1 . For general µ-integrable f , kf ∗ νk1 ≤ kf + ∗ νk1 + kf − ∗ νk1 = kf + k1 + kf − k1 = kf k1 .

444N

Convolutions

321

(ii) If p = ∞, then directly from the formula (f ∗ ν)(x) = f : X → R is a bounded Borel measurable function then |(f ∗ ν)(x)| ≤

R

R

f (xy −1 )∆(y −1 )ν(dy) we see that if

∆(y)−1 ν(dy) supy∈X |f (y)| = γ supy∈X |f (y)|

for every x. Since changing f on a µ-negligible set changes f ∗ ν on a µ-negligible set, we can argue as in (a) above to see that f • 7→ (f ∗ ν)• defines a linear operator from L∞ to itself of norm at most γ. R p , so that γ = ∆(y −1 )1/q ν(dy). Suppose for the (iii) Now suppose that 1 < p < ∞. Set q = p−1 moment that f ∈ Lp is a non-negativeRBorel measurable function, and let h : X → R be another non-negative Borel measurable function such that hq dµ ≤ 1. In this case Z

ZZ h(x)f (xy −1 )∆(y −1 )ν(dy)µ(dx)

(f ∗ ν) × h dµ = ZZ =

h(x)f (xy −1 )∆(y −1 )µ(dx)ν(dy)

(by 417Ha, because (x, y) 7→ h(x)f (xy −1 )∆(y −1 ) is Borel measurable and {x : h(x) 6= 0} is a countable union of sets of finite measure for µ, while ν is totally finite) ZZ = h(xy)f (x)µ(dx)ν(dy) by 442Kc, as usual, at least if the last integral is finite. But, for any y ∈ X, Z Z ¡ ¢1/q h(xy)f (x)µ(dx) ≤ kf kp |h(xy)|q µ(dx) Z ¡ ¢1/q −1 = kf kp ∆(y ) |h(x)|q µ(dx) ≤ kf kp ∆(y −1 )1/q . So

R

(f ∗ ν) × h dµ =

RR

h(xy)f (x)µ(dx)ν(dy) ≤

R

kf kp ∆(y −1 )1/q ν(dy) = γkf kp .

Because µ (being a quasi-Radon measure) is semi-finite, this means that f ∗ν ∈ Lp and that kf ∗νkp ≤ γkf kp (366D-366E, or 244Xf and 244Fa). (Once again, we need to know that every member of Lq can be represented by a Borel measurable function; this is a consequence of R 443J or 412Xc.) For general Borel measurable f : X → R such that |f |p dµ < ∞, we know that from 444G that f ∗ ν is Borel measurable, while |f ∗ ν| ≤ |f | ∗ ν (and f ∗ ν is defined wherever |f | ∗ ν is finite), so that kf ∗ νkp ≤ k|f | ∗ νkp ≤ γk|f |kp = γkf kp . Finally, if f ∈ Lp is arbitrary, then there is a Borel measurable g : X → R such that f =a.e. g, so that f ∗ ν =a.e. g ∗ ν and kf ∗ νkp = kg ∗ νkp ≤ γkgkp = γkf kp . It follows at once that we have a bounded linear operator f • 7→ (f ∗ ν)• : Lp → Lp , of norm at most γ. 444N The following lemma on exchanging the order of repeated integrals will be fundamental to the formulae in the rest of the section. Lemma Let X be a topological group and µ a left Haar measure on X. Suppose that f , g, h ∈RL0 (µ) (the space of measurable real-valued functions defined µ-a.e. in X) are non-negative. Then, writing d(x, y) to denote integration with respect to the quasi-Radon product measure µ × µ,

RR

f (x)g(y)h(xy)dxdy =

RR

f (x)g(y)h(xy)dydx =

R

f (x)g(y)h(xy)d(x, y)

in [0, ∞]. proof Following the standard pattern in results of this type, I deal with successively more complicated functions f , g and h. Evidently the situation is symmetric, so that it will be enough if I can show that RR R f (x)g(y)h(xy)dxdy = f (x)g(y)h(xy)d(x, y).

322

Topological groups

444N

(a) Suppose first that f = χF , g = χG and h = χH, where F , G, H are Borel subsets of X. In this case

RR

R R

f (x)g(y)h(xy)dxdy = supU,V ∈Σf

V

U

f (x)g(y)h(xy)dxdy,

where Σf is the ideal of measurable sets of finite measure for µ. P P For y ∈ X, n ∈ N and U ∈ Σf write q(y) =

R

f (x)h(xy)dx = µ(F ∩ Hy −1 ),

R

qU (y) =

f (x)h(xy)dx = µ(U ∩ F ∩ Hy −1 ),

U

(n)

q (n) (y) = min(n, q(y)),

qU (y) = min(n, qU (y)).

Then every qU is continuous, by 443C (with a little help from 323Cc), while supU ∈Σf qU (y) = q(y) for every (n) y, because µ is semi-finite. Because µ is τ -additive and effectively locally finite, (q (n) )• = supU ∈Σf (qU )• in 0 f L (µ) for every n (414Ab); because Σ is upwards-directed, Z

Z q (n) (y)g(y)dy Z Z (n) qU (y)g(y)dy, qU (y)g(y)dy = sup sup

q(y)g(y)dy = sup

n∈N

=

n∈N,U ∈Σf

that is,

RR

U ∈Σf

RR

f (x)g(y)h(xy)dxdy = supU ∈Σf

U

On the other hand, for any U ∈ Σf , we surely have

RR

f (x)g(y)h(xy)dxdy.

R R

f (x)g(y)h(xy)dxdy = supV ∈Σf U

V

U

f (x)g(y)h(xy)dxdy,

again because µ is semi-finite. Putting these together, we have the result. Q Q R Looking at the other side of the equation, f (x)g(y)h(xy)d(x, y) = (µ × µ)W , where W = (F × G) ∩ {(x, y) : xy ∈ H} is a Borel set; so that ZZ f (x)g(y)h(xy)dxdy =

sup (µ × µ)((U × V ) ∩ W ) U,V ∈Σf

=

Z

sup


U,V ∈Σf

U ×V

(417C(iii)). But now we can apply 417Ha to see that, for any U , V ∈ Σf ,

R

U ×V

f (x)g(y)h(xy)d(x, y) =

Taking the supremum over U and V , we get

R


(b) Clearly both sides of our equation

R


R R V

RR

RR

U

f (x)g(y)h(xy)dxdy.

f (x)g(y)h(xy)dxdy.

f (x)g(y)h(xy)dxdy

are additive in f , g and h separately (subtraction, of course, will be another matter, as I am allowing ∞ to appear without restriction); and also behave identically if f or g or h is multiplied by a non-negative scalar. So the identity will be valid if f , g and h are all finite sums of non-negative multiples of characteristic functions of Borel sets. Moreover, by repeated use of B.Levi’s theorem, we see that if hfn in∈N , hgn in∈N and hhn in∈N are non-decreasing sequences of such functions with suprema f , g and h, then Z

Z

fn (x)gn (y)hn (xy)d(x, y) ZZ ZZ fn (x)gn (y)hn (xy)dxdy = f (x)g(y)h(xy)dxdy. = sup

f (x)g(y)h(xy)d(x, y) = sup

n∈N

n∈N

So the identity is valid for all non-negative Borel functions f , g and h.

444O

Convolutions

323

(c) Finally, suppose only that f , g and h are non-negative, measurable and defined almost everywhere. In this case, by 443J(b-iv), there are Borel measurable functions f0 , g0 and h0 , non-negative, defined everywhere on X and equal almost everywhere to f , g and h respectively. Let E be the conegligible set {x : x ∈ dom f ∩ dom h ∩ dom g, f (x) = f0 (x), g(x) = g0 (x), h(x) = h0 (x)}. R R We find that f (x)h(xy)dx = f0 (x)h0 (xy)dx for every y ∈ X. P P E ∩ Ey −1 is conegligible (see 443A), −1 and f (x)h(xy) = f0 (x)h0 (xy) for every x ∈ E ∩ Ey . Q Q Consequently

RR

f (x)g(y)h(xy)dxdy =

RR

f0 (x)g0 (y)h0 (xy)dxdy.

Secondly, f (x)g(y)h(xy) = f0 (x)g0 (y)h0 (xy) (µ × µ)-a.e. P P Set W = {(x, y) : x ∈ E, y ∈ E, xy ∈ E}. Fubini’s theorem, applied to (U × V ) \ W where U , V ∈ Σf , shows that W is conegligible; but of course f (x)g(y)h(xy) = f0 (x)g0 (y)h0 (xy) whenever (x, y) ∈ W . Q Q Accordingly

R


R

f0 (x)g0 (y)h0 (xy)d(x, y).

Combining this with the result of (b), applied to f0 , g0 and h0 , we see that once again

R


RR

f (x)g(y)h(xy)dxdy,

as required. 444O Convolutions of functions: Theorem R Let X be a topological group and µ a left Haar measure on X. For f , g ∈ L0 = L0 (µ), write (f ∗ g)(x) = f (y)g(y −1 x)dy whenever this is defined in R, taking the integral with respect to µ. (a) Writing ∆ for the left modular function of X, Z Z (f ∗ g)(x) = f (y)g(y −1 x)dy = f (xy)g(y −1 )dy Z Z −1 −1 = ∆(y )f (xy )g(y)dy = ∆(y −1 )f (y −1 )g(yx)dy whenever any of these integrals is defined in R. (b) If f =a.e. f1 (that is, f = f1 a.e.) and g =a.e. g1 , then f ∗ g = f1 ∗ g1 . (c)(i) |(f ∗ g)(x)| ≤ (|f | ∗ |g|)(x) whenever either is defined. (ii) ((f1 + f2 ) ∗ g)(x) = (f1 ∗ g)(x) + (f2 ∗ g)(x), (f ∗ (g1 + g2 ))(x) = (f ∗ g1 )(x) + (f ∗ g2 )(x), ((αf ) ∗ g)(x) = (f ∗ (αg))(x) = α(f ∗ g)(x) whenever the right-hand expressions are defined. (d) If f , g and h belong to L0 and any of

R

(|f | ∗ |g|)(x)|h|(x)dx,

RR is defined in [0, ∞[ (writing on X × X), then

R

|f (x)g(y)h(xy)|dydx,

RR R

|f (x)g(y)h(xy)|dxdy, |f (x)g(y)h(xy)|d(x, y)

d(x, y) for integration with respect to the quasi-Radon product measure µ × µ

R RR

(f ∗ g)(x)h(x)dx, f (x)g(y)h(xy)dydx,

RR

f (x)g(y)h(xy)dxdy,

R


are all defined, finite and equal, provided that in the expression (f ∗ g)(x)h(x) we interpret the product as 0 when h(x) = 0 and (f ∗ g)(x) is undefined. (e) If f , g and h belong to L0 , f ∗g and g∗h are defined a.e. and x ∈ X is such that either (|f |∗(|g|∗|h|))(x) or ((|f | ∗ |g|) ∗ |h|)(x) is defined in R, then (f ∗ (g ∗ h))(x) and ((f ∗ g) ∗ h)(x) are defined and equal.

324

Topological groups

444O

(f) If a ∈ X and f , g ∈ L0 , a•l (f ∗ g) = (a•l f ) ∗ g,

a•r (f ∗ g) = f ∗ (a•r g),

(a•r f ) ∗ g = ∆(a−1 )f ∗ (a−1 •l g). (g) If X is abelian then f ∗ g = g ∗ f for all f and g. proof (a) Use 441J and 442Kc to see that the four formulae for f ∗ g coincide. (b) Setting E = {y : y ∈ dom f ∩ dom f1 ∩ dom g ∩ dom g1 , f (y) = f1 (y), g(y) = g1 (y)}, E is conegligible. If x ∈ X, then f (y)g(y −1 x) = f1 (y)g1 (y −1 x) for every y ∈ E ∩ xE −1 , which is also conegligible, by 443A; so (f ∗ g)(x) = (f1 ∗ g1 )(x) if either is defined. (c) These are all elementary. (d) First consider non-negative f , g and h. The point is that, if any of the integrals is defined and finite, Z

ZZ (f ∗ g)(x)h(x)dx =

(by 442Kb)

ZZ f (x)g(x

−1

y)h(y)dxdy =

∆(x−1 )f (x−1 )g(xy)h(y)dxdy

ZZ =

∆(x−1 )f (x−1 )g(xy)h(y)dydx

(by 444N, recalling that x 7→ ∆(x−1 )f (x−1 ) will belong to L0 if f does, by 442J and 442H) ZZ ZZ −1 = f (x)g(x y)h(y)dydx = f (x)g(y)h(xy)dydx (substituting xy for y in the inner integral, as permitted by 441J). (The ‘and finite’ at the beginning of the last sentence is there because I have changed the rules since the last paragraph, and f ∗ g is not allowed to take the value ∞. So we have to take care that {y : h(y) > 0,

R

f (x)g(x−1 y)dx = ∞}

is negligible.) Now applying 444N again, we get Z ZZ (f ∗ g)(x)h(x)dx = f (x)g(y)h(xy)dxdy ZZ Z = f (x)g(y)h(xy)dydx = f (x)g(y)h(xy)d(x, y) if any of these integrals is finite. For the general case, the hypothesis on |f |, |g| and |h| is sufficient to ensure that the four expressions are equal for any combination of f ± , g ± and h± ; adding and subtracting these combinations appropriately, we get the result. (e) The point is that, for non-negative f , g and h, Z ((f ∗ g) ∗ h)(x) = (setting h0 (z) = h(z −1 x))

Z (f ∗ g)(z)h(z

−1

x)dz =

(f ∗ g)(z)h0 (z)dz

ZZ =

f (y)g(z)h0 (yz)dzdy

(using (d); to see that h0 is measurable, refer to 443A as usual) ZZ = f (y)g(z)h(z −1 y −1 x)dzdy Z = f (y)(g ∗ h)(y −1 x)dy = (f ∗ (g ∗ h))(x)

444R

Convolutions

325

at least as long as one of the expressions here Ris finite. (Note that, as inRR255J, we need to suppose that f ∗ g and a.e. when Rmoving from (f ∗ g)(z)h(z −1 x)dz to f (y)g(z)h(z −1 y −1 )dydz and from RR g ∗ h are defined −1 −1 −1 f (y)g(z)h(z y x)dzdy to f (y)(g ∗ h)(y x)dy, since in part (d) I am more tolerant of infinities in the repeated integrals than I was in the definition of f ∗ g.) Once again, subject to the inner integrals implicit in the formulae f ∗ (g ∗ h) and (f ∗ g) ∗ h being adequately defined, we can use addition and subtraction to obtain the result for general f , g and h. (f ) These are immediate from the formulae in (a), using 442K if necessary. (g) If X is abelian, then ∆(y) = 1 for every y, so (g ∗ f )(x) =

R

g(y)f (y −1 x)dy =

R

g(y)∆(y −1 )f (xy −1 )dy = (f ∗ g)(x)

if either (f ∗ g)(x) or (g ∗ f )(x) is defined. 444P Proposition Let X be a topological group and µ a left Haar measure on X. (a) If f ∈ L1 (µ)+ and g ∈ L0 (µ) then f ∗ g, as defined in 444O, is equal to (f µ) ∗ g as defined in 444H. (b) If f ∈ L0 (µ) and g ∈ L1 (µ)+ then f ∗ g = f ∗ (gµ). proof Again, these are immediate from the formulae above: (f ∗ g)(x) =

R

R

g(y −1 x)f (y)µ(dy) =

R

R

f (xy −1 )∆(y −1 )(gµ)(dy) = (f ∗ gµ)(x) R R whenever these are defined, using 235M, as usual, to calculate d(f µ), d(gµ). (Note that as we assume throughout that f and g are defined µ-almost everywhere, all the functions y 7→ g(y −1 x), y 7→ f (xy −1 ) are also defined µ-a.e., by the results set out in 443A.) (f ∗ g)(x) =

f (xy −1 )∆(y −1 )g(y)µ(dy) =

g(y −1 x)(f µ)(dy) = (f µ ∗ g)(x),

444Q Proposition Let X be a topological group and µ a left Haar measure on X. (a) Let f , g be non-negative µ-integrable functions. Then, defining f ∗ g as in 444O, we have f ∗ g ∈ L1 = L1 (µ) and (f µ) ∗ (gµ) = (f ∗ g)µ 1

for all f , g ∈ L . (b) For any f , g ∈ L1 , f ∗ g ∈ L1 and

R

f ∗ g dµ =

R

f dµ

R

g dµ,

kf ∗ gk1 ≤ kf k1 kgk1 .

proof (a) Putting 444K and 444P together, f µ ∗ gµ = (f µ ∗ g)µ, so that f ∗ g = f µ ∗ g is µ-integrable, and (f ∗ g)µ = (f µ ∗ g)µ = f µ ∗ gµ. R R R (b) Taking h = χX in 444Od, we get f ∗ g dµ = f dµ g dµ. Now kf ∗ gk1 =

R

|f ∗ g| ≤

R

|f | ∗ |g| =

R

|f |

R

|g| = kf k1 kgk1 .

444R Proposition Let X be a topological group and µ a left Haar measure on X. Take any p ∈ [1, ∞]. (a) kf ∗ gkp ≤ kf k1 kgkp whenever f ∈ L1 (µ), g ∈ Lp (µ). (b) For a real-valued function f defined on a subset of X, set f˜(x) = f (x−1 ) whenever this is defined. Then f ∗ g˜ = (g ∗ f˜)∼ for all f , g ∈ L0 . If X is unimodular then kf˜kp = kf kp for every f ∈ L0 . (c) Set q = ∞ if p = 1, p/(p−1) if 1 < p < ∞, 1 if p = ∞. If f ∈ Lp (µ) and g ∈ Lq (µ), then f ∗ g˜ is defined everywhere in X and is continuous, and kf ∗ g˜k∞ ≤ kf kp kgkq . If X is unimodular, then f ∗ g ∈ Cb (X) and kf ∗ gk∞ ≤ kf kp kgkq for every f ∈ Lp (µ), g ∈ Lq (µ). Remark In the formulae above, interpret kgk∞ as kg • k∞ = ess sup |g| for g ∈ L∞ = L∞ (µ), and as ∞ for g ∈ L0 \ L∞ . Because µ is strictly positive, this agrees with the usual definition kgk∞ = supx∈X |g(x)| when g is continuous. proof (a) If f ≥ 0, then

326

Topological groups

444R

kf ∗ gkp = kf µ ∗ gkp ≤ (f µ)(X)kgkp = kf k1 kgkp by 444Pa and 444Ma. Generally, kf ∗ gkp ≤ kf + ∗ gkp + kf − ∗ gkp ≤ (kf + k1 + kf − k1 )kgkp = kf k1 kgkp . (b)(i) By 443A, as usual, f˜ ∈ L0 whenever f ∈ L0 . For any x ∈ X, (f ∗ g˜)(x) =

R

f (y)˜ g (y −1 x)dy =

R

f˜(y −1 )g(x−1 y)dy = (g ∗ f˜)(x−1 ) = (g ∗ f˜)∼ (x)

if any of these are defined. (ii) If X is unimodular, then, for any f ∈ L0 and p ∈ [1, ∞[, kf˜kpp =

R

|f (x−1 )|p dx =

R

∆(x−1 )|f (x−1 )|p dx = kf kpp ;

while ess sup |f˜| = ess sup |f | because E −1 is conegligible whenever E ⊆ X is conegligible. (c)(i) For any x ∈ X, (f ∗ g˜)(x) =

R

f (y)g(x−1 y)dy =

R

f × (x•l g) R in the language of 443G and 444O. By 443Gb, x•l g ∈ Lq , so (f ∗ g˜)(x) = f × (x•l g) is defined. (ii) If p > 1, so that q < ∞, then x 7→ (x•l g)• : X → Lq is continuous (443Ge), so x 7→ (f ∗ g˜)(x) =

R

f • × (x•l g)•

is continuous, because f • ∈ Lp ∼ = (Lq )∗ . If p = 1, then (f ∗ g˜)(x) =

R

f (xy)g(y)dy =

R

(x−1 •l f ) × g

for every x; since x 7→ (x−1 •l f )• : X → L1 is continuous, so is f ∗ g˜. (iii) If X is unimodular then f ∗ g = f ∗ g˜˜ is continuous, because g˜ ∈ Lq by (b), and kf ∗ gk∞ ≤ kf kp k˜ g kq = kf kp kgkq . 444S Remarks (a) From 444Ob we see that ∗, regarded as a function from L1 × Lp to Lp (444Ra), not merely descends to a bilinear operator from L1 × Lp to Lp but can actually be regarded as a function from L1 × Lp to Lp . Similarly, if the group is unimodular, and p1 + 1q = 1, the map ∗ : Lp × Lq → Cb (X) (444Rc) descends to a map from Lp × Lq to Cb (X). (b) 444Qb tells us that if µ is a left Haar measure on a topological group then L1 (µ), with the operation f ∗ g • = (f ∗ g)• , is a normed algebra (therefore, of course, a Banach algebra); it is commutative if X is abelian (444Og). If, for f ∈ L1 (µ), we write f µ for the countably additive functional defined by setting R (f µ)(E) = E f dµ for every Borel set E, then the formula f • 7→ f µ defines an embedding of L1 (µ) as a subalgebra of the Banach algebra of τ -additive Borel measures described in 444E; 444Qa tells us that the embedding respects the multiplication. •

444T Proposition Let X be a topological group and µ a left Haar measure on X. Then for any p ∈ [1, ∞[, f ∈ Lp (µ) and ² > 0 there is a neighbourhood U of the identity e in X such that kν ∗ f − f kp ≤ ² and kf ∗ ν − f kp ≤ ² whenever ν is a quasi-Radon measure on X such that νU = νX = 1. proof (a) To begin with, suppose that f is non-negative, continuous and bounded, and that G = {x : f (x) > 0} has finite measure; set M = supx∈X f (x). Write U for the family of neighbourhoods of e. Take δ > 0, η ∈ ]0, 1] such that (2δ + (1 + δ)p − 1)1/p kf kp ≤ ², (1 − η)p

R

((f − ηχX)+ )p dµ − M p η p ≥ (1 − δ)

R

f p dµ,

(1 − η)(1−p)/p ≤ 1 + δ.

For each U ∈ U , set HU = int{x : f (y) ≥ f (x) − η for every y ∈ xU −1 ∪ U −1 x}.

444T

Convolutions

327

Then HU is open and for every x ∈ X there is a U ∈ U such that |f (y) − f (x)| ≤ 21 η whenever y ∈ xU U −1 ∪ U −1 xU , so that x ∈ int xU ⊆ HU . Thus {HU : U ∈ U } is an upwards-directed family of open sets with union X, and there is a U ∈ U such that µ(G \ HU ) ≤ η; moreover, because ∆ is continuous, we can suppose that ∆(y −1 ) ≥ 1 − η for every y ∈ U . Now suppose that ν is a quasi-Radon measure on X such that νU = νX = 1. Then, for any x ∈ G ∩ HU , (ν ∗ f )(x) =

R

f (y −1 x)ν(dy) =

R

U

f (y −1 x)ν(dy) ≥ f (x) − η

because x ∈ HU and y −1 x ∈ U −1 x whenever y ∈ U . Similarly, (f ∗ ν)(x) =

R

U

f (xy −1 )∆(y −1 )ν(dy) ≥ (f (x) − η)(1 − η)

for every x ∈ HU . Now this means that, setting h1 = ν ∗ f , h2 = f ∗ ν we have (f ∧ hi )(x) ≥ (f (x) − η)(1 − η) for every x ∈ G ∩ HU , both i. Accordingly Z Z p p (f ∧ hi ) dµ ≥ (1 − η) ((f − ηχX)+ )p dµ G∩HU Z Z p + p fp ≥ (1 − η) ((f − ηχX) ) dµ − G\HU Z Z p + p p p ≥ (1 − η) ((f − ηχX) ) dµ − M η ≥ (1 − δ) f p dµ. Now, just because f and hi are non-negative, and p ≥ 1, |f − hi |p + 2(f ∧ hi )p ≤ f p + hpi . Also, writing γ=

R

∆(y)(1−p)/p ν(dy) =

R U

∆(y)(1−p)/p ν(dy) ≤ (1 − η)(1−p)/p ≤ 1 + δ,

we have (444M), so that

R

R

hpi dν

kh1 kp = kν ∗ f kp ≤ kf kp , kh2 kp = kf ∗ νkp ≤ γkf kp R ≤ (1 + δ)p f p dµ for both i, and

|f − hi |p dµ ≤

R

f p dµ +

R

R

hpi dµ − 2 (f ∧ hi )p dµ ≤ (2δ + (1 + δ)p − 1)

R

f p dµ

for both i. But this means that max(kf − f ∗ νkp , kf − ν ∗ f kp ) ≤ (2δ + (1 + δ)p − 1)1/p kf kp ≤ ². As ν is arbitrary, we have found a suitable U . (b) For any continuous bounded function f such that µ{x : f (x) 6= 0} < ∞, we can find neighbourhoods U1 , U2 of e such that kf + − ν ∗ f + kp ≤ 12 ²,

kf + − f + ∗ νkp ≤ 12 ²

kf − − ν ∗ f − kp ≤ 12 ²,

kf − − f − ∗ νkp ≤ 12 ²

whenever νU1 = νX = 1, whenever νU2 = νX = 1. So we shall have kf − ν ∗ f kp ≤ ²,

kf − f ∗ νkp ≤ ²

whenever ν(U1 ∩ U2 ) = νX = 1. (c) For general f ∈ Lp (µ), there is a bounded continuous function g : X → R such that µ{x : g(x) 6= 0} < ∞ and kf − gkp ≤ 41 ² (415Pa). Now there is a neighbourhood U1 of e such that kg − ν ∗ gkp ≤ 14 ²,

kg − g ∗ νkp ≤ 14 ²

whenever νU1 = νX = 1. There is also a neighbourhood U2 of e such that ∆(y −1 )(1−p)/p ≤ 2 for every y ∈ U2 , so that kg ∗ ν − f ∗ νkp ≤ 2kg − f kp ≤ 21 ²

328

Topological groups

444T

whenever νU2 = νX = 1. Since we have kν ∗ g − ν ∗ f kp ≤ kg − f kp ≤ 14 ² whenever νX = 1, we get kf − ν ∗ f kp ≤ ², kf = f ∗ νkp ≤ ² whenever ν(U1 ∩ U2 ) = νX = 1. This completes the proof. 444U Corollary Let X be a topological group S and µ a left Haar measure on X. For any Haar measurable E ⊆ X such that 0 < µE < ∞, and any f ∈ 1≤p<∞ Lp (µ), write fE (x) =

1 µE

R

fE0 (x) =

f dµ, xE

1 µ(Ex)

R

Ex

f dµ

for x ∈ X. Then, for any p ∈ [1, ∞[, f ∈ Lp and ² > 0, there is a neighbourhood U of the identity in X such that kfE − f kp ≤ ² and kfE0 − f kp ≤ ² whenever E ⊆ U is a non-negligible Haar measurable set. proof Take δ ∈ ]0, 1[ such that δ(1 − δ)(1−p)/p kf kp ≤ 21 ². By 444T, there is a neighbourhood U of the identity such that kf − f ∗ νkp ≤ 21 ², kf − ν ∗ f kp ≤ ² whenever ν is a quasi-Radon measure such that νU = νX = 1. Shrinking U if necessary, we may suppose also that U = U −1 , that µU < ∞ and that |∆(x) − 1| ≤ δ for every x ∈ U , where ∆ is the left modular function of X. If E ⊆ U and µE > 0, consider the quasi-Radon measures ν, ν 0 , ν˜ and ν˜ 0 on X defined by setting νF =

1 µE −1

R

E∩F

∆(x−1 )µ(dx),

ν 0F =

µ(E∩F ) , µE

ν˜F = νF −1 ,

ν˜ 0 F = ν 0 F −1

whenever these are defined. (They are quasi-Radon measures because ν and ν 0 are totally finite indefiniteintegral measures over µ and the map x 7→ x−1 is a homeomorphism.) Because E ⊆ U = U −1 , we have ν˜U = νU −1 =

1 µE −1

R E

∆(x−1 )µ(dx) = 1 = ν˜X

by 442Ka, while ν˜ 0 X = ν˜ 0 U = ν 0 U −1 = 1. Now consider f ∗ ν˜ and ν˜ 0 ∗ f . For any x ∈ X, Z (f ∗ ν˜)(x) =

Z f (xy −1 )∆(y −1 )˜ ν (dy) =

f (xy)∆(y)ν(dy)

(because ν˜ is the image of ν under the map y 7→ y −1 ) Z 1 = χE(y)∆(y −1 )f (xy)∆(y)µ(dy) −1 µE

(noting that ν is an indefinite-integral measure over µ, and using 235M) Z 1 = χE(x−1 y)f (y)µ(dy) µE −1 Z 1 µE = f (y)µ(dy) = fE (x), µE −1 xE µE −1 Z Z (˜ ν 0 ∗ f )(x) = f (y −1 x)˜ ν 0 (dy) = f (yx)ν 0 (dy) (because ν˜ 0 is the image of ν 0 under the map y 7→ y −1 ) Z Z 1 ∆(x−1 ) = χE(y)f (yx)µ(dy) = χE(yx−1 )f (y)µ(dy) µE

(by 442Kc) =

1 µ(Ex)

µE

Z Ex

f (y)µ(dy) = fE0 (x).

So ¯ µE −1 ¯ kf − fE kp ≤ kf − f ∗ ν˜kp + ¯ − 1¯kf ∗ ν˜kp . µE

444V

Now µE −1 =

Convolutions

R E

329

∆(y −1 )µ(dy), so that |µE − µE −1 | ≤

R E

also kf ∗ ν˜kp ≤ kf kp

|∆(y −1 ) − 1|µ(dy) ≤ δµE,

R

|

µE −1 µE

− 1| ≤ δ;

∆(y)(1−p)/p ν(dy) ≤ kf kp (1 − δ)(1−p)/p

by 444Mb, so ¯ µE −1 ¯ 1 ¯ − 1¯kf ∗ ν˜kp ≤ δ(1 − δ)(1−p)/p kf kp ≤ ², µE

2

and kf − fE kp ≤ ². On the other hand, kf − fE0 kp = kf − ν˜ 0 ∗ f kp ≤ ²; as E is arbitrary, we have found a suitable U . 444V So far I have not emphasized the special properties of compact groups. But of course they are the centre of the subject, and for the sake of a fundamental theorem in §446 I give the following result. Theorem Let X be a compact topological group and µ a left Haar measure on X. (a) For any u, v ∈ L2 = L2 (µ) we can interpret their convolution u ∗ v either as a member of the space C(X) of continuous real-valued functions on X, or as a member of the space L2 . (b) If w ∈ L2 , then u 7→ u ∗ w is a compact linear operator whether regarded as a map from L2 to C(X) or as a map from L2 to itself. (c) If w ∈ L2 and w = w ˜ (as defined in 443Af), then u 7→ u ∗ w : L2 → L2 is a self-adjoint operator. proof (a) Being compact, X is unimodular (442Ic). As noted in 444Sa, ∗ can be regarded as a bilinear map from L2 × L2 to Cb (X) = C(X). Because µX must be finite, we now have a natural map f 7→ f • from C(X) to L2 , so that we can think of u ∗ v as a member of L2 for u, v ∈ L2 . (b)(i) Evidently (u, v) 7→ u ∗ v : L2 × L2 → C(X) is bilinear. So u 7→ u ∗ w : L2 → C(X) is linear, for any w ∈ L2 . (ii) Let B be the unit ball of L2 , and give it the topology induced by the weak topology Ts (L2 , L2 ), so that B is compact (4A4Ka). Let •l be the left action of G on L2 as in 443G, so that a•l f • = (a•l f )• for every a ∈ G, f ∈ L2 = L2 (µ). If f , g ∈ L2 and a ∈ X, then (f ∗ g)(a) =

R

f (x)g(x−1 a)dx =

R

f (x)˜ g (a−1 x)dx =

R

f × a•l g˜,

where g˜(x) = g(x−1 ) whenever this is defined. So if u, w ∈ L2 and a ∈ X, (u∗w)(a) = (u|a•l w), ˜ where w ˜ = g˜• • 2 whenever w = g , as in 443Af. It follows that, for any w ∈ L , the function (a, u) 7→ (u ∗ w)(a) : X × B → R is continuous. P P We know that •l : X × L2 → L2 is continuous when L2 is given its norm topology (443Ge). Given a0 ∈ X and u0 ∈ B and ² > 0, set G = {a : a ∈ X, ka•l w ˜ − a0 •l wk ˜ 2 < 12 ²} and 1 H = {u : u ∈ B, |(u|a0 •l w) ˜ − (u0 |a0 •l w)| ˜ < 2 ²}, so that G and H are open sets (for the topologies under consideration) containing a0 and u0 respectively. Then, for any a ∈ G and u ∈ H, |(u ∗ w)(a) − (u0 ∗ w)(a0 )| = |(u|a•l w) ˜ − (u0 |a0 •l w)| ˜ ≤ |(u|a•l w) ˜ − (u|a0 •l w)| ˜ + |(u|a0 •l w) ˜ − (u0 |a0 •l w)| ˜ 1 2

≤ ka•l w ˜ − a0 •l wk ˜ 2 + ² ≤ ². As a0 , u0 and ² are arbitrary, we have the result. Q Q Because X is compact, this means that u 7→ u ∗ w : B → C(X) is continuous when C(X) is given its norm topology and B is given the weak topology (4A2G(g-ii)). Because B is compact in the weak topology, {u ∗ w : u ∈ B} is compact in C(X). But this implies that u 7→ u ∗ w is a compact linear operator (definition: 3A5Ka).

330

Topological groups

444V

√ (iii) Because X is compact, µ is totally finite, so, for f ∈ C(X), kf k2 ≤ kf k∞ µX, and the natural map f 7→ f • : C(X) → L2 is a bounded linear operator. Consequently the map u 7→ (u ∗ w)• : L2 → L2 is a compact operator, by 4A4La. (c) Now suppose that w = w. ˜ In this case (u ∗ w|v) = (u|v ∗ w) for all u, v ∈ L2 . P P Express u, v and w • • • as f , g and h where f , g and h are square-integrable Borel measurable functions defined everywhere on X. We have Z

ZZ

(f ∗ h)(x)g(x)dx = f (y)h(y −1 x)g(x)dydx ZZ = f (y)h(y −1 x)g(x)dxdy RR (because (x, y) 7→ f (y)h(y −1 x)g(x) is Borel measurable, µ is totally finite and |f (y)h(y −1 x)g(x)|dydx = (|u| ∗ |w|||v|) is finite) ZZ Z ˜ −1 y)g(x)dxdy = f (y)(g ∗ h)(y)dy ˜ = f (y)h(x (u ∗ w|v) =

= (u|v ∗ w) ˜ = (u|v ∗ w). Q Q As u and v are arbitrary, this shows that u 7→ u ∗ w : L2 → L2 is self-adjoint. 444X Basic exercises > (a) Let X be a Hausdorff topological group. Show that if λ and ν are totally finite Radon measures on X then λ ∗ ν is the image measure (λ × ν)φ−1 , where φ(x, y) = xy for x, y ∈ X, and in particular is a Radon measure. > (b) Let X be a topological group and λ, ν two totally finite quasi-Radon measures on X. Writing supp λ for the support of λ, show that supp(λ ∗ ν) = (supp λ)(supp ν). + (c) Let X be a topological group and MqR the family of totally finite quasi-Radon measures on X. Show + + + + that (λ, ν) 7→ λ ∗ ν : MqR × MqR → MqR is continuous for the narrow topology on MqR . (Hint: 437Ma.)

(d) Let X be a Hausdorff topological group. Show that X is abelian iff its Banach algebra of signed τ -additive Borel measures is commutative. (e) Let X be a topological group, and Mτ its Banach algebra of signed τ -additive Borel measures. (i) Show that we have actions •l , •r of X on Mτ defined by writing (a•l ν)(E) = ν(aE), (a•r ν)(E) = ν(Ea−1 ). (ii) Show that (a•l λ) ∗ ν = a•l (λ ∗ ν), λ ∗ (a•r ν) = a•r (λ ∗ ν) for all a ∈ X and λ, ν ∈ Mτ . (f ) Let X be a compact topological group, and B a norm-bounded subset of the Banach algebra Mτ of signed τ -additive Borel measures on X. Show that (λ, ν) 7→ λ ∗ ν : B × B → Mτ is continuous for the vague topology on Mτ . (g) Let X be a topological group, and ν a totally finite quasi-Radon measure on X. Show that for any Borel sets E, F ⊆ X, the function (g, h) 7→ ν(gE ∩ F h) is Borel measurable. (Hint: for Borel sets W ⊆ X × X, set ν 0 W = ν{x : (x, x) ∈ W }. Consider the action of X × X on itself defined by writing (g, h)•(x, y) = (gx, yh−1 ).) (h) Let X be a topological group and f a real-valued function defined on a subset of X. (i) Show that a•r (ν ∗ f ) = ν ∗ (a•r f ) (definition: 441Ac) whenever a ∈ X and ν is a measure on X. (ii) Show that if X carries Haar measures, then a•l (f ∗ ν) = (a•l f ) ∗ ν whenever a ∈ X and ν is a measure on X. (i) Let X be a topological group carrying Haar measures, f : X → R a bounded continuous function and ν a totally finite quasi-Radon measure on X. Show that f ∗ ν is continuous.

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(j) Let X be a topological group carrying Haar measures, f a real-valued function defined on a subset of X, and λ, ν totally finite quasi-Radon measures on X. Show that ((f ∗ ν) ∗ λ)(x) = (f ∗ (ν ∗ λ))(x) whenever the right-hand side is defined. (See also 444Yi.) (k) Let X be an abelian topological group carrying Haar measures. Show that f ∗ ν = ν ∗ f for every measure ν on X and every real-valued function f defined on a subset of X. > (l) Let X be a topological group and µ a left Haar measure on X. (i) Let ν be a totally finite quasiRadon measure on X such that x 7→ ν(xF ) is continuous for every closed set F ⊆ X. Show that ν is truly continuous with respect to µ. (Hint: if µF = 0, apply 444K to ν ∗ χF −1 to see that ν(xF ) = 0 for µ-almost every x.) (ii) Let ν be a totally finite Radon measure on X such that x 7→ ν(xK) is continuous for every compact set K ⊆ X. Show that ν is truly continuous with respect to µ. (m) Let X be a topological group carrying Haar measures, and ν a non-zero totally finite quasi-Radon measure on X such that ν(xE) = 0 whenever x ∈ X and νE = 0. (i) Show that ν is strictly positive, so that X is ccc. (ii) Show that a subset of X is ν-negligible iff it is Haar negligible. (Hint: if µ is a left Haar measure on X, consider the convolution χG ∗ ν for a suitable set G.) (n) Use the method of part (b) of the proof of 444M to prove part (a) there. > (o) Let X be the group S 1 × S 1 , with the topology defined by giving the first coordinate the usual topology of S 1 and the second coordinate the discrete topology, so that X is a locally compact abelian group. Let µ be a Haar measure on X. (i) Find a Borel measurable function f : X × X → {0, 1} such that RR RR f (x, y)µ(dx)µ(dy) 6= f (x, y)µ(dy)µ(dx). (ii) Let ν be the Radon measure on X defined by setting νE = #({s : (s, s−1 ) ∈ E}) if this RR is finite, ∞ otherwise. RRDefine g : X → {0, 1} by setting g(s, t) = 1 if s = t, 0 otherwise. Show that g(xy)ν(dy)µ(dx) = ∞, g(xy)µ(dx)ν(dy) = 0. (iii) Find a closed set F ⊆ X such that x 7→ ν(xF ) is not Haar measurable. > (p) Let X be a Hausdorff topological group and for a ∈ X write δa for the Radon measure on X defined by saying that δa (E) = (χE)(a) whenever E ⊆ X. (i) Show that δa ∗ δb = δab for all a, b ∈ X. (ii) Show that, in the notation of 444Xe, δa ∗ νˆ is the completion of a−1 •l ν and νˆ ∗ δa is the completion of a•r ν for every a ∈ X and every totally finite τ -additive Borel measure ν on X with completion νˆ. (iii) Show that δa ∗ f = a•l f for every a ∈ X and every real-valued function f defined on a subset of X. (iv) Show that if X carries Haar measures, and has left modular function ∆, f ∗ δa = ∆(a−1 )Sa0 −1 (f ) for every a ∈ X and every real-valued function f defined on a subset of X. (v) Use these formulae to relate 444Of to 444B. (q) Let X be a topological group and µ a left Haar measure on X. For a real-valued function f defined on a subset of X set f˜(x) = f (x−1 ) whenever x ∈ X and x−1 ∈ dom f . Show that if f , g ∈ L0 (X) then (f ∗ g)∼ = g˜ ∗ f˜. (r) Let X be a locally compact Hausdorff topological group and µ a left Haar measure on X. Show that if f , g : X → R are continuous functions with compact support, then f ∗ g is a continuous function with compact support. (s) In 444Rc, show that f ∗ g˜ is uniformly continuous for the bilateral uniformity. (Hint: in 443Ge, x 7→ x•l u is uniformly continuous.) (t) Let X be a topological group with a totally finite Haar measure µ. Show that (i) (u∗w|v) = (u|v∗w) ˜ for any u, v, w ∈ L2 = L2 (µ), where w ˜ and u∗v are defined as in 443Af and 444V (ii) the map u 7→ u∗w : L2 → L2 is a compact linear operator for any w ∈ L2 . (Hint: for (ii), use 443L.) (u) Let X be a topological group with a Haar probability measure µ. Show that L2 (µ) with convolution is a Banach algebra. 444Y Further exercises (a) Find a subgroup X of {0, 1}N and quasi-Radon probability measures λ, ν on X and a set A ⊆ X such that (λ ∗ ν)∗ (A) = 1 but (λ × ν){(x, y) : x, y ∈ X, x + y ∈ A} = 0.

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444Yb

(b) Let X be a topological semigroup, that is, a semigroup with a topology such that multiplication is continuous. (i) For totally finite τ -additive Borel measures λ, ν on X, show that there is a τ -additive Borel measure λ ∗ ν defined by saying that (λ ∗ ν)(E) = (λ × ν){(x, y) : xy ∈ R E} for everyR Borel set E ⊆ X. (ii) Show that in this context (λ1 ∗ λ2 ) ∗ λ3 = λ1 ∗ (λ2 ∗ λ3 ). (iii) Show that f d(λ ∗ ν) = f (xy)λ(dx)ν(dy) whenever f is (λ ∗ ν)-integrable. (iv) Show that if the topology is Hausdorff and λ and ν are tight (that is, inner regular with respect to the compact sets) so is λ ∗ ν. (v) Show that we have a Banach algebra of signed τ -additive Borel measures on X, as in 444E. (C)

(c) Let X be a topological group, and write Mτ for the complexification of the L-space Mτ of 444E, as (C) described in 354Yk. Show that Mτ , with the natural extension of the convolution operator of 444E, is a (C) complex Banach algebra, and that we still have |λ ∗ ν| ≤ |λ| ∗ |ν| for λ, ν ∈ Mτ . (d) Find a locally compact Hausdorff topological group X, a Radon probability measure ν on X and an open set G ⊆ X such that {(xGx−1 )• : x ∈ X} is not a separable subset of the measure algebra of ν. (e) Let X be a Polish group. We say that a subset A of X is Haar null if there are a universally measurable set E ⊇ A and a non-zero Radon measure ν on X such that ν(xEy) = 0 for every x, y ∈ X. (i) Show that the family of Haar null sets is a translation-invariant σ-ideal of subsets of X. (Hint: if hEn in∈N is a sequence of universally measurable Haar null sets, we can find Radon probability measures νn concentrated near the identity such that νn (xEn y) = 0 for every x, y and n; now construct an infinite convolution product ν = ν0 ∗ ν1 ∗ . . . and show that ν(xEn y) = 0 for every x, y and n.) (ii) Show that if Y is another Polish group, φ : X → Y is a surjective continuous homomorphism and B ⊆ Y is Haar null, then φ−1 [B] is Haar null in X. (iii) Show that if X is a locally compact Polish group then a subset of X is Haar null iff it is Haar negligible in the sense of 442H. (See Solecki 01.) (f ) Let X be a topological group and µ a left Haar measure on X. Let τ be a T -invariant extended Fatou norm on L0 (µ) (§374). Show that if ν is any totally finite quasi-Radon measure on X, then we have a linear operator f • 7→ (ν ∗ f )• from Lτ to itself, of norm at most νX. (g) Let X be a topological group with a left Haar measure µ, Mτ the Banach algebra of signed τ -additive Borel measures on X, and p ∈ [1, ∞]. (i) Show that we have a multiplicative linear operator T from Mτ to the Banach algebra B(Lp (µ); Lp (µ)) defined by writing (T ν)(f • ) = (ˆ ν ∗ f )• whenever ν is a totally finite p τ -additive Borel measure on X with completion νˆ and f ∈ L (µ). (Hint: Use 444K and 444B to show that (λ ∗ ν) ∗ f =a.e. λ ∗ (ν ∗ f ) for enough λ, ν and f . See also 444Yi.) (ii) Show that kT νk = kνk for every ν ∈ Mτ+ . (h) Let X be a unimodular topological group with left Haar measure µ. Suppose that p, q, r ∈ [1, ∞] 1 are such that p1 + 1q = 1 + 1r , interpreting ∞ as 0. Show that if f ∈ Lp (µ) and g ∈ Lq (µ) then f ∗ g ∈ Lr (µ) and kf ∗ gkr ≤ kf kp kgkq . (Hint: 255Ym. Take care to justify any changes in order of integration.) (i) Let X be a topological group carrying Haar measures. Investigate conditions under which the associative laws λ ∗ (ν ∗ f ) = (λ ∗ ν) ∗ f ,

λ ∗ (f ∗ ν) = (λ ∗ f ) ∗ ν,

f ∗ (λ ∗ ν) = (f ∗ λ) ∗ ν,

f ∗ (g ∗ ν) = (f ∗ g) ∗ ν,

f ∗ (ν ∗ g) = (f ∗ ν) ∗ g,

ν ∗ (f ∗ g) = (ν ∗ f ) ∗ g

will be valid, where λ and ν are quasi-Radon measures on X and f , g are real-valued functions. Relate your results to 444Xp. (j) Let X be a topological group and µ a left Haar measure on X. Let τ be a T -invariant extended Fatou norm on L0 (µ) such that τ ¹Lτ is an order-continuous norm. For a totally finite quasi-Radon measure ν on X, let Tν : Lτ → Lτ be the corresponding linear operator (444Yf). Show that for any u ∈ Lτ and ² > 0 there is a neighbourhood U of the identity in X such that τ (Tν u − u) ≤ ² whenever νU = νX = 1. (k) Let X be a topological group with a left HaarRmeasure µ. For u ∈ L2 = L2 (µ), set A = {a•l u : a ∈ X} (443G) in L2 , and D = {v ∗ u : v ∈ L1 (µ), v ≥ 0, v = 1}. (i) Show that the closed convex hull of A in

444 Notes

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333

L2 is the closure of D. (Hint: (α) use 444Od to show that if w ∈ L2 and (w0 |w) ≥ γ for every w0 ∈ A, then (w0 |w) ≥ γ for every w0 ∈ D (β) use 444U to show that A ⊆ D.) (ii) Show that the closed linear subspace Wu generated by A is the closure of {v ∗ u : v ∈ L1 }. (iii) Show that if w ∈ L2 and w ∈ A⊥ , that is, (u0 |w) = 0 for every u0 ∈ A, then Ww ⊆ Wu⊥ . (iv) Show that if X is σ-compact, then Wu is separable. (Hint: A is σ-compact, by 443Ge.) (v) Set C = {f • : f ∈ L2 ∩ C(X)}. Show that C ∩ Wu is dense in Wu . (Hint: v ∗ u ∈ C for many v, by 444Rc.) (vi) Show that if X is σ-compact, then Wu has an orthonormal basis in C. (vii) Show that L2 has an orthonormal basis in C. (Hint: if X is σ-compact, take a maximal orthogonal family of subspaces Wu , find a suitable orthonormal basis of each, and use (iii) to see that these assemble to form a basis of L2 . For a general locally compact Hausdorff group, start with a σ-compact open subgroup, and then deal with its cosets. For a general topological group with a Haar measure, use 443L.) (Compare 416Yg.) (l) Let X be a topological group with a left Haar measure µ. Let λ be the quasi-Radon product measure on X × X. Let U be the set of those h ∈ L1 (λ) such that (X × X) \ {(x, y) : (x, y) ∈ dom h, h(x, y) = 0} can be covered by a sequence of open sets of finite measure. (i) Show that if hR ∈ U, then (x, y) 7→ h(y, y −1 x) belongs to U. (Hint: 443Xb.) (ii) Show that if h ∈ U, then (T h)(x) = h(y, y −1 x)µ(dy) is defined for almost every x ∈ X and T h is µ-integrable, with kT hk1 ≤ khk1 . (Hint: 255Xd.) (ii) Show that if h1 , h2 ∈ U are equal λ-a.e. then T h1 = T h2 µ-a.e. (iii) Show that every member of L1 (λ) can be represented by a member of U. (Hint: 443Xl.) (iv) Show that if f , g ∈ L1 (µ) and both are zero outside some countable union of open sets of finite measure, then T (f ⊗ g) = f ∗ g, where (f ⊗ g)(x, y) = f (x)g(y). (v) Show that if we set T˜(h• ) = (T h)• for h ∈ U, then T˜ : L1 (λ) → L1 (µ) is the unique continuous linear operator such that T˜(u ⊗ v) = u ∗ v for all u, v ∈ L1 (µ), where u ∗ v is defined in 444S and ⊗ : L1 (µ) × L1 (µ) → L1 (λ) is the canonical bilinear map (253E). (m) In 444Yl, suppose that µX = 1. (i) Show that the map T˜ belongs to the class Tλ,¯ ¯ µ of §373. (ii) Show that if p ∈ [1, ∞] then kT hkp ≤ khkp whenever h ∈ U ∩ Lp (λ). (n) Rewrite this section in terms of right Haar measures instead of left Haar measures. 444 Notes and comments The aim of this section and the next is to work through ideas from the second half of Chapter 25, and Chapter 28, in forms natural in the context of general topological groups. (It is of course possible to go farther; see 444Yb. It is the glory and confusion of twentieth-century mathematics that it has no firm stopping points.) The move from R to an arbitrary topological group is a large one, and I think it is worth examining the various aspects of this leap as they affect the theorems here. The most conspicuous change, and the one which most greatly affects the forms of the results, is the loss of commutativity. We are forced to re-examine every formula to determine exactly which manipulations can still be justified. Multiplications must be written the right way round, and inversions especially must be watched. But while there are undoubtedly some surprises, we find that in fact (provided we take care over the definitions) the most important results survive. Of course I wrote the earlier results out with a view to what I expected to do here, but no dramatic manoeuvers are needed to turn the fundamental results 255G, 255H, 255J, 257B, 257E, 257F into the new versions 444Od, 444Qb, 444Oe, 444C, 444B, 444Qa. (The changed order of presentation is an indication of the high connectivity of the web here, not of any new pattern.) In fact what makes the biggest difference is not commutativity, as such, but unimodularity. In groups which are not unimodular we do have new phenomena, as in 444Mb and 444Of, and these lead to complications in the proofs of such results as 444U, even though the result there is exactly what one would expect. In this section I ignore right Haar measures entirely. I do not even put them in the exercises. If you wish to take this theory farther, you may some day have to work out the formulae appropriate to right Haar measures. (You can check your results in Hewitt & Ross 63, 20.32.) But for the moment, I think that they are likely to be just a source of confusion. There is one point which you may have noticed. The theory of groups is essentially symmetric. In the definition of ‘group’ there is no distinction between left and right. In the formulae defining group actions, we do have such a distinction, because they must reflect the fact that we write g •x rather than x•g. With •l and •r , for instance (444Of), if we want them to be actions in the standard sense we have to put an −1 into the definition of •l but not into the definition of •r . But we

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still expect that, for instance, λ ∗ ν and ν ∗ λ will be related in some transparent way. However there is an exception to this rule in the definitions of ν ∗ f and f ∗ ν (444H, 444J). The modular function appears in the latter, so in fact the definition applies only in a more restricted class of groups. In abelian groups we assume that f ∗ ν and ν ∗ f will be equal, and they are (444Xk), but strictly speaking, on the definitions here, we can write f ∗ ν = ν ∗ f only for abelian topological groups carrying Haar measures. From the point of view of the proofs in this section, the principal change is that the Haar measures here are no longer assumed to be σ-finite. I am well aware that non-σ-finite measures are a minority interest, especially in harmonic analysis, but I do think it interesting that σ-finiteness is not relevant to the main results, and the techniques required to demonstrate this are very much in the spirit of this treatise (see, in particular, the proof of 444N, and the repeated applications of 443Jb). The basic difficulty is that we can no longer exchange repeated integrals, even of non-negative Borel measurable functions, quite automatically. Let me emphasize that the result in 444N is really rather special. If we try to generalize it to other measures or other types of function we encounter the usual obstacles (444Xo). A difficulty of a different kind arises in the proof of 444Fc. Here I wish to show that the function g 7→ (g •E)• : G → A is Borel measurable for every Borel measurable set E. The first step is to deal with open sets E, and it would be nice if we could then apply the Monotone Class Theorem. But the difficulty is that even though the map (a, b) 7→ a \ b : A × A → A is continuous, it does not quite follow that the map g 7→ (g •(E \ F ))• = (g •E)• \ (g •F )• is Borel measurable whenever g 7→ (g •E)• and g 7→ (g •F )• are, because the map g 7→ ((g •E)• , (g •F )• ) : G → A × A might conceivably fail to be Borel measurable, if the metric space A is not separable, that is, if the Maharam type of the measure ν is uncountable. Of course the difficulty is easily resolved by an extra twist in the argument. I use different techniques for the two parts of 444M as an excuse to recall the ideas of §371; in fact part (a) is slightly easier than part (b) when proved by the method of the latter (444Xn). 444U is a kind of density theorem. Compared with the density theorems in §§223 and 261, it is a ‘mean’ rather than ‘pointwise’ density theorem; if E is concentrated near the identity, then fE• approximates f • in Lp , but there is no suggestion that we can be sure that fE (x) l f (x) for any particular xs unless we know much more about the set E. In fact this is to be expected from the form of the results concerning Lebesgue measure. The sets E considered there are generally intervals or balls, and even in such a general form as 223Ya we need a notion of scalar multiplication separate from the group operation.

445 The duality theorem In this section I present a proof of the Pontryagin-van Kampen duality theorem (445U). As in Chapter 28, and for the same reasons, we need to use complex-valued functions; the relevant formulae in §444 apply unchanged, and I shall not repeat them here, but you may wish to re-read that section taking all functions to be complex- rather than real-valued. (It is possible to avoid complex-valued measures, which I relegate to the exercises.) The duality theorem itself applies only to abelian locally compact Hausdorff groups, and it would be reasonable, on first reading, to take it for granted that all groups here are of this type, which simplifies some of the proofs a little. My exposition is based on that of Rudin 67. I start with the definition of ‘dual group’, including a description of a topology on the dual (445A), and the simplest examples (445B), with a mention of FourierStieltjes transforms of measures (445C-445D). The elementary special properties of dual groups of groups carrying Haar measures are in 445E-445G; in particular, in these cases, the bidual of a group begins to make sense, and we can start talking about Fourier transforms of functions. Serious harmonic analysis begins with the identification of the dual group with the maximal ideal space of L1 (445H-445K). The next idea is that of ‘positive definite’ function (445L-445M). Putting these together, we get the first result here which asserts that the dual group of an abelian group X carrying Haar measures is sufficiently large to effectively describe functions on X (Bochner’s theorem, 445N). It is now easy to establish that X can be faithfully embedded in its bidual (445O). We also have most of the machinery necessary to describe the correctly normalized Haar measure of the dual group, with a first step towards identifying functions whose Fourier transforms will have inverse Fourier transforms (the Inversion Theorem, 445P). This leads directly to the Plancherel Theorem, identifying the L2 spaces of X and its dual (445R).

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335

At this point it is clear that the bidual X cannot be substantially larger than X, since they must have essentially the same L2 spaces. A little manipulation of shifts and convolutions in L2 (445S-445T) shows that X must be dense in X, and a final appeal to local compactness shows that X is closed in X. 445A Dual groups Let X be any topological group. (a) A character on X is a continuous group homomorphism from X to S 1 = {z : z ∈ C, |z| = 1}. It is easy to see that the set X of all characters on X is a subgroup of the group (S 1 )X , just because S 1 is an abelian topological group. (If χ, θ ∈ X , then x 7→ χ(x)θ(x) is continuous, and (χθ)(xy) = χ(xy)θ(xy) = χ(x)χ(y)θ(x)θ(y) = χ(x)θ(x)χ(y)θ(y) = (χθ)(x)(χθ)(y).) So X itself is an abelian group. (b) Give X the topology of uniform convergence on subsets of X which are totally bounded for the bilateral uniformity on X (4A5Hb, 4A5O). (If E is the set of totally bounded subsets of X, then the topology of X is generated by the pseudometrics ρE , where ρE (χ, θ) = supx∈E |χ(x) − θ(x)| for E ∈ E and χ, θ ∈ X . It will be useful, in this formula, to interpret sup ∅ = 0, so that ρ∅ is the zero pseudometric. Note that E is closed under finite unions, so {ρE : E ∈ E} is upwards-directed, as in 2A3Fe.) Then X is a Hausdorff topological group. (If x ∈ E ∈ E and χ, χ0 , θ, θ0 ∈ X , |(χθ)(x) − (χ0 θ0 )(x)| = |χ(x)(θ(x) − θ0 (x)) + θ0 (x)(χ(x) − χ0 (x))| ≤ |θ(x) − θ0 (x)| + |χ(x) − χ0 (x)|, |χ−1 (x) − χ−1 0 (x)| = |χ(x) − χ0 (x)|, so ρE (χθ, χ0 θ0 ) ≤ ρE (χ, χ0 ) + ρE (θ, θ0 ),

ρE (χ−1 , χ−1 0 ) = ρE (χ, χ0 ).

If χ 6= θ then there is an x ∈ X such that χ(x) 6= θ(x), and now {x} ∈ E and ρ{x} (χ, θ) > 0.) (c) Note that if X is locally compact, then its totally bounded sets are just its relatively compact sets (4A5Oe), so the topology of X is the topology of uniform convergence on compact subsets of X. (d) If X is compact, then X is discrete. P P X itself is totally bounded, so U = {χ : |χ(x) − 1| ≤ 1 for every x ∈ X} is a neighbourhood of the identity ι in X . But if χ ∈ U , x ∈ X then |χ(x)n − 1| ≤ 1 for every n ∈ N, so χ(x) = 1. Thus U = {ι} and ι is an isolated point of X ; it follows that every point of X is isolated. Q Q (e) If X is discrete then X is compact. P P The only totally bounded sets in X are the finite sets, so the topology of X is just that induced by its embedding in (S 1 )X . On the other hand, every homomorphism from X to S 1 is continuous, so X is a closed set in (S 1 )X , which is compact by Tychonoff’s theorem. Q Q 445B Examples (a) If X = R with addition, then X can also be identified with the additive group R, if we write χy (x) = eiyx for x, y ∈ R. P P It is easy to check that every χy , so defined, is a character on R, and that y 7→ χy : R → X is a homomorphism. On the other hand, if χ is a character, then (because it is continuous) there is a δ ≥ 0 such that |χ(x) − 1| ≤ 1 whenever |x| ≤ δ. χ(δ) is uniquely expressible as eiα where |α| ≤ π2 . Set y = α/δ, so that χ(δ) = χy (δ). Now χ( 21 δ) must be one of the square roots of χ(δ), so is ±χy ( 12 δ); but as |χ( 21 δ) − 1| ≤ 1, it must be +χy ( 12 δ). Inducing on n, we see that χ(2−n δ) = χy (2−n δ) for every n ∈ N, so that χ(2−n kδ) = χy (2−n kδ) for every k ∈ Z, n ∈ N; as χ and χy are continuous, χ = χy . Thus the map y 7→ χy is surjective and is a group isomorphism between R and X . As for the topology of X , R is a locally compact topological group, so the totally bounded sets are just the relatively compact sets (4A5Oe), that is, the bounded sets in the usual sense (2A2F). Now a straightforward calculation shows that for any α ≥ 0 in R and ² ∈ ]0, 2[, ² 2

ρ[−α,α] (χy , χz ) ≤ ² ⇐⇒ α|y − z| ≤ 2 arcsin , so that the topology of X agrees with that of R. Q Q

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Topological groups

445Bb

(b) Let X be the group Z with its discrete topology. Then we may identify its dual group X with S 1 itself, writing χζ (n) = ζ n for ζ ∈ S 1 , n ∈ Z. P P Once again, it is elementary to check that every χζ is a character, and that ζ 7→ χζ is an injective group homomorphism from S 1 to X . If χ ∈ X , set ζ = χ(1); then χ = χζ . So X ∼ = S 1 . And because the only totally bounded sets in X are finite, χ 7→ χζ is continuous, therefore a homeomorphism. Q Q (c) On the other hand, if X = S 1 with its usual topology, then we may identify its dual group X with Z, writing χn (ζ) = ζ n for n ∈ Z, ζ ∈ S 1 . P P The verification follows the same lines as in (a) and (b). As usual, the key step is to show that the map n 7→ χn : Z → X is surjective. We can do this by applying (a). If χ ∈ X , then x 7→ χ(eix ) is a character of R, so there is a y ∈ R such that χ(eix ) = eiyx for every x ∈ R. In particular, e2iyπ = χ(1) = 1, so y ∈ Z, and χ = χy . Concerning the topology of X , we know from 445Ad that it must be discrete, so that also matches the usual topology of Z. Q Q (d) Let hXj ij∈J be any family of topological groups, and X their product (4A5G). For each jQ∈ J let Xj be the dual group of Xj . Then the dual group of X can be identified with the subgroup X of j∈I Xj Q consisting of those χ ∈ j∈J Xj such that {j : χ(j) is not the identity} is finite; the action of X on X is defined by the formula Q χ•x = j∈J χ(j)(x(j)). (This is well-defined because only finitely many terms in the product are not equal to 1.) If I is finite, so Q that X = j∈I Xj , the topology of X is the product topology. P P As usual, it is easy to check that •, as defined above, defines an injective homomorphism from X to the dual group of X. If θ is any character on X, then for each j ∈ I we have a continuous group homomorphism εj : Xj → X defined by setting εj (ξ)(j) Q = ξ, εj (ξ)(k) = ek , the identity of Xk , for every k 6= j. Setting χ(j) = θεj for each j, we obtain χ ∈ j∈I Xj . Now there is a neighbourhood U of the identity of X such that |θ(x) − 1| ≤ 1 for every x ∈ U , and we may suppose that U is of the form {x : x(j) ∈ Gj for every j ∈ J}, where J ⊆ I is finite and Gj is a neighbourhood of ej for every j ∈ J. If k ∈ I \ J, εk (ξ) ∈ U for every ξ ∈ Xk , so that |χ(k)(ξ) − 1| ≤ 1 for every ξ, and χ(k) must be the identity character on Xk ; n this shows that χ ∈ X . If x ∈ X and x(j) = ej for j ∈ J, then again |θ(x Q ) − 1| ≤ 1 for every n ∈ N, so • θ(x) = χ x = 1. For any x ∈ X, we can express it as a finite product y j∈J εj (x(j)) where y(j) = ej for every j ∈ J, so that Q Q θ(x) = θ(y) j∈J θεj (x(j)) = j∈J χ(j)(x(j)) = χ•x. Thus • defines an isomorphism between X and the dual group of X. As for the topology Q of X , a subset of X is totally bounded iff it is included in a product of totally bounded sets (4A5Od). If E = j∈I Ej is such a product (and not empty), then for χ, θ ∈ X P supj∈I ρEj (χ(j), θ(j)) ≤ ρE (χ, θ) ≤ j∈I ρEj (χ(j), θ(j)), so (if I is finite) the topology on X is just the product topology. Q Q 445C Fourier-Stieltjes transforms Let X be a topological group, and X its dual group. For any totally finite topological measure ν on X,R we can form its ‘characteristic function’ or Fourier-Stieltjes ∧ ∧ transform ν : X → C by writing ν(χ) = χ(x)ν(dx) (cf. 285A). 445D Theorem Let X be a topological group, and X its dual group. If λ and ν are totally finite ∧ ∧ quasi-Radon measures on X, then (λ ∗ ν)∧ = λ × ν. proof If χ ∈ X , then, by 444C, Z ZZ ∧ (λ ∗ ν) (χ) = χ d(λ ∗ ν) = χ(xy)λ(dx)ν(dy) ZZ Z Z ∧ ∧ = χ(x)χ(y)λ(dx)ν(dy) = χ(x)λ(dx) · χ(y)ν(dy) = λ(χ)ν(χ).

445E

The duality theorem

337

445E Let us turn now to groups carrying Haar measures. I start with three welcome properties. Proposition Let X be a topological group with a neighbourhood of the identity which is totally bounded for the bilateral uniformity on X, and X its dual group, with its dual group topology. (a) The map (χ, x) 7→ χ(x) : X × X → S 1 is continuous. (b) Let X be the dual group of X , again with its dual group topology, the topology of uniform convergence on totally bounded subsets of X . Then we have a continuous homomorphism x 7→ x ˆ : X → X defined by setting x ˆ(χ) = χ(x) for x ∈ X, χ ∈ X . ∧ (c) For any totally finite quasi-Radon measure ν on X, its Fourier-Stieltjes transform ν : X → C is uniformly continuous. Remark Note that the condition here is satisfied by any topological group X carrying Haar measures (443H). proof Fix an open totally bounded set U0 containing the identity. (a) Let χ0 ∈ X , x0 ∈ X and ² > 0. Then x0 U0 is totally bounded, so 1 2

V = {χ : |χ(y) − χ0 (y)| ≤ ² for every y ∈ x0 U0 } is a neighbourhood of χ0 . Also 1 2

U = {x : x ∈ x0 U0 , |χ0 (x) − χ0 (x0 )| ≤ ²} is a neighbourhood of x0 . And if χ ∈ V , x ∈ U we have |χ(x) − χ0 (x0 )| ≤ |χ(x) − χ0 (x)| + |χ0 (x) − χ0 (x0 )| ≤ ². As χ0 , x0 and ² are arbitrary, (χ, x) 7→ χ(x) is continuous. (b)(i) It is easy to check that x ˆ, as defined above, is always a homomorphism from X to S 1 , and that x 7→ x ˆ : X → (S 1 )X is a homomorphism. Because ρ{x} is always one of the defining pseudometrics for the topology of X (445Ab), x ˆ is always continuous, so belongs to X. (ii) To see that b is continuous, I argue as follows. Take an open set H ⊆ X and x0 ∈ X such that x ˆ0 ∈ H. Then there are a totally bounded set F ⊆ X and an ² > 0 such that x ∈ H whenever x ∈ X and ρF (x, x ˆ0 ) ≤ ². Now x0 U0 is a totally bounded neighbourhood of x0 , so 1 2

V = {θ : θ ∈ X , |θ(y) − 1| ≤ ² for every y ∈ x0 U0 } is a neighbourhood of the identity in X . There are therefore χ0 , . . . , χn ∈ X such that F ⊆

S k≤n

χk V . Set

1 2

U = {x : x ∈ x0 U0 , |χk (x) − χk (x0 )| < ² for every k ≤ n}. Then U is an open neighbourhood of x0 in X. If x ∈ U and χ ∈ F then there is a k ≤ n such that θ = χ−1 k χ ∈ V , so that |χ(x) − χ(x0 )| = |χk (x)θ(x) − χk (x0 )θ(x0 )| 1 2

1 2

≤ |χk (x) − χk (x0 )| + |θ(x) − θ(x0 )| ≤ ² + ² = ². But this shows that ρF (ˆ x, x ˆ0 ) ≤ ², so x ˆ ∈ H. So we have x0 ∈ U ⊆ {x : x ˆ ∈ H}. As x0 is arbitrary, {x : x ˆ ∈ H} is open; as H is arbitrary, x 7→ x ˆ is continuous. S (c) Let ² > 0. Because ν is τ -additive, there are x0 , . . . , xn ∈ X such that ν(X \ k≤n xk U0 ) ≤ 31 ². Set S E = k≤n xk U0 ; then E is totally bounded. So V = {θ : |θ(x) − 1| ≤

² 1+3νX

for every x ∈ E}

is a neighbourhood of the identity in X . If χ, χ0 ∈ X are such that χ−1 χ0 ∈ V , then

338

Topological groups

445E

Z ∧

0

∧

|χ(x) − χ0 (x)|ν(dx)

|ν(χ) − ν(χ )| ≤

≤ 2ν(X \ E) +

²νE 1+3νX

≤ ².

∧

Since ² is arbitrary (and X is abelian), this is enough to show that ν is uniformly continuous. 445F Fourier transforms of functions Let X be a topological group with a left Haar measure µ. ∧ For any µ-integrable complex-valued function f , define its Fourier transform f : X → C by setting ∧ R ∧ f (χ) = f (x)χ(x)µ(dx) for every character χ of X. (Compare 283A. If f is non-negative, then f = (f µ)∧ ∧

∧

as defined in 445C, where f µ is the indefinite-integral measure, as in 444K.) Note that f = g whenever ∧

∧

f =a.e. g, so we can equally well write u(χ) = f (χ) whenever u = f • in L1C (µ). 445G Proposition Let X be a topological group with a left Haar measure µ. Then for any µ-integrable ∧ ∧ ∧ ∧ complex-valued functions f and g, (f ∗ g)∧ = f × g; that is, (u ∗ v)∧ = u × v for all u, v ∈ L1C (µ). proof For any character χ on X, ZZ Z χ(x)(f ∗ g)(x)dx = χ(xy)f (x)g(y)dxdy ZZ ∧ ∧ = χ(x)χ(y)f (x)g(y)dxdy = f (χ)g(χ) (using 444Od). 445H Theorem Let X be a topological group with a left Haar measure µ; let X be its dual group and let Φ be the set of non-zero multiplicative linear functionals on the complex Banach algebra L1C =L1C (µ). Then there is a one-to-one correspondence between X and Φ, defined by the formulae φ(f • ) =

R

∧

f × χ dµ = f (χ) for every f ∈ L1C = L1C (µ),

φ(a•l u) = χ(a)φ(u) for every u ∈ L1C , a ∈ X, for χ ∈ X and φ ∈ Φ. Remark I follow 443G in writing a•l f • = (a•l f )• , where (a•l f )(x) = f (a−1 x) for f ∈ L1C and a, x ∈ X as in 441Ac. ∞ proof (a) If χ ∈ X then we can think of its equivalence class χ• as a member of L∞ C = LC (µ), so that we R R ∧ can define φχ ∈ (L1C )∗ by writing φχ (u) = χ• × u for every u ∈ L1C ; that is, φχ (f • ) = f × χ = f (χ) for every f ∈ L1C . 445G tells us that φχ is multiplicative. To see that it is non-zero, recall that µ is strictly positive (442Aa) and that χ is continuous. Let G be an open set containing the identity e of X such that |χ(x) − 1| ≤ 12 for every x ∈ G; then Re(χ(x)) ≥ 21 for every x ∈ G, so

|

R

χ(x)dx| ≥ Re G

R

χ(x)dx = G

R

G

Re(χ(x))dx ≥ 12 µG > 0.

Accordingly φχ (χG)• 6= 0 and φχ 6= 0. (I hope that no confusion will arise if I continue occasionally to write χE for the characteristic function of a set E, even if the symbol χ is already active in the sentence.) (b) Now suppose that φ is a non-zero multiplicative linear functional on L1C . Fix on some g0 ∈ L1C such that φ(g0• ) = 1. Let ∆ be the left modular function of X. (If you are reading this proof on the assumption that X is abelian, then a•r f = a−1 •l f and ∆ ≡ 1, so the argument below simplifies usefully.) (i) For any f ∈ L1C and a ∈ X, φ(a−1 •l f )• = ∆(a)φ(a•r f )• . P P Let ² > 0. Then for any sufficiently small open neighbourhood U of the identity, if we set h = k(a•r f ) ∗ h − a•r f k1 ≤ ²,

1 χU , µU

we shall have

kh ∗ f − f k1 ≤ ²

445H

The duality theorem

339

(444T, with 444P). Now ∆(a)(a•r f ) ∗ h = f ∗ (a−1 •l h), by 444Of, so |φ(a−1 •l f )• − ∆(a)φ(a•r f )• | ≤ |φ(a−1 •l f )• − φ(f ∗ (a−1 •l h))• | + |∆(a)φ((a•r f ) ∗ h)• − ∆(a)φ(a•r f )• | = |φ(a−1 •l f )• − φ(f )• φ(a−1 •l h)• | + ∆(a)|φ((a•r f ) ∗ h)• − φ(a•r f )• | ≤

|φ(a−1 •l f )•

− φ(a−1 •l h)• φ(f )• |

+ ∆(a)k(a•r f ) ∗ h − a•r f k1 (because kφk ≤ 1 in

(L1C )∗ ,

by 4A6F) ≤ |φ(a−1 •l f )• − φ((a−1 •l h) ∗ f )• | + ²∆(a) ≤ ka−1 •l f − (a−1 •l h) ∗ f k1 + ²∆(a) = ka−1 •l (f − h ∗ f )k1 + ²∆(a)

(444Of) = kf − h ∗ f k1 + ²∆(a) (443Gb) ≤ (1 + ∆(a))². As ² is arbitrary, we have the result. Q Q (ii) For any f , g ∈ L1C and a ∈ X, φ(a•l f )• φ(g • ) = φ(f • )φ(a•l g)• . P P φ(a•l f )• φ(g • ) = φ((a•l f ) ∗ g)• = φ(a•l (f ∗ g))• = ∆(a−1 )φ(a−1 •r (f ∗ g))• = ∆(a−1 )φ(f ∗ (a−1 •r g))• = φ(f • ∗ (a•l g)• ) = φ(f • )φ(a•l g)• , using the other two formulae in 444Of for the second and fourth equalities. Q Q (iii) Translating this into the language of L1C , we have φ(a•l u)φ(v) = φ(a•l f )• φ(g • ) = φ(f • )φ(a•l g)• = φ(u)φ(a•l v) whenever a ∈ X and u = f • , v = g • ∈ L1C . (iv) Let v0 be g0• , so that φ(v0 ) = 1, and set χ(a) = φ(a•l v0 ) = φ(a•l g0 )• for every a ∈ X. Then if a, b ∈ X, χ(ab) = φ(ab•l v0 ) = φ(a•l (b•l v0 )) is an action of X on L1C , as noted in 443Gd) = φ(a•l (b•l v0 ))φ(v0 ) = φ(b•l v0 )φ(a•l v0 ) (by (iii) above)

(because

•

l

= χ(a)χ(b). So χ : X → C is a group homomorphism. Moreover, because continuous (indeed, of norm at most 1), χ is continuous. Finally,

•

l

is continuous (443Ge), and φ also is

|χ(a)| ≤ ka•l v0 k1 = kv0 k1 for every a ∈ X, by 443Gb; it follows at once that {χ(a)n : n ∈ Z} is bounded, so that |χ(a)| = 1, for every a ∈ X. Thus χ ∈ X . Moreover, for any u ∈ L1C , φ(a•l u) = φ(a•l u)φ(v0 ) = φ(u)φ(a•l v0 ) = χ(a)φ(u). R • (v) Now φ = φχ . P P Because φ ∈ (L1C )∗ , there is some h ∈ L∞ h(x)f (x)dx for C (µ) such that φ(f ) = 1 every f ∈ LC (243Gb/243K; recall that by the rules of 441D, µ is suppose to be a quasi-Radon measure, therefore strictly localizable, by 415A). In this case, for any f ∈ L1C ,

340

Topological groups

445H

Z φ(f ) = φ(f ∗ g0 ) = •

(444Od)

ZZ h(x)(f ∗ g0 )(x)dx =

•

ZZ

h(xy)f (x)g0 (y)dydx

Z −1

=

h(y)f (x)g0 (x

y)dydx =

φ(x•l g0 )• f (x)dx

Z =

χ(x)f (x)dx = φχ (f • ). Q Q

(c) Thus we see that the formulae announced do define a surjection from X onto Φ. We have still to confirm that it is injective. But if χ, θ are distinct members of X , then {x : χ(x) 6= θ(x)} is a non-empty open set, so has positive measure, because µ is strictly positive; because µ is semi-finite, they represent different linear functionals on L1C , and φχ 6= φθ . This completes the proof. 445I The topology of the dual group: Proposition Let X be a topological group with a left Haar measure µ, and X its dual group. For χ ∈ X , let χ• be its equivalence class in L0C = L0C (µ), and φχ ∈ (L1C )∗ = (L1C (µ))∗ the multiplicative linear functional corresponding to χ, as in 445H. Then the maps χ 7→ χ• and χ 7→ φχ are homeomorphisms between X and its images in L0C and (L1C )∗ , if we give L0C the topology of convergence in measure (245A/245M) and (L1C )∗ the weak* topology (2A5Ig). proof (a) Note that χ 7→ χ• is injective because µ is strictly positive, so that if χ, θ are distinct members of X then the non-empty open set {x : χ(x) 6= θ(x)} has non-zero measure; and that χ 7→ φχ is injective by 445H. So we have one-to-one correspondences between X and its images in L0C and (L1C )∗ . Write T for the topology of X as defined in 445Ab, Tm for the topology induced by its identification with its image in L0C , and Tw for the topology induced by its identification with its image in (L1C )∗ . Let E be the family of non-empty totally bounded subsets of X and Σf the set of measurable sets of finite measure; for E ∈ E, F ∈ Σf and f ∈ L1C = L1C (µ) set ρE (χ, θ) = supx∈E |χ(x) − θ(x)|, ρ0F (χ, θ) = ρ00f (χ, θ) = |

R

R

F

min(1, |χ(x) − θ(x)|)µ(dx),

f (x)χ(x)µ(dx) −

R

f (x)θ(x)µ(dx)|

for χ, θ ∈ X . Then T is generated by the pseudometrics {ρE : E ∈ E}, Tm is generated by {ρ0F : F ∈ Σf } and Tw is generated by {ρ00f : f ∈ L1C }. (b) Tm ⊆ T. P P Suppose that F ⊆ X is a measurable set of finite measure, and ² > 0. There is a non-empty totally bounded open set U ⊆ X (443H). Since S {xU : x ∈ X} is an open S cover of X and µ is τ -additive, there are y0 , . . . , yn ∈ X such that µ(F \ j≤n yj U ) ≤ 13 ²; set E = j≤n yj U . Then E is totally bounded, and ρ0F (χ, θ) ≤ ² whenever ρE (χ, θ) ≤

² . 1+3µE

As F and ² are arbitrary, the identity map

(X , T) → (X , Tm ) is continuous (2A3H), that is, Tm ⊆ T. Q Q (c) Tw ⊆ Tm . P P If f ∈ L1C and ² > 0 let F ∈ Σf , M > 0 be such that θ ∈ X and ρ0F (χ, θ) ≤

² , 4M

R

(|f | − M χF )+ dµ ≤ 14 ². If χ,

then Z ¯ (χ − θ) × f ¯ ≤ |χ − θ| × |f | Z Z ≤M |χ − θ| + 2 (|f | − M χF )+ F Z 1 1 ≤ ² + 2M min(1, |χ − θ|) = ² + 2M ρ0F (χ, θ) ≤ ².

¯ ρ00f (χ, θ) = ¯

Z

2

F

2

445K

The duality theorem

341

As f and ² are arbitrary, this shows that Tw ⊆ Tm . Q Q (d) Finally, T ⊆ Tw . P P Fix χ ∈ X , E ∈ E and ² > 0. Let u ∈ L1C be such that φχ (u) = 1, and represent • 1 u as f where f ∈ LC (µ). Set U = {a : a ∈ X, ka•l u − uk1 < 14 ²}; then U is an open neighbourhood of the identity e of X, becauseSa 7→ a•l u is continuous (443Ge). Because E is totally bounded, there are y0 , . . . , yn ∈ X such that E ⊆ k≤n yk U . Set fk = yk−1 •l f , so that fk• = yk •l u for each k ≤ n. Now suppose that θ ∈ X is such that ² 4

ρ00f (θ, χ) ≤ ,

ρ00fk (θ, χ) ≤

² 4

for every k ≤ n.

Take any x ∈ E. Then there is a k ≤ n such that x ∈ yk U , so that yk−1 x ∈ U and kx•l u − yk •l uk1 = kyk •l (yk−1 x•l u − u)k1 = kyk−1 x•l u − uk1 ≤

² 4

(using 443Gd for the second equality). Now φχ (x•l u) = χ(x) (445H), so |φθ (x•l u) − χ(x)| ≤ |φθ (x•l u − yk •l u)| + |φθ (yk •l u) − φχ (yk •l u)| + |φχ (yk •l u − x•l u)| 3 4

≤ 2kx•l u − yk •l uk1 + ρ00fk (θ, χ) ≤ ². On the other hand, ² 4

|θ(x) − φθ (x•l u)| = |θ(x)||1 − φθ (u)| = ρ00f (θ, χ) ≤ . So |θ(x) − φ(x)| ≤ ². As x is arbitrary, ρE (θ, χ) ≤ ². As χ, E and ² are arbitrary, this shows that T ⊆ Tw . Q Q 445J Corollary For any topological group X carrying Haar measures, its dual group X is locally compact and Hausdorff. proof Let Φ be the set of non-zero multiplicative linear functionals on L1C = L1C (µ), for some left Haar measure µ on X, and give it its weak* topology. Then Φ ∪ {0} ⊆ (L1C )∗ is the set of all multiplicative linear functionals on L1C , and is closed for the weak* topology, because {φ : φ ∈ (L1C )∗ , φ(u ∗ v) = φ(u)φ(v)} is closed for all u, v ∈ L1C . Because the unit ball of (L1C )∗ is a compact Hausdorff space for the weak* topology (3A5F), so is Φ ∪ {0}. So Φ itself is an open subset of a compact Hausdorff space and is a locally compact Hausdorff space (3A3Bg). Since the topology on X can be identified with the weak* topology on Φ (445I), X also is locally compact and Hausdorff. 445K Proposition Let X be a topological group and µ a left Haar measure on X. Let X be the dual group of X, and write C0 = C0 (X ; C) for the Banach algebra of continuous functions h : X → C such that {χ : |h(χ)| ≥ ²} is compact for every ² > 0. ∧ (a) For any u ∈ L1C = L1C (µ), its Fourier transform u belongs to C0 . ∧ (b) The map u 7→ u : L1C → C0 is a multiplicative linear operator, of norm at most 1. (c) Suppose that X is abelian. For f ∈ L1C = L1C (µ), set f˜(x) = f (x−1 ) whenever this is defined. Then ∧ f˜ ∈ L1 and kf˜k1 = kf k1 . For u ∈ L1 , we may define u ˜ ∈ L1 by setting u ˜ = f˜• whenever u = f • . Now u ˜ is C

∧

C

∧

C

the complex conjugate of u, so (u ∗ u ˜)∧ = |u|2 . ∧ ∧ (d) Still supposing that X is abelian, {u : u ∈ L1C } is a norm-dense subalgebra of C0 , and kuk∞ = r(u), 1 the spectral radius of u (4A6K), for every u ∈ LC . proof (a) As in 445H and 445J, let Φ be the set of non-zero multiplicative linear functionals on L1C , so that ∧ Φ ∪ {0} is compact for the weak* topology of (L1C )∗ , and u(χ) = φχ (u) for every χ ∈ X . By the definition of the weak* topology, φ 7→ φ(u) is continuous; since we can identify the weak* topology on Φ with the dual ∧ group topology of X (445I), u is continuous. Also, for any ² > 0, ∧ {χ : χ ∈ X , |u(χ)| ≥ ²} ∼ = {φ : φ ∈ Φ, |φ(u)| ≥ ²},

342

Topological groups

445K

which is a closed subset of Φ ∪ {0}, therefore compact. (b) It is immediate from the definition of ∧ that it is a linear operator from L1C to C0 ; it is multiplicative ∧ by 445G, and of norm at most 1 because all the multiplicative linear functionals u 7→ u(χ) must be of norm at most 1 (4A6F). (c) Now suppose that X is abelian. If f ∈ L1C , then

R

f˜(x)dx =

R

f (x−1 )dx =

R

f (x)dx

by 442Kb, so f˜ ∈ L1C ; the same formulae tell us that kf˜k1 = kf k1 . If f =a.e. g then f˜ =a.e. g˜ (442G, or otherwise), so u ˜ is well-defined. If χ ∈ X , and u = f • , then Z Z Z ∧ −1 ˜ u ˜(χ) = f (x)χ(x)dx = f (x )χ(x)dx = f (x)χ(x−1 )dx Z Z ∧ = f (x)χ(x)dx = f (x)χ(x)dx = u(χ), ∧

∧

so u ˜ is the complex conjugate of u, and ∧

∧

∧

(u ∗ u ˜)∧ = u × u ˜ = |u|2 . ∧

(d) To see that A = {u : u ∈ L1C } is dense in C0 , we may use the Stone-Weierstrass theorem in the form 4A6B. A is a subalgebra of C0 ; it separates the points (because the canonical map from X to (L1C )∗ is injective); if χ ∈ X , there is an h ∈ A such that h(χ) 6= 0 (because elements of X act on L1C as non-zero functionals); and the complex conjugate of any function in A belongs to A, by (c) above. Accordingly A is dense in C0 , by 4A6B. ∧ The calculation of kuk∞ is an immediate consequence of the characterization of r(u) as sup{|φ(u)| : φ ∈ Φ} (4A6K) and the identification of Φ with X . Remark This is the first point in this section where we really need to know whether or not our group is abelian. 445L Positive definite functions Let X be a group. (a) A function h : X → C is called positive definite if Pn −1 ¯ j,k=0 ζj ζk h(xk xj ) ≥ 0 for all ζ0 , . . . , ζn ∈ C and x0 , . . . , xn ∈ X. (b) Suppose that h : X → C is positive definite. Then, writing e for the identity of X, (i) |h(x)| ≤ h(e) for every x ∈ X; (ii) h(x−1 ) = h(x) for every x ∈ X. P P If ζ ∈ C and x ∈ X, take n = 1, x0 = e, x1 = x, ζ0 = 1 and ζ1 = ζ in the definition in (a) above, and observe that ¯ −1 ) = h(e−1 e) + ζh(e−1 x) + ζh(x ¯ −1 e) + ζ ζh(x ¯ −1 x) ≥ 0. (1 + |ζ|2 )h(e) + ζh(x) + ζh(x Taking ζ = 0, x = e we get h(e) ≥ 0. Taking ζ = 1 we see that h(x) + h(x−1 ) is real, and taking ζ = i, we see that h(x) − h(x−1 ) is purely imaginary; that is, h(x−1 ) = h(x), for any x. Taking ζ such that |ζ| = 1, ζh(x) = −|h(x)| we get 2h(e) − 2|h(x)| ≥ 0, that is, |h(x)| ≤ h(e) for every x ∈ X. Q Q (c) If h : X → C is positive definite and χ : X → S 1 is a homomorphism, then h × χ is positive definite. P P If ζ0 , . . . , ζn ∈ C and x0 , . . . , xn ∈ X then Pn Pn −1 −1 ¯ Q j,k=0 ζj ζk (h × χ)(xk xj ) = j,k=0 ζj χ(xj )ζk χ(xk )h(xk xj ) ≥ 0. Q (d) If X is an abelian topological group and µ a Haar measure on X, then for any f ∈ L2C (µ) the convolution f ∗ f˜ : X → C is continuous and positive definite, where f˜(x) = f (x−1 ) whenever this is defined.

445M

The duality theorem

343

P P As in 444Rc, f ∗ f˜ is defined everywhere on X and is continuous. (The definition of ∼ has shifted since §444, but the argument there applies unchanged to the present situation.) Now, if x0 , . . . , xn ∈ X and ζ0 , . . . , ζn ∈ C, n X

ζj ζ¯k (f ∗ f˜)(x−1 k xj ) =

j,k=0

=

n X

Z ζj ζ¯k

j,k=0 Z X n

f (y)f˜(y −1 x−1 k xj )dy

ζj ζ¯k f (y)f (x−1 j xk y)dy

j,k=0

=

Z X n

ζj ζ¯k f (xj y)f (x−1 j xk xj y)dy

j,k=0

=

Z X n Z

=

j,k=0 n X

|

ζj ζ¯k f (xj y)f (xk y)dy ζj f (xj y)|2 dy ≥ 0.

j=0

So f ∗ f˜ is positive definite. Q Q 445M Proposition Let X be a topological and ν a quasi-Radon measure on X. If h : X → C is a RR group continuous positive definite function, then h(y −1 x)f (x)f (y)ν(dx)ν(dy)≥ 0 for every ν-integrable function f. proof (a) Extend f , if necessary, to the whole of X; since the hypothesis implies that dom f is conegligible, this does not affect the integrals. Let λ be the product quasi-Radon measure on X × X; because h is continuous (by hypothesis) and bounded (by 445L(b-i)), the function (x, y) 7→ h(y −1 x)f (x)f (y) is λ-integrable, and (because {x : f (x) 6= 0} can be covered by a sequence of sets of finite measure) I=

RR

h(y −1 x)f (x)f (y)ν(dx)ν(dy) =

R

h(y −1 x)f (x)f (y)λ(d(x, y))

(417H). (b) Let ² > 0. Set γ = supx∈X |h(x)| = h(e) (445L(b-i)). Let F ⊆ X be a non-empty measurable set R of finite measure for ν such that γ (X×X)\(F ×F ) |f (x)f (y)|λ(d(x, y)) ≤ 12 ² and f is bounded on F ; say |f (x)| ≤ M for every x ∈ F . Let δ > 0 be such that δ(M 2 + 2M γ)(νF )2 + 2M 2 γδ ≤ 21 ². Let G be the set {G × H : G, H ⊆ X are open, |h(y −1 x) − h(y1−1 x1 )| ≤ δ whenever x, x1 ∈ G, y, y1 ∈ H}. Because h is continuous, G is a cover of X × X. Because λ is τ -additive, there is a finiteSset G0 ⊆ G such S that λ((F × F ) \ G0 ) ≤ δ; we may suppose that G0 is non-empty. Set W = (F × F ) ∩ G0 . Enumerate G0 as hGi × Hi ii≤n . Let F be a finite partition of F into measurable sets such that |f (x) − f (x0 )| ≤ δ whenever x, x0 belong to the same member of F. Let E be the partition of F generated by F ∪ {F ∩ Gj : j ≤ n} ∪ {F ∩ Hj : j ≤ n}. EnumerateSE as hEk ik≤m ; for each k ≤ m choose xk ∈ Ek . Set J = {(j, k) : j ≤ m, k ≤ m, Ej × Ek ⊆ W }; then W = (j,k)∈J Ej × Ek . (c) If (j, k) ∈ J, x ∈ Ej and y ∈ Ek then 2 |h(y −1 x)f (x)f (y) − h(x−1 k xj )f (xj )f (xk )| ≤ δ(M + 2M γ),

because there must be some r ≤ n such that Ej × Ek ⊆ Gr × Hr , so that |h(y −1 x) − h(x−1 k xj )| ≤ δ, while there are members of F including Ej and Ek , so that |f (x) − f (xj )| ≤ δ and |f (y) − f (xk )| ≤ δ; at the same time,

344

Topological groups

|f (x)f (y)| ≤ M 2 ,

445M

|h(x−1 k xj )||f (y)| ≤ M γ,

|h(x−1 k xj )||f (xj )| ≤ M γ

because x, y, xj and xk all belong to F . (d) Set ζj = f (xj )νEj for j ≤ m, so that ζ¯j = f (xj )νEj . Now consider Z X ¯ ¯ ¯ ¯ h(y −1 x)f (x)f (y)λ(d(x, y)) − ζj ζ¯k h(x−1 k xj ) W

(j,k)∈J

Z ¯ X ¯ = (j,k)∈J

X Z

≤

(j,k)∈J

X

≤

Ej ×Ek

Ej ×Ek

¯ ¯ h(y −1 x)f (x)f (y) − h(x−1 k xj )f (xj )f (xk )λ(d(x, y))

|h(y −1 x)f (x)f (y) − h(x−1 k xj )f (xj )f (xk )|λ(d(x, y))

δ(M 2 + 2M γ)νEj νEk

(j,k)∈J

≤ δ(M 2 + 2M γ)λW ≤ δ(M 2 + 2M γ)(νF )2 . On the other hand, Z ¯ ¯

¯ h(y −1 x)f (x)f (y)λ(d(x, y))¯ ≤ γ

Z |f (x)f (y)|λ(d(x, y)) (X×X)\(F ×F )

(X×X)\W

Z

+γ

|f (x)f (y)|λ(d(x, y)) (F ×F )\W

1 2

≤ ² + γδM 2 , and X

¯ ¯

X

¯ 2 ˜ ¯ h(x−1 k xj )f (xj )f (xk ) ≤ γM

j≤m,k≤m,(j,k)∈J /

νEj νEk

j≤m,k≤m,(j,k)∈J /

= γM 2 λ((F × F ) \ W ) ≤ γM 2 δ. Putting these together, |I −

m X

1 2

2 2 2 2 ζj ζ¯k h(x−1 k xj )| ≤ δ(M + 2M γ)(νF ) + ² + γδM + γM δ ≤ ².

j,k=0

But

Pm

−1 ¯ j,k=0 ζj ζk h(xk xj )

≥ 0, because h is positive definite. As ² is arbitrary, I ≥ 0, as required.

445N Bochner’s Theorem (Herglotz 11, Bochner 33, Weil 40) Let X be an abelian topological group with a Haar measure µ, and X its dual group. Then for any continuous positive definite function h : X → C there is a unique totally finite Radon measure ν on X such that

R

h(x)f (x)µ(dx) = h(x) =

R

R

∧

f (χ)ν(dχ) for every f ∈ L1C = L1C (µ),

χ(x)ν(dχ) for every x ∈ X.

proof (a) If h(e) = 0, where e is the identity in X, then h = 0, by 445L(b-i), and the result is trivial. Otherwise, since multiplying h by a positive scalar leaves h positive definite and does not affect the result, we may suppose that h(e) = 1. For f , g ∈ L1C = L1C (µ) set (f |g) =

RR

f (x)g(y)h(y −1 x)µ(dx)µ(dy) =

RR

f (x)g(y −1 )h(yx)µ(dx)µ(dy)

(by 442Kb, since X is unimodular). Then, by 445M, (f |f ) ≥ 0 for every f ∈ L1C . Also (f1 + f2 |g) = (f1 |g) + (f2 |g), (ζf |g) = ζ(f |g) and (g|f ) = (f |g) for all f , g, f1 , f2 ∈ L1C and ζ ∈ C. P P Only the last is anything but trivial, and for this we have

445N

The duality theorem

345

ZZ g(x)f (y)h(y −1 x)µ(dx)µ(dy)

(g|f ) = ZZ

g(x)f (y)h(y −1 x)µ(dy)µ(dx)

=

(by 417Ha, because (x, y) 7→ g(x)f (y)h(y −1 x) is integrable for the product measure and zero off the square of a countable union of sets of finite measure) ZZ = g(x)f (y)h(x−1 y)µ(dy)µ(dx) (using 445L(b-ii))

ZZ f (y)g(x)h(x−1 y)µ(dy)µ(dx)

=

= (f |g). Q Q (b) If f , g ∈ L1C , |(f |g)|2 ≤ (f |f )(g|g). P P (Really this is just Cauchy’s inequality.) For any α, β ∈ C, 2 ¯ |g)) + |β|2 (g|g) = (αf + βg|αf + βg) ≥ 0. |α| (f |f ) + 2 Re(αβ(f If (f |f ) = 0 we have 2 Re(α(f |g)) + (g|g) ≥ 0 for every α ∈ C so in this case (f |g) = 0; similarly (f |g) = 0 if ¯ |g) = −|αβ(f |g)|, (g|g) = 0; otherwise we can find non-zero α, β such that |α|2 = (g|g), |β|2 = (f |f ) and αβ(f in which case the inequality simplifies to |(f |g)| ≤ |αβ| and |(f |g)|2 ≤ (f |f )(g|g), as required. Q Q R (c) Now consider the functional ψ ∈ (L1C )∗ = (L1C (µ))∗ corresponding to h, so that ψ(f • ) = h × f dµ for every f ∈ L1C . Then |ψ(f • )|2 ≤ (f |f ) for every f ∈ L1C . P P Let ² > 0. Then there is an open neighbourhood U of e such that U = U −1 and |h(y −1 x) − h(e)| ≤ ² whenever x, y ∈ U , ka•l f − f k1 ≤ ² for every a ∈ U where (a•l f )(x) = f (a−1 x) whenever this is defined, as usual (443Ge). Shrinking U if need be, we may 1

χU ∈ L1C . Then suppose that µU < ∞, and of course µU > 0. Set g = µU Z Z Z Z ¯ 1 ¯¯ • −1 |(f |g) − ψ(f )| = f (x)h(y x)µ(dx)µ(dy) − f (x)h(x)µ(dx)µ(dy)¯ µU U X Z Z Z ZU X ¯ 1 ¯¯ = f (yx)h(x)µ(dx)µ(dy) − f (x)h(x)µ(dx)µ(dy)¯ µU U X U X Z Z ¯ 1 ¯¯ ((y −1 •l f )(x) − f (x))h(x)µ(dx)µ(dy)¯ = µU U X Z Z 1 ≤ |(y −1 •l f )(x) − f (x)||h(x)|µ(dx)µ(dy) µU U X Z 1 ≤ ky −1 •l f − f k1 µ(dy) ≤ ². µU

U

Also ¯ 1 |(g|g) − 1| = ¯

µU

≤

Z Z 2

1 µU 2

Z UZ

¯ (h(y −1 x) − 1)µ(dx)µ(dy)¯ U

|h(y −1 x) − 1|µ(dx)µ(dy) ≤ ². U

U

So max(0, |ψ(f • )| − ²)2 ≤ |(f |g)|2 ≤ (f |f )(g|g) ≤ (1 + ²)(f |f ). Letting ² ↓ 0 we have the result. Q Q

346

Topological groups

445N

(d) If we look at (f |f ), however, and apply 444Od, we see that ZZ (f |f ) = f (x)f (y)h(y −1 x)µ(dx)µ(dy) ZZ Z = f (x)f (y −1 )h(yx)µ(dx)µ(dy) = h(x)(f ∗ f˜)(x)µ(dx), where f˜(x) = f (x−1 ) whenever this is defined; that is, (f |f ) = ψ(f ∗ f˜)• . (Note that f˜ ∈ L1C because X is unimodular, as in part (c) of the proof of 445K.) So (c) tells us that |ψ(f • )|2 ≤ ψ(f ∗ f˜)• for every f ∈ L1C , that is, |ψ(u)|2 ≤ ψ(u ∗ u ˜) for every u ∈ L1C , defining u ˜ as in 445Kc. (e) In fact ∧

|ψ(u)| ≤ kuk∞ for every u ∈ L1C . P P Set u0 = u and uk+1 = uk ∗ u ˜k for every k ∈ N. We need to know that uk = u ˜k for k ≥ 1. To see this, represent uk−1 as f • where f ∈ L1C , so that uk = (f ∗ f˜)• . Now Z ∼ −1 ˜ ˜ (f ∗ f ) (x) = (f ∗ f )(x ) = f (x−1 y)f˜(y −1 )dy Z Z = f (x−1 y)f (y)dy = f˜(y −1 x)f (y)dy = (f ∗ f˜)(x) k−1 for every x, so (f ∗ f˜)∼ = f ∗ f˜ and u ˜k = uk . Accordingly uk+1 = uk ∗uk for k ≥ 1 and we have uk = (u1 )2 for every k ≥ 1. At the same time, we have |ψ(uk )|2 ≤ ψ(uk+1 ) for every k, by (d), so that, for k ≥ 1, k

|ψ(u)|2 ≤ ψ(uk ) ≤ kuk k1 = ku12 |ψ(u)| ≤ ku21

p

k−1

1/2k

k1

k−1

k1 ,

.

Letting k → ∞, |ψ(u)| ≤ r(u1 ), where r(u1 ) is the spectral radius of u1 . ∧ ∧ ∧ ∧ At this point, recall that r(u1 ) = ku1 k∞ (445Kd), while |u|2 = u1 (445Kc), so r(u1 ) = kuk2∞ and ∧ |ψ(u)| ≤ kuk∞ . Q Q ∧

(f ) Now consider ∧ as a linear operator from L1C to C0 = C0 (X ; C), as in 445K. If u = 0 then ψ(u) = 0, by ∧ ∧ (e), so setting A = {u : u ∈ L1C } we have a linear functional ψ0 : A → C defined by saying that ψ0 (u) = ψ(u) 1 for every u ∈ LC . By (e), kψ0 k ≤ 1. Because A is norm-dense in C0 (445Kd), ψ0 has a unique extension to a bounded linear operator ψ1 , still of norm at most 1, from C0 to C (2A4I). (g) Suppose that 0 ≤ q ≤ 1 in C0 , writing 1 for the constant function with value 1, and set α = ψ1 (q). Then for any ζ ∈ C, γ ≥ 0 we have |ζ −γα| ≤ max(|ζ|, |γ|, |ζ −γ|). P P Let ² > 0. Set V = {x : |1−h(x)| < ²}; 1 then V is an open neighbourhood of e; set f = µV χV and u = f • , so that ∧

kuk∞ = r(u) ≤ kuk1 = 1, |1 − ψ(u)| = |

1 µV

R V

(1 − h(x))dx| ≤ ².

Set v = u ∗ u ˜; then ∧

ψ1 (v) = ψ(v) ≥ |ψ(u)|2 ≥ (1 − ²)2 , ∧

∧

∧

∧

using part (d) for the central inequality. But v = |u|2 , so that 0 ≤ v ≤ 1 and ψ1 (v) ≤ 1. ∧ ∧ Now consider kζ v − γqk∞ . If χ ∈ X , then ζ v(χ) and γq(χ) both lie in the triangle with vertices 0, ζ and ∧ γ, because 0 ≤ v ≤ 1, 0 ≤ q ≤ 1. So

445O

The duality theorem

347

∧

|ζ v(χ) − γq(χ)| ≤ max(|γ|, |ζ|, |γ − ζ|). As χ is arbitrary, ∧

kζ v − γqk∞ ≤ max(|γ|, |ζ|, |γ − ζ|). Accordingly ∧

∧

|ζ − γα| ≤ |ζ − ζψ1 (v)| + |ψ1 (ζ v − γq)| ≤ |ζ|(1 − (1 − ²)2 ) + kζ v − γqk∞ ≤ 2²|ζ| + max(|ζ|, |γ|, |ζ − γ|). ∧

Q As ² is arbitrary, we have the result. Q p Taking ζ = γ = 1 we see that |1−α| ≤ 1, so that Re α ≥ 0. Taking ζ = ±i, we see that |i±γα| ≤ 1 + γ 2 for every γ ≥ 0, so that Im α = 0. Thus ψ1 (q) ≥ 0; and this is true whenever 0 ≤ q ≤ 1 in C0 . (h) It follows at once that ψ1 (q) ≥ 0 whenever q ≥ 0 in C0 . Applying the Riesz Representation Theorem, in the form 436K, to the restriction R of ψ1 to C0 (X ; R), we see that there is a unique totally finite Radon R measure ν on X such that ψ1 (q) = q dν for every real-valued q ∈ C0 ; of course it follows that ψ1 (q) = q dν for every q ∈ C0 . Unwrapping the definition of ψ1 , we see that

R

∧

h(x)f (x)µ(dx) = ψ(f • ) = ψ1 (f ) =

R

∧

f (χ)ν(dχ)

for every f ∈ L1C (µ), and that this uniquely defines ν. (i) For the second formula, argue as follows. Given f ∈ L1C (µ), consider the function (x, χ) 7→ f (x)χ(x) : X × X → C. Because (χ, x) 7→ χ(x) is continuous (445Ea), this is measurable for the product quasi-Radon measure µ × ν on X × X . It is integrable because νX < ∞ and |χ(x)| = 1 for every χ, x; moreover, it is zero off the set {x : f (x) 6= 0} × X , which is a countable union of products of sets R of finite measure. Note also that because χ 7→ χ(x) is continuous and bounded for every x ∈ X, h0 (x) = χ(x)ν(dχ) is defined, and |h0 (x)| ≤ νX, for every x ∈ X. What is more, h0 is continuous. P P Let X be the dual group of X , and for x ∈ X let x ˆ be the corresponding member of X. Then, in the language of 445C, applied to the topological group X , h0 (x) =

R

∧

x ˆ dν = ν(ˆ x)

∧

for every x ∈ X. But ν : X → C is continuous, by 445Ec, and x 7→ x ˆ : X → X is continuous, by 445Eb; so h0 also is continuous. Q Q We may therefore apply Fubini’s theorem (417H) to see that Z ZZ ZZ 0 f (x)h (x)µ(dx) = f (x)χ(x)ν(dχ)µ(dx) = f (x)χ(x)µ(dx)ν(dχ) Z Z ∧ = f (χ)ν(dχ) = f (x)h(x)µ(dx). Since this is true for every f ∈ L1C , h0 =a.e. h; since both are continuous, h0 = h, as required. 445O Proposition Let X be a Hausdorff abelian topological group carrying Haar measures. Then the map x 7→ x ˆ from X to its bidual group X is a homeomorphism between X and its image in X. In particular, the dual group X of X separates the points of X. proof We already know that b is continuous (445Eb). Now let U be any neighbourhood of the identity e of X. Let V ⊆ U be an open neighbourhood of e such that V V −1 ⊆ U and µV < ∞. Then f = χV ∈ L2C (µ), so f ∗ f˜ is positive definite (445Ld) and there is a totally finite Radon measure ν on X such that R ∧ (f ∗ f˜)(x) = χ(x)ν(dχ) for every x ∈ X (445N). Note that, writing e for the identity of X and ν : X → C for the Fourier transform of ν, R R R R ∧ ν(e) = e(χ)ν(dχ) = χ(e)ν(dχ) = (f ∗ f˜)(e) = f (y)f˜(y −1 )µ(dy) = |f (y)|2 µ(dy) 6= 0. ∧

∧

Now ν is continuous (445Ec), so W = {x : ν(x) 6= 0} is a neighbourhood of e. If x ∈ X and x ˆ ∈ W , then (f ∗ f˜)(x) =

R

χ(x)ν(dχ) =

R

∧

x ˆ(χ)ν(dχ) = ν(ˆ x) 6= 0,

348

Topological groups

445O

so there is some y ∈ X such that f (y)f˜(y −1 x) 6= 0, that is, f (y) 6= 0 and f (x−1 y) 6= 0, that is, y and x−1 y both belong to V ; in which case x ∈ V V −1 ⊆ U . Thus U ⊇ {x : x ˆ ∈ W }. This means that, writing S for {{x : x ˆ ∈ H} : H ⊆ X is open}, every neighbourhood of e for the original topology T of X is a neighbourhood of e for S. But (it is easy to check) (X, S) is a topological group because X is a topological group and bis a homomorphism. So T ⊆ S (4A5Fb). As we know already that S ⊆ T, the two topologies are equal. It follows at once that if T is Hausdorff, then (because S is Hausdorff) the map bis an injection and is a homeomorphism between X and its image in X. 445P The Inversion Theorem Let X be an abelian topological group and µ a Haar measure on X. Then there is a unique Haar measure λ on the dual group X of X such that whenever f : X → C is ∧ continuous, µ-integrable and positive definite, then f : X → C is λ-integrable and f (x) =

R

∧

f (χ)χ(x)λ(dχ)

for every x ∈ X. proof (a) Write P for the set of µ-integrable positive definite continuous functions h : X → C. For h ∈ P , let νh be the corresponding totally finite Radon measure on X defined in 445N, so that

R

f (x)h(x)µ(dx) =

R

∧

f (χ)νh (dχ)

for every f ∈ L1C = L1C (µ). (b) The basis of the argument is the following fact. If f ∈ L1C and h1 , h2 ∈ P , then

R

Z P P

∧

R

∧

∧

¯ 1 dνh . f ×h 2

Z

∧

∧

∧

¯ 2 dνh = f ×h 1

¯ 1 dνh = f ×h 2

(445G)

¯ 1 )∧ dνh (f ∗ h 2 Z

=

Z ¯ 1 )(x)µ(dx) = h2 (x)(f ∗ h

¯ 2 (x−1 )(f ∗ h ¯ 1 )(x)µ(dx) h

(by 445Lb) ¯1) ∗ h ¯ 2 )(e) = ((f ∗ h ¯2) ∗ h ¯ 1 )(e) = ((f ∗ h (because ∗ is associative and commutative, by 444Oe and 444Og) Z ∧ ∧ ¯ 2 dνh . Q = f ×h Q 1 ∧

∧

¯ 1 and h ¯ 2 are both bounded (by Now because h

R

|h1 |dµ and

R

|h2 |dµ respectively), and νh1 and νh2 are

∧

both totally finite measures, and {f : f ∈ L1C (µ)} is k k∞ -dense in C0 = C0 (X ; C) (445Kd), we must have

R

∧

¯ 2 dνh = p×h 1

R

∧

¯ 1 dνh p×h 2

for every p ∈ C0 . (c) Let K be the family of compact subsets of X . For K ∈ K set ∧

¯ PK = {h : h ∈ P, h(χ) > 0 for every χ ∈ K}. Then PK is non-empty. P P Set U = {x : x ∈ X, |1 − χ(x)| ≤

1 2

for every χ ∈ K}.

Then U is a neighbourhood of the identity e of X, by 445Eb. Let V be an open neighbourhood of e, of finite 1 measure, such that V V −1 ⊆ U , set g = µV χV , and try h = g ∗ g˜. Then h is real-valued and non-negative

445P

The duality theorem

349

(because g and g˜ are), continuous and positive definite (445Ld), zero outside U (because V V −1 ⊆ U , as in the proof of 445O), and

R

R

h dµ =

R

g dµ ·

g˜ dµ = 1

(444Qb). Next, ∧

∧

∧ 2 ¯ = h = |g| h

(445Kc) is non-negative, and if χ ∈ K then ∧

|1 − h(χ)| = | because |1 − χ(x)| ≤

1 2

R

R

h(x) − h(x)χ(x)µ(dx)| ≤

h(x)|1 − χ(x)|µ(dx) ≤

1 2

if x ∈ U and h(x) = 0 if x ∈ X \ U . So ∧

∧

¯ h(χ) = h(χ) ≥

1 2

for every χ ∈ K, and h ∈ PK . Q Q (d) Because K is upwards-directed, {PK : K ∈ K} is downwards-directed and generates a filter F on P . Let Ck = Ck (X ; C) be the space of continuous complex-valued functions on X with compact support. If q ∈ Ck , then φ(q) = limh→F

R

q ∧

¯ h

dνh

∧

¯ we interpret 0/0 as 0 if necessary. P is defined in C, where in the division q/h P Setting K = {χ : q(χ) 6= 0}, we see in fact that for any h1 , h2 ∈ PK we may define a function p ∈ Ck by setting p(χ) =

q(χ) ∧

∧

¯ 1 (χ)h ¯ 2 (χ) h

if χ ∈ K,

= 0 if q(χ) = 0, so that Z

q ∧

¯1 h

Z

(by (b) above)

Z =

So this common value must be φ(q). Q Q If q, q 0 ∈ Ck and α ∈ C, then Z q + q0 ∧

¯ h

Z

Z

∧

¯ 2 dνh = p×h 1

dνh1 =

q ∧

¯2 h

dνh2 .

Z dνh = αq ∧

¯ h

∧

¯ 1 dνh p×h 2

q ∧

¯ h

Z dνh +

dνh = α

Z

q ∧

¯ h

q0 ∧

¯ h

dνh ,

dνh

whenever h ∈ PK , where K = {χ : |q(χ)| + |q 0 (χ)| > 0}; so φ(q + q 0 ) = φ(q) + φ(q 0 ) and φ(αq) = αφ(q). ∧ ¯ ≥ 0 for every h ∈ PK , so φ(q) ≥ 0. Moreover, if q ≥ 0, then q/h R (e) By the Riesz Representation Theorem (in the form 436J) there is a Radon measure λ on X such that q dλ = φ(q) for any continuous function q of compact support. (As in part (h) of the proof of 445M, the shift from real-valued q to complex-valued q is elementary.) (f ) Now λ is translation-invariant. P P Take θ ∈ X and q ∈ Ck . Set K = {χ : q(χ) 6= 0} and L = θ−1 K, and take any h ∈ PK . Set h1 (x) = h(x)θ−1 (x) for x ∈ X. Then h1 is positive definite (445Lc); of course it is continuous and µ-integrable; and for any χ ∈ L,

350

Topological groups

R

∧

¯ 1 (χ) = h

445P ∧

¯ h(x)θ(x)χ(x)µ(dx) = h(θχ) > 0.

So h1 ∈ PL . To relate νh1 to νh , observe that if f ∈ L1C then ∧

f (θχ) = so

R

∧

f (θχ)νh1 (dχ) =

R

R

f (x)θ(x)χ(x)µ(dx) = (f × θ)∧ (χ),

(f × θ)(x)h1 (x)µ(dx) =

So we see that the equation

R

p(θχ)νh1 (dχ) =

R

R

f (x)h(x)µ(dx) =

R

∧

f (χ)νh (dχ).

p(χ)νh (dχ)

∧

is valid whenever p is of the form f , for some f ∈ L1C , and therefore for every p ∈ C0 . Set q1 (χ) = q(θχ) for every χ ∈ X , so that q1 ∈ Ck and L = {χ : q1 (χ) 6= 0}. Accordingly Z Z q1 (χ) q(θχ) ν (dχ) = νh1 (dχ) φ(q1 ) = h1 ∧ ∧ ¯ 1 (χ) ¯ h h(θχ) Z q(χ) = νh (dχ) = φ(q). ∧ ¯ h(χ) So

R

q(θχ)λ(dχ) =

R

q(χ)λ(dχ).

As q and θ are arbitrary, λ is translation-invariant (441L). Q Q (g) Thus λ is a Haar measure on X . I have still to confirm that f (x) =

R

∧

f (χ)χ(x)λ(dχ)

whenever f is continuous, positive definite and µ-integrable, and x ∈ X. But recall the formula from (b) above. If q ∈ Ck , K = {χ : q(χ) 6= 0} and h ∈ PK , then we must have Z

Z

∧

Z

∧

q × f¯

q × f¯dλ =

∧

¯ h

Z

∧

¯ q×h

dνh =

∧

¯ h

dνf =

q dνf .

∧ R q × f¯dλ = q dνf for every q ∈ Ck . ∧ ∧ R In particular, q × f¯dλ ≥ 0 whenever q ≥ 0; since f¯ is continuous (445Ka), and λ, being a Radon ∧ measure, is strictly positive, this shows that f¯ ≥ 0. Also

Thus we have

R

R

R

∧

∧

f¯dλ = sup{ q × f¯dλ : q ∈ Ck , 0 ≤ q ≤ 1} = νf (X ) < ∞, ∧

so we have an indefinite-integral measure f¯λ; since this is a Radon measure (416S), and acts on Ck in the same way as νf , it is actually equal to νf (436J). In particular, for any x ∈ X, f (x) =

R

χ(x)νf (dχ) =

R

∧

χ(x) f¯(χ)λ(dχ)

by the second formula in 445N, and 235M. But Z

∧

f¯(χ) = (by 445Lb)

Z f (x)χ(x)µ(dx) =

Z =

f (x)χ(x−1 )µ(dx)

(because X is abelian, therefore unimodular)

f (x−1 )χ(x)µ(dx)

445R

The duality theorem

351

∧

= f (χ−1 ) for every χ ∈ X . So Z (because X is abelian)

Z

∧

χ(x)f (χ−1 )λ(dχ) =

f (x) = Z =

∧

χ−1 (x)f (χ)λ(dχ)

∧

f (χ)χ(x)λ(dχ).

(h) We should check that λ is unique. Since all Haar measures on X are multiples of each other (442B), R ∧ it is enough to observe that the formula f (e) = f dλ can be valid for at most one such measure, if we can find any f ∈ P such that f (e) 6= 0; and such functions are available in abundance, by the construction in part (c) of the proof. 445Q Remark We can extract the following useful fact from part (g) of the proof above. If h : X → C ∧ ¯ is non-negative and λ-integrable, and the Radon is µ-integrable, continuous and positive definite, then h ∧ ¯ measure νh of 445N is just the indefinite-integral measure hλ. Note also that λ is actually a Radon measure; of course it has to be, because X is locally compact and Hausdorff (445J). 445R The Plancherel Theorem Let X be an abelian topological group with a Haar measure µ, and X its dual group. Let λ be the Haar measure on X corresponding to µ (445P). Then there is a normed ∧

space isomorphism T : L2C (µ) → L2C (λ) defined by setting T (f • ) = f • whenever f ∈ L1C (µ) ∩ L2C (µ). ∧

∧

proof (a) Since f = g whenever f =a.e. g, the formula certainly defines an operator T from L1C (µ) ∩ L2C (µ) to L0C (λ), and of course it is linear. If f ∈ L1C (µ) ∩ L2C (µ), set f˜(x) = f (x−1 ) whenever this is defined; then h = f ∗ f˜ is integrable, continuous ∧

∧

and positive definite (445Ld), and h = |f |2 (445Kc). Now Z

Z

∧

|f |2 dλ = (445P)

∧

h dλ = h(e) Z

Z f (x)f˜(x−1 )µ(dx) =

=

|f (x)|2 µ(dx).

Thus kT uk2 = kuk2 whenever u ∈ L1C (µ) ∩ L2C (µ); since L1C (µ) ∩ L2C (µ) is k k2 -dense in L2C (µ) (244Ha/244Ob, or otherwise), we have a unique isometry T : L2C (µ) → L2C (λ) agreeing with the given formula on L1C (µ)∩L2C (µ) (2A4I). (b) The meat of the theorem is of course the proof that T is surjective. P P?? Suppose, if possible, that W = T [L2C (µ)] is not equal to L2C (λ). Because T is linear, W is a linear subspace of L2C (λ); because T is an isometry, W is isometric to L2C (µ), and in particular is complete, therefore closed in L2C (λ) (3A4Fd). There is therefore a Rnon-zero continuous linear functional on L2C (λ) which is zero on W (3A5Ad), and this is of the form u 7→ u × v for some v ∈ L2C (λ) (244J/244Oc). What this means is that there is a g ∈ L2C (λ) such R ∧ that g • 6= 0 in L2C (λ) but f × g dλ = 0 for every f ∈ L1C (µ) ∩ L2C (µ). Suppose that f ∈ L1C (µ) and that h is any µ-integrable continuous positive definite function on X. Then ¯ ∈ L1 (µ) ∩ L2 (µ) (444Ra). Accordingly h is bounded (445Lb), so |h|2 is also µ-integrable, and f ∗ h C C

R

∧

∧

¯ dλ = g×f ×h

R

¯ ∧ dλ = 0. g × (f ∗ h)

352

Thus

Topological groups

R

445R

∧

∧

¯ is the Radon measure on X corresponding to h constructed in 445N g × f dνh = 0, where νh = hλ ∧

(see 445Q). And this is true for every f ∈ L1C (µ). But as {f : f ∈ L1C (µ)} is k k∞ -dense in C0 (X ; C) ∧ ∧ R ¯ ∈ L2 (λ), by (a), so |g × h|dλ ¯ (445Kd), and g is νh -integrable (because h < ∞), g must be zero νh -a.e., C ∧ ¯ = 0 λ-a.e. Now recall (from part (c) of the proof of 445P, for instance) that for every compact that is, g × h ∧

¯ set K ⊆ X there is a µ-integrable continuous positive definite h such that h(χ) 6= 0 for every χ ∈ K, so g = 0 a.e. on K. Since λ is tight (that is, inner regular with respect to the compact sets), g = 0 a.e. (412Jc), which is impossible. X XQ Q Thus T is surjective and we have the result. ∧

445S While we do not have a direct definition of f for f ∈ L2C \ L1C , the map T : L2C (µ) → L2C (λ) does ∧

correspond to the map f 7→ f in many ways. In particular, we have the following useful properties. Proposition Let X be an abelian topological group with a Haar measure µ, X its dual group, λ the associated Haar measure on X and T : L2C (µ) → L2C (λ) the standard isometry described in 445R. Suppose that f0 , f1 ∈ L2C (µ) and g0 , g1 ∈ L2C (ν) are such that T f0• = g0• and T f1• = g1• , and take any θ ∈ X . Then (a) setting f2 = f¯0 , g2 (χ) = g0 (χ−1 ) whenever this is defined, T f2• = g2• ; (b) setting f3 = f1 × θ, g3 (χ) = g1 (θχ) whenever this is defined, T f3• = g3• ; ∧

(c) setting f4 = f0 × f1 ∈ L1C (µ), f 4 (θ) = (g0 ∗ g1 )(θ). proof (a) We have isometries R1 : L2C (µ) → L2C (µ) and R2 : L2C (λ) → L2C (λ) defined by setting R1 f • = (f¯)• for every f ∈ L2C (µ) and R2 g • = (˜ g )• for every g ∈ L2C (λ), where g˜(χ) = g(χ−1 ) whenever this is defined. 1 Now if f ∈ LC (µ), then ∧

f¯(χ) =

R

f (x)χ(x)µ(dx) =

R

˜∧ f (x)χ−1 (x)µ(dx) = f (χ)

for every χ ∈ X . So T R1 u = R2 T u for every u ∈ L1C (µ) ∩ L2C (µ); as L1C ∩ L2C is dense in L2C , T R1 = R2 T , which is what we need to know. (b) This time, set R1 f • = (f × θ)• for every f ∈ L2C (µ) and R2 g • = (θ−1 •l g)• for g ∈ L2C (λ), where (θ•l g)(χ) = g(θ−1 χ) whenever this is defined. Once again, R1 : L2C (µ) → L2C (µ) and R2 : L2C (λ) → L2C (λ) are isometries. If f ∈ L1C (µ), then (f × θ)∧ (χ) =

R

∧

f (x)θ(x)χ(x)µ(dx) = f (θχ)

for every χ, so T R1 f • = R2 T f • ; once again, this is enough to prove that T R1 = R2 T , as required. (c) We have Z (g0 ∗ g1 )(θ) =

Z −1

g0 (χ

)g1 (θχ)λ(dχ) =

g2 (χ)g3 (χ)λ(dχ) = (g3• |g2• )

(where ( | ) is the standard inner product of L2C (λ)) = (T f3• |T f2• ) (using (a) and (b)) = (f3• |f2• ) (because linear isometries of Hilbert space preserve inner products, see 4A4Jb) Z Z = f3 (x)f2 (x)µ(dx) = f1 (x)θ(x)f0 (x)µ(dx) = (f0 × f1 )∧ (θ).

445T Corollary Let X be an abelian topological group with a Haar measure µ, and λ the corresponding Radon measure on the dual group X of X (445P). Then for any non-empty open set H ⊆ X , there is an ∧

∧

f ∈ L1C (µ) such that f 6= 0 and f (χ) = 0 for χ ∈ X \ H.

445Xf

The duality theorem

353

proof Let V1 and V2 be non-empty open sets of finite measure such that V1 V2 ⊆ H, and let g1 , g2 be their characteristic functions. Then there are f1 , f2 ∈ L2C (µ) such that T fj• = gj• for both j, where T : L2C (µ) → L2C (λ) is the isometry of 445R. In this case, by 445Sc, (f1 × f2 )∧ = g1 ∗ g2 . But it is easy to check that g1 ∗ g2 is non-zero and zero outside H. 445U The Duality Theorem (Pontryagin 34, Kampen 35) Let X be a locally compact Hausdorff abelian topological group. Then the canonical map x 7→ x ˆ from X to its bidual X (445E) is an isomorphism between X and X as topological groups. ˆ ⊆ X. Accordingly X ˆ is itself, with its proof By 445O, b is a homeomorphism between X and its image X subspace topology, a locally compact topological group, and is closed in X (4A5Mc). ˆ 6= X. Let µ be a Haar measure on X (441E) and λ the associated Radon ?? Suppose, if possible, that X ∧ ˆ but not zero measure on the dual X of X (445P). By 445T, there is a g ∈ L1C (λ) such that g is zero on X everywhere, so that g is not zero a.e. We may suppose that g is defined everywhere on X . But in this case we have ∧

0 = g(ˆ x) =

R

g(χ)ˆ x(χ)λ(dχ) =

R

g(χ)χ(x)λ(dχ)

for every x ∈ X. R By 418J, 3gR: X → C is almost continuous. Let K ⊆ {χ : χ ∈ X , g(χ) 6= 0} be a compact set such that |g|dλ ≥ 4 X |g|dλ and g¹K is continuous. Set q(χ) = g(χ)/|g(χ)| for χ ∈ K. Now consider the linear K span A of {ˆ x : x ∈ X} as a linear space of complex-valued functions on X . Since x cy = x ˆ × yˆ for all x, y ∈ X, −1 = x ¯ˆ for every x ∈ X, h ¯ ∈ A for every h ∈ A; the constant A is a subalgebra of Cb = Cb (X ; C); since xd function eˆ belongs to A; and A separates the points of X . By the Stone-Weierstrass theorem, in the form 1 281G, there is an h ∈ A such that |h(χ) − q(χ)| R R ≤ 2 for every χ ∈ K and |h(χ)| ≤ 1 for every χ ∈ X . Of course g × h dλ = 0 for every h ∈ A because g × x ˆ dλ = 0 for every x ∈ X. Now, however, Z Z Z Z ¯ ¯ |g|dλ = g × q dλ ≤ ¯ g × h dλ¯ + |g| × |h − q|dλ K K K ZK Z Z Z Z ¯ ¯ 1 1 ¯ ¯ ≤ g × h dλ + |g|dλ ≤ |g|dλ + |g|dλ < |g|dλ, X \K

2

K

X \K

2

K

K

which is impossible. X X ˆ = X and the proof is complete. Thus X 445X Basic exercises (a) Consider the additive group Q with its usual topology. Show that its dual group can be identified with the additive group R with its usual topology. (b) Let X be any topological group, and X its dual group. Show that if ν is any totally finite Radon ∧ measure on X, then ν : X → C is continuous. (c) Let X be a topological group carrying Haar measures and X its dual group. For a totally finite R ∧ ∧ quasi-Radon measure ν on X set ν(x) = χ(x)ν(dχ) for every x ∈ X. (i) Show that ν is continuous. (ii) R ∧ R ∧ Show that for any totally finite quasi-Radon measure µ on X, µ dν = ν dµ. ¯1 > (d) Let X be a group and h1 , h2 : X → C positive definite functions. Show that h1 + h2 , αh1 and h are also positive definite for any α ≥ 0. (e) (i) Let X be a group, Y a subgroup of X and h : Y → C a positive definite function. Set h1 (x) = h(x) if x ∈ Y , h1 (x) = 0 if x ∈ X \ Y . Show that h1 is positive definite. (ii) Let X and Y be groups, φ : X → Y a group homomorphism and h : Y → C a positive definite function. Show that hφ : X → C is positive definite. (f ) Let X be a topological group and X its dual group. (i) Let ν be any totally finite topological measure R on X and set h(x) = χ(x)ν(dχ) for x ∈ X. Show thatR h : X → C is positive definite. (ii) Let ν be any totally finite topological measure on X. Show that χ 7→ χ dν : X → C is positive definite.

354

Topological groups

445Xg

> (g) Let X be a topological group withRa left Haar measure, and h : X → C a bounded continuous function. Show that h is positive definite iff h(x−1 y)f (y)f (x)dxdy ≥ 0 for every integrable function f . > (h) Let φ : R r → C be a function. Show that it is the characteristic function of a probability distribution on Rr iff it is continuous and positive definite and φ(0) = 1. > (i) Let X be a compact Hausdorff abelian topological group, and µ the Haar probability measure on X. Show that the corresponding Haar measure on the dual group X of X is just counting measure on X . > (j) Let X be an abelian group with its discrete topology, and µ counting measure on X. Let X be the dual group and λ the corresponding Haar measure on X . Show that λX = 1. 1 (k) Let X be the topological group R, and µ = √ µL , where µL is Lebesgue measure. (i) Show that if 2π

we identify the dual group X of X with R, writing χ(x) = eiχx for x, χ ∈ R, then the Haar measure on X corresponding to the Haar measure µ on X is µ itself. (ii) Show that if we change the action of R on itself by setting χ(x) = e2πiχx , then the Haar measure on X corresponding to µL is µL . (l) Let X0 , . . . , Xn be abelian topological groups with Haar measures µ0 , . . . , µn , and let X = X0 × . . . × Xn be the product group with its Haar measure µ = µ0 × . . . × µn . For each k ≤ n let Xk be the dual group of Xk and λk the Haar measure on Xk corresponding to µk . Show that if we identify X = X0 × . . . × Xn with the dual group of X, then the Haar measure on X corresponding to µ is just the product measure λ0 × . . . × λn . > (m) Let X be a compact Hausdorff Rabelian topological group, with dual group X , and µ the Haar probability measure on X. (i) Show that χ(x)µ(dx) = 0 for every χ ∈ X except the identity. (ii) Show ∧• that {χ• : χ ∈ X } is an orthonormal basis of the Hilbert space L2C (µ). (Hint: (u|χ• ) = (T u|χ ) where T is the operator of 445R.) (n) Let X be an abelian topological group with a Haar measure µ, λ the associated Haar measure on the dual group X and T : L2C (µ) → L2C (λ) the standard isometry. Suppose that u, v ∈ L2C (µ). Suppose that f0 , f1 ∈ L2C (µ), g0 , g1 ∈ L2C (ν) are such that T f0• = g0• and T f1• = g1• . Show that (f0 ∗ f1 )(x) = R g0 (χ)g1 (χ)χ(x)λ(dχ) for any x ∈ X. (o) Let X be a locally compact Hausdorff abelian topological group and X its dual group. Show that a function h : X → C is the Fourier-Stieltjes transform of a totally finite Radon measure on X iff it is continuous and positive definite. (p) (i) Show that we can define a binary operation +2adic on X = {0, 1}N by setting x+2adic y = z whenever Pk x, y, z ∈ X and i=0 2i (x(i) + y(i) − z(i)) is divisible by 2k+1 for every k. (ii) Show that if we give X its usual topology then (X, +2adic ) is an abelian topological group. (iii) Show that the usual measure on X is n the Haar probability measure for this group operation. (iv) Show that G = {ζ : ζ ∈ C, ∃ n ∈ N, ζ 2 = 1} is Q∞ 2i x(i) a subgroup of C. (v) Show that the dual of (X, +2adic ) is {χζ : ζ ∈ G} where χζ (x) = i=0 ζ for ζ ∈ G and x ∈ X. (vi) Show that the functions f , g : X → X described in 388E are of the form x 7→ x ±2adic x0 for a certain x0 ∈ X. b be its completion 445Y Further exercises (a) Let X be any Hausdorff topological group. Let X b under its bilateral uniformity (4A5N). Show that the dual groups of X and X can be identified as groups. Show that they can be identified as topological groups if either X is metrizable or X has a totally bounded neighbourhood of the identity. (b) Let hXj ij∈I be a countable family of topological groups, with product X; let Xj be the dual group of each Q Xj , and X the dual group of X. Show that the topology of X is generated by sets of the form X ∩ j∈I Hj where Hj ⊆ Xj is open for each j.

445 Notes

The duality theorem

355

(c) Let X be a real linear topological space, with addition as its group operation. Show that its dual group is just the set of functionals x 7→ eif (x) where f : X → R is a continuous linear functional. Hence show that there are abelian groups with trivial duals. (d) Let X be the group of rotations of R 3 , that is, the group of orthogonal real 3 × 3 matrices with determinant 1, and give X its usual topology, corresponding to its embedding as a subspace of R 9 . Show that the only character on X is the constant function 1. (Hint: (i) show that two rotations through the same angle are conjugate in X; (ii) show that if 0 < θ ≤ π2 then the product of two rotations through the angle θ about orthogonal axes is not a rotation through an angle 2θ.) (e) Let X be a finite abelian group, endowed with its discrete topology. Show that its dual is isomorphic to X. (Hint: X is a product of cyclic groups.) (f ) Show that if I is any uncountable set, then there is a quasi-Radon probability measure ν on the ∧ topological group RI such that ν is not continuous. (Hint: take ν to be a power of a suitably widely spread probability distribution on R.) (g) Let X be an abelian topological group with a Haar measure µ, and λ the associated Haar measure on the dual group X of X. Let T : L2C (µ) → L2C (λ) be the standard isomorphism. Suppose that f ∈ L2C (µ), R g ∈ L1C (λ) ∩ L2C (λ) are such that T f • = g • . Show that f (x) = g(χ)χ(x)λ(dχ) for almost every x ∈ X. (Hint: look first at locally compact Hausdorff X.) (h) Let X be a locally compact Hausdorff abelian topological group with dual group X , hνn in∈N a sequence of Radon probability measures on X Rand ν another Radon probability measure on X. Show R that the following are equiveridical: (i) limn→∞ f dνn = f dν for every bounded continuous real-valued ∧ ∧ function on X; (ii) limn→∞ ν n (χ) = ν(χ) for every χ ∈ X . (Hint: compare 285L. For the critical step, showing that {νn : n ∈ N} is uniformly tight, use the formulae in 445N to show that there is an integrable ∨ R ∨ f : X → C such that 0 ≤ f ≤ 1 and f (x)ν(dx) ≥ 1 − 12 ².) (i) Let X be a topological group carrying Haar measures and X its dual group. Let Mτ be the complex Banach space of signed totally finite τ -additive Borel measures on X (put the ideas of 437F and 437Yb together). Show Rthat X separates the points of Mτ in the sense that if ν ∈ Mτ is non-zero, there is an x ∈ X such that χ(x)ν(dχ) 6= 0, if the integral is appropriately interpreted. (Hint: use the method in the proof of 445U.) (j) Let X be an abelian topological group carrying Haar measures. Let Mτ be the complex Banach space of signed totally finite τ -additive Borel measures on X. Show that the dual R X of X separates the points of Mτ in the sense that if ν ∈ Mτ is non-zero, there is a χ ∈ X such that χ(x)ν(dx) 6= 0. (Hint: use 443L to reduce to the case in which X is locally compact and Hausdorff; now use 445U and 445Yi.) (k) Let X be an abelian topological group and µ a Haar measure on X. Show that the spectral radius of any non-zero element of L1C (µ) is non-zero. (Hint: 445Yj, 445Kd.) (l) Show that for any integer p ≥ 2 there is an operation +padic on {0, . . . , p − 1}N with properties similar to those of the operation +2adic of 445Xp. 445 Notes and comments I repeat that this section is intended to be a more or less direct attack on the duality theorem. At every point the clause ‘let X be a locally compact Hausdorff abelian topological group’ is present in spirit. The actual statement of each theorem involves some subset of these properties, purely in accordance with the principle of omission of irrelevant hypotheses, not because I expect to employ the results in any more general setting. In 445Ab I describe a topology on the dual group in a context so abstract that we have rather a lot of choice. For groups carrying Haar measures, the alternative descriptions of the topology on the dual (445I) make it plain that this must be the first topology to study. By 445E it is already becoming fairly convincing. But it is not clear that there is any such pre-eminent topology in the general case.

356

Topological groups

445 Notes

Fourier-Stieltjes transforms hardly enter into the arguments of this section; I mention them mostly because they form the obvious generalization of the ideas in §285. But I note that the principal theorem of §285 (that sequential convergence of characteristic functions corresponds to sequential convergence of distributions, 285L) generalizes directly to the context here (445Yh). I have tried to lay this treatise out in such a way that we periodically return to themes from past chapters at a higher level of sophistication. There seem to be four really important differences between this section and the previous treatment in Chapter 28. (i) The first is the obvious one; we are dealing with general locally compact Hausdorff abelian groups, rather than with R and S 1 and Z. Of course this puts much heavier demands on our technique, and, to begin with, leaves our imaginations unfocused. (ii) The second concerns differentiation, or rather its absence; since we no longer have any differential structure on our groups, a substantial part of the theory evaporates, and we are forced to employ new tactics in the rest. (iii) The third concerns the normalization of the measure on the dual group. As soon as we know that X is a locally compact group (445J) we know that it carries Haar measures. The problem is to describe the particular one we need in appropriate terms. In the case of the dual pairs (R, R) or (S 1 , Z), we have measures already presented (counting measure on Z, Lebesgue measure on ]−π, π] and R). (They are not in fact dual in the sense we need here, at least not if we use the simplest formulae for the duality, and have to be corrected in each case by a factor of 2π. See 445Xk.) But since we do have dual pairs already to hand, we can simultaneously develop theories of Fourier transforms and inverse Fourier transforms (for the pair (S 1 , Z) the inverse Fourier transform is just summation of trigonometric series), and the problem is to successfully match operations which have independent existences. (iv) The final change concerns interesting feature R a of Pan n Z and R. Repeatedly, in §§282-283, the formulae invoked symmetric limits limn→∞ k=−n or lima→∞ −a to approach some conditionally convergent sumR or integral. Elsewhere one sometimes deals R −a R 1 with singularities 1 by examining ‘Cauchy principal values’; if −1 f is undefined, try lima↓0 ( −1 f + a f ). This particular method seems to disappear in the general context. But the general challenge of the subject remains the ∧ same: to develop a theory of the transform u 7→ u which will apply to the largest possible family of objects u and will enable us to justify, in the widest possible contexts, the manipulations listed in the notes to §284. The calculations in 445S and 445Xn, treating ‘shift’ and ‘convolution’ in L2 , are typical. In terms of the actual proofs of the results here, ‘test functions’ (284A) have gone, and in their place we take a lot more trouble over the Banach algebra L1 . This algebra is the key to one of the magic bits, which turns up in rather undignified corners in 445Kd and part (e) of the proof of 445N. Down to 445O, the dominating problem is that we do not know that the dual group X of a group X is large enough to tell us anything interesting about X. (After that, the problem reverses; we have to show that X is not too big.) We find that (under rather specially arranged circumstances) we are able to say something useful about the spectral radius of a member of L1 , and we use this to guarantee that it has a non-trivial Fourier transform. If we identify X with the maximal ideal space of L1 (445H), then the Fourier transform on L1 becomes the ‘Gelfand map’, a general construction of great power in the theory of commutative Banach algebras. There is one similarity between the methods of this section and those of §284. In both cases we have isomorphisms between L2C (µ) and L2C (λ) (the Plancherel Theorem), but cannot define the Fourier transform of a function in L2C in any direct way; indeed, while the Fourier transform of a function in L1C , or even of a (totally finite) measure, can really be thought of as a (continuous) function, the transform of a function in L2 is at best a member of L2 , not a function at all. We manoeuvre around this difficulty by establishing that our (genuine) Fourier transforms match dense subspaces isometrically. In §284 I used test functions, and in the present section I use L1 ∩ L2 . Test functions are easier partly because the Fourier transform of a test function is again a test function, and all the formulae we need are easy to establish for such functions. Searching for classes of functions which will be readily manageable in general locally compact abelian groups, we come to the ‘positive definite’ functions. The phrase is unsettling, since the functions themselves are in no obvious sense positive (nor even, as a rule, real-valued). Also their natural analogues in the theory of bilinear forms are commonly called ‘positive semi-definite’. However, their Fourier transforms, whether regarded as measures (445N) or as functions (445Q), are positive, and, as a bonus, we get a characterization of the Fourier transforms of measures (445Xf, 445Xh), answering a question left hanging in 285Xr.

446B

The structure of locally compact groups

357

446 The structure of locally compact groups I develop those fragments of the structure are needed for the main theorem of the next both itself important and with a proof which but the rest of the section, from 446D on, is not expected to be abelian.

theory of locally compact Hausdorff topological groups which section. Theorem 446B here is of independent interest, being uses the measure theory of this chapter in an interesting way; starred. Note that in this section, unlike the last, groups are

446A Finite-dimensional representations (a) Definitions (i) For any r ∈ N, write Mr = Mr (R) for the space of r × r real matrices. If we identify it with the space B(R r ; R r ), where R r is given its Euclidean norm, then Mr becomes a unital Banach algebra (4A6C), with identity I, the r × r identity matrix. Write GL(r, R) for the group of invertible elements of Mr . (ii) Let X be a topological group. A finite-dimensional representation of X is a continuous homomorphism from X to a group of the form GL(r, R) for some r ∈ N. If the homomorphism is injective the representation is called faithful (cf. 4A5Be). (b) Observe that if X is any topological group and φ is a finite-dimensional representation with kernel Y , then X/Y has a faithful finite-dimensional representation ψ defined by writing ψ(x• ) = φ(x) for every x ∈ X (4A5La). 446B Theorem Let X be a compact Hausdorff topological group. Then for any a ∈ X, other than the identity, there is a finite-dimensional representation φ of X such that φ(a) 6= I. proof (a) Let U be a symmetric neighbourhood of the identity e in X such that a ∈ / U U . Because X is completely regular (3A3Bb), there is a non-zero continuous function h : X → [0, ∞[ such that h(x) = 0 for every x ∈ X \ U ; replacing h by x 7→ h(x) + h(x−1 ) if necessary, we may suppose that h(x) = h(x−1 ) for every x. Let µ be a (left) Haar measure on X (441E), and set w = h• in L0 (µ). (b) Define an operator T from L2 = L2 (µ) to itself by setting T u = u ∗ w for every u ∈ L2 , where ∗ is convolution; that is, T f • = (f ∗ h)• for every f ∈ L2 = L2 (µ). Then T is a compact self-adjoint operator on the real Hilbert space L2 (444V). (c) For any z ∈ X, define Sz : L2 → L2 by setting Sz u = z •l u for u ∈ L2 , where •l is defined as in 443G. Then Sz is a norm-preserving linear operator. Also Sz commutes with T . P P If f ∈ L2 , then Sz T f • = (z •l (f ∗ h))• = ((z •l f ) ∗ h)• (444Of) = T Sz f • . Q Q (d) Now Sa T w 6= T w. P P Set g = h ∗ h, so that T w = g • and g, a•l g are both continuous functions (444Rc). Then (a•l g)(e) = g(a−1 ) =

R

h(y)h(y −1 a−1 )dy = 0

because if y ∈ U then y −1 a−1 ∈ / U , while g(e) =

R

h(y)h(y −1 )dy =

R

h(y)2 dy > 0.

So the open set {x : g(x) 6= (a•l g)(x)} is non-empty; because µ is strictly positive (442Aa), it is nonnegligible, and Sa T w = (a•l g)• 6= g • = T w. Q Q (e) The closed linear subspace {u : Sa u = u} therefore does not include T [L2 ]. But T [L2 ] is the closed linear span of {T v : v is an eigenvector of T } (4A4M), so there is an eigenvector v ∗ of T such that

358

Topological groups

446B

Sa T v ∗ 6= T v ∗ . Let γ ∈ R be such that T v ∗ = γv ∗ ; since T v ∗ 6= 0, γ 6= 0, and V = {u : T u = γu} is finite-dimensional (4A4Lb). (f ) Sz [V ] ⊆ V for every z ∈ X. P P T Sz u = Sz T u = Sz (γu) = γSz u for every u ∈ V . Q Q We therefore have a map z 7→ Sz ¹ V : X → B(V ; V ). As observed in 443Gc, this is actually a semigroup homomorphism, and of course Se ¹ V is the identity of B(V ; V ), so Sz ¹ V is always invertible, and we have a group homomorphism from X to the group of invertible elements of B(V ; V ). Taking an orthonormal basis (v1 , . . . , vr ) of V , we have a homomorphism φ from X to GL(r, R), defined by setting φ(z) = h(Sz vi |vj )i1≤i,j≤r for every z ∈ X. Moreover, φ is continuous. P P For any u ∈ L2 , z 7→ Sz u : X → L2 is continuous, by 443Ge. But this means that all the maps z 7→ (Sz vi |vj ) are continuous; since the topology of GL(r, R) can be defined in terms of these functionals (see the formulae in 262H), φ is continuous. Q Q Thus φ is a finite-dimensional representation of X. But V was chosen to contain v ∗ ; of course T v ∗ ∈ V , while Sa T v ∗ 6= T v ∗ ; so that φ(a) is not the identity, as required. 446C Corollary Let X be a compact Hausdorff topological group. Then for any neighbourhood U of the identity of X there is a finite-dimensional representation of X with kernel included in U . T proof Let Φ be the set of finite-dimensional representations of X. By 446B, φ∈Φ ker(φ) = {e}, where T e is the identity of X. Because X \ int U is compact and disjoint from φ∈Φ ker(φ) (and ker(φ) is closed T for every φ), there must be φ0 , . . . , φn ∈ Φ such that i≤n ker(φi ) ⊆ U . For each i ≤ n, let ri ∈ N be the Pn integer such that φi is a continuous homomorphism from X to GL(ri , R). Set r = i=0 ri . Then we have a map φ : X → GL(r, R) given by the formula 

φ0 (x) 0 φ1 (x)  0 φ(x) =  ... ... 0 0

... ... ... ...

 0 0   ... φn (x)

for every x ∈ X. It is easy to check that φ is a continuous homomorphism, and ker(φ) = So we have an appropriate representation of X.

T i≤n

ker(φi ) ⊆ U .

*446D Notation (a) It will help to be clear on an elementary point of notation. If X is a group and A ⊆ X I will write A0 = {e}, where e is the identity of X, and An+1 = AAn for n ∈ N, so that A3 = {x1 x2 x3 : x1 , x2 , x3 ∈ A}, etc. Now we find that Am+n = Am An and Amn = (Am )n for all m, n ∈ N. Writing A−1 = {x−1 : x ∈ A} as usual, we have (Ar )−1 = (A−1 )r . But note that if we also continue to write A−1 = {x−1 : x ∈ A}, then AA−1 is not in general equal to A0 ; and that there is no simple relation between Ar , B r and (AB)r , unless X is abelian. (b) In the rest of this section, I shall make extensive use of the following device. If X is a group with identity e, e ∈ A ⊆ X and n ∈ N, write Dn (A) = {x : x ∈ X, xi ∈ A for every i ≤ n}. Then (i) D0 (A) = X. (ii) D1 (A) = A. (iii) Dn (A) ⊆ Dm (A) whenever m ≤ n. (iv) Dmn (A) ⊆ Dm (Dn (A)) for all m, n ∈ N. P P If x ∈ Dmn (A) then (xi )j ∈ A whenever j ≤ n, i ≤ m. Q Q (v) If r ∈ N and Ar ⊆ B, then Dn (A) ⊆ Dnr (B) for every n ∈ N; in particular, A ⊆ Dr (B). P P For r = 0 this is trivial. Otherwise, take x ∈ Dn (A) and i ≤ nr. Then we can express i as i1 + . . . + ir where ij ≤ n for each j, so that Qr xi = j=1 xij ∈ Ar ⊆ B. Q Q (vi) If A = A−1 then Dn (A) = Dn (A)−1 for every n ∈ N.

*446F


359

(vii) If Dm (A) ⊆ B for some m ∈ N, then Dmn (A) ⊆ Dn (B) for every n ∈ N. P P If x ∈ Dmn (A) and i ≤ n, then xij ∈ A for every j ≤ m, so xi ∈ Dm (A) ⊆ B. Q Q (c) In (b), if X is a topological group and A is closed, then every Dn (A) is closed; if moreover A is compact, then Dn (A) is compact for every n ≥ 1. If A is a neighbourhood of e, then so is every Dn (A), because the map x 7→ xi is continuous for every i ≤ n. *446E ‘Groups with no small subgroups’ (a) Definition Let X be a topological group. We say that X has no small subgroups if there is a neighbourhood U of the identity e of X such that the only subgroup of X included in U is {e}. (b) If X is a Hausdorff topological group and U is a compact symmetric neighbourhood of the identity e such that the only subgroup of X included in U is {e}, then {Dn (U ) : n ∈ N} is a base of neighbourhoods of e, where Dn (U ) = {x : x ∈ X, xi ∈ U for every iT≤ n}. P P By 446Dc, hDn (U )in≥1 is a non-increasing sequence of compact neighbourhoods of e, and if x ∈ n∈N Dn (U ) then xi ∈ U for every i ∈ N; as U −1 = U , T U includes the subgroup {xi : i ∈ Z}, so x = e. Thus n∈N Dn (U ) = {e} and {Dn (U ) : n ∈ N} is a base of neighbourhoods of e (4A2Gd). Q Q (c) In particular, a locally compact Hausdorff topological group with no small subgroups is metrizable (4A5Q). *446F Lemma Let X be a group, and U ⊆ X. Let f : X → [0, ∞[ be a bounded function such that f (x) = 0 for x ∈ X \ U ; set α = supx∈X f (x). Let A ⊆ X be a symmetric set containing e, and K a set including Ak , where k ≥ 1. Define g : X → [0, ∞[ by setting 1 Pk−1 i g(x) = i=0 sup{f (yx) : y ∈ A } k

for x ∈ X. Then (a) f (x) ≤ g(x) ≤ α for every x ∈ X, and g(x) = 0 if x ∈ / K −1 U . (b) |g(ax) − g(x)| ≤

jα k

if j ∈ N, a ∈ Aj and x ∈ X.

(c) For any x, z ∈ X, |g(x) − g(z)| ≤ supy∈K |f (yx) − f (yz)|. proof (a) Of course f (x) =

1 k

Pk−1 i=0

f (ex) ≤ g(x) ≤

1 k

Pk−1 i=0

α=α

for every x. Suppose that g(x) 6= 0. Then there must be an i < k and a y ∈ Ai such that f (yx) 6= 0, so that yx ∈ U . But also, because e ∈ A, y ∈ Ak ⊆ K, so x ∈ y −1 U ⊆ K −1 U . (b) Suppose first that j = 1, so that a ∈ A. If ² > 0 there are yi ∈ Ai , for i < k, such that g(ax) ≤ 1 Pk−1 i+1 for each i, so i=0 f (yi ax) + ². Now yi a ∈ A

k

g(x) ≥

1 k

Pk−2 i=0

α k

f (yi ax) ≥ g(ax) − ² − .

As ² is arbitrary, g(x) ≥ g(ax) − αk . Similarly, as a−1 ∈ A (because A is symmetric), g(ax) ≥ g(x) − |g(ax) − g(x)| ≤ αk . For the general case, induce on j. (If j = 0, then a = e and the result is trivial.)

α k

and

(c) Set γ = supy∈K |f (yx) − f (yz)|. If ² > 0, there are yi ∈ Ai , for i < k, such that g(ax) ≤ 1 Pk−1 i=0 f (yi x) + ². Now every yi belongs to K, so

k

g(z) ≥

1 k

Pk−1 i=0

f (yi z) ≥

1 k

Pk−1 i=0

(f (yi x) − γ) ≥ g(x) − ² − γ.

As ² is arbitrary, g(z) ≥ g(x) − γ; similarly, g(x) ≥ g(z) − γ.

360

Topological groups

*446G

*446G Lemma Let X be a locally compact Hausdorff topological group and hAn in∈N a sequence of symmetric subsets of X all containing the identity e of X. Suppose that for every neighbourhood W of e there is an n0 ∈ N such that An ⊆ W for every n ≥ n0 . Let U be a compact neighbourhood of e and k(n) k(n)+1 suppose that for each n ∈ N we have k(n) ∈ N such that An ⊆ U , An 6⊆ U . Let F be a non-principal k(n) ultrafilter on N and set Q = lim supn→F An (definition: 4A2A). (i) If Q2 = Q then Q is a compact subgroup of X included in U and meeting the boundary of U . (ii) If Q2 6= Q then there are a neighbourhood W of e and an infinite set I ⊆ N such that for every n ∈ I there are an x ∈ An and an i ≤ k(n) such that xi ∈ / W. k(n)

proof (a) Because every An is included in U , so is Q, and Q, being closed (4A2Ta), is compact. Because x 7→ x−1 is a homeomorphism, k(n) −1

Q−1 = lim supn→F (An

)

k(n)

k(n) = lim supn→F (A−1 = lim supn→F An n )

= Q.

k(n) Q because e ∈ An for every n. k(n) we have an an ∈ An and an xn ∈

And of course e ∈ For each n ∈ N, An such that an xn ∈ / U . Now a = limn→F an is defined (because U is compact), and belongs to Q (4A2Ta). Also limn→∞ xn = e, because every neighbourhood of e includes all but finitely many of the An , so a = limn→F an xn ∈ / int U , and a belongs to the boundary of U . Thus Q meets the boundary of U . (b) From (a) we see that if Q2 = Q then Q is a compact subgroup of X, included in U and meeting the boundary of U . So henceforth let us suppose that Q2 6= Q and seek to prove (ii). Let w ∈ Q2 \ Q. Let W0 ⊆ U be an open neighbourhood of e such that W0 wW02 ∩ QW02 = ∅ (4A5Ee). (c) Fix a left HaarR measure µ on X. Let f : X → [0, ∞[ be R a continuous function such that {x : f (x) > 0} ⊆ W0 and f (x)dx = 1. Set α = supx∈X f (x), β = f (x)2 dx, so that α is finite and β > 0. W0 U 2 W0 ⊆ U 4 is compact, so has finite measure, and there is an η > 0 be such that 2η(1 + αµ(W0 U 2 W0 )) < β. Let W be a neighbourhood of e such that W ⊆ W0 and |f (yax) − f (ybx)| ≤ η whenever y ∈ (U −1 )2 ∪ U , x ∈ X and ab−1 ∈ W (4A5Pa). (d) Express w as w0 w00 where w0 , w00 ∈ Q. Then k(n)

{n : An

k(n)

{n : An

∩ W0 w0 6= ∅},

k(n)

{n : An k(n)

⊆ QW0 } = {n : An

∩ W0 w00 6= ∅},

∩ (U \ QW0 ) = ∅}

all belong to F, by 4A2Tb and 4A2Tc. Also {n : k(n) ≥ 1} is cofinite in N, because An ⊆ U for all n large enough. Let I be the intersection of these four sets, so that I ∈ F and must be infinite. (e) Let n ∈ I. ?? Suppose, if possible, that xi ∈ W for every x ∈ An , i ≤ k(n). (The rest of the proof will be a search for a contradiction.) Note that k(n) ≥ 1. Choose xj ∈ An , for j < 2k(n), such that the products x2k(n)−1 x2k(n)−2 . . . xk(n) , xk(n)−1 . . . x0 belong to W0 w0 , W0 w00 respectively; set w ˜ = x2k(n)−1 . . . x0 , so that 2k(n)

w ˜ ∈ An k(n)

Since An

∩ W0 w0 w00 W0 ⊆ W0 wW0 . k(n)

⊆ QW0 and W0 wW02 ∩ QW02 is empty, wW ˜ 0 does not meet An

W0 .

(f ) Define g : X → [0, ∞[ by setting g(x) =

1 k(n)

Pk(n)−1 i=0

sup{f (yx) : y ∈ Ain }. k(n)

Then g(wx)f ˜ (x) = 0 for every x ∈ X. P P If f (x) 6= 0, then x ∈ W0 , so wx ˜ ∈ / An 446Fa. Q Q Accordingly

W0 , and g(wx) ˜ = 0, by

*446G


R

(g(x) − g(wx))f ˜ (x)dx =

R

361

g(x)f (x)dx ≥ β

since g ≥ f (446Fa). Set y0 = e, yi+1 = xi yi for i < 2k(n), so that y2k(n) = w ˜ and 2k(n)

yi ∈ Ain ⊆ An

k(n) 2

= (An

) ⊆ U2

for every i ≤ 2k(n). Then g(x) − g(wx) ˜ =

P2k(n)−1 i=0

g(yi x) − g(yi+1 x)

for every x ∈ X. Let i < 2k(n) be such that

R

(g(yi x) − g(yi+1 x))f (x)dx ≥

1 2k(n)

R

(g(x) − g(wx))f ˜ (x)dx ≥

Set u = xi , v = yi , so that u ∈ An , v ∈ U 2 and Z Z (g(x) − g(ux))f (v −1 x)dx = (g(vx) − g(uvx))f (x)dx Z = (g(yi x) − g(yi+1 x))f (x)dx ≥

β . 2k(n)

β . 2k(n)

(g) We have Z (g(x) − g(uk(n) x))f (v −1 x)dx k(n)−1 Z

=

X

(g(uj x) − g(uj+1 x))f (v −1 x)dx

j=0 k(n)−1 Z

=

X j=0

= k(n)

(g(x) − g(ux))f (v −1 u−j x)dx Z (g(x) − g(ux))f (v −1 x)dx k(n)−1 Z

X

+

(g(x) − g(ux))(f (v −1 u−j x) − f (v −1 x))dx,

j=0

that is, Z k(n)

(g(x) − g(ux))f (v −1 x)dx Z = (g(x) − g(uk(n) x))f (v −1 x)dx k(n)−1 Z

−

X

(g(x) − g(ux))(f (v −1 u−j x) − f (v −1 x))dx.

j=0

Set β1 =

Pk(n)−1 R j=0

β2 = then

(g(x) − g(ux))(f (v −1 u−j x) − f (v −1 x))dx,

R

(g(x) − g(uk(n) x))f (v −1 x)dx;

R

β2 − β1 = k(n) (g(x) − g(ux))f (v −1 x)dx ≥ 21 β. (h)(i) Examine β1 . We know that, because u ∈ An , |g(x) − g(ux)| ≤

α k(n)

for every x (see 446Fb). On

the other hand, we are supposing that xj ∈ W for every j ≤ k(n) and every x ∈ An , so, in particular,

362

Topological groups

*446G

uj ∈ W ⊆ W0 for every j ≤ k(n). Also, as noted in (f), v ∈ U 2 . So for any j < k(n) we must have |f (v −1 u−j x) − f (v −1 x)| ≤ η for every x ∈ X, by the choice of W , while f (v −1 u−j x) − f (v −1 x) = 0 unless x ∈ W0 U 2 W0 . So k(n)−1 Z

|β1 | ≤

X

|g(x) − g(ux)||f (v −1 u−j x) − f (v −1 x)|dx

j=0 k(n)−1

≤

X j=0

α ηµ(W0 U 2 W0 ) k(n)

= αηµ(W0 U 2 W0 ).

(ii) Now consider β2 . As uk(n) ∈ W , |f (zuk(n) x) − f (zx)| ≤ η for every z ∈ U , by the choice of W , so k(n) (because An ⊆ U ) |g(uk(n) x) − g(x)| ≤ η for every x (446Fc). Accordingly |β2 | ≤ η

R

f (v −1 x)dx = η

R

f (x)dx = η.

(i) But this means that β ≤ 2(|β1 | + |β2 |) ≤ 2η(1 + αµ(W0 U 2 W0 )) < β, which is absurd. X X (j) Thus for every n ∈ I there are an x ∈ An and an i ≤ k(n) such that xi ∈ / W , and (ii) is true. This completes the proof. *446H Lemma Let X be a locally compact Hausdorff topological group. For A ⊆ X, n ∈ N set Dn (A) = {x : xi ∈ A for every i ≤ n}. If U ⊆ X is a compact symmetric neighbourhood of the identity such that there is no subgroup of X included in U other than {e}, then there is an r ≥ 1 such that Drn (U )n ⊆ U for every n ∈ N. proof ?? Suppose, if possible, that for every r ≥ 1 there is an nr ≥ 1 such that Drnr (U )nr 6⊆ U . Of course nr ≥ 1. Set Ar = Drnr (U ) for each r ≥ 1, A0 = U , and apply 446G to the sequence hAr ir∈N . Note that the k(r) k(r) there are such that Ar ⊆ U , so k(r) < nr for r ≥ 1; also k(r) ≥ 1, because Ar ⊆ U . Since U includes no non-trivial subgroup, (i) of 446G is impossible, and we are left with (ii). Let W , I be as declared there, so that Ar 6⊆ Dk(r) (W ) for every r ∈ I. There must be some m ≥ 1 such that Dm (U ) ⊆ W (446Eb). Take r ∈ I such that r ≥ m; then rnr ≥ mk(r), so

Ar = Drnr (U ) ⊆ Dmk(r) (U ) ⊆ Dk(r) (Dm (U )) (446D(b-iv)) ⊆ Dk(r) (W ), which is impossible. X X *446I Lemma Let X be a locally compact Hausdorff topological group and U a compact symmetric neighbourhood of the identity in X such that U does not include any subgroup of X other than {e}. For n ∈ N, set Dn (U ) = {x : xi ∈ U for every i ≤ n}, and let F be any non-principal ultrafilter on N. Suppose that hxn in∈N is a sequence in X such that xn ∈ Dn (U ) for every n ∈ N. Then we have a continuous i(n) homomorphism q : R → X defined by setting q(t) = limn→F xn whenever hi(n)in∈N is a sequence in Z such that limn→F i(n)/n = t in R. Remark See 2A3S for the elementary properties of limn→F . i(n)

proof (a) If hi(n)in∈N is any sequence in N such that limn→F i(n)/n is defined in R, then limn→F xn is defined in X. P P There is some m ∈ N such that m > limn→F i(n)/n, so that J = {n : i(n) ≤ mn} ∈ F ; but if n ∈ J, then

*446J


363

xn ∈ Dn (U ) ⊆ Dmn (U m ) ⊆ Di(n) (U m ), i(n)

i(n)

by 446D(b-v), and xn ∈ U m . But this means that F contains {n : xn ∈ U m }; as U m is compact, i(n) limn→F xn is defined in X. Q Q i(n) More generally, if hi(n)in∈N is any sequence in Z such that limn→F i(n)/n is defined in R, then limn→F xn is defined in X. P P At least one of {n : i(n) ≥ 0}, {n : i(n) ≤ 0} belongs to F. In the former case, i(n) max(0,i(n)) limn→F xn = limn→F xn is defined; in the latter case, i(n)

limn→F xn

max(0,−i(n)) −1

= limn→F (xn

)

max(0,−i(n)) −1

= (limn→F xn

)

is defined. Q Q i(n)

(b) If V is any neighbourhood of e, there is a δ > 0 such that limn→F xn ∈ V whenever hi(n)in∈N is a sequence in Z such that limn→F |i(n)/n| ≤ δ. P P By 446Eb, there is an m ≥ 1 such that Dm (U ) ⊆ V . Take 1 δ
i(n)

and ∈ Dm (U ); since Dm (U ), like U , is symmetric (446D(b-vi)), xn i(n) n ∈ J, so limn→F xn ∈ Dm (U ) ⊆ V , as required. Q Q

∈ Dm (U ). This is true for every i(n)

(c) It follows at once that if hi(n)in∈N is a sequence in Z such that limn→F i(n)/n = 0, then limn→F xn = e. Consequently, if hi(n)in∈N , hj(n)in∈N are sequences in Z such that limn→F i(n)/n and limn→F j(n)/n j(n) i(n) P Set k(n) = i(n) − j(n). Then both exist in R and are equal, then limn→F xn = limn→F xn . P k(n) limn→F xn = e because limn→F k(n)/n = 0. But now i(n)

limn→F xn

j(n) k(n) xn

= limn→F xn

j(n)

= limn→F xn

.Q Q

(d) We do therefore have a function q : R → X defined by the given formula. Now q(s + t) = q(s)q(t) for all s, t ∈ R. P P Take sequences hi(n)in∈N , hj(n)in∈N such that s = limn→∞ i(n)/n, t = limn→∞ j(n)/n; then s + t = limn→∞ (i(n) + j(n))/n, so i(n)+j(n)

q(s + t) = limn→F xn

i(n) j(n)

= limn→F xn xn

= q(s)q(t). Q Q

(e) Thus q is a homomorphism. Finally, (b) shows that it is continuous at 0, so it must be continuous. *446J Lemma Let X be a locally compact Hausdorff topological group with no small subgroups. Then there is a neighbourhood V of the identity e such that x = y whenever x, y ∈ V and x2 = y 2 . proof Let U be a symmetric compact neighbourhood of e not including any subgroup of X except {e}. Let hUn in∈N be a non-increasing sequence of neighbourhoods of e comprising a base of neighbourhoods of e (446Ec), and with U0 = U . ?? Suppose, if possible, that for each n ∈ N there are distinct xn , yn ∈ Un such that x2n = yn2 . Set an = x−1 n yn ; then −2 −1 −1 x−1 n an xn = xn yn xn = yn xn = an .

Accordingly m −1 m −m x−1 n an xn = (xn an xn ) = an

for every m ∈ N. Since U includes no non-trivial subgroup, and an 6= e, there is a k(n) ∈ N such that ain ∈ U for i ≤ k(n) k(n)+1 k(n) and an ∈ / U . Let F be any non-principal ultrafilter on N; then a = limn→F an is defined and belongs to U . Also, because hxn in∈N and hyn in∈N both converge to e, so does han in∈N , and k(n)+1

a = limn→F an thus a cannot be e. k(n) −k(n) However, x−1 xn = an for each n, so n an

∈ X \ int U ;

364

Topological groups k(n)

e−1 ae = limn→F x−1 n an

*446J k(n) −1

xn = limn→F (an

)

= a−1 .

So a = a−1 and {e, a} is a non-trivial subgroup of X included in U , which is supposed to be impossible. X X Thus some Un serves for V . *446K Lemma Let X be a locally compact Hausdorff topological group with no small subgroups. For A ⊆ X set Dn (A) = {x : xi ∈ A for every i ≤ n}. Then there is a compact symmetric neighbourhood U of the identity e such that whenever V is a neighbourhood of e there are an n0 ∈ N and a neighbourhood W of e such that whenever n ≥ n0 , x ∈ Dn (U ), y ∈ Dn (U ) and xn y n ∈ W , then xy ∈ Dn (U ). proof (a) Let U0 be a compact symmetric neighbourhood of e such that (α) U03 includes no subgroup of X other than {e} (β) whenever x, y ∈ U0 and x2 = y 2 , then x = y; such a neighbourhood exists by 446J. Let r ≥ 1 be such that Drn (U0 )n ⊆ U0 for every n ∈ N (446H). Let U be a compact symmetric neighbourhood of e such that U r ⊆ U0 . In this case Dn (U ) ⊆ Drn (U0 ) for every n, by 446D(b-v). So Dn (U )n ⊆ U0 for every n. R (b) Fix a left Haar measure µ on X. Let f : X → [0, ∞[ beR a continuous function such that f (x)dx = 1 and f (x) = 0 for x ∈ X \ U0 . Set α = supx∈X |f (x)|, β = f (x)2 dx, so that α is finite and β > 0. For n ≥ 1, set 1 Pn−1 i fn (x) = i=0 sup{f (yx) : y ∈ Dn (U ) } n

for x ∈ X. Because Dn (U )n ⊆ U0 , we can apply 446F to see that, for each n, (i) fn ≥ f , (ii) fn (x) = 0 if x ∈ / U02 , (iii) |fn (ax) − fn (x)| ≤

jα n

if j ∈ N, a ∈ Dn (U )j and x ∈ X,

(iv) for any x, z ∈ X, |fn (x) − fn (z)| ≤ supy∈U0 |f (yx) − f (yz)|. It follows that (v) for any ² > 0 there is a neighbourhood W of e such that |fn (ax) − fn (bx)| ≤ ² whenever a, b, x ∈ X, n ∈ N and ab−1 ∈ W (4A5Pa). (c) It will help to have the following fact available. Suppose we are given sequences hxn in∈N , hyn in∈N such that xn and yn belong to Dn (U ) for every n ∈ N and limn→∞ xnn ynn = e. Write γn = sup{|fn (ynj x) − fn (x−j n x)| : j ≤ n, x ∈ X} for each n ∈ N. Then limn→∞ γn = 0. P P?? Otherwise, there is an η > 0 such that J = {n : γn > η} is infinite. Let W be a neighbourhood of e such that |fn (ax) − fn (bx)| ≤ η whenever n ∈ N, x, a, b ∈ X and ab−1 ∈ W ((b-v) above); let W 0 be a neighbourhood of e such that ab ∈ W whenever a, b ∈ U and ba ∈ W 0 j(n) j(n) j(n) j(n) / W , while xn and yn (4A5Ej). Then for each n ∈ J there must be a j(n) ≤ n such that yn xn ∈ j(n) j(n) 0 both belong to U , so that x y ∈ / W . Let F be a non-principal ultrafilter on N containing J. By 446I, −i(n) i(n) there are continuous homomorphisms q, q˜ from R to X such that q(t) = limn→F xn , q˜(t) = limn→F yn −1 whenever hi(n)in∈N is a sequence in Z such that limn→F i(n)/n = t in R. (Of course xn ∈ Dn (U ) for every n, by 446D(b-vi).) Setting t0 = limn→F j(n)/n ∈ [0, 1], j(n) j(n) yn

q(−t0 )˜ q (t0 ) = limn→F xn

∈ / int W 0 ,

so q(t0 ) 6= q˜(t0 ) and q 6= q˜. But q(−1)˜ q (1) = limn→F xnn ynn = e, −i(n)

so q(1) = q˜(1). Now if 0 ≤ i(n) ≤ n, then xn ∈ Dn (U )n ⊆ U0 ; so if 0 ≤ t ≤ 1, q(t) ∈ U0 . Similarly, q˜(t) ∈ U0 whenever t ∈ [0, 1]. But recall that U0 was chosen so that if x, y ∈ U0 and x2 = y 2 then x = y. In particular, since q( 21 ) and q˜( 12 ) both belong to U0 , and their squares q(1), q˜(1) are equal, q( 12 ) = q˜( 21 ). Repeating this argument, we see that q(2−k ) = q˜(2−k ) for every k ∈ N, so that q(2−k i) = q˜(2−k i) for every k ∈ N, i ∈ Z; since q and q˜ are supposed to be continuous, they must be equal; but q(t0 ) 6= q˜(t0 ). X XQ Q

*446K


365

(d) Now let V be any neighbourhood of e. ?? Suppose, if possible, that for every neighbourhood W of e and n0 ∈ N there are n ≥ n0 , x, y ∈ Dn (U ) such that xn y n ∈ W but xy ∈ / Dn (V ). For k ∈ N choose nk ∈ N and x ˜k , y˜k ∈ Dnk (U ) such that ˜k y˜k ∈ / Dnk (V ), and nk > nk−1 if k ≥ 1. Now we know that hDk (U )ik∈N is a x ñk k y˜knk ∈ Dk (U ) but x non-increasing sequence constituting a base of neighbourhoods of e (446Eb), so limk→∞ x ñk k y˜knk = e. Set J = {nk : k ∈ N}. For n = nk , set xn = x ˜k , yn = y˜k ; for n ∈ N \ J, set xn = yn = e. Then xn , yn ∈ Dn (U ) for every n ∈ N and hxnn ynn in∈N → e as n → ∞, while xn yn ∈ / Dn (V ) for n ∈ J. We know from (d) that γn = sup{|fn (ynj x) − fn (x−j n x)| : j ≤ n, x ∈ X} → 0 as n → ∞; for future reference, take a sequence hj(n)in≥1 such that 1 ≤ j(n) ≤ n for every n ≥ 1, limn→∞ j(n)/n = 0 and limn→∞

nγn j(n)

= 0.

(e) Let l ≥ 1 be such that Dl (U03 ) ⊆ V (446Eb again), and set K = U02l+1 . Then Dnl (U03 ) ⊆ Dn (V ) for every n (446D(b-vii)). So xn yn ∈ / Dnl (U03 ) for n ∈ J. For each n ∈ J choose m(n) ≤ nl such that m(n) 3 (xn yn ) ∈ / U0 ; for n ∈ N \ J, set m(n) = 1. Then (xn yn )−m(n) x ∈ / U02 for any x ∈ U0 , n ∈ J, and −m(n) fn ((xn yn ) x)f (x) = 0 for any x ∈ X, n ∈ J. So

R

| (fn ((xn yn )−m(n) x) − fn (x))f (x)dx| =

R

fn (x)f (x)dx ≥ β

for n ∈ J, because fn ≥ f ((b-i) above). (f ) Of course m(n) > 0 for every n ∈ J, therefore for every n. So we can set 1 Pm(n)−1 f ((xn yn )i x) gn (x) = i=0 m(n) R R for x ∈ X, n ∈ N. Note that gn , like f , is non-negative, and also that gn (x)dx = f (x)dx = 1. We need to know that gn (x) = 0 if x ∈ / K; this is because (xn yn )i ∈ Dn (U )2nl ⊆ U02l for every i ≤ m(n), while f (x) = 0 if x ∈ / U0 , and U0 is symmetric. We also have |gn (ax) − gn (x)| ≤ sup{|f (wax) − f (wx)| : w ∈ U02l } for every n ∈ N, a, x ∈ X (cf. 446Fc), so for every η > 0 there must be a neighbourhood W of e such that |gn (ax) − g(x)| ≤ η whenever n ∈ N, a ∈ W and x ∈ X, by 4A5Pb. Since also gn (ax) = gn (x) = 0 if a ∈ U0 and x ∈ / U0R−1 K, and U0−1 K has finite measure, we see that for every η > 0 there is a neighbourhood W of e such that |gn (ax) − g(x)|dx ≤ η for every a ∈ W , n ∈ N. (g) Returning to the formula in (e), we see that, for any n ∈ J, ¯ ¯

Z

¡

¯ ¢ fn ((xn yn )−m(n) x) − fn (x) f (x)dx¯ X ¯m(n)−1 =¯

Z

¯ ¡ ¢ fn ((xn yn )−i−1 x) − fn ((xn yn )−i x) f (x)dx¯

i=0 m(n)−1 Z

¯ X =¯ i=0

¯ = m(n)¯ ¯ ≤ ln¯ ¯ ≤ ln¯

Z Z

Z

¯ ¡ ¢ fn ((xn yn )−1 x) − fn (x) f ((xn yn )i x)dx¯

¯ ¡ ¢ fn ((xn yn )−1 x) − fn (x) gn (x)dx¯

¡

¯ ¢ ¯ fn (yn−1 x−1 n x) − fn (x) gn (x)dx

¡

¯ ¢ −1 ¯ fn (yn−1 x) − fn (x) − fn (yn−1 x−1 n x) + fn (xn x) gn (x)dx Z ¯ ¡ ¯ ¢ −1 ¯ + ln¯ 2fn (x) − fn (x−1 n x) − fn (yn x) gn (x)dx .

366

Topological groups

*446K

Next, ¡ ¢ −1 j(n) 2fn (x) − fn (x−1 n x) − fn (yn x) ¡ ¢ −j(n) = j(n) fn (x) − fn (x−1 x) n x) − fn (x) + fn (xn ¡ ¢ −1 + j(n) fn (x) − fn (yn x) − fn (x) + fn (yn−j(n) x) + 2fn (x) − fn (x−j(n) x) − fn (yn−j(n) x) n for every x, and finally Z ¡ ¢ 2fn (x) − fn (x−j(n) x) − fn (yn−j(n) x) gn (x)dx n Z Z ¡ ¢ ¡ ¢ −j(n) = fn (x) − fn (yn x) gn (x)dx − fn (ynj(n) x) − fn (x) gn (x)dx Z ¢ ¡ + fn (ynj(n) x) − fn (x−j(n) x) gn (x)dx n Z Z ¡ ¢ ¡ ¢ j(n) j(n) = fn (yn x) − fn (x) gn (yn x)dx − fn (ynj(n) x) − f (x) gn (x)dx Z ¡ ¢ + fn (ynj(n) x) − fn (x−j(n) x) gn (x)dx n Z ¡ ¢¡ ¢ = fn (ynj(n) x) − fn (x) gn (ynj(n) x) − gn (x) dx Z ¢ ¡ x) gn (x)dx. + fn (ynj(n) x) − fn (x−j(n) n So if we write β1n = ln β2n =

ln j(n)

β3n =

ln j(n)

R¡

¢

−1 fn (yn−1 x) − fn (x) − fn (yn−1 x−1 n x) + fn (xn x) gn (x)dx,

R¡

R¡

x) gn (x)dx,

−j(n)

x) gn (x)dx,

j(n)(fn (x) − fn (yn−1 x)) − fn (x) + fn (yn

β4n =

ln j(n)

β5n =

R¡

j(n)

fn (yn

ln j(n)

R¡

¢¡

x) − fn (x) j(n)

fn (yn

j(n)

gn (yn

−j(n)

x) − fn (xn

¢

−j(n)

j(n)(fn (x) − fn (x−1 n x)) − fn (x) + fn (xn

¢

¢

x) − gn (x) dx,

¢

x) gn (x)dx

for n ≥ 1, we have |β1n | + |β2n | + |β3n | + |β4n | + |β5n | ≥ β for every n ∈ J. (h) Now β1n → 0 as n → ∞. P P Z ¯ ¡ ¯ ¢ −1 ¯ fn (yn−1 x) − fn (x) − fn (yn−1 x−1 ¯ n x) + fn (xn x) gn (x)dx ¯ =¯ ¯ =¯ ¯ =¯

Z Z Z

¡ ¢ fn (yn−1 x) − fn (x) gn (x)dx − ¡ ¢ fn (yn−1 x) − fn (x) gn (x)dx −

Z Z

¯ ¡ ¢ −1 ¯ fn (yn−1 x−1 n x) − fn (xn x) gn (x)dx ¯ ¡ ¢ fn (yn−1 x) − fn (x) gn (xn x)dx¯

¡ ¢¡ ¢ ¯ fn (yn−1 x) − fn (x) gn (x) − gn (xn x) dx¯ Z α |gn (x) − gn (xn x)|dx ≤ n

*446K


by (b-iii). So |β1n | ≤ αl

R

367

|gn (x) − gn (xn x)|dx → 0

as n → ∞, by (f), since surely hxn in∈N → e. Q Q (i) Now look at β2n . If we set

R

γn0 = sup{ |gn (ax) − g(x)|dx : a ∈ Dn (U )j(n) },

R then γn0 → 0 as n → ∞. P P Given ² > 0, there is a neighbourhood W of e such that |gn (ax) − gn (x)|dx ≤ ² whenever n ≥ 1 and a ∈ W , as noted at the end of (f). Let p be such that Dp (U0 ) ⊆ W . Then, for all )pj(n) ⊆ U0 and Dn (U )j(n) ⊆ Dp (U0 ) ⊆ W (446D(b-v)) and R n large enough, pj(n) ≤ n, so that Dn (Uj(n) |gn (ax) − g(x)|dx ≤ ² for every a ∈ Dn (U ) .Q Q We have Z ¯ ¡ ¯ ¢ −j(n) ¯ j(n)(fn (x) − fn (x−1 x) gn (x)dx¯ n x)) − fn (x) + fn (xn X ¯j(n)−1 =¯

Z

¯ ¡ ¢ −i −i−1 fn (x) − fn (x−1 x) gn (x)dx¯ n x) − fn (xn x) + fn (xn

i=0 j(n)−1 Z

¯ X =¯

¢ ¡ fn (x) − fn (x−1 n x) gn (x)dx −

Z

¯ ¢ ¡ −i−1 x) gn (x)dx¯ fn (x−i n x) − fn (xn

i=0 j(n)−1 Z

¯ X =¯

¡ ¢ fn (x) − fn (x−1 n x) gn (x)dx −

Z

¯ ¡ ¢ i ¯ fn (x) − fn (x−1 n x) gn (xn x)dx

i=0 j(n)−1 Z

¯ X =¯

¡ ¢¡ ¢ i fn (x) − fn (x−1 n x) gn (x) − gn (xn x) dx|

i=0 j(n)−1 Z

≤

X

α n

i 0 |fn (x) − fn (x−1 n x)||gn (x) − gn (xn x)|dx ≤ j(n) γn .

i=0

So |β2n | ≤ lαγn0 → 0 as n → ∞. Similarly, hβ3n in≥1 → 0. (j) As for β4n , we have

R

j(n)

|fn (yn

j(n)

x) − fn (x)||gn (yn

x) − gn (x)|dx ≤

αj(n) 0 γn , n

putting the last remark in (c) and the definition of γn0 together. So |β4n | ≤ lαγn0 → 0 as n → ∞. (k) We come at last to β5n . Here, for every n ≥ 1, Z Z ¯ ¡ ¯ ¯ ¯ ¢ ¯ fn (ynj(n) x) − fn (xn−j(n) x) gn (x)dx¯ ≤ ¯fn (ynj(n) x) − fn (x−j(n) x)¯gn (x)dx n Z ≤ γn gn (x)dx = γn by the definition of γn in (c) above. So |β5n | ≤ l by the choice of the j(n).

n γn j(n)

→0

368

Topological groups

*446K

P5 (l) Thus βin → 0 as n → ∞ for every i. But this is impossible, because 0 < β ≤ i=1 |βin | for every n ∈ J. X X This contradiction shows that we must be able to find a neighbourhood W of e and an n0 ∈ N such that xy ∈ Dn (V ) whenever n ≥ n0 , x, y ∈ Dn (U ) and xn y n ∈ W ; as V is arbitrary, U has the property required. *446L Definition Let X be a topological group. A B-sequence in X is a non-increasing sequence hVn in∈N of closed neighbourhoods of the identity, constituting a base of neighbourhoods of the identity, such that there is some M such that for every n ∈ N the set Vn Vn−1 can be covered by at most M left translates of Vn . *446M Proposition Let X be a locally compact Hausdorff topological group with no small subgroups. Then it has a B-sequence. proof (a) For A ⊆ X set Dn (A) = {x : xi ∈ A for every i ≤ n}. We know from 446K that there is a compact symmetric neighbourhood U of the identity e such that whenever V is a neighbourhood of e there are an n0 ∈ N and a neighbourhood W of e such that whenever n ≥ n0 , x ∈ Dn (U ), y ∈ Dn (U ) and xn y n ∈ W , then xy ∈ Dn (V ). Shrinking U if necessary, we may suppose that U includes no subgroup of X other than e, so that there is an r ≥ 1 such that Drn (U )n ⊆ U for every n ∈ N (446H). Let V be a closed symmetric neighbourhood of e such that V 2r ⊆ U . Then Dn (V )2 ⊆ Dn (U ) for every n ∈ N. P P Dn (V ) ⊆ D2rn (U ), by 446D(b-v), so (Dn (V )2 )n ⊆ D2rn (U )2n ⊆ U , and Dn (V )2 ⊆ Dn (U ) (446D(b-v) again). Q Q Take n0 ∈ N and a neighbourhood W of e such that whenever n ≥ n0 , x, y ∈ Dn (U ) and xn y n ∈ W −1 W , then xy ∈ Dn (V ). (b) Let M be so large that U can be covered by M left translates of W . Then for any n ≥ n0 , Dn (V )Dn (V )−1 = Dn (V )2 can be covered S by M left translates of Dn (V ). P P Let z0 , . . . , zM −1 be such that U ⊆ i≤M zi W . For each i < M , set Ai = {x : x ∈ Dn (U ), xn ∈ zi W }; if Ai 6= ∅ choose xi ∈ Ai ; otherwise, set xi = e. For any y ∈ Dn (V )2 , y ∈ Dn (U ), so y n ∈ U and there is some i < M such that y ∈ Ai . In this case xi also n −1 belongs to Ai . Now zi−1 y n and zi−1 xni both belong to W , so x−n W , and S x−1 i y belongs to W i y ∈ Dn (V ), 2 by the choice of W and n0 . But this means that y ∈ xi Dn (V ). As y is arbitrary, Dn (V ) ⊆ i
kT −Ik 1−kT −Ik

≤1

*446O


369

(c) Now let V be a compact neighbourhood of the identity e of X. Let V1 be a neighbourhood of e such that (V1 V1−1 )−1 V1 V1−1 ⊆ V . For δ > 0, set Uδ = {x : x ∈ V , kφ(x) − Ik ≤ δ}. Then each Uδ is a compact neighbourhood of e, because φ is continuous. Also T δ>0 Uδ = {x : x ∈ V, φ(x) = I} = {e}. So {Uδ : δ > 0} is a base of neighbourhoods of e (4A2Gd), and there is a δ1 > 0 such that Uδ1 ⊆ V1 ; of course we may suppose that δ1 ≤ 81 . (d) If δ ≤ 12 and x ∈ Uδ Uδ−1 , then kφ(x) − Ik ≤ 4δ. P P Express x as yz −1 where y, z ∈ Uδ . Then 1 −1 kφ(z) − Ik ≤ 2 , so kφ(z )k ≤ 2 and kφ(x) − Ik = k(φ(y) − φ(z))φ(z −1 )k ≤ 2kφ(y) − φ(z)k ≤ 4δ. Q Q 2

(e) Now if δ ≤ δ1 , Uδ Uδ−1 can be covered by at most m = 17r left translates of Uδ . P P?? Suppose, if possible, otherwise. Then we can choose x0 , . . . , xm ∈ Uδ Uδ−1 such that xj ∈ / xi Uδ whenever i < j ≤ m. If i < j ≤ m, then −1 −1 x−1 Uδ Uδ−1 ) ⊆ (V1 V1−1 )−1 V1 V1−1 ⊆ V , i xj ∈ (Uδ Uδ )

and x−1 / Uδ , so kφ(x−1 i xj ∈ i xj ) − Ik > δ. Set Ti = φ(xi ) for each i ≤ m; then kTi − Ik ≤ 4δ ≤

1 2

for each i, by (d), while −1 −1 δ < kφ(x−1 i xj ) − Ik = kTi Tj − Ik ≤ kTi kkTj − Ti k ≤ 2kTj − Ti k

whenever i < j ≤ m. Write Bi = B(Ti , 4δ ) for each i; then all the Bi are disjoint. But also they are all included in B(I, 17δ 4 ), so we have ¡ δ ¢r 2 ¡ 17δ ¢r2 2 (17r + 1) γ≤ γ, 4

4

which is impossible. X XQ Q (f ) Accordingly, setting Wn = U2−n δ1 , hWn in∈N is a B-sequence in X. *446O Theorem Let X be a locally compact Hausdorff topological group. Then it has an open subgroup Y which has a compact normal subgroup Z such that Y /Z has no small subgroups. proof (a) Let U be a compact neighbourhood of the identity e of X. Then there is a subgroup Y0 of X, included in U , and a neighbourhood W0 of e such that every subgroup of X included in W0 is also included in Y0 . P P(i) To begin with (down to the end of (iii)) let us suppose that X is metrizable. Let hVn in∈N be a non-increasing sequence of symmetric neighbourhoods of e running over a base of neighbourhoods of e, and such that V12 ⊆ V0 ⊆ U . For each n ∈ N, set An = {x : xi ∈ Vn for every i ∈ N}. S (ii)?? Suppose, if possible, that k∈N Akn 6⊆ U for every n ∈ N. For each n ∈ N let k(n) ∈ N be such k(n) k(n)+1 that An ⊆ U , An 6⊆ U . Then hAn in∈N and U satisfy the conditions of 446G (because Am ⊆ Vn whenever m ≥ n, and {Vn : n ∈ N} is a base of neighbourhoods of e). Let F be a non-principal ultrafilter k(n) on N and set Q = lim supn→F An . IfSW is any neighbourhood of e, there is an n ∈ N such that Vn ⊆ W , so that xi ∈ W whenever x ∈ m≥n Am , i ∈ N. Thus (ii) of 446G is not true, and Q must be a closed subgroup of X included in U and meeting the boundary of U . By 446C, there are an r ∈ N and a continuous homomorphism φ : Q → GL(r, R) such that the kernel Z of φ is included in int V1 . Let G ⊆ X be an open set including Z and with closure disjoint from (X \ int V1 ) ∪ {x : x ∈ Q, kφ(x) − Ik ≥ 61 }; such can be found because Z is compact and Hausdorff, therefore normal (4A2Gb). Then Z ⊆ G, and any subgroup Z 0 of Q included in G has kφ(x)i − Ik ≤ 61 for every x ∈ Z 0 and i ∈ N, so that Z 0 ⊆ Z, by 4A6N. Set V = G.

370

Topological groups k(n)+1

*446O j(n)

j(n)+1

6⊆ U for every n, we can find j(n) ≤ k(n) such that An ⊆ V , An 6⊆ V Since V ⊆ U and An j(n) 0 0 0 for every n. Set Q = lim supn→F An . As before, (ii) of 446G cannot be true of Q , and Q must be a j(n) k(n) closed subgroup of X meeting the boundary of V . Because e ∈ An , An ⊆ An for every n, and Q0 ⊆ Q 0 0 (4A2Ta); also Q ⊆ V , so Q ⊆ Z. But Z does not meet the boundary of V . X X S k k (iii) So there is some n ∈ N such that An ⊆ U for every k ∈ N. Because A−1 n = An , Y0 = k∈N An is a subgroup of X. Any subgroup of X included in Vn is a subset of An so is included in Y . Thus we have a pair Y0 , W0 = Vn of the kind required, at least when X is metrizable. (iv) Now suppose that X is σ-compact. Let U1 be a neighbourhood of e such that U12 ⊆ U . Then there is a closed normal subgroup X0 of X such that X0 ⊆ U1 and X 0 = X/X0 is metrizable (4A5S). By (i)-(iii), there are a subgroup Y00 of X 0 , included in the image of U1 in X 0 , and a neighbourhood W00 of the identity in X 0 such that any subgroup of X 0 included in W00 must also be included in Y00 . Write π : X → X 0 for the canonical homomorphism and consider Y0 = π −1 [Y00 ], W0 = π −1 [W00 ]. Then W0 is a neighbourhood of e and Y0 is a subgroup of X included in π −1 [π[U1 ]] = U1 X0 ⊆ U12 ⊆ U . And if Z is any subgroup of X included in W0 , then π[Z] ⊆ W00 so π[Z] ⊆ Y00 and Z ⊆ Y0 . Thus in this case also we have the result. (v) Finally, for the general case, observe that X has a σ-compact open subgroup X1 (4A5El). So we can find a subgroup Y0 of X1 , included in U ∩ X1 , and a neighbourhood W0 of the identity in X1 such that any subgroup of X1 included in W0 is also included in Y0 . But of course Y0 and W0 also serve for X and U . This completes the proof of (a). Q Q (b) By 446C, there is a finite-dimensional representation φ : Y0 → GL(r, R), for some r ∈ N, such that the kernel Z of φ is included in int W0 . Let W1 be a neighbourhood of e in X such that kφ(x) − Ik ≤ 61 for every x ∈ W1 ∩ Y0 , and set W = W1 Z ∩ W0 . Note that if x ∈ W ∩ Y0 , there is a z ∈ Z such that xz ∈ W1 ∩ Y0 , so that kφ(x) − Ik = kφ(xz) − Ik ≤ 16 . Of course Z ⊆ int W1 Z, so Z ⊆ int W . If Y 0 is a subgroup of Y0 included in W , then kφ(x)i − Ik ≤ 16 for every i ∈ N and x ∈ Y 0 , so Y 0 ⊆ Z. Consequently any subgroup of X included in W is a subgroup of Z, since by the choice of Y0 and W0 it is a subgroup of Y0 . Now let Y be the normalizer of Z in X. Z is compact, so G = {x : xZx−1 ⊆ int W } is open, and contains e; but also G ⊆ Y , because if x ∈ G then xZx−1 is a subgroup of X included in W , and must be included in Z. Accordingly Y = GY is open. Since any subgroup of Y included in W is a subgroup of Z, we see that any subgroup of Y /Z included in the image of W is the trivial subgroup, and Y /Z has no small subgroups, as required. *446P Corollary Let X be a locally compact Hausdorff topological group. Then it has a chain hXξ iξ≤κ of closed subgroups, where κ is an infinite cardinal, such that (i) X0 is open, (ii) Xξ+1 is a normal subgroup of Xξ for every ξ < κ, (iii) Xξ is T compact for ξ ≥ 1, (iv) Xξ = η<ξ Xη for non-zero limit ordinals ξ ≤ κ, (v) Xξ /Xξ+1 has a B-sequence for every ξ < κ, (vi) Xκ = {e}, where e is the identity of X. proof (a) By 446O, X has an open subgroup X0 with a compact normal subgroup X1 such that X0 /X1 has no small subgroups. By 446M, X0 /X1 has a B-sequence. (b) Let Φ be the set of finite-dimensional representations of X1 ; if we distinguish the trivial homomorphisms from X1 to each GL(r, R), Φ is infinite. Set κ = #(Φ) and let hφξ i1≤ξ<κ run over Φ. For 1 ≤ ξ ≤ κ, set Xξ = {x : x ∈ X1 , φη (x) = I for 1 ≤ η < ξ}. Then hXξ iξ≤κ satisfies conditions (i)-(iv). As for (v), I have already checked the case ξ = 0, and if 1 ≤ ξ < κ, then φξ ¹Xξ is a finite-dimensional representation of Xξ with kernel Xξ+1 , so Xξ /Xξ+1 has a faithful finitedimensional representation (446Ab), and therefore has a B-sequence (446N).

446 Notes


371

Finally, Xκ = {e} by 446B; if x ∈ X1 and x 6= e, there is a φ ∈ Φ such that φ(x) 6= φ(e), so that x ∈ / Xκ . 446X Basic exercises (a) Show that the constructions of 446B-446C produce representations which are homomorphisms from X into orthogonal groups. (b) Let X be a locally compact Hausdorff abelian topological group. Show that for every element a of X, other than the identity, there is a finite-dimensional representation φ of X such that φ(a) 6= I. (Hint: 445O.) (c) Q Let hXn in∈N be any sequence of groups with their discrete topologies, and X the product topological group n∈N Xn . Show that X has a B-sequence. (Hint: set Vn = {x : x(i) = e(i) for i < n}.) 446Y Further exercises (a) Let X be the countable group of all bijections from N to itself which are products of an even number of transpositions. Give X the discrete topology, so that it is a locally compact topological group. Show that any finite-dimensional representation of X is trivial. (Hint: X is simple.) (b) Let X be a compact Hausdorff topological group, f ∈ C(X) and ² > 0. Show that there are a finite-dimensional representation φ : X → GL(r, R) and a, b ∈ R r such that |f (x) − (φ(x)(a)|b)| ≤ ² for every x ∈ X. 446 Notes and comments The ideas above are extracted from the structure theory for locally compact groups, as described in Montgomery & Zippin 55. (A brisker and sometimes neater, but less complete, exposition can be found in Kaplansky 71.) The full theory goes very much deeper into the analysis of groups with no small subgroups. One of the most important ideas, hidden away in 446I and part (c) of the proof of 446K, is that of ‘one-parameter subgroup’; if X is a group with no small subgroups, there are enough continuous homomorphisms from R to X not only to provide a great deal of information on the topological group structure of X, but even to set up a differential structure (Kaplansky 71, §II.3). For our purposes here, however, all we need to know is that groups with no small subgroups have ‘B-sequences’ (446L-446M), which can form the basis of a theory corresponding to Vitali’s theorem and Lebesgue’s Density Theorem in R r (447C-447D below). There are four essential elements in the argument here. Working from the outside, the first step is 446O: starting from a locally compact Hausdorff group X, we can find an open subgroup X0 of X and a compact normal subgroup X1 of X0 such that X0 /X1 has no small subgroups. This depends on a subtle argument based on the first key lemma, the dichotomy in 446G, which in turn uses the ‘smoothing’ construction in 446F and a careful analysis of inequalities involving integrals. (Naturally enough, the translation-invariance of the Haar integral is a leitmotiv of this investigation.) Note the remarkable transition in 446H. The sets Dn (U ) are defined solely in terms of powers, while the sets Dn (U )n involve products. We are able to obtain information about products x1 . . . xn from information about the powers xij for i, j ≤ n. Next, we need to find a chain of closed subnormal subgroups of X1 , decreasing to {e}, such that the quotients all have faithful finite-dimensional representations (in this context, this means that they are isomorphic to compact subgroups of GL(r, R)). This step depends on the older ideas in 446B-446C, where we use the theory of compact operators on Hilbert spaces to show that a compact group has many representations as actions on finite-dimensional subspaces of its L2 space. (Observe that in this section I revert to real-valued, rather than complex-valued, functions.) This can be thought of as a development of the result of 445O. If X is a locally compact abelian group, its characters separate its points (cf. 446Xb); if X is compact but not necessarily abelian, its finite-dimensional representations separate its points. (But if X is neither compact nor abelian, there are further difficulties; see 446Ya.) The other two necessary facts are that both groups with no small subgroups, and groups with faithful finite-dimensional representations, have B-sequences. The latter is reasonably straightforward (446N); any 2 complications are due entirely to the fact that the natural measure on GL(r, R), inherited from R r , is not quite invariant under multiplication, so we have to manipulate some inequalities. For groups with no small subgroups (446M) we have much more to do. The proof I give here depends on a second key lemma, 446K, refining the methods of 446G; a slightly stronger version of this result is the basis of the analysis of one-parameter subgroups in the general theory (compare Montgomery & Zippin 55, §3.8).

372

Topological groups

§447 intro.

447 Translation-invariant liftings I devote a section to the main theorem of Ionescu Tulcea & Ionescu Tulcea 67: a group carrying Haar measures has a translation-invariant lifting (447J). The argument uses an inductive construction of the same type as that used in §341 for the ordinary Lifting Theorem. It depends on the structure theory for locally compact groups described in §446. On the way I describe a Vitali theorem for certain metrizable groups (447C), with a corresponding density theorem (447D). 447A Liftings and lower densities Let X be a group carrying Haar measures, Σ its algebra of Haar measurable sets and A its Haar measure algebra (442H, 443A). (a) Recall that a lifting of A is either a Boolean homomorphism θ : A → Σ such that (θa)• = a for every a ∈ A, or a Boolean homomorphism φ : Σ → Σ such that E4φE is Haar negligible for every E ∈ Σ and φE = ∅ whenever E is Haar negligible (341A). Such a lifting θ or φ is left-translation-invariant if θ((xE)• ) = x(θE • ) or φ(xE) = x(φE) for every E ∈ Σ and x ∈ X. (In the notation of 443C, a lifting θ : A → Σ is left-translation-invariant if θ(x•l a) = xθa for every x ∈ X, a ∈ A.) The language of 341A demanded a named measure; I spoke there of a lifting of a measure space (X, Σ, µ) or of a measure µ. But (as noted in 341Lh) what the concept really depends on is a triple (X, Σ, I), where Σ is an algebra of subsets of X and I is an ideal of Σ. Variations in the measure which do not affect the algebra of measurable sets or the null ideal are irrelevant. So, in the present context, we can speak of a ‘lifting of Haar measure’ without declaring which Haar measure we are using, nor even whether it is a left or right Haar measure. (b) Now suppose that Σ0 is a σ-subalgebra of Σ. In this case, a partial lower density on Σ0 is a function φ : Σ0 → Σ such that φE = φF whenever E, F ∈ Σ0 and E4F is negligible, E4φE is negligible for every E ∈ Σ0 , φ∅ = ∅ and φ(E ∩ F ) = φE ∩ φF for all E, F ∈ Σ0 . (See 341C-341D.) As in (a), such a function is left-translation-invariant if xE ∈ Σ0 and φ(xE) = x(φE) for every x ∈ X, E ∈ Σ0 . 447B Lemma Let X be a group carrying Haar measures and Y a subgroup of X. Write ΣY for the algebra of Haar measurable subsets E of X such that EY = E, and suppose that φ : ΣY → ΣY is a lefttranslation-invariant partial lower density. Then G ⊆ φ(GY ) for every open set G ⊆ X. proof Of course GY is open (4A5Ed), so belongs to ΣY . Let a ∈ G and let U be an open neighbourhood of the identity in X such that U −1 U a ⊆ G. Then U aY is a non-empty open set, therefore not negligible (442Aa), and there is an x ∈ U aY ∩ φ(U aY ). Express x as uay where u ∈ U and y ∈ Y ; then ua = xy −1 ∈ U a ∩ φ(U aY ), because φ(U aY ) ∈ ΣY . So a = u−1 ua ∈ u−1 φ(U aY ) = φ(u−1 U aY ) ⊆ φ(GY ). As a is arbitrary, G ⊆ φ(GY ). 447C Vitali’s Theorem Let X be a topological group with a left Haar measure µ, and hVn in∈N a B-sequence in X (definition: 446L). If A ⊆ X is any S set and Kx is an infinite subset of N for every x ∈ A, then there is a disjoint family V of sets such that A \ V is negligible and every member of V is of the form xVn for some x ∈ A, n ∈ Kx . proof (a) There is surely some r such that Vr is totally bounded for the bilateral uniformity on X (443H); replacing Vi by Vr for i < r and Kx by Kx \ r for each x, we may suppose that V0 is totally bounded. (b) Choose hIn in∈N inductively, as follows. Given Ij ⊆ X for j < n, choose a set In ⊆ A which is maximal subject to the conditions n ∈ Kx for every x ∈ In , xVn ∩ yVj = ∅ whenever x ∈ In , j < n and y ∈ Ij , xVn ∩ yVn = ∅ whenever x, y ∈ In are distinct. On completing the induction, set V = {xVn : n ∈ N, x ∈ In }; this is a disjoint family.

447D

Translation-invariant liftings

373

S S (c) ?? Suppose, if possible, that A \ V is not negligible. By 415B, the subspace measure on A \ V is τ -additive and has a non-empty support.STake any a belonging to this support and set G = int(aV0 ); then G is totally bounded and δ = µ∗ (G ∩ A \ V) is non-zero. Let M be such that every Vn Vn−1 can be covered by M left translates of Vn , so that µ(Vn Vn−1 ) ≤ M µVn for every n. Set Jn = In ∩ GV0 V0−1 ,

En = Jn Vn ,

ñ = En Vn−1 E

for each n. If n ∈ N and x ∈ Jn , then xVn ⊆ GV0 V0−1 V0 . Accordingly (because hxVn ix∈Jn is disjoint) P #(Jn )µVn = x∈Jn µ(xVn ) ≤ µ(GV0 V0−1 V0 ) < ∞ because GV0 V0−1 V0 is totally bounded (4A5Ob). So Jn is finite and Note that if x ∈ In and S En is closed. S G ∩ xVn 6= ∅, then x ∈ GVn−1 ⊆ GV0 V0−1 and x ∈ Jn ; so G ∩ V = G ∩ n∈N En . Also, hEn in∈N is a P∞ −1 disjoint P∞ sequence of subsets of GV0 V0 V0 ; accordingly n=0 µEn is finite, and there is an m ∈ N such that M n=m µEn < δ. Observe next that, for any x ∈ X and n ∈ N, µ(xVn Vn−1 ) = µ(Vn Vn−1 ) ≤ M µVn = M µ(xVn ). So ñ ≤ µE

P

µ(xVn Vn−1 ) ≤ M

P

x∈Jn µ(xVn ) = M µEn S ñ < δ. for each n, and µ( n≥m )E S ñ cannot include A ∩ G \ S V, and there is a z belonging to This means that n≥m E [ [ [ [ ñ ) = A ∩ G \ ( ñ ) A∩G\( V ∪ E En ∪ E x∈Jn

n≥m

n≥m

n∈N

=A∩G\(

[

En ∪

n
[

ñ ). E

n≥m

S Now there must be a first k ≥ m such that k ∈ Kz and zVk ⊆ G \ n
for almost every x ∈ X. proof (a) Let α < 1, and set A = {x : x ∈ E, lim inf n→∞

µ(E ∩ xVn ) < α}. µVn

?? Suppose, if possible, that A is not negligible. Then there is an open set G of finite measure such that µ∗ (G ∩ A) = γ > 0. Let δ > 0 be such that γ > α(γ + δ) + δ. Take a Borel set F which is a measurable envelope of G ∩ A and a closed set F1 ⊆ G \ F such that µF1 ≥ µ(G \ F ) − δ. Writing H = G \ F1 , we see that H ∩ A = G ∩ A and µH ≤ µ∗ (H ∩ A) + δ = γ + δ. For each x ∈ H ∩ A, set

374

Topological groups

447D

Kx = {n : xVn ⊆ H, µ(E ∩ xVn ) ≤ αµVn }. Then Kx is infinite. By Vitali’s theorem in the form 447C, there is a disjoint family V ⊆ {xVn : x ∈ S H ∩ A, n ∈ Kx } such that (H ∩ A) \ SV is negligible. Since every member of V has non-zero measure, while S µH is finite, V is countable. Now µ( V) ≥ µ∗ (H ∩ A), so µ(H \ V) ≤ δ; also, because µ(E ∩ V ) ≤ αµV for every V ∈ V, and V is disjoint, S S µ(E ∩ V) ≤ αµ( V) ≤ αµH and γ = µ∗ (A ∩ H) ≤ µ(E ∩ H) ≤ αµH + δ ≤ α(γ + δ) + δ, which is impossible, by the choice of δ. X X (b) As α is arbitrary, lim inf n→∞

µ(E ∩ xVn ) =1 µVn

for almost every x ∈ E. Similarly, lim inf n→∞

µ(xVn \ E) = 1, µVn

lim sup n→∞

µ(E ∩ xVn ) =0 µVn

for almost every x ∈ X \ E, so µ(E ∩ xVn ) = χE(x) n→∞ µVn lim

for almost every x ∈ X. 447E We need to recall some results from 443O-443Q. Let X be a locally compact Hausdorff topological group, and Y a closed subgroup of X such that the modular function of Y is the restriction to Y of the modular function of X. Let µ be a left Haar measure on X and µY a left Haar measure on Y . (a) Writing Ck (X) for the space of continuous real-valued functions on X with compact support, and X/Y for the set of left cosets of Y in X with R the quotient topology, we have a linear operator T : Ck (X) → Ck (X/Y ) defined by writing (T f )(x• ) = Y f (xy)µY (dy) for every x ∈ X, f ∈ Ck (X) (443O); Rmoreover, + + T R [Ck (X) ] = Ck (X/Y ) (443Oa), and we have an invariant Radon measure λ on X/Y such that T f dλ = f dµ for every f ∈ Ck (X) (see part (b) of the proof of 443Q). If Y is a normal subgroup of X, so that X/Y is the quotient group, λ is a left Haar measure. If Y is compact and µY is the Haar probability measure on Y , then λ is the image measure µπ −1 , where π(x) = x• = xY for every x ∈ X (443Pd). ˜ = {x• : x ∈ E} belongs to the domain of λ, (b) If E ⊆ X and EY = Y , then E is Haar measurable iff E ˜ is λ-negligible (443Pc). and E is Haar negligible iff E R (c) Now suppose that X is σ-compact. Then for any Haar measurable E ⊆ X, µE = g dλ in [0, ∞], where g(x• ) = µY (Y ∩ x−1 E) is defined for almost every x ∈ X (443Pe). In particular, E is Haar negligible iff µY (Y ∩ x−1 E) = 0 for almost every x ∈ X. (d) Again suppose that X is σ-compact. Then we can extend the operator T of part (a) to an operator R • 1 from L1 (µ) to LR1 (λ) by writing (T f )(x ) = f (xy)µ Y (dy) whenever f ∈ L (µ), x ∈ X and the integral R 1 1 is defined, and T f dλ = f dµ for every f ∈ L (µ) (443Pe). If f ∈ L (µ), and we set fx (y) = f (xy) for all those x ∈ X, y ∈ Y for which xy ∈ dom f , then Q = {x : fx ∈ L1 (µY )} is µ-conegligible, and x 7→ fx• : Q → L1 (µY ) is almost continuous (443Pf). (e) If X is σ-compact, Y is compact and µY is the Haar probability measure on Y , so that λ is the image measure µπ −1 , then we can apply 235I to the formula in (d) to see that

RR

f (xy)µY (dy)µ(dx) =

R

(T f )(x• )µ(dx) =

R

T f dλ =

R

f dµ

R for every µ-integrable function f , and therefore (because µ is σ-finite) for every function f such that f dµ R is defined in [−∞, ∞]. In particular, µE = ν(Y ∩ x−1 E)µ(dx) for every Haar measurable set E ⊆ X.

447F


375

447F Lemma Let X be a σ-compact locally compact Hausdorff topological group and Y a closed subgroup of X such that the modular function of Y is the restriction to Y of the modular function of X. Let Z be a compact normal subgroup of Y such that the quotient group Y /Z has a B-sequence. Let ΣY be the σ-algebra of those Haar measurable subsets E of X such that EY = E, and ΣZ the algebra of Haar measurable sets E ⊆ X such that EZ = E. Let φ : ΣY → ΣY be a left-translation-invariant partial lower density. Then there is a left-translation-invariant partial lower density ψ : ΣZ → ΣZ extending φ. proof (a) Let µ be a left Haar measure on X, µY a left Haar measure on Y and µRZ the Haar probability R measure on Z; then there is a left Haar measure ν on Y /Z such that g(y)µ Y (dy) = (T g)(u)ν(du) for every R g ∈ Ck (Y ), where (T g)(y • ) = g(yz)µZ (dz) for every y ∈ Y (447Ea). We are supposing R that Y /Z has a B-sequence hVn in∈N . It follows that there is a sequence hhn in∈N in Ck (Y )+ such that (i) hn (y)µY (dy) = 1 for every n (ii) whenever F ⊆ Y is Haar measurable (regarded as a subset of Y , that is), and F Z = Z, then limn→∞

R

χF (by)hn (y)µY (dy) = χF (b)

for µY -almost every b ∈ Y . P P Since any subsequence of hVn in∈N is a B-sequence, and Y /Z is locally compact, we may suppose that every Vn is compact. For each n ∈ N, choose a non-negative h0n ∈ Ck (Y /Z) such that

R

R

h0n dν = 1,

|h0n −

1 χVn |dν νVn

≤ 2−n .

R (This is possible by 416I, or otherwise.) Let hn ∈ Ck (Y )+ be such that T hn = h0n ; then hn dµY = R 0 hn dν = 1. Now if F ⊆ Y is Haar measurable and F Z = Z, there is a Haar measurable F˜ ⊆ Y /Z such that F = {y : y • ∈ F˜ } (447Eb). Take b ∈ Y and n ∈ N. Because

R

χF (byz)hn (yz)µZ (dz) =

R

χF˜ (b• y • )hn (yz)µZ (dz) = χF˜ (b• y • )h0n (y • )

for every y ∈ Y , Z

ZZ χF (by)hn (y)µY (dy) =

(447Ed)

χF (byz)hn (yz)µZ (dz)µY (dy) Z

=

Z χF˜ (b• y • )h0n (y • )µY (dy) =

χF˜ (b• u)h0n (u)ν(du)

because y 7→ y • is inverse-measure-preserving for µY and ν (447Ee). So Z Z • ˜ |χF (b) − χF (by)hn (y)µY (dy)| = |χF (b ) − χF˜ (b• u)h0n (u)ν(du)| Z 1 ≤ |χF˜ (b• ) − χF˜ (b• u)χVn (u)ν(du)| νVn Z 1 + |h0n (u) − χVn (u)|ν(du) νVn

≤ |χF˜ (b• ) −

ν(F˜ ∩b• Vn ) | + 2−n . νVn

Since {v : v ∈ Y /Z, limn→∞

ν(F˜ ∩vVn ) νVn

6= χF˜ (v)}

R is ν-negligible (447D), its inverse image in Y is µY -negligible, so χF (b) = limn→∞ χF (by)hn (y)µY (dy) for almost every b, as claimed. Q Q R (b) We find now that if E ∈ ΣZ , then limn→∞ χE(xy)hn (y)µY (dy) = χE(x) for µ-almost every x ∈ X. P P Set Ex = Y ∩x−1 E for x ∈ X. Because X is σ-compact, we can express E as the union of a non-decreasing (k) sequence hF (k) ik∈N where each F (k) is Haar measurable and relatively compact; set Fx = Y ∩ x−1 F (k) for (k) (k) each x. In this case, for any k ∈ N, Qk = {x : Fx ∈ dom(µY )} is conegligible, and x 7→ (χFx )• : Qk →

376

Topological groups

447F

R (k) L1 (µY ) is almost continuous (447Ed), so that T x 7→ χFx (y)hn (y)µY (dy) : Qk → [0, 1] is measurable, for each n. But this means that, setting Q = k∈N Qk , x 7→

R

χEx (y)hn (y)µY (dy) = limk→∞

R

(k)

χFx (y)hn (y)µY (dy) : Q 7→ [0, 1] (k)

(k)

is measurable, for every n ∈ N. Note that if x ∈ X and y ∈ Y then Fxy = y −1 Fx , so that Qk Y = Qk for every k ∈ N, and QY = Q. R ˜ = {x : x ∈ Q, limn→∞ χE(xy)hn (y)µY (dy) = χE(x)}. This is a Haar measurable Now consider Q subset of X. If a ∈ Q, then ˜ = {y : y ∈ Y , limn→∞ χE(ays)hn (s)µY (ds) = χE(ay)} Y ∩ a−1 Q ˜ is µ-negligible is µY -conegligible, by the choice of the hn in (a) above. Because Q is µ-conegligible, Q \ Q ˜ is conegligible, as required. Q (447Ec) and Q Q (c) We are now ready for the formulae at the centre of this proof. For any Haar measurable set E ⊆ X, n ∈ N and γ < 1, set [ ψnγ (E) = {G ∩ φF : G ⊆ X is open, F ∈ ΣY , Z χE(xy)hn (y)µY (dy) is defined and at least γ for every x ∈ G ∩ F }, ψE =

T

S γ<1

T n∈N

m≥n

ψmγ (E).

The rest of the proof is devoted to checking that ψ¹ΣZ is a left-translation-invariant partial lower density extending φ. (d) I had better make one remark straight away. If G ⊆ X is open, then G ⊆ φ(GY ) (447B). It follows that if G ⊆ X is open, F ∈ ΣY and G ∩ φF 6= ∅, then G ∩ F 6= ∅. P P If a ∈ G ∩ φF , then −1 a ∈ φ(GY ) ∩ φF = φ(GY ∩ F ), so GY ∩ F 6= ∅, that is, G ∩ F = G ∩ F Y is non-empty. Q Q I mention this now because we need to know that the condition

R

χE(xy)hn (y)µY (dy) is defined and at least γ for every x ∈ G ∩ F

in the definition of ψnγ (E) is never vacuously satisfied if G ∩ φF 6= ∅. In particular, ψnγ (∅) = ∅ whenever n ∈ N and 0 < γ < 1, so that ψ∅ = ∅. (e) If E ⊆ X is Haar measurable, n ∈ N and ² > 0, Rthen for almost every a ∈ X there are an open set G ⊆ X and an F ∈ ΣY such that a ∈ G ∩ φF and |χE(xy) − χE(ay)|hn (y)µY (dy) ≤ ² whenever x ∈ G ∩ F. P P Take δ > 0 such that δ(1 + 2khn k∞ ) ≤ ². Because X is σ-compact, locally compact and Hausdorff, therefore Lindelöf and R regular (4A2Hd, 3A3Bb), there is a sequence hfr ir∈N of continuous functions from X to [0, 1] such that |χE(x) − fr (x)|µ(dx) ≤ 2−r for every r (415Pb). Set gr = |χE − fr | R ˜ by writing (T˜f )(x• ) = f (xy)µY (dy) whenever this is defined for each r. For f ∈ L1 (µ), define T˜f ∈ L1 (λ) R R ˜ = f dµ. (447Ed); we have T˜f dλ Set Q = {x : Y ∩ x−1 E ∈ dom µY }, so that Q ∈ ΣY is conegligible (447Ec). For each r, set F˜r = {u : u ∈ X/Y , T˜gr (u) is defined and at least δ}; then R R ˜ F˜r ≤ 1 T gr dλ ˜ = 1 gr dµ ≤ 2−r /δ. λ δ

δ

T

˜ So r∈N F˜r is T λ-negligible. Set Fr = {x : x ∈ X, x• ∈ F˜r } for each r; then Fr Y = Fr , Fr is Haar measurable (447Eb) and S r∈N Fr is µ-negligible (also by 447Eb). Since (X \ Fr )4φ(X \ Fr ) is negligible for each r, Q1 = Q ∩ r∈N φ(X \ Fr ) \ Fr is conegligible. Note that Q1 Y = Q1 . Suppose that a ∈ Q1 . Then there is an r ∈ N such that a ∈ Q1 ∩ φ(X \ Fr ) \ Fr = (Q1 \ Fr ) ∩ φ(Q1 \ Fr ). R Set F = Q1 \Fr ∈ ΣY . Consider the function x 7→ fr (xy)hn (y)µY (dy). We chose hn with compact support L ⊆ Y say. If V is a compact neighbourhood of a in X, then fr is uniformly continuous on V L for the right uniformity on X (4A5Ha, 4A2Je). There is therefore an open neighbourhood U of the identity of X such

447F


377

that |fr (x0 ) − fr (x)| ≤ δ whenever x, x0 ∈ V L and x0 x−1 ∈ U ; of course we may suppose that G = U a is a subset of V . Take any x ∈ G ∩ F . Then if y ∈ L, we have ay, xy both in V L, while xy(ay)−1 ∈ U , so that |fr (xy) − fr (ay)| ≤ δ. Accordingly |fr (xy) − fr (ay)|hn (y) ≤ δhn (y) for every y ∈ Y , and

R

|fr (ay) − fr (xy)|hn (y)µY (dy) ≤ δ.

At the same time, because both x and a belong to F = Q1 \ Fr ,

R

R Putting these together,

R

|χE(ay) − fr (ay)|hn (y)µY (dy) ≤ δkhn k∞ , |χE(xy) − fr (xy)|hn (y)µY (dy) ≤ δkhn k∞ .

|χE(ay) − χE(xy)|hn (y)µY (dy) ≤ δ(1 + 2khn k∞ ) ≤ ².

Thus G and F witness that a has the property required; as a is any member of the conegligible set Q1 , we have the result. Q Q (f ) If E ∈ ΣZ then E4ψE is negligible. P P By (e), applied in turn to every n and every ² of the form 2 , there is a conegligible set Q1 ⊆ X such that whenever R a ∈ Q1 , n ∈ N and ² > 0 there are an open set G containing a and an F ∈ ΣY such that a ∈ φF and |χE(xy) − RχE(ay)|hn (y)µY (dy) ≤ ² for every x ∈ G ∩ F . By (b), there is a conegligible set Q2 ⊆ X such that limn→∞ χE(ay)hn (y)µY (dy) = χE(a) for every a ∈ Q2 . 1 R Suppose that a ∈ Q1 ∩ Q2 ∩ E. Let γ < 1; set ² = 2 (1 − γ). Because a ∈ Q2 , there is an n ∈ N such that χE(ay)hm (y)µY (dy) ≥ 1 − ² for every m R≥ n. Take any m ≥ n. Because a ∈ Q1 , there are an open set G and an F ∈ ΣY such that a ∈ G ∩ φF and |χE(ay) − χE(xy)|hm (y)µY (dy) ≤ ² whenever x ∈ G ∩ F . But now −i

R

χE(xy)hm (y)µY (dy) ≥ 1 − 2² = γ

for every x ∈ G ∩ F , so a ∈ ψmγ (E). This is true for every m ≥ n; as γ is arbitrary, a ∈ ψE. As a is arbitrary, Q1 ∩ Q2 ∩ E ⊆ ψE. Now suppose that a ∈ Q1 ∩ Q2 ∩ ψE. Then there is an n ∈ N such that a ∈ ψm,3/4 (E) for every m ≥ n. R 1 There is an m ≥ n such that | χE(ay)h m (y)µY (dy) − χE(a)| ≤ 4 . There are an open set G1 and an R 3 F1 ∈ ΣY such that a ∈ G1 ∩ φF1 and χE(xy)hm (y)µY (dy) ≥ 4 for every x ∈ G1 ∩ F1 . There are also an R open set G2 and an F2 ∈ ΣY such that a ∈ G2 ∩ φF2 and |χE(ay) − χE(xy)|hm (y)µY (dy) ≤ 14 for every x ∈ G2 ∩ F2 . Set G = G1 ∩ G2 , F = F1 ∩ F2 ; then a ∈ G ∩ φF , so G ∩ F is not empty ((d) above). Take x ∈ G ∩ F . Then

R

R

|

3 4

χE(xy)hm (y)µY (dy) ≥ , 1 4

|χE(ay) − χE(xy)|hm (y)µY (dy) ≤ ,

R

1 4

χE(ay)hm (y)µY (dy) − χE(a)| ≤ ,

so χE(a) ≥ 41 and a ∈ E. This shows that Q1 ∩ Q2 ∩ ψE ⊆ E. Accordingly E4ψE ⊆ X \ (Q1 ∩ Q2 ) is negligible, as required. Q Q In particular, ψE is Haar measurable for every E ∈ ΣZ . (g) If E ∈ ΣZ then ψE ∈ ΣZ . P P We have just seen that ψE is Haar measurable. Take z ∈ Z, n ∈ N, γ < 1 and a ∈ ψ (E). Then there are an open G ⊆ X and an F ∈ ΣY such that a ∈ G ∩ φF and nγ R χE(xy)hn (y)µY (dy) ≥ γ for every x ∈ G ∩ F . Because F and φF belong to ΣY , az ∈ φF . Of course Gz R is an open set containing az. If x ∈ Gz ∩ F , then xz −1 ∈ G ∩ F and χE(xz −1 y)hn (y)µY (dy) ≥ γ. But χE(xz −1 y) = χE(xy · y −1 z −1 y) = χE(xy)

378

Topological groups

447F

for every y ∈ Y , because Z C Y (so z 0 = y −1 z −1 y ∈ Z) and we are supposing that E ∈ ΣZ (so xyz 0 ∈ E iff xy ∈ E). So

R

χE(xy)hn (y)µY (dy) =

R

χE(xz −1 y)hn (y)µY (dy) ≥ γ.

As x is arbitrary, Gz and F witness that az ∈ ψnγ E. This shows that, for any n and γ, az ∈ ψnγ (E) whenever a ∈ ψnγ (E) and z ∈ Z. It follows at once that az ∈ ψE whenever a ∈ ψE and z ∈ Z, as claimed. Q Q P Suppose that n ∈ N, γ < 1 (h) For any Haar measurable E ⊆ X and c ∈ X, ψ(cE) = cψE. P and a ∈ ψ (E). Then there are an open set G ⊆ X and an F ∈ Σ nγ Y such that a ∈ G ∩ φF and R χE(xy)hn (y)µY (dy) ≥ γ for every x ∈ G ∩ F . Now cG is an open set containing ca, cF ∈ ΣY and φ(cF ) = cφF contains ca, and if x ∈ cG ∩ cF we have

R

χ(cE)(xy)hn (xy)µY (dy) =

R

χE(c−1 xy)hn (xy)µY (dy) ≥ γ

because c−1 x ∈ G ∩ F . But this means that cG, cF witness that ca ∈ ψnγ (cE). Since a is arbitrary, cψnγ (E) ⊆ ψnγ (cE); as n and γ are arbitrary, cψE ⊆ ψ(cE). Similarly, of course, c−1 ψ(cE) ⊆ ψE, so in Q fact ψ(cE) = cψE, as claimed. Q (i) If E1 , E2 ⊆ X are Haar measurable and E1 \ E2 is µ-negligible, ψE1 ⊆ ψE2 . P P Take n ∈ N, γ < 1 and a ∈ ψ (E ). Then there are an open set G ⊆ X and an F ∈ Σ such that a ∈ G ∩ φF and nγ 1 Y R χE1 (xy)hn (y)µ R Y (dy) ≥ γ for every x ∈ G ∩ F . Let Q be the set of those x ∈ X such that µY measures Y ∩ x−1 E and χ(E1 \ E2 )(xy)µY (dy) = 0; then QY = Q is conegligible (447Ec). Now Q ∩ F ∈ ΣY , and φ(Q ∩ F ) = φF contains a. But if x ∈ G ∩ Q ∩ F , χE1 (xy) ≤ χE2 (xy) for µY -almost every y. So

R

χE2 (xy)hn (y)µY (dy) ≥

R

χE1 (xy)hn (y)µY (dy) ≥ γ.

Thus G and Q∩F witness that a ∈ ψnγ (E2 ). As a is arbitrary, ψnγ (E1 ) ⊆ ψnγ (E2 ); as n and γ are arbitrary, ψE1 ⊆ ψE2 . Q Q In particular, (i) ψE1 = ψE2 whenever E1 4E2 is negligible (ii) ψE1 ⊆ ψE2 whenever E1 ⊆ E2 . (j) If E1 , E2 ⊆ X are Haar measurable, ψ(E1 ∩ E2 ) = ψE1 ∩ ψE2 . P P By (i), ψ(E1 ∩ E2 ) ⊆ ψE1 ∩ ψE2 . So take a ∈ ψE1 ∩ ψE2 . Let γ < 1. Set δ = 12 (1 + γ) < 1. Then there are n1 , n2 ∈ N such that a ∈ ψmδ (E1 ) for every m ≥ n1 and a ∈ ψmδ (E2 ) for every m ≥ n2 . Set n = max(n1 , n2 ) and takeRany m ≥ n. Then there are open sets G1 , G2 ⊆ X and R F1 , F2 ∈ ΣY such that a ∈ G1 ∩ G2 ∩ φF1 ∩ φF2 and χE1 (xy)hm (y)µY (dy) ≥ δ for every x ∈ G1 ∩ F1 , χE2R(xy)hm (y)µY (dy) ≥ δ for every x ∈ G2 ∩ F2 . Let Q ∈ ΣY be the conegligible set of those x ∈ X such that χ(E1 ∩ E2 )(xy)hm (y)µY (dy) is defined. Set G = G1 ∩ G2 , F = F1 ∩ F2 ∩ Q; then G is open, F ∈ ΣY and φF = φ(F1 ∩ F2 ) = φF1 ∩ φF2 , so that a ∈ G ∩ φF . Now take any x ∈ G ∩ F . We have Z (1 − χ(E1 ∩ E2 ))(xy)hm (y)µY (dy) Z Z ≤ (1 − χE1 (xy))hm (y)µ(dy) + (1 − χE2 (xy))hm (y)µ(dy) ≤ 2(1 − δ) = 1 − γ, R

R and χ(E1 ∩ E2 )(xy)hm (y)µY (dy) ≥ γ, because hm (y)µY (dy) = 1. As x is arbitrary, G and F witness that a ∈ ψmγ (E1 ∩ E2 ). And this is true for every m ≥ n. As γ is Q arbitrary, a ∈ ψ(E1 ∩ E2 ). As a is arbitrary, ψE1 ∩ ψE2 ⊆ ψ(E1 ∩ E2 ) and the two are equal. Q (i) If E ∈ ΣY , ψE = φE. P P (i) Suppose a ∈ φE, n ∈ N and γ < 1. Set F = E ∩ φE ∈ ΣY , G = X. Then G ∩ φF = φ(E ∩ φE) = φE contains a. Take any x ∈ F . Then

R

χE(xy)hn (y)µY (dy) =

R

hn (y)µY (dy) = 1;

as x is arbitrary, so a ∈ ψnγ (E). As n and γ are arbitrary, and a ∈ ψE; as a is arbitrary, φE ⊆ ψE. (ii) Suppose a ∈ ψE. Then there must be some open G ⊆ X and F ∈ ΣY and n ∈ N such that a ∈ G ∩ φF

447G


379

R and χE(xy)hn (y)µY (dy) > 0 for every x ∈ G ∩ F . This surely implies that G ∩ F ⊆ EY = E, so that GY ∩ F ⊆ E. But a ∈ G ⊆ φ(GY ), by 447B, so a ∈ φ(GY ) ∩ φF = φ(GY ∩ F ) ⊆ φE. This shows that ψE ⊆ φE. Q Q (l) Thus we have assembled all the facts required to establish that ψ¹ΣZ is a left-translation-invariant partial lower density extending φ. 447G Lemma Let X be a σ-compact locally compact Hausdorff topological group, and hYn in∈N a nonincreasing sequence of compact subgroups of X with intersection Y . Let Σ be the algebra of Haar measurable subsets of X; set ΣYn = {E : E ∈ Σ, EYn = E} for each n, and ΣY = {E : E ∈ Σ, EY = E}. Suppose that for each n ∈ N we are given a left-translation-invariant partial lower density φn : ΣYn → ΣYn , and that φn+1 extends φn for every n. Then there is a left-translation-invariant partial lower density φ : ΣY → ΣY extending every φn . proof (a) Fix a left Haar measure µ on X, and for each n ∈ N let νn be the Haar probability measure on Yn (442Ie). As noted in 443Rb, the modular function of X must be equal to 1, and equal to the modular function of Yn , everywhere on every Yn . S (b) We need to know that for any E ∈ ΣY there is an F in the σ-algebra Λ generated by n∈N ΣYn such that E4F is negligible. P P Because X is σ-compact and µ is a Radon S measure (442Ac), there is a sequence hK i of compact sets such that K ⊆ E for every i and E \ i i∈N i S T i∈N Ki is negligible. For each T Y (4A5Eh), so is included in E. Set F = i∈N n∈N Ki Yn ; then F belongs to the i ∈ N, n∈N Ki Yn = KiS Q σ-algebra generated by n∈N ΣYn , and Ki ⊆ F ⊆ E for every i, so E4F is negligible. Q (c) For each E ∈ ΣY , n ∈ N set gEn (x) = νn (Yn ∩ x−1 E) whenever this is defined. By 447Ee, gEn is defined µ-almost everywhere and is Σ-measurable. In fact gEn is ΣY -measurable, because gEn (xy) = gEn (x) whenever x ∈ X, y ∈ Y and either is defined. If F ∈ Σ R Yn then gEn (x)χF (x) = νn (Yn ∩ x−1 (E ∩ F )) whenever this is defined, which is almost everywhere; so F gEn dµ = µ(E ∩ F ), by 447Ee. If E, E 0 ∈ ΣY and E4E 0 is negligible, then gEn =a.e. gE 0 n , because gE4E 0 ,n = 0 a.e. It follows that hgEn in∈N → gE µ-a.e. for every E ∈ ΣY . P P Let G ⊆ X be any non-empty relatively compact open set, and set U = GY0 , so that U is also a non-empty relatively compact open set and U Yn = U for every n, U Y = U . Set µU (F ) = (µU )−1 µF whenever F ∈ Σ and F ⊆ U , so that µU is a probability measure on U . (U ) (U ) Writing ΣYn , ΣY and Λ(U ) for the subspace σ-algebras on U generated by ΣYn , ΣY and Λ, we see that (U )

if F ∈ ΣYn then

R F

gEn dµU = (µU )−1

R F

gEn dµ = µU (E ∩ F ). (U )

So gEn ¹U is a conditional expectation of χ(E ∩ U ) on ΣYn . By Lévy’s martingale theorem (275I), hgEn in∈N converges almost everywhere on U to a conditional expectation g of χ(E ∩ U ) on Λ(U ) , because of course S (U ) Λ(U ) is the σ-algebra of subsets of U generated by n∈N ΣYn . But as there is an F ∈ Λ such that E4F is negligible, by (b) above, g must be equal to χE almost everywhere on U . Thus gEn → χE almost everywhere on U and therefore almost everywhere on G. As G is arbitrary, gEn → χE almost everywhere on X, by 412Jb (applied to the family K of subsets of relatively compact open sets). Q Q (d) Now we can use the method of 341G, as follows. For E ∈ ΣY , k ≥ 1 and n ∈ N set Hkn (E) = {x : x ∈ dom(gEn ), gEn (x) ≥ 1 − 2−k } ∈ ΣYn , ˜ kn (E) = φn (Hkn (E)), H φE =

T

S k≥1

T n∈N

m≥n

˜ km (E). H

380

Topological groups

447G

By the arguments of parts (e)-(i) of the proof of 341G, φ is a lower density on ΣY extending every φn . (e) To see that φ is left-translation-invariant, we may argue as follows. Let E ∈ ΣY and a ∈ X. Then, for any n, gaE,n (x) = νn (Yn ∩ x−1 aE) = gEn (a−1 x) for almost every x, so aHkn (E)4Hkn (aE) is negligible, for every k, and ˜ kn (aE) = φn (Hkn (aE)) = φn (aHkn (E)) = aφn (Hkn (E)) = aH ˜ kn (E). H Accordingly φ(aE) =

\ [ \

˜ km (aE) H

k≥1 n∈N m≥n

= a(

\ [ \

˜ km (E)) = aφ(E), H

k≥1 n∈N m≥n

as required. 447H Lemma Let X be a locally compact Hausdorff topological group, and Σ the algebra of Haar measurable sets in X. Then there is a left-translation-invariant lower density φ : Σ → Σ. proof (a) To begin with (down to the end of (c) below) let us suppose that X is σ-compact. By 446P, there is a family hXξ iξ≤κ of closed subgroups of X, where κ is an infinite cardinal, such that X0 is an open subgroup of X, for every ξ < κ, Xξ+1 is a normal subgroup of Xξ and Xξ /Xξ+1 has a B-sequence, T for every non-zero limit ordinal ξ ≤ κ, Xξ = η<ξ Xη , X1 is compact, Xκ = {e}, where e is the identity of X. Note that for every ξ ≤ κ, the modular function ∆ξ of Xξ is just the restriction to Xξ of the modular function ∆ of X. P P For ξ = 0 this is because X0 is an open subgroup of X (443Rd). For ξ ≥ 1, Xξ is compact, so ∆ξ and ∆¹Xξ are both constant with value 1, as noted in 443Rb. Q Q (b) For each ξ ≤ κ, write Σξ for the σ-algebra {E : E ∈ Σ, EXξ = E}. I seek to choose inductively a family hφξ iξ≤κ such that each φξ : Σξ → Σξ is a left-translation-invariant partial lower density, and φξ extends φη whenever η < ξ. (i) Start Since X0 is an open subgroup of X, every member of Σ0 is open, and we can start the induction by setting φ0 E = E for every E ∈ Σ0 . (ii) Inductive step to a successor ordinal If we have defined hφη iη≤ξ , where ξ < κ, then Xξ+1 C Xξ is compact and Xξ /Xξ+1 has a B-sequence. So the conditions of 447F are satisfied and φξ has an extension to a left-translation-invariant partial lower density φξ+1 : Σξ+1 → Σξ+1 . Of course φξ+1 extends φη for every η ≤ ξ because φξ does. (iii) Inductive step to a limit ordinal of countable cofinality If we have defined hφη iη<ξ , where ξ ≤ κ is a non-zero limit ordinal of countable cofinality, let hξn in∈N be a strictly increasing sequence with supremum T ξ; we may suppose that ξ0 ≥ 1, so that every Xξn is compact. Then Xξ = n∈N Xξn , so by 447G there is a left-translation-invariant partial lower density φ : Σξ → Σξ extending every φξn , and therefore extending φη whenever η < ξ. (iv) Inductive step to a limit ordinal of uncountable cofinality Suppose we have defined hφη iη<ξ , where ξ ≤ κ is a limit ordinal of uncountable cofinality. Then for every E ∈ Σξ there are an η < ξ and an F ∈ Ση such that E4F is negligible. P P (Cf. part (b) of the proof of 447G.) Because X is σ-compact, there S are non-decreasingS sequences hKi ii∈N , hLi ii∈N of compact subsets of E, X \ E respectively such that E \ i∈I Ki and (X \ E) \ i∈N Li are negligible. For each i ∈ N, Ki Xξ ∩ Li S ⊆ EXξ \ E is empty; by 4A5Eh, there is an ηi < ξ such that Ki Xηi ∩ Li is empty. Set η = supi∈N ηi , F = i∈N Ki Xη ; this works. Q Q

447I


381

Accordingly we have a function φξ : Σξ → Σξ defined by writing φξ (E) = φη (F ) whenever E ∈ Σξ , η < ξ, F ∈ Ση and E4F is negligible. P P If η ≤ η 0 < ξ and F ∈ Ση , F 0 ∈ Ση0 are such that E4F and E4F 0 are 0 both negligible, then F 4F is negligible so φη (F ) = φη0 (F ) = φη0 (F 0 ). Q Q It is easy to check that φξ is a left-translation-invariant partial lower density (cf. part (d) of the proof of 341H), and of course it extends φη for every η < ξ. (c) On completing the induction, we see that Σκ = Σ, so that φκ : Σ → Σ is a left-translation-invariant lower density. (d) For the general case, recall that X certainly has an open σ-compact subgroup Y say (4A5El). If Σ is the algebra of Haar measurable subsets of X, and T is the algebra of Haar measurable subsets of Y , then T is just Σ ∩ PY = {E ∩ Y : E ∈ Σ}, and the Haar negligible subsets of Y are just sets of the form E ∩ Y where E is a Haar negligible subset of X (443F). Let ψ : T → T be a left-translation-invariant lower density on Y . For E ∈ Σ set φE = {x : x ∈ X, e ∈ ψ(Y ∩ x−1 E)}, where e is the identity of X. It is easy to check that φ∅ = ∅, φE = φF if E4F is negligible, φ(E ∩ F ) = φE ∩ φF for all E, F ∈ Σ, directly from the corresponding properties of ψ. If E ∈ Σ and a ∈ X, then x ∈ φE ⇐⇒ e ∈ ψ(Y ∩ x−1 E) ⇐⇒ e ∈ ψ(Y ∩ (ax)−1 aE) ⇐⇒ ax ∈ φ(aE), so φ(aE) = aφE. I have not yet checked that E4φE is always negligible. But if E ∈ Σ, then E4φE = {x : e ∈ ψ(Y ∩ x−1 E)4(Y ∩ x−1 E)}, so (E4φE) ∩ Y = {x : x ∈ Y, e ∈ ψ(x−1 (E ∩ Y ))4x−1 (E ∩ Y )} = {x : x ∈ Y, e ∈ x−1 ψ(E ∩ Y )4x−1 (E ∩ Y )} = ψ(E ∩ Y )4(E ∩ Y ) is negligible. Moreover, for any a ∈ X, (E4φE) ∩ aY = a((a−1 E4φ(a−1 E)) ∩ Y ) because φ is translation-invariant, so (E4φE) ∩ aY is negligible. Since {aY : a ∈ X} is an open cover of X, E4φE is negligible (412Jc). In particular, φE ∈ Σ. So φ : Σ → Σ is a left-translation-invariant lower density, as required. 447I Theorem (Ionescu Tulcea & Ionescu Tulcea 67) Let X be a locally compact Hausdorff topological group. Then it has a left-translation-invariant lifting for its Haar measures. proof (Cf. 345B-345C.) Write Σ for the algebra of Haar measurable subsets of X, and let φ : Σ → Σ be a left-translation-invariant lower density (447H). Let φ0 : Σ → Σ be any lifting such that φ0 E ⊇ φE for every E ∈ Σ (341Jb). For E ∈ Σ, set φE = {x : e ∈ φ0 (x−1 E)}, where e is the identity of X. It is easy to check that φ : Σ → PX is a Boolean homomorphism. Also x ∈ φE =⇒ e ∈ φ(x−1 E) =⇒ e ∈ φ0 (x−1 E) =⇒ x ∈ φE. So φ is a lifting (341I). Finally, φ is left-translation-invariant by the argument used in (d) of the proof of 447H (and also in (e) of the proof of 345B).

382

Topological groups

447J

447J Corollary Let X be any topological group carrying Haar measures. Then it has a left-translationinvariant lifting for its Haar measures. proof Let µ be a left Haar measure on X. By 443L, we have a locally compact Hausdorff topological group Z and a continuous homomorphism f : X → Z, inverse-measure-preserving for µ and an appropriate left Haar measure ν on Z, such that for every E in the domain Σ of µ there is an F in the domain T of ν such that f −1 [F ] ⊆ E and E \ f −1 [F ] is negligible. Let ψ be a left-translation-invariant lifting for ν. For E ∈ Σ, we may define φE ∈ Σ by saying that φE = f −1 [ψF ] whenever F ∈ T and E4f −1 [F ] is negligible. P P By the construction of Z and f , there is always an F ∈ T such that E4f −1 [F ] is negligible, so that there is a candidate for φE. If F , F 0 ∈ T are such that E4f −1 [F ] and E4f −1 [F ] are both negligible, then f −1 [F 4F 0 ] is negligible, so F 4F 0 is negligible (because f is inverse-measure-preserving), ψF = ψF 0 and f −1 [ψF ] = f −1 [ψF 0 ]. Thus φE is uniquely defined for every E ∈ Σ. Q Q It is now easy to check that φ is a lifting for µ (cf. 341Xd), and φ is left-translation-invariant because f is a group homomorphism and ψ is left-translation-invariant. 447X Basic exercises > (a) Let X = R × {−1, 1}, with the usual topology, and define a multiplication on X by setting (x, δ)(y, ²) = (x + δy, δ²). Show that X is a locally compact topological group. Show that there is no lifting of the Haar measure algebra of X which is both left- and right-translation-invariant. (Hint: 345Xc.) (b) Let X be a topological group carrying Haar measures which has a B-sequence. Show that it has a B-sequence hVn in∈N such that supn∈N µ(Vn Vn−1 )/µVn is finite for any Haar measure µ on X, whether left or right. (c) Let X be a topological group with a left Haar measure µ, and hVn in∈N a B-sequence for X. Show 1 R that if f ∈ L0 (µ) is locally integrable, then f (x) = limn→∞ f dµ for almost every x. xVn µVn

447Y Further exercises (a) Describe a compact Hausdorff topological group such that its Haar measure has no lifting which is both left- and right-translation-invariant. (b) Let (X, Σ, µ) be a measure space, with measure algebra A, and θ :SA → Σ a lower density. Show that we have a function q : L∞ (A)+ → L∞ (Σ) such that {x : q(u)(x) > α} = β>α θ[[u > β]] for every α ≥ 0 and u ∈ L∞ (A)+ . Show that q(u)• = u, q(αu) = αq(u), q(u ∧ v) = q(u) ∧ q(v) and q(χa) = χ(θa) for every u, v ∈ L∞ (A)+ , α ≥ 0 and a ∈ A. 447 Notes and comments The structure of the proof of 447I is exactly that of the proof of the ordinary Lifting Theorem in §341; the lifting is built from a lower density which is constructed inductively on a family of sub-σ-algebras. To get a translation-invariant lifting it is natural to look for a translation-invariant lower density, and a simple trick (already used in §345) ensures that this is indeed enough. The refinements we need here are dramatic but natural. To make the final lower density φ (in 447H) translation-invariant, it is clearly sensible (if we can do it) to keep all the partial lower densities φξ translation-invariant. This means that their domains Σξ should be translation-invariant. It does not quite follow that they have to be of the form Σξ = {E : EXξ = Xξ } for closed subgroups Xξ , but if we look at the leading example of {0, 1}I (345C) this is also a reasonable thing to try first. So now we have to consider what extra hypotheses will be needed to make the induction work. The inductive step to limit ordinals of uncountable cardinality remains elementary, at least if the Xξ are compact (part (b-iv) of the proof of 447H). The inductive step to limit ordinals of countable cofinality (447G) is harder, but can be managed with ideas already presented. Indeed, compared with the version in 341G, we have the advantage of a formula for the auxiliary functions gEn , which is very helpful when we come to translation-invariance. We have to do something about the fact that we are no longer working with a probability space – that is the point of the µU in part (c) of the proof of 447G. (Another expression of the manoeuvre here is in 369Xq.) Where we do need a new idea is in the inductive step to a successor ordinal. If Σξ+1 is to be translationinvariant, it must be much bigger than the σ-algebra generated by Σξ ∪ {E}, as discussed in 341F. To make the step a small one (and therefore presumably easier), we want Xξ+1 to be a large subgroup of Xξ in some

448A

Invariant measures on Polish spaces

383

sense; as it turns out, a helpful approach is to ask for Xξ+1 to be a normal subgroup of Xξ and for Xξ /Xξ+1 to be small. At this point we have to know something of the structure theory of locally compact topological groups. The right place to start is surely the theory of compact Hausdorff groups. Such a group X actually has a continuous decreasing chain hXξ iξ≤κ of closed normal subgroups, from X0 = X to Xκ = {e}, such that all the quotients Xξ /Xξ+1 are Lie groups. I do not define ‘Lie group’ here, because for our purposes it is enough to know that the quotients have faithful finite-dimensional representations, and therefore have ‘B-sequences’ in the sense of 446L. Having identified this as a relevant property, it is not hard to repeat arguments from §221 and §261 to prove versions of Vitali’s theorem and Lebesgue’s Density Theorem in such groups (447C-447D). This will evidently provide translation-invariant lower densities for groups of this special type, just as Lebesgue lower density is a translation-invariant lower density on Rr (345B). Of course we still have to find a way of combining this construction with a translation-invariant lower density on Σξ to produce a translation-invariant lower density on Σξ+1 , and this is what I do in 447F. The argument I offer is essentially that of Ionescu Tulcea & Ionescu Tulcea 67, §7, and is the deepest part of this section. For compact groups, these ideas are all we need, and indeed the step to a limit ordinal of countable cofinality is a little easier, since we have a Haar probability measure on the whole group. The next step, to general σ-compact locally compact groups, demands much deeper ideas from the structure theory, but from the point of view of the present section the modifications are minor. The subgroups Xξ are now not always normal subgroups of X, which means that we have to be more careful in the description of the quotient spaces X/Xξ (they must consist of left cosets), and we have to watch the modular functions of the Xξ in order to be sure that there are invariant measures on the quotients. An extra obstacle at the beginning is that we may have to start the chain with a proper subgroup X0 of X, but since X0 can be taken to be open, it is pretty clear that this will not be serious, and in fact it gives no trouble (part (b-i) of the proof of 447H). For ξ ≥ 1, the Xξ are compact, so the inductive steps to limit ordinals are nearly the same. The step to a general locally compact Hausdorff topological group (part (d) of the proof of 447H) is essentially elementary. And finally I note that the whole thing applies to general topological groups with Haar measures (447J), for the usual reasons. There is an implicit challenge here: find expressions of the arguments used in this section which will be valid in the more general context. The measure-theoretic part of such a programme might be achievable, but I do not see any hope of a workable structure theory to match that of §446 which does not use 443L or something like it.

448 Invariant measures on Polish spaces In this section I set out an interesting result concerning measures on Polish spaces which are invariant under actions of Polish groups. In contrast to §441, we no longer have a strong general existence theorem for such measures, but instead have a natural necessary and sufficient condition in terms of countable dissections: there is an invariant probability measure on X iff there is no countable dissection of X into Borel sets which can be rearranged, by the action of the group, into two copies of X (448P). The principal ideas needed here have already been set out in §394, and in many of the proofs I allow myself to direct you to the corresponding arguments there rather than write the formulae out again. I do not think you need read through §394 before embarking on this section; I will try to give sufficiently detailed references so that you can take them one paragraph at a time, and many of the arguments referred to are in any case elementary. But unless you are already familiar with this topic, you will need a copy of §394 to hand to fully follow the proofs below. 448A Definition (Compare 394A.) Let A be a Dedekind σ-complete Boolean algebra, and G a subgroup of Aut A. For a, b ∈ A I will say that an isomorphism φ : Aa → Ab between the corresponding principal ideals belongs to the countably full local semigroup generated by G if there are a countable partition of unity hai ii∈I in Aa and a family hπi ii∈I in G such that φc = πi c whenever i ∈ I and c ⊆ ai . If such an isomorphism exists I will say that a and b are G-σ-equidecomposable. I write a 4σG b to mean that there is a b0 ⊆ b such that a and b0 are G-σ-equidecomposable. As in §394, I will say that a function f with domain A is G-invariant if f (πa) = f (a) for every a ∈ A, π ∈ G.

384

Topological groups

448A

I have expressed these definitions, and most of the work below, in terms of abstract Dedekind σ-complete Boolean algebras. The applications I have in mind for this section are to σ-algebras of sets. If you have already worked through §394, the version here should come very easily; but even if you have not, I think that the extra abstraction clarifies some of the ideas. 448B I begin with results corresponding to 394B-394D; there is hardly any difference, except that we must now occasionally pause to check that a partition of unity is countable. Lemma Let A be a Dedekind σ-complete Boolean algebra and G a subgroup of Aut A. Write G∗σ for the countably full local semigroup generated by G. (a) If a, b ∈ A and φ : Aa → Ab belongs to G∗σ , then φ−1 : Ab → Aa also belongs to G∗σ . (b) Suppose that a, b, a0 , b0 ∈ A and that φ : Aa → Aa0 , ψ : Ab → Ab0 belong to G∗σ . Then ψφ ∈ G∗σ ; its domain is Ac where c = φ−1 (b ∩ a0 ), and its set of values is Ac0 where c0 = ψ(b ∩ a0 ). (c) If a, b ∈ A and φ : Aa → Ab belongs to G∗σ , then φ¹ Ac ∈ G∗σ for any c ⊆ a. (d) Suppose that a, b ∈ A and that ψ : Aa → Ab is an isomorphism such that there are a countable partition of unity hai ii∈I in Aa and a family hφi ii∈I in G∗σ such that ψc = φi c whenever i ∈ I and c ⊆ ai . Then ψ ∈ G∗σ . proof (a) As 394Bb. (b) As 394Bc. (c) As 394Bd. (d) For each i ∈ I, let haij ij∈J(i) , hπij ij∈J(i) witness that φi ∈ G∗σ ; then hai ∩ aij ii∈I,j∈J(i) and hπij ii∈I,j∈J(i) witness that ψ ∈ G∗σ . 448C Lemma Let A be a Dedekind σ-complete Boolean algebra and G a subgroup of Aut A. Write G∗σ for the countably full local semigroup generated by G. (a) For a, b ∈ A, a 4σG b iff there is a φ ∈ G∗σ such that a ∈ dom φ and φa ⊆ b. (b)(i) 4σG is transitive and reflexive; (ii) if a 4σG b and b 4σG a then a and b are G-σ-equidecomposable. (c) G-σ-equidecomposability is an equivalence relation on A. (d) If hai ii∈I and hbi ii∈I are countable families in A, of which hbi ii∈I is disjoint, and ai 4σG bi for every i ∈ I, then supi∈I ai 4σG supi∈I bi . proof The arguments of 394C apply unchanged, calling on 448B in place of 394B. 448D Theorem Let A be a Dedekind σ-complete Boolean algebra and G a subgroup of Aut A. Then the following are equiveridical: (i) there is an a 6= 1 such that a is G-σ-equidecomposable with 1; (ii) there is a disjoint sequence han in∈N of non-zero elements of A which are all G-σ-equidecomposable; (iii) there are non-zero G-σ-equidecomposable a, b, c ∈ A such that a ∩ b = 0 and a ∪ b ⊆ c; (iv) there are G-σ-equidecomposable a, b ∈ A such that a ⊂ b. proof As 394D. 448E Definition (Compare 394E.) Let A be a Dedekind σ-complete Boolean algebra and G a subgroup of Aut A. I will say that G is countably non-paradoxical if the statements of 448D are false; that is, if one of the following equiveridical statements is true: (i) if a is G-σ-equidecomposable with 1 then a = 1; (ii) there is no disjoint sequence han in∈N of non-zero elements of A which are all G-σ-equidecomposable; (iii) there are no non-zero G-σ-equidecomposable a, b, c ∈ A such that a ∩ b = 0 and a ∪ b ⊆ c; (iv) if a ⊆ b ∈ A and a, b are G-σ-equidecomposable then a = b.

448G


385

448F We now come to one of the points where we need to find a new path because we are looking at algebras which need not be Dedekind complete. Provided the original group G is countable, we can still follow the general line of §394, as follows. Lemma (Compare 394G.) Let A be a Dedekind σ-complete Boolean algebra and G a countable subgroup of Aut A. Let C be the fixed-point subalgebra of G. (a) For any a ∈ A, upr(a, C) (314V) is defined, and is given by the formula upr(a, C) = sup{πa : π ∈ G}. G∗σ

(b) If is the countably full local semigroup generated by G, then φ(a ∩ c) = φa ∩ c whenever φ ∈ G∗σ , a ∈ dom φ and c ∈ C. (c) upr(φa, C) = upr(a, C) whenever φ ∈ G∗σ and a ∈ dom φ; consequently, upr(a, C) ⊆ upr(b, C) whenever a 4σG b. (d) If a 4σG b and c ∈ C then a ∩ c 4σG b ∩ c. So a ∩ c and b ∩ c are G-σ-equidecomposable whenever a and b are G-σ-equidecomposable and c ∈ C. proof (a) As remarked in 394Ga, C is order-closed. Because G is countable and A is Dedekind σ-complete, c∗ = sup{πa : π ∈ G} is defined in A. If c ∈ C, then a ⊆ c ⇐⇒ πa ⊆ πc for every π ∈ G ⇐⇒ πa ⊆ c for every π ∈ G ⇐⇒ c∗ ⊆ c, so c∗ = inf{c : a ⊆ c ∈ C}, taking the infimum in A, as required by the definition of upr(a, C). (b) Suppose that hai ii∈I , hπi ii∈I witness that φ ∈ G∗σ . Then φ(a ∩ c) = supi∈I πi (ai ∩ a ∩ c) = supi∈I πi (ai ∩ a) ∩ c = φa ∩ c. (c) For c ∈ C, a ⊆ c ⇐⇒ a ∩ c = a ⇐⇒ φ(a ∩ c) = φa ⇐⇒ φa ∩ c = φa ⇐⇒ φa ⊆ c. (d) There is a φ ∈ G∗σ such that φa ⊆ b; now a ∩ c 4σG φ(a ∩ c) = φa ∩ c ⊆ b ∩ c. 448G With this support, we can now continue with the ideas of 394H-394L, adding at each step the hypothesis ‘G is countable’ to compensate for the weakening of the hypotheses ‘A is Dedekind complete, G is fully non-paradoxical’ to ‘A is Dedekind σ-complete, G is countably non-paradoxical’. Lemma (Compare 394H.) Let A be a Dedekind σ-complete Boolean algebra and G a countable countably non-paradoxical subgroup of Aut A. Write C for the fixed-point subalgebra of G. Take any a, b ∈ A. Then c0 = sup{c : c ∈ C, a ∩ c 4σG b} is defined in A and belongs to C; a ∩ c0 4σG b and b \ c0 4σG a. proof Let hπn in∈N be a sequence running over G. Define han in∈N , hbn in∈N inductively, setting an = (a \ supi
bn = πn an .

Then han in∈N is a disjoint sequence in Aa and hbn in∈N is a disjoint sequence in Ab , and supn∈N an is G-σequidecomposable with supn∈N bn . Set a0 = a \ supn∈N an , b0 = b \ supn∈N bn , c0 = 1 \ upr(a0 , C). Then a ∩ c0 ⊆ supn∈N an 4σG b. Now b0 ⊆ c0 . P P?? Otherwise, because c0 = 1 \ supn∈N πn a0 , there must be an n ∈ N such that πn a0 ∩ b0 6= 0. But in this case d = a0 ∩ πn−1 b0 6= 0, and we have d ⊆ (a \ supi
386

Topological groups

448G

Now take any c ∈ C such that a ∩ c 4σG b, and consider c0 = c \ c0 . Then b0 ∩ c0 = 0, that is, b ∩ c0 = supn∈N bn ∩ c0 , which is G-σ-equidecomposable with supn∈N an ∩ c0 = (a \ a0 ) ∩ c0 . But now a ∩ c0 4σG b ∩ c0 4σG (a ∩ c0 ) \ (a0 ∩ c0 ); because G is countably non-paradoxical, a0 ∩ c0 must be 0, that is, c0 ⊆ c0 and c0 = 0. So c0 has the required properties. 448H Lemma (Compare 394I.) Let A be a Dedekind σ-complete Boolean algebra, not {0}, and G a countable countably non-paradoxical subgroup of Aut A. Let C be the fixed-point subalgebra of G. Suppose that a, b ∈ A and that upr(a, C) = 1. Then there are non-negative u, v ∈ L0 = L0 (C) such that [[u ≥ n]] = max{c : c ∈ C, there is a disjoint family hdi ii
for every n ∈ N. Moreover, we have (i) [[u ∈ N]] = [[v ∈ N]] = 1, (ii) [[v > 0]] = upr(b, C), (iii) u ≤ v ≤ u + χ1. proof The argument of 394I applies unchanged, except that every 4τG must be replaced with a 4σG , and we use 448F and 448G in place of 394G and 394H. C is Dedekind σ-complete because it is order-closed in the Dedekind σ-complete algebra A (314Eb). 448I Notation (Compare 394J.) In the context of 448H, I will write bb : ac for u, db : ae for v. 448J Lemma (Compare 394K-394L.) Let A be a Dedekind σ-complete Boolean algebra, not {0}, and G a countable countably non-paradoxical subgroup of Aut A with fixed-point subalgebra C. Suppose that a, a1 , a2 , b, b1 , b2 ∈ A and that upr(a, C) = upr(a1 , C) = upr(a2 , C) = 1. Then (a) b0 : ac = d0 : ae = 0 and b1 : ac ≥ χ1. (b) If b1 4σG b2 then bb1 : ac ≤ bb2 : ac and db1 : ae ≤ db2 : ae. (c) db1 ∪ b2 : ae ≤ db1 : ae + db2 : ae. (d) If b1 ∩ b2 = 0, then bb1 : ac + bb2 : ac ≤ bb1 ∪ b2 : ac. (e) bb : a2 c ≥ bb : a1 c × ba1 : a2 c, db : a2 e ≤ db : a1 e × da1 : a2 e. proof As in 394K-394L. 448K For the result corresponding to 394Mb, we again need to find a new approach; I deal with it by adding a further hypothesis to the list which has already accreted. Definition Let A be a Dedekind σ-complete Boolean algebra and G a subgroup of Aut A with fixed-point algebra C. I will say that G has the σ-refinement property if for every a ∈ A there is a d ⊆ a such that d 4σG a \ d and a0 = a \ upr(d, C) is a relative atom over C, that is, every b ⊆ a0 is expressible as a0 ∩ c for some c ∈ C. (If we replace 4σG with 4τG , as used in §394, we see that 394Ma could be read as ‘if A is a Dedekind complete Boolean algebra, then any subgroup of Aut A has the τ -refinement property’.) 448L I give the principal case in which the ‘σ-refinement property’ just defined arises. Proposition Let A be a Dedekind σ-complete Boolean algebra with countable Maharam type (definition: 331F). Then any countable subgroup of Aut A has the σ-refinement property.

448N


387

proof (a) Let E be a countable subset of A which τ -generates A, and E the subalgebra of A generated by E; then E is countable (331Gc), and the smallest order-closed subset of A including E is a subalgebra of A (313Fc), so must be A itself. (b) Suppose that b ∈ A \ {0} and π ∈ Aut A are such that b ∩ πb = 0. Then there is an e ∈ E such that b ∩ e \ πe 6= 0. P P?? Otherwise, set D = {d : d ∈ A, b ∩ d \ πd = 0}. Then E ⊆ D, but b ∈ / D. So D cannot be order-closed. case 1 If D0 ⊆ D is a non-empty upwards-directed set with supremum d0 ∈ / D, then b ∩ d0 \ πd0 6= 0, so there is a d ∈ D0 such that b ∩ d \ πd0 6= 0; but now d∈ / D, which is impossible. case 2 If D0 ⊆ D is a non-empty downwards-directed subset of D with infimum d0 ∈ / D, then b ∩ d0 \ πd0 6= 0. But π is order-continuous, so there is a d ∈ D0 such that b ∩ d0 \ πd 6= 0; and now d ∈ / D, which is impossible. Thus in either case we have a contradiction. X XQ Q (c) Now let G be a countable subgroup of Aut A, with fixed-point algebra C, and let h(πn , en )in∈N be a sequence running over G × E. Take any a ∈ A. For k ∈ N set ak = a ∩ ek ∩ πk−1 (a \ ek ), a0k = ak \ supj
πj a0j ∩ a0k = 0, πj a0j ∩ πk a0k = πk (a0k ∩ πk−1 πj a0j ) = 0.

So, setting d = supk∈N a0k and d0 = supk∈N πk a0k , d and d0 are disjoint and G-σ-equidecomposable and included in a, and d 4σG a \ d. Consider a0 = a \ upr(d, C). Since a0k = ak \ supj 0 there is an a∗ ∈ A such that upr(a∗ , C) = 1 and db : a∗ e ≤ bb : a∗ c + ²b1 : a∗ c for every b ∈ A. proof As part (b) of the proof of 394M. 448N Theorem (Compare 394N.) Let A be a Dedekind σ-complete Boolean algebra and G a countable countably non-paradoxical subgroup of Aut A with fixed-point algebra C. Suppose that G has the σ-refinement property of 448K. Then there is a function θ : A → L∞ (C) such that (i) θ is additive, non-negative and sequentially order-continuous; (ii) θa = 0 iff a = 0, θ1 = χ1; (iii) θ(a ∩ c) = θa × χc for every a ∈ A, c ∈ C; in particular, θc = χc for every c ∈ C;

388

Topological groups

448N

(iv) If a, b ∈ A are G-σ-equidecomposable, then θa = θb; in particular, θ is G-invariant. proof The arguments of the proof of 394N apply here also, though we have to take things in a slightly different order. As in 394N, set θa (b) =

db : ae ∈ L0 (C) b1 : ac

whenever upr(a, C) = 1 and b ∈ A. This time, turn immediately to part (c) of the proof to see that if en is chosen (using 448M) such that upr(en , C) = 1 and db : en e ≤ bb : en c + 2−n b1 : en c for every b ∈ A, then θen b ≤ θa b + 2−n d1 : ae whenever upr(a, C) = 1 and b ∈ A. So we can write θb = inf n∈N θen b = inf upr(a,C)=1 θa b for every b ∈ A, and we have a function θ : A → L0 as before. The rest of the proof is unchanged, except that we have a simplification in (h), since we need consider only the case κ = ω. 448O This concludes the adaptations we need from §394. I now return to the specific problem addressed in the present section. The first step is a version of 448N with weaker hypotheses. Theorem Let A be a Dedekind σ-complete Boolean algebra, not {0}, and G a countable subgroup of Aut A with the σ-refinement property. Let C be the fixed-point algebra of G. Then the following are equiveridical: (i) there are a Dedekind σ-complete Boolean algebra D, not {0}, and a G-invariant sequentially ordercontinuous non-negative additive function θ : A → L∞ (D) such that θ1 = χ1; (ii) if a ∈ A and 1 4σG a, then upr(1 \ a, C) 6= 1; (iii) if a ∈ A and 1 4σG a, then 1 64σG 1 \ a. proof (a)(i)⇒(iii) Take θ : A → L∞ (D) as in (i). If 1 4σG a, then there is a b ⊆ a which is G-σequidecomposable with 1, so that θb = χ1, just as in 394N(v)/448N(iv). But this means that θa = χ1; so that θ(1 \ a) 6= χ1 and 1 64σG 1 \ a. (b) not-(ii)⇒not-(iii) Suppose that (ii) is false; that there is an a ∈ A such that 1 4σG a and upr(1 \ a, C) = 1. Let G∗σ be the countably full local semigroup generated by G; then there is a ψ ∈ G∗σ such that ψ1 ⊆ a. Set b0 = 1 \ ψ1 and bn = ψ n b0 for every n ≥ 1; then b0 ∩ bn ⊆ b0 ∩ ψ1 = 0 for every n ≥ 1, so bm ∩ bn = ψ m (b0 ∩ bn−m ) = 0 whenever m < n. Let hπi ii∈N be a sequence running over G; then supi∈N πi b0 = upr(b0 , C) ⊇ upr(1 \ a, C) = 1. Set aj = πj b0 \ supi
ψ2 d = supi∈N ψ 2i+1 πi−1 (d ∩ ai )

for every d ∈ A. Because ψ 2i πi−1 ai ⊆ b2i for every i, hψ 2i πi−1 ai ii∈N is disjoint, so ψ1 ∈ G∗σ ; similarly, ψ2 ∈ G∗σ . Thus 1 4σG ψ1 1 ⊆ supi∈N b2i , 1 4σG ψ2 1 ⊆ supi∈N b2i+1 ⊆ 1 \ ψ1 1 and (iii) is false. (c) For the rest of this proof, therefore, I suppose that (ii) is true and seek to prove (i). Let I be the σ-ideal of A generated by {1 \ a : a ∈ A, 1 4σG a}. Then 1 ∈ / I. P P?? Otherwise, there is a sequence han in∈N such that 1 4σG an for every n and supn∈N 1 \ an = 1. Choose ψn ∈ G∗σ such that

448O


389

ψn 1 ⊆ an , and set cn = upr(1 \ ψn 1, C) for each n, so that supn∈N cn = 1. Set c0n = cn \ supi
belongs to I, while han \ din∈N , hπn (an \ d)in∈N are disjoint. Because I is a σ-ideal and G is countable, d∗ = upr(d, C) = supπ∈G πd ∈ I. Define ψ ∈ Gσ∗ by setting ψa = supn∈N πn (a ∩ an \ d∗ ) ∪ (a ∩ d∗ ) for every a ∈ A. Then 1 \ ψ1 ∈ I, by the definition of I, so (ψ1)• = 1 in B. But (ψ1)• = sup(πn (an \ d∗ ))• = sup(πn an )• n∈N

n∈N

= sup π ñ bn = b. n∈N

˜ is countably non-paradoxical. Q So b = 1. As b is arbitrary, G Q ˜ has the σ-refinement property. P (f ) G P Let b ∈ B. Then there is an a ∈ A such that a• = b. Because G is supposed to have the σ-refinement property, there is a d ⊆ a such that d 4σG a \ d and a \ upr(d, C) is a relative atom over C. Set e = d• ⊆ b. We know that there are a partition of unity hdn in∈N in Ad and a sequence hπn in∈N in G such that πn dn ⊆ a \ d for every n and hπn dn in∈N is disjoint. Now hd•n in∈N is a partition of unity in Be , π ñ d•n ⊆ b \ e • σ for every n, and h˜ πn dn in∈N is disjoint; so e 4G˜ b \ e. Suppose that b0 ⊆ b \ upr(e, D). Then it is expressible as a•0 where a0 ⊆ a and (a0 ∩ πd)• = b0 ∩ π ˜e = 0 for every π ∈ G. So if we set a1 = a0 \ supπ∈G πd, we shall have a1 ⊆ a \ upr(d, C) and a•1 = b0 . Now a \ upr(d, C) is supposed to be a relative atom over C, so a1 = a ∩ c for some c ∈ C. In this case, π ˜ c• = (πc)• = c• for every π ∈ G, so c• ∈ D, while b0 = b ∩ c• . As b0 is arbitrary, b \ upr(e, D) is a relative atom over D.

390

Topological groups

448O

˜ has the σ-refinement Thus e has both the properties required by the definition 448K. As b is arbitrary, G property. Q Q (g) 448N now tells us that there is a sequentially order-continuous non-negative additive functional θ0 : B → L∞ (D) such that θ0 1 = χ1 and θ0 (˜ π b) = θ0 b for every b ∈ B, π ∈ G. If we set θa = θ0 a• , it is easy to see that θ has all the properties required by (i) of this theorem. Thus (ii)⇒(i), and the proof is complete. 448P At last we come to Polish spaces. Theorem (Nadkarni 90, Becker & Kechris 96) Let G be a Polish group acting on a non-empty Polish space (X, T) with a Borel measurable action •. For Borel sets E, F ⊆ X say that E 4σG F if there are a countable partition hEi ii∈I of E into Borel sets, and a family hgi ii∈I in G, such that gi •Ei ⊆ F for every i and hgi •Ei ii∈I is disjoint. Then the following are equiveridical: (i) there is a G-invariant Radon probability measure T µ on X; (ii) if F ⊆ X is a Borel set such that X 4σG F , then n∈N gn •F 6= ∅ for any sequence hgn in∈N in G; (iii) there are no disjoint Borel sets E, F ⊆ X such that X 4σG E and X 4σG F . proof (a) Let us start with the easy parts. (i)⇒(ii) Let µ be a G-invariant Radon probability measure on X, and suppose that X 4σG F . Let hEi ii∈I be a countable partition of X into Borel sets and hhi ii∈I a family in G such that hhi •Ei ii∈I is disjoint and hi •Ei ⊆ F for every i. Then P P µF ≥ i∈I µ(hi •Ei ) = i∈I µEi = µX, T so F is conegligible. Consequently n∈N gn •F must be conegligible and cannot be empty, for any sequence hgn in∈N in G. As F and hgn in∈N are arbitrary, (ii) is true. (ii)⇒(iii) Assume (ii). ?? If there are disjoint E, F such that X 4σG E and X 4σG F , then we have a countable partition hE of X into Borel sets and a family hgi ii∈I in G such that gi •Ei ⊆ E for every i ∈ I. Ti ii∈I −1 But there is an x ∈ i∈I gi •F , by (ii). In this case there is a j ∈ I such that x ∈ Ej and gj •x ∈ E ∩ F , which is impossible. X X So (iii) must be true. (b) For the rest of the proof, therefore, I shall assume (iii) and seek to prove (i). Let T be the topology of X, and B its Borel σ-algebra. For g ∈ G define πg : B → B by writing πg E = g •E ˜ = {πg : g ∈ G} is a subgroup of Aut B. Observe for every E ∈ B. Then πgh = πg πh for all g, h ∈ G, so G σ σ that for E, F ∈ B, E 4G F , in the sense here, iff E 4G˜ F in the sense of 448A. By the Becker-Kechris theorem (424H), there is a Polish topology T1 on X, giving rise to the same Borel σ-algebra B as the original topology, for which the action of G is continuous. Let U be a countable base for T1 ; replacing U by {U0 ∪ . . . ∪ Un : U0 , . . . , Un ∈ U} if necessary, we may suppose that U ∪ U 0 ∈ U for all U , U 0 ∈ U. (We are going to have three Polish topologies on X in this proof, so watch carefully.) ˜ are (c) For the time being (down to the end of (f) below) let us suppose that G, and therefore G, ˜ has the σ-refinement property, by 448L. We can therefore apply 448O to see that countable. In this case G (iii) implies that ˜ (i)0 there are a Dedekind σ-complete Boolean algebra D, not {0}, and a G-invariant sequentially ∞ order-continuous non-negative additive functional θ : B → L (D) such that θX = χ1. To see that (i)0 ⇒(i), argue as follows. Express D as Σ/J where Σ is a σ-algebra of subsets of a set Z and J is a σ-ideal of Σ (314N). Then we can identify L∞ (D) with the quotient L∞ /W , where L∞ is the space of bounded Σ-measurable real-valued functions on Z and W is the set {f : f ∈ L∞ , {z : f (z) 6= 0} ∈ J } (363H). For each E ∈ B, let fE ∈ L∞ be a representative of θE ∈ L∞ (D); because θ(πE) = θE whenever ˜ we may suppose that fπE = fE whenever E ∈ B and π ∈ G. ˜ E ∈ B and π ∈ G, ˜ Then B is countable and πE ∈ B for Let B be the subalgebra of B generated by {πU : U ∈ U , π ∈ G}. ˜ By 4A3I, there is yet another Polish topology S on X which is zero-dimensional and every E ∈ B, π ∈ G. such that every member of B is open-and-closed for S. Of course S ⊇ U , so B is still the algebra of S-Borel sets (423Fb). Let V be a countable base for S consisting of sets which are open-and-closed for S, and let B1 be the subalgebra of B generated by V ∪ B; then B1 is countable and consists of open-and-closed sets

448P


391

for S. Let hWn in∈N be a sequence running over V. Let ρ be a complete metric on X defining the topology S, and for m, n ∈ N set S Wmn = {Wi : i ≤ n, diamρ (Wi ) ≤ 2−m }; then for each m ∈ N, hWmn in∈N is a non-decreasing sequence in B1 with union X. (d) Consider the subsets of Z of the following types: PE = {z : fE (z) < 0}, where E ∈ B1 , QEF = {z : fE∪F (z) 6= fE (z) + fF (z)}, where E, F ∈ B1 and E ∩ F = ∅, R = {z : fX (z) 6= 1}, Sm = {z : supn∈N fWmn (z) 6= 1}, where m ∈ N. Because fE• = θE ≥ 0 for every E ∈ B, • = θ(E ∪ F ) = θE + θF = fE• + fF• whenever E ∩ F = ∅, fE∪F

• supn∈N fW mn

• = θX = χ1, fX S = supn∈N θWmn = θ( n∈N Wmn ) = θX = χ1

for every m ∈ N, all the sets PE , QEF , R and Sm belong to J . Since D 6= {0}, Z ∈ / J ; so there is a z0 ∈ Z not belonging to R or PE or QEF or Sm whenever m ∈ N and E, F ∈ B1 are disjoint. Set νE = fE (z0 ) for every E ∈ B1 . If E, F ∈ B1 are disjoint, then ν(E ∪ F ) = νE + νF because z0 ∈ / QEF ; thus ν : B1 → R is additive. If E ∈ B1 then νE ≥ 0 because z0 ∈ / PE , so ν is non-negative. νX = 1 because z0 ∈ / R. For each m ∈ N, supn∈N νWmn = 1 because z0 ∈ / Sm . (e) For any ² > 0 there is an S-compact set K ⊆ X such that νE ≥ 1 − ² whenever E ∈ B T1 and E ⊇ K. P P For each m ∈ N we have a k(m) ∈ N such that νWm,k(m) ≥ 1 − 2−m−1 ². Set K = m∈N Wm,k(m) . Because every Wm,k(m) is S-closed, K is S-closed, therefore ρ-complete; because every Wm,k(m) is a finite union of sets of diameter at most 2−m , K is ρ-totally bounded, therefore S-compact (4A2Jd). ?? Suppose, if possible, that E ∈ B1 is such that K ⊆ E and νE < 1 − ². For every m ∈ N, Pm Pm T ν( i≤m Wi,k(i) ) ≥ 1 − i=0 ν(X \ Wi,k(i) ) ≥ 1 − i=0 2−i−1 ² > 1 − ² > νE T because ν is non-negative and finitely additive. So i≤m Wi,k(i) \ E must be non-empty. There is therefore an ultrafilter F on X containing Wi,k(i) \ E for every i ∈ N. Now for each i there must be a j ≤ k(i) such that diam Wj ≤ 2T−i and Wj ∈ F, so F is a ρ-Cauchy filter, and S-converges to x say. Because every Wi,k(i) / E; but K is supposed to be included in E. X X is S-closed, x ∈ i∈N Wi,k(i) = K; because E ∈ S, x ∈ Thus inf{νE : K ⊆ E ∈ B1 } ≥ 1 − ². As ² is arbitrary, we have the result. Q Q (f ) By 416O, there is an S-Radon measure µ on X extending ν. Because µ is just the completion of its restriction to B, it is also T-Radon and T1 -Radon (433Cb). Now µ is G-invariant. P P Take any g ∈ G. Set µg E = µ(g •E) whenever E ⊆ X and g •E ∈ dom µ. The map x 7→ g •x is a homeomorphism for T1 , so µg is also a T1 -Radon measure. (Setting φ(x) = g −1 •x, µg is the image measure µφ−1 .) Again because T and T1 have the same Borel σ-algebras, µg is T-Radon. If E ∈ B, then E and g •E belong to B ⊆ B1 , so µg E = µ(g •E) = ν(g •E) = ν(πg E) = fπg E (z0 ) = fE (z0 ) (because fπg E = fE , as declared in (c) above) = νE = µE. In particular, µg E = µE for every E in the algebra generated by U. But µg and µ are both T-Radon measures, and U is a base for T closed under finite unions, so µg = µ (415H(iv)). As g is arbitrary, µ is G-invariant. Q Q

392

Topological groups

448P

Thus we have found a G-invariant Radon probability measure, and (i) is true. (g) Thus (iii)⇒(i) if G is countable. Now let us consider the general case. Because G is a Polish group, it has a countable dense subgroup H. (Take H to be the subgroup generated by any countable dense subset of G.) Of course there can be no disjoint E, F ∈ B such that X 4σH E and X 4σH F , so there must be an H-invariant Radon measure µ on X, by the arguments of (b)-(f). (H need not be a Polish group in its subspace topology. But if we give it its discrete topology, then x 7→ h•x is still a T1 -homeomorphism for every h ∈ H, so the action of H on X is still continuous if H is given the discrete topology and X is given T1 .) Now µ is G-invariant. P P For any g ∈ G, let µg be the Radon probability measure defined by setting µg E = µ(g •E) whenever this is defined. (As in (f) above, this formula does define a probability measure which is Radon for either T or T1 .) Let f : X → R be any bounded T1 -continuous function. Then

R

f dµg =

R

f (g −1 •x)µ(dx)

(applying 235I with φ(x) = g −1 •x). Now there is a sequence hhn in∈N in H converging to g. In this case, because G is a topological group, g −1 = limn→∞ h−1 n . Because the action of G on X is T1 -continuous, −1 •x, for T , for every x ∈ X. Because f is T -continuous, f (g −1 •x) = lim g −1 •x = limn→∞ h−1 1 1 n→∞ f (hn •x) n in R for every x ∈ X. By Lebesgue’s Dominated Convergence Theorem,

R

f dµg =

R

• f (g −1 •x)µ(dx) = limn→∞ f (h−1 n x)µ(dx) = limn→∞

R

f dµhn =

R

f dµ

because µ is H-invariant, so µhn = µ for every n. As f is arbitrary, µg = µ, by 415I. As g is arbitrary, µ is G-invariant. Q Q Thus (iii)⇒(i) in all cases, and the proof is complete. 448X Basic exercises (a) Show that the results in 448Fb and 448Fd remain true if G is not assumed to be countable. (b) In part (c) of the proof of 448O, show that I is just the set of those d ∈ A such that d ⊆ upr(1 \ a, C) for some a such that 1 4σG a. (c) Show that, in part (c) of the proof of 448P, we can if we wish take Z = X and Σ = C. >(d) Let (X, Σ) be a standard Borel space and Σ0 a countable subalgebra of Σ. Show that there is a sequence hhEni ii∈N iP n∈N of partitions of unity in Σ such that whenever ν : Σ → R is a finitely additive ∞ functional and νX = i=0 νEni for every n ∈ N, then ν¹Σ0 is countably additive. (e) Let (X, Σ) be a standard Borel space, Y any set, T a σ-algebra of subsets of Y and J a σ-ideal of subsets of T. Suppose that (Y, T, J ) is complete in the sense that H ∈ T whenever H ⊆ F ∈ J . Let θ : Σ → L∞ (T/J ) be a non-negative, sequentially order-continuous additive function. Show that there is a family hµy iy∈Y of probability measures with domain Σ such that, for every E ∈ Σ, the function y 7→ µy E is T-measurable and represents θE. > (f ) Set X = [0, 1] \ Q, G = Q and define • : G × X → X by requiring that g •x − g − x ∈ Z for g ∈ G, x ∈ X. Show that this is a Borel measurable action and that Lebesgue measure on X is G-invariant. Find a metric on X, inducing its topology, for which all the maps x 7→ g •x are isometries. > (g) Show that a Polish group carries Haar measures iff it is locally compact. (Hint: 443E.) (h) Give ZN its usual (product) topology and abelian group structure. Show that it is a Polish group, and has no Haar measure. >(i) Let (X, ρ) be a metric space, G a group and • an action of G on X such that x 7→ g •x is an isometry for every g ∈ G. (i) Show that if µ is a G-invariant quasi-Radon probability measure on X then {g •x : g ∈ G} is totally bounded for every x in the support of µ. (ii) Show that if the action is transitive and there is a non-zero G-invariant quasi-Radon measure on X, then X is covered by totally bounded open sets. (iii) Suppose that X has measure-free weight (see §438; for instance, X could be separable). Show that if the action is transitive and there is a G-invariant Borel probability measure on X then X is totally bounded.

449B

Amenable groups

393

(j) Let (A, µ ¯) be a measure algebra, with the operation 4 and the measure-algebra topology. (i) Show that A is a topological group. (ii) Show that if µ ¯ is σ-finite and A has countable Maharam type, it is a Polish group. (iii) Show that if (A, µ ¯) is semi-finite and not purely atomic, then A has no Haar measures. (k) Let (A, µ ¯) be the measure algebra of Lebesgue measure on [0, 1], with its measure metric, and G = Autµ¯ A the group of measure-preserving automorphisms on A. (i) Show that if we give G the topology induced by the pointwise topology of the isometry group of A, then it is a Polish group. (Hint: 441Xm(iv).) (ii) Show that if ν is a G-invariant Borel probability measure on A, then ν{0, 1} = 1. (Hint: 448Xi(i).) 448Y Further exercises (a) Let A be a Dedekind σ-complete Boolean algebra, and G a subgroup of Aut A; let G∗σ be the countably full local semigroup generated by G, and write H for the union of all the full local semigroups generated by countable subgroups of G (following the definition in 394A as written, without troubling about whether A is Dedekind complete). (i) Show that G∗σ ⊆ H. (ii) Find an example in which H 6= G∗σ . (iii) Show that if A is Dedekind complete then G∗σ = H. (iv) Show that if A is ccc then G∗σ = H is the full local semigroup generated by G. (b) In 448N, show that θ is uniquely defined. 448 Notes and comments The keys to the work here are in 448F, 448G and 448L. Even though we no longer have a Dedekind complete algebra, the fact that we are working with countable groups means that the suprema we actually need are defined. The final step, however, uses yet another idea. In a standard Borel space, given a finitely additive functional on the σ-algebra, we can sometimes confirm an adequate approximation to countable additivity by looking at only countably many sequences (448Xd). This enables us to pass from a G-invariant map θ : A → L∞ (D) to a G-invariant Radon measure (448Xe, parts (d)-(f) of the proof of 448P), without needing to know anything about the algebra D except that it is Dedekind σ-complete. In particular (and in contrast to the corresponding step in 394P) we do not need to suppose that D is a measurable algebra. I do not know whether there is a useful ergodicity condition which could be added to the hypotheses of 448O to ensure that D there becomes {0, 1}. 448P was proved in the case G = Z by Nadkarni 90; the extension to general Borel actions by Polish groups is due to Becker & Kechris 96. (See Nadkarni 90 for notes on the history of the problem, and Kechris 95 for the basic general theory of Polish groups and Borel actions.) It is a remarkable result, but its application is limited by the difficulty of determining whether either condition (ii) or condition (iii) is satisfied. Much commoner situations are those of 448Xf-448Xk, where either there is no invariant measure or we can find one easily.

449 Amenable groups I end this chapter with a brief introduction to ‘amenable’ topological groups. I start with the definition (449A) and straightforward results assuring us that there are many amenable groups (449C). At a slightly deeper level we have a condition for a group to be amenable in terms of a universal object constructible from the group, not invoking ‘all compact Hausdorff spaces’ (449E). I give some notes on amenable locally compact groups, concentrating on a list of properties equivalent to amenability (449I), and end with a version of Tarski’s theorem characterizing amenable discrete groups (449K). 449A Definition A topological group G is amenable if whenever X is a non-empty compact Hausdorff space and • is a continuous action of G on X, then there is a G-invariant Radon probability measure on X. Warning: other definitions have been used, commonly based on conditions equivalent to amenability for locally compact Hausdorff groups, such as those listed in 449I(ii)-449I(x). In addition, many authors use the phrase ‘amenable group’ to mean a group which is amenable in its discrete topology. The danger of this to the non-specialist is that many theorems concerning amenable discrete groups do not generalize in the ways one might expect.

394

Topological groups

449B

449B Lemma Let G be a topological group, X a locally compact Hausdorff space, and • a continuous action of G on X. (a) Writing C0 for the Banach space of continuous real-valued functions on X vanishing at ∞ (436I), the map a 7→ a−1 •f : C0 → C0 (definition: 441Ac) is uniformly continuous for the right uniformity on G and the norm uniformity of C0 , for any f ∈ C0 . (b) If µ is a G-invariant Radon measure on X and 1 ≤ p < ∞, then a 7→ a−1 •u : G → Lp (definition: 441K) is uniformly continuous for the right uniformity on G and the norm uniformity of Lp = Lp (µ), for any u ∈ Lp (µ). proof (a)(i) Note first that if a ∈ G and f ∈ C0 , then x 7→ a•x : X → X is a homeomorphism (4A5Bd), so x 7→ f (a•x) belongs to C0 ; but this is just the function a−1 •f . (ii) For any ² > 0 and f ∈ C0 there is a neighbourhood V of the identity e of G such that kf − P?? Suppose, if possible, otherwise. For each symmetric neighbourhood V a−1 •l f k∞ ≤ ² for every a ∈ V . P of e set ² 2

QV = {(a, x) : a ∈ V, x ∈ X, |f (x)| ≥ , |f (x) − f (a•x)| ≥ ²}. We are supposing that there are a ∈ V and x ∈ X such that |f (x) − f (a•x)| ≥ ². If |f (x)| ≥ 21 ² then (a, x) ∈ QV . Otherwise, |f (a•x)| ≥ 21 ², a−1 ∈ V and |f (a•x) − f (a−1 a•x)| ≥ ², so (a−1 , a•x) ∈ QV . Thus QV is never empty. Since QV ⊆ QV 0 whenever V ⊆ V 0 , there is an ultrafilter F on G × X such that QV ∈ F for every neighbourhood V of e. Setting π1 (a, x) = a and π2 (a, x) = x for (a, x) ∈ G × X, we see that the image filter π1 [[F]] contains every neighbourhood of e, so converges to e, while π2 [[F]] contains the compact set {x : |f (x)| ≥ 12 ²}, so must have a limit x0 in X. So F → (e, x0 ) in G × X. Next, because the action is continuous, •[[F]] → e•x0 = x0 , and there must be an F ∈ F such that |f (x0 ) − f (a•x)| ≤ 13 ² for every (a, x) ∈ F . Also, of course, there is an F 0 ∈ F such that |f (x0 ) − f (x)| ≤ 13 ² whenever (a, x) ∈ F 0 . But now there is an (a, x) ∈ QG ∩ F ∩ F 0 , and we have |f (x) − f (a•x)| ≥ ²,

|f (x0 ) − f (a•x)| ≤ 31 ²,

|f (x0 ) − f (x)| ≤ 13 ²

simultaneously, which is impossible. X XQ Q Now we find that if a, b ∈ G, ab−1 ∈ V and x ∈ X, then |(a−1 •f )(x) − (b−1 •f )(x)| = |f (a•x) − f (b•x)| = |f (ab−1 •(b•x) − f (b•x)| ≤ ². As x is arbitrary, ka−1 •f − b−1 •f k∞ ≤ ²; as ² is arbitrary, a 7→ a−1 •f is uniformly continuous for the right uniformity. (b)(i) Suppose that f : X → R is continuous and has compact support K. Let H ⊇ K be an open set of finite measure. Then V0 = {a : a ∈ G, a•x ∈ H for every x ∈ K} is a neighbourhood of e. P P If we take a continuous function f0 with compact support such that χK ≤ f0 ≤ χH (4A2G(e-i)), then Q Let ² > 0. By (a) again, there is V0 ⊇ {a : kf0 − a−1 •f0 k∞ < 1}, which is a neighbourhood of e by (a). Q a symmetric neighbourhood V1 of e such that (kf − a−1 •f k∞ )p µH ≤ ²p for every a ∈ V1 ; we may suppose that V1 ⊆ V0 . If a ∈ V1 , f (x) = f (a•x) = 0 for every x ∈ X \ H, so that kf − a−1 •f kpp =

R

H

|f − a−1 •f |p dµ ≤ (kf − a−1 •f k∞ )p µH ≤ ²p .

Now suppose that a, b ∈ G and that ab−1 ∈ V1 . Then Z ka−1 •f − b−1 •f kpp =

|f (a•x) − f (b•x)|p µ(dx) Z |f (a•(b−1 •x)) − f (b•(b−1 •x))|p µ(dx)

= (because µ is G-invariant, see 441L)

Z =

|ba−1 •f − f )|p dµ ≤ ²p ,

and ka−1 •f − b−1 •f kp ≤ ². As ² is arbitrary, a 7→ (a−1 •f )• is uniformly continuous for the right uniformity.

449C

Amenable groups

395

(ii) In general, given u ∈ Lp (µ) and ² > 0, there is an f ∈ Ck (X) such that ku − f • kp ≤ ² (416I). Let V be a neighbourhood of e such that ka−1 •f − a−1 •f k ≤ ² whenever ab−1 ∈ V ; then ka−1 •u − (a−1 •f )• kp = ku − f • kp (because µ is G-invariant), so ka−1 •u − b−1 •ukp ≤ ka−1 •u − a−1 •f • kp + ka−1 •f − b−1 •f kp + kb−1 •f • − b−1 •ukp ≤ 3² whenever ab−1 ∈ V (using 441Kc). As ² is arbitrary, a 7→ a−1 •u is uniformly continuous for the right uniformity. This completes the proof. 449C Theorem (a) Let G and H be topological groups such that there is a continuous surjective homomorphism from G onto H. If G is amenable, so is H. (b) Let G be a topological group and suppose that there is a dense subset A of G such that every finite subset of A is included in an amenable subgroup of G. Then G is amenable. (c) Let G be a topological group and H a normal subgroup of G. If H and G/H are both amenable, so is G. (d) Let G be a topological group with two amenable subgroups H0 and H1 such that H0 is normal and H0 H1 = G. Then G is amenable. (e) The product of any family of amenable topological groups is amenable. (f) Any abelian topological group is amenable. (g) Any compact Hausdorff topological group is amenable. proof (a) Let φ : G → H be a continuous surjective homomorphism. Let X be a non-empty compact Hausdorff space and • : H × X → X a continuous action. For a ∈ G and x ∈ X, set a•1 x = φ(a)•x. Then • 1 is a continuous action of G on X, so there is a G-invariant Radon probability measure µ on X. Because φ[G] = H, µ is also H-invariant; as X and • are arbitrary, H is amenable. (b)(i) Let X be a non-empty compact Hausdorff space and • a continuous action of G on X. Let P be the set of Radon probability measures on X with the narrow topology, so that P is a compact Hausdorff space (437O). For a ∈ G and x ∈ X, set Ta (x) = a•x, so that Ta : X → X is a homeomorphism. For a ∈ G and µ ∈ P write a•µ for the image measure µTa−1 , so that a•µ ∈ P (418I); it is easy to check that (a, µ) 7→ a•µ : G × P → P is an action. The point is that it is continuous. P P Let f ∈ C(X), a0 ∈ G, µ0 ∈ P 1 •f k and ² > 0. By 449Ba, there is a neighbourhood V of a0 in G such that ka−1 •f − a−1 ∞ ≤ 2 ² for every R −1 R −10 1 a ∈ V . Next, there is a neighbourhood W of µ0 in P such that | a0 •f dµ − a0 •f dµ0 | ≤ 2 ² for every µ ∈ W . But now, if a ∈ V and µ ∈ W , Z Z Z Z • • | f d(a µ) − f d(a0 µ0 )| = | f Ta dµ − f Ta0 dµ0 | Z Z • = | a−1 •f dµ − a−1 0 f dµ0 | Z Z Z Z −1 • −1 • −1 • • ≤ | a f dµ − a0 f dµ| + | a0 f dµ − a−1 0 f dµ0 | 1

• ≤ ka−1 •f − a−1 0 f k∞ + 2 ² ≤ ². R As ², a0 and µ0 are arbitrary, (a, µ) 7→ f d(a•µ) is continuous; as f is arbitrary,

•

is continuous. Q Q

(ii) Because the topology of P is Hausdorff, it follows that Qa = {µ : µ ∈ P, a•µ = µ} is closed in P for any a ∈ G, and that Gµ = {a : a ∈ G, a•µ = µ} is closed in G for any µ ∈ P . Now for any finite subset I of A there is an amenable subgroup HI of G including I. The restriction of the action to HIT× X is a T continuous action of HI on X, so has an HI -invariant Radon probability measure, and a∈I Qa ⊇ a∈HI Qa T is non-empty. Because P is compact, there is a µ ∈ a∈A Qa . Since Gµ includes the dense set A, it is the whole of G, and µ is G-invariant. As X and • are arbitrary, G is amenable. (c) Let X be a compact Hausdorff space and • a continuous action of G on X. Let P be the space of Radon probability measures on X with its vague topology. Define a•µ, for a ∈ G and µ ∈ P , as in (b-i) above, so that this is a continuous action of G on P . Set Q = {µ : µ ∈ P, a•µ = µ for every a ∈ H}; then Q is a closed subset of P and, because H is amenable, is non-empty, since it is the set of H-invariant Radon probability measures on X. Next, b•µ ∈ Q for every µ ∈ Q and b ∈ G. P P If a ∈ H, then

396

Topological groups

449C

a•(b•µ) = (ab)•µ = (bb−1 ab)•µ = b•((b−1 ab)•µ) = b•µ, because H is normal, so b−1 ab ∈ H. As a is arbitrary, b•µ ∈ Q. Q Q Accordingly we have a continuous action of G on the compact Hausdorff space Q. If a ∈ H and b ∈ G, then b•µ = (ba)•µ for every µ ∈ Q. We therefore have a map • : G/H × Q → Q defined by setting b• •µ = b•µ for every b ∈ G, µ ∈ Q. Moreover, this is a continuous action, because if W ⊆ Q is relatively open then {(b, µ) : b•µ ∈ W } is open in G × Q, so its image {(b• , µ) : b• •µ ∈ W } is open in (G/H) × Q. Because G/H is amenable, λ on Q. R R there is a (G/H)-invariant Radon probability measure R Now consider the formula p(f ) = Q ( X f (x)µ(dx))λ(dµ). If f ∈ C(X), then µ 7→ X f (x)µ(dx) is continuous for the vague topology on Q, so p(f ) is well-defined. Clearly p is a linear functional, p(f ) ≥ 0 if R f ≥ 0, and p(χX) = 1; so there is a Radon probability measure ν on X such that p(f ) = f dν for every f ∈ C(X) (436J/436K). If b ∈ G, then Z

Z

Z ¡Z

f d(b ν) = •

¢

f Tb dν = p(f Tb ) = f Tb dµ λ(dµ) Q X Z Z Z Z Z Z • • • = f d(b µ)λ(dµ) = f d(b µ)λ(dµ) = f dµλ(dµ) Q

X

Q

X

Q

X

(because λ is G/H-invariant) Z = f dν for every f ∈ C(X), so that b•ν = ν. Thus ν is G-invariant. As X and

•

are arbitrary, G is amenable.

(d) The canonical map from H1 to G/H0 is a continuous surjective homomorphism. By (a), G/H0 is amenable; by (c), G is amenable. (e) By (c) or (d), the product of two amenable topological groups is amenable, since each can be regarded as a normal subgroup of the product. It follows that the product of finitely many amenable topological groups is amenable. Now let hGi ii∈I be any family of amenable topological groups with product G. For finite J ⊆ I let HJ be the set of those aQ∈ G such that a(i) is the identity in Gi for every i ∈ I \J. Then HJ is isomorphic (as topological group) to i∈J Gi , so is amenable. Since {HJ : J ∈ [I]<ω } is an upwards-directed family of subgroups of G with dense union, (b) tells us that G is amenable. (f )(i) The first step is to observe that the group Z, with its discrete topology, is amenable. P P Let X be a compact Hausdorff space and • a continuous action of Z on X. Set φ(x) = 1•x for x ∈ X. Then φ : X → X is continuous, so by 437S there is a Radon probability measure µ on X such that µ is equal to the image measure µφ−1 . Because φ is bijective, we see that, for E ⊆ X, E ∈ dom µ iff φ[E] ∈ dom(µφ−1 ) = dom µ, and in this case µφ[E] = µE; that is, φ−1 , like φ, is inverse-measure-preserving. Now we can induce on n to see that µ(φn )−1 and µ(φ−n )−1 are equal to µ for every n. Since n•x = φn (x) for every x ∈ X and n ∈ Z, µ is Z-invariant. As X and • are arbitrary, Z is amenable. Q Q (ii) Now let G be any abelian topological group. Q For each finite set I ⊆ G let φI : ZI → G be the continuous homomorphism defined by setting φI (z) = a∈I az(a) for z ∈ ZI . By (i) just above and (e), we know that ZI (with its discrete topology) is amenable, so (a) tells us that the subgroup GI = φI [ZI ] is amenable. But now {GI : I ∈ [G]<ω } is an upwards-directed family of amenable subgroups of G with union G, so from (b) we see that G is amenable. (g) Let G be a compact Hausdorff topological group, X a compact Hausdorff space and • a continuous action of G on X. Fix x0 ∈ X and set φ(a) = a•x0 for a ∈ G; then φ is continuous. Let µ be the Haar probability measure on G, and ν the Radon probability measure µφ−1 on X. If F ∈ dom ν and a ∈ G, then ν(a•F ) = µ{b : φ(b) ∈ a•F } = µ{b : b•x0 ∈ a•F } = µ{b : a−1 •b•x0 ∈ F } = µ{b : a−1 b ∈ φ−1 [F ]} = µφ−1 [F ] = νF. So ν is G-invariant. As X and

•


449D

Amenable groups

397

449D Theorem Let G be a topological group. (a) Write U for the set of bounded real-valued functions on G which are uniformly continuous for the right uniformity of G. Then U is an M -space, and we have an action •l of G on U defined by the formula (a•l f )(y) = f (a−1 y) for a, y ∈ G and f ∈ U . (b) Let Z ⊆ RU be the set of Riesz homomorphisms z : U → R such that z(χG) = 1. Then Z is a compact Hausdorff space, and we have a continuous action of G on Z defined by the formula (a•l z)(f ) = z(a−1 •l f ) for a ∈ G, z ∈ Z and f ∈ U . (c) Setting a ˆ(f ) = f (a) for a ∈ G and f ∈ U , the map a 7→ a ˆ : G → Z is a continuous function from G b onto a dense subset of Z. If a, b ∈ G then a•lˆb = ab. (d) Now suppose that X is a compact Hausdorff space, (a, x) 7→ a•x is a continuous action of G on X, and x0 ∈ X. Then there is a continuous function φ : Z → X such that φ(ˆ e) = x0 and φ(a•l z) = a•φ(z) for every a ∈ G and z ∈ Z. (e) If G is Hausdorff then the action of G on Z is faithful and the map a 7→ a ˆ is a homeomorphism between G and its image in Z. proof (a) Because U is a norm-closed Riesz subspace of Cb (G) containing the constant functions (4A2Jg), it is an M -space. To see that the given formula defines an action, we need to check that a•l f belongs to U whenever a ∈ G and f ∈ U . Of course a•l f is continuous and ka•l f k∞ = kf k∞ is finite. If ² > 0 there is a neighbourhood V of the identity e in G such that |f (b) − f (c)| ≤ ² whenever b, c ∈ G and bc−1 ∈ V ; now a−1 V a is a neighbourhood of e, and if bc−1 ∈ aV a−1 then (a−1 b)(a−1 c)−1 ∈ V , so |(a•l f )(b) − (a•l f )(c)| = |f (a−1 b) − f (a−1 c)| ≤ ². As ² is arbitrary, a•l f is uniformly continuous with respect to the right uniformity. Now •l is an action, just as in 441Ac. (b)(i) Because U is an M -space with standard order unit χG, Z is a compact Hausdorff space and U can be identified, as normed Riesz space, with C(Z) (354L). For any a ∈ G, the map f 7→ a•l f : U → U is a Riesz homomorphism leaving the constant functions fixed. So we can define a•l z, for z ∈ Z, by saying that (a•l z)(f ) = z(a−1 •l f ) for any f ∈ U , and a•l z will belong to Z for any a ∈ G and z ∈ Z. As usual, it is easy to check that this formula defines an action of G on Z. P Take a0 ∈ G, z0 ∈ Z and any neighbourhood W of a0 •l z0 in Z. (ii) (a, z) 7→ a•l z is continuous. P Because U corresponds to the whole of C(Z), and Z is completely regular, there is an f ∈ U + such that −1 • 1 • (a0 •l z0 )(f ) = 1 and z(f ) = 0 for every z ∈ Z \ W . Set W0 = {z : z(a−1 0 l f ) > 2 }. Observe that z0 (a0 l f ) = −1 1, so W0 is an open subset of Z containing z0 . Next, set V0 = {a : a ∈ G, ka−1 •l f − a0 •l f k∞ ≤ 21 }. There is a neighbourhood V of e such that |f (b) − f (c)| ≤ 12 whenever b, c ∈ G and bc−1 ∈ V . If a ∈ V a0 then ab(a0 b)−1 = aa−1 0 ∈ V , so • |(a−1 •l f )(b) − (a−1 0 l f )(b)| = |f (ab) − f (a0 b)| ≤

1 2

for every b ∈ G, and a ∈ V0 . Thus V0 ⊇ V a0 is a neighbourhood of a0 . Now if a ∈ V0 and z ∈ W0 we shall have • (a•l z)(f ) = z(a−1 •l f ) ≥ z(a−1 0 lf ) −

1 2

>0

and a•l z ∈ W . As a0 , z0 and W are arbitrary, the action of G on Z is continuous. Q Q (c) Of course a ˆ, as defined, is a Riesz homomorphism taking the correct value at χG, so belongs to Z. Because the topology on G is compatible with the right uniformity (4A5Ha), the map a 7→ a ˆ is continuous. ?? If {ˆ a : a ∈ G} is not dense in Z, there is a non-zero h ∈ C(Z) such that h(ˆ a) = 0 for every a ∈ G; but as U is identified with C(Z), there is an f ∈ U such that z(f ) = h(z) for every z ∈ Z. In this case, f cannot be the zero function, but f (a) = a ˆ(f ) = h(ˆ a) = 0 for every a ∈ G. X X Thus the image of G is dense, as claimed. If a, b ∈ G and f ∈ U then b ), (a•lˆb)(f ) = ˆb(a−1 •l f ) = (a−1 •l f )(b) = f (ab) = ab(f b so a•lˆb = ab.

398

Topological groups

449D

(d) We have a Riesz homomorphism T : C(X) → RG defined by setting (T g)(a) = g(a•x0 ) for every g ∈ C(X) and a ∈ G. Now T g ∈ U for every g ∈ C(X). P P T g(a) = (a−1 •g)(x0 ); since the map a 7→ a−1 •g is uniformly continuous (449Ba), so is T g. Q Q Of course T (χX) = χG. So if z ∈ Z, zT : C(X) → R is a Riesz homomorphism such that (zT )(χX) = 1. There is therefore a unique φ(z) ∈ X such that (zT )(g) = g(φ(z)) for every g ∈ C(X) (354L). Since the function z 7→ g(φ(z)) = z(T g) is continuous for every g ∈ C(X), φ is continuous. Now suppose that a ∈ G. Then φ(ˆ a) = a•x0 . P P If g ∈ C(X), then g(φ(ˆ a)) = a ˆ(T g) = (T g)(a) = g(a•x0 ). Q Q So if a, b ∈ G, then b = (ab)•x0 = a•(b•x0 ) = a•φ(b). φ(a•lˆb) = φ(ab) Since {ˆb : y ∈ G} is dense in Z, and all the functions here are continuous, φ(a•l z) = a•φ(z) for all a ∈ G and z ∈ Z. (e) Now suppose that the topology of G is Hausdorff. Then it is defined by the bounded uniformly continuous functions (4A2Ja); the map a 7→ a ˆ is therefore injective and is a homeomorphism between G and its image in Z. If a, b ∈ G are distinct, then a•eˆ = a ˆ 6= ˆb = b•eˆ, so the action is faithful. Remark Following Brook 70, the space Z, together with the canonical action of G on it and the map a 7→ a ˆ : G → Z, is called the greatest ambit of the topological group G. 449E Corollary Let G be a topological group. Then the following are equiveridical: (i) G is amenable; (ii) there is a G-invariant Radon probability measure on the greatest ambit of G; (iii) writing U for the space of bounded real-valued functions on G which are uniformly continuous for the right uniformity, there is a positive linear functional p : U → R such that p(χG) = 1 and p(a•l f ) = p(f ) for every f ∈ U and a ∈ G. proof Let Z be the greatest ambit of G. (i)⇒(ii) As soon as we know that Z is a compact Hausdorff space and the action of G on Z is continuous (449Db), this becomes a special case of the definition of ‘amenable topological group’. (ii)⇒(i) Let µ be a G-invariant Radon probability measure on Z. Given any continuous action of G on a non-empty compact Hausdorff space X, fix x0 ∈ X and let φ : Z → X be the corresponding continuous function such that φ(a•l z) = a•φ(z) for every a ∈ G and z ∈ Z, as in 449Dd. Let ν be the image measure µφ−1 . Then ν is a Radon probability measure on X (418I). If F ∈ dom ν and a ∈ G, then ν(a•F ) = µφ−1 [a•F ] = µ(a•l φ−1 [F ]) = µφ−1 [F ] = νF . As a and F are arbitrary, ν is G-invariant; as X and

•


(ii) ⇐⇒ (iii) The identification of U with C(Z) (see (b-i) of the proof of 449D) means that we have a one-to-one correspondence between Radon probabilityR measures µ on Z and positive linear functionals p on U such that p(χG) = 1, given by the formula p(f ) = z(f )µ(dz) for f ∈ U (436J/436K). Now µ is G-invariant Z Z ⇐⇒ (a•l z)(f )µ(dz) = z(f )µ(dz) for every f ∈ U, a ∈ G (441L)

Z

Z z(a−1 •l f )µ(dz) = z(f )µ(dz) for every f ∈ U, a ∈ G Z Z • ⇐⇒ z(a l f )µ(dz) = z(f )µ(dz) for every f ∈ U, a ∈ G ⇐⇒

⇐⇒ p(a•l f ) = p(f ) for every f ∈ U, a ∈ G. So there is a G-invariant µ, as required by (ii), iff there is a G-invariant p as required by (iii).

449H

Amenable groups

399

449F Corollary If G is an amenable topological group, then every open subgroup of G is amenable. proof (a) Let H be an open subgroup of G. Write U for the set of bounded real-valued functions on G which are uniformly continuous for the right uniformity of G, and V for the set of bounded real-valued functions on H which are uniformly continuous for the right uniformity of H. Let A ⊆ G be a set meeting each right coset of H in just one point, so that each member of G is uniquely expressible as ya where y ∈ H and a ∈ A. Define T : V → RG by setting (T f )(ya) = f (y) whenever f ∈ V , y ∈ H and a ∈ A. Then T is a positive linear operator. Also T [V ] ⊆ U . P P Let f ∈ V . Of course T f is bounded. If ² > 0, there is a neighbourhood W of the identity in H such that |f (x) − f (y)| ≤ ² whenever x, y ∈ H and xy −1 ∈ W . Because H is open, W is also a neighbourhood of the identity in G. Now suppose that x, y ∈ G and xy −1 ∈ W . Express x as x0 a and y as y0 b where x0 , y0 ∈ H and a, b ∈ A. Then x0 ab−1 y0−1 ∈ W ⊆ H, so ab−1 ∈ H and a ∈ Hb and a = b and x0 y0−1 ∈ W and |(T f )(x) − (T f )(y)| = |f (x0 ) − f (y0 )| ≤ ². As ² is arbitrary, T f is uniformly continuous and belongs to U . Q Q (b) Next, b•l (T f ) = T (b•l f ) whenever f ∈ V and b ∈ H. P P If x ∈ G, express it as ya where y ∈ H and a ∈ H. Then (b•l T f )(x) = (T f )(b−1 x) = (T f )(b−1 ya) = f (b−1 y) = (b•f )(y) = T (b•l f )(x). Q Q (c) By 449E, there is a positive linear functional p : U → R such that p(χG) = 1 and p(a•l f ) = p(f ) whenever f ∈ U and a ∈ G. Set q(f ) = p(T f ) for f ∈ V ; then q is a positive linear operator, q(χH) = 1 and q is H-invariant, by (b). So by 449E in the other direction, H is amenable. 449G Example Let F2 be the free group on two generators, with the discrete topology. Then F2 is a σ-compact unimodular locally compact Polish group. But it is not amenable. P P Let a and b be the generators of F2 . Then every element of F2 is uniquely expressible as a word (possibly empty) in the letters a, b, a−1 , b−1 in which the letters a, a−1 are never adjacent and the letters b, b−1 are never adjacent. Write A for the set of elements of F2 for which the canonical word does not begin with either b or b−1 , and B for the set of elements of F2 for which the canonical word does not begin with either a or a−1 . Then A ∪ B = F2 and A ∩ B = {e}. ?? Suppose, if possible, that F2 is amenable. Every member of `∞ (F2 ) is uniformly continuous with respect to the right uniformity. So there is an F2 -invariant positive linear functional p : `∞ (F2 ) → R such that p(χF2 ) = 1. Let ν be the corresponding non-negative additive functional on PF2 , so that νC = p(χC) for every C ⊆ F2 . For c ∈ F2 and C ⊆ F2 , c•l χC = χ(cC), so ν(cC) = νC for every C ⊆ F2 and c ∈ F2 . In particular, ν(bn A) = νA for every n ∈ Z; but as all the bn A, for n ∈ Z, are disjoint, νA = 0. Similarly νB = 0 and 0 = ν(A ∪ B) = νF2 = p(χF2 ) = 1, which is absurd. X X Thus F2 is not amenable, as claimed. Q Q 449H In this section so far, I have taken care to avoid assuming that groups are locally compact. Some of the most interesting amenable groups are very far from being locally compact (e.g., 449Xc). But of course a great deal of work has been done on amenable locally compact groups. In particular, there is a remarkable list of equivalent properties, some of which I will now present. Lemma Let G be a locally compact Hausdorff topological group, and U the space of bounded real-valued functions on G which are uniformly continuous for the right uniformity, as in 449D-449E. Let µ be a left Haar measure on G, and ∗ the corresponding convolution on L0 (µ) (444O). (a) If h ∈ L1 (µ) and f ∈ L∞ (µ) then h ∗ f ∈ U . (b) Let p : U → R be a positive linear functional such that p(a•l f ) = p(f ) for every f ∈ U and every R a ∈ G. Then p(h ∗ f ) = p(f ) h dµ for every h ∈ L1 (µ) and every f ∈ U . proof (a) For any x ∈ G, (h ∗ f )(x) =

R

h(xy)f (y −1 )µ(dy) =

R

(x−1 •l h) × f˜,

400

Topological groups

449H

where f˜(y) = f (y −1 ) whenever y −1 ∈ dom f . By 449Bb, applied to the left action of G on itself, x 7→ (x−1 •l h)• : G → L1R(µ) is uniformly continuous for the right uniformity of G and the norm uniformity of L1 (µ). Since u 7→ u × v : L1 (µ) → R is uniformly continuous for every v ∈ L∞ (µ), x 7→ (h ∗ f )(x) = R −1 (x •l h)• × f˜• is uniformly continuous for the right uniformity (3A4Cb). Of course supx∈G |(h ∗ f )(x)| ≤ khk1 kf k∞ is finite, so h ∗ f ∈ U . R (b) Let ² > 0. Then there are a compact set K ⊆ G such that G\K |h|dµ ≤ ² (412Jd) and a symmetric neighbourhood V0 of e such that |f (x) xy −1 ∈ V0 . Let a0 , . . . , aSn ∈ G be such that R S S − f (y)| ≤ ² whenever K ⊆ i≤n ai V0 , and set Ei = ai V0 \ j
i=0

¯ =¯ ¯ =¯

Z

h(y)f (y −1 x)µ(dy) −

n X

Z h(y)f (y −1 x)µ(dy) +

i=0 n X

F

Z ≤ kf k∞

|h|dµ + F

¯ ¯

i=0

Z ≤ kf k∞

n X

|h|dµ + ² X\K

Thus kh ∗ f −

i=0

Z (

i=0

Z Ei

Ei

¯ ¯ h(y)f (y −1 x)µ(dy) − αi f (a−1 i x))

¯ ¯ h(y)(f (y −1 x) − f (a−1 i x))µ(dy)

n Z X i=0

Pn

¯ ¯ αi f (a−1 i x)

|h|dµ ≤ ²(kf k∞ + khk1 ).

Ei

αi ai •l f k∞ ≤ ²(kf k∞ + khk1 ).

Since p(ai •l f ) = p(f ) for every i, it follows that Z |p(h ∗ f ) − p(f )

h dµ| ≤ ²(kf k∞ + khk1 )p(χG) + |

n X

Z αi −

h dµ||p(f )|

i=0

Z ≤ ²(kf k∞ + khk1 )p(χG) + |

h dµ|kf k∞ p(χG) F

≤ ²(2kf k∞ + khk1 )p(χG). As ² is arbitrary, p(h ∗ f ) = p(f )

R

h dµ, as claimed.

+ 449I Theorem Let G be a locally compact Hausdorff group, and µ a left Haar measure on G. Let Ck1 R be the set of continuous functions h : G → [0, ∞[ with compact supports such that h dµ = 1. Then the following are equiveridical: (i) G is amenable; (ii) there is a positive linear functional p : Cb (X) → R such that p(χG) = 1 and p(a•l f ) = p(f ) for every u ∈ Cb (X) and every a ∈ G, where a•l f is defined as in 441Ac; (iii) there is a positive linear functional p˜ : L∞ (µ) → R such that p˜(χG• ) = 1 and p˜(a•l u) = p˜(u) for every u ∈ L∞ (µ) and every a ∈ G, where a•l u is defined as in 441K and 443G; R (iv) there is a positive linear functional p˜ : L∞ (µ) → R such that p˜(χG• ) = 1 and p˜(g ∗ f )• = p˜(f • ) g dµ for every g ∈ L1 (µ) and f ∈ L∞ (µ); (v) there is a positive linear functional p˜ : L∞ (µ) → R such that p˜(χG• ) = 1 and p˜(ν ∗ u) = νG · p˜(u) for every u ∈ L∞ (µ) and every totally finite Radon measure ν on G, where ν ∗ u is defined as in 444Ma; + ∞ R (vi) for every finite R set I ⊆ G, finite set J ⊆ L (µ) and ² > 0, there is an h ∈ Ck1 such that | f (ax)h(x)µ(dx) − f (x)h(x)µ(dx)| ≤ ² for every a ∈ I and f ∈ J;

449I

Amenable groups

401

R + (vii) for every finite set J ⊆ L1 (µ) and ² > 0, there is an h ∈ Ck1 such that kg ∗ h − ( g dµ)hk1 ≤ ² for every g ∈ J; + (viii) for every compact set K ⊆ G and ² > 0, there is an h ∈ Ck1 such that ka•l h − hk1 ≤ ² for every −1 a ∈ K, where (a•l h)(x) = h(a x) for every x ∈ G; (ix) for every compact set K ⊆ G and ² > 0, there is a compact set L ⊆ G with non-zero measure such that µ(L4aL) ≤ ²µL for every a ∈ K; (x) (Emerson & Greenleaf 67) for every compact set K ⊆ G and ² > 0, there is a compact set L ⊆ G with non-zero measure such that µ(KL) ≤ (1 + ²)µL. proof (a)(i)⇒(v) Write U for the space of bounded real-valued functions on G which are uniformly continuous for the right uniformity. Then we have a positive linear functional p : U → R such that p(χG) = 1 and p(a•l f R) = p(f ) for every f ∈ U and a ∈ G. Now if f ∈ L∞ = L∞ (µ), h1 , h2 ∈ L1 = L1 (µ) and R h1 dµ = h2 dµ, then p(h1 ∗ f ) = p(h2 ∗ f ). P P By 449Ha, both h1 ∗ f and h2 ∗ f belong to U . Set h = h1 − h2 . By 444T, there is a neighbourhood V of the identity e of G such that kh ∗ ν − hk1 ≤ ² whenever ν is a quasi-Radon measure on G such that νV = νG = 1. In particular, taking ν to be the indefinite-integral measure over µ defined from g =

1 µV

χV , kh ∗ g − hk1 ≤ ² (using 444Pb). Now

|p(h1 ∗ f ) − p(h2 ∗ f )| = |p(h ∗ f )| ≤ |p((h ∗ g) ∗ f )| + |p((h ∗ g − h) ∗ f )| ≤ |p(h ∗ (g ∗ f ))| + k(h ∗ g − h) ∗ f k∞ (because ∗ is associative, 444Oe)

Z ≤ |p(g ∗ f )

h dµ| + kh ∗ g − hk1 kf k∞ ≤ ²kf k∞

by 449Hb. As ² is arbitrary, p(h1 ∗ f ) = p(h2 ∗ f ), as claimed. Q Q Of course p(h ∗ f ) = 0 whenever h ∈ L1 , f ∈ L∞ and f = 0 a.e. We can therefore define a functional R ∞ 1 p˜ : L∞ = L∞ (µ) → R by saying that p˜(f • ) = p(h ∗ f ) whenever f ∈ L , h ∈ L and h dµ = 1. p˜ is R • 1 ∞ positive and linear because p is. It follows that p(h ∗ f ) = p ˜ (f ) h dµ whenever h ∈ L and f ∈ L . Also R p˜(χG• ) = p(χG) = 1 because h ∗ χG = ( h)χG for every h ∈ L1 . If u ∈ L∞ and ν is a totally finite Radon measure onR G, express u as f • where f ∈ L∞ , so that ν ∗ u = (ν ∗ f )• . Taking any non-negative h ∈ L1 such that h dµ = 1, we have h ∗ (ν ∗ f ) = hµ ∗ (ν ∗ f ) (444Pa; here hµ is the indefinite-integral measure, as in 444J) = (hµ ∗ ν) ∗ f (444Ic) = (h ∗ ν)µ ∗ f (444K) = (h ∗ ν) ∗ f (444Pa again). So p˜(ν ∗ u) = p˜((ν ∗ f )• ) = p(h ∗ (ν ∗ f )) Z = p((h ∗ ν) ∗ f ) = (h ∗ ν)dµ · p˜(u) = νG · p˜(u) (444K). As ν and u are arbitrary, p˜ has the required properties. (b)(v)⇒(iii) Take p˜ from (v), a ∈ G and u ∈ L∞ . Let δa be the point-supported probability measure concentrated at a, and f : G → R a function such that f • = u. For any x ∈ G, R (δa ∗ f )(x) = f (y −1 x)δa (dy) = f (a−1 x) = (a•l f )(x),

402

Topological groups

449I

so δa ∗ f = a•l f . Accordingly δa ∗ u = a•l u. Now, of course, p˜(a•l u) = p˜(δa ∗ u) = δa (G)˜ p(u) = p˜(u), as required by (iii). (c)(iii)⇒(vi) Suppose that p˜ has the properties described in (iii). If I ⊆ G and J ⊆ L∞ (µ) are ¤ finite, £ and ² > 0, let J ∗ ⊆ L∞ be the finite set {a−1 •l f • : a ∈ I ∪ {e}, f ∈ J ∪ {χG}}. Let η ∈ 0, 12 be such that 2η(1 + 5 supv∈J ∗ kvk∞ ) ≤ ². By 243Gb, we can identify L∞ with (L1 )∗ , because Haar measures are (by definition) quasi-Radon, and quasi-Radon measure spaces are (strictly) localizableR (415A). Because p˜ ∈ (L∞ )∗ ∼ p(v) − v × u0 | ≤ η for = (L1 )∗∗ has norm 1, there is a u0 ∈ L1R such that ku0 k ≤ 1 and |˜ ∗ every v ∈ J (4A4If). In particular, taking v = χG• , u0 ≥ 1 − η. By 416I, there is a continuous function R R h0 : G → R with compact support such that ku0 − h•0 k1 ≤ η. Then h0 dµ ≥ 1 − 2η and |h0 |dµ ≤ 1 + η. R + 1 So if we set h+ h0 dµ, h = h+ 0 = h0 ∨ 0, γ = 0 we have γ

kh+ 0 − h0 k1 =

1 2

R

|h0 | − h0 ≤ 2η

+ • and kh − h+ 0 k1 = |γ − 1| ≤ 2η, so ku0 − h k1 ≤ 5η, while h ∈ Ck1 . This means that

|˜ p(v) −

R

1 2

v × h• k ≤ η + 5ηkvk∞ ≤ ²

for every v ∈ J ∗ . Now suppose that f ∈ J and a ∈ I. Then ¯ ¯

Z

Z f (ax)h(x)µ(dx) − ¯ =¯ ¯ ≤¯

¯ f (x)h(x)µ(dx)¯ Z

Z (a−1 •l f • ) × h• −

¯ f • × h• ¯

Z (a

−1 •

lf

•

) × h − p˜(a •

−1 •

¯ ¯ • ¯ ¯ ˜(f ) − lf ) + p •

Z

¯ f • × h• ¯

(because p˜ is G-invariant) ≤² because f • and a−1 •l f • belong to J ∗ . So h has the property required in (vi). + (d)(vi)⇒(ii) If I ⊆ G and RJ ⊆ Cb (X) are finite, and ² > 0, then the set A(I, J, ²) of those h ∈ Ck1 R such that | f (ax)h(x)µ(dx) − f (x)h(x)µ(dx)| ≤ ² for every a ∈ IRand f ∈ J is non-empty. So there is a + filter F on Ck1 containing every A(I, J, ²). If we set p(f ) = limh→F f (x)h(x)µ(dx) for f ∈ Cb (X), then p witnesses that (ii) is true.

(e)(ii)⇒(i) Given p as in (ii), its restriction to the space of bounded right-uniformly-continuous functions is positive, linear and G-invariant, so G is amenable. (f )(v)⇒(iv) Take p˜ as in (v). If g ∈ L1 , f ∈ L∞ and g ≥ 0, then p˜(g ∗ f )• = p˜(gµ ∗ f )• = p˜(gµ ∗ f • ) Z • = (gµ)(G)˜ p(f ) = g dµ · p˜(f • ) as required by (iv). (g)(iv)⇒(vii) Suppose that (iv) is true. α) As in (d), just because p˜ ∈ (L∞ )∗ is a positive linear functional and p˜(χG)• = 1, we see that if (α J ⊆ L∞ is finite and ² > 0, then R + A(J, ²) = {h : h ∈ Ck1 , | f × h dµ − p˜(f • )| ≤ ² for every f ∈ J}

449I

Amenable groups

403

R + is non-empty. So we have a filter F on Ck1 containing every A(J, ²), and p˜(f • ) = limh→F f × h dµ for every f ∈ L∞ . R β ) Now 0 = limh→F (g ∗ h)• − ( g dµ)h• for the weak topology of L1 , for every g ∈ L1 . P P Set R(β −1 −1 γ = g dµ. Let f ∈ L∞ . Define g 0 by setting g 0 (x) = ∆(x )g(x ) whenever this is defined, where ∆ is R + the left modular function of G; then g 0 ∈ L1 and g 0 dµ = γ (442K(b-ii)). If h ∈ Ck1 , then Z

ZZ f × (g ∗ h)dµ =

(444Od) =

f (xy)g(x)h(y)µ(dx)µ(dy) Z Z ¡

¢ ∆(x−1 )g 0 (x−1 )f (xy)µ(dx) h(y)µ(dy)

Z (g 0 ∗ f )(y)h(y)µ(dy)

= (444Oa)

Z =

(g 0 ∗ f ) × h dµ.

As we have p˜(g 0 ∗ f )• = p˜(f • )

R

g 0 dµ = γ p˜(f • ),

we get Z lim

h→F

Z f × (g ∗ h − γh)dµ = lim

h→F 0

(g 0 ∗ f − γf ) × h dµ

= p˜(g ∗ f )• − γ p˜(f • ) = 0. As f is arbitrary, and (L1 )∗ can be identified with L∞ , this is all we need. Q Q (γγ ) Now take any finite set J ⊆ L1 and ² > 0. On (L1 )J let T be the locally convex linear space topology which is the product topology when each copy of L1 is given the norm topology, and S the corresponding weak topology. Define T : Ck (G) → (L1 )J by setting ® R T h = (g ∗ h)• − ( g dµ)h• g∈J for h ∈ Ck (G), where Ck (G) is the linear space of continuous real-valued functions on G with compact support. Then T is linear. Moreover, by (β), limh→F T h = 0 in (L1 )J for the product topology, if each copy of L1 is given its weak topology. But this is just S, by 4A4Be. In particular, 0 belongs to the S-closure + + + of T [Ck1 ]. But Ck1 is convex and T is linear, so T [Ck1 ] is convex; by 4A4Ed, 0 belongs to the T-closure of R + + T [Ck1 ]. There is therefore an h ∈ Ck1 such that k(g ∗ h)• − ( g dµ)h• k1 ≤ ² for every g ∈ J. As J and ² are arbitrary, (vii) is true. (h)(vii)⇒(viii) Suppose that (vii) is true and that we have a compact set K ⊆ G and ² > 0. Set η = 31 ². + Fix any h0 ∈ Ck1 . Let V be a neighbourhood of e such that kc•l h0 − h0 k1 ≤ η whenever c ∈ V (443Ge). + Let I ⊆ G be a finite set such that K ⊆ IV . By (vii), there is an h1 ∈ Ck1 such that kb•l h0 ∗ h1 − h1 k1 ≤ η for every b ∈ I,

kh0 ∗ h1 − h1 k1 ≤ η.

+ (I omit brackets because (b•l h0 ) ∗ h1 = b•l (h0 ∗ h1 ), see 444Of.) Set h = h0 ∗ h1 . Then h ∈ Ck1 . P P h is continuous, by 444Rc (or otherwise). If we set Mi = {x : hi (x) 6=R 0} for both i, then h(x) = 0 for every R R x ∈ G \ M0 M1 , so h has compact support. Of course h ≥ 0, and h dµ = h0 dµ h1 dµ = 1 (444Qb), so + h ∈ Ck1 .Q Q If a ∈ K, there are b ∈ I, c ∈ V such that a = bc, so that

ka•l h − hk1 = kb•l (c•l h0 ∗ h1 ) − h0 ∗ h1 k1 ≤ kb•l (c•l h0 − h0 ) ∗ h1 k1 + kb•l h0 ∗ h1 − h1 k1 + kh1 − h0 ∗ h1 k1 ≤ kc•l h0 − h0 k1 + η + η ≤ 3η = ².

404

Topological groups

449I

So this h will serve. (i)(viii)⇒(ix) Suppose that (viii) is true. α) I show first that for any ², δ > 0 and any compact set K ⊆ G there are Borel sets E, F ⊆ G such (α that 0 < µE < ∞, µF ≤ δ and µ(aE4E) ≤ ²µE whenever a ∈ K \ F . P P Of course it is enough to deal + with the case in which µK > 0. Set η = ²δ/µK. Let h ∈ Ck1 be such that ka•l h − hk1 < η for every a ∈ K. Set K0 = {x : h(x) 6= 0} and K ∗ = K0 ∪ KK0 , so that K ∗ ⊆ G is compact. Set Q = {(a, x, t) : a ∈ K, x ∈ G, t ∈ R, either h(x) ≤ t < h(a−1 x) or h(a−1 x) ≤ t < h(x)}. Then Q is a Borel subset of G × G × R included in the compact set K × K ∗ × [0, khk∞ ]. Let µL be Lebesgue measure on R, and let µ × µ × µL be the τ -additive product measure on G × G × R (417D). (Of course this is actually a Radon measure.) For t ∈ R let Et be the open set {x : h(x) > t}. Now 417H tells us that Z µL {t : (a, x, t) ∈ Q}(µ × µ)(d(a, x))

(µ × µ × µL )(Q) = G×G

(where µ×µ is the τ -additive product measure on G×G, so that we can identify µ×µ×µL with (µ×µ)×µL ), as in 417Db)

Z Z |h(a−1 x) − h(x)|µ(dx)µ(da)

= K

G

(we can use 417H again because {x : h(a−1 x) 6= h(x)} ⊆ K ∗ if a ∈ K, and µK ∗ is finite) Z = ka•l h − hk1 µ(da) < ηµK K

(by the choice of h)

Z = ηµK

h = ηµK(µ × µL ){(x, t) : 0 ≤ t < h(x)} Z

∞

= ηµK

µEt µL (dt) 0

as in 252N. (The c.l.d. and τ -additive product measures on G × R coincide, by 417T.) On the other hand, Z (µ × µ × µL )(Q) =

µ{x : (a, x, t) ∈ Q}(µ × µL )(d(a, t)) K×R

(again, we can use 417H because x ∈ K ∗ whenever (a, x, t) ∈ Q) Z = µ(Et 4aEt )(µ × µL )(d(a, t)) K×R Z ∞Z = µ(Et 4aEt )µ(da)µL (dt) 0

K

because Et = aEt = G whenever t < 0 and a ∈ G. So there must be some t > 0 such that R µ(Et 4aEt )µ(da) < ηµKµEt = ²δµEt , K in which case µEt > 0 and F = {a : a ∈ K, µ(Et 4aEt ) ≥ ²µEt } has measure at most δ. So we can set E = Et . Q Q β ) Now let K ⊆ G be a compact set and ² > 0, as in the statement of (ix); enlarging K and lowering (β ² if necessary, we may suppose that µK > 0 and ² ≤ 1. Set K1 = K ∪ KK, so that K1 is still compact. By (α), we have a Borel set E ⊆ G of finite measure such that F = {a : a ∈ K1 , µ(aE4E) > 31 ²µE} has measure less than 21 µK. If a ∈ K, then F ∪ a−1 F cannot cover K, so there is a b ∈ K \ F such that ab ∈ / F; thus b and ab both belong to K1 \ F , and 1 3

2 3

µ(aE4E) ≤ µ(aE4abE) + µ(abE4E) ≤ µ(E4bE) + ²µE ≤ ²µE.

449I

Amenable groups

405

If we now take a compact set L ⊆ E such that 8µ(E \ L) ≤ ²µL, we shall have µ(aL4L) ≤ µ(aE \ aL) + µ(aE4E) + µ(E \ L) 2 3

≤ 2µ(E \ L) + ²(µL + µ(E \ L)) ≤ ²µL for every a ∈ K, as required. (j)(ix)⇒(x) Suppose that (ix) is true, and that K ⊆ G is compact and ² > 0. Enlarging K if necessary, we may suppose that it includes a neighbourhood of e. Of course we may also suppose that ² ≤ 1. α) The first thing to note is that there is a set I ⊆ G such that KI = G and m = supy∈G #({x : (α x ∈ I, y ∈ Kx}) is finite. P P Let V be a neighbourhood of e such that V V −1 ⊆ K. Let I ⊆ G be −1 −1 maximal subject to x V ∩ y V = ∅ for all distinct x, y ∈ I. If x ∈ G, there must be a y ∈ I such that x−1 V ∩ y −1 V 6= ∅, so that x−1 ∈ y −1 V V −1 and x ∈ V V −1 y ⊆ Ky ⊆ KI; as x is arbitrary, G ⊆ KI. If y ∈ G and Iy = {x : x ∈ I, y ∈ Kx}, then Iy ⊆ K −1 y, so that {x−1 V : x ∈ Iy } is a disjoint family of subsets of y −1 KV . But this means that #(Iy )µV ≤ µ(y −1 KV ) = µ(KV ). Accordingly supy∈G #(Iy ) ≤

µ(KV ) µV

is

finite. Q Q β ) Set γ = supa∈K ∆(a), K ∗ = KKK −1 . Let δ > 0 be such that (β δγm < ²(µK − δγ), and let η > 0 be such that 1+

δγm µK−δγ

η δ

≤ (1 + ²)(1 − µK ∗ ).

By (ix), there is a compact set L∗ ⊆ G such that µ(aL∗ 4L∗ ) ≤ ηµL∗ for every a ∈ K ∗ . Set L = {x : x ∈ L∗ , µ(K ∗ \ L∗ x−1 ) ≤ δ}; note that L is closed (because x 7→ µ(K ∗ ∩ L∗ x−1 ) is continuous, see 443C), therefore compact. η δ

(γγ ) µL ≥ (1 − µK ∗ )µL∗ . P P Set W = {(x, y) : x ∈ K ∗ , y ∈ L∗ \ L, xy ∈ / L∗ }. Then W is a relatively compact Borel subset of G × G, so we may apply Fubini’s theorem (in the form 417H) to see that Z Z ∗ ∗ ∗ −1 δµ(L \ L) ≤ µ(K \ L y )µ(dy) = µW −1 [{y}]µ(dy) L∗ \L Z Z = µW [{x}]µ(dx) = µ((L∗ \ L) \ x−1 L∗ )µ(dx) K∗ Z Z ≤ µ(L∗ \ x−1 L∗ )µ(dx) = µ(xL∗ \ L∗ )µ(dx) ≤ ηµK ∗ µL∗ K∗

K∗

by the choice of L∗ . Accordingly η δ

µL = µL∗ − µ(L∗ \ L) ≥ (1 − µK ∗ )µL∗ . Q Q In particular, µL > 0. (δδ ) Set J = {x : x ∈ I, L ∩ Kx 6= ∅}. Then ∆(x)(µK − γδ) ≤ µ(Kx ∩ L∗ ) for every x ∈ J. P P Take y ∈ L ∩ Kx. Then yx−1 ∈ K so ∆(y) = ∆(yx−1 )∆(x) ≤ γ∆(x). Next, µ(K ∗ y \ L∗ ) = ∆(y)µ(K ∗ \ L∗ y −1 ) ≤ δ∆(y). Since x ∈ K −1 y, Kx ⊆ KK −1 y ⊆ K ∗ y and µ(Kx \ L∗ ) ≤ µ(K ∗ y \ L∗ ) ≤ δ∆(y) ≤ δγ∆(x) and µ(Kx ∩ L∗ ) = µ(Kx) − µ(Kx \ L∗ ) ≥ ∆(x)µK − δγ∆(x) = ∆(x)(µK − δγ). Q Q (²²) Now recall that

P x∈I

χ(Kx) ≤ mχG, by the definition of m, so that

406

Topological groups

(µK − δγ)

P x∈J

∆(x) ≤

P

449I

µ(Kx ∩ L∗ ) ≤ mµL∗ .

x∈J

For each x ∈ J, choose zx ∈ Kx ∩ L. Since KI = G, L ⊆ KJ and [ [ [ KL ⊆ KKx ⊆ KKK −1 zx = K ∗ zx , x∈J

X

µ(KL \ L∗ ) ≤

x∈J

x∈J

µ(K ∗ zx \ L∗ ) =

x∈J

≤δ

X

X

∆(zx )µ(K ∗ \ L∗ zx−1 )

x∈J

∆(zx ) ≤ δγ

x∈J

X

∆(x) ≤

x∈J

δγm µL∗ . µK−δγ

Accordingly

µ(KL) ≤ µL∗ (1 +

δγm ) µK−δγ

≤ µL ·

1+

δγm µK−δγ

η δ

1 − µK ∗

(by (γ) above) ≤ (1 + ²)µL, by the choice of δ and η. Thus we have found an appropriate set L. (k)(x)⇒(ii) Write L for the family of all compact subsets of G with non-zero measure. For L ∈ L, define 1 R pL : Cb (X) → R by setting pL (f ) = f dµ for f ∈ Cb (X). For finite I ⊆ G, ² > 0 set L µL

1 2

A(I, ²) = {L : L ∈ L, µ(aL \ L) ≤ ²µL for every a ∈ I}. By (x), no A(I, ²) is empty, since there is always an L ∈ L such that µ((I ∪ {e})L) ≤ (1 + 12 ²)µL. So we have an ultrafilter F on L containing every A(I, ²). Set p(f ) = limL→F pL (f ) for f ∈ Cb (X); then p : Cb (X) → R is a positive linear functional and p(χG) = 1. If a ∈ G, f ∈ Cb (X) and L ∈ L, then Z Z ¯ 1 ¯¯ |pL (a•l f ) − pL (f )| = f (a−1 x)µ(dx) − f (x)µ(dx)¯ µL L L Z Z ¯ 1 ¯¯ = f dµ − f dµ¯ µL

≤

aL

L

1 kf k∞ µ(aL4L) µL

=

1 kf k∞ (µ(aL \ L) + µ(a−1 L \ L)). µL

Since F contains A({a, a−1 }, ²) for every ² > 0, |p(a•l f ) − p(f )| = limL→F |pL (a•l f ) − pL (f )| = 0. So p witnesses that (ii) is true. 449J If we make a further step back towards the origin of this topic, and suppose that our group is discrete, then we have a striking further condition to add to the list above. I give this as a corollary of a general result on group actions recalling the main theorems of §§394 and 448. Tarski’s theorem Let G be a group acting on a non-empty set X. Then the following are equiveridical: (i) there is an additive functional ν : PX → [0, 1] such that νX = 1 and ν(a•A) = νA whenever A ⊆ X and a ∈ G; (ii) there are no A0 , . . . , An , a0 , . . . , an , b0 , . . . , bn such that A0 , . . . , An are subsets of X covering X, a0 , . . . , an , b0 , . . . , bn all belong to G, and a0 •A0 , b0 •A0 , a1 •A1 , b1 •A1 , . . . , bn •An are all disjoint.

449K

Amenable groups

407

proof (a)(i)⇒(ii) This is elementary, for if ν : PX → [0, 1] is a non-zero additive functional and A0 , . . . , An cover X and a0 , . . . , bn ∈ G, then Pn Pn • • i=0 ν(ai Ai ) + i=0 ν(bi Ai ) ≥ 2νX > νX, and a0 •A0 , . . . , bn •An cannot all be disjoint. (b)(ii)⇒(i) Now suppose that (ii) is true. α) Suppose that c0 , . . . , cn ∈ G. Then there is a finite set I ⊆ X such that #({ci •x : i ≤ n, x ∈ I}) < (α 2#(I). P P?? Otherwise, by the Marriage Lemma in the form 4A1H, applied to the set R = {((x, j), ci •x) : x ∈ X, j ∈ {0, 1}, i ≤ n} ⊆ (X × {0, 1}) × X, there is an injective function φ : X × {0, 1} → X such that φ(x, j) ∈ {ci •x : i ≤ n} for every S x ∈ X and j ∈ {0, 1}. Now set Bij = {x : φ(x, 0) = ci •x, φ(x, 1) = cj •x} for i, j ≤ n, so that X = i,j≤n Bij . Let Aij ⊆ Bij be such that hAij ii,j≤n is a partition of X, and set aij = ci , bij = cj for i, j ≤ n; then aij •Aij ⊆ φ[Aij × {0}], bij •Aij ⊆ φ[Aij × {1}] are all disjoint, which is supposed to be impossible. X XQ Q β ) Suppose that J ⊆ G is finite and ² > 0. Then there is a non-empty finite set I ⊆ X such that (β #(I4c•I) ≤ ²#(I) for every c ∈ J. P P It is enough to consider the case in which the identity e of G belongs to J. ?? Suppose, if possible, that there is no such set I. Let m ≥ 1 be such that (1 + 21 ²)m ≥ 2. Set K = J m = {c1 c2 . . . cm : c1 , . . . , cm ∈ J}. By (α), there is a finite set I0 ⊆ X such that #(I0∗ ) < 2#(I0 ), where I0∗ = {a•x : a ∈ K, x ∈ I0 }. Choose c1 , . . . , cm and I1 , . . . , Im inductively such that given that Ik is a non-empty finite subset of X, where k < m, take ck+1 ∈ J such that #(Ik 4ck+1 •Ik ) > ²#(Ik ) and set Ik+1 = Ik ∪ ck+1 •Ik . Then Ik ⊆ {a•x : a ∈ J k , x ∈ I0 } for each k ≤ m, and in particular Im ⊆ I0∗ . But also #(Ik+1 ) = #(Ik ) + #((ck+1 •Ik ) \ Ik ) 1 2

1 2

= #(Ik ) + #((ck+1 •Ik )4Ik ) ≥ (1 + ²)#(Ik ) for every k < m, so 1 2

#(I0∗ ) ≥ #(Im ) ≥ (1 + ²)m #(I0 ) ≥ 2#(I0 ), contrary to the choice of I0 . X XQ Q (γγ ) There is therefore an ultrafilter F on [X]<ω \ {∅} such that Ac² = {I : I ∈ [X]<ω \ {∅}, #(I4c•I) ≤ ²#(I)} belongs to F for every c ∈ G and ² > 0. For I ∈ [X]<ω \ {∅} and A ⊆ X set νI (A) = #(A ∩ I)/#(I), and set νA = limI→F νI A for every A ⊆ X, so that ν : PX → [0, 1] is an additive functional and νX = 1. Now ν is G-invariant. P P If A ⊆ X and c ∈ G and ² > 0, then Ac−1 ,² ∈ F. If I ∈ Ac−1 ,² , then |νI (c•A) − νI (A)| =

1 |#(I #(I)

∩ (c•A)) − #(I ∩ A)|

=

1 |#((c−1 •I) ∩ A) − #(I #(I)

≤

1 |#((c−1 •I)4I| #(I)

∩ A)|

≤ ².

As ² is arbitrary, limI→F νI (c•A) − νI (A) = 0, and ν(c•A) = νA. As A and c are arbitrary, ν is G-invariant. Q Q So ν witnesses that (i) is true, and the proof is complete. 449K Corollary Let G be a group with its discrete topology. Then the following are equiveridical: (i) G is amenable; (ii) there are no A0 , . . . , An , a0 , . . . , an , b0 , . . . , bn such that A0 , . . . , An are subsets of G covering G, a0 , . . . , an , b0 , . . . , bn all belong to G, and a0 A0 , b0 A0 , a1 A1 , b1 A1 , . . . , bn An are all disjoint.

408

Topological groups

449K

proof All we have to observe is that (α) every function from G to R is uniformly continuous for the right uniformity of G, so that G is amenable iff there is an invariant positive linear functional p : `∞ (G) → R such that p(χG) = 1 (β) that a positive linear functional on `∞ (G) is G-invariant iff the corresponding additive functional on PG is G-invariant. So (i) of 449J is equivalent to amenability of G as defined in 449A. 449L Theorem Let G be a group which is amenable in its discrete topology, X a set, and • an action of G on X. Let Σ be a subring of PX and ν : Σ → [0, ∞[ a finitely additive functional which is G-invariant in the sense that g •E ∈ Σ and ν(g •E) = νE whenever E ∈ Σ and g ∈ G. Then there is an extension of ν to a G-invariant non-negative finitely additive functional ν˜ defined on the ideal I of subsets of X generated by Σ. proof (a) There is a non-negative finitely additive functional θ : I → R extending ν. P P Let V be the linear subspace of `∞ (X) generated by {χE : E ∈ Σ}, so that V can be identified with the Riesz space R S(Σ) (361L). Let U be the solid linear subspace of `∞R(X) generated by V . For u ∈ U set q(u) = inf{ v dν : v ∈ V , |u| ≤ v}. Then q is a seminorm, and | v dν| ≤ q(v) for every v ∈ V . So there is a linear functional R f : U → R such that f (v) = v dν for every v ∈ V and |f (u)| ≤ q(u) for every u ∈ U (4A4D(a-i)). Set θA = f (χA) for A ∈ I. Then θ : I → R is additive and extends ν. If A ∈ I, there is an E ∈ Σ including A. Now we have R θ(E \ A) = f (χE − χA) ≤ q(χE − χA) ≤ χE dν = νE = θE, so θA ≥ 0. So θ is non-negative. Q Q (b) As in the proof of 449K, we have a positive G-invariant linear functional p : `∞ (G) → R such that p(χG) = 1. For A ∈ I, set fA (a) = θ(a−1 •A) for a ∈ G, and νÃ = p(fA ). Then ν˜ : I → [0, ∞[ is additive. If E ∈ Σ then fA (a) = νE for every a, so ν˜ extends ν. If A ∈ I and a, b ∈ G, then faA (b) = θ(b−1 aA) = fA (a−1 b), so faA = a•l fA and ν˜(aA) = p(faA ) = p(fA ) = νÃ. Thus ν˜ is G-invariant, as required. 449M Corollary (Banach 23) If r = 1 or r = 2, there is a functional µ ˜ : PRr → [0, ∞] such that (i) µ ˜(A ∪ B) = µ Ã + µ ˜B whenever A, B ⊆ Rr are disjoint (ii) µ ˜E is the Lebesgue measure of E whenever E ⊆ Rr is Lebesgue measurable (iii) µ ˜(g[A]) = µ Ã whenever A ⊆ Rr and g : Rr → Rr is an isometry. proof (a) The point is that the group G of all isometries of Rr , with the discrete topology, is amenable. P P Let G0 ⊆ G be the subgroup consisting of rotations about 0; because r ≤ 2, this is abelian, therefore amenable (449Cf). Let G1 ⊆ G be the subgroup consisting of isometries keeping 0 fixed; then G0 is a normal subgroup of G1 , and G1 /G0 is abelian, so G1 is amenable (449Cc). Let G2 ⊆ G be the normal subgroup consisting of translations; then G2 is abelian, therefore amenable. Now G = G1 G2 , so G is amenable (449Cd). Q Q (b) Let Σ be the ring of subsets of Rr with finite Lebesgue measure, and let ν be the restriction of Lebesgue measure to Σ. Then ν is G-invariant. By 449L, there is a G-invariant extension ν˜ of ν to the ideal I generated by Σ. Setting µ Ã = νÃ for A ∈ I, ∞ for A ∈ PRr \ I, we have a suitable functional µ ˜. 449X Basic exercises > (a) Let G be a topological group. On G define a binary operation ¦ by saying that x ¦ y = yx for all x, y ∈ G. Show that (G, ¦) is a topological group isomorphic to G, so is amenable iff G is. (b) Show that any finite topological group is amenable. >(c) Let X be a set and G the group of all permutations of X. (i) Give X the metric ρ such that ρ(x, y) = 1 for all distinct x, y ∈ X, so that G is the isometry group of X. Show that G, with the topology of pointwise convergence (441G), is amenable. (Hint: for any I ∈ [X]<ω , {a : a ∈ G, a(x) = x for every x ∈ X \ I} is amenable.) (ii) Show that if X is infinite then G, with its discrete topology, is not amenable. (Hint: the left action of F2 on itself can be regarded as an embedding of F2 in G; use 449F.)

449Xq

Amenable groups

409

> (d) Show that, for any r ∈ N, the isometry group of Rr , with the topology of pointwise convergence, is amenable. (Hint: 443Xx, 449Cd.) (e) Find a metrizable σ-compact locally compact Hausdorff topological group which is amenable but not unimodular. (Hint: 442Xf, 449Cd.) (f ) Let G be a topological group and U the space of bounded real-valued functions on G which are uniformly continuous for the right uniformity of G. Show that •r , as defined in 441Ac, gives an action of G on U . (g) Show that a dense subgroup of an amenable topological group is amenable. (Hint: 449E(iii).) ˆ its completion with respect to its bilateral uniformity (h) Let G be a Hausdorff topological group, and G ˆ (definition: 4A5Hb). Show that G is amenable iff G is. (i) Let • be an action of a group G on a set X, and U a Riesz subspace of `∞ (X), containing the constant functions, such that a•f ∈ U for every f ∈ U , a ∈ G. Show that the following are equiveridical: (i) there is a Pn G-invariant positive linear functional p : U → R such that p(χX) = 1; (ii) supx∈X i=0 f (xi ) − f (ai •xi ) ≥ 0 whenever f0 , . . . , fn ∈ U and a0 , . . . , an ∈ G. (Hint: if (ii) is true, let V be the linear subspace generated by {f − a•f : f ∈ U , a ∈ G} and show that inf g∈V kg − χXk∞ = 1.) (j) Let G be the group with generators a, b and relations a2 = b3 = e; that is, the quotient of the free group on two generators a and b by the normal subgroup generated by {a2 , b3 }. Show that G, with its discrete topology, is not amenable. > (k) Let G be a locally compact Hausdorff group, and µ a left Haar measure on G. Show that the following are equiveridical: (i) G is amenable; (ii) there is a positive linear functional p# : L∞ (µ) → R such that p# (χG• ) = 1 and p# (a•r u) = p# (u) for every u ∈ L∞ (µ) and every a ∈ G, where •r is defined as in + ∞ 443G; (iii) for R every finite set I R⊆ G, finite set J ⊆ L (µ) and ² > 0, there is an h ∈ Ck1 (definition: 449I) such that | f (xa)h(x)µ(dx) − f (x)h(x)µ(dx)| ≤ ² for every a ∈ I and f ∈ J. > (l) Let G be a unimodular locally compact Hausdorff group, and µ a Haar measure on G. Show that G is amenable iff whenever K ⊆ G is compact and ² > 0 there is a compact set L ⊆ G with non-zero measure such that µ(LK) ≤ (1 + ²)µL. (m) Let G be a locally compact Hausdorff group, and µ a left Haar measure on G. Show that the following are equiveridical: (α) G is amenable; (β) for every finite set I ⊆ G and ² > 0 there is a u ∈ L2 (µ) such that kuk2 = 1 and ka•l u − uk2 ≤ ² for every a ∈ I; (γ) for every I ∈ [G]<ω and ² > 0 there is an h ∈ Ck (G) such that khk2 = 1 and ka•l h − hk2 ≤ ² for every a ∈ I; (δ) for every I ∈ [G]<ω and ² > 0 there is an h ∈ Ck (G)+ such that khk2 = 1 and ka•l h − hk2 ≤ ² for every a ∈ I. (Hint: 449I(vii)⇒ (β) ⇒ (γ) ⇒ (δ) ⇒449I(iv).) (n) Let G be a locally compact Hausdorff topological group with left Haar measure µ. Show that the following are equiveridical: (i) G is amenable; (viii)0 for every compact set K ⊆ G and ² > 0, there is an + h ∈ Ck1 (definition: 449I) such that h(x−1 ) = h(x) for every x ∈ G and ka•l h − hk1 ≤ ² for every a ∈ K; 0 (ix) for every compact set K ⊆ G and ² > 0 there is a symmetric compact set L ⊆ G with non-empty interior such that µL > 0 and µ(L4aL) ≤ ²µL for every a ∈ K. (o) Let G be a unimodular locally compact Hausdorff topological group with Haar measure µ. Show that the following are equiveridical: (i) G is amenable; (iii)0 there is a positive linear functional p˜ : L∞ (µ) → R such that p˜(χG• ) = 1 and p˜(a•l u) = p˜(a•r u) = p˜(u) for every u ∈ L∞ (µ) and a ∈ G. (p) Let G be a locally compact Hausdorff group and µ a left Haar measure on G. Suppose that G is exponentially bounded, that is, limn→∞ (µ(K n ))1/n ≤ 1 for every compact set K ⊆ G. (i) Show that G is amenable. (ii) Show that (G, ¦), as defined in 449Xa, is exponentially bounded. (q) Let G be a locally compact Hausdorff group and B(G) its Borel σ-algebra. Show that G is amenable iff there is a non-zero finitely additive ν : B(G) → [0, 1] such that ν(aE) = νE for every a ∈ G and E ∈ B(G).

410

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449Xr

(r) Let X be a non-empty set, G a group and • an action of G on X. Suppose that G is an amenable group when given its discrete topology. Show that there is an additive functional ν : PX → [0, 1] such that νX = 1 and ν(a•A) = νA for every A ⊆ X and every a ∈ G. > (s) Let G be a group which is amenable when given its discrete topology. Let ν : PG → [0, 1] be an additive functional such that νG = 1 and ν(aE) = νE for every E ⊆ G, a ∈ G. For E ⊆ G set R ν˜E = ν(Ex)ν(dx) (notation: 363Lf). Show that ν˜ : PG → [0, 1] is an additive functional, that ν˜G = 1 and that ν˜(aE) = ν˜(Ea) = ν˜E for every E ⊆ G and a ∈ G. (t) Let G be a locally compact Hausdorff group and µ a left Haar measure on G. Show that if G, with its discrete topology, is amenable, then there is a functional λ : PG → [0, ∞], extending µ, such that λ(A ∪ B) = λA + λB whenever A, B ⊆ G are disjoint and λ(xA) = λA for every x ∈ G and A ⊆ G. (u) Let X be a compact metrizable space, φ : X → X a continuous function and µ a Radon probability 1 Pn i measure on X such that µφ−1 = µ. Show that for µ-almost every x ∈ X, limn→∞ i=0 f (φ (x)) is n+1

defined for every f ∈ C(X). (Hint: 4A2Pe, 372K.) 449Y Further exercises (a) If S is a semigroup with identity e and X is a set, an action of S on X is a map (s, x) 7→ s•x : S × X → X such that s•(t•x) = (st)•x and e•x = x for every s, t ∈ S and x ∈ X. A topological semigroup S with identity is amenable if for every non-empty compact Hausdorff space X and every R R continuous action of S on X there is a Radon probability measure µ on X such that f (s•x)µ(dx) = f (x)µ(dx) for every s ∈ S and f ∈ C(X). Show that (i) (N, +), with the discrete topology, is amenable; (ii) if S is a topological semigroup and S is an upwards-directed family of amenable sub-semigroups of S with dense union in S, then S is amenable; (iii) if hSi ii∈I is a family of amenable topological semigroups with product S then S is amenable; (iv) if S is an amenable topological semigroup, S 0 is a topological semigroup, and there is a continuous multiplicative surjection from S onto S 0 , then S 0 is amenable; (v) if S is an abelian topological semigroup, then it is amenable. (b) Let G be the group with generators a, b and relations a2 = b2 = e. Show that G, with its discrete topology, is amenable. (c) Let G be a topological group and U the space of bounded real-valued functions on G which are + uniformly continuous for the right uniformity. Let MqR be the space of totally finite quasi-Radon measures + on G. (i) Show that if ν ∈ MqR then ν ∗ f (definition: 444H) belongs to U for every f ∈ U . (ii) Show that + + (ν, f ) 7→ ν ∗ f : MqR × U → U is continuous if MqR is given its narrow topology (437Jd) and U is given its norm topology. (iii) Show that if p : U → R is a continuous linear functional such that p(a•l f ) = p(f ) for every f ∈ U and a ∈ G, then p(ν ∗ f ) = νG · p(f ) for every f ∈ U and every totally finite quasi-Radon measure ν on G. (d) Re-work 449I for general groups carrying Haar measures. (e) Let X be a set, A a subset of X, and G a group acting on X. Show that the following are equiveridical: (i) there is a functional θ : PX → [0, ∞] such that θA = 1, θ(B ∪ C) = θB + θC and θ(a•B) = θB for all disjoint B, C ⊆ X and a ∈ G; (ii) there are no A0 , . . . , An , a0 , . . . , an , b0 , . . . , bn such that A0 , . . . , An are disjoint subsets of A covering A, a0 , . . . , bn all belong to G, and a0 •A0 , b0 •A0 , a1 •A1 , b1 •A1 , . . . , bn •An are all disjoint subsets of A.  3  4 0 5 5 (f ) (Swierczkowski 58) Let G be the group of orthogonal 3 × 3 matrices. Set S =  − 54 35 0  0 0 1   1 0 0 4  3 . Show that S and T are free in G (that is, no non-trivial product of the form and T =  0 5 5 4 0 − 5 35

449 Notes

Amenable groups

411

in its discrete topology. (Hint: S n0 T n1 S n2 T n3 . . . S n2k T n2k+1 can be the identity), so that G is not amenable     3 4 0 0 0 0 let R be the ring of 3 × 3 matrices over the field Z5 . In R set σ =  −4 3 0 , τ =  0 3 4 . 0 0 0 0 −4 3 Show that σ 2 = σ. Now suppose ρ ∈ R is defined as a non-trivial product of the elements σ, τ and their 0 0 0 0 transposes  σ , τ in  which  σ and σ are never adjacent, τ and τ are never adjacent, and the last term is a 1 σ. Set  b  = ρ  0 . Show that if the first term in the product is σ or σ 0 , then c = 0 and b 6= 0, and c 0 otherwise a = 0 and b 6= 0.) (g) Let F2 be the free group on two generators a, b. (i) Show that there is a set W ⊆ F2 such that aW = bW ∪ b2 W = F2 \ W . (ii) Let S2 be the unit sphere in R3 . Show that if S, T are the matrices of 449Yf, there is a set A ⊆ S2 such that (A, S[A]) and (A, T [A], T 2 [A]) are both partitions of a cocountable subset of S2 . (iii) Show that there is no non-zero rotation-invariant additive functional from PS2 to [0, 1]. (iv) Show that there is no rotation-invariant additive functional on the ball {x : x ∈ R3 , kxk ≤ 1} which extends Lebesgue measure. (h) Give an example of a topological semigroup S with identity such that S is amenable in the sense of 449Ya but (S, ¦) is not, where a ¦ b = ba for a, b ∈ S. 449 Notes and comments The general theory of amenable groups is outside the scope of this book. Here I have tried only to indicate some of the specifically measure-theoretic arguments which are used in the theory. Primarily we have the Riesz representation theorem, enabling us to move between linear functionals and measures. Since the invariant measures considered in the definition of ‘amenable group’ are all Radon measures on compact Hausdorff spaces, they can equally well be thought of as linear functionals on spaces of continuous functions. What is striking is that the definition in terms of continuous actions on arbitrary compact Hausdorff spaces can be reduced to a question concerning an invariant mean on a single space easily constructed from the group (449E). The first part of this section deals with general topological groups. It is a remarkable fact that some of the most important non-locally-compact topological groups are amenable. For most of these we shall have to wait until we have done ‘concentration of measure’ (§§476, 492) and can approach ‘extremely amenable’ groups (§493). But there is an easy example in 449Xc which already indicates one of the basic methods. 449I is taken from Greenleaf 69, where you will find many references to its development. Historically the subject was dominated by the case of discrete groups, in which combinatorial rather than measuretheoretic formulations seem more appropriate; we also have an extra method of attack, suggested in 449Yb. In 449I, conditions (i)-(iii) relate to ‘invariant means’ of one kind or another. Note that the invariant means p and p˜ of 449E(iii) and 449I(ii) are normalized by the condition p(χG) = p˜(χG• ) = 1, while the left Haar measure µ of 449I has a degreeR of freedom; so that when they come together in 449I(iii) the two sides of the equation p˜(g ∗ f )• = p˜(f • ) g dµ must move together if we change µ by a scalar factor; of course this happens through the hidden dependence of the convolution operation on µ. Between 449I(i) and 449I(iii) there is a double step. First we note that a convolution g ∗ f is a kind of weighted average of left translates of f , so that if we have a mean which is invariant under translations we can hope that it will be invariant under convolutions (449H, 449Yc). What is more remarkable is that an invariant mean on the space U of bounded uniformly continuous functions should necessarily have an extension to an invariant mean on the space L∞ of (equivalence classes of) bounded Haar measurable functions. Condition (vi) in 449I looks at a different aspect of the phenomenon. In effect, it amounts to saying that not only is there an invariant mean, but that there is an invariant mean defined by the formula 1 R p(f ) = limL→F f for a suitable filter F on the family of sets of non-zero finite measure. This may be L µL

called a ‘FølnerR condition’, following Følner 55. Condition (iv) looks for an invariant mean of the form p(f ) = limh→F f × h dµ, where F is a suitable filter on L1 (µ). The techniques developed in §444 to handle Haar measures on groups which are not locally compact can also be used in 449H-449I, using ‘totally bounded for the bilateral uniformity’ in place of ‘compact’ when appropriate (449Yd). 443L provides another route to the same generalization.

412

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449 Notes

The words ‘right’ and ‘left’ appear repeatedly in this section, and it is not perhaps immediately clear which of the ordinary symmetries we can expect to find. The fact that the operation x 7→ x−1 : G → G always gives us an isomorphism between a group and the same set with the multiplication reversed (449Xa) means that we do not have to distinguish between ‘left amenable’ and ‘right amenable’ groups, at least if we start from the definition in 449A. In 449C also there is nothing to break the symmetry between left and right. In 449B and 449D-449E, however, we must commit ourselves to the left action of the group on the space of functions which are uniformly continuous with respect to the right uniformity. If we wish to change one, we must also change the other. In the list of conditions in 449I, some can be reflected, but in groups which are not unimodular many cannot; see 449Xk-449Xl. For semigroups we do have a difference between ‘left’ and ‘right’ amenability (449Yh). In the case of discrete groups, in which all considerations of measurability and continuity evaporate, we have a completely different technique available, as in 449J. Here we can go directly from a non-paradoxicality condition, a weaker version of conditions already introduced in 394E and 448E, to a Følner condition ((β) in part (b) of the proof of 449J) which easily implies amenability. I remind you that I still do not know whether these ideas can be applied to other algebras than PX (394Z). The difficulty is that the unscrupulous use of the axiom of choice in the infinitary Marriage Lemma (4A1H) seems to give us no control over the nature of the sets Aij described in (b-α) of the proof of 449J; moreover, the structure of the proof depends on having a suitable invariant measure (counting measure on X) to begin with. For more on amenable discrete groups and their connexions with measure theory see Laczkovich 02.

Chap. 45 intro.

Introduction

413

Chapter 45 Perfect measures and disintegrations One of the most remarkable features of countably additive measures is that they provide us with a framework for probability theory, as described in Chapter 27. The extraordinary achievements of probability theory since Kolmogorov are to a large extent possible because of the rich variety of probability measures which can be constructed. We have already seen image measures (112E) and product measures (§254). The former are elementary, as the paragraph number declares, but a glance at the index will confirm that they have many surprises to offer; the latter are obviously fundamental to any idea of what probability theory means. In this chapter I will look at some further constructions. The most important are those associated with ‘disintegrations’ or ‘regular conditional probabilities’ (§§452-453) and methods for confirming the existence of measures on product spaces with given images on subproducts (§§454-455). We find that these constructions have to be based on measure spaces of special types; the measures involved in the principal results are the Radon measures of Chapter 41 (of course), the compact and perfect measures of Chapter 34, and an intermediate class, the ‘countably compact’ measures of Marczewski 53 (451B). So the first section of this chapter is a systematic discussion of compact, countably compact and perfect measures. A ‘disintegration’, when present, is likely to provide us with a particularly effective instrument for studying a measure, analogous to Fubini’s theorem for product measures (see 452F). §§452-453 therefore concentrate on theorems guaranteeing the existence of disintegrations compatible with some pre-existing structure, typically an inverse-measure-preserving function (452I, 452O, 453K) or a product structure (452B-452D, 452M). Both depend on the existence of suitable liftings, and for the topological version in §453 we need a ‘strong’ lifting, so much of that section is devoted to the study of such liftings. One of the central concerns of probability theory is to understand ‘stochastic processes’, that is, models of systems evolving randomly over time. If we think of our state space as consisting of functions, so that a whole possible history isQdescribed by a random function of time, it is natural to think of our functions as Q members of some set n∈N Zn (if we think of observations as being taken at discrete time intervals) or t∈[0,∞[ Zt (if we regard our system as evolving continuously), where Zt represents the set of possible states of the system at time t. We are therefore led to consider measures on such product spaces, and the new idea is that we may have some definite intuition concerning the joint distribution of finite strings (f (t0 ), . . . , f (tn )) of values of our random Q function, that is to say, we may think we know something about the image measures on finite products i≤n Zti . So we come immediately to a fundamental question: given Q Q a (probability) measure µJ on i∈J Zi for each finite J ⊆ T , when will there be a measure on i∈T Zi compatible with every µJ ? In §454 I give the most important generally applicable existence theorems for such measures, and in §455 I show how they can be applied to a general construction for models of Markov processes. The abstract theory of §454 yields measures on product spaces which, from the point of view of a probabilist, are unnaturally large, often much larger than intuition suggests. General techniques for determining which such measures can be transferred to more amenable spaces are beyond the scope of this book, but I show how the most important example, Brownian motion, can be interpreted as a Radon measure on C([0, ∞[) rather than as a Baire measure on R[0,∞[ (455D). One of the defining characteristics of Brownian motion is the fact that all its finite-dimensional marginals are Gaussian distributions. Stochastic processes with this property form a particularly interesting class, which I examine in §456. From the point of view of this volume, one of their most striking properties is Talagrand’s theorem that, regarded as measures on powers RI , they are τ -additive (456K). The next two sections look again at some of the ideas of the previous sections when interpreted as answers to questions of the form ‘can all the measures in such-and-such a family be simultaneously extended to a single measure?’ If we seek only a finitely additive common extension, there is a reasonably convincing general result (457A); but countably additive measures remain puzzling even in apparently simple circumstances (457Z). In §458 I introduce ‘relatively independent’ families of σ-algebras, with the associated concept of ‘relative product’ of measures. Finally, in §459, I give some basic results on symmetric measures and exchangeable random variables, with de Finetti’s theorem (459C) and corresponding theorems on representing permutation-invariant measures on products as mixtures of product measures (459E, 459G).

414

Perfect measures, disintegrations and processes

§451 intro.

451 Perfect, compact and countably compact measures In §§342-343 I introduced ‘compact’ and ‘perfect’ measures as part of a study of the representation of homomorphisms of measure algebras by functions between measure spaces. An intermediate class of ‘countably compact’ measures (the ‘compact’ measures of Marczewski 53) has appeared in the exercises. It is now time to collect these ideas together in a more systematic way. In this section I run through the standard properties of compact, countably compact and perfect measures (451A-451J), with a couple of simple examples of their interaction with topologies (451L-451O). An example of a perfect measure space which is not countably compact is in 451T. Some new ideas, involving non-trivial set theory, show that measurable functions from compact totally finite measure spaces to metrizable spaces have ‘essentially separable ranges’ (451Q); consequently, any measurable function from a Radon measure space to a metrizable space is almost continuous (451S). 451A Let me begin by recapitulating the principal facts already covered. T (a) A family K of sets is a compact class if K0 6= ∅ whenever K0 ⊆ K is a non-empty downwardsdirected family. If K ⊆ PX, then K is a compact class iff there is a compact topology on X for which every member of K is closed (342D). A subfamily of a compact class is compact (342Ab). (b) A measure on a set X is compact if it is inner regular with respect to some compact class of sets; equivalently, if it is inner regular with respect to the closed sets for some compact topology on X (342F). All Radon measures are compact measures (416Wa). If (X, Σ, µ) is a semi-finite compact measure space with measure algebra A, (Y, T, ν) is a complete strictly localizable measure space with measure algebra B, and π : A → B is an order-continuous Boolean homomorphism, there is a function g : Y → X such that g −1 [E] ∈ T and g −1 [E]• = π(E • ) for every E ∈ Σ (343B). T (c) A family T K of sets is a countably compact class if n∈N Kn 6= ∅ whenever hKn in∈N is a sequence in K such that i≤n Ki 6= ∅ for every n ∈ N. Any subfamily of a countably compact class is countably compact. If K is a countably compact class, then there is a countably compact class K∗ ⊇ K which is closed under finite unions and countable intersections (413R). (d) A measure space (X, Σ, µ) is perfect if whenever f : X → R is measurable, E ∈ Σ and µE > 0, there is a compact set K ⊆ f [E] such that µf −1 [K] > 0. A countably separated semi-finite measure space is compact iff it is perfect (343K). A measure space (X, Σ, µ) is isomorphic to the unit interval with Lebesgue measure iff it is an atomless complete countably separated perfect probability space (344K). 451B Now for the new class of measures. Definition Let (X, Σ, µ) be a measure space. Then (X, Σ, µ), or µ, is countably compact if µ is inner regular with respect to some countably compact class of sets. Evidently compact measures are also countably compact. A simple example of a countably compact measure which is not compact is the countable-cocountable measure on an uncountable set (342M). For an example of a perfect measure which is not countably compact, see 451T. Note that if µ is inner regular with respect to a countably compact class K, then it is also inner regular with respect to K ∩ Σ (411B), and K ∩ Σ is still countably compact. 451C Proposition (Ryll-Nardzewski 53) Any semi-finite countably compact measure is perfect. proof The central idea is the same as in 342L, but we need to refine the second half of the argument. (a) Let (X, Σ, µ) be a countably compact measure space, f : X → R a measurable function, and E ∈ Σ a set of positive measure. Let K be a countably compact class such that µ is inner regular with respect to K; by 451Ac, we may suppose that K is closed under finite unions and countable intersections. Because µ is semi-finite, there is a measurable set F ⊆ E such that 0 < µF < ∞; replacing F by a set of the form F ∩ f −1 [[−n, n]] if necessary, we may suppose that f [F ] is bounded; P finally, we may suppose that F ∈ K. Let h²q iq∈Q be a family of strictly positive real numbers such that q∈Q ²q < 21 µF . For each q ∈ Q,

451E

Perfect, compact and countably compact measures

415

0 set Eq = {x : x ∈ F, f (x) ≤ q}, Eq0 = {x : x ∈ F, f (x) > q}, and T choose Kq0, Kq ∈ K ∩ Σ such that Kq ⊆ Eq , 0 0 0 0 Kq ⊆ Eq and µ(Eq \ Kq ) ≤ ²q , µ(Eq \ Kq ) ≤ ²q . Then K = q∈Q (Kq ∪ Kq ) ∈ K ∩ Σ, K ⊆ F and P µ(F \ K) ≤ q∈Q µ(Eq \ Kq ) + µ(Eq0 \ Kq0 ) < µF ,

so µK > 0. (b) Take any t ∈ f [K]. Enumerate Q as hqn in∈N and choose hLn in∈N in K by the rule Ln = Kqn if t < qn , = Kq0 n if t > qn , T

= F if t = qn .

Now i≤n Li 6= ∅ for every n ∈ N. P P Because t ∈ f [K], there must be some s ∈ f [K] such that s < qi whenever i ≤ n and t < qi , while s > qi whenever i ≤ n and t > qi . Let x ∈ K be such that f (x) = s. Then, for any i ≤ n, either t < qi , f (x) < qi so x ∈ / Kq0 i and x ∈ Kqi = Li or t > qi , f (x) > qi so x ∈ / Kqi and x ∈ Kq0 i = Li , or t = qi and x ∈ F = Li . T So x ∈ i≤n Li . Q Q T As K is a countably compact class, there must be some x ∈ n∈N Ln . But this means that, for any n ∈ N, if t > qn then x ∈ Kq0 n and f (x) > qn , if t < qn then x ∈ Kqn and f (x) < qn . So in fact f (x) = t. Accordingly t ∈ f [K]. (c) What this shows is that f [K] ⊆ f [K] and f [K] is closed. Because (by the choice of F ) it is also bounded, it is compact (2A2F). Of course we now have f [K] ⊆ f [E], while µf −1 [f [K]] ≥ µK > 0. As f and E are arbitrary, µ is perfect. 451D Proposition Let (X, Σ, µ) be a measure space, and E ∈ Σ; let µE be the subspace measure on E. (a) If µ is compact, so is µE . (b) If µ is countably compact, so is µE . (c) If µ is perfect, so is µE . proof (a)-(b) Let K be a (countably) compact class such that µ is inner regular with respect to K. Then µE is inner regular with respect to K (412Oa), so is (countably) compact. (c) Suppose that f : E → R is ΣE -measurable, where ΣE = Σ ∩ PE is the subspace σ-algebra, and F ⊆ E is such that µF > 0. Set g(x) = arctan f (x) if x ∈ E, = 2 if x ∈ X \ E. Then g is Σ-measurable, so there is a compact set K ⊆ g[F ] such that µg −1 [K] > 0. Set L = {tan t : t ∈ K}; then L ⊆ f [F ] is compact and f −1 [L] = g −1 [K] has non-zero measure. As f and F are arbitrary, µE is perfect. 451E Proposition Let (X, Σ, µ) be a perfect measure space. (a) If (Y, T, ν) is another measure space and f : X → Y is an inverse-measure-preserving function, then ν is perfect. (b) In particular, µ¹ T is perfect for any σ-subalgebra T of Σ. proof (a) Suppose that g : Y → R is T-measurable and F ∈ T is such that νF > 0. Then gf : X → R is Σ-measurable and µf −1 [F ] > 0. So there is a compact set K ⊆ (gf )[f −1 [F ]] such that µ(gf )−1 [K] > 0. But now K ⊆ g[F ] and νg −1 [K] > 0. As g and F are arbitrary, ν is perfect. (b) Apply (a) to Y = X, ν = µ¹ T and f the identity function. Remark We shall see in 452R that there is a similar result for countably compact measures; but for compact measures, there is not (342Xf, 451Xg).

416


451F

451F Lemma (Sazonov 66) Let (X, Σ, µ) be a semi-finite measure space. Then the following are equiveridical: (i) µ is perfect; (ii) µ¹ T is compact for every countably generated σ-subalgebra T of Σ; (iii) µ¹ T is perfect for every countably generated σ-subalgebra T of Σ; (iv) for every countable set E ⊆ Σ there is a σ-algebra T ⊇ E such that µ¹ T is perfect. proof (a)(i)⇒(ii) Suppose that µ is perfect, and that T is a countably generated σ-subalgebra of Σ. Let P∞ hEn in∈N be a sequence in T which σ-generates it, and define f : X → R by setting f (x) = n=0 3−n χEn (x) for every x ∈ X. Then f is measurable. Set K = {f −1 [L] : L ⊆ f [X] is compact}. Then K is a compact class. P P If K0 ⊆ K is non-empty and has the finite intersection property, then L0 = {L : L ⊆ f [X] is compact, f −1 [L] ∈ K0 } is also a non-empty family with the finite intersection property. So there is an T T α ∈ L0 ; since α ∈ f [X], there is an x such that f (x) = α, and now x ∈ K0 . As K0 is arbitrary, K is a compact class. Q Q Observe next that, for any n ∈ N, P P En = {x : ∃ I ⊆ n, i∈I 3−i + 3−n ≤ f (x) < i∈I 3−i + 3−n+1 }. So T0 = {f −1 [F ] : F ⊆ R} contains every En ; as it is a σ-algebra of subsets of X, it includes T. Now µ¹ T is inner regular with respect to K. P P If E ∈ T and µE > 0, there is a set F ⊆ R such that E = f −1 [F ]. Because f is Σ-measurable and µ is perfect, there is a compact set L ⊆ f [E] such that µf −1 [L] > 0. But now f −1 [L] ∈ K ∩ T, and f −1 [L] ⊆ E because L ⊆ F . Because K is closed under finite unions, this is enough to show that µ¹ T is inner regular with respect to K. Q Q Thus K witnesses that µ¹ T is a compact measure. (b)(ii)⇒(i) Now suppose that µ¹ T is compact for every countably generated σ-algebra T ⊆ Σ, that f : X → R is a measurable function, and that µE > 0. Let F ⊆ E be a measurable set of non-zero finite measure, and T the σ-algebra generated by {F } ∪ {f −1 [ ]−∞, q[ ] : q ∈ Q}, so that T is countably generated and f is T-measurable. Because µ¹ T is compact, so is the subspace measure (µ¹ T)F (451Da); but this is now perfect (342L or 451C), while F ∈ T and µF > 0, so there is a compact set L ⊆ f [F ] ⊆ f [E] such that µf −1 [L] > 0. As f and E are arbitrary, µ is perfect. (c)(i)⇒(iv) is trivial. (d)(iv)⇒(iii) If (iv) is true, and T is a countably generated σ-subalgebra of Σ, let E be a countable set generating it. Then there is a σ-algebra T1 ⊇ E such that µ¹ T1 is perfect. By 451Eb, µ¹ T = (µ¹ T1 )¹ T is compact, therefore perfect. (e)(iii)⇒(ii) If (iii) is true, and T is a countably generated σ-subalgebra of Σ, then µ¹ T is perfect; but as (i)⇒(ii), and T is a countably generated σ-subalgebra of itself, µ¹ T is compact. ˆ µ ˜ µ 451G Proposition Let (X, Σ, µ) be a measure space. Let (X, Σ, ˆ) be its completion and (X, Σ, ˜) its c.l.d. version. Then (a)(i) if µ is compact, so are µ ˆ and µ ˜; (ii) if µ is semi-finite and either µ ˆ or µ ˜ is compact, then µ is compact. (b)(i) If µ is countably compact, so are µ ˆ and µ ˜; (ii) if µ is semi-finite and either µ ˆ or µ ˜ is countably compact, then µ is countably compact. (c)(i) If µ is perfect, so are µ ˆ and µ ˜; (ii) if µ ˆ is perfect, then µ is perfect; (iii) if µ is semi-finite and µ ˜ is perfect, then µ is perfect. ˇ for proof (a)-(b) The arguments for µ ˆ and µ ˜ run very closely together. Write µ ˇ for either of them, and Σ its domain. (i) If µ is inner regular with respect to K, so is µ ˇ (412Ha). So if µ is (countably) compact, so is µ ˇ. (ii) Now suppose that µ is semi-finite. The point is that if K is closed under countable intersections and µ ˇ is inner regular with respect to K, so is µ. P P Suppose that E ∈ Σ and that µE > γ. Choose sequences hEn in∈N in Σ and Kn in K inductively, as follows. E0 is to be such that E0 ⊆ E and γ < µE0 < ∞.

451I


417

ˇ be such that Kn ⊆ En and µ Given that γ < µEn < ∞, let Kn ∈ K ∩ Σ ˇKn > γ; now take En+1 ∈ Σ such that E ⊆ K and µE = µ ˇ K (212Cb or 213Fc), and continue. At the end of the induction, n+1 n n+1 n T T K = E is a member of Σ ∩ K included in E and of measure at least γ. As E and γ are n n n∈N n∈N arbitrary, µ is inner regular with respect to K. Q Q It follows that if µ ˇ is compact or countably compact, so is µ. P P Let K be a (countably) compact class such that µ ˇ is inner regular with respect to K; by 451Aa or 451Ac, there is a (countably) compact class K∗ , including K, which is closed under countable intersections, so that µ is inner regular with respect to K∗ , and is itself (countably) compact. Q Q ˆ ˆ such that µ α) Let f : X → R be Σ-measurable, (c)(i)(α and E ∈ Σ Ê > 0. Then there are a µ-conegligible set F0 ∈ Σ such that f ¹F0 is Σ-measurable (212Fa), and an F1 ∈ Σ such that F1 ⊆ E and µ ˆ(E \ F1 ) = 0. Set F = F0 ∩ F1 . By 451Dc, the subspace measure µF is perfect, while f ¹F is ΣF -measurable; so there is a compact set K ⊆ f [F ] such that µ(F ∩ f −1 [K]) > 0. But now K ⊆ f [E] and µ ˆf −1 [K] > 0. As f and E are arbitrary, µ ˆ is perfect. ˜ ˜ is the domain of µ ˜ such that µ β ) Let f : X → R be Σ-measurable, (β where Σ ˜, and E ∈ Σ ˜E > 0. Then there is a set F ∈ Σ such that µF < ∞ and µ ˆ(F ∩ E) is defined and greater than 0 (213D). In this ˆ case, µ ˆ and µ ˜ induce the same subspace measure µ ˆF on F . Accordingly f ¹F is Σ-measurable. Because µ ˆ is perfect (by (α) just above), so is µ ˆF (451Dc), and there is a compact set K ⊆ f [F ∩ E] such that µ ˆF (f ¹F )−1 [K] > 0. But now, of course, K ⊆ f [E] and µ ˜f −1 [K] > 0. As f and E are arbitrary, µ ˜ is perfect. (ii) Suppose that µ ˆ is perfect. Since µ = µ ˆ¹Σ, µ is perfect, by 451Eb. (iii) Similarly, if µ ˜ is perfect and µ is semi-finite, then µ = µ ˜¹Σ, by 213Hc, so µ is perfect. 451H Lemma Let hXi ii∈I be a family of sets with product X. Suppose that Ki ⊆ PXi for each i ∈ I, and set K = {πi−1 [K] : i ∈ I, K ∈ Ki }, where πi : X → Xi is the coordinate map for each i ∈ I. Then (a) if every Ki is a compact class, so is K; (b) if every Ki is a countably compact class, so is K. proof (a) For each i ∈ I, let Ti be a compact topology on Xi such that every member of Ki is closed. Then the product topology T on X is compact (3A3J), and every member of K is T-closed, so K is a compact class. T (b) If hKn in∈N is a sequence in K such that k≤n Kk 6= ∅ for every n ∈ N, then we must be able to express each Kn as πj−1 [Ln ], where jn ∈ I, Ln ∈ Kjn for every n. Now, for i ∈ I, Li = {Kjn : n ∈ N, jn = i} n is a countable subset of Ki , and any finite T subfamily of Li has non-empty intersection. Since K0 6= ∅, Xi 6= ∅; so, whether Li is empty or not, Xi ∩ Li is non-empty. So T Q T Li ) k∈N Kk = i∈I (Xi ∩ is not empty. As hKn in∈N is arbitrary, K is countably compact. 451I Theorem Let (X, Σ, µ) and (Y, T, ν) be measure spaces, with c.l.d. product (X × Y, Λ, λ). (a) If µ and ν are compact, so is λ. (b) If µ and ν are countably compact, so is λ. (c) If µ and ν are perfect, so is λ. proof (a)-(b) Let K ⊆ PX, L ⊆ PY be (countably) compact classes such that µ is inner regular with respect to K and ν is inner regular with respect to L. Set M0 = {K × Y : K ∈ K} ∪ {X × L : L ∈ L}. Then M0 is (countably) compact, by 451H. By 451Aa/451Ac, there is a (countably) compact class M ⊇ M0 which is closed under finite unions and countable intersections. By 412R, λ is inner regular with respect to M, so is (countably) compact. (c)(i) Let f : X × Y → R be Λ-measurable, and V ∈ Λ a set of positive measure. Then there are G ∈ Σ, H ∈ T such that µG, νH are both finite and λ(V ∩ (G × H)) > 0. Recall that the subspace measure λG×H on G × H is just the product of the subspace measures µG and µH (251P(ii-α)), and is the completion of its b H generated by {E × F : E ∈ ΣG , F ∈ TH }, where ΣG and TH are the restriction θ to the σ-algebra ΣG ⊗T b H, subspace σ-algebras on G, H respectively, the domains of µG and µH (251K). Next, for any W ∈ ΣG ⊗T

418


451I

there are countable families E ⊆ ΣG , F ⊆ TH such that W belongs to the σ-algebra of subsets of G × H generated by {E × F : E ∈ E, F ∈ F} (331Gd). b H ; let (ii) The point is that θ is perfect. P P Let Λ0 be any countably generated σ-subalgebra of ΣG ⊗T 0 hWn in∈N be a sequence in Λ generating it. Then there are countable families E ⊆ ΣG , F ⊆ TH such that every Wn belongs to the σ-algebra generated by {E × F : E ∈ E, F ∈ F}. Let Σ0 , T0 be the σ-algebras of b 0 , so Λ0 ⊆ Σ0 ⊗T b 0. subsets of G and H generated by E and F respectively; then every Wn belongs to Σ0 ⊗T 0 Let λ0 be the product of the measures µ¹Σ0 = µG ¹Σ0 and ν¹ T . Then λ0 is the completion of its restriction b 0. to Σ0 ⊗T Now trace through the results above. µG and νH are perfect (451Dc), so µG ¹Σ0 and νH ¹ T0 are compact (451F), so λ0 is compact ((a) of this theorem), so λ0 is perfect (342L or 451C). But θ must agree with λ0 on Λ0 , by Fubini’s theorem (252D), or otherwise, so θ¹Λ0 is a restriction of λ0 , and is perfect (451Eb). Thus θ¹Λ0 is perfect for every countably generated σ-subalgebra Λ0 of dom θ. By 451F, θ is perfect. Q Q (iii) By 451G(c-i), λG×H is perfect. Now f ¹G × H is measurable, and λG×H (V ∩ (G × H)) > 0, so there is a compact set K ⊆ f [V ∩ (G × H)] such that λG×H ((G × H) ∩ f −1 [K]) > 0; in which case K ⊆ f [V ] and λf −1 [K] > 0. As f and V are arbitary, λ is perfect. 451J Theorem Let h(Xi , Σi , µi )ii∈I be a family of probability spaces with product (X, Σ, µ). (a) If every µi is compact, so is µ. (b) (Marczewski 53) If every µi is countably compact, so is µ. (c) If every µi is perfect, so is µ. proof The same strategy as in 451I is again effective. (a)-(b) For each i ∈ I, let Ki ⊆ PXi be a (countably) compact class such that µi is inner regular with respect to Ki . Set M0 = {πi−1 [K] : i ∈ I, K ∈ Ki }, so that M0 is (countably) compact. Let M ⊇ M0 be a (countably) compact class which is closed under finite unions and countable intersections. By 412T, µ is inner regular with respect to M, so is (countably) compact. N (c) Let Λ0 be a countably generated σ-subalgebra of c i∈I Σi , the σ-algebra of subsets of X generated by N the sets {x : x(i) ∈ E} for i ∈ I and E ∈ Σi . Then λ¹Λ0 is perfect. P P For every W ∈ c Σi , we must be i∈I

able to find countable subsets Ei of Σi such that W is in the σ-algebra generated by {πi−1 [E] : i ∈ I, E ∈ Σi }; so there are in fact countable sets Ei ⊆ Σi such that the σ-algebra generated by {πi−1 [E] : i ∈ I, E ∈ Σi } includes Λ0 . Let Ti be the σ-subalgebra of Σi generated by Ei , so that µi ¹ Ti is compact. Let λ0 be the product of hµi ¹ Ti ii∈I ; then λ0 is compact, by (a) above, therefore perfect. Now λ is an extension of λ0 , by N 254G or otherwise, so λ0 is an extension of λ¹Λ0 , and λ¹Λ0 is perfect. Q Q As Λ0 is arbitrary, λ¹ c i∈I Σi is perfect, and its completion λ (254Ff) is also perfect. Remark This theorem is generalized in 454Ab. *451K The following result is interesting because it can be reached from an unexpectedly weak hypothesis, and has occasional applications. Proposition Let h(Xi , Σi , µi )ii∈I be a family of perfect probability spaces, and (X, Λ, λ) their product. For N J ⊆ I, x ∈ X set πJ (x) = x¹J. Let K be the set {V : V ⊆ X, πJ [V ] ∈ c i∈J Σi for every J ⊆ I}. Then λ is inner regular with respect to K. proof (a) Take any W ∈ Λ and γ < λW . Then there is a family hEni in∈N,i∈I such that Eni ∈ Σi for every n ∈ N, i ∈ I, Jn = {i : i ∈ I, Eni 6= Xi } is finite for every n ∈ N, S Q X \ W ⊆ n∈N i∈I Eni , P∞ Q n=0

i∈I

µi Eni =

P∞ Q n=0

i∈Jn

µi Eni < 1 − γ

451M


419

(254A). For each i ∈ I, let Σ0i be the σ-subalgebra of Σi generated by {Eni : n ∈ N}. Then µi ¹Σ0i is a compact measure (451F); let Ki be a compact class such that µi ¹Σ0i is inner P∞regular Q with respect to Ki . Let h²ni in∈N,i∈Jn be a family of strictly positive real numbers such that n=0 i∈Jn (µi Eni + ²ni ) ≤ 1 − γ. For n ∈ N, i ∈ I choose Kni ∈ Ki ∩ Σ0i such that Kni ⊆ Xi \ Eni and µi Kni ≥ µi (Xi \ Eni ) − ²ni , that is, µi (Xi \ Kni ) ≤ µi Eni + ²ni . T S (b) Set V = n∈N i∈Jn {x : x ∈ X, x(i) ∈ Kni }. Then [ {x : x(i) ∈ Xi \ Kni for every i ∈ Jn } X \V = n∈N

⊇

[

{x : x(i) ∈ Eni for every i ∈ Jn } =

µ(X \ V ) ≤

P∞ Q n=0

Eni ⊇ X \ W,

n∈N i∈I

n∈N

so V ⊆ W . Next,

[Y

i∈Jn

µi (Xi \ Kni ) ≤

P∞ Q n=0

i∈Jn (µi Eni

+ ²ni ) ≤ 1 − γ,

so µV ≥ γ.

T S (c) Take any J ⊆ I. For m ∈ N, set Vm = n≤m i∈Jn {x : x ∈ X, x(i) ∈ Kni }. Then πJ [V ] = P Of course πJ [V ] ⊆ πJ [Vm ] for every m, just because V ⊆ Vm . On the other hand, suppose m∈N πJ [V Tm ]. P Q that y ∈ m∈N πJ [Vm ]. Then for every m ∈ N there is a zm ∈ i∈I\J Xi such that y ∪ zm ∈ Vm , writing y ∪zm ∈ X for the common extension of the functions y and zm . Now for n ≤ m ∈ N there is an i(m, n) ∈ Jn such that (y ∪ zm )(i(m, n)) ∈ Kn,i(m,n) . Let F be a non-principal ultrafilter on N. For every n ∈ N there is an in ∈ Jn such that {m : m ≥ n, i(m, n) = in } ∈ F, because Jn is finite. Fix i ∈ I \ J for a moment. If we set Ki0 = {Kni : nT∈ N, in = i}, then for any m ∈ N there is an m0 ≥ m such that i(m0 , n) = in for every n ≤ m, so that n≤m,in =i Kni contains zm0 (i) and is therefore non-empty. Thus Ki0 has the finite T intersection property; because K is a compact class, we can choose a point z(i) ∈ Ki0 . Doing this for each Q i ∈ I \ J, we have z ∈ i∈I\J Xi . Consider y ∪ z. Let n ∈ N. If in ∈ J, then there is an m ≥ n such that i(m, n) = in , in which case

T

(y ∪ z)(in ) = y(in ) = (y ∪ zm )(i(m, n)) ∈ Kn,i(m,n) = Knin . S If in ∈ I \ J, then Knin ∈ Ki0 n so (y ∪ z)(in ) = z(in ) ∈ Knin . Thus T y ∪ z ∈ i∈Jn {x : x(i) ∈ Kni }. As n is Q arbitrary, y ∪ z ∈ V and y ∈ πJ [V ]. As y is arbitrary, πJ [V ] = m∈N πJ [Vm ]. Q Now observe that every Vm is in the algebra of subsets of X generated by sets of the form {x : x(i) ∈ H} N where i ∈ I and H ∈ Σi , which we can identify with the free product Qi∈I Σi (315La). This means that Vm can be expressed as a finite union of cylinder sets of the form C = i∈I HQ i where Hi ∈ Σi for every i and {i : Hi 6= Xi } is finite (315Jb). But in this case πJ [C] is either empty or i∈J Hi , and in either case N N belongs to c i∈J Σi . So πJ [Vm ], being a finite union of such sets, also belongs to c i∈J Σi . As this is true N T for every m ∈ N, πJ [V ] = m∈N πJ [Vm ] belongs to c i∈J Σi . (d) This is true for every J ⊆ I, so V ∈ K. As W and γ are arbitrary, λ is inner regular with respect to K. 451L Proposition Let (X, Σ) be a standard Borel space. Then any semi-finite measure µ with domain Σ is compact, therefore perfect. proof If T is a Polish topology on X with respect to which Σ is the Borel σ-algebra, then µ is inner regular with respect to the family K of T-compact sets (433Ca), which is a compact class. 451M Proposition Let (X, Σ, µ) be a perfect measure space and T a T0 topology on X with a countable network consisting of measurable sets. (For instance, µ might be a topological measure on a regular space with a countable network (4A2Ng), or a second-countable space. In particular, X might be a separable metrizable space.) Then µ is inner regular with respect to the compact sets. proof This is a refinementPof 343K. Let hEn in∈N be a sequence in Σ running over a network for T. Define ∞ g : XP → R by setting g = n=0 3−n χHn (cf. 343E). Then g is measurable, because every χEn is. Writing αI = i∈I 3−i for I ⊆ N, and

420


Hn =

S I⊆n

451M

¤ £ αI + 21 3−n , αI + 3−n+1 ,

we see that En = g −1 [Hn ] for each n ∈ N. This shows that g is injective, because if x, y are distinct points in X there is an open set containing one but not the other, and now there is an n ∈ N such that En contains that one and not the other, so that just one of g(x), g(y) belongs to Hn . Also g −1 : g[X] → X is continuous, since (g −1 )−1 [En ] = g[En ] = Hn ∩ g[X] is relatively open in g[X] for every n ∈ N (4A2B(a-ii)). Now suppose that E ∈ Σ and µE > 0. Then there is a compact set K ⊆ g[E] such that µg −1 [K] > 0. But as g is injective, g −1 [K] ⊆ E, and as g −1 is continuous, g −1 [K] is compact. By 412B, this is enough to show that µ is inner regular with respect to the compact sets. 451N Corollary Let (X, Σ, µ) be a complete perfect measure space, Y a Hausdorff space with a countable network consisting of Borel sets and f : X → Y a measurable function. If the image measure µf −1 is locally finite, it is a Radon measure. proof Because f is measurable, µf −1 is a topological measure; by 451Ea, it is perfect; by 451M, it is tight; and it is complete because µ is. Because Y has a countable network, it is Lindelöf (4A2Nb), and µf −1 is σ-finite (411Ge), therefore locally determined. So it is a Radon measure. 451O Corollary Let (X, Σ, µ) be a perfect measure space, Y a separable metrizable space, and f : X → Y a measurable function. (a) If E ∈ Σ and γ < µE, there is a compact set K ⊆ f [E] such that µ(E ∩ f −1 [K]) ≥ γ. (b) If ν = µφ−1 is the image measure, then µ∗ f −1 [B] = ν∗ B for every B ⊆ Y . (c) If moreover µ is σ-finite, then µ∗ f −1 [B] = ν ∗ B for every B ⊆ Y . proof (a) Consider the subspace measure µE , the measurable function f ¹E from E to the separable metrizable space f [E], and the image measure ν 0 = µE (f ¹E)−1 on f [E]. By 451Ea and 451M, this is tight, while ν 0 f [E] = µE; so there is a compact set K ⊆ f [E] such that ν 0 K ≥ γ, and this serves. (b)(i) If F ∈ dom ν and F ⊆ B, then νF = µf −1 [F ] ≤ µ∗ f −1 [B]; as F is arbitrary, µ∗ f −1 [B] ≥ ν∗ B. (ii) If E ∈ Σ and E ⊆ f −1 [B] and γ < µE, then (a) tells us that there is a compact set K ⊆ f [E] such that µ(E ∩ f −1 [K]) > γ, in which case ν∗ B ≥ νK > γ. As E and γ are arbitrary, ν∗ B ≥ µ∗ f −1 [B]. (c)(i) If F ∈ dom ν and F ⊇ B, then νF = µf −1 [F ] ≥ µ∗ f −1 [B]; as F is arbitrary, µ∗ f −1 [B] ≤ ν ∗ B. (ii) If µ∗ f −1 [B] = ∞, then of course µ∗ f −1 [B] = ν ∗ B. Otherwise, because µ is σ-finite, we can find a disjoint sequence hEn in∈N of subsets of X of finite measure, covering X, such that E0 ⊇ f −1 [B] and µE0 = µ∗ f −1 [B]. Let ² > 0. For eachSn ≥ 1, (a) tells us that there is a compact set Kn ⊆ f [En ] such that µf −1 [En \ Kn ] ≤ 2−n ². Set H = Y \ n≥1 Kn ; then νH ≤ µE + ², and B ⊆ H. So ν ∗ B ≤ νH ≤ µE + ² = µ∗ f −1 [B] + ². As ² is arbitrary, ν ∗ B ≤ µ∗ f −1 [B]. 451P I turn now to a remarkable extension of the idea above to general metric spaces Y . Lemma Let S (X, Σ, µ) be a semi-finite compact measure space,Pand hEi ii∈I a disjoint family of subsets of S X such that i∈J Ei ∈ Σ for every J ⊆ I. Then µ( i∈I Ei ) = i∈I µEi . proof (a) To begin with (down to S the end of part (d) of the proof) assume that µ is totally finite and that every Ei is negligible. Set X0 = i∈I Ei , and let µ0 be the subspace measure on X0 . Define f : X0 → I by S setting f (x) = i if i ∈ I, x ∈ Ei , and let ν be the image measure µ0 f −1 , so that νJ = µ( i∈J Ei ) for J ⊆ I; then (I, PI, ν) is a totally finite measure space.

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(b) ν is purely atomic. P P?? Suppose, if possible, otherwise; that there is a K ⊆ I such that νK > 0 and the subspace measure ν¹ PK is atomless. In this case there is an inverse-measure-preserving function g : K → [0, γ], where S [0, γ] is given Lebesgue measure (343Cc); write λ for Lebesgue measure on [0, γ]. Set X1 = f −1 [K] = i∈K Ei and let µ1 be the subspace measure on X1 . Now gf : X1 → [0, γ] is inversemeasure-preserving for µ1 and λ. Because µ is compact, so is µ1 (451Da), so µ1 is perfect (342L or 451C). By 451N, the image measure λ1 = µ1 (gf )−1 is a Radon measure. But λ1 must be an extension of Lebesgue measure λ, because gf is inverse-measure-preserving for µ1 and λ, and λ1 and λ must agree on all compact sets. By 416E(b-ii), λ1 S and λ are identical, and, in particular, have the same domains. Now for any set A ⊆ [0, γ], (gf )−1 [A] = i∈J Ei ∈ Σ, where J = g −1 [A] ⊆ I; so A ∈ dom λ1 = dom λ. But we know from 134D or 419J that not every subset of [0, γ] can be Lebesgue measurable. X XQ Q (c) But ν is also atomless. P P?? Suppose, if possible, that M ⊆ I is an atom for ν. Set γ = νM = S µ( i∈M Ei ), F = {F : F ⊆ M, ν(M \ F ) = 0}. Because νF is defined for every F ⊆ M , and M is an atom, F is an ultrafilter on M ; and because ν is countably additive, the intersection of any sequence in F belongs to F, that is, F is ω1 -complete (definition: 4A1Ib). Also F must be non-principal, because we are supposing that ν{i} = 0 for every i ∈ M . By 4A1K, there are a regular uncountable cardinal κ and a function h : M → κ such that the image filter H = h[[F]] is normal. For each ξ < κ, κ \ ξ ∈ H, so S Gξ = (hf )−1 [κ \ ξ] = {Ei : h(i) ≥ ξ} ∈ Σ, µGξ = νh−1 [κ \ ξ] = γ > 0. At this point I apply the full strength of the hypothesis that µ is a compact measure. Let K ⊆ Σ be a compact class such that µ is inner regular with respect to K, and for each ξ < κ choose Kξ ∈TKξ such that Kξ ⊆ Gξ and µKξ ≥ 21 γ. Let S ⊆ [κ]<ω be the family of those finite sets L ⊆ κ such that ξ∈L Kξ = ∅. Because H is a normal filter, there is an H ∈ H such that, for every n ∈ N, [H]n is either a subset of S or disjoint from S (4A1L). If we look at {GξT: ξ ∈ H}, we see that it has empty intersection, because h(f (x)) ≥ ξ for every x ∈ Gξ , and sup H = κ. So ξ∈H Kξ = ∅. Because all the Kξ belong to the compact class K, there must be a finite T set L0 ⊆ H such that ξ∈L0 Kξ = ∅, that is, L0 ∈ S. But this means that [H]n ∩ S 6= ∅, where n = #(L0 ), so that [H]n ⊆ S, by the choice of H. However, H is surely infinite, so we can find distinct ξ0 , . . . , ξ2n in H. If we now look at Kξ0 , . . . , Kξ2n , we see that #({i : i ≤ 2n, x ∈ Kξi }) < n for every x ∈ X, so P2n P2n R 1 χKξi ≥ γ(2n + 1), i=0 χKξi ≤ (n − 1)χG0 , i=0 2

which is impossible, because µG0 = γ. X XQ Q (d) Thus ν is P simultaneously atomless and purely atomic, which means that νI = 0, that is, that S µ( i∈I Ei ) = 0 = i∈I µEi . (e) Now let us return to the general case. Of course P P S i∈I µEi = supJ⊆I is finite i∈J µEi ≤ µ( i∈I Ei ). P S S ?? Suppose, Pif possible, that i∈I µEi < µ( i∈I Ei ). Because µ is semi-finite, there is a set F S⊆ i∈I Ei such L must be countable, so µ( i∈J Ei ) = P that i∈I µEi < µF < ∞. Set L = {i : i ∈SI, µEi > 0}; then 0 i∈J µEi < µF , and µG > 0, where G = F \ i∈L Ei . Set Ei = G ∩ Ei for every i ∈ I, and let µG be the measure on G. Then µG is compact (451Da) and totally finite, µG Ei0 = 0 for S subspace S S every0 i ∈ I, 0 0 i∈J Ei = G ∩ i∈J Ei is measured by µG for every J ⊆ I, every Ei is µG -negligible, but µG ( i∈I Ei ) = µG is not zero; which contradicts the result of (a)-(d) above. X X P S So i∈I µEi = µ( i∈I Ei ), as required. 451Q Lemma Let (X, Σ, µ) be a totally finite compact measure space, Y a metrizable space, and f : X → Y a measurable function. Then there is a closed separable subspace Y0 of Y such that f −1 [Y \ Y0 ] is negligible.

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proof (a) (Cf. 438D.) By 4A2L(h-ii), there is a σ-disjoint base U for thePtopology of Y . Express U as S −1 U [U ]iU ∈Un is disjoint, so U ∈Un µf −1 [U ] ≤ µX is finite, n∈N n where Un is disjoint for each n. Then hf −1 and Vn = {V : V ∈ Un , µf [V ] > 0} is countable for each n. If W ⊆ Un \ Vn , then S S µ( U ∈W f −1 [U ]) = f −1 [ W] is measurable. By 451P,

S S P µf −1 [ (Un \ Vn )] = µ( U ∈Un \Vn f −1 [U ]) = U ∈Un \Vn µf −1 [U ] = 0.

Set V=

S n∈N

Vn ,

Then Y0 is closed, and f −1 [Y \ Y0 ] ⊆

Y0 = Y \ S n∈N

S

(U \ V).

S f −1 [ (Un \ Vn )]

is negligible, so f −1 [Y0 ] is conegligible. On the other hand, Y0 is separable. P P Because U is a base for the topology of X, {Y ∩ U : U ∈ U} is a base for the topology of Y (4A2B(a-iv)). But this is included in the countable family {Y ∩ V : V ∈ V} ∪ {∅}, so Y is second-countable, therefore separable (4A2Oc). Q Q So we have found an appropriate Y0 . 451R Proposition Let (X, Σ, µ) be a semi-finite compact measure space, Y a metrizable space and f : X → Y a measurable function. (a) The image measure ν = µf −1 is tight. (b) If ν is locally finite and µ is complete and locally determined, ν is a Radon measure. proof (a) Take F ⊆ Y such that νF > 0. Then µf −1 [F ] > 0. Because µ is semi-finite, there is an E ∈ Σ such that E ⊆ f −1 [F ] and 0 < µE < ∞. Consider the subspace measure µE and the restriction f ¹E. µE is a totally finite compact measure and f ¹E is measurable, so 451Q tells us that there is a closed separable subspace Y0 ⊆ Y such that µ(E \ f −1 [Y0 ]) = 0. Set E1 = E ∩ f −1 [Y0 ], so that µE1 > 0. Again, the subspace measure µE1 is a totally finite compact measure, therefore perfect, while f [E1 ] ⊆ Y0 . So the image measure µE1 (f ¹E1 )−1 on Y0 is perfect (451Ea), therefore tight (451M), and there is a compact set K ⊆ Y0 ∩F such that νK = µf −1 [K] > 0. By 412B, this is enough to show that ν is tight. (b) ν is complete because µ is. Now suppose that H ⊆ Y is such that H ∩ F belongs to the domain T of ν whenever µF < ∞. In this case µ is inner regular with respect to E = {E : E ∈ Σ, E ∩ f −1 [H] ∈ Σ}. P P Suppose that E ∈ Σ and that µE > 0. Applying (a) to µE and f ¹E, there is a compact set K ⊆ f [E] such that µf −1 [K] > 0. Now νK < ∞, because ν is locally finite, so K ∩ H ∈ T and f −1 [K] ∩ f −1 [H] ∈ Σ. Thus f −1 [K] is a non-negligible member of E included in E. Since E is closed under finite unions, this is enough to show that µ is inner regular with respect to E. Q Q Accordingly f −1 [H] ∈ Σ, by 412Ja. As H is arbitrary, ν is locally determined, therefore a Radon measure. 451S Theorem (Fremlin 81, Koumoullis & Prikry 83) Let (X, T, Σ, µ) be a Radon measure space and Y a metrizable space. Then a function f : X → Y is measurable iff it is almost continuous. proof If f is almost continuous it is surely measurable, by 418E. Now suppose that f is measurable and that E ∈ Σ and γ < µE. Let E0 ⊆ E be such that E0 ∈ Σ and γ < µE0 < ∞. Applying 451Q to the subspace measure µE0 and the restricted function f ¹E0 , we see that there is a closed separable subspace Y0 of Y such that µ(E0 \ f −1 [Y0 ]) = 0. Set E1 = E0 ∩ f −1 [Y0 ]; then µE1 > γ. Applying 418J to µE1 and f ¹E1 : E1 → Y0 , we can find a measurable set F ⊆ E1 such that f ¹F is continuous and µF ≥ γ. As E and γ are arbitrary, f is almost continuous. 451T Example (Vinokurov & Makhkamov 732 , Musial 76) There is a perfect completion regular quasi-Radon probability space which is not countably compact. 2I

am indebted to J.Pachl for the reference.

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proof (a) Let Ω be the set of non-zero countable limit ordinals. For each ξ ∈ Ω, let hθξ (n)in∈N be a strictly increasing sequence in ξ with supremum ξ, and set Qξ = {x : x ∈ {0, 1}ω1 , x(θξ (n)) = 0 for every n ∈ N}. Write X = {0, 1}ω1 \

S ξ∈Ω

Qξ .

ω1

Let ν be the usual measure on {0, 1} , and T its domain; let µ be the subspace measure on X, and Σ = dom µ. (b) It is convenient to note immediately the following fact: for every countable set J ⊆ ω1 , the set πJ [X] is conegligible in {0, 1}J , where πJ (x) = x¹J for x ∈ {0, 1}ω1 . P P Set A = {ξ : ξ ∈ Ω, θξ (n) ∈ J for every n ∈ N}. Then A is countable, because ξ ≤ sup J for every ξ ∈ A. So S D = ξ∈A {y : y ∈ {0, 1}J , y(θξ (n)) = 0 for every n ∈ N} is negligible in {0, 1}J , being a countable union of negligible sets. If y ∈ {0, 1}J \ D, define x ∈ {0, 1}ω1 by setting x(η) = y(η) for η ∈ J, x(η) = 1 for η ∈ ω1 \ J. Then x ∈ / Qξ for any ξ ∈ A, because x¹J = y¹J, while x ∈ / Qξ for any ξ ∈ Ω \ A by the definition of A. So x ∈ X. As y is arbitrary, πJ [X] ⊇ {0, 1}J \ D is conegligible. Q Q (c) µ is a completion regular quasi-Radon measure because ν is (415E, 415B, 412Pd). Also µX = 1. P P Let F ∈ T be a measurable envelope for X. Then there is a countable J ⊆ ω1 such that νJ πJ [F ] is defined and equal to νF (254Od), where νJ is the usual measure on {0, 1}J . But we know that νJ πJ [X] = 1, so µX = ν ∗ X = νF = νJ πJ F = 1. Q Q (d) µ is perfect. P P Take E ∈ Σ such that µE > 0, and a measurable function f : E → R. Set f1 (x) =

f (x) 1+|f (x)|

for x ∈ E, 1 for x ∈ X \ E; then f1 : X → R is measurable. Let g : {0, 1}ω1 → R

be a measurable function extending f1 . By 254Pb, there are a countable set J ⊆ ω1 , a conegligible set W ⊆ {0, 1}J , and a measurable h : W → R such that g extends hπJ . By (b), W 0 = W ∩ πJ [X] is conegligible, while W 00 = W 0 ∩ πJ [E] = {z : z ∈ W 0 , h1 (z) < 1} is measurable and not negligible. Because W 00 is a non-negligible measurable subset of the perfect measure space {0, 1}J , there is a compact set K1 ⊆ h[W 00 ] such that νJ h−1 [K1 ] > 0. Set K = {

t 1−|t|

: t ∈ K1 }; then K is compact, and we have

K1 ⊆ h[W 00 ] = hπJ [X ∩ πJ−1 [W ]] = g[E ∩ πJ−1 [W ]] ⊆ f1 [E], K ⊆ f [E], while f1 , g and hπJ all agree on the µ-conegligible set X ∩ πJ−1 [W ], so µf −1 [K] = µf1−1 [K1 ] = µ(X ∩ (hπJ )−1 [K1 ]) = ν ∗ (X ∩ (hπJ )−1 [K1 ]) = ν(hπJ )−1 [K1 ] (because ν ∗ X = 1 and (hπJ )−1 [K1 ] is measurable) = νJ h−1 [K1 ] > 0. As f is arbitrary, µ is perfect. Q Q (e) ?? Suppose, if possible, that µ is countably compact. Let K be a countably compact class of sets such that µ is inner regular with respect to K; we may suppose that K ⊆ Σ. (i) For I ⊆ ω1 set U (I) = {x : x ∈ X, x(η) = 0 for every η ∈ I}.

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It will be helpful to know that if E ∈ Σ and µE > 0, there is a γ < ω1 such that µ(E ∩ U (I)) > 0 for every finite I ⊆ ω1 \ γ. P P Express E as X ∩ F where F ∈ T. Let J ⊆ ω1 be a countable set such that ν(F 0 \ F ) = 0, where F 0 = πJ−1 [πJ [F ]] (254Od), and γ < ω1 such that J ⊆ γ. If I ⊆ ω1 \ γ is finite, then I ∩ J = ∅, while U (I) is determined by coordinates in I and F 0 is determined by coordinates in J; so µ(E ∩ U (I)) = ν ∗ (X ∩ F ∩ U (I)) = ν(F ∩ U (I)) = ν(F 0 ∩ U (I)) = νF 0 · νU (I) = µE · νU (I) > 0. Thus this γ serves. Q Q (ii) Let M be the family of countable subsets M of ω1 ∪ K such that (α) if I ⊆ M ∩ ω1 is finite there is a K ∈ M ∩ K such that K ⊆ U (I) and µK > 0; (β) if K ∈ M ∩ K, I ⊆ M ∩ ω1 is finite and µ(K ∩ U (I)) > 0, then there is a K 0 ∈ M ∩ K such that K 0 ⊆ K ∩ U (I) and µK 0 > 0; (γ) if γ ∈ M ∩ ω1 then γ ⊆ M ; (δ) if K ∈ M ∩ K and µK > 0 then there is a γ ∈ M ∩ ω1 such that µ(K ∩ U (I)) > 0 whenever I ⊆ ω1 \ γ is finite. Then every countable M ⊆ ω1 ∪ K is included in some member of M. P P Choose hNn in∈N as follows. N0 = M . Given that Nn is a countable subset of ω1 ∪ K then let Nn+1 ⊆ ω1 ∪ K be a countable set such that (α) if I ⊆ Nn ∩ ω1 is finite there is a K ∈ Nn+1 ∩ K such that K ⊆ U (I) and µK > 0; (β) if K ∈ Nn ∩ K, I ⊆ Nn ∩ ω1 is finite and µ(K ∩ U (I)) > 0, then there is a K 0 ∈ Nn+1 ∩ K such that K 0 ⊆ K ∩ U (I) and µK 0 > 0; (γ) if γ ∈ Nn ∩ ω1 then γ ⊆ Nn+1 ; (δ) if K ∈ Nn ∩ K and µK > 0 then there is a γ ∈ Nn+1 ∩ ω1 such that µ(K ∩ U (I)) > 0 whenever I ⊆ ω1 \ γ is finite; (²) Nn ⊆ Nn+1 . S On completing the induction, set M 0 = n∈N Nn ; this serves (because every finite subset of M 0 is a subset of some Nn ). Q Q (iii) Choose a sequence hMn in∈N in M such that, for each n, Mn ∪ {sup(Mn ∩ ω1 ) + 1} ⊆ Mn+1 . Set γn = sup(Mn ∩ ω1 ) for each n. Note that γn ⊆ Mn , because if η < γn then there is some ξ ∈ Mn such that η < ξ; now ξ ⊆ Mn because Mn ∈ M, so η ∈ Mn . Also γn + 1 ∈ Mn+1 for each n, so hγn in∈N is strictly increasing, and ξ = supn∈N γn belongs to Ω. Set J = {θξ (n) : n ∈ N}. Then J ∩ η is finite for every η T < ξ, and in particular J ∩ γn is finite for every n. Set I0 = J ∩ γ0 and In = J ∩ γn \ γn−1 for n ≥ 1. Then n∈N U (In ) = Qξ is disjoint from X. Choose a sequence hKn in∈N in K as follows. Because I0 is a finite subset of M0 ∩ω1 , there is a K0 ∈ M0 ∩K such that K0 ⊆ U (I0 ) and µK0 > 0. Given that Kn ∈ Mn ∩K and µKn > 0, then there is a β ∈ Mn ∩ω1 such that µ(Kn ∩ U (I)) > 0 for every finite I ⊆ ω1 \ β; now β ≤ γn and In+1 ∩ γn = ∅, so µ(Kn ∩ U (In+1 )) > 0. But Kn ∈ Mn+1 ∩ K and In+1 is a finite subset of Mn+1 ∩ ω1 , so there is a Kn+1 ∈ Mn+1 ∩ K such that Kn+1 ⊆ Kn ∩ U (In+1 ) and µKn+1 > 0. Continue. In this way we find a non-increasing sequence hKn in∈N in K such that Kn T ⊆ U (In ) for every T T n and no Kn is empty. But in this case i≤n Ki = Kn is non-empty for every n, while n∈N Kn ⊆ X ∩ n∈N U (In ) is empty. So K is not a countably compact class. X X (f ) Thus µ is not countably compact, and has all the properties claimed. *451U Weakly α-favourable spaces There is an interesting variation on the concept of ‘countably compact’ measure space, as follows. For any measure space (X, Σ, µ) we can imagine an infinite game for two players, whom I will call ‘Empty’ and ‘Nonemepty’. Empty chooses a non-negligible measurable set E0 ; Nonempty chooses a non-negligible measurable set F0 ⊆ E0 ; Empty chooses a non-negligible measurable set E1 ⊆ F0 ; Nonempty chooses T T a non-negligible measurable set F1 ⊆ E1 , and so on. At the end of the game, Empty wins if n∈N En = n∈N Fn is empty; otherwise Nonempty wins. If you have seen ‘Banach-Mazur’

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games, you will recognise this as a similar construction, in which open sets are replaced by non-negligible measurable sets. A strategy for Nonempty is a rule to determine his moves in terms of the preceding moves for Empty; S that is, a function σ : n∈N (Σ \ N )n+1 → Σ \ N , where N is the ideal of negligible sets, such that σ(E0 , E1 , . . . , En ) ⊆ En , at least whenever E0 , . . . , En ∈ Σ \ N are such that Ek+1 ⊆ σ(E0 , . . . , Ek ) for every k < n; since it never matters what Nonempty does if Empty has already broken the rules, T we usually just demand that σ(E0 , . . . , En ) ⊆ En for all E0 , . . . , En ∈ Σ \ N . σ is a winning strategy if n∈N En 6= ∅ whenever hEn in∈N is a sequence in Σ \ N such that En+1 ⊆ σ(E0 , . . . , En ) for every n ∈ N. In terms of the game, we interpret this as saying that Nonempty will win if he plays Fn = σ(E0 , . . . , En ) whenever faced with the position (E0 , F0 , E1 , F1 , . . . , Fn−1 , En ). (Since it is supposed that Nonempty will use the same strategy throughout the game, the moves F0 , . . . , Fn−1 are determined by E0 , . . . , En−1 and there is no advantage in taking them separately into account when choosing Fn .) Now we say that the measure space (X, Σ, µ) is weakly α-favourable if there is such a winning strategy for Nonempty. It turns out that the class of weakly α-favourable spaces behaves in much the same way as the class of countably compact spaces. For the moment, however, I leave the details to the exercises (451Yg-451Yq). See Fremlin 00. 451X Basic exercises (a) (i) Show that any purely atomic measure space is perfect. (ii) Show that any strictly localizable purely atomic measure space is countably compact. (iii) Show that the space of 216E/342N is not countably compact. > (b) Show that a compact measure space in which singleton sets are negligible is atomless. > (c) Let (X, Σ, µ) be a measure space, and ν an indefinite-integral measure over µ (§234). Show that ν is compact, or countably compact, or perfect if µ is. (d) In 413Xn, show that µ is a countably compact measure. (Hint: show that the algebra Σ in 413Xn is a countably compact class.) (e) Let h(Xi , Σi , µi )ii∈I be a family of measure spaces, with direct sum (X, Σ, µ). Show that µ is compact, or countably compact, or perfect iff every µi is. (f ) Let (X, Σ, µ) be a measure space. For A ⊆ X, write µA for the subspace measure on A. Suppose that whenever E ∈ Σ and µE > 0 there is a set A ⊆ X such that µA is perfect and µ∗ (A ∩ E) > 0. Show that µ is perfect. (g) (i) Give an example of a compact probability space (X, Σ, µ) and a σ-subalgebra T of Σ such that µ¹ T is not compact. (ii) Give an example of a compact probability space (X, Σ, µ), a set Y and a function f : X → Y such that the image measure µf −1 is not compact. (Hint: 342M, 342Xf, 439Xa.) (h)QLet hXi ii∈I be a family of sets, with product X. Suppose that Ki ⊆ PXi for each i, and set K = { i∈I Ki : Ki ∈ Ki for each i}. (i) Show that if Ki is a compact class for each i, so is K. (ii) Show that if Ki is a countably compact class for each i, so is K. (i) Let X be a Polish space and E a subset of X. Show that the following are equiveridical: (i) E is universally measurable; (ii) every Borel probability measure on E is perfect; (iii) every σ-finite Borel measure on E is compact; (iv) f [E] is universally measurable in R for every Borel measurable function f : X → R. (j) In 451M, show that µ is a compact measure. (k) Find a Radon measure space (X, T, Σ, µ), a continuous function f : X → [0, 1] and a set B ⊆ [0, 1] such that µ∗ (f −1 [B]) < (µf −1 )∗ B. (l) Let (X, Σ, µ) be a σ-finite measure space. Show that it is perfect iff whenever f : X → R is measurable there is a Kσ set H ⊆ f [X] such that f −1 [H] is conegligible.

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451Xm

(m) Let X be a metrizable space, and µ a semi-finite topological measure on X which (regarded as a measure) is compact. Show that µ is τ -additive. (n) Let (X, Σ, µ) be a compact strictly localizable measure space (e.g., any Radon measure space), (Y, T, ν) a σ-finite measure space, and f : X → L0 (ν) a function. Show that the following are equiveridical: (i) f is measurable, when L0 (ν) is given its topology of convergence in measure; (ii) there is a function h ∈ L0 (λ), where λ is the c.l.d. product measure on X × Y , such that f (x) = h•x for almost every x ∈ X, where hx (y) = h(x, y). (Hint: 418R.) > (o) Let (X, T, Σ, µ) be a Radon measure space. Show that Σ = PX iff µ is purely atomic. (Hint: if Σ = PX, apply 451S with Y = X, the discrete topology on Y and the identity function from X to Y .) (p) Let (X, T, Σ, µ) be a Radon measure space and U a normed space. Show that if f , g : X → U are measurable functions, then f + g is measurable. (Cf. 418Xj.) 451Y Further exercises (a) Show that for any probability space (X, Σ, µ), there is a compact probability space (Y, T, ν) with a subspace isomorphic to (X, Σ, µ). (b) Let hXi ii∈I be a family of sets, and Σi a σ-algebra of subsets ofQXi for each i. Suppose that for each finite N J ⊆ I we are given a finitely additive functional νJ on XJ = i∈J Xi , with domain the algebra TJ = i∈J Σi generated by sets of the form {x : x ∈ XJ , x(i) ∈ E} for i ∈ J, E ∈ Σi , and that (α) νK {x : x ∈ XK , x¹J ∈ W } = νJ W whenever J ⊆ K ∈ [I]<ω and W ∈ TJ (β) µi = νi is a countably compact probability measure for every i ∈ I. Show that there is a countably compact measure µ on X = XI such that µ{x : x ∈ X, x¹J ∈ W } = νJ W whenever J ∈ [I]<ω and W ∈ TJ . (Hint: 454D.) (Compare 418M.) (c) Describe µ in the case of 451Yb in which I = [0, 1], Xi = [0, 1]N \ {i}, Σi is the algebra of Lebesgue measurable subsets of Xi , and νJ E = µL {t : zJt ∈ E} for every E ∈ i∈J Σi , where zJt (i) = t for i ∈ J, t ∈ [0, 1]. Contrast this with the difficulty encountered in 418Xv. (d) Let (X, Σ, µ) be aSsemi-finite compact measure space, and hEi iS i∈I a point-finite family of measurable S subsets of X such that i∈J Ei ∈ Σ for every J ⊆ I. Show that µ( i∈I Ei ) = supJ⊆I is finite µ( i∈J Ei ). (Hint: 438Ya.) (e) Let X be a metacompact space, and µ a semi-finite topological measure on X which (regarded as a measure) is compact. Show that µ is τ -additive. (f ) Let (X, Σ, µ) be a compact measure space, V a Banach space and f : X → V a measurable function such that kf k : X → [0, ∞[ is integrable. Show that f is Bochner integrable (253Yf). (g) Show that any purely atomic measure space is weakly α-favourable, so that the space of 216E/342N is weakly α-favourable but not countably compact. (h) Show that the direct sum of a family of weakly α-favourable measure spaces is weakly α-favourable. (i) Show that an indefinite-integral measure over a weakly α-favourable measure is weakly α-favourable. (j) (i) Show that a countably compact measure space is weakly α-favourable. (ii) Show that a semi-finite weakly α-favourable measure space is perfect. (k) Show that any measurable subspace of a weakly α-favourable measure space is weakly α-favourable. (l) Let (X, Σ, µ) be a weakly α-favourable measure space, (Y, T, ν) a semi-finite measure space, and f : X → Y a (Σ, T)-measurable function such that f −1 [F ] is negligible whenever F ⊆ Y is negligible. Show that (Y, T, ν) is weakly α-favourable.

§452 intro.

Integration and disintegration of measures

427

(m) (i) Show that a measure space is weakly α-favourable iff its completion is weakly α-favourable. (ii) Show that a semi-finite measure space is weakly α-favourable iff its c.l.d. version is weakly α-favourable. (n) Show that the c.l.d. product of two weakly α-favourable measure spaces is weakly α-favourable. (o) Show that the product of any family of weakly α-favourable probability measures is weakly αfavourable. (p) Show that the space of 451T is not weakly α-favourable. (q) Let (X, Σ, µ) be a complete locally determined measure space and φ a lower density on X such that φX = X; let T be the corresponding density topology (414P). Show that (X, Σ, µ) is weakly α-favourable iff (X, T) is weakly α-favourable (definition: 4A2A). 451 Notes and comments For a useful survey of results on countably compact and perfect measures, with historical notes, see Ramachandran 02. The concepts of ‘compact’, ‘countably compact’ and ‘perfect’ measure space can all be regarded as attempts to understand and classify the special properties of Lebesgue measure on [0, 1], regarded as a measure space. Because a countably separated perfect probability space is very nearly isomorphic to Lebesgue measure (344K), we can think of a perfect measure space as one in which the countably-generated σ-subalgebras look like Lebesgue measure (451F). The arguments of 451Ic and 451Jc already hint at the kind of results we can hope for. When we form a product measure, each measurable set in the product will depend, in effect, on sequences of measurable sets in the factors, and therefore can be studied in terms of countably generated subalgebras; so that many results about products of perfect measures will be derivable, if we wish to take that route, from results about products of copies of Lebesgue measure. Of course my normal approach in this treatise is to go straight for the general result; but like anyone else I often start from a picture based on the familiar special case. In the next section we shall have some theorems for which countable compactness, rather than perfectness, seems to be the relevant property. The first half of the section (down to 451O) is essentially a matter of tidying up the theory of compact and perfect measures, and showing that the same ideas will cover the new class of countably compact measures. (You may like to go back to 342G, in which I worked through the basic properties of compact measures, and contrast the arguments used there with the slightly more sophisticated ones above.) In 451P-451S I enter new territory, showing that for compact measures (and therefore for Radon measures) the theory of measurable functions into metric spaces is particularly simple, without making any assumptions about measure-free cardinals.

452 Integration and disintegration of measures A standard method of defining measures is through a formula µE =

R

µy E ν(dy)

where (Y, T, ν) is a measure space and hµy iy∈Y is a family of measures on another set X. In practice these constructions commonly involve technical problems concerning the domain of µ (as in 452Xi), which is why I have hardly used them so far in this treatise. There are not-quite-trivial examples in 417Yb, 434R and 436F, and the indefinite-integral measures of §234 can also be expressed in this way (452Xf); for a case in which this approach is worked out fully, see 453N. But when a formula of this kind is valid, as in Fubini’s theorem, it is likely to be so useful that it dominates further investigation of the topic. In this section I give one of the two most important theorems guaranteeing the existence of appropriate families hµy iy∈Y when µ and ν are given (452I); another will follow in the next section (453K). They both suppose that we are provided with a suitable function f : X → Y , and rely heavily on the Lifting Theorem (§341) and on considerations of inner regularity from Chapter 41. A variant of the argument can be applied to certain measures on product spaces, with a projection onto a factor in place of the function f (452M).

428


§452 intro.

The formal definition of a ‘disintegration’ (which is nearly the same thing as a ‘regular conditional probability’) is in 452E. The main theorem depends, for its full generality, on the concept of ‘countably compact measure’ (451B). It can be strengthened when µ is actually a Radon measure (452O). The greater part of the section is concerned with general disintegrations, in which the measures µy are supposed to be measures on X and are not necessarily related to any particular structure on X. However a natural, and obviously important, class of applications has X = Y × Z and each µy based on the section {y} × Z, so that it can be regarded as a measure on Z. I begin the section, therefore, with an outline of the elementary theory of measures defined by ‘integrating’ such a family (452A-452D). For such measures, we R can derive the fundamental requirement, that µy E ν(dy) should be defined for all measurable sets E ⊆ X, from conditions expressed in terms of the factors Y and Z. 452A Lemma Let (Y, T, ν) be a probability space, X a set, Rand hµy iy∈Y a family of probability measures on X. Let A be the family of subsets A of X such that θE = µy E ν(dy) is defined. (a) A is a Dynkin class. (b) If Σ is any σ-subalgebra of A then µ = θ¹Σ is a probability measure on X. ˆ its domain, then (c) Suppose now that every µy is complete. If, in (b), µ ˆ is the completion of µ and Σ ˆ ⊆ A and µ ˆ Σ ˆ = θ¹ Σ. proof For (a) and (b), we have only to look at the definitions of ‘Dynkin class’ and ‘measure’ and apply the ˆ then there are E 0 , E 00 ∈ Σ such that E 0 ⊆ E ⊆ E 00 elementary properties of the integral. For (c), if E ∈ Σ, 0 00 0 00 and θE = θE . So µy E = µy E for ν-almost every y; since all the µy are supposed to be complete, µy E is defined and equal to µy E 0 for almost every y, and θE is defined and equal to θE 0 = µE 0 = µ Ê. 452B Theorem (a) Let X be a set, (Y, T, ν) a probability space, and hµy iy∈Y a family ofRprobability measures on X. Let E be a family of subsets of X, closed under finite intersections, such that µy E ν(dy) is defined for every E ∈ E. (i) If Σ is the σ-algebra of subsetsR of X generated by E, we have a probability measure µ on X, with domain Σ, given by the formula µE = µy E ν(dy) for every E ∈ Σ. R ˆ its domain, then µ (ii) If µ ˆ is the completion of µ and Σ ˆy E ν(dy) is defined and equal to µ Ê for every ˆ where µ E ∈ Σ, ˆy is the completion of µy for each y ∈ Y . (b) Let Z be a set, (Y, T, ν) a probability space, and hµy iy∈Y a family ofRprobability measures on Z. Let H be a family of subsets of Z, closed under finite intersections, such that µy H ν(dy) is defined for every H ∈ H. (i) If Υ is the σ-algebra of subsets of ZRgenerated by H, we have a probability measure µ on Y × Z, b defined by setting µE = µy E[{y}]ν(dy) for every E ∈ T⊗Υ. b with domain T⊗Υ, R ˆ (ii) If µ ˆ is the completion of µ and Σ its domain, then µ ˆy E[{y}]ν(dy) is defined and equal to µ Ê for ˆ where µ every E ∈ Σ, ˆy is the completion of µy for each y ∈ Y . proof (a) Define A ⊆ P(Y × Z) as in 452A. Then E ⊆ A, so by the Monotone Class Theorem (136B) Σ ⊆ A and we have (i). Applying 452Ac to hˆ µy iy∈Y we have (ii). (b) Set X = Y × Z. For y ∈ Y , let µ0y be the measure on X defined by setting µ0y E = µy E[{y}] whenever this is defined; that is, µ0y is the image of µy under the function z 7→ (y, z) : Z → X. Set E = {F ×H : F ∈ T, H ∈ H}. Then E is a family of subsets of X closed under finite intersections, and

R

µ0y (F × H)ν(dy) =

R

χF (y)µy H ν(dy)

is defined for every F ∈ T, H ∈ H. By (a), we have a measure µ on X, with domain the σ-algebra Σ generated by E, defined by writing µE =

R

µ0y E ν(dy) =

R

µy E[{y}]ν(dy)

b (the set {H : Y × H ∈ Σ} is a σ-algebra of subsets of Z for every E ∈ Σ. Of course Σ includes T⊗Υ b including H, so includes Υ) and is therefore equal to T⊗Υ. ˆ This proves (i). If now E ∈ Σ, (a-ii) tells us that µ Ê =

R

µ ˆ0y E ν(dy) =

R

µ ˆy E[{y}]ν(dy).

452D


429

452C Theorem(a) Let Y be a topological space, ν a τ -additive topological probability on Y , (X, T) a topological space, and hµy iy∈Y a family of τ -additive topological probability measures on X. Suppose that there is a base U for T, closed under finite unions, such that y 7→ µy U is lower semi-continuous for every U ∈ U. R (i) We can define a τ -additive Borel probability measure µ on X by writing µE = µy E ν(dy) for every Borel set E ⊆ X. R ˆ its domain, then µ (ii) If µ ˆ is the completion of µ and Σ ˆy E ν(dy) is defined and equal to µ Ê for every ˆ where µ E ∈ Σ, ˆy is the completion of µy for each y ∈ Y . (b) Let Y be a topological space, ν a τ -additive topological probability on Y , (Z, U) a topological space, and hµy iy∈Y a family of τ -additive topological probability measures on Z. Suppose that there is a base V for U, closed under finite unions, such that y 7→ µy V is lower semi-continuous for every V ∈ R V. (i) We can define a τ -additive Borel probability measure µ on Y × Z by writing µE = µy E[{y}]ν(dy) for every Borel set E ⊆ Y × Z. R ˆ its domain, then µ (ii) If µ ˆ is the completion of µ and Σ ˆy E[{y}]ν(dy) is defined and equal to µ Ê for ˆ where µ every E ∈ Σ, ˆy is the completion of µy for each y ∈ Y . proof (a) For A ⊆ X, set fA (y) = µy A when this is defined. We may suppose that ∅ ∈ U. If W ⊆ U is a non-empty upwards-directed set with union G, hfW iW ∈W is an upwards-directed family of lower semicontinuous fG , because every µy is τ -additive. So fG is lower semi-continuous, R R functions with supremum and also fG dν = supW ∈W fW dν, by 414Ba. Taking E to be the family Rof open subsets of X in 452Ba, we see that we have a τ -additive Borel measure µ on X such that µE = µy E ν(dy) for every Borel set E ⊆ X. Moreover, if G is a non-empty upwards-directed family of open subsets of X with union G∗ , then W = {W : W ∈ U , W ⊆ G for some G ∈ G} is an upwards-directed family with union G∗ , so µG∗ =

R

fG∗ dν = supW ∈W

R

fW dν ≤ supG∈G µG ≤ µG∗ .

As G is arbitrary, µ is τ -additive. This proves (i); (ii) follows immediately, as in 452Ba. S (b) Let U be the family of sets expressible as i≤n Hi × Vi where Hi ⊆ Y is open and Vi ∈ V for every i ≤ n. Because V is a base for U, U is a base for the topology of X = Y × Z. For y ∈ Y let µ0y be the measure on X defined by saying that µ0y E = µy E[{y}] whenever this is defined. Then µ0y is a τ -additive topological probability measure on X, by 418Ha or otherwise. If U ∈ U , y 7→ µ0y U is lower semi-continuous. P P Express S U as i≤n Hi × Vi where Hi ⊆ Y is open and Vi ∈ V for each i. Suppose that y ∈ Y and γ < µy U . Set S T I = {i : i ≤ n, y ∈ Hi }, H = Y ∩ i∈I Hi and V = i∈I Vi . Then U [{y}] = V ⊆ U [{y 0 }] for every y 0 ∈ H. Also H 0 = {y 0 : µy0 V > γ} is a neighbourhood of y. So H ∩ H 0 is a neighbourhood of y, and µ0y0 G[{y 0 }] > γ for every y 0 ∈ H ∩ H 0 . As y and γ are arbitrary, we have the result. Q Q Now applying (a) to hµ0y iy∈Y we see that (b) is true. 452D Theorem (a) Let (Y, S, T, ν) be a Radon probability space, (X, T) a topological space, and hµy iy∈Y a uniformly tight (definition: 437Tb) family of Radon probability measures on X. Suppose that there is a base U for T, closed under finite unions, such that y 7→ µy U is lower semi-continuous for every R U ∈ U. Then we have a Radon probability measure µ ˜ on X such that µ ˜E = µy E ν(dy) whenever µ ˜ measures E. (b) Let (Y, S, T, ν) be a Radon probability space, (Z, U) a topological space, and hµy iy∈Y a uniformly tight family of Radon probability measures on Z. Suppose that there is a base V for U, closed under finite unions, such that y 7→ µy V is lower semi-continuous for every V ∈ V. Then we have a Radon probability R measure µ ˜ on X = Y × Z such that µ ˜E = µy E[{y}]ν(dy) whenever µ ˜ measures E. proof I take the two parts together. By 452C we have a τ -additive Borel probability measure µ satisfying the appropriate formula. Now for any ² > 0 there is a compact set K ⊆ X such that µK ≥ 1 − 2². P P In (a), we just have to take a K such that µy K ≥ 1 − 2² for every y ∈ Y . In (b), start with a compact set K1 ⊆ Y such that νK1 ≥ 1 − ². Now there is a compact set K2 ⊆ Z such that µy K2 ≥ 1 − ² for every y ∈ Y . So K = K1 × K2 is compact and µK ≥ (1 − ²)2 ≥ 1 − 2². Q Q Since µ is a probability measure it is surely locally finite and effectively locally finite, so the conditions of 416F(iv) are satisfied and the c.l.d. version µ ˜ of µ is a Radon measure on X. But of course µ ˜ is just the completion of µ, so 452C(a-ii) or 452C(b-ii) tells us that the declared formula also applies to µ ˜.

430


452E

452E All the constructions above can be thought of as special cases of the following. Definition Let (X, Σ, µ) andR (Y, T, ν) be measure spaces. A disintegration of µ over ν is a family hµy iy∈Y of measures on X such that µy E ν(dy) is defined in [0, ∞] and equal to µE for every E ∈ Σ. If f : X → Y is an inverse-measure-preserving function, a disintegration hµy iy∈Y of µ over ν is consistent with f if, for each F ∈ T, µy f −1 [F ] = 1 for ν-almost every y ∈ Y . hµy iy∈Y is strongly consistent with f if, for almost every y ∈ Y , µy is a probability measure for which f −1 [{y}] is conegligible. A trivial example of a disintegration is when ν is a probability measure and µy = µ for every y. Of course this is of little interest. The archetypal disintegration is 452Bb when all the µy are the same, in which case Fubini’s theorem tells us that we are looking at a product measure on X = Y × Z. Clearly this disintegration is consistent with the function (y, z) 7→ y : X → Y , and if µ is a probability measure then it is strongly consistent. The phrase regular conditional probability is used for special types of disintegration; typically, when µ and ν and every µy are probabilities, and sometimes supposing that every µy has the same domain as µ. I have seen the word decomposition used for what I call a disintegration. 452F RR Proposition Let (X, Σ, µ) and (Y, T, ν) be measure spaces and hµy iy∈Y a disintegration of µ over R ν. Then f (x)µ (dx)ν(dy) is defined and equal to f dµ for every [−∞, ∞]-valued function f such that y R f dµ is defined in [−∞, ∞]. proof (a) Suppose first that f is non-negative. Let H ∈ Σ be a conegligible set such that f ¹H is Σmeasurable. For n ∈ N set P4n Enk = {x : x ∈ H, 2−n k ≤ f (x)} for k ≥ 1, fn = 2−n k=1 χEnk . RThen hfn in∈N is a non-decreasing sequence of functions with limn→∞ fn (x) = f (x) for every x ∈ H. Now µy (X \ H)ν(dy) = 0, so X \ H is µy -negligible for almost every y. Set V = {y : µy (X \ H) = 0, Enk ∈ dom µy for every n ∈ N, k ≥ 1}; then V is ν-conegligible. For y ∈ V ,

R

R

P 4n fn dµy = limn→∞ 2−n k=1 µy Enk , R is ν-virtually measurable, so y 7→ f dµy is ν-virtually measurable and

f dµy = limn→∞

while each function y 7→ µy Enk ZZ

ZZ f dµy ν(dy) = lim

fn dµy ν(dy) = lim 2

n→∞

n→∞

= lim 2−n n→∞

4 Z X n

−n

∞ X

Z µEnk = lim

k=1

n→∞

k=1

fn dµ =

µy Enk ν(dy)

Z f dµ.

(b) For general f we now have ZZ

ZZ

ZZ f + (x)µy (dx)ν(dy) − f − (x)µy (dx)ν(dy) Z Z Z = f + dµ − f − dµ = f dµ,

f (x)µy (dx)ν(dy) =

where f + , f − are the positive and negative parts of f . Remark When X = Y × Z and our disintegration is a family hµ0y iy∈Y of measures on RX defined from a family hµRy iy∈Y of probability measures on Z, as in 452Bb, we can more naturally write f (y, z)µy (dz) in place of f (x)µ0y (dx), and we get

RR

as in 252B.

f (y, z)µy (dz)ν(dy) =

R

f dµ whenever the latter is defined in [−∞, ∞]

452H


431

452G The most useful theorems about disintegrations of course involve some restrictions on their form, most commonly involving consistency with some kind of projection. I clear the path with statements of some elementary facts. Proposition Let (X, Σ, µ) and (Y, T, ν) be measure spaces, f : X → Y an inverse-measure-preserving function, and hµy iy∈Y a disintegration of µ over ν. (a) If hµy iy∈Y is consistent with f , and F ∈ T, then µy f −1 [F ] = χF (y) for ν-almost every y ∈ Y ; in particular, almost every µy is a probability measure. (b) If hµy iy∈Y is strongly consistent with f it is consistent with f . (c) If ν is countably separated (definition: 343D) and hµy iy∈Y is consistent with f , then it is strongly consistent with f . proof (a) We have µy f −1 [F ] = 1 for almost every y ∈ F . Since also µy (X \ f −1 [F ]) = µy f −1 [Y \ F ] = 1,

µy X = µy f −1 [Y ] = 1

for almost every y ∈ Y \ F , µy f −1 [F ] = 0 for almost every y ∈ X \ F . (b) If F ∈ T, then f −1 [F ] ⊇ f −1 [{y}] is µy -conegligible for almost every y ∈ F ; since we are also told that µy X = 1 for almost every y, µy f −1 [F ] = 1 for almost every y ∈ F . (c) There is a countable F ⊆ T separating the points of Y ; we may suppose that Y \ F ∈ F for every F ∈ F. Now HF = F \ {y : µy f −1 [F ] is defined and equal to 1} are negligible for every F ∈ F, so that Z = {y : µy X = 1} \

S F ∈F

HF

is conegligible. For y ∈ Z, set Fy = {F : y ∈ F ∈ F }; then T T {y} = Fy , f −1 [{y}] = {f −1 [F ] : F ∈ Fy }, while µy f −1 [F ] = 1 for every F ∈ Fy . Because Fy is countable, µy f −1 [{y}] = 1. This is true for almost every y, so hµy iy∈Y is strongly consistent with f . 452H Lemma Let (X, Σ, µ) and (Y, T, ν) be spaces, and T : L∞ (µ) → L∞ (ν) a positive R probability R • • linear operator such that T (χX ) = χY and T u = u whenever u ∈ L∞ (µ)+ . Let K be a countably compact class of subsets of X, closed under finite unions and countable intersections, such that µ is inner regular with respect to K. Then there is a disintegration hµy iy∈Y of µ over ν such that (i) µy X = 1, K ⊆ domR µy and µy is inner regular with respect to K, for every y ∈ Y ; (ii) setting hg (y) = g dµy whenever g ∈ L∞ (µ) and y ∈ Y are such that the integral is defined, hg ∈ L∞ (ν) and T (g • ) = h•g for every g ∈ L∞ (µ). proof (a) Completing ν does not change its L∞ (ν) or L∞ (ν), nor does it change the families which are disintegrations over ν; so we may assume throughout that ν is complete. It therefore has a lifting θ : B → T, where B is the measure algebra of ν, which gives rise to a Riesz homomorphism S from L∞ (ν) ∼ = L∞ (B) ∞ • to the space L (T) of bounded T-measurable real-valued functions on Y such that (Sv) = v for every v ∈ L∞ (ν) (363I, 363F, 363Ha). (b) For y ∈ Y and E ∈ Σ, set ψy E = (ST (χE • ))(y). Because 0 ≤ T (χE • ) ≤ χY • in L∞ (ν), 0 ≤ ψy E ≤ 1. The maps E 7→ χE 7→ χE • 7→ T (χE • ) 7→ ST (χE • ) are all additive, so ψy : Σ → [0, 1] is additive for each y ∈ Y . For fixed E ∈ Σ, µE =

R

χE dµ =

R

(χE • ) =

R

T (χE • ) =

R

ST (χE • ) =

R

ψy E ν(dy).

(c) At this point, recall that µ is supposed to be inner regular with respect to the countably compact class K. By 413Sa, there is for every y ∈ Y a complete measure µ0y on X such that µ0y X ≤ ψy X ≤ 1, K ⊆ dom µ0y , and µ0y K ≥ ψy K for every K ∈ K ∩ Σ.

432


452H

(d) Now, for any fixed E ∈ Σ, µ0y E is defined and equal to ψy E for almost every y ∈ Y . P P Let hKn in∈N , 0 hKn0 in∈N be sequences in K ∩ Σ such that K ⊆ E and K ⊆ X \ E for every n, while µE = supn∈N µKn n S T n and µ(X \ E) = supn∈N µKn0 . Set L = n∈N Kn , L0 = n∈N (X \ Kn0 ). Then both L and L0 belong to the domain of every µ0y , and sup ψy Kn ≤ sup µ0y Kn ≤ µ0y L ≤ µ0y L0

n∈N

n∈N

≤ inf µ0y (X \ Kn0 ) = µ0y X − sup µ0y Kn0 ≤ 1 − sup ψy Kn0 n∈N

n∈N

n∈N

for every y. On the other hand, Z Z 0 (1 − sup ψy Kn )ν(dy) ≤ νY − sup ψy (Kn0 )ν(dy) = µX − sup µKn0 = µE n∈N n∈N n∈N Z Z = sup µKn = sup ψy Kn ν(dy) ≤ sup ψy Kn ν(dy). n∈N

n∈N

n∈N

So supn∈N ψy Kn = µ0y L = µ0y L0 = 1 − supn∈N ψy Kn0 for almost every y. Because L ⊆ E ⊆ L0 and µ0y is complete, E ∈ dom µ0y and µ0y E = 1 − supn∈N ψy Kn0 ≥ 1 − ψy (X \ E) ≥ ψy E for almost every y ∈ Y . Similarly, µ0y (X \ E) ≥ ψy (X \ E) for almost every y. But as µ0y E + µ0y (X \ E) = µ0y X ≤ ψy X ≤ 1 whenever the left-hand side is defined, we must have µ0y E = ψy E for almost every y, as claimed. Q Q It follows at once that

R

for every E ∈ Σ, and

hµ0y iy∈Y

µ0y E ν(dy) =

ψy E ν(dy) = µE

is a disintegration of µ over ν.

(e) At this point observe that

R

R

µ0y X ν(dy) = µX =

R

χX • =

R

T (χX • ) = νY ,

so F0 = {y : µ0y X < 1} is negligible. Taking any y0 ∈ Y \ F0 and setting µy = µ0y0 for y ∈ F0 , = µy for y ∈ Y \ F0 , we find ourselves with a disintegration hµy iy∈Y of µ over ν with the same properties as hµ0y iy∈Y , but now consisting entirely of probability measures. R (f ) For g ∈ L∞ (µ), set hg (y) = g dµy whenever y ∈ Y is such that the integral is defined. Consider the set V of those g ∈ L∞ (µ) such that hg ∈ L∞ (ν) and T g • = h•g in L∞ (ν). If E ∈ Σ, then hχE (y) = ψy E for almost every y, so h•χE = (ST (χE • ))• = T (χE • ); accordingly χE ∈ V . It is easy to check that V is closed under addition and scalar multiplication, so it contains all simple functions. Next, if hgn in∈N is a non-decreasing sequence of simple functions with limit g ∈ L∞ (ν), then hg = supn∈N hgn wherever the right-hand side is defined. Also T is order-continuous, because it preserves integrals, so T g • = supn∈N T gn• = supn∈N h•gn = h•g and g ∈ V . Finally, if g ∈ L∞ (µ) is zero almost everywhere,R there is a negligible E ∈ Σ such that g(x) = 0 for every x ∈ X \ E; µy E = 0 for almost every y, so hg (y) = g dµy = 0 for almost every y and again g ∈ V . Putting these together, we see that V = L∞ (ν), as required by (ii) as stated above.

452J


433

452I Theorem (Pachl 78) Let (X, Σ, µ) be a countably compact measure space, (Y, T, ν) a σ-finite measure space, and f : X → Y an inverse-measure-preserving function. Then there is a disintegration hµy iy∈Y of µ over ν, consistent with f , such that µy X ≤ 1 for every y ∈ Y . Moreover, (i) if K is a countably compact class of subsets of X such that µ is inner regular with respect to K, then we can arrange that K ⊆ dom µy for every y ∈ Y ; (ii) if, in (i), K is closed under finite unions and countable intersections, then we can arrange that K ⊆ dom µy and µy is inner regular with respect to K for every y ∈ Y . proof (a) Consider first the case in which ν and µ are probability measures and we are provided with a R class K as in (ii). In this case, for each u ∈ L∞ (µ), F 7→ f −1 [F ] u is countably additive. So we have an R R operator T : L∞ (µ) → L∞ (ν) defined by saying that F T u = f −1 [F ] u whenever u ∈ L∞ (µ) and F ∈ T. R R Of course T is linear and positive and T u = u whenever u ∈ L∞ (µ). By 452H, there is a disintegration hµy iy∈Y of µ over ν such that (α) for every y ∈ Y , µy X = 1, K ⊆ dom µy and µyR is inner regular with respect to K; (β) T (g • ) = h•g whenever g ∈ L∞ (µ) and hg (y) = g dµy when the integral is defined. If now F ∈ T, set g = χf −1 [F ] in (β); then T g • is defined by saying that

R

H

T g• =

R

f −1 [H]

g = µf −1 [F ∩ H] = ν(F ∩ H)

for every H ∈ T, so that T g • = χF • and we must have µy f −1 [F ] = 1 for almost every y ∈ F . (b) The theorem is stated in a way that makes it quotable in parts without committing oneself to a particular class K. But if we are given a class satisfying (i), we can extend it to one satisfying (ii), by 413Sb; and if we are told only that µ is countably compact, we know from the definition that we shall be able to choose a countably compact class satisfying (i). (c) This proves the theorem on the assumption that µ and ν are probability measures. If µX = νY = 0 then the result is trivial, as we can take every µy to be the zero measure. Otherwise, because ν is σ-finite, there is a partition hYn in∈N of Y P into measurable sets of finite measure. Let hγn in∈N be a sequence of strictly ∞ positive real numbers such that n=0 γn νYn = 1, and write P∞ ν 0 F = n=0 γn ν(F ∩ Yn ) for F ∈ T, P∞ µ0 E = n=0 γn µ(E ∩ Xn ) for E ∈ Σ. It is easy to check (α) that ν 0 and µ0 are probability measures (β) that f is inverse-measure-preserving for µ0 and ν 0 (γ) that if µ is inner regular with respect to K so is µ0 . Note that ν 0 and ν have the same negligible sets. By (a)-(b), µ0 has a disintegration hµy iy∈Y over ν 0 which is consistent with f , and (if appropriate) has the properties demanded in (i) or (ii). Now, if E ∈ Σ,

µE = =

∞ X n=0 ∞ X

γn−1 µ0 (E ∩ Xn ) = Z γn−1

n=0

∞ X

Z γn−1

µy (E ∩ Xn )ν 0 (dy)

n=0

µy E ν 0 (dy) Yn

(because µy X = 1, µy Xn = (χYn )(y) for ν 0 -almost every y, every n) Z ∞ Z ∞ Z X X = µy E ν(dy) = µy (E ∩ Xn )ν(dy) = µy E ν(dy). n=0

Yn

n=0

So hµy iy∈Y is a disintegration of µ over ν. If F ∈ T, then µy f −1 [F ] = 1 for ν 0 -almost every y, that is, for ν-almost every y, so hµy iy∈Y is still consistent with f with respect to the measure ν. 452J Remarks (a) In the theorem above, I have carefully avoided making any promises about the domains of the µy beyond that in (i). If Σ0 is the σ-algebra generated by K ∩ Σ, then whenever E ∈ Σ there are E 0 , E 00 ∈ Σ0 such that E 0 ⊆ E ⊆ E 00 and µ(E 00 \ E 0 ) = 0. (For µ, like ν, must be σ-finite, so

434


452J

we can choose E 0 to be a countable union of members of K ∩ Σ, and E 00 to be the complement of such a union.) Thus we shall have a σ-algebra on which every µy is defined and which will be adequate to describe nearly everything about µ. The example of Lebesgue measure on the square (452E) shows that we cannot ordinarily expect the µy to be defined on the whole of Σ itself. In many important cases, of course, we can say more (452Xl). (b) Necessarily (as remarked in the course of the proof) µy X = 1 for almost every y. In some applications it seems right to change µy for a negligible set of y’s so that every µy is a probability measure. Of course this cannot be done if X = ∅ 6= Y , but this case is trivial (we should have to have νY = 0). In other cases, we can make sure that any new µy is equal to some old one, so that a property required by (i) or (ii) remains true of the new disintegration. If we want to have ‘µy f −1 [{y}] = µy X = 1 for every y ∈ Y ’, strengthening ‘strongly consistent’, we shall of course have to begin by checking that f is surjective. (c) The question of whether ‘σ-finite’ can be weakened to ‘strictly localizable’ in the hypotheses of 452I is related to the Banach-Ulam problem (452Yb). See also 452O. 452K Example The hypothesis ‘countably compact’ in 452I is in fact essential (452Ye). To see at least that it cannot be omitted, we have the following elementary example. Set Y = [0, 1], and let ν be Lebesgue measure on Y , with domain T. Let X ⊆ [0, 1] have outer measure 1 and inner measure 0 (134D, 419J); let µ be the subspace measure on X. Set f (x) = x for x ∈ X. Then there is no disintegration hµy iy∈Y of µ over ν which is consistent with f . P P?? Suppose, if possible, that hµy iy∈[0,1] is such a disintegration. Then, in particular, the sets Hq = [0, q] \ {y : X ∩ [0, q] ∈ dom µy , µy (X ∩ [0, q]) = 1}, Hq0 = [q, 1] \ {y : X ∩ [q, 1] ∈ dom µy , µy (X ∩ [q, 1]) = 1} S are negligible for every q ∈ [0, 1]. Set G = [0, 1] \ q∈Q∩[0,1] (Hq ∪ Hq0 ), so that G is ν-conegligible. Then there must be some y ∈ G\X. Now µy (X ∩[0, q 0 ]) = µy (X ∩[q, 1]) = 1 whenever q, q 0 ∈ Q and 0 ≤ q < y < q 0 ≤ 1, so that µy (X ∩ {y}) = 1. But X ∩ {y} = ∅. X XQ Q 452L The same ideas as in 452I can be used to prove a result on the disintegration of measures on product spaces. It will help to have a definition. Q Definition Let hXi ii∈I be a family of sets, and λ a measure on X = i∈I Xi . For each i ∈ I set πi (x) = x(i) for x ∈ X. Then the image measure λπi−1 is the marginal measure of λ on Xi . 452M I return to the context of 452B-452D. Theorem Let Y and Z be sets and T ⊆ PY , Υ ⊆ PZ σ-algebras. Let µ be a non-zero totally finite measure b with domain T⊗Υ, and ν the marginal measure of µ on Y . Suppose that the marginal measure λ of µ on Z is inner regular with respect to a countably compact class K ⊆ PZ which is closed under finite unions and countable intersections. Then there is a family hµy iy∈Y of complete probability measures on Z, all measuring every member of K and inner regular with respect to K, such that µE = b and for every E ∈ T⊗Υ,

R

f dµ =

R

µy E[{y}]ν(dy)

RR

f (y, z)µy (dz)ν(dy) R whenever f is a [−∞, ∞]-valued function such that f dµ is defined in [−∞, ∞]. proof (a) To begin with, assume that µ is a probability measure and that ν is complete. Let B be the measure algebra of ν and θ : B → T a lifting. For H ∈ Υ, F ∈ T set νH F = µ(F × H); then νRH : T → [0, 1] is countably additive and νH F ≤ νF for every F ∈ T, so there is a vH ∈ L1 (ν) such that F vH = νH F for every F ∈ T and 0 ≤ vH ≤ χ1. We can therefore think of vH as a member of L∞ (ν) ∼ = L∞ (B). Let ∞ ∞ T : L (B) → L (T) be the Riesz homomorphism associated with θ, and set ψy H = (T vH )(y) for every y ∈Y.

452M


435

Each ψy : Υ → [0, ∞[ is finitely additive. So we have a complete measure µy on Z such that µy Z ≤ ψy Z = 1, K ⊆ dom µy , µy is inner regular with respect to K and µy K ≥ ψy K for every K ∈ K. For H ∈ Υ, F ∈ T we have

R

F

So

R

ψy H ν(dy) =

R

F

T vH =

µy K · χF (y)ν(dy) ≥

R F

R

F

vH = µ(F × H).

ψy K ν(dy) = µ(F × K)

for every K ∈ K. Now note that, for any H ∈ Υ and F ∈ T, µ(F × H) −

sup

µ(F × K) =

K∈K,K⊆H

≤

inf

µ(F × (H \ K))

inf

λ(H \ K) = 0

K∈K,K⊆H K∈K,K⊆H

because λ is inner regular with respect to K (and, like µ, is a probability measure). So Z Z (µy )∗ H · χF (y)ν(dy) ≥ sup µy K · χF (y)ν(dy) K∈K,K⊆H

≥

sup

µ(F × K) = µ(F × H).

K∈K,K⊆H

In particular,

R R

(µy )∗ H ν(dy) ≥ µ(Y × H) = λH,

and similarly (µy )∗ (Z \ H)ν(dy) ≥ λ(Z \ H). functions Rg1 , g2 such RTaking ν-integrable R R that g1 (y) ≤ (µy )∗ H and g2 (y) ≤ (µy )∗ (Z \ H) for almost every y, g1 dν = (µy )∗ H ν(dy) and g2 dν = (µy )∗ (Z \ H)ν(dy) (133J(a-ii)), we must have for almost every y, while

R

g1 (y) + g2 (y) ≤ (µy )∗ H + (µy )∗ (Z \ H) ≤ µy Z ≤ 1 g1 + g2 dν ≥ 1; so that, for almost all y, g1 (y) + g2 (y) = (µy )∗ H + (µy )∗ (Z \ H) = µy Z = 1,

and (because µy is complete) µy H is defined and equal to g1 (y) (413Ef). It now follows that

R

F

µy H ν(dy) =

for every F ∈ T. But since also

R

F

R F

R F

R

g1 (y)ν(dy) =

R

(µy )∗ H · χF (y)ν(dy) ≥ µ(F × H)

µy (Z \ H)ν(dy) ≥ µ(F × (Z \ H)),

µy H + µy (Z \ H)ν(dy) ≤ νF = µ(F × H) + µ(F × (Z \ H)),

we must actually have F µy H ν(dy) = µ(F × H). All this is true for every F ∈ T, H ∈ Υ. But now, setting

b Υ, µE = E = {E : E ∈ T⊗

R

µy E[{y}]ν(dy)},

we see that E is a Dynkin class and includes I = {F × H : F ∈ T, H ∈ Υ}, which is closed under finite intersections; so that the Monotone Class Theorem tells us that E includes the σ-algebra generated by I, b and is the whole of T⊗Υ. (b) The rest is just tidying up. First I should note that the construction in (a) allows µy Z to be less than 1 for a ν-negligible set of y; but of course all we have to do, if that happens, is to amend µy arbitrarily on that set to any of the ‘ordinary’ values of µy . Next, if the original measure ν is not complete, just apply the argument of (a) to its completion. If the original measure µ is not a probability measure, apply the arguments so far to suitable scalar multiples of µ and ν. (c) Thus we have the formula µE =

R

µy E[{y}]ν(dy)

436


452M

b for every E ∈ T⊗Υ. The second formula announced follows as in the remark following 452F. 452N Corollary Let Y and Z be sets and T ⊆ PY , Υ ⊆ PZ σ-algebras. Let µ be a probability measure b and ν the marginal measure of µ on Y . Suppose that with domain T⊗Υ, either Υ is the Baire σ-algebra with respect to a compact Hausdorff topology on Z or Υ is the Borel σ-algebra with respect to an analytic Hausdorff topology on Z or (Z, Υ) is a standard Borel space. Then there is a family hµy iy∈Y of probability measures on Z, all with domain Υ, such that µE = b and for every E ∈ T⊗Υ,

R

f dµ =

R

µy E[{y}]ν(dy)

RR

f (y, z)µy (dz)ν(dy) R whenever f is a [−∞, ∞]-valued function such that f dµ is defined in [−∞, ∞]. proof In each case, the marginal measure of µ on Z is tight, (that is, inner regular with respect to the compact sets) for a Hausdorff topology on Z. (Use 412D when Υ is the Baire σ-algebra on a compact Hausdorff space and 433Ca when it is the Borel σ-algebra on an analytic Hausdorff space; when (Z, Υ) is a standard Borel space, take any appropriate Polish topology on Z and use 423Ba.) So 452M tells us that we can achieve the formulae sought with Radon probability measures µy . Since (in all three cases) dom µy will include Υ for every y, we can get the result as stated by replacing each µy by µy ¹Υ. 452O Proposition Let (X, T, Σ, µ) be a Radon measure space, (Y, T, ν) a strictly localizable measure space, and f : X → Y an inverse-measure-preserving function. Then there is a disintegration hµy iy∈Y of µ over ν, consistent with f , such that every µy is a Radon measure on X. proof (a) Let hYi ii∈I be a decomposition of Y . For each i ∈ I, let νi be the subspace measure on Yi and λi the subspace measure on Xi = f −1 [Yi ]. Then fi = f ¹Xi is inverse-measure-preserving for λi and νi . Let Ki be the family of compact subsets of Xi ; of course Ki is a (countably) compact class and λi is inner regular with respect to Ki (412Oa). By 452I, we can choose, for each i ∈ I, a disintegration h˜ µy iy∈Yi of λi over νi , consistent with f ¹Xi , such that (α) every µ ˜y is complete and totally finite (β) µ ˜y measures every compact subset of Xi and is inner regular with respect to Ki for every y ∈ Yi . By 412Ja, µ ˜y measures every relatively closed subset of Xi for every y ∈ Yi . Now, for i ∈ I and y ∈ Yi , set µy E = µ ˜y (E ∩ Xi ) whenever E ⊆ X and E ∩ Xi is measured by µ ˜y . Then µy is a complete totally finite measure on X; it is inner regular with respect to Ki and measures every closed subset of X. It follows at once that it is tight and measures every Borel set, that is, is a Radon measure on X. R S S (b) Now µE = µy E ν(dy) for every E ∈ Σ. P P i∈J E ∩R Xi = E ∩ f −1 [ i∈J Yi ] belongs to Σ for every P ˜y (E ∩ Xi )νi (dy) = µ(E ∩ Xi ). So J ⊆ I. By 451P, µE = i∈I µ(E ∩ Xi ). For i ∈ I, we have Yi µ X XZ µ ˜y (E ∩ Xi )νi (dy) µE = µ(E ∩ Xi ) = i∈I

=

XZ i∈I

i∈I

µy E ν(dy) =

Z

Yi

µy E ν(dy)

Yi

by 214M. Q Q Thus hµy iy∈Y is a disintegration of µ over ν. (c) Finally, if F ∈ T and i ∈ I, then Yi ∩ F \{y : µy f −1 [F ] is defined and equal to 1} = (F ∩ Yi ) \ {y : y ∈ Yi , µ ˜y f −1 [F ∩ Yi ] = 1} is negligible for every i, so µy f −1 [F ] = 1 for almost every y. Thus hµy iy∈Y is consistent with f .

*452R


437

452P Corollary (cf. Blackwell 56) Let (X, T, Σ, µ) be a Radon measure space, (Y, S, T, ν) an analytic Radon measure space and f : X → Y an inverse-measure-preserving function. Then there is a disintegration hµy iy∈Y of µ over ν, strongly consistent with f , such that every µy is a Radon measure on X. proof By 433B, ν is countably separated; now put 452O and 452Gc together. 452Q Disintegrations and conditional expectations Fubini’s theorem provides a relatively concrete description of the conditional expectation of a function on a product of probability R spaces with respect to the σ-algebra defined by one of the factors, by means of the formula g(x, y) = f (x, z)dz (253H). This generalizes straightforwardly to measures with disintegrations, as follows. Proposition Let (X, Σ, µ) and (Y, T, ν) be probability spaces and f : X → Y an inverse-measure-preserving function. Suppose that hµy iy∈Y is a disintegration of µ over ν which is consistent with f , and that g is a µ-integrable real-valuedRfunction. (a) Setting h0 (y) = g dµy whenever y ∈ Y and the integral is defined in R, h0 is a Radon-Nikod´ ym R derivative of the functional F 7→ f −1 [F ] g dµ : T → R. R (b) Now suppose that ν is complete. Setting h1 (x) = g dµf (x) whenever x ∈ X and the integral is defined in R, then h is a conditional expectation of g on the σ-algebra Σ0 = {f −1 [F ] : F ∈ T}. proof (a) If F ∈ T,R then f −1 [F ] is µy -conegligible for almost every y ∈ F , and µy -negligible for almost every y ∈ Y \ F , so g × χf −1 [F ]dµy = h0 (y) × χF (y) for almost every y, and

R

F

h0 dν =

RR

g × χf −1 [F ]dµy ν(dy) =

R

f −1 [F ]

g dµ

(452F). As F is arbitrary, we have the result. (b) Of course Σ0 is a σ-algebra (111Xd), and it is included in Σ because f is inverse-measure-preserving. By 452F, Y0 = {y : g is µy -integrable} is conegligible, so dom h1 = f −1 [Y0 ] is conegligible. If α ∈ R, then because y 7→

R

F = {y : y ∈ Y0 ,

R

g dµy ≥ α} ∈ T

g dµy is ν-virtually measurable and ν is complete. So {x : x ∈ dom h1 , h1 (x) ≥ α} = f −1 [F ]

belongs to Σ0 , and h1 is Σ0 -measurable. If F ∈ T, then Z f −1 [F ]

Z h1 dµ =

(235I)

Z f −1 [F ]

Z Z g dµf (x) µ(dx) =

g dµy ν(dy) F

Z (g × χ(f −1 [F ]))dµy ν(dy)

=

(because µy f −1 [F ] = µy X = 1 for almost every y ∈ F , µy f −1 [F ] = 0 for almost every y ∈ Y \ F ) Z = g × χ(f −1 [F ])dµ (452F)

Z =

g dµ. f −1 [F ]

As F is arbitrary, h1 is a conditional expectation of g on Σ0 , as claimed. *452R I take the opportunity to interpolate an interesting result about countably compact measures. It demonstrates the power of 452I to work in unexpected ways. Theorem (Pachl 79) Let (X, Σ, µ) be a countably compact measure space, (Y, T, ν) a strictly localizable measure space, and f : X → Y an inverse-measure-preserving function. Then ν is countably compact. proof (a) For most of the proof (down to the end of (b) below) I suppose that µ and ν are totally finite.

438


*452R

Let Z be the Stone space of the Boolean algebra T. (I am not using the measure algebra here!) For F ∈ T, let F ∗ be the corresponding open-and-closed subset of Z. For each y ∈ Y , the map F 7→ χF (y) is a Boolean homomorphism from T to {0, 1}, so belongs to Z; define g : Y → Z by saying that g(y)(F ) = χF (y) for y ∈ Y , F ∈ T, that is, that g −1 [F ∗ ] = F for every F ∈ T. Let Z be the family of zero sets in Z, and Λ the Baire σ-algebra of Z. The set {W : W ⊆ Z, g −1 [W ] ∈ T} is a σ-algebra of subsets of Z containing all the open-and-closed sets, so contains every zero set (4A3Oe) and includes Λ. Set λW = νg −1 [W ] for W ∈ Λ. Then λ is a Baire measure on Z, so is inner regular with respect to Z (412D). Set h = gf : X → Z. Then h is a composition of inverse-measure-preserving functions, so is inversemeasure-preserving. By 452I, there is a disintegration hµz iz∈Z of µ over λ which is consistent with h. (b) Let K be the family of sets {g −1 [V ] : V ∈ Z, µz h−1 [V ] = µz X = 1 for every z ∈ V }. T (i) K is a countably compact class of sets. P P Let hKn in∈N be a sequence in K such that i≤n Ki 6= ∅ for each n ∈ N. For each n ∈ N, let Vn ∈ Z be such that Kn = g −1 [Vn ] and µz h−1 Vn = µz X = 1 for every z ∈ Vn . Then T T g −1 [ i≤n Vi ] = i≤n Ki 6= ∅ for every n ∈ N, so {Vn : n ∈ N} has the finite intersection property and (because Z is compact) there is a T z ∈ n∈N Vn . Now µz h−1 [Vn ] = µz X = 1 for every n ∈ N, so Thus

∅ 6=

T n∈N

T n∈N

h−1 [Vn ] = f −1 [

T n∈N

Kn ].

Kn is non-empty. As hKn in∈N is arbitrary, K is a countably compact class. Q Q

(ii) ν is inner regular with respect to K. P P Suppose that F ∈ T and γ < νF . Choose a sequence hVn in∈N in Z as follows. Start with V0 = F ∗ , so that λV0 = νg −1 [V0 ] = νF > γ. Given that Vn ∈ Z and λVn > γ, then we know that µz h−1 [Vn ] = µz X = 1 for λ-almost every z ∈ Vn ; because λ is inner regular with respect to Z, there is a Vn+1 ∈ Z such that Vn+1 ⊆ Vn , λVn+1 > γ and µz h−1 [Vn ] = µz X = 1 for every z ∈ Vn+1 . Continue. T At the end of the induction, set V = n∈N Vn . Then V ∈ Z. If z ∈ V , then µz h−1 [V ] = limn→∞ µz h−1 [Vn ] = 1 = µz X, so g −1 [V ] ∈ K. Because V ⊆ V0 = F ∗ , g −1 [V ] ⊆ F , and νg −1 [V ] = λV = limn→∞ λVn ≥ γ. As F and γ are arbitrary, ν is inner regular with respect to K. Q Q Thus K witnesses that ν is countably compact. (c) For the general case, let hYi ii∈I be a decomposition of Y . For each i ∈ I, set Xi = f −1 [Yi ]; let µi be the subspace measure on Xi and νi the subspace measure on Yi . Then µi is countably compact (451Db) and f ¹Xi : Xi → Yi is inverse-measure-preserving for µi and νi , so νi is countably compact, by (a)-(b) above. S Let Ki ⊆ PYi be a countably compact class such that νi is inner regular with respect to Ki . Then K = i∈I Ki is a countably compact class (because any sequence in K with the finite intersection property must lie within a single Ki ). By 413R, there is a countably compact class K∗ ⊇ K which is closed under finite unions; by 412Aa, ν is inner regular with respect to K∗ , so is countably compact. This completes the proof.

452Xl


439

*452S Corollary (Pachl 78) If (X, Σ, µ) is a countably compact totally finite measure space, and T is any σ-subalgebra of Σ, then µ¹ T is countably compact. 452X Basic exercises (a) Let Y be a first-countable topological space, ν a topological probability measure on Y , Z a topological space, and hµy iy∈Y a family of topological probability measures on Z such that y 7→ µy V is lower semi-continuous R for every open set V ⊆ Z. Show that there is a Borel probability measure µ on Y × Z such that µE = µy E[{y}]ν(dy) for every Borel set E ⊆ Y × Z. (Hint: 434R.) (b) Let (Y, T, ν) be a probability space, Z a topological space and P the set of topological probability measures on Z with its narrow topology (437Jd). Let y 7→ µy : Y → P be a function which is measurable in the sense of 411L. Show that, writing R B(Z) for the Borel σ-algebra of Z, we have a probability measure b b µ defined on T⊗B(Z) such that µE = µy E[{y}]ν(dy) for every E ∈ T⊗B(Z). (c) Let (Y, T, ν) be a probability space, Z a topological space and PBa the set of Baire probability measures on Z with its vague topology (437Jc). Let y 7→ µy : Y → PBa be a measurable function. Show b that, writing Ba(Z) for the Baire σ-algebra of Z, we have a probability measure µ defined on T⊗Ba(Z) R b such that µE = µy E[{y}]ν(dy) for every E ∈ T⊗Ba(Z). (d) Let (Y, S, T, ν) be a Radon probability space, (X, T) a topological space, and hµy iy∈Y a family of Radon probability measures on X. Suppose that (i) there is a base U for T, closed under finite unions, such that y 7→ µy U is lower semi-continuous for every U ∈ U (ii) ν is inner regular with respect to the family {K : K ⊆ Y , {µyR : y ∈ K} is uniformly tight}. Show that we have a Radon probability measure µ ˜ on Y × Z such that µ ˜E = µy E[{y}]ν(dy) whenever µ ˜ measures E. (e) Let (Y, S, T, ν) be a Radon probability space, (Z, U) a Prokhorov Hausdorff space (437V), and P the space of Radon probability measures on Z with its narrow topology. Suppose that y 7→ µy : Y R → P is almost continuous. Show that we have a Radon probability measure µ ˜ on Y × Z such that µ ˜E = µy E[{y}]ν(dy) whenever µ ˜ measures E. (f ) Let (X, T, ν) be a measure space, and µ an indefinite-integral measure over ν. Show that there is a disintegration hµx ix∈X of µ over ν such that µx {x} = µx X for every x ∈ X. (g) Let (X, Σ, µ) and (Y, T, ν) be measure spaces and hµy iy∈Y a disintegration of µ over ν. Show that hˆ µy iy∈Y is a disintegration of µ ˆ over ν, where µ ˆy and µ ˆ are the completions of µy and µ respectively. > (h) Let (X, Σ, µ) and (Y, T, ν) be measure spaces, and ν 0 an indefinite-integral measure over ν, defined from a ν-virtually measurable function g : Y → [0, ∞[ (234B). Suppose that hµy iy∈Y is a disintegration of µ over ν 0 . Show that hg(y)µy iy∈Y is a disintegration of µ over ν. (i) Let (Y, T, ν) be a probability space, X a set and hµy iy∈Y a family of probability measures on X. Set R θA = µ∗y (A) ν(dy) for every A ⊆ X. (i) Show that θ is an outer measure on X. (ii) Let µ be the measure on X defined from θ by Carathéodory’s construction. Show that hµy iy∈Y is a disintegration of µ over ν. (iii) Suppose that X = [0, 1]2 , ν is Lebesgue measure on [0, 1] = Y and µy E = ν{x : (x, y) ∈ E} whenever this is defined. Show that, for any E measured by µ, µy E ∈ {0, 1} for ν-almost every y. R RR (j) Explore connexions between 452F and the formula f dµ = f dνz λ(dz) of 443Pe. (k) Let (X, Σ, µ) be a countably compact σ-finite measure space, (Y, T, ν) a σ-finite measure space, and f : X → Y a (Σ, T)-measurable function such that f −1 [F ] is µ-negligible whenever F ⊆ Y is ν-negligible. Show that there is a disintegration hµy iy∈Y of µ over ν such that, for each F ∈ T, µy (X \ f −1 [F ]) = 0 for almost every y ∈ F . (Hint: Reduce to the case in which µ is totally finite, and disintegrate µ over ν 0 = (µf −1 )¹ T.) > (l) Let (X, Σ, µ) be a non-empty countably compact measure space such that Σ is countably generated (as σ-algebra), (Y, T, ν) a σ-finite measure space, and f : X → Y an inverse-measure-preserving function. (i) Show that there is a disintegration hµy iy∈Y of µ over ν, consistent with f , such that every µy is a probability measure with domain Σ. (ii) Show that if hµ0y iy∈Y is any other disintegration of µ over ν which is consistent with f , then µy = µ0y ¹Σ for almost every y.

440


452Xm

(m) Let (X, Σ) be a non-empty standard Borel space, µ a measure with domain Σ, (Y, T, ν) a σ-finite measure space, and f : X → Y an inverse-measure-preserving function. (i) Show that there is a disintegration hµy iy∈Y of µ over ν, consistent with f , such that every µy is a probability measure with domain Σ. (ii) Show that if hµ0y iy∈Y is any other disintegration of µ over ν which is consistent with f , then µy = µ0y ¹Σ for almost every y. (n) Let (X, Σ, µ) be a totally finite countably compact measure space and T ⊆ Σ a countably-generated σ-algebra; set ν = µ¹ T. Show T that there is a disintegration hµx ix∈X of µ over ν such that µx Hx = µx X = 1 for every x ∈ X, where Hx = {F : x ∈ F ∈ T} for every x. (Hint: apply 452I with Y = {Hx : x ∈ X}.) (o) Show that 452I can be deduced from 452M. (Hint: start with the case νY = 1; set λW = µ{x : b (x, f (x)) ∈ W } for W ∈ Σ⊗T.) (p) Show that, in 452M, we shall have µ Ê =

R

µy E −1 [{y}]ν(dy) whenever the completion µ ˆ of µ measures

E. > (q) Let T be the Borel σ-algebra of [0, 1], ν the restriction of Lebesgue measure to T, Z ⊆ [0, 1] a set with inner measure 0 and outer measure 1, and Υ the Borel σ-algebra of Z. Show that there is a probability b measure µ on [0, 1] × Z defined by setting µE = ν ∗ {y : (y, y) ∈ E} for E ∈ T⊗Υ. Show that there is no disintegration of µ over ν which is consistent with the projection (y, z) 7→ y. > (r) Let (X, Σ, µ) be a complete totally finite countably compact measure space and T a σ-subalgebra of Σ containing all negligible sets. Show Rthat there is a family hµx ix∈X of probability measures on X such that (i) x 7→ µx E is T-measurable and µx E µ(dx) = µE for every E ∈ Σ (ii) if F ∈ T, then µx F = 1 for almost every x ∈ F . Show that if g is any µ-integrable real-valued function, then g is µx -integrable for R almost every x, and x 7→ g dµx is a conditional expectation of g on T. 452Y Further exercises (a) Let Z be a set, (Y, T, ν) a measure space, and hµy iy∈Y a family of measures on Z. Let Υ be a σ-algebra of subsets of Z such that, for every H ∈ Υ, y 7→ µy H : Y → [0, ∞] is defined ν-a.e. and is ν-virtually measurable. For F ∈ T, set HF = {H : H ∈ Υ, µy H is defined for every y ∈ F and b defined by setting supy∈F µy H < ∞}. Show that there is a measure µ on Y × Z, with domain T⊗Υ, n Z X

µE = sup{

i=0

µ(E[{y}] ∩ Hi )ν(dy) : F0 , . . . , Fn ∈ T are disjoint,

Fi

νFi < ∞ and Hi ∈ HFi for every i ≤ n} b for E ∈ T⊗Υ. (b) Let (X, Σ, µ) be a semi-finite countably compact measure space, (Y, T, ν) a strictly localizable measure space, and f : X → Y an inverse-measure-preserving function. Suppose that the magnitude of ν (definition: 332Ga) is finite or a measure-free cardinal (definition: 438A). Show that there is a disintegration hµy iy∈Y of µ over ν which is consistent with f . (c) Give an example to show that the phrase ‘strictly localizable’ in the statements of 452O and 452Yb cannot be dispensed with. (d) Give an example to show that, in 452M, we cannot always arrange that Υ ⊆ dom µy for ν-almost every y ∈ Y . (e) Let (X, Σ, µ) be a probability space such that whenever (Y, T, ν) is a probability space and f : X → Y is an inverse-measure-preserving function, there is a disintegration hµy iy∈Y of µ over ν which is consistent with f . Show that µ is countably compact. (Hint: 452R, or Pachl 78.)

453A

Strong liftings

441

452 Notes and comments 452B and 452C correspond respectively to the ordinary and τ -additive product measures of §§251 and 417. I have not attempted to find a suitable general formulation for the constructions when the measures involved are not probability measures. In 452Ya I set out a possible version which at least agrees with the c.l.d. product measure when all the µy are the same. Any product measure which has an associated Fubini theorem can be expected to be generalizable in the same way; for instance, 434R becomes 452Xa. The hypotheses in 452B are closely matched with the conclusion, and clearly cannot be relaxed substantially if the theorem is to remain true. 452C and 452D are a rather different matter. While the condition ‘y 7→ µy V is lower semi-continuous’ is a natural one, and plainly necessary for the argument given, the integrated measure µ can be τ -additive or Radon for other reasons. In particular, the most interesting specific example in this book of a Radon measure constructed through these formulae (453N below) does not satisfy the lower semi-continuity condition for the section measures. Early theorems on disintegrations concentrated on cases in which all the measure spaces involved were ‘standard’ in that the measures were defined on standard Borel algebras, or were the completions of such measures. Theorem 452I here is the end (so far) of a long search for ways to escape from topological considerations. As usual, of course, the most important applications (in probability theory) are still rooted in the standard case. Being countably separated, such spaces automatically yield disintegrations which are concentrated on fibers, in the sense that µy f −1 [{y}] = µy X = 1 for almost every y (452P). The general question of when we can expect to find disintegrations of this type is an important one to which I will return in the next section. 452I and 452O, as stated, assume that the functions f : X → Y controlling the disintegrations are inverse-measure-preserving. In fact it is easy to weaken this assumption (452Xk). Note the constructions for conditional expectations in 452Q and 452Xr. Obviously 452I and 452M are nearly the same theorem; but I write out formally independent proofs because the constructions needed to move between them are not quite trivial. In fact I think it is easier to deduce 452I from 452M than the other way about (452Xo). The point of 452N is that the spaces (X, Σ) there have the ‘countably compact measure property’, that is, any totally finite measure with domain Σ is countably compact. I will return to this in the exercises to §454 (454Xe et seq.). The method of 452R, due to J.Pachl, may have inspired the proof of (vi)⇒(i) in 343B. In the general introduction to this work I wrote ‘I have very little confidence in anything I have ever read concerning the history of ideas’. We have here a case indicating the difficulties a historian faces. I proved 343B in the winter of 1996-97, while a guest of the University of Wisconsin at Madison. Around that time I was renewing my acquaintance with Pachl 78. I know I ran my eye over the proof of 452R, without, I may say, understanding it, as became plain when I came to write the first draft of the present section in the summer of 1997; whether I had understood it twenty years earlier I do not know. It is entirely possible that a subterranean percolation of Pachl’s idea was what dislodged an obstacle to my attempts to prove 343B, but I was not at the time conscious of any connexion.

453 Strong liftings The next step involves the concept of ‘strong’ lifting on a topological measure space (453A); I devote a few pages to describing the principal cases in which strong liftings are known to exist (453B-453J). When we have Radon measures µ and ν, with an almost continuous inverse-measure-preserving function between them, and a strong lifting of ν, we can hope for a disintegration hµy iy∈Y such that (almost) every µy lives on the appropriate fiber. This is the content of 453K. I end the section with a note on the relation between strong liftings and Stone spaces (453M) and with V.Losert’s example of a space with no strong lifting (453N). Much of the work here is based on ideas in Ionescu Tulcea & Ionescu Tulcea 69. 453A The proof of the first disintegration theorem I presented, 452I, depended on two essential steps: the use of a lifting of (Y, T, ν) to define the finitely additive functionals λy , and the use of a countably compact class to convert these into countably additive functionals. In 452O I observed that if our countably compact class is the family of compact sets in a Hausdorff space, we can get Radon measures in our disintegration. Similarly, if we have a lifting of a special type, we can hope for special properties of the disintegration. A particularly important kind of lifting, in this context, is the following.

442


453A

Definition Let (X, T, Σ, µ) be a topological measure space. A lifting φ : Σ → Σ is strong or of local type if φG ⊇ G for every S open set G ⊆ X, that is, if φF ⊆ F for every closed set F ⊆ X. I will say that φ is almost strong if G∈T G \ φG is negligible. Similarly, if A is the measureSalgebra of µ, a lifting θ : A → Σ is strong if θG• ⊇ G for every open set G ⊆ X, and almost strong if G∈T G \ θG• is negligible. Obviously a strong lifting is almost strong. 453B We already have the machinery to describe a particularly striking class of strong liftings. Theorem Let X be a topological group with a Haar measure µ, and Σ its algebra of Haar measurable sets. (a) If φ : Σ → Σ is a left-translation-invariant lifting, in the sense of 447A, then φ is strong. (b) µ has a strong lifting. proof (a) Apply 447B with Y = {e} and φ = φ. (b) For there is a left-translation-invariant lifting (447J). Remark In particular, translation-invariant liftings on Rr or {0, 1}I (§345) are strong. 453C Proposition Let (X, T, Σ, µ) be a topological measure space and φ : Σ → Σ a lifting. Write L∞ for the space of bounded Σ-measurable real-valued functions on X, so that L∞ can be identified with L∞ (Σ) (363H) and the Boolean homomorphism φ : Σ → Σ gives rise to a Riesz homomorphism T : L∞ → L∞ (363F). (a) If φ is a strong lifting, then T f = f for every bounded continuous function f : X → R. (b) If (X, T) is completely regular and T f = f for every f ∈ Cb (X), then φ is strong. proof (a) Suppose first that f ≥ 0. For α ∈ R, set Gα = {x : x ∈ X, f (x) > α}; then Gα is open, so φGα ⊇ Gα . We have f ≥ αχGα , so T f ≥ αT (χGα ) = αT (χ(φGα )) ≥ αT (χGα ), that is, (T f )(x) ≥ α whenever f (x) > α. As α is arbitrary, T f ≥ f . At the same time, setting γ = kf k∞ , we have T (γχX − f ) ≥ γχX − f ,

T (γχX) = γχ(φX) = γχX,

so T f ≤ f and T f = f . For general f ∈ Cb (X), T f = T (f + − f − ) = T f + − T f − = f + − f − = f , where f + and f − are the positive and negative parts of f . (b) Let G ⊆ X be open and x any point of G. Then there is an f ∈ Cb (X) such that f ≤ χG and f (x) = 1. In this case f = T f ≤ T (χG) = χ(φG), so x ∈ φG. As x is arbitrary, G ⊆ φG; as G is arbitrary, φ is strong. 453D Proposition Let (X, T, Σ, µ) be a topological measure space. (a) If µ has a strong lifting it is strictly positive (definition: 411Nf). (b) If µ is strictly positive and complete, and has an almost strong lifting, it has a strong lifting. (c) If µ has an almost strong lifting it is τ -additive, so has a support. (d) If µ is complete and µX > 0 and the subspace measure µE has an almost strong lifting for some conegligible set E ⊆ X, then µ has an almost strong lifting. proof (a) If φ : Σ → Σ is a strong lifting, then G ⊆ φG = ∅ whenever G is a negligible open set, so µ is strictly positive. S (b) If µ is strictly positive and complete and φ : Σ → Σ is an almost strong lifting, set A = G∈T G \ φG. For each x ∈ A, let Ix be the ideal of subsets of X generated by

453E

Strong liftings

443

{F : F ⊆ X is closed, x ∈ / F } ∪ {B : B ⊆ X is negligible}. Then X ∈ / Ix , because µ is strictly positive, so a closed set not containing x cannot be conegligible. There is therefore a Boolean homomorphism ψx : PX → {0, 1} such that ψx F = 0 for every F ∈ Ix (311D). Set ˜ = (φE \ A) ∪ {x : x ∈ A, ψx E = 1} φE for E ∈ Σ. It is easy to check that φ˜ : Σ → PX is a Boolean homomorphism. (Compare the proof of 341J.) If E ∈ Σ, then ˜ ⊆ (E4φE) ∪ A E4φE ˜ ∈ Σ. If E is negligible, then E ∈ Ix and ψx E = 0 for every is negligible, so (because µ is complete) φE ˜ = φE = ∅. Thus φ˜ is a lifting. Now suppose that x ∈ G ∈ T. If x ∈ A, then X \ G ∈ Ix , so x ∈ A, so φE ˜ If x ∈ ˜ As x and G are arbitrary, ψx (X \ G) = 0, ψx G = 1 and x ∈ φG. / A, then x ∈ φG and again x ∈ φG. ˜ φ is a strong lifting. (c) Suppose that φ : Σ → Σ is an almost strong lifting. Let G be a non-empty upwards-directed family of open sets with union H. If supG∈G µG = ∞, this is surely equal to µH. Otherwise, S there is a non-decreasing sequence hGn in∈N in G such that G \ H0 is negligible for every G ∈ G, where H0 = n∈N Gn (215Ab). Then φG ⊆ φH0 for every G ∈ G. This means that S H \ φH0 ⊆ G∈G G \ φG is negligible, because φ is almost strong, and µH ≤ µ(φH0 ) = µH0 = limn→∞ µGn = supG∈G µG. As G is arbitrary, µ is τ -additive. By 411Nd, it has a support. (d) Now suppose that µ is complete, that µX > 0 and that there is a conegligible E ⊆ X such that µE has an almost strong lifting φ. Let ψ : PX → {∅, X} be any Boolean homomorphism such that ψA = ∅ whenever A is negligible. (This is where I use the hypothesis that X is not negligible.) Define φ˜ : Σ → PX by setting ˜ = φ(E ∩ F ) ∪ (ψF \ E). φF Then φ˜ is a Boolean homomorphism because φ and ψ are; ˜ ⊆ ((E ∩ F )4φ(E ∩ F )) ∪ (X \ E) F 4φF ˜ ∈ Σ, for every F ∈ Σ, because µ is complete; and if F is negligible, then φ(E ∩ F ) = is negligible, so φF ˜ ψF = ∅ so φF = ∅. Thus φ˜ is a lifting. Finally, S S ˜ G∈T G \ φG ⊆ (X \ E) ∪ G∈T ((G ∩ E) \ φ(G ∩ E)) is negligible because φ is almost strong and E is conegligible. 453E Proposition Let (X, T, Σ, µ) be a complete strictly localizable topological measure space with an almost strong lifting, and A ⊆ X a non-negligible set. Then the subspace measure µA has an almost strong lifting. proof Let φ : Σ → Σ be an almost strong lifting. Because µ is strictly localizable, A has a measurable envelope W say (213L). Write ΣA for the subspace σ-algebra on A. Let ψ : ΣA → {∅, A} be any Boolean homomorphism such that ψH = ∅ for every negligible set H ⊆ A. If E, F ∈ Σ and E ∩ A = F ∩ A, then φE ∩ φW = φF ∩ φW . P P µ((E4F ) ∩ W ) = µ∗ ((E4F ) ∩ A) = 0, so (φE ∩ φW )4(φF ∩ φW ) = φ((E4F ) ∩ W ) = ∅. Q Q We can therefore define a function φ˜ : ΣA → PA by setting ˜ = (φE ∩ φW ∩ A) ∪ (ψH \ φW ) φH

444


453E

whenever E ∈ Σ and H = E ∩ A. It is easy to check that φ˜ is a Boolean homomorphism. If E ∈ Σ then ˜ ∩ A) ⊆ (E4φE) ∪ (A \ φW ) ⊆ (E4φE) ∪ (W \ φW ) (E ∩ A)4φ(E ˜ ∩ A) ∈ ΣA (because µ and µA are complete). If H ∈ ΣA is negligible, then is negligible, so φ(E ˜ ⊆ φH ∪ ψH = ∅, φH so φ˜ is a lifting. S Now set B = (A \ φW ) ∪ G∈T G \ φG. Because φ is almost strong, B is negligible. If H ⊆ A is relatively ˜ ⊆ B. P ˜ open, then H \ φH P Take x ∈ H \ φH. Express H as G ∩ A where G ⊆ X is open. If x ∈ φW , then x∈ / φG so x ∈ B; if x ∈ / φW , then of course x ∈ B. Q Q Thus S ˜ : H ⊆ A is relatively open} ⊆ B {H \ φH is negligible and φ˜ is almost strong. 453F Proposition Let (X, T, Σ, µ) be a complete strictly localizable topological measure space. (a) If T has a countable network consisting of measurable sets, any lifting for µ is almost strong. (b) Suppose that µX > 0 and µ is inner regular with respect to K = {K : K ∈ Σ, µK has an almost strong lifting}, where µK is the subspace measure on K. Then µ has an almost strong lifting. proof (a) Let E ⊆ Σ be a countable network for T, and φ : Σ → Σ a lifting. Then [

G \ φG =

G∈T

[

E \ φG

G∈T,E∈E,E⊆G

⊆

[

G∈T,E∈E,E⊆G

E \ φE =

[

E \ φE

E∈E

is negligible, so φ is almost strong. P (b) Let L ⊆ K be a disjoint family such that µ∗ A = L∈L µ∗ (A ∩ L) for every A ⊆ X (412Ib). For each L ∈ L, S let ΣL be the corresponding subspace σ-algebra and φL : ΣL → ΣL an almost strong lifting. Set E = L; then P µ∗ (X \ E) = L∈L µ(L \ E) = 0, S so E is conegligible. For F ∈ ΣE set φF = L∈L φL (F ∩ L); then φ is a Boolean homomorphism from ΣE to PE. If F ∈ ΣE , then P P µ∗ (F 4φF ) = L∈L µ∗ (L ∩ (F 4φF )) = L∈L µ∗ ((F ∩ L)4φL (F ∩ L)) = 0, while if µF = 0 then φL (F ∩ L) = ∅ for every L, so φF = ∅. Thus φ is a lifting. Now set S A = {H \ φH : H ⊆ E is relatively open}. If L ∈ L, then A∩L=

S

{(H ∩ L) \ φL (H ∩ L) : H ⊆ E is relatively open}

is negligible, because φL is almost strong; thus φ is an almost strong lifting for µE . By 453Dd, µ also has an almost strong lifting. 453G Corollary (a) A non-zero quasi-Radon measure on a separable metrizable space has an almost strong lifting. (b) A non-zero Radon measure µ on an analytic Hausdorff space X has an almost strong lifting. proof (a) A quasi-Radon measure is complete and strictly localizable (415A), so, if non-zero, has a lifting (341K). A separable metrizable space has a countable network consisting of open sets (4A2P(a-i)), so this lifting must be almost strong.

453I

Strong liftings

445

(b) If K ⊆ X is compact and non-negligible, it is metrizable (423Dc), so that the subspace measure µK has an almost strong lifting, by (a); as µ is tight (that is, inner regular with respect to the closed compact sets), it has an almost strong lifting, by 453Fb. Remark In particular, Lebesgue measure on Rr has an almost strong lifting and therefore, by 453Db, a strong lifting, as already noted in 453B. 453H Lemma Let (X, Σ, µ) be a complete strictly localizable measure space and T a topology on X generated by a family U ⊆ Σ. Suppose that φ : Σ → Σ is a lifting such that φU ⊇ U for every U ∈ U . Then µ is a τ -additive topological measure, and φ is a strong lifting. proof Of course φ is a lower density, and φX = X, so by 414P we have a density topology Td = {E : E ∈ Σ, E ⊆ φE} with respect to which µ is a τ -additive topological measure. But our hypothesis is that U ⊆ Td , so T ⊆ Td and µ is a τ -additive topological measure with respect to T. Also, of course, φG ⊇ G for every G ∈ T, so φ is a strong lifting. 453I Proposition Let h(Xi , Ti , Σi , µi )ii∈I be a family of topological probability spaces such that every Ti has a countable network consisting of measurable sets and every µi is strictly positive. Let λ be the Q (ordinary) complete product measure on X = i∈I Xi . Then λ is a τ -additive topological measure and has a strong lifting. proof (a) The strategy of the proof is as follows. We may suppose that I = κ is a cardinal. Write Λ for the domain of λ, and for each ξ ≤ κ let Λξ be the σ-algebra of members of Λ determined by coordinates in ξ; write πξ : X → Xξ for the canonical map. I seek to define a lifting φ : Λ → Λ such that φW ⊇ W for every open set W ∈ Λ. This will be the last in a family hφξ iξ≤κ of partial liftings, constructed inductively as in the proof of 341H, with dom φξ = Λξ for each ξ. The inductive hypothesis will be that φξ extends φη whenever η ≤ ξ, and φξ πη−1 [G] ⊇ πη−1 [G] for every η < ξ and every open G ⊆ Xη . The induction starts with Λ0 = {∅, X}, φ0 ∅ = ∅, φ0 X = X. For ξ ≤ κ, set Bξ = {W • : W ∈ Λξ }. (b) Inductive step to a successor ordinal ξ + 1 Suppose that φξ has been defined, where ξ < κ. (i) By 341Nb, there is a lifting φ0ξ : Λ → Λ extending φξ . Let Eξ ⊆ Σξ be a countable network for Tξ , and πξ : X → Xξ the canonical map. Set S Q = {πξ−1 [E] \ φ0ξ (πξ−1 [E]) : E ∈ Eξ }; then Q is negligible. (ii) For x ∈ Q, let Ix ⊆ Λ be the ideal generated by {W : W ∈ Λξ , x ∈ / φξ W } ∪ {πξ−1 [F ] : F ⊆ Xξ is closed, πξ (x) ∈ / F } ∪ {W : λW = 0}. Then X ∈ / Ix . P P?? Otherwise, there are a W ∈ Λξ , a closed F ⊆ Xξ and a negligible W 0 such that −1 0 W ∪ W ∪ πξ [F ] = X while x ∈ / φξ W ∪ πξ−1 [F ]. But in this case 0 = λW 0 ≥ λ((X \ W ) ∩ (X \ πξ−1 [F ])) = λ(X \ W ) · λ(X \ πξ−1 [F ]) = λ(X \ W ) · µξ (Xξ \ F ) > 0 because µξ is strictly positive and φξ W 6= X. X XQ Q There is therefore a Boolean homomorphism ψx : Σ → {0, 1} which is zero on Ix . (iii) Set φξ+1 W = (φ0ξ W \ Q) ∪ {x : x ∈ Q, ψx W = 1} for every W ∈ Λξ+1 . Then φξ+1 is a Boolean homomorphism from Λξ+1 to PX. Because φξ+1 W 4φ0ξ W ⊆ Q is negligible, φξ+1 W ∈ Λ for every W ∈ Λξ+1 . If λW = 0 then φ0ξ W = ∅ and ψx W = 0 for every x ∈ Q, so φξ+1 W = ∅; thus φξ+1 : Λξ+1 → Λ is a partial lifting. If W ∈ Λξ , then, for x ∈ Q,

446


453I

x ∈ φξ W =⇒ x ∈ / φξ (X \ W ) =⇒ X \ W ∈ Ix =⇒ ψx (X \ W ) = 0 =⇒ ψx W = 1 ⇐⇒ x ∈ φξ+1 W =⇒ W ∈ / Ix =⇒ x ∈ φξ W, so φξ+1 W = φξ W . Thus φξ+1 extends φξ . (iv) Suppose that η ≤ ξ and G ⊆ Xη is open. If η < ξ then φξ+1 (πη−1 [G]) = φξ (πη−1 [G]) ⊇ πη−1 [G] by the inductive hypothesis. If η = ξ, take any x ∈ πξ−1 [G]. If x ∈ Q, then X\πξ−1 [G] ∈ Ix , so ψx (πξ−1 [G]) = 1 and x ∈ φξ+1 (πξ−1 [G]). If x ∈ / Q, there is an E ∈ Eξ such that x(ξ) ∈ E ⊆ G. In this case, x ∈ πξ−1 [E] \ Q ⊆ φ0ξ (πξ−1 [E]) \ Q ⊆ φξ+1 (πξ−1 [E]) ⊆ φξ+1 (πξ−1 [G]). As x is arbitary, πη−1 [G] ⊆ φξ+1 (πη−1 [G]) in this case also. Thus the induction continues. (c) Inductive step to a non-zero limit ordinal ξ of countable cofinality Suppose that 0 < ξ ≤ κ, that cf ξ = ω and that φη has been defined for every η < ξ. Let S hζn in∈N be a non-decreasing sequence in ξ with limit ξ. Then Bξ is the closed subalgebra of A generated by n∈N Bζn (using 254N and 254Fe, or otherwise). By 341G, there is a partial lower density φ : Λξ → Λ extending every φζn , and therefore extending φη for b ξ , where Λ b ξ is the σ-subalgebra of Λ generated by Λξ ∪{W : λW = 0}), every η < ξ. By 341J (applied to λ¹ Λ there is a partial lifting φξ : Λξ → Λ such that φW ⊆ φξ W for every W ∈ Λξ . If η < ξ and W ∈ Λη , then φη W = φW ⊆ φξ W ,

X \ φη W = φη (X \ W ) ⊆ φξ (X \ W ) = X \ φξ W ,

so φξ extends φη . If η < ξ and G ⊆ Xη is open, φξ (πη−1 [G]) = φη+1 (πη−1 [G]) ⊇ πη−1 [G]. So again the induction continues. S (d) Inductive step to a limit ordinal ξ of uncountable cofinality In this case, Bξ = η<ξ Bη , as in the proof of 341H; so there will be a unique partial lifting φξ : Λξ → Λ extending φη for every η < ξ (set φξ W = φη W 0 whenever W ∈ Λξ , η < ξ, W 0 ∈ Λη and W 4W 0 is negligible). As in (c), we again have φξ (πη−1 [G]) = φη+1 (πη−1 [G]) ⊇ πη−1 [G] whenever η < ξ and G ⊆ Xη is open. (e) At the end of the induction, we have a lifting φ = φκ of Λ such that φU ⊇ U for every U ∈ U , where U = {πξ−1 [G] : ξ < κ, G ∈ Tξ }. By 453H, λ is a τ -additive topological measure and φ is a strong lifting. 453J Corollary Let h(Xi , Ti , Σi , µi )ii∈I be a family of quasi-Radon probability spaces such that every Ti has a countable network consisting of measurable sets and every µi is strictly positive. Then the ordinary Q product measure λ on X = i∈I Xi is quasi-Radon and has a strong lifting. If every Xi is compact and Hausdorff, then λ is a Radon measure. proof We have just seen that λ is a τ -additive topological measure; but also it is inner regular with respect to the closed sets, by 412Ua, so it is a quasi-Radon measure. If all the Xi are compact and Hausdorff, so is X, so λ is a Radon measure (416G). 453K We come now to the construction of disintegrations from strong liftings. Theorem Let (X, T, Σ, µ) and (Y, S, T, ν) be Radon measure spaces and f : X → Y an almost continuous inverse-measure-preserving function. Suppose that ν has an almost strong lifting. Then there is a disintegration hµy iy∈Y of µ over ν such that every µy is a Radon measure and µy X = µy f −1 [{y}] = 1 for almost every y ∈ Y .

453K

Strong liftings

447

proof (a) Suppose first that X is compact and that f is continuous. Turn back to the proof of 452I, parts (a)-(g). In that argument, suppose that θ : B → T corresponds to S an almost strong lifting φ : T → T (see 341Ba). Set B = H∈S H \ φH, so that B is negligible. Take K to be the family of compact subsets of X. Then all the µy will necessarily be Radon measures, and almost all of them will be probability measures, as in 452O. For every y, f −1 [{y}] is a closed set, so is necessarily measured by µy . But also it is µy -conegligible for every y ∈ Y \ B. P P Let K ⊆ X \ f −1 [{y}] be a compact set. Then f [K] is a compact set not containing y. Because Y is Hausdorff, there is an open set H containing y such that H ∩ f [K] = ∅ (4A2F(h-i)). Now y ∈ H \ B ⊆ φH ⊆ φH. Set E = f −1 [H]. Then vE = (χH)• , as in (f) of the proof of 452I, and T vE = χ(φH), so λy E = T vE (y) = 1. Because E ∈ K, µy E ≥ λy E; since we always have µy X ≤ 1, we must have µy E = µy X = 1. But K ∩ E = ∅, so µy K = 0. As K is arbitrary, µy (X \ f −1 [{y}]) = 0. Q Q Thus µy f −1 [{y}] = µy X = 1 whenever y ∈ Y \ B and µy X = 1, which is almost always. (b) Now suppose that µ and ν are totally finite. (i) Let K be the family of subsets K ⊆ X such that K is compact and f ¹K is continuous, whenever F ∈ T and ν(F ∩ f [K]) > 0 then µ(K ∩ f −1 [F ]) > 0, either K = ∅ or µK > 0. Take any E ∈ Σ such that µE > 0. Then there is a K ∈ K such that K ⊆ E and µK > 0. P P Let K0 ⊆ E be a compact set such that f ¹K0 is continuous and µK0 > 0. Let δ > 0 be such that µK0 − δνY > 0. For compact sets K ⊆ K0 set q(K) = µK − δνf [K]. Choose hαn in∈N , hKn in≥1 as follows. Given that Kn is a compact subset of K0 , where n ∈ N, set αn = sup{q(K) : K ⊆ Kn is compact}, and choose aTcompact subset Kn+1 of Kn such that q(Kn+1 ) ≥ max(q(Kn ), αn − 2−n ). Continue. Set K = n∈N Kn . We have q(K) = µK − δνf [K] ≥ lim µKn − δ inf νf [Kn ] = lim q(Kn ) = sup q(Kn ) n→∞

n∈N

n→∞

n∈N

because hq(Kn )in∈N is non-decreasing. Of course K ⊆ E, µK ≥ q(K) ≥ q(K0 ) > 0, and f ¹K is continuous because K ⊆ K0 . ?? If there is an F ∈ T such that ν(F ∩ f [K]) > 0 but µ(K ∩ f −1 [F ]) = 0, take a compact set K 0 ⊆ K \ f −1 [F ] such that µK 0 > µK − δν(F ∩ f [K]). Then f [K 0 ] ⊆ f [K] \ F , so q(K 0 ) = µK 0 − δνf [K 0 ] ≥ µK 0 − δ(νf [K] − ν(F ∩ f [K])) > µK − δνf [K] = q(K). Let n ∈ N be such that q(K 0 ) > q(K) + 2−n ; then K 0 is a compact subset of Kn , so αn ≥ q(K 0 ) > q(K) + 2−n ≥ q(Kn+1 ) + 2−n ≥ αn , which is impossible. X X Thus K belongs to K and will serve. Q Q S (ii) By 342B, there is a countable disjoint set K0 ⊆ K such that µ(X \ K0 ) = 0. Enumerate K0 as hKn in<#(K0 ) ; for convenience of notation, if K0 is finite, set Kn = ∅ for n ≥ #(K0 ), so that every Kn P∞ belongs to K and µE = n=0 µ(E ∩ Kn ) for every E ∈ Σ. (iii) For each n ∈ N, define λn : T → R by setting λn F = µ(Kn ∩ f −1 [F ]) for every F R∈ T. Then λn is a measure dominated by ν, so there is a T-measurable gn : Y → [0, 1] such that λn F = F gn for every F ∈ T, by the Radon-Nikod´ ym theorem. Because λn (Y \ f [Kn ]) = 0, we may suppose that gn (y) = 0 for y∈ / f [Kn ]. We have

448


R P∞ F

n=0 gn

=

P∞ R n=0 F

gn =

P∞ n=0

453K

µ(Kn ∩ f −1 [F ]) = µf −1 [F ] = νF

P∞ for every F ∈ T, so n=0 gn (y) =P1 for ν-almost every y. Reducing the gn further on a set of measure zero, ∞ if need be, we may suppose that n=0 gn (y) ≤ 1 for every y. ˜ n be the subspace measure on f [Kn ] induced by λn , and µ (iv) For each n ∈ N, let λ ñ the subspace ˜ n . Also, λ ˜ n has an measure on Kn induced by µ. Then f ¹Kn is inverse-measure-preserving for µ ñ and λ almost strong lifting. P P If Kn = ∅, this is trivial. Otherwise, νf [Kn ] ≥ µKn > 0, so the subspace measure ˜ n have the same domain νñ induced by ν on f [Kn ] has an almost strong lifting, by 453E. But νñ and λ T ∩ P(f [Kn ]) and the same null ideal, because Kn ∈ K; so an almost strong lifting for νñ is an almost strong ñ. Q lifting for λ Q ˜ n such that every µny is a Radon By (a) above, we can find a disintegration hµny iy∈f [Kn ] of µ ñ over λ measure on Kn , µny Kn ≤ 1 for every y and µny {x : x ∈ Kn , f (x) = y} = µny Kn = 1 ˜ n -almost every y ∈ f [Kn ], that is, for ν-almost every y ∈ f [Kn ]. For y ∈ Y \ f [Kn ], let µny be the zero for λ measure on Kn . (v) Now, for y ∈ Y , set µy E =

P∞

n=0 gn (y)µny (E

∩ Kn )

for all those E ⊆ X such that the sum is defined. Then µy is a Radon measure and µy X ≤ 1. P P Because every µny is a complete measure, so is µy . We have P∞ P∞ µy X = n=0 gn (y)µny Kn ≤ n=0 gn (y) ≤ 1 by the choice of the gn . If G ⊆ X is open then µny measures G ∩ Kn for every n, so µy measures G; accordingly µy measures every compact set. If µy E > 0, there is some n ∈ N such that gn (y) > 0 and µny (E ∩ Kn ) > 0; now there is a compact set K ⊆ E ∩ Kn such that µny K > 0, in which case µy K > 0. By 412B, µy is tight, and is a Radon measure. Q Q (vi) hµy iy∈Y is a disintegration of µ over ν. P P If E ∈ Σ then

µE =

∞ X

µ(E ∩ Kn )

n=0

(by the choice of the Kn in (ii) above) ∞ ∞ Z X X = µ ñ (E ∩ Kn ) = n=0

n=0

˜ n (dy) µny (E ∩ Kn )λ f [Kn ]

ñ) (because hµny iy∈f [Kn ] is a disintegration of µ ñ over λ Z ∞ ∞ Z X X = µny (E ∩ Kn )λn (dy) = µny (E ∩ Kn )λn (dy) n=0

f [Kn ]

n=0

(because λn (Y \ f [Kn ]) = 0) ∞ Z X gn (y)µny (E ∩ Kn )ν(dy) = n=0

(235A) =

Z X ∞

Z gn (y)µny (E ∩ Kn )ν(dy) =

µy E ν(dy). Q Q

n=0

R (vii) In particular, µy X ν(dy) = νY ; since µy X ≤ 1 for every y, µy X = 1 for almost every y. And µy f −1 [{y}] = 1 for almost every y. P P

453N

Strong liftings

449

{y : µy f −1 [{y}] 6= 1} ⊆ {y : µy X 6= 1} ∪ {y : µ∗y (X \ f −1 [{y}]) > 0} [ ⊆ {y : µy X 6= 1} ∪ {y : y ∈ f [Kn ], µny (Kn \ f −1 [{y}]) > 0} n∈N

is negligible. Q Q (c) Now let us turn to the general case. This proceeds just as in 452O. Let hYi ii∈I be a decomposition of Y . For each i ∈ I, take Xi , λi and νi as in the proof of 452O. Note that λi and νi are Radon measures, so that we can apply (b) above to find a disintegration h˜ µy iy∈Yi of λi over νi such that every µ ˜y is a Radon measure and µ ˜y Xi = µ ˜y f −1 [{y}] = 1 for νi -almost every y ∈ Yi . Just as in 452O, we can set µy E = µ ˜y (E ∩ Xi ) whenever y ∈ Yi and µy measures E ∩ Xi , to obtain a disintegration hµy iy∈Y of µ over ν in which every µy is a Radon measure and µy X = 1 for almost every y; and this time S {y : y ∈ Y, µy f −1 [{y}] 6= 1} = i∈I {y : y ∈ Yi , µ ˜y f −1 [{y}] 6= 1} is negligible. So we have a disintegration of the required type. 453L Remark If f is surjective, we can arrange that every µy is a Radon probability measure for which f −1 [{y}] is µy -conegligible, just by changing some of the µy to point masses. If f is not surjective, then we can still (if X itself is not empty) arrange that every µy is a Radon probability measure; but it might be more appropriate to make some of the µy the zero measure, so that f −1 [{y}] is always µy -conegligible. I have continued to express this theorem in terms of measures µy on the whole space X. Of course, if we take it that Xy = f −1 [{y}] is to be conegligible for every y, it will sometimes be easier to think of µy as a measure on Xy ; this is very much what we do in the case of Fubini’s theorem, where all the Xy are, in effect, the same. 453M Strong liftings and Stone spaces Let (X, T, Σ, µ) be a quasi-Radon measure space, and (Z, S, T, ν) the Stone space of the measure algebra (A, µ ¯) of µ. For E ∈ Σ let E ∗ ⊆ Z be the open• and-closed set corresponding to the equivalence class E ∈ A. Let R be the relation T ∗ F ⊆X is closed {(z, x) : z ∈ Z \ F or x ∈ F } ⊆ Z × X (415Q). For every lifting φ : Σ → Σ we have a unique function gφ : X → Z such that φE = gφ−1 [E ∗ ] for every E ∈ Σ (see 341P). Now we have the following easy facts. (a) φ is strong iff (gφ (x), x) ∈ R for every x ∈ X. P P (gφ (x), x) ∈ R for every x ∈ X ⇐⇒ x ∈ F whenever F is closed and gφ (x) ∈ F ∗ ⇐⇒ F ⊆ gφ−1 [F ∗ ] for every closed set F ⊆ X ⇐⇒ F ⊆ φF for every closed set F ⊆ X ⇐⇒ φ is strong. Q Q (b) If T is Hausdorff, so that R is the graph of a function f (415R), then φ is strong iff f gφ (x) = x for every x ∈ X. (For (gφ (x), x) ∈ R iff f gφ (x) = x.) 453N Losert’s example (Losert 79) There is a compact Hausdorff space with a strictly positive completion regular Radon probability measure which has no strong lifting. proof (a) Let ν be the usual measure on {0, 1}N = Y . Let M ⊆ Y be a closed nowhere dense set such that νM > 0 (cf. 419B), and ν1 a Radon probability measure on Y such that ν1 M = 1 (e.g., a point mass at some point of M ).

450


453N

Let I be any set of cardinal at least ω2 such that I ∩ (I × I) = ∅. Let λ be the product measure on Y I , giving each factor the measure ν; of course λ can be identified with the usual measure on {0, 1}N×I (254N). Note that λ and ν are both strictly positive. For i ∈ I write Mi = {z : z ∈ Y I , z(i) ∈ M }; then Mi is closed in Y I . (z) Set A = {(i, j) : i, j ∈ I, i 6= j}. For z ∈ Y I and (i, j) ∈ A let νij be the Radon probability measure on Y given by setting (z)

νij = ν1 if z ∈ Mi ∩ Mj , = ν otherwise. (z)

Now, for z ∈ Y I , let λz be the Radon product measure of hνij i(i,j)∈A on Y A . (b) Let U be the family of sets U ⊆ Y A of the form {u : u(i, j) ∈ Uij for (i, j) ∈ B}, where B ⊆ A is finite and Uij ⊆ Y is open-and-closed for every (i, j) ∈ B. Then the function z 7→ λz U : Y I → [0, 1] is Borel measurable for every U ∈ U . P P Express U in the given form. For C ⊆ B set EC = {z : z ∈ Y I , C = {(i, j) : (i, j) ∈ B, z ∈ Mi ∩ Mj }}, so that hEC iC⊆B is a partition of Y I into Borel sets. For any C ⊆ B, λz U =

Y

(z)

νij (Uij ) =

(i,j)∈B

Y

ν1 Uij ·

(i,j)∈C

Y

νUij

(i,j)∈B\C

is constant for z ∈ EC . Q Q (c) There is a Radon measure on X = Y I × Y A given by the formula µE =

R

λz E[{z}]λ(dz)

R for every Baire set E ⊆ X. P P Let E be the class of those sets E ⊆ X such that λz E[{z}]λ(dz) is defined. Then E is closed under monotone limits of sequences, and E \ E 0 ∈ E whenever E, E 0 ∈ E and E 0 ⊆ E; also E contains all the basic open-and-closed sets in X of the form V × U , where V ⊆ Y I is open-and-closed and U ∈ U . By the Monotone Class Theorem (136B), E includes the σ-algebra generated by such sets, which is R the Baire σ-algebra Ba of X (4A3Of). Of course E 7→ λz E[{z}]λ(dz) is countably additive on Ba, so is a Baire measure on X, and has a unique extension to a Radon measure, by 432F. Q Q µ is strictly positive. P P Let W ⊆ X be any non-empty open set. Then it includes an open set of the form U × V where U = {z : z ∈ Y I , z(i) ∈ Vi for every i ∈ J}, V = {u : u ∈ Y A , u(j, k) ∈ Ujk for every (j, k) ∈ B}, J ⊆ I and B ⊆ A are finite sets, and Vi , Ujk ⊆ Y are non-empty open sets for every i ∈ J and (j, k) ∈ B. Now ν is strictly positive, so λU 0 > 0, where U 0 = {z : z ∈ U, z ∈ / Mj whenever (j, k) ∈ B}. (z)

(This is where we need to know that the Mj are nowhere dense.) But if z ∈ U 0 then νjk = ν for every (j, k) ∈ B, so Q λz V = (j,k)∈B νUjk > 0. Accordingly µW ≥

R U0

λz V λ(dz) > 0.

As W is arbitrary, µ is strictly positive. Q Q Write Σ for the domain of µ. (d) Fix on a self-supporting compact set K ⊆ X. I seek to show that, regarded as a subset of Y I∪A , K is determined by coordinates in some countable set. (i) There is a zero set L ⊇ K such that µL = µK. P P Let hKn in∈N be a sequence of compact subsets of X \ K such that limn→∞ µKn = µ(X \ K). For each n ∈ N there is a continuous function fn : X → [0, 1] which is zero on K and 1 on Kn ; now L = {x : fn (x) = 0 for every n ∈ N} is a zero set including K and of the same measure as K. Q Q

453N

Strong liftings

451

(ii) By 4A3Nc, L is determined by coordinates in a countable subset of I ∪A, that is, there are countable sets J0 ⊆ I, B0 ⊆ A such that whenever (z, u) ∈ L, (z 0 , u0 ) ∈ X, z¹J0 = z 0 ¹J0 and u¹B0 = u0 ¹B0 we shall have (z 0 , u0 ) ∈ L. Set J = J0 ∪ {i : (i, j) ∈ B0 } ∪ {j : (i, j) ∈ B0 },

B = A ∩ (J × J);

then J ⊇ J0 and B ⊇ B0 are still countable, and L is determined by coordinates in J ∪ B. (iii) Take any (z0 , u0 ) ∈ X \ K. Because K is closed, we can find finite sets J1 ⊆ I and B1 ⊆ A, open-and-closed sets Vi ⊆ Y for i ∈ J1 , and open-and-closed sets Uij ⊆ Y for (i, j) ∈ B1 , such that W = {(z, u) : z(i) ∈ Vi for every i ∈ J1 , u(i, j) ∈ Uij for every (i, j) ∈ B1 } contains (z0 , u0 ) and is disjoint from K. Set W1 = {(z, u) : (z, u) ∈ X, z(i) ∈ Vi for every i ∈ J1 ∩ J, u(i, j) ∈ Uij for every (i, j) ∈ B1 ∩ B}, Q = {z : z ∈ Y I , λz ((L ∩ W1 )[{z}]) > 0}, so that W1 is an open-and-closed set in X and Q is a Borel set in Y I ((b) above). Now Q is determined by coordinates in J. P P Suppose that z ∈ Q, z 0 ∈ Y I and z¹J = z 0 ¹J. Because both L and W1 are determined by coordinates in J ∪ B, (L ∩ W1 )[{z}] = (L ∩ W1 )[{z 0 }] = H say, and H is determined by coordinates in B. At the same time, for any (i, j) ∈ B, Mi ∩ Mj is determined by coordinates in J, so contains z iff it contains (z) (z 0 ) (z) (z 0 ) z 0 , and νij = νij . This means that, writing λ0z and λ0z0 for the products of hνij i(i,j)∈B and hνij i(i,j)∈B on Y B , λ0z = λ0z0 . So λz0 ((L ∩ W1 )[{z 0 }]) = λz0 H = λ0z0 H 0 = λ0z H 0 = λz H = λz ((L ∩ W1 )[{z}]) > 0, where H 0 = {u¹B : u ∈ H} (254Ob), and z 0 ∈ Q. Q Q (iv) Set J2 = ({i : (i, j) ∈ B1 \ B} ∪ {j : (i, j) ∈ B1 \ B}) \ J. Then J2 is a finite subset of I \ J, and B1 ⊆ (J ∪ J2 ) × (J ∪ J2 ). Set G = {z : z ∈ Y I , z(i) ∈ / M for every i ∈ J2 }, so that G is a dense open subset of Y I . Set G1 = {z : z ∈ Y I , z(i) ∈ Vi for every i ∈ J1 \ J}. Then G1 is a non-empty open set, so G ∩ G1 6= ∅ and λ(G ∩ G1 ) > 0. (v) Set U = {u : u ∈ Y A , u(i, j) ∈ Uij for every (i, j) ∈ B1 \ B}. (z)

If z ∈ G, then z ∈ / Mi ∩ Mj whenever (i, j) ∈ B1 \ B, so νij = ν for every (i, j) ∈ B1 \ B, and Q λz U = (i,j)∈B1 \B νUij > 0. (vi) ?? Suppose, if possible, that λQ > 0. Because Q is determined by coordinates in J and G ∩ G1 is determined by coordinates in J2 ∪ (J1 \ J), λ(Q ∩ G ∩ G1 ) = λQ · λ(G ∩ G1 ) > 0. If z ∈ Q ∩ G ∩ G1 ,

λz ((L ∩ W )[{z}]) = λz (U ∩ (L ∩ W1 )[{z}]) I

(because W = W1 ∩ (Y × U ) ∩ (G1 × Y A ), and z ∈ G1 )

452


453N

= λz U · λz ((L ∩ W1 )[{z}]) (because (L ∩ W1 )[{z}] is determined by coordinates in B, while U is determined by coordinates in B1 \ B, and λz is a product measure) >0 because z ∈ G ∩ Q. But this means that 0<

R

λz ((L ∩ W )[{z}])λ(dz) = µ(L ∩ W ) = µ(K ∩ W ) = µ∅,

which is absurd. X X Thus λQ must be zero. (vii) Consequently µ(K ∩ W1 ) = µ(L ∩ W1 ) =

R

λz ((L ∩ W1 )[{z}])λ(dz) = 0;

because K is self-supporting, K ∩ W1 = ∅. And W1 contains (z0 , u0 ) and is determined by coordinates in J ∪ B. (viii) What this means is that there can be no (z, u) ∈ K such that z¹J = z0 ¹J and u¹B = u0 ¹B. At this point, recall that (z0 , u0 ) was an arbitrary point of X \ K. So what must be happening is that K is determined by coordinates in the countable set J ∪ B. By 4A3Nc again, in the other direction, K is a zero set. (e) Part (d) shows that every self-supporting compact subset of X is a zero set. Since µ is certainly inner regular with respect to the self-supporting compact sets, it is inner regular with respect to the zero sets, that is, is completion regular. It follows that whenever E ∈ Σ there is an E 0 ⊆ E, determined by coordinates in a countable subset of I ∪ A, such that E \ E 0 is negligible. (Take E 0 to be a countable union of self-supporting compact sets.) (f ) ?? Now suppose, if possible, that we could find a strong lifting φ for µ. For each i ∈ I, take a set Ei ⊆ φ(Mi × Y A ) such that µEi = µφ(Mi × Y A ) and Ei is determined by coordinates in Ji ∪ Bi , where Ji ⊆ I and Bi ⊆ A are countable. Set Ji∗ = {j : (j, k) ∈ Bi } ∪ {k : (j, k) ∈ Bi }, so that Ji∗ is also countable. Because #(I) ≥ ω2 , there are distinct i, j ∈ I such that i ∈ / Jj∗ and j ∈ / Ji∗ (4A1Ea). So neither (i, j) nor (j, i) belongs to Bi ∪ Bj . Set F = {u : u ∈ Y A , u(i, j) ∈ M, u(j, i) ∈ M }. Then µ((Mi ∩ Mj ) × (Y A \ F )) = 0. P P If z ∈ Mi ∩ Mj , then (z)

(z)

λz F = νij M · νji M = (ν1 M )2 = 1, so λz (Y A \ F ) = 0. But (Mi ∩ Mj ) × (Y A \ F ) is a Baire set, so Z µ((Mi ∩ Mj ) × (Y

A

λz ((Mi ∩ Mj ) × (Y A \ F ))[{z}]λ(dz)

\ F )) = Z

λz (Y A \ F )λ(dz) = 0. Q Q

= Mi ∩Mj

Accordingly

Ei ∩ Ej ⊆ φ(Mi × Y A ) ∩ φ(Mj × Y A ) = φ((Mi ∩ Mj ) × Y A ) ⊆ φ(Y I × F ) (because ((Mi ∩ Mj ) × Y A ) \ (Y I × F ) is negligible) ⊆YI ×F

453Y

Strong liftings

453

because Y I × F is closed and φ is supposed to be strong. However, Ei ∩ Ej is determined by coordinates in Ji ∪ Jj ∪ Bi ∪ Bj , while Y I × F is determined by coordinates in {(i, j), (j, i)}, which does not meet Bi ∪ Bj . So either Ei ∩ Ej is empty or F = Y A . But F 6= Y A because M 6= Y , while µ(Ei ∩ Ej ) = µ(φ(Mi × Y A ) ∩ φ(Mj × Y A )) = µ((Mi ∩ Mj ) × Y A ) = λ(Mi ∩ Mj ) = (νM )2 > 0, so Ei ∩ Ej 6= ∅. X X (g) Thus µ has no strong lifting, as claimed. 453X Basic exercises > (a) Let (A, µ ¯) be a measure algebra and (Z, T, Σ, µ) its Stone space. Show that the canonical lifting of µ (341O) is strong. (b) Let S be the right-facing Sorgenfrey topology on R (415Xc). Show that there is a lifting of Lebesgue 1 δ

measure on R which is strong with respect to S. (Hint: set φE = {x : limδ↓0 µ(E ∩ [x, x + δ]) = 1}, and use 341J.) > (c) Let µ be the usual measure on the split interval (343J, 419L). Show that µ has a strong lifting. (d) Let (X, T, Σ, µ) be a complete locally determined topological measure space such that µ is inner regular with respect to the closed sets, and φ : Σ → Σ a strong lifting. Show that µ is a quasi-Radon measure with respect to the lifting topology Tl (414Q). Show that if T is regular then Tl ⊇ T. (e) Let (X, T, Σ, µ) be a topological measure space which has a strong lifting. Show that any non-zero indefinite-integral measure over µ (§234) has a strong lifting. (f ) Let (X, T, Σ, µ) be a Radon measure space such that (X, Σ, µ) is countably separated and µX > 0; for example, (X, T) could be an analytic space (433B). Show that µ is inner regular with respect to the compact metrizable subsets of X, so has an almost strong lifting. (Hint: there is an injective measurable f : X → R, which must be almost continuous.) (g) Let (X, T, Σ, µ) be a complete locally determined topological measure space such that µ is effectively locally finite and inner regular with respect to the closed sets, and φ : Σ → Σ a lower density such that φG ⊇ G for every open G ⊆ X. Show that µ is a quasi-Radon measure with respect to both T and the density topology associated with φ. (h) Let (X, T, Σ, µ) be a quasi-Radon measure space and φ : Σ → Σ a lower density such that φG ⊇ G for every open G ⊆SX. Let hGx ix∈X be a family of open sets in X such that x ∈ / φ(X \ Gx ) for every x ∈ X. (i) Show that A \ x∈A (Gx ∩ Ux ) is negligible whenever A ⊆ X and Ux is a neighbourhood of x for every x ∈ A. (ii) Let S be the topology on X generated by T ∪ {{x} ∪ Gx : x ∈ X}. Show that µ is quasi-Radon with respect to S. (i) Let X and Y be Hausdorff spaces, and µ a Radon probability measure on X × Y ; set π(x, y) = y for x ∈ X, y ∈ Y , and let ν be the image measure µπ −1 . Suppose that ν has an almost Rstrong lifting. Show that there is a family hµy iy∈Y of Radon probability measures on X such that µE = µy (E −1 [{y}])ν(dy) for every E ∈ dom µ. (j) Use 453Xe to simplify part (b) of the proof of 453K. R (k) In 453N, show that λz E[{z}]λ(dz) is defined and equal to µE whenever µ measures E. 453Y Further exercises (a) Let (Y, S, T, ν) be a Radon measure space such that νY > 0 and whenever (X, T, Σ, µ) is a Radon measure space and f : X → Y is an almost continuous inverse-measure-preserving function, then there is a disintegration hµy iy∈Y of µ over ν such that µy f −1 [{y}] = 1 for almost every y. Show that ν has an almost strong lifting. (Hint: Start with the case in which Y is compact. Take f : X → Y to be the function described in 415R, 416V and 453Mb. Set φE = {y : µy E ∗ = µy X = 1}.)

454


453Z

453Z Problems (a) If (X, T, Σ, µ) and (Y, S, T, ν) are compact Radon measure spaces with strong liftings, does their product necessarily have a strong lifting? What if they are both Stone spaces? (b) If (X, T, Σ, µ) is a Radon probability space with countable Maharam type, must it have an almost strong lifting? 453 Notes and comments As I noted in §452, early theorems on disintegrations concentrated on cases in which all the measure spaces involved were ‘standard’ in that the measures were defined on standard Borel spaces (§424), or were the completions of such measures. Under these conditions the distinction between 452I and 453K becomes blurred; measures (when completed) have to be Radon measures (433Cb), liftings have to be almost strong (453F) and disintegrations have to be concentrated on fibers (452Gc). Theorem 453K provides disintegrations concentrated on fibers without any limitation on the size of the spaces involved, though making strong topological assumptions. The strength of 453K derives from the remarkable variety of the (Radon) measure spaces which have strong liftings, as in 453F, 453G, 453I and 453J. For some ten years there were hopes that every selfsupporting Radon measure had an strong lifting, which were finally dashed by Losert 79; I give a version of the example in 453N. This is a special construction, and it remains unclear whether some much more direct approach might yield another example (453Za). I should perhaps remark straight away that if the continuum hypothesis is true, then any self-supporting Radon measure space of Maharam type at most ω1 has a strong lifting; I will present the argument in Volume 5. In particular, subject to the continuum hypothesis, Z × Z has a strong lifting, where Z is the Stone space of the Lebesgue measure algebra, and we have a positive answer to 453Zb.

454 Measures on product spaces A central concern of probability theory is the study of ‘processes’, that is, families hXt it∈T of random variables thought of as representing the evolution of a system in time. The representation of such processes as random variables in the modern sense, that is, measurable functions on an abstract probability space, was one of the first challenges faced by Kolmogorov. In this section I give a version of Kolmogorov’s theorem on the extension of consistent families of measures on subproducts to a measure on the whole product (454D). It turns out that some restriction on the marginal measures is necessary, and ‘perfectness’ seems to be an appropriate hypothesis, necessarily satisfied if the factor spaces are standard Borel spaces or the marginal measures are Radon measures. If we have marginal measures with stronger properties then we shall be able to infer corresponding properties of the measure on the product space (454A, generalizing 451J). The apparatus here makes it easy to describe joint distributions of arbitrary families of real-valued random variables (454J-454M), extending the ideas of §271. For the sake of the theorem that almost all Brownian paths are continuous (455D) I briefly investigate measures on C(T ), where T is a Polish space (454N-454P). 454A Theorem Let h(Xi , Σi , µi )ii∈I be a non-empty family of totally finite measure spaces. Set X = N c Σi generated i∈I Xi and let µ be a measure on X which is inner regular with respect to the σ-algebra i∈I −1 by {πi [E] : i ∈ I, E ∈ Σi }, where πi : X → Xi is the coordinate map for each i ∈ I. Suppose that every πi is inverse-measure-preserving. (a) If K ⊆ PX is a family of sets which is closed under finite unions and countable intersections, and µi is inner regular with respect to Ki = {K : K ⊆ Xi , πi−1 [K] ∈ K} for every i ∈ I, then µ is inner regular with respect to K. (b)(i) If every µi is a compact measure, so is µ; (ii) if every µi is a countably compact measure, so is µ; (iii) if every µi is a perfect measure, so is µ. Q

proof If X is empty this is all trivial, so we may suppose that X 6= ∅. (a) Set A = {πi−1 [E] : i ∈ I, E ∈ Σi }. If A ∈ A, V ∈ Σ and µ(A ∩ V ) > 0, there is a K ∈ K ∩ A such that K ⊆ A and µ(K ∩ V ) > 0. P P Express A as πi−1 [E], where E ∈ Σi ; take L ∈ Ki such that L ⊆ E and µi L > µi E − µ(A ∩ V ), and set K = πi−1 [L]. Q Q

454C

Measures on product spaces

455

N By 412C, µ¹ c i∈I Σi is inner regular with respect to K; by 412Ab, so is µ. (b)(i)-(ii) Suppose that every µi is (countably) compact. Then for each i ∈ I we can find a (countably) compact class Ki ⊆ PXi such that µi is inner regular with respect to Ki . Set L = {πi−1 [K] : i ∈ I, K ∈ Ki }. Then L is (countably) compact (451H). So there is a (countably) compact K ⊇ L which is closed under finite unions and countable intersections (342D, 413R). Now µ is inner regular with respect to K, by (a), and therefore (countably) compact. N (iii) Let T0 be a countably generated σ-subalgebra of c i∈I Σi . Then there must be some countable subfamily E of {πi−1 [E] : i ∈ I, E ∈ Σi } such that T0 is included in the σ-algebra generated by E (use 331Gd). Set Ei = {E : E ∈ Σi , πi−1 [E] ∈ E} for each i, so that Ei is countable, and let Σ0i be the σ-algebra N generated by Ei . Then µi ¹Σ0i is compact (451F). Applying (i), we see that µ¹ c i∈I Σ0i is compact, therefore N N perfect; while T0 ⊆ c i∈I Σ0i . As T0 is arbitrary, µ¹ c i∈I Σi is perfect (451F). But as the completion of µ is N exactly the completion of µ¹ c i∈I Σi , µ also is perfect, by 451Gc. 454B Corollary Let hXi ii∈I be a family of Polish spaces with product X. Then any totally finite Baire measure on X is a compact measure. N proof If µ is a Baire measure on X, then its domain Ba(X) is c i∈I B(Xi ), where B(Xi ) is the Borel σ-algebra of Xi for each i ∈ I (4A3Na). So each image measure µi on Xi is a Borel measure, therefore tight (that is, inner regular with respect to the compact sets, 433Ca), and by 454A(b-i) µ is compact. 454C Theorem(Marczewski & Ryll-Nardzewski 53) Let (X, Σ, µ) be a perfect totally finite measure space and (Y, T, ν) any totally finite measure space. Let Σ ⊗ T be the algebra of subsets of X × Y generated by {E × F : E ∈ Σ, F ∈ T}. If λ0 : Σ ⊗ T → [0, ∞[ is a non-negative finitely additive functional such that λ0 (E ×Y ) = µE and λ0 (X ×F ) = νF whenever E ∈ Σ and F ∈ T, then λ0 has a unique extension b generated by Σ ⊗ T. to a measure defined on the σ-algebra Σ⊗T proof (a) By 413K, it will be enough to show that limn→∞ λ0 Wn = 0 for every non-increasing sequence hWn in∈N in Σ ⊗ T with empty intersection. Take such a sequence. Each Wn must belong to the algebra generated by some finite subset of {E × F : E ∈ Σ, F ∈ T}, so there must be a countable set E ⊆ Σ such that every Wn belongs to the algebra generated by {E × F : E ∈ E, F ∈ T}; let Σ0 be the σ-subalgebra of b Σ generated by E, so that every Wn belongs to Σ0 ⊗T. (b) By 451F, µ¹Σ0 is a compact measure; let K ⊆ PX be a compact class such that µ¹Σ0 is inner regular with respect to K. We may suppose that K is the family of closed sets for a compact topology on X (342Da). Let W be the family of those elements W of Σ0 ⊗ T such that every horizontal section W −1 [{y}] belongs to K. Then W is closed under finite unions and intersections. (c) S If W ∈ Σ0 ⊗ T and ² > 0, then there is a W 0 ∈ W such that W 0 ⊆ W and λ0 (W \ W 0 ) ≤ ². P P Express W as i≤n Ei × Fi , where Ei ∈ Σ0 , Fi ∈ T for each i ≤ n. (Cf. 315Jb.) For each i ≤ n, take Ki ∈ K ∩ Σ0 S 1 such that µ(Ei \ Ki ) ≤ n+1 ², and set W 0 = i≤n Ki × Fi . Then W 0 ∈ W, W 0 ⊆ W and λ0 (W \ W 0 ) ≤ =

n X i=0 n X

λ0 ((Ei × Fi ) \ (Ki × Fi )) ≤

n X

λ0 ((Ei \ Ki ) × Y )

i=0

µ0 (Ei \ Ki ) ≤ ². Q Q

i=0

(d) Take anyT² > 0. Then for each n ∈ N we can find Wn0 ∈ W such that Wn0 ⊆ Wn and λ0 (Wn \ Wn0 ) ≤ 2 ². Set Vn = i≤n Wi0 , so that Vn ∈ W and Pn λ0 (Wn \ Vn ) ≤ i=0 λ0 (Wi \ Wi0 ) ≤ 2² −n

for each n, and hVn in∈N is non-increasing, with empty intersection. Because Vn ∈TΣ0 ⊗ T, its projection Hn = Vn [X] belongs to T, for each n. Of course hHn in∈N is nonincreasing; also n∈N Hn = ∅. P P If y ∈ Y , then hVn−1 [{y}]in∈N is a non-increasing sequence in K with

456


454C

T T empty intersection, because n∈N Vn ⊆ n∈N Wn is empty. But K is a compact class, so there must be some n such that Vn−1 [{y}] is empty, that is, y ∈ / Hn . Q Q Accordingly limn→∞ νHn = 0. But as Vn ⊆ X × Hn , limn→∞ λ0 Vn = 0. This means that limn→∞ λ0 Wn ≤ 2². But as ² is arbitrary, limn→∞ λ0 Wn = 0, as required. 454D Theorem (Kolmogorov 33, §III.4) Q N Let h(Xi , Σi , µi )ii∈I be a family of totally finite perfect −1 measure spaces. Set X = i∈I Xi , and write i∈I Σi for the algebra of subsets of X generated N by {πi [E] : i ∈ I, E ∈ Σi }, where πi : X → Xi is the coordinate map for each i ∈ I. Suppose that λ0 : i∈I Σi → [0, ∞[ is a non-negative finitely additive functional such that λ0 πi−1 [E] = µi E whenever i ∈ I and E ∈ Σi . Then N λ0 has a unique extension to a measure λ with domain c i∈I Σi , and λ is perfect. proof (a) The Nargument follows the same pattern as that of 454C. This time, take a non-increasing sequence hWn in∈N in i∈I Σi with empty intersection. Each Wn belongs to the algebra generated by some finite −1 subset of {πi [E] : i ∈ I, E ∈ Σi }, so we can find countable sets Ei ⊆ Σi such that every Wn belongs to −1 the subalgebra generated by N {πi [E] : i ∈ I, E ∈ Ei }. Let Ti be the σ-subalgebra of Σi generated by Ei , so that every Wn belongs to i∈I Ti . (b) For each i ∈ I, µi ¹ Ti is compact (451F); let Ti be a compact topology on Xi such that µi ¹ Ti is inner regular with respect to the closed sets (342F). Let T be the product topology on X, so that T is compact N (3A3J). Let W be the family of closed sets in X belonging to i∈I Ti . N P We can (c) If W ∈ i∈I Ti and ² > 0, there is a W 0 ∈ W such that W 0 ⊆ W and λ0 (W \ W 0 ) ≤ ². P S T express W as k≤n i∈Jk πi−1 [Eki ] where each Jk is a finite subset of I and Eki ∈ Σi for k ≤ n, i ∈ Jk (cf. 315Jb). Let h²ki ik≤n,i∈Jk be a family of strictly positive numbers with sum at most ². For each k ≤ n, S T i ∈ Jk take Kki ∈ Ki ∩ Ti such that Kki ⊆ Eki and µi (Eki \ Kki ) ≤ ²ki , and set W 0 = k≤n i∈Jk πi−1 [Kki ]. Q Q (d) Take any T ² > 0. Then for each n ∈ N we can find Wn0 ∈ W such that Wn0 ⊆ Wn and λ0 (Wn \ Wn0 ) ≤ 2−n ². Set Vn = i≤n Wi0 . Then hVn in∈N is a non-increasing sequence of closed sets in the compact space X, and has empty intersection, so there is some n such that Vn is empty, and Pn λ0 Wn ≤ i=0 λ0 (Wi \ Wi0 ) ≤ 2². As ² is arbitrary, limn→∞ λ0 Wn = 0. N (e) As hWn in∈N is arbitrary, λ0 has a unique countably additive extension to c i∈I Σi , by 413K, as before. Of course the extension is perfect, by 454A(b-iii). 454E Corollary measure spaces. Let C be the family of Q Let h(Xi , Σi , µi )ii∈I be a family Tof perfect −1 subsets of X = i∈I Xi expressible in the form X ∩ i∈J πi [Ei ] where J ⊆ I is finite and Ei ∈ Σi for every i ∈ I, writing πi (x) = x(i) for x ∈ X, i ∈ I. Suppose that λ0 : C → R is a functional such that (i) λ0 πi−1 [E] = µi E whenever i ∈ I and E ∈ Σi (ii) λ0 C = λ0 (C ∩ πi−1 [E]) + λ0 (C \ πi−1 [E]) whenever C ∈ C, N i ∈ I and E ∈ Σi . Then λ0 has a unique extension to a measure on c i∈I Σi , which is necessarily perfect. N proof By 326Q, λ0 has an extension to an additive functional on i∈I Σi , so we can apply 454D. Q 454F Corollary Let h(Xi , Σi )ii∈I be a family of standard Borel spaces. Set X = i∈I Xi , and let N −1 i∈I Σi be the algebra of subsets of X generated by {πi [E] : i ∈ I, E ∈ Σi }, where πi : X → Xi is the coordinate map for each i. Let λ0 : Σ → [0, ∞[ be a non-negative finitely additive functional such that all the marginal functionals E 7→ λ0 πi−1 [E] : Σi → [0, ∞[ are countably additive. Then λ0 has a unique N extension to a measure defined on c i∈I Σi , which is a compact measure. proof This follows immediately from 454D and 454A if we note that all the measures λ0 πi−1 are necessarily compact, therefore perfect (451L).

454H


457

454G Corollary Let hXi ii∈I be a family of sets, and Σi a σ-algebra of subsets Q of Xi for each i ∈ I. Suppose that for each finite set J ⊆ I we are given a totally finite measure µJ on ZJ = i∈J Xi with domain N c Σi such that (i) whenever J, K are finite subsets of I and J ⊆ K, then the canonical projection from i∈J ZK to ZJ is inverse-measure-preserving (ii) every marginal measure µ{i} on Z{i} ∼ = Xi is perfect. Then N Q c there is a unique measure µ defined on Σi such that the canonical projection π ˜J : Xi → ZJ is i∈I

i∈I

inverse-measure-preserving for every finite J ⊆ I. proof All we need to observe is that N N πJ−1 [V ] : J ∈ [I]<ω , V ∈ i∈J Σi }. i∈I Σi = {˜

Because all the canonical projections from XK ontoNXJ are inverse-measure-preserving, we have µJ V = N −1 µK V 0 whenever J, K are finite subsets of I, V ∈ Σi , V 0 ∈ Σi and π ˜J−1 [V ] = π ˜K [V 0 ]. So i∈J i∈K N −1 we have ˜J [V ] = µJ V whenever J ⊆ I is finite and i∈I Σi → [0, ∞[ such that λ0 π N a functional λ0 : V ∈ i∈J Σi . It is easy to check that λ0 is finitely additive and satisfies the conditions of 454D. So λ0 can N be extended to a measure µ defined on c i∈I Σi . N If J ⊆ I is finite, then µJ and µ˜ πJ−1 agree on i∈J ΣJ and therefore (by the Monotone Class Theorem, N 136C) on c i∈J Σi ; that is, π ˜J is inverse-measure-preserving. To see that µ itself is unique, observe that N N the conditions define its values on i∈I Σi and therefore on c i∈I Σi , by the Monotone Class Theorem once more. 454H Corollary Let h(Xn , Σn )in∈N be a sequence of standard Borel spaces. For each n ∈ N set N Q Zn = i
R

νz W [{z}]˜ µn (dz)

for every n ∈ N and W ∈ Tn+1 , and Z ZZ ZZ f d˜ µn+1 = ... f (ξ0 , . . . , ξn )ν(ξ0 ,... ,ξn−1 ) (dξn ) ν(ξ0 ,... ,ξn−2 ) (dξn−1 ) . . . νξ0 (dξ1 )ν∅ (dξ0 ) for every n ∈ N and µ ñ+1 -integrable real-valued function f . proof (a) We must check that the first formula given actually defines µ ñ+1 (W ) for every W ∈ Tn+1 . Of course this is an induction on n. µ ˜0 will be the unique probability measure on the singleton set Z0 . Given R that dom µ ñQ = Tn , the class W of sets W ⊆ Zn+1 for which νz W [{z}]˜ µn (dz) is defined will contain all sets of the form i≤n Ei where Ei ∈ Σi for every i ≤ n, just because the function z 7→ νz En is Tn -measurable. Since W is closed under increasing sequential unions and differences of comparable sets, the Monotone Class Theorem (136B) tells us that it includes the σ-algebra generated by the cylinder sets, which is Tn+1 . (b) Accordingly hνz0 iz∈Zn is a disintegration of µ ñ+1 over µ ñ , where νz0 is the measure with domain Tn+1 0 defined by writing νz (W ) = νz W [{z}] for W ∈ Tn+1 , z ∈ Zn . By 452F, Z Z Z f d˜ µn+1 = f (w)νz0 (dw)˜ µn (dz) Zn+1 Zn Zn+1 Z Z = f (z, ξn+1 )νz (dξn+1 )˜ µn (dz) Zn

Xn

for every µ ñ+1 -integrable function f . (The second equality can be regarded as an application of the changeof-variable formula 235Ib applied to the (νz , νz0 )-inverse-measure-preserving function ξ 7→ (z, ξ) : Xn+1 → R Zn+1 .) Now a direct induction yields the general formula for f d˜ µn+1 in the statement of this corollary.

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454H

(c) The canonical maps from Zn+1 to Zn are all inverse-measure-preserving, just because every νz is a N probability measure. We therefore have a well-defined functional λ0 : n∈N Σn → [0, 1] defined by setting N λ0 π ñ−1 [W ] = µn W whenever n ∈ N and W ∈ i
454N


459

(e) The only point I have not covered is the uniqueness of ν. But suppose that ν 0 is another measure on RI with the properties described in (i). If i0 , . . . , in ∈ I and π ˜ (x) = (x(i0 ), . . . , x(in )) for x ∈ RI , then −1 0 −1 n+1 the image measures ν π ˜ and ν π ˜ on R are both the distribution of Xi0 , . . . , Xin , by the argument of (c) above. This means that ν and ν 0 agree on the algebra of subsets of RI generated by sets of the form {x : x(i) ∈ E} where i ∈ I and E ⊆ R is Borel. By 454D, they agree on all zero sets, and must be equal (412L). 454K Definition In the context of 454J, I will call ν the (joint) distribution of the process hXi ii∈I . Note that if I = n ∈ N \ {0}, then ν is a Radon measure on Rn , so is the distribution of hXi ii
j∈J

Z =

exp(i

X j∈J

Z 0

αj x(j))ν (dx) =

exp(i

X

αj z(j))µ0J (dz),

j∈J

µ0J

so µJ and have the same characteristic function, therefore are equal. This is true for every J, so ν and ν 0 are equal, by 454D. (b) Taking ν and ν 0 to be the two distributions, (a) (with 454L) tells us that their restrictions to the Baire σ-algebra of RI are the same, so they must be identical. 454N Continuous processes The original, and still by farQthe most important, context for 454D is when every (Xi , Σi ) is R with its Borel σ-algebra, so that X = i∈I Xi can be identified with RI . In the discussion so far, the set I has been an abstract set, except in the very special case of 454H. But some of the most important applications (to which I shall come in the next section) involve index sets carrying a topological structure; for instance, I could be the unit interval [0, 1] or the half-line [0, ∞[. In such a case, we have a wide variety of subspaces of RI (for instance, the space of continuous functions) marked out as

460


454N

special, and it is important to know when, and in what sense, our measures on the product space RI can be regarded as, or replaced by, measures on the subspace of interest. In the next few paragraphs I look briefly at spaces of continuous functions on Polish spaces. Lemma Let T be a separable metrizable space and (X, Σ, µ) a semi-finite measure space. Let T be a topology on X such that µ is inner regular with respect to the closed sets. (a) Let φ : X × T → R be a function such that (i) for each x ∈ X, t 7→ φ(x, t) is continuous (ii) for each t ∈ T , x 7→ φ(x, t) is Σ-measurable. Then µ is inner regular with respect to K = {K : K ⊆ X, φ¹K × T is continuous}. (b) Let θ : X → C(T ) be a function such that x 7→ θ(x)(t) is Σ-measurable for every t ∈ T . Give C(T ) the topology Tc of uniform convergence on compact subsets of T . Then θ is almost continuous. proof (a) Take E ∈ Σ and γ < µE; take F ∈ Σ such that F ⊆ E and γ < µF < ∞. Let U be a countable base for the topology of T consisting of non-empty sets, D a countable dense subset of T and V a countable base for the topology of R. For U ∈ U , V ∈ V set EU V = {x : φ(x, t) ∈ V for every t ∈ U ∩ D}; then EU V ∈ Σ. Let h²U V iU ∈U ,V ∈V be a family of strictly positive numbers with sum at most µF − γ. For each U ∈ U , V ∈ V take a closed set FU V ⊆ F \ EU V such that µFU V ≥ µ(F \ EU V ) − ²U V . Consider T K = U ∈U ,V ∈V FU V ∪ (F ∩ EU V ). Then K ⊆ E and µK ≥ γ. If x ∈ K, t ∈ T and φ(x, t) ∈ V0 ∈ V, let V ∈ V be such that φ(x, t) ∈ V and V ⊆ V0 . Then {t0 : φ(x, t0 ) ∈ V } is an open set containing t, so there is some U ∈ U such that t ∈ U and φ(x, t0 ) ∈ V for every t0 ∈ U . This means that x ∈ EU V , so that (K \ FU V ) × U contains (x, t), and is a relatively open set in K × T . If (x0 , t0 ) ∈ (K \ FU V ) × U , then x0 ∈ EU V , so φ(x0 , t00 ) ∈ V whenever t00 ∈ U ∩ D; as D is dense, φ(x0 , t00 ) ∈ V whenever t00 ∈ U ; in particular, φ(x0 , t0 ) ∈ V ⊆ V0 . This shows that (K × T ) ∩ φ−1 [V0 ] is relatively open in K × T ; as V0 is arbitrary, φ¹K × T is continuous. So K ∈ K. As E and γ are arbitrary, µ is inner regular with respect to K. (b) Set φ(x, t) = θ(x)(t) for x ∈ X, t ∈ T . Because θ(x) ∈ C(T ) for every x, φ is continuous in the second variable; and the hypothesis on θ is just that φ is measurable in the first variable. So µ is inner regular with respect to K as described in (a). But θ¹K is continuous for every K ∈ K, by 4A2G(g-ii). So θ is almost continuous. 454O Proposition Let T be an analytic metrizable space (e.g., a Polish space, or any Souslin-F subset of a Polish space), and µ a probability measure on C(T ) with domain the σ-algebra Σ generated by the evaluation functionals f 7→ f (t) : C(T ) → R for t ∈ T . Give C(T ) the topology Tc of uniform convergence on compact subsets of T . Then the completion of µ is a Tc -Radon measure. proof If T is empty this is trivial, so let us suppose henceforth that T 6= ∅. (a) Let D be a countable dense subset of T . Let π : C(T ) → RD be the restriction map. Set X = π[C(T )] ⊆ RD ; then X, with the topology it inherits from RD , is a separable metrizable space. Note that, because D is dense, π is injective. We need to know that π is an isomorphism between (C(T ), Σ) and (X, B), where B is the Borel σ-algebra of X. P P Since the Borel σ-algebra of RD is just the σ-algebra generated by the functionals g 7→ g(t) : RD → R as t runs over D (4A3Dc/4A3E), B is the σ-algebra of subsets of X generated by the functionals g 7→ g(t) : X → R for t ∈ D. So π is surely (Σ, B)-measurable. On the other hand, if t ∈ X, there is a sequence htn in∈N in D converging to t, so that π −1 (g)(t) = limn→∞ g(tn ) for every g ∈ X, and g 7→ π −1 (g)(t) : X → R is B-measurable. Accordingly π −1 is (B, Σ)-measurable. Q Q (b) Let hUn in∈N be a sequence running over a base for the topology of T , with no Un empty. For each n ∈ N, g ∈ RD set ωn (g) = supt,u∈Un ∩D min(1, g(t) − g(u)), so that ωn : RD → [0, 1] is T-measurable, where T is the Borel (or Baire) algebra of RD . For g ∈ RD , g ∈ X iff g has an extension to a continuous function on T , that is,

454Xb


461

for every t ∈ T , k ∈ N there is an n ∈ N such that t ∈ Un and ωn (g) ≤ 2−k . Turning this round, RD \ X is the projection onto the first coordinate of the set S T Q = k∈N n∈N {(g, t): either t ∈ / Un or ωn (g) > 2−k } ⊆ RD × T . But (because every Un is an open set and every ωn is Borel measurable) Q is a Borel set in the analytic space RD × T . So Q is analytic (423Eb) and RD \ X is analytic (423Bb). Since RD , being Polish (4A2Qc), is a Radon space (434Kb), X is a Radon space (434F). (c) The image measure ν = µπ −1 on X is a Borel probability measure. Because X is a Radon space, ν is tight, and its completion νˆ is a Radon measure. By 454Nb, π −1 : X → C(T ) is almost continuous if we give C(T ) the topology Tc . So the image measure λ = νˆ(π −1 )−1 is a Radon measure for Tc (418I). But of course λ is the completion of µ, just because π is a bijection and νˆ is the completion of ν. 454P Corollary Let T be an analytic metrizable space. (a) C(T ), with either the topology Tp of uniform convergence on finite subsets of T or the topology Tc of uniform convergence on compact subsets of T , is a measure-compact Radon space. (b) Let µ be a Baire probability measure on RT such that µ∗ C(T ) = 1. Then µ is τ -additive, so has a unique extension µ ˜ which is a Radon measure on RT . µ ˜C(T ) = 1, and the subspace measure on C(T ) induced by µ ˜ is a Radon measure if C(T ) is given the topology Tc . proof (a) Let µ be a probability measure on C(T ) which is either a Baire measure or a Borel measure with respect to either Tp or Tc . Let µ ˜ be the completion of µ¹Σ, where Σ is the σ-algebra generated by the functionals f 7→ f (t); because all these are Tp -continuous, Σ is certainly included in the Baire σ-algebra for Tp , so that Σ ⊆ dom µ. 454O tells us that µ ˜ is a Radon measure for Tc . Because Tp is a coarser Hausdorff topology, µ ˜ is also a Radon measure for Tp . Also µ ˜ must extend µ, because its domain includes that of µ and the completion of µ must extend µ ˜ (in fact, of course, this means that µ ˜ is actually the completion of µ). Now µ ˜ is τ -additive (for either topology), so µ also is; as µ is arbitrary, C(T ) is measure-compact (for either topology). On the other hand, if µ is a Borel measure (for either topology), it must be tight for that topology; so that C(T ) is a Radon space. (b) Write µC for the subspace measure on C(T ). Recall that the domain Σ of µ is just the σ-algebra generated by the functionals f 7→ f (t) : RT → R, as t runs over T (4A3Na), so that the domain ΣC of µC is the σ-algebra of subsets of C(T ) generated by the functionals f 7→ f (t) : C(T ) → R. By 454O, the completion µ ˆC of µC is a Radon measure on C(T ) if we give C(T ) the topology Tc of uniform convergence on compact subsets of T . The embedding C(T ) ⊆ RT is of course continuous for Tc and the product topology on RT , so we have a Radon image measure µ ˜ on RT defined by saying that µ ˜E = µ ˆC (E ∩ C(T )) whenever E ∩ C(T ) is measured by µ ˆC . If E ∈ Σ, then µ ˜E = µ ˆC (E ∩ C(T )) = µC (E ∩ C(T )) = µ∗ (E ∩ C(T )) = µE because µ∗ C(T ) = 1, so µ ˜ extends µ. Of course µ ˜C(T ) = 1 and the subspace measure on C(T ) induced by µ ˜ is just µ ˆC , so is a Radon measure with respect to Tc . Finally, because µ has an extension to a Radon measure, it must itself be τ -additive. Because Σ includes a base for the topology of RT , µ can have only one extension to a Radon measure on RT (415H(iv)). 454X Basic exercises > (a) Let µ be Lebesgue measure on [0, 1]. Let X0 ⊆ [0, 1] be a set of inner measure zero and outer measure 1. Set X1 = [0, 1] \ X0 ; for each i, let µi be the subspace measure on Xi and Σi its domain. Show that we have a finitely additive functional λ defined on Σ0 ⊗ Σ1 by the formula λ((E ∩ X0 ) × (F ∩ X1 )) = µ(E ∩ F ) for all E, F ∈ dom µ, and that λ has no extension to a measure on X0 × X1 . (b) Adapt the example of 419K to provide a counter-example for 454G if we omit the hypothesis that the marginal measures µ{i} must be perfect.

462


454Xc

(c) Adapt the example of 419K/454Xb to provide a counter-example for 454H if we omit the hypothesis Q that the (Xn , Σn ) must be standard Borel spaces. (Hint: if z ∈ i≤n Xi , try νz (E) = 1 if z(n) ∈ E, 0 otherwise.) > (d) Let X be a set and Σ a σ-algebra of subsets of X. Let us say that (X, Σ) has the perfect measure property if every totally finite measure with domain Σ is perfect. Show that (i) if (X, Σ) has the perfect measure property, so does (E, ΣE ) for any E ∈ Σ, where ΣE is the subspace σ-algebra on E Q N (ii) if h(Xi , Σi )ii∈I is a family of spaces with the perfect measure property, then ( i∈I Xi , c i∈I Σi ) has the perfect measure property. (e) Let X be a set and Σ a σ-algebra of subsets of X. Let us say that (X, Σ) has the countably compact measure property if every totally finite measure with domain Σ is countably compact. Show that (i) if (X, Σ) has the countably compact measure property it has the perfect measure property (ii) if (X, Σ) has the countably compact measure property so does (E, ΣE ) for every E ∈ Σ, where ΣE is the subspace σ-algebra on E (iii) if h(Xi , Σi )ii∈I is a family of spaces with the countably compact measure Q N property, then ( i∈I Xi , c i∈I Σi ) has the countably compact measure property. (f ) Suppose that (X, Σ) has the countably compact measure property. (i) Let µ be a totally finite measure with domain Σ, (Y, T, ν) a measure space, and f : X → Y an inverse-measure-preserving function. Show that µ has a disintegration hµy iy∈Y over ν which is consistent with f . (ii) Let Y be any set, T any σ-algebra b of subsets of Y , and λ a probability measure with R domain Σ⊗T. Show that there is a family hµy iy∈Y b of probability measures on X such that λW = µy W −1 [{y}]ν(dy) for every W ∈ Σ⊗T, where ν is the marginal measure of λ on Y . (Hint: 452M.) (g) (i) Let X be any set, and Σ the countable-cocountable algebra on X. Show that (X, Σ) has the countably compact measure property. (ii) Show that any standard Borel space has the countably compact measure property. (h) Let (X, Σ) be a space with the perfect measure property, and T the smallest σ-algebra including Σ and closed under Souslin’s operation. Show that (X, T) has the perfect measure property. (i) Let X be a Radon Hausdorff space, and Σ the algebra of universally measurable sets in X (434D). Show that (X, Σ) has the countably compact measure property. > (j) Let hXi ii∈I be a family of real-valued random variables. Show that hXi ii∈I is independent iff its distribution is the product of the distributions of the Xi , and is then a quasi-Radon measure on RI . (Hint: 272G, 415E.) R R (k) Let I be a set and ν, ν 0 two quasi-Radon measures on RI such that eif (x) ν(dx) = eif (x) ν 0 (dx) for every continuous linear functional f : RI → R. Show that ν = ν 0 . > (l) Let Σ be the σ-algebra of subsets of C([0, ∞[) generated by the functionals f 7→ f (t) for t ≥ 0. Give C([0, ∞[) the topology Tc of uniform convergence on compact sets. (i) Show that Tc is Polish, and that Σ ∩ Tc is a base for Tc which generates Σ as σ-algebra. (ii) Use this to give a quick proof of 454O in this case. (m) Let T be a Polish space, and Tc the topology on C(T ) of uniform convergence on compact sets. Show that if T is any Hausdorff topology on C(T ), coarser than Tc , such that all the functionals f 7→ f (t), for t ∈ T , are Baire measurable for T, then (C(T ), T) is a measure-compact Radon space. 454Y Further exercises (a) In 454A, show that µ is weakly α-favourable (definition: 451U) if every µi is. (b) Let Σ be the algebra of Lebesgue measurable subsets of R. Show that (R, Σ) has the perfect measure property (454Xd) iff c is measure-free.

454 Notes


463

(c) Let Σ be the Borel σ-algebra of ω1 with its order topology. Show that (ω1 , Σ) has the perfect measure property. (Hint: 439Yf.) (d) Let (X, Σ, µ) be a semi-finite measure space with a topology such that µ is inner regular with respect to the closed sets, T a second-countable space and Y a separable metrizable space. Suppose that φ : X × T → Y is continuous in the second variable and measurable in the first, as in 454N. Show that µ is inner regular with respect to K = {K : K ⊆ X, φ¹K × T is continuous}. 454 Notes and comments 454A generalizes Theorem 451J, which gave the same result (with essentially the same proof) for product measures. One of the themes of this section is the idea that we can deduce properties of measures on product spaces from properties of their marginal measures, that is, the image measures on the factors. The essence of ‘compactness’, ‘countable compactness’ and ‘perfectness’ is that we can find enough points in the measure space to do what we want. (See, for instance, the characterization of compactness in 343B, or Pachl’s characterization of countable compactness in 452Ye.) Since the canonical feature of a product space is that we put in every point the Axiom of Choice provides us with, it’s perhaps not surprising that such properties can be inherited by measures on product spaces. Theorems 454C and 454D can be regarded as further variations on the same theme. A finitely additive non-negative functional on an algebra of sets will have an extension to a measure if, and only if, it is sequentially smooth in the sense that the measures of a decreasing sequence of sets with empty intersection converge to zero (413K). If we have a decreasing sequence of sets, with measures bounded away from zero, but with empty intersection, one interpretation of the phenomenon is that some points which ought to have been present got left out of the sets. What 454D tells us is that perfectness (and countable additivity) of the marginal measures is enough to ensure that there are enough points in the product to stop this happening. In effect, 454C tells us that it will be enough if every marginal but one is perfect. These results are of course associated with the projective limit constructions in 418M-418Q. In the theorems there we had Radon measures, so that they were actually compact rather than perfect; in return for the stronger hypothesis on the measures, we could handle projective limits corresponding to rather small subsets of the product spaces (see the formulae in 418O-418Q). Just as in §418, the patterns change when we have countable rather than uncountable families to deal with (418P-418Q, 454H). In 454J-454M, I insist rather arbitrarily that ‘the’ joint distribution of a family hXi ii∈I of real-valued random variables is the completion of a Baire measure on RI . Of course all the ideas can also be expressed in terms of the Baire measure itself, but I have sought a formulation which is consistent with the rules set out in §271. When I is countable, we get a Radon measure (454J(iii)), as in the finite-dimensional case. There are other cases in which the distribution is a quasi-Radon measure (454Xj). As always, we can ask whether the distribution is τ -additive; in this case it will have a canonical extension to a quasi-Radon measure (415N). Important examples of this phenomenon are described in 455D and 456K. Because RI has a linear topological space structure, we have a notion of ‘characteristic function’ for any probability measure on RI measuring the zero sets, and the characteristic function of a Baire measure determines that measure (454M, 454Xk). In 454O, C(T ), with Tc , has a countable network (4A2Oe), so the subspace measure µC induced by µ on C(T ) must be a τ -additive topological measure with respect to Tc (414Xc) and has a unique extension to a quasi-Radon measure on C(T ) (415M). The hard bit is the next step, showing that C(T ), under Tc , is a Radon space; this is the real point of 454N-454O. For the most important case, in which T = [0, ∞[, we have a useful simplification, because Tc is actually Polish (454Xl). Even in this case, however, we need to observe that the measure we are seeking is a little more complicated than a simple completion of a measure on RT . We must complete the subspace measure on C(T ), and C(T ) is far from being a measurable set. The measure µ ˜ of 454P will not as a rule be completion regular, for instance. Spaces of continuous functions are so important that it is worth noticing that the results here will be valid for various topologies on C(T ) (454Xm).

464


§455 intro.

455 Markov processes and Brownian motion For a ‘Markov’ process, in which the evolution of the system after a time t depends only on the state at time t, the general theory of §454 leads to a straightforward existence theorem (at least for random variables taking values in standard Borel spaces) dependent only on a natural consistency condition on the conditional distributions (455A). I apply this to the most important example, Brownian motion, in 455C. 455A Theorem Let T be a totally ordered set with least element a, and for each t ∈ T let (Xt , Σt ) be Q a standard Borel space. Set X = t∈T Xt and for each t ∈ T let πt : X → Xt be the coordinate map. Let µa be a probability measure with domain Σa . Suppose that we are given, for each pair s < t in T , a family (s,t) hνx ix∈Xs of probability measures on Xt all with domain Σt , and suppose that (s,t) (i) whenever s < t in T and E ∈ Σt , x 7→ νx (E) is Σs -measurable, (ii) whenever s < t < u in T , x ∈ Xs and E ∈ Σu , then R (t,u) (s,u) (s,t) νx (E) = νy (E)νx (dy). N Then there is a unique probability measure µ on X, with domain c t∈T Σt , such that µa is the image measure R (s,t) µπa−1 and, writing µt for the image measure on Xt , µt (E) = νx (E)µs (dx) for every E ∈ Σt whenever s < t in T . Q Moreover, if J ⊆ T is a finite set containing a, and we write π ˜J for the canonical map from X onto ZJ = t∈J Xt and µ ˜J for the image measure µ˜ πJ−1 , then for any µ ˜J -integrable real-valued function f we have Z ZZ ZZ n−1 ,tn ) f d˜ µJ = ... f (x0 , . . . , xn )νx(tn−1 (dxn ) n−2 ,tn−1 ) 1) νx(tn−2 (dxn−1 ) . . . νx(a,t (dx1 )µa (dx0 ) 0

where t0 = a, t1 , . . . , tn are the elements of J written in ascending order. N proof (a) For each finite set I ⊆ T , write TI = c t∈I Σt . If I = {t0 , t1 , . . . , tn } is a finite subset of T with a = t0 < t1 < . . . < tn , then we have a probability measure λI on ZI with domain TI such that Z ZZ ZZ n−1 ,tn ) f dλI = ... f (x0 , . . . , xn )νx(tn−1 (dxn ) n−2 ,tn−1 ) 1) νx(tn−2 (dxn−1 ) . . . νx(a,t (dx1 )µa (dx0 ) 0

for every λI -integrable function f . P P Use 454H on the finite sequence (Xt0 , . . . , Xtn ). The measures νz required by 454H must be constructed by the rule (t ,t

νz = νz(tmm )m+1

)

Q for m < n, z ∈ i≤m Xti , while of course ν∅ = µa . (Having a finite sequence rather than an infinite one clearly makes things easier; we can stop the argument at the end of part (b) of the proof of 454H.) Q Q (b) Of course the point of this is that these measures λI form a consistent family; if a ∈ I ⊆ J ∈ [T ]<ω , then the canonical projection πIJ : ZJ → ZI is inverse-measure-preserving. P P It is enough to consider the case in which J has just one more point than I, since then we can induce on #(J \ I). In this case, express J as {t0 , . . . , tn } where a = t0 < . . . < tn , and suppose that I = J \ {tm }. If W ∈ TI , then Z −1 λJ πIJ [W ] =

ZZ ...

Z =

ZZ ...

Z ...

n−1 ,tn ) χW (x0 , . . . , xm−1 , xm+1 , . . . , xn )νx(tn−1 (dxn ) m−1 ,tm ) . . . νx(tmm ,tm+1 ) (dxm+1 )νx(tm−1 (dxm ) . . . µa (dx0 )

g(x0 ,... ,xm−1 ) (xm+1 )νx(tmm ,tm+1 ) (dxm+1 ) m−1 ,tm ) νx(tm−1 (dxm ) . . . µa (dx0 )

where

(∗ )

455A

Markov processes and Brownian motion

Z

465

Z

g(x0 ,... ,xm−1 ) (xm+1 ) =

...

χW (x0 , . . . , xm−1 , xm+1 , . . . , xn ) n−1 ,tn ) m+1 ,tm+2 ) νx(tn−1 (dxn ) . . . νx(tm+1 (dxm+2 ).

Here, of course, we use the hypothesis (ii); since (t

m−1 νxm−1

,tm+1 )

(E) =

R

(t ,tm+1 )

νxmm

(t

m−1 (E)νxm−1

(t

,t

,tm )

(dxm )

)

m−1 m+1 for every E ∈ Σtm+1 , and since g(x0 ,... ,xm−1 ) is bounded and νxm−1 -integrable, Z m−1 ,tm+1 ) (dxm+1 ) g(x0 ,... ,xm−1 ) (xm+1 )νx(tm−1 ZZ m−1 ,tm ) = g(x0 ,... ,xm−1 ) (xm+1 )νx(tmm ,tm+1 ) (dxm+1 )νx(tm−1 (dxm )

(452F). Substituting this into (*) above, Z

ZZ

−1 λJ πIJ [W ] =

... Z

g(x0 ,... ,xm−1 ) (xm+1 ) m−1 ,tm ) νx(tmm ,tm+1 ) (dxm+1 )νx(tm−1 (dxm ) . . . µa (dx0 )

Z

m−1 ,tm+1 ) g(x0 ,... ,xm−1 ) (xm+1 )νx(tm−1 (dxm+1 ) . . . µa (dx0 ) Z Z Z = . . . . . . χW (x0 , . . . , xm−1 , xm+1 , . . . , xn )

=

...

n−1 ,tn ) m−1 ,tm+1 ) νx(tn−1 (dxn ) . . . νx(tm−1 (dxm+1 ) . . . µa (dx0 )

= λI W, applying the formula in (a) again. Q Q (Some of the formulae here are inappropriate if m = n. In this case, of course, Z −1 λJ πIJ [W ] =

Z ...

Z

=

Z ...

n−1 ,tn ) χW (x0 , . . . , xn−1 )νx(tn−1 (dxn ) . . . µa (dx0 ) n−2 ,tn−1 ) χW (x0 , . . . , xn−1 )νx(tn−2 (dxn−1 ) . . . µa (dx0 )

= λI W.) (c) We also see from the formulae in (a) that if J ⊆ T is of the form {t0 , . . . , tn } with n ≥ 1 and a = t0 < . . . < tn , then for any λJ -integrable function f Z ZZ ZZ n−1 ,tn ) f dλJ = ... f (x0 , . . . , xn )νx(tn−1 (dxn ) ZZ =

n−2 ,tn−1 1) νx(tn−2 (dxn−1 ) . . . νx(a,t (dx1 )µa (dx0 ) 0 n−1 ,tn ) f (x0 , . . . , xn )νx(tn−1 (dxn )λI (d(x0 , . . . , xn−1 ))

where I = {t0 , . . . , tn−1 }. (d) Part (b) tells us that of measures on the finite products ZJ , and therefore N we have a consistent family −1 have a functional λ on Σ defined by setting λ˜ π t J [W ] = λJ W for every finite J ⊆ T containing a t∈T N and W ∈ Σ , writing π ˜ : X → Z for the canonical maps. λ is finitely additive, and its images t J J t∈J µt = λπt−1 are all countably additive because they are also images of measures λ{a,t} . So we have a measure µ extending λ, by 454F. We have to check that each λJ is the image measure µ˜ πJ−1 ; but this is true because

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455A

N they agree on t∈J Σt (using the Monotone Class Theorem in the form 136C, as always). So the integral formula sought for µ ˜J is just that obtained in part (a) for λJ . (e) Accordingly, if s < t, we can calculate µt E, for E ∈ Σt , as follows: setting J = {a, s, t}, I = J \ {t}, we have µt E = µπt−1 [E] = µ˜ πJ−1 [F ] (where F = {z : z ∈ ZJ , z(t) ∈ E}) Z (s,t) = λJ F = νz(s) (E)λI (dz) (by (c))

Z =

Z (s,t) νw(s) (E)µ(dw)

=

νx(s,t) (E)µs (dx).

455B Remarks (a) I noted in 454I that in theorems of this type we do not really need to know that the factors (Xt , Σt ) are standard Borel spaces; what counts is that they should have the ‘perfect measure property’ (454Xd). In the present context, we may be able to do even better, since all that matters is that the marginal measures µt should be perfect, and we have a direct description of them from the formula R (a,t) µt = νx µa (dx). So if, for instance, we can show from this presentation that all the µt are tight (that is, inner regular with respect to compact sets), then we need no further control on the algebras Σt . (b) The formula µt (E) =

R

(s,t)

νx

(E)µs (dx) for every E ∈ Σt whenever s < t in T

in 455A is the defining property of a Markov process. In something closer to probabilists’ language, it comes out as ¡ (s,t) ¢ (s,t) Pr(Xt ∈ E) = E νXs (E) where νx (E) = Pr(Xt ∈ E|Xs = x). The idea is that hXt it∈T is a family of random variables, and that we start from the assurance that ‘history is irrelevant’; if, at time s, we wish to make guesses about the behaviour of Xt , the state of the system at a future time t, then we expect that it will be useful to look at the current state Xs , but once we know the value of Xs then any further information about Xs0 for s0 < s will tell us nothing more about Xt . (s,t) We are given the distributions νx , which can be thought of as the conditional distributions of Xt given that Xs = x, or (in the language of §452) as a disintegration of the joint distribution of (Xs , Xt ) over the distribution of Xs . The condition (ii) of 455A is plainly necessary if the system is going to make sense at all; the content of the theorem is that this condition is also sufficient (at least when the random variables Xt are taking values in standard spaces) to ensure that the system as a whole can indeed be represented as a family of random variables, in Kolmogorov’s sense, on a suitable probability space. 455C Brownian motion As an example of the use of 455A I give a first existence proof for Brownian motion or Wiener measure. Theorem There are a probability space (Ω, Σ, µ) and a family hXt it≥0 of real-valued random variables on Ω such that (i) X0 = 0 almost everywhere; (ii) whenever 0 ≤ s < t then Xt − Xs is normally distributed with expectation 0 and variance t − s; (iii) whenever 0 ≤ t1 < . . . < tn , then Xt2 − Xt1 , Xt3 − Xt2 , . . . , Xtn − Xtn−1 are independent. proof (a) In 455A, take T = [0, ∞[, so that a = 0; take R for each of the factor spaces, so that X = hXt it≥0 becomes a name for the identity function on the product Ω = RT (instead of the name of the product itself!), and each Σt is the Borel σ-algebra B of R. Let µ0 be the Borel probability measure on R which gives (s,t) measure 1 to {0}, and whenever 0 ≤ s < t, x ∈ R let νx be the distribution of a normal random variable with expectation x and variance t − s. (Strictly speaking, since elsewhere I insist that a distribution is a (s,t) Radon measure, and that a Radon measure is complete, νx must be the restriction of such a distribution to B.)

455C


467

(b) The condition (s,t)

whenever 0 ≤ s < t and E ∈ B, x 7→ νx

(E) is B-measurable,

in 455A is elementary; (s,t)

νx

1 (E) = √

R

2π(t−s)

2

E

e−(y−x)

/2(t−s)

dy

(274Ad) is actually a continuous function of x, because the integrand is continuous as a function of two variables (and is dominated by an integrable function as x varies over any bounded interval). As for the condition R (t,u) (s,u) (s,t) whenever 0 ≤ s < t < u, x ∈ R and E ∈ B, then νx (E) = νy (E)νx (dy), this is really just the theorem that the sum of independent normal random variables is normal (274B). For if U , V are independent normal random variables, U has expectation x and variance t − s, and V has expectation 0 and variance u − t, then U + V will be normal with expectation x and variance u − s, so that (s,u) (s,t) (t,u) νx (E) = Pr(U + V ∈ E). On the other side of the equation, νx is the distribution of U , while νy is the distribution of V + y; so that

(t,u)

νx(s,u) (E) = Pr(U + V ∈ E) = (νx(s,t) ∗ ν0 (272S)

)(E)

Z (t,u)

= (257Xb with 257D, or 444A)

ν0

(E − y)νx(s,t) (dy)

Z =

νy(t,u) (E)νx(s,t) (dy),

as required. (c) Thus the conditions of 455A are satisfied, and we get a probability measure µ on Ω = R[0,∞[ . The conclusion µa is the image measure µπa−1 says just that Pr(X0 = 0) = 1. If s < t and E ∈ B, then ZZZ χE(z − y)νy(s,t) (dz)νx(0,s) (dy)µ0 (dx)

Pr(Xt − Xs ∈ E) = Z

(0,s)

νy(s,t) (E + y)ν0

=

(dy)

Z =

(s,t)

(s,t)

ν0

(0,s)

(E)ν0

(s,t)

(dy) = ν0

(E),

so Xt − Xs has distribution ν0 , that is, is normally distributed with expectation 0 and variance t − s. Similarly, if 0 ≤ t1 < . . . < tn and E1 , . . . , En−1 ∈ B, then

468


455C

Pr(Xti+1 − Xti ∈ Ei for 1 ≤ i < n) ZZ Z = . . . χE1 (x2 − x1 )χE2 (x3 − x2 ) . . . χEn−1 (xn − xn−1 ) ZZ =

n−1 ,tn ) 1) (dxn ) . . . νx(0,t (dx1 )µ0 (dx0 ) νx(tn−1 0

Z ...

ZZZ =

χ(E1 + x1 )(x2 )χ(E2 + x2 )(x3 ) . . . χ(En−1 + xn−1 )(xn ) ZZ

...

n−1 ,tn ) 1) νx(tn−1 (dxn ) . . . νx(0,t (dx1 )µ0 (dx0 ) 0 n−1 ,tn ) (dxn ) χ(En−1 + xn−1 )(xn )νx(tn−1 n−2 ,tn−1 ) χ(En−2 + xn−2 )(xn−1 )νx(tn−2 (dxn−1 ) 1) . . . χ(E1 + x1 )(x2 )νx(t11 ,t2 ) (dx2 )νx(0,t (dx1 )µ0 (dx0 ) 0

(t

= ν0 n−1 =

n−1 Y

,tn )

(t

(En−1 )ν0 n−2

,tn−1 )

(t ,t2 )

(En−2 ) . . . ν0 1

(E1 )

Pr(Xi+1 − Xi ∈ Ei ).

i=1

So Xt2 − Xt1 , . . . , Xtn − Xtn−1 are independent. 455D This construction of Brownian motion is sufficient to show that there is a the defining conditions (i)-(iii), which can be studied with the tools of measure theory. the most important property of Brownian motion, that ‘sample paths are continuous’. inequalities from Chapter 27 and the ideas of 454N-454P, we can find a proof of this, as

process, satisfying It gives no hint of With some simple follows.

Theorem In 455C, we can take Ω to be C([0, ∞[) and µ to be a Radon measure when C([0, ∞[) is given the topology Tc of uniform convergence on compact sets, while Xt (f ) = f (t) for f ∈ C([0, ∞[). proof (a) The construction in 455C in fact gives us Ω = R[0,∞[ , Xt (f ) = f (t) for f ∈ R[0,∞[ , and a Baire measure µ. The main part of the argument here (down to the end of (e)) is devoted to showing that µ∗ C([0, ∞[) = 1; the result will then follow easily from 454Pb. (b) ?? Suppose, if possible, that µ∗ C([0, ∞[) < 1. Then there is a non-negligible Baire set H ⊆ R[0,∞[ \ C([0, ∞[). There is a countable set D ⊆ [0, ∞[ such that H is determined by coordinates in D (4A3Nb); we may suppose that D includes Q ∩ [0, ∞[. (c) (The key.) Let q, q 0 be rational numbers such that 0 ≤ q < q 0 , and ² > 0. Then Pr(supt∈D∩[q,q0 ] |Xt − Xq | ≥ ²) ≤

√ 18 q 0 −q −²2 /18(q 0 −q) √ e . ² 2π

P P If q = t0 < t1 < . . . < tn = q 0 , set Yi = Xti − Xti−1 for 1 ≤ i ≤ n, so that Xtm − Xq = 1 ≤ m ≤ n, and Y1 , . . . , Yn are independent. By Etemadi’s lemma (272U), 1 3

Pr(sup |Xti − Xq | ≥ ²) ≤ 3 max Pr(|Xti − Xq | ≥ ²) i≤n

i≤n

1

²

|Xti − Xq | ≥ √ ) = 3 max Pr( √ ti −q 3 ti −q 1≤i≤n Z ∞ 2 1 = 6 max √ e−x /2 dx √ 1≤i≤n

1 Xti ti −q

(because √

2π

²/3 ti −q

− Xq is standard normal) Z

= ≤

∞

2 6 √ e−x /2 dx 2π ²/3√q 0 −q √ 18 q 0 −q −²2 /18(q 0 −q) √ e ² 2π

Pm i=1

Yi for

455D


469

by 274Ma. Thus if I ⊆ [q, q 0 ] is any finite set containing q and q 0 , Pr(supt∈I |Xt − Xq | ≥ ²) ≤

√ 18 q 0 −q −²2 /18(q 0 −q) √ e . ² 2π

Taking hIn in∈N to be a non-decreasing sequence of finite sets with union D ∩[q, q 0 ], starting from I0 = {q, q 0 }, we get Pr(

sup

t∈D∩[q,q 0 ]

|Xt − Xq | ≥ ²) = lim Pr(sup |Xt − Xq | ≥ ²) n→∞

≤

t∈In

√ 18 q 0 −q −²2 /18(q 0 −q) √ e , ² 2π

as required. Q Q (d) If ² > 0 and n ≥ 1, then Pr(there are t, u ∈ D ∩ [0, n] such that |t − u| ≤ 2

18n ≤ √ e−n

2 2

² /18

² 2π

1 n2

and |Xt − Xu | ≥ 3²)

.

P P Divide [0, n] into n3 intervals [qi , qi+1 ] of length 1/n2 . For each of these, Pr(supt∈D∩[qi ,qi+1 ] |Xt − Xqi | ≥ ²) ≤

2 2 18 √ e−n ² /18 . n² 2π

So Pr(there are i < n3 , t ∈ D ∩ [qi , qi+1 ] such that |Xt − Xqi | ≥ ²) 2

18n is at most √ e−n ² 2π

2 2

² /18

.

But if t, u ∈ [0, n] and |t − u| ≤ 1/n2 and |Xt − Xu | ≥ 3², there must be an i < n3 such that both t and u belong to [qi , qi+2 ], so that either there is a t0 ∈ D ∩ [qi , qi+1 ] such that |Xt0 − Xqi | ≥ ² or there is a t0 ∈ D ∩ [qi+1 , qi+2 ] such that |Xt0 − Xqi+1 | ≥ ². So Pr(there are t, u ∈ D ∩ [0, n] such that |t − u| ≤

1 n2

and |Xt − Xu | ≥ 3²)

≤ Pr(there are i < n3 , t ∈ D ∩ [qi , qi+1 ] such that and |Xt − Xqi | ≥ ²) 2

18n ≤ √ e−n ² 2π

2 2

² /18

,

as required. Q Q (e) So if we take G²n to be the Baire set {f : f ∈ R[0,∞[ , there are t, u ∈ D ∩ [0, n] such that |t − u| ≤

1 n2

and |f (t) − f (u)| ≥ 3²}, we have 2

18n µG²n ≤ √ e−n ² 2π

2 2

² /18

,

and limn→∞ µG²n = 0. We can therefore find a Sstrictly increasing sequence hnk ik∈N in N such that P ∞ k≥1 G1/k,nk . k=1 µ(G1/k,nk ) < µH, so that there is an f ∈ H \ What this means is that if k ≥ 1 and t, u ∈ D ∩ [0, nk ] are such that |t − u| ≤

1 , n2k

3 k

then |f (t) − f (u)| < .

Since nk → ∞ as k → ∞, there is a continuous function f 0 such that f 0 ¹D = f ¹D. But H is determined by coordinates in D, so f 0 belongs to H ∩ C([0, ∞[), which is supposed to be empty. X X (f ) Thus µ∗ C([0, ∞[) = 1. By 454Pb, we have an extension of µ to a measure µ ˜ such that µ ˜C([0, ∞[) = 1 and the subspace measure µ ˜C is a Radon measure for Tc . If we interpret the random variables Xt (f ) = f (t)

470


455D

as functions defined on the probability space (R[0,∞[ , µ ˜), they still have the properties (i)-(iii) of 455C, because µ ˜ extends µ; and since C([0, 1[) is µ ˜-conegligible, their restrictions to C([0, ∞[), regarded as random variables on the probability space (C([0, ∞[), µ ˜C ) will also have these properties, as required. 455X Basic exercises (a) In 455A, set T = [0, ∞[, let each (Xt , Σt ) be R with its Borel σ-algebra, let (s,t) µ0 be any atomless Borel probability measure on R, and for x ∈ R, 0 ≤ s < t let νx be the Borel measure on R giving measure 1 to {ψ(x, t − s)}, where ψ(x, t) =

x 1−xt

if xt 6= 1 and x 6= 0,

= 0 if xt = 1, 1 t

= − if x = 0. Show that the conditions of 455A are satisfied, that the measure µ constructed in 455A is a Baire measure on RT , and that µ is not τ -additive. (Hint: setting φ(x)(t) = ψ(x, t), show that φ : R → RT is inversemeasure-preserving for µ0 and µ, and that every point in RT has a neighbourhood of zero measure.) R (s,t) (b) Suppose, in 455A, that every (Xt , Σt ) is R Rwith its Borel σ-algebra, and that yνx (dy) = x R whenever s < t in T and x ∈ R. Suppose also that yµ0 (dy) = α. Show that πt dµ = α for every t ∈ T (that is, that all the random variables πt : X → R have the same expectation). (s,t)

(c) Suppose, in 455C,P we set νx (E) = λt−s (E − x), where λt is a Poisson distribution with expectation t, that is, λt (E) = e−t m∈E∩N tm /m! (cf. 285Q, 285Xo). Show that we can again use 455A to build a measure on Ω = R[0,∞[ , and that as in 455C the increments Xt − Xs over non-overlapping intervals are independent. (s,t)

(d) Show that, in 455A, we can set νx (E) = λt−s (E − x), where λt is a Cauchy distribution with centre 0 and scale parameter t (285Xm). Show that we obtain a measure on Ω = R[0,∞[ , such that, as in 455C, the increments Xt − Xs over non-overlapping intervals are independent. (e) (i) The standard gamma distribution with parameter γ is the probability distribution λγ on R with probability density function t 7→

1 γ−1 −t t e Γ(γ)

for t > 0. Show that its expectation is γ. (Hint:

225Xj(iv).) Show that its variance is γ. (ii) Show that if X and Y are independent random variables with standard gamma distributions, then X + Y has a standard gamma distribution. (Hint: 272T, 252Yk.) (iii) R∞ 1 1 Show that limγ↓0 γΓ(γ) = 1, so that limγ↓0 λγ [1, ∞[ = 1 e−t > 0. (iv) Show that, in 455A, we can set γ

t

(s,t) νx (E)

= λt−s (E − x) to get a process in which increments over non-overlapping intervals are independent, the set of non-decreasing functions has full outer measure, and the set of continuous functions is negligible. (v) Show that this process can be represented by a Radon measure on R[0,∞[ . (Hint: 438P, 432D.) (f ) Use 272Ye to simplify the formulae in the proof of 455D. 455Y Further exercises (a) Suppose, in 455A, that T = N or T = [0, ∞[, that every (Xt , Σt ) is R with (s,t) (0,t−s) its Borel σ-algebra, and that νQ (E) = ν0 (E − x) whenever x ∈ R, E ∈ B and s < t. Take t ∈ T and x let µ ˜[0,t] be the image of µ on 0≤s≤t Xs . Show that we have a disintegration of µ over µ ˜[0,t] defined by the formula µW = where for g ∈

Q 0≤s≤t

Xs , f ∈

Q s∈T

R

µ{f : g a f ∈ W }˜ µ[0,t] (dg),

Xs we write (g a f )(s) = g(s) if s ≤ t, g(t) + f (s − t) if s ≥ t.

(b) Let hXn in∈Z be a double-ended sequence of real-valued random variables such that (i) for each n ∈ Z, Yn = Xn+1 − Xn is independent of {Xi : i ≤ n} (ii) hYn in∈Z is identically distributed. Show that the Yn are essentially constant. (Hint: 285Yc.)

455 Notes


471

S (c) Write Dn for {2−n i : i ∈ N} and D = n∈N Dn , the set of dyadic rationals in [0, ∞[. For d ∈ D define fd ∈ C([0, ∞[) as follows. If n ∈ N, fn (t) = 0 if t ≤ n, t − n if n ≤ t ≤ n + 1, 1 if t ≥ n + 1. If 1 (1 − 2n |t − d|)) 2n+1 P

d = 2−n k where n ≥ 1 and k ∈ N is odd, fd (t) = max(0, √

for t ≥ 0. Now let hZd id∈D be

an independent family of standard normal distributions, and set Gn (t) = d∈Dn fd (t)Zd for t ≥ 0, so that each Gn is a random continuous function on [0, ∞[. Show that for any n ∈ N, ² > 0, 2 2π

Pr(supt∈[0,n] |Gn+1 (t) − Gn (t)| ≥ ²) ≤ 2n n · √

R∞

√ 2² 2n

2

e−x

/2

dx,

and hence that hGn in∈N converges almost surely to a continuous function. Explain how to interpret this as a construction of the measure on C([0, ∞[) described in 455D. (d) Let hλt it∈]0,∞[ be any family of probability distributions on R such that λs ∗ λt = λs+t for all s, (s,t) t > 0. Show that we can set νx (E) = λt−s (E − x) in 455A to obtain a measure on R[0,∞[ such that the increments over non-overlapping intervals are independent. (e) In 455Xc, show that the measure µ obtained on R[0,∞[ is a Baire measure, and has an extension to a Radon measure µ ˜ on R[0,∞[ such that µ ˜Q = 1, where Q is the set of non-decreasing functions f : [0, ∞[ → N such that limt↓0 f (t) = 0, lims↓t f (s) − lims↑t f (s) ≤ 1 for every t ≥ 0, and limt→∞ f (t) = ∞. (Hint: show that Q is a K-analytic space with metrizable compact sets, so is a Radon space.) (f ) In 455Xd, show that the measure µ obtained on R[0,∞[ is a Baire measure, and that C([0, ∞[) is negligible. (Hint: estimate Pr(|X(i+1)/n − Xi/n | ≤ ² for every i < n).) (g) Let µ be the Radon measure on C([0, ∞[) representing Brownian motion, and γ > limt→∞ t−γ f (t) = 0 for µ-almost every f ∈ C([0, ∞[).

1 2.

Show that

455 Notes and comments Obviously the principal applications of 455A will be with the totally ordered sets N and [0, ∞[, as in 454H and 455C. I use the letters t, T to suggest the probabilistic intuitions behind these results; we think of the spaces Xt in 455A as being the sets of possible states of a system at ‘time’ t, (s,t) are descriptions of how we believe the system is likely to evolve between times s so that the measures νx and t, given that it is in state x at time s. In the case of ‘discrete time’, when we observe the system only at clearly separated moments, it is easy to handle non-Markov processes, in which evolution between times (n−1,n) n − 1 and n can depend on the whole history up to time n − 1; thus in 454H the measures νz = νz Q (n−1,m) for any m > n. are defined for every z ∈ i
472


455 Notes

Perhaps it is worth recalling here that a Baire measure on a completely regular space has at most one quasi-Radon measure extending it (415G). The step from 455C to 455D amounts to showing that Brownian motion, interpreted as a Baire probability measure on R[0,∞[ , has an extension to a Radon measure µ ˜ on R[0,∞[ for which C([0, ∞[) is conegligible, so that we have the option of regarding µ ˜ as a Radon measure on C([0∞[); by 454P or 454Xl, we can give C([0, ∞[) either the topology of pointwise convergence or the topology of uniform convergence on compact sets. The point I wish to make is that the step from µ to µ ˜ is a canonical one and worth performing as soon as we know that µ is τ -additive, whether or not we come to a measure living on the continuous functions. In 455Xe, for instance, µ ˜ gives measure 1 to the set of non-decreasing functions, and this is clearly an essential property of the process. There are of course many routes to the fundamental theorems 455C-455D, and I do not claim that the one described here is the ‘best’. I use it because it is an instructive example of the class of problems which led to the theorems of this section. The excursion through a measure on R[0,∞[ is by no means necessary; in 455Yc I sketch an alternative method (‘Lévy’s construction’) which avoids it. Generally it is true that interesting stochastic processes have much stronger properties than those described in 455A. Any construction using these abstract methods should be regarded as provisional; it may be a useful, even necessary, first confirmation that there is something there to work on, but we are likely to need much more penetrating techniques to form a proper conception of what we have found. The general argument above does however provide a clear description of what is necessary for the ‘independent increments’ property of Brownian motion: the fact that if t1 < t2 < . . . < tn then Xt2 − Xt1 , . . . , Xtn − Xtn−1 are independent. We need the Markov property (so that, given Xtn−1 , Xtn is independent of previous history) (s,t) (s,t) (s,t) and the translation property of the νx , encapsulated in the formula νx (E) = ν0 (E −x), which assures (s,t) us that the distribution of Xt − Xs is ν0 (so that Xs , Xt − Xs are independent). It also has a place for (s0 ,t0 ) (s,t) whenever t0 − s0 = t − s (455Ya). And the = νx ‘temporal homogeneity’, that is, the fact that νx method does at least make it clear which property of normal distributions we need (part (b) of the proof of 455C), so leads us to examine other important processes (455Xc-455Xe, 455Yd). These are all based on `ve 77 or families of ‘infinitely divisible’ distributions; for the general theory of such distributions, see Loe Fristedt & Gray 97.

456 Gaussian distributions Uncountable powers of R are not as a rule measure-compact (439P, 455Xa). Accordingly distributions, in the sense of 454K, need not be τ -additive. But, as we have seen in 455D, some, at least, of the distributions most important to us are indeed τ -additive, and therefore have interesting canonical extensions. This section is devoted to a remarkable result, taken from Talagrand 81, concerning a class of distributions which are of great importance in probability theory. It demands a combination of techniques from classical probability theory and from the topological measure theory of this volume. I begin with the definition and fundamental properties of what I call ‘centered Gaussian distributions’ (456A-456E). These are minor adaptations of the classical finite-dimensional theory. Another relatively easy idea is that of ‘universal’ Gaussian distribution (456F-456H). In 456I we come to a much deeper result, a step towards classifying the ways in which a Gaussian family of n-dimensional random variables can accumulate at 0. The ideas are combined in 456J-456K to complete the proof of Talagrand’s theorem that Gaussian distributions on powers of R are τ -additive. 456A Definitions (a) Write µG for the Radon probability measure on R which is the distribution of a 1 standard normal random variable, that is, the probability distribution with density function x 7→ √ e−x

2

/2

2π

(I)

(274A). For any set I, write µG for the measure on RI which is the product of copies of µG ; this is always quasi-Radon (415E/453J); if I is countable, it is Radon (417Q); if I = n ∈ N \ {0}, it is the probability distribution with density function x 7→ (2π)−n/2 e−x . x/2 (272I); if I = ∅, it is the unique probability measure on the singleton set R∅ . (b) I will use the phrase centered Gaussian distribution to mean a measure µ on a power RI of R such that µ is the completion of a Baire measure (that is, is a distribution in the sense of 454K) and every

456B

Gaussian distributions

473

continuous linear functional f : RI → R is either zero almost everywhere or is a normal random variable with zero expectation. (Note that I call the distribution concentrated on the point 0 in RI a ‘Gaussian distribution’.) I (c) If I is a set and µ R is a centered Gaussian distribution on R , its covariance matrix is the family hσij ii,j∈I where σij = x(i)x(j)µ(dx) for i, j ∈ I. (The integral is always defined and finite because each function x 7→ x(i) is either essentially constant or normally distributed, and in either case is squareintegrable.)

456B I start with some fundamental facts about Gaussian distributions. Proposition (a) Suppose that I and J are sets, µ is a centered Gaussian distribution on RI , and T : RI → RJ is a continuous linear operator. Then there is a unique centered Gaussian distribution on RJ for which T is inverse-measure-preserving; if J is countable, this is the image measure µT −1 . (b) Let I be a set, and µ, ν two centered Gaussian distributions on RI . If they have the same covariance matrices they are equal. (I) (c) For any set I, µG is the centered Gaussian distribution on RI with the identity matrix for its covariance matrix. (d) Suppose that I is a countable set. Then a measure µ on RI is a centered Gaussian distribution iff it (N) is of the form µG T −1 where T : RN → RI is a continuous linear operator. (e) Suppose hIj ij∈J is a disjoint family of sets with union I, and that for each j ∈ J we have a centered Gaussian distribution νj on RIj . Then the product ν of the measures νj , regarded as a measure on RI , is a centered Gaussian distribution. (f) Let I be any set, µ a centered Gaussian distribution on RI and E ⊆ RI a set such that µ measures E. Writing −E = {−x : x ∈ E}, µ(−E) = µE. proof (a)(i) For Baire sets F ⊆ RJ , set νF = µT −1 [F ]; this is always defined because T is continuous (4A3Kc). This makes ν a Baire probability measure on RJ for which T is inverse-measure-preserving. Because µ is complete, T is still inverse-measure-preserving for the completion νˆ of ν (235Hc). If g : RJ → R is a continuous linear functional, so is gT : RI → R; now ν{y : g(y) ≤ α} = µ{x : gT (x) ≤ α} for every α, so g and gT have the same distribution, and are both either zero a.e. or normal random variables. As g is arbitrary, νˆ is a centered Gaussian distribution as defined in 456Ab. Of course it is the only such distribution on RJ for which T is inverse-measure-preserving. (ii) Now suppose that J is countable. Then RJ is is Polish (4A2Qc), so ν is a Borel measure and νˆ is a Radon measure (433Cb). RJ has a countable network consisting of Borel sets, µ is perfect (454A(b-iii)) and totally finite, and T is measurable (418Bd), so µT −1 is a Radon measure (451N). Thus νˆ and µT −1 are Radon measures agreeing on the Borel sets and must be equal. (b) The point is that µf −1 = νf −1 for every continuous linear functional f : RI → R. P P By (a), µf −1 and νf −1 are Radon measures on R, and by the definition of ‘Gaussian distribution’ each is either a normal P distribution with expectation zero, or is concentrated at 0. By 4A4Be, we can express f in the form f (x) = i∈I βi x(i) for every x ∈ RI , where {i : βi 6= 0} is finite. In this case Z Z Z X t2 (µf −1 )(dt) = f (x)2 µ(dx) = βi βj x(i)x(j)µ(dx) =

X

Z βi βj

i,j∈I

x(i)x(j)ν(dx) =

Z t2 (νf −1 )(dt)

i,j∈I

because µ and ν have the same covariance matrices. But this means that µf −1 and νf −1 have the same variance; if this is zero, they both give measure 1 to {0}; otherwise, they are normal distributions with the same expectation and the same variance, so again are equal. Q Q By 454M, µ = ν. (I)

(c) Being a completion regular quasi-Radon probability measure (415E), µG is the completion of a Baire probability measure on RI . If f : RI → R is a continuous linear functional, then it is expressible in the

474


456B

P form f (z) = i∈I βi z(i), where J = {i : βi 6= 0} is finite. I need to show that f is either zero a.e. or a normal random variable with expectation 0. If J = ∅ then f = 0 everywhere and we can stop. Otherwise, P (I) f = i∈J βi πi , where πi (x) = x(i) for i ∈ I, x ∈ RI . Now, with respect to the measure µG , hπi ii∈J is an independent family of normal random variables with zero expectation (272G). So hβi πi ii∈J is independent P (272E), and βi πi is normal for i ∈ J (274Ae). By 274B, f = i∈J βi πi is normal, and of course it has zero (I) expectation. As f is arbitrary, µG is a centered Gaussian distribution. We have R R (I) x(i)x(i)µG (dx) = t2 µG (dt) = 1 for i ∈ I, and

R

(I)

x(i)x(j)µG (dx) =

R

tµG (dt) ·

R

tµG (dt) = 0

(J)

if i, j ∈ I are distinct. So the covariance matrix of µG is the identity matrix. By (b), it is the only centered Gaussian distribution with this covariance matrix. (N)

(d)(i) It follows from (c) and (a) that if µ = µG T −1 , where T : RN → RI is a continuous linear operator, then µ is a centered Gaussian distribution. (ii) Now suppose that µ is a centered Gaussian distribution on RI . Set πi (x) = x(i) for i ∈ I and x ∈ RI ; for i ∈ I, set ui = πi• in L2 = L2 (µ). By 4A4Jg, there is a countable orthonormal family hvj ij∈J in L2 such that every vj is a linear combination of the uP i , and every ui is a linear combination of the vj . We may suppose that J ⊆ N. For i ∈ I, express ui as j∈J αij vj , where {j : αij 6= 0} is finite. Define P T : RN → RI by setting (T z)(i) = j∈J αij z(j) for every z ∈ RN , i ∈ I. Then T is a continuous linear (N)

functional. Set ν = µG T −1 , so that ν is a centered Gaussian distribution on RI , by (a). Because RI is Polish, both µ and ν must be Radon measures. (iii) Now µ and ν have the same covariance matrices. P P If i, i0 ∈ I then Z X x(i)x(i0 )µ(dx) = (ui |ui0 ) = αij αi0 j 0 (vj |vj 0 ) j,j 0 ∈J

=

X j∈J

Z =

X

αij αi0 j =

Z αij αi0 j 0

j,j 0 ∈J

(N)

z(j)z(j 0 )µG (dz) Z

(N)

(T z)(i)(T z)(i0 )µG (dz) =

x(i)x(i0 )ν(dx). Q Q

By (b), µ = ν is of the required form. (e) We must first confirm that ν is the completion of a Baire measure. P P If we write Ba(RIj ) for the Baire Ij σ-algebra of R , then each νj is the completion of its restriction νj ¹Ba(RIj ), so ν is also the product of the N measures νj ¹Ba(RIj ) (254I), and is therefore the completion of its restriction to c j∈J Ba(RIj ) (254Ff). But N N N as Ba(RIj ) = c Ba(R) for every j (4A3Na), c Ba(RIj ) can be identified with c Ba(R) = Ba(RI ), i∈Ij

j∈J

i∈I

so that ν is indeed the completion of ν¹Ba(RI ). Q Q Now P suppose that f : RI → R is a continuous linear functional. Then we can express f in the form f (x) = i∈K αi x(i) for every x ∈ RI , where K ⊆ I is finite. Set L = {j : K ∩ Ij 6= ∅} and Kj = K ∩ Ij for P P j ∈ L, so that L and every Kj are finite; for j ∈ L, x ∈ RI set fj (x) = i∈Kj αi x(i). Now f = j∈L fj . P If we set gj (y) = i∈Kj αi y(i) for y ∈ RIj , then gj is either zero a.e. or a normal random variable with respect to the probability measure νj . Since ν{x : fj (x) ≤ α} = ν{x : gj (x¹Ij ) ≤ α} = νj {y : gj (y) ≤ α} for every α ∈ R, fj (regarded as a random variable on (RI , ν)) has the same distribution as gj (regarded as a random variable on (RIjP , νj )). This is true for every j ∈ L. Moreover, the different fj , as j runs over L, are independent. So f = j∈L fj is the sum of independent random variables which are all either normal or essentially constant. By 274B again, f also is either normal or essentially constant. And of course its expectation is zero. As f is arbitrary, this shows that ν is a centered Gaussian distribution.

456C


475

(f ) Set T x = −x for x ∈ RI , so that T is a continuous linear operator and we have a unique centered Gaussian distribution ν on RI such that T is inverse-measure-preserving for µ and ν, by (a). For any i, j ∈ I, R R R x(i)x(j)ν(dx) = (T x)(i)(T x)(j)µ(dx) = x(i)x(j)µ(dx), so µ and ν have the same covariance matrices and are equal, by (b). Accordingly µ(−E) = µT −1 [E] = νE = µE whenever µ measures E. 456C Lemma Let I be a finite set and µ a centered Gaussian distribution on RI . Suppose that γ ≥ 0 and α = µ{x : supi∈I |x(i)| ≥ γ}. Then µ{x : supi∈I |x(i)| ≥ 12 γ} ≥ 2α(1 − α)3 . proof The case γ = 0, α = 1 is trivial; suppose that γ > 0. We may suppose that I has a total order ≤. Give RI×4 ∼ = (RI )4 the product λ of four copies of µ; then λ is a centered Gaussian distribution (456Be). P3 Define T : RI×4 → RI by setting (T y)(i) = 21 r=0 y(i, r) for i ∈ I, y ∈ RI×4 ; then λT −1 is a centered Gaussian distribution on RI (456Ba). Now λT −1 has the same covariance matrix as µ. P P If i, j ∈ I then Z

Z x(i)x(j)(λT −1 )(dx) = =

(because the map y 7→ hy(i, r)ii∈I : R

(T y)(i)(T y)(j)λ(dy) = 1 4

3 Z X

=

3 X 3 Z X

y(i, r)y(j, s)λ(dy)

r=0 s=0

y(i, r)y(j, r)λ(dy) =

r=0 I×4

Z

1 4

1 4

3 Z X

x(i)x(j)µ(dx)

r=0

→ RI is inverse-measure-preserving for each r) x(i)x(j)µ(dx). Q Q

So λT −1 = µ, by 456Bb. Define Eir = {y : y ∈ RI×4 , |y(i, r)| ≥ γ} for r < 4 and i ∈ I, and Er =

S i∈I

Eir

for r < 4, so that Set

Er0

= Er \

λEr = λ{y : supi∈I |y(i, r)| ≥ γ} = µ{x : supi∈I |x(i)| ≥ γ} = α.

S s6=r

Es , so that λEr0 = λ{y : sup |y(i, r)| ≥ γ, sup |y(i, s)| < γ for s 6= r) i∈I i∈I Y λ{y : sup |y(i, s)| < γ} = λ{y : sup |y(i, r)| ≥ γ} · i∈I

i∈I

s6=r

(because these are independent events) = α(1 − α)3 . 0 Next, for i ∈ I and r < 4, set Eir = Eir \ ( 0 hEir ii∈I,r<4 is disjoint. Set

S j
Ejr ∪

0 Fir = {y : y ∈ Eir , y(i, r)

S s6=r

P s6=r

Es ), so that Er0 =

S i∈I

0 Eir . Observe that

y(i, s) ≥ 0}.

0 Then νFir ≥ 12 νEir . P P We can think of RI×4 as a product RJ × RK , where J = I × {r} and K = I × (4 \ {r}). In this case, λ becomes identified with a product λJ × λK , where λJ and λK are centered 0 Gaussian distributions on RJ and RK respectively, and Eir is of the form V × W , where

476


456C

V = {v : v ∈ RJ , |v(i, r)| ≥ γ, |v(j, r)| < γ for j < i}, W = {w : w ∈ RK , |w(j, s)| < γ for every j ∈ I, s 6= r}. In the same representation, Fir becomes (V + × W + ) ∪ (V − × W − ), where

W+

V + = {v : v ∈ V , v(i, r) ≥ γ}, V − = {v : v ∈ V , v(i, r) ≤ −γ}, P P = {w : w ∈ W , r6=s w(i, s) ≥ 0}, W − = {w : w ∈ W , r6=s w(i, s) ≤ 0}.

By 456Bf, λJ V + = λJ V − and λK W + = λK W − ; since V − = V \ V + , while W + ∪ W − = W , we have 1 2

1 2

λJ V + = λJ V − = λJ V ,

λK W + = λK W − ≥ λK W .

But this means that

λFir = λ(V + × W + ) + λ(V − × W − ) 1 2

1 2

0 , = λJ V + · λK W + + λJ V − · λK W − ≥ λJ V · λK W = λEir

as claimed. Q Q P4 At this point, observe that if y ∈ Fir then | s=0 y(i, s)| ≥ |y(i, r)| ≥ γ. So 1 2

1 2

µ{x : sup |x(i)| ≥ γ}) = λT −1 [{x : sup |x(i)| ≥ γ})] i∈I

i∈I 3

1X y(i, r)| i∈I 2 r=0

= λ{y : sup | = λ{y : sup | i∈I

≥ λ(

[

3 X

y(i, r)| ≥ γ)

r=0

Fir ) =

i∈I,r<4

≥

1 2

XX r<4 i∈I

1 2

≥ γ})

X

λFir

i∈I,r<4 0 λEir ≥

1 2

X

λEr0 = 2α(1 − α)3 ,

r<4

which is what we set out to prove. 456D The support of a Gaussian distribution: Proposition Let I be a set and µ a centered Gaussian distribution on RI . Write Z for the set of those x ∈ RI such that f (x) = 0 whenever f : RI → R is a continuous linear functional and f = 0 a.e. Then Z is a self-supporting closed linear subspace of RI with full outer measure. If I is countable it is the support of µ. proof (a) Being the intersection of a family of closed linear subspaces, of course Z is a closed linear subspace. (b) Z is of full outer measure. P P Let F ⊆ RI be a non-negligible zero set. Let J ⊆ I be a countable set such that F is determined by coordinates in J. For i ∈ I, x ∈ X set πi (x) = x(i); then each πi is either normally distributed or zero almost everywhere, so is square-integrable; set ui = πi• in L2 = L2 (µ). Let hvk ik∈K be a countable orthonormal family in L2 such that every vk is a linear combination of the ui , for i ∈ J, and every ui , for i ∈ J, is a linear combination P of the vk (4A4Jg). Extend hvk ik∈K to a Hamel basis hvl il∈L of L2 . For every i ∈ I, we can express ui as l∈L αil vl , where {l : αil 6= 0} is finite; and the construction ensures that αil = 0 if i ∈ J and l ∈ L \ K. P Consider the linear operator T0 : RK → RJ defined by setting (T0 z)(i) = k∈K αik z(k) for z ∈ RK and (K) (K) i ∈ J. If we give RK the product measure µG , then the image measure µG T0−1 is a Gaussian distribution (456Ba), with covariance matrix

456G


477

Z

Z (K) (K) x(i)x(i0 )(µG T0−1 )dx = (T0 z)(i)(T0 z)(i0 )µG (dz) X X (K) = αik αi0 k0 z(k)z(k 0 )µG (dz) = αik αi0 k

σii0 =

k,k0 ∈K

=

X

k∈K

Z αik αi0 k (vk |vk0 ) = (ui |ui0 ) =

x(i)x(i0 )µ(dx).

k,k0 ∈K (K)

But this means that µG T0−1 has the same covariance matrix as µ˜ πJ−1 , where π ˜J x = x¹J for x ∈ RI . Since this also is a centered Gaussian distribution, the two measures must be equal (456Bb). We know that π ˜J [F ] has non-zero measure, so there is a z0 ∈ RK such that T0 z0 ∈ π ˜J [F ]. Extend z0 arbitrarily to z1 ∈ RL . P Set x1 (i) = l∈L αil z1 (l) for i ∈ I. Then π ˜J x1 = T0 z0 ∈ π ˜J [F ], so x1 ∈ F , because F is determined by I coordinates in J. If a continuous linear functional f : R → R is zero a.e., it can be expressed in the form P f (x) = i∈I βi x(i) where {i : βi 6= 0} is finite (4A4Be). In this case, P P P 0 = f • = i∈I βi ui = l∈L i∈I βi αil vl P in L2 . Since hvl il∈L is linearly independent, i∈I βi αil = 0 for every l ∈ L. But this means that P f (x1 ) = i∈I,l∈L βi αil z1 (l) = 0. As f is arbitrary, x1 ∈ Z and Z ∩ F 6= ∅. As F is arbitrary, and µ is inner regular with respect to the zero sets, Z has full outer measure. Q Q (c) Z is self-supporting. P P If W ⊆ RI is an open set meeting Z, there is an open set V , depending on coordinates in a finite set J ⊆ I, such that V ⊆ W and V ∩ Z 6= ∅. Write π ˜J (x) = x¹J for x ∈ RI , and νJ −1 J for the image measure µ˜ πJ on R ; by 456Ba, this is a centered Gaussian distribution. By 456Bd, there (N) is a continuous linear operator T : RN → RJ such that νJ = µG T −1 . Since the support of µG is R, the (N) support of µG is RN (417E(iv), or otherwise), and the support of νJ is Z1 = T [RN ] (411Ne). Write Q for the set of linear functionals g : RJ → R (necessarily continuous) which are zero on Z1 . If g ∈ Q, then gT = 0, so g = 0 νJ -a.e. and g˜ πJ = 0 µ-a.e. This means that g˜ πJ (x) = 0 for every x ∈ Z, that is, g(y) = 0 for every y ∈ π ˜J [Z]. Because Z1 is a linear subspace of RJ , this is enough to show that π ˜J [Z] ⊆ Z1 . Now recall that V ∩ Z 6= ∅ so Z1 ∩ π ˜J [V ] 6= ∅, while V = π ˜J−1 [˜ πJ [V ]]. Since π ˜J is an open map (4A2B(f-i)), π ˜J [V ] is open and µ∗ (W ∩ Z) = µW ≥ µV = νJ π ˜J [V ] > 0, because Z1 is the support of νJ . Q Q (d) If I is countable, µ is a topological measure so measures Z, and Z is the support of µ. 456E Remarks (a) In the context of 456D, I will call Z the support of the centered Gaussian distribution µ, even though µ need not be a topological measure, so the definition 411Nb is not immediately applicable. In 456L we shall see that Z really is the support of a canonical extension of µ. (b) It is worth making one elementary point at once. If I and J are sets, µ and ν are centered Gaussian distributions on RI and RJ respectively with supports Z and Z 0 , and T : RI → RJ is an inverse-measurepreserving continuous linear operator, then T z ∈ Z 0 for every z ∈ Z. P P If g : RJ → R is a continuous linear I functional which is zero ν-a.e., then gT : R → R is a continuous linear functional which is zero µ-a.e., so g(T z) = (gT )(z) = 0. Q Q 456F Universal Gaussian distributions: Definition A centered Gaussian distribution on RI is universal if its covariance matrix hσij ii,j∈I is the inner product for a Hilbert space structure on I. (See 456Xd.)

478


456G

456G Proposition Let I be any set, and µ a centered Gaussian distribution on I. Then there are a set J, a universal centered Gaussian distribution ν on RJ , and a continuous inverse-measure-preserving linear operator T : RJ → RI . proof (a) Set RL2 = L2 (µ). Then for any finite K ⊆ L2 there is a centered Gaussian distribution µK on RK such that x(u)x(v)µ(dx) = (u|v) for all u, v ∈ K. P P If K = ∅ or K = {0} this is trivial, as we take µ to be the trivial distribution concentrated at 0. Otherwise, let hwi ii
Z (n)

x(u)x(v)µK (dx) = =

(T y)(u)(T y)(v)µG (dy) n−1 X n−1 X

Z αui αvj

(n)

y(i)y(j)µG (dy)

i=0 j=0

=

n−1 X n−1 X

αui αvj (wi |wj ) = (u|v)

i=0 j=0

for all u, v ∈ K. So µK is the distribution we seek. Q Q −1 (b) If K ⊆ L2 is finite and L ⊆ K, then µL = µK πKL , where πKL (x) = x¹L for x ∈ RK . P P Since −1 µK πKL is a centered Gaussian distribution, all we have to do is to check its covariance matrix. But if u, v ∈ L then

Z

Z −1 y(u)y(v)(µK πKL )(dy) =

=

Z

(πKL x)(u)(πKL x)(v)µK (dx) Z x(u)x(v)µK (dx) = (u|v) = y(u)y(v)µL (dy).

−1 By 456Bb, µL = µK πKL .Q Q 2

−1 (c) By 454G, there is a Baire measure ν 0 on RL such that ν 0 π ˜K [E] = µK E for every finite K ⊆ L2 and 2 every Borel set E ⊆ RK , where π ˜K (x) = x¹K for x ∈ RL . Take ν to be the completion of ν 0 . Then π ˜K 2 is inverse-measure-preserving for ν and µK , for every finite K ⊆ L2 . If f : RL → R is a continuous linear functional, there are a finite K ⊆ L2 and a linear functional g : RK → R such that f = π ˜K g, so that

ν{x : f (x) ≤ α} = µK {x : g(x) ≤ α} for every α, and f and g have the same distribution; as g is either normal with zero expectation or zero a.e., so is f . As f is arbitrary, ν is a centered Gaussian distribution. (d) ν is universal. P P If u, v ∈ L2 set K = {u, v}. Then Z

Z x(u)x(v)ν(dx) =

(˜ πK x)(u)(˜ πK x)(v)ν(dx) Z

=

y(u)y(v)µK (dy) = (u|v).

Thus the covariance matrix of ν is just the inner product of the standard Hilbert space structure of L2 . Q Q (e) For i ∈ I, let ui ∈ L2 be the equivalence class of the square-integrable function x 7→ x(i) : RI → R. 2 2 Define T : RL → RI by setting (T y)(i) = y(ui ) for every i ∈ I, y ∈ RL . Then there is a centered Gaussian distribution µ0 on RI such that T is inverse-measure-preserving for ν and µ0 . Now the covariance matrix of µ0 is defined by

456H


Z

479

Z

Z (T y)(i)(T y)(j)ν(dy) = y(ui )y(uj )ν(dy) Z = (ui |uj ) = x(i)x(j)µ(dx)

x(i)x(j)µ0 (dx) =

for all i, j ∈ I. So µ and µ0 are equal and T is inverse-measure-preserving for ν and µ. 456H Lemma Let µ beRa universal centered Gaussian distribution on RI ; give I a corresponding Hilbert space structure such that x(i)x(j)µ(dx) = (i|j) for all i, j ∈ I. Let F ⊆ RI be a set determined by coordinates in J, where J ⊆ I is a closed linear subspace for the Hilbert space structure of I. Let W be the union of all the open subsets of RI which meet F in a negligible set, and W 0 the union of the open subsets of RI which meet F in a negligible set and are determined by coordinates in J. If F ⊆ W then F ⊆ W 0 . proof (a) Let Z be the support of µ in the sense of 456D. We need to know thatP Z is just the set of all linear functionals from I to R. P P If K ⊆ I is finite and hαi ii∈K ∈ RK and f (x) = i∈I αi x(i) for x ∈ RI , then P P kf k22 = i,j∈K αi αj (i|j) = k i∈K αi ik2 . So, for x ∈ RI , x ∈ Z ⇐⇒ f (x) = 0 whenever f ∈ (RI )∗ and kf k2 = 0 X X ⇐⇒ αi x(i) = 0 whenever K ⊆ I is finite and αi i = 0 in I i∈K

i∈K

⇐⇒ x : I → R is linear. Q Q ⊥

(b) Let J ⊥ be the orthogonal complement of J in I, so that I = J ⊕ J ⊥ (4A4Je). Give RJ and RJ the centered Gaussian distributions µJ , µJ ⊥ induced by µ and the projections x 7→ x¹J, x 7→ x¹J ⊥ . Then the ⊥ ⊥ product measure λ on RJ ×RJ is also a centered Gaussian distribution (456Be). Define T : RJ ×RJ → RI ⊥ by setting T (u, v)(j + k) = u(j) + v(k) whenever j ∈ J, k ∈ J ⊥ , u ∈ RJ and v ∈ RJ . Then T is inversemeasure-preserving for λ and µ. P P T is a continuous linear operator so we have a centered Gaussian distribution µ0 on RI such that T is inverse-measure-preserving for λ and µ0 . If j, j 0 ∈ J and k, k 0 ∈ J ⊥ , Z

Z 0

0

0

T (u, v)(j + k)T (u, v)(j 0 + k 0 )λ(d(u, v))

x(j + k)x(j + k )µ (dx) = Z

(u(j) + v(k))(u(j 0 ) + v(k 0 ))λ(d(u, v)) Z Z 0 = u(j)u(j )µJ (du) + v(k)v(k 0 )µJ ⊥ (dv) Z Z = x(j)x(j 0 )µ(dx) + x(k)x(k 0 )µ(dx)

=

= (j|j 0 ) + (k|k 0 ) = (j + k|j 0 + k 0 ) Z = x(j + k)x(j 0 + k 0 )µ(dx). Thus µ and µ0 have the same covariance matrix and are equal, and T is inverse-measure-preserving for λ and µ. Q Q (c) Take any z ∈ W ∩ Z. Then there is an open set V , determined by coordinates in a finite set K0 ⊆ I, such that z ∈ V and µ(V ∩ F ) = 0. Let ² > 0 be such that y ∈ V whenever x ∈ RI and |x(i) − z(i)| < 2² for every i ∈ K0 . Express each k ∈ K0 as k 0 + k 00 where k 0 ∈ J and k 00 ∈ J ⊥ . Set V 0 = {x : x ∈ RI , |x(k 0 ) − z(k 0 )| < ² for every k ∈ K0 }.

480


456H

Then V 0 is an open set, determined by coordinates in J, and contains z. Also µ(V 0 ∩ F ) = 0. P P Set V 00 = {x : x ∈ RI , |x(k 00 ) − z(k 00 )| < ² for every k ∈ K0 }. Then V ⊇ V 0 ∩ V 00 ∩ Z. Since Z has full outer measure (456D), µ(V 0 ∩ V 00 ∩ F ) = µ∗ (V 0 ∩ V 00 ∩ F ∩ Z) ≤ µ(V ∩ F ) = 0. Now 0 = µ(V 0 ∩ V 00 ∩ F ) = λT −1 [V 0 ∩ F ∩ V 00 ] (because V 0 ∩ F ∩ V 00

= λ{(x¹J, x¹J ⊥ ) : x ∈ V 0 ∩ F ∩ V 00 } is determined by coordinates in J ∪ J ⊥ ) = µJ {x¹J : x ∈ V 0 ∩ F } · µJ ⊥ {x¹J ⊥ : x ∈ V 00 }

because V 0 ∩ F is determined by coordinates in J, while V 00 is determined by coordinates in J ⊥ . However, z ∈ V 00 , and z¹J ⊥ belongs to the support Z 0 of µJ ⊥ , by 456Eb; since Z 0 is self-supporting, and {x¹J ⊥ : x ∈ V 00 } is open, µJ ⊥ {x¹J ⊥ : x ∈ V 00 } > 0. We conclude that 0 = µJ {x¹J : x ∈ V 0 ∩ F } = µ(V 0 ∩ F ). Q Q (d) This shows that z ∈ W 0 . As z is arbitrary, W ∩ Z ⊆ W 0 . ?? Suppose, if possible, that there is a point z0 ∈ F \ W 0 . If i, j ∈ J and α ∈ R, then {x : x(i + j) 6= x(i) + x(j)},

{x : x(αi) 6= αx(i)}

are negligible open sets determined by coordinates in J, so are included in W 0 and do not contain z0 . Thus z0 ¹J : J → R is linear. Let z : I → R be a linear functional extending z0 ¹J. Then z ∈ Z and z¹J = z0 ¹J; as both F and W 0 are determined by coordinates in J, z ∈ F \ W 0 . But this means that z ∈ Z ∩ W \ W 0 , which is impossible. X X So F ⊆ W 0 , as claimed. 456I Cluster sets: Lemma Let I be a countable set, n ≥ 1 an integer and µ a centered Gaussian distribution on RI×n . For ² > 0 set R I² = {i : i ∈ I, |x(i, r)|2 µ(dx) ≤ ²2 for every r < n}; suppose that no I² is empty. (a) There is a closed set F ⊆ Rn such that T F = ²>0 {hx(i, r)ir 0 such that supi∈I² ,r0 {Si x : i ∈ I² }, so that Fx is a closed subset of Rn . For A ⊆ Rn set EA = {x : x ∈ RI×n , A ∩ Fx 6= ∅}. (N)

(ii) By 456Bd, there is a continuous linear operator T : RN → RI×n such that µ = µG T −1 . Set Ti = Si T for i ∈ I; then Ti : RN → Rn is a continuous linear operator. For y ∈ RN , set T {Ti (y) : i ∈ I² }. F˜y = FT (y) = ²>0

For A ⊆ Rn set Ã = T −1 [EA ] = {y : y ∈ RN , A ∩ F˜y 6= ∅}. E ˜K is a Borel subset of RN . P (iii) If K ⊆ Rn is compact, then E P Let V be a countable base for the topology of Rn , and for k ≥ 1 let Vk be the set of members of V with diameter at most 1/k which meet K. Then Ti−1 [V ] is a Borel set for every V ∈ V and i ∈ I, so

456I


E0 =

T

S k≥1

S V ∈Vk

i∈I1/k

481

Ti−1 [V ]

is a Borel set. ˜K , take k ≥ 1. There is a z ∈ F˜y ∩ K. Let V ∈ V be such that z ∈ V and diam V ≤ 1/k; in this If y ∈ E case V ∈ Vk . Because z ∈ {Ti (y) : i ∈ I1/k }, there is an i ∈ I1/k such that Ti (y) ∈ V . As k is arbitrary, this ˜K ⊆ E 0 . shows that y ∈ E 0 ; thus E ˜ If y ∈ / EK , then K is a compact set disjoint from the closed set F˜y . There is therefore some ² > 0 such that K ∩ {Ti (y) : i ∈ I² } = ∅ (since these form a downwards-directed family of compact sets with empty intersection). Next, there is a δ > 0 such that B(z, δ) ∩ {Ti (y) : i ∈ I² } = ∅ for every z ∈ K (2A2Ed). Let k ≥ 1 be such that 1/k ≤ min(², δ). If V ∈ Vk and i ∈ I1/k , there is some z ∈ K ∩ V so V ⊆ B(z, δ) and ˜K . Ti (y) ∈ / V . This shows that y ∈ / E 0 . As y is arbitrary, E 0 ⊆ E 0 ˜K = E is a Borel set. Q So E Q ˜ It follows at once that EH is a Borel set for every Kσ set H, in particular, for any open or closed set H. (iv)P We need a simple estimate on the coefficients of the linear operators Ti . Let αirj be such that ∞ Ti (y) = h j=0 αirj y(j)ir
˜K > 0. Then (v) Now suppose that H ⊆ Rn is open, that K ⊆ H is compact, and that µG E P∞ h²j ij∈N be a sequence of strictly positive real numbers such that j=0 ²j ≤

(N) ˜ µG E P Let ² > 0. Let H = 1. P (N) ˜ 1 2 min(², µG EK ), and for each

j ∈ N let γj ≥ 0 be such that µG [−γj , γj ] ≥ 1 − ²j . Then

˜ = {y : y ∈ E ˜K , |y(j)| ≤ γj for every j ∈ N} E P∞ (N) ˜ ˜0 has measure at least µG E K − j=0 ²j > 0. By 254Sb, there are an m ∈ N and a set E , of measure at ˜ 0 there is a y ∈ E ˜ such that y(j) = y 0 (j) whenever j > m. Set least 1 − 12 ², such that for every y 0 ∈ E (N) 00 0 ˜ = {y : y ∈ E ˜ , |y(j)| ≤ γj for every j ∈ N}, so that µ E ˜ 00 ≥ 1 − ². E G √ Pm Let δ > 0 be such that z 0 ∈ H whenever z ∈ K and kz −z 0 k ≤ 2δ. Let η > 0 be such that 2η n j=0 γj ≤ ˜ 00 and i ∈ Iη , there is a y ∈ E ˜ such that y(j) = y 0 (j) for j > m. Also |y(j) − y 0 (j)| ≤ 2γj for δ. If y 0 ∈ E j ≤ m, so P∞ Pm δ |Ti (y)(r) − Ti (y 0 )(r)| ≤ j=0 |αirj ||y(j) − y 0 (j)| ≤ j=0 2ηγj ≤ √ n

0

for every r < n, and kTi (y) − Ti (y )k ≤ δ. Now F˜y ∩ K 6= ∅; take z ∈ F˜y ∩ K. For every ζ > 0, there is an i ∈ Imin(η,ζ) such that kz − Ti (y)k ≤ δ, so that kz − Ti (y 0 )k ≤ 2δ. This means that B(x, 2δ) ∩ {Ti (y 0 ) : i ∈ Iζ } is not empty. As B(x, 2δ) is compact, ˜ 00 ⊆ E ˜H , while it must meet F˜y0 . But this means that H ∩ F˜y0 6= ∅, by the choice of δ. As y 0 is arbitrary, E (N) ˜ 00 µG E ≥ 1 − ². ˜H is conegligible, as claimed. Q This works for every ² > 0. So E Q (N) ˜ (vi) If H ⊆ Rn is open and µG E H > 0, then (because H is σ-compact) there is a compact set K ⊆ H (N) ˜ (N) ˜ such that µG EK > 0, and (e) tells us that µG E H = 1. Set (N) ˜ (N) ˜ V0 = {V : V ∈ V, µ E EV < 1}. V = 0} = {V : V ∈ V, µ G

Then we see that V0 = {V : V ∈ V, F˜y ∩ V = ∅} for almost every y ∈ RN , that is, V0 = {V : V ∈ V, Fx ∩ V = ∅}

G

482


456I

S for almost every xS∈ RI×n . But as every Fx is closed, we have Fx = Rn \ V0 for almost every x. So we can set F = Rn \ V0 . ∼ (RI×n )2 the measure λ corresponding to the product measure µ × µ; by 456Be, this (b)(i) Give RI×2n = is a centered Gaussian distribution. For (x1 , x2 ) ∈ (RI×n )2 , set T Fx0 1 x2 = ²>0 {(Si x1 , Si x2 ) : i ∈ I² }. set F 0 ⊆ R2n such that F 0 = Fx0 1 x2 for almost all x1 , x2 . (Of course I² = {i : RBy (a), we2 have a closed 2 |xj (i, r)| λ(dx) ≤ ² for every j ∈ {1, 2}, r < n} whenever ² > 0.) (ii) Now (z, 0) ∈ F 0 . P P Take x1 ∈ RI×n such that F = Fx1 and E = {x2 : F 0 = Fx0 1 x2 } is conegligible. I×n (Almost has these properties.) For k ∈ N, z ∈ {Si x1 : i ∈ I2−k }; let ik ∈ I2−k be such P every point of R that r 0 there is an i ∈ I² such that |z(r) − x1 (i, r)| ≤ ² and |x2 (i, r)| ≤ ² for every r < n. But now |αz(r) − T˜(x1 , x2 )(r)| ≤ 2² for every r < n. Q Q So T˜−1 [{x : αz ∈ Fx }] = {(x1 , x2 ) : αz ∈ F ˜ } ⊇ {(x1 , x2 ) : (z, 0) ∈ Fx0 x } T (x1 ,x2 )

1

2

is λ-conegligible, and αz ∈ Fx for µ-almost every x, that is, αz ∈ F , as claimed. (c) Suppose now that F is bounded. (i) For L ⊆ I, α ≥ 0 set Q(L, α) =

S

i∈L,r
: |x(i, r)| ≥ α}.

By 456C, applied to the image of µ under the map x 7→ x¹L : RI → RL , µQ(L, 21 α) ≥ 2µQ(L, α)(1 − µQ(L, α))3 for every finite L ⊆ I and every α ≥ 0. Let β > 0 be such that δ = 2β(1 − (n + 1)β)3 − β > 0, and let α0 > 0 be such that α02 β ≥ 1 and kzk < 12 α0 for every z ∈ F , so that µ{x : |x(i, r)| ≥ α0 } ≤ β whenever i ∈ I1 and r < n, and µQ({i}, α0 ) ≤ nβ for every i ∈ I1 . Set K = {z : z ∈ Rn , 21 α0 ≤ maxr
456J


483

{x : Si (x) ∈ K for some i ∈ I1/k } form a non-increasing sequence of measurable sets with negligible intersection, there is a k ≥ 1 such that µ{x : Si (x) ∈ K for some i ∈ I1/k } < δ. (ii) ?? Suppose, if possible, that µQ(I1/k , α0 ) > β. Let L ⊆ I1/k be a finite set of minimal size such that γ = µQ(L, α0 ) ≥ β. Since µQ({i}, α0 ) ≤ nβ for any i ∈ L, and L is minimal, we must have β ≤ γ ≤ (n + 1)β. Now this means that 1 2

µQ(L, α0 ) ≥ 2γ(1 − γ)3 (see (i) above) δ β

≥ 2γ(1 − (n + 1)β)3 = γ(1 + ) ≥ γ + δ, so that µ(Q(L, 12 α0 ) \ Q(L, α0 )) ≥ δ. But if x ∈ Q(L, 12 α0 ) \ Q(L, α0 ) there is some i ∈ L such that maxr
δ ≤ µ(Q(L, α0 ) \ Q(L, α0 )) ≤ µ({x : Si (x) ∈ K for some i ∈ L} ≤ µ({x : Si (x) ∈ K for some i ∈ I1/k } < δ which is absurd. X X (iii) Thus µQ(I1/k , α0 ) ≤ β. For α ≥ 0, set f (α) = µQ(I1/k , α); then f is non-increasing. Also f ( 12 α) ≥ 2f (α)(1 − f (α))3 for every α. P P?? Otherwise, because f (α) = sup{µQ(L, α) : L ⊆ I1/k is finite}, there is a finite L ⊆ I1/k such that f ( 12 α) < 2γ(1 − γ)3 , where γ = µQ(L, α). But in this case µQ(L, 12 α) ≤ f ( 21 α) < 2γ(1 − γ)3 , which is impossible, as remarked in (i). X XQ Q Set ζ = limα→∞ f (α). Then ζ = limα→∞ f ( 12 α) ≥ 2ζ(1 − ζ)3 . But we also know, from (ii), that ζ ≤ f (α0 ) ≤ β. So (1 − ζ)3 ≥ (1 − β)3 > What this means is that if we set ² = k1 then

1 2

and ζ must be 0.

limα→∞ µ{x : supi∈I² ,r α} = 0, that is, supi∈I² ,r
S (a) There is an upwards-directed family G of open Baire sets in RJ such that W0 = G is a Baire set and µW0 > supG∈G µG. S Let G0 ⊆ G be a countable upwards-directed set such that supG∈G0 µG = supG∈G µG, and set W1 = W0 \ G0 ; then µW1 > 0 and µ(W1 ∩ G) = 0 for every G ∈ G. Let W be a non-negligible zero set included in W1 . For each n ∈ N, let Vn be a countable base for the topology of Rn consisting of open balls. Let Gn∗ be the family of open sets of RJ of the form T −1 [V ], where T : RJ → Rn is a continuous linear operator, V ⊆ Rn is open and µ(W ∩ T −1 [V ]) = 0. Of course G0∗ = ∅.

484


456J

J n (b) For n ≥ 1 and V ∈ Vn , let TnV be the family S ∗ of continuous linear operators T : R → R such that −1 W ∩ T [V ] is negligible, but not included in Gn−1 . Index Tnv as hTi ii∈I(n,V ) ; it will be convenient to do this in such a way that all the I(n, V ) are disjoint. Define fir , for i ∈ I(n, V ) and r < n, by saying S • that Ti (x)S= hfir (x)ir
(c) If j ∈ I(n, V ), then there is a δ > 0 such that {Ti (x) : i ∈ I 0 (n, V ), ρn (φn (i), φn (j)) ≤ δ} 0 is bounded for almost every x ∈ RJ . P P For i ∈ I 0 (n, V ) and r < n, set fir = fir − fjr , Ti0 = Ti − Tj ; for δ > 0, write Iδ0 for 0 {i : i ∈ I 0 (n, V ), ρn (φn (i), φn (j)) ≤ δ} = {i : i ∈ I 0 (n, V ), kfir k2 ≤ δ for every r < n}.

Because φn (j) ∈ {φn (i) : i ∈ I 0 (n, V )}, no Iδ0 is empty. So 456Ia tells us that there is a closed set F ⊆ Rn such that T F = Fx = δ>0 {Ti0 (x) : i ∈ Iδ0 } for almost every x ∈ RJ . By 456Ib, αz ∈ F whenever z ∈ F and |α| ≤ 1. ?? If F is not bounded, then it 1 n

must include a line L through 0. (The sets { z : z ∈ F , kzk = n}, for n ≥ 1, form a non-increasing sequence of non-empty compact sets, so there is a point z0 belonging to them all; take L to be the set of multiples of z0 .) Let D ⊆ L be a countable dense set. For z ∈ D and k ∈ N we know that for every i ∈ I(n, V ), Ti (x) ∈ / V for almost every x ∈ W , for almost every x ∈ RJ there is an i ∈ I 0 (n, V ) such that kTi (x) − Tj (x) − zk ≤ 2−k and therefore for almost every x ∈ W , Ti (x) ∈ / V for every i ∈ I 0 (n, V ), but there is an i ∈ I 0 (n, V ) such −k that kTi (x) − Tj (x) − zk ≤ 2 so that for almost every x ∈ W , the distance from Tj (x) + z to the closed set Rn \ V is at most 2−k . This is true for every k ∈ N, so we get Tj (x) + z ∈ / V for almost every x ∈ W . And this is true for every z ∈ D, so we get for almost every x ∈ W , Tj (x) + z ∈ / V for every z ∈ D, so Tj (x) ∈ / V + L. n n−1 0 Let S : R → R be a linear operator with kernel L, and set V = S[V ]. Then V 0 ⊆ Rn−1 is open, and −1 0 ∗ W ∩ (STj ) [V ] = W ∩ Tj−1 [V + L] is negligible. But this means that Tj−1 [V ] ⊆ (STj )−1 [V 0 ] ∈ Gn−1 and S ∗ −1 Tj [V ] is included in Gn−1 ; which contradicts the definition of TnV . X X 0 So F is bounded. By 456Ic, there is some δ > 0 such that supi∈Iδ0 ,r
456K


485

P Let h²j ij∈J0 and h²0nV k in≥1,V ∈Vn ,k∈N be families of strictly positive real numbers such that j∈J0 ²j and P∞ P P∞ 0 1 0 n=1 V ∈Vn k=0 ²nV k are both at most 3 µW . Let hγj ij∈J0 and hγnV k in≥1,V ∈Vn ,k∈N be such that µ{x : x ∈ RJ , |x(j)| ≥ γj } ≤ ²j for every j ∈ J0 , 0 0 µ{x : x ∈ RJ , supi∈InV k kTi (x)k ≥ γnV k } ≤ ²nV k for every n ≥ 1, V ∈ Vn , k ∈ N.

Set W 0 = {x : x ∈ W , |x(j)| ≤ γj for every j ∈ J0 }; then µW 0 ≥ 23 µW and W 0 is of the form C × RJ\J0 , where C ⊆ RJ0 is compact. Set 0 W 00 = {x : x ∈ W 0 , kTi (x)k ≤ γnV k whenever n ≥ 1, V ∈ Vn , k ∈ N and i ∈ InV k };

then µW 00 ≥ 31 µW . S (f ) Set I = n≥1,V ∈Vn I(n, V ) × n. If K ⊆ I is finite, then 0 0 WK = {x : x ∈ W 0 , kfir (x)k ≤ γnV k whenever n ≥ 1, V ∈ Vn , k ∈ N,

(i, r) ∈ K and φn (i) ∈ K ∩ φn [InV k ]} P For each quintuple (i, r, n, V, k) with n ≥ 1, V ∈ Vn , k ∈ N, (i, r) ∈ K and has measure at least 31 µW . P φn (i) ∈ φn [InV k ], there is a sequence him im∈N in InV k such that ρn (φn (i), φn (im )) ≤ 2−m for every m; so that kfir − fim r k2 ≤ 2−m for every m. But this means that fir (x) = limm→∞ fim r (x) for almost every 0 00 x ∈ RJ . Accordingly |fir (x)| ≤ γnV k for almost every x ∈ W . Since there are only countably many such 00 0 0 Q quintuples (i, r, n, V, k), we see that W \ WK is negligible, so µWK ≥ µW 00 ≥ 31 µW . Q (g) For x ∈ RJ , define T x ∈ RI by setting (T x)(i, r) = fir (x) for i ∈ I(n, V ) and r < n. Then T : RJ → RI is a continuous linear operator. By 4A4H, T [W 0 ] is closed. For finite K ⊆ I, let HK be the family of open subsets H of RK such that µ{x : x ∈ W , T x¹K ∈ H} = 0. Then HK is closed under countable unions so has a largest member HK . Now there is a K ∈ [I]<ω such 0 0 that T x¹K ∈ HK for every x ∈ WK . P P?? Otherwise, choose for each K ∈ [I]<ω an xK ∈ WK such <ω <ω that T xK ¹K ∈ / HK . Let F be an ultrafilter on [I] containing {K : L ⊆ K ∈ [I] } for every finite L ⊆ I. If (i, r) ∈ I, there are n ≥ 1, V ∈ Vn and k ∈ N such that r < n and φn (i) ∈ φn [InV k ], in 0 <ω which case |fir (xK )| ≤ γnV . This means that limK→F fir (xK ) must be k whenever (i, r) ∈ K ∈ [I] 0 0 ∗ defined in [−γnV k , γnV k ]; consequently y = limK→F T xK is defined in RI . Since xK ∈ W 0 for every K, y ∗ ∈ T [W 0 ] = T [W 0 ]. Let x∗ ∈ W 0 be such that T x∗ = y ∗ . Since x∗ ∈ W , there are n ≥ 1, V ∈ Vn and i ∈ I(n, V ) such that Ti (x∗ ) ∈ V . Set L = {(i, r) : r < n}, H = {z : z ∈ RL , hz(i, r)ir 0 and 0 0 that T x¹K ∈ HK for every x ∈ WK and that WK ⊆ W and that {x : x ∈ W , T x¹K ∈ HK } is negligible. Faced with this we have to abandon the original supposition that µ is not τ -additive.

456K We now have all the ideas needed for the main theorem of this section. Theorem (Talagrand 81) Every centered Gaussian distribution is τ -additive. proof ?? Suppose, if possible, that µ is a centered Gaussian distribution on a set RI which is not τ -additive. (a) By 456G, there are a set J and a universal centered Gaussian distribution ν on RJ and a continuous linear operator T : RJ → RI which is inverse-measure-preserving for ν and µ. By 418Ha, ν is not τ -additive. (b) As in part (a) of the proof of 456J, there are a non-negligible zero set W ⊆ RJ and a family G of open sets, covering W , such that ν(W ∩ G) = ∅ for every G ∈ G. Give J a Hilbert space structure such

486


456K

R that x(i)x(j)ν(dx) = (i|j) for all i, j ∈ J. Let K0 ⊆ J be a countable set such that W is determined by coordinates in K0 , and let K be the closed linear subspace of J generated by K0 . Let S G 0 be the family of open sets determined by coordinates in K which meet W in negligible sets. Then W ⊆ G 0 , by 456H. Let λ be the centered Gaussian distribution on RK for which the map π ˜K = x 7→ x¹K : RJ → RK is K inverse-measure-preserving. Then π ˜K [W ] is a zero set in R , λ˜ πK [W ] = νW > 0, {˜ πK [G] : G ∈ G 0 } is K a family of open sets in R covering π ˜K [W ], and λ(˜ πK [W ] ∩ π ˜K [G]) = ν(W ∩ G) = 0 for every G ∈ G 0 ; so λ is not τ -additive. However, K, regarded as a normed space, is separable (see 4A4Bi); and if we set πj (y) = y(j) for y ∈ RK and j ∈ K, then kπi• − πj• k2 = ki − jk for all i, j ∈ K. So {πj• : j ∈ K} is separable in L2 (λ). And this is impossible, by 456J. X X Thus every centered Gaussian distribution must be τ -additive. 456L Corollary If µ is a centered Gaussian distribution on RI , there is a unique quasi-Radon measure µ ˜ on RI extending µ. The support of µ as defined in 456D is the support of µ ˜ as defined in 411N. proof By 415L, µ has a unique extension to a quasi-Radon measure µ ˜. Now the support Z of µ is a closed set, so µ ˜Z = µ∗ Z (415L(i)). Also Z is self-supporting for µ. If G ⊆ RI is an open set meeting Z, then there is a cozero set H ⊆ G which also meets Z, and µ∗ (Z ∩ H) > 0. It follows that µ∗ (Z \ H) < 1; as µ ˜ extends µ, µ ˜(Z \ H) < 1 and µ ˜(Z ∩ G) > 0. This shows that Z is self-supporting for µ ˜, so must be the support of µ ˜ in the standard sense. 456M Gaussian processes I take half a page to spell out the connexion between centered Gaussian distributions, as defined in 456A, and the corresponding processes considered in 454J-454K. Proposition Let hXi ii∈I be a family of random variables on a probability space (Ω, Σ, µ). Then the following are equiveridical: (i) the distribution of hXi ii∈I , in the sense of 454K, isPa centered Gaussian distribution; n (ii) whenever i0 , . . . , in ∈ I and α0 , . . . , αn ∈ R then r=0 αr Xir is either zero a.e. or a normal random variable with zero expectation; (iii) whenever i0 , . . . , in ∈ I then the joint distribution of Xi0 , . . . , Xin , in the sense of 271C, is a centered Gaussian distribution; (iv) whenever J ⊆ I is finite then there is an independent family hYk ik∈K of standard normal random variables on Ω such that each Xi , for i ∈ J, is almost everywhere equal to a linear combination of the Yk . proof (ii) ⇐⇒ (i) is immediate from the definition in 456Ab and 454L. (i) ⇐⇒ (iii) is also direct from 456Ab and the identification of the two concepts of ‘distribution’ (454K). (iv)⇒(ii) is direct from 274A-274B. (i)⇒(iv) For i ∈ J set ui = Xi• in L2 (µ). By 4A4Jg there is an orthonormal family hvk ik∈K in L2 (µ) such that each vk is a linear combination of the ui and each ui is a linear combination of the vk . Take Yk such that Yk• = vk for each k; then each Xi is equal almost everywhere to a linear combination of the Yk , while each Yk is equal almost everywhere to a linear combination of the Xi . As #(K) must be the dimension of the linear span of {ui : i ∈ J}, K is finite. Any linear combination of the Yk is equal almost everywhere to a linear combination of the Xk , so is either zero a.e. or a normal random variable with zero expectation. Because (ii)⇒(iii), hYk ik∈K has a centered Gaussian distribution ν say. Each Yk has variance (vk |vk ) = 1, so is a standard normal random variable. The covariance matrix of ν is given by Z y(j)y(k)ν(dy) = E(Yj × Yk ) = (vj |vk ) = 1 if j = k = 0 otherwise. (K)

But this means that ν has the same covariance matrix as the product measure µG , which is also a centered (K) Gaussian distribution (456Bc). So ν = µG (456Bb). But this means that hYk ik∈K is independent (272G). Thus we have found a suitable family.

456Yb


487

456N Definition A family of random variables satisfying the equivalent conditions in 456M is a centered Gaussian process. 456X Basic exercises > (a) Let I be a set and hσij ii,j∈I a family of real numbers. Show that the following are equiveridical: (i) hσij ii,j∈I is the covariance matrix of a centered Gaussian distribution on RI ; (ii) there are a Hilbert space U and a family hui ii∈I in U such that (ui |uj ) = σij for all i, j ∈ I; (iii) hσij ii,j∈J is the covariance matrix of a centered Gaussian distribution on RJ for every finite J ⊆ I; (iv) P σij = σji for all i, j ∈ I and i,j∈J αi αj σij ≥ 0 whenever J ⊆ I is finite and hαi ii∈J ∈ RJ . (Hint: 4A4Jg, for finite I, and 454G. For (iv)⇒(iii) use the finite-dimensional version of 4A4M.) (n)

(n)

(b) Let n ≥ 1 be an integer. (i) Show that µG T −1 = µG for any orthogonal linear operator T : Rn → Rn . (ii) Set p(x) =

1 x kxk

(n)

for x ∈ Rn \ {0}; take p(0) to be any point of S n−1 . Show that µG p−1 is a multiple

of (n − 1)-dimensional Hausdorff measure on S n−1 . (Hint: 443T.) (c) Let I be a countable set, µ a centered Gaussian distribution on RI , and γ ≥ 0. Set α = µ{x : supi∈I |x(i)| ≥ γ). Show that µ{x : supi∈I |x(i)| ≥ 12 γ) ≥ 2α(1 − α)3 . (d) (i) Let U and V be Hilbert spaces, I ⊆ U a set and f : I → V a function such that (f (u)|f (v)) = (u|v) for all u, v ∈ I. Show that there is a linear operator T : U → V , extending f , such that (T u|T v) = (u|v) for all u, v ∈ U . (ii) Let I be a set and hσij ii,j∈I a family of real numbers. Show that there is at most one inner product space structure on I for which σij = (i|j) for all i, j ∈ I. (e) Let hXn in∈N be an independent sequence of standard normalPrandom variables. Show that for any P∞ ∞ square-summable sequence hαn in∈N and any bijection π : N → N, n=0 αn Xn and n=0 απ(n) Xπ(n) are finite and equal a.e. (Hint: 273B.) > (f ) For any set I, I will say that a centered Gaussian quasi-Radon measure on RI is a quasi-Radon measure µ on RI such that every continuous linear functional f : RI → R is either zero a.e. or is normally distributed with zero expectation. Show that (i) there is a one-to-one correspondence between centered Gaussian quasi-Radon measures µ on RI and centered Gaussian distributions ν on RI obtained by matching µ with ν iff they agree on the zero sets of RI ; R R (ii) if µ, ν are centered Gaussian quasi-Radon measures on RI and x(i)x(j)µ(dx) = x(i)x(j)ν(dx) for all i, j ∈ I, then µ = ν; (iii) the support of a centered Gaussian quasi-Radon measure on RI is a linear subspace of RI ; (iv) if hIj ij∈J is a disjoint family of sets with union I, and µj is a centered Gaussian quasi-Radon measure on RIj for each j ∈ J, then the quasi-Radon product of hµj ij∈J , regarded as a measure on RI , is a centered Gaussian quasi-Radon measure. >(g) Show that the completion of the measure on Ω = R[0,∞[ described in 455C is a centered Gaussian distribution with covariance matrix σ = hmin(s, t)is,t∈[0,∞[ . > (h) Let hXi ii∈I be a Gaussian process. Show that hXi ii∈I is independent iff E(Xi × Xj ) = 0 for all distinct i, j ∈ I. 456Y Further exercises (a) Let (Ω, Σ, µ) be a probability space with measure algebra (A, µ ¯), and hui ii∈I a family in L2 (µ) ∼ ¯) which is a Gaussian process in the sense that whenever Xi ∈ L2 (µ) = L2 (A, µ is such that Xi• = ui for every i, then hXi ii∈I is a Gaussian process. Suppose that γ ≥ 0 and that α=µ ¯(supi∈I [[|ui | ≥ γ]]). Show that µ ¯(supi∈I [[|ui | ≥ 21 γ]]) ≥ 2α(1 − α)3 . (b) Let U be a Hilbert space with an orthonormal basis huj ij∈J , and µ the universal centered Gaussian distribution on RU with covariance matrix defined by the inner product of U . Show that there is a function (J) T : RJ → RU , inverse-measure-preserving for µG and µ, such that whenever hj in∈N is a sequence Pn∞ P∞ of distinct elements of J and hαn in∈N is a square-summable sequence in R, then (T x)( n=0 αn ujn ) = n=0 αn x(jn ) for almost every x ∈ RJ .

488


456Yc

(c) Let U be an infinite-dimensional Hilbert space and µ the universal centered Gaussian distribution on RU with covariance matrix defined by the inner product of U . Show that µC = 0 for every compact set C ⊆ RU . (Hint: there is an m ∈ N such that {u : |x(u)| ≤ m for every x ∈ C} is somewhere dense in U .) (d) Let µ be a centered Gaussian distribution on a set RI . Show that the following are equiveridical: (i) µ hasqcountable Maharam type; (ii) L2 (µ) is separable; (iii) I is separable under the pseudometric R (i, j) 7→ (x(i) − x(j))2 µ(dx). ¤ £ (e) Fix p ∈ − 12 , 12 \ {0}. (i) For α ∈ R set fα (t) = |t − α|p − |t|p when this is defined. Show that fα ∈ L2 (µL ), where µL is Lebesgue measure, that kfα k22 = |α|2p+1 kf1 k22 and that kfα − fβ k2 = kfα−β k2 for all α, β ∈ R. (ii) Show that there is a centered Gaussian process hXα iα∈R such that E(Xα × Xβ ) = |α|2p+1 + |β|2p+1 − |α − β|2p+1 for all α, β ∈ R. (iii) Show that such a process can be represented by a Radon measure on C(R). (Hint: 455Yc.) (This is fractional Brownian motion.) 456 Notes and comments This section has taken a direct route to Talagrand’s theorem 456K, leaving all the real reasons for studying Gaussian processes (see Fernique 97) to one side. I hope that it will nevertheless be clear from such fragments as 252Xh, 456Bb, 456C and the exercises here that they are one of the many concepts of probability theory which are both important and delightful.

457 Simultaneous extension of measures The questions addressed in §§451, 454 and 455 can all be regarded as special cases of a general class of problems: given a set X and a family hνi ii∈I of (probability) measures on X, when can we expect to find a measure on X extending every νi ? An alternative formulation, superficially more general, is to ask: given a set X, a family h(Yi , Ti , νi )ii∈I of probability spaces, and functions φi : X → Yi for each i, when can we find a measure Q on X for which every φi is inverse-measure-preserving? Even the simplest non-trivial case, when X = i∈I Yi and every φi is the coordinate map, demands a significant construction (the product measures of Chapter 25). In this section I bring together a handful of important further cases which are accessible to the methods of this chapter. I begin with a discussion of extensions of finitely additive measures (457A-457D), which are much easier, before considering the problems associated with countably additive measures (457E-457H). 457A It is helpful to start with a widely applicable result on common extensions of finitely additive measures. Lemma Let A be a Boolean algebra and hBi ii∈I a non-empty family of subalgebras of A. For each i ∈ I, we may identify L∞ (Bi ) with the closed linear subspace of L∞ (A) generated by {χb : b ∈ Bi } (363Ga). Suppose R that for each i ∈ I we are given a finitely additive functional νi : Bi → [0, 1] such that νi 1 = 1. Write . . . dνi for the corresponding positive linear functional on L∞ (Bi ) (363L). Then the following are equiveridical: (i) there is an additive functional µ : A → [0, 1] extending Pn every νi ; (ii) whenever i , . . . , i ∈ I, a ∈ B for k ≤ n, and 0 n k ik k=0 χak ≥ mχ1 in S(A), where m ∈ N, then Pn ν a ≥ m; k=0 ik k Pn k=0 χak ≤ mχ1, where m ∈ N, then Pn (iii) whenever i0 , . . . , in ∈ I, ak ∈ Bik for k ≤ n, and ν a ≤ m; k=0 ik k Pn Pn R (iv) whenever i0 , . . . , in ∈ I, uk ∈ L∞ (Bik ) for every k ≤ n, and P uk dνik ≥ 1; k=0 uk ≥ χ1, then i=0 P n ∞ R (v) whenever i0 , . . . , in ∈ I, uk ∈ L∞ (Bik ) for every k ≤ n, and k=0 uk ≤ χ1, then i=0 uk dνik ≤ 1. proof (a) It is elementary R to check that if (i) is true then R(ii)-(v) are all true, simply because we have a positive linear functional dµ extending all the functionals dνi . Pn (b)(ii)⇒(iii) Given that ak ∈ Bik and k=0 χak ≤ mχ1, then Pn Pn k=0 χ(1 \ ak ) = (n + 1)χ1 − k=0 χak ≥ (n + 1 − m)χ1,

457C

Simultaneous extension of measures

so

Pn k=0

νik ak = n + 1 −

Pn k=0

489

νik (1 \ ak ) ≤ n + 1 − (n + 1 − m) = m,

as required by (iii). (c)(iii)⇒(i) Assume (iii). Set ψa = sup{νi a : i ∈ I, a ∈ Bi } for a ∈ A (interpreting sup ∅ as 0, as usual in such contexts). Then (iii) tells us that ψ satisfies the condition (ii) of 391F. So there is a non-negative finitely additive functional µ such that µ1 = 1 and µa ≥ ψa for every a ∈ A, that is, µb ≥ νi b whenever i ∈ I and b ∈ Bi . But observe now that, because µ1 = ν1 and µ(1 \ b) ≥ νi (1 \ b), we actually have µb = νi b for every b ∈ Bi , so that µ extends νi , for every i ∈ I. (d)(iv)⇒(ii) & (v)⇒(iii) are elementary; if m > 0, apply (iv) or (v) to uk =

1 m χak .

457B Corollary Let X be a set and hYi ii∈I a family of sets. Suppose that for each i ∈ I we have an algebra Ei of subsets of Yi , an additive functional νi : Ei → [0, 1] such that νi Yi = 1, and a function fi : X → Yi . Then the following are equiveridical: (i) there is an additive functional µ : PX → [0, 1] such that µfi−1 [E] = νi E whenever i ∈P I and E ∈ Ei ; n (ii) whenever i0 , . . . , in ∈ I and Ek ∈ Eik for k ≤ n, then there is an x ∈ X such that k=0 νik Ek ≤ #({k : k ≤ n, fik (x) ∈ Ek }). Pn Pn Pn proof (i)⇒(ii) is elementary; if m = d k=0 νik Ek e − 1, then k=0 µfi−1 [Ek ] > mµX, so k=0 χfi−1 Ek 6≤ k k mχX, that is, there is an x ∈ X such that Pn Pn #({k : fik (x) ∈ Ek }) = k=0 (χfi−1 [Ek ])(x) ≥ m + 1 ≥ k=0 νik Ek . k (ii)⇒(i) Now suppose that (ii) is true. For i ∈ I set Bi = {fi−1 [E] : E ∈ Ei }. Note that if E ∈ Ei and νi E > 0, then (applying (ii) with n = 0, i0 = i and E0 = E) fi−1 [E] cannot be empty; accordingly we have 0 an additive functional νi0 : Bi → [0, 1] defined by setting νi0 f −1 [E] = Pνni E for every E ∈ Ei , and νi X = 1. If i0 , . . . , in ∈ I, H0 ∈ Bi0 , . . . , Hn ∈ Bin and m ∈ N are such that k=0 χHk ≤ mχX, express each Hk as fi−1 [Ek ], where Ek ∈ Eik ; then there is an x ∈ X such that k Pm Pn Pn 0 k=0 χHk (x) ≤ m. k=0 νik Ek ≤ #({k : fk (x) ∈ Ek }) = k=0 νik Hk = But this means that the condition of 457A(iii) is satisfied, with A = PX, so 457A(i) and (i) here are also true. 457C Corollary (a) Let A be a Boolean algebra and B1 , B2 two subalgebras of A with finitely additive functionals νi : Bi → [0, 1] such that ν1 1 = ν2 1 = 1. Then the following are equiveridical: (i) there is an additive functional µ : A → [0, 1] extending both the νi ; (ii) whenever b1 ∈ B1 , b2 ∈ B2 and b1 ∪ b2 = 1, then ν1 b1 + ν2 b2 ≥ 1; (iii) whenever b1 ∈ B1 , b2 ∈ B2 and b1 ∩ b2 = 0, then ν1 b1 + ν2 b2 ≤ 1. (b) Let X, Y1 , Y2 be sets, and for i ∈ {1, 2} let Ei be an algebra of subsets of Yi , νi : Ei → [0, 1] an additive functional such that νi Yi = 1, and fi : X → Yi a function. Then the following are equiveridical: (i) there is an additive functional µ : PX → [0, 1] such that µfi−1 [E] = νi E whenever i ∈ {1, 2} and E ∈ Ei ; (ii) f1−1 [E1 ] ∩ f2−1 [E2 ] 6= ∅ whenever E1 ∈ E1 , E2 ∈ E2 and ν1 E1 + ν2 E2 > 1; (iii) νE1 ≤ νE2 whenever E1 ∈ E1 , E2 ∈ E2 and f1−1 [E1 ] ⊆ f2−1 [E2 ]. proof (a)(i)⇒(iii) is elementary (and is a special case of 457A(i)⇒457A(iii)). (iii)⇒(ii) If (iii) is true, and b1 ∈ B1 , b2 ∈ B2 are such that b1 ∪ b2 = 1, then (1 \ b1 ) ∩ (1 \ b2 ) = 0, so ν1 b1 + ν2 b2 = 2 − ν1 (1 \ b1 ) − ν2 (1 \ b2 ) ≥ 1. (ii)⇒(i) The P Suppose that i0 , . . . , in ∈ {1, 2}, ak ∈ Bik Pnpoint is that (ii) here implies (ii) of 457A. P for kP≤ n and χa ≥ mχ1 in S(A), where m ∈ N. Set Kj = {k : kP≤ n, ik = j} for each j, k k=0 P m1 u = k∈K1 χak ∈ S(B1 ), v = k∈K2 χak ∈ S(B2 ). Then we can express u as j=0 χcj where cj ∈ B1 for each j ≤ m1 and c0 ⊇ c1 ⊇ . . . ⊇ cm1 (see the proof of 361Ec). Taking cj = 0 for m1 < j ≤ m if necessary,

490


457C

Pm2 we may suppose that m1 ≥ m. Similarly, v = j=0 χdj where m2 ≥ m, dj ∈ B2 for each j ≤ m2 and d0 ⊇ . . . ⊇ dm2 . For j < m, set bj = 1 \ (cj ∪ dm−j−1 ). Then, because bj ∩ cj = 0, Pm1 Pj−1 u × χbj = r=0 χ(cr ∩ bj ) = r=0 χ(cr ∩ bj ) ≤ jχbj , and similarly v × χbj ≤ (m − j − 1)χbj , so mχbj ≤ (u + v) × χbj = u × χbj + v × χbj ≤ (m − 1)χbj , and bj must be 0. Thus cj ∪ dm−j−1 = 1 for every j < m. But this means that ν1 cj + ν2 dm−j−1 ≥ 1 for every j < m, so that n X

νik ak =

k=0

X

ν1 ak +

k∈K1

=

m1 X j=0

X

Z ν2 ak =

Z u dν1 +

v dν2

k∈K2

ν1 cj +

m2 X

ν2 dj ≥

j=0

m−1 X

ν1 cj + ν2 dm−1−j ≥ m,

j=0

as required. Q Q Because 457A(ii) implies 457A(i), we have the result. (b) We can convert (i) and (ii) here into (a-i) and (a-iii) just above by the same translation as in 457B. So (i) and (ii) are equiveridical. As for (iii), this corresponds exactly to replacing E2 by Y2 \ E2 in (ii). *457D The proof of 457A is based, at some remove, on the Hahn-Banach theorem, as applied in the proof of 391E-391F. An alternative proof uses the max-flow min-cut theorem of graph theory. To show the power of this method I apply it to an elaboration of 457C, as follows. Proposition (Strassen 65) Let A be a Boolean algebra and B1 , B2 two subalgebras of A. Suppose that νi : Bi → [0, 1] are finitely additive functionals such that ν1 1 = ν2 1 = 1, and θ : A → [0, ∞[ another additive functional. Then the following are equiveridical: (i) there is an additive functional µ : A → [0, ∞[ extending both the νi , and such that µa ≤ θa for every a ∈ A; (ii) ν1 b1 + ν2 b2 ≤ 1 + θ(b1 ∩ b2 ) for every b1 ∈ B1 , b2 ∈ B2 . proof (a) As usual in this context, (i)⇒(ii) is elementary; if µ ≤ θ extends both νj , and bj ∈ Bj for both j, then ν1 b1 + ν2 b2 = µb1 + µb2 = µ(b1 ∪ b2 ) + µ(b1 ∩ b2 ) ≤ 1 + θ(b1 ∩ b2 ). (b) For the reverse implication, suppose to begin with (down to the end of (d) below) that A is finite. Let I, J and K be the sets of atoms of B1 , B2 and A respectively. Consider the transportation network (V, E, γ) where V = {(0, 0)} ∪ {(b, 1) : b ∈ I} ∪ {(d, 2) : d ∈ K} ∪ {(c, 3) : c ∈ J} ∪ {(1, 4)}, E = {e0b : b ∈ I} ∪ {e1d : d ∈ K} ∪ {e2d : d ∈ K} ∪ {e3c : c ∈ J}, where for b ∈ I, e0b runs from (0, 0) to (b, 1), for d ∈ K, e1d runs from (b, 1) to (d, 2), where b is the member of I including d, for d ∈ K, e2d runs from (d, 2) to (c, 3), where c is the member of J including d, for c ∈ J, e3c runs from (c, 3) to (1, 4). Define the capacity γ(e) of each link by setting γ(e0b ) = ν1 b for b ∈ I, γ(e1d ) = γ(e2d ) = θd for d ∈ K,

*457D


491

γ(e3c ) = ν2 c for c ∈ J. By the max-flow min-cut theorem (4A4N), there are a flow φ and a cut X of the same value; that is, we have a function φ : E → [0, ∞[ and a set X ⊆ E such that P P e starts from v φ(e) = e ends at v φ(e) for every v ∈ V \ {(0, 0), (1, 4)}, φ(e) ≤ γ(e) for every e ∈ E,

P e starts from (0,0)

φe =

P e ends at (1,4)

φe =

P e∈X

γ(e),

and there is no path from (0, 0) to (1, 4) using only links in E \ X. Now, for any d ∈ K, there is exactly one link e1d ending at d and exactly one link e2d starting from d. So φ(e1d ) = φ(e2d ), and we may define an additive functional µ on A by setting P P µa = d∈K,d ⊆ a φ(e1d ) = d∈K,d ⊆ a φ(e2d ) for every a ∈ A. (c)(i) µb ≤ ν1 b for every b ∈ B1 . P P Because I is the set of atoms of the finite Boolean algebra B1 , it is enough to show that µb ≤ ν1 b for every b ∈ I. Now, for such b, X X X µb = µd = φ(e1d ) = φ(e) d∈K,d ⊆ b

d∈K,d ⊆ b

X

=

φ(e) =

φ(e0b )

e starts from (b,1)

≤ γ(e0b ) = ν1 b,

e ends at (b,1)

because the only link ending at (b, 1) is e0b . Q Q (ii) Similarly, because the only link starting at (c, 3) has capacity ν2 c, µc ≤ ν2 c for every c ∈ J. But this means that µc ≤ ν2 c for every c ∈ B2 . (iii) In third place, because µd = φ(e1d ) ≤ γ(e1d ) = θd for every d ∈ K, µa ≤ θa for every a ∈ A. (d) (The key.) µ1 ≥ 1. P P We have X X µ1 = µd = φ(e1d ) d∈K

=

X

X

d∈K

φ(e1d ) =

b∈I d∈K,d ⊆ b

=

X

X

b∈I e ends at (b,1)

X

X

φ(e)

b∈I e starts from (b,1)

φ(e) =

X

φ(e) =

e starts from (0,0)

X

γ(e).

e∈X

Set b∗ = sup{b : b ∈ I, e0b ∈ X} ∈ B1 , a∗1 = sup{d : d ∈ K, e1d ∈ X}, a∗2 = sup{d : d ∈ K, e2d ∈ X}, c∗ = sup{c : c ∈ J, e3c ∈ X} ∈ B2 . For any d ∈ K, we have a four-link path e0b , e1d , e2d , e3c from (0, 0) to (1, 4), where b ∈ I, c ∈ J are the atoms of B1 , B2 including d. At least one of the links in this path must belong to X, so that d is included in b∗ ∪ a∗1 ∪ a∗2 ∪ c∗ . Thus, writing a = (1 \ b∗ ) ∩ (1 \ c∗ ), a ⊆ a∗1 ∪ a∗2 and θa ≤ θa∗1 + θa∗2 . But this means that

492


µ1 =

X

γ(e)

e∈X

X

=

X

γ(e0b ) +

b∈I,e0b ∈X

X

=

*457D

d∈K,e1d ∈X

X

ν1 b +

b∈I,e0b ∈X

X

γ(e1d ) +

d∈K,e2d ∈X

X

θd +

d∈K,e1d ∈X

X

θd +

d∈K,e2d ∈X

X

γ(e2d ) +

γ(e3c )

c∈J,e3c ∈X

ν2 c

c∈J,e3c ∈X

= ν1 b∗ + θa∗1 + θa∗2 + ν2 c∗ ≥ ν1 b∗ + θ((1 \ b∗ ) ∩ (1 \ c∗ )) + ν2 c∗ ≥ ν1 b∗ + ν1 (1 \ b∗ ) + ν2 (1 \ c∗ ) − 1 + ν2 c∗ (applying the hypothesis (ii)) = 1, as claimed. Q Q Since we already know that ν1 1 = 1 and that µb ≤ ν1 b for every b ∈ B1 , we must have µ1 = 1 and µb = ν1 b for every b ∈ B, so that µ extends ν1 . Similarly, µ extends ν2 . (e) Thus the proposition is proved in the case in which A is finite. In the general case, for each finite subset K of A write AK for the subalgebra of A generated by K. Then (b)-(d) tell us that there is a nonnegative additive functional µK on AK , dominated by θ on AK , agreeing with ν1 on AK ∩ B1 and agreeing with ν2 on AK ∩ B2 . Let µ be any cluster point of the µK in [0, 1]A as K increases through the finite subsets of A; then µ will be a non-negative additive functional on A, dominated by θ, and extending ν1 and ν2 . This proves the result. 457E Proposition Let X be a non-empty set and hνi ii∈I a family of probability measures on X satisfying the conditions of Lemma 457A, taking A = PX and Bi = dom νi for each i. Suppose that there is a countably compact class K ⊆ PX such that every νi is inner regular with respect to K. Then there is a probability measure µ on X extending every νi . proof If I = ∅ this is trivial. Otherwise, by 457A, there is a finitely additive functional ν on PX extending every νi . Now 413S tells us that there is a complete measure µ on X such that µX ≤ νX and µK ≥ νK for every K ∈ K. In this case, for any i ∈ I and E ∈ Ti = dom νi , we must have µ∗ E ≥

sup

µK ≥

K∈K,K⊆E

≥

sup K∈K∩dom νi ,K⊆E

sup

µK

K∈K∩dom νi ,K⊆E

νK =

sup K∈K∩dom νi ,K⊆E

νi K = νi E.

In particular, µX ≥ νi X = 1. Now µ∗ E = 1 − µ∗ (X \ E) ≤ 1 − νi (X \ E) = νi E for any E ∈ Ti ; as µ is complete, µE is defined and equal to νi E for every E ∈ Ti , and µ extends νi , as required. 457F Proposition (a) Let (X, Σ, µ) be a perfect probability space and (Y, T, ν) any probability space. Write Σ ⊗ T for the algebra of subsets of X × Y generated by {E × F : E ∈ Σ, F ∈ T}. Suppose that Z ⊆ X × Y is such that (i) Z is expressible as the intersection of a sequence in Σ ⊗ T, (ii) Z ∩ (E × F ) 6= ∅ whenever E ∈ Σ, F ∈ T are such that µE + νF > 1. Then there is a probability measure λ on Z such that the maps (x, y) 7→ x : Z → X and (x, y) 7→ y : Z → Y are both inverse-measure-preserving. N (b) Let Q h(Xi , Σi , µi )ii∈I be a family of perfect probability spaces. Write i∈I Σi for the algebra of subsets of X = i∈I Xi generated by {{x : x ∈ X, x(i) ∈ E} : i ∈ I, E ∈ Σi }. Suppose that Z ⊆ X is such that N (i) Z is expressible as the intersection of a sequence in i∈I Σi ,

457H


493

(ii) whenever i0P , . . . , in ∈ I and Ek ∈ Σik for k ≤ n, there is a z ∈ Z such that #({k : k ≤ n n, z(ik ) ∈ Ek }) ≥ k=0 µik Ek . Then there is a perfect probability measure λ on Z such that z 7→ z(i) : Z → Xi is inverse-measurepreserving for every i ∈ I. proof (a) Apply 457Cb to the coordinate maps f1 : Z → X and f2 : Z → Y . The condition (ii) here shows that 457C(b-ii) is satisfied, so there is an additive functional θ : PZ → [0, 1] such that θf1−1 [E] = µE for every E ∈ Σ and θf2−1 [F ] = νF for every F ∈ T. Define θ0 : Σ ⊗ T → [0, 1] by setting θ0 W = θ(Z ∩ W ) for every W ∈ Σ ⊗ T. Then θ0 (E × Y ) = µE for every E ∈ Σ and θ0 (X × F ) = νF for every F ∈ T. Because µ is perfect, θ0 has an extension to a measure T ˜ defined on Σ⊗T b (454C). Now Z is supposed to be expressible as n∈N Wn where Wn ∈ Σ ⊗ T for every λ n; since ˜ n = θ0 Wn = θ(Z ∩ Wn ) = θZ = 1 λW ˜ = 1. So if we take λ to be the subspace measure on Z induced by λ, ˜ λ will be a probability for every n, λZ measure on Z. If E ∈ Σ, then ˜ ∩ (E × Y )) = λ(E ˜ ×Y) λ(Z ∩ (E × Y )) = λ(Z = θ0 (E × Y ) = θ(Z ∩ (E × Y )) = µE. So f1 : Z → X is inverse-measure-preserving for λ and µ. Similarly, f2 : Z → Y is inverse-measurepreserving for λ and ν. (b) We use the same ideas, but appealing to 457B and 454D instead of 457Cb and 454C. Taking fi : X → Xi to be the coordinate map for each i ∈ I, (ii) here, with 457B, tells us that there is an additive functional θ : N PZ → [0, 1] such that θfi−1 [E] = µi E whenever i ∈ I and E N ∈ Σi . 0 Define θ : i∈I Σi → [0, 1] by setting θ0 W = θ(Z ∩ W ) for every W ∈ i∈I Σi . Then θ0 {x : x ∈ X, x(i) ∈ E} = θ{z : z ∈ Z, z(i) ∈ E} = µi E ˜ defined whenever i ∈ I and E ∈ Σi . Because every µi is perfect, θ0 has an extension to a perfect measure λ N T N on c i∈I Σi (454D). Now Z is supposed to be expressible as n∈N Wn where Wn ∈ i∈I Σi for every n; since ˜ n = θ0 Wn = θ(Z ∩ Wn ) = θZ = 1 λW ˜ = 1. So if we take λ to be the subspace measure on Z induced by λ, ˜ λ will be a probability for every n, λZ measure on Z; by 451Dc, λ is perfect. If i ∈ I and E ∈ Σi , then ˜ : x ∈ X, x(i) ∈ E} λ{z : z ∈ Z, z(i) ∈ E} = λ{x = θ0 {x : x ∈ X, x(i) ∈ E} = θ{z : z ∈ Z, z(i) ∈ E} = µi E. So z 7→ z(i) : Z → Xi is inverse-measure-preserving for λ and µi for every i ∈ I, as required. 457G Proposition Let X be a set and hµi ii∈I a family of probability measures on X which is upwardsdirected in the sense that for any i, j ∈ I there is a k ∈ I such that µk extends both µi and µj . Suppose that for any countable J ⊆ I there is a measure on X extending µi for every i ∈ J. Then there is a measure on X extending µi for every i ∈ I. S proof Set Σi = dom µi for each i ∈ I. Because hµi ii∈I is upwards-directed, T = i∈I Σi is an algebra of subsets of X, and we have a finitely additive functional ν : T → [0, 1] defined by saying that νE = µi E whenever i ∈ I and E ∈ Σi . Now if hEn in∈N is S any non-increasing sequence in T with empty intersection, there is a countable set J ⊆ I such that En ⊆ i∈J Σi for every n ∈ N. We are told that there is a measure λ on X extending µi for every i ∈ J; now νEn = λEn for every n ∈ N, so limn→∞ νEn = 0. By 413K, ν has an extension to a measure on X, which of course extends every µi . 457H Example Set X = {(x, y) : 0 ≤ x < y ≤ 1} ⊆ [0, 1]2 . Write π1 , π2 : X → R for the coordinate maps, and µL for Lebesgue measure on [0, 1], with ΣL its domain.

494


457Ha

(a) There is a finitely additive functional ν : PX → [0, 1] such that νπi−1 [E] = µL E whenever i ∈ {1, 2} and E ∈ ΣL . P P If E1 , E2 ∈ ΣL and µL E1 + µL E2 > 1, then neither is empty and inf E1 < sup E2 , so there are x ∈ E1 , y ∈ E2 such that x < y, and (x, y) ∈ π1−1 [E1 ] ∩ π2−1 [E2 ]. So the result follows by 457Cb. Q Q (b) However, there is no measure µ on X for which both π1 and π2 are inverse-measure-preserving. P P?? If there were,

R

π1 (x, y)µ(d(x, y)) =

R

xµL (dx) =

R

yµL (dy) =

R

π2 (x, y)µ(d(x, y))

by 235I; but π1 (x, y) < π2 (x, y) for every (x, y) ∈ X, so this is impossible. X XQ Q (c) If we write Ti = {πi−1 [E] : E ⊆ [0, 1] is Borel} for each i, then we have a measure νi with domain Ti defined by setting νi πi−1 [E] = µL E for each Borel set E ⊆ [0, 1]. Now ν1 and ν2 have no common extension to a Borel measure on X, even though X is a Polish space and each νi is a compact measure, being inner regular with respect to the compact class Ki = {πi−1 [K] : K ⊆ ]0, 1[ is compact}. (The trouble is that K1 ∪ K2 is not compact, so we cannot apply 457E.) 457I Example Let µL be Lebesgue measure on [0, 1] and ΣL its domain. Set P3 3 P3 X = {(ξ1 , ξ2 , ξ3 ) : 0 ≤ ξi ≤ 1 for each i, i=1 ξi ≤ , i=1 ξi2 ≤ 1}. 2

For 1 ≤ i ≤ 3 set πi (x) = ξi for x = (ξ1 , ξ2 , ξ3 ) ∈ X. P3 P Set (a) If Ei ∈ ΣL for i ≤ 3, then there is an x ∈ X such that #({i : πi (x) ∈ Ei }) ≥ i=1 µL Ei . P αi = inf(Ei ∪ {1}) for each i, and set P3 P3 P3 m = d i=1 µL Ei e ≤ d i=1 1 − αi e = 3 − b i=1 αi c, P3 P3 so that i=1 αi < 4 − m. Take ξi ∈ Ei ∪ {1} such that i=1 ξi < 4 − m. It will be enough to consider the case in which ξ1 ≤ ξ2 ≤ ξ3 . P3 (i) If m = 1, then i=1 ξi < 3 so ξ1 < 1 and ξ1 ∈ E1 . Set x = (ξ1 , 0, 0); then x ∈ X and P3 #({i : πi (x) ∈ Ei }) ≥ 1 ≥ i=1 µL Ei . (ii) If m = 2, then ξ1 + ξ2 ≤ 34 ≤ 32 . Also

P3

i=1 ξi

< 2 so ξ2 < 1 and ξ1 ∈ E1 , ξ2 ∈ E2 . Set x = (ξ1 , ξ2 , 0). We have 1 2

1 2

ξ2 ≤ (ξ2 + ξ3 ) ≤ 1 − ξ1 , so 1 2

5 4

ξ12 + ξ22 ≤ ξ12 + (1 − ξ1 )2 = 1 − ξ1 + ξ12 ≤ 1 because ξ1 ≤

2 3

≤ 45 . So x ∈ X and #({i : πi (x) ∈ Ei }) ≥ 2 ≥

(iii) If m = 3 then x ∈ X and

P3

i=1 ξi

P3 i=1

µL Ei .

< 1 so ξi ∈ Ei for every i; set x = (ξ1 , ξ2 , ξ3 ). Since #({i : πi (x) ∈ Ei }) = 3 ≥

P3 i=1

P3

2 i=1 ξi

≤

P3

i=1 ξi

≤ 1,

µL Ei .

Putting these together, we have the result. Q Q (b) There is no finitely additive functional ν on X such that νπi−1 [E] = µE for each i and every E ∈ ΣL . P P?? Suppose there were. Set Ti = {πi−1 [E] : E ∈ ΣL } and νi = ν¹ Ti for each i. Then νi is a probability measure on X; moreover, because X is compact, πi−1 [K] is compact for every compact K ⊆ [0, 1], so νi is inner regular with respect to the compact subsets of X. By 457E, the νi have a common extension to a countably additive measure µ. Now

457Xe


R X

so we must have ξ1 + ξ2 + ξ3 =

3 2

ξ1 + ξ2 + ξ3 µ(dx) = 3

R1 0

495 3 2

t dt = ,

for µ-almost every x; similarly,

R X

ξ12 + ξ22 + ξ32 µ(dx) = 3

R1 0

t2 dt = 1,

so we must have ξ12 + ξ22 + ξ32 = 1 for µ-almost every x. Since 3 2

( − ξ3 )2 = (ξ1 + ξ2 )2 ≤ 2(ξ12 + ξ22 ) ≤ 2(1 − ξ32 ) for almost every x, ξ3 − ξ32 ≥

1 12

for almost every x, which is impossible, since µ{x : ξ3 ≤

1 2

1 6

− √ } > 0. X XQ Q

457J Example There are a set X and a family hµi ii∈I of probability measures on X such that (i) for every countable set J ⊆ I there is a measure on X extending µi for every i ∈ J (ii) there is no measure on X extending µi for every i ∈ I. proof By 439Fc, there is an uncountable universally negligible subset of [0, 1]. Because [0, 1] and PN are uncountable Polish spaces, they have isomorphic Borel structures (424Cb), so there is an uncountable universally negligible set X0 ⊆ PN. The map a 7→ N \ a is an autohomeomorphism of PN, so X1 = {N \ a : a ∈ X0 } is universally negligible, and X = X0 ∪ X1 is universally negligible (439Cb). For n ∈ N, set En = {a : n ∈ a ∈ X} and Σn = {∅, En , X \ En , X}; note that, because X is closed under complementation, neither En nor X \ En is empty, and we have a probability measure µn with domain Σn defined by setting µn En = µn (X \ En ) = 12 . Next, for a ∈ X, set Σ0a = {∅, {a}, X \ {a}, X), and let µ0a be the probability measure with domain Σ0a defined by setting µ0a {a} = 0. If J ⊆ X is countable, then there is a probability measure on X extending µn for every n ∈ N and µ0a for every a ∈ J. P P Because X0 is uncountable, there is a b ∈ X0 such that neither b nor b0 = PN \ b belongs to J. Let µ be the probability measure with domain PX defined by setting µ{b} = µ{b0 } = 21 ; this extends all the µn and all the µ0a for a ∈ J. Q Q ?? Suppose, if possible, that µ is a measure on X extending every µn and every µ0a . In this case, because µ extends every µn , its domain includes the Borel σ-algebra B of X, and µ¹B is a Borel probability measure on X. Since X is universally negligible, there is a point a ∈ X such that µ{a} > 0; in which case µ cannot extend µ0a . X X Thus the µn , µ0a constitute a family of the kind required. 457X Basic exercises (a) Let X be a non-empty set and hνi ii∈I a family of probability measures on X satisfying the conditions of Lemma 457A, taking A = PX and Bi = dom νi for each i. Suppose that there is a totally finite measure θ on X such that θE is defined and greater than or equal to νi E whenever i ∈ I and νi measures E. Show that there is a measure on X extending every νi . (b) Find a set X and non-negative additive functionals µ1 , µ2 defined on subalgebras of PX which agree on dom µ1 ∩ dom µ2 but have no common extension to a non-negative additive functional. (Hint: take #(X) = 3.) (c) Let A be a Boolean algebra and hνi ii∈I a family of non-negative finitely additive functionals, each νi being defined on a subalgebra Bi of A. Show that if any finite number of the νi have a common extension to an additive functional on a subalgebra of A, then the whole family has a common extension to an additive functional on the whole algebra A. (d) Set X = {0, 1, 2} and in the algebra PX let Bi be the subalgebra {0, {i}, X \ {i}, X} for each i. Let νi : Bi → [0, 1] be the additive functional such that νi {i} = 21 , νi X = 1. Show that any pair of ν0 , ν1 , ν2 have a common extension to an additive functional on PX, but that the three together have no such extension. (e) Let A be a Boolean algebra, B a subalgebra of A, and ν : B → [0, ∞[, θ : A → [0, ∞[ additive functionals such that νb ≤ θb for every b ∈ B. Show directly, without using either 457D or 391F, that there

496


457Xe

is an additive functional µ : A → [0, ∞[, extending ν, such that µa ≤ θa for every a ∈ A. (Hint: first consider the case in which A is the algebra generated by B ∪ {c}.) >(f ) Let (Y1 , S1 , T1 , ν1 ) and (Y2 , S2 , T2 , ν2 ) be Radon probability spaces and X ⊆ Y1 × Y2 a closed set. Show that the following are equiveridical: (i) there is a measure on X such that the coordinate map from X to Yi is inverse-measure-preserving for both i; (ii) there is a Radon measure on X such that the coordinate map from X to Yi is inverse-measure-preserving for both i; (iii) for every compact K ⊆ Y1 , ν1 K ≤ ν2∗ (X[K]). (Hint: for (iii)⇒(ii), use 457C to show that there is a finitely additive functional ν on PX of the required type; now observe that ν must give large mass to compact subsets of X, and apply 413S.) (g) Let X ⊆ [0, 1]2 be a Lebesgue measurable set such that X ∩ (E × F ) is not negligible for any nonnegligible sets E, F ⊆ [0, 1]. (For the construction of such sets, see the notes to §325.) Show that there is a Radon measure on X such that both the coordinate projections from X to [0, 1] are inverse-measurepreserving, where [0, 1] is given Lebesgue measure. (Hint: show that there is a measure-preserving bijection φ between conegligible subsets of [0, 1] which is covered by X; φ can be taken to be of the form φ(x) = x−αn for x ∈ En .) (h) Set X = {(t, 2t) : 0 ≤ t ≤ 12 } ∪ {(t, 2t − 1) : 12 ≤ t ≤ 1}. Show that there is a Radon measure on X for which both the coordinate maps onto [0, 1] are inverse-measure-preserving, but that X does not include the graph of any measure-preserving bijection between conegligible subsets of [0, 1]. (i) Let X be the eighth-sphere {x : x ∈ [0, 1]3 , kxk = 1}. Show that there is a measure on X such that all three coordinate maps from X onto [0, 1] are inverse-measure-preserving. (Hint: 265Xe.) (j) Set X = {x : x ∈ [0, 1]3 , ξ1 + ξ2 + ξ3 = 32 }. Show that there is a measure on X such that all the coordinate maps from X onto [0, 1] are inverse-measure-preserving. (Hint: note that X is a regular hexagon; try one-dimensional Hausdorff measure on its boundary.) (k) Explain how to adapt the example in 457J to provide a family hµi ii∈I of probability measures on a set X such that (i) hµi ii∈I is upwards-directed (iii) there is no measure on X extending µi for every i ∈ I. 457Y Further exercises (a) Show that for any n ≥ 2 there are a finite set X and a family hµi ii≤n of measures on X such that {µi : i ≤ n, i 6= j} have a common extension to a measure on X for every j ≤ n, but the whole family {µi : i ≤ n} has no such extension. (b) RShow thatRthe example in 457H has the property: if fi is a νi -integrable real-valued function for each i, and f1 dν1 + f2 dν2 < 1, then there is an (x, y) ∈ dom f1 ∩ dom f2 such that f1 (x, y) + f2 (x, y) < 1. (c) Suppose we replace the set X in 457H with X 0 = X ∪ {(x, x) : x ∈ [0, 21 ]}, and write νi0 for the measures on X 0 defined by theR coordinate Show that (i) if fi is a νi0 -integrable real-valued R projections. 0 0 0 function on X for each i, and f1 dν1 + f2 dν2 ≤ 1, then there is an (x, y) ∈ dom f1 ∩ dom f2 such that f1 (x, y) + f2 (x, y) ≤ 1 (ii) there is no measure on X 0 extending both νi0 . (d) In 457Xj, show that there are many Radon measures on X such that all the coordinate maps from X onto [0, 1] are inverse-measure-preserving. 457Z Problems (a) Characterize the sets X ⊆ [0, 1]2 for which there is a measure on X such that both the projections from X to [0, 1] are inverse-measure-preserving. (b) Set X = {x : x ∈ [0, 1]3 , kxk = 1}. Is there more than one Radon measure on X for which all the coordinate maps from X onto [0, 1] are inverse-measure-preserving? (See 457Xi, 457Yd.) 457 Notes and comments In the context of this section, as elsewhere (compare 391E-391G and 391J), finitely additive extensions, as in 457A-457D, generally present easier problems than countably additive extensions. So techniques for turning additive functionals into measures (391D, 454C, 454D, 457E, 457G)

458B

Relative independence and relative products

497

are very valuable. Note that 457D offers possibilities in this direction: if θ there is countably additive, µ also will be (457Xa). 457H-457J demonstrate obstacles which can arise when seeking countably additive extensions even when finitely additive extensions give no difficulty. For finitely additive extensions a problem can arise at any finite number of measures (see 457Ya), but there is no further obstruction with infinite families (457Xc). For countably additive measures we have a positive result (457G) only under very restricted circumstances; relaxing any of the hypotheses can lead to failure (457J, 457Xk). Even in the apparently concrete case in which we have an open or closed set X ⊆ [0, 1]2 and we are seeking a measure on X with prescribed image measures on each coordinate, there can be surprises (457H, 457Xg, 457Xh), and I know of no useful description of the sets for which such a measure can be found (457Za). The two-dimensional case has a special feature: when verifying the conditions (ii) or (iii) in 457A, or the condition (ii) of 457B, it is enough to consider only one set associated with each coordinate (457C). Put another way, in conditions (iv) and (v) of 457A it is enough to examine characteristic functions. This is not the case as soon as we have three coordinates (457I).

458 Relative independence and relative products Stochastic independence is one of the central concepts of probability theory, and pervades measure theory. We come now to a generalization of great importance. If X1 , X2 and Y are random variables, we may find that X1 and X2 are ‘relatively independent over Y ’, or ‘independent when conditioned on Y ’, in the sense that if we know the value of Y , then we learn nothing further about one of the Xi if we are told the value of the other. For any stochastic process, where information comes to us piecemeal, this idea is likely to be fundamental. In this section I set out a general framework for discussion of relative independence (458A), introducing relative distributions (458C) and relative independence in measure algebras (458H). In the second half of the section I look at ‘relative product measures’ (458I, 458L), giving the basic existence theorems (458J, 458N, 458O). 458A Relative independence Let (X, Σ, µ) be a probability space and T a σ-subalgebra of Σ. (a) I say that a family hEi ii∈I in Σ is relatively (stochastically) independent over T if whenever J T ⊆ I is finiteR Q and not empty, and giR isQa conditional expectation of χEi on T for each i ∈ J, then µ( i∈J Ei ) = i∈J gi dµ. (Note that i∈J gi dµ does not depend on which conditional expectation gi we take.) A family hΣi ii∈I of σ-subalgebras of Σ is relatively independent over T if hEi ii∈I is relatively independent over T whenever Ei ∈ Σi for every i ∈ I. (b) I say that a family hfi ii∈I in L0 (µ) (the space of almost-everywhere-defined virtually measurable real-valued functions, or ‘random variables’) is relatively independent over T if hΣi ii∈I is relatively independent over T (with respect to the completion of µ), where Σi is the σ-algebra defined by fi in the sense of 272C, that is, the σ-algebra generated by {fi−1 [F ] : F ⊆ R is a Borel set}. (c) I remark at once that a family of σ-algebras or random variables is relatively independent iff every finite subfamily is (cf. 272Bb). 458B It is sometimes useful to know that ‘relative independence’ can be defined without using the apparatus of conditional expectations. Lemma Let (X, Σ, µ) be a probability space, T0 a σ-subalgebra of Σ, and hΣi ii∈I a family of σ-subalgebras of Σ. Let T be the family of finite subalgebras ofQ T. For Λ ∈ T write AΛ for the set of non-negligible atoms in Λ. For non-empty finite J ⊆ I and hEi ii∈J ∈ i∈J Σi , set φΛ (hEi ii∈J ) =

P H∈AΛ

µH ·

Q i∈J

µ(Ei ∩H) . µH

Then hΣi ii∈I is relatively independent over T iff limΛ∈T,Λ↑ φΛ (hEi ii∈J ) = µ( finite and not empty and Ei ∈ Σi for every i ∈ J.

T i∈J

Ei ) whenever J ⊆ I is

498


458B

proof (a) The point is just that if J ⊆ I is Rfinite Q and not empty, Ei ∈ Σi for i ∈ J, and gi is a conditional expectation of χEi on T for each i, then P Adjusting each gi on a i∈J gi dµ = limΛ↑ φΛ (hEi ii∈J ). P negligible set if necessary, we may suppose that it is T-measurable, defined everywhere on X and takes values between 0 and 1. Fix n ∈ N for the moment. Let Λn be the finite subalgebra of T generated by sets of the form {x : gi (x) ≤ 2−n k} for i ∈ J and 1 ≤ k ≤ 2n + 1, and Λ any finite subalgebra of Σ0 including Λn . If H is an atom of Λ and µH > 0, then there are integers ki , for i ∈ J, 2−n ki ≤ gi (x) < 2−n (ki + 1) for every i ∈ J and x ∈ H. So 2−n ki ≤

µi (E∩H) µH

< 2−n (ki + 1) P Q P Q for each i. Accordingly between H∈AΛ µH · i∈J 2−n ki and H∈AΛ µH · i∈J 2−n (ki +1), R Q φΛ0(hEi ii∈J ) Rlies Q −n 00 0 00 (x) = min(1, 2−n (k + 1)) when k, gin that is, between i∈J gin dµ, where gin (x) = 2 i∈J gin dµ and −n −n 2 k ≤ gi (x) < 2 (k + 1). But this means that

|φΛ (hEi ii∈J ) −

Z Y

Z gi dµ| ≤ max(

i∈J

|

Y

0 gin

i∈J

Z X

≤ max(

−

Y

(because all the

00 gin

gi |dµ,

i∈J

0 |gin − gi |dµ,

i∈J 0 gi , gin ,

Z Z X

|

Y

00 gin −

i∈J

Y

gi |dµ)

i∈J

00 |gin − gi |dµ)

i∈J

take values in [0, 1], see 285O) ≤ 2−n #(J).

Since this is true for every Λ ⊇ Λn and every n ∈ N, limΛ↑ φΛ (hEi ii∈J ) =

RQ i∈J

gi dµ. Q Q

(b) Accordingly the condition given exactly matches the definition in 458A. 458C All the fundamental theorems concerning stochastic independence have relativized forms. For the first we need a concept of ‘relative probability distribution’, as follows. Definition Let (X, Σ, µ) be a probability space, T a σ-subalgebra of subsets of X, and f ∈ L0 (µ). Then a relative distribution of f over T will be a family hνx ix∈X of Radon probability measures on R such that R x 7→ νx (H) : X → [0, 1] is T-measurable and F νx (H)µ(dx) = µ(F ∩ f −1 [H]) for every Borel set H ⊆ R and every F ∈ T, that is, x 7→ νx H is a conditional expectation of χf −1 [H] on T. 458D Theorem Let (X, Σ, µ) be a probability space, T a σ-subalgebra of Σ, and f ∈ L0 (µ). Then there is a relative distribution of f over T, which is essentially unique in the sense that if hνx ix∈X and hνx0 ix∈X are two such families, then νx = νx0 for µ¹ T-almost every x. ˆ for the domain of µ proof (a) Write µ0 for the restriction of µ to T, µ ˆ for the completion of µ, Σ ˆ, and B for ˆ T⊗B)-measurable, b the Borel σ-algebra of R. Then the function x 7→ (x, f (x)) : dom f → X × R is (Σ, just ˆ for every F ∈ T and H ∈ B. So we have a probability measure λ on X × R defined because F ∩ f −1 [H] ∈ Σ b by setting λW = µ{x : (x, f (x)) ∈ W } for every W ∈ T⊗B. The marginal measure on R is tight (that is, inner regular with respect to the compact sets) just because it is a Borel probability measure (433Ca). By R 452M, we have a family hλx ix∈X of Radon probability measures on R such that λW = λx W −1 [{x}]µ0 (dx) b for every W ∈ T⊗B. (b) The functions x 7→ λx H need not be T-measurable. However, if we set gq (x) = λx (]−∞, q]) for q ∈ Q, then every gq is µ0 -virtually measurable, so there is a µ0 -conegligible set G such that every gq ¹G is T-measurable. By the Monotone Class Theorem (136B), the family {H : H ∈ B, x 7→ λx H : G → [0, 1] is T-measurable} is the whole of B. So if we set νx = λx for x ∈ G, and take λx to be the point mass at 0 (for instance) for x ∈ X \ G, we shall have a relative distribution of f over T as defined in 458C.

458E


499

hνx0 ix∈X is another relative distribution of f over T. Then for each H ∈ B we have R (c) Now suppose R that 0 ν Hµ(dx) = F νx Hµ(dx) for every F ∈ T, so that νx H = νx0 H for µ0 -almost every x. But this means F x that for µ0 -almost every x, we have νx H = νx0 H for every interval H with rational endpoints; and for such x we must have νx = νx0 (416Eb). 458E Now we can state and prove a result corresponding to 272G. Theorem Let (X, Σ, µ) be a probability space, T a σ-subalgebra of Σ, and hfi ii∈I a family in L0 (µ). For each i ∈ I, let hνix ix∈X be a relative distribution of fi over T, and f˜i : X → R an arbitrary extension of fi to the whole of X. Then the following are equiveridical: (i) hfi ii∈I is relatively independent over T; (ii) for any Baire set W ⊆ RI and any F ∈ T, µ ˆ(F ∩ f −1 [W ]) =

R

F

λx W µ(dx),

where µ ˆ is the completion of µ, f (x) = hf˜i (x)ii∈I for x ∈ X, and λx is the product of hνix ii∈I for each x. R proof (a) Note first that if i ∈ I and H ⊆ R is a Borel set, then F νix Hµ(dx) = µ ˆ(F ∩ fi−1 [H]) for every −1 F ∈ T, so x 7→ νix H is a conditional expectation of χfi [H] on T. Suppose that hfi ii∈I is relatively independent, and F ∈ T. Let C be the family of subsets of RI expressible in the form C = {z : z ∈ RI , z(i) ∈ Hi for every i ∈ J} where J ⊆ I is finite and Hi ⊆ R is a Borel set for each i ∈ J. For such a set C,

µ ˆ(F ∩ f −1 [C]) = µ ˆ(F ∩

\

f˜i−1 [Hi ]) =

Z Y n

(interpreting an empty product as χX)

νix Hi µ(dx)

F i=0

i∈J

Z =

λx Cµ(dx). F

So W = {W : W ⊆ RI , µ(F ∩ f −1 [W ]) =

R

λx W µ(dx)}

includes C; since it is a Dynkin class, it contains every Baire subset of RI (by the Monotone Class Theorem, 136B), and (ii) is true. (b) Now suppose that (ii) is true. Let Σi be the σ-algebra defined by fi for each i. If J ⊆ I is finite and Ei ∈ Σi for each i ∈ J, then there are Borel sets Hi ⊆ R such that Ei 4fi−1 [Hi ] is negligible for each i, so that x 7→ νix Hi is a conditional expectation of χEi on T. Now by the same equations as before, in the opposite direction, Z Y

Z νix Hi µ(dx) =

F i∈J

λx Cµ(dx) F

(where C = {z : z(i) ∈ Hi for i ∈ J}) =µ ˆ(F ∩ f −1 [C]) = µ ˆ(F ∩

\ i∈J

fi−1 [Hi ]) = µ ˆ(F ∩

\

Ei )

i∈J

for every F ∈ T. As hEi ii∈J is arbitrary, hΣi ii∈I and hfi ii∈I are relatively independent. ˜ Remarks Of T course the ungainly shift to fi is unnecessary if I is countable; but for uncountable I the intersection i∈I dom fi , which is the only suitable domain for f , may not be conegligible. I said that λx should be ‘the product of hνix ii∈I ’. Since the νix are Radon probability measures, we have two possible interpretations of this: either the ‘ordinary’ product measure of §254 or the ‘quasi-Radon’ product measure of §417. But as we are interested only in the values of λx W for Baire sets W , it makes no difference which we choose.

500


458F

458F Proposition (Compare 272K.) Let (X, Σ, µ) be a probability space, T a σ-subalgebra of Σ, and hΣi ii∈I a family of σ-subalgebras of Σ which is relatively independent over T. Let hIj ij∈J be a partition of ˜ j be the σ-algebra generated by S ˜ I, and for each j ∈ J let Σ i∈Ij Σi . Then hΣj ij∈J is relatively independent over T. proof For each E ∈ Σ let gE be a conditional expectation of χE on T. Take T any finite K R⊆ J, Q and let W ˜ j for each j ∈ K and µ(F ∩ be the set of families hWj ij∈K such that Wj ∈ Σ j) = F j∈K W j∈K fWj dµ T for every F ∈ T. For each j ∈ K, let Cj be the family of sets expressible as W = X ∩ i∈L Ei where L ⊆ Ij is finite and Ei ∈ Σi for i ∈ L. Note that in this case

RQ

T

µ(F ∩ W ) = µ(F ∩ i∈L Ei ) = f dµ i∈L Ei Q for every F ∈ T, so fW =a.e. i∈L fEi , taking the product to be χX if L isTempty. If Wj ∈ Cj for each j ∈ K, then hWj ij∈J ∈ W . P P Express Wj as X ∩ i∈Lj Ei where Lj ⊆ Ij is finite and Ei ∈ Σi whenever j ∈ K and i ∈ Lj . Then µ(F ∩ (where L =

\

\

Wj ) = µ(F ∩

j∈K

S j∈K

Ei )

i∈L

Lj ) =

Z Y F i∈L

(because hΣi ii∈I is relatively independent)

Z =

fEi dµ

Y Y

F j∈K i∈L j

Z fEi dµ =

Y F j∈K

fWi dµ

for every F ∈ T. Q Q ˜ k such that Observe next that if we fix k ∈ K, and a family hWj ij∈K\{k} , then the set of those Wk ∈ Σ hWj ij∈K ∈ W is a Dynkin class, so if it includes Ck it must include the σ-algebra generated by Ck , viz., ˜ ˜ k . Now an easy induction on n shows that if hWj ij∈K ∈ Q / Cj }) = n, then Σ j∈K Σj and #({j : Wj ∈ Q ˜j ⊆ W . hWj ij∈K ∈ W . Taking n = #(K) we see that j∈K Σ ˜ j ij∈J is relatively independent over T, as claimed. As this is true for every finite K ⊆ J, hΣ 458G Example The simplest examples of relatively independent σ-algebras arise as follows. Let (X, Σ, µ) be a probability space, hTi ii∈I an independent family of σ-subalgebras of Σ, as in 272Ab, and T a σS subalgebra of Σ which is independent of the σ-algebra generated by i∈I Ti . For each i ∈ I, let Σi be the σ-algebra generated by T ∪ Ti . Then hΣi ii∈I is relatively independent over T. proof For i ∈ I, set Σi0 = Ti and Σi1 = T. Then hΣi² i(i,²)∈I×{0,1} is relatively independent over T. P P Let J ⊆ I × {0, 1} be a non-empty finite set and Ei² ∈ Σi² for each (i, ²) ∈ J. Set J0 = {i : (i, 0) ∈ J}, J1 = {i : (i, 1) ∈ J}. Then gi0 = µEi0 χX is a conditional expectation of χEi0 on T for each i ∈ J0 , and gi1 = χEi1 is a conditional expectation of χEi1 on T for each i ∈ J1 . If F ∈ T, then µ(F ∩

\

\

Ei² ) = µ((F ∩

Ei² ) ∩

i∈J1

(i,²)∈J

= µ(F ∩

\

Ei² ) ·

i∈J1

(taking the product to be 1 if J0 is empty)

Z =

Y

\

Ei² )

i∈J0

Y

µEi²

i∈J0

gi² dµ.

F (i,²)∈J

As F and hEi² i(i,²)∈J are arbitrary, hΣi² i(i,²)∈I×{0,1} is relatively independent over T. Q Q Since each Σi is the σ-algebra generated by Σi0 ∪ Σi1 , 458F tells us that hΣi ii∈I is also relatively independent over T.

458J


501

458H Measure algebras We can look at the same ideas in the context of measure algebras. (a) Suppose that (A, µ ¯) is a probability algebra, and that C is a closed subalgebra of A. If a ∈ A, then R we can say that u ∈ L∞ (C) is the conditional expectation of χa on C if c u = µ ¯(c ∩ a) for every c ∈ C (365R). Now we can say that a family hbi ii∈I in A is relatively (stochastically) independent over C if R Q µ ¯(c ∩ inf i∈J bi ) = c i∈J ui whenever J ⊆ I is a non-empty finite set and ui is the conditional expectation of χbi on C for every i ∈ J; while a family hBi ii∈I of subalgebras of A is relatively (stochastically) independent over C if hbi ii∈I is relatively independent over C whenever bi ∈ Bi for every i ∈ I. Following 458F, we have the result that if hBi ii∈I is relatively independent over C, and hIj ij∈J is a ˜ j is the closed subalgebra of A generated by S ˜ partition of I, and B i∈Ij Bi for every j ∈ J, then hBj ij∈J is relatively independent over C. The most natural proof, from where we are now standing, is to express (A, µ ¯) as the measure algebra of a probability space (X, Σ, µ), set T = {F : F • ∈ C} and Σi = {E : E • ∈ Bi } for each i ∈ I, and use 458F.) (b) Note that if a ∈ A and u is the conditional expectation of a on C, then [[u > 0]] = upr(a, C), by 365Rc. So if hBi ii∈I is a family of subalgebras of A which is relatively independent over C, and J ⊆ I is finite, and bi ∈ Bi for each i ∈ J, then inf i∈J bi = 0 iff inf i∈J upr(bi , C) = 0. (If ui is a conditional expectation of bi on C for each i, then Q inf i∈J upr(bi , C) = inf i∈J [[ui > 0]] = [[ i∈J ui ]] RQ is zero iff µ ¯(inf i∈J bi ) = i∈J ui = 0.) (c) We have a straightforward version of 458G, as follows. If (A, µ ¯) is a probability algebra, hCi ii∈I is a (stochastically) independent family of closed subalgebras of A, C is a closed subalgebra of A independent S of the algebra generated by i∈I Ci , and Bi is the closed subalgebra of A generated by C ∪ Ci for each i, then hBi ii∈I is relatively independent over C. (Either repeat the proof of 458G, looking at Bi0 = Ci and Bi1 = C for each i, or move to a measure space representing A and quote 458G.) 458I Relative free products of probability algebras: Definition Let h(Ai , µ ¯i )ii∈I be a family of probability algebras and (C, ν¯) a probability algebra, and suppose that we are given a measure-preserving Boolean homomomorphism πi : C → Ai for each i ∈ I. A relative free product of h(Ai , µ ¯i , πi )ii∈I over (C, ν¯) is a probability algebra (A, µ ¯), together with a measure-preserving Boolean homomorphism φi : Ai → A for each i ∈ I, such that φi πi = φj πj : C → A for all i, j ∈ I, writing D for the common value of the φi [πi [C]], S hφi [Ai ]ii∈I is relatively independent over D, A is the closed subalgebra of itself generated by i∈I φi [Ai ]. 458J Theorem Let h(Ai , µ ¯i )ii∈I be a family of probability algebras, (C, ν¯) a probability algebra and πi : C → Ai a measure-preserving Boolean homomomorphism for each i ∈ I. Then h(Ai , µ ¯i , πi )ii∈I has an essentially unique relative free product over (C, ν¯). proof (a)(i) Let B be the free product of hAi ii∈I (315H); write εi : Ai → B for the canonical embedding of Ai in B. For each i ∈ I, a ∈ A we have a completely additive functional c 7→ µ ¯i (a ∩ πi c) : C → [0, 1]; let R uia ∈ L∞ (C) be a corresponding Radon-Nikod´ ym derivative, so that c uia d¯ ν=µ ¯i (a ∩ πi c) for every c ∈ C (365E). By 326Q, there is a unique additive functional λ : B → [0, 1] such that λ(inf i∈J εi ai ) =

RQ

i∈J

ui,ai d¯ ν

whenever J ⊆ I is a non-empty finite set and ai ∈ Ai for every i ∈ J. Of course ui1 = χ1 in L∞ (C) for every i ∈ I, interpreting the ‘1’ in ui1 in the Boolean algebra Ai , and the ‘1’ in χ1 in the Boolean algebra C; so (this time interpreting ‘1’ in B) λ1 = 1 (the final ‘1’ being a real number, of course). Note also that ui,a∩πi c = χc × uia for every c ∈ C and a ∈ Ai , because

R

c0

for every c0 ∈ C.

ui,a∩πi c d¯ ν=µ ¯(a ∩ πi c ∩ πi c0 ) =

R

c∩c0

uia d¯ ν=

R

c0

χc × uia d¯ ν

502


458J

(ii) By 393C, there are a probability algebra (A, µ ¯) and a Boolean homomorphism φ : B → A such that λ = µ ¯φ. We can of course suppose that A is the order-closed subalgebra of itself generated by φ[B] (which is in fact automatically the case if we use the construction in the proof of 393C). For each i ∈ I, set φi = φεi : Ai → A. It is a Boolean homomorphism because φ and εi are. If a ∈ Ai , then µ ¯φi a = λεi a =

R

uia d¯ ν=µ ¯i (a ∩ πi 1) = µ ¯i a,

so φi is measure-preserving. If i, j ∈ I and c ∈ C then µ ¯(φi πi c 4 φj πj c) = λ(εi πi c 4 εj πj c) = λ(εi πi c) + λ(εj πj c) − 2λ(εi πi c ∩ εj πj c) Z Z Z = ui,πi c d¯ ν + uj,πj c d¯ ν − 2 ui,πi c × uj,πj c d¯ ν Z Z Z = χc d¯ ν + χc d¯ ν − 2 χc × χc d¯ ν = 0. So φi πi = φj πj ; write π for this common value. Set D = π[C], so that π is a measure-preserving Boolean isomorphism between C and D. Let T : L0 (C) → L0 (D) be the corresponding f -algebra isomorphism. (iii) If i ∈ I and a ∈ Ai , then T uia is the conditional expectation of χ(φi a) on D. P P For any d ∈ D, set c = π −1 d; then µ ¯(d ∩ φi a) = µ ¯(φi (πi c ∩ a)) = λ(εi (πi c ∩ a)) Z Z Z = ua∩πi c d¯ ν = uia d¯ ν = T uia c

d

by 365H. Q Q (iv) Suppose that J ⊆ I is finite and not empty and that di ∈ φi [Ai ] for each i ∈ J. For each i ∈ I let vi ∈ L∞ (D) be the conditional expectation of χdi on D, and let ai ∈ Ai be such that φi a = di . Set c = π −1 d. Then µ ¯(d ∩ inf di ) = µ ¯(inf (di ∩ d)) = λ(inf εi (ai ∩ πi c)) i∈J i∈J i∈J Z Y Z Y = ui,ai ∩πi c d¯ ν= ui,ai d¯ ν Z

i∈J

=

T( πc

Y

c i∈J

Z Y

ui,ai ) d¯ µ=

d i∈J

i∈J

T ui,ai d¯ µ=

Z Y

vi d¯ µ.

d i∈J

As hdi ii∈J and d are arbitrary, hφi [Ai ]ii∈I is relatively independent over D. This shows that (A, µ ¯, hφi ii∈I ) is a relative free product of h(Ai , µ ¯i , πi )ii∈I over (C, ν¯). (b) Now suppose that (A0 , µ ¯0 , hφ0i ii∈I ) is another relative free product of h(Ai , µ ¯i , πi )ii∈I over (C, ν¯). Then we have a Boolean homomorphism ψ : B → A0 such that φ0i = ψεi for every i ∈ I (315I). In this case, µ ¯0 ψ = λ. P P Let π 0 be the common value of φ0i πi for i ∈ I, set D0 = π[C], and let T 0 : L0 (C) → L0 (D0 ) be the isomorphism corresponding to π 0 : C → D. If a ∈ A and c ∈ C, then µ ¯0 (φ0i a ∩ π 0 c) = µ ¯(a ∩ πc ) =

R

c

uia d¯ ν=

R

π0 c

T 0 uia d¯ µ0

by 365H (compare (a-iii) above); it follows that T 0 uia is the conditional expectation of χ(φ0i a) on D0 . If J ⊆ I is finite and not empty, and ai ∈ Ai for i ∈ J, then Z Y µ ¯0 (ψ(inf εi ai )) = µ ¯0 (inf φ0i ai ) = T 0 ui,ai d¯ µ0 i∈J

i∈J

Z =

T 0(

Y

i∈J

i∈J

ui,ai )d¯ µ0 =

Z Y i∈J

ui,ai d¯ ν = λ(inf εi ai ). i∈J

458K


503

Because λ is the only additive functional on B taking the right values on elements of this form, µ ¯0 ψ = λ. Q Q 0 In particular, ψb = 0 whenever b ∈ B and λb = 0. It follows that ψb = ψb whenever b, b0 ∈ B and φb = φb0 , since in this case λ(b 4 b0 ) = µ ¯(φb 4 φb0 ) is zero. So we have a function θ : φ[B] → A0 defined by setting θ(φb) = ψb for every b ∈ B, and of course θ is a Boolean homomorphism; moreover, µ ¯0 θ(φb) = µ ¯0 ψb = λb = µ ¯φb so θ is measure-preserving and an isometry for the measure metrics of A and A0 . If i ∈ I and a ∈ A, then θφi a = θφεi a = ψεi a = φ0 a,

S S so θφi = φ0i for every i. Because A and A0 are the closed subalgebras generated by i∈I φi [Ai ] and i∈I φ0i [Ai ] respectively, φ[B] and ψ[B] are dense (323J). The isometry θ therefore extends uniquely to a measure ˆ i = φ0 for every i. algebra isomorphism θˆ : A → A0 which must be the unique isomorphism such that θφ i Thus (A, µ ¯, hφi ii∈I ) and (A0 , µ ¯0 , hφ0i ii∈I ) are isomorphic, and the relative free product is essentially unique. 458K Developing the argument of the last part of the proof of 458J, we have the following. ¯0i )ii∈I be two families of probability algebras, and ψi : Ai → A0i a Theorem Let h(Ai , µ ¯i )ii∈I , h(A0i , µ measure-preserving Boolean homomorphism for each i. Let (C, ν¯), (C0 , ν¯0 ) be probability algebras and πi : C → Ai , πi0 : C0 → A0i measure-preserving Boolean homomomorphisms for each i ∈ I; suppose that we have a measure-preserving isomorphism ψ : C → C0 such that πi0 ψ = ψi πi : C → A0i for each i. Let (A, µ ¯, hφi ii∈I ) and (A0 , µ ¯0 , hφ0i ii∈I ) be relative free products of h(Ai , µ ¯i , πi )ii∈I , h(A0i , µ ¯0i , πi0 )ii∈I over (C, ν¯), (C0 , ν¯0 ) respectively. Then there is a unique measure-preserving Boolean homomorphism ψˆ : A → A0 such ˆ i = φ0 πi : Ai → A0 for every i ∈ I. that ψφ i proof By the uniqueness assertion of 458J, we may suppose that (A, µ ¯, hφi ii∈I ) has been constructed by the method in the proof of 458J. (a) For i ∈ A, a ∈ A, a0 ∈ A0 let uia ∈ L∞ (C), u0ia0 ∈ L∞ (C0 ) be defined as in the proof of 458J, so that R R 0 u d¯ ν=µ ¯i (a ∩ πi c), u d¯ ν0 = µ ¯0i (a0 ∩ πi0 c0 ) c ia c0 ia0 whenever c ∈ C, c0 ∈ C0 . Let T : L0 (C) → L0 (C0 ) be the sequentially order-continuous Riesz homomorphism such that T (χc) = χ(ψc) for every c ∈ C (364R). Now u0i,ψi a = T uia whenever i ∈ I and a ∈ Ai . P P If c ∈ C, then Z

Z T uia d¯ ν0 =

ψc

uia d¯ ν c

(365H)

Z 0

0

=µ ¯(a ∩ πi c) = µ ¯ ψi (a ∩ πi c) = µ ¯

(ψi a ∩ πi0 ψc)

= ψc

ui,ψi a .

Because ψ is surjective, it follows that T uia = ui,ψi (a) . Q Q (b) Let B be the free product of hAi ii∈I εi : Ai → B the canonical embedding for each i; let λ be the functional on B defined by the process of (a-i) in the proof of 458J. By 315I, there is a Boolean homomorphism θ : B → A0 such that θεi = φ0i ψi : Ai → A0 for every i. Now µ ¯0 θ = λ. P P If J ⊆ I is finite and ai ∈ Ai for every i ∈ J, then

µ ¯0 θ(inf εi ai ) = µ ¯0 (inf φi ψi ai ) = i∈J

i∈J

=

Z Y i∈J

∞

(because T ¹L (C) is multiplicative)

Z Y

u0i,ψi ai d¯ ν0

i∈J

Z T ui,ai d¯ ν0 =

T(

Y

i∈J

ui,ai )d¯ ν0

504


=

Z Y

458K

ui,ai )d¯ ν = λ(inf εi ai ). i∈J

i∈J

As λ, θ and ν¯ are all additive, λ = ν¯θ0 (using 315Jb). Q Q (c) Let φ : B → A be the map described in (a-ii) of the proof of 458J. Then µ ¯(φb) = λb = µ ¯0 (θb) for every b ∈ B; in particular, φb = 0 ⇒ µ ¯(φb) = 0 ⇒ µ ¯0 (θb) = 0 ⇒ θb = 0. ˜ = θ, and θ˜ is measure-preserving on There is therefore a Boolean homomorphism θ˜ : φ[B] → A0 such that θφ ˜ φ[A]. Since φ[B] is topologically dense in A (use 323H), θ has an extension to a measure-preserving Boolean homomorphism ψˆ : A → A0 (324O). Now, for i ∈ I and a ∈ Ai , ˆ i a = ψφε ˆ i a = θφε ˜ i a = θεi a = φ0 ψi a, ψφ i as required. (d) To see that ψˆ is unique, we need observe only that the given formula defines in on the subalgebra ˆ being measure-preserving, must be continuous. φ[B] and that this is topologically dense in A, while ψ, 458L Relative product measures: Definitions (a) Let hXi ii∈I be a family of sets, Y a Q set, and πi : Xi → Y a function for each i ∈ I. The fiber product of h(Xi , πi )ii∈I is the set ∆ = {x : x ∈ i∈I Xi , πi x(i) = πj x(j) for all i, j ∈ I}. (b) Let h(Xi , Σi , µi )ii∈I be a family of probability spaces and (Y, T, ν) a probability space, and suppose that we are given an inverse-measure-preserving function πi : Xi → Y for each i ∈ I; let ∆ be the fiber product of hXi , πi ii∈I . A relative product measure on ∆ is a probability measure µ on ∆ such that (†) whenever J ⊆ I is finite and not empty and Ei ∈ Σi for i ∈ J, and gi is a Radon-Nikod´ ym derivative of the functional F 7→ µi (E ∩ πi−1 [F ]) :R TQ→ [0, 1] for each i ∈ J, then µ{x : x ∈ ∆, x(i) ∈ Ei for every i ∈ J} is defined and equal to i∈J gi dν; (‡) for every W ∈ Σ there is a W 0 in the σ-algebra generated by {{x : x ∈ ∆, x(i) ∈ E} : i ∈ I, E ∈ Σi } such that µ(W 4W 0 ) = 0. Remark If µ is a relative product measure of h(µi , πi )ii∈I over ν, then all the functions x 7→ x(i) : ∆ → Xi are inverse-measure-preserving. P P The condition (†) tells us that if E ∈ Σi and g is any Radon-Nikod´ ym derivative of F 7→ µi (E ∩ πi−1 [F ]), then µ{x : x(i) ∈ E} =

R

g dν = µi E. Q Q

It follows that if I is not empty then we have an inverse-measure-preserving function π : ∆ → Y defined by setting πx = πi x(i) whenever x ∈ ∆ and i ∈ I. Note that when verifying (†) we need check the equality µ{x : x ∈ ∆, x(i) ∈ Ei for every i ∈ J} = RQ ym derivatives for any given hEi ii∈J . i∈J gi dν for only one representative family hgi ii∈J of Radon-Nikod´ 458M Proposition Suppose that h(Xi , Σi , µi )ii∈I is a family of probability spaces, (Y, T, ν) a probability space, πi : Xi → Y an inverse-measure-preserving function for each i ∈ I, ∆ the fiber product of hXi , πi ii∈I and µ a relative product measure on ∆ with domain Σ. Let (Ai , µ ¯i ), (C, ν¯) and (A, µ ¯) be the measure algebras of (Xi , Σi , µi ), (Y, T, ν) and (∆, Σ, µ) respectively, and for i ∈ I define π ¯i : C → Ai and φ¯i : Ai → A by setting π ¯i F • = πi−1 [F ]• , φ¯i E • = {x : x ∈ ∆, x(i) ∈ E}• for every F ∈ T, E ∈ Σi . Then (A, µ ¯, hφ¯i ii∈I ) is a relative free product of h(Ai , µ ¯i , π ¯i )ii∈I over (C, ν¯). Q proof The case I = ∅ is trivial (if you care to follow through the definitions to the letter, ∆ = i∈I Xi = {∅} and A is the two-point algebra). So I will take it that I is not empty. (a) Of course we have to check that all the π ¯i and φ¯i are measure-preserving Boolean homomorphisms between the appropriate algebras, but in view of the remark following S the definition 458L, this is elementary. The condition that A should be the closed subalgebra generated by i∈I φ¯i [Ai ] is just a translation of the condition (‡).

458N


505

(b) As I is not empty, we have a well-defined inverse-measure-preserving map π : ∆ → Y given by the formula π(x) = πi x(i) whenever x ∈ ∆ and i ∈ I. Let π ¯ : C → A be the corresponding measure-preserving homomorphism, so that π ¯ = φ¯i π ¯i for every i. Set D = π ¯ [C] ⊆ A, and let T : L∞ (C) → L∞ (D) be the • f -algebra isomorphism corresponding to π ¯ (363F). For i ∈ I and E ∈ Σi , set uiE = T giE ∈ L∞ (D). Then • uiE is the conditional expectation of χ{x : x(i) ∈ E} on D. P P If d ∈ D, it is of the form φ¯i π ¯i F • where F ∈ T, so that Z Z Z • uiE d¯ µ = uiE × χd d¯ µ = T (giE × χF • )d¯ µ d Z Z • = giE × χF • d¯ ν= giE dν F

−1 −1 • = µi (E ∩ πi−1 [F ]) = µ(φ−1 ¯(d ∩ φ−1 i [E] ∩ φi [πi [F ]]) = µ i [E] ).

As d is arbitrary, we have the result. Q Q ¯ (c) It follows that if J ⊆ I is finite and not empty, conditional expectation R Q and ai ∈ φi [Ai ] and vi is the −1 • of χai on D for each i ∈ J, then µ ¯(inf i∈J ai ) = v d¯ µ . P P Express a as φ [E i i ] , where Ei ∈ Σi , so i i∈J i that vi = ui,Ei for each i. Then Z Z Y Z Y Z Y Y • • • µ= T gi,E d¯ µ = T ( g )d¯ µ = gi,E d¯ ν vi d¯ i,Ei i i i∈J

i∈J

=

Z Y

gi,Ei dν = µ(

i∈J

\

i∈J

i∈J

φ−1 ¯(inf ai ). Q Q i [Ei ]) = µ i∈J

i∈J

But this is exactly what we need to know to see that hφ¯i [Ai ]ii∈I is relatively independent over D, completing the proof that (A, µ ¯, hφ¯i ii∈I ) is a relative free product of (h(Ai , µ ¯i , π ¯i )ii∈I over C, ν¯). 458N There is no general result on relative product measures to match 458J (see 458Xg, 458Xh). The general question of when we can expect relative product measures to exist seems interesting. Here I give a couple of sample results. Proposition Let h(Xi , Σi , µi )ii∈I be a family of probability spaces, (Y, T, ν) a probability space, and πi : Xi → Y an inverse-measure-preserving function for each i. Suppose that for Q each i we have a disintegration hµiy iy∈Y of µi such that µ∗iy πi−1 [{y}] = µiy Xi = 1 for every y ∈ Y . Let ∆ ⊆ i∈I Xi be the fiber product of N h(Xi , πi )ii∈I , and Υ the subspace σ-algebra on ∆ induced by c i∈I Σi . For y ∈ Y , letR λy be the product of hµiy ii∈I , (λy )∆ the subspace measure on ∆ and λ0y its restriction to Υ. Then µW = λ0y W ν(dy) is defined for every W ∈ Υ, and µ is a relative product measure of h(µi , πi )ii∈I over ν. proof If y ∈ Y , then λ0y ∆ = λ∗y ∆ == (λy )∗

Q i∈I

πi−1 [{y}] = 1

(254Lb). For i ∈ I and E ∈ Σi set giE (y) = µiy E when this is defined; then giE is a Radon-Nikod´ ym Q −1 derivative of F 7→ µi (E ∩ πi [F ]) : T → [0, 1] (452Qa). Write X for i∈I Xi ; for i ∈ I and x ∈ X set φi (x) = x(i). If J ⊆ I is finite and Ei ∈ Σi for each i ∈ J, then Z λ0y (∆ ∩

\

Z φ−1 i [Ei ])ν(dy) =

i∈J

Z =

\

(λy )∆ (∆ ∩ λy (X ∩

\

φ−1 i [Ei ])ν(dy)

i∈J

φ−1 i [Ei ])ν(dy)

i∈J

(because

λ∗y ∆

= 1 and λy measures every

φ−1 i [Ei ]

for almost every y) Z Y Z Y = µiy Ei ν(dy) = gi,Ei dν. i∈J

i∈J

506


458N

R T In particular, λ0y (∆ ∩ i∈J φ−1 [Ei ])ν(dy) is defined. Ri The set {W : W ⊆ X, λ0y (W ∩ ∆)ν(dy) is defined} is a Dynkin class of subsets of X containing T −1 i∈J φi [Ei ] whenever J ⊆ I is finite and not empty and Ei ∈ Σi for each i ∈ J; by the Monotone Class R N Theorem, it includes c i∈I Σi . So µW = λ0y W ν(dy) is defined for every W ∈ Υ. Moreover, the formula RQ T displayed above tells us that µ(∆ ∩ i∈I φ−1 i [Ei ]) = i∈J gi,Ei dν whenever J ⊆ I is finite and Ei ∈ Σi for each i ∈ I. Thus (†) of 458L is satisfied. And (‡) is true by the choice of Υ. 458O The latitude I have permitted in the definition of ‘relative product’ makes it possible to look for relative product measures with further properties, as in the following. Proposition Let h(Xi , Ti , Σi , µi )ii∈I be a family of compact Radon probability spaces, (Y, S, T, ν) a Radon probability space, and πi : Xi → Y a continuous inverse-measure-preserving function for each i. Then h(µi , πi )ii∈I has a relative product measure µ over ν which is Q a Radon measure for the topology on the fiber product of h(Xi , πi )ii∈I induced by the product topology on i∈I Xi . proof (a) For each i ∈ I, E ∈ Σi let giE be a Radon-Nikod´ ym derivative of F 7→ µi (E ∩πi−1 [F ]) : T → [0, 1]. Q T Let C be the family of subsets of X = i∈I Xi expressible in the form C = X ∩ i∈J φ−1 i [Ei ] where J ⊆ I is finite and Ei ∈ Σi for each i ∈ J, writing φi (x) = x(i) for x ∈ X, i ∈ I. Because the expression here is unique unless one of the Ei is empty or Xi , we have a functional λ0 : C → [0, 1] defined by setting RQ T λ0 ( i∈J φ−1 i∈J gi,Ei dν i [Ei ]) = whenever J ⊆ I is finite and not empty and Ei ∈ Σi for every i ∈ I. It is easy to check that λ0 is additive in the sense of 454E so (because every µi is perfect, by 416Wa and 343K) it has an extension to a measure N λ on X with domain c i∈I Σi . By 454A, with K the family of compact subsets of X, λ is inner regular with ˜ on X extending λ. respect to the compact sets. By 416O, there is a Radon measure λ ˜ (b) Let ∆ be the fiber product of h(Xi , πi )ii∈I . Now the point is that ∆ is λ-conegligible. P P Because ˜ every πi is continuous, ∆ is closed. ?? If it is not conegligible, then, because λ isQ τ -additive, there must be a basic open set of non-zero measure disjoint from ∆; express such a set as W = i∈J φ−1 i [Gi ] where J ⊆ I ˜ is inner regular with respect to the compact sets, is finite and Gi ⊆ Xi is open for each i ∈ J. Because λ ˜ > 0; setting Ki = φi [K], Ki ⊆ Gi is compact for each i and there is set K ⊆ W such that λK Qa compact −1 0 W = i∈J φi [Ki ] is non-negligible. Now we have RQ Q −1 −1 ˜ Q 0 < λ( i∈J φi [Ki ]) = λ0 ( i∈J φi [Ki ]) = i∈J gi,Ki dν, so F = {y : y ∈ Y , gi,Ki (y) > 0 for every i ∈ J} is non-negligible. On the other hand, for each i ∈ J we have

R

g dν Y \πi [Ki ] i,Ki

= µi (Ki ∩ πi−1 [Y \ Ki ]) = 0

T so that F \ πi [Ki ] is negligible. Accordingly i∈J πi [Ki ] is non-negligible, and must meet the support Y0 of Y ; let y be any point of the intersection. For i ∈ J, choose x(i) ∈ Ki such that πi x(i) = y. For i ∈ I \ J, πi [Xi ] is a compact subset of Y , and νπi [Xi ] = µi πi−1 [πi [Xi ]] = 1, so Y0 ⊆ πi [Xi ] and we can therefore choose x(i) ∈ Xi with πi x(i) = y. This defines x ∈ ∆. But as x(i) ∈ Ki for i ∈ J, we also have T T −1 x ∈ i∈J φ−1 i [Ki ] ⊆ i∈J φi [Gi ] ⊆ X \ ∆, which is impossible. X X ˜ Thus ∆ is λ-conegligible, as claimed. Q Q ˜ and Σ its domain, so that µ is a Radon probability (c) Let µ be the subspace measure on ∆ induced by λ, measure on ∆ with its subspace topology (416Rb). Concerning (†) of 458L, if J ⊆ I is finite and not empty and Ei ∈ Σi for i ∈ J, then RQ T T ˜ T µ( φ−1 [Ei ]) = λ( φ−1 [Ei ]) = λ0 ( φ−1 [Ei ]) = gi,E dν, i∈J

i

i∈J

i

i∈J

i

{φ−1 i [E]

i∈J

i

as required. Finally, for (‡), the σ-algebra Υ generated by : i ∈ I, E ∈ Σi } is just the subspace N N c ˜ σ-algebra induced by i∈I Σi . Let A be the measure algebra of λ and B ⊆ A the set {W • : W ∈ c i∈I Σi }.

458Xg


507

N Then B is a closed subalgebra of A. If W ⊆ X is open, then for every ² > 0 there is a W0 ∈ c i∈I Σi such ˜ that W0 ⊆ W and λ(W \ W0 ) ≤ ², so W • ∈ B; accordingly {W : W • ∈ B} contains every open set and every Borel set and must be the whole of dom λ. Returning to the measure µ, we see that if W ∈ Σ there N ˜ must be a W0 ∈ c i∈I Σi such that λ(W 4W0 ) = 0; now W0 ∩ ∆ ∈ Υ and µ(W 4(W0 ∩ ∆)) = 0. So (‡) also is true, and we have a relative product measure of the declared type. 458P We can of course make a general search through theorems about product measures, looking for ways of re-presenting them as theorems about relative product measures. There is an associative law, for instance (458Xn). To give an idea of what is to be expected, I offer a result corresponding to 253D. Proposition Let (X1 , Σ1 , µ1 ), (X2 , Σ2 , µ2 ) and (Y, T, ν) be probability spaces, and π1 : X1 → Y , π2 : X2 → Y inverse-measure-preserving functions. Let ∆ be the fiber product of (X1 , π1 ) and (X2 , π2 ), and suppose that µ is a relative product measure of (µ1 , π1 ) and (µ2 , π2 ) over ν; set πx = π1 x(1) = π2 x(2) for x ∈ ∆. Take f1 ∈ L1 (µ1 ) and f2 ∈ L2 (µ2 ), and set (f1 ⊗ f2 )(x) = f1 (x(1))f2 (x(2)) when x ∈ ∆ ∩ (dom f1 × R R dom f2 ). For i = 1, 2 let gi ∈ L1 (ν) be a Radon-Nikod´ ym derivative of F 7→ π−1 [F ] fi dµi : T → R. Then F g1 × g2 dν = i R f ⊗ f2 dµ for every F ∈ T. π −1 [F ] 1 proof When f1 and f2 are characteristic functions of measurable sets, this is just the definition of ‘relative product measure’. The formula for gi corresponds to a linear operator from L1 (µi ) to L1 (ν), so the result is true for simple functions f1 and f2 . If f1 and f2 are almost everywhere limits of non-decreasing sequences hf1n in∈N , hf2n in∈N of non-negative simple functions, then the corresponding sequences hg1n in∈N , hg2n in∈N will also be non-decreasing and non-negative and convergent to g1 , g2 ν-a.e.; moreover, because x 7→ x(1) and x 7→ x(2) are inverse-measure-preserving, f1 ⊗ f2 = limn→∞ f1n ⊗ f2n µ-a.e. So in this case we shall have Z Z f1 ⊗ f2 dµ = lim f1n ⊗ f2n dµ n→∞ π −1 [F ] π −1 [F ] Z Z = lim g1n × g2n dν = g1 × g2 dν n→∞

F

F

for every F ∈ T. Finally, considering positive and negative parts, we can extend the result to general integrable f1 and f2 . 458X Basic exercises > (a) Find an example of a probability space (X, Σ, µ) with σ-subalgebras Σ1 , Σ2 and T of Σ such that Σ1 and Σ2 are independent but are not relatively independent over T. R (b) In 458C, show that (writing µ ˆ for the completion of µ) µ ˆ(F ∩ f −1 [H]) = F νx Hµ(dx) for every F ∈ T and every universally measurable H ⊆ R. (c) Let (X, Σ, µ) be a probability space, hTi ii∈I an independent family of σ-subalgebras of Σ, as in 272Ab, S and T a σ-subalgebra of Σ which is independent of the σ-algebra generated by i∈I Ti . For each i ∈ I, let T0i be a σ-subalgebra of T, and Σi the σ-algebra generated by T0i ∪ Ti . Show that hΣi ii∈I is relatively independent over T. ˜ j directly from 313G, without appealing to 458F. (d) In 458H, prove the result concerning the B (e) Let h(Ai , µ ¯i )ii∈I be a family of probability algebras. Show that their probability algebra free product (325K) can be identified with their relative free product over (C, ν¯) if C is the two-element Boolean algebra, ν¯ its unique probability measure, and πi : C → Ai the trivial Boolean homomorphism for every i. > (f ) Let Y be a set, hZi ii∈I a family of sets, and πi : Y × Zi → YQthe canonical map for each i. Show that the fiber product of h(Y × Zi , πi )ii∈I can be identified with Y × i∈I Zi . (g) Let ν be Lebesgue measure on [0, 1], and X1 , X2 ⊆ [0, 1] disjoint sets with outer measure 1. For each i ∈ {1, 2} let µi be the subspace measure on Xi and πi : Xi → [0, 1] the identity map. Show that (µ1 , π1 ) and (µ2 , π2 ) have no relative product measure over ν.

508


458Xh

(h) Let ν be the usual measure on the split interval I k (343J), and µ Lebesgue measure on [0, 1]. Set π1 (t) = t+ , π2 (t) = t− for t ∈ [0, 1]. Show that (µ, π1 ) and (µ, π2 ) have no relative product measure over ν. (i) Let ν be Lebesgue measure on [0, 1]. For each t ∈ [0, 1], set Xt = [0, 1] \ {t}; let µt be the subspace measure on Xt and πt : Xt → [0, 1] the identity map. Show that h(µt , πt )it∈[0,1] has no relative product measure over ν. (j) Let (X, Σ, µ) be a probability space, T a σ-subalgebra of Σ, and hΣi ii∈I a family of σ-subalgebras of Σ, all including T. Set πi (x) = x for every x ∈ X. Show that hΣi ii∈I is relatively independent over T iff µ¹Σ∗ is a relative product measure of h(µ¹Σi , πi )ii∈I over µ¹ T, where Σ∗ is the σ-algebra generated by S i∈I Σi . (k) (i) Let h(Xi , Σi , µi )ii∈I be a family of probability spaces and (X, Σ, µ) their ordinary probability space product. Show that µ is a relative product measure of h(µi , πi )ii∈I over ν where Y is a singleton set, ν its unique probability measure, and πi : Xi → Y the unique function for each i. (ii) Let h(Xi , Ti , Σi , µi )ii∈I be a family of quasi-Radon probability spaces and (X, T, Σ, µ) their quasi-Radon probability space product. Show that µ is a relative product measure of hµi ii∈I in the same sense as in (i). (l) Suppose that h(Xi , Σi , µi )ii∈I is a family of probability spaces and (Y, T, ν) is a probability space, and that for each i ∈ I we are given an inverse-measure-preserving function πi : Xi → Y . Write µ î and νˆ for the completions of µi , ν respectively. Show that h(µi , πi )ii∈I has a relative product measure over ν iff h(ˆ µi , πi )ii∈I has a relative product measure over νˆ. (m) Let h(Xi , Σi , µi )ii∈I be a family of probability spaces, (Y, T, ν) a probability space, and πi : Xi → Y an inverse-measure-preserving function for each i ∈ I. Show that if h(µi , πi )ii∈I has a relative product measure over ν, so does h(µi , πi )ii∈J for any J ⊆ I. (n) Let h(Xi , Σi , µi )ii∈I be a family of probability spaces, (Y, T, ν) a probability space, and πi : Xi → Y an inverse-measure-preserving function for each i ∈ I. Let hJk ik∈K be a partition of I into non-empty sets. For each k ∈ K, let ∆k be the fiber product of h(Xi , πi )ii∈Jk ; suppose that µ ˜k is a relative product ˜ k its domain. Define π measure of h(µi , πi )ii∈Jk and Σ ˜k : ∆k → Y by setting π ˜k (x) = πi x(i) whenever x ∈ ∆k and i ∈ Jk , so that π ˜k is inverse-measure-preserving. Suppose that µ is a relative product measure of h(˜ µk , π ˜k )ik∈K over ν. Show that µ can be regarded as a relative product measure of h(µi , πi )ii∈I over ν. (o) Let h(Xi , Σi , µi )ii∈I and h(Xi0 , Σ0i , µ0i )ii∈I be two families of probability spaces, (Y, T, ν) and (Y 0 , T0 , ν 0 ) probability spaces, and πi : Xi → Y , πi0 : Xi0 → Y 0 inverse-measure-preserving functions for each i. Suppose that we have a measure space isomorphism g : Y → Y 0 and inverse-measure-preserving functions fi : Xi → Xi0 , for i ∈ I, such that gπi = πi0 fi for every i. Show that if there is a relative product measure of h(µi , πi )ii∈I over ν, then there is a relative product measure of h(µ0i , πi0 )ii∈I over ν 0 . (p) Let h(Xi , Σi , µi )ii∈I be a countable family of probability spaces, (Y, T, ν) a probability space, and πi : Xi → Y an inverse-measure-preserving function for each i. Suppose that for each i we have a disintegration of µi over ν which is strongly consistent with πi . Show that h(µi , πi )ii∈I has a relative product measure over ν. (q) Let (X, Σ, µ), (X 0 , Σ0 , µ0 ) and (Y, T, ν) be probability spaces. Suppose that π : X → Y and π 0 : X 0 → Y are inverse-measure-preserving functions, and that µ0 has a disintegration hµ0y iy∈Y over (Y, T, ν) which is stronglyRconsistent with π 0 . Show that (µ, π) and (µ0 , π 0 ) have a relative product measure over ν. (Hint: b 0 .) set λW = µ0πx W [{x}]µ(dx) for every W ∈ Σ⊗Σ > (r) Let Y be a Hausdorff space, hZi ii∈I a family of Hausdorff spaces, µi a Radon probability measure on Zi × Y and πi : Y × Zi → Y the canonical map for each i. Suppose that all the image measures µi πi−1 on Y are the same, and that all the Zi are compact. Show that there is a Radon Q but countably many of −1 probability measure µ on Y × Z such that µ = µφ for each i, where φi (y, z) = (y, z(i)) for y ∈ Y , i i i∈I i Q z ∈ j∈I Zj .

458 Notes


509

(s) Let h(Xi , Ti , Σi , µi )ii∈I be a countable family of Radon probability spaces, (Y, S, T, ν) a Radon probability space, and πi : Xi → Y an almost continuous inverse-measure-preserving function for each i. Show that h(µi , πi )ii∈I has a relative product measure over ν which is aQRadon measure for the topology on the fiber product of h(Xi , πi )ii∈I induced by the product topology on i∈I Xi . (t) Let (X, Σ, µ) be a probability space and T a σ-subalgbra of Σ. Let f1 , f2 be µ-integrable real-valued functions which are relatively independent over T, and suppose that f1 × f2 is integrable. Let g1 , g2 be conditional expectations of f1 , f2 on T. Show that g1 × g2 is a conditional expectation of f1 × f2 on T. (Hint: 458P.) 458Y Further exercises (a) Let (X, Σ, µ) be a complete probability space, T a σ-algebra of subsets of Σ, and hfi ii∈I a family in L0 (µ) which is relatively independent over T. Show that there is a family hνx ix∈X of countably compact quasi-Radon probability measures on RI such that whenever F ∈ T and J ⊆ I is countable and H ⊆ RJ is a Borel set, then R T µ{x : x ∈ F ∩ i∈J dom fi , hfi (x)ii∈J ∈ H} = F νz {x : x¹J ∈ H}µ(dz). (b) Let (X, Σ, µ) be a probability space, T a σ-subalgebra of Σ, and hΣn in∈N a sequence of σ-subalgebras S of Σ which is relatively independent over T. For each n ∈ N let Σ∗n be the σ-algebra generated by m≥n Σm , T and set Σ∞ = n∈N Σ∗n . Show that for every E ∈ Σ∞ there is an F ∈ T such that E4F is negligible. (Compare 272O.) (c) Let (X, Σ, µ) be a probability space, T a σ-subalgebra of Σ, and f , g ∈ L0 (µ) relatively independent over T; suppose that hνx ix∈X and hνx0 ix∈X are relative distributions of f and g over T. Show that hνx ∗νx0 ix∈X is a relative distribution of f + g over T. (Compare 272S.) (d) Let (X, Σ, µ) be a probability space, T a σ-subalgebra of Σ, and hfn in∈N a sequence in L2 (µ) such R that hfn in∈N is relatively independent over T and F fn dµ = 0 for every n ∈ N and every F ∈ T. (i) Suppose P∞ 1 that hβn in∈N is a non-decreasing sequence in ]0, ∞[, diverging to ∞, such that n=0 2 kfn k22 < ∞. Show βn Pn 1 Pn 1 that limn→∞ i=0 fi = 0 a.e. (ii) Suppose that supn∈N kfn k2 < ∞. Show that limn→∞ n+1 i=0 fi = 0 βn

a.e. (Compare 273D.) (e) Let h(Xi , Σi , µi )ii∈I be a family of measure spaces, (Y, T, ν) a measure space, and πi : Xi → Y an inverse-measure-preserving function for each i ∈ I. For i ∈ I and E ∈ Σi let giE be a Radon-Nikod´ ym Q derivative of the functional F 7→ µi (E ∩ πi−1 [F ]). Let C be the family of subsets of X = i∈I Xi of the T form X ∩ i∈J φ−1 X, i ∈ I. If i [Ei ] where J ⊆ I is finite, Ei ∈ Σi for i ∈ J and φi (x) = x(i) for x ∈ RQ T [E ] where J ⊆ I is finite and not empty and E ∈ Σ for i ∈ J, set λ C = C = i∈J φ−1 i i i 0 i i∈J gi,Ei dν. Let ∆ ⊆ X be the fiber product of h(Xi , πi )ii∈I . Show that the following are equiveridical: P (i) h(µi , πi )ii∈I ∞ has a relative product measure over ν (ii) whenever hCn in∈N is a sequence in C covering ∆, n=0 λ0 Cn ≥ 1 T 0 ∗ (iii) ν ( i∈I πi [Xi ]) = 1 and whenever T is a countably generated σ-subalgebra of T, J ⊆ I is countable, and Σ0i is a countably generated σ-subalgebra of Σi including {πi−1 [F ] : F ∈ T0 } for every i ∈ J, then h(µi ¹Σ0i , πi )ii∈J has a relative product over ν¹ T0 . (f ) Let X be a Hausdorff space and PR the set of Radon probability measures on X. For ν1 , ν2 ∈ PR , write M(ν1 , ν2 ) for the set of Radon probability measures µ on X × X which have marginal measures ν1 on the R first copy of X, ν2 on the second. (i) For a continuous pseudometric ρ on X, set dρ (ν1 , ν2 ) = inf{ ρ(x, y)µ(d(x, y)) : µ ∈ M(ν1 , ν2 )}. Show that dρ is a pseudometric on PR . (ii) Show that if X is completely regular and P is a family of pseudometrics defining the topology of X, then {dρ : ρ ∈ P} defines the narrow topology of PR . (Compare 274Ya, 437Yi.) 458 Notes and comments At a couple of points in Volume 3 (Dye’s theorem, in §388, and Kawada’s theorem, in §394) I took the trouble to generalize standard theorems to ‘non-ergodic’ forms. In both 388L and 394P the results are complicated by potentially non-trivial closed subalgebras of the probability algebra

510


458 Notes

we are studying. I remarked on both occasions that the generalization is only a matter of technique, but I do not suppose that it was obvious just why this must be so. It is however a fundamental theorem of the topic of ‘random reals’ in the theory of Forcing that any theorem about probability algebras must have a relativized form as a theorem about probability algebras with arbitrary closed subalgebras. The concept of ‘relative Maharam type’ from §333, for instance, is what matches ‘Maharam type’ for simple algebras; the concept of ‘exchangeable’ sequence is what matches ‘independent identically distributed’ sequence. (In probability theory, the keyword is ‘mixture’.) In this section I present another example in the idea of ‘relatively independent’ closed subalgebras (458H). I should emphasize that the theory of random reals, when we eventually come to it in Volume 5, will not as a rule apply directly to measure spaces; it deals with measure algebras. But of course the ideas generated by this theory can often be profitably applied to constructions in measure spaces, and this is what I am seeking to do with relatively independent σ-algebras and relative product measures. Just as independent σ-algebras are associated with product spaces (272J), relatively independent algebras are associated with relative products (458Xj). The archetype of a relative product measure is 458N; it is a kind of disintegrated product. It is frequently profitable to express the ‘relative’ concepts of measure theory in terms of disintegrations. I introduce ‘relative free products’ of probability algebras before proceeding to measure spaces because the uniqueness property proved in 458J shows that we have an unambiguous definition. For measure spaces it seems for the moment better to leave ourselves a bit of freedom, not (for instance) favouring one product construction over another (458Xk). The requirement that a relative product measure be carried by the fiber product is seriously limiting (458Xg-458Xi, 458Ye), and forces us to seek strictly consistent disintegrations (458N), at least for uncountable products (see 458Xp). However, as we might hope, the special case of compact spaces with Radon measures and continuous functions is amenable to a different approach (458O); and we have a one-sided method for the product of two spaces (458Xq) which is reminiscent of 454C and 457F. There are corresponding complications when we come to look at maps between different relative products. For measure algebras, we have a natural theorem (458K), based on the same algebraic considerations as the corresponding theorems in §§315 and 325; the only possibly surprising feature is the need to assume that ψ : C → C0 is actually an isomorphism. For measure spaces there is a similar result (458Xo).

459 Symmetric measures and exchangeable random variables Among the relatively independent families of random variables discussed in 458A-458E, it is natural to give extra attention to those which are ‘relatively identically distributed’. It turns out that these have a particularly appealing characterization as the ‘exchangeable’ families (459C). In the same way, among the measures on a product space X I there is a special place for those which are invariant under permutations of coordinates (459E, 459G). 459A The following elementary fact seems to have gone unmentioned so far. Lemma Let (X, Σ, µ) and (Y, T, ν) be probability spaces and f : X → Y an inverse-measure-preserving function; set Σ0 = {f −1 [F ] : F ∈ T}. Let T1 be a σ-subalgebra of T and Σ1 = {f −1 [F ] : F ∈ T1 }. If g ∈ L1 (ν) and h is a conditional expectation of g on T1 , then hf is a conditional expectation of gf on Σ1 . proof h is ν¹T1 -integrable and f is inverse-measure-preserving for µ¹Σ1 and ν¹T1 , so hf is µ¹Σ1 -integrable. If E ∈ Σ1 then there is an F ∈ T1 such that E = f −1 [F ], and now

R

gf dµ = E

R

f −1 [F ]

gf dµ =

R

F

g dν =

R

As E is arbitrary, hf is a conditional expectation of hg on Σ1 .

F

h dν =

R

E

hf dµ.

459B

Symmetric measures and exchangeable random variables

511

459B Theorem Let (X, Σ, µ) be a probability space, Z a set, Υ a σ-algebra of subsets of Z and hfi ii∈I an infinite family of (Σ, Υ)-measurable functions from X to Z. For each i ∈ I, set Σi = {fi−1 [H] : H ∈ Υ}. Then the following are equiveridical: (i) whenever i0 , . . . , ir ∈ T I are distinct, j0 , . . . , jr ∈ I are distinct, and Hk ∈ Υ for each k ≤ r, T −1 then µ( k≤r fi−1 [H ]) = µ( k k≤r fjk [Hk ]); k (ii) there is a σ-subalgebra T of Σ such that (α) hΣi ii∈I is relatively independent over T, (β) whenever i, j ∈ I, H ∈ Υ and F ∈ T, then µ(F ∩ fi−1 [H]) = µ(F ∩ fj−1 [H]). Moreover, if I is totally ordered by ≤, we can add (iii) whenever i0
S α) For each n, r ∈ N, let Σnr be the σ-subalgebra of Σ generated by n≤i≤n+r Σi ; let Tn be the (α T S σ-algebra generated by r∈N Σnr , and T = n∈N Tn . For n ∈ N and H ∈ Υ, let gnH : X → R be a T-measurable function which is a conditional expectation of χfn−1 [H] on T. β ) (The key.) For any n ∈ N and Borel set H ⊆ R, gnH is a conditional expectation of χfn−1 [H] on (β Tn+1 . P P For m, r ∈ N, let hmr : X → [0, 1] be a Σmr -measurable function which is a conditional expectation of χfn−1 [H] on Σmr ; for m ∈ N, set hm = limr→∞ hmr where this is defined. By Lévy’s martingale theorem (275I) hm is defined almost everywhere and is a conditional expectation of χfn−1 [H] on Tm . For m, r ∈ N, define Fmr : X → Z r+2 by setting Fmr (x) = (fn (x), fm (x), fm+1 (x), . . . , fm+r (x)) for x ∈ X. At this point, examine the hypothesis (iii). This implies that if m > n and r ∈ N then \ −1 −1 fm+k [Hk ]) µ(Fmr [H 0 × H0 × . . . × Hr ]) = µ(fn−1 [H 0 ] ∩ k≤r

=

µ(fn−1 [H 0 ]

=

−1 µ(Fm+1,r [H 0

∩

\

−1 fm+1+k [Hk ])

k≤r

× H0 × . . . × Hr ])

−1 for all H 0 , H0 , . . . , Hr ∈ Υ. By the Monotone Class Theorem (136C), the image measures µFmr and N −1 r+2 µFm+1,r agree on the σ-algebra c r+2 Υ of subsets of Z generated by measurable rectangles; set N N −1 −1 c λ = µFmr ¹ r+2 Υ = µFm+1,r ¹ c r+2 Υ. N Let Λ be the σ-subalgebra of c r+2 Υ generated by sets of the form Z × H0 × . . . × Hr , and let h be a conditional expectation of χ(H × Z r+1 ) on Λ with respect to λ. Then 459A tells us that hFmr is a conditional expectation of χ(fn−1 [H]) on Σmr , and is therefore equal almost everywhere to hmr . Similarly, hFm+1,r =a.e. hm+1,r , and this is true for every r ∈ N. But as Fmr and Fm+1,r are both inverse-measurepreserving for R 2 R µ and λ, this means that hmr , h and hm+1,r all have the same distribution. In particular, hmr dµ = h2m+1,r dµ. Now hhmr ir∈N and hhm+1,r ir∈N converge almost everywhere to hm and hm+1 respectively, so

R

h2m dµ = limr→∞

R

h2mr dµ = limr→∞

R

h2m+1,r dµ =

R

h2m+1 dµ.

On the other hand, hm+1 is a conditional expectation of hm on Tm+1 (233Eh). This means that

512


R

459B

R

hm × hm+1 dµ = hm+1 × hm+1 dµ R (233K). A direct calculation tells us that (hm − hm+1 )2 dµ = 0, so that hm =a.e. hm+1 . Inducing on r, we see that hm =a.e. hr whenever n < m ≤ r. Now the reverse martingale theorem (275K) tells us that limm→∞ hm is defined almost everywhere and is a conditional expectation of χfn−1 [H] on T, that is, is equal almost everywhere to gnH . Since the hm , for m > n, are equal almost everywhere, they are all equal to gnH a.e. In particular, gnH is equal a.e. to hn+1 , and is a conditional expectation of χfn−1 [H] on Tn+1 . Q Q Qr T −1 (γγ ) If n ∈ N and H0 , . . . , Hr ∈ Υ, then i=0 gn+i,Hi is a conditional expectation of χ( i≤r fn+i [Hi ]) on T. P P Induce on r. For r = 0 this is just the definition of g . For the inductive step to r ≥ 1, observe nH0 Qr Q r −1 −1 that gnH0 × i=1 χfn+i [Hi ] is a conditional expectation of i=0 χfn+i [H ] on T , by 233Eg or 233K, Qr i −1 n+1 because gnH0 is a conditional expectation of χfn−1 [H0 ] on Tn+1 and i=1 χfn+i [Hi ] is Tn+1 -measurable. Qr Qr −1 But as (by the inductive hypothesis) i=1 gn+i,Hi is a conditional expectation of i=1 χfn+i [Hi ] on T, Qr Qr −1 Q i=0 gn+i,Hi is a conditional expectation of i=0 χfn+i [Hi ] on T, by 233Eg/233K again. Q Qr T (δδ ) In particular, i=0 giHi is a conditional expectation of χ( i≤r fi−1 [Hi ]) on T for every r ∈ N and H0 , . . . , Hr ∈ Υ. This shows that hΣn in∈N is relatively independent over T. (²²) Now consider part (β) of the condition (ii). For this, observe that if m > 0, H ∈ Υ and Hi ∈ Υ for i ≤ r, then T T −1 −1 −1 µ(f0−1 [H] ∩ i≤r fm+i+1 [Hi ]) = µ(fm [H] ∩ i≤r fm+i+1 [Hi ]). By the Monotone Class Theorem, −1 µ(F ∩ f0−1 [H]) = µ(F ∩ fm [H])

for any F ∈ Tm+1 and in particular for any F ∈ T. Thus (ii) is true. (c) Now suppose that sequence hjk ik∈N in I. For each n, let Tn be the σT S there is a strictly increasing algebra generated by k≥n Σjk , and set T = n∈N Tn . Then (b), applied to hfjk ik∈N , tells us that hΣjk ik∈N is relatively independent over T and that for each H ∈ Υ there is a function gH which is a conditional expectation of χ(fj−1 [H]) on T for every k ∈ N. k T T α) If i0 , . . . , ir ∈ I are distinct and H0 , . . . , Hr ∈ Υ, then µ( k≤r fi−1 (α [Hk ]) = µ( k≤r fj−1 [Hk ]). P P k k Let ρ be the permutation of {0, . . . , r} such that iρ(0) < iρ(1) < . . . < iρ(r) . Then µ(

\ k≤r

fi−1 [Hk ]) = µ( k

\

fi−1 [Hρ(k) ]) = µ( ρ(k)

k≤r

\

fj−1 [Hρ(k) ]) k

k≤r

Z Y Z Y r r \ = ( gHρ(k) )dµ = ( gHk )dµ = µ( fj−1 [Hk ]). Q Q k k=0

k=0

k≤r

β ) Now suppose that i0 , . . . , ir ∈ I are distinct. Then there is some m ∈ N such that jk 6= il for (β any l ≤ r and k ≥ m. In this case, consider the sequence fi0 , . . . , fir , fjm , fjm+1 , . . . . By (α) here, this sequence satisfies the condition (iii). We can therefore apply the construction of (c). But observe that the tail σ-algebra obtained from fi0 , . . . , fir , fjm , . . . is precisely T, as defined from hfjk ik∈N just above. So hΣik ik≤r is relatively independent over T. As i0 , . . . , ir are arbitrary, hΣi ii∈I is relatively independent over [H]) have the same conditional expectations T. At the same time we see that if H ∈ Υ then all the χ(fi−1 k over T as χ(fj−1 [H]). So (ii-β) is satisfied. m (d) Finally, if there is no strictly increasing sequence in I, then (I, ≥) is well-ordered; since I is infinite, the well-ordering starts with an initial segment ofTorder type ω, thatTis, a sequence hjk ik∈N such that j0 > [Hk ]) whenever i0 > . . . > ir [Hk ]) = µ( k≤r fj−1 j1 > . . . . But note now that (iii) tells us that µ( k≤r fi−1 k k and j0 > . . . > jr and Hk ∈ Υ for every k. So we can apply (c) to (I, ≥) to get the result in this case also. 459C Exchangeable random variables I spell out the leading special case of this theorem.

459D


513

de Finetti’s theorem Let (X, Σ, µ) be a probability space, and hfi ii∈I an infinite family in L0 (µ). Then the following are equiveridical: (i) the joint distribution of (fi0 , fi1 , . . . , fir ) is the same as the joint distribution of (fj0 , fj1 , . . . , fjr ) whenever i0 , . . . , ir ∈ I are distinct and j0 , . . . , jr ∈ I are distinct; (ii) there is a σ-subalgebra T of Σ such that hfi ii∈I is relatively independent over T and all the fi have the same relative distributions over T. Moreover, if I is totally ordered by ≤, we can add (iii) the joint distribution of (fi0 , fi1 , . . . , fir ) is the same as the joint distribution of (fj0 , fj1 , . . . , fjr ) whenever i0 < . . . < ir and j0 < . . . < jr in I. Remark Families of random variables satisfying the condition in (i) are called exchangeable. The equivalence of (i) and (ii) can be expressed by saying that ‘an exchangeable family of random variables is a mixture of independent identically distributed families’. proof Changing each fi on a negligible set will not change either their joint distributions (271De) or their relative distributions over T or their relative independence; so we may suppose that every fi is a Σ-measurable function from X to R. Now look at 459B, taking (Z, Υ) to be R with its Borel σ-algebra. The condition 459B(i) reads whenever i0 , . . . , ir ∈ ITare distinct, j0 , . . . , jr ∈ I are distinct, and Hk ∈ Υ for each k ≤ r, then T [Hk ]), µ( k≤r fi−1 [Hk ]) = µ( k≤r fj−1 k k matching (i) here, by 271B; similarly, (iii) of 459B matches (iii) here. Equally, condition (ii) here is just a re-phrasing of 459B(ii) in the language of 458A. So 459B gives the result. 459D Specializing 459B in another direction, we have the case in which X is actually the product Z I . In this case, the condition 459B(i) corresponds to a strong kind of symmetry in the measure µ. It now makes sense to look for subsets of X = Z I which are essentially invariant under permutations, and we have the following result. Proposition Let Z be a set, Υ a σ-algebra of subsets of Z, I an infinite set and µ a measure on Z I with N domain the σ-algebra c I Υ generated by {πi−1 [H] : i ∈ I, H ∈ Υ}, taking πi (x) = x(i) for x ∈ Z I and i ∈ I. For each permutation ρ of I, define ρˆ : Z I → Z I by setting ρˆ(x) = xρ for x ∈ Z I . Suppose that N ρ−1 [E]) = 0 for every µ = µˆ ρ−1 for every ρ. Let E be the family of those sets E ∈ c I Υ such that µ(E4ˆ N permutation ρ of I, and V the family of those sets E ∈ c I Υ such that E is determined by coordinates in I \ {i} for every i ∈ I. N (a) E is a σ-subalgebra of c I Υ. (b) V is a σ-subalgebra of E. (c) If E ∈ E and J ⊆ I is infinite, then there is a V ∈ V, determined by coordinates in J, such that µ(E4V ) = 0. (d) Setting Σi = {πi−1 [H] : H ∈ Υ} for each i ∈ I, (α) hΣi ii∈I is relatively independent over E, (β) for every H ∈ Υ there is a E-measurable function gH : Z I → [0, 1] which is a conditional expectation of χ(πi−1 [H]) on E for every i ∈ I. proof (a) is elementary. (b) Let V ∈ V. Suppose that ρ : I → I is a permutation, i0 , . . . , ir ∈ I are distinct and Hk ∈ Υ for every k ≤ r. Then there is a permutation σ : I → I such that σ(ik ) = ρ(ik ) for every k ≤ r and J = {i : σ(i) 6= i} is finite. By 254Ta, V is determined by coordinates in I \ J, so σ ˆ −1 [V ] = V . Set jk = ρ−1 (ik ) for each k ≤ r. Now \ \ [Hk ]) πj−1 [H ]) = µ(V ∩ µ(ˆ ρ−1 [V ] ∩ πi−1 k k k k≤r

k≤r

= µ(ˆ σ

−1

[V ] ∩

\ k≤r

[Hk ]) = µ(V ∩ πi−1 k

\ k≤r

[Hk ]). πi−1 k

514


459D

N By the Monotone Class Theorem, as usual, µ(E ∩ ρˆ−1 [V ]) = µ(E ∩ V ) for every E ∈ c I Υ. In particular, taking E = V and E = Z I \ V , we see that V 4ˆ ρ−1 [V ] is negligible. As ρ is arbitrary, V ∈ E. This shows that V ⊆ E. Of course V is a σ-algebra, since it is just the intersection of the σ-algebras N {V : V ∈ c I Σ, V is determined by coordinates in I \ {i}}. N (c) For each n ∈ N, there is a set En ∈ c I Σ, determined by a finite set Jn of coordinates, such that µ(E4En ) ≤ 2−n . Choose permutations ρn of I such that hρn [Jn ]in∈N is a disjoint sequence T of subsets S of J. Set Fn = ρˆ−1 [E ]; then F is determined by coordinates in ρ [J ] for each n ∈ N, so V = n n n n n n∈N m≥n Fm belongs to V and is determined by coordinates in J. Also µ(E4Fn ) = µ(ˆ ρ[E]4En ) = µ(E4En ) ≤ 2−n for each n, so µ(E4V ) = 0, as required. (d) Let hjn in∈N be any sequence of distinct points S T of I. Set J = {jn : n ∈ N}. For n ∈ N let Tn be the σ-algebra generated by k≥n Σjk , and set T = n∈N Tn , so that T = {V : V ∈ V, V is determined by coordinates in J}. P P Of course T ⊆ V and every member of T is determined by coordinates in J, because every member of T0 is. On the other hand, if V ∈ V is determined by coordinates in J, then fix some w ∈ Z I\J . In this case, identifying Z I with Z J × Z I\J , the set V1 = {z : z ∈ Z J , (z, w) ∈ V } must belong N to c J Υ, so V = V1 × Z I\J belongs to T0 . Applying the same idea to J \ {jk : k < n}, we see that V ∈ Tn for every n, so that V ∈ T. Q Q Part (c) of the proof of 459B tells us that hΣi ii∈I is relatively independent over T and that for every H ∈ Υ there is a T-measurable gH which is a conditional expectation of χ(πi−1 [H]) on T for every i ∈ I. Now (c) here tells us that gH is a conditional expectation of χ(πi−1 [H]) on E; and examining the definition in 458Aa, we see that hΣi ii∈I is relatively independent over E, as claimed. 459E If µ is countably compact, we have a strong disintegration theorem, as follows. Theorem Let Z be a set, Υ a σ-algebra of subsets of Z, I an infinite set, and µ a countably compact N probability measure on Z I with domain the σ-algebra c I Υ generated by {πi−1 [H] : i ∈ I, H ∈ Υ}, taking πi (x) = x(i) for x ∈ Z I and i ∈ I. Then the following are equiveridical: (i) for every permutation ρ of I, x 7→ xρ : Z I → Z I is inverse-measure-preserving for µ; (ii) for every transposition ρ of two elements of I, x 7→ xρ : Z I → Z I is inverse-measurepreserving for µ; (iii) for each n ∈ N and any two injective functions p, q : n → I the maps x 7→ xp : Z I → Z n , x 7→ xq : Z I → Z n induce the same measure on Z n ; (iv) there are a probability space (Y, T, ν) and a family hλy iy∈Y of probability measures on Z such that hλIy iy∈Y is a disintegration of µ over ν, writing λIy for the product of copies of λy . Moreover, if I is totally ordered, we can add (v) for each n ∈ N and any two strictly increasing functions p, q : n → I the maps x 7→ xp : Z I → Z n , x 7→ xq : Z I → Z n induce the same measure on Z n . If the conditions (i)-(v) are satisfied, then there is a countably compact measure λ, with domain Υ, which is the common marginal measure of µ on every coordinate; and if K is a countably compact class of subsets of Z, closed under finite unions and countable intersections, such that λ is inner regular with respect to K, then (iv)0 there are a probability space (Y, T, ν) and a family hλy iy∈Y of complete probability measures on Z, all with domains including K and inner regular with respect to K, such that hλIy iy∈Y is a disintegration of µ over ν. proof (a) Since any set I can be totally ordered, we may suppose from the outset that we have been given a total ordering ≤ on I. I start with the easy bits. (iv)0 ⇒(iv) is trivial, at least if there is a common countably compact marginal measure on Z. N (iv)⇒(i) If (iv) is true and ρ : I → I is a permutation, take any E ∈ c I Σ and set E 0 = {x : x ∈ Z I , xρ ∈ E}. For any y ∈ Y , x 7→ xρ is an isomorphism of the measure space (Z I , λIy ), so

459F


µE 0 =

R

λIy E 0 ν(dy) =

R

515

λIy E ν(dy) = µE.

As E is arbitrary, (i) is true. (i)⇒(ii) is trivial. (ii)⇒(iii) There is a permutation ρ of I such that q = ρp and ρ moves only finitely many points of I, that is, ρ is a product of transpositions. By (ii), x 7→ xρ and x 7→ xρ−1 are inverse-measure-preserving for µ, that is, are isomorphisms of (Z I , µ). But this means that x 7→ xp and x 7→ xρp = xq must induce the same measure on Z n . (iii)⇒(v) is trivial. (b) So for the rest of the proof I assume that (v) is true. Taking n = 1 in the statement of (v), we see that there is a common image measure λ = µπi−1 for every i ∈ I. By 452R, λ is countably compact. Let K ⊆ PZ be a countably compact class, closed under finite unions and countable intersections, such that λ is inner regular with respect to K. N In 459B, set X = Z I and Σ = c I Υ and fi = πi : X → Z for i ∈ I. Then (v) here corresponds to (iii) of N 459B, so (translating (ii) of 459B) we have a σ-subalgebra T of c I Υ and a family hgH iH∈Υ of T-measurable functions from Z I to [0, 1] such that R Q T µ( i∈J πi−1 [Hi ]) = ( i∈J gHi )dµ whenever J ⊆ I is finite and not empty and Hi ∈ Υ for i ∈ J. In particular, gH is a conditional expectation of χ(πi−1 [H]) on T whenever H ∈ Υ and i ∈ I. Fix i∗ ∈ I for the moment. Set ν = µ¹T. The inverse-measure-preserving function πi∗ from (X, µ) to (Z, λ) gives us an integral-preserving Riesz homomorphism T0 : L∞ (λ) → L∞ (µ) defined by setting T0 h• = (hπi∗ )• for every h ∈ L∞ (λ). Let P : L1 (µ) → L1 (ν) be the conditional expectation operator; then T = P T0 : L∞ (λ) → L∞ (ν) is an integral-preserving positive linear operator, and T (χZ • ) = χX • . By 452H, we have a family hλx ix∈X of completeR probability Rmeasures on Z, all with domains including R K and inner regular with respect to K, such that F hπi∗ dµ = F Z h dλx ν(dx) for every h ∈ L∞ (λ) and 0 F ∈ T. In particular, setting gH (x) = λx H whenever H ∈ Υ and x ∈ X are such that H ∈ dom λx , then 0 gH will be a conditional expectation of χπi−1 ∗ [H] on T, and will be equal ν-almost everywhere to gH . This means that if J ⊆ I is finite and not empty and Hi ∈ Υ for i ∈ J, Z Z Y Z Y \ −1 I 0 λx ( πi [Hi ])ν(dx) = λx Hi ν(dx) = gH dν i X

X i∈J

i∈J

=

Z Y X i∈J

R

X i∈J

gHi dν =

Z Y X i∈J

gHi dµ = µ(

\

πi−1 [Hi ]).

i∈J

λIx E

Thus the family W of sets E ⊆ X such that ν(dx) and µE are defined and equal contains all N measurable cylinders. As W is a Dynkin class it includes c I Υ. But this says exactly that hλIx ix∈X is a disintegration of µ over ν, as required by (iv)0 . Thus (v)⇒(iv)0 and the proof is complete. ˜ τ+ (X) the set of τ -additive totally finite topological 459F Lemma Let X be a topological space, M ˜ τ+ (X) a measure which is inner regular with respect to the Borel sets. Let Y be measures on X and µ ∈ M a topological space, ν a totally finite τ -additive topological measure on Y and hµy iy∈Y a family of complete ˜ τ+ (X) such that y 7→ µy is continuous for the given topology on Y and the narrow topology measures in M + ˜ on Mτ (X) (definition: 437Jd). (a) Suppose that R U is a base for the topology of X, closed under finite intersections and containing X, such that µU = µy U ν(dy) for every U ∈ U. Then hµyRiy∈Y is aRR disintegration of µ over ν. (b) Suppose that X is completely regular and that u dµ = u dµy ν(dy) for every u ∈ Cb (X). Then hµy iy∈Y is a disintegration of µ over ν. R proof (a) Let E be the family of those sets E ⊆ X such that µE and µy E ν(dy) are defined and equal. Then E is a Dynkin class including U , so includes the σ-algebra generated by U; in particular, it includes

516


459F

U ∗ = {U0 ∪ . . . ∪ Un : U0 , . . . , Un ∈ U }. For A ⊆ X, set fA (y) = µy A whenever this is defined; note that if G ⊆ X is open, then fG is lower semi-continuous, being the composition of the continuous function y 7→ µy and the semi-continuous function θ 7→ θG. If H ⊆ X is open, set G = {G : G ∈ U ∗ , G ⊆ H}; then G is upwards-directed and has union H, so fH (y) = supG∈G fG (y) for every y ∈ Y , because every µy is τ -additive. Accordingly Z

Z µy H ν(dy) =

Z fH dν = sup

G∈G

fG dν

(414Ba) = sup µG = µH, G∈G

and H ∈ E. Again using the Monotone Class Theorem, we see that every Borel set belongs to E. Now if E ∈ dom µ, there are Borel sets E 0 , E 00 such that E 0 ⊆ E ⊆ E 00 and µE = µE 0 = µE 00 , because µ is inner regular with respect to the Borel sets. Accordingly

R

fE 0 dν = µE 0 = µE = µE 00 =

R

fE 00 dν,

and F = {y : fE 0 (y) = fE 00 (y)} is conegligible. But we are supposing that every µy is complete, so µy E is defined and equal to µy E 0 for every y ∈ F , and

R

µy E ν(dy) =

R

µy E 0 ν(dy) = µE 0 = µE.

As E is arbitrary, hµy iy∈Y is a disintegration of µ over ν. (b) In this case, the set U of cozero sets is a base for the topology of X closed under finite intersections. If U ∈ U , there is a continuous u : X → R such that U = {x : u(x) 6= 0}; setting un = χX ∧ n|u| for n ∈ N, hun in∈N is a non-decreasing sequence in Cb (X) with supremum χU , so Z ZZ µU = sup un dµ = sup un dµy ν(dy) n∈N n∈N ZZ Z = (sup un )dµy ν(dy) = µy U ν(dy). n∈N

Thus the condition of (a) is satisfied and hµy iy∈Y is a disintegration of µ over ν. 459G Theorem Let Z be a Hausdorff space, I an infinite set, and µ ˜ a quasi-Radon probability measure on Z I such that the marginal measures on each copy of Z are Radon measures. Write PR (Z) for the set of Radon probability measures on Z with its narrow topology. Then the following are equiveridical: (i) for every permutation ρ of I, w 7→ wρ : Z I → Z I is inverse-measure-preserving for µ ˜; (ii) for every transposition ρ of two elements of I, w 7→ wρ : Z I → Z I is inverse-measurepreserving for µ ˜; (iii) for each n ∈ N and any two injective functions p, q : n → I the maps w 7→ wp : Z I → Z n and w 7→ wq : Z I → Z n induce the same measure on Z n ; (iv) there are a probability space (Y, T, ν) and a family hµy iy∈Y of τ -additive Borel probability measures on Z such that h˜ µIy iy∈Y is a disintegration of µ ˜ over ν, writing µ ˜Iy for the τ -additive product of copies of µy ; (v) there is a Radon probability measure ν˜ on PR (Z) such that hθ˜I iθ∈PR (Z) is disintegration of µ ˜ over ν˜, writing θ˜I for the quasi-Radon product of copies of θ. Moreover, if I is totally ordered, we can add (vi) for each n ∈ N and any two strictly increasing functions p, q : n → I the maps w 7→ wp : Z I → Z n and w 7→ wq : Z I → Z n induce the same measure on Z n . proof (a) As in 459E, we need consider only the case in which I is totally ordered, and the implications (v)⇒(iv)⇒(i)⇒(ii)⇒(iii)⇒(vi)

459G


517

are elementary. So henceforth I will suppose that (vi) is true and seek to prove (v). N (b) Write µ for µ ˜¹ c I B(Z), where B(Z) is the Borel σ-algebra of Z. Then (vi) is also true of µ. (Strictly speaking, we ought to check that the different images of µ all have the same domain. But this is true, N because the image of µ corresponding to a strictly increasing function p : r → I has domain c r B(Z).) The ˜ which is the image of µ (unique) marginal measure λ of µ is the restriction to B(Z) of the Radon measure λ ˜, so is a tight Borel measure, therefore countably compact. By 454A, µ is countably compact. So 459E, with K the family of compact subsets of Z, tells us that there are a probability space (Y0 , T, ν0 ) and a family hµy iy∈Y0 in PR (Z) such that hµIy iy∈Y0 is a disintegration of µ over ν0 , writing µIy for the ordinary product of copies of µy . It follows that the family h˜ µIy iy∈Y0 of Radon product measures is a disintegration of µ over ν0 , just because µ ˜Iy extends µIy . Note also that hµy iy∈Y0 is a disintegration of λ; this is clearly achieved by the proof of 459E, and it is necessarily true if hµIy iy∈Y0 is to be a disintegration of µ. It follows that hµy iy∈Y0 is ˜ of λ over ν0 (452B(a-ii)). a disintegration of the completion λ P∞ ˜ S Let hKn in∈N be a disjoint sequence of compact subsets of Z such that n=0 λKn = 1, and set X0 = n∈N Kn . Since ˜ 0 = λX0 = 1 = λX

R

Y0

µy (X0 )ν0 (dy),

µy X0 = 1 for ν0 -almost every y, and Y = {y : y ∈ Y0 , µy X0 = 1} is ν0 -conegligible. It follows that, writing µy iy∈Y is a disintegration of ν for the subspace measure on Y , h˜ µIy iy∈Y is a disintegration of µ over ν, and h˜ ˜ λ over ν. (c) Give X0 the disjoint union topology corresponding to the partition {Kn : n ∈ N}, so that X0 is σ-compact and locally compact; note that the identity map from X0 to Z is continuous, and that the Borel σ-algebra B(X0 ) is just {X0 ∩ E : E ∈ B(Z)}. Consequently the subspace measure µy ¹PX0 induced on X0 by µy is a Radon measure on X0 for every y ∈ Y . Let X = X0 ∪ {x∗ } be the one-point compactification of X0 , and for y ∈ Y let λ0y be the Radon probability measure on X extending µy ¹PX0 (so that λ0y {x∗ } = 0). In ˜ 0 be the Radon probability measure on X extending λ¹PX ˜ ˜0 ˜ the same way, let λ 0 ; of course λ X0 = λX0 = 1. (d) Let ν1 be the image measure on PR (X) defined by setting ν1 Q = ν{y : y ∈ Y , λ0y ∈ Q} whenever this is defined. Then ν1 is a complete probability measure. For each n ∈ N, Kn is an open-and-closed set in X, so θ 7→ θKn is continuous for the narrow topology on PR (X), and P∞ Q∗ = {θ : θ ∈ PR (X), θX0 = 1} = {θ : θ ∈ PR (X), n=0 θKn = 1} is a Baire set; of course ν1 Q∗ = 1. If u ∈ C(X), then Z

Z

Z u dθ ν1 (dθ) =

PR (X)

Z

Q∗

X

Z

u dθ ν1 (dθ) X

Z

= Q∗

is defined and equal to

R R Y

X0

Z

Z

u dθ ν1 (dθ) = X0

u dθ ν1 (dθ) PR (X)

u dµy ν(dy) =

R X0

X0

u dλ

R

by 235L and 452F. In particular (because ν1 is complete), θ 7→ X u dθ is dom ν1 -measurable. So ν1 measures R every Baire subset of PR (X). P P Setting u ˆ(θ) = X u dθ for u ∈ C(X) and θ ∈ PR (X), the functions u ˆ are continuous and separate the points of PR (X), which is a compact Hausdorff space (437O). By the StoneWeierstrass theorem (281E), every continuous real-valued function on PR (X) is uniformly approximated by linear combinations of products of such functions u ˆ, and is therefore dom ν1 -measurable. It follows at once that all zero sets and all Baire sets are measured by ν1 . Q Q By 432F, there is a Radon measure ν˜1 on PR (X) agreeing with ν1 on the Baire sets. In particular, Q∗ is ν˜1 -conegligible. Now R R R R R R ˜= ˜0 u dθ ν˜1 (dθ) = u ˆ dν1 = u dλ = u dλ u dλ PR (X)

X

PR (X)

X0

X0

X

518


459G

˜ 0 over ν˜1 . for every u ∈ C(X). By 459Fb, hθiθ∈PR (X) is a disintegration of λ (e) Because the identity map ι : X0 → Z is continuous, we have a map ψ : Q∗ → PR (Z) defined by saying that ψ(θ) = (θ¹PX0 )ι−1 (416Rb, 418I). If G ⊆ Z is open and α > 0, then {θ : ψ(θ)(G) > α} = {θ : θ(G ∩ X0 ) > α} is open in PR (X), so ψ is continuous. We therefore have a Radon probability measure ν˜ = (˜ ν1 ¹PQ∗ )ψ −1 on PR (Z). (Perhaps I should remark here that PR (Z) is a Hausdorff space, by 437Je.) (f ) Suppose that i0 , . . . , ir ∈ I are distinct and Gk ⊆ Z is open for each k ≤ r. For each k ≤ r, set Ak = {u : u ∈ C(X), 0 ≤ u ≤ χ(X0 ∩ Gk )}. For u ∈ C(X), define u ¯ : Z → R by setting u ¯(z) = u(z) if z ∈ X0 , 0 if z ∈ Z \ X0 . Then u ¯ is Borel measurable. Now, writing πik (w) = w(ik ) for w ∈ Z I and k ≤ r, R Qr T µ( k≤r πi−1 [Gk ]) = supuk ∈Ak ∀k≤r ( k=0 u ¯k πik )dµ. k P P Of course

R Qr T µ( k≤r πi−1 [G ]) ≥ sup ( k=0 u ¯k πik )dµ k u ∈A ∀k≤r k k k

because u ¯k ≤ χGk whenever uk ∈ Ak . On the other hand, given ² > 0, then for each k ≤ r we have ˜ 0 (Gk ∩ X0 ) = λGk , so there is a uk ∈ Ak such that λ λGk ≤ ² + Now

R

˜0 = ² + u dλ X k

R

R

˜0 = ² + uk dλ

X0

R

χGk πik − u ¯k πik dµ =

R

X0

˜ =²+ uk dλ

R

Z

u ¯k dλ.

χGk − u ¯k dλ ≤ ²

for each k, so µ(

\

πi−1 [Gk ]) − k

Z Y r

k≤r

u ¯k πik dµ =

k=0

Z Y ¡ r

χGk πik −

k=0

≤

Z X r

r Y

¢ u ¯k πik dµ

k=0

(χGk πik − u ¯k πik )dµ ≤ (r + 1)².

k=0

Thus

R Qr T µ( k≤r πi−1 [Gk ]) ≤ (r + 1)²+ ¯k πik dµ. k=0 u k

As ² is arbitrary,

R Qr T [G ]) = sup ¯k πik dµ. Q Q µ( k≤r πi−1 k u ∈A ∀k≤r k=0 u k k k

Next observe that if uk ∈ Ak for k ≤ r, then R Qr R ¯k πik dµ = P k=0 u ZI

¡Qr R R (X)

k=0

¢

uk dθ ν˜1 (dθ).

P P Z

(452F, because

Qr

r Y

Z I k=0

k=0

Z Z u ¯k πik dµ =

r Y

Z I k=0

Y

u ¯k πik dµIy ν(dy)

N u ¯k πik is c I B(Z)-measurable) Z Y Z Z Y Z ¡ r ¢ ¡ r = u ¯k dµy ν(dy) = Y k=0 Z r Z Y

Z =

¡

Y k=0

Z =

¡

X0 r Y

PR (X) k=0

(235L)

¢ uk dλ0y ν(dy) = Z X

¢ uk dθ ν1 (dθ)

¢ u ¯k dµy ν(dy)

Y k=0 X0 r Z Y

Z

¡

Y k=0

X

¢ uk dλ0y ν(dy)

459H


Z

Z

r ¡Y

= Qr

R

k=0 X

¢ uk dθ ν˜1 (dθ)

Z

PR (X) k=0

because θ 7→

519

uk dθ is continuous, and ν˜1 agrees with ν1 on Baire sets. Q Q Accordingly

µ ˜(

\

πi−1 [Gk ]) k

= µ(

k≤r

\

Z πi−1 [Gk ]) k

k≤r

=

Z

uk ∈Ak ∀k≤r

Z

PR (X) k=0 X r Z Y

Z =

sup

r ¡Y

sup uk ∈Ak ∀k≤r

(because ν˜1 is τ -additive and the

=

sup

¡

Z

r ¡Y

= Z =

Z sup

PR (X) k=0 uk ∈Ak r Y

Z I k=0

u ¯k πik dµ

¢ uk dθ ν˜1 (dθ) ¢ uk dθ ν˜1 (dθ)

PR (X) uk ∈Ak ∀k≤r k=0 X Qr R functions θ 7→ k=0 X uk dθ are

directed)

r Y

continuous and the sets Ak are upwards-

¢ uk dθ ν˜1 (dθ)

X

θ(Gk ∩ X0 )˜ ν1 (dθ)

PR (X) k=0

(because every θ is τ -additive, and Gk ∩ X0 is an open set in the completely regular space X) Z Y r = θ(Gk ∩ X0 )˜ ν1 (dθ) Z =

Q∗ k=0 r Y Q∗ k=0

(235L again)

Z ψ(θ)(Gk )˜ ν1 (dθ) =

Z θ˜I (

= PR (Z)

r Y

θ(Gk )˜ ν (dθ)

PR (Z) k=0

\

πi−1 [Gk ])˜ ν (dθ), k

k≤r

where I write θ˜I to indicate that I intend to use the Radon product measure, even though in this formula it makes no difference which product measure is employed. R (g) What this means is that µ ˜W = θ˜I W ν˜(dθ) for every open measurable cylinder W . Since the map θ 7→ θ˜I : PR (Z) → PR (Z I ) is continuous (437Mb), 459Fa tells us that hθ˜I iθ∈PR (Z) is a disintegration of µ ˜ over ν˜, as required. 459H Following the results of §452 (especially 452Ye), we do not generally expect to find disintegrations of measures which are not countably compact. It may however illuminate the constructions here if I give a specific example related to the contexts of 459E and 459G. Example (Dubins & Freedman 79) There are a separable metrizable space Z and a quasi-Radon measure on Z N , invariant under permutations of coordinates, which cannot be disintegrated into powers of measures on Z. proof (a) Let λ be Lebesgue measure on [0, 1]. Q = [0, 1] × [0, 1]N , with its usual topology, is a compact metrizable space, so has just c Borel sets (4A3F). Let hWξ iξ
520


459H

(b) Set X = ([0, 1]2 )N and define φ : Q → X by setting φ(t, htn in∈N ) = h(t, tn )in∈N for t, tn ∈ [0, 1]. Then φ is a homeomorphism between Q and φ[Q], so there is a unique Radon measure µ# on X such that φ is inverse-measure-preserving for λ × λN and µ# . Now µ# is invariant under permutations of coordinates, because if ρ : N → N is a permutation and ρˆ(x) = xρ for x ∈ X, then ρˆφ = φ¯ ρ, where ρ¯(t, htn in∈N ) = (t, htρ(n) in∈N ); and as ρ¯ : Q → Q is inverse-measure-preserving, so is ρˆ : X → X. Also Z N has full outer measure for µ# . P P If µ# W > 0, then (λ × λN )φ−1 [W ] > 0, so there is some ξ < c −1 such that Wξ ⊆ φ [W ]. Now h(tξ , tξn )in∈N ∈ Z N ∩ W . Q Q Accordingly the subspace measure µ ˜ on Z is a probability measure. Because µ# is invariant under permutations of coordinates, so is µ ˜; because µ# is a Radon measure, µ ˜ is a quasi-Radon measure (416Ra). (c) ?? Suppose, if possible, that R there are a probability space (Y, T, ν)Nand a family hµy iy∈Y of probability measures on Z such that µ ˜E = µN y E ν(dy) for every Borel set E ⊆ Z . (The argument to follow will not depend on which product measure is used in forming the µN y .) Looking at sets of the form (Z ∩ H) × Z × Z × . . . , where H ⊆ [0, 1]2 is a Borel set, we see that µy (Z ∩ H) must be defined for almost every y; as Z is second-countable, µy must be a topological measure for almost every y. Looking at sets of the form (Z ∩ (G0 × [0, 1])) × (Z ∩ (G1 × [0, 1])) × Z × . . . , where G0 and G1 are disjoint Borel subsets of [0, 1], we see that µy (Z ∩ (G0 × [0, 1])) · µy (Z ∩ (G1 × [0, 1])) = 0 for almost every y; as [0, 1] is second-countable and Hausdorff, there must be, for almost every y ∈ Y , an sy ∈ [0, 1] such that µy (Z ∩ ({sy } × [0, 1])) = 1. Next, if G ⊆ [0, 1] is a Borel set, then µy (Z ∩ ([0, 1] × G)) = λG for almost every y. P P h(y) = µN y ((Z ∩ ([0, 1] × G)) × Z × . . . ) = µy (Z ∩ ([0, 1] × G)) is defined for almost every y, and h is ν-integrable, with

R

h dν = µ ˜((Z ∩ ([0, 1] × G)) × Z × . . . ) = µ# (([0, 1] × G) × [0, 1]2 × . . . ) = λG.

At the same time, Z h(y)(1 − h(y))ν(dy) = µ ˜((Z ∩ ([0, 1] × G)) × (Z ∩ ([0, 1] × ([0, 1] \ G))) × Z × . . . ) = µ# (([0, 1] × G) × ([0, 1] × ([0, 1] \ G)) × [0, 1]2 × . . . ) R

= λG(1 − λG). R R R h dν = ( h)2 . But this means that (h(y) − h)2 ν(dy) = 0 and h(y) = λG for 2

Rearranging, we see that almost every y. Q Q It follows that, for at least some y, µy (Z ∩ ({sy } × G)) = λG for every interval G ⊆ [0, 1] with rational endpoints. But this is impossible, because all the vertical sections of Z are countable. X X Thus there is no such disintegration, as claimed. 459X Basic exercises (a) Let (X, Σ, µ) be a probability space and hfn in∈N an exchangeable sequence of real-valued random variables on X all with finite expectation. Use 459C and 273I to show 1 Pn that h i=0 fi in∈N converges a.e. (Compare 276Xe.) n+1

(b) Let (X, Σ, µ) be a probability space and hfn in∈N an exchangeable sequence of real-valued random Pn 1 Pn variables on X all with finite variance, such that limn→∞ i=0 fi = 0 a.e. Show that hPr( i=0 fi ≥ n+1 √ α n + 1)in∈N is convergent for every α ∈ R. (Hint: 274I.) > (c) (Diaconis & Freedman 80) Let X be a non-empty compact Hausdorff space and I an infinite set ˜ be a Radon probability measure on X I invariant under permutations of I. For k ≤ n let including N. Let λ k Dnk ⊆ n be the set of injective functions from k to n and Ωnk the set X I × nk × Dnk , endowed with the ˜ and the uniform probability measures on the finite sets nk and Dnk . Define fnk : Ωnk → X k product νnk of λ and gnk : Ωnk → X k by setting fnk (w, p, q) = wp, gnk (w, p, q) = wp if p ∈ Dnk = wq otherwise.

459 Notes


521

−1 ˜ where each µnw (i) Show that there is a disintegration hµknw iw∈X I of the image measure νnk fnk over λ −1 is a suitable point-supported measure. (ii) Show that the image measure νnk gnk is the image measure R k(k−1) ˜ k = λ˜ ˜ π −1 , where π ˜ k W − µk W λ(dw)| ˜ λ ˜k (w) = w¹k for w ∈ X I . (iii) Show that |λ ≤ for every nw

k

2n

k

Baire set W ⊆ X . (iv) Show that there is a Radon probability measure θn on PR (X) for which w 7→ µnw is inverse-measure-preserving. (v) Show that if θ is any cluster point of hθn in∈N in PR (X) then h˜ µI iµ∈PR (X) I ˜ over θ, writing µ is a disintegration of λ ˜ for the Radon product of copies of any µ ∈ PR (X). > (d) (Hewitt & Savage 55) Let X be a non-empty compact Hausdorff space and I an infinite set. Let Q be the set of Radon probability measures on X I which are invariant under permutations of I. Show that (i) Q is a closed convex subset of the set PR (X I ) of all Radon probability measures on X I with its narrow topology; (ii) Q is isomorphic, as topological convex structure, to PR (PR (X)); (iii) the extreme points of Q are just the powers of Radon probability measures on X. (e) Let X be a non-empty Hausdorff space. Let Q be the set of Radon probability measures on X N which are invariant under permutations of N. Show that (i) Q is a closed convex subset of the set PR (X N ) of all Radon probability measures on X N with its narrow topology; (ii) Q is isomorphic, as topological convex structure, to PR (PR (X)); (iii) the extreme points of Q are just the powers of Radon probability measures on X. 459Y Further exercises (a) Discuss the problems which arise in 459B, 459C, 459E and 459G if the index set I is finite. 459 Notes and comments As I have presented this material, the centre of the argument is the martingales in part (b) of the proof of 459B. We are trying to resolve the functions fi into ‘common’ and ‘independent’ parts. The ‘common’ part is given by the conditional expectations of the fi over an appropriate σ-algebra T, and we approach these by looking at the conditional expectations of each fi on σ-algebras Tn generated by ‘distant’ fj . All the most important ideas are already exhibited when the index set I is equal to N. Note in particular that in the basic hypothesis that all finite strings (fi0 , . . . , fir ) have the same joint distribution, it is enough to look at increasing strings. But there is a striking phenomenon which appears in sharper relief with uncountable sets I: any sequence hjk ik∈N of distinct elements of I can be used to generate an adequate σ-algebra, because while the tail σ-algebra of sets depends on the choice of the jk , they all lead to the same closed subalgebra of the measure algebra (459D). Perhaps I should emphasize at this point that I really does have to be infinite, though for large finite I there are approximations to the results here. The proof of 459B is one of the standard proofs of de Finetti’s theorem, with trifling modifications. In the case of real-valued random variables we have a notion of relative distribution (458C) which gives a quick way of saying that all the fi have the same conditional expectations over T, as in 459C(ii). For variables taking values in other spaces the situation may be different (459H), unless (as in §452) we have a countably compact measure (459E). Specializing to the case X = Z I in 459B, we find ourselves examining symmetric measures on infinite product spaces, which are of great interest in themselves. Note that while in the hypothesis of 459E I have asked for the measure µ on the product space Z I to be countably compact, what is actually necessary is that the marginal measure on Z should be countably compact. By 454A, this comes to the same thing. As in 452O, we can look for a disintegration consisting of Radon measures, provided of course that the marginalR measure is a Radon measure. What we have to work harder for is a direct expression in terms of an integral θ˜I ν˜(dθ) where ν˜ is itself a Radon probability measure on the space of Radon probability measures θ (459G). But most of the extra work consists of finding the correct reduction to the case of compact spaces. For compact spaces we can approach by a completely different route (459Xc). I will not go farther with this idea here, but I note that the method can be used in a wide variety of problems involving symmetric structures.

522

Compact sets of measurable functions

Chapter 46 Pointwise compact sets of measurable functions This chapter collects results inspired by problems in functional analysis. §§461 and 466 look directly at measures on linear topological spaces. The primary applications are of course to Banach spaces, but as usual we quickly find ourselves considering weak topologies. In §461 I look at ‘barycenters’, or centres of mass, of probability measures, with the basic theorems on existence and location of barycenters of given measures and the construction of measures with given barycenters. In §466 I examine topological measures on linear spaces in terms of the classification developed in Chapter 41. A special class of normed spaces, those with ‘Kadec norms’, is particularly important, and in §467 I sketch the theory of the most interesting Kadec norms, the ‘locally uniformly rotund’ norms. In the middle sections of the chapter, I give an account of the theory of pointwise compact sets of measurable functions, as developed by A.Bellow, M.Talagrand and myself. The first step is to examine pointwise compact sets of continuous functions (§462); these have been extensively studied because they represent an effective tool for investigating weakly compact sets in Banach spaces, but here I give only results which are important in measure theory, with a little background material. In §463 I present results on the relationship between the two most important topologies on spaces of measurable functions, not identifying functions which are equal almost everywhere: the pointwise topology and the topology of convergence in measure. These topologies have very different natures but nevertheless interact in striking ways. In particular, we have important theorems giving conditions under which a pointwise compact set of measurable functions will be compact for the topology of convergence in measure (463F, 463K). The remaining two sections are devoted to some remarkable ideas due to Talagrand. The first, ‘Talagrand’s measure’ (§464), is a special measure on PI (or `∞ (I)), extending the usual measure of PI in a canonical way. In §465 I turn to the theory of ‘stable’ sets of measurable functions, showing how a concept arising naturally in the theory of pointwise compact sets led to a characterization of Glivenko-Cantelli classes in the theory of empirical measures.

461 Barycenters and Choquet’s theorem One of the themes of this chapter will be the theory of measures on linear spaces, and the first fundamental concept is that of ‘barycenter’ of a measure, its centre of mass (461Aa). The elementary theory (461B461D) uses non-trivial results from the theory of locally convex spaces (§4A4), but is otherwise natural and straightforward. It is not always easy to be sure whether a measure has a barycenter in a given space, and I give a representative pair of results in this direction (461E, 461G). Deeper questions concern the existence and nature of measures on a given compact set with a given barycenter. The Riesz representation theorem is enough to tell us just which points can be barycenters of measures on compact sets (461H). A new idea (461J) shows that the measures can be moved out towards the boundary of the compact set. We need a precise definition of ‘boundary’; the set of extreme points seems to be the appropriate concept (461N). 461A Definitions (a) Let X be a Hausdorff locally convex linear topological space, R and µ a probability measure on a subset A of X. Then x∗ ∈ X is the barycenter or resultant of µ if A f dµ is defined and equal to f (x∗ ) for every f ∈ X ∗ . Because X ∗ separates the points of X (4A4Ec), µ can have at most one barycenter. (b) Let X be any real linear space, and C ⊆ X a convex set (definition: 2A5E). Then a function g : C → R is convex if g(tx + (1 − t)y) ≤ tg(x) + (1 − t)g(y) for all x, y ∈ C and t ∈ [0, 1]. (Compare 233G, 233Xd.) (c) The following elementary remark is useful. Let X be a linear space over R, C ⊆ X a convex set, and g : C → R a function. Then g is convex iff the set {(x, α) : x ∈ C, α ≥ g(x)} is convex in X × R (cf. 233Xd).

461E

Barycenters and Choquet’s theorem

523

461B Proposition Let X and Y be Hausdorff locally convex linear topological spaces, and T : X → Y a continuous linear operator. Suppose that A ⊆ X, B ⊆ Y are such that T [A] ⊆ B, and let µ be a probability measure on A which has a barycenter x∗ in X. Then T x∗ is the barycenter of the image measure µT −1 on B. proof All we have to observe is that if g ∈ Y ∗ then gT ∈ X ∗ (4A4Bd), so that g(T x∗ ) =

R

A

g(T x)µ(dx) =

R

B

g(y)ν(dy)

by 235I. 461C Theorem Let X be a Hausdorff locally convex linear topological space, C ⊆ X a convex set and µ a topological probability measure on a subset A of C. Suppose that µ has a barycenter x∗ in X which R ∗ belongs to C. Then g(x ) ≤ A g dµ for every lower semi-continuous convex function g : C → R. proof Take any γ < g(x∗ ). Let D be the convex set {(x, α) : x ∈ C, α ≥ g(x)} in X × R (461Ac). Then the closure D of D is also convex (2A5Eb). Now (x∗ , γ) ∈ / D. P P Set δ = 12 (g(x∗ ) − γ) > 0. As g is lower ∗ semi-continuous, there is an open neighbourhood G of x such that g(x) ≥ g(x∗ ) − δ for every x ∈ C ∩ G, so that G × ]−∞, γ + δ[ is an open set containing (x∗ , γ) and disjoint from D. Q Q Consequently there is a continuous linear functional h : X × R → R such that h(x∗ , γ) < inf w∈D h(w) (4A4Eb). Now there are f ∈ X ∗ , β ∈ R such that h(x, α) = f (x) + αβ for every x ∈ X, α ∈ R (4A4Be). So we have f (x∗ ) + βγ = h(x∗ , γ) < h(x, g(x)) = f (x) + βg(x) for every x ∈ C. In particular, f (x∗ ) + βγ < f (x∗ ) + βg(x∗ ), so β > 0. Integrating with respect to µ, we have f (x∗ ) + βγ <

R

f dµ + β R R and γ < A g dµ. As γ is arbitrary, g(x∗ ) ≤ A g dµ. R Remark Of course A g dµ might be infinite. A

R

A

g dµ = f (x∗ ) + β

R

A

g dµ,

461D Theorem Let X be a Hausdorff locally convex linear topological space, and µ a probability measure on X such that (i) the domain of µ includes the cylindrical σ-algebra of X (ii) there is a compact convex set K ⊆ X such that µ∗ K = 1. Then µ has a barycenter in X, which belongs to K. proof If f ∈ X ∗ , then γf = supx∈K |fR(x)| is finite, and {x : |f (x)| ≤ γf } is a measurable set including K, so must be conegligible, and φ(f ) = f dµ is defined and finite. Now φ : X ∗ → R is a linear functional and φ(f ) ≤ supx∈K f (x) for every f ∈ X ∗ ; because K is compact and convex, there is an x0 ∈ K such that φ(f ) = f (x0 ) for every f ∈ X ∗ (4A4Ef), so that x0 is a barycenter of µ in X. 461E Theorem Let X be a complete locally convex linear topological space, and A ⊆ X a bounded set. Let µ be a τ -additive topological probability measure on A. Then µ has a barycenter in X. proof (a) If f ∈ X ∗ , then f ¹A is continuous and bounded (4A4Bg), therefore µ-integrable. For each open neighbourhood G of 0 in X, set FG = {y : y ∈ X, |f (y) −

R

A

f dµ| ≤ 2τG (f ) for every f ∈ X ∗ }

where τG (f ) = supx∈G f (x) for f ∈ X ∗ . Then FG is non-empty. P P Set H = int(G ∩ (−G)), so that H is an open neighbourhood of 0. Because A is bounded, there is an m ≥ 1 such that A ⊆ mH. The set {x + H : x ∈ A} is an open cover of A, and µ is a τ -additive topological measure, so there are x0 , . . . , xn ∈ A S 1 such that µ(A \ i≤n (xi + H)) ≤ . Set m S Ei = A ∩ (xi + H) \ j
524


Z |f (y) −

f dµ| ≤ A

≤

n X

Z

i=0

Z

|µEi f (xi ) −

f dµ| + Ei

i=0 n Z X

461E

|f |dµ E

|f (x) − f (xi )|µ(dx) + mτG (f )µE Ei

(because E ⊆ mH, so f (x) ≤ mτG (f ) and f (−x) ≤ mτG (f ) for every x ∈ E) n X ≤ τG (f )µEi + mτG (f )µE i=0

(because if i ≤ n and x ∈ Ei , then x − xi and xi − x belong to G, so |f (x) − f (xi )| = |f (x − xi )| ≤ τG (f )) ≤ 2τG (f ). As f is arbitrary, y ∈ FG and FG 6= ∅. Q Q (c) Since FG∩H ⊆ FG ∩ FH for all neighbourhoods G and H of 0, {FG : G is a neighbourhood of 0} is a filter base and generates a filter F on X. Now F is Cauchy. P P If G is any neighbourhood of 0, let G1 ⊆ G be a closed convex neighbourhood of 0, and set H = 14 G1 . ?? If y, y 0 ∈ FH and y − y 0 ∈ / G, then there is an f ∈ X ∗ such that f (y − y 0 ) > τG1 (f ) (4A4Eb). But now R R 1 1 1 τH (f ) = τG1 (f ) < (|f (y) − A f dµ| + |f (y 0 ) − A f dµ|) ≤ (2τH (f ) + 2τH (f )). X X 4

4

4

This means that FH − FH ⊆ G; as G is arbitrary, F is Cauchy. Q Q (d) Because X is complete, F has a limit x∗ say. Take any f ∈ X ∗ . ?? If f (x∗ ) 6= R G = {x : |f (x)| ≤ 13 |f (x∗ ) − A f dµ|}. Then G is a neighbourhood of 0 in X, and Z Z Z |f (x∗ ) − f dµ| = lim |f (x) − f dµ| ≤ sup |f (x) − f dµ| x→F x∈FG A A A Z 2 ≤ 2τG (f ) ≤ |f (x∗ ) − f dµ|. X X 3

∗

So f (x ) =

R A

R A

f dµ, set

A

∗

f dµ; as f is arbitrary, x is a barycenter of µ.

461F Lemma Let X be a normed space, R and µ a probability measure on X such that every member of the dual X ∗ of X is integrable. Then f 7→ f dµ : X ∗ → R is a bounded linear functional on X ∗ . proof Replacing µ by its completion if necessary, we may suppose that µ is complete, so that every member of X ∗ is Σ-measurable, where Σ is the domain of µ. (The point is that µ and its completion give rise to the same integrals, T by 212Fb.) For each n ∈ N let En be a measurable envelope of Bn = {x : kxk ≤ n}; replacing En by i≥n Ei if necessary, we may suppose that hEn in∈N is non-decreasing. If n ∈ N, f ∈ X ∗ then be negligible, and R {x : x ∈ En , |f (x)| > nkf k} is a measurable subset of En disjoint from Bn , so must R | En f | ≤ nkf k. We therefore have an element φn of X ∗∗ defined by setting φn (f ) = En f for every f ∈ X ∗ . But also φ(f ) = limn→∞ φn (f ) for every f , because hEn in∈N is a non-decreasing sequence of measurable sets with union X. By the Uniform Boundedness Theorem (3A5Ha), {φn : n ∈ N} is bounded in X ∗∗ , and φ ∈ X ∗∗ , with kφk ≤ supn∈N kφn k. 461G Proposition Let X be a reflexive Banach space, and µ a probability measure on X such that every member of X ∗ is µ-integrable. Then µ has a barycenter in X. R proof By 461F, f 7→ f is a bounded linear functional on X ∗ ; but this means that it is represented by a member of X, which is a barycenter of µ. 461H Theorem Let X be a Hausdorff locally convex linear topological space, and K ⊆ X a compact set. Then the closed convex hull of K in X is just the set of barycenters of Radon probability measures on K.

461J


525

proof (a) If µ is a Radon probability measure on K with barycenter x∗ , then f (x∗ ) =

R

K

f (x)µ(dx) ≤ supx∈K f (x) ≤ supx∈Γ(K) f (x);

because Γ(K) is closed and convex, it must contain x∗ (4A4Eb). (b) Now suppose that x∗ ∈ Γ(K). Let W ⊆ C(X) be the set of functionals of the form f + αχX, where f ∈ X ∗ and α ∈ R. Set U = {f ¹K : f ∈ W }, so that U is a linear subspace of C(K) containing χK. If f1 , f2 ∈ W and f1 ¹K = f2 ¹K, then {x : f1 (x) = f2 (x)} is a closed convex set including K, so contains x∗ , and f1 (x∗ ) = f2 (x∗ ); accordingly we have a functional φ : U → R defined by setting φ(f ¹K) = f (x∗ ) for every f ∈ W . Of course φ is linear; moreover, φ(g) ≤ supx∈K g(x) for every g ∈ U , by 4A4Eb. Applying this to ±g, we see that |φ(g)| ≤ kgk∞ for every g ∈ U . We therefore have an extension of φ to a continuous linear functional ψ on C(K) such that kψk ≤ 1, by the Hahn-Banach theorem (3A5Ab). Now ψ(χK) = φ(χK) = χX(x∗ ) = 1; so if 0 ≤ h ≤ χK then |1 − ψ(h)| = |ψ(χK − h)| ≤ kχK − hk∞ ≤ 1, and ψ(h) ≥ 0. It follows that ψ(h) ≥ 0 forR every h ∈ C(K)+ . But this means that there is a Radon probability measure µ on K such that ψ(h) = h dµ for every h ∈ C(K) (436J/436K). As µK = ψ(χK) = 1, µ is a probability measure; and for any f ∈ X ∗

R

K

f dµ = ψ(f ¹K) = φ(f ¹K) = f (x∗ ),

so x∗ is the barycenter of µ, as required. 461I Corollary: Kreˇın’s theorem Let X be a complete Hausdorff locally convex linear topological space, and K ⊆ X a weakly compact set. Then the closed convex hull Γ(K) of K is weakly compact. proof Give K the weak topology induced by Ts (X, X ∗ ). Let P be the set of Radon probability measures on K (or, if you prefer, the set of tight Borel probability measures) with the narrow topology, R so that P is compact (437O). By 461E, every µ ∈ P has a barycenter b(µ) in K. If f ∈ X ∗ , f (b(µ)) = K f dµ, while f ¹K is continuous; so b : P → X is continuous for the vague topology on P and the weak topology on X. Accordingly b[P ] is weakly compact. But b[P ] is the weakly closed convex hull of K, by 461H applied to the weak topology on X. By 4A4Ed, Γ(K) has the same closure for the original topology of X as it has for the weak topology, and Γ(K) = b[P ] is weakly compact. 461J Lemma Let X be a Hausdorff locally convex linear topological space, and K a compact convex subset of X. Then for any x∗ ∈ K there is a Radon probability measure µ on K such that x∗ is the barycenter of µ and µ{ 12 (x + y) : x ∈ M1 , y ∈ M2 } = 0 for all disjoint closed convex subsets M1 , M2 of K. proof (a) Let P ⊆ C(K)∗ be the set of linear functionals φ such that (i) φ(g) ≥ 0 for every non-negative g ∈ C(K) (ii) φ(χK) = 1 (iii) φ(f ¹K) = f (x∗ ) for every f ∈ X ∗ . Then P is a weak*-closed subset of the unit ball of C(K)∗ , so is compact for the weak* topology on C(K)∗ (3A5F). In the proof of 461H we found a Radon probability measure on K with barycenter x∗ precisely by finding a member of P , so P is not empty. (b) For φ, ψ ∈ P , say that φ 4 ψ if φ(g) ≤ ψ(g) for every convex continuous function g : K → R. Then 4 is reflexive and transitive, and Rφ = {ψ : ψ ∈ P , φ 4 ψ} is closed for everyTφ ∈ P . Let Q ⊆ P be a maximal set such that {Rφ : φ ∈ Q} has the finite intersection property; then φ∈Q Rφ is not empty; fix φ˜ in the intersection. Of course φ˜ ∈ P (because Q is not empty). Note that if ψ ∈ P and φ˜ 4 ψ, then ˜ {Rφ : φ ∈ Q ∪ {ψ}} has non-empty intersection, so that ψ ∈ Q and ψ 4 φ. ˜ Let µ ˜ be the Radon probability measure on K corresponding to φ, so that x∗ is the barycenter of µ ˜. (c) Now take disjoint closed convex sets M1 , M2 ⊆ K, and set M = { 21 (x + y) : x ∈ M1 , y ∈ M2 }. ?? Suppose, if possible, that µ ˜M > 0. Set q(x, y) = 12 (x+y) for x ∈ M1 , y ∈ M2 , so that q : M1 ×M2 → M is continuous. Let µ ˜M be the subspace measure on M induced by µ ˜, so that µ ˜M is a Radon measure on M (416Rb). Let λ be a Radon measure on M1 × M2 such that µ ˜M = λq −1 (418L), and define ψ : C(K) → R by writing

526


R

ψ(g) =

K\M

g d˜ µ+

R

1 (g(x) M1 ×M2 2

461J

+ g(y))λ(d(x, y)).

Then ψ is linear, ψ(g) ≥ 0 whenever g ∈ C(K)+ , and ψ(χK) = µ ˜(K \ M ) + λ(M1 × M2 ) = µ ˜(K \ M ) + µ ˜M (M ) = 1. If g : K → R is continuous and convex, then Z

Z

ψ(g) =

g d˜ µ+ K\M

Z ≥

Z g d˜ µ+

K\M

M1 ×M2

g d˜ µ+

=

gq dλ M1 ×M2

K\M

Z =

1 2

g( (x + y))λ(d(x, y)) Z

Z

Z g d˜ µ+

K\M

(235I)

1 (g(x) + g(y))λ(d(x, y)) 2 M1 ×M2

g d˜ µM M

Z

Z

=

g × χ(K \ M )d˜ µ+ K

(131Fa)

g × χM d˜ µ K

Z ˜ g d˜ µ = φ(g).

= K

˜ ¹K) = f (x∗ ) for every f ∈ X ∗ , so that ψ ∈ P . Also Applying this to ±f ¹K, we see that ψ(f ¹K) = φ(f ˜ ˜ φ 4 ψ, so (as remarked in (b)) ψ 4 φ. Because M1 and M2 are disjoint non-empty compact convex sets in the Hausdorff locally convex space X, there are an f ∈ X ∗ and an α ∈ R such that f (x) < α < f (y) whenever x ∈ M1 , y ∈ M2 (4A4Ee). Set g(x) = |f (x) − α| for x ∈ K; then g is a continuous convex function. If x ∈ M1 and y ∈ M2 , then 1 2

1 2

1 2

g( (x + y)) = |f ( (x + y)) − α| = |(f (x) − α) + (f (y) − α)| 1 2

1 2

< (|f (x) − α| + |f (y) − α|) = (g(x) + g(y)). Looking at the formulae above for ψ(g), we see that we have Z Z 1 ψ(g) = g d˜ µ+ (g(x) + g(y))λ(d(x, y)) 2 K\M M1 ×M2 Z Z 1 ˜ > g d˜ µ+ g( (x + y))λ(d(x, y)) = φ(g) K\M

M1 ×M2

2

˜ which is impossible. X (because λ(M1 × M2 ) = µ ˜M > 0). But this means that ψ 64 φ, X Thus µ ˜M = 0. As M1 and M2 are arbitrary, µ ˜ has the properties required. 461K Choquet’s theorem Let X be a Hausdorff locally convex linear topological space, and K ⊆ X a metrizable compact convex set. Let E be the set of extreme points of K. Then for any x∗ ∈ K there is a Radon probability measure µ on K such that x∗ is the barycenter of µ and µE = 1. proof By 461J, there is a Radon probability measure µ on K such that x∗ is the barycenter of µ and µ( 21 (M + M 0 )) = 0 whenever M , M 0 are disjoint closed convex subsets of K, writing 21 (M + M 0 ) for { 12 (x + y) : x ∈ M , y ∈ M 0 }. Now µE = 1. P P Let U be the set of convex open subsets of X. Because the topology of X is locally convex, U is a base for the topology of X, and V = {K ∩ U : U ∈ U} is a base for the topology of K (4A2B(a-iv)); because K is compact and metrizable, it is separable (4A2Od), and there

461L


527

is a countable base V0 ⊆ V (4A2O(a-iii)). Set M = {V : V ∈ V0 }, so that M is a countable family of closed convex subsets of K. If x ∈ K \ E, then there are distinct y, z ∈ K such that x = 21 (y + z). Now there must be disjoint M , 0 M ∈ M such that y ∈ M and z ∈ M 0 , so that x ∈ 12 (M + M 0 ). Thus S K \ E ⊆ { 12 (M + M 0 ) : M , M 0 ∈ M, M ∩ M 0 = ∅}. As M is countable, µ(K \ E) = 0 and µE = 1, as required. Q Q 461L In the form above, Choquet’s theorem really does depend on the metrizability of X (see 461Yb). But there is a version which applies more generally, as follows. Lemma Let X be a Hausdorff locally convex linear topological space and K a compact convex subset of X. Let T : X → RN be a continuous linear operator. Let E be the set of extreme points of K and x0 a point of K such that T x0 ∈ / T [E]. Then x0 is expressible as 12 (y1 + y2 ) where y1 , y2 ∈ K and T y1 6= T y2 . proof (a) For each n ∈ N, set γn = supx∈K |(T x)(n)|, and consider P∞ 1 g(x) = n=0 n |(T x)(n) − (T x0 )(n)| 2 (γn +1)

for x ∈ K. Then g : K → [0, ∞[ is a continuous convex function and g −1 [{0}] = {x : x ∈ K, T x = T x0 }, so g(x) > 0 for every x ∈ E, while g(x0 ) = 0. Let M be the closed convex hull of {(x, g(x)) : x ∈ K} in X × R. (b) ?? Suppose, if possible, that (x0 , α) ∈ / M for every α > 0. Then for every n ∈ N there is a continuous linear functional hn : X × R → R such that hn (x0 , 2−n ) > supx∈K hn (x, g(x)). Let fn ∈ X ∗ , βn ∈ R be such that hn (x, α) = fn (x) + βn α for every x ∈ X, α ∈ R; because hn (x0 , 2−n ) > hn (x0 , 0), βn > 0. Set Y = RN × RN and define S : X → Y by setting Sx = (T x, hfn (x)in∈N ) for x ∈ X; then S is continuous and linear, so S[K] is compact and convex in Y . Set P∞ 1 g˜(w, z) = n=0 n |w(n) − (T x0 )(n)| 2 (γn +1)

for (w, z) ∈ S[K], so that g˜ is a continuous convex function and g = g˜S. Let F be the set of extreme points of S[K]. By Choquet’s theorem, there is a Radon probability measure ν on S[K] such that Sx0 is the barycenter of ν and νF = 1. For each n ∈ N, we have fn (x) + βn g(x) = hn (x, g(x)) ≤ hn (x0 , 2−n ) = fn (x0 ) + 2−n βn for each x ∈ K. So for (w, z) ∈ S[K] we get z(n) + βn g˜(w, z) ≤ fn (x0 ) + 2−n βn . Integrating with respect to ν, and recalling that Sx0 = (T x0 , hfn (x0 )in∈N ) is the barycenter of ν, we get

R

g˜ dν ≤ fn (x0 ) + 2−n βn , R R and (because βn > 0) g˜ dν ≤ 2−n . As n is arbitrary, g˜ dν = 0. R But F ⊆ S[E] (4A4Gc), and g(x) > 0 for every x ∈ E, so g˜(w, z) > 0 for every (w, z) ∈ F . Since νF = 1, g˜ dν > 0, which is impossible. X X fn (x0 ) + βn

(c) There is therefore some α0 > 0 such that (x0 , α0 ) ∈ M . Of course L = {(x, g(x)) : x ∈ K} is compact (being a continuous image of K), so there is a Radon probability measure λ on L such that (x0 , α0 ) is the barycenter of λ (461H). Let µ be the image measure λπ1−1 on K, where π1 (x, α) = x; then µ is a Radon probability measure on K, and x0 is the barycenter of µ (461B). At the same time, Z

Z

Z

g dµ = K

gπ1 (x, α)λ(d(x, α)) = L

(because if (x, α) ∈ L then α = g(x)) = α0 > 0

αλ(d(x, α)) L

528


461L

(because (x, α) 7→ α is a continuous linear functional on X × R andR (x0 , α0 ) is the barycenter of λ). Looking back at the construction of g, there must be some n ∈ N such that K |(T x)(n) − (T x0 )(n)|µ(dx) is non-zero. R (d) Since x 7→ (T x)(n) is a continuous linear functional, K (T x)(n)µ(dx) = (T x0 )(n) = β say, and M1 = {x : (T x)(n) ≤ β},

M2 = {x : (T x)(n) > β}

must both have non-zero measure. We therefore have Radon probability measures µ1 , µ2 on K defined by saying that µ1 H = µ(H ∩ M1 )/µM1 , µ2 H = µ(H ∩ M2 )/µM2 whenever these are defined. For each i, we have a barycenter yi0 of µi ; since (T y10 )(n) =

R

(T x)(n)µ1 (dx) ≤ β < K

R

K

(T x)(n)µ2 (dx) = (T y20 )(n),

T y10 6= T y20 . On the other hand, for any f ∈ X ∗ , f ((µM1 )y10 + (µM2 )y20 ) = µM1 so x0 =

(µM1 )y10

+

R

f dµ1 + µM2 K

R

f dµ2 = K

R K

f dµ = f (x0 ),

(µM2 )y20 .

(e) Thus x0 is a proper convex combination of y10 and y20 , which are members of K distinguished by T . There must therefore be distinct y1 , y2 in the line segment Γ({y10 , y20 }) such that x0 = 12 (y1 + y2 ), and in this case T y1 6= T y2 , as required. 461M Theorem Let X be a Hausdorff locally convex linear topological space and K a compact convex subset of X. Let E be the set of extreme points of K. Then for any x∗ in K there is a Baire probability measure µ on K such that x∗ is the barycenter of µ and µ∗ E = 1. proof (a) By 461J, there is a Radon probability measure µ ˜ on K such that x∗ is the barycenter of µ ˜ and 1 0 0 of µ ˜( 2 (M + M )) = 0 whenever M , M are disjoint closed convex subsets of K. Let µ be the restriction R ∗ µ ˜ to the Baire σ-algebra of K; since f ¹K is surely Baire measurable for every f ∈ X , we have f dµ = R f d˜ µ = f (x∗ ) for every f ∈ X ∗ , and x∗ is still the barycenter of µ. (b) The point is that µ ˜Z = 0 for every zero set P Since Z ⊆ K is a closed S Z in K which is disjoint from E. P Gδ set in K, K \ Z is expressible as a union n∈N Fn of compact sets. For any n ∈ N, z ∈ Z and y ∈ Fn , ∗ there is an f ∈ X ∗ such that f (z) 6= f (y); since Z and Fn are compact, there is a finite S set Φn ⊆ X such that whenever z ∈ Z and y ∈ Fn there is an f ∈ Φn such that f (z) 6= f (y). Set Φ = n∈N Φn ∪ {0}; then Φ is countable; let hfn in∈N be a sequence running over Φ, and define T : X → RN by setting (T x)(n) = fn (x) for n ∈ N, x ∈ X. Then T is a continuous linear operator and T [Z] is disjoint from T [Fn ] for every n ∈ N, so T [Z] is disjoint from T [E]. Lemma 461L now tells us that for every z ∈ Z there are x, y ∈ K such that z = 21 (x + y) and T x 6= T y. 0 0 For n ∈ N and q < q 0 in Q set Mnqq0 = {x : x ∈ K, fn (x) ≤ q}, Mnqq 0 = {x : x ∈ K, fn (x) ≥ q }. 1 0 0 Then Mnqq0 and Mnqq ˜( 2 (Mnqq0 + Mnqq 0 are disjoint closed convex subsets of K, so µ 0 )) = 0; but as Z ⊆ S 1 0 0 ˜Z = 0. Q Q n∈N,q
R

K

f dµ =

R

E

f dν

for every f ∈ X ∗ , by 214F. 461O It is a nearly universal rule that when we encounter a compact convex set we should try to identify its extreme points. I look now at some sets which arose naturally in §437.

461P


529

Proposition Let X be a set, Σ a σ-algebra of subsets of X, and φ : X → X a (Σ, Σ)-measurable function. Let Mσ be the L-space of countably additive real-valued functionals on Σ, and Q ⊆ Mσ the set of probability measures with domain Σ for which φ is inverse-measure-preserving. If µ ∈ Q, then µ is an extreme point of Q iff φ is ergodic with respect to µ. proof (a) Suppose that φ is µ-ergodic, and that µ = 12 (µ1 + µ2 ) where µ1 , Rµ2 ∈ Q. Then 0 ≤ µ1 E ≤ 2µE for every E ∈ Σ, so there is a Σ-measurable f : X → [0, 2] such that µ1 E = E f dµ for every E ∈ Σ (232F). For α ≥ 0, set Gα = {x : x ∈ X, f (x) > α}. Then, for any E ∈ Σ, µ1 E =

R

R

f × χE dµ =

R∞ 0

µ(E ∩ Gα )dα

where the integral dα is taken with respect to Lebesgue measure (252O). We are supposing that µ1 φ−1 [E] = µ1 E for every E ∈ Σ. In particular, for any α ≥ 0, Z ∞ Z ∞ αµGα + µGβ dβ = µ(Gα ∩ Gβ )dβ = µ1 Gα α 0 Z ∞ = µ1 φ−1 [Gα ] = µ(φ−1 [Gα ] ∩ Gβ )dβ 0 Z ∞ −1 ≤ αµφ [Gα ] + µ(φ−1 [Gα ] ∩ Gβ )dβ α Z ∞ = αµGα + µ(φ−1 [Gα ] ∩ Gβ )dβ α

because µ also belongs to Q. So

R∞ α

µGβ dβ ≤

R∞ α

µ(φ−1 [Gα ] ∩ Gβ )dβ.

It follows that µ(φ−1 [Gα ] ∩ Gβ ) = µGβ , that is, Gβ \ φ−1 [Gα ] is µ-negligible, for almost every β >Sα. Take a sequence hβn in∈N with infimum α such that Gβn \ φ−1 [Gα ] is µ-negligible for every n; then Gα = n∈N Gβn , so Gα \ φ−1 [Gα ] is µ-negligible. As µGα = µφ−1 [Gα ], Gα 4φ−1 [Gα ] is µ-negligible. Since φ is µ-ergodic, µGα must be 0 or 1 (372J). As this is true for every α, f must be µ-essentially constant, and µ1 = µ. As µ1 and µ2 are arbitrary, µ is an extreme point of Q. (b) Suppose that φ is not ergodic. Let E ∈ Σ be such that φ−1 [E] = E and 0 < µE < 1. Set α = µE, β = 1 − µE and let µ1 , µ2 be the uncompleted indefinite-integral measures over µ defined by α1 χE and 1 β χ(X \ E) (234Cc). Then both are probability measures with domain Σ. For any F ∈ Σ, 1 α

1 α

1 α

1 α

µ1 (φ−1 [F ]) = µ(E ∩ φ−1 [F ]) = µ(φ−1 [E] ∩ φ−1 [F ]) = µ(φ−1 [E ∩ F ]) = µ(E ∩ F ) = µ1 F. So µ1 and µ1 φ−1 agree on Σ and µ1 ∈ Q. Similarly, µ2 ∈ Q. As neither µ1 nor µ2 is equal to µ, µ is not extreme in Q. 461P Corollary Let X be a compact Hausdorff space and φ : X → X a continuous function. Let Qφ be the non-empty compact convex set of Radon probability measures µ on X such that φ is inverse-measurepreserving for µ, with its narrow topology and the convex structure inherited from its identifications with subsets of C(X)∗ and of the L-space of countably additive functionals on the Borel σ-algebra of X (437S). Then the extreme points of Qφ are those for which φ is inverse-measure-preserving and ergodic. proof (a) If µ0 ∈ Qφ is not extreme, let B = B(X) be the Borel σ-algebra of X, so that φ is (B, B)measurable, and write Q0φ for the set of φ-invariant Borel probability measures on X. Then 461O tells us that the extreme points of Q0φ are just the measures for which φ is ergodic. If µ ∈ Qφ , then µ¹B ∈ Qφ0 , and of course the function µ 7→ µ¹B is injective (416Eb) and preserves convex combinations. So µ0 ¹B is not extreme in Q0φ . By 461O, φ is not µ0 ¹B-ergodic, and therefore not µ0 -ergodic.

530


461P

(b) If µ0 ∈ Qφ and φ is not µ0 -ergodic, let E ∈ dom µ0 be such that 0 < µ0 E < 1 and φ−1 [E] = E. Set α = µ0 E and β = 1 − α, and let µ1 , µ2 be the indefinite-integral measures over µ0 defined by 1 χ(X β

1 χE α

and

\ E). Then µ1 is a Radon probability measures on X (416S), so the image measure µ1 φ−1 is also a

Radon measure (418I). The argument of part (b) of the proof of 461O tells us that µ1 φ−1 agrees with µ1 on Borel sets, so µ1 = µ1 φ−1 (416Eb) and µ1 ∈ Qφ . Similarly, µ2 ∈ Qφ , and µ0 = αµ1 + βµ2 , so µ0 is not extreme in Qφ . 461X Basic exercises > (a) Let X be a Hausdorff locally convex linear topological space, C ⊆ X a convex set, and g : C → R a function. Show that the following are equiveridical: (i) g is convex and lower semi-continuous; (ii) there are a non-empty set D ⊆ X ∗ and a family hβf if ∈D in R such that g(x) = supf ∈D f (x) + βf for every x ∈ C. (Compare 233Hb.) > (b) Let X be a Hausdorff locally convex linear topological space, K ⊆ X a compact convex set, and x an extreme point of K. Let µ be a probability measure on X such that µ∗ K = 1 and x is the barycenter of µ. (i) Show that {y : y ∈ K, f (y) 6= f (x)} is µ-negligible for every f ∈ X ∗ . (Hint: 461D.) (ii) Show that if µ is a Radon measure then µ{x} = 1. (c) For each n ∈ N, define dn ∈ c 0 by saying that dn (n) = 1, dn (i) = P 0 if i 6= n. Let µ be the point∞ supported Radon probability measure on c 0 defined by saying that µE = n=0 2−n−1 χE(2n dn ) for every ∗ ∗ E ⊆ c 0 . (i) Show that every member of c 0 is µ-integrable. (Hint: c 0 can be identified with `1 .) (ii) Show that µ has no barycenter in c 0 . (d) Let I be an uncountable set, and X = {x : x ∈ `∞ (I), {i : x(i) 6= 0} is countable}. Show that X is a closed linear subspace of `∞ (I). (i) Show that there is a probability measure µ on X such that R (α) µ{x : kxk ≤ 1} is defined and equal to 1 (β) µ{x : x(i) = 1} = 1 for every i ∈ I. (ii) Show that f dµ is defined for every f ∈ X ∗ . (Hint: for any f ∈ X ∼ , there is a countable set J ⊆ X such that f (x) = 0 whenever x¹J = 0, so that f =a.e. f (χJ).) (iii) Show that µ has no barycenter in X. > (e) Let X be a complete Hausdorff locally convex linear topological space, and K ⊆ X a compact set. Show that every extreme point of Γ(K) belongs to K. (Hint: show that it cannot be the barycenter of any measure on K which is not supported by a single point.) > (f ) Let X be a Hausdorff locally convex linear topological space, and K ⊆ X a metrizable compact set. Show that Γ(K) is metrizable. (Hint: we may suppose that X is complete, so that Γ(K) is compact. Show that Γ(K) is a continuous image of the space P of Radon probability measures on K, and use 437Ra.) (g) Let X be a Hausdorff locally convex linear topological space and K ⊆ X a compact set. Show that the Baire σ-algebra of K is just the subspace σ-algebra induced by the cylindrical σ-algebra of X. (Hint: disjoint closed subsets of K can be separated by finite subsets of X ∗ .) >(h) Let X be a Hausdorff locally convex linear topological space, and K ⊆ X a compact convex set; let E be the set of extreme points of K. Let hfn in∈N be a sequence in X ∗ such that (i) supx∈K,n∈N |fn (x)| is finite (ii) limn→∞ fn (x) = 0 for every x ∈ E. Show that limn→∞ fn (x) = 0 for every x ∈ K. > (i) Let X be a Hausdorff locally convex linear topological space, and K ⊆ X a metrizable compact convex set. Show that the set E of extreme points of K is a Gδ set, therefore Polish. Show that the algebra of Borel subsets of K is just the subspace algebra of the cylindrical σ-algebra of X. (j) Let X be a Hausdorff locally convex linear topological space, and K ⊆ X a compact set. Let us say that a point x of K is extreme if the only Radon probability measure on K with barycenter x is the one for which {x} has measure 1. (Cf. 461Xb.) (i) Show that if X is complete, then x ∈ X is an extreme point of K iff it is an extreme point of Γ(K). (ii) Writing E for the set of extreme points of K, show that any point of K is the barycenter of some probability measure on E. (iii) Show that if K is metrizable then E is a Gδ subset of K and any point of K is the barycenter of some Radon probability measure on E.

461 Notes


531

(k) Let X be a set, Σ a σ-algebra of subsets of X, and P the set of probability measures with domain Σ, regarded as a convex subset of the linear space of countably additive functionals on Σ. Show that µ ∈ P is an extreme point in P iff it takes only the values 0 and 1. (l) Let X be a set, Σ a σ-algebra of subsets of X, φ : X → X a (Σ, Σ)-measurable function and Mσ the L-space of countably additive functionals with domain Σ. (i) Show that the set V = {µ : µ ∈ Mσ , µφ−1 [E] = µE for every E ∈ Σ} is a norm-closed Riesz subspace of Mσ , so is an L-space in its own right. (ii) Let Qφ be the set of probability measures in V . Show that if µ, ν are distinct extreme points of Qφ then there is an E ∈ Σ which is negligible for one and conegligible for the other. (Hint: show that supn∈N µ ∧ nν = 0 and apply 326I to µ − ν.) > (m) Set S 1 = {z : z ∈ C, |z| = 1}, and let w ∈ S 1 be such that wn 6= 1 for any integer 1. Define φ : S 1 → S 1 by setting φ(z) = wz for every z ∈ S 1 . Show that the only Radon probability measure on S 1 for which φ is inverse-measure-preserving is the Haar probability measure µ of S 1 . (Hint: use 281N to show R Pn 1 k that limn→∞ n+1 k=0 f (w z) = f dµ for every f ∈ C(S 1 ); now put 461P and 372I together.) (n) Set φ(x) = 2 min(x, 1 − x) for x ∈ [0, 1] (cf. 372Xm). Show that there are many point-supported Radon measures on [0, 1] for which φ is inverse-measure-preserving. 461Y Further exercises (a) Let X be a Hausdorff locally convex linear topological space. (i) Show that if M0 , . . . , Mn are non-empty compact convex subsets Pnempty intersection, then there is a Pn of X with continuous convex function g : X → R such that g( i=0 αi xi ) < i=0 αi g(xi ) whenever xi ∈ Mi and αi > 0 for every i ≤ n. (ii) Show that if K ⊆ X is compact and x∗ ∈ Γ(K) then there is a Radon probability measure µ on K, with barycenter x∗ , such that µ(α0 M0 + . . . + αn Mn )P= 0 whenever M0 , . . . , Mn are n compact convex sets with empty intersection, αi ≥ 0 for every i ≤ n, and i=0 αi = 1. (b) In R[0,2[ let K be the set of those functions u such that (α) 0 ≤ u(s) ≤ u(t) ≤ 1 whenever 0 ≤ s ≤ t ≤ 1 (β) |u(t + 1)| ≤ u(s0 ) − u(s) whenever 0 ≤ s < t < s0 ≤ 1. (i) Show that K is a compact convex set. (ii) Show that the set E of extreme points of K is just the set of functions of the types 0, χ[0, 1], χ[s, 1] ± χ{1 + s} and χ ]s, 1] ± χ{1 + s} for 0 < s < 1. (iii) Set w(s) = s for s ∈ [0, 1], 0 for s ∈ ]1, 2[. Show that if µ is any Radon probability measure on K with barycenter w then µE = 0. (c) Let G be a topological group. Show that it is amenable iff whenever X is a Hausdorff linear topological space, and • is a continuous action of G on X such that x 7→ a•x is a linear operator for every a ∈ G, and K ⊆ X is a compact convex set such that a•x ∈ K for every a ∈ G and x ∈ K, then there is an x ∈ K such that a•x = x for every a ∈ G. Use this to simplify parts of the proof of 449C. (Hint: 493B.) (d) Let X be a set, Σ a σ-algebra of subsets of X, and Φ a set of (Σ, Σ)-measurable functions from X to itself. Let Mσ be the L-space of countably additive real-valued functionals on Σ, and Q ⊆ Mσ the set of probability measures with domain Σ for which every member of Φ is inverse-measure-preserving. (i) Show that if µ ∈ Q, then µ is an extreme point of Q iff µE ∈ {0, 1} whenever E ∈ Σ and µ(E4φ−1 [E]) = 0 for every φ ∈ Φ. (ii) Show that if µ ∈ Q and Φ is countable and commutative, then µ is an extreme point of Q iff µE ∈ {0, 1} whenever E ∈ Σ and E = φ−1 [E] for every φ ∈ Φ. (e) Let X be a non-empty Hausdorff space, and define φ : X N → X N by setting φ(x)(n) = x(n + 1) for x ∈ X N , n ∈ N. Let Qφ be the set of Radon measures on X N for which φ is inverse-measure-preserving. Show that a Radon probability measure λ on X N is an extreme point of Qφ iff it is a Radon product measure µN for some Radon probability measure µ on X. 461 Notes and comments The results above are a little unusual in that we have studied locally convex spaces for several pages without encountering two topologies on the same space more than once (461I). In fact some of the most interesting properties of measures on locally convex spaces concern their relationships with strong and weak topologies, but I defer these ideas to later parts of the chapter. For the moment, we just have the basic results (i) affirming that barycenters exist (461D, 461E, 461G) (ii) that points can be represented as barycenters (461H, 461K, 461M). The last two can be thought of as refinements of the

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461 Notes

Kreˇın-Milman theorem. Any compact convex set K (in a locally compact Hausdorff space) is the closed convex hull of the set E of its extreme points. By 461N, given x ∈ K, we can actually find a measure on E with barycenter x; and if K is metrizable we can do this with a Radon measure. Of course the second part of 461N is a straightforward consequence of the first, because in the metrizable case E is a Borel set, the cylindrical σ-algebra (on K or E) is the Borel algebra, and any Borel probability measure on E must be tight (461Xi, 433Ca). But I do not know of any proof of 461N which does not pass through 461K. Kreˇın’s theorem (461I) is a fundamental result in the theory of linear topological spaces. The proof here, using the Riesz representation theorem and vague topologies, is a version of the standard one (e.g., Bourbaki 87, II.4.1), written out to be a little heavier in the measure theory and a little lighter in the topological linear space theory than is usual. There are of course proofs which do not use measure theory. In §437 I have already looked at an archetypal special case of 461H and 461M. If X is a compact Hausdorff space and P the compact convex set of Radon probability measures on X with the vague topology, then the set of extreme points of P can be identified with ∆ = {δ(x) : x ∈ X}, where δ(x) is the Radon probability measure concentrated at the point x (437P, 437Xo). If we think of P as a subset of C(X)∗ with the weak* topology, so that the dual of the linear topological space C(X)∗ can be identified with C(X), then any µ ∈ P is the barycenter of a Baire measure ν on ∆. In fact (because ∆ here is compact) µ is the barycentre of a Radon measure on ∆, and this is just the image measure µδ −1 . Another class of examples arising in §437 is explored in 461O-461P, 461Xk-461Xn and 461Yd-461Ye. It is when we have an explicit listing of the extreme points, as in 461Yb and 461Ye, that we begin to feel that we understand a compact convex set.

462 Pointwise compact sets of continuous functions In preparation for the main work of this chapter, beginning in the next section, I offer a few pages on spaces of continuous functions under their ‘pointwise’ topologies (462Ab). There is an extensive general theory of such spaces, described in Arkhangel’skii 92; here I present only those fragments which seem directly relevant to the theory of measures on normed spaces and spaces of functions. In particular, I star the paragraphs 462B-462D, which are functional analysis rather than measure theory. They are here because although this material is well known, and may be found in many places, I think that the ideas, as well as the results, are essential for any understanding of measures on linear topological spaces. Measure theory enters the section in the proof of 462E, in the form of an application of the Riesz representation theorem, though 462E itself remains visibly part of functional analysis. In the rest of the section, however, we come to results which are pure measure theory. For (countably) compact spaces X, the Radon measures on C(X) are the same for the pointwise and norm topologies (462G). This fact has extensive implications for the theory of separately continuous functions (462I) and for the theory of convex hulls in linear topological spaces (462J). 462A Definitions (a) A regular Hausdorff space X is angelic if whenever A is a subset of X which is relatively countably compact in X, then (i) its closure A is compact (ii) every point of A is the limit of a sequence in A. (A Fr´ echet-Urysohn space is a topological space in which, for any set A, every point of the closure of A is a limit of a sequence in A. So (ii) here may be written as ‘every compact subspace of X is a Fréchet-Urysohn space’.) (b) If X is any set and A a subset of RX , then the topology of pointwise convergence on A is that inherited from the usual product topology of RX ; that is, the coarsest topology on A for which the map f 7→ f (x) : A → R is continuous for every x ∈ X. I shall commonly use the symbol Tp for such a topology. In this context, I will say that a sequence or filter is pointwise convergent if it is convergent for the topology of pointwise convergence. Note that if A is a linear subspace of RX then Tp is a linear space topology on A (4A4Ba).

*462B

Pointwise compact sets of continuous functions

533

*462B Theorem (Pryce 71) Let X be a topological space such that there is a sequence hXn in∈N of countably compact subsets of X, covering X, with the property that a function f : X → R is continuous whenever f ¹Xn is continuous for every n ∈ N. Then the space C(X) of continuous real-valued functions on X is angelic in its topology of pointwise convergence. proof Of course C(X) is regular and Hausdorff under Tp , because RX is, so we need attend only to the rest of the definition in 462Aa. Let A ⊆ C(X) be relatively countably compact for Tp . (a) Since {f (x) : f ∈ A}, being a continuous image of A, must be relatively countably compact in R (4A2G(f-iv)), therefore relatively compact (4A2Le), for every x ∈ X, the closure A of A in RX is compact, by Tychonoff’s theorem. ?? Suppose, if possible, that A 6⊆ C(X); let g ∈ A be a discontinuous function. By the hypothesis of the theorem, there is an n ∈ N such that g¹Xn is not continuous; take x∗ ∈ Xn such that g¹Xn is discontinuous at x∗ . Let ² > 0 be such that for every neighbourhood U of x∗ in Xn there is a point x ∈ U such that |g(x) − g(x∗ )| ≥ ². Choose sequences hfi ii∈N in A, hxi ii∈N in Xn as follows. Given hfi ii 0 and x ∈ Xn , there is a y ∈ D ∩ Xn such that |f (y) − f (x)| ≤ ² for every f ∈ I; whenever J ⊆ D is finite and ² > 0 there is an f ∈ B such that |f (x) − g(x)| ≤ ² for every x ∈ J. P P For any finite set I ⊆ RX and n ∈ N, the set QIn = {hf (x)if ∈I : x ∈ Xn } is a subset of the separable metric space RI , so is itself separable, and there is a countable dense set DIn ⊆ Xn such that Q0In = {hf (x)if ∈I : x ∈ DIn } is dense in QIn . Similarly, because g ∈ A, we can choose for any finite set J ⊆ X a sequence hfJi ii∈N in A such that limi→∞ fJi (x) = g(x) for every x ∈ J. Now construct hDm im∈N , hBm im∈N inductively by setting S S Dm = {DIn : n ∈ N, I ⊆ {g} ∪ i 0 such that J = {i : |g(y) − fi (y)| ≥ ²} is infinite. Let n be such that y ∈ Xn . For each m ∈ N, Im = {fi : i ≤ m} is a finite subset of B, so there is an xm ∈ DIm n such that |f (xm ) − f (y)| ≤ 2−m for every f ∈ Im ∪ {g}. Let x∗ ∈ Xn be a cluster point of hxm im∈N , and h ∈ C(X) a cluster point of hfi ii∈J . Then because g(x) = limi→∞ fi (x) for every x ∈ D, g(xm ) = h(xm ) for every m ∈ N, and g(x∗ ) = h(x∗ ); because |g(y) − fi (y)| ≥ ² for every i ∈ J, |g(y) − h(y)| ≥ ²; because |fi (xm ) − fi (y)| ≤ 2−m whenever i ≤ m, fi (x∗ ) = fi (y) for every i ∈ N, and h(x∗ ) = h(y); because |g(xm ) − g(y)| ≤ 2−m for every m, g(x∗ ) = g(y); but this means that g(y) = g(x∗ ) = h(x∗ ) = h(y) 6= g(y), which is absurd. X XQ Q

534


*462B

So g = limi→∞ fi for Tp . As g is arbitrary, A has both properties required in 462Aa; as A is arbitrary, C(X) is angelic. Remark For a slight strengthening of this result, see 462Ya. *462C Proposition (Pryce 71) Let (X, T) be an angelic regular Hausdorff space. (a) Any subspace of X is angelic. (b) If S is a regular topology on X finer than T, then S is angelic. (c) Any countably compact subset of X is compact and sequentially compact. proof (a) Let Y be any subset of X. Then of course the subspace topology on Y is regular and Hausdorff. If A ⊆ Y is relatively countably compact in Y , then A is relatively countably compact in X, so A, the closure of A in X, is compact. Now if x ∈ A, there is a sequence hxn in∈N in A converging to x; but hxn in∈N must have a cluster point in Y , and (because T is Hausdorff) this cluster point can only be x. Accordingly A ⊆ Y and is the closure of A in Y . Thus A is relatively compact in Y . Moreover, any point of A is the limit of a sequence in A. As A is arbitrary, Y is angelic. (b) By hypothesis, S is regular, and it is Hausdorff because it is finer than T. Now suppose that A ⊆ X is S-relatively countably compact. Because the identity map from (X, S) to (X, T) is continuous, A is T-relatively countably compact (4A2G(f-iv)), and the T-closure A of A is T-compact. Let F be any ultrafilter on X containing A. Then F has a T-limit x ∈ X. ?? If F is not S-convergent to x, there is an H ∈ S such that x ∈ H and X \ H ∈ F , so that A \ H ∈ F. Now x belongs to the T-closure of A \ H, because A \ H ∈ F and F is T-convergent to x; because T is angelic, there is a sequence hxn in∈N in A \ H which T-converges to x. But now hxn in∈N has a S-cluster point x0 . x0 must also be a T-cluster point of hxn in∈N , so x0 = x; but every xn belongs to the S-closed set X \ H, so x0 ∈ / H, which is impossible. X X Thus every ultrafilter on X containing A is S-convergent. Because S is regular, the S-closure A˜ of A is S-compact (3A3De). Again because S is finer than T, and T is Hausdorff, the two topologies must agree on A˜ = A. But now every point of A is the T-limit of a sequence in A, so every point of A˜ is the S-limit of a sequence in A. As A is arbitrary, S is angelic. (c) If K ⊆ X is countably compact, then of course it is relatively countably compact in its subspace topology, so (being angelic) must be compact in its subspace topology. If hxn in∈N is a sequence in K, let x be any cluster point of hxn in∈N . If {n : xn = x} is infinite, then this immediately provides us with a subsequence converging to x. Otherwise, take n such that x 6= xi for i ≥ n. Since x ∈ {xi : i ≥ n}, and {xi : i ≥ n} is relatively countably compact, there is a sequence hyi ii∈N in {xi : i ≥ n} converging to x. The topology of X being Hausdorff, {yi : i ∈ N} must be infinite, and hyi ii∈N , hxi ii∈N have a common subsequence which converges to x. As hxi ii∈N is arbitrary, K is sequentially compact. *462D Theorem Let U be any normed space. Then it is angelic in its weak topology. proof Write X for the unit ball of the dual space U ∗ , with its weak* topology. Then X is compact (3A5F). We have a natural map u 7→ u ˆ : U → RX defined by setting u ˆ(x) = x(u) for x ∈ X, u ∈ U . By the definition of the weak* topology, u ˆ ∈ C(X) for every u ∈ U . The weak topology of U is normally defined in terms of all functionals u 7→ f (u), for f ∈ U ∗ ; but as every member of U ∗ is a scalar multiple of some x ∈ X, we can equally regard the weak topology of U as defined just by the functionals u 7→ x(u) = u ˆ(x), for x ∈ X. But b this means that the map u 7→ u ˆ is a homeomorphism between U , with its weak topology, and its image U in C(X), with the topology of pointwise convergence. b (462Ca), and U is angelic in its weak topology. Since C(X) is Tp -angelic (462B), so is U 462E Theorem Let X be a locally compact Hausdorff space, and C0 (X) the Banach lattice of continuous real-valued functions on X which vanish at infinity (436I). Write Tp for the topology of pointwise convergence on C0 (X). (i) A sequence hun in∈N in C0 (X) is weakly convergent to u ∈ C0 (X) iff it is Tp -convergent and normbounded. (ii) A subset K of C0 (X) is weakly compact iff it is norm-bounded and Tp -compact.

462G


535

proof (a) Since all the maps f → 7 f (x), where x ∈ X, are bounded linear functionals on C0 (X), the Tp is coarser than the weak topology Ts ; so a Ts -convergent sequence is Tp -convergent, and a Ts -compact set is Tp -compact. (b) If K ⊆ C0 (X) is Ts -compact, then f [K] ⊆ R must be compact, therefore bounded, for every f ∈ C0 (X)∗ ; by the Uniform Boundedness Theorem (3A5Hb), K is norm-bounded. Applying this to {u} ∪ {un : n ∈ N}, we see that any Ts -convergent sequence hun in∈N with limit u is norm-bounded. (c) Suppose that hun iRn∈N is norm-bounded and Tp -convergent to u. By Lebesgue’s Dominated ConverR gence Theorem, limn→∞ un dµ = u dµ for every totally finite Radon measure µ on X. But by the Riesz Representation Theorem (in the form 436K), this says just that limn→∞ f (un ) = f (u) for every positive linear functional f on C0 (X). Since every member of C0 (X)∗ is expressible as the difference of two positive linear functionals (356Dc), limn→∞ f (un ) = f (u) for every f ∈ U ∗ , that is, hun in∈N is Ts -convergent to u. Putting this together with (a) and (b), we see that (i) is true. (d) Now suppose that K ⊆ C0 (X) is norm-bounded and Tp -compact. Then it is Tp -closed, therefore Ts closed. Because Tp is angelic (462B), any sequence hun in∈N in K has a Tp -convergent subsequence (462Cc), which is also Ts -convergent, by (c). This means that K is (relatively) countably compact in C0 (X) for the topology Ts . Since Ts is angelic (462D), the Ts -closure of K is Ts -compact; but this is just K, so K is Ts -compact, as claimed. 462F Proposition Let X be a countably compact topological space. Then a subset of Cb (X) is weakly compact iff it is norm-bounded and compact for the topology Tp of pointwise convergence. proof I should perhaps begin by remarking that, because X is countably compact, every continuous realvalued function on X is bounded (4A2G(f-v)), so that we have a natural norm k k∞ on C(X) = Cb (X) (354Hb). A weakly compact subset of C(X) is norm-bounded and Tp -compact by the same arguments as in (a)-(b) of the proof of 462E. Suppose that K ⊆ Cb (X) is norm-bounded and Tp -compact. We have a natural map x 7→ x ˆ : X → RK defined by writing x ˆ(u) = u(x) for every u ∈ K, x ∈ X. By the definition of Tp , x ˆ ∈ C(K) for every x ∈ X, if we interpret C(K) as the space of Tp -continuous real-valued functions on K; and x 7→ x ˆ : X → C(K) is continuous for the given topology on X and the topology of pointwise convergence on C(K) because K ⊆ C(X). It follows that Z = {ˆ x : x ∈ X} is countably compact for the topology of pointwise convergence on C(K) (4A2G(f-vi)). But now Z must be actually compact for the topology of pointwise convergence on C(K), by 462B. Next, consider the natural map u 7→ u ˆ : K → RZ defined by setting u ˆ(f ) = f (u) for f ∈ Z, u ∈ K. Just as in the last paragraph, this is a continuous function from K to C(Z), if we give K, Z and C(Z) their topologies of pointwise convergence. So L = {ˆ u : u ∈ K} is compact for the topology of pointwise convergence on C(Z). Moreover, it is norm-bounded, because supφ∈L kφk∞ = supu∈K,f ∈Z |ˆ u(f )| = supu∈K,x∈X |u(x)| = supu∈K kuk∞ is finite. So 462E tells us that L is weakly compact in C(Z). (Note that C(Z) = C0 (Z) because Z is compact.) Now we have an operator T : C(Z) → RX defined by setting T (φ)(x) = φ(ˆ x) for φ ∈ C(Z), x ∈ X. Because x 7→ x ˆ : X → Z is continuous, T φ ∈ C(X) for every φ ∈ C(Z), and of course T is norm-preserving. So T [L] is weakly compact in C(X) (2A5If). But if u ∈ K and x ∈ X, (T u ˆ)(x) = u ˆ(ˆ x) = x ˆ(u) = u(x), so T u ˆ = u. Accordingly K = T [L], and K is weakly compact, as claimed. 462G Theorem Let X be a countably compact topological space. Then the totally finite Radon measures on C(X) are the same for the topology of pointwise convergence and the norm topology. proof (a) As in 462F, C(X) = Cb (X) is a Banach space under its norm k k∞ . Write Tp for the topology of pointwise convergence on C(X) and T∞ for the norm topology. Because Tp ⊆ T∞ , every totally finite T∞ -Radon measure is Tp -Radon (418I).

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(b) For the rest of the proof, therefore, I take a totally finite Tp -Radon measure µ on C(X) and seek to prove that it is T∞ -Radon. By 418L, applied to the identity map from (C(X), T∞ ) to (C(X), Tp ), it will be enough to show that whenever E ∈ dom µ and µE > 0 there is some T∞ -compact set K such that µ(E ∩ K) > 0. (c) I show first that if E ∈ dom µ and µE > 0, and ² > 0, there is some g ∈ C(X) such that µ(E ∩ B(g, ²)) > 0, where B(g, ²) = {f : kf − gk∞ ≤ ²}. P P?? Suppose, if possible, otherwise. Let K ⊆ E be a non-empty Tp -compact set such that µK > 0. Since all the balls B(0, k) are Tp -closed, we may suppose that K is k k∞ -bounded, therefore weakly compact (462F). Choose hKn in∈N and hfn in∈N as follows. K0 = K. Given that Kn ⊆ E is non-negligible and Tp -compact, and that hfi ii ² for every g ∈ Dn } is a Tp -open set and E \ Hn is negligible, so µ(Kn ∩ Hn ) > 0 and we can find a non-negligible Tp -closed set Kn+1 ⊆ Kn ∩ Hn ; choose fn ∈ Kn+1 . Continue. At theSend of the induction, let f ∗ be any cluster point of hfn in∈N for the weak topology on C(X). The set Γ = n∈N Γn is convex, so its norm-closure Γ is also convex (2A5Eb), therefore closed for the weak topology (3A5Ee), and contains f ∗ . So there is a g ∈ Γ such that kf ∗ − gk∞ ≤ 21 ². Now there is some n such that g ∈ Γn . Let g 0 ∈ Dn be such that kg − g 0 k∞ ≤ 12 ², so that kf ∗ − g 0 k ≤ ². Since fi ∈ Kn+1 for every i ≥ n, and Kn+1 is Tp -closed, therefore closed for the weak topology, f ∗ ∈ Kn+1 ⊆ Hn , and kf ∗ − g 0 k > ², which is impossible. X XQ Q (d) What this means is that if we take Kn to be the family of subsets of C(X) which can be covered −n by finitely many balls of radius T at most 2 , then µ is inner regular with respect to Kn (see 412Aa), and therefore with respect to K = n∈N Kn (412Ac). But K is just the set of subsets of C(X) which are totally bounded for the norm-metric ρ on C(X). By 412Ac again, µ must be inner regular with respect to the family of Tp -closed ρ-totally bounded sets, which are surely T∞ -closed therefore ρ-complete and T∞ -compact. By 418L, µ is T∞ -Radon. 462H Corollary Let X be a countably compact Hausdorff space, and give C(X) its topology of pointwise convergence. If µ is any Radon measure on C(X), it is inner regular with respect to the family of compact metrizable subsets of C(X). proof In the language of 462G, µ is inner regular with respect to the family of T∞ -compact sets; but as Tp ⊆ T∞ , the two topologies agree on all such sets, and they are compact and metrizable for Tp . 462I Proposition Let X be a topological space, Y a Hausdorff space, f : X × Y R→ R a bounded separately continuous function, and ν a totally finite Radon measure on Y . Set φ(x) = f (x, y)ν(dy) for every x ∈ X. Then φ¹C is continuous for any relatively countably compact set C ⊆ X. proof (a) To begin with, suppose that X itself is countably compact. For y ∈ Y , set uy (x) = f (x, y) for every x ∈ X. Then every uy is continuous, because f is continuous in the first variable, and y 7→ uy : Y → C(X) is continuous, if we give C(X) the topology Tp of pointwise convergence, because f is continuous in the second variable. We therefore have a Tp -Radon image measure µ on C(X), by 418I. By 462G, µ is a Radon measure for the norm topology of C(X). Now recall that f is bounded. If |f (x, y)| ≤ M for all x ∈ X, y ∈ Y , then kuy k∞ ≤ M for every y ∈ Y , and the ball B(0, M ) in C(X) is µ-conegligible. By 461E, applied to the subspace measure on B(0, M ), µ has a barycenter h in C(X). Now we can compute h by the formulae Z h(x) = (because g 7→ g(x) belongs to C(X)∗ )

Z =

(by 235I)

g(x)µ(dg)

uy (x)ν(dy)

462Y


537

Z =

f (x, y)ν(dy) = φ(x),

for every x ∈ X. So φ = h is continuous. (b) For the general result, we have only to apply (a) to f ¹C for each countably compact C ⊆ X. 462J Corollary Let X be a topological space such that whenever h ∈ RX is such that h¹C is continuous for every relatively countably compact C ⊆ X, then h is continuous. Write Tp for the topology of pointwise convergence on C(X). Let K ⊆ C(X) be a Tp -compact set such that suph∈K,x∈C |h(x)| is finite for any countably compact set C ⊆ X. Then the Tp -closed convex hull of K, taken in C(X), is Tp -compact. proof The closed convex hull Γ(K) of K, taken in RX , is closed and included in a product of closed bounded intervals, therefore compact. If h ∈ Γ(K), then there is a Radon probability measure µ on K such that h is R the barycenter of µ (461H), so that h(x) = f (x)µ(df ) for every x ∈ X. If C ⊆ X is countably compact, the map (x, f ) 7→ f (x) : C × K → R is separately continuous, because K ⊆ C(X) is being R given the topology of pointwise convergence, and is bounded by hypothesis. So 462I tells us that x 7→ f (x)µ(df ) : C → R is continuous, that is, that h¹C is continuous. By the hypothesis on X here, h is continuous. Thus the Tp -compact set Γ(K) is included in C(X), and must be the closed convex hull of K in C(X). Remark The hypothesis whenever h ∈ RX is such that h¹C is continuous for every relatively countably compact C ⊆ X, then h is continuous is clumsy, but seems the best way to cover the large number of potential applications of the ideas here. Besides the obvious case of countably compact spaces X, we have all first-countable spaces (for which, of course, the other hypotheses can be relaxed, as in 462Xc), and all k-spaces. (A k-space is a topological space X such that a set G ⊆ X is open iff G ∩ K is relatively open in K for every compact set K ⊆ X; see Engelking 89, 3.3.18 et seq. In particular, all locally compact spaces are k-spaces.) Perhaps the most important case of all is when X is the the dual U ∗ of a complete locally convex Hausdorff linear topological space U , endowed with the topology Ts (U ∗ , U ), so that U , with its weak topology Ts (U, U ∗ ), can be regarded as a linear subspace of C(X) with its topology of pointwise convergence (see 4A4Ch). 462X Basic exercises (a) (i) Show that R, with the right-facing Sorgenfrey topology, is angelic. (ii) Show that any metrizable space is angelic. (iii) Show that the one-point compactification of an angelic locally compact Hausdorff space is angelic. (iv) Find a first-countable regular Hausdorff space which is not angelic. >(b) Let X be any countably compact topological space. Show that a norm-bounded sequence in Cb (X) which is pointwise convergent is weakly convergent. (c) Let X be a first-countable topological space, (Y, T, ν) a totally finite measure space, and f : X×Y → R a bounded function such that y 7→ f (x, R y) is measurable for every x ∈ X, and x 7→ f (x, y) is continuous for almost every y ∈ Y . Show that x 7→ f (x, y)ν(dy) is continuous. (d) Give an example of a Tp -compact subset K of C([0, 1]) such that the convex hull of K is not relatively compact in C([0, 1]). 462Y Further exercises (a) Let X be a topological space such that there is a sequence hXn in∈N of countably compact subsets of X, covering X, with the property that a function f : X → R is continuous whenever f ¹Xn is continuous for every n ∈ N. Let Tp be the topology of pointwise convergence on C(X). Show that, for a set K ⊆ C(X), the following are equiveridical: (i) φ[K] is bounded for every Tp -continuous function φ : C(X) → R; (ii) whenever hfn in∈N is a sequence in K and A ⊆ X is countable, there is a cluster point of hfn ¹Ain∈N in C(A) for the topology of pointwise convergence on C(A); (iii) K is relatively compact in C(X) for Tp . (See Asanov & Velichko 81.)

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462Yb

(b) Let U be a metrizable locally convex linear topological space. Show that it is angelic in its weak topology. (Hint: start with the case in which U is complete, using Grothendieck’s theorem and the full strength of 462B, with X = U ∗ .) (c) In 462I, show that the conclusion remains valid for any totally finite τ -additive topological measure ν on Y which is inner regular with respect to the relatively countably compact subsets of Y . (d) Show that if X is any compact topological space (more generally, any topological space such that X n is Lindelöf for every n ∈ N), then C(X), with its topology of pointwise convergence, is countably tight. 462Z Problem Let K be a compact Hausdorff space. Is C(K), with the topology of pointwise convergence, necessarily a pre-Radon space? 462 Notes and comments The theory of pointwise convergence in spaces of continuous functions is intimately connected with the theory of separately continuous functions of two variables. For if X and Y are topological spaces, and f : X × Y → R is any separately continuous function, then we have natural maps x 7→ fx : X → C(Y ) and y 7→ f y : Y → C(X), writing fx (y) = f y (x) = f (x, y), which are continuous if C(X) and C(Y ) are given their topologies of pointwise convergence; and if X is a topological space and Y is any subset of C(X) with its topology of pointwise convergence, the map (x, y) 7→ y(x) : X × Y → R is separately continuous. I include a back-and-forth shuffle between C(X) and separately continuous functions in 462G-462I-462J largely as a demonstration of the principle that all the theorems here can be expressed in both languages. ˇ 462Yb is a compendium of Smulian’s theorem with part of Eberlein’s theorem; 462E and 462J can be thought of as the centre of Krein’s theorem. There are many alternative routes to these results, which may ¨ the 69 or Grothendieck 92. In particular, 462E can be proved without using measure be found in Ko theory; see, for instance, Fremlin 74, A2F. Topological spaces homeomorphic to compact subsets of C(X), where X is some compact space and C(X) is given its topology of pointwise convergence, are called Eberlein compacta; see 467O-467P. A positive answer to A.Bellow’s problem (463Za below) would imply a positive answer to 462Z; so if the continuum hypothesis, for instance, is true, then C(K) is pre-Radon in its topology of pointwise convergence for any compact space K.

463 Tp and Tm We are now ready to start on the central ideas of this chapter with an investigation of sets of measurable functions which are compact for the topology of pointwise convergence. Because ‘measurability’ is, from the point of view of this topology on RX , a rather arbitrary condition, we are looking at compact subsets of a topologically irregular subspace of RX ; there are consequently relatively few of them, and (under a variety of special circumstances, to be examined later in the chapter and also in Volume 5) they have some striking special properties. The presentation here is focused on the relationship between the two natural topologies on any space of measurable functions, the ‘pointwise’ topology Tp and the topology Tm of convergence in measure (463A). In this section I begin with results which apply to any σ-finite measure space (463B-463G) before turning to some which apply to perfect measure spaces (463H-463K) – in particular, to Lebesgue measure. These lead to some interesting properties of separately continuous functions (463L-463M). 463A Notation Let (X, Σ, µ) be a measure space, and L0 = L0 (Σ) the space of all Σ-measurable functions from X to R, so that L0 is a linear subspace of RX . On L0 we shall be concerned with two very different topologies. The first is the topology Tp of pointwise convergence (462Ab); the second is the topology Tm of (local) convergence in measure (245Ab). Both are linear space topologies. P P For Tp I have already R noted this in 462Ab. For Tm , repeat the argument of 245D; Tm is defined by the functionals f 7→ F min(1, |f |)dµ, where µF < ∞, and these satisfy the criterion of 2A5B. Q Q Tp is Hausdorff (3A3Id) and locally convex (4A4Ce); only in exceptional circumstances is either true of Tm . However, Tm can easily

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be pseudometrizable (if, for instance, µ is σ-finite, as in 245Eb), while Tp is not, except in nearly trivial cases. Associated with the topology of pointwise convergence on RX is the usual topology of PX (definition: 4A2A); the map χ : PX → RX is a homeomorphism between PX and its image {0, 1}X ⊆ RX . Tm is intimately associated with the topology of convergence in measure on L0 = L0 (µ) (§245). A subset of L0 is open for Tm iff it is of the form {f : f • ∈ G} for some open set G ⊆ L0 ; consequently, a subset K of L0 is compact, or separable, for Tm iff {f • : f ∈ K} is compact or separable for the topology of convergence in measure on L0 . It turns out that the identity map from (L0 , Tp ) to (L0 , Tm ) is sequentially continuous (463B). Only in nearly trivial cases is it actually continuous (463Xa(i)), and it is similarly rare for the reverse map from (L0 , Tm ) to (L0 , Tp ) to be continuous (463Xa(ii)). If, however, we relativise both topologies to a Tp compact subset of L0 , the situation becomes very different, and there are many important cases in which the topologies are comparable. 463B Lemma Let (X, Σ, µ) be a measure space, and L0 the space of Σ-measurable real-valued functions on X. Then every pointwise convergent sequence in L0 is convergent in measure to the same limit. proof 245Ca. 463C Proposition (Ionescu Tulcea 73) Let (X, Σ, µ) be a measure space, and L0 the space of Σmeasurable real-valued functions on X. Write Tp , Tm for the topologies of pointwise convergence and convergence in measure on L0 ; for A ⊆ L0 , write TpA , TmA for the corresponding subspace topologies. (a) If A ⊆ L0 and TpA is metrizable, then the identity map from A to itself is (TpA , TmA )-continuous. (b) Suppose that µ is semi-finite. Then, for any A ⊆ L0 , TmA is Hausdorff iff whenever f , g are distinct members of A the set {x : f (x) 6= g(x)} is non-negligible. (c) Suppose that K ⊆ L0 is such that TpK is compact and metrizable. Then TpK = TmK iff TmK is Hausdorff. (d) Suppose that µ is σ-finite, and that K ⊆ L0 is Tp -sequentially compact. Then TpK = TmK iff TmK is Hausdorff, and in this case TpK is compact and metrizable. (e) Suppose that K ⊆ L0 is such that TpK is compact and metrizable. Then whenever ² > 0 and E ∈ Σ is a non-negligible measurable set, there is a non-negligible measurable set F ⊆ E such that |f (x) − f (y)| ≤ ² whenever f ∈ K and x, y ∈ F . proof (a) All we need is to remember that sequentially continuous functions from metrizable spaces are continuous (4A2Ld), and apply 463B. R (b) TmA is Hausdorff iff for any distinct f , g ∈ A there is a measurable set F of finite measure such that min(1, |f − g|)dµ > 0, that is, µ{x : x ∈ F, f (x) 6= g(x)} > 0; because µ is semi-finite, this happens iff F µ{x : f (x) 6= g(x)} > 0. (c) If TpK = TmK then of course TmK is Hausdorff, because TpK is. If TmK is Hausdorff then the identity map (K, TpK ) → (K, TmK ) is an injective function from a compact space to a Hausdorff space and (by (a)) is continuous, therefore a homeomorphism, so the two topologies are equal. (d) If TpK = TmK then TmK must be Hausdorff, just as in (c). So let us suppose that TmK is Hausdorff. Note that, by 245Eb, the topology of convergence in measure on L0 is metrizable; in terms of L0 , this says just that the topology of convergence in measure on L0 is pseudometrizable. So TmK is Hausdorff and pseudometrizable, therefore metrizable (4A2La). We are told that any sequence in K has a TpK -convergent subsequence. But this subsequence is now TmK convergent (463B), so TmK is sequentially compact; being metrizable, it is compact (4A2Lf). Moreover, the same is true of any TpK -closed subset of K, so every TpK -closed set is TmK -compact, therefore TmK closed. Thus the identity map from (K, TmK ) to (K, TpK ) is continuous. Since TmK is compact and TpK is Hausdorff, the two topologies are equal; and, in particular, TpK is compact and metrizable. (e) Let ρ be a metric on K inducing the topology TpK . Let D ⊆ K be a countable dense set. For each n ∈ N, set

540


463C

1

Gn = {x : |f (x) − g(x)| ≤ ² whenever f , g ∈ D and ρ(f, g) ≤ 2−n }. 3 S S Because D is countable, Gn is measurable. Now n∈N Gn = X. P P?? If x ∈ X \ n∈N Gn , then for each n ∈ N we can find fn , gn ∈ D such that ρ(fn , gn ) ≤ 2−n and |fn (x) − gn (x)| ≥ 31 ². Because K is compact, there is a strictly increasing sequence hnk ik∈N such that hfnk ik∈N and hgnk ik∈N are both convergent to f , g say. Now ρ(f, g) = limk→∞ ρ(fnk , gnk ) = 0,

1 3

|f (x) − g(x)| = limk→∞ |fnk (x) − gnk (x)| ≥ ²,

so f = g while f (x) 6= g(x), which is impossible. X XQ Q There is therefore some n ∈ N such that µ(E ∩ Gn ) > 0. Since K, being compact, is totally bounded for ρ, there is a finite set D0 ⊆ D such that every member of D is within a distance of 2−n of some member of D0 . Now there is a measurable set F ⊆ E ∩ Gn such that µF > 0 and |g(x) − g(y)| ≤ 31 ² whenever g ∈ D0 and x, y ∈ F . So |f (x) − f (y)| ≤ ² whenever f ∈ D and x, y ∈ F . But as D is dense in K, |f (x) − f (y)| ≤ ² whenever f ∈ K and x, y ∈ F , as required. 463D Proposition Let (X, Σ, µ) be a measure space, and L0 the space of Σ-measurable real-valued functions on X. Write Tp , Tm for the topologies of pointwise convergence and convergence in measure on L0 . Suppose that K ⊆ L0 is Tp -compact and that there is no Tp -continuous surjection from any closed subset of K onto ω1 +1 with its order topology. Then the identity map from (K, Tp ) to (K, Tm ) is continuous. proof (a) It is worth noting straight away that ξ 7→ χξ : ω1 + 1 → {0, 1}ω1 is a homeomorphism between ω1 + 1 and a subspace of {0, 1}ω1 . So our hypothesis tells us that there is no continuous surjection from any closed subset of K onto {0, 1}ω1 , and therefore none onto {0, 1}A for any uncountable A. (b) For the moment, fix on a sequence hfn in∈N in K and a set E ∈ Σ of finite measure. Set q(x) = max(1, 2 supn∈N |fn (x)|) for each x ∈ X. For any infinite I ⊆ N, set gI = supn∈N inf i∈I,i≥n fi ,

hI = inf n∈N supi∈I,i≥n fi ;

because supfR∈K |f (x)| is surely finite for each x ∈ X, gI and hI are defined in L0 , and gI ≤ hI . For f ∈ L0 set τ 0 (f ) = E min(1, |f |/q),R and for I ∈ [N]ω (the set of infinite subsets of N) set ∆(I) = τ 0 (hI − gI ). Since hI − gI ≤ q, ∆(I) = E (hI − gI )/q. If I, J ∈ [N]ω and J \ I is finite, then gI ≤ gJ ≤ hJ ≤ hI , so ∆(J) ≤ ∆(I), with equality only when gI = gJ a.e. on E and hI = hJ a.e. on E. (i) There is a J ∈ [N]ω such that ∆(I) = ∆(J) for every I ∈ [J]ω . P P For J ∈ [N]ω , set ∆(J) = ω ω inf{∆(I) : I ∈ [J] }. Choose hIn in∈N in [N] inductively in such a way that In+1 ⊆ In and ∆(In+1 ) ≤ ∆(In ) + 2−n for every n. If we now set J = {min{i : i ∈ In , i ≥ n} : n ∈ N}, J ⊆ N will be an infinite set and J \ In is finite for every n. If I ∈ [J]ω then, for every n, ∆(J) ≤ ∆(In+1 ) ≤ ∆(In ) + 2−n ≤ ∆(In ∩ I) + 2−n = ∆(I) + 2−n ; as n is arbitrary, ∆(J) = ∆(J), as required. Q Q Now for any I ∈ [J]ω we have ∆(I) = ∆(J), so that gI = gJ a.e. on E and hI = hJ a.e. on E. (ii) Now ∆(J) = 0. P P?? Otherwise, F = {x : x ∈ E, gJ (x) < hJ (x)} has positive measure. Set T K0 = n∈N {fi : i ∈ J, i ≥ n}, the closure being taken for Tp , so that K0 is Tp -compact. Let A be the family of sets A ⊆ F such that whenever L, M ⊆ A are finite and disjoint there is an f ∈ K0 such that f (x) = gJ (x) for x ∈ L and f (x) = hJ (x) for x ∈ M . Then A has a maximal member A0 say. If we define φ : L0 → [0, 1]A0 by setting φ(f )(x) = min(hJ (x), max(f (x), gJ (x)))/(hJ (x) − gJ (x)) for x ∈ A0 and f ∈ L0 , φ[K0 ] is a compact subset of [0, 1]A0 , and whenever L, M ⊆ A0 are finite there is a g ∈ φ[K0 ] such that g(x) = 0 for x ∈ L and g(x) = 1 for x ∈ M . This means that φ[K0 ] ∩ {0, 1}A0 is dense in {0, 1}A0 and must therefore be the whole of {0, 1}A0 . So {0, 1}A0 is a continuous image of a closed subset of K. By (a), A0 is countable. For each pair L, M of disjoint finite subsets of A0 , we have a cluster point fLM of hfj ij∈J such that fLM (x) = gJ (x) for x ∈ L and fLM (x) = hJ (x) for x ∈ M . Let I(L, M ) be an infinite subset of J such that limi→∞,i∈I(L,M ) fi (x) = fLM (x) for every x ∈ A0 . Then gI(L,M ) = gJ and hI(L,M ) = hJ almost everywhere on E. Because µF > 0 and A0 has only countably many finite subsets,

463E

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541

there is a y ∈ F such that gI(L,M ) (y) = gJ (y) and hI(L,M ) (y) = hJ (y) whenever L and M are disjoint finite subsets of A0 . What this means is that if L and M are disjoint finite subsets of A0 , then there are infinite sets I 0 , 00 I ⊆ I(L, M ) such that limi→∞,i∈I 0 fi (y) = gJ (y) and limi→∞,i∈I 00 fi (y) = hJ (y); so that there are f 0 , f 00 ∈ K0 such that f 0 (x) = gJ (x) for x ∈ L ∪ {y}, f 00 (x) = gJ (x) for x ∈ L,

f 0 (x) = hJ (x) for x ∈ M ,

f 00 (x) = hJ (x) for x ∈ M ∪ {y}.

But this means that A0 ∪ {y} ∈ A; and A0 was chosen to be maximal. X XQ Q (iii) So ∆(J) = 0 and gJ = hJ almost everywhere on E. But if we enumerate J in ascending order as hni ii∈N , gJ = lim inf i→∞ fni and hJ = lim supi→∞ fni , so hfni ii∈N converges almost everywhere on E. Thus we see that for every sequence in K, and every set of finite measure, there is a subsequence converging almost everywhere on that set. (c) ?? Suppose, if possible, that the identity map from (K, Tp ) to (K, Tm ) is not continuous at f0 ∈ K. Then there are an E ∈ Σ, of finite measure, and R an ² > 0 such that C = {f : f ∈ K, τE (f − f0 ) ≥ ²} meets every Tp -neighbourhood of f , where τE (f ) = E min(1, |f |) for every f ∈ L0 , and there is an ultrafilter F on L0 which contains C and converges to f0 for Tp . Consider the map ψ : L0 → L0 (µE ), where µE is the subspace measure on E, defined by setting ψ(f ) = (f ¹E)• for f ∈ L0 . We know from (b) that every sequence in K has a subsequence convergent almost everywhere on E, so every sequence in ψ[K] has a subsequence which is convergent for the topology of convergence in measure on L0 (µE ). Since this is metrizable, ψ[K] is relatively compact in L0 (µE ) (4A2Le), and the image filter ψ[[F]] has a limit v ∈ L0 (µE ). Let f1 ∈ L0 be such that ψ(f1 ) = v. For any countable set A ⊆ X there is a g ∈ C such that g¹A = f0 ¹A and g = f1 almost everywhere on E. P P If X = ∅ this is trivial, so we may, if necessary, enlarge A by one point so that it is not empty. Let hxn in∈N be a sequence running over A. Then for each n ∈ N the set {g : g ∈ C, |g(xi ) − f0 (xi )| ≤ 2−n for every i ≤ n, τE (g, f1 ) ≤ 2−n } belongs to F, so is not empty; take gn in this set. Let g ∈ K be any cluster point of hgn in∈N . Since hgn in∈N converges to f1 almost everywhere on E, g = f1 a.e. on E. Since τE (gn − f0 ) ≥ ² for every n, τE (g − f0 ) ≥ ² and g ∈ C. Since hgn (xi )in∈N converges to f0 (xi ) for every i, g¹A = f0 ¹A. So we have the result. Q Q In particular, there is a g ∈ C such that g =a.e. f1 , so F = {x : x ∈ E, f0 (x) 6= f1 (x)} has non-zero measure. Now choose hgξ iξ<ω1 in K and hxξ iξ<ω1 in F inductively so that gξ ∈ C,

gξ = f1 almost everywhere on E, gξ (xη ) = f0 (xη ) for η < ξ

(choosing gξ ), xξ ∈ F ,

gη (xξ ) = f1 (xξ ) for η < ξ

(choosing xξ ). If we now set A = {xξ : ξ < ω1 }, T K1 = ξ<η<ω1 {f : f ∈ K, either f (xξ ) = f0 (xξ ) or f (xξ ) = f (xη ) = f1 (xη )}, then K1 is a closed subset of K containing every gξ and also f0 . But if we look at {f ¹A : f ∈ K1 }, this is homeomorphic to ω1 + 1; which is supposed to be impossible. X X So we conclude that the identity map from (K, Tp ) to (K, Tm ) is continuous. 463E Corollary Let (X, Σ, µ) be a measure space, and L0 the space of Σ-measurable real-valued functions on X. Write Tp , Tm for the topologies of pointwise convergence and convergence in measure on L0 . Suppose that K ⊆ L0 is compact and countably tight for Tp . Then the identity map from (K, Tp ) to (K, Tm ) is continuous. If Tm is Hausdorff on K, the two topologies coincide on K. proof Since ω1 + 1 is not countably tight (the top point ω1 is not in the closure of any countable subset of ω1 ), ω1 + 1 is not a continuous image of any closed subset of K (4A2Kb), and we can apply 463D to see that the identity map is continuous. It follows at once that if Tm is Hausdorff on K, then the topologies coincide.

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463F Theorem (Ionescu Tulcea 74) Let (X, Σ, µ) be a σ-finite measure space, and K a convex set of measurable functions from X to R such that (i) K is compact for the topology Tp of pointwise convergence (ii) {x : f (x) 6= g(x)} is not negligible for any distinct f , g ∈ K. Then K is metrizable for Tp , which agrees with the topology of convergence in measure on K. proof (a) Let hfn in∈N be a sequence in K. Then it has a pointwise convergent subsequence. P P Because K is compact, we surely have supf ∈K |f (x)| < ∞ for every x ∈ X, so that q0 = χX + supn∈N |fn | is defined in RX , and is measurable. Let hXk ik∈N be a sequence of measurable sets of finite measure covering X, and set P∞ 1 χXk , q1 = k=0 k 2 (1+µXk )

so that q1 is a strictly positive measurable function and is square-integrable. Set K 0 = {f × q1 /q0 : f ∈ K}, so that K 0 is another convex pointwise compact set of measurable functions, this time all dominated by q1 , so that K 0 ⊆ L2 (µ). Setting gn = fn × q1 /q0 , the sequence hgn• in∈N of equivalence classes is a norm-bounded sequence in the Hilbert space L2 = L2 (µ). It therefore has a weakly • convergent subsequence hgn(i) ii∈N say (4A4Kb), with limit v. ?? Suppose, if possible, that hfn(i) ii∈N is not pointwise convergent. Then there must be some x0 ∈ X such that lim inf i→∞ fn(i) (x0 ) < lim supi→∞ fn(i) (x0 ); let α < β in R be such that I = {i : fn(i) (x0 ) ≤ α}, • :i∈ I 0 = {i : fn(i) (x0 ) ≥ β} are both infinite. In this case v belongs to the weak closures of both D = {gn(i) 0 0 • I} and D = {gn(i) : i ∈ I }. It must therefore belong to the norm closures of their convex hulls Γ(D), Γ(D0 ) (4A4Ed). Accordingly we can find vn ∈ Γ(D), vn0 ∈ Γ(D0 ) such that kv − vn k2 ≤ 3−n , kv − vn0 k2 ≤ 3−n for every n ∈ N. Setting A = {fn(i) : i ∈ I}, A0 = {fn(i) : i ∈ I 0 }, we see that there must be hn ∈ Γ(A), h0n ∈ Γ(A0 ) such that vn = (hn × q1 /q0 )• , vn0 = (h0n × q1 /q0 )• for every n ∈ N. Now if g : X → R is a measurable function ˜ = g × q0 /q1 , we have such that g • = v, and h

R

|g − (hn × q1 /q0 )|2 ≤ 9−n ,

˜ µ{x : |h(x) − hn (x)| ≥ 2−n q0 (x)/q1 (x)} = µ{x : |g(x) − (hn × q1 /q0 )(x)| ≥ 2−n } ≤ 4n kv − vn k22 ≤ 2−n ˜ a.e. Similarly, h0 → h ˜ a.e. for every n ∈ N, and hn → h n At this point, recall that K is supposed to be convex, so all the hn , h0n belong to K. Let F be any non-principal ultrafilter on N. Because K is pointwise compact, h = limn→F hn and h0 = limn→F h0n are ˜ both defined in K for the topology of pointwise convergence. For any x such that limn→∞ hn (x) = h(x), we 0 0 ˜ ˜ Similarly, h =a.e. h, ˜ and h =a.e. h . surely have h(x) = h(x); so h =a.e. h. Now at last we apply the hypothesis that distinct members of K are not equal almost everywhere, to see that h = h0 . But if we look at what happens at the distinguished point x0 above, we see that f (x0 ) ≤ α for every f ∈ A, so that f (x0 ) ≤ α for every f ∈ Γ(A), hn (x0 ) ≤ α for every n ∈ N, and h(x0 ) ≤ α; and similarly h0 (x0 ) ≥ β. So h 6= h0 , which is absurd. X X Q This contradiction shows that hfn(i) ii∈N is pointwise convergent, and is an appropriate subsequence. Q (b) Now 463Cd tells us that K is metrizable for Tp , and that Tp agrees on K with the topology of convergence in measure. 463G Corollary Let (X, T, Σ, µ) be a σ-finite topological measure space in which µ is strictly positive. Suppose that whenever h ∈ RX is such that h¹C is continuous for every relatively countably compact C ⊆ X, then h is continuous. If K ⊆ Cb (X) is a norm-bounded Tp -compact set, then it is Tp -metrizable. proof By 462J, the Tp -closed convex hull Γ(K) of K in Cb (X) is Tp -compact. Because µ is strictly positive, µ{x : f (x) 6= g(x)} > 0 whenever f and g are distinct continuous real-valued functions on X. So the result is immediate from 463F.

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463H Lemma Let (X, Σ, µ) be a perfect probability space, and hEn in∈N a sequence in Σ. Suppose that there is an ² > 0 such that ²µF ≤ lim inf n→∞ µ(F ∩ En ) ≤ lim supn→∞ µ(F ∩ En ) ≤ (1 − ²)µF for every F ∈ Σ. Then hEn in∈N has a subsequence hEnk ik∈N such that µ∗ A = 0 and µ∗ A = 1 for any cluster point A of hEnk ik∈N in PX; in particular, hEnk ik∈N has no measurable cluster point. proof (a) If we replace µ by its completion, we do not change µ∗ and µ∗ (212Ea, 413Eg), so we may suppose that µ is already complete. (b) Suppose that hEn in∈N is actually stochastically independent, with µEn = is an integer. In this case µ∗ A = 1 for any cluster point A of hEn in∈N .

1 k

for every n, where k ≥ 2

P P (i) There is a non-principal ultrafilter F on N such that A = limn→F En in PX (4A2Bc); that is, χA(x) = limn→F χEn (x) for every x ∈ X; that is, A = {x : x ∈ X, {n : x ∈ En } ∈ F}. (ii) We have a measurable function φ : X → Y = {0, 1}N defined by setting φ(x)(n) = (χEn )(x) for every x ∈ X, n ∈ N. Because µ is complete and perfect and totally finite, and {0, 1}N is compact and metrizable, the image measure ν = µφ−1 is a Radon measure (451N). By 451Oc, µ∗ φ−1 [B] = ν ∗ B for every B ⊆ Y . Moreover, for any basic open set of the form H = {y : y(i) = ²i for every i ≤ n}, µφ−1 [H] = ν˜H, where ν˜ is the product measure corresponding to the measure ν0 on {0, 1} for which ν0 {1} = k1 , ν0 {0} = k−1 k . Since ν˜ is also a Radon measure (416U), ν˜ = ν (415H(v)). Set B = {y : y ∈ Y , {n : y(n) = 1} ∈ F}, so that φ−1 [B] = A and B is determined by coordinates in {n, n + 1, . . . } for every n ∈ N. By 451Oc, µ∗ A = ν ∗ B; by the zero-one law (254Sa), ν ∗ B must be either 0 or 1. So µ∗ A is either 0 or 1. (iii) To see that µ∗ A cannot be 0, we repeat the arguments of (ii) from the other side, as follows. Let λ0 be the uniform probability measure on {0, 1, . . . , k − 1}, giving measure k1 to each point; let λ be the corresponding product measure on Z = {0, . . . , k − 1}N . Let ψ : Z → Y be defined by setting ψ(z)(n) = 1 if z(n) = 0, 1 otherwise; then ψ is inverse-measure-preserving (254G). Since λ is a Radon measure and ψ is continuous, λψ −1 is a Radon measure on Y and must be equal to ν. Accordingly ν ∗ B = λ∗ ψ −1 [B], by 451Oc yet again or otherwise. (iv) We have a measure space automorphism θ : Z → Z defined by setting θ(z)(n) = z(n) +k 1 for every z ∈ Z, n ∈ N, where +k is addition mod k. So, writing C = ψ −1 [B], λ∗ C = λ∗ θi [C] for every i ∈ N. Now, for z ∈ Z, {n : z(n) = 0} ∈ F ⇐⇒ {n : ψ(z)(n) = 1} ∈ F ⇐⇒ ψ(z) ∈ B ⇐⇒ z ∈ C. But for any z ∈ Z, there is some i < k such that {n : z(n) = i} ∈ F, so that θk−i (z) ∈ C. Thus S Pk−1 ∗ i i ∗ ∗ ∗ ∗ i=0 λ θ [C] ≥ 1 and λ C > 0. But this means that µ A = ν B = λ C is non-zero, i 0 there is some n ∈ N such that 0 < µ(F ∩ En ) < µF . Let k ≥ 2 be such that k1 < ². Then there are a strictly increasing sequence hm(i)ii∈N in N and a stochastically independent sequence hFi ii∈N in Σ such that Fi ⊆ Em(i) and µFi = k1 for every i ∈ N. P P Choose hm(i)ii∈N , Fi inductively, as follows. Let Σi be the (finite) algebra generated by {Fj : j < i}. Choose m(i) such that m(i) > m(j) for any j < i and µ(F ∩ Em(i) ) ≥ k1 µF for every F ∈ Σi . List the atoms of Σi as Gi0 , . . . , Gipi , and choose Fir ⊆ Em(i) ∩ Gir such that µFir = k1 µGir , for each r ≤ pi ; this is possible S by 215D. Set Fi = r≤pi Fir ; then µ(Fi ∩ F ) = k1 µF for every F ∈ Σi . Continue. It is easy to check that µ(Fi1 ∩ . . . ∩ Fir ) = 1/k r whenever i1 < . . . < ir , so that hFi ii∈N is stochastically independent. Q Q If A is a cluster point of hEm(i) ii∈N , then there is a non-principal ultrafilter F on N such that A = limi→F Em(i) in PX. In this case, A ⊇ A0 , where A0 = limi→F Fi . But (b) tells us that µ∗ A0 must be 1, so µ∗ A = 1. (d) Thus we have a subsequence hEm(i) ii∈N of hEn in∈N such that any cluster point of hEm(i) ii∈N has outer measure 1. But the same argument applies to hX \ Em(i) ii∈N to show that there is a strictly increasing sequence hik ik∈N such that every cluster point of hX\Em(ik ) ik∈N has outer measure 1. Since complementation

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is a homeomorphism of PX, µ∗ (X \ A) = 1, that is, µ∗ A = 0, for every cluster point A of hEm(ik ) ik∈N . So if we set n(k) = m(ik ), any cluster point of hEnk ik∈N will have inner measure 0 and outer measure 1, as claimed. 463I Lemma Let (X, Σ, µ) be a perfect probability space, and hEn in∈N a sequence in Σ. Then either hχEn in∈N has a subsequence which is convergent almost everywhere or hEn in∈N has a subsequence with no measurable cluster point in PX. proof Consider the sequence hχEn• in∈N in the Hilbert space L2 = L2 (µ). This is a norm-bounded sequence, so has a weakly convergent subsequence hχEn• i ii∈N with limit v say (4A4Kb). Express v as g • where g : X → R is Σ-measurable. case 1 Suppose that g(x) ∈ {0, 1} for almost every x ∈ X; set F = {x : g(x) = 1}. Then limi→∞

R

F

χEni =

R

F

g = µF ,

limi→∞

R

X\F

χEni =

R

X\F

g = 0.

So, replacing hχEni ii∈N with a sub-subsequence if necessary, we may suppose that |µF −

R

F

χEni | ≤ 2−i ,

|

R

X\F

χEni | ≤ 2−i

for every i. But as 0 ≤ χEni ≤ 1 everywhere, we have

R

|χF − χEni | ≤ 2−i+1

for every i, so that χEni → χF a.e., and we have a subsequence of hχEn in∈N which is convergent almost everywhere. R case 2 Suppose that {x : g(x) ∈ / {0, 1}} has positive measure. Note that because F g = limi→∞ µ(F ∩ Eni ) lies between 0 and µF for every F ∈ Σ, 0 ≤ g ≤ 1 a.e., and µ{x : 0 < g(x) < 1} > 0. There is therefore an ² > 0 such that µG > 0, where G = {x : ² ≤ g(x) ≤ 1 − ²}. Write µG for the subspace measure on G, and ΣG for its domain; set ν = (µG)−1 µG , so that ν is a probability measure. We know that µG is perfect (451Dc), so ν also is (see the definition in 451Ad). Now if F ∈ ΣG , limi→∞ ν(F ∩ Eni ) = (µG)−1

R

F

g

lies between ²µF/µG = ²νF and (1 − ²)νF . By 463H, there is a strictly increasing sequence hi(k)ik∈N such that B ∈ / ΣG whenever B is a cluster point of hG ∩ Eni(k) ik∈N in PG. If A is any cluster point of hEni(k) ik∈N in RX , then A ∩ G is a cluster point of hG ∩ Eni(k) ik∈N in PG, so cannot belong to ΣG . Thus A ∈ / Σ. So in this case we have a subsequence hEni(k) ik∈N of hEn in∈N which has no measurable cluster point. 463J Fremlin’s Alternative (Fremlin 75a) Let (X, Σ, µ) be a perfect σ-finite measure space, and hfn in∈N a sequence of real-valued measurable functions on X. Then either hfn in∈N has a subsequence which is convergent almost everywhere or hfn in∈N has a subsequence with no measurable cluster point in RX . proof (a) If µX = 0 then of course hfn in∈N itself is convergent a.e., so we may suppose that µX > 0. If there is any x ∈ X such that supn∈N |fn (x)| = ∞, then hfn in∈N has a subsequence with no cluster point in RX , measurable or otherwise; so we may suppose that hfn in∈N is bounded at each point of X. (b) Let λ be the c.l.d. product of µ with Lebesgue measure on R, and Λ its domain. Then λ is perfect (451Ic) and also σ-finite (251K). There is therefore a probability measure ν on Y with the same domain and the same negligible sets as λ (215B(vii)), so that ν also is perfect. For any function h ∈ RX , write Ω(h) = {(x, α) : x ∈ X, α ≤ h(x)} ⊆ X × R (compare 252N). (c) By 463I, applied to the measure space (X × R, Λ, ν) and the sequence hχΩ(fn )in∈N , we have a strictly increasing sequence hn(i)ii∈N such that either hχΩ(fn(i) )ii∈N is convergent ν-a.e. or hΩ(fn(i) )ii∈N has no cluster point in Λ.

463K

Tp and Tm

545

case 1 Suppose that hχΩ(fn(i) )ii∈N is convergent ν-a.e. Set W = {(x, α) : limi→∞ χΩ(fn(i) )(x, α) is defined}. Then W is λ-conegligible, so W −1 [{α}] = {x : (x, α) ∈ W } is µ-conegligible for almost every T α (252D). Set D = {α : W −1 [{α}] is µ-conegligible}, and let Q ⊆ D be a countable dense set; then G = α∈Q W −1 [{α}] is µ-conegligible. But if x ∈ G, then for any α ∈ Q the set {i : fn(i) (x) ≥ α} = {i : χΩ(fn(i) )(x, α) = 1} is either finite or has finite complement in N, so hfn(i) (x)ii∈N must be convergent in [−∞, ∞]. Since hfn (x)in∈N is supposed to be bounded, hfn(i) (x)ii∈N is convergent in R. Thus in this case we have an almost-everywhereconvergent subsequence of hfn in∈N . case 2 Suppose that hΩ(fn(i) )ii∈N has no cluster point in Λ. Let h be any cluster point of hfn(i) ii∈N in R . Then there is a non-principal ultrafilter F on N such that h = limi→F fn(i) in RX . Set A = limi→F Ω(fn(i) ), so that A ∈ / Λ. If x ∈ X and α ∈ R, then X

α < h(x) =⇒ {i : α < fn(i) (x)} ∈ F =⇒ (x, α) ∈ A, h(x) < α =⇒ {i : α < fn(i) (x)} ∈ / F =⇒ (x, α) ∈ / A. Thus Ω0 (h) ⊆ A ⊆ Ω(h), where Ω0 (h) = {(x, α) : α < h(x)}. ?? If h is Σ-measurable, then S Ω0 (h) = q∈Q {x : h(x) > q} × ]−∞, q], Ω(h) = (X × R) \

S

q∈Q {x

: h(x) < q} × [q, ∞[

0

belong to Λ, and λ(Ω(h) \ Ω (h)) = 0 (because every vertical section of Ω(h) \ Ω0 (h) is negligible). But as Ω0 (h) ⊆ A ⊆ Ω(h), A ∈ Λ (remember that product measures in this book are complete), which is impossible. X X Thus h is not Σ-measurable. As h is arbitrary, hfn(i) ii∈N has no measurable cluster point in RX . So at least one of the envisaged alternatives must be true. 463K Corollary Let (X, Σ, µ) be a perfect σ-finite measure space. Write L0 ⊆ RX for the space of real-valued Σ-measurable functions on X. (a) If K ⊆ L0 is relatively countably compact for the topology Tp of pointwise convergence on L0 , then every sequence in K has a subsequence which is convergent almost everywhere. Consequently K is relatively compact in L0 for the topology Tm of convergence in measure. (b) If K ⊆ L0 is countably compact for Tp , then it is compact for Tm . (c) Suppose that K ⊆ L0 is countably compact for Tp and that µ{x : f (x) 6= g(x)} > 0 for any distinct f , g ∈ K. Then the topologies Tm , Tp agree on K, so both are compact and metrizable. proof (a) Since every sequence in K must have a Tp -cluster point in L0 , 463J tells us that every sequence in K has a subsequence which is convergent almost everywhere, therefore Tm -convergent. Now K is relatively sequentially compact in the pseudometrizable space (L0 , Tm ), therefore relatively compact (4A2Le). (b) As in (a), every sequence hfn in∈N in K has a subsequence hgn in∈N which is convergent almost everywhere. But hgn in∈N has a Tp -cluster point g in K, and now g(x) = limn→∞ gn (x) for every x for which the limit is defined; accordingly gn → g a.e., and g is a Tm -limit of hgn in∈N in K. Thus every sequence in K has a Tm -cluster point in K, and (because Tm is pseudometrizable) K is Tm -compact. (c) The point is that K is sequentially compact under Tp . P P Note that as K is countably compact, supf ∈K |f (x)| is finite for every x ∈ K. (I am passing over the trivial case K = ∅.) If hfn in∈N is a sequence in K, then, by (b), it has a subsequence hgn in∈N which is convergent for the pseudometrizable topology Tm . Now hgn in∈N has a further subsequence hhn in∈N which is convergent a.e., by 245K. ?? If hhn in∈N is not Tp -convergent, then there are a point x0 ∈ X and two further subsequences hh0n in∈N , hh00n in∈N of hhn in∈N such that limn→∞ h0n (x0 ), limn→∞ h00n (x0 ) exist and are different. Now hh0n in∈N , hh00n in∈N must have cluster points h0 , h00 ∈ K with h0 (x0 ) 6= h00 (x0 ). However, h0 (x) = limn→∞ hn (x) = h00 (x)

546


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whenever the limit is defined, which is almost everywhere; so h0 =a.e. h00 . And this contradicts the hypothesis that if two elements of K are equal a.e., they are identical. X X Thus hhn in∈N is a Tp -convergent subsequence of hfn in∈N . As hfn in∈N is arbitrary, K is Tp -sequentially compact. Q Q Now 463Cd gives the result. 463L Proposition Let X0 , . . . , Xn be compact Hausdorff spaces, each carrying a strictly positive Radon measure. Let X be their product and Ba(X) the Baire σ-algebra of X. Then any separately continuous function f : X → R is Ba(X)-measurable. proof For i ≤ n let µi be a strictly positive Radon measure on Xi ; let µ be the product Radon measure on X. (a) The proof relies on the fact that (∗) if g, g 0 : X → R are distinct separately continuous functions, then µ{x : g(x) 6= g 0 (x)} > 0; I seek to prove this, together with the stated result, by induction on n. The induction starts easily with n = 0, so that X can be identified with X0 , a separately continuous function on X is just a continuous function on X0 , and (∗) is true because µ = µ0 is strictly positive. (b) For the inductive step to n + 1, given a separately continuous function f : X0 × . . . × Xn+1 → R, set ft (y) = f (y, t) for every y ∈ Y = X0 × . . . × Xn and t ∈ Xn+1 . Then every ft is separately continuous, therefore Ba(Y )-measurable, by the inductive hypothesis. Moreover, again because f is separately continuous, the function t 7→ ft (y) is continuous for every y, that is, t 7→ ft : Xn+1 → RY is continuous; and finally, by the inductive hypothesis (∗), ν{y : ft (y) 6= ft0 (y)} > 0 whenever t, t0 ∈ Xn+1 and ft 6= ft0 , where ν is the product Radon measure on Y . Since ν, being a Radon measure, is perfect (416W), we can apply 463Kc to see that K = {ft : t ∈ Xn+1 } is metrizable for the topology of pointwise convergence. Let ρ be a metric on K inducing its topology, and hgi ii∈N a sequence running over a dense subset of K. (I am passing over the trivial case K = ∅ = Xn+1 .) For m, i ∈ N set Emi = {t : ρ(ft , gi ) ≤ 2−m , ρ(ft , gj ) > 2−m for j < i}, so that every Emi belongs to Ba(Xn+1 ). Set f (m) (y, t) = gi (y) for t ∈ Emi . Then f (m) : X → R is Ba(X)-measurable because every gi is Ba(Y )-measurable, every Emi belongs to Ba(Xm+1 ), and Ba(X) = (m) (m) b Ba(Y )⊗Ba(X im∈N → f at every point, because ρ(ft , ft ) ≤ 2−m for every n+1 ) (4A3Od). Also hf m ∈ N and t ∈ Xn+1 . So f is Ba(X)-measurable. (c) We still have to check that (∗) is true at the new level. But if h, h0 : X → R are distinct separately continuous functions, then there are t0 ∈ Xn+1 , y0 ∈ Y such that h(y0 , t0 ) 6= h0 (y0 , t0 ). Let G be an open set containing t0 such that h(y0 , t) 6= h0 (y0 , t) whenever t ∈ G. Then ν{y : h(y, t) 6= h0 (y, t)} > 0 for every t ∈ G, by the inductive hypothesis, so µ{(y, t) : h(y, t) 6= h0 (y, t)} =

R

ν{y : h(y, t) 6= h0 (y, t)}µn+1 (dt) > 0

because µn+1 is strictly positive. Thus the induction continues. 463M Corollary Let X0 , . . . , Xn be Hausdorff spaces with product X, and µ any Radon measure on X; let Σ be the domain of µ. Then every separately continuous function f : X → R is Σ-measurable. proof (a) Suppose first that there is a conegligible compact set C ⊆ X. For each i ≤ n, let πi : X → Xi −1 be the Q coordinate projection, and µi = µπi the image Radon measure; let Zi be the support of µi and Z = i≤n Zi . Note that every Zi is a closed subset of πi [C], so that Zi is compact and C ⊆ Z. By 463L, f ¹Z is Ba(Z)-measurable; because Z is conegligible, f is Σ-measurable. (b) In general, if C ⊆ X is compact, then we can apply (a) to the measure E 7→ µ(E ∩ C) to see that f ¹C is measurable. As µ is complete and locally determined and tight (that is, inner regular with respect to the compact sets), f is measurable (see 412Ja). *463N There are important open questions concerning the possibility of extending the results of 463K to non-perfect measure spaces (see 463Za-463Zb). I give here one curious fact which is relevant to a question which arises naturally in the next section.

*463O

Tp and Tm

547

Proposition Let (X, Σ, µ) be a σ-finite measure space, and L0 the space of Σ-measurable real-valued functions on X. Suppose that K ⊆ L0 is an infinite set, compact for the topology Tp of pointwise convergence, and that µ{x : f (x) 6= g(x)} > 0 for any distinct f , g ∈ K. Then there is a sequence of distinct elements of K which is Tp -convergent. proof (a) To begin with (down to the end of (c) below) I will suppose that µX = 1 and that 0 ≤ f ≤ χX for every f ∈ K. R ∆(A) = 0. Otherwise, set ∆(A) = R For any set A ⊆ K, define ∆(A) as follows. If A =X ∅, then (sup A − inf A)dµ, where sup A and inf A are taken in R , and dµ is the lower integral (133I). (There is no suggestion at this point that sup A or inf A should be measurable.) Clearly ∆(A) ≤ ∆(B) whenever A ⊆ B ⊆ K. If A has more than one member, then ∆(A) > 0, because two elements of K which are equal a.e. are identical; this is where I use the hypothesis that K is separated by the topology of convergence in measure. (b) ?? Now suppose, if possible, that there is no non-trivial convergent sequence in K. (Here, and henceforth, all topological terms will refer to the topology Tp .) Then there is an infinite closed set K1 ⊆ K such that ∆(L) = ∆(K1 ) for every infinite closed set L ⊆ K1 . P P Write L for the family of infinite closed subsets of K. For L ∈ L, set ∆(L) = inf{∆(L0 ) : L0 ∈ L, L0 ⊆ L}. Choose hLn in∈N in L inductively, as follows. L0 = K. Given Ln , let Ln+1 ∈TL be such that Ln+1 ⊆ Ln and ∆(Ln+1 ) ≤ ∆(Ln ) + 2−n ; continue. At the end of the induction, set K1 = n∈N Ln . Because there are supposed to be no convergent sequences in K, K1 must be infinite (4A2G(h-i)). Now ∆(K1 ) ≤ inf n∈N ∆(Ln ) ≤ supn∈N ∆(Ln ) ≤ ∆(K1 ) ≤ ∆(K1 ), so ∆(L) = ∆(K1 ) whenever L ⊆ K1 and L ∈ L, as required. Q Q (c) Now there is no non-trivial convergent sequence in K1 , so K1 cannot be scattered (4A2G(h-ii)), and it must have an infinite closed subset K2 without isolated points which has a countable π-base U say (4A2G(iiii)). We may suppose that ∅ ∈ / U . Set g0 = inf K2 , g1 = sup K2 in RX . Then g0 and g1 are measurable. P P For each non-empty U ∈ U, choose hU ∈ U . Then g00 = inf U ∈U hU is measurable, and g0 ≤ g00 . But, for any t ∈ X, {f : f ∈ K2 , f (t) < g00 (t)} is a relatively open set in K2 , not containing any hU and therefore not including any member of U, and must accordingly be empty. This means that g00 ≤ f for every f ∈ K2 and Q g0 = g00 is measurable. Similarly, g1 = supU ∈U hU is measurable. Q Since

R

g1 − g0 =

R

g1 − g0 = ∆(K2 ) > 0,

there is a rational number α such that E = {t : g0 (t) < α < g1 (t)} is non-negligible. For t ∈ E, the set {f : f ∈ K2 , f (t) < α} is a non-empty relatively open set in K2 , so includes some member Ut of U. Because U is countable, there is a U ∈ U such that D = {t : t ∈ E, Ut = U } is non-negligible. But now observe that f (t) < α whenever f ∈ U and t ∈ D, so f (t) R≤ α whenever f ∈ U and t ∈ D. Now there is a measurable function h such that h ≤ sup U − inf U and h = ∆(U ) (133J(a-ii)). In this case h ≤ g1 − g0 and also {t : h(t) < g1 (t) − g0 (t)} ⊇ D is non-negligible, so that ∆(U ) =

R

h<

R

g1 − g0 = ∆(K2 ).

On the other hand, U is infinite, because K2 has no isolated points, so U is an infinite closed subset of K2 with ∆(U ) < ∆(K2 ); which contradicts the choice of K2 . X X (d) This shows that, at least in the case in which µX = 1 and K ⊆ [0, 1]X , K has a non-trivial convergent sequence. For the general case, observe that µX is certainly non-zero (since K contains functions which are not equal a.e.), so there is a probability measure ν with the same measurable sets and the same negligible sets 1 2

(215B(vii)). Moreover, if we set q(f )(t) = (1 +

f (t) ) 1+|f (t)|

for f ∈ RX and t ∈ X, then K 0 = {q(f ) : f ∈ K}

is homeomorphic to K, and it is still true that ν{t : f (t) 6= g(t)} > 0 for all distinct f , g ∈ K 0 , while K 0 is bounded by the constant functions 0 and χX. So, applying (a)-(c) to ν and K 0 , we see that K 0 has a non-trivial convergent sequence and therefore K also does. *463O Corollary (Talagrand 80) Under the hypotheses of 463N, K cannot be homeomorphic to βN, ˇ the Stone-Cech compactification of N.

548


*463O

proof For there are no non-trivial convergent sequences in βN (4A2I(b-iv)). 463X Basic exercises > (a) Let (X, Σ, µ) be a measure space, L0 the space of Σ-measurable real-valued functions on X, Tp the topology of pointwise convergence on L0 and Tm the topology of convergence in measure on L0 . (i) Show that Tm ⊆ Tp iff for every measurable set E of finite measure there is a countable set D ⊆ E such that µ∗ D = µE. (ii) Show that Tp ⊆ Tm iff 0 < µ∗ {x} < ∞ for every x ∈ X. (b) Let (X, Σ, µ) be a σ-finite measure space, and K ⊆ L0 (Σ) a Tp -countably compact set. Show that the following are equiveridical: (i) every sequence in K has a subsequence which converges almost everywhere; (ii) K is Tm -compact; (iii) K is totally bounded for the uniformity associated with the linear space topology Tm . Show that if moreover the topology on K induced by Tm is Hausdorff, then K is Tp -metrizable. (c) Let (X, Σ, µ) be a complete locally determined measure space, and K ⊆ L0 (Σ) a set such that whenever E ∈ Σ is a non-negligible measurable set and ² > 0, there is a non-negligible measurable set F ⊆ E such that |f (x) − f (y)| ≤ ² whenever f ∈ K and x, y ∈ F . Show that Tp and Tm agree on K, and that if supf ∈K |f (x)| is finite for every x ∈ X then K is relatively compact in L0 for both topologies. (d) (i) Show that there is a set of Borel measurable functions on [0, 1] which countably tight, compact and non-metrizable for the topology of pointwise convergence. (ii) Show that there is a strictly localizable measure space (X, Σ, µ) with a set K of measurable functions which is countably tight, compact, Hausdorff and non-metrizable for both the topology of pointwise convergence and the topology of convergence in measure. (Hint: the one-point compactification of any discrete space is countably tight.) (e) Let X be a topological space and K ⊆ C(X) a convex Tp -compact set. Show that if there is a strictly positive σ-finite topological measure on X, then K is Tp -metrizable. (f ) Use Komlós’ theorem (276H) to shorten the proof of 463F. (g) Let (X, Σ, µ) be any complete σ-finite measure space. Show that if K ⊆ L0 (Σ) is Tm -relatively compact, and supf ∈K |f (x)| is finite for every x ∈ X, then K is Tp -relatively countably compact in L0 (Σ). (h) Let K be the set of non-decreasing functions from [0, 1] to {0, 1}. Show that K, with its topology of pointwise convergence, is homeomorphic to the split interval (419L). Show that the identity map from (K, Tp ) to (K, Tm ) (for any Radon measure µ on [0, 1]) is continuous. > (i) Let K be the set of non-decreasing functions from ω1 to {0, 1}. Show that if µ is the countablecocountable measure on ω1 then K is a Tp -compact set of measurable functions and is also Tm -compact, but the identity map from (K, Tp ) to (K, Tm ) is not continuous. > (j) Let K be the set of functions f : [0, 1] → R such that max(kf k∞ , Var[0,1] f ) ≤ 1, where Var[0,1] f is the variation of f (224A). Show that K is Tp -compact and that (for any Radon measure on [0, 1]) the identity map from (K, Tp ) to (K, Tm ) is continuous. R > (k) Let A be the set of functions f : [0, 1] → [0, 1] such that |f |dµ · Var[0,1] f ≤ 1, where µ is Lebesgue measure. Show that every member of A is measurable and that R every sequence in A has a subsequence which converges almost everywhere to a member of A, but that : A → [0, 1] is not Tp -continuous, while A is Tp -dense in [0, 1][0,1] . (l) Let X be a Hausdorff space and K ⊆ C(X) a Tp -compact set. Show that if there is a strictly positive σ-finite Radon measure on X then K is Tp -metrizable. > (m) Let (X, Σ, µ) be any perfect measure space and K ⊆ L0 (Σ) a Tp -compact set. Show that the identity map on K is (Tp , Tm )-continuous, so that K is Tm -compact. (n) Let (X, Σ, µ) be a localizable measure space and K ⊆ L0 (Σ) a non-empty Tp -compact set. Show that sup{f • : f ∈ K} is defined in L0 (µ).

463 Notes

Tp and Tm

549

463Y Further exercises (a) Let (X, Σ, µ) be a probability space and V a Banach space. A function φ : X → V is scalarly measurable (often called weakly measurable) if hφ : X → R isR Σ-measurable for every h ∈ V ∗ . φ is Pettis integrable, with indefinite Pettis integral θ : Σ → V , if E hφ dµ is defined and equal to h(θE) for every E ∈ Σ and every h ∈ V ∗ . (i) Show that if φ is scalarly measurable, then K = {hφ : h ∈ V ∗ , khk ≤ 1} is a Tp -compact subset of L0 (Σ). (ii) Show R that if φ is scalarly measurable, then it is Pettis integrable iff every function in K is integrable and f 7→ E f : K → R is Tp -continuous for every E ∈ Σ. (Hint: 4A4Cg.) (iii) In particular, if φ is bounded and scalarly measurable and the identity map from (K, Tp ) to (K, Tm ) is continuous, then φ is Pettis integrable. (See Talagrand 84, chap. 4.) (b) Show that any Bochner integrable function (253Yf) is Pettis integrable. (c) Let µ be Lebesgue measure on [0, 1], and define φ : [0, 1] → L∞ (µ) by setting φ(t) = χ[0, t]• for every t ∈ [0, 1]. (i) Show that if h ∈ L∞ (µ)∗ and khk ≤ 1, then hφ has variation at most 1. (ii) Show that K = {hφ : h ∈ L∞ (µ)∗ , khk ≤ 1} is a Tp -compact set of Lebesgue measurable functions, and that the identity map from (K, Tp ) to (K, Tm ) is continuous, so that φ is Pettis integrable. (iii) Show that φ is not Bochner integrable. (d) Let (X, Σ, µ) be a σ-finite measure space and suppose that µ is inner regular with respect to some family E ⊆ Σ of cardinal at most ω1 . (Subject to the continuum hypothesis, this is true for any subset of R, for instance.) Show that if K ⊆ L0 (Σ) is Tp -compact then it is Tm -compact. (See Talagrand 84, 9-3-3.) (e) Assume that the continuum hypothesis is true; let 4 be a well-ordering of [0, 1] with order type ω1 (4A1Ad). Let (Z, ν) be the Stone space of the measure algebra of Lebesgue measure on [0, 1], and q : Z → [0, 1] the canonical inverse-measure-preserving map (416V). Let g : [0, 1] → [0, ∞[ be any function. Show that there is a function f : [0, 1]R× Z → [0, ∞[ such that (α) f is continuous in the second variable (β) f (t, z) = 0 whenever q(z) 4 t (γ) f (t, z)ν(dz) = g(t) for every t ∈ [0, 1]. Show that f is universally measurable in the first variable, but need not be measurable for the product Radon measure on [0, 1] × Z. Setting fz (t) = f (t, z), show that K = {fz : z ∈ Z} is a Tp -compact set of Lebesgue measurable functions and that g belongs to the Tp -closed convex hull of K in R[0,1] . 463Z Problems (a) A.Bellow’s problem Let (X, Σ, µ) be a probability space, and K ⊆ L0 (Σ) a Tp -compact set such that {x : f (x) 6= g(x)} is non-negligible for any distinct functions f , g ∈ K, as in 463F and 463Kc. Does it follow that K is metrizable for Tp ? A positive answer would displace several of the arguments of this section, and have other consequences (see 462Z, for instance). It is known that under any of a variety of special axioms (starting with the continuum hypothesis) there is indeed a positive answer; see Talagrand 84, chap. 12. (b) Let X ⊆ [0, 1] be a set of outer Lebesgue measure 1, and µ the subspace measure on X, with Σ its domain. Let K be a Tp -compact subset of L0 (Σ). Must K be Tm -compact? (c) Let X0 , . . . , Xn be compact Hausdorff spaces and f : X0 × . . . × Xn → R a separately continuous function. Must f be universally measurable? 463 Notes and comments The relationship between the topologies Tp and Tm is complex, and I do not think that the results here are complete; in particular, we have a remarkable outstanding problem in 463Za. Much of the work presented here has been stimulated by problems concerning the integration of vector-valued functions. I am keeping this theory firmly in the ‘further exercises’ (463Ya-463Yc), but it is certainly the most important source of examples of pointwise compact sets of measurable functions. In particular, since the set {hφ : h ∈ V ∗ , khk ≤ 1} is necessarily convex whenever V is a Banach space and φ : X → V is a function, we are led to look at the special properties of convex sets, as in 463F. There are obvious connexions with the theory of measures on linear topological spaces, which I will come to in §466. The dichotomy in 463J shows that sets of measurable functions on perfect measure spaces are either ‘good’ (relatively countably compact for Tp , relatively compact for Tm ) or ‘bad’ (with neither property). It is known that the result is not true for arbitrary σ-finite measure spaces (see §464 below), but it is not clear whether there are important non-perfect spaces in which it still applies in some form; see 463Zb.

550


463 Notes

Just as in §462, many questions concerning the topology Tp on RX can be re-phrased as questions about real-valued functions on products X × K which are continuous in the second variable. For the topology of pointwise convergence on sets of measurable functions, we find ourselves looking at functions which are measurable in the first variable. In this way we are led to such results as 463L-463M and 463Ye. Concerning 463L and 463Zc, R.Pol has recently (June 2000) proved that if X and Y are any compact Hausdorff spaces, and f : X × Y → R is separately continuous, then f is Borel measurable. A substantial proportion of the questions which arise naturally in this topic are known to be undecidable without using special axioms. I am avoiding such questions in this volume, but it is worth noting that the continuum hypothesis, in particular, has many striking consequences here, of which 463Ye is a sample. It also decides 463Za and 463Zb (see 463Yd).

464 Talagrand’s measure An obvious question arising from 463H and its corollaries is, do we really need the hypothesis that the measure involved is perfect? A very remarkable construction of M.Talagrand (464C) shows that these results are certainly not true of all probability spaces (464D). Investigating the properties of this measure we are led to some surprising facts about additive functionals on algebras PI and the duals of `∞ spaces (464M, 464R). 464A The usual measure on PI Recall from 254J and 416U that for any set I we have a standard measure ν, a Radon measure for the usual topology on PI, defined by saying that ν{a : a ⊆ I, a ∩ J = c} = 2−#(J) whenever J ⊆ I is a finite set and c ⊆ J, or by copying from the usual product measure on {0, 1}I by means of the bijection a 7→ χa : PI → {0, 1}I . We shall need a couple of simple facts about these measures. (a) If hIj ij∈J is any partition of I, then ν can be identified Q with the product of the family hνj ij∈J , where νj is the usual measure on PIj , and we identify PI with j∈J PIj by matching a ⊆ I with ha ∩ Ij ij∈J ; this is the ‘associative law’ 254N. It follows that if we have any family hAj ij∈J of subsets of PI, and if for each j the by coordinates in Ij ’ in the sense that, for a ⊆ I, a ∈ Aj iff a ∩ Ij ∈ Aj , then Q T set Aj is ‘determined ν ∗ ( j∈J Aj ) = j∈J ν ∗ Aj (use 254Lb). (b) Similarly, if f1 , f2 are non-negative real-valued functions on PI, and if there are disjoint sets I1 , R I2 ⊆ I such that fj (a) = fj (a ∩ Ij ) for every a ⊆ I and both j, then the upper integral f1 + f2 dν is R R f1 dν + f2 dν. P P We may suppose that I2 = I \ I1 . For each j, define gj : PIj → [0, ∞[ by setting gj = fj ¹ PIj , so that fj (a) = gj (a ∩ Ij ) for every a ⊆ I. Let νj be the usual measure on PIj , so that we can identify ν with the produce measure ν1 × ν2 , if we identify PI with PI1 × PI2 ; that is, we think of a subset of I as a pair (a1 , a2 ) where aj ⊆ Ij for both j. Now we have Z

Z f1 + f2 dν =

(by 253K)

Z g1 dν1 +

Z =

Z g1 dν1 ·

Z =

g2 dν2 Z 1dν2 +

Z 1dν1 ·

g2 dν2

Z f1 dν +

f2 dν

by 253J, because we can think of f1 (a1 , a2 ) as g1 (a1 )1(a2 ) for all a1 , a2 . Q Q (c) If A ⊆ PI is such that b ∈ A whenever a ∈ A, b ⊆ I and a4b is finite, then ν ∗ A must be either 0 or 1; this is the zero-one law 254Sa, applied to the set {χa : a ∈ A} ⊆ {0, 1}I and the usual measure on {0, 1}I .

464C

Talagrand’s measure

551

464B Lemma Let I be any set, and ν the usual measure Q∞ on PI. (a)(i) There is a sequence hm(n)in∈N in N such that n=0 1 − 2−m(n) = 21 . Q (ii) Given such a sequence, write X for n∈N (PI)m(n) , and let λ be the product measure on X. We have a function φ : X → PI defined by setting S T φ(hhani ii 21 . k=0 (ii) If t ∈ I, then {x : x ∈ X, t ∈ / φ(x)} = {hhani ii
Q∞ n=0

λn {hai ii
S

T i
T n∈N

i
ani }

ai },

where λn is the product measure on (PI)m(n) for each n. But this is just Q∞ 1 −m(n) = , n=0 1 − 2 2

1 2

by the choice of hm(n)in∈N . Accordingly λ{x : t ∈ φ(x)} = for every t ∈ I. Next, each set Et = {x : t ∈ φ(x)} is determined by coordinates in Jt = {(t, n, i) : n ∈ N, i < m(n)}. Since the sets Jt are disjoint, the sets Et , for different t, are stochastically independent (464Aa), so if J ⊆ I is finite, Q λ{x : J ⊆ φ(x)} = t∈J λEt = 2−#(J) = ν{a : J ⊆ a}. This shows that λφ−1 [F ] = νF whenever F is of the form {a : J ⊆ a} for some finite J ⊆ I. By the Monotone Class Theorem (136B), λφ−1 [F ] = νF for every F belonging to the σ-algebra generated by sets of this form. But this σ-algebra certainly contains all sets of the form {a : a ∩ J = c} where J ⊆ I is finite and c ⊆ J, which are the sets corresponding to the basic cylinder sets in the product {0, 1}I . By 254G, φ is inverse-measure-preserving. (b) This uses the same idea as (a-ii). Writing ν 3 for the product measure on (PI)3 , then, for any t ∈ I, ν 3 {(a, b,c) : t ∈ (a ∩ b) ∪ (a ∩ c) ∪ (b ∩ c)} = ν 3 {(a, b, c) : t ∈ a ∩ b} + ν 3 {(a, b, c) : t ∈ a ∩ c} + ν 3 {(a, b, c) : t ∈ b ∩ c} − 2ν 3 {(a, b, c) : t ∈ a ∩ b ∩ c} =3·

1 4

−2·

1 8

1 2

= .

Once again, these sets are independent for different t, and this is all we need to know in order to be sure that the map is inverse-measure-preserving. 464C Lemma Let I be any set, and let ν be the usual measure on PI. ´ski 45) If F ⊆ PI is any filter containing every cofinite set, then ν∗ F = 0 and ν ∗ F is (a) (see Sierpin either 0 or 1. If F is a non-principal ultrafilter then ν ∗ F = 1. T (b) (Talagrand 803 ) If hFn in∈N is a sequence of filters on I, all of outer measure 1, then n∈N Fn also has outer measure 1. 3 The

date of this paper is misleading, as there was an unusual backlog in the journal; in reality it preceded Fremlin & Talagrand 79.

552


464C

proof (a) That ν ∗ F ∈ {0, 1} is immediate from 464Ac. If F is an ultrafilter, then PI = F ∪ {I \ a : a ∈ F}; but as a 7→ I \ a is a measure space automorphism of (PI, ν), ν ∗ {I \ a : a ∈ F } = ν ∗ F, and both must be at least 21 , so ν ∗ F = 1. Equally, ν ∗ (PI \ F) = ν ∗ {I \ a : a ∈ F} = 1, so ν∗ F = 0. Returning to a general filter containing every cofinite set, this is included in a non-principal ultrafilter, so also has inner measure 0. Q∞ (b) Let hm(n)in∈N be a sequence in N such that n=0 1 − 2−m(n) = 12 , and let X, λ and φ : X → PI Q m(n) be asTin 464Ba. Consider the set D = n∈N Fn as a subset of X. By 254Lb, λ∗ D = 1. If we set F = n∈N Fn , then we see that whenever x = hhani ii
464G


553

(b) In §§361-363 I examined some of the bands in M . The most significant ones for our present purposes are the band Mτ of completely additive functionals (362Bb) and its complement Mτ⊥ (352P); because M is ⊥ Dedekind P complete, we have M = Mτ ⊕⊥Mτ (353I). In fact Mτ is just the set of those θ ∈ M such that θa = t∈a θ{t} for every a ⊆ I, while Mτ is the set of those θ ∈ M such that θ{t} = 0 for every t ∈ I. P P For any θ ∈ M , we can set αt = θ{t} for each t ∈ I; in this case, P | t∈J αt | = |θJ| ≤ kθk P P for every finite set J ⊆ I, so t∈I |αt | is finite, and we have a functional θ1 defined by setting θ1 a = t∈a αt for every a ⊆ I. It is easy to check that θ1 ∈ Mτ . If we write θ2 = θ − θ1 then θ2 {t} = 0 for every t ∈ I. So if φ ∈ Mτ and 0 ≤ φ ≤ |θ2 |, we still have φ{t} = 0 for every t ∈ I, and φa P= 0 for every finite a ⊆ I and φ = 0; thus θ2 ∈ Mτ⊥ . Now θ ∈ Mτ iff θ2 = 0, that is, iff θ = θ1 and θa = t∈a θ{t} for every a ⊆ I; while θ ∈ Mτ⊥ iff θ1 = 0, that is, θ{t} = 0 for every t ∈ I. Q Q Observe that if θ ∈ Mτ⊥ and a, b ⊆ I are such that a4b is finite, then θa = θb, because θ(a \ b) = θ(b \ a) = 0. (c) It will be useful to have an elementary fact out in the open. If θ ∈ M + , then {a : a ⊆ I, θa = θI} is a filter; this is because {a : θa = 0} is an ideal in PI. 464G We also need a new result not exactly covered by those in Chapters 35 and 36. Lemma Let A be any Boolean algebra. Write M for the L-space of bounded additive functionals on A, and M + for its positive cone, the set of non-negative additive functionals. Suppose that ∆ : M + → [0, ∞[ is a functional such that (α) ∆ is non-decreasing, (β) ∆(αθ) = α∆(θ) whenever θ ∈ M + , α ≥ 0, (γ) ∆(θ1 + θ2 ) = ∆(θ1 ) + ∆(θ2 ) whenever θ1 , θ2 ∈ M + are such that, for some e ⊆ I, θ1 (1 \ e) = θ2 e = 0, (δ) |∆(θ1 ) − ∆(θ2 )| ≤ kθ1 − θ2 k for all θ1 , θ2 ∈ M + . Then there is a non-negative h ∈ M ∗ extending ∆. proof (a) If θ1 , θ2 ∈ M + are such that θ1 ∧ θ2 = 0 in M + , then ∆(θ1 + θ2 ) = ∆(θ1 ) + ∆(θ2 ). P P Let ² > 0. Then there is an e ∈ A such that θ1 (1 \ e) + θ2 e ≤ ² (362Ba). Set θ10 a = θ1 (a ∩ e), θ20 (a) = θ2 (a \ e) for a ∈ A; then θ10 and θ20 belong to M + and θ10 (1 \ e) = θ20 e = 0, so ∆(θ10 + θ20 ) = ∆(θ10 ) + ∆(θ20 ), by hypothesis (γ). On the other hand, 0 ≤ θ1 a − θ10 a = θ(a \ e) ≤ θ(1 \ e) ≤ ² for every a ∈ A, so kθ1 − θ10 k ≤ ². Similarly, kθ2 − θ20 k ≤ ² and k(θ1 + θ2 ) − (θ10 − θ20 )k ≤ 2². But this means that we can use the hypothesis (δ) to see that |∆(θ1 + θ2 ) − ∆(θ1 ) − ∆(θ2 )| ≤ |∆(θ10 + θ20 ) − ∆(θ10 ) − ∆(θ20 )| + 4² = 4². As ² is arbitrary, ∆(θ1 + θ2 ) = ∆(θ1 ) + ∆(θ2 ). Q Q (b) Now recall that M , being a Dedekind complete Riesz space, can be identified with an order-dense solid linear subspace of L0 (C) for some Boolean algebra C (368H). Inside L0 (C) we have the order-dense Riesz subspace S(C) (364K). Write S1 for M ∩ S(C), so that S1 is an order-dense Riesz subspace of M (352N, 353A). Pm If θ1 , θ2 ∈ S1+ = S1 ∩ M + , then ∆(θ1 + θ2 ) = ∆(θ1 ) + ∆(θ2 ). P P We can express θ1 as i=0 αi χci , where c0 , . . . , cm are disjoint members of C, and αi ≥ 0 for each i (361Ec); adding a term 0 · χ(1 \ supi≤m ci ) Pn if necessary, we may suppose that supi≤m ci = 1. Similarly, we can express θ2 as β χd where P j=0 j j d0 , . . . , dn ∈ C are disjoint, βj ≥ 0 for every j and supj≤n dj = 1. In this case, θ1 = i≤m,j≤n αi χ(ci ∩ dj ) P and θ2 = i≤m,j≤n βj χ(ci ∩ dj ). Re-enumerating {ci ∩ dj : i ≤ m, j ≤ n} as hei : i ≤ k} we have expressions P P of θ1 , θ2 in the form i≤k γi χei , i≤k δi χei , while e0 , . . . , ek are disjoint. Pr (r) (r) Setting θ1 = i=0 γi χei for r ≤ k, we see that θ1 ∧ γr+1 χer+1 = 0 for r < k, so (a) above, together with the hypothesis (β), tell us that

554

Pointwise compact sets of measurable functions (r+1)

(r)

464G

(r)

) = ∆(θ1 ) + ∆(γr+1 χer+1 ) = ∆(θ1 ) + γr+1 ∆(χer+1 ) Pk Pk for r < k. Accordingly ∆(θ1 ) = i=0 γi ∆(χei ). Similarly, ∆(θ2 ) = i=0 δi ∆(χei ) and Pk ∆(θ1 + θ2 ) = i=0 (γi + δi )∆(χei ) = ∆(θ1 ) + ∆(θ2 ), ∆(θ1

as claimed. Q Q (c) Consequently, ∆(θ1 +θ2 ) = ∆(θ1 )+∆(θ2 ) for all θ1 , θ2 ∈ M + . P P Let ² > 0. Because S1 is order-dense in M , and the norm of M is order-continuous (354N), S1 is norm-dense (354Ef), and there are θ10 , θ20 ∈ S1+ such that kθj − θj0 k ≤ ² for both j (354Be). But now, just as in (a), |∆(θ1 + θ2 ) − ∆(θ1 ) − ∆(θ2 )| ≤ |∆(θ10 + θ20 ) − ∆(θ10 ) − ∆(θ20 )| + 4² = 4². As ² is arbitrary, we have the result. Q Q (d) Now (c) and the hypothesis (β) are sufficient to ensure that ∆ has an extension to a positive linear functional (355D). 464H The next lemma contains the key ideas needed for the rest of the section. Lemma Let I be any set, and M the L-space of bounded additive functionals on PI; let ν be the usual measure on PI. For θ ∈ M + , set R ∆(θ) = θ dν. (a) For every θ ∈ M + , 12 θI ≤ ∆(θ) ≤ θI. (b) There is a non-negative h ∈ M ∗ such that h(θ) = ∆(θ) for every θ ∈ M + . (c) If θ ∈ (Mτ⊥ )+ , where Mτ ⊆ M is the band of completely additive functionals, then θ ≤ ∆(θ) ν-a.e., and ν ∗ {a : α ≤ θa ≤ ∆(θ)} = 1 for every α < ∆(θ). (d) Suppose that θ ∈ (Mτ⊥ )+ and β, γ ∈ [0, 1] are such that θI = 1 and βθ0 I ≤ ∆(θ0 ) ≤ γθ0 I whenever θ0 ≤ θ in M + . Then, for any α < β, (i) for any finite set K ⊆ PI, the set QK = {a : a ⊆ I, αθb ≤ θ(a ∩ b) ≤ γθb for every b ∈ K} has outer measure 1 in PI; (ii) if α ≥ 21 , the set R = {(a, b, c) : a, b, c ⊆ I, θ((a ∩ b) ∪ (a ∩ c) ∪ (b ∩ c)) ≥ 2α2 + (1 − 2α)γ 2 } has outer measure 1 in (PI)3 ; (iii) if α ≥ 12 , then 2α2 + (1 − 2α)γ 2 ≤ γ. (e) Any θ ∈ M + can be expressed as θ1 + θ2 where ∆(θ1 ) = 21 θ1 I and ∆(θ2 ) = θ2 I. (f) Suppose that 0 ≤ θ0 ≤ θ in M . (i) If ∆(θ) = 21 θI, then ∆(θ0 ) = 12 θ0 I. (ii) If ∆(θ) = θI, then ∆(θ0 ) = θ0 I. proof (a) Of course ∆(θ) ≤

R

θI dν = θI.

On the other hand, because a 7→ I \ a : PI → PI is an automorphism of the measure space (PI, ν), R R f (I \ a)ν(da) = f (a)ν(da) for any real-valued function f . In particular, R R ∆(θ) = θ(a)ν(da) = θ(I \ a)ν(da). So Z 2∆(θ) = (133J(b-ii))

Z θ(a)ν(da) +

Z θ(I \ a)ν(da) ≥

θ(a) + θ(I \ a)ν(da)

464H


555

Z =

θI ν(da) = θI,

and ∆(θ) ≥ 12 θI. (b) I use 464G with A = PI. Examine the conditions (α)-(δ) there. α) Of course ∆ is non-decreasing (133Jc). (α β ) ∆(αθ) = α∆(θ) for every θ ∈ M + and every α ≥ 0, by 133J(b-iii). (β (γγ ) If θ1 , θ2 ∈ M + then ∆(θ1 + θ2 ) ≤ ∆(θ1 ) + ∆(θ2 ) by 133J(b-ii), as in (a). If θ1 , θ2 ∈ M + and e ⊆ I are such that θ1 (I \ e) = θ2 e = 0, then θ1 a = θ1 (a ∩ e), θ2 (a) = θ2 (a \ e) for every a ⊆ I. So ∆(θ1 + θ2 ) = ∆(θ1 ) + ∆(θ2 ) by 464Ab. (δδ ) For every a ⊆ I, θ2 a ≤ θ1 a + kθ2 − θ1 k, so ∆(θ2 ) ≤ ∆(θ1 ) + kθ2 − θ1 k. Similarly, ∆(θ1 ) ≤ ∆(θ2 ) + kθ1 − θ2 k. So |∆(θ1 ) − ∆(θ2 )| ≤ kθ1 − θ2 k, and ∆ satisfies condition (δ) of 464G. (²²) Accordingly 464G tells us that ∆ has an extension to a member of M ∗ . (c) If θ ∈ Mτ⊥ , then θa = θb whenever a4b is finite (464Fb), so all the sets {a : θa > α} have outer measure either 0 or 1, by 464Ac. But this means that if f is a measurable function and θ ≤a.e. f and R f = ∆(θ), as in 133Ja, {a : f (a) ≤ ∆(θ)} has positive measure and meets A = {a : θa > ∆(θ)} in a negligible set; so A cannot have full outer measure and is negligible. On the other hand, if α < ∆(θ), then θ cannot be dominated a.e. by α1, so {a : θa > α} is not negligible and has outer measure 1. (d)(i) The point is just that for any b ⊆ I, the functional θb , defined by saying that θb (a) = θ(a ∩ b) for every a ⊆ I, belongs to M + and is dominated by θ, so that βθb I ≤ ∆(θb ) ≤ γθb I, and {a : αθb I ≤ θb a ≤ γθb I} has outer measure 1, by (c). (If αθb I = ∆(θb ), this is because θb I = 0, and the result is trivial; otherwise, αθb I < ∆(θb ) ≤ γθb I.) But this just says that, for any b ⊆ I, {a : αθb ≤ θ(a ∩ b) ≤ γθb} has outer measure 1. Now, given any finite set K ⊆ PI, let B be the subalgebra of PI generated by K, and b0 , . . . , bn the atoms of B. Then all the sets Ai = {a : αθbi ≤ θ(a ∩ bi ) ≤ γθbi } have T outer measure 1; because each Ai is determined by coordinates in bi , and b0 , . . . S , bn are disjoint, A = i≤n Ai still has outer measure 1, by 464Aa. But if a ∈ A and b ∈ K, then b = i∈J bi for some J ⊆ {0, . . . , n} and X X αθb = αθbi ≤ θ(a ∩ bi ) i∈J

= θ(a ∩ b) ≤

i∈J

X

γθbi = γθb.

i∈J

So {a : αθb ≤ θ(a ∩ b) ≤ γθb for every b ∈ K} includes A and has outer measure 1. (ii) ?? Suppose, if possible, otherwise; then ν∗3 (PI \ R) > 0, where ν 3 is the product measure on (PI)3 , and there is a measurable set W ⊆ (PI)3 \R such that ν 3 W > 0. For a ⊆ I, set Wa = {(b, c) : (a, b, c) ∈ W }; R 2 3 then ν W = ν Wa ν(da), where ν 2 is the product measure on (PI)2 , by Fubini’s theorem (252D). Set E = {a : ν 2 Wa is defined and not zero}; then νE > 0. Since {a : α ≤ θa ≤ γ} has outer measure 1, there is an a ∈ E such that α ≤ θa ≤ γ. R For b ⊆ I, set Wab = {c : (b, c) ∈ Wa } = {c : (a, b, c) ∈ W }. Then 0 < ν 2 Wa = νWab ν(db), so F = {b : νWab is defined and not 0} has non-zero measure. But also, by (i), {b : θb ≥ α, θ(a ∩ b) ≤ γθa} has outer measure 1, so we can find a b ∈ F such that θb ≥ α and θ(a ∩ b) ≤ γθa. By (i) again,

556


464H

{c : θ(c ∩ (a4b)) ≥ αθ(a4b)} has outer measure 1, so meets Wab ; accordingly we have a c ⊆ I such that (a, b, c) ∈ W while θ(c ∩ (a4b)) ≥ αθ(a4b). Now calculate, for this triple (a, b, c), θ((a ∩ b) ∪ (a ∩ c) ∪ (b ∩ c)) = θ(a ∩ b) + θ(c ∩ (a4b)) ≥ θ(a ∩ b) + αθ(a4b) (by the choice of c) = α(θa + θb) + (1 − 2α)θ(a ∩ b) ≥ α(θa + α) + (1 − 2α)γθa (by the choice of b, recalling that 1 − 2α ≤ 0) ≥ 2α2 + (1 − 2α)γ 2 by the choice of a. But this means that (a, b, c) ∈ W ∩ R, which is supposed to be impossible. X X (iii) Now recall that the map (a, b, c) 7→ (a ∩ b) ∪ (a ∩ c) ∪ (b ∩ c) is inverse-measure-preserving (464Ac). Since θa ≤ ∆(θ) ≤ γ for ν-almost every a, we must have θ((a ∩ b) ∪ (a ∩ c) ∪ (b ∩ c)) ≤ γ for ν 3 -almost every (a, b, c). But as R is not negligible, there must be some (a, b, c) ∈ R such that θ((a ∩ b) ∪ (a ∩ c) ∪ (b ∩ c)) ≤ γ, and 2α2 + (1 − 2α)γ 2 ≤ γ. (e)(i) M ∗ , being the dual of an L-space, is an M -space (356P), so can be represented as C(Z) for some compact Hausdorff space Z (354L). The functional h of (b) above therefore corresponds to a function w ∈ C(Z). Any θ ∈ M + acts on M ∗ as a positiveR linear functional, so corresponds to a Radon measure µθ on Z R(436J/436K); we have ∆(θ) = h(θ) = w dµθ . The inequalities 21 θI ≤ ∆(θ) ≤ θI become 1 w dµθ ≤ µθ Z, because the constant function χZ corresponds to the standard order unit of M ∗ 2 µθ Z ≤ (356P), so that µθ Z =

R

χZ dµθ = kθk = θI

for every θ ≥ 0. Since 0 ≤ h(θ) ≤ kθk for every θ ≥ 0, kwk∞ = khk ≤ 1 and 0 ≤ w ≤ χZ. (ii) Now suppose that β < γ and that G = {z : z ∈ Z, β < w(z) < γ} is non-empty. In this case there is a non-zero θ0 ∈ M + such that βθ0 I ≤ ∆(θ0 ) ≤ γθ0 I whenever 0 ≤ θ0 ≤ θ0 . α) We have a solid linear subspace V = {v : v ∈ C(Z), v(z) = 0 for every z ∈ G} of C(Z). Consider P P(α U = {θ : θ ∈ M, (θ|v) = 0 for every v ∈ V }, where I write ( | ) for the duality between M and C(Z) corresponding to the identification of C(Z) with M ∗ . R β ) If θ ∈ U ∩ M + , then βθI ≤ ∆(θ) ≤ γθI. To see this, observe that v dµθ = (θ|v) = 0 for every (β v ∈ V , so µθ (Z \ G) = sup{µθ K : K ⊆ Z \ G is compact} Z = sup{ v dµθ : v ∈ C(Z), 0 ≤ v ≤ χ(Z \ G)} (because Z is normal, so that whenever K ⊆ Z \ G is compact there is a v ∈ C(Z) such that χK ≤ v ≤ χ(Z \ G), by Urysohn’s Lemma) = 0. R Accordingly β < w < γ µθ -a.e. and w dµθ must lie between βµθ Z = βθI and γµθ Z = γθI. (γγ ) Because M is an L-space, it is perfect (356P), so its dual M ∗ = M × is also perfect, and C(Z) is perfect; moreover, the duality ( | ) identifies M with C(Z)× . Now V ⊥ , taken in C(Z), contains any continuous function zero on Z \ G, so is not {0}; since V ⊥ , like C(Z), must be perfect (356L), (V ⊥ )× is non-trivial. Take any ψ > 0 in (V ⊥ )× . Being perfect, C(Z) is Dedekind complete (356K), so there is a band projection P : C(Z) → V ⊥ (353I). Now ψP is a positive element of C(Z)× which is zero on V , and must correspond to a non-zero element θ0 of U ∩ M + .

464Ja


557

(δδ ) If 0 ≤ θ0 ≤ θ0 in M , then, for any v ∈ V , |(θ0 |v)| ≤ (θ0 ||v|) ≤ (θ0 ||v|) = 0, Q because |v| ∈ V . So θ0 ∈ U and βθ0 I ≤ ∆(θ0 ) ≤ γθ0 I, by (β). Thus θ0 has the required property. Q (iii) It follows at once that w(z) ≥ 12 for every z ∈ Z. P P?? If w(z0 ) < 12 , then we can apply (ii) with ¤ £ 1 β = −1, γ ∈ w(z0 ), 2 to see that there is a non-zero θ ∈ M + such that ∆(θ) ≤ γθI < 12 θI, which is impossible, by (a). X XQ Q £ ¤ £ ¤ (iv) But we find also that w(z) ∈ / 12 , 1 for any z ∈ Z. P P?? If w(z0 ) = δ ∈ 12 , 1 , then 2δ 2 +(1−2δ)δ 2 > δ (because δ(2δ − 1)(1 − δ) > 0). We can therefore find α, β and γ such that 21 ≤ α < β < δ < γ and 2α2 + (1 − 2α)γ 2 > γ. But now {z : β < w(z) < γ} is non-empty, so by (ii) there is a non-zero θ ∈ M + such that βθ0 I ≤ ∆(θ0 ) ≤ γθ0 I whenever 0 ≤ θ0 ≤ θ. Multiplying θ by a suitable scalar if necessary, we can arrange that θI should be 1. But this is impossible, by (d-iii). X XQ Q (v) Thus w takes only the values 21 and R1; let H1 and H2 be the R corresponding open subsets of Z. Take θ ∈ M + . For u ∈ C(Z), set φ(u) = u dµθ and φj (u) = Hj u dµθ for each j. Then each φj is a positive linear functional on C(Z) and φj ≤ φ. But φ is the image of θ under the canonical isomorphism from M to C(Z)× ∼ = M ×× , and C(Z)× is solid in C(Z)∼ (356B), so both φ1 and φ2 belong to the image of M , and correspond to θ1 , θ2 ∈ M . For any u ∈ C(Z), (θ1 + θ2 |u) = φ1 (u) + φ2 (u) = (θ|u), so θ = θ1 + θ2 . We have Z ∆(θj ) = φj (w) =

w dµθ Hj

1 2

1 2

= µθ (H1 ) = θ1 I if j = 1, = µθ (H2 ) = θ2 I if j = 2. So we have a suitable decomposition θ = θ1 + θ2 . (f ) This is easy. Set θ00 = θ − θ0 ; then 1 0 2θ I

≤ ∆(θ0 ) ≤ θ0 I,

1 00 2θ I

≤ ∆(θ00 ) ≤ θ00 I

by (a), while ∆(θ0 ) + ∆(θ00 ) = ∆(θ) by (b), and of course θ0 I + θ00 I = θI. But this means that ∆(θ0 ) − 12 θ0 I ≤ ∆(θ) − 12 θI,

θ0 I − ∆(θ0 ) ≤ θI − ∆(θ),

and the results follow. 464I Measurable and purely non-measurable functionals As before, let I be any set, ν the usual measure on PI, T its domain, and M the L-space of bounded additive functionals on PI. Following Fremlin & Talagrand 79, I say that θ ∈ M is measurable if it is T-measurable when regarded as a real-valued function on PI, and purely non-measurable if {a : a ⊆ I, |θ|(a) = |θ|(I)} has outer measure 1. (Of course the zero functional is both measurable and purely non-measurable.) 464J Examples Before going farther, I had better offer some examples of measurable and purely nonmeasurable functionals. Let I, ν and M be as in 464I. (a) Any θ ∈ Mτ is measurable, where Mτ is the space of completely additive functionals on PI. P P By P 464Fb, θ can be expressed as a sum of point masses; say θa = α for some family hα i in R. Since t t t∈I t∈a P |α | must be finite, {t : α = 6 0} is countable, and we can express θ as the limit of a sequence of finite t t t∈I P sums t∈K αt tˆ, where tˆ(a) = 1 if t ∈ a, 0 otherwise. But of course every tˆ is a measurable function, so P ˆ Q t∈K αt t is measurable for every finite set K, and θ is measurable. Q

558


464Jb

(b) For a less elementary measurable functional, consider the following construction. Let htn in∈N be any 1 n

sequence of distinct points in I. Then limn→∞ #({i : i < n, ti ∈ a}) =

1 2

for ν-almost every a ⊆ I. P P Set

fn (a) = 1 if tn ∈ a, 0 otherwise. Then hfn in∈N is an independent sequence of random variables. By any 1 Pn−1 1 of the versions of the Strong Law of Large Numbers in §273 (273D, 273H, 273I), limn→∞ i=0 fi = n

a.e., which is what was claimed. Q Q

2

So if we take any non-principal ultrafilter F on N, and set θa =

1 n

limn→F #({i : i < n, ti ∈ a}) for a ⊆ I, θ will be constant ν-almost everywhere, and measurable; and it is easy to check that θ is additive. Note that θ{t} = 0 for every t, so θ ∈ Mτ⊥ , by 464Fb. (c) If F is any non-principal ultrafilter on I, and we set θa = 1 for a ∈ F , 0 otherwise, then θ is an additive functional which is purely non-measurable, by 464Ca. For further remarks on where to look for measurable and purely non-measurable functionals, see 464P464Q below. 464K The space Mm : Lemma Let I be any set, ν the usual measure on PI, and M the L-space of R bounded additive functionals on PI. Write Mm for the set of measurable θ ∈ M , and ∆(θ) = θ dν for θ ∈ M + , as in 464H. (a) |θ| ∈ Mm for every θ ∈ Mm . (b) A functional θ ∈ M + is measurable iff ∆(θ) = 12 θI. (c) Mm is a solid linear subspace of M . proof (a)(i) Take any θ ∈ Mm ∩ Mτ⊥ , where Mτ is the space of completely additive functionals on PI. α) θ =a.e. 21 θI. P (α P For any α ∈ R, Aα = {a : θa < α} is measurable; but also b ∈ A whenever a ∈ A and a4b is finite, by 464Fb, so νA must be either 1 or 0, by 464Ac. Setting δ = sup{α : νAα = 0}, we see that νAδ = 0, νAδ+2−n = 1 for every n ∈ N, so that θ = δ a.e. Also, because a 7→ I \ a is a measure space automorphism, θ(I \ a) = δ for almost every a, so there is some a such that θa = θ(I \ a) = δ, and δ = 12 θI. Q Q β ) For any b ⊆ I, θ(a ∩ b) = 12 θb for almost every a. P (β P We know that θa = 12 θI for almost every a, by (α). But a 7→ a4b : PI → PI is inverse-measure-preserving, so θ(a4b) = 12 θI for almost every a. This means that θa = θ(a4b) for almost every a, and 1 2

1 2

θ(a ∩ b) = (θb + θa − θ(a4b)) = θb for almost every a. Q Q −n

2

(γγ ) θ+ , taken in M , is measurable. P P For any n ∈ N we can find bn ⊆ I such that θ− bn +θ+ (I \bn ) ≤ , so that |θ+ a − θ(a ∩ bn )| = |θ+ a − θ+ (a ∩ bn ) + θ− (a ∩ bn )| ≤ θ+ (I \ bn ) + θ− bn ≤ 2−n

for every a ⊆ I. But as θ(a ∩ bn ) is constant a.e. for every n, by (β), so is θ+ , and θ+ is measurable. Q Q Consequently |θ| = 2θ+ − θ is measurable. (ii) Now take an arbitrary θ ∈ Mm . Because M is Dedekind complete (354N, 354Ee), M = Mτ + Mτ⊥ (353I), and we can express θ as θ1 +θ2 where θ1 ∈ Mτ and θ2 ∈ Mτ⊥ ; moreover, |θ| = |θ1 |+|θ2 | (352Fb). Now θ1 is measurable, by 464Ja, so θ2 = θ − θ1 is measurable; as θ2 ∈ Mτ⊥ , (i-γ) tells us that |θ2 | is measurable. On the other hand, |θ1 | belongs to Mτ and is measurable, so |θ| = |θ1 | + |θ2 | is measurable. Thus (a) is true. R (b) Let f be a ν-integrable function such that ∆(θ) ≤ f a.e. and f dν = ∆(θ). Then θI − f (a) ≤ θI − θa = θ(I \ a) for almost every a, so

R R θI − ∆(θ) ≤ θ(I \ a)ν(da) = θ(a)ν(da)

464M


559

because a 7→ I \ a is a measure space automorphism, as in the proof of 464Ha. So if ∆(θ) = 21 θI then R R θ dν = θ dν and θ is ν-integrable (133Jd), therefore (because ν is complete) (dom ν)-measurable. On the other hand, if θ is measurable, then ∆(θ) =

R

θ dν =

R

θ(I \ a)ν(da) = θI −

R

θ dν = θI − ∆(θ),

so surely ∆(θ) = 21 θI. (c) Of course Mm is a linear subspace. If θ0 ∈ Mm and |θ| ≤ |θ0 |, then |θ0 | ∈ Mm , by (a), so ∆(|θ0 |) = by (b). Because θ+ ≤ |θ| ≤ |θ0 |, ∆(θ+ ) = 12 θ+ I (464H(f-i)), and θ+ is measurable, by (b) in the reverse direction. Similarly, θ− is measurable, and θ = θ+ − θ− is measurable. As θ and θ0 are arbitrary, Mm is solid. 1 2 |θ0 |(I),

464L The space Mpnm : Lemma Let I be any set, ν the usual measure on PI, and M the L-space of bounded additive functionals on PI. This time, write Mpnm for the set of those members of M which are purely non-measurable in the sense of 464I. (a) If θ ∈ M + , then θ is purely non-measurable iff ∆(θ) = θI. (b) Mpnm is a solid linear subspace of M . proof (a)(i) If θ is purely non-measurable, and f R≥ θ is integrable, then {a : f (a) ≥ θI} is a measurable set including {a : θa = θI}, so has measure 1, and f ≥ θI; as f is arbitrary, ∆(θ) = θI. (ii) If ∆(θ) = θI, then ∆(θ0 ) = θ0 I whenever 0 ≤ θ0 ≤ θ, by 464Hf. But this means that θ0 cannot be measurable whenever 0 < θ0 ≤ θ, by 464Kc above, so that θ0 ∈ / Mτ whenever 0 < θ0 ≤ θ, by 464Ja. Thus ⊥ θ ∈ Mτ . By 464Hc, ν ∗ {a : α ≤ θa ≤ ∆(θ)} = 1 for every α < ∆(θ). Let hm(n)in∈N be a sequence in N such that Q∞ −m(n) = 12 , and define X, λ and φ as in 464Ba. Set ηn = 2−n /m(n) > 0 for each n. Consider n=0 1 − 2 Q m(n) the sets An = {a : θa ≥ (1 − ηn )θI} for each n ∈ N. Then ν ∗ An = 1 for each n, and λ∗ ( n∈N An ) = 1. Q m(n) Because φ is inverse-measure-preserving, ν ∗ (φ[ n∈N An ]) = 1 (413Eh). But if x = hhani ii
T θ(φ(x)) ≥ supn∈N θ( i
560


464M

+ + (b) Now recall that every element of M + is expressible in the form θ1 +θ2 where θ1 ∈ Mm and θ2 ∈ Mpnm ; this is 464He, using 464Kb and 464La again. Because Mm and Mpnm are linear subspaces, with intersection ⊥ ⊥ ⊥⊥ {0}, M = Mm ⊕ Mpnm . Now Mpnm ⊆ Mm , so Mm + Mm = M and Mm = Mm is a complemented ⊥ band (352R); similarly, Mpnm is a complemented band. Since (Mm + Mpnm ) = {0}, Mm and Mpnm are complementary bands (see 352S).

464N Corollary (Fremlin & Talagrand 79) Let I be any set, and let µ be Talagrand’s measure on PI; write Σ for its domain. Then every bounded additive functional on PI is Σ-measurable. proof Defining M , Mm and Mpnm as in 464K and 464L, we see that every functional in Mm is Σ-measurable + because it is (by definition) (dom ν)-measurable, where ν is the usual measure on PI. If θ ∈ Mpnm , then F = {a : θa = θI} is a non-measurable filter; but this means that µF = 1, by the construction of µ, so that θ = θI µ-a.e. So if θ is any member of Mpnm , both θ+ and θ− are Σ-measurable, and θ = θ+ − θ− also is. 464O Remark Note that we have a very simple description of the behaviour of additive functionals as seen by the measure µ. Since Mτ ⊆ Mm , we have a three-part band decomposition M = Mτ ⊕ (Mm ∩ Mτ⊥ ) ⊕ Mpnm . (i) Functionals in Mτ are T-measurable, where T is the domain of ν, therefore Σ-measurable, just because they can be built up from the functionals a 7→ χa(t), as in 464Fb. (ii) A functional in Mm is T-measurable, by definition; but a functional θ in Mm ∩ Mτ⊥ is actually constant, with value 21 θI, ν-almost everywhere. P P For θ ≥ 0, this is 464Kb; the general case follows because Mm ∩ Mτ⊥ is solid in M . Q Q Thus the almost-constant nature of the functionals described in 464Jb is typical of measurable functionals in Mτ⊥ . (iii) Finally, a functional θ ∈ Mpnm is equal to θI µ-almost everywhere; once again, this follows from the definition of ‘purely non-measurable’ and the construction of µ for θ ≥ 0, and from the fact that Mpnm is solid for other θ. R (iv) Thus we see that any θ ∈ Mτ⊥ = (Mm ∩ Mτ⊥ ) ⊕ Mpnm is constant µ-a.e. We also have θ dµ = ∆(θ) for every θ ≥ 0 (look at θ ∈ Mm and Rθ ∈ Mpnm separately, using 464La for the latter), so that if h ∈ M ∗ is the linear functional of 464Hb, then θ dµ = h(θ) for every θ ∈ M . 464P More on purely non-measurable functionals (a) We can discuss non-negative additive funcˇ tionals on PI in terms of the Stone-Cech compactification βI of I, as follows. For any set A ⊆ βI set HA = {a : a ⊆ I, A ⊆ b a}, where b a ⊆ βI is the open-and-closed set corresponding to a ⊆ I. If A 6= ∅, HA is a filter on I. Write A for the family of those sets A ⊆ βI such that ν ∗ HA = 1, where ν is the usual measure on PI. Then A is a σ-ideal. P P Of course T A ∈ A whenever A ⊆ B ∈ A, since then T HA ⊇ HB . If hAn in∈N Q Note that is a sequence in A with union A, then ν ∗ ( n∈N HAn ) = 1, by 464C; but HA ⊇ n∈N HAn . Q if A ∈ A then A ∈ A, because HA = HA . We see also that {z} ∈ A for every z ∈ βI \ I (since H{z} is a non-principal ultrafilter, as in 464Jb), while {t} ∈ / A for any t ∈ I (since H{t} = {a : t ∈ a} has measure 12 ). Because βI can be identified with the Stone space of PI (4A2I(b-i)), we have a one-to-one correspondence between non-negative additive functionals θ on PI and Radon measures µθ on βI, defined by writing µθ (b a) = θa whenever a ⊆ I and θ ∈ M + (416Qb). (This is not the same as the measure µθ of part (e) of the proof of 464H, which is on a much larger space.) Now suppose that θ is a non-negative additive functional T on PI. Then Fθ = {a : θa = θI} is either PI or a filter on I. If we set Fθ = {b a : a ∈ Fθ }, then Fθ = HFθ (4A2I(b-iii)). Since a ∈ Fθ iff θa = θI, \ \ Fθ = {b a : a ∈ Fθ } = {b a : θa = θI} \ [ = {b a : µθ b a = µθ (βI)} = βI \ {b a : µθ b a = 0}. But this is just the support of µθ (411Nd), because {b a : a ⊆ I} is a base for the topology of βI. Thus we see that θ ∈ M + is purely non-measurable iff the support Fθ of the measure µθ belongs to A. If you like, θ is purely non-measurable iff the support of µθ is ‘small’. (b) Yet another corollary of 464C is the following. Since M is a set of real-valued functions on PI, it has the corresponding topology Tp of pointwise convergence as a subspace of RPI . Now if C ⊆ Mpnm

464Qc


561

is countable, its Tp -closure C is included in Mpnm . P P It is enough to consider the case in which C is non-empty and 0 ∈ / C. For each θ ∈ C, F|θ| = {a : |θ|(a) = |θ|(I)} is a filter with outer measure 1, so F = {a : |θ|(a) = |θ|(I) for every θ ∈ C} also has outer measure 1, by 464Cb. Now suppose that θ0 ∈ C. If a ∈ F, then |θ|(I \ a) = 0 for every θ ∈ C, that is, θb = 0 whenever θ ∈ C and b ⊆ I \ a (362Ba). But this means that θ0 b = 0 whenever b ⊆ I \ a, so |θ0 |(I \ a) = 0 and |θ0 |(a) = |θ0 |(I). Thus F|θ0 | ⊇ F, so has outer measure 1, and θ0 is also purely non-measurable. Q Q (c) If θ ∈ M is such that θa = 0 for every countable set a ⊆ I, then θ ∈ Mpnm . P P ν is inner regular with respect to the family W of sets which are determined by coordinates in countable subsets of I, by 254Ob. But if W ∈ W and νW > 0, let J ⊆ I be a countable set such that W is determined by coordinates in J; then |θ|(J) = 0, so if a is any member of W we shall have a ∪ (I \ J) ∈ W ∩ F|θ| . As W is arbitrary, ν ∗ F|θ| = 1 and θ is purely non-measurable. Q Q In particular, if θ ∈ Mσ ∩ Mτ⊥ , where Mσ is the space of countably additive functionals on PI (362B), then θ ∈ Mpnm . (For ‘ordinary’ sets I, Mσ = Mτ ; see 438Xa. But this observation is peripheral to the concerns of the present section.) S In the language of (a) above, we have a closed set in βI, being F = βI \ {b a : a ∈ [I]≤ω }; and if θ is such that the support of µθ is included in F , then θ is purely non-measurable. 464Q More on measurable functionals (a) We know that Mm is a band in M , and that it includes the band Mτ . So it is natural to look at the band Mm ∩ Mτ⊥ . I showed in part (a) of the proof of 464K that if θ ∈ Mm ∩ Mτ⊥ , and b ⊆ I, we have θ(a ∩ b) = 21 θb for ν-almost every a. (b) It follows that if θ is any non-zero non-negative functional in Mm ∩ Mτ⊥ , we can find a family haξ iξ<ω1 T in PI which is independent in the sense that θ( ξ∈K aξ ) = 2−#(K) θI for every non-empty finite K ⊆ I. P P Choose the aξ inductively, observing that at the inductive step we have to satisfy only countably many conditions of the form θ(aξ ∩ b) = 21 θb, where b runs over the subalgebra generated by {aη : η < ξ}; so that ν-almost any a will serve for aξ . Q Q In terms of the associated measure µθ on βI, this means that h(b aξ )• iξ<ω1 is an uncountable independent family in the measure algebra of µθ , and µθ has Maharam type at least ω1 . Turning this round, we see that if λ is a Radon measure on βI, of countable Maharam type, and λI = 0, then the corresponding functional on PI is purely non-measurable. [For a stronger result in this direction, see §531 in Volume 5.] (c) Another striking property of measurable additive functionals is the following. If θ ∈ Mm ∩ Mτ⊥ , n+1 and n ∈ N, then θ(a0 ∩ a1 ∩ . . . ∩ an ) = 2−n−1 θI for ν n+1 -almost every a0 , . . . , aT is the n ⊆ I, where S ν n+1 product measure on (PI) . P P For K ⊆ {0, . . . , n} write ψK (a0 , . . . , an ) = I ∩ i∈K ai \ i≤n,i∈K / ai for S a0 , . . . , an ⊆ I; for S ⊆ P{0, . . . , n} write φS (x) = K∈S ψK (x) for x ∈ (PI)n+1 . Then, for any t ∈ I, P ν n+1 {x : t ∈ φS (x)} = K∈S ν n+1 {x : t ∈ φK (x)} = 2−n−1 #(S). Moreover, for different t, these sets are independent. So if #(S) = 2n , that is, if S is just half of P{0, . . . , n}, then φS will be inverse-measure-preserving, by the arguments in 464B. (In fact 464Bb is the special case n = 2, S = {K : #(K) ≥ 2}.) Accordingly we shall have θ(φS (x)) = 21 θI for almost every x whenever S ⊆ P{0, . . . , n} has 2n members. Since there are only finitely many sets S, the set E is ν n+1 -conegligible, where 1 2

E = {x : x ∈ (PI)n+1 , θ(φS (x)) = θI whenever S ⊆ P{0, . . . , n} and #(S) = 2n }. But given x = (a0 , . . . , an ) ∈ E, let K1 , . . . , K2n+1 be a listing of P{0, . . . , n} in such an order that θ(ψKi (x)) ≤ θ(ψKj (x)) whenever i ≤ j, and consider S = {K1 , . . . , K2n }; since P 2n i=1

1 2

θ(ψKi (x)) = θ(φS (x)) = θI =

1 2

P2n+1 i=0

θ(ψKi (x)),

we must have θ(ψKi (x)) = 2−n−1 θI for every i. In particular, θ(ψ{0,... ,n} (x)) = 2−n−1 θI, that is, θ( 2−n−1 θI. As this is true whenever a0 , . . . , an ∈ E, we have the result. Q Q

T i≤n

ai ) =

562


464R

464R A note on `∞ (I) As already noted in 464F, we have, for any set I, a natural additive map χ : PI → ` (I) ∼ = L∞ (PI), giving rise to an isomorphism between the L-space M of bounded additive functionals on PI and `∞ (I)∗ . If we write µ ˜ for the image measure µχ−1 on `∞ (I), where µ is Talagrand’s measure on PI, ˜ for the domain of µ ˜ ˜ includes the and Σ ˜, then every member of `∞ (I)∗ is Σ-measurable, by 464N. Thus Σ ∞ cylindrical σ-algebra of ` (I) (4A3T). We also have a band decomposition `∞ (I)∗ = `∞ (I)∗m ⊕ `∞ (I)∗pnm corresponding to the decomposition M = Mm ⊕ Mpnm . In this context, Mτ corresponds to `∞ (I)× (363K). Since we can identify `∞ (I) with `1 (I)∗ = `1 (I)× (243Xl), and `1 (I), like any L-space, is perfect, `∞ (I)× is the canonical image of `1 (I) in `∞ (I)∗ . Because any functional in Mτ⊥ is µ-almost constant (464O), any functional in (`∞ (I)× )⊥ will be µ ˜-almost constant. ∞

464X Basic exercises (a) Let I be any set and λ a Radon measure on βI. Show that if the support of λ is a separable subset of βI \ I, then the corresponding additive functional on PI is purely non-measurable. (b) Let I be any set, and µ ˜ the image of Talagrand’s measure on `∞ (I), as in 464R. Show that µ ˜ has a ∞ barycenter in ` (I) iff I is finite. 464Y Further exercises (a) Show that there is a sequence hFn in∈N of distinct non-principal ultrafilters on N with the following property: if we define h(a) = {n : a ∈ Fn } for a ⊆ N, then {h(a) : a ⊆ N} is νnegligible. (b) Let µ be Talagrand’s measure on PN, and let λ be the corresponding product measure on X = PN × PN. Define Φ : X → `∞ by setting Φ(a, b) = χa−χb for all a, b ⊆ N. R Show that Φ is Pettis integrable (463Ya) with indefinite Pettis integral Θ defined by setting (ΘE)(n) = E fn dλ, where fn (a, b) = χa(n) − χb(n). Show that K = {hΦ : h ∈ `∞∗ , khk ≤ 1} contains every fn , and in particular is not Tm -compact, so the identity map from (K, Tp ) to (K, Tm ) is not continuous. 464Z Problem Let I be an infinite set, and µ ˜ the image in `∞ (I) of Talagrand’s measure on PI (464R). Is µ ˜ a topological measure for the weak topology of `∞ (I)? 464 Notes and comments The central idea of this section appears in 464B: the algebraic structure of PI leads to a variety of inverse-measure-preserving functions φ from powers (PI)K to PI. The simplest of these is the measure space automorphism a 7→ I \ a, as used in the proofs of 464Ca, 464Hf, 464Ka and 464Kb. Then we have the map (a, b, c) 7→ (a ∩ b) ∪ (a ∩ c) ∪ (b ∩ c), as in 464Bb, and theSgeneralization of T this in the argument of 464Qc; and, most important of all, the map hhani ii
465Ae

Stable sets

563

Once having noted this remarkable dichotomy between ‘measurable’ and ‘purely non-measurable’ functionals, it is natural to look for other ways in which they differ. The position seems to be that ‘simple’ Radon measures on βI \ I (e.g., all measures with separable support (464Xa) or countable Maharam type (464Qb)) have to correspond to purely non-measurable functionals. Of course the simplest possible measures on βI are those concentrated on I, which give rise to functionals which belong to Mτ and are therefore measurable; it is functionals in Mm ∩ Mτ⊥ which have to give rise to ‘complicated’ Radon measures. The fact that every element of Mτ⊥ is almost constant for Talagrand’s measure leads to an interesting Pettis integrable function (464Yb). The suggestion that Talagrand’s measure on `∞ might be a topological measure for the weak topology (464Z) is a bold one, but no more outrageous than the suggestion that it might measure every continuous linear functional once seemed. Talagrand’s measure does of course measure every Baire set for the weak topology (4A3U). I note here that the usual measure on {0, 1}I , when transferred to `∞ (I), is actually a Radon measure for the weak* topology Ts (`∞ , `1 ), because on {0, 1}I this is just the ordinary topology.

465 Stable sets The structure of general pointwise compact sets of measurable functions is complex and elusive. One particular class of such sets, however, is relatively easy to describe, and has a variety of remarkable properties, some of them relevant to important questions arising in the theory of empirical measures. In this section I outline the theory of ‘stable’ sets of measurable functions from Talagrand 84 and Talagrand 87. The first steps are straightforward enough. The definition of stable set (465B) is not obvious, but given this the basic properties of stable sets listed in 465C are natural and easy to check, and we come quickly to the fact that (for complete locally determined spaces) pointwise bounded stable sets are relatively pointwise compact sets of measurable functions (465D). A less transparent, but still fairly elementary, argument leads to the next reason for looking at stable sets: the topology of pointwise convervence on a stable set is finer than the topology of convergence in measure (465G). At this point we come to a remarkable fact: a uniformly bounded set A of functions on a complete probability space is stable if and only if certain laws of large numbers apply ‘nearly uniformly’ on A. These laws are expressed in conditions (ii), (iv) and (v) of 465M. For singleton sets A, they can be thought of as versions of the strong law of large numbers described in §273. To get the full strength of 465M a further idea in this direction needs to be added, described in 465H here. The theory of stable sets applies in the first place to sets of true functions. There is however a corresponding notion applicable in function spaces, which I explore briefly in 465O-465R. Finally, I mention the idea of ‘R-stable’ set (465S-465U), obtained by using τ -additive product measures instead of c.l.d. product measures in the definition. 465A Notation Throughout this section, I will use the following notation. (a) If X is a set and Σ a σ-algebra of subsets of X, I will write L0 (Σ) for the space of Σ-measurable functions from X to R, as in §463. (b) I will identify N with the set of finite ordinals, so that each n ∈ N is the set of its predecessors, and a power X n becomes identified with the set of functions from {0, . . . , n − 1} to X. (c) If (X, Σ, µ) is any measure space, then for finite sets I (in particular, if I = k ∈ N) I write µI for the c.l.d. product measure on X I , as defined in 251W. (For definiteness, let us take µ0 to be the unique probability measure on X 0 = {∅}.) If (X, Σ, µ) is a probability space, then for any set I µI is to be the product probability measure on X I , as defined in §254. (d) We shall have occasion to look at Nfree powers of algebras of sets. If X is a set and Σ is an algebra of subsets of X, then for any set I write I Σ for the algebra of subsets of X I generated by sets of the form N N {w : w(i) ∈ E} where i ∈ I and E ∈ Σ, and c I Σ for the σ-algebra generated by I Σ. (e) Now for a new idea, which will be used in almost every paragraph of the section. If X is a set, A ⊆ RX a set of real-valued functions defined on X, E ⊆ X, α < β in R and k ≥ 1, write

564


Dk (A, E, α, β) =

[

465Ae

{w : w ∈ E 2k , f (w(2i)) ≤ α,

f ∈A

f (w(2i + 1)) ≥ β for i < k}. We are now ready for the main definition. 465B Definition Let (X, Σ, µ) be a semi-finite measure space. Following Talagrand 84, I say that a set A ⊆ RX is stable if whenever E ∈ Σ, 0 < µE < ∞ and α < β in R, there is some k ≥ 1 such that (µ2k )∗ Dk (A, E, α, β) < (µE)2k . Remark I hope that the next few results will show why this concept is important. It is worth noting at once that these sets Dk need not be measurable, and that some of the power of the definition derives precisely from the fact that quite naturally arising sets A can give rise to non-measurable sets Dk (A, E, α, β). If, however, the set A is countable, then all the corresponding Dk will be measurable; this will be important in the results following 465R. 465C I start with a list of the ‘easy’ properties of stable sets, derivable more or less directly from the definition. Proposition Let (X, Σ, µ) be a semi-finite measure space. (a) If A ⊆ RX is stable, then any subset of A is stable. (b) If A ⊆ RX is stable, then A, the closure of A in RX for the topology of pointwise convergence, is stable. (c) Suppose that A ⊆ RX , E ∈ Σ, n ≥ 1 and α < β are such that 0 < µE < ∞ and (µ2n )∗ Dn (A, E, α, β) < (µE)2n . Then 1 (µ2k )∗ Dk (A, E, α, β) (µE)2k

→0

as k → ∞. (d) If A, B ⊆ RX are stable, then A ∪ B is stable. (e) If A ⊆ RX is stable, then γA = {γf : f ∈ A} is stable, for any γ ∈ R. (f) If A ⊆ RX is stable and g ∈ L0 = L0 (Σ), then A + g = {f + g : f ∈ A} is stable. (g) If A ⊆ L0 is finite it is stable. (h) If A ⊆ RX is stable so is {|f | : f ∈ A}. (i) If A ⊆ RX is stable and g ∈ L0 , then A × g = {f × g : f ∈ A} is stable. (j) If µ ˆ, µ ˜ are the completion and c.l.d. version of µ, then A ⊆ RX is stable with respect to one of the measures µ, µ ˆ, µ ˜ iff it is stable with respect to the others. (k) Let ν be an indefinite-integral measure over µ (§234). If A ⊆ RX is stable with respect to µ, it is stable with respect to ν and with respect to ν¹Σ. (l) Let h : R → R be a continuous non-decreasing function. If A ⊆ RX is stable, so is {hf : f ∈ A}. (m) In particular, if A ⊆ RX is stable, so is {f + : f ∈ A}. proof (a) This is immediate from the definition in 465B, since Dk (B, E, α, β) ⊆ Dk (A, E, α, β) for all k, E, α and β. (b) Given E such that 0 < µE < ∞, and α < β, take α0 , β 0 such that α < α0 < β 0 < β. Then it is easy to see that Dk (A, E, α, β) ⊆ Dk (A, E, α0 , β 0 ), so (µ2k )∗ Dk (A, E, α, β) ≤ (µ2k )∗ Dk (A, E, α0 , β 0 ) < (µE)2k for some k ≥ 1. As E, α and β are arbitrary, A is stable. (c) For any m ≥ 1 and l < n, if we identify X 2(mn+l) with (X 2n )m × X 2l , we see that Dmn+l (A, E, α, β) becomes identified with a subset of Dn (A, E, α, β)m × E 2l . (If w ∈ Dmn+l (A, E, α, β), there is an f ∈ A such that f (w(2i)) ≤ α, f (w(2i + 1)) ≥ β for i < mn + l. Now (w(2rn), w(2rn + 1), . . . , w(2rn + 2n − 1)) ∈ Dn (A, E, α, β) for r < m.) So

465C

Stable sets

565

1 (µ2(mn+l) )∗ Dmn+l (A, E, α, β) (µE)2(mn+l)

≤

¡ 2n ∗ ¢m 1 (µ ) Dn (A, E, α, β) (µE)2l (µE)2(mn+l)

(251Wm) =

¡

¢m 1 (µ2n )∗ Dn (A, E, α, β) (µE)2n

→0

as m → ∞. (d) Note that Dk (A ∪ B, E, α, β) = Dk (A, E, α, β) ∪ Dk (B, E, α, β) for all E, k, α and β. Now, given that 0 < µE < ∞ and α < β, there are m, n ≥ 1 such that (µ2m )∗ Dm (A, E, α, β) < (µE)2m and (µ2n )∗ Dn (A, E, α, β) < (µE)2n . So, by (c) above, 1 (µ2k )∗ Dk (A ∪ B, E, α, β) (µE)2k

≤

¡ 2k ∗ ¢ 1 (µ ) Dk (A, E, α, β) + (µ2k )∗ Dk (B, E, α, β) 2k (µE)

→0

as k → ∞, and there is some k such that (µ2k )∗ Dk (A ∪ B, E, α, β) < (µE)2k . As E, α and β are arbitrary, A ∪ B is stable. (e)(i) If γ > 0, Dk (γA, E, α, β) = Dk (A, E, α/γ, β/γ) for all k, E, α and β, so the result is elementary. Similarly, if γ = 0, then Dk (γA, E, α, β) = ∅ whenever k ≥ 1, E ∈ Σ and α < β, so again we see that γA is stable. (ii) If γ = −1 then, for any k, E, α and β, [ Dk (−A, E, α, β) = {w : w ∈ E 2k , f (w(2i)) ≥ −α, f ∈A

f (w(2i + 1)) ≤ −β ∀ i < k} = φ[Dk (A, E, −β, −α)], where φ : X 2k → X 2k is the measure space automorphism defined by setting φ(w) = (w(1), w(0), w(3), w(2), . . . , w(2k − 1), w(2k − 2)) 2k

for w ∈ X . So, given E ∈ Σ and α < β, there is a k ≥ 1 such that (µ2k )∗ Dk (−A, E, α, β) = (µ2k )∗ Dk (A, E, −β, −α) < (µE)2k . So −A is stable. (iii) Together with (i) this shows that γA is stable for every γ ∈ R. (f ) Take E such that 0 < µE < ∞, and α < β. Set η = 21 (β − α) > 0. Then there is a γ ∈ R such that F = {x : x ∈ E, γ ≤ g(x) ≤ γ + η} has non-zero measure. Set α0 = α − γ, β 0 = β − γ − η. Then Dk (A + g, F, α, β) ⊆ Dk (A, F, α0 , β 0 ), while α0 < β 0 . So if we take k ≥ 1 such that (µ2k )∗ Dk (A, F, α0 , β 0 ) < (µF )2k , then (µ2k )∗ Dk (A + g, E, α, β) ≤ (µ2k )∗ Dk (A + g, F, α, β) + µ2k (E 2k \ F 2k ) < (µE)2k . (g) If A is empty, or contains only the constant function 0 with value 0, this is trivial. Now (f) tells us that {g} = {0} + g is stable for every g ∈ L0 , and from (d) it follows that any finite subset of L0 is stable. (h) Writing |A| for {|f | : f ∈ A}, then Dk (|A|, E, α, β) ⊆ Dk (A, E, α, β) ∪ Dk (−A, E, α, β) = Dk (A ∪ (−A), E, α, β)

566


465C

whenever 0 ≤ α < β. So (d) and (e) tell us that in this case there is some k such that Dk (|A|, E, α, β) has outer measure less than (µE)2k . (i) Let E ∈ Σ, α < β be such that 0 < µE < ∞. Set E0 = {x : x ∈ E, g(x) = 0},

E1 = {x : x ∈ E, g(x) > 0},

E2 = {x : x ∈ E, g(x) < 0}. (i) Suppose that µE0 > 0. Then D1 (A × g, E, α, β) does not meet E0 , so µ∗ D1 (A × g, E, α, β) < µE. (ii) Suppose that µE1 > 0. Let η > 0 be such that max(α,

α ) 1+η

= α0 < β 0 = min(β,

β ). 1+η

Let γ > 0 be such that µF > 0, where F = {x : x ∈ E, γ ≤ g(x) ≤ γ(1 + η)}. If x ∈ F then f (x)g(x) ≤ α =⇒ f (x) ≤

α g(x)

≤

α0 , γ

f (x)g(x) ≥ β =⇒ f (x) ≥

β g(x)

≥

β0 . γ

So Dk (A × g, F, α, β) ⊆ Dk (A, F,

α0 β 0 , ) γ γ

for every k. Now, because A is stable, there is some k ≥ 1 such that (µ2k )∗ Dk (A, F,

α0 β 0 , ) γ γ

< (µF )2k ,

and in this case (µ2k )∗ Dk (A × g, E, α, β) < (µE)2k , just as in the argument for (f) above. (iii) If µE2 > 0, then we know from (e) that −A is stable, so (ii) tells us that there is a k ≥ 1 such that (µ2k )∗ Dk ((−A) × (−g), E, α, β) < (µE)2k . Since one of these three cases must occur, and since E, α and β are arbitrary, A × g is stable. (j) The product measures µ2k , µ ˆ2k and µ ˜2k are all the same (251Wn), so this follows immediately from the definition in 465B. (k) Let h be a Radon-Nikod´ ym derivative of ν with respect to µ (234B). Suppose that 0 < νE < ∞ and α < β. Then there is an F ∈ Σ such that F ⊆ E ∩ dom h, h(x) > 0 for every x ∈ F , and 0 < µF < ∞. There is a k ≥ 1 such that (µ2k )∗ Dk (A, F, α, β) < (µF )2k , that is, there is a W ⊆ F 2k \ Dk (A, F, α, β) Q2k−1 ˜ such that µ2k W > 0. In this case, ν 2k W > 0. P P Set h(w) = i=0 h(w(i)) for w ∈ (dom h)2k . Then ˜ with respect to µ2k (253I), and h(w) ˜ ν 2k is the indefinite integral of h > 0 for every w ∈ F 2k . Q Q Since 2k 2k ∗ 2k W ⊆ E \ Dk (A, E, α, β), (ν ) Dk (A, E, α, β) < (νE) ; as E, α and β are arbitrary, A is stable with respect to ν. Since ν is the completion of its restriction ν¹Σ (234Db), A is also stable with respect to ν¹Σ, by (j). (l) Write B = {hf : f ∈ A}. Suppose that 0 < µE < ∞ and α < β. If either α < h(γ) for every γ ∈ R or h(γ) < β for every γ ∈ R, µ∗ D1 (B, E, α, β) = µ∅ = 0 < µE. Otherwise, because h is continuous, the Intermediate Value Theorem tells us that there are α0 < β 0 such that α < h(α0 ) < h(β 0 ) < β. In this case Dk (B, E, α, β) ⊆ Dk (A, E, α0 , β 0 ) for every k. Because A is stable, there is some k ≥ 1 such that (µ2k )∗ Dk (A, E, α0 , β 0 ) < (µE)2k , so that (µ2k )∗ Dk (B, E, α, β) < (µE)2k . As E, α and β are arbitrary, B is stable. (m) follows at once. 465D Now for the first result connecting the notion of ‘stable’ set with the concerns of this chapter.

465F

Stable sets

567

Proposition Let (X, Σ, µ) be a complete locally determined measure space, and A ⊆ RX a stable set. (a) A ⊆ L0 = L0 (Σ) (that is, every member of A is Σ-measurable). (b) If {f (x) : f ∈ A} is bounded for each x ∈ X, then A is relatively compact in L0 for the topology of pointwise convergence. proof (a) ?? Suppose, if possible, that there is a non-measurable f ∈ A. Then there is an α ∈ R such that D0 = {x : f (x) > α} ∈ / Σ. Because µ is locally determined, there is an F0 ∈ Σ such that µF0 < ∞ and D0 ∩ F0 ∈ / Σ. Let F1 ⊆ F0 be a measurable envelope of D0 ∩ F0 (132Ee). Then S D0 ∩ F1 = D0 ∩ F0 is not measurable; because µ is complete, F1 \ D0 cannot be negligible. Now D0 = n∈N {x : f (x) ≥ α + 2−n }, so there is some β > α such that D1 = F1 ∩ {x : f (x) ≥ β} is not negligible. Let E be a measurable envelope of D1 . Then, setting P = {x : x ∈ E, f (x) ≤ α}, Q = {x : x ∈ E, f (x) ≥ β} we have µ∗ P = µ∗ Q = µE > 0. Now suppose that k ≥ 1. Then Dk ({f }, E, α, β) ⊇ (P × Q)k , so (µ2k )∗ Dk ({f }, E, α, β) = (µ∗ P · µ∗ Q)k = (µE)2k (251Wm again). Since this is true for every k, {f } is not stable, and (by 465Ca) A cannot be stable; which contradicts our hypothesis. X X (b) Because {f (x) : x ∈ A} is bounded for each x, A, the closure of A in RX , is compact for the topology of pointwise convergence. But A is stable, by 465Cb, so is included in L0 , by (a). 465E The topology Ts (L2 , L2 ) Some of the arguments below will rely on ideas of compactness in function spaces. There are of course many ways of expressing the method, but a reasonably accessible one uses the Hilbert space L2 , as follows. Let (X, Σ, µ) be any measure space. Then L2 = L2 (µ) is a Hilbert space with a corresponding weak topology Ts (L2 , L2 ) defined by the functionals u 7→ (u|v) for v ∈ L2 . In the 2 2 present section it will be more convenient to regard this as a topology Ts (L L2 = L2 (µ) R , L ) on the space 2 of square-integrable real-valued functions, defined by the functionals f 7→ f × g for g ∈ L . The essential fact we need is that norm-bounded sets are relatively weakly compact. In L2 , this is because Hilbert spaces are reflexive (4A4Ka). In L2 , given an ultrafilter F containing a k k2 -bounded set B ⊆ L2 , v = limf →F f • must be defined in L2 for Ts (L2 , L2 ), and now there is a g ∈ L2 such that g • = v; in which case limf →F

R

f × h = limf →F (f • |h• ) = (g • |h• ) =

R

g×h

for every h ∈ L2 . Note that we are free to take g to be a Σ-measurable function with domain X (241Bk). 465F Lemma Let (X, Σ, µ) be a measure space, and B ⊆ L2 = L2 (µ) a k k2 -bounded set. Suppose that h ∈ L2 belongs to the closure of B for Ts (L2 , L2 ). Then for any δ > 0, k ≥ 1 the set [ W = {w : w ∈ X k , w(i) ∈ dom f ∩ dom h f ∈B

and f (w(i)) ≥ h(w(i)) − δ for every i < k} is µk -conegligible in X k . proof (a) Since completing the measure µ does not change the space L2 (244Xa) nor the product measure µk (251Wn), we may suppose that µ is complete. (b) The first substantive fact to note is that there is a sequence hfn in∈N in B converging to h for Ts = Ts (L2 , L2 ). P P Setting C = {f • : f ∈ B}, C is a bounded set in L2 = L2 (µ) and h• belongs to the 2 2 Ts (L , L )-closure of C. But L2 , being a normed space, is angelic in its weak topology (462D), and C is relatively compact in L2 , so there is a sequence in C converging to h• . We can represent this sequence as hfn• in∈N where fn ∈ B for every n, and now hfn in∈N → h for Ts . Q Q (c) The second T component of the proof is the following simple idea. Suppose that hEn in∈N is a sequence in Σ such that n∈I En is negligible for every infinite set I ⊆ N. For m ≥ 1, I ⊆ N set T Vm (I) = n∈I {w : w ∈ X m , ∃ i < m, w(i) ∈ En }. Then µm Vm (I) = 0 for every infinite I ⊆ N. P P Induce on m. For m = 1 this is just the original hypothesis on hEn in∈N . For the inductive step to m + 1, identify µm+1 with the product of µm and µ, and observe that

568


465F

Vm+1 (I)−1 [{x}] = {w : (w, x) ∈ Vm+1 (I)} = Vm ({n : n ∈ I, x ∈ / En }) for every x ∈ X. Now, setting Ix = {n : n ∈ I, x ∈ / En }, F = {x : Ix is finite}, we have S T F = r∈N n∈I\r En , so F is negligible, while if x ∈ / F then Vm (Ix ) is negligible, by the inductive hypothesis. But this means that almost every horizontal section of Vm+1 (I) is negligible, and Vm+1 (I), being measurable, must be negligible, by Fubini’s theorem (252F). Thus the induction proceeds. Q Q (d) Now let us return to the main line of the argument T from (b). For each n ∈ N, set En =T{x : x ∈ dom fn ∩ dom h, fn (x) < h(x) − δ}. If I ⊆ N is infinite, then n∈I En is negligible. P P Setting G = n∈I En , R R µG is finite (becauseR δχEn R≤ |h − fn | a.e., so µEn < ∞ for every n) and G fn ≤ G h − δµG for every n ∈ I. But limn→∞ G fn = G h and I is infinite, so δµG ≤ inf n∈I | By (c), it follows that V = T

T

n∈N {w

R

h− G

R

G

fn | = 0. Q Q

: w ∈ X k , ∃ i < k, w(i) ∈ En }

is negligible. But if we set Y = n∈N dom fn ∩ dom h, Y is a conegligible subset of X, Y k is a conegligible subset of X k , and Y k \ V ⊆ W , so W is conegligible, as required. 465G Theorem Let (X, Σ, µ) be a semi-finite measure space, and A ⊆ L0 = L0 (Σ) a stable set of measurable functions. Let Tp and Tm be the topologies of pointwise convergence and convergence in measure, as in §463. Then the identity map from A to itself is (Tp , Tm )-continuous. proof ?? Suppose, if possible, otherwise. (a) We must have an f0 ∈ A, a set F ∈ Σ of finiteR measure, and an ² > 0 such that for every Tp neighbourhood U of f0 there is an f ∈ A ∩ U such that F min(1, |f − f0 |)dµ ≥ ². Set B = {min(χF, |f − f0 |) : f ∈ A}. Then B = {χF − (χF − |f − f0 |)+ : f ∈ A}

R is stable, by 465Cf, 465Ch, 465Cm and 465Ce, used repeatedly. Setting B 0 = {f : f R∈ B, f ≥ ²}, B 0 is again stable (465Ca). Our hypothesis is that f0 is in the Tp -closure of A0 = {f : f ∈ A, min(χF, |f −f0 |) ≥ ²}; since f 7→ min(χF, |f − f0 |) is Tp -continuous, 0 belongs to the Tp -closure of the image of A0 under this map, which is B 0 . (b) Let F be an ultrafilter on B 0 which Tp -converges to 0. Because B 0 is k k2 -bounded (since 0 ≤ f ≤ χF for every f ∈ B 0 ), F also has a Ts (L2 , L2 )-limit h say, as noted in 465E; and we can suppose that h is measurable and defined everywhere. We must have

R

F

h = limf →F

R

F

f ≥ ² > 0.

So there is a δ > 0 such that E = {x : x ∈ F, h(x) ≥ 3δ} has measure greater than 0. (c) Because B 0 is stable, there must be some k ≥ 1 such that (µ2k )∗ Dk (B 0 , E, δ, 2δ) < (µE)2k . Let W ⊆ E 2k \ Dk (B 0 , E, δ, 2δ) be a measurable set of positive measure. By Fubini’s theorem, there must be x0 , . . . , xk−1 such that µk V is defined and greater than 0, where V = {v : v ∈ X k , (x0 , v(0), x1 , v(1), . . . , xk−1 , v(k − 1)) ∈ W }. Set C = {f : f ∈ B 0 , f (xi ) ≤ δ for every i < k}; then C ∈ F , because F → 0 for Tp . Accordingly h belongs to the Ts (L2 , L2 )-closure of C. But now 465F tells us that there must be a v ∈ V and an f ∈ C such that f (v(i)) ≥ h(v(i)) − δ for every i < k. Consider w = (x0 , v(0), . . . , xk−1 , v(k − 1)). We know that w ∈ W (because v ∈ V ), so, in particular, w ∈ E 2k and h(w(i)) ≥ 3δ for every i < 2k; accordingly f (w(2i + 1)) = f (v(i)) ≥ h(v(i)) − δ ≥ 2δ

465H

Stable sets

569

for every i < k. On the other hand, f (w(2i)) = f (xi ) ≤ δ for every i < k, because f ∈ C. But this means that f witnesses that w ∈ Dk (B 0 , E, δ, 2δ), which is supposed to be disjoint from W . X X This contradiction shows that the theorem is true. 465H We shall need some interesting and important general facts concerning powers of measures. I start with an important elaboration of the strong law of large numbers. Theorem Let (X, Σ, µ) be any probability space. For n ∈ N, write Λn for the domain of the product measure µn . For w ∈ X N , k ≥ 1, n ≥ 1 write νwk for the probability measure with domain PX defined by writing 1 k

νwk (E) = #({i : i < k, w(i) ∈ E}) n for E ⊆ X, and νwk for the corresponding product measure on X n . R n exists, and is Then whenever n ≥ 1 and f : X n → R is bounded and Λn -measurable, limk→∞ f dνwk R n N N equal to f dµ , for µ -almost every w ∈ X .

proof (a) For k ≥ n and w ∈ X N , set hk (w) =

(k−n)! k!

P π:n→k is injective

Setting M = supv∈X n |f (v)|, we have Z ¯ ¯ ¯ (k−n)! n ¯ ¯hk (w) − f dνwk =¯ k!

≤M

X

f (wπ).

f (wπ) −

π:n→k is injective

¡ (k−n)! k!

−

1 ¢ M¡ + n kn kn k

−

1 kn

X

¯ f (wπ)¯

π:n→k

k! ¢ (k−n)!

→0

as k → ∞, for every w ∈ X N . (b) Write Λ for the domain of µN , and for k ≥ n set Tk = {W : W ∈ Λ, wπ ∈ W whenever w ∈ W and π ∈ Sk }, where Sk is the set of bijections ψ : N → N such that ψ(i) = i for i ≥ k. Then hTk ik≥n is a non-increasing sequence of σ-subalgebras of Λ. For any injective function π : n → k there are just (k − n)! extensions of π to a member of Sk . So 1 P hk (w) = π∈Sk f (wπ¹n) k!

for every w. Observe that hk (wψ) = hk (w) for every ψ ∈ Sk , w ∈ X N , so hk is Tk -measurable. (Of course we need to look back at the definition of hk to confirm that it is Λ-measurable.) (c) For any k ≥ n, hk is a conditional expectation of hn on Tk . P P If W ∈ Tk , then R R R P P 1 1 h (w)dw = π∈Sk W f (wπ¹n)dw = π∈Sk W g(wπ)dw W k k!

k!

where g(w) = f (w¹n) for w ∈ X N . Now observe that for every π ∈ Sk the map R w 7→ wπ is aRmeasure space N automorphism of X which leaves W unchanged, because W ∈ T ; so that by k R R R RW g(wπ)dw R = W g(w)dw, R 235Ic. So W hk = W g. But (since W ∈ Tk ⊆ Tn ) W hn is also equal to W g, and W hk = W hn . As W is arbitrary, hk is a conditional expectation of hn on Tk . Q Q (d) By the reverse martingale theorem (275K), h∞ (w) = limk→∞ hk (w) is defined for almost every R n w ∈ X N . Accordingly limk→∞ f dνwk is defined for almost every w. R T n (e) To see that the limit is f dµ , observe that if W ∈ T∞ = k∈N Tk then µN W must be either 0 or N 1. P P Set γ = µN W . Let ² > 0. Then there is a V ∈ N Σ (notation: 465Ad) such that µN (W 4V ) ≤ ² (254Fe). There is some k such that V is determined by coordinates in k. If we set π(i) = 2k − i for i < 2k, i for i ≥ 2k, then V 0 = {wπ : w ∈ V } is determined by coordinates in 2k \ k, so µN (V ∩ V 0 ) = (µN V )2 .

570


465H

On the other hand, because W ∈ T2k , the measure space automorphism w 7→ wπ does not move W , and µN (W \ V 0 ) = µN (W \ V ). Accordingly γ = µN (W ∩ W ) ≤ µN (V ∩ V 0 ) + 2µN (W \ V ) ≤ (µN V )2 + 2² ≤ (γ + ²)2 + 2². As ² is arbitrary, γ ≤ γ 2 and γ ∈ {0, 1}. Q Q (e) Now 275K tells us that h∞ is T∞ -measurable, therefore essentially constant, to R Rand must be equal R its expectation almost everywhere. But, setting W = X N in (c), we see that hk = f (w¹n)dw = f dµn for every k, so limk→∞

R

n f dνwk = h∞ (w) =

R

f dµn

for almost every w, as claimed. 465I Now for a string of lemmas, working towards the portmanteau Theorem 465M. The first is elementary. Lemma Let X be a set, and Σ a σ-algebra of subsets of X. For w ∈ X N , k ≥ 1, n ≥ 1 write νwk for the probability measure with domain PX defined by writing 1 k

νwk (E) = #({i : i < k, w(i) ∈ E}) N I for E ⊆ X. Then for any k ∈ N and any set I, w 7→ νwk (W ) is c N Σ-measurable (notation: 465Ad) for N every W ∈ c I Σ. N I proof Write W for the set of subsets W of X I such that w 7→ νwk (W ) is c N Σ-measurable. Then X I ∈ W, S W 0 \W ∈ W whenever W , W 0 ∈ W and W ⊆ W 0 , and n∈N Wn ∈ W whenever hWn in∈N is a non-decreasing sequence in W. Write V for the set of Σ-cylinders in X I , that is, sets expressible in the form {v : v(i) ∈ Ei for every i ∈ J}, where J ⊆ I is finite and Ei ∈ Σ for i ∈ J. Then V ⊆ W. P P If J ⊆ I is finite and Ei ∈ Σ for i ∈ J, then 1 Pk−1 w 7→ νwk Ei = j=0 χEi (w(j)) k

N is c N -measurable for every i ∈ J. So I w 7→ νwk {v : v(i) ∈ Ei ∀ i ∈ J} =

Q i∈J

νwk Ei

is also measurable. Q Q Because V ∩ V 0 ∈ V for all V , V 0 ∈ V, the Monotone Class Theorem (136B) tells us that W must include N the σ-algebra generated by V, which is c I Σ. 465J The next three lemmas are specifically adapted to the study of stable sets of functions. Lemma Let (X, Σ, µ) be a probability space. For any n ∈ N, W ⊆ X n I say that W is symmetric if wπ ∈ W whenever w ∈ W and π : n → n is a bijection. Give each power X n its product measure µn . (a) Suppose that for each n ≥ 1 we have a measurable set Wn ⊆ X n , and that Wm+n ⊆ Wm × Wn for all m, n ≥ 1, identifying X m+n with X m × X n . Then limn→∞ (µn Wn )1/n is defined and equal to δ = inf n≥1 (µn Wn )1/n . (b) Now suppose that each Wn is symmetric. Then there is an E ∈ Σ such that µE = δ and E n \ Wn is negligible for every n ∈ N. (c) Next, let hDn in≥1 be a sequence of sets such that Dn ⊆ X n is symmetric for every n ≥ 1, whenever 1 ≤ m ≤ n, v ∈ Dn then v¹m ∈ Dm . Then δ = limn→∞ ((µn )∗ Dn )1/n is defined and there is an E ∈ Σ such that µE = δ and (µn )∗ (Dn ∩ E n ) = (µE)n for every n ∈ N. proof (a) For any η > 0, there is an m ≥ 1 such that µm Wm ≤ (δ + η)m . If n = mk + i, where k ≥ 1 and i < m, then (identifying X n with (X m )k × X i ) Wn ⊆ (Wm )k × X i , so

465J

Stable sets

571

µn Wn ≤ (δ + η)mk ≤ γ(δ + η)mk+i , where γ = maxi
1 i ). δ+η

So lim supn→∞ (µn Wn )1/n ≤ (δ + η) lim supn→∞ γ 1/n = δ + η. As η is arbitrary, δ ≤ lim inf n→∞ (µn Wn )1/n ≤ lim supn→∞ (µn Wn )1/n ≤ δ, and limn→∞ (µn Wn )1/n = δ. (b) It is enough to consider the case δ > 0. (i) Consider the family V of sequences hVn in≥1 such that for each n ≥ 1, Vn is a symmetric measurable subset of X n and µn Vn ≥ δ n , if 1 ≤ m ≤ n then v¹m ∈ Vm for every v ∈ Vn . Observe that W = hWn in≥1 ∈P V. Order V by saying that hVn in≥1 ≤ hVn0 in≥1 if Vn ⊆ Vn0 for every n. For −n n V)= ∞ V = hVnT in≥1 in V, set θ(V µ Vn . Any non-increasing sequence hhVkn in≥1 ik∈N in V has a lower n=1 2 V ) = θ(W W 0) bound h k∈N Vkn in≥1 in V, so there must be a W 0 = hWn0 in≥1 ∈ V such that W 0 ≤ W and θ(V 0 0 whenever V ∈ V and V ≤ W ; that is, whenever hVn in≥1 ∈ V and Vn ⊆ Wn for every n ≥ 1, then µn Vn = µn Wn0 for every n. (x)

(x)

0 }. Then Vn is measurable for almost every x; (ii) For x ∈ X, n ≥ 1 set Vn = {w : (x, w) ∈ Wn+1 (x) let X1 ⊆ X be a conegligible set such that Vn is measurable for every x ∈ X1 and every n ≥ 1. Every (x) (x) (x) (x) Vn is symmetric, and if 1 ≤ m ≤ n and v ∈ Vn then v¹m ∈ Vm . It follows that if m, n ≥ 1 then Vm+n (x) (x) becomes identified with a subset of Vm × Vn . (x) (x) From (a) we see that δx = limn→∞ (µn Vn )1/n is defined for every x ∈ X1 . The map x 7→ µn Vn : X1 → [0, ∞[ is measurable for each n, by Fubini’s theorem (252D), so x 7→ δx is also measurable. Since (x) Vn ⊆ Wn0 ⊆ Wn for every x and n, δx ≤ δ for every x ∈ X1 .

(iii) Set E = {x : x ∈ X1 , δx = δ}. Then µE ≥ δ. P P?? Otherwise, there is some β < δ such that µF < β, where F = {x : x ∈ X1 , δx ≥ β}. Now X1 \ F = {x : x ∈ X1 , lim (µn Vn(x) )1/n < β} n→∞ [ \ ⊆ {x : x ∈ X1 , µn Vn(x) ≤ β n }, m∈N n≥m

so there is some m ∈ N such that µH ≥ 1 − β, where T (x) H = n≥m {x : x ∈ X1 , µn Vn ≤ β n }. Set γn = µn Wn0 ≥ δ n for each n. Then, for any n ≥ m, Z n+1 0 γn+1 = µ Wn+1 = µn Vn(x) µ(dx) X1 Z Z n (x) = µ Vn µ(dx) + µn Vn(x) µ(dx) ≤ β n + βγn H

X1 \H

(x) Vn

because ⊆ Wn0 for every x, and µ(X1 \ H) ≤ β. An easy induction shows that γm+k ≤ kβ m+k−1 + β k γm for every k ∈ N. But this means that δ k = δ −m δ m+k ≤ δ −m γm+k ≤ β k δ −m (kβ m−1 + γm ) for every k; setting η = (δ − β)/β > 0, k(k−1) η 2

≤ (1 + η)k =

¡ δ ¢k β

≤ δ −m (kβ m−1 + γm )

572


465J

for every k, which is impossible. X XQ Q (x)

(x)

(iv) Next, for any x ∈ E, V (x) = hVn in≥1 ∈ V and V (x) ≤ W 0 , so µn (Wn0 \ Vn ) = 0 for every n ≥ 1. 0 This means that, given n ≥ 1, every vertical section of (E × Wn0 ) \ Wn+1 (regarded as a subset of X × X n ) 0 0 is negligible; so (E × Wn ) \ Wn+1 is negligible. We are assuming that δ > 0, so (x)

E ⊆ {x : x ∈ X1 , V1

6= ∅} ⊆ {x : {w : (x, w) ∈ W20 } 6= ∅} ⊆ W10 .

Now a simple induction shows that E n \ Wn0 is negligible for every n ≥ 1, so that E n \ Wn is negligible for every n, and we have an appropriate E. (Of course µE = δ exactly, because (µE)n ≤ µWn for every n.) (c) For each n ∈ N let Vn be a measurable envelope of Dn in X n . Define hWn in≥1 inductively by saying W1 = V1 , Wn+1 = {w : w ∈ X n+1 , wπ¹n ∈ Wn , wπ ∈ Vn+1 for every bijection π : n + 1 → n + 1} for each n ≥ 1. Then an easy induction on n shows that Wn is measurable and symmetric and that Dn ⊆ Wn ⊆ Vn , so that Wn is a measurable envelope of Dn and µn Wn = (µn )∗ Dn . Now hWn in≥1 satisfies the hypotheses of (b), so δ = limn→∞ (µn Wn )1/n = limn→∞ ((µn )∗ Dn )1/n is defined and there is a set E ∈ Σ, of measure δ, such that E n \ Wn is negligible for every n; but this means that (µn )∗ (E n ∩ Dn ) = µn (E n ∩ Wn ) = δ n for every n, as required. 465K Lemma Let (X, Σ, µ) be a complete probability space, and A ⊆ [0, 1]X a stable set. Suppose that R N ² > 0 is such that f dµ ≤ ²R2 for every f ∈ A. Then there are an n ≥ 1 and a W ∈ c n Σ (notation: 465Ad) and a γ > µn W such that f dν ≤ 3² whenever f ∈ A and ν is a probability measure on X with domain including Σ such that ν n W ≤ γ. S proof (a) For n ∈ N write Cñ = f ∈A {x : f (x) ≥ ²}n . Then hCñ in∈N satisfies the conditions of 465Jc, so n ∗ n n ˜ δ = limn→∞ ((µn )∗ Cñ )1/n is defined, and there is an E ∈ Σ such that S µE = δ and (µ ) (E ∩ nCn ) = δ X for every n ∈ N. Now, for B ⊆ [0, 1] and n ∈ N, write Cn (B) = f ∈B {x : x ∈ E, f (x) ≥ ²} , so that Cn (A) = E n ∩ Cñ , and (µn )∗ Cn (A) = δ n for every n. For any B ⊆ [0, 1]X , hCn (B)in∈N also satisfies the conditions of 465Jc, and δB = limn→∞ ((µn )∗ Cn (B))1/n is defined; we have δA = δ. (b) If B, B 0 ⊆ [0, 1]X , then (µn )∗ Cn (B) ≤ (µn )∗ Cn (B ∪ B 0 ) ≤ (µn )∗ Cn (B) + (µn )∗ Cn (B 0 ) for every n, so δB ≤ δB∪B 0 ≤ max(δB , δB 0 ). It follows that if G = {G : G ⊆ [0, 1]X is Tp -open, δG∩A < δ}, where Tp is the usual topology of [0, 1]X , no finite subfamily of G can cover A. Accordingly, since the Tp -closure A of A is Tp -compact, there is an h ∈ A such that δG∩A = δ for every Tp -open set G containing h. R (c) At this point recall Rthat every function in A is measurable (465Cb, 465Da) and that : A → [0, 1] is Tp -continuous (465G). So h ≤ ²2 and µ{x : h(x) ≥ ²} ≤ ². ?? Suppose, if possible, that δ > ². Then there is some η > 0 such that µF > 0, where F = {x : x ∈ E, h(x) < ² − η}. For k ∈ N, u ∈ F k set Gu = {f : f ∈ [0, 1]X , f (u(i)) < ² − η for every i < k}. Then Gu is an open neighbourhood of h, so δGu ∩A = δ and (µk )∗ (Ck (Gu ∩ A)) ≥ δ k . But because F ⊆ E and Ck (Gu ∩ A) ⊆ E k , this means that (µk )∗ (F k ∩ Ck (Gu ∩ A)) = (µF )k . For u, v ∈ X k write u#v for (u(0), v(0), u(1), v(1), . . . , u(k − 1), v(k − 1)) ∈ X 2k . Then

465L

Stable sets

573

Ck (Gu ∩ A) ∩ F k ⊆ {v : u#v ∈ Dk (A, F, ² − η, ²)} for any u ∈ F k . So (µk )∗ {v : u#v ∈ Dk (A, F, ² − η, ²)} = (µF )k for every u ∈ F k . But this means that (µ2k )∗ Dk (A, F, ² − η, ²) = (µF )2k . Since this is so for every k ≥ 1, A is not stable. X X N (d) Thus δ ≤ ². There is therefore some n ≥ 1 such that (µn )∗ Cñ < (2²)n . Let W ∈ c n Σ be a measurable envelope of Cñ , and try γ = (2²)n . If ν is any probability measure on X with domain including Σ such that ν n W ≤ γ, then for any f ∈ A we have {x : f (x) ≥ ²}n ⊆ Cñ ⊆ W , (ν{x : f (x) ≥ ²})n ≤ ν n W ≤ (2²)n , R so that ν{x : f (x) ≥ ²} ≤ 2². As 0 ≤ f (x) ≤ 1 for every x ∈ X, f dν ≤ 3², as required. 465L Lemma (Talagrand 87) Let (X, Σ, µ) be a complete probability space, and RA ⊆ [0, 1]X R a set which is not stable. Then there are measurable functions h0 , h1 : X → [0, 1] such that h0 dµ < h1 dµ ˜ k = 1 for every k ≥ 1, where and (µ2k )∗ D ˜k = D

[

{w : w ∈ X 2k , f (w(2i)) ≤ h0 (w(2i)),

f ∈A

f (w(2i + 1)) ≥ h1 (w(2i + 1)) for every i < k}.

(∗)

proof The proof divides into two cases. case 1 Suppose that there is an ultrafilter F on A such that the Tp -limit g0 of F is not measurable. R R R R Let h00 , h01 be measurable functions such that h00 ≤ g0 ≤ h01 and h00 = g0 , h01 = g0 (133Ja). Then R R R δ = 15 h01 − h00 > 0 (133Jd). Set h0 = h00 + 2δχX, h1 = h01 − 2δχX, so that h0 < h1 . Set Q0 = {x : x ∈ X, g0 (x) ≤ h0 (x) − δ}. Then µ∗ Q0 = 1. P P?R? Otherwise, there is a non-negligible R measurable set F ⊆ X \ Q0 . But in this case h00 + δχF ≤ g0 and g0 > h00 , which is impossible. X XQ Q Similarly, µ∗ Q1 = 1, where Q1 = {x : g0 (x) ≥ h1 (x) + δ}. If k ≥ 1, x0 , . . . , xk−1 ∈ Q0 and y0 , . . . , yk−1 ∈ Q1 , then there is an f ∈ A such that |f (xi ) − g0 (xi )| ≤ δ, |f (yi ) − g0 (yi )| ≤ δ for every i < k. But this means that f (xi ) ≤ h0 (xi ) and f (yi ) ≥ h1 (yi ) for every i < k, ˜ k . Thus and (x0 , y0 , . . . , xk−1 , yk−1 ) ∈ D ˜ k ⊇ Q0 × Q1 × Q0 × Q1 × . . . × Q0 × Q1 , D which has full outer measure, by 254L. As k is arbitrary, we have found appropriate h0 , h1 in this case. case 2 Now suppose that for every ultrafilter F on A, the Tp -limit of F is measurable. (i) We are supposing that A is not stable, so there are E ∈ Σ and α < β such that µE > 0 and (µ2k )∗ Dk (A, E, α, β) = (µE)2k for every k ≥ 1. The first thing to note is that if I is the set of those B ⊆ A for which there is some k ∈ N such that (µ2k )∗ Dk (B, E, α, β) < (µE)2k , then I is an ideal of subsets of A. P P Of course ∅ ∈ I and B ∈ I whenever B ⊆ B 0 ∈ I. Also 465Cc tells us that, if B ∈ I, then limk→∞

1 (µ2k )∗ Dk (B, E, α, β) (µE)2k

= 0. It follows easily (as in the proof of 465Cd) that B ∪ B 0 ∈ I for all

B, B 0 ∈ I. Q Q

(ii) I is a proper ideal of subsets of A (by the choice of E, k, α and β), so there is an ultrafilter F on A such that F ∩ I = ∅. Let g0 be the Tp -limit of F. Then g0 is measurable. Set δ = 31 (β − α)µE > 0, and define h0 , h1 by setting h0 (x) = α if x ∈ E, = g0 (x) + δ if x ∈ X \ E, h1 (x) = β if x ∈ E, = g0 (x) − δ if x ∈ X \ E. Then

574


R

h1 −

R

465L

h0 ≥ (β − α)µE − 2δµ(X \ E) > 0.

(iii) We shall need to know a little more about sets of the form Dk (B, E, α, β) for B ∈ F. In fact, if B ⊆ A and B ∈ / I, then for any finite sets I and J (µI × µJ )∗ DI,J (B, E, α, β) = (µE)#(I)+#(J) , where DI,J (B, E, α, β) is S I J f ∈B {(u, v) : u ∈ E , v ∈ E , f (u(i)) ≤ α for i ∈ I, f (v(j)) ≥ β for j ∈ J}. P P We may suppose that I = k and J = l where k, l ∈ N. Take m = max(1, k, l). Then we have an inversemeasure-preserving map φ : X 2m → X I × X J defined by saying that φ(w) = (u, v) where u(i) = w(2i) for i < k and v(i) = w(2i + 1) for i < l. ?? If (µI × µJ )∗ DI,J (B, E, α, β) < (µE)k+l , there is a non-negligible measurable set V ⊆ (E I × E J ) \ DI,J (B, E, α, β). Now φ−1 [V ] is non-negligible and depends only on coordinates in {2i : i < k} ∪ {2i + 1 : i < l}, so µ2m (E 2m ∩ φ−1 [V ]) = µ2m (φ−1 [V ]) · (µE)2m−k−l > 0. But φ−1 [V ] ∩ Dm (B, E, α, β) = ∅, so (µ2m )∗ Dm (B, E, α, β) < (µE)2m , and B ∈ I. X X µJ )∗ DI,J (B, E, α, β) = (µE)k+l , as required. Q Q

So (µI ×

˜ k < 1, where D ˜ k is defined from h0 and h1 (iv) ?? Suppose, if possible, that k ≥ 1 is such that (µ2k )∗ D by the formula (*) in the statement of the lemma. Let W ⊆ X 2k be a measurable set of positive measure ˜ k . For I, J ⊆ k write WIJ for disjoint from D {w : w ∈ W, w(2i) ∈ E for i ∈ I, w(2i) ∈ / E for i ∈ k \ I, w(2i + 1) ∈ E for i ∈ J, w(2i + 1) ∈ / E for i ∈ k \ J}. Then there are I, J ⊆ K such that µ2k WIJ > 0. We can identify X 2k with X I × X J × X k\I × X k\J , matching any w ∈ X 2k with (w0 , w1 , w2 , w3 ) where w0 (i) = w(2i) for i ∈ I, w1 (i) = w(2i + 1) for i ∈ J, w2 (i) = w(2i) for i ∈ k \ I, w3 (i) = w(2i + 1) for i ∈ k \ J. ˜ for the image of WIJ under this matching. The condition WIJ ∩ D ˜ k = ∅ translates into Write W ˜ , f ∈ A, (†) whenever (w0 , w1 , w2 , w3 ) ∈ W either there is an i ∈ I such that f (w0 (i)) > α or there is an i ∈ J such that f (w1 (i)) < β or there is an i ∈ k \ I such that f (w2 (i)) > g0 (w2 (i)) + δ or there is an i ∈ k \ J such that f (w3 (i)) < g0 (w3 (i)) − δ. (v) By Fubini’s theorem, applied to (X I × X J ) × (X k\I × X k\J ), we can find w2 ∈ X k\I , w3 ∈ X k\J ˜ }. Set such that (µI × µJ )(V ) is defined and greater than 0, where V = {(w0 , w1 ) : (w0 , w1 , w2 , w3 ) ∈ W B = {f : f ∈ A, |f (w2 (i)) − g0 (w2 (i))| ≤ δ for i ∈ k \ I, |f (w3 (i)) − g0 (w3 (i))| ≤ δ for i ∈ k \ J}. Then B ∈ F, because F → g0 for Tp . So (µI × µJ )∗ DI,J (B, E, α, β) = (µE)#(I)+#(J) , by (iii) above. Since WIJ ⊆ W ⊆ E 2k , V is included in E I × E J and meets DI,J (B, E, α, β); that is, there are f ∈ B and (w0 , w1 ) ∈ V such that f (w0 (i)) ≤ α for i ∈ I and f (w1 (i)) ≥ β for i ∈ J. But because f ∈ B we also have f (w2 (i)) ≤ g0 (w2 (i)) + δ for i ∈ k \ I and f (w3 (i)) ≥ g0 (w3 (i)) − δ for i ∈ k \ J; which contradicts the list in (†) above. X X ˜ k = 1 for every k, and in this case also we have an appropriate pair h0 , h1 . (vi) Thus (µ2k )∗ D

465M

Stable sets

575

465M Theorem (Talagrand 82, Talagrand 87) Let (X, Σ, µ) be a complete probability space, and A a non-empty uniformly bounded set of real-valued functions defined on X. Then the following are equiveridical: (i) A is stable. R 1 Pk−1 (ii) Every function in A is measurable, and limk→∞ supf ∈A | f | = 0 for almost every i=0 f (w(i)) − k

w ∈ X N. (iii) Every function in A is measurable, and for every ² > 0 there are a finite subalgebra T of Σ in which every atom of T is non-negligible and a sequence hhk ik≥1 of measurable functions on X N such that ¯ 1 Pk−1 ¯¯ ¯ hk (w) ≥ supf ∈A i=0 f (w(i)) − E(f |T)(w(i)) k

for every w ∈ X N , k ≥ 1, and lim supk→∞ hk (w) ≤ ² N

for almost every w ∈ X . (Here E(f |T) is the (unique) conditional expectation of f on T.) 1 Pk−1 1 Pl−1 N (iv) limk,l→∞ supf ∈A | i=0 f (w(i)) − l i=0 f (w(i))| = 0 for almost every w ∈ X . k R 1 Pk−1 1 Pl−1 N (v) limk,l→∞ supf ∈A | i=0 f (w(i)) − i=0 f (w(i))|µ (dw) = 0. k

l

proof All the statements (i)-(v) are unaffected by translations (by constant functions) and scalar multiplications of the set A, so it will be enough to consider the case in which A ⊆ [0, 1]X . As in 465I-465H, I write νwk (E) = k1 #({i : i < k, w(i) ∈ E}) for w ∈ X N , k ≥ 1 and E ⊆ X; so for any R 1 Pr−1 function f : X → R, f dνwk = i=0 f (w(i)). k

α) If A is stable, then every function in A is measurable, by 465Da. Let ² > 0. Set (a)(i)⇒(iii)(α 1 2 η = 27 ² > 0. By 465Db, the Tp -closure A of A in RX is a Tp -compact set of measurable functions, and by 465G it is Tm -compact; because A is uniformly bounded, it must be totally bounded for the norm k k1 . So there are f0 , . . . , fm ∈ A such that for every f ∈ A there is an i ≤ m such that kf − fi k1 ≤ η. Let T0 be the 1 η

finite subalgebra of Σ generated by the sets {x : jη ≤ fi (x) < (j + 1)η} for i ≤ m and j ≤ . Then T0 may have negligible atoms, but if we absorb these into non-negligible atoms we get a finite subalgebra T of Σ such that |fi (x) − E(fi |T)(x)| ≤ η for almost every x ∈ X, every i ≤ m. (Because T is a finite algebra with non-negligible atoms, two T-measurable functions which are equal almost everywhere must be identical, and we have unique conditional expectations with respect to T.) Since kE(f |T) − E(g|T)k1 ≤ kf − gk1 for all integrable functions f and g (242Je), kf − E(f |T)k1 ≤ 3η for every f ∈ A. β ) Set A0 = {f − E(f |T) : f ∈ A}. Then A0 is stable. P (β P Suppose that µE > 0 and α < β. The set B = {E(f |T) : f ∈ A} is a uniformly bounded subset S of a finite-dimensional space of functions, so is k k∞ -compact. So there are g0 , . . . , gr ∈ B such that B ⊆ i≤r {g : kg − gi k∞ ≤ 31 (β − α)}. By 465Cf and S 465Cd, C = i≤r A − gi is stable. So there is a k ≥ 1 such that (µ2k )∗ Dk (C, E, 23 α + 13 β, 13 α + 23 β) < (µE)2k . But for every g ∈ A0 there is an h ∈ C such that kg − hk∞ ≤ 13 (β − α), so Dk (A0 , E, α, β) ⊆ Dk (C, E, 23 α + 13 β, 13 α + 23 β), (µ2k )∗ Dk (A0 , E, α, β) < (µE)2k . As E, α and β are arbitrary, A0 is stable. Q Q N 0 n c (γγ ) By R 465Ch, {|g| √ : g ∈ A } is stable. By 0465K there are an n ≥ 1, a W ∈ n Σ and a γ > µ W such that |g|dν ≤ 3 3η = ² whenever g ∈ A and ν is a probability measure on X, with domain Σ, n such that ν n W ≤ γ. Now νwk is a probability measure on X; set qk (w) = νwk (W ). Applying 465H to the n characteristic function of W , we see that limk→∞ qk (w) = µ W for almost every w ∈ X N . Also, because N W ∈ c n Σ, every qk is measurable, by 465I.

576


465M

Set hk (w) = 1 if qk (w) > γ, ² if qk (w) ≤ γ. Then every hk is measurable and limk→∞ hk (w) = ² for almost every w. N 0 R For any w ∈ X and any f ∈ A,ng = f − E(f |T)R∈ A and kgk∞ ≤ 1. So either hk (w) = 1 and certainly |g|dνwk ≤ hk (w), or hk (w) = ², νwk (W ) ≤ γ and |g|dνwk ≤ ². Thus we have ¯ ¯ R¯ 1 Pk−1 ¯¯ ¯ ¯f − E(f |T)¯dνwk ≤ hk (w) i=0 f (w(i)) − E(f |T)(w(i)) = k

N

for every w ∈ X and every f ∈ A, as required by (iii). (b)(iii)⇒(ii) & (v) Set gk (w) = supf ∈A | 0 gkl (w) = supf ∈A |

1 k

1 k

Pk−1 i=0

Pk−1 i=0

f (w(i)) −

f (w(i)) −

1 l

R

f dµ|,

Pl−1 i=0

f (w(i))|,

for w ∈ X N and k, l ≥ 1. Let ² > 0. Let T, hhk ik≥1 be as in (iii), and let E0 , . . . , Er be the atoms of T. For w ∈ X N , k ≥ 1, j ≤ r set qkj (w) = |µEj − νwk Ej |. Then for any f ∈ A, E(f |T) is expressible as P r X j=0 αj χEj where αj ∈ [0, 1] for every j (remember that A ⊆ [0, 1] ), so Z Z Z X ¯ ¯ ¯ ¯ 1 k−1 ¯ ¯ ¯ f (w(i)) − f dµ = f dνwk − f dµ¯ k

i=0

¯ ≤¯

Z

1 k

≤

Z f dνwk −

k−1 X

¯ ¯ E(f |T)dνwk ¯ + ¯

Z

Z E(f |T)dνwk −

¯ E(f |T)dµ¯

r ¯ ¯ X ¯ ¯ ¯f (w(i)) − E(f |T)(w(i))¯ + αj ¯µEj − νwk Ej ¯

i=0

j=0

≤ hk (w) +

r X

qkj (w),

j=0 l−1 X ¯ 1 k−1 ¯ 1X ¯ f (w(i)) − f (w(i))¯

k

l

i=0

i=0

≤ hk (w) +

r X

qkj (w) + hl (w) +

j=0

qlj (w).

j=0

Taking the supremum over f , we have gk (w) ≤ hk (w) + 0 gkl (w) ≤ hk (w) +

r X

Pr

Pr

j=0 qkj (w)

j=0 qkj (w),

+

Pr

j=0 qlj (w)

+ hl (w).

But, for each j ≤ r, limk→∞ qkj (w) = 0 for almost every w, by 465H (or 273J) applied to the characteristic function of Ej . So lim supk→∞ gk (w) ≤ lim supk→∞ hk (w) ≤ ² for almost every w. At the same time, lim supk,l→∞

R

0 gkl ≤ lim supk,l→∞

R

R Pr R Pr R hk + j=0 qkj + j=0 qlj + hl ≤ 2².

As ² is arbitrary, {w : lim supk→∞ gk (w) ≥ 2−i } is negligible for every i ∈ N, and limk→∞ gk = 0 almost R 0 everywhere, as required, while equally limk,l→∞ gkl = 0. Thus (ii) and (v) are true. (c)(ii)⇒(iv) is trivial, since Z Z l−1 l−1 X X ¯ 1 k−1 ¯ ¯ 1 k−1 ¯ ¯1 X ¯ 1X ¯ f (w(i)) − f (w(i))¯ ≤ ¯ f (w(i)) − f ¯ + ¯ f (w(i)) − f ¯. k

i=0

l

i=0

k

i=0

l

i=0

465M

Stable sets

577

(d) not-(i)⇒not-(iv) & not-(v) Now suppose that A is not stable. α) In this case, by 465L, there are measurable functions h0 , h1 : X → [0, 1] such that (α ˜ k = 1 for every k ∈ N, where and (µ2k )∗ D [ ˜k = {w : w ∈ X 2k , f (w(2i)) ≤ h0 (w(2i)), D

R

h0 dµ <

R

h1 dµ

f ∈A

f (w(2i + 1)) ≥ h1 (w(2i + 1)) for every i < k}. β ) Set δ = (β

1 4

R

h1 − h0 > 0. Let k0 ≥ 1 be so large that

R

µk {w : w ∈ X k , |

hj −

1 k

Pk i=0

hj (w(i))| ≤ δ} ≥

1 2

for both j and for every k ≥ k0 (273J or 465H. The point is that Z µk {w : w ∈ X k , |

hj −

1 k

k X

hj (w(i))| ≤ δ}

i=0

Z

N

N

= µ {w : w ∈ X , |

hj −

1 k

k X

hj (w(i))| ≤ δ}

i=0

→ 1 as k → ∞.) Let hkn in≥1 be such that kn ≥

2 δ

Pn−1 i=0

ki for every n ≥ 1. Since δ ≤ 14 , every kn is at least as large as k0 .

Q P (γγ ) Set mn = 2 i
φ(w)(mn + kn + i) = w(n)(2i + 1)

˜ ˜ k has outer measure 1 in X 2kn , so D ˜ =Q for n ∈ N and i < kn . For each n ∈ N, D n n∈N Dkn has outer Q 2kn N ˜ ˜ measure 1 in n∈N X , and φ[D] has outer measure 1 in X . Note that φ[D] is just the set of w ∈ X N such that, for every n ∈ N, there is an f ∈ A such that f (w(i)) ≤ h0 (w(i)) for mn ≤ i < mn + kn , f (w(i)) ≥ h1 (w(i)) for mn + kn ≤ i < mn + 2kn = mn+1 . If we set Vn0 = {w : w ∈ X N , | Vn1 = {w : w ∈ X N , |

1 kn

1 kn

Pmn +kn −1 i=mn

Pmn +2kn −1 i=mn +kn

h0 (w(i))−

R

h1 (w(i)) −

R

h0 | ≤ δ}, h1 | ≤ δ},

every Vn0 and Vn1 has measure at least 21 , because kn ≥ k0 . ˜ Then (δδ ) Now suppose that n ∈ N and w ∈ Vn0 ∩ Vn1 ∩ φ[D]. ¯ sup ¯

1 f ∈A mn +2kn

mn +2k Xn −1 i=0

f (w(i)) −

1 mn +kn

mn X +kn −1

¯ f (w(i))¯ ≥

i=0

δ . (2+δ)(1+δ)

˜ there must be an f ∈ A such that P P Since w ∈ φ[D], f (w(i)) ≤ h0 (w(i)) for mn ≤ i < mn + kn , f (w(i)) ≥ h1 (w(i)) for mn + kn ≤ i < mn + 2kn = mn+1 . Pmn −1 Pmn +kn Pmn +2kn −1 Set s = i=0 f (w(i)), t = i=m f (w(i)) and t0 = i=m f (w(i)). Then n n +kn

578


1 mn +2kn

mn +2k Xn −1

f (w(i)) −

i=0

1 mn +kn

mn X +kn −1

465M

f (w(i))

i=0

s + t + t0 s+t (t0 − t − s)kn + t0 mn − = mn + 2kn mn + kn (mn + 2kn )(mn + kn ) 0 (t − t − mn )kn t0 − t − δkn ≥ ≥ (mn + 2kn )(mn + kn ) (2 + δ)(1 + δ)kn

=

because mn ≤ δkn , by the choice of kn . To estimate t and t0 , we have R Pmn +kn −1 Pmn +kn −1 t = i=m f (w(i)) ≤ i=m h0 (w(i))≤ kn ( h0 + δ) n n because w ∈ Vn0 . On the other hand, R Pmn +2kn −1 Pmn +2kn −1 t0 = i=m f (w(i)) ≥ i=m h1 (w(i))≥ kn ( h1 − δ) n +kn n +kn because w ∈ Vn1 . So

R

t0 − t ≥ kn ( h1 − h0 − 2δ) = 2kn δ, and 1 mn +2kn

mn +2k Xn −1

f (w(i)) −

i=0

≥

1 mn +kn

mn X +kn −1

f (w(i))

i=0

2δkn − δkn δ = .Q Q (2 + δ)(1 + δ)kn (2 + δ)(1 + δ)

(²²) The Vn0 and Vn1 are all independent. So µN (Vn0 ∩ Vn1 ) ≥

1 4

for every n, and

W = {w : w ∈ X N , w ∈ Vn0 ∩ Vn1 for infinitely many n} ˜ has outer measure has measure 1 (by the Borel-Cantelli lemma, 273K, or otherwise). Accordingly W ∩ φ[D] N ˜ then (δ) tells us that 1 in X . But if w ∈ W ∩ φ[D], l−1 X ¯ 1 k−1 ¯ 1X lim sup sup ¯ f (w(i)) − f (w(i))¯ ≥ k,l→∞ f ∈A k i=0

l

i=0

δ , (2+δ)(1+δ)

˜ So (iv) must be false. because there are infinitely many n such that w ∈ Vn0 ∩ Vn1 ∩ φ[D]. (ζζ ) We see also that, for any n ∈ N, Z 0 gm ≥ n +kn ,mn +2kn

≥

δ µN (Vn0 (2+δ)(1+δ)

∩ Vn1 )

δ , 4(2+δ)(1+δ)

writing 0 (w) = supf ∈A | gkl

1 k

Pk−1 i=0

f (w(i)) −

1 l

Pl−1 i=0

f (w(i))|

R 0 for k, l ∈ N. So lim supk,l→∞ gkl > 0, and (v) is false. 465N Corollary (Talagrand 84) Let (X, Σ, µ) be a measure space. (a) Let A ⊆ RX be a stable set. Suppose that there is a measurable function g : X → [0, ∞[ such that |f (x)| ≤ g(x) for every x ∈ X, f ∈ A. Then the convex hull Γ(A) of A in RX is stable. (b) Let A, B ⊆ RX be two stable sets. Suppose that that there is a measurable function g : X → [0, ∞[ such that |f (x)| ≤ g(x) for every x ∈ X, f ∈ A ∪ B. Then A + B = {f1 + f2 : f1 ∈ A, f2 ∈ B} is stable.

465P

Stable sets

579

proof (a)(i) Consider first the case in which µX = 1 and A ⊆ [−1, 1]X . In this case, R R 1 Pk−1 1 Pk−1 supf ∈Γ(A) | f |= supf ∈A | f| i=0 f (w(i))− i=0 f (w(i))− k

k

N

for every k ≥ 1, w ∈ X . So Γ(A) satisfies condition (ii) of 465M whenever A does, and we have the result. (ii) Now suppose just that µX = 1. Set A0 = {

f 1+g

: f ∈ A}. Then A0 is stable (465Ci), so Γ(A0 ) is

stable, by (i), and Γ(A) = {f × (1 + g) : f ∈ A0 } is stable. (iii) For the general case, take E ∈ Σ and α < β such that 0 < µE < ∞. Let ν be the indefinite-integral measure defined by µ and the function (µE)−1 χE. Then A is stable with respect to ν (465Ck), so Γ(A) is stable with respect to ν, by (ii). There is therefore some k ≥ 1 such that (ν 2k )∗ Dk (Γ(A), E, α, β) < 1. But (because ν 2k is the indefinite integral of (µE)−2k χE 2k ) (µ2k )∗ Dk (Γ(A), E, α, β) < (µE)2k . As E, α and β are arbitrary, Γ(A) is stable with respect to µ. (b) A + B ⊆ 2Γ(A ∪ B), so the result follows from 465Cd, (a) and 465Ce. 465O Stable sets in L0 The notion of ‘stability’ as defined in 465B is applicable only to true functions; in such examples as 465Xj, the irregularity of the set A is erased entirely if we look at its image in the space L0 of equivalence classes of measurable functions. We do, however, have a corresponding concept for subsets of function spaces, which can be expressed in the language of §325. If (A, µ ¯) is a semi-finite measure N ¯k ) for the localizable measure algebra free product of k copies of (A, µ ¯), algebra, and k ≥ 1, I write ( c k A, µ as described in 325H. If Q ⊆ L0 (A), k ≥ 1, a ∈ A has finite measure and α < β in R, set dk (Q, a, α, β) = sup (a ∩ [[v ≤ α]]) ⊗ (a ∩ [[v ≥ β]]) ⊗ . . . v∈Q

⊗ (a ∩ [[v ≤ α]]) ⊗ (a ∩ [[v ≥ β]]) N in c 2k A, taking k repetitions of the formula (a ∩ [[v ≤ α]])⊗(a ∩ [[v ≥ β]]) to match the corresponding formula S Dk (A, E, α, β) = f ∈A ((E ∩ {x : f (x) ≤ α}) × (E ∩ {x : f (x) ≥ β))k . (Note that the supremum supv∈Q . . . is defined because a⊗2k = a ⊗ . . . ⊗ a has finite measure in the measure N ¯2k ). Of course I mean to take dk (Q, a, α, β) = 0 if Q = ∅.) Now we can say that Q is algebra ( c A, µ 2k

stable if whenever 0 < µ ¯a < ∞ and α < β there is a k ≥ 1 such that µ ¯2k dk (Q, a, α, β) < (¯ µa)2k ; that is, dk (Q, a, α, β) 6= a ⊗ . . . ⊗ a. We have the following relationships between the two concepts of stability. 465P Theorem Let (X, Σ, µ) be a semi-finite measure space, with measure algebra (A, µ ¯). (a) Suppose that A ⊆ L0 (Σ) and that Q = {f • : f ∈ A} ⊆ L0 (µ), identified with L0 = L0 (A) (364Jc). Then Q is stable in the sense of 465O iff every countable subset of A is stable in the sense of 465B. (b) If µ is strictly localizable and Q ⊆ L0 (µ) is stable, then there is a stable set B ⊆ L0 such that Q = {f • : f ∈ B}. proof (a)(i) Suppose that all countable subsets of A are stable, and take a ∈ A such that 0 < µ ¯a < ∞ and α < β in R. For each k ∈ N there is a countable set Qk ⊆ Q such that dk (Qk , a, α, β) = dk (Q, a, α, β), because N S a⊗2k is of finite measure in c 2k A. Now there is a countable set A0 ⊆ A such that {f • : f ∈ A0 } = k∈N Qk . Let E ∈ Σ be such that E • = a. As µE = µ ¯a ∈ ]0, ∞[ and A0 is stable, there is a k ≥ 1 such that 2k ∗ 0 2k 0 (µ ) Dk (A , E, α, β) < (µE) . Because A is countable, Dk (A0 , E, α, β) is measurable. But 325He tells us that we have an order-continuous measure-preserving Boolean homomorphism π from the measure algebra N Q • of µ2k to c 2k A, such that π( i<2k Fi )• = F0• ⊗ . . . ⊗ F2k−1 for all F0 , . . . , F2k−1 ∈ Σ; accordingly [ µ ¯2k dk (Q, a, α, β) = µ ¯2k dk (Qk , a, α, β) ≤ µ ¯2k dk ( Qi , a, α, β) 2k

0

i∈N 2k

=µ ¯ (πDk (A , E, α, β) ) = µ Dk (A0 , E, α, β) < (µE)2k = µ ¯2k a⊗2k .

•

580


465P

As a, α and β are arbitrary, Q is stable. (ii) Now suppose that Q is stable and that A0 is any countable subset of A. Take E ∈ Σ such that 0 < µE < ∞, and α < β in R. Set a = E • ∈ A. This time, writing Q0 for {f • : f ∈ A0 }, we have πDk (A0 , E, α, β)• = dk (Q0 , a, α, β) ⊆ dk (Q, a, α, β) for every k ≥ 1. There is some k such that dk (Q, a, α, β) 6= a⊗2k , and in this case µ2k Dk (A0 , E, α, β) < (µE)2k ; as E, α and β are arbitrary, A0 is stable. (b)(i) If µX = 0 this is trivial; suppose that µX > 0. Replacing µ by its completion does not change either L0 (µ) or the stable subsets of RX (241Xb, 465Cj), and leaves µ strictly localizable (212Ga), so we may suppose that µ is complete. Let hEi ii∈I be a decomposition of X into sets of finite measure. Amalgamating any negligible Ei into other non-negligible ones, we may suppose that µEi > 0 forSeach i. Writing µi for the subspace measure on Ei , we have a consistent lifting φi for µi (346J). Set φE = i∈I φi (E ∩ Xi ) for E ∈ Σ; then φ is a lifting of µ. Let θ be the corresponding lifting from A to Σ (341B) and T : L∞ (A) → L∞ (Σ) the associated linear operator, defined by saying that T (χa) = χ(θa) for every a ∈ A (363F). Since θ(a)• = a for every a ∈ A, (T v)• = v for every v ∈ L∞ . (ii) We need to know that if v ∈ L∞ and α < α0 , then {x : (T v)(x) ≤ α} ⊆ φ({x : (T v)(x)P ≤ α0 }). P P n 1 1 0 0 0 0 0 Let v ∈ S(A) be such that kv − v k∞ ≤ 2 (α − α) (363C), and set γ = (α + α ). Express v as α χa i i i=0 2 Pn where a0 , . . . , an ∈ A are disjoint. Then T v 0 = i=0 αi χ(θai ). Now kT v − T v 0 k∞ ≤ kv − v 0 k∞ ≤ γ − α, so

{x : (T v)(x) ≤ α} ⊆ {x : (T v 0 )(x) ≤ γ} [ [ = {θai : i ≤ n, αi ≤ γ} = φ( {θai : i ≤ n, αi ≤ γ}) (because φ(θa) = θa for every a ∈ A) = φ({x : (T v 0 )(x) ≤ γ}) ⊆ φ({x : (T v)(x) ≤ α0 }), as claimed. Q Q Similarly, if β 0 < β then {x : (T v)(x) ≥ β} ⊆ φ({x : (T v)(x) ≥ β 0 }). (iii) For the moment, suppose that Q ⊆ L∞ (A), which we may identify with L∞ (µ) (363I). Set B = T [Q], so that Q = {f • : f ∈ A}. Then B is stable. P P Let E ∈ Σ be such that 0 < µE < ∞, and α < β. Let i ∈ I be such that µ(E ∩ Ei ) > 0, and α0 , β 0 ∈ R such that α < α0 < β 0 < β. Setting a = (E ∩ Ei )• , we have 0 < µ ¯a < ∞, so there is some k ∈ N such that dk (Q, a, α0 , β 0 ) 6= a⊗2k . Let π be the N measure-preserving Boolean homomorphism from the measure algebra of µ2k to c 2k A described in part (a) of this proof; as noted in 325He, the present context is enough to ensure that π is an isomorphism. So there is a W ∈ dom µ2k such that πW • = dk (Q, a, α0 , β 0 ); since dk (Q, a, α0 , β 0 ) ⊆ a⊗2k ⊆ (Ei2k )• , we may suppose that W ⊆ Ei2k . If f ∈ B, then πDk ({f }, E, α0 , β 0 ) = dk ({f • }, a, α0 , β 0 ) ⊆ dk (Q, a, α0 , β 0 ), so Dk ({f }, E, α0 , β 0 ) \ W is negligible. 0 2k At this point, Q recall that Qφi was supposed to be a consistent lifting of µi . So we have a lifting φ of µi 0 such that φ ( j<2k Fj ) = j<2k φi Fj for all F0 , . . . , F2k−1 ∈ Σi . In particular, if f ∈ B and we set F2j = {x : x ∈ E ∩ Ei , f (x) ≤ α},

F2j+1 = {x : x ∈ E ∩ Ei , f (x) ≥ β},

0 F2j = {x : x ∈ E ∩ Ei , f (x) ≤ α0 },

0 F2j+1 = {x : x ∈ E ∩ Ei , f (x) ≥ β 0 },

for j < k, we shall have

465R

Stable sets

Y

Fj ⊆

j<2k

Y

=

Y j<2k

Q j<2k

φFj0

j<2k

(by (ii) above, because f = T v for some v)

because

581

φi Fj0 = φ0 (

Y

Fj0 ) ⊆ φ0 W

j<2k

Fj0 = Dk ({f }, E, α0 , β 0 ). As f is arbitrary, Dk (B, E ∩ Ei , α, β) ⊆ φ0 W . But now 0 (µ2k )∗ Dk (B, E ∩ Ei , α, β) ≤ µ2k (φ0 W ) = µ2k i (φ W )

(251Wl) 2k = µ2k i W =µ W

=µ ¯2k dk (Q, a, α0 , β 0 ) < µ ¯2k a⊗2k = µ(E ∩ Ei )2k . As usual, it follows that (µ2k )∗ Dk (B, E, α, β) < (µE)2k ; as E, α and β are arbitrary, B is stable, as claimed. Q Q (iv) Thus the result is true if Q is included in the unit ball of L∞ . In general, set h(α) = tanh α for α ∈ R, and consider A0 = {hf : f ∈ A}, Q0 = {(hf )• : f ∈ A}. Since every countable subset of A is stable, so is every countable subset of A0 (465Cl) and Q0 is stable (by (a) of this theorem). But Q0 is included in the unit ball of L∞ , so there is a stable set B 0 ⊆ L0 such that Q0 = {g • : g ∈ B 0 }. Setting B = {h−1 g : g ∈ B 0 }, B is stable and {f • : f ∈ B} = Q. So we have the general theorem. 465Q Remarks Using 465Pa, we can work through the first part of this section to get a list of properties of stable subsets of L0 . For instance, the convex hull of an order-bounded stable set in L0 is stable, as in 465Na. It is harder to relate such results as 465M to the idea of stability in L0 , but the argument of 465Pb gives a line to follow: if (X, Σ, µ) is complete and strictly localizable, there is a linear operator T : L∞ (µ) → L∞ (Σ), defined from a lifting, such that, for Q ⊆ L∞ , T [Q] is stable iff Q is stable. So when Pk−1 µ is a complete probability measure, we can look at the averages ψwk (v) = k1 i=0 (T v)(w(i)) for v ∈ Q, w ∈ X N to devise criteria for stability of Q in terms of the linear functionals ψwk . Working in L1 , however, we can look for results of a different type, as follows. 465R Theorem (Talagrand 84) Let (A, µ ¯) and (B, ν¯) be measure algebras, and T : L1 (A, µ ¯) → L (B, ν¯) a bounded linear operator. If Q is stable and order-bounded in L1 (A, µ ¯), then T [Q] ⊆ L1 (B, ν¯) is stable. 1

proof (a) To begin with (down to (d) below) let us suppose that (A, µ ¯) and (B, ν¯) are the measure algebras of measure spaces (X, Σ, µ) and (Y, T, ν), so that we can identify L1µ¯ , L1ν¯ with L1 (µ) and L1 (ν) (365B), Q is countable, so that Q can be expressed as {f • : f ∈ A}, where A ⊆ L0 (Σ) is countable and stable (465P), T is positive, µ and ν are totally finite, T (χX • ) = χY • , Q ⊆ L∞ (A) is k k∞ -bounded, so that we may take A ⊆ L∞ to be k k∞ -bounded, νY = 1, µY = 1, and that kT k ≤ 1. (b) The idea of the argument is that for any n ≥ 1 we have a positive linear operator Un : L1 (µn ) → L (ν n ) defined as follows. 1

582


465R

Qn Q If f0 , . . . , fn−1 ∈ L1 (µ), set (f0 ⊗ . . . ⊗ fn−1 )(w) = i=0 fi (w(i)) whenever w ∈ i
for all u0 , . . . , un−1 ∈ L (µ). Moreover, kUn+1 k ≤ kUn kkT k for every n (see 253F), so kUn k ≤ 1 for every n. For i < n ∈ N, we have a natural operator Rni : L1 (µ) → L1 (µn ), defined by saying that Rni f • = (f πni )• for every f ∈ L1 (µ), where πni (w) = w(i) for w ∈ X n . Similarly, we have an operator Sni : L1 (ν) → L1 (ν n ). Observe that Rni u = e ⊗ . . . ⊗ e ⊗ u ⊗ e ⊗ . . . ⊗ e where e = χX • and the u is put in the position corresponding to the coordinate i. Since T e = (χY )• = e0 say, Un Rni u = e0 ⊗ . . . ⊗ T u ⊗ . . . ⊗ e0 = Sni T u, for every u ∈ L1 (µ). (c) Let B ⊆ L∞ (T) be a countable k k∞ -bounded set such that T [Q] = {g • : g ∈ B}. (T [Q] is k k∞ bounded because T is positive and T (χX)• = (χY )• .) IRseek to show that B is stable by using the criterion 465M(v). Let ² > 0. Then there is an m ≥ 1 such that fkl (w)µN (dw) ≤ ² for any k, l ≥ m, writing 1 Pk−1 1 Pl−1 fkl (w) = supf ∈A | i=0 f (w(i)) − i=0 f (w(i))| k

l

for w ∈ X N ; note that fkl is measurable because A is countable. Take any k, l ≥ m and consider 1 Pk−1 1 Pl−1 gkl (z) = supg∈B | i=0 g(z(i)) − i=0 g(z(i))| for z ∈ Y N . I claim that

R

k

l

R R gkl dν N ≤ ². P P Set n = max(k, l). Then gkl dν N = g˜ dν n , where 1 Pk−1 1 Pl−1 g˜(z) = supg∈B | i=0 g(z(i)) − i=0 g(z(i))| k

n

1

l

n

for z ∈ Y . If we look at g˜ in L (ν ), we see that it is 1 Pk−1 1 Pl−1 • • supg∈B | i=0 Sni g − i=0 Sni g | •

k

1

1

l

n

where Sni : L (ν) → L (ν ) is defined in (b) above. Thus 1 Pk−1 1 Pl−1 g˜• = supv∈T [Q] | i=0 Sni v − i=0 Sni v|. k

l

Similarly, setting 1 Pk−1 1 Pl−1 f˜(w) = supf ∈A | i=0 f (w(i)) − i=0 f (w(i))| k

l

n

for w ∈ X , 1 Pl−1 1 Pk−1 f˜• = supu∈Q | i=0 Rni u − i=0 Rni u|. k

l

Now consider Un f˜• . For any v ∈ T [Q], we can express v as T u where u ∈ Q, so

465R

Stable sets

583

l−1 l−1 X X ¯ 1 k−1 ¯ ¯ 1 k−1 ¯ 1X 1X ¯ Sni v − Sni v ¯ = ¯ Sni T u − Sni T u¯

k

l

i=0

k

i=0

¯1 =¯ k

l

i=0 k−1 X

Un Rni u −

i=0

i=0

1 l

l−1 X

¯ Un Rni u¯

i=0

(because Un Rni = Sni T , as noted in (b) above) k−1 l−1 ¯ ¯ 1X 1X = ¯Un ( Rni u − Rni u)¯

k

≤ Un (|

1 k

l

i=0 k−1 X

Rni u −

i=0

1 l

i=0 l−1 X

Rni u|)

i=0

(because Un is positive) ≤ Un f˜• . As v is arbitrary, g˜• ≤ Un f˜• , and Z

Z N

gkl dν =

g˜ dν n = k˜ g • k1 ≤ kUn f˜• k1 ≤ kf˜• k1

(because kUn k ≤ 1) ≤² because k, l ≥ m. Q Q (d) As ² is arbitrary, B satisfies the criterion 465M(v), and is stable. So T [Q] is stable, by 465Pa in the other direction. (e) Now let us seek to unwind the list of special assumptions used in the argument above. Suppose we drop the last two, and assume only that (A, µ ¯) and (B, ν¯) are the measure algebras of measure spaces (X, Σ, µ) and (Y, T, ν), Q is countable, T is positive, µ and ν are totally finite, T (χX • ) = χY • , Q ⊆ L∞ (A) is k k∞ -bounded, νY = 1. R Then T [Q] is stable. P P Define µ1 : Σ → [0, ∞[ by setting µ1 E = T (χE • ) for every E ∈ Σ. Then µ1 is countably additive because T is (sequentially) order-continuous (355Ka). If µE = 0 then χE • = 0 in L1 (µ) and µ1 E = 0, so µ1 is truly continuous with respect to µ (232Bb) and has a Radon-Nikod´ ym derivative (232E). By 465Ck, A is stable with respect to µ1 , while µ1 X = νY = 1, because T (χX)• = χY • . Let (A1 , µ ¯1 ) be the measure algebra of µ1 . If E ∈ Σ and µ1 E = 0, then T (χE • ) = 0. Accordingly we can define an additive function θ : A1 → L1 (ν) by setting θE • = T (χE • ) for every E ∈ Σ. (Note that the two • s here must be interpreted differently. In the formula θE • , the equivalence class E • is to be taken in A1 . In the formula χE • = (χE)• , the equivalence class is to be taken in L0 (µ). In the rest of this proof I will pass over such points without comment; I hope the context will always make it clear how each • is to be read.) Because T is positive, θ is non-negative, and by the definition of µ1 we have kθak1 = µ ¯1 a for every a ∈ A1 . So we have a positive linear operator T1 : L1 (A1 , µ ¯1 ) → L1 (ν) defined by setting T1 (χE • ) = θE • = T (χE • ) for every E ∈ Σ (365I). If f : X → R is simple (that is, a linear combination of characteristic functions of sets in Σ), then T1 f • = T f • . So this is also true for every f ∈ L∞ (Σ); in particular, it is true for every f ∈ A, so that

584


465R

T [Q] = T1 [Q1 ], where Q1 = {f • : f ∈ A} ⊆ L1 (µ1 ). But µ1 , Q1 and T1 satisfy all the conditions of (a), so (b)-(d) tell us that T1 [Q1 ] is stable, and T [Q] is stable, as required. Q Q (f ) The next step is to drop the condition ‘νY = 1’. But this is elementary, since we are still assuming that ν is totally finite, and multiplying ν by a non-zero scalar doesn’t change L0 (ν) or the stability of any of its subsets, while the case νY = 0 is trivial. So we conclude that if (A, µ ¯) and (B, ν¯) are the measure algebras of measure spaces (X, Σ, µ) and (Y, T, ν), Q is countable, T is positive, µ and ν are totally finite, T (χX • ) = χY • , Q ⊆ L∞ (A) is k k∞ -bounded, then T [Q] is stable. (g) We can now attack what remains. We find that if (A, µ ¯) and (B, ν¯) are the measure algebras of measure spaces (X, Σ, µ) and (Y, T, ν), Q is countable, T is positive, then T [Q] is stable. P P At this point recall that we are supposing that Q is order-bounded. Let u0 ∈ L1 (µ) be such that |u| ≤ u0 for every u ∈ Q; let f0 ∈ L0 (Σ)+ be such that f0• = u0 . Setting A0 = {(f ∧f0 )∨(−f0 ) : f ∈ A}, A0 is still stable, because its image in L1 (µ) is still Q, or otherwise. Set v0 = T u0 , and let g0 ∈ L0 (T)+ be such that g0• = v0 . Because T is positive, |T u| ≤ T |u| ≤ v0 for every u ∈ Q. So we can represent T [Q] as {g • : g ∈ B}, where B ⊆ L0 (T) is a countableR set and |g| ≤ g0 for every gR ∈ B. Set F0 = {y : g0 (y) 6= 0}. Define measures µ1 , ν1 by setting µ1 E = E f0 dµ for E ∈ Σ, ν1 F = F g0 dν for F ∈ T. Then both µ1 and ν1 are totally finite. By 465Ck, A0 is stable with respect to µ1 . Set A1 = { f (x) f0

f f0

: f ∈ A0 }, interpreting

as 0 if f0 (x) = 0; then A1 is stable with respect to µ1 , by 465Ci, and kf k∞ ≤ 1 for every f ∈ A1 .

Take Q1 = {f • : f ∈ A1 } ⊆ L1 (µ1 ), so that Q1 is stable. We have a norm-preserving positive linear operator R : L1 (µ1 ) → L1 (µ) defined by setting Rf • = (f ×f0 )• for every f ∈ L1 (µ1 ) (use 235A). Observe that R[Q1 ] = Q and R(χX)• = u0 . Similarly, we have a normpreserving positive linear operator S : L1 (ν1 ) → L1 (ν) defined by setting Sg • = (g × g0 )• for g ∈ L1 (ν1 ). The set of values of S is just {g • : g ∈ L1 (ν), g(y) = 0 whenever g0 (y) = 0}, which is the band of L1 (ν) generated by v0 . So {u : u ∈ L1 (µ1 ), T R|u| ∈ S[L1 (ν1 )]} is a band in L1 (ν1 ) containing χX • , and must be the whole of L1 (µ1 ). Thus we have a positive linear operator T1 = S −1 T R : L1 (µ1 ) → L1 (ν1 ), and T1 (χX)• = χY • in L1 (ν1 ). By (f), T1 [Q1 ] is stable in L1 (ν1 ). Observe that T1 [Q1 ] = {S −1 g • : g ∈ B} = {g • : g ∈ B1 }, where B1 = {

g g0

: g ∈ B}, interpreting

g (y) g0

as 0 if y ∈ Y \ F0 . Consequently B1 and B = {g × g0 : g ∈ B1 } are

stable with respect to ν1 . By 465Ck, once more, B1 is stable in L1 (ν0 ), where ν0 F =

R

1

F g0

dν1 = ν(F ∩ F0 )

for any F ∈ T. But because g(y) = 0 whenever g ∈ B and y ∈ Y \ F0 , (ν 2k )∗ Dk (B, F, α, β) = (ν 2k )∗ Dk (B, F ∩ F0 , α, β) = (ν02k )∗ Dk (B, F, α, β) whenever F ∈ T, α < β and k ≥ 1; so B is also stable with respect to ν, and Q = {g • : g ∈ B} is stable in L1 (ν). Q Q (h) The worst is over. If we are not told that T is positive, we know that it is expressible as the difference of positive linear operators T1 and T2 (371D); now T1 [Q] and T2 [Q] will be stable, by the work above, so T [Q] ⊆ T1 [Q] − T2 [Q] is stable, by 465Nb. If we are not told that Q is countable, we refer to 465P to see that we need only check that countable subsets of T [Q] are stable, and these are images of countable subsets of

*465U

Stable sets

585

Q. Finally, the identification of the abstract measure algebras (A, µ ¯1 ) and (B, ν¯1 ) with the measure algebras of measure spaces is Theorem 321J. *465S R-stable sets The theory above has been developed in the context of general measure (or probability) spaces and the ‘ordinary’ product measure of measure spaces. For τ -additive measures – in particular, for Radon measures – we have an alternative product measure, as described in §417. If (X, T, Σ, µ) is a semi-finite τ -additive topological measure space such that µ is inner regular with respect to the Borel sets, write µ ˜I for the τ -additive product measure on X I , as described in 417D (for the product of two spaces) and 417F (for the product of any family of probability spaces); just as in 251W, we can extend the construction of 417D to arbitrary finite products. Now say that A ⊆ RX is R-stable if whenever 0 < µE < ∞ and α < β there is a k ≥ 1 such that (˜ µ2k )∗ Dk (A, E, α, β) < (µE)2k . Because we have a version of Fubini’s theorem for the products of τ -additive topological measures (417H), all the arguments of this section can be applied to R-stable sets, yielding criteria for R-stability exactly like those in 465M. Because the τ -additive product measure extends the c.l.d. product measure, stable sets are always Rstable. (We must have (˜ µ2k )∗ Dk (A, E, α, β) ≤ (µ2k )∗ Dk (A, E, α, β) for all k, A, E, α and β.) For an example of an R-stable set which is not stable, see 465U. The concept of ‘R-stability’ is used in Talagrand 84 in applications to the integration of vector-valued functions. I give one result, however, to show how it is relevant to a question with a natural expression in the language of this chapter. *465T Proposition (Talagrand 84) Let (X, T, Σ, µ) be a semi-finite τ -additive topological measure space such that µ is inner regular with respect to the Borel sets. If A ⊆ C(X) is such that every countable subset of A is R-stable, then A is R-stable. proof For any α < β and k ≥ 1, Gk (A, X, α, β) =

[

{w : w ∈ X 2k , f (w(2i)) < α,

f ∈A

f (w(2i + 1)) > β for i < k} is open. Suppose that 0 < µE < ∞. Because all the product measures µ ˜2k are τ -additive, we can find a 0 countable set A ⊆ A such that µ ˜2k Gk (A, E, α, β) = µ ˜2k Gk (A0 , E, α, β) for every k ≥ 1 and all rational α, β. Now, if α < β, there are rational α0 , β 0 such that α < α0 < β 0 < β, and a k ≥ 1 such that µ ˜2k Dk (A0 , E, α0 , β 0 ) < (µE)2k ; in which case (˜ µ2k )∗ Dk (A, E, α, β) ≤ µ ˜2k Gk (A, E, α0 , β 0 ) = µ ˜2k Gk (A0 , E, α0 , β 0 ) ≤µ ˜2k Dk (A0 , E, α0 , β 0 ) < (µE)2k . As E, α and β are arbitrary, A is R-stable. *465U I come now to the promised example of an R-stable set which is not stable. I follow the construction in Talagrand 88, which displays an interesting characteristic related to 465O-465P above. Example There is a Radon probability space with an R-stable set of continuous functions which is not stable. proof (a) Let (X, Σ, µ) be an atomless probability space (e.g., the unit interval with Lebesgue measure). Define hrn in∈N by setting r0 = 1, r1 = 2, rn+1 = 2nrn for n ≥ 1; then 2n ≤ rn < rn+1 for every n. Let hΣn in∈N be an increasing sequence of finite subalgebras of Σ, each Σn having rn atoms of the same size; this is possible because rn+1 is always a multiple of rn . Write Hn for the set of atoms of Σn . Next, for each n ∈ N, let hGH iH∈Hn be an independent family in Σn+1 of sets of measure 2−n ; such a family exists because rn+1 is a multiple of 2nrn .

586


*465U

S Let E be the family of all sets expressible in the form E = i∈N Hi where, for some strictly increasing sequence hni ii∈N in N, Hi ∈ Hni and Hj ⊆ GHi whenever i < j in N. Set A = {χE : E ∈ E} ⊆ L0 (Σ). (b) A is stable. P P Suppose that F ∈ Σ and that µF > 0. Take n ∈ N so 2−n < (µF )2 . Set Slarge that 3 · Q H = {H : H ∈ Hn , µ(H ∩ F ) > 0}; enumerate H as hHi ii
S k>n

1 rk2

+

1 2k rk

+

1 ) 2k rk

Uk ) ≥ (µF )2 − 3

≤ 3 · 2−k ,

P∞ k=n+1

2−k > 0.

Set V = {w : w ∈ F12m , w(2i) ∈ Hi for every i < m, (w(1), w(3)) ∈ / Then µ2m (V ) ≥

Qm−1 i=0

µ(F1 ∩ Hi ) · µ2 (F12 \

S k>n

S k>n

Uk }.

Uk ) · (µF )m−2 > 0.

?? Suppose, if possible, that there is a point w ∈ S V ∩ Dm (A, F, 0, 1). Then there is an E ∈ E such that w(2i) ∈ / E, w(2i + 1) ∈ E for i < m. Express E as j∈N Ej where Ej ∈ Hnj for every j ∈ N, where hnj ij∈N is strictly increasing, and Ek ⊆ GEj whenever j < k. Because w(2i) ∈ F1 ∩ Hi \ E, Ej cannot include Hi for any i < m, j ∈ N; so Ej ∩ Hi = ∅ whenever i < m and nj ≤ n. Because w(1), w(3) ∈ S E, there are j0 , j1 ∈ N such that w(1) ∈ Ej0 and w(3) ∈ Ej1 . Since both w(1) and w(3) belong to F1 ⊆ i n, which is impossible. X X Thus Dm (A, F, 0, 1) does not meet V , and (µ2m )∗ Dm (A, F, 0, 1) < (µF )2m . Now if α < β, then Dm (A, F, α, β) = Dm (A, F, 0, 1) if 0 ≤ α < β ≤ 1, ∅ otherwise; so in all cases (µ2m )∗ Dm (A, F, α, β) < (µF )2m . As F also is arbitrary, A is stable, as claimed. Q Q (c) Now let (Z, S, T, ν) be the Stone space of the measure algebra of (X, Σ, µ) (321K), so that ν is a Radon measure (411Pe). For E ∈ Σ, write E ∗ for the corresponding open-and-closed set in Z, so that E 7→ E ∗ : Σ → T is a measure-preserving Boolean homomorphism. Set A∗ = {χE ∗ : E ∈ E} ⊆ C(Z). Write ν m for the c.l.d. product measure on Z m for m ≥ 1. We already know that ν 2 is not a topological measure (419Xc). (d) The point is that (ν 2m )∗ Dm (A∗ , Z, 0, 1) = 1 for every m ≥ 1. P P?? Suppose, if possible, otherwise. ˜ ⊆ Z 2m such that ν 2m (W ˜ ) > 0 and W ˜ ∩ Dm (A∗ , Z, 0, 1) > 0. For u, v ∈ Z m , write Then there is a set W u#v = (u(0), v(0), u(1), v(1), . . . , u(m − 1), v(m − 1)) ∈ Z 2m . Then there is an ² > 0 such that ˜ } is defined and greater than m²} V˜ = {v : v ∈ Z m , ν m {u : u ∈ Z m , u#v ∈ W has non-zero inner measure for ν m . Now there are sets F˜ij ∈ T, for i ∈ N and j < m, such that S Q P∞ Qm ˜ Z m \ V˜ ⊆ i∈N j 0. I seek to choose hnk ik∈N in N, hHk ik∈N and hEkj ik∈N,j
*465U

Stable sets

587

Ekj ⊆ Eij ∩ GHi for i < k, j < m, Q Q j
Q j
Ekj ) > 0,

ni < nk for every i < k, Hk ∈ Hnk , if k = ik m + jk , where ik ∈ N, jk < m, then µ(Hk ∩ Ekjk ) > 0, 0 for every k ∈ N. The induction proceeds as follows. Set Ekj = X if k = 0, GHk−1 ∩ Ek−1,j otherwise, so Q Q 0 0 that µ(V ∩ j 0. Because V ∩ j 0 and j
(if k = 0) 2−nk ≤ 12 ²,

nk > ni for i < k,

(1 − 2−mnk )ηrnk < δ.

(This is possible because limn→∞ 2−mn rn = ∞.) Set H = {H : H ∈ Hnk , µ(H ∩ Ekjk ) > 0}; then −mnk #(H) ≥ ηrnk . Consider the family hGm , H iH∈H . These are stochastically independent sets of measure 2 −mnk #(H) m m so their union has measure 1 − (1 − 2 ) > 1 − δ, and there is an H ∈ H such that µ (V ∩ G ∩ k Hk Q j 0. Thus the induction continues. Look at the sequence hHk ik∈N and its union E. We have Hk ∈ Hnk for every k; moreover, if i < k, then µ(Hk ∩ Ekjk ) > 0, while Ekjk ⊆ GHi ; since Hk is an atom of Σnk , while GHi ∈ Σnk , Hk ⊆ GHi . Thus E ∈ E. Next, whenever i ≤ k ∈ N, Q Q Q Q j 0. Moving back to Z, this translates into S Q ν m ((E ∗ )m \ i≤k j 0. S Q But this means that (E ∗ )m is not included in i≤k j m² ≥ m ν m {u : u#v ∈ W

∞ X k=0

=m

∞ X

2−n0 −k ≥ m

∞ X

2−nk

k=0

µHk ≥ mµE = mνE ∗ .

k=0

˜ and u(j) ∈ So there must be some u such that u#v ∈ W / E ∗ for every j < m. But now, setting w = u#v, ∗ ∗ ˜ ; which is supposed to be we have w(2j) ∈ / E , w(2j + 1) ∈ E for j < m, and w ∈ Dm (A∗ , Z, 0, 1) ∩ W impossible. X XQ Q (e) This shows that A∗ is not stable. It is, however, R-stable. P P We have a measure algebra isomorphism between the measure algebras of µ and ν defined by the map E 7→ E ∗ : Σ → T. The corresponding isomorphism between L0 (µ) and L0 (ν) takes {f • : f ∈ A} to {h• : h ∈ A∗ }. By 465Pa and (b) above, {f • : f ∈ A} is stable in L0 (µ), so {h• : h ∈ A∗ } is stable in L0 (ν), and every countable subset of A∗ is stable. Since A∗ ⊆ C(X), it follows that A∗ is stable (465T). Q Q

588


*465V

*465V Remark This example is clearly related to 419E. The argument here is significantly deeper, but it does have an idea in common with that in 419E, besides the obvious point that both involve the Stone spaces of atomless probability spaces. Suppose that, in the context of 465U, we take E0 to be the family of sets E expressible as the union of a finite chain H0 , . . . , Hk where Hi ∈ Hni for i ≤ k and Hj ⊆ GHi for i < j ≤ k. Then we find, on repeating the argument of (b) in the proof of 465U, that the countable set A∗0 = {χE ∗ : E ∈ E0 } is stable, so that, setting Wm = Dm (A∗0 , Z, 0, 1), ν 2m Wm is small for large m. On the other hand, setting ˜ m = S{V : V ⊆ Z m is open, V \ Wm is negligible}, W ˜ m , so that (ν 2m )∗ W ˜ m = 1 for all m ≥ 1. Of course, writing ν˜2m for the we see that Dm (A∗ , Z, 0, 1) ⊆ W 2m 2m ˜ Radon product measure on Z , we have ν˜ (Wm ) = ν 2m Wm < 1 for large m, just as in 419E. Both A ⊆ L0 (Σ) and A∗ ⊆ L0 (T) are relatively pointwise compact. Note that while I took (X, Σ, µ) and (Z, T, ν) to be quite separate, it is entirely possible for them to be actually the same space. In this case it is natural to take every Σn to consist of open-and-closed sets, so that every member of E is open, and E ∗ becomes identified with the closure of E for E ∈ E. 465X Basic exercises (a) Let (X, Σ, µ) be a measure space, and hfn in∈N a sequence of measurable real-valued functions on X which converges a.e. Show that {fn : n ∈ N} is stable. (b) Let C be the family of convex sets in Rr . Show that {χC : C ∈ C} is stable with respect to Lebesgue measure on Rr , but that if r ≥ 2 there is a Radon probability measure ν on Rr such that {χC : C ∈ C, C is closed} is not stable with respect to ν. (c) Show that, for any M ≥ 0, the set of functions f : R → R of variation at most M is stable with respect to any Radon measure on R. (Hint: show that if µ is a Radon measure on R and E ∈ dom µ has non-zero finite measure, and (2k − 1)(β − α) > M , then (µ2k )∗ Dk (A, E, α, β) < (µE)2k .) (d) Let (X, Σ, µ) be a measure space, and A ⊆ RX . Show that A is stable iff {f + : f ∈ A} and {f : f ∈ A} are both stable. −

(e) Let (X, Σ, µ) and (Y, T, ν) be measure spaces, and φ : X → Y an inverse-measure-preserving function. Show that if B ⊆ RY is stable with respect to ν, then {gφ : g ∈ B} is stable with respect to µ. (f ) Let (X, Σ, µ) be a semi-finite measure space and A ⊆ RX . (i) Suppose that µ is inner regular with respect to the family {F : F ∈ Σ, {f × χF : f ∈ A} is stable}. Show that A is stable. (ii) Use this together with 465Ce, 465Cf and 465Cl to prove 465Ci. (g) Let (X, Σ, µ) be a totally finite measure space and T a σ-subalgebra of Σ. Let A ⊆ L0 (T) be any set. (i) Show that if A is µ¹ T-stable then it is µ-stable. (ii) Give an example to show that A can be µ-stable and pointwise compact without being µ¹ T-stable. (Hint: take µ to be Lebesgue measure on [0, 1] and T the countable-cocountable algebra.) (h) Let (X, Σ, µ) be a measure space, and A ⊆ RX . For g : X → [0, ∞[ set Ag = {(f ∧ g) ∨ (−g) : f ∈ A}. Show that A is stable iff Ag is stable for every integrable g : X → [0, ∞[. (i) Let (X, Σ, µ) be a locally determined measure space and A ⊆ RX a stable set. Let A be the closure of A for the topology of pointwise convergence. Show that {f • : f ∈ A} is just the closure of {f • : f ∈ A} ⊆ L0 (µ) for the topology of convergence in measure. (j) Show that there is a disjoint family I of finite subsets of [0, 1] such that A = {χI : I ∈ I} is not stable, though A is pointwise compact and the identity map on A is continuous for the topology of pointwise convergence and the topology of convergence in measure. (k) Let (X, Σ, µ) be a probability space. Show that A ⊆ RX is stable iff ¡ ¢1/m inf m∈N (µ2m )∗ Dm (A, X, α, β) =0 whenever α < β in R.

465 Notes

Stable sets

589

(l) Let (X, Σ, µ) be a measure space. Show that a countable set A ⊆ L0 (Σ) is not stable iff there are ˜ m (A, E, α, β) = (µE)m for every m ≥ 1, where D ˜ m (A, E, α, β) E ∈ Σ and α < β such that µE > 0 and µm D m is the set of those w ∈ E such that for every I ⊆ m there is an f ∈ A such that f (w(i)) ≤ α for i ∈ I, f (w(i)) ≥ β for i ∈ m \ I. (Hint: see part (iii) of case 2 of the proof of 465L.) (m) Let (X, Σ, µ) be a semi-finite measure space, and A ⊆ L0 (Σ) a set which is compact and metrizable for the topology of pointwise convergence. Show that A is stable. (Hint: apply the ideas of case 2 in the proof of 465L to a countable dense subset of A.) (n) Let (X, Σ, µ) be a probability space and A ⊆ L0 (Σ) a uniformly bounded set of functions. Show that A is stable iff ¯ N ¯ 1 Pk−1 R 1 Pl−1 ¯ limn→∞ supf ∈A,k,l≥n ¯ i=0 f (w(i)) − i=0 f (w(i)) µ (dw) = 0. k

l

(o) Show that there is a set A ⊆ R[0,1] such that g(x) = supf ∈A |f (x)| is finite for every x, and A is stable with respect to Lebesgue measure, but its convex hull Γ(A) is not. (Hint: take A = {g(x)χ{x} : x ∈ [0, 1]}, where g is such that for every m ≥ 1, every non-negligible compact set K ⊆ [0, 1]m there is a w ∈ K such that g(w(i)) = 2i+1 for every i < m.) (p) Let (X, Σ, µ) be a strictly localizable measure space and Q ⊆ L0 (µ) a set which is stable in the sense of 465R. Show that the closure of Q (for the topology of convergence in measure) is stable. (Hint: 465Xi.) (q) Let (X, Σ, µ) be a probabilityR space, T a σ-subalgebra of Σ, and A ⊆ RX a countable stable set of Σ-measurable functions such that supf ∈A |f (x)|µ(dx) < ∞. Show that if for each f ∈ A we choose a conditional expectation gf of f on T (requiring each gf to be T-measurable and defined everywhere on X), then {gf : f ∈ A} is stable. (r) Give an example of a probability algebra (A, µ ¯), a conditional expectation operator P : L1µ¯ → L1µ¯ 1 (365R), and a uniformly integrable stable set A ⊆ Lµ¯ such that P [A] is not stable. (Hint: start by picking a sequence hvn in∈N in P [L1µ¯ ] which is norm-convergent to 0 but not stable, and express this as hP un in∈N where µ ¯[[un > 0]] ≤ 2−n for every n.) 2

465Y Further exercises (a) RFind an integrable continuous function f : [0, 1[ → [0, ∞[ such that, in 2 the notation of 465H, lim supk→∞ f dνwk = ∞ for almost every w ∈ [0, 1]N . (Hint: arrange for f (x, x) to grow very fast as x ↑ 1.) (b) Show that in 465M we may replace the condition ‘A is uniformly bounded’ by the condition ‘|f | ≤ f0 for every f ∈ A, where f0 is integrable’. (Hint: Talagrand 87.) (c) Let (X, T, Σ, µ) be a τ -additive topological measure space such that µ is inner regular with respect to the Borel sets. Show that a countable R-stable subset of RX is stable. (d) Let (X, T, Σ, µ) be a complete τ -additive topological probability space such that µ is inner R regular with respect to the Borel sets, and A ⊆ [0, 1]X an R-stable set. Suppose that ² > 0 is such that f dµ ≤ ²2 for every f ∈ A. Show that there are an n ≥ 1 and aR Borel set W ⊆ X n and a γ > µ ñ W (writing µ ñ n for the τ -additive product measure on X ) such that f dν ≤ 3² whenever f ∈ A, ν : PX → [0, 1] is a point-supported probability measure and ν n W ≤ γ. 465 Notes and comments The definition in 465B arose naturally when M.Talagrand and I were studying pointwise compact sets of measurable functions; we found that in many cases a set of functions was relatively pointwise compact because it was stable (465Db). Only later did it appear that the concept was connected with ‘Glivenko-Cantelli’ classes in the theory of empirical measures, as explained in Talagrand 87. It is not the case that all pointwise compact sets of measurable functions are stable. In fact I have already offered examples in 463Xi and 464E above. In both cases it is easy to check from the definition in 465B that they are not stable, as can also be deduced from 465G. Another example is in 465Xj. You will observe

590


465 Notes

however that all these examples are ‘pathological’ in the sense that either the measure space is irregular (from some points of view, indeed, any measure space not isomorphic to Lebesgue measure on the unit interval can be dismissed as peripheral), or the set of functions is uninteresting. It is clear from 465R and 465T, for instance, that we should start with countable sets. So it is natural to ask: if we have a separable pointwise compact set of real-valued measurable functions on the unit interval, must it be stable? It turns out that this is undecidable in Zermelo-Fraenkel set theory (Shelah & Fremlin 93); I hope to return to the question in Volume 5. (If we ask for ‘metrizable’, instead of ‘separable’, we get a positive answer; see 465Xm.) The curious phrasing of the statement of 465M(iii), with the auxiliary functions hk , turns on the fact that all the expressions ‘supf ∈A . . . ’ here give rise to functions which need not be measurable. Thus the simple pointwise convergence described in (ii) and (iv) is not at all the same thing as the convergence in (v), which R may be thought of as a kind of k k1 -convergence if we write kgk1 = |g| for arbitrary real-valued functions g. (Since the sets A here are uniformly bounded, it may equally be thought of as convergence in measure.) Similarly, 465M(iii) is saying much more than just ¯ 1 Pk−1 ¯¯ ¯ limk→∞ supf ∈A i=0 f (w(i)) − E(f |T)(w(i)) = 0 a.e., k

though of course for countable sets A these distinctions disappear. On the other hand, since the convergence is certainly not monotonic in k, k k1 -convergence does not imply pointwise convergence. So we can look for something stronger than P either (iv) or (v) of 465M, P as in 465Xn. But this is still nowhere near the strength of 465M(iii), in which | . . . | is replaced by | . . . |. For further variations, see Talagrand 87 and Talagrand 96. If we wish to adapt the ideas here to spaces of equivalence classes of functions rather than spaces of true functions, we find that problems of measurability evaporate, and that (because the definition of stability looks only at sets of finite measure) all the relevant suprema can be interpreted as suprema of countable sets. Consequently a subset of L0 (µ) or L0 (A) is stable if all its countable subsets are stable (465Pa). It is remarkable that, for strictly localizable measures µ, we can lift any stable set in L0 to a stable set in L0 (465Pb, 465Q). By moving to function spaces we get a language in which to express a new kind of permanence property of stable sets (465R, 465Xq). See also Talagrand 89. The definition of ‘stable set’ of functions seems to be utterly dependent on the underlying measure space. But 465R tells us that in fact the property of being an order-bounded stable subset of L1 (¯ µ) is invariant under normed space automorphisms. (‘Order-boundedness’ is a normed space invariant in L1 spaces by the Chacon-Krengel theorem, 371D.) Since stability can be defined in terms of order-bounded sets (465Xh), we could, for instance, develop a theory of stable sets in abstract L-spaces. The theory of stable sets is of course bound intimately to the theory of product measures, and such results as 465J have independent interest as theorems about sets in product spaces. So any new theory of product measures will give rise to a new theory of stable sets. In particular, the τ -additive product measures in §417 lead to R-stable sets (465S). It is instructive to work through the details, observing how the properties of the product are employed. Primarily, of course, we need ‘associative’ and ‘commutative’ laws, and Fubini’s theorem; but some questions of measurability need to be re-examined, as in 465Yd. I have starred 465U because it involves the notion of R-stability. In fact this appears only in the final stage, and the construction, as set out in part (a) of the proof, is an instructive challenge to any intuitive concept of what stable sets are like.

466 Measures on linear topological spaces In this section I collect a number of results on the special properties of topological measures on linear topological spaces. The most important is surely Phillips’ theorem (466A-466B): on any Banach space, the weak and norm topologies give rise to the same totally finite Radon measures. This is not because the weak and norm topologies have the same Borel σ-algebras, though this does happen in interesting cases (466C-466E, §467). When the Borel σ-algebras are different, we can still ask whether the Borel measures are ‘essentially’ the same, that is, whether every (totally finite) Borel measure for the weak topology extends to a Borel measure for the norm topology. A construction due to M.Talagrand (466H) gives a negative answer.

466B

Measures on linear topological spaces

591

Just as in Rr , a totally finite quasi-Radon measure on a locally convex linear topological space is determined by its characteristic function (466K). I end the section with a note on measurability conditions sufficient to ensure that a linear operator between Banach spaces is continuous (466L-466M). 466A Theorem Let (X, T) be a metrizable locally convex linear topological space and µ a σ-finite measure on X which is quasi-Radon for the weak topology Ts (X, X ∗ ). Then the support of µ is separable, so µ is quasi-Radon for the original topology T. If X is complete and µ is locally finite with respect to T, then µ is Radon for T. proof (a) Let hVn in∈N be a sequence running over a base of neighbourhoods of 0 in X, and for n ∈ N set |f (x)| ≤ 1 for every x ∈ Vn }. Then each Vn◦ is a convex Ts -compact subset of X ∗ Vn◦ = {f : f ∈ X ∗ , S ∗ (4A4Bh), and X = n∈N Vn◦ . Let Z ⊆ X be the support of µ, and for n ∈ N set Kn = {f ¹Z : f ∈ Vn◦ }. Since the map f 7→ f ¹Z : X ∗ → C(Z) is linear and continuous for Ts and the topology Tp of pointwise convergence on Z, each Kn is convex and Tp -compact. Next, the subspace measure µZ on Z is σ-finite and strictly positive, so by 463E each Kn is Tp -metrizable. Let U be the base for Tp consisting of sets of the form U (I, h, ²) = {f : f ∈ C(Z), |f (x) − h(x)| < ² for every x ∈ I}, where I ⊆ Z is finite, h ∈ RI and ² > 0. For each n ∈ N, write Vn = {Kn ∩ U : U ∈ U }, so that Vn is a base for the subspace topology of Kn (4A2B(a-iv)). Since this is compact and metrizable, therefore second-countable, there is a countable base Vn0 ⊆ Vn (4A2Ob). Now there is a countable set U 0 ⊆ U such that Vn0 ⊆ {Kn ∩ U : U ∈ U 0 } for every n ∈ N, and a countable set D ⊆ Z such that U 0 ⊆ {U (I, h, ²) : I ⊆ D is finite, h ∈ RI , ² > 0}. Let D0 ⊆ X be the linear span of D over the rationals. Then D0 is again countable, and its T-closure Y is a linear subspace of X (4A4Bi). ?? If Z 6⊆ Y , then µ(X \ Y ) > 0. But, writing Y ◦ = {f : f ∈ X ∗ , f (x) ≤ 0 for every x ∈ Y } = {f : f ∈ X ∗ , f (x) = 0 for every x ∈ Y }, S X \ Y must be f ∈Y ◦ {x : |f (x)| > 0}, by 4A4Eb. Because µ is τ -additive, there must be an f ∈ Y ◦ such that µ{x : |f (x)| > 0} > 0. Let n ∈ N be such that f ∈ Vn◦ . Since f (x) = 0 for every x ∈ D, f ¹Z belongs to the same members of Vn0 as the zero function; since Vn0 is a base for the Hausdorff subspace topology of Kn , f ¹Z actually is the zero function, and {x : |f (x)| > 0} ⊆ X \ Z, so µZ < µX, contrary to the choice of Z. X X Thus Z ⊆ Y . Because Y = D0 is T-separable, and T is metrizable, Z is also T-separable (4A2P(a-iv)). (b) The subspace topology TY induced on Y by T is a separable metrizable locally convex linear space topology, so the Borel σ-algebras of TY and the associated weak topology Ts (Y, Y ∗ ) are equal (4A3V). But Ts (Y, Y ∗ ) is just the subspace topology induced on Y by Ts (4A4Ea), Accordingly the subspace measure µY on Y is Ts (Y, Y ∗ )-quasi-Radon (415B). Since every TY -open set is Ts (Y, Y ∗ )-Borel, µY is a topological measure for TY . Since TY is finer than Ts (Y, Y ∗ ), µY is effectively locally finite for TY and inner regular with respect to the TY -closed sets, and therefore is a quasi-Radon measure for TY (415D(i)). By 415J, there is a measure µ ˜ on X, quasi-Radon for T, such that µ ˜E = µY (E ∩ Y ) whenever µ ˜ measures E. But as Y is T-closed and µ-conegligible, and µ is complete, we have µ = µ ˜, and µ is quasi-Radon for T. (c) If X is complete and µ is locally finite with respect to T, then (X, T) is a pre-Radon space (434Jg), so µ is a Radon measure for T (434Jb). 466B Corollary (Compare 462G.) If X is a Banach space and µ is a totally finite measure on X which is quasi-Radon for the weak topology of X, it is a Radon measure for both the norm topology and the weak topology. proof By 466A, µ is a Radon measure for the norm topology; by 418I, or otherwise, it is a Radon measure for the weak topology. Remark Thus Banach spaces, with their weak topologies, are pre-Radon.

592


466C

466C Definition A normed space X has a Kadec norm (also called Kadec-Klee norm) if the norm and weak topologies coincide on the sphere {x : kxk = 1}. Example For any set I, any p ∈ [1, ∞[, the Banach space `p (I) norm. P P If x ∈ `p (I) P has a Kadec p and kxkp = 1 and ² > 0, let J ⊆ I be a finite set such that i∈I\J |x(i)| ≤ ². Set H = {y : y ∈ P `p (I), i∈J |y(i) − x(i)|p < ²}; then H is open for the weak topology of `p (I). Q Q For further examples, see §467. 466D Proposition (Hansell 01) Let X be a normed space with a Kadec norm. Then there is a network S for the norm topology on XSexpressible in the form n∈N Vn , where for each n ∈ N Vn is an isolated family for the weak topology and Vn is the difference of two closed sets for the weak topology. S proof Let U be a σ-disjoint base for the norm topology of X (4A2L(h-ii)); express it as n∈N Un where every Un is disjoint. For rational numbers q, q 0 with 0 < q < q 0 , set Sqq0 = {x : q < kxk ≤ q 0 }, and for A ⊆ X write W (A, q, q 0 ) for the interior of A ∩ Sqq0 taken in the subspace weak topology of Sqq0 . Set Vnqq0 = {W (U, q, q 0 ) : U ∈ Un }, so that Vnqq0 is a disjoint family of relatively-weakly-open subsets of Sqq0 , and S is an isolated family for the weak topology. Now n∈N,q,q0 ∈Q,0 0, then take n ∈ N, U ∈ Un such that x ∈ U ⊆ {y : ky − xk ≤ ²}. Let δ > 0 be such that {y : ky − xk ≤ δ} ⊆ U . Next, because k k is a Kadec norm, there is a weak neighbourhood V of 0 such that ky − xk ≤ 21 δ whenever y ∈ x − V and kyk = kxk. Let V 0 be an open weak neighbourhood of 0 such that V 0 + V 0 ⊆ V . Let η > 0 be such that η < 1, ηkxk ≤ 12 δ and y ∈ V 0 whenever kyk ≤ ηkxk. If y ∈ x − V 0 and (1 − η)kxk ≤ kyk ≤ (1 + η)kxk, then ky − x− so kx −

kxk yk kyk

≤

1 2δ

kxk y kyk

kxk yk kyk

kxk y kyk

= |1 −

∈ x + V . Since

kyk |kxk kxk

= (x − y) + (y −

kxk y) kyk

≤ ηkxk,

∈V0+V0 ⊆V,

and kx − yk ≤ δ and y ∈ U . This means that if we take q, q 0 ∈ Q such that

(1 − η)kxk ≤ q ≤ kxk ≤ q 0 ≤ (1 + η)kxk, (x − V 0 ) ∩ Sqq0 ⊆ U and x ∈ W (U, q, q 0 ) ∈ Vnqq0 . Since of course W (U, q, q 0 ) ⊆ U ⊆ {y : ky − xk ≤ ²}, and x and ² are arbitrary, we have the result. Q Q To get a σ-isolated family for the weak topology which is a network for the norm topology on the whole of X, we just have to add the singleton set {0}. To S see that the union of each of our isolated families is the difference of two weakly open sets, observe that Vnqq0 is a relatively weakly open subset of Sqq0 , which is the difference of the weakly open sets {x : kxk > q} and {x : kxk > q 0 }. 466E Corollary Let X be a normed space with a Kadec norm. (a) The norm and weak topologies give rise to the same Borel σ-algebras. (b) The weak topology has a σ-isolated network, so is hereditarily weakly θ-refinable. proof (a) Write Bk k , BTs for the Borel σ-algebras for the weak and norm topologies. Of course BTs ⊆ Bk k . Let hVn in∈N be a sequence covering a network for the S norm topology as in 466D. Because Vn is (for the weak S topology) isolated and its union S belongs to BTs , S W ∈ BTs for every W ⊆ Vn , n ∈ N. But this means that W ∈ BTs for every W ⊆ V = n∈N Vn ; and as n∈N Vn is a network for the norm topology, every norm-open set belongs to BTs , and Bk k ⊆ BTs . Thus the two Borel σ-algebras are equal. (b) Of course V is also a network for the weak topology, so the weak topology has a σ-isolated network; by 438Ld, it is hereditarily weakly θ-refinable. 466F Proposition Let X be a Banach space with a Kadec norm. Then the following are equiveridical: (i) X is a Radon space in its norm topology; (ii) X is a Radon space in its weak topology; (iii) the weight of X (for the norm topology) is measure-free in the sense of §438.

466H


593

proof (a)(i) ⇐⇒ (ii) By 466Ea, the norm and weak topologies give rise to the same algebra B of Borel sets. If X is a Radon space in its norm topology, then any totally finite measure with domain B is inner regular with respect to the norm-compact sets, therefore inner regular with respect to the weakly compact sets, and X is Radon in its weak topology. If X is a Radon space in its weak topology, then any totally finite measure µ with domain B has a completion µ ˆ which is a Radon measure for the weak topology, therefore also for the norm topology, by 466B; as µ is arbitrary, X is a Radon space for the norm topology. (b)(i) ⇐⇒ (iii) is 438H. 466G Definition A partially ordered set X has the σ-interpolation property or countable separation property if whenever A, B are non-empty countable subsets of X and x ≤ y for every x ∈ A, y ∈ B, then there is a z ∈ X such that x ≤ z ≤ y for every x ∈ A, y ∈ B. A Dedekind σ-complete partially ordered set (314Ab) always has the σ-interpolation property. 466H Proposition (Jayne & Rogers 95) Let X be a Riesz space with a Riesz norm, given its weak topology Ts = Ts (X, X ∗ ). Suppose that (α) X has the σ-interpolation property (β) there is a strictly increasing family hpξ iξ<ω1 in X. Then there is a Ts -Borel probability measure µ on X such that (i) µ is not inner regular with respect to the Ts -closed sets; (ii) µ is not τ -additive for the topology Ts ; (iii) µ has no extension to a norm-Borel measure on X. Accordingly (X, Ts ) is not a Radon space (indeed, is not Borel-measure-complete). proof (a) Let K be the set {f : f ∈ X ∗ , f ≥ 0, kf k ≤ 1} = {f : kf k ≤ 1} ∩

T

x∈X + {f

: f (x) ≥ 0},

∗

so that K is a weak*-closed subset of the unit ball of X and is weak*-compact. Because X ∗ is a solid linear subspace of the order-bounded dual X ∼ of X (356Da), every member of X ∗ is the difference of two non-negative members of X ∗ , and K spans X ∗ . We shall need to know that if x < y in X, there is an f ∈ K such that f (x) < f (y); set f =

1 |g| kgk

where g ∈ X ∗ is such that g(x) 6= g(y). (Recall that the norm of X ∗

is a Riesz norm, as also noted in 356Da.) Set A=

S

ξ<ω1 {x

: x ∈ X, x ≤ pξ };

then every sequence in A has an upper bound in A, but A has no greatest member. It follows that if B ⊆ K is countable there is an x ∈ A such that f (x) = supy∈A f (y) for every f ∈ B, so that f (x) = f (y) whenever x ≤ y ∈ A and f ∈ B. Let E be the family of those setsSE ⊆ X such that A ∩ E is cofinal with A. Observe that if hEn in∈N is P?? Otherwise, there is for any sequence of subsets of X, and n∈N En ∈ E, then some En belongs to E. P each n ∈ N an xn ∈ A such that xn 6≤Sy for every y ∈ A ∩ En . Now there is an x ∈ A such that xn ≤ x for every n ∈ N, and x 6≤ y for every y ∈ n∈N En . X XQ Q (b) If G ∈ E is a Ts -open set, there is a k ∈ N such that for every z ∈ A there are x ∈ A and f0 , . . . , fk ∈ K such that x ≥ z and {y : y ∈ X, |fi (y−x)| ≤ 2−k for every i ≤ k} is included in G. P P For each k ∈ N let Hk be the set of those x ∈ X such that {y : |fi (y − x)| ≤ 2−k for every S i ≤ k} is included in G for some f0 , . . . , fk ∈ K. Because the linear span of K is X ∗ , and G is Ts -open, G = k∈N Hk . So, as remarked in (a), there is some k ∈ N such that Hk ∈ E, and this k will serve. Q Q T (c) (The key.) If hGn in∈N is any sequence of Ts -open sets belonging to E, then n∈N Gn ∈ E. P P Start from any z ∗ ∈ A. For each n ∈ N, choose kn ∈ N as in (b), so that there is a family hfnzi iz∈A,i≤kn in K such that for every z ∈ A there is an x ∈ A such that x ≥ z and {y : |fnzi (y − x)| ≤ 2−kn ∀ i ≤ kn } ⊆ Gn .

594


466H

Let F be any ultrafilter on A containing {x : x ∈ A, x ≥ z} for every z ∈ A, and write fni = limz→F fnzi for n ∈ N and i ≤ kn , the limit being taken for the weak* topology on K. Let z1∗ > z ∗ be such that z1∗ ∈ A and fni (x − z1∗ ) = 0 whenever x ∈ A, x ≥ z1∗ , n ∈ N and i ≤ kn . Choose sequences hxn in∈N , hyn in∈N and hzn in∈N in A inductively, as follows. Set y0 = z1∗ . Given yn , the set Cn = {z : z ∈ A, z ≥ yn , |(fnzi − fni )(yn − z1∗ )| ≤ 2−kn for every i ≤ kn } belongs to F, so is not empty; take zn ∈ Cn . Because yn ≥ z1∗ , we have fni (yn −z1∗ ) = 0 and |fnzn i (yn −z1∗ )| ≤ 2−kn , for every i ≤ kn . Let xn ∈ A be such that xn ≥ zn and {y : |fnzn i (y − xn )| ≤ 2−kn for every i ≤ kn } ⊆ Gn . Now let yn+1 ∈ A be such that yn+1 ≥ xn and fnzn i (y − yn+1 ) = 0 whenever y ∈ A and y ≥ yn+1 . Continue. At the end of the induction, let z2∗ be an upper bound for {yn : n ∈ N} in A. For n ∈ N, set Pn un = z1∗ + j=0 xj − yj , vn = un + z2∗ − yn+1 . Then hun in∈N is non-decreasing and un ≤ vn ≤ z2∗ for every n ∈ N; moreover, vn − vn+1 = yn+1 − xn+1 − yn+1 + yn+2 ≥ 0 for every n, so hvn in∈N is non-increasing, and um ≤ vn for all m, n ∈ N. Because X has the σ-interpolation property, there is an x ∈ X such that un ≤ x ≤ vn for every n ∈ N. Since z1∗ ≤ x ≤ z2∗ , x ∈ A and x > z ∗ . Take any n ∈ N. Then fnzn i (x − un ) ≤ fnzn i (vn − un ) = fnzn i (z2∗ − yn+1 ) = 0 for every i ≤ kn . On the other hand, xn − un = (yn − z1∗ ) −

Pn−1

j=0 (xj

− yj )

lies between 0 and yn − z1∗ , so 0 ≤ fnzn i (xn − un ) ≤ 2−kn

T for every i ≤ kn , and |fnzn i (x − xn )| ≤ 2−kn for every i. Thus x ∈ Gn . As n is arbitrary, x ∈ n∈N Gn . T T As x > z ∗ , this shows that z ∗ is not an upper bound of n∈N Gn ∩ A. As z ∗ is arbitrary, n∈N Gn ∈ E. Q Q (d) Set T

K0 = {G \ H : G, H ∈ Ts , G ∈ E, H ∈ / E}.

P Express each En as GnT\ Hn where Gn , Hn are Then n∈N En 6= ∅ for any sequence hEnSin∈N in K0 . P T -open, G ∈ E and H ∈ / E. Then H ∈ / E, as noted in (a), while n∈N Gn ∈ E, by (c); so n n∈N n Ts T Sn G \ H is non-empty. Q Q E = n∈N n n∈N n n∈N n It follows that K = K0 ∪ {∅} is a countably compact class in the sense of 413L. Moreover, E ∩ E 0 ∈ K for all E, E 0 ∈ K (using (b) and (c) again), so if we define φ0 : K → {0, 1} by writing φ0 (E) = 1 for E ∈ K0 , φ0 (∅) = 0, then K and φ0 will satisfy all the conditions of 413M. There is therefore a measure µ ˆ on X extending φ0 and inner regular with respect to Kδ , the family of sets expressible as intersections of sequences in K. The domain of µ ˆ must include every member of K; but if G ∈ Ts then either G or X \ G belongs to K0 , so is measured by µ ˆ, and µ ˆ is a topological measure. We need to observe that, because φ0 takes only the values 0 and 1, µ Ê ≤ 1 for every E ∈ Kδ , and µ ˆX ≤ 1; since φ0 X = 1, µ ˆX = 1 and µ ˆ is a probability measure. (e) We may therefore take µ to be the restriction of µ ˆ to the algebra B of Borel sets, and µ is a Ts -Borel probability measure. Now µ is not inner regular with respect to the Ts -closed sets. P P For each ξ < ω1 , pξ < pξ+1 < pξ+2 , so there are gξ , hξ ∈ K such that gξ (pξ ) < gξ (pξ+1 ) and hξ (pξ+1 ) < hξ (pξ+2 ). Let D ⊆ ω1 be any set such that D and ω1 \ D are both uncountable, and set S G = ξ∈D {x : gξ (pξ ) < gξ (x), hξ (x) < hξ (pξ+2 )}. Then G ∈ Ts , and pξ+1 ∈ G for every ξ ∈ D, so G ∈ E and µG = φ0 G = 1. On the other hand, if η ∈ ω1 \ D, then for every ξ ∈ D either ξ < η and hξ (pξ+2 ) ≤ hξ (pη+1 ), or ξ > η and gξ (pη+1 ) ≤ gξ (pξ ); thus pη+1 ∈ /G

466J


595

for any η ∈ ω1 \ D, and X \ G ∈ E. But this means that if F ⊆ G is closed then X \ F ∈ K0 and µF = 0. Thus µG > supF ⊆G is closed µF . Q Q (f ) Because Ts is regular, µ cannot be τ -additive, by 414Mb. It follows at once that (X, Ts ) is not Borelmeasure-complete, and in particular is not a Radon space. To see that µ has no extension to a norm-Borel measure, we need to look again at the set A. For each ξ < ω1 , Fξ = {x : x ≤ pξ }. Then every Fξ is norm-closed (354Bc) and hFξ iξ<ω1 is an increasing family with union A. Consequently, A is norm-closed (4A2Ld, 4A2Ka). At the same time, every Fξ is convex (cf. 351Ce), so A is also convex. It follows that A, like every Fξ , is Ts -closed (3A5Ee). So µ measures A and every Fξ . Because X \ Fξ is a Ts -open set belonging to E, µFξ = 0 for every ξ < ω1 ; because X \ A is a Ts -open set not belonging to E, µA = 1. But ω1 is a measure-free cardinal (438Cd), so 438I tells us that λA = supξ<ω1 λFξ for any semi-finite norm-Borel measure λ on X. Thus µ has no extension to a norm-Borel measure, and the proof is complete. 466I Examples The following spaces satisfy the hypotheses of 466H. (a) (Talagrand 78a, or Talagrand 84, 16-1-2) X = `∞ (I) or {x : x ∈ `∞ (I), {i : x(i) 6= 0} is countable}, where I is uncountable. P P X has the σ-interpolation property because it is Dedekind complete, and if hiξ iξ<ω1 is any family of distinct elements of I, we can set pξ (iη ) = 1 for η ≤ ξ, pξ (i) = 0 for all other i ∈ I to obtain a strictly increasing family hpξ iξ<ω1 in X. Q Q (b) (de Maria & Rodriguez-Salinas 91) X = `∞ /cc0 . P P (i) To see that X has the σ-interpolation property, let A, B ⊆ X be non-empty countable sets such that u ≤ v for all u ∈ A, v ∈ B. Let hxn in∈N , hyn in∈N be sequences in `∞ such that A = {x•n : n ∈ N} and B = {yn• : n ∈ N}. Set x ñ = supi≤n xi , yñ = inf i≤n yi for n ∈ N; then x ˜•n ≤ yñ• , so (˜ xn − yñ )+ ∈ c0 . Set kn = max({n} ∪ {i : x ñ (i) ≥ yñ (i) + 2−n }) for n ∈ N, and define x ∈ `∞ by writing x(i) = 0 if i ≤ k0 , =x ñ (i) if kn < i ≤ kn+1 . Then it is easy to check that u ≤ x ≤ v for every u ∈ A, v ∈ B; as A and B are arbitrary, X has the σ-interpolation property. •

(ii) To see that X has the other property, recall that there is a family hIξ iξ<ω1 of infinite subsets of N such that Iξ \ Iη is finite if η ≤ ξ, infinite if ξ < η (4A1Fa). Setting pξ = χ(N \ Iξ )• , we have a strictly increasing family hpξ iξ<ω1 in X. Q Q 466J Proposition Let X be a linear topological space andRΣ its cylindrical σ-algebra. If µ and ν are R probability measures with domain Σ such that eif (x) µ(dx) = eif (x) ν(dx) for every f ∈ X ∗ , then µ = ν. ∗

proof Define T : X → RX by setting (T x)(f ) = f (x) for f ∈ X ∗ , x ∈ X. Then T is linear and continuous ∗ ∗ for the weak topology of X. So if F ⊆ RX is a Baire set for the product topology of RX , T −1 [F ] is a Baire set for the weak topology of X (4A3Kc) and belongs to Σ (4A3U). We therefore have Baire measures ∗ ∗ µ0 , ν 0 on RX defined by saying that µ0 F = µT −1 [F ] and ν 0 F = νT −1 [F ] for every Baire set FP⊆ RX . ∗ n If h : RX → R is a continuous linear functional, it can be expressed in the form h(z) = i=0 αi z(fi ) ∗ where f0 , . . . , fn ∈ X and α0 , . . . , αn ∈ R. So Pn Pn h(T x) = i=0 αi (T x)(fi ) = i=0 αi fi (x) = f (x) Pn for every x ∈ X, where f = i=0 αi fi . This means that

R

eih(z) µ0 (dz) =

R

eih(T x) µ(dx) =

R

eif (x) µ(dx) =

R

eif (x) ν(dx) =

R

eih(z) ν 0 (dz).

As h is arbitrary, µ0 = ν 0 (454Ma). ∗ Now let Σ0 be the family of subsets of X of the form T −1 [F ] where F ⊆ RX is a Baire set. This is a σ-algebra and contains all sets of the form {x : f (x) ≥ α} where f ∈ X ∗ and α ∈ R. So every member of X ∗ is Σ0 -measurable and Σ0 must include the cylindrical σ-algebra of X. Since µ and ν agree on Σ0 they must be identical.

596


466K

466K Proposition If X linear topological space and µ, ν are quasi-Radon probability R is a locally convex R measures on X such that eif (x) µ(dx) = eif (x) ν(dx) for every f ∈ X ∗ , then µ = ν. proof Write T for the given topology on X and Ts = Ts (X, X ∗ ) for the weak topology. By 466J, µ and ν must agree on the cylindrical σ-algebra Σ of X. Since Σ includes a base for Ts , every weakly open set G is the union of an upwards-directed family of open sets belonging to Σ; as µ and ν are τ -additive, µG = νG. Consequently µ and ν agree on Ts -closed sets, and therefore on T-closed convex sets (4A4Ed). Write H for the family of T-open sets which are expressible as the union of a non-decreasing sequence of T-closed convex sets; then µ and ν agree on H. If τ is a T-continuous seminorm on X, x0 ∈ X and α > 0, then {x : τ (x − x0 ) < α} ∈ H; and sets of this kind constitute a base for T (4A4Cb). Also the intersection of two members of H belongs to H. By 415H(v), µ = ν. Remark This generalizes 285M and 454Xk, which are the special cases X = Rr (for finite r) and X = RI . 466L Proposition Suppose that X and Y are Banach spaces and that T : X → Y is a linear operator such that gT : X → R is universally Radon-measurable, in the sense of 434Ec, for every g ∈ Y ∗ . Then T is continuous. proof ?? Suppose, if possible, otherwise. Then there is a g ∈ Y ∗ such that gT is not continuous (4A4Ib). For each ∈ N, take xn ∈ X such that kxn k = 2−n and g(T xn ) > 2n. Define h : {0, 1}N → X by setting Pn P∞ ∞ 0 h(t) = n=0 t(n)xn (4A4Ie). Then h is continuous, because kh(t) − h(t )k ≤ n=0 2−n |t(n) − t0 (n)| for all t, t0 ∈ {0, 1}N . Let ν be the usual measure on {0, 1}N ; then the image measure µ = νh−1 is a Radon measure on X (418I), so gT must be dom µ-measurable, and φ = gT h is dom ν-measurable. In this case there is an m ∈ N such that E = {t : |φ(t)| ≤ m} has measure greater than 21 . But as g(T xm ) > 2m, we see that if t ∈ E then t0 ∈ / E, where t0 differs from t at the mth coordinate only, so that |φ(t) − φ(t0 )| = g(T xm ). Since the map t 7→ t0 is an automorphism of the measure space ({0, 1}N , ν), νE ≤ 12 , which is impossible. X X 466M Corollary If X is a Banach space, Y is a separable Banach space, and T : X → Y is a linear operator such that the graph of T is a Souslin-F set in X × Y , then T is continuous. proof It is enough to show that T ¹Z is continuous for every separable closed linear subspace Z of X (because then it must be sequentially continuous, and we can use 4A2Ld). Write Γ ⊆ X × Y for the graph of T . If H ⊆ Y is open, then Γ ∩ (Z × H) is a Souslin-F set in the Polish space Z × Y , so is analytic (423E), and its projection (T ¹Z)−1 [H] is also analytic (423Ba), therefore universally measurable (434Dc). Thus T ¹Z : Z → Y is a universally measurable function, and gT ¹Z must be universally measurable for any g ∈ Y ∗ (434Df). By 466L, T ¹Z is continuous; as Z is arbitrary, T is continuous. 466X Basic exercises (a) Let (X, T) be a metrizable locally convex linear topological space and µ a totally finite measure on X which is quasi-Radon for the topology T. Show that µ is quasi-Radon for the weak topology Ts (X, X ∗ ). > (b) Let hen in∈N be the usual orthonormal basis of `2 . Give `2 the Radon probability measure ν such that ν{en } = 2−n−1 for every n. Let I be an uncountable set, and set X = (`2 )I with the product linear structure and the product topology T, each copy of `2 being given its norm topology; write Ts for the weak topology of X. (i) Let λ be the τ -additive product of copies of ν (417G). Show that λ is quasi-Radon for T but is not inner regular with respect to the Ts -closed sets. (ii) Let µs be the τ -additive product measure of copies of µ when each copy of `2 is given its weak topology instead of its norm topology. Show that µs is quasi-Radon for Ts but does not measure every T-Borel set. (Hint: setting E = {en : n ∈ N}, µ(E I ) = 1 and E I is relatively Ts -compact.) > (c) Let X be a metrizable locally convex linear topological space and µ a τ -additive totally finite measure on X with domain the cylindrical σ-algebra of X. Show that µ has an extension to a quasi-Radon measure on X. (Hint: 4A3U, 415N.) (d) Let X be a metrizable locally convex linear topological space which is Lindelöf in its weak topology, and Σ the cylindrical σ-algebra of X. Show that any totally finite measure with domain Σ has an extension to a quasi-Radon measure on X.

466Zb


597

> (e) Let X be a separable Banach space. Show that it is a Radon space when given its weak topology. (Hint: 4A3V, 467E.) (f ) Let K be a compact metric space, and C(K) the Banach space of continuous real-valued functions on K. Show that the σ-algebra of subsets of C(K) generated by the functionals x 7→ x(t) : C(K) → R, for t ∈ K, is just the cylindrical σ-algebra of C(K). (Hint: 4A2Pe.) Show that C(K) is a pre-Radon space when given its weak topology. (See 462Z.) (g) Re-write part (d) of the proof of 466H to avoid any appeal to results from §413. (h) Let f : R → R be a function such that f (1) = 1 and f (x + y) = f (x) + f (y) for all x, y ∈ R. Show that the following are equiveridical: (i) f (x) = x for every x ∈ R; (ii) f is continuous at some point; (iii) f is bounded on some non-empty open set; (iv) f is bounded on some Lebesgue measurable set of non-zero measure; (v) f is Lebesgue measurable; (vi) f is Borel measurable; *(vii) f is bounded on some non-meager b Gδ set; *(viii) f is B-measurable, where Bb is the Baire property algebra of R. (Hint: 443Db.) (i) Set X = {x : x ∈ RN , {n : x(n) 6= 0} is finite}, and give X any norm. Show that any linear operator from X to any normed space is universally measurable. (j) Let X be a locally convex linear topological space and µ, ν two totally finite quasi-Radon measures on X. Show that if µ and ν give the same measure to every half-space {x : f (x) ≥ α}, where f ∈ X ∗ and α ∈ R, then µ = ν. (k) Let X be a Hilbert space and µ, ν two totally finite Radon measures on X. Show that if µ and ν give the same measure to every ball B(x, δ), where x ∈ X and δ ≥ 0, then µ = ν. (Hint: every open half-space is the union of a non-decreasing sequence of balls.) (l) Let µ be a Radon measure on `∞ (I), where I is any set. Set fi (x) = x(i) for x ∈ `∞ (I), i ∈ I. Show that {fi• : i ∈ I} is relatively compact in L0 (µ). (Hint: 463Xc.) > (m) Show that Talagrand’s measure, intepreted as a measure on `∞ (I) (464R), is not τ -additive for the weak topology. 466Y Further exercises (a) Give an example of a Hausdorff locally convex linear topological space (X, T) with a probability measure µ on X which is a Radon measure for the weak topology Ts (X, X ∗ ) but not for the topology T. (Hint: take C = C([0, 1]) and X = C ∗ with the Mackey topology for the dual pair (X, C), that is, the topology of uniform convergence on weakly compact subsets of C.) (b) In 466H, show that there is a Borel set for the norm topology on X which is not a Borel set for the weak topology. (Hint: we can adjust the family hpξ iξ<ω1 to ensure that, for someS² > 0, kpξ − pη k ≥ ² whenever η < ξ. For ξ < ω1 set Fξ = {x : x ≤ pξ , x 6≤ pη for any η < ξ}. Show that ξ∈D Fξ is norm-Borel S for every D ⊆ ω1 , while µ∗ ( ξ<ω1 Fξ ) = 1.) (c) Find Banach spaces X and Y and a linear operator from X to Y which is not continuous but whose graph is an Fσ set in X × Y . 466Z Problems (a) Does every probability measure defined on the Ts (`∞ , (`∞ )∗ )-Borel sets of `∞ = `∞ (N) extend to a measure defined on the k k∞ -Borel sets? It is by no means obvious that the Borel sets of `∞ are different for the weak and norm topologies; for a proof see Talagrand 78b. (b) Assume that c is measure-free. Does it follow that `∞ , with its weak topology, is a Radon space? Note that a positive answer to 464Z (with I = N) would settle this, since Talagrand’s measure, when interpreted as a measure on `∞ , cannot agree on the weakly Borel sets with any Radon measure on `∞ (466Xl, 466Xm).

598


466 Notes

466 Notes and comments I have given a proof of 466A using the machinery of §463; when the measure µ is known to be a Radon measure for the weak topology, rather than just a quasi-Radon measure or a τ -additive measure on the cylindrical algebra (466Xc), the theorem is older than this, and for an instructive alternative approach see Talagrand 84, 12-1-4. Another proof is in Jayne & Rogers 95. On any Banach space we have at least three important σ-algebras: the norm-Borel σ-algebra (generated by the norm-open sets), the weak-Borel algebra (generated by the weakly open sets) and the cylindrical algebra (generated by the continuous linear functionals). (Note that the Baire σ-algebras corresponding to the norm and weak topologies are the norm-Borel algebra (4A3Kb) and the cylindrical algebra (4A3U).) If our Banach space is naturally represented as a subspace of some RI (e.g., because it is a space of continuous functions), then we have in addition the σ-algebra generated by the functionals x 7→ x(i) for i ∈ I, and the Borel algebra for the topology of pointwise convergence. We correspondingly have natural questions concerning when these algebras coincide, as in 466E and 4A3V and 466Xf, and when a measure on one of the algebras leads to a measure on another, as in 466A-466B. The question of which Banach spaces are Radon spaces in their norm topologies is, if not exactly ‘solved’, at least reducible to a classical problem in set theory by the results in §438. It seems much harder to decide which non-separable Banach spaces are Radon spaces in their weak topologies. We have a simple positive result for spaces with Kadec norms (466F), and after some labour a negative result for a couple of standard examples (466I). The case of `∞ seems still to be open in ‘ordinary’ set theories (466Zb). `∞ is of particular importance in this context because the dual of any separable Banach space is isometrically isomorphic to a linear subspace of `∞ (4A4Id). So a positive answer to either question in 466Z would have very interesting consequences – and would be correspondingly surprising. 466L and 466M belong to a large family of results of the general form: if, between spaces with both topological and algebraic structures, we have a homomorphism (for the algebraic structures) which is not continuous, then it is wildly irregular. I hope to return to some of these ideas in Volume 5. For the moment I just give a version of the classical result that an additive function f : R → R which is Lebesgue measurable must be continuous (466Xh). The definition of ‘universally measurable’ function which I gave in §434 has a number of paradoxical aspects. I have already remarked that in some contexts we might prefer to use the notion of ‘universally Radon-measurable’ function; this is also appropriate for 466L. But when our space X, for any reason, has few Borel measures, as in 466Xi, there are correspondingly many universally measurable functions defined on X. Of course 466M can also be thought of as a generalization of the closed graph theorem; but note that, unlike the closed graph theorem, it needs a separable codomain (466Yc). A curious geometric question concerning measures on metric spaces is the following. If two totally finite Radon measures on a metric space agree on balls, must they be identical? It is known that (even for compact spaces) the answer, in general, is ‘no’ (Davies 71); in Hilbert spaces the answer is ‘yes’, for nontrivial reasons (466Xk); in any finite-dimensional normed space, the answer is again ‘yes’ (Jackson & Mauldin 99).

*467 Locally uniformly rotund norms In the last section I mentioned Kadec norms. These are interesting in themselves, but the reason for including them in this book is that in a normed space with a Kadec norm the weak topology has the same Borel sets as the norm topology (466Ea). The same will evidently be true of any space which has an equivalent Kadec norm. Now Kadec norms themselves are not uncommon, but equivalent Kadec norms appear in a striking variety of cases. Here I describe the principal class of spaces (the ‘weakly K-countably determined’ Banach spaces, 467H) which have equivalent Kadec norms. In fact they have ‘locally uniformly rotund’ norms, which are much easier to do calculations with. Almost everything here is pure functional analysis, mostly taken from Deville Godefroy & Zizler 93, which is why I have starred the section. The word ‘measure’ does not appear until 467P. At that point, however, we find ourselves with a striking result (Schachermayer’s theorem) which appears to need the structure theory of weakly compactly generated Banach spaces developed in 467C-467M. 467A Definition Let X be a linear space with a norm k k. k k is locally uniformly rotund or locally uniformly convex if whenever kxk = 1 and ² > 0, there is a δ > 0 such that kx − yk ≤ ² whenever kyk = 1 and kx + yk ≥ 2 − δ.

467Cb

Locally uniformly rotund norms

599

If X has a locally uniformly rotund norm, then every subspace of X has a locally uniformly rotund norm. 467B Proposition A locally uniformly rotund norm is a Kadec norm. proof Let X be a linear space with a locally uniformly rotund norm k k. Set SX = {x : kxk = 1}. Suppose that G is open for the norm topology and that x ∈ G ∩ SX . Then there is an ² > 0 such that G ⊇ B(x, ²) = {y : ky − xk ≤ ²}. Let δ > 0 be such that kx − yk ≤ ² whenever kyk = 1 and kx + yk ≥ 2 − δ. Now there is an f ∈ X ∗ such that f (x) = kf k = 1 (3A5Ac). So V = {y : f (y) > 1 − δ} is open for the weak topology. But if y ∈ V ∩ SX , then kx + yk ≥ f (x + y) ≥ 2 − δ, so kx − yk ≤ ² and y ∈ G. As x is arbitrary, G ∩ SX is open for the weak topology on SX ; as G is arbitrary, the norm and weak topologies agree on SX . 467C A technical device (a) I will use the following notation for the rest of the section. Let X be a linear space and p : X → [0, ∞[ a seminorm. Define qp : X × X → [0, ∞[ by setting qp (x, y) = 2p(x)2 + 2p(y)2 − p(x + y)2 = (p(x) − p(y))2 + (p(x) + p(y))2 − p(x + y)2 for x ∈ X. (b) A norm k k on X is locally uniformly rotund iff whenever x ∈ X and ² > 0 there is a δ > 0 such that kx − yk ≤ ² whenever qk k (x, y) ≤ δ. P P(i) Suppose that k k is locally uniformly rotund, x ∈ X and ² > 0. (α) If x = 0 then qk k (x, y) = kyk2 = kx − yk2 for every y so we can take δ = ²2 . (β) If x 6= 0 set x0 = 0

0

kx − y k ≤

1 2 ²kxk

0

0

1 x. kxk

Let η > 0 be such that

0

whenever ky k = 1 and kx + y k ≥ 2 − η. Let δ > 0 be such that √ √ √ 1 δ + 2 δkxk ≤ ηkxk2 , δ < kxk, δ ≤ ²kxk2 . 2

Now if qk k (x, y) ≤ δ, we must have (kxk − kyk)2 ≤ δ < kxk2 , so that y 6= 0 and

(kxk + kyk)2 − kx + yk2 ≤ δ,

√ ¯ 1 δ 1 ¯¯ |kyk−kxk| ¯ − kyk = ≤ , kxk

kyk

kxk

kxk + kyk − kx + yk ≤ Set y 0 =

1 y, kyk

y 00 =

1 y. kxk

kxk

δ kxk+kyk+kx+yk

≤

δ . kxk

Then ky 0 k = 1, and √ ¯ 1 1 ¯¯ δ 1 − kyk ≤ ≤ ²kxk. ky 0 − y 00 k = ¯ kyk

kxk

kxk

2

Accordingly √ 1 δ kx + yk − kxk kxk √ √ 1 δ kyk δ δ δ (kxk + kyk − )− =1+ − − kxk kxk kxk kxk kxk2 kxk √ √ √ kxk− δ δ δ δ 2 δ 1+ − = 2 − ≥ 2 − η. − − kxk kxk2 kxk kxk2 kxk

kx0 + y 0 k ≥ kx0 + y 00 k − ky 0 − y 00 k ≥ ≥ ≥

But this means that kx0 − y 0 k ≤ 12 ²kxk, so that kx0 − y 00 k ≤ ²kxk and kx − yk ≤ ². As x and ² are arbitrary, the condition is satisfied. (ii) Suppose the condition is satisfied. If kxk = 1 and ² > 0, take δ ∈ ]0, 2] such that kx − yk ≤ ² whenever q(x, y) ≤ 4δ; then if kyk = 1 and kx + yk ≥ 2 − δ, q(x, y) = 4 − kx + yk2 ≤ 4δ and kx − yk ≤ ². As x and ² are arbitrary, k k is locally uniformly rotund. Q Q

600


467Cc

(c) We have the following elementary facts. Let X be a linear space. (i) For any seminorm p on X, qp (x, y) ≥ 0 for all x, y ∈ X. P P qp (x, y) ≥ (p(x) + p(y))2 − p(x + y)2 ≥ 0 because p(x + y) ≤ p(x) + p(y). Q Q P 2 (ii) Suppose that hpi ii∈I is a family of seminorms on X such that is finite for every i∈I pi (x) pP P 2 p (x) for x ∈ X; then p is a seminorm on X and q = P Of course x ∈ X. Set p(x) = i p i∈I i∈I qpi . P p(αx) = |α|p(x) for α ∈ R and x ∈ X. If x ∈ X, then p(x) = kφ(x)k2 , where φ(x) = hpi (x)ii∈I ∈ `2 (I). Now for x, y ∈ X, 0 ≤ φ(x + y) ≤ φ(x) + φ(y) in `2 (I), so p(x + y) = kφ(x + y)k2 ≤ kφ(x) + φ(y)k2 ≤ kφ(x)k2 + kφ(y)k2 = p(x) + p(y). P Thus p is a seminorm. Now the calculation of qp = i∈I qpi is elementary. Q Q In particular, qp ≥ qpi for every i ∈ I. (iii) If k k is an inner product norm on X, then qk k (x, y) = kx − yk2 for all x, y ∈ X. P P 2kxk2 + 2kyk2 − kx + yk2 = 2(x|x) + 2(y|y) − (x + y|x + y) = (x|x) + (y|y) − (x|y) − (y|x) = (x − y|x − y). Q Q 467D Lemma Let X be a normed space. Suppose that there are a normed space Y with a locally uniformly rotund norm and a bounded linear operator T : Y → X such that T [Y ] is dense in X and, for every x ∈ X and γ > 0, there is a z ∈ Y such that kx − T zk2 + γkzk2 = inf y∈Y kx − T yk2 + γkyk2 . Then X has an equivalent locally uniformly rotund norm. proof (a) For each n ∈ N, x ∈ X set pn (x) =

p inf y∈Y kx − T yk2 + 2−n kyk2 .

Then pn : X → [0, ∞[ is a norm on X, equivalent to k k. P P (i) The functionals (x, y) 7→ kx − T yk, −n/2 (x, y) → 7 2 kyk from X × Y to [0, ∞[ are both seminorms, so the functional (x, y) 7→ φ(x, y) = p 2 −n 2 kx − T yk + 2 kyk also is, by 467C(c-ii). (ii) If x ∈ X and α ∈ R, take z ∈ Y such that pn (x) = φ(x, z); then pn (αx) ≤ φ(αx, αz) = |α|φ(x, z) = |α|pn (x). If α 6= 0, apply the same argument to see that pn (x) ≤ |α|−1 pn (αx), so that pn (αx) = |α|pn (x). (iii) Now take any x1 , x2 ∈ X. Let z1 , z2 ∈ Y be such that pn (xi ) = φ(xi , zi ) for both i. Then 1 2

1 2

1 2

1 2

1 2

1 2

pn (x1 + x2 ) = 2pn ( x1 + x2 ) ≤ 2φ( x1 + x2 , z1 + z2 ) 1 2

1 2

≤ 2( φ(x1 , z1 ) + φ(x2 , z2 )) = pn (x1 ) + pn (x2 ). Thus pn is a seminorm. (iv) pn (x) ≤ φ(x, 0) = kxk for every x ∈ X. (vi) For any x ∈ X and y ∈ Y , either kT yk ≥ 21 kxk and φ(x, y) ≥ 2−n/2 kyk2 ≥

1 kxk, 2·2n/2 kT k

or kT yk ≤ 21 kxk and 1 2

φ(x, y) ≥ kx − T yk ≥ kxk; this shows that pn (x) ≥ min( 21 , 12 2−n/2 kT k−1 )kxk. (I am passing over the trivial case X = {0}, kT k = 0.) In particular, pn (x) = 0 only when x = 0. Thus pn is a norm equivalent to p. Q Q

467E


601

(c) For any x ∈ X, limn→∞ pn (x) = 0. P P Let ² > 0. Let y ∈ Y be such that kx − T yk ≤ ²; then lim supn→∞ pn (x)2 ≤ lim supn→∞ kx − T yk2 + 2−n kyk22 ≤ ²2 . Q As ² is arbitrary, limn→∞ pn (x) = 0. Q p P ∞ −n p (x)2 for x ∈ X. The sum is always finite because p (x) ≤ kxk for every (d) Set kxk0 = n n n=0 2 0 n, so k k is a seminorm; and it is a norm equivalent to k k because p0 is. Now k k0 is locally uniformly rotund. P P Take x ∈ X and ² > 0. Let n ∈ N be such that pn (x) ≤ 14 ². Choose y ∈ Y such that 2 pn (x) = kx−T yk2 +2−n kyk2 . (This is where we really use the hypothesis that the infimum in the definition of pn is attained.) Let δ > 0 be such that 2n δ ≤ ( 14 ²)2 and kT kky 0 − yk ≤ 14 ² whenever qk k (y 0 , y) ≤ 22n δ. If qk k0 (x, x0 ) ≤ δ, then q2−n/2 pn (x, x0 ) ≤ δ, by 467C(c-ii), that is, qpn (x, x0 ) ≤ 2n δ. Let y 0 ∈ Y be such that pn (x0 )2 = kx0 − T y 0 k2 + 2−n ky 0 k22 . Then pn (x + x0 )2 ≤ kx + x0 − T y − T y 0 k2 + 2−n ky + y 0 k2 ≤ (kx − T yk + kx0 − T y 0 k)2 + 2−n ky + y 0 k2 , so qpn (x, x0 ) = 2pn (x)2 + 2pn (x0 )2 − pn (x + x0 )2 ≥ 2(kx − T yk2 + 2−n kyk2 ) + 2(kx0 − T y 0 k2 + 2−n ky 0 k2 ) − (kx − T yk + kx0 − T y 0 k)2 − 2−n ky + y 0 k2 = (kx − T yk − kx0 − T y 0 k)2 + 2−n (2kyk2 + 2ky 0 k22 − ky + y 0 k2 ). This means that qk k (y, y 0 ) ≤ 2n qpn (x,x0 ) ≤ 22n δ, so kT kky − y 0 k ≤ 41 ², while kx − T yk + kx0 − T y 0 k ≤ 2kx − T yk +

√ p 3 qpn (x, x0 ) ≤ 2pn (x) + 2n/2 δ ≤ ². 4

Finally 1 4

3 4

kx − x0 k ≤ kT kky − y 0 k + kx − T yk + kx0 − T y 0 k ≤ ² + ² = ². As x and ² are arbitrary, this shows that k k0 is locally uniformly rotund. Q Q This completes the proof. 467E Theorem Let X be a separable normed space. Then it has an equivalent locally uniformly rotund norm. proof (a) It is enough to show that the completion of X has an equivalent locally uniformly rotund norm; since the completion of X is separable, we may suppose that X itself is complete. Let hxi ii∈N be a sequence in X running over a dense subset of X. Define T : `2 → X by setting P∞ y(i) xi T y = i=0 i 2 (1+kxi k)

for y ∈ `2 = `2 (N); then T y is always defined (4A4Ie); T is a linear operator and pP∞ P∞ −2i kyk kT yk ≤ i=0 2−i |y(i)| ≤ 2 i=0 2 for every y ∈ `2 , by Cauchy’s inequality (244E). So T is a bounded linear operator. (b) T satisfies the conditionspof 467D. P P T [`2 ] is dense because it contains every xi . Given x ∈ X and γ > 0, α ≥ 0, the function y 7→ kx − T yk2 + γkyk22 is a seminorm and norm-continuous, so the set Cα (x) = {y : y ∈ `2 , kx − T yk2 + γkyk22 ≤ α2 } is convex and norm-closed. Consequently, Cα (x) is weakly closed (4A4Ed); since kyk2 ≤ γ −1/2 α for every y ∈ Cα (x), Cα (x) is weakly compact (4A4Ka). Set β = inf y∈`2 kx − T yk2 + γkyk22 . Then {Cα (x) : α > β}

602


467E

is a T downwards-directed set of non-empty weakly compact sets, so has non-empty intersection; taking any z ∈ α>β Cα (x), β = kx − T zk2 + γkzk22 . Q Q (c) So 467D gives the result. 467F Lemma Let (X, k k) be a Banach space, and hTi ii∈I a family of bounded linear operators from X to itself such that (i) for each i ∈ I, the subspace Ti [X] has an equivalent locally uniformly rotund norm, P (ii) for each x ∈ X, ² > 0 there is a finite set J ⊆ I such that kx − i∈J Ti xk ≤ ², (iii) for each x ∈ X, ² > 0 the set {i : i ∈ I, kTi xk ≥ ²} is finite. Then X has an equivalent locally uniformly rotund norm. proof (a) Let k ki be a locally uniformly rotund norm on Xi = Ti [X] equivalent to k k on Xi . Reducing k ki by a scalar multiple if necessary, we may suppose that kTi xki ≤ kxk for every x ∈ X, i ∈ I. By (iii), supi∈I kTi xk is finite for every x ∈ X; by the Uniform Boundedness Theorem (3A5Ha), M = supi∈I supkxk≤1 kTi xk is finite. For finite sets J ⊆ I and k ≥ 1, set qP P P 1 2 2 pJk (x) = K⊆J kx − i∈K Ti xk ; i∈J kTi xki + k for n ∈ N and k ≥ 1 set (n)

pk (x) = sup{pJk (x) : J ⊆ I, #(J) ≤ n}. (n)

By 467C(c-ii), as usual, all the pJk are seminorms, and it follows at once that the pk are seminorms. P Observe that if K ⊆ I is finite, then kx − i∈K Ti xk ≤ (1 + M #(K))kxk for every x, so if J ⊆ I is finite then p pJk (x) ≤ #(J) + 2#(J) (1 + M #(J))kxk, p (n) and pk (x) ≤ n + 2n (1 + M n)kxk for every n ∈ N, k ≥ 1. Setting βnk = 22n+k for n, k ∈ N, q P∞ P∞ −1 (n) 2 0 kxk = n=0 k=1 βnk pk (x) is finite for every x ∈ X, so that k k0 is a seminorm on X; moreover, kxk0 ≤ βkxk for every x ∈ X, where qP ∞ P∞ −1 n β= n=0 k=1 βnk (n + 2 (1 + M n)) is finite. Since we also have 1 1 1 (0) kxk0 ≥ √ p1 (x) = √ p∅1 (x) = √ kxk 2

2

2

0

for every x, k k is a norm on X equivalent to k k. (b) NowPk k0 is locally uniformly rotund. P P Take x ∈ X and ² > 0. Let K ⊆ I be a finite set such that kx − i∈K Ti xk ≤ 14 ²; we may suppose that Ti x 6= 0 for every i ∈ K. Set α1 = mini∈K kTi xki , J = {i : i ∈ I, kTi xki ≥ α1 } and α0 = supi∈I\J kT xi ki . (For completeness, if K = ∅, take J = ∅, α0 = supi∈I kTi xki and α1 = α0 + 1.) Then J is finite and α0 < α1 , by hypothesis (iii) of the lemma. Set n = #(J). Let k be so large that (βnk + 1)η ≤ min( kTi x − zk ≤

² 1+4n

²2 , α12 16k

2n (M n+1) kxk2 k

< 12 (α12 − α02 ). Let η > 0 be such that

− α02 ),

whenever i ∈ K, z ∈ Xi and qk ki (Ti x, z) ≤ (βnk + 1)η;

this is where we use the hypothesis that every k ki is locally uniformly rotund (and equivalent to k k on Xi ). Now suppose that y ∈ X and qk k0 (x, y) ≤ η. Then qp(n) (x, y) ≤ βnk η, by 467C(c-ii). Let L ∈ [I]n be such k

(n)

that pk (x + y)2 ≤ pLk (x + y)2 + η. Then qpLk (x, y) = 2pLk (x)2 + 2pLk (y)2 − pLk (x + y)2 (n)

(n)

(n)

≤ 2pk (x)2 + 2pk (y)2 − pk (x + y)2 + η ≤ (βnk + 1)η.

467G


603

We also have (n)

(n)

2pLk (x)2 ≥ 2pLk (x)2 + 2pLk (y)2 − 2pk (y)2 ≥ pLk (x + y)2 − 2pk (y)2 (n)

(n)

(n)

≥ pk (x + y)2 − η − 2pk (y)2 ≥ 2pk (x)2 − (βnk + 1)η, so X

1 2

(n)

kT xi k2i ≤ pJk (x)2 ≤ pk (x)2 ≤ pLk (x)2 + (βnk + 1)η

i∈J

≤

X

kT xi k2i +

i∈L

<

X

2n (M n+1) kxk2 k

1 2

+ (βnk + 1)η

kT xi k2i + α12 − α02 .

i∈L

Since #(L) = #(J) and kT xi k2i ≤ α02 < α12 ≤ kT xj k2j whenever j ∈ J and i ∈ I \ J, we must actually have L = J. In particular, K ⊆ L. But this means that (by 467C(c-ii) again) qk ki (Ti x, Ti y) ≤ qpLk (x, y) ≤ (βnk + 1)η P P for every i ∈ K, so that k i∈K Ti x − i∈K Ti yk ≤ 14 ². P The last element we need is that, setting p˜(z) = √1k kz − i∈K Ti zk, p˜ is a seminorm on X and is one of the constituents of pLk ; so that

and (by the choice of η) kTi x − Ti yk ≤

1 (kx − k

X

² 1+4n

Ti xk − ky −

i∈K

X

Ti yk)2 ≤ qp˜(x, y) ≤ qpLk (x, y)

i∈K

≤ (βnk + 1)η ≤

²2 , 16k

¯ ¯ P P and ¯kx − i∈K Ti xk − ky − i∈K Ti yk¯ ≤ 14 ². This means that P P ky − i∈K Ti yk ≤ 41 ² + kx − i∈K Ti xk ≤ 12 ². Putting these together, kx − yk ≤ kx −

P i∈K

Ti xk +

P i∈K

kTi x − Ti yk + ky −

P i∈K

Ti yk ≤ ².

0

And this is true whenever qk k0 (x, y) ≤ η. As x and ² are arbitrary, k k is locally uniformly rotund. Q Q 467G Theorem Let X be a Banach space. Suppose that there are an ordinal ζ and a family hPξ iξ≤ζ of bounded linear operators from X to itself such that (i) if ξ ≤ η ≤ ζ then Pξ Pη = Pη Pξ = Pξ ; (ii) P0 (x) = 0 and Pζ (x) = x for every x ∈ X; (iii) if ξ ≤ ζ is a non-zero limit ordinal, then limη↑ξ Pη (x) = Pξ (x) for every x ∈ X; (iv) if ξ < ζ then Xξ = {(Pξ+1 − Pξ )(x) : x ∈ X} has an equivalent locally uniformly rotund norm. Then X has an equivalent locally uniformly rotund norm. Remark A family hPξ iξ≤ζ satisfying (i), (ii) and (iii) here is called a projectional resolution of the identity. proof For ξ < ζ set Tξ = Pξ+1 − Pξ . From condition (i) we see easily that Tξ Tη = Tξ if ξ = η, 0 otherwise; and that Tξ Pη = Tξ if ξ < η, 0 otherwise. I seek to show that the conditions of 467F are satisfied by hTξ iξ<ζ . Condition (i) of 467F is just condition (iv) here. Let Z be the set of those x ∈ X for which conditions (ii) and (iii) of 467F are satisfied; that is,

604


467G

P for each ² > 0 there is a finite set J ⊆ ζ such that kx − ξ∈J Tξ xk ≤ ², and {ξ : kTξ xk ≥ ²} is finite. Then Z is a linear subspace of X. For ξ ≤ ζ, set Yξ = Pξ [X]. Then Yξ ⊆ Z. P P Induce on ξ. Since Y0 = {0}, the induction starts. For the inductive step to a successor ordinal ξ + 1 ≤ ζ, Yξ+1 = Yξ + Xξ ⊆ Z. For the inductive step to a non-zero limit ordinal ξ ≤ ζ, given x ∈ Yξ and ² > 0, we know that there is a ξ 0 < ξ such that kPη x − Pξ xk ≤ 31 ² whenever ξ 0 ≤ η ≤ ξ. So kTη xk ≤ 23 ² whenever ξ 0 ≤ η < ξ, and {η : kTη xk ≥ ²} = {η : η < ξ 0 , kTη xk ≥ ²} = {η : kTη Pξ0 xk ≥ ²}

P is finite, by the inductive hypothesis. Moreover, there is a finite set J ⊆ ξ 0 such that kPξ0 x− η∈J Tη Pξ0 xk ≤ P 2 Q η∈J Tη xk ≤ ². As x and ² are arbitrary, Yξ ⊆ Z. Q 3 ², and now kx − In particular, X = Yζ ⊆ Z and conditions (ii) and (iii) of 467F are satisfied. So 467F gives the result. Σ space if 467H Definitions (a) A topological space X is K-countably determined or a Lindel¨ of-Σ N there are a subset A of N and an usco-compact relation R ⊆ A × X such that R[A] = X. Observe that all K-analytic Hausdorff spaces (§422) are K-countably determined. (b) A normed space X is weakly K-countably determined if it is K-countably determined in its weak topology. (c) Let X be a normed space and Y , W closed linear subspaces of X, X ∗ respectively. I will say that (Y, W ) is a projection pair if X = Y ⊕ W ◦ and ky + zk ≥ kyk for every y ∈ Y , z ∈ W ◦ , where W ◦ = {z : z ∈ X, f (z) ≤ 1 for every f ∈ W } = {z : z ∈ X, f (z) = 0 for every f ∈ W }. 467I Lemma (a) If X is a weakly K-countably determined normed space, then any closed linear subspace of X is weakly K-countably determined. (b) If X is a weakly K-countably determined normed space, Y is a normed space, and T : X → Y is a continuous linear surjection, then Y is weakly K-countably determined. (c) If X is a Banach space and Y ⊆ X is a dense linear subspace which is weakly K-countably determined, then X is weakly K-countably determined. proof (a) Let A ⊆ N N , R ⊆ A × X be such that R is usco-compact (for the weak topology on X) and R[A] = X. Let Y be a (norm-)closed linear subspace of X; then Y is closed for the weak topology (3A5Ee). Also the weak topology on Y is just the subspace topology induced by the weak topology of X (4A4Ea). Set R0 = R ∩ (A × Y ). Then R0 is usco-compact whether regarded as a subset of A × X or as a subset of A × Y (422Da, 422Db, 422Dg). Since Y = R0 [A], Y is weakly K-countably determined. (b) Let A ⊆ N N , R ⊆ A × X be such that R is usco-compact for the weak topology on X and R[A] = X. Because T is continuous for the weak topologies on X and Y (3A5Ec), R1 = {(φ, y): there is some x ∈ X such that (φ, x) ∈ R and T x = y} is usco-compact in A × Y (422Db, 422Df). Also R1 [A] = T [R[A]] = Y . So Y is weakly K-countably determined. (c) Let A ⊆ N N , R ⊆ A × Y be such that R is usco-compact (for the weak topology on Y ) and R[A] = Y . Then, as in (a), R is usco-compact when regarded as a subset of A × X. By 422Dd, the set R1 = {(hφn in∈N , hyn in∈N ) : (φn , yn ) ∈ R for every n ∈ N} N

is usco-compact in A × Y N . Now examine S = {(hyn in∈N , x) : x ∈ X, yn ∈ Y and kyn − xk ≤ 2−n for every n ∈ N}. Then S is usco-compact in Y N × X. P P (Remember that we are using weak topologies on X and Y throughout.) If y = hyn in∈N is a sequence in Y and (yy , x) ∈ S, then x = limn→∞ yn ; so S[{yy }] has at most one member and is certainly compact. Let F ⊆ X be a weakly closed set and y ∈ Y N \ S −1 [F ]. case 1 If there are m, n ∈ N such that kym − yn k > 2−m + 2−n , let f ∈ Y ∗ be such that kf k ≤ 1 and f (ym − yn ) > 2−m + 2−n . Then G = {zz : f (zm − zn ) > 2−m + 2−n } is an open set in Y N containing y and disjoint from S −1 [F ], because S[G] is empty.

467J


605

case 2 Otherwise, y is a Cauchy sequence and (because X is a Banach space) has a limit x ∈ X, which does not belong to F . Let δ > 0 and f0 , . . . , fr ∈ X ∗ be such that {w : |fi (w) − fi (x)| ≤ δ for every i ≤ r} does not meet F . Let n ∈ N be such that 2−n kfi k ≤ 13 δ for every i ≤ r. Then G = {zz : |fi (zn ) −fi (yn )| < 13 δ for every i ≤ r} is an open set in Y N containing Y . If z ∈ G and (zz , w) ∈ S, then kzn − wk ≤ 2−n so |fi (w) − fi (x)| ≤ |fi (w) − fi (zn )| + |fi (zn ) − fi (yn )| + |fi (yn ) − fi (x)| 1 3

≤ 2−n kfi k + δ + 2−n kfi k ≤ δ for every i ≤ r, and w ∈ / F . Thus again G ∩ S −1 [F ] is empty. This shows that there is always an open set containing y and disjoint from S −1 [F ]. As y is arbitrary, −1 S [F ] is closed. As F is arbitrary, S is usco-compact. Q Q It follows that SR1 ⊆ AN × X is usco-compact (422Df), while (SR1 )[AN ] = S[R1 [AN ]] = S[Y N ] = X because Y is dense in X. Finally, AN is homeomorphic to a subset of N N because it is a subspace of (N N )N ∼ = N N . So X is weakly K-countably determined. 467J Lemma Let X be a weakly K-countably determined Banach space. Then there is a family M of subsets of X ∪ X ∗ such that (i) whenever B ⊆ X ∪ X ∗ there is an M ∈ M such S that B ⊆ M and #(M ) ≤ max(ω, #(B)); (ii) whenever M0 ⊆ M is upwards-directed, then M0 ∈ M; (iii) whenever M ∈ M then (M ∩ X, M ∩ X ∗ ) (where the closures are taken for the norm topologies) is a projection pair of subspaces of X and X ∗ . proof A ⊆ N N , R ⊆ A × X be such that R is usco-compact in A × X and R[A] = X. Set S (a) Let n S = n∈N N and for σ ∈ S set Fσ = R[Iσ ], where Iσ = {φ : σ ⊆ φ ∈ N N }; set S0 = {σ : σ ∈ S, Fσ 6= ∅}. Let M be the family of those sets M ⊆ X ∪ X ∗ such that (α) whenever x, y ∈ M ∩ X and q ∈ Q then x + y and qx belong to M (β) whenever f , g ∈ M ∩ X ∗ and q ∈ Q then f + g and qf belong to M (γ) kxk = max{f (x) : f ∈ M ∩ X ∗ , kf k ≤ 1} for every x ∈ M ∩ X (δ) supx∈Fσ f (x) = supx∈Fσ ∩M f (x) for every f ∈ M ∩ X ∗ , σ ∈ S0 . (b) For each x ∈ X, choose hx ∈ X ∗ such that khx k ≤ 1 and hx (x) = kxk; for each f ∈ X ∗ and σ ∈ S0 choose a countable set Cf σ ⊆ Fσ such that sup{f (x) : x ∈ Cf σ } = sup{f (x) : x ∈ Fσ }. Given B ⊆ X, define hBn in∈N by setting Bn+1 = Bn ∪ {x + y : x, y ∈ Bn ∩ X} ∪ {qx : q ∈ Q, x ∈ Bn ∩ X} ∪ {f + g : f, g ∈ Bn ∩ X} ∪ {qf : q ∈ Q, f ∈ Bn ∩ X} [ ∪ {hx : x ∈ Bn ∩ X} ∪ {Cf σ : f ∈ Bn ∩ X ∗ , σ ∈ S0 },

S

Bn belongs to M and has cardinal at most max(ω, #(B)). S (c) The definition of M makes it plain that if M0 ⊆ M is upwards-directed then M0 belongs to M.

for each n ∈ N. Then M =

n∈N

(d) Now take M ∈ M and set Y = M ∩ X, W = M ∩ X ∗ . These are linear subspaces (2A5Ec). If y ∈ M ∩ X and z ∈ W ◦ , then there is an f ∈ M ∩ X ∗ such that kf k ≤ 1 and f (y) = kyk, so that ky + zk ≥ f (y + z) = f (y) = kyk. Because the function y 7→ ky+zk−kyk is continuous, ky+zk ≥ kyk for every y ∈ Y and z ∈ W ◦ . In particular, if y ∈ Y ∩ W ◦ , kyk ≤ ky − yk = 0 and y = 0, so Y + W ◦ = Y ⊕ W ◦ . If x ∈ Y + W ◦ , then there are sequences hyn in∈N in Y and hzn in∈N in W ◦ such that x = limn→∞ yn +zn ; now kym −yn k ≤ k(ym +zm )−(yn +zn )k → 0 as n → ∞, so (because X is a Banach space) hyn in∈N is convergent to y say; in this case, y ∈ Y and x − y = limn→∞ zn belongs to W ◦ , so x ∈ Y + W ◦ . This shows that Y ⊕ W ◦ is a closed linear subspace of X. (e) ?? Suppose, if possible, that Y ⊕ W ◦ 6= X. Then there is an x0 ∈ X \ (Y ⊕ W ◦ ). By 4A4Eb, there is an f ∈ X ∗ such that f (x0 ) > 0 and f (y) = f (z) = 0 whenever y ∈ Y and z ∈ W ◦ ; multiplying f by a scalar if necessary, we can arrange that f (x0 ) = 1. By 4A4Eg, f belongs to the weak* closure of W in X ∗ .

606


467J

Let φ ∈ A be such that (φ, x0 ) ∈ R. Then K = R[{φ}] is weakly compact. Now the weak* closure of W is also its closure for the Mackey topology of uniform convergence on weakly compact subsets of X (4A4F). So there is a g ∈ W such that |g(x) − f (x)| ≤ 17 for every x ∈ K. Next, because K is bounded, and g belongs to the norm closure of M ∩ X ∗ , there is an h ∈ M ∩ X ∗ such that |h(x) − g(x)| ≤ 71 for every x ∈ K. This means that |h(x) − f (x)| ≤ 27 for every x ∈ K, and K is included in the weakly open set G = {x : h(x) − f (x) < 37 }, that is, φ does not belong to R−1 [X \ G], which is relatively closed in A, because R is usco-compact regarded as a relation between A and X with the weak topology. There is therefore a σ ∈ S such that φ ∈ Iσ and Iσ ∩ R−1 [X \ G] = ∅, that is, Fσ ⊆ G. In this case, x0 ∈ Fσ , so σ ∈ S0 , while h(x) − f (x) < 37 for every x ∈ Fσ . But, because M ∈ M, there is a y ∈ M ∩ X ∩ Fσ such that h(y) ≥ h(x0 ) − 17 , and as y ∈ Y we must now have 0 = f (y) = h(y) − (h(y) − f (y)) > h(x0 ) − = f (x0 ) − (f (x0 ) − h(x0 )) −

4 7

≥1−

3 7

−

1 7

−

3 7

4 7

= 0,

which is absurd. X X Thus X = Y ⊕ W ◦ and (Y, W ) is a projection pair. This completes the proof. 467K Theorem Let X be a weakly K-countably determined Banach space. Then it has an equivalent locally uniformly rotund norm. proof Since the completion of X is weakly K-countably determined (467Ic), we may suppose that X is complete. The proof proceeds by induction on the weight of X. (a) The induction starts by observing that if w(X) ≤ ω then X is separable (4A2Li) so has an equivalent locally uniformly rotund norm by 467E. (b) So let us suppose that w(X) = κ > ω and that any weakly K-countably determined Banach space of weight less than κ has an equivalent locally uniformly rotund norm. Let M be a family of subsets of X ∪ X ∗ as in 467J. Then there is a non-decreasing family hMξ iξ≤κ in S M such that #(Mξ ) ≤ max(ω, #(ξ)) for every ξ ≤ κ, Mκ is dense in X, and Mξ = η<ξ Mη for every limit ordinal ξ ≤ κ. P P By 4A2Li, there is a dense subset of X of cardinal κ; enumerate it as hxξ iξ<κ . Choose Mξ inductively, as follows. M0 = ∅. GIven Mξ with #(Mξ ) ≤ max(ω, #(ξ)), then by 467J(i) there is an Mξ+1 ∈ M such that Mξ+1 ⊇ Mξ ∪ {xξ } and #(Mξ+1 ) ≤ max(ω, #(Mξ ∪ {xξ }) ≤ max(ω, #(ξ + 1)). Given that S hMη iη<ξ is a non-decreasing family in M with #(Mη ) ≤ max(ω, #(η)) for every η < ξ, set Mξ = η<ξ Mη ; then 467J(ii) tells us that Mξ ∈ M, while #(Mξ ) ≤ max(ω, #(ξ)), as required by the inductive hypothesis. Q Q At the end of the induction, Mκ ⊇ {xξ : ξ < κ} will be dense in X. (c) For each ξ < κ, set Yξ = Mξ ∩ X, Wξ = Mξ ∩ X ∗ . Then (Yξ , Wξ ) is a projection pair, by 467J(iii). Since X = Yξ ⊕ Wξ◦ , we have a projection Pξ : X → Yξ defined by saying that Pξ (y + z) = y whenever P The point is that Yξ ⊆ Yη and Wξ ⊆ Wη , y ∈ Yξ and z ∈ Wξ◦ . Now Pξ Pη = Pη Pξ = Pξ whenever ξ ≤ η. P so Wη◦ ⊆ Wξ◦ . If x ∈ X, express it as y + z1 where y ∈ Yξ and z1 ∈ Wξ◦ , and express z1 as y 0 + z where y 0 ∈ Yη and z 0 ∈ Wη◦ . Then Pξ x = y ∈ Yξ ⊆ Yη so Pη Pξ x = Pξ x. On the other hand, x = y + y 0 + z, y + y 0 ∈ Yη and z ∈ Wη◦ , so Pη x = y + y 0 ; and as Q y 0 = z1 − z belongs to Wξ◦ , Pξ (y + y 0 ) = y, so Pξ Pη x = Pξ x. Q Note that the condition ky + zk ≥ kyk whenever y ∈ Yξ , z ∈ Wξ◦ ensures that kPξ k ≤ 1 for every ξ. Next, if ξ ≤ κ is a non-zero limit ordinal, Pξ x = limη↑ξ Pη x. P P We know that

467N


607

S Pξ x ∈ Yξ = Mξ ∩ X = η<ξ Mη ∩ X. S So, given ² > 0, there is an x0 ∈ η<ξ Mη such that kPξ x − x0 k ≤ 21 ². Let η < ξ be such that x0 ∈ Mη . If η ≤ η 0 ≤ ξ, then kPξ x − Pη0 xk = kPξ (Pξ x − x0 ) − Pη0 (Pξ x − x0 )k (because x0 ∈ Yη , so Pξ x0 = Pη0 x0 = x0 ) ≤ 2kPξ x − x0 k ≤ ². As ² is arbitrary, Pξ x = limη↑ξ Pη x. Q Q (d) Now observe that every Yξ is weakly K-countably determined (467Ia), while w(Yξ ) ≤ max(ω, #(ξ)) < κ for every ξ < κ (using 4A2Li, as usual). So the inductive hypothesis tells us that Yξ = Pξ [X] has an equivalent locally uniformly rotund norm for every ξ < κ. By 467G, X = Pκ [X] has an equivalent locally uniformly rotund norm. Thus the induction proceeds. 467L Weakly compactly generated Banach spaces The most important class of weakly K-countably determined spaces is the following. A normed space X is weakly compactly generated if there is a S sequence hKn in∈N of weakly compact subsets of X such that n∈N Kn is dense in X. 467M Proposition (Talagrand 75) A weakly compactly generated Banach space is weakly K-countably determined. proof Let X be a Banach space with a sequence hKn in∈N of weakly compact subsets of X such that S n∈N Kn is dense in X. Set S Pn Ln = { i=0 αi xi : |αi | ≤ n, xi ∈ j≤n Kj for every i ≤ n} S for S n ∈ N. Then every Ln is weakly compact, and Y = n∈N Ln is a linear subspace of X including n∈N Kn , therefore dense. Now Y is a countable union of weakly compact sets, therefore K-analytic for its weak topology (422Gc, 422Hc); in particular, it is weakly K-countably determined. By 467Ic, X is weakly K-countably determined. 467N Theorem Let X be a Banach lattice with an order-continuous norm (§354). Then it has an equivalent locally uniformly rotund norm. proof (a) Consider first the case in which X has a weak order unit e. Then the interval [0, e] is weakly compact. P P We have X ∗ = X × and X ∗∗ = X ×∼ (356D). The canonical identification of X with its image ∗∗ in X is an order-continuous Riesz homomorphism from X onto a solid order-dense Riesz subspace of X ×× (356I) and therefore of X ×∼ = X ∗∗ . In particular, [0, e] ⊆ X is matched with an interval [0, eˆ] ⊆ X ∗∗ . But [0, eˆ] = {θ : θ ∈ X ∗∗ , 0 ≤ θ(f ) ≤ f (e) for every f ∈ (X ∗ )+ } is weak*-closed and norm-bounded in X ∗∗ , therefore weak*-compact; as the weak* topology on X ∗∗ corresponds to the weak topology of X, [0, e] is weakly compact. Q Q Now, for each n ∈ N, Kn = [−ne, ne] = n[0, e] − n[0, e] is weakly compact. If x ∈ X + , then hx ∧ nein∈N + − converges to X, because S the norm of X is order-continuous; so for any x ∈ X, hx ∧ ne − x ∧ nein∈N converges to x. Thus n∈N Kn is dense in X and X is weakly compactly generated. By 467M and 467K, X has an equivalent locally uniformly rotund norm. (b) For the general case, let hxi ii∈I be a maximal disjoint family in X + . For each i ∈ I let Xi be the band in X generated by xi , and Ti : X → Xi the band projection onto Xi (354Ee, 353Hb). Then hTi ii∈I satisfies the conditions of 467F. P P (i) Each Ti is a continuous linear operator because the given norm k k of X is a Riesz norm. Next, each Xi has a weak order unit xi , and the norm on Xi is order-continuous, so (a) tells us that there is an equivalent locally uniformly rotund norm on Xi = Ti [X]. (ii) If x ∈ X, set x0 = supi∈I Ti |x|; then (|x| − x0 ) ∧ xi = 0 for every i, so, by the maximality of hxi ii∈I , x0 = |x|. If ² > 0 then, because the norm of X is order-continuous, there is a finite J ⊆ I such that P kx − j∈J Ti xk = kx0 − supj∈J Ti |x|k ≤ ².

608


Moreover, if i ∈ I \ J, then kTi xk ≤ kx −

P j∈J

467N

Tj xk ≤ ².

Thus conditions (ii) and (iii) of 467F are satisfied. Q Q Accordingly 467F tells us that X has an equivalent locally uniformly rotund norm. 467O Eberlein compacta: Definition A topological space K is an Eberlein compactum if it is homeomorphic to a weakly compact subset of a Banach space. 467P Proposition Let K be a compact Hausdorff space. (a) The following are equiveridical: (i) K is an Eberlein compactum; (ii) there is a set L ⊆ C(K), separating the points of K, which is compact for the topology of pointwise convergence. (b) Suppose that K is an Eberlein compactum. (i) K has a σ-isolated network, so is hereditarily weakly θ-refinable. (ii) (Schachermayer 77) If w(K) is measure-free, K is a Radon space. proof (a)(i)⇒(ii) If K is an Eberlein compactum, we may suppose that it is a weakly compact subset of a Banach space X. Set L = {f ¹K : f ∈ X ∗ , kf k ≤ 1}; since the map f 7→ f ¹K : X ∗ → C(K)} is continuous for the weak* topology of X ∗ and the topology Tp of pointwise convergence on C(K), L is Tp -compact; and L separates the points of K because X ∗ separates the points of X. (ii)⇒(i) If L ⊆ C(K) is Tp -compact and separates the points of K,Sset Ln = {f : f ∈ L, kf k∞ ≤ n} for each n ∈ N. Then Ln is Tp -compact for each n. Set L0 = {0} ∪ n∈N 2−n Ln ; then L0 ⊆ C(K) is 0 norm-bounded and Tp -compact and separates the points of K. Now define x 7→ x ˆ : K → RL by setting x ˆ(f ) = f (x) for x ∈ K, f ∈ L0 . Then x ˆ ∈ C(L0 ) for every x and x 7→ x ˆ is continuous for the given topology ˆ = {ˆ of K and the topology of pointwise convergence on C(L0 ); so the image K x : x ∈ K} is Tp -compact. Since it is also bounded, it is weakly compact (462E). But x 7→ x ˆ is injective, because L0 separates the points ˆ and is an Eberlein compactum. of K; so K is homeomorphic to K, (b) Again suppose K is actually a weakly compact subset of a Banach space X. As in 467M, set Pn S Ln = { i=0 αi xi : |αi | ≤ n, xi ∈ K for every i ≤ n} for each n ∈ N. Then Y = n∈N Ln is a weakly compactly generated Banach space. (I am passing over the trivial case K = ∅.) So Y has an equivalent locally uniformly rotund norm (467M, 467K), which is a Kadec norm (467B), and Y , with the weak topology, has a σ-isolated network (466Eb). It follows at once that K has a σ-isolated network (4A2B(a-ix)), so is hereditarily weakly θ-refinable (438Ld); and if w(K) is measure-free, K is Borel-measure-complete (438M), therefore Radon (434Jf, 434Ka). 467X Basic exercises (a) (i) Show that a continuous image of a K-countably determined space is K-countably determined. (ii) Show that the product of a sequence of K-countably determined spaces is K-countably determined. (iii) Show that any K-countably determined topological space is Lindelöf. (Hint: 422De.) (iv) Show that any Souslin-F subset of a K-countably determined topological space is K-countably determined. (Hint: 422Hc.) (b) Let X be a σ-compact Hausdorff space. Show that a subspace T Y of X is K-countably determined iff there is a countable family K of compact subsets of X such that {K : x ∈ K ∈ K} ⊆ Y for every x ∈ Y . (c) Show that if X and Y are weakly K-countably determined normed spaces, then X × Y , with an appropriate norm, is weakly K-countably determined. (d) Show that a normed space X is weakly compactly generated iff there is a weakly compact set K ⊆ X such that the linear subspace of X generated by K is dense in X. (e) Show that any separable normed space is weakly compactly generated.

467 Notes


609

(f ) Show that any reflexive Banach space is weakly compactly generated. > (g) Show that if X is a weakly compactly generated Banach space, then it is K-analytic in its weak topology. (Hint: in 467M, use the proof of 467Ic.) (h) Show that if X is a Banach space and there is a set A ⊆ X such that A is K-countably determined for the weak topology and the linear subspace generated by A is dense, then X is weakly K-countably determined. (i) Show that the one-point compactification of any discrete space is an Eberlein compactum. 467Y Further exercises (a) Let hX Qn in∈N be a sequence of weakly K-countably determined normed spaces. Investigate normed subspaces of n∈N Xn which will be weakly K-countably determined. (b) (i) Show that if (Ω, Σ, µ) is a probability space, then L1 (µ) has a locally uniformly rotund Riesz norm. (Hint: apply the construction of 467D with Y = L2 (µ) and T the identity operator; show that all the norms pn are Riesz norms.) (ii) Show that if X is any L-space then it has a locally uniformly rotund Riesz norm. (Hint: apply the construction of 467F/467N, noting that if the Ti in 467F are band projections and k k and all the k ki are Riesz norms, then all the norms in the proof of 467F are Riesz norms.) 467 Notes and comments The purpose of this section has been to give an idea of the scope of Proposition 466F. ‘Local uniform rotundity’ has an important place in the geometrical theory of Banach spaces, but for the many associated ideas I refer you to Deville Godefroy & Zizler 93. From our point of view, Theorem 467E is therefore purely accessory, since we know by different arguments that on separable Banach spaces the weak and norm topologies have the same Borel σ-algebras (4A3V). We need it to provide the first step in the inductive proof of 467K. Since the concept of ‘K-analytic’ space is one of the fundamental ideas of Chapter 43, it is natural here to look at ‘K-countably determined’ spaces, especially as many of the ideas of §422 are directly applicable (467Xa). But the goal of this part of the argument is Schachermayer’s theorem 467P(b-ii), which uses ‘weakly compactly generated’ spaces (467L). ‘Eberlein compacta’ are of great interest in other ways; they are studied at length in Arkhangel’skii 92. I mention order-continuous norms here (467N) because they are prominent in the theory of Banach lattices in Volume 3. Note that the methods here do not suffice in general to arrange that the locally uniformly rotund norm found on X is a Riesz norm; though see 467Yb. It is in fact the case that every Banach lattice with an order-continuous norm has an equivalent locally uniformly rotund Riesz norm, but this requires further ideas (see Davis Ghoussoub & Lindenstrauss 81).

610

Geometric measure theory

Chapter 47 Geometric measure theory I offer a chapter on geometric measure theory, continuing from Chapter 26. The greater part of it is directed specifically at a version of the Divergence Theorem (475N), and I do not attempt to provide a balanced view of the subject, for which I must refer you to Mattila 95, Evans & Gariepy 92 and Federer 69. However §472, at least, deals with something which must be central to any approach, Besicovitch’s Density Theorem for Radon measures on Rr . In §473 R I examine Lipschitz functions, and give crude forms of some fundamental ineqalities relating integrals k grad f kdµ with other measures of the variation of a ∂ function f . In §474 I introduce perimeter R measures λE R and outward-normal functions ψE as those for which the Divergence Theorem, in the form E div φ dµ = φ. ψE dλ∂E , will be valid, and give the geometric description of ψE (x) as the Federer exterior normal to E at x. In §475 I show that λ∂E can be identified with normalized Hausdorff r − 1-dimensional measure on the essential boundary of E. §471 is devoted to Hausdorff measures on general metric spaces, extending the ideas introduced in §264 for Euclidean space, up to basic results on densities (471P) and Howroyd’s theorem (471S). In §476 I turn to a different topic, the problem of finding the subsets of Rr on which Lebesgue measure is most ‘concentrated’ in some sense. I present a number of classical results, the deepest being the Isoperimetric Theorem (476L): among sets with a given measure, those with the smallest perimeters are the balls.

471 Hausdorff measures I begin the chapter by returning to a class of measures which we have not examined since Chapter 26. The primary importance of these measures is in studying the geometry of Euclidean space; in §265 I looked briefly at their use in describing surface measures, which will reappear in §475. Hausdorff measures are also one of the basic tools in the study of fractals, but for such applications I must refer you to Falconer 90 and Mattila 95. All I shall attempt to do here is to indicate some of the principal ideas which are applicable to general metric spaces, and to look at some special properties of Hausdorff measures related to the concerns of this chapter and of §261. 471A Definition Let (X, ρ) be a metric space and r ∈ ]0, ∞[. For δ > 0 and A ⊆ X, set ∞ X

θrδ A = inf{

(diam Dn )r : hDn in∈N is a sequence of subsets of X covering A,

n=0

diam Dn ≤ δ for every n ∈ N}. (As in §264, take diam ∅ = 0 and inf ∅ = ∞.) It will be useful to note that every θrδ is an outer measure. Now set θr A = supδ>0 θrδ A for A ⊆ X; θr also is an outer measure on X, as in 264B; this is r-dimensional Hausdorff outer measure on X. Let µHr be the measure defined by Carathéodory’s method from θr ; µHr is r-dimensional Hausdorff measure on X. Notation It may help if I list some notation already used elsewhere. Suppose that (X, ρ) is a metric space. I write B(x, δ) = {y : ρ(y, x) ≤ δ},

U (x, δ) = {y : ρ(y, x) < δ}

for x ∈ X, δ ≥ 0; recall that U (x, δ) is open (2A3G). For x ∈ X and A, A0 ⊆ X I write ρ(x, A) = inf y∈A ρ(x, y),

ρ(A, A0 ) = inf y∈A,z∈A0 ρ(y, z);

for definiteness, take inf ∅ to be ∞, as before. 471B Definition Let (X, ρ) be a metric space. An outer measure θ on X is a metric outer measure if θ(A1 ∪ A2 ) = θA1 + θA2 whenever A1 , A2 ⊆ X and ρ(A1 , A2 ) > 0.

471D

Hausdorff measures

611

471C Proposition Let (X, ρ) be a metric space and θ a metric outer measure on X. Let µ be the measure on X defined from θ by Carathéodory’s method. Then µ is a topological measure. proof (Compare 264E, part (b) of the proof.) Let G ⊆ X be open, and A any subset of X such that θA < ∞. Set An = {x : x ∈ A, ρ(x, A \ G) ≥ 2−n }, B0 = A0 ,

Bn = An \ An−1 for n > 1. S Observe that An ⊆ An+1 for every n and n∈N An = n∈N Bn = A ∩ G. The point is that if m, n ∈ N and n ≥ m + 2, and if x ∈ Bm and y ∈ Bn , then there is a z ∈ A \ G such that ρ(y, z) < 2−n+1 ≤ 2−m−1 , while ρ(x, z) must be at least 2−m , so ρ(x, y) ≥ ρ(x, z) − ρ(y, z) ≥ 2−m−1 . Thus ρ(Bm , Bn ) > 0 whenever n ≥ m + 2. It follows that for any k ≥ 0 Pk S m=0 θB2m = θ( m≤k B2m ) ≤ θ(A ∩ G) < ∞, Pk m=0

S

S θB2m+1 = θ( m≤k B2m+1 ) ≤ θ(A ∩ G) < ∞.

P∞ Consequently n=0 θBn < ∞. P∞ But now, given ² > 0, there is an m such that n=m θBm ≤ ², so that

θ(A ∩ G) + θ(A \ G) ≤ θAm +

∞ X

θBn + θ(A \ G)

n=m

≤ ² + θAm + θ(A \ G) = ² + θ(Am ∪ (A \ G)) (since ρ(Am , A \ G) ≥ 2−m ) ≤ ² + θA. As ² is arbitrary, θ(A ∩ G) + θ(A \ G) ≤ θA. As A is arbitrary, G is measured by µ; as G is arbitrary, µ is a topological measure. 471D Theorem Let (X, ρ) be a metric space and r > 0. Let µHr be r-dimensional Hausdorff measure on X, and Σ its domain; write θr for r-dimensional Hausdorff outer measure on X, as defined in 471A. (a) µHr is a topological measure. (b) For every A ⊆ X there is a Gδ set H ⊇ A such that µHr H = θr A. (c) θr is the outer measure defined from µHr (that is, θr is a regular outer measure). (d) Σ is closed under Souslin’s operation. (e) µHr E = sup{µHr F : F ⊆ E is closed} whenever E ∈ Σ and µHr E < ∞. (f) If A ⊆ X and θr A < ∞ then A is separable and the set of isolated points of A is µHr -negligible. (g) µHr is atomless. (h) If µHr is totally finite it is a quasi-Radon measure. proof (a) The point is that θr , as defined in 471A, is a metric outer measure. P P (Compare 264E, part (a) of the proof.) Let A1 , A2 be subsets of X such that ρ(A1 , A2 ) > 0. Of course θ(A ∪ B) ≤ θr A + θr B, because θr is an outer measure. For the reverse inequality, we may suppose that θr (A ∪ B) < ∞, so that θr A and θr B are both finite. Let ² > 0 and let δ1 , δ2 > 0 be such that θr A1 + θr A2 ≤ θrδ1 A1 + θrδ2 A2 + ², defining the θrδi as in 471A. Set δ = min(δ1 , δ2 , 21 ρ(AP 1 , A2 )) > 0 and let hDn in∈N be a sequence of sets of ∞ diameter at most δ, covering A1 ∪ A2 , and such that n=0 (diam Dn )r ≤ θrδ (A1 ∪ A2 ) + ². Set K = {n : Dn ∩ A1 6= ∅},

L = {n : Dn ∩ A2 6= ∅}.

Because S ρ(x, y) > diam Dn whenever x ∈ A1 , y ∈ A2 and n ∈ N, K ∩ L = ∅; and of course A1 ⊆ A2 ⊆ n∈L Dn . Consequently

S n∈K

Dk ,

612


X

θr A1 + θr A2 ≤ ² + θrδ1 A1 + θrδ2 A2 ≤ ² +

(diam Dn )r +

n∈K

≤²+

∞ X

471D

X

(diam Dn )r

n∈L

(diam Dn )r ≤ 2² + θrδ (A1 ∪ A2 ) ≤ 2² + θr (A1 ∪ A2 ).

n=0

As ² is arbitrary, θr (A1 ∪ A2 ) ≥ θr A1 + θr A2 . The reverse inequality is true just because θr is an outer measure, so θr (A1 ∪ A2 ) = θr A1 + θr A2 . As A1 and A2 are arbitrary, θr is a metric outer measure. Q Q Now 471C tells us that µHr must be a topological measure. (b) (Compare 264Fa.) If θr A = ∞ this is trivial. S Otherwise, P∞for each n ∈ N, let hDni ii∈N be a sequence of sets of diameter at most 2−n such that A ⊆ i∈N Di and i=0 (diam Dni )r ≤ θr,2−n (A) + 2−n , defining r −n−i θr,2−n as in 471A. Let ηni ∈ ]0, 2−n ] be such that (2ηni + (diam Dni )r , and set S + diam Dni ) ≤ 2 Gni = {x : ρ(x, Dni ) < ηni } for all n, i ∈ N; then Gni = x∈Dni U (x, ηni ) is an open set including Dni and (diam Gni )r ≤ 2−n−i + (diam Dni )r . Set T S H = n∈N i∈N Gni , so that H is a Gδ set including A. For any δ > 0, there is an n ∈ N such that 3 · 2−n ≤ δ, so that diam Gmi ≤ diam Dmi + 2ηmi ≤ δ for every i ∈ N and m ≥ n, and θrδ (H) ≤

∞ X

(diam Gmi )r ≤

i=0

∞ X

2−m−i + (diam Dmi )r

i=0

≤ 2−m+1 + θr,2−m (A) + 2−m ≤ 2−m+2 + θr (A) for every m ≥ n. Accordingly θrδ H ≤ θr A for every δ > 0, so θr H ≤ θr A. Of course this means that θr H = θr A; and since, by (a), µHr measures H, we have µHr H = θr A, as required. (c) (Compare 264Fb.) If A ⊆ X, θr A ≥ µ∗Hr A (by (b)) = inf{θr E : A ⊆ E ∈ Σ} ≥ θr A. (d) Use 431C. (e) By (b), there is a Borel set H ⊇ E such that µHr H = µHr E, and now there is a Borel set H 0 ⊇ H \ E such that µHr H 0 = µHr (H \ E) = 0, so that G = H \ H 0 is a Borel set included in E and µHr G = µHr E. Now G is a Baire set (4A3Kb), so is Souslin-F (421L), and µHr G = supF ⊆G is closed µHr F , by 431E. (f ) For every n ∈ N, there must be a sequence hDni ii∈N of sets of diameter at most 2−n covering A; now if D ⊆ A is a countable set which meets Dni whenever i, n ∈ N and A ∩ Dni 6= ∅, D will be dense in A. Now if A0 is the set of isolated points in A, it is still separable (4A2P(a-iv)); but as the only dense subset of A0 is itself, it is countable. Since θrδ {x} = (diam{x})r = 0 for every δ > 0, µHr {x} = 0 for every x ∈ X, and A0 is negligible. (g) In fact, if A ⊆ X and θr A > 0, there are disjoint A0 , A1 ⊆ A such that θr Ai > 0 for both i. P P (i) Suppose first that A is not separable. For each n ∈ N, let D ⊆ A be a maximal set such that ρ(x, y) ≥ 2−n n S for all distinct x, y ∈ Dn ; then n∈N Dn is dense in A, so there is some n ∈ N such that Dn is uncountable; if we take A1 , A2 to be disjoint uncountable subsets of Dn , then θr A1 = θr A2 = ∞. (ii) If A is separable, then set G = {G : G ⊆ X is open, θr (A ∩ G) A is hereditarily Lindel¨ S = 0}. Because S S of (4A2P(a-iii)), there S is a countable subset G0 of G such that A ∩ G = A ∩ G0 (4A2H(c-i)), so A ∩ G is negligible and A \ G has at least two points x0 , x1 . If we set Ai = U (xi , 21 ρ(xi , x1−i )) for each i, these are disjoint subsets of A of non-zero outer measure. Q Q (h) If µHr is totally finite, then it is inner regular with respect to the closed sets, by (e). Also, because X must be separable, by (f), therefore hereditarily Lindelöf, µHr must be τ -additive (414O). Finally, µHr is complete just because it is defined by Carathéodory’s method. So µHr is a quasi-Radon measure.

471G

Hausdorff measures

613

471E Corollary If (X, ρ) is a metric space, r > 0 and Y ⊆ X then r-dimensional Hausdorff measure (Y ) (X) (X) µHr on Y extends the subspace measure (µHr )Y on Y induced by r-dimensional Hausdorff measure µHr (X) on X; and if either Y is measured by µHr or Y has finite r-dimensional Hausdorff outer measure in X, then (Y ) (X) µHr = (µHr )Y . (X)

(Y )

proof Write θr and θr for the two r-dimensional Hausdorff outer measures. If A ⊆ Y and hDn in∈N is anyPsequence of subsets of X P covering A, then hDn ∩ Y in∈N is a sequence ∞ ∞ of subsets of Y covering A, and n=0 (diam(Dn ∩ Y ))r ≤ n=0 (diam Dn )r ; moreover, when calculating diam(Dn ∩ Y ), it doesn’t matter whether we use the metric ρ on X or the subspace metric ρ¹ Y × Y on Y . What this means is that, for any δ > 0, θrδ A is the same whether calculated in Y or in X, so that (Y ) (X) θr A = supδ>0 θrδ A = θr A. (Y ) (X) (X) Thus θr = θr ¹ PY . Also, by 471Db, θr is a regular outer measure. So 214Hb gives the results. 471F Corollary Let (X, ρ) be an analytic metric space (that is, a metric space in which the topology is analytic in the sense of §423), and write µHr for r-dimensional Hausdorff measure on X. Suppose that ν is a locally finite indefinite-integral measure over µHr . Then ν is a Radon measure. proof Since dom ν ⊇ dom µ, ν is a topological measure. Because X is separable, therefore hereditarily Lindelöf, ν is σ-finite and τ -additive, therefore locally determined and effectively locally finite. Next, it is inner regular with respect to the closed sets. P P Let f be a Radon-Nikod´ ym derivative of ν. If νE > 0, there is an E 0 ⊆ E such that 0 < νE 0 =

R

f × χE 0 dµHr < ∞.

R There is a µHr -simple function g such that g ≤ f × χE 0 µHr -a.e. and g dµHr > 0; setting H = E 0 ∩ {x : g(x) >R 0}, νHr H < ∞. Now there is a closed set F ⊆ H such that µHr F > 0, by 471De, and in this case νF ≥ F g dµHr > 0. By 412B, this is enough to show that ν is inner regular with respect to the closed sets. Q Q Since ν is complete (234A), it is a quasi-Radon measure, therefore a Radon measure (434Jf, 434Jb). 471G Increasing Sets Lemma(Davies 70) Let (X, ρ) be a metric space and r > 0. (a) Suppose that δ > 0 and that hAn in∈N is a non-decreasing sequence of subsets of X with union A. Then θr,6δ (A) ≤ (5r + 2) supn∈N θrδ (An ). (b) Suppose that δ > 0 and that hAn in∈N is a non-decreasing sequence of subsets of X with union A. Then θrδ A ≤ supn∈N θrδ An . proof (a) If supn∈N θrδ (An ) = ∞ this is trivial; suppose otherwise.

¤ ¤ (i) Take any γ > γ 0 > supn∈N θrδ (An ). For each i ∈ N, let ζi ∈ 0, 61 δ be such that (α + 2ζi )r ≤ αr + 2−i−1 (γ −γ 0 ) whenever 0 ≤ α P ≤ δ. For each n ∈ N, there is aPsequence hCni ii∈N of sets covering An such that ∞ ∞ diam Cni ≤ δ for every i and i=0 (diam Cni )r ≤ γ 0 . Since i=0 (diam Cni )r is finite, limi→∞ diam Cni = 0; rearranging the sequence, we may suppose that diam Cni + ζi ≥ diam Cn,i+1 for every i. Pk In this case, limi→∞ supn∈N diam Cni = 0. P P Given ² > 0, take k such that j=0 max(0, ² − ζj )r > γ. Pk For any i ≥ k and n ∈ N, diam Cnj ≥ max(0, diam Cni − ζj ) for every j ≤ k, and j=0 (diam Cnj )r ≤ γ, so diam Cni ≤ ². Q Q (ii) By Ramsey’s theorem (4A1G, with n = 2), there is an infinite set I ⊆ N such that for all i, j ∈ N there is an s ∈ N such that either Cmi ∩ Cnj = ∅ whenever m, n ∈ I and s ≤ m < n or Cmi ∩ Cnj 6= ∅ whenever m, n ∈ I and s ≤ m < n, for each i ∈ N, αi = limn∈I,n→∞ diam Cni is defined in R. (Apply 4A1Fb with the families Jij = {J : J ∈ [N]ω , either Cmi ∩ Cnj = ∅ whenever m, n ∈ J and m < n or Cmi ∩ Cnj 6= ∅ whenever m, n ∈ J and m < n} Jiq0

ω

= {J : J ∈ [N] , either diam Cni ≤ q for every n ∈ J or diam Cni ≥ q for every n ∈ J}

614


471G

for i, j ∈ N and q ∈ Q.) Of course αj − ζi ≤ αi ≤ δ whenever i ≤ j, because diam S Cnj T− ζi ≤ diam Cni ≤ δ for every n. Set Dni = {x : ρ(x, Cni ) ≤ 2αi + 3ζi } for all n, i ∈ N, and Di = s∈N n∈I\s Dni for i ∈ N. (I am identifying each s ∈ N with the set of its predecessors.) (iii) Set L = {(i, j) : i, j ∈ N, ∀ s ∈ N ∃ m, n ∈ I, s ≤ m < n and Cmi ∩ Cnj 6= ∅}. If (i, j) ∈ L then there is an s ∈ N such that Cmi ⊆ Dmin(i,j) whenever m ∈ I and m ≥ s. P P By the choice of I, we know that there is an s0 ∈ N such that Cmi ∩ Cnj 6= ∅ whenever m, n ∈ I and s0 ≤ m < n. Let s1 ≥ s0 be such that diam Cmi ≤ αi + min(ζi , ζj ),

diam Cmj ≤ αj + min(ζi , ζj )

whenever m ∈ I and m ≥ s1 . Take m0 ∈ I such that m0 ≥ s1 , and set s = m0 + 1. Let m ∈ I be such that m ≥ s. α) Suppose that i ≤ j and x ∈ Cmi . Take any n ∈ I such that m ≤ n. Then there is an n0 ∈ I such (α that n < n0 . We know that Cmi ∩ Cn0 j and Cni ∩ Cn0 j are both non-empty. So ρ(x, Cni ) ≤ diam Cmi + diam Cn0 j ≤ αi + ζi + αj + ζi ≤ 2αi + 3ζi and x ∈ Dni . This is true for all n ∈ I such that n ≥ m, so x ∈ Di . As x is arbitrary, Cmi ⊆ Di . β ) Suppose that j ≤ i and x ∈ Cmi . Take any n ∈ I such that n > m. Then Cmi ∩ Cnj is not (β empty, so ρ(x, Cnj ) ≤ diam Cmi ≤ αi + ζj ≤ αj + 2ζj and x ∈ Dnj . As x and n are arbitrary, Cmi ⊆ Dj . Thus Cmi ⊆ Dmin(i,j) in both cases. Q Q (iv) Set D=

S i∈N

Di ,

J = {i : i ∈ N, ∃ s ∈ N, Cni ⊆ D whenever n ∈ I and n ≥ s}.

If i ∈ N \ J and j ∈ N, then (iii) tells us that (i, j) ∈ / L, so there is some s ∈ N such that Cmi ∩ Cnj = ∅ whenever m, n ∈ I and s ≤ m < n. (v) For l ∈ N, µ∗Hr (Al \ D) ≤ 2γ. P P Let ² > 0. Then there is a k ∈ N such that diam Cni ≤ ² whenever n ∈ N and i > k. Next, there is an s ∈ N such that Cni ⊆ D whenever i ≤ k, i ∈ J, n ∈ I and s ≤ n, Cmi ∩ Cnj = ∅ whenever i, j ≤ k, i ∈ / J, m, n ∈ I and s ≤ m < n. Take m, n ∈ I such that max(l, s) ≤ m < n. Then [ Al \ D = Al ∩ Cmi \ D i∈N

⊆

[

(An ∩ Cmi \ D) ∪

i≤k

⊆

[

(An ∩ Cmi ) ∪

[

[

j>k

Cnj ∪

[

Cmi

i>k

(Cmi ∩ Cnj ) ∪ [

[ j>k

i≤k,i∈J,j≤k /

=

Cmi

i>k

i≤k,i∈J /

⊆

[

Cnj ∪

[

Cmi

i>k

Cmi .

i>k

Since diam Cnj ≤ ² and diam Cmi ≤ ² for all i, j > k, P∞ P∞ θr² (Al \ D) ≤ i=k+1 (diam Cni )r + i=k+1 (diam Cmi )r ≤ 2γ.

471G

Hausdorff measures

615

This is true for every ² > 0, so µ∗Hr (Al \ D) ≤ 2γ, as claimed. Q Q (vi) This is true for each l ∈ N. But this means that µ∗Hr (A \ D) ≤ 2γ (132Ae). Now θr,6δ D ≤ 5r γ. P P For each i ∈ N, diam Di ≤ ≤ Next, for any k ∈ N,

lim

diam Dni

lim

diam Cni + 4αi + 6ζi = 5αi + 6ζi ≤ 6δ.

n∈I,n→∞ n∈I,n→∞

Pk i=0

αir = limn∈I,n→∞

Pk

r i=0 (diam Cni )

≤ γ0,

so k X

(diam Di )r ≤ 5r

i=0

k X

k X (αi + 2ζi )r ≤ 5r ( αir + 2−i−1 (γ − γ 0 ))

i=0

i=0

(by the choice of the ζi ) ≤ 5r γ. Letting k → ∞, θr,6δ D ≤

P∞

r i=0 (diam Di )

≤ 5r γ. Q Q

Putting these together, θr,6δ A ≤ θr,6δ D + θr,6δ (A \ D) ≤ θr,6δ D + µ∗Hr (A \ D) ≤ (5r + 2)γ. As γ is arbitrary, we have the preliminary result (a). (b) Now let us turn to the sharp form (b). Once again, we may suppose that supn∈N θrδ An is finite. (i) Take γ, γ 0 such that supn∈N θrδ An < γ 0 < γ. As in (a)(i) above, we can find a family hCni in,i∈N such that S An ⊆ i∈N Cni , diam Cni ≤ δ for every i ∈ N, P∞ r 0 i=0 (diam Cni ) ≤ γ for each n, and limi→∞ supn∈N diam Cni = 0. Replacing each Cni by its closure if necessary, we may suppose that every Cni is a Borel set. Let Q ⊆ X be a countable set which meets Cni whenever n, i ∈ N and Cni is not empty. This time, let I ⊆ N be an infinite set such that αi = limn∈I,n→∞ diam Cni is defined in [0, δ] for every i ∈ N, limn∈I,n→∞ ρ(z, Cni ) is defined in [0, ∞] for every i ∈ N and every z ∈ Q. (Take ρ(z, ∅) = ∞ if any of the Cni are empty.) It will be helpful to note straight away that the limit limn∈I,n→∞ ρ(x, Cni ) is defined in [0, ∞] for every i ∈ N and x ∈ Q. P P If limn∈I,n→∞ ρ(y, Cni ) = ∞ for any z ∈ Q, then limn∈I,n→∞ ρ(x, Cni ) = ∞, and we can stop. Otherwise, for any ² > 0, there are a z ∈ Q such that ρ(x, z) ≤ ² and an s ∈ N such that Cmi is not empty and |ρ(z, Cmi ) − ρ(z, Cni )| ≤ ² whenever m, n ∈ I \ s; in which case |ρ(x, Cmi ) − ρ(x, Cni )| ≤ 3² whenever m, n ∈ I \ s. As ² is arbitrary, limn∈I,n→∞ ρ(x, Cni ) is defined in R. Q Q Let F be a non-principal ultrafilter on N containing I, and for i ∈ N set Set D =

Di = {x : limn→F ρ(x, Cni ) = 0}.

S i∈N

Di . (Actually it is easy to check that every Di is closed.)

616


(ii) Set A∗ =

S

T m∈N

note that A∗ is a Borel set. For k, m ∈ N, set A∗km =

S n∈I\m

T

i∈N

Cni \ D;

i≥k

Cni .

S n∈I\m

471G

For fixed k, hA∗km im∈N is a non-decreasing sequence of sets. Also its union includes A∗ . P P Take x ∈ A∗ . α)?? If x ∈ (α / Q, let ² > 0 be such that Q ∩ B(x, ²) = ∅. Let k ∈ N be such that diam SCni ≤ ² whenever n ∈ N and i ≥ k; then x ∈ / CS ni whenever n ∈ N and i ≥ k. Let m ∈ N be such that x ∈ i∈N Cni whenever n ∈ I and n ≥ m. Then x ∈ i 0. Let k ∈ N be such that diam Cni ≤ ² whenever i ≥ k and Pany ∞ n ∈ N. For any m ∈ I, θr² A∗km ≤ i=k (diam Cmi )r ≤ γ. By (a), θr,6² A∗ ≤ (5r + 2) sup θr² A∗km m∈N

= (5r + 2) sup θr² A∗km ≤ (5r + 2)γ. m∈I

As ² is arbitrary, µHr A∗ ≤ (5r + 2)γ < ∞. Q Q P∞ r ∗ 0 P?? Suppose, if possible, otherwise. Take β such that γ 0 − Actually, µHr A ≤ γ − i=0 αi . P P∞ (iv) r ∗ ∗ ∗ i=0 αi < β < µHr A . For x ∈ A and k ∈ N, set fk (x) = min{n : n ∈ I, x ∈ Akn }; then hfk ik∈N is a ∗ non-decreasing sequence of Borel measurable functions from A to N. Choose hsk ik∈N inductively so that µHr {x : x ∈ A∗ , fj (x) ≤ sj for every j ≤ k} ≥ β for every k ∈ N. Set A˜ = {x : x ∈ A∗ , fj (x) ≤ sj for every j ∈ N}; because µHr A∗ is finite, µHr A˜ ≥ β. Pk−1 P∞ Take ² > 0 such that θr² A˜ > γ 0 − i=0 αir . Let k ∈ N be such that θr² A˜ + i=0 αir > γ 0 and diam Cni ≤ ² ˜ Pk−1 (diam Cni )r > γ 0 . If whenever n ∈ N and i ≥ k. Take n ∈ I such that n ≥ sj for every j < k and θr² A+ i=0 P∞ P S ∞ r 0 r ˜ ˜ then x ∈ A∗ ⊆ x ∈ A, kn i=0 (diam Cni ) > γ , i=k (diam Cni ) ; but this means that i≥k Cni , so θr² A ≤ contrary to the choice of the Cni . X XQ Q (v) Now observe that A⊆

S

T m∈N

S n≥m

i∈N

Cni ⊆ A∗ ∪ D.

Moreover, for any i ∈ N, diam Di ≤ αi ≤ δ. P P If x, y ∈ Di then for every ² > 0 ρ(x, Cni ) ≤ ²,

ρ(y, Cni ) ≤ ²,

diam Cni ≤ αi + ²

for all but finitely many n ∈ I. So ρ(x, y) ≤ αi + 3². As x, y and ² are arbitrary, diam Di = diam Di ≤ αi . Of course αi ≤ δ because diam Cni ≤ δ for every n. Q Q Now P∞ P∞ θrδ D ≤ i=0 (diam Di )r ≤ i=0 αir . Putting this together with (iv), θrδ A ≤ θrδ D + θrδ A∗ ≤ θrδ D + µHr A∗ ≤ γ. As γ and γ 0 are arbitrary, θrδ A ≤ supn∈N θrδ An , as required.

471I

Hausdorff measures

617

471H Corollary Let (X, ρ) be a metric space, and r > 0. For A ⊆ X, set P∞ θr∞ A = inf{ n=0 (diam Dn )r : hDn in∈N is a sequence of subsets of X covering A}. Then θr∞ is a capacity on X. proof (a) Of course 0 ≤ θr∞ A ≤ θr∞ B whenever A ⊆ B ⊆ X. (b) Suppose that hAn in∈N is a non-decreasing sequence of subsets of A with union A. By (a), γ = limn→∞ θr∞ An is defined and less than or equal to θr∞ A. If γ = ∞, of course it is equal to θr∞ A. Otherwise, take δ = (γ + 1)1/r . For n, k ∈ N there is a sequence hDnki ii∈N of sets, covering An , such that P∞ r −k . But in this case diam Dnki ≤ δ for all n, k and i, so the Dnki witness that i=0 (diam Dnki ) ≤ γ + 2 θrδ An ≤ γ. By 471Gb, γ ≥ θrδ A ≥ θr∞ A and again we have γ = θr∞ A. (c) Let K P ⊆ X be any set, and suppose that γ > θr∞ K. Let hDn in∈N be a sequence of sets, covering ∞ r K, such that n=0 (diam Dn ) < γ. Let h²n in∈N be a sequence of strictly positive real numbers such that P∞ r ≤ γ. Set Gn = {x : ρ(x, Dn ) < ²n } for each n; then Gn is open and diam Gn ≤ n=0 (diam Dn + 2²n ) S diam Dn + 2²n . So G = n∈N Gn is an open set including K, and hGn in∈N witnesses that θr∞ G ≤ γ. As K and γ are arbitrary, condition (iii) of 432J is satisfied and θr∞ is a capacity. Remark θr∞ is r-dimensional Hausdorff capacity on X. 471I Theorem Let (X, ρ) be a metric space, and r > 0. Write µHr for r-dimensional Hausdorff measure on X. If A ⊆ X is analytic, then µHr A is defined and equal to sup{µHr K : K ⊆ A is compact}. proof (a) Before embarking on the main line of the proof, it will be convenient to set out a preliminary result. For δ > 0, n ∈ N, B ⊆ X set S Pn (n) θrδ (B) = inf{ i=0 (diam Di )r : B ⊆ i≤n Di , diam Di ≤ δ for every i ≤ n}, (n)

(n)

(n)

taking inf ∅ = ∞ as usual. Then θrδ (B) ≤ θrδ (B) for every n. Now the point is that θrδ (B) = sup{θrδ (I) : (n) (n) I ⊆ B is finite}. P P Set γ = supI∈[B]<ω θrδ (I). Of course γ ≤ θrδ (B). If γ = ∞ there is nothing more to say. γ 0 > γ. For each I ∈ [B]<ω , we have a function fI : I → {0, . . . , n} such P Otherwise, rtake any 0 that i∈J ρ(xi , yi ) ≤ γ whenever J ⊆ {0, . . . , n} and xi , yi ∈ I and fI (xi ) = fI (yi ) = i for every i ∈ J, while ρ(x, y) ≤ δ whenever x, y ∈ I and fI (x) = fI (y). Let F be an ultrafilter on [B]<ω such that {I : x ∈ I ∈ [B]<ω } ∈ F for every x ∈ B (4A1Ia). Then for every x ∈ B there is an f (x) ∈ {0, . . . , n} such that {I : x ∈ I ∈ [B]<ω , fI (x) = f (x)} ∈ F. Set Di = f −1 [{i}] for i ≤ n. If x, y ∈ B and f (x) = f (y), there is an I ∈ [B]<ω containing both x and y such that fI (x) = f (x) = f (y) = fI (y), so that ρ(x, y) ≤ δ; thus diam Di ≤ δ for each i. If J ⊆ {0, . . . , n} and for each i ∈PJ we take xi , yi ∈ Di , then there is an I ∈ [B]<ω such that fI (xi ) = fI (yi ) = i for every i ∈ J, so i∈J ρ(xi , yi )r ≤ γ 0 . This means that P (n) (n) r 0 0 0 Q i≤n (diam Di ) ≤ γ , so that θrδ (B) ≤ γ . As γ is arbitrary, θrδ (B) ≤ γ, as claimed. Q (b) Now let us turn to the set A. Because A is Souslin-F (422Ha), µHr measures A (471Da, 471Dd). Set γ = sup{µHr (K) : K ⊆ A is compact}. ?? Suppose, if possible, that µHr (A) > γ. Take γ 0 ∈S ]γ, µHr (A)[. Let δ > 0 be such that γ 0 < θrδ (A). Let N ∗ f : N → A be a continuous surjection. For σ ∈ S = n∈N N n , set Fσ = {φ : φ ∈ N N , φ(i) ≤ σ(i) for every i < #(σ)}, so that f [F∅ ] = A. Now choose ψ ∈ N N and a sequence hIn in∈N of finite subsets of N N inductively, as follows. (n) Given that Ij ⊆ Fψ¹n for every j < n and that θrδ (f [Fψ¹n ]) > γ 0 , then θrδ (f [Fψ¹n ]) > γ 0 , so by (a) above (n) there is a finite subset In of Fψ¹n such that θrδ (f [In ]) ≥ γ 0 . Next, lim θrδ f [F(ψ¹n)a i ] = θrδ

i→∞

[

f [F(ψ¹n)a i ]

i∈N

(by 471G) = θrδ f [Fψ¹n ] > γ 0 ,

618


471I

S so we can take ψ(n) such that j≤n Ij ⊆ Fψ¹n+1 and θrδ f [Fψ¹n+1 ] > γ 0 , and continue. At the end of the induction, set K = {φ : φ ≤ ψ}. Then f [K] is a compact subset of A, and In ⊆ K for every n ∈ N, so (n)

(n)

θrδ (f [K]) ≥ θrδ (f [In ]) ≥ γ 0 for every n ∈ N. On the other hand, µHrP (f [K]) ≤ γ, so there is a sequence hDi ii∈N of sets, covering ∞ f [K], all of diameter less than δ, such that i=0 (diam Di )r < γ 0 . Enlarging the Di slightly ifSneed be, we may suppose that they are all open. But in this case there is some finite n such that K ⊆ i≤n Di , and Pn (n) θrδ (K) ≤ i=0 (diam Di )r < γ 0 ; which is impossible. X X This contradiction shows that µHr (A) = γ, as required. 471J Proposition Let (X, ρ) and (Y, σ) be metric spaces, and f : X → Y a function which is γ(X) Lipschitz, that is, such that σ(f (x), f (y)) ≤ γρ(x, y) for all x, y ∈ X, where γ ≥ 0. If r > 0 and θr , (Y ) (Y ) (X) θr are the r-dimensional Hausdorff outer measures on X and Y respectively, then θr f [A] ≤ γ r θr A for every A ⊆ X. (X)

proof (Compare 264G.) Let δ > 0. Set η = δ/(1 + γ) and consider θrη : PX → [0, ∞], defined as in 471A. (X) (X) We know that θr A ≥ θrη A, so there is a sequence hDn in∈N of sets, all of diameter at most η, covering P∞ S (X) A, with n=0 (diam Dn )r ≤ θr A + δ. Now φ[A] ⊆ n∈N φ[Dn ] and diam φ[Dn ] ≤ γ diam Dn ≤ γη ≤ δ for every n. Consequently (Y )

θrδ (φ[A]) ≤

P∞

r n=0 (diam φ[Dn ])

≤

P∞ n=0

(X)

γ r (diam Dn )r ≤ γ r (θr

+ δ),

and (Y )

(Y )

(X)

θr (φ[A]) = limδ↓0 θrδ (φ[A]) ≤ γ r θr

A,

as claimed. 471K Lemma Let (X, ρ) be a metric space, and r > 0. Let µHr be r-dimensional Hausdorff measure on X. µHr A = 0 iff for every ² > 0 there is a countable family D of sets, covering A, such that P If A ⊆ X, then r D∈D (diam D) ≤ ². proof If µHr A = 0 and ²P > 0, then, in the language of 471A, θr1 A ≤ ², so there is a sequence hDn in∈N of ∞ sets covering A such that n=0 (diam Dn )r ≤ ². If the condition is satisfied, then for any ², δ > 0 there is a countable family D of sets, covering A, P it as hDn in∈N ; if it is finite, enumerate such that D∈D (diam D)r ≤ min(², δ r ). If D is infinite, enumerate S it as hDn in 0. Let δ > 0 be suchSthat δ s−r (1 + there is a sequence hAn in∈N of sets of Pµ∞Hr A) ≤ ². Then diameter at most δ such that A ⊆ n∈N An and n=0 (diam An )r ≤ 1 + µ∗Hr A. But now, by the choice of P∞ δ, n=0 (diam An )s ≤ ². As ² is arbitrary, µHs A = 0, by 471K.

471M There is a generalization of the density theorems of §§223 and 261 for general Hausdorff measures, which (as one expects) depends on a kind of Vitali theorem. I will use the following notation for the next few paragraphs. Definition If (X, ρ) is a metric space and A ⊆ X, write A∼ for {x : x ∈ X, ρ(x, A) ≤ 2 diam A}, where ρ(x, A) = inf y∈A ρ(x, y). (Following the conventions of 471A, ∅∼ = ∅.)

471P

Hausdorff measures

619

471N Lemma Let (X, ρ) be a metric space. Let F be a family of subsets of X such that {diam F : F ∈ F} is bounded. Set T S Y = δ>0 {F : F ∈ F, diam F ≤ δ}. Then S there isSa disjoint family I ⊆ F such that (i) F ⊆ F ∈I F ∼ ; S S (ii) Y ⊆ J ∪ F ∈I\J F ∼ for every finite J ⊆ I. proof (a) Let γ be an upper bound for {diam F : F ∈ F }. Choose S hIn in∈N , hJn in∈N inductively, as follows. I0 = ∅. Given In , set Fn0 = {F : F ∈ F, diam F ≥ 2−n γ, F ∩ In = ∅}, and let Jn ⊆ Fn0 be a maximal disjoint set; now set In+1 = In ∪ Jn , and continue. At the end of the induction, set S S I 0 = n∈N In , I = I 0 ∪ {{x} : x ∈ F \ I 0 , {x} ∈ F}. The construction ensures that every In is a disjoint subset of F, so I 0 and I are also disjoint subfamilies of F. S S (b) ?? Suppose, if possible, that there is a point x in F \ F ∈I F ∼ . Let F ∈ F be such that x ∈ F . S 0 Since x ∈ / I and {x} ∈ / I, {x} ∈ / F, and diam F > 0; let n ∈ N be such that 2−n γ ≤ diam F ≤ 2−n+1 γ. 0 If F ∈ / Fn , there is a D ∈ In such that F ∩ D 6= ∅; otherwise, since Jn is maximal and F ∈ / Jn , there is a D ∈ Jn such that F ∩ D 6= ∅. In either case, we have a D ∈ I such that F ∩ D 6= ∅ and diam F ≤ 2 diam D. But in this case ρ(x, D) ≤ diam F ≤ 2 diam D and x ∈ D∼ , which is impossible. X X S if possible, that there are a point x ∈ Y and a finite set J ⊆ I such that x∈ / J ∪ S S S (c) ?? Suppose, ∼ J ), so that F ∩ J = ∅. As F ∈I\J F . Then there is an F ∈ F such that x ∈ F and diam F < ρ(x, in (b), F cannot be {x}, and there must be an n ∈ N such that 2−n γ < diam F ≤ 2−n+1 γ. As in (b), there must be a D ∈ In+1 such that F ∩ D 6= ∅, so that x ∈ D∼ ; and as D cannot belong to J , we again have a contradiction. X X 471O Lemma Let (X, ρ) be a metric space, and r > 0. Let µHr be r-dimensional Hausdorff measure on X. Suppose that A, F are such that P∞ (i) F is a family of closed subsets of X such that n=0 (diam Fn )r is finite for every disjoint sequence hFn in∈N in F, (ii) for every x ∈ A, δ > 0 there is an F ∈ F such that x ∈ F and 0 < diam F ≤ δ. S Then there is a countable disjoint family I ⊆ F such that A \ I has zero r-dimensional Hausdorff measure. proof Replacing F by {F : F ∈ F , 0 < diam F ≤ 1} if necessary, we may suppose that supF ∈F diam F is finite and thatSdiam F > 0 for every F ∈ F. Take a disjoint family I ⊆ F as in 471N. If I is finite, then A ⊆ Y ⊆ I, where Y is defined as in 471N, so we can stop. Otherwise, hypothesis (i) tells us that {F : FP∈ I, diam F ≥ δ} is finite for every δ > 0, so I is countable;Penumerate it as hFn in∈N ; we ∞ ∞ Fn )r < ∞. Since diam Fn∼ ≤ 5 diam Fn for every n, n=0 (diamSFn∼ )r is must have S finite, and P∞ n=0 (diam ∼ r inf n∈N i=n (diam Fi ) = 0. But now observe that the construction ensures that A \ I ⊆ i≥n Fi∼ for S every n ∈ N. By 471K, µHr (A \ I) = 0, as required. 471P Theorem Let (X, ρ) be a metric space, and r > 0. Let µHr be r-dimensional Hausdorff measure on X. Suppose that A ⊆ X and µ∗Hr A < ∞. µ∗Hr (A∩D) :x (diam D)r ∗ µ (A∩B(x,δ)) lim supδ↓0 Hr r δ

(a) limδ↓0 sup{

∈ D, 0 < diam D ≤ δ} = 1 for µHr -almost every x ∈ A.

(b)

≥ 1 for µHr -almost every x ∈ A. So 2−r ≤ lim supδ↓0

for µHr -almost every x ∈ A. (c) If A is measured by µHr , then

µ∗Hr (A∩B(x,δ)) (diam B(x,δ))r

≤1

620


limδ↓0 sup{

µ∗Hr (A∩D) (diam D)r

471P

: x ∈ D, 0 < diam D ≤ δ} = 0

for µHr -almost every x ∈ X \ A. proof (a)(i) Note first that as the quantities sup{


: x ∈ D, 0 < diam D ≤ δ}

decrease with δ, the limit is defined in [0, ∞] for every x ∈ X. Moreover, since diam D = diam D and µ∗Hr (A ∩ D) ≥ µ∗Hr (A ∩ D) for every D, sup{


: x ∈ D ⊆ X, 0 < diam D ≤ δ}

= sup{

µ∗Hr (A∩F ) (diam F )r

: F ⊆ X is closed, x ∈ F, 0 < diam F ≤ δ}

for every x and δ. (ii) Fix ² for the moment, and set A² = {x : x ∈ A, limδ↓0 sup{ Then θrη (A) ≤ µ∗Hr A −

² µ∗ A² 1+² Hr


: x ∈ D, 0 < diam D ≤ δ} > 1 + ²}.

for every η > 0, where θrη is defined in 471A. P P Let F be the family

{F : F ⊆ X is closed, 0 < diam F ≤ η, (1 + ²)(diam F )r ≤ µ∗Hr (A ∩ F )}. Then every member of A² belongs to sets in F of arbitrarily small diameter. Also, if hFn in∈N is any disjoint sequence in F, P∞ P∞ r ∗ ∗ n=0 (diam Fn ) ≤ n=0 µHr (A ∩ Fn ) ≤ µHr A < ∞ because every Fn , being closed, is measured by µHr . (If you like, Fn ∩ A is measured by the subspace measure S on A for every n.) So 471O tells usSthat there is a countable disjoint family I ⊆ F such that A² \ I is negligible, and µ∗Hr A² = µ∗Hr (A² ∩ I). Because θrη is an outer measure and θrη ≤ µ∗Hr , θrη A ≤ θrη (A ∩

[

I) + θrη (A \

[

I) ≤

X

(diam F )r + µ∗Hr (A \

[

I)

F ∈I

(because I is countable) ≤

[ [ 1 µ∗Hr (A ∩ I) + µ∗Hr (A \ I) 1+²

≤ µ∗Hr A −

² µ∗ (A² 1+² Hr

∩

[

= µ∗Hr A −

I) = µ∗Hr A −

[ ² µ∗Hr (A ∩ I) 1+²

² µ∗ A² , 1+² Hr

as claimed. Q Q (iii) Taking the supremum as η ↓ 0, µ∗Hr A ≤ µ∗Hr A −

² µ∗ A² 1+² Hr

and µHr A² = 0.

This is true for any ² > 0. But {x : x ∈ A, limδ↓0 sup{ is just

S n∈N


: x ∈ D, 0 < diam D ≤ δ} > 1}

A2−n , so is negligible.

(iv) Next, for 0 < ² ≤ 1, set A0² = {x : x ∈ A, µ∗Hr (A ∩ D) ≤ (1 − ²)(diam D)r whenever x ∈ D and 0 < diam D ≤ ²}.

471P

Hausdorff measures

621

Then A0² is negligible. P P Let hDn in∈N be any sequence of sets of diameter at most ² covering A0² . Set 0 K = {n : Dn ∩ A² 6= ∅}. Then X µ∗Hr A0² ≤ µ∗Hr (A ∩ Dn ) n∈K

≤ (1 − ²)

X

(diam Dn )r ≤ (1 − ²)

∞ X

(diam Dn )r .

n=0

n∈K

As hDn in∈N is arbitrary, µ∗Hr A0² ≤ (1 − ²)θr² A0² ≤ (1 − ²)µ∗Hr A0² , Q and µ∗Hr A0² (being finite) must be zero. Q This means that {x : x ∈ A, limδ↓0 sup{


: x ∈ D, 0 < diam D ≤ δ} < 1} ⊆

S n∈N

A02−n

is also negligible, and we have the result. (b) We need a slight modification of the argument in (a)(iv). This time, for 0 < ² ≤ 1, set A˜0² = {x : x ∈ A, µ∗Hr (A ∩ B(x, δ)) ≤ (1 − ²)δ r whenever 0 < δ ≤ ²}. Then µ∗Hr A˜0² ≤ ². P P Note first that, as µHr {x} = 0 for every x, µ∗Hr (A ∩ B(x, δ)) ≤ (1 − ²)δ r whenever 0 x ∈ A˜² and 0 ≤ δ ≤ ². Let hDn in∈N be a sequence of sets of diameter at most ² covering A˜0² . Set K = {n : Dn ∩ A˜0² 6= ∅}, and for n ∈ K choose xn ∈ Dn ∩ A˜0² and set δn = diam Dn . Then Dn ⊆ B(xn , δn ) S and δn ≤ ² for each n, so A˜0² ⊆ n∈K B(xn , δn ) and X µ∗Hr (A˜0² ∩ B(xn , δn )) µ∗Hr A˜0² ≤ n∈K

X

≤

(1 − ²)δnr ≤ (1 − ²)

As hDn in∈N is arbitrary, Now

≤ (1 −

(diam Dn )r .

n=0

n∈K

µ∗Hr A˜²0

∞ X

²)µ∗Hr A˜0²

{x : x ∈ A, lim supδ↓0

and A˜0² must be negligible. Q Q µ∗Hr (A∩B(x,δ)) δr

< 1} =

S n∈N

A˜02−n

is negligible. As for the second formula, we need note only that diam B(x, δ) ≤ 2δ for every x ∈ X, δ > 0 to obtain the first inequality, and apply (a) to obtain the second. (c) Let ² > 0. This time, write A˜² for {x : x ∈ X, limδ↓0 sup{


: x ∈ D, 0 < diam D ≤ δ} > ²}.

Let E ⊆ A be a closed set such that µ(A \ E) ≤ ²2 (471De). For η > 0, let Fη be the family {F : F ⊆ X \ E is closed, 0 < diam F ≤ η, µHr (A ∩ F ) ≥ ²(diam F )r }. Just as in (a) above, every point in A˜² \E belongs to members of Fη of arbitrarily small diameter. If hFn in∈N is a disjoint sequence in Fη , P∞ 1 r n=0 (diam F ) ≤ µHr (A \ E) ≤ ² ²

S is finite. There is therefore a countable disjoint family Iη ⊆ Fη such that µHr ((A˜² \ E) \ Iη ) = 0. If θrη is the outer measure defined in 471A, we have [ [ θrη (A˜² \ A) ≤ θrη ( Iη ) + θrη (A˜² \ (E ∪ Iη )) X [ ≤ (diam F )r + µ∗Hr (A˜² \ (E ∪ Iη )) ≤ ². F ∈Iη

622


471P

As η is arbitrary, µ∗Hr (A˜² \ A) ≤ ². But now {x : x ∈ X \ A, limδ↓0 sup{ is

S n∈N


: x ∈ D, 0 < diam D ≤ δ} > 0}

A˜2−n \ A, and is negligible.

471Q I now come to a remarkable fact about Hausdorff measures on analytic spaces: their Borel versions are semi-finite (471S). We need some new machinery. Lemma Let (X, ρ) be a metric space, and r > 0, δ > 0. Suppose that θrδ (X), as defined in 471A, is finite. (a) There is a non-negative additive functional ν on PX such that νX = 5−r θrδ (X) and νA ≤ (diam A)r whenever A ⊆ X and diam A ≤ 51 δ. (b) If X is compact, there is a Radon measure µ on X such that µX = 5−r θrδ (X) and µG ≤ (diam G)r whenever G ⊆ X is open and diam G ≤ 51 δ. proof (a) I use 391E. If θrδ (X) = 0 the result is trivial. Otherwise, set γ = 5r /θrδ (X) and define φ : PX → [0, 1] by setting φA = min(1, γ(diam A)r ) if diam A ≤ 15 δ, 1 for other A ⊆ X. Now P i∈I χAi ≥ mχX, then P whenever hAi ii∈I is a finite family of subsets of X, m ∈ N and φA ≥ m. i i∈I P P Discarding any Ai for which φAi = 1, if necessary, we may suppose that diam Ai ≤ 51 δ and φAi = γ(diam Ai )r for every i. Choose hIj ij≤m , hJj ij
i∈I

i∈I

≥ 5−r mγθrδ (X) = m. Q Q By 391E, there is an additive functional ν0 : PX → [0, 1] such that ν0 X = 1 and ν0 A ≤ φA for every A ⊆ X. Setting ν = 5−r θrδ (X)ν0 , we have the result. (b) Now suppose that X is compact. By 416N, there is a Radon measure µ on X such that µX ≤ νX and µK ≥ νK for every compact K ⊆ X. Because X itself is compact, µX = νX = 5−r θrδ (X). If G is open and diam G ≤ 51 δ, µG = µX − µ(X \ G) ≤ νX − ν(X \ G) (because X \ G is compact) = νG ≤ (diam G)r , as required. 471R Lemma (Howroyd 95) Let (X, ρ) be a compact metric space and r > 0. Let µHr be r-dimensional Hausdorff measure on X. If µHr (X) > 0, there is a Borel set H ⊆ X such that 0 < µHr H < ∞. proof (a) Let δ > 0 be such that θr,5δ (X) > 0, where θr,5δ is defined as in 471A. Then there is a family V of open subsets of X such that (i) diam V ≤ δ for every V ∈ V (ii) {V : V ∈ V, diam V ≥ ²} is finite for every ² > 0 (iii) whenever A ⊆ X and 0 < diam A < 41 δ there is a V ∈ V such that A ⊆ V and S diam V ≤ 8 diam A. P P For each k ∈ N, let Ik be a finite subset of X such that X = x∈Ik B(x, 2−k−2 δ); now set V = {U (x, 2−k−1 δ) : k ∈ N, x ∈ Ik }. Then V is a family of open sets and (i) and (ii) are satisfied. If A ⊆ X and 0 < diam A < 41 δ, let k ∈ N be such that 2−k−3 δ ≤ diam A < 2−k−2 δ. Take x ∈ Ik such that B(x, 2−k−2 δ) ∩ A 6= ∅; then A ⊆ U (x, 2−k−1 δ) ∈ V and diam U (x, 2−k−1 δ) ≤ 2−k δ ≤ 8 diam A. Q Q

471R

Hausdorff measures

623

In particular, {V : V ∈ V, diam V ≤ ²} covers X for every ² > 0. (b) Set P = {µ : µ is a Radon measure on X, µV ≤ (diam V )r for every V ∈ V}. P is non-empty (it contains the zero measure, for instance). Now if G ⊆ X is open, µ 7→ µG is lower semi-continuous for the narrow topology (437Jd), so P is a closed set in the narrow topology on the set of Radon measures on X, which may be identified with a compact subset of C(X)∗ with its weak* topology (437Kc). Moreover, since there is a finite subfamily of V covering X, γ = sup{µX : µ ∈ P } is finite, and P is compact (437O). Because µ 7→ µX is continuous, P0 = {µ : µ ∈ P, µX = γ} is non-empty. Of course P and P0 are both convex, and P0 , like P , is compact. By the Kreˇın-Mil’man theorem (4A4Gb), applied in C(X)∗ , P has an extreme point ν say. Note next that θr,5δ (X) is certainly finite, again because X is compact. By 471Qb, γ > 0, and ν is non-trivial. For any ² > 0, there is a finite cover of X by sets in V of diameter at most ², which have measure at most ²r (for ν); so ν is atomless. In particular, ν{x} = 0 for every x ∈ X. (c) For ² > 0, set G² =

S

{V : V ∈ V, 0 < diam V ≤ ² and νV ≥ 12 (diam V )r }.

Then G² is ν-conegligible. P P?? Otherwise, ν(X \ G² ) > 0. Because V²0 = {V : V ∈ V, diam V > ²} is finite, there is a Borel set E ⊆ X \ G² such that νE > 0 and, for every V ∈ V²0 , either E ⊆ V or E ∩ V = ∅. Because ν is atomless, there is a measurable set E0 ⊆ E such that νE0 = 12 νE (215D); set E1 = E \ E0 . Define Radon measures ν1 , ν2 on X by setting νi (F ) = 2ν(F ∩ Ei ) + ν(F \ E) whenever ν measures F \ E1−i , for each i (use 416S if you feel the need to check that this defines a Radon measure on the definitions of this book). If V ∈ V, then, by the choice of E, either E ⊆ V and νi V = νV ≤ (diam V )r or E ∩ V = ∅ and νi V = νV ≤ (diam V )r or 0 < diam V ≤ ² and νV < 21 (diam V )r , in which case νi V ≤ 2νV ≤ (diam V )r or diam V = 0 and νi V = νV = 0 = (diam V )r . So both νi belong to P and therefore to P0 , since νi X = νX = γ. But ν = 21 (ν1 + ν2 ) and ν1 6= ν2 , so this is impossible, because ν is supposed to be an extreme point of P0 . X XQ Q T (d) Accordingly, setting H = n∈N G2−n , νH = νX = γ. Now examine µHr H. (i) µHr H ≥ 8−r γ. P P Let hAn in∈N An ≤ 81 δ for every n. S be a sequence of sets0 covering H with diam 0 0 Set K = {n : diam An > 0}, H = H ∩ n∈K An ; then H \ H is countable, so νH = νH. For each n ∈ K, let Vn ∈ V be such that An ⊆ Vn and diam Vn ≤ 8 diam An ((a) above). Then ∞ X

(diam An )r =

n=0

X n∈K −r

≥8

(diam An )r ≥ 8−r X

X

(diam Vn )r

n∈K

νVn ≥ 8

−r

νH 0 = 8−r γ.

n∈K

As hAn in∈N is arbitrary, 8−r γ ≤ θr,δ/8 (H) ≤ µ∗Hr H = µHr H. Q Q (ii) µHr H ≤ 2γ. P P Let η > 0. Set F = {V : V ∈ V, 0 < diam V ≤ η, νV ≥ 12 (diam V )r }. Then F is a family of closed subsets of X, and (by the definition of G² ) every member of H belongs to members of F of arbitrarily small diameter. Also νF ≥ 21 (diam F )r for every F ∈ F, so P∞ P∞ r n=0 (diam Fn ) ≤ 2 n=0 νFn < ∞ for any disjoint sequence hFn in∈N in F. By 471O, there is a countable disjoint family I ⊆ F such that S µHr (H \ I) = 0. Accordingly

624


θrη (H) ≤ As η is arbitrary, µHr H =

P

µ∗Hr H

r F ∈I (diam F )

+ θrη (H \

S

I) ≤

471R

P F ∈I

2νF ≤ 2γ.

≤ 2γ. Q Q

(e) But this means that we have found a Borel set H with 0 < µHr H < ∞, as required. 471S Theorem (Howroyd 95) Let (X, ρ) be an analytic metric space, and r > 0. Let µHr be rdimensional Hausdorff measure on X, and B the Borel σ-algebra of X. Then the Borel measure µHr ¹B is semi-finite and tight (that is, inner regular with respect to the compact sets). proof Suppose that E ∈ B and µHr E > 0. Since E is analytic (423E), 471I above tells us that there is a compact set K ⊆ E such that µHr K > 0. Next, by 471R, there is a Borel set H ⊆ K such that 0 < µHr H < ∞. (Strictly speaking, µHr H here should be calculated as the r-dimensional Hausdorff measure of H defined by the subspace metric ρ¹K × K on K. By 471E we do not need to distinguish between this and the r-dimensional measure calculated from ρ itself.) By 471I again (applied to the subspace metric on H), there is a compact set L ⊆ H such that µHr L > 0. Thus E includes a non-negligible compact set of finite measure. As E is arbitrary, this is enough to show both that µHr ¹B is semi-finite and that it is tight. 471X Basic exercises (a) Define a metric ρ on X = {0, 1}N by setting ρ(x, y) = 2−n if x¹n = y¹n and x(n) 6= y(n). Show that the usual measure µ on X is one-dimensional Hausdorff measure. (Hint: diam F ≥ µF for every closed set F ⊆ X.) (b) Let (X, ρ) be a metric space and r > 0; let µHr , θr∞ be r-dimensional Hausdorff measure and capacity on X. (i) Show that, for A ⊆ X, µHr A = 0 iff θr∞ A = 0. (ii) Suppose that E ⊆ X and δ > 0 are such that µHr E < ∞ and δ < θr∞ E. Show that there is a closed set F ⊆ E such that µHr F > 0 and δµHr (F ∩ G) ≤ (diam G)r whenever µHr measures G. (Hint: show that {G• : θr∞ G ≤ δµHr (E ∩ G)} cannot be order-dense in the measure algebra of µHr . This is a version of ‘Frostman’s Lemma’.) (c) Let (X, ρ) be an analytic metric space, (Y, σ) a metric space, and f : X → Y a Lipschitz function, that is, a function which is γ-Lipschitz for some γ ≥ 0. Show that if r > 0 and A ⊆ X is measured by Hausdorff r-dimensional measure on X, with finite measure, then f [A] is measured by Hausdorff r-dimensional measure on Y . (d) Let (X, ρ) be a metric space and r > 0. Show that a set A ⊆ X is negligible P∞for Hausdorffr rdimensional measure on X iff there is a sequence hA i of subsets of X such that n n∈N n=0 (diam An ) is T S finite and A ⊆ n∈N m≥n Am . (e) Let (X, ρ) be a metric space. (i) Show that there is a unique dimH (X) ∈ [0, ∞] such that the r-dimensional Hausdorff measure of X is infinite if 0 < r < dimH (X), zero if r > dimH (X). (dimH (X) is the S Hausdorff dimension of X.) (ii) Show that if hAn in∈N is any sequence of subsets of X, then dimH ( n∈N An ) = supn∈N dimH (An ). (f ) Let (X, ρ) be a metric space, and µ any topological measure on X. Suppose that E ⊆ X and that µE is defined and finite. (i) Show that (x, δ) 7→ µ(E ∩ B(x, δ)) : X × [0, ∞[ → R is upper semi-continuous. (ii) Show that x 7→ lim supδ↓0

1 µ(E δr

∩ B(x, δ)) : X → [0, ∞] is Borel measurable. (iii) Show that if X is

separable, then µB(x, δ) > 0 for every δ > 0, for almost every x ∈ X. (g) Give R its usual metric. Let C ⊆ R be the Cantor set, and r = ln 2/ ln 3. Show that lim inf δ↓0

µHr (C∩B(x,δ)) (diam B(x,δ))r

≤ 2−r

for every x ∈ R. (h) Let (X, ρ) be a metric space and r > 0. Let µHr be r-dimensional Hausdorff measure on X and µ ˜Hr its c.l.d. version (213D-213E). Show that µ ˜Hr is inner regular with respect to the closed sets, and that µ ˜Hr A = µHr A for every analytic set A ⊆ X.

471Yh

Hausdorff measures

625

471Y Further exercises (a) The next few exercises (down to 471Yd) will all be based on the following context. Let (X, ρ) be a metric space and ψ : PX → [0, ∞] a function such that ψ∅ = 0 and ψA ≤ ψA0 whenever A ⊆ A0 ⊆ X. Set ∞ X

θψδ A = inf{

ψDn : hDn in∈N is a sequence of subsets of X covering A,

n=0

diam Dn ≤ δ for every n ∈ N} for δ > 0, and θψ A = supδ>0 θψδ A for A ⊆ X Show that θψ is a metric outer measure. Let µψ be the measure defined from θψ by Carathéodory’s method. (b) Suppose that ψA = inf{ψE : E is a Borel set including A} for every A ⊆ X. Show that θψ = µ∗ψ and that µψ E = sup{µψ F : F ⊆ E is closed} whenever µψ E < ∞. (c) Suppose that X is separable and that there is a β ≥ 0 such that ψA∼ ≤ βψA for every A ⊆ X, where A∼ is defined in 471M. (i) Suppose that A ⊆ X and F is a family of closed subsets of X such that P ∞ n=0 ψFn is finite for every disjoint sequence hFn in∈N in F and for every x ∈ A, δ > 0 there is an S F ∈F such that x ∈ F and 0 < diam F ≤ δ. Show that there is a disjoint family I ⊆ F such that µψ (A \ I) = 0. (ii) Suppose that δ > 0 and that θψδ (X) < ∞. Show that there is a non-negative additive functional ν on PX such that νX = β1 θψδ (X) and νA ≤ ψA whenever A ⊆ X and diam A ≤ 15 δ. (iii) Now suppose that for every x ∈ X, ² > 0 there is a δ > 0 such that ψB(x, δ) ≤ ². Show that if X is compact and µψ X > 0 there is a compact set K ⊆ X such that 0 < µψ K < ∞. (d) State and prove a version of 471P appropriate to this context. (e) Give an example of a set A ⊆ R2 which is measured by Hausdorff 1-dimensional measure on R2 but is such that its projection onto the first coordinate is not measured by Hausdorff 1-dimensional measure on R. (f ) Show that the space (X, ρ) of 471Xa can be isometrically embedded as a subset of a metric space (Y, σ) in such a way that (i) diam B(y, δ) = 2δ for every y ∈ Y and δ ≥ 0 (ii) Y \ X is countable. Show that if µH1 is one-dimensional Hausdorff measure on Y , then µH1 B(y, δ) ≤ δ for every y ∈ Y , δ ≥ 0, so that lim supδ↓0

µH1 B(y,δ) diam B(x,δ)

=

1 2

for every y ∈ Y . (g) Let hkn in∈N be a sequence in N \ {0, 1, 2, 3} such that

P∞

1

n=0 k n

< ∞. Set X =

Q n∈N

kn . Set m0 = 1,

mn+1 = k0 k1 . . . kn for n ∈ N. Define a metric ρ on X by saying that ρ(x, y) = 1/2mn if n = min{i : x(i) 6= y(i)} and min(x(n), y(n)) = 0, = 1/mn if n = min{i : x(i) 6= y(i)} and min(x(n), y(n)) > 0. Let ν be the product measure on X obtained by giving each factor kn the uniform probability measure in which each singleton set has measure 1/kn . (i) Show that S if A ⊆ X then ν ∗ A ≤ diam A. (ii) Show that ν is one-dimensional Hausdorff measure on X. (iii) Set E = n∈N {x : x ∈ X, x(n) = 0}. Show that νE < 1. (iv) Show that lim supδ↓0

ν(E∩B(x,δ)) νB(x,δ)

≥

1 2

for every x ∈ X. (v) Show that there is a familySF of closed balls in X such that every point of X is the centre of arbitrarily small members of F, but ν( I) < 1 for any disjoint subfamily I of F. (h) Let ρ be the metric on {0, 1}N defined in 471Xa. Show that for any integer s ≥ 1 there is a bijection f : [0, 1]s → {0, 1}N such that whenever 0 < r ≤ s, µ∗Hr is Hausdorff r-dimensional outer measure on [0, 1]s (for its usual metric) and µ ˜∗H,r/s is Hausdorff rs -dimensional measure on {0, 1}ω , then there is an α > 0 such ∗ −1 ∗ that µHr f [A] ≤ µH,r/s A ≤ αµ∗Hr f −1 [A] for every A ⊆ {0, 1}N .

626


471Yi (X)

(X×R s )

(i) Let (X, ρ) be a metric space, and r > 0; let s ≥ 1 be an integer. Write µHr and µH,r+s for Hausdorff r-dimensional measure on X and Hausdorff r + s-dimensional measure on X × Rs respectively, and µLs for (X×R s ) Lebesgue measure on Rs . (i) Show that there are non-zero constants c, c0 such that cµH,r+s (E × F ) ≤ (X)

(X×R s )

µHr (E) · µLs (F ) ≤ c0 µH,r+s (E × F ) for all Borel sets E ⊆ X, F ⊆ Rs . (Hint: Federer 69, 2.10.45.) (ii) (X×R s )

(X)

Write λ for the c.l.d. product of µHr and µLs , and µ ˜H,r+s (X×R s )

(X×R s )

for the c.l.d. version of µH,r+s . Show that (X×R s )

these have the same domain Λ and that c˜ µH,r+s (W ) ≤ λW ≤ c0 µ ˜H,r+s (W ) for every W ∈ Λ. 471 Notes and comments In the exposition above, I have worked throughout with simple r-dimensional measures for r > 0. As noted in 264Db, there are formulae in which it is helpful to interpret µH0 as counting measure. More interestingly, when we come to use Hausdorff measures to give us information about the geometric structure of an object (e.g., in the observation that the Cantor set has ln 2/ ln 3-dimensional Hausdorff measure 1, in 264J), it is sometimes useful to refine the technique by using other functionals than A 7→ (diam A)r in the basic formulae of 264A or 471A. The most natural generalization is to functionals of the form ψA = h(diam A) where h : [0, ∞] → [0, ∞] is a non-decreasing function (264Yo). But it is easy to see that many of the arguments are valid in much greater generality, as in 471Ya-471Yc. For more in these directions see Rogers 70 and Federer 69. In the context of this book, the most conspicuous peculiarity of Hausdorff measures is that they are often very far from being semi-finite. (This is trivial for non-separable spaces, by 471Df. That Hausdorff one-dimensional measure on a subset of R2 can be purely infinite is not I think obvious; I gave an example in 439H.) The response I ordinarily recommend in such cases is to take the c.l.d. version. But then of course we need to know just what effect this will have. In geometric applications, one usually begins by checking that the sets one is interested in have σ-finite measure, and that therefore no problems arise; but it is a striking fact that Hausdorff measures behave relatively well on analytic sets, even when not σ-finite, provided we ask exactly the right questions (471I, 471S, 471Xh). The geometric applications of Hausdorff measures, naturally, tend to rely heavily on density theorems; it is therefore useful to know that we have effective versions of Vitali’s theorem available in this context (471N-471O), leading to a general density theorem (471P) similar to that in 261D; see also 472D below. I note that 471P is useful only after we have been able to concentrate our attention on a set of finite measure. And traps remain. For instance, the formulae of 261C-261D cannot be transferred to the present context without re-evaluation (471Yg).

472 Besicovitch’s Density Theorem The first step in the program of the next few sections is to set out some very remarkable properties of Euclidean space. We find that in Rr , for geometric reasons (472A), we have versions of Vitali’s theorem (472B-472C) and Lebesgue’s Density Theorem (472D) for arbitrary Radon measures. I add a version of the Hardy-Littlewood Maximal Theorem (472F). Throughout the section, r ≥ 1 will be a fixed integer. As usual, I write B(x, δ) for the closed ball with centre x radius δ. k k will represent the Euclidean norm, and x .y the scalar product of x and y, so that Pand r x . y = i=1 ξi ηi if x = (ξ1 , . . . , ξr ) and y = (η1 , . . . , ηr ). 472A Besicovitch’s Covering Lemma Suppose that ² > 0 is such that (5r + 1)(1 − ² − ²2 )r > (5 + ²)r . Let x0 , . . . , xn ∈ Rr , δ0 , . . . , δn > 0 be such that kxi − xj k > δi ,

δj ≤ (1 + ²)δi

whenever i < j ≤ n. Then #({i : i ≤ n, kxi − xn k ≤ δi + δn }) ≤ 5r . proof Set I = {i : i ≤ n, kxi − xn k ≤ δn + δi }.

472A

Besicovitch’s Density Theorem

627

(a) It will simplify the formulae of the main argument if we suppose for the time being that δn = 1; in this case 1 ≤ (1 + ²)δi , so that δi ≥

1 1+²

for every i ≤ n, while we still have δi < kxi − xn k for every i < n,

and kxi − xn k ≤ 1 + δi for every i ∈ I. For i ∈ I, define x0i by saying that – if kxi − xn k ≤ 2 + ², x0i = xi ; – if kxi − xn k > 2 + ², x0i is to be that point of the closed line segment from xn to xi which is at distance 2 + ² from xn . (b) The point is that kx0i − x0j k > 1 − ² − ²2 whenever i, j are distinct members of I. P P We may suppose that i < j. case 1 Suppose that kxi − xn k ≤ 2 + ² and kxj − xn k ≤ 2 + ². In this case kx0i − x0j k = kxi − xj k ≥ δi ≥

1 1+²

≥ 1 − ².

case 2 Suppose that kxi − xn k ≥ 2 + ² ≥ kxj − xn k. In this case kx0i − x0j k = kx0i − xj k ≥ kxi − xj k − kxi − x0i k ≥ δi − kxi − xn k + 2 + ² ≥ δi − δi − 1 + 2 + ² = 1 + ². case 3 Suppose that kxi − xn k ≤ 2 + ² ≤ kxj − xn k. Then kx0i − x0j k = kxi − x0j k ≥ kxi − xj k − kxj − x0j k > δi − kxj − xn k + 2 + ² ≥ δi − δj − 1 + 2 + ² ≥ δi − δi (1 + ²) + 1 + ² ≥ 1 + ² − ²(2 + ²) (because δi < kxi − xn k ≤ 2 + ²) = 1 − ² − ²2 . case 4 Suppose that 2 + ² ≤ kxj − xn k ≤ kxi − xn k. Let y be the point on the line segment between xi and xn which is the same distance from xn as xj . In this case ky − xj k ≥ kxi − xj k − kxi − yk ≥ δi − kxi − xn k + kxj − xn k ≥ kxj − xn k − 1. Because the triangles (xn , y, xj ) and (xn , x0i , x0j ) are similar, kx0i − x0j k =

2+² ky kxj −xn k

− xj k ≥ (2 + ²)

kxj −xn k−1 kxj −xn k

≥1+²

because kxj − xn k ≥ 2 + ². case 5 Suppose that 2 + ² ≤ kxi − xn k ≤ kxj − xn k. This time, let y be the point on the line segment from xn to xj which is the same distance from xn as xi is. We now have ky − xi k ≥ kxi − xj k − kxj − yk > δi − kxj − xn k + kxi − xn k ≥ δj − ²δi − (δj + 1) + kxi − xn k = kxi − xn k − 1 − ²δi ≥ kxi − xn k(1 − ²) − 1, so that kx0i − x0j k =

2+² ky kxi −xn k

≥ (2 + ²)

− xi k > (2 + ²)

(2+²)(1−²)−1 2+²

So we have the required inequality in all cases. Q Q

kxi −xn k(1−²)−1 kxi −xn k

= 1 − ² − ²2 .

628


472A

2

) for i ∈ I. These are disjoint, all have Lebesgue measure (c) Now consider the balls B(x0i , 1−²−² 2 −r 2 βr (1 − ² − ²2 )r where βr is the measure of the unit ball B(0, 1), and are all included in the ball B(xn , 2 + −r ² + 1−² βr (5 + ²)r . So we must have 2 ), which has measure 2 2−r βr (1 − ² − ²2 )r #(I) ≤ 2−r βr (5 + ²)r . But ² was declared to be so small that this implies that #(I) ≤ 5r , as claimed. (e) This proves the lemma in the case δn = 1. For the general case, replace each xi by δn−1 xi and each δi by δi /δn ; the change of scale does not affect the hypotheses or the set I. 472B Theorem Let A ⊆ Rr be a bounded set, and I a family of non-trivial closed balls in Rr such that every point of A is the centre of aSmember of I. Then there is a family hIk ik<5r of countable subsets of I such that each Ik is disjoint and k<5r Ik covers A. proof (a) For each x ∈ A let δx > 0 be such that B(x, δx ) ∈ I. If either A is empty or supx∈A δx = ∞, the result is trivial. (In the latter case, take x ∈ A such that δx ≥ diam A and set I0 = {B(x,S δx )}, Ik = ∅ for k > 0.) So let us suppose henceforth that {δx : x ∈ A} is bounded in R. In this case, C = x∈A B(x, δx ) is bounded in Rr . Fix ² > 0 such that (5r + 1)(1 − ² − ²2 )r > (5 + ²)r . S (b) Choose inductively a sequence hBn in∈NSin I ∪ {∅} as follows. Given S hBi ii δxi , δxj ≤ αi ≤ (1 + ²)δxi . But now 472A gives the result at once. Q Q (c) We may therefore define a function f : N → {0, 1, . . . , 5r − 1} by setting f (n) = min{k : 0 ≤ k < 5r , f (i) 6= k for every i ∈ In } for every n ∈ N. Set Ik = {Bi : i ∈ N, f (i) = k, Bi 6= ∅} for each k < 5r . By the choice of f , i ∈ / Ij , so that B ∩ B = ∅, whenever i < j and f (i) = f (j); thus every I is disjoint. Since B ⊆ C for every i, i j k i P P∞ {µBi : f (i) = k} ≤ µ∗ C for every k < 5r , and i=0 µBi ≤ 5r µ∗ C is finite. (d) ?? Suppose, if possible, that

S S S A 6⊆ k<5r Ik = n∈N Bn . S S Take x ∈ A \ n∈N Bn . Then, first, A 6⊆ i
is impossible. X X (e) Thus A ⊆

S

S k<5r

Ik , as required.

472C Theorem Let λ be a Radon measure on Rr , A a subset of Rr and I a family of non-trivial closed balls in Rr such that every point of A is the centre of arbitrarily small members of I. Then S (a) there is a countable disjoint I0 ⊆ I such that λ(A \ I0 ) S = 0; P (b) for every ² > 0 there is a countable I1 ⊆ I such that A ⊆ I1 and B∈I1 λB ≤ λ∗ A + ². proof (a)(i) S The first step is to show that if A0 ⊆ A is bounded then there is a finite disjoint set J ⊆ I such ∗ 0 that λ (A ∩ J ) ≥ 6−r λ∗ A0 . P P If λ∗ A0 =S0 take J = ∅. Otherwise, by 472B, there is a family hJk ik<5r of disjoint countable subsets of I such that k<5r Jk covers A0 . Accordingly P5r −1 S λ∗ A0 ≤ k=0 λ∗ (A0 ∩ Jk )

472D


629

S and there is some k < 5r such that λ∗ (A0 ∩ Jk ) ≥ 5−r λ∗ A0 . Let hBi ii∈N be a sequence running over Jk ; then S S limn→∞ λ∗ (A0 ∩ i≤n Bi ) = λ∗ (A0 ∩ Jk ) ≥ 5−r λ∗ A0 , S so there is some n ∈ N such that λ∗ (A0 ∩ i≤n Bi ) ≥ 6−r λ∗ A0 , and we can take J = {Bi : i ≤ n}. Q Q (ii) Now choose hKn in∈N inductively, as follows. Start by fixing on a sequence hmn in∈N in N such that every member of N appears infinitely often as an mn . Take K0 = S S ∅. Given that Kn is a finite disjoint subset of I, set I 0 = {B : B ∈ I, B ∩ Kn = S ∅}, An = A ∩ B(0, mn ) \ Kn . Because every point of A is the centre of arbitrarily small members of I, and Kn is closed, every member of An is the centre ofS (arbitrarily small) ∗ members of I 0 , and (i) tells us that there is a finite disjoint set Jn ⊆ I 0 such that λ (A ∩ Jn ) ≥ 6−r λ∗ An . n S Set Kn+1 = K ∪ Jn , and continue. At the end of the induction, set I0 = n∈N Kn ; because hKn in∈N is non-decreasing and every Kn is disjoint, I0 is disjoint, and of course I0 ⊆ I. The effect of this construction is to ensure that

(because

S

λ∗ (A ∩ B(0, mn ) \

[

Kn+1 ) = λ∗ (An \

[

Jn ) = λ∗ An − λ∗ (An ∩

[

Jn )

Jn is a closed set, therefore measured by λ) ≤ (1 − 6−r )λ∗ An = (1 − 6−r )λ∗ (A ∩ B(0, mn ) \

[

Kn )

for every n. So, for any m ∈ N,

S λ∗ (A ∩ B(0, m) \ Kn ) ≤ λ∗ (A ∩ B(0, m))(1 − 6−r )#({j:j
(b)(i) Let E ⊇ A be such that λE = λ∗ A, and H ⊇ E an open set such that λH ≤ λE + 21 ² (256Bb). Set I 0 = {B : B ∈ I, B ⊆ H}. Then every point of AS is the centre of arbitrarily small members of I 0 , so by 0 (a) there is a disjoint family I0 ⊆ I such that λ(A \ I0 ) = 0. Of course S P 1 ∗ B∈I0 λB = λ( I0 ) ≤ λH ≤ λ A + 2 ². S (ii) For m ∈ N set Am = A ∩ B(0, m) \ I0 . Then there is a Jm ⊆ I, covering Am , such that −m−2 ². P P There is an open set G ⊇ Am such that λG ≤ 5−r 2−m−2 ². Now I 00 = {B : B∈Jm λB ≤ 2 B ∈ I,SB ⊆ G} covers Am , so there is a family hJmk ik<5r of disjoint countable subfamilies of I 00 such that Jm = k<5r Jmk covers Am . For each k, P S B∈Jmk λB = λ( Jmk ) ≤ λG,

P

so

P B∈Jm

(iii) Setting I1 = I0 ∪

S m∈N

λB ≤ 5r λG ≤ 2−m−2 ². Q Q

Jm we have a cover of A by members of I, and

X B∈I1

λB ≤

X B∈I0

λB +

∞ X X

λB

m=0 B∈Jm ∞ X −m−2

1 ≤ λ∗ A + ² + 2 2 m=0

² = λ∗ A + ².

472D Besicovitch’s Density Theorem Let λ be any Radon measure on Rr . Then, for any locally λ-integrable real-valued function f , R 1 (a) f (y) = limδ↓0 f dλ, λB(y,δ) B(y,δ) R 1 (b) limδ↓0 |f (x) − f (y)|λ(dx) = 0 B(y,δ) λB(y,δ)

630


472D

for λ-almost every y ∈ Rr . Remark The theorem asserts that, for λ-almost every y, limits of the form limδ↓0

1 λB(y,δ)

. . . are defined;

in my usage, this includes the assertion that λB(y, δ) > 0 for all sufficiently small δ > 0. proof (Compare 261C and 261E.) (a) Let Z be the support of λ (411Nd); then Z is λ-conegligible and λB(y, δ) > 0 whenever y ∈ Z and δ > 0. For q < q 0 in Q, n ∈ N set Anqq0 = {y : y ∈ Z ∩ dom f, kyk ≤ n, f (y) ≤ q, lim supδ↓0

1 λB(y,δ)

R

B(y,δ)

f dλ > q 0 }.

R Then λAnqq0 = 0. P P Let ² > 0. Then there is an η ∈ ]0, ²] such that F |f |dλ ≤ ² whenever F ⊆ B(0, n) and λF ≤ η (225A). Let E be a measurable envelope of Anqq0 included in {y : y ∈ Z ∩ dom f, f (y) ≤ q}, and take an open set G R⊇ E such that λ(G \ E) ≤ η (256Bb). Let I be the family of non-singleton closed 0 balls B ⊆ G such that B f ≥ q 0 λB. Then every point of A Snqq belongs to arbitrarily small S members of 0 I, so there is a disjoint family I ⊆ I such that λ(A \ I ) = 0 (472C). Now λ(E \ I0 ) = 0 and 0 nqq 0 S λ(( I0 ) \ E) ≤ η ≤ ², so [ X q 0 λE ≤ q 0 λ( I0 ) + ²|q 0 | = q 0 λB + ²|q 0 | X Z

≤

B∈I0

Z

B∈I0

Z

f dλ + ²|q 0 | =

B

S

f dλ + ²|q 0 | I0

f dλ + ²(1 + |q 0 |) ≤ qλE + ²(1 + |q 0 |),

≤ E

and (q 0 − q)λ∗ Anqq0 = (q 0 − q)λE ≤ (1 + |q 0 |)². As ² is arbitrary, λ∗ Anqq0 = 0. Q Q As n, q and q 0 are arbitrary, 1 λB(y,δ)

lim supδ↓0

R B(y,δ)

f ≤ f (y)

for λ-almost every y ∈ Z, therefore for λ-almost every y ∈ Rr . Similarly, or applying the same argument to −f , lim inf δ↓0

1 λB(y,δ)

for λ-almost every y, and limδ↓0

1 λB(y,δ)

R

B(y,δ)

R B(y,δ)

f ≥ f (y)

f exists = f (y)

for λ-almost every y. (b) Now, for each q ∈ Q set gq (x) = |f (x) − q| for x ∈ dom f . By (a), we have a λ-conegligible set D such that limδ↓0

1 λB(y,δ)

R

g dλ B(y,δ) q

= gq (y)

for every y ∈ D and q ∈ Q. Now, if y ∈ D and ² > 0, there is a q ∈ Q such that |f (y) − q| ≤ ², and a δ0 > 0 such that | whenever 0 < δ ≤ δ0 . But in this case

1 λB(y,δ)

R

g dλ B(y,δ) q

− gq (y)| ≤ ²

*472F


1 λB(y,δ)

631

Z |f (x) − f (y)|λ(dx) Z 1 |f (x) − q|λ(dx) +

B(y,δ)

≤

λB(y,δ)

1 λB(y,δ)

B(y,δ)

Z |q − f (y)|λ(dx) B(y,δ)

≤ 2². As ² is arbitrary, limδ↓0

1 λB(y,δ)

R B(y,δ)

|f (x) − f (y)|λ(dx) = 0;

as this is true for every y ∈ D, the theorem is proved. *472E Proposition Let λ, λ0 be Radon measures on Rr , and G ⊆ Rr an open set. Let Z be the support of λ, and for x ∈ Z ∩ G set M (x) = sup{

λ0 B λB

: B ⊆ G is a non-trivial ball with centre x}.

Then 5r 0 λG t

λ{x : x ∈ Z, M (x) ≥ t} ≤ for every t > 0.

proof The function M : Z → [0, ∞] is lower semi-continuous. P P If M (x) > t ≥ 0, there is a δ > 0 such that B(x, δ) ⊆ G and λ0 B(x, δ) > tλB(x, δ). Because λ is a Radon measure, there is an open set V ⊇ B(x, δ) such that V ⊆ G and λ0 B(x, δ) > tλV ; because B(x, δ) is compact, there is an η > 0 such that B(x, δ + 2η) ⊆ V . Now if y ∈ Z and ky − xk ≤ η, B(x, δ) ⊆ B(y, δ + η) ⊆ V , 0

so λ B(y, δ + η) > tλB(y, δ + η) and M (y) > t. Q Q In particular, Ht = {x : x ∈ Z ∩ G, M (x) > t} is always measured by λ. Now, given t > 0, let I be the set of non-trivial closed balls BS ⊆ G such that λ0 B > tλB. By 472B, there is a family hIk ik<5r of countable disjoint subsets of I such that k<5r Ik covers Ht . So λHt ≤

P5r −1 P k=0

B∈Ik

λB ≤

1 t

P5r −1 P k=0

B∈Ik

λ0 B ≤

5r 0 λ G, t

as claimed. *472F Theorem Let λ be a Radon measure on Rr , and f ∈ Lp (λ) any function, where 1 < p < ∞. R 1 |f |dλ. Then f ∗ is lower Let Z be the support of λ, and for x ∈ Z set f ∗ (x) = supδ>0 λB(x,δ) B(x,δ) ¡ 5r p ¢1/p semi-continuous, and kf ∗ kp ≤ 2 kf kp . p−1

proof (a) Replacing f by |f | if necessary, we may suppose that f ≥ 0. Z is λ-conegligible, so that f ∗ is defined λ-almost everywhere. Next, f ∗ is lower P I repeat an idea from the proof of 472E. R semi-continuous. P If f ∗ (x) > t ≥ 0, there is a δ > 0 such that B(x,δ) |f |dλ > tλB(x, δ). Because λ is a Radon measure, there R is an open set V ⊇ B(x, δ) such that B(x,δ) |f |dλ > tλV ; because B(x, δ) is compact, there is an η > 0 such that B(x, δ + 2η) ⊆ V ; and now f ∗ (y) > t for every y ∈ Z ∩ B(x, η). Q Q (b) For t > 0, set Ht = {x : x ∈ Z, f ∗ (x) > t}, Ft = {x : x ∈ dom f, f (x) ≥ t}. Then λHt ≤

2·5r t

R

Ft/2

f dλ

for every t > 0. P P Set g = f × χFt/2 . Because ( 2t )p λFt/2 ≤ kf kpp is finite, λFt/2 is finite, χFt/2 ∈ Lq (λ) 1 1 (where p + q = 1) and g ∈ L1 (λ) (244E). Let λ0 be the indefinite-integral measure defined by g over λ (§234); then λ0 is totally finite, and is a Radon measure (416S). Set

632


M (x) = sup{ for x ∈ Z. Then f ∗ (x) ≤ M (x) +

t 2

λ0 B λB

*472F

: B ⊆ Rr is a non-trivial ball with centre x}

for every x ∈ Z, just because

R

t 2

f dλ ≤ λB +

B

R B

t 2

g dλ = λB + λ0 B

for every closed ball B. Accordingly t 2

λHt ≤ λ{x : M (x) > } ≤

2·5r 0 r λR t

=

2·5r t

R Ft/2

f dλ. Q Q

(c) As in part (c) of the proof of 286A, we now have Z

Z (f ∗ )p dλ =

∞

Z λ{x : f ∗ (x)p > t}dt = p

0

Z

≤ 2 · 5r p

∞

0

Z = 2 · 5r p

tp−1 λ{x : f ∗ (x) > t}dt

0

Z tp−2

∞

Ft/2

2p−1 f (x)p λ(dx) p−1 r R

=

2p 5r p p−1

Z

Z

f dλdt = 2 · 5r p Z

f (x) Rr

2f (x)

tp−2 dtλ(dx)

0

f p dλ.

Taking pth roots, we have the result. 472X Basic exercises (a) Show that if λ, λ0 are Radon measures on Rr which agree on closed balls, they are equal. (b) Let λ be a Radon measure on Rr . Let A ⊆ Rr be a non-empty set, and ² > 0. Show that there is a S sequence hBn in∈N of closed balls in Rr , all of radius at most ² and with centres in A, such that A ⊆ n∈N Bn P∞ and n=0 λBn ≤ λ∗ A + ². (c) Let λ be a non-zero Radon measure on Rr and F its support. Show that we have a lower density φ (definition: 341C) for the subspace measure λF defined by setting φE = {x : x ∈ F , limδ↓0

λ(E∩B(x,δ)) λB(x,δ)

= 1}

whenever λF measures E. 472Y Further exercises (a) (i) Let I be a finite family of intervals (open, closed S or half-open) in R. Show that there are subfamilies I0 , I1 ⊆ I, both disjoint, such that I0 ∪ I1 covers I. (Hint: induce on #(I).) Show that this remains true if any totally ordered set is put in place of R. (ii) Show that if I is any family of non-trivial intervals in R such that none contains the centre of any other, then I is expressible as I0 ∪ I1 where both I0 and I1 are disjoint. (b) Let m = m(r) be the largest number such that there are u1 , . . . , um ∈ Rr such that kui k = 1 for every i, kui − uj k ≥ 1 for all i 6= j. Let A ⊆ Rr be a bounded set and x 7→ δx : A → ]0, ∞[¤ a bounded ¤ 1 such function; set Bx = B(x, δx ) for x ∈ A. (i) Show that m < 3r . (ii) Show that there is an ² ∈ 0, 10 that whenever ku0 k = . . . = kum k = 1 there are distinct i, j ≤ m such that ui . uj > 12 (1 + ²). (iii) Suppose b has that u, v ∈ Rr are such that 31 ≤ kuk ≤ 1, kvk ≤ 1 + ² and ku − vk > 1. Show that the angle u0v cosine at most 21 (1 + ²). (Hint: maximise

a2 +b2 −c2 2ab

subject to

1 3

≤ a ≤ 1, b ≤ 1 + ² and c ≥ 1.) (iv) Suppose S that hxn in∈N is a sequence in A / Bxi for i < n and (1 + ²)δxn ≥ sup{δx : x ∈ A \ i
473A

Poincar´ e’s inequality

633

(c) Use 472Yb to prove an alternative version of 472B, but with the constant 9r in place of 5r . (d) Let λ be a Radon measure on Rr , and f a λ-integrable real-valued function. Show that f ∗ (x) = R 1 supδ>0 |f |dλ is defined and finite for λ-almost every x ∈ Rr . B(x,δ) λB(x,δ)

(e) Let λ, λ0 be Radon measures on Rr . (i) Show that g(x) = limδ↓0 0

λ0 Br (x) λBr (x)

is defined in R for λ-almost

every x. (ii) Setting λ0 = supn∈N λ ∧ nλ in the cone of Radon measures on Rr (437Yj), show that g is a Radon-Nikod´ ym derivative of λ0 with respect to λ. (Hint: show that if λ and λ0 are mutually singular then g = 0 λ-a.e.) 472 Notes and comments I gave primacy to the ‘weak’ Vitali’s theorem in 261B because I think it is easier than the ‘strong’ form in 472C, it uses the same ideas as the original one-dimensional theorem in 221A, and it is adequate for the needs of Volume 2. Any proper study of general measures on Rr , however, will depend on the ideas in 472A-472C. You will see that in 472B, as in other forms of Vitali’s theorem, there is a key step in which a sequence is chosen greedily. This time we must look much more carefully at the geometry of Rr because we can no longer rely on a measure to tell us what is happening. (Though you will observe that I still use the elementary properties of Euclidean volume in the argument of 472A.) Once we have reached 472C, however, we are in a position to repeat all the arguments of 261C-261E in much greater generality (472D), and, as a bonus, can refine 261F (472Xb). For more in this direction see Mattila 95 and Federer 69, §2.8. It is natural to ask whether the constant ‘5r ’ in 472A is best possible. The argument of 472A is derived from Sullivan 94, where a more thorough analysis is given. It seems that even for r = 2 the best constant is unknown. (For r = 1, the best constant is 2; see 472Ya.) Note that even for finite families I we should have to find the colouring number of a graph (counting two balls as linked if they intersect), so it may well be a truly difficult problem. The method in 472B amounts to using the greedy colouring algorithm after ordering the balls by size, and one does not expect such approaches to give exact colouring numbers. Of course the questions addressed here depend only on the existence of some function of r to do the job. An alternative argument runs through a kind of pointwise version of 472A (472Yb-472Yc). It gives a worse constant but is attractive in other ways. For many of the applications of 472C, the result of 472Yb is already sufficient. ¡ 5r p ¢1/p The constant 2 in 472F makes no pretence to be ‘best’, or even ‘good’. The only reason for p−1

giving a formula at all is to emphasize the remarkable fact that it does not depend on the measure λ. The theorems of this section are based on the metric geometry of Euclidean space, not on any special properties of Lebesgue measure. The constants do depend on the dimension, so that even in Hilbert space (for instance) we cannot expect any corresponding results.

473 Poincar´ e’s inequality In this section I embark on the main work of the chapter, leading up to the Divergence Theorem in §475. I follow the method in Evans & Gariepy 92. The first step is to add some minor results on differentiable and Lipschitz functions to those already set out in §262 (473B-473C). Then we need to know something about convolution products (473D), extending ideas in §§256 and 444; in particular, it will be convenient to ˜ n in∈N of smoothing functions with some useful special properties (473E). have a fixed sequence hh The new ideas of the section begin with the Gagliardo-Nirenberg-Sobolev inequality, relating kf kr/(r−1) to k grad f k1 . In its simplest form (473H) it applies only to Rfunctions with compactRsupport; we need R to work much harder to get results which we can use to estimate B |f |r/(r−1) in terms of B k grad f k and B f for balls B (473I, 473K). 473A Notation For the next three sections, r ≥ 2 will be a fixed integer. For x ∈ Rr and δ ≥ 0, B(x, δ) = {y : ky − xk ≤ δ} will be the closed ball with centre x and radius δ. I will write ∂B(x, δ) for the

634


473A

boundary of B(x, δ), the sphere {y : ky − xk = δ}. Sr−1 = ∂B(0, 1) will be the unit sphere. As in Chapter 26, I will use Greek letters to represent coordinates of vectors, so that x = (ξ1 , . . . , ξr ), etc. µ will always be Lebesgue measure on Rr . βr = µB(0, 1) will be the r-dimensional volume of the unit ball, that is, βr = =

πk k!

if r = 2k is even,

22k+1 k!π k (2k+1)!

if r = 2k + 1 is odd

(252Q). ν will be normalized Hausdorff r − 1-dimensional measure on Rr , that is, ν = 2−r+1 βr−1 µH,r−1 , where µH,r−1 is r − 1-dimensional Hausdorff measure on Rr as described in §264, and βr−1 = =

22k k!π k−1 (2k)! πk k!

if r = 2k is even,


is the Lebesgue measure of a ball of diameter 1 in Rr−1 (264I). Recall from 265F and 265H that νSr−1 = 2πβr−2 = rβr (counting β0 as 1). 473B Differentiable functions (a) Recall from §262 that a function φ from a subset of Rr to Rs (where s ≥ 1) is differentiable at x ∈ Rr , with derivative an s × r matrix T , if for every ² > 0 there is a δ > 0 such that kφ(y) − φ(x) − T (y − x)k ≤ ²ky − xk whenever ky − xk ≤ δ; this includes the assertion that B(x, δ) ⊆ dom φ. In this case, the coefficients of T are the partial derivatives are the coordinate functions of φ, and

∂ ∂ξi

∂φj (x) ∂ξi

at x, where φ1 , . . . , φs

represents partial differentiation with respect to the ith coordinate

(262Ic). (b) When s = 1, so that we have a real-valued function f defined on a subset of Rr , I will write (grad f )(x) for the derivative of f at x, ´the gradient of f . If we strictly adhere to the language of (a), grad f is a ³ 1 × r matrix r

∂f ∂ξ1

...

∂f ∂ξr

; but it is convenient to treat it as a vector, so that grad f (x) (when defined)

belongs to R , and we can speak of y . grad f (x) rather than (grad f (x))(y), etc. (c) Chain rule for functions of many variables I find that I have not written out the following basic fact. Let φ : A → Rs and ψ : B → Rp be functions, where A ⊆ Rr and B ⊆ Rs . If x ∈ A is such that φ is differentiable at x, with derivative S, and ψ is differentiable at φ(x), with derivative T , then the composition ψφ is differentiable at x, with derivative T S. P P Recall that if we regard S and T as linear operators, they have finite norms (262H). Given ² > 0, let η > 0 be such that ηkT k+η(kSk+η) ≤ ². Let δ1 , δ2 > 0 be such that φ(y) is defined and kφ(y)−φ(x)−S(y − x)k ≤ ηky − xk whenever ky − xk ≤ δ1 , and ψ(z) is defined and kψ(z) − ψφ(x) − T (z − φ(x))k ≤ ηkz − φ(x)k whenever kz − φ(x)k ≤ δ2 . Set δ = min(δ1 ,

δ2 ) η+kSk

> 0. If ky − xk ≤ δ, then φ(y) is defined and

kφ(y) − φ(x)k ≤ kS(y − x)k + kφ(y) − φ(x) − S(y − x)k ≤ (kSk + η)ky − xk ≤ δ2 , so ψφ(y) is defined and kψφ(y) − ψφ(x) − T S(y − x)k ≤ kψφ(y) − ψφ(x) − T (φ(y) − φ(x))k + kT kkφ(y) − φ(x) − S(y − x)k ≤ ηkφ(y) − φ(x) − S(y − x)k + kT kηky − xk ≤ η(kSk + η)ky − xk + kT kηky − xk ≤ ²ky − xk; as ² is arbitrary, ψφ is differentiable at x with derivative T S. Q Q

473Cc


635

(d) It follows that if f and g are real-valued functions defined on a subset of Rr , and x ∈ dom f ∩ dom g is such that (grad f )(x) and (grad g)(x) are both then grad(f × g)(x) is defined and equal to µ defined, ¶ f (y) f (x) grad g(x) + g(x) grad f (x). P P Set φ(y) = for y ∈ dom f ∩ dom g; then φ is differentiable at µ ¶ g(y) grad f (x) x with derivative the 2 × r matrix (262Ib). Set ψ(z) = ζ1 ζ2 for z = (ζ1 , ζ2 ) ∈ R2 ; then ψ is grad g(x) differentiable everywhere, with derivative the 1 × 2 matrix ( ζ2 ζ1 ). So f × g = ψφ is differentiable at x with derivative ¶ µ grad f (x) ( g(x) f (x) ) = g(x) grad f (x) + f (x) grad g(x). Q Q grad g(x) (e) Let D be a subset of Rr and φ : D → Rs any function. Set D0 = {x : x ∈ D, φ is differentiable at x}. Then the derivative of φ, regarded as a function from D0 to Rrs , is (Lebesgue) measurable. P P Use 262P; the point is that, writing T (x) for the derivative of φ at x, T (x) is surely a derivative of φ¹D0 , relative to D0 , at every point of D0 . Q Q (See also 473Ya.) (f ) A function φ : Rr → Rs is smooth if it is differentiable arbitrarily often; that is, if all its repeated partial derivatives ∂ m φj ∂ξi1 ...∂ξim

are defined and continuous everywhere on Rr . I will write D for the family of real-valued functions on Rr which are smooth and have compact support. 473C Lipschitz functions (a) If f and g are bounded real-valued Lipschitz functions, defined on any subsets of Rr , then f × g, defined on dom f ∩ dom g, is Lipschitz. P P Let γf , Mf , γg and Mg be such that |f (x)| ≤ Mf and |f (x)−f (y)| ≤ γf kx−yk for all x, y ∈ dom f , while |g(x)| ≤ Mg and |g(x)−g(y)| ≤ γg kx−yk for all x, y ∈ dom g. Then for any x, y ∈ dom f ∩ dom g, |f (x)g(x) − f (y)g(y)| ≤ |f (x)||g(x) − g(y)| + |g(y)||f (x) − f (y)| ≤ (Mf γg + Mg γf )kx − yk. So Mf γg + Mg γf is a Lipschitz constant for f × g. Q Q (b) Suppose that F1 , F2 ⊆ Rr are closed sets with convex union C. Let f : C → R be a function such that f ¹F1 and f ¹F2 are both Lipschitz. Then f is Lipschitz. P P For each j, let γj be a Lipschitz constant for f ¹Fj , and set γ = max(γ1 , γ2 ), so that γ is a Lipschitz constant for both f ¹F1 and f ¹F2 . Take any x, y ∈ C. If both belong to the same Fj , then |f (x) − f (y)| ≤ γkx − yk. If x ∈ Fj and y ∈ / Fj , then y must belong to F3−j , and (1 − t)x + ty ∈ F1 ∪ F2 for every t ∈ [0, 1], because C is convex. Set t0 = sup{t : t ∈ [0, 1], (1 − t)x + ty ∈ Fj }, z = (1 − t0 )x + t0 y; then z ∈ F1 ∩ F2 , because both are closed, so |f (x) − f (y)| ≤ |f (x) − f (z)| + |f (z) − f (y)| ≤ γkx − zk + γkz − yk = γkx − yk. As x and y are arbitrary, γ is a Lipschitz constant for f . Q Q (c) Suppose that f : Rr → R is Lipschitz. Recall that by Rademacher’s theorem (262Q), grad f is defined almost everywhere. All the partial derivatives of f are (Lebesgue) measurable, by 473Be, so grad f is (Lebesgue) measurable on its domain. If γ is a Lipschitz constant for f , k grad f (x)k ≤ γ whenever grad f (x) is defined. P P If z ∈ Rr , then 1 t

limt↓0 |f (x + tz) − f (x) − tz . grad f (x)| = 0, so 1 t

|z . grad f (x)| = limt↓0 |f (x + tz) − f (x)| ≤ γkzk; as z is arbitrary, k grad f (x)k ≤ γ. Q Q

636


473Cd

(d) Conversely, if f : Rr → R is differentiable and k grad f (x)k ≤ γ for every x, then γ is a Lipschitz constant for f . P PTake x, y ∈ Rr . Set g(t) = f ((1−t)x+ty) for t ∈ R. The function t 7→ (1−t)x+ty : R → Rr is everywhere differentiable, with constant derivative y − x, so by 473Bc g is differentiable, with derivative g 0 (t) = y − x . gradf (g(t)) for every t; in particular, |g 0 (t)| ≤ γky − xk for every t. Now, by the Mean Value Theorem, there is a t ∈ [0, 1] such that g(1) − g(0) = g 0 (t), so that |f (y) − f (x)| = |g 0 (t)| ≤ γky − xk. As x and y are arbitrary, f is γ-Lipschitz. Q Q (e) Note that if f ∈ D then all its partial derivatives are continuous functions with compact support, so are bounded (436Ia), and f is Lipschitz as well as bounded, by (d) here. (f )(i) If D ⊆ Rr is bounded and f : D → R is Lipschitz, then there is a Lipschitz function g : Rr → R, with compact support, extending f . P P By 262Bb there is a Lipschitz function f1 : Rr → R which extends f . Let γ > 0 be such that D ⊆ B(0, γ) and γ is a Lipschitz constant for f1 ; set M = |f1 (0)| + γ 2 ; then |f1 (x)| ≤ M for every x ∈ D, so if we set f2 (x) = max(−M, min(f (x), M )) for x ∈ Rr , f2 is a bounded Lipschitz function extending f . Let f3 (x) = max(0, 1 −

kxk ) γ

for x ∈ Rr ; then f3 is a bounded Lipschitz

function with compact support. By (a), g = f3 × f2 is Lipschitz, and g : Rr → R is a function with compact support extending f . Q Q (ii) It follows that if D ⊆ Rr is bounded and f : D → Rs is Lipschitz, then there is a Lipschitz function g : Rr → Rs , with compact support, extending f . P P By 262Ba, we need only apply (i) to each coordinate of f . Q Q 473D Smoothing by convolution We shall need a miscellany of facts, many of them R special cases of results in §§255 and 444, concerning convolutions on Rr . Recall that I write (f ∗ g)(x) = f (y)g(x − y)µ(dy) whenever f and g are real-valued functions defined almost everywhere in Rr and the integral is defined. Recall that f ∗ g = g ∗ f (255Fb, 444Og). Now we have the following. Lemma Suppose that f and g are Lebesgue measurable real-valued functions defined µ-almost everywhere in Rr . (a) If f is integrable and g is essentially bounded, then their convolution f ∗ g is defined everywhere in Rr and uniformly continuous, and supx∈R r |(f ∗ g)(x)| ≤ kf k1 kgk∞ . (b) If f is locally integrable and g is bounded and has compact support, then f ∗ g is defined everywhere in Rr and is continuous. (c) If f and g are defined everywhere in Rr and x ∈ Rr \ ({y : f (y) 6= 0} + {z : g(z) 6= 0}), then (f ∗ g)(x) is defined and equal to 0. (d) If f is integrable and g is bounded, Lipschitz and defined everywhere, then f ∗ grad g and grad(f ∗ g) are defined everywhere and equal, where f ∗ grad g = (f ∗

∂g ,... ∂ξ1

,f∗

∂g ). ∂ξr

Moreover, f ∗ grad g = grad(f ∗ g)

is Lipschitz. (e) If f is locally integrable, and g ∈ D, then f ∗ g is defined everywhere and is smooth. (f) If f is essentially bounded and g ∈ D, then f ∗ g is Lipschitz as well as smooth. (g) If f is integrable and φ : Rr → Rr is a bounded measurable function with components φ1 , . . . , φr , and we write (f ∗ φ)(x) = ((f ∗ φ1 )(x), . . . , (f ∗ φr )(x)), then k(f ∗ φ)(x)k ≤ kf k1 supy∈R r kφ(y)k for every x ∈ Rr . proof (a) See 255K. (b) Suppose that g(y) = 0 when kyk ≥ n. Given x ∈ Rr , set f˜ = f × χB(x, n + 1). Then f˜ ∗ g is defined everywhere and continuous, by (a), while (f ∗ g)(z) = (f˜ ∗ g)(z) whenever z ∈ B(x, 1); so f ∗ g is defined everywhere in B(x, 1) and is continuous at x. (c) We have only to note that f (y)g(x − y) = 0 for every y. (d) Let γ be a Lipschitz constant for g. We know that grad g is defined almost everywhere, is measurable, and that k grad g(x)k ≤ γ whenever it is defined (473Cc); so (f ∗ grad g)(x) is defined for every x, by (a) here. Fix x ∈ Rr . If y, z ∈ Rr set

473E


θ(y, z) =

1 ¡ g(x − y kzk

637

¢ + z) − g(x − y) − z . grad g(x − y)

whenever this is defined. Then |θ(y, z)| ≤ 2γ whenever it is defined. Now suppose that hzn in∈N is a sequence in Rr \ {0} converging to R 0. Then limn→∞ θ(y, zn ) = 0 whenever grad g(x − y) is defined, which almost everywhere. So limn→∞ f (y)θ(y, zn )µ(dy) = 0, by Lebesgue’s Dominated Convergence Theorem. But this means that ¢ 1 ¡ (f ∗ g)(x + zn ) − (f ∗ g)(x) − ((f ∗ grad g)(x)). zn → 0 kzn k

as n → ∞. As hzn in∈N is arbitrary, grad(f ∗ g)(x) is defined and is equal to (f ∗ grad g)(x). Now grad g is bounded, because g is Lipschitz, so grad(f ∗ g) = f ∗ grad g is also bounded, by (a), and f ∗ g must be Lipschitz (473Cd). (e) By (b), f ∗ g is defined everywhere and is continuous. Now, for any i ≤ r,

∂ (f ∂ξi

∗ g) = f ∗

∂g ∂ξi

everywhere. P P Let n ∈ N be such that g(y) = 0 if kyk ≥ n. Given x ∈ Rr , set f˜ = f × χB(x, n + 1). Then (f ∗ g)(z) = (f˜ ∗ g)(z) for every z ∈ B(x, 1), so that ∂f ∗g (x) ∂ξi

=

∂ f˜∗g (x) ∂ξi

∂g = (f˜ ∗ )(x) ∂ξi

(by (d)) = (f ∗ (because of course

∂g ∂ξi

∂g )(x) ∂ξi

is also zero outside B(0, n)). Q Q Inducing on k, ∂k ∂kg (f ∗ g)(x) = (f ∗ )(x) ∂ξi1 . . . ∂ξik ∂ξi1 . . . ∂ξik

for every x ∈ Rr and every i1 , . . . , ik ; so we have the result. (f ) The point is just that all the partial derivatives of g, being smooth functions with compact support, are integrable, and that |

∂ (f ∂ξi

∗ g)(x)| = |(f ∗

∂g )(x)| ∂ξ

≤ kf k∞ k

∂g k1 ∂ξi

for every x and every i ≤ r. This means that grad(f ∗ g) is bounded, so that f ∗ g is Lipschitz, by 473Cd. (g) If x, z ∈ Rr , then z .(f ∗ φ)(x) =

r X

Z ζi (f ∗ φi )(x) =

i=1

Z ≤

|f (y)|| Z

≤

f (y)

r X

ζi φi (x − y)µ(dy)

i=1 r X

ζi φi (x − y)|µ(dy)

i=1

|f (y)|kzkkφ(x − y)kµ(dy) ≤ kzkkf k1 sup kφ(y)k. y∈R r

As z is arbitrary, k(f ∗ φ)(x)k ≤ kf k1 supy∈R r kφ(y)k. 473E Lemma (a) Define h : R → [0, 1] by setting h(t) = exp( 0

1 ) t2 −1

for |t| < 1, 0 for |t| ≥ 1. Then h is

smooth, and h (t) ≤ 0 for t ≥ 0. ˜ n : Rr → R by setting (b) For n ≥ 1, define h αn =

R

h((n + 1)2 kxk2 )µ(dx),

˜ n (x) = 1 h((n + 1)2 kxk2 ) h αn

638


473E

R ˜ n ∈ D, h ˜ n (x) ≥ 0 for every x, h ˜ n (x) = 0 if kxk ≥ 1 , and h ˜ n dµ = 1. for every x ∈ Rr . Then h n+1

˜ n )(x) = f (x) for every x ∈ dom f at which f is continuous. (c) If f ∈ L0 (µ), then f (x) = limn→∞ (f ∗ h (d) If f : Rr → R is uniformly continuous (in particular, if it is either Lipschitz or a continuous function ˜ n k∞ = 0. with compact support), then limn→∞ kf − f ∗ h 0 ˜ n )(x) for µ-almost every x ∈ Rr . (e) If f ∈ L (µ) is locally integrable, then f (x) = limn→∞ (f ∗ h p ˜ (f) If f ∈ L (µ), where 1 ≤ p < ∞, then limn→∞ kf − f ∗ hn kp = 0. proof (a) Set h0 (t) = exp(− 1t ) for t > 0, 0 for t ≤ 0. A simple induction on n shows that the nth derivative (n) h0 of h0 is of the form 1 t

1 t

(n)

h0 (t) = qn ( ) exp(− ) for t > 0, = 0 for t ≤ 0, where each qn is a polynomial of degree 2n; the inductive hypothesis depends on the fact that lims→∞ q(s)e−s = 0 for every polynomial q. So h0 is smooth. Now h(t) = h0 (1 − t2 ) so h also is smooth. If 0 ≤ t < 1 then h0 (t) = − exp(

1 2t )· 2 2 t2 −1 (t −1)

< 0;

if t > 1 then h0 (t) = 0; since h0 is continuous, h0 (t) ≥ 0 for every t ≥ 0. (b) We need only observe that x 7→ (n + 1)2 kxk2 = (n + 1)2

Pr

2 i=1 ξi

is smooth and that the composition of smooth functions is smooth (by 473Bc). (c) If f is continuous at x and ² > 0, let n0 ∈ N be such that |f (y) − f (x)| ≤ ² whenever y ∈ dom f and ky − xk ≤

1 . n0 +1

Then for any n ≥ n0 ,

Z Z ˜ ˜ ˜ n (y)µ(dy)| |(f ∗ hn )(x) − f (x)| = | f (x − y)hn (y)µ(dy) − f (x)h Z Z ˜ n (y)µ(dy) ≤ ²h ˜ n (y)µ(dy) = ². ≤ |f (x − y) − f (x)|h As ² is arbitrary, we have the result. (d) Repeat the argument of (c), but ‘uniformly in x’; that is, given ² > 0, take n0 such that |f (y)−f (x)| ≤ 1 ˜ n )(x) − f (x)| ≤ ² for every n ≥ n0 and every ² whenever x, y ∈ Rr and ky − xk ≤ , and see that |(f ∗ h n0 +1 x. (e) We know from 472Db or 261E that, for almost every x ∈ Rr , 1 µB(x,δ)

limδ↓0

R B(x,δ)

|f (y) − f (x)|µ(dy) = 0.

Take any such x. Set γ = f (x), Set g(y) = |f (y) − γ| for every y ∈ dom f . Let ² > 0. Then there is some δ > 0 such that

q(t) βr tr

≤ ² whenever 0 < t ≤ δ, where q(t) =

by 265G, so q 0 (t) =

R ∂B(y,t)

R B(x,t)

g dµ =

RtR 0

∂B(x,s)

1 δ

g dν for almost every t ∈ [0, δ], by 222E. If n + 1 ≥ , then Z

˜ n )(x) = (g ∗ h = (265G again)

g(y)ν(dy)dt

Z ˜ n (x − y)µ(dy) = g(y)h

1 αn

Z 0

δ

Z

˜ n (x − y)µ(dy) g(y)h B(x,δ)

g(y)h((n + 1)2 t2 )ν(dy)dt ∂B(x,t)

473F


Z

=

1 αn

=

1 − αn

δ

639

h((n + 1)2 t2 )q 0 (t)dt

0

Z

δ

2(n + 1)2 h0 ((n + 1)2 t2 )q(t)dt

0

(integrating by parts (225F), because q(0) = h((n + 1)2 δ 2 ) = 0 and both q and h are absolutely continuous) Z δ ² ≤− 2(n + 1)2 h0 ((n + 1)2 t2 )βr tr dt αn

r

0

0

(because 0 ≤ q(t) ≤ ²βr t and h ((n + 1)2 t2 ) ≤ 0 for 0 ≤ t ≤ δ) =² ˜ n )(x) = γ for every n, (applying the same calculations with χRr in place of g). But now, since (γχRr ∗ h

R

˜ n )(x) − γ| = | (f (y) − γ)h ˜ n (x − y)µ(dy)| ≤ |(f ∗ h

R

˜ n (x − y)µ(dy) ≤ ² |f (y) − γ|h

1 ˜ n )(x); and this is true for µ-almost every x. whenever n + 1 ≥ . As ² is arbitrary, f (x) = γ = limn→∞ (f ∗ h δ

˜ n ; use 444P for the identification (f ) Apply 444T with ν the indefinite-integral measure over µ defined by h ñ ∗ f = f ∗ h ˜ n. of ν ∗ f with h 473F Lemma For any measure space (X, Σ, λ) and any non-negative f1 , . . . , fk ∈ L0 (λ),

R Qk

1/k

i=1

fi

dλ ≤

Qk ¡R i=1

¢1/k

fi dλ

.

R proof Induce on k. Note that we can suppose R that every fi is integrable; for if any fi is zero, then fi = 0 a.e. and the result is trivial; and if all the fi are greater than zero and any of them is infinite, the result is again trivial. The induction starts with the trivial case k = 1. For the inductive step to k ≥ 2, we have Z Y k

1/k

fi

dλ ≤ k

i=1

k−1 Y

1/k

fi

1/k

kk/(k−1) kfk kk

i=1

(by Hölder’s inequality, 244E) =

¡

Z k−1 Y

¡

k−1 Y

1/(k−1)

fi

dλ

Z ¢(k−1)/k ¡

i=1

≤

Z

(

fi dλ)1/(k−1)

¢1/k fk dλ

Z ¢(k−1)/k ¡

¢1/k fk dλ

i=1

(by the inductive hypothesis) =

k Z Y ( fi dλ)1/k , i=1

as required. 473F Lemma Let (X, Σ, λ) be a σ-finite measure space and k ≥ 2 an integer. Write λk for the product measure on X k . For x = (ξ1 , . . . , ξk ) ∈ X k , t ∈ X, 1 ≤ i ≤ k set Si (x, t) =R (ξ10 , . . . , ξk0 ) where ξi0 = t and ξj0 = ξj for j 6= i. Then if h ∈ L1 (λk ) is non-negative, and we set hi (x) = Si (x, t)λ(dt) whenever this is defined in R, we have

R Qk (

proof Induce on k.

R

h )1/(k−1) dλk ≤ ( h dλk )k/(k−1) . i=1 i

640


473F

(a) If k = 2, we have Z ZZ Z Z ¡ ¢¡ ¢ h1 × h2 dλk = h(τ1 , ξ2 )dτ1 h(ξ1 , τ2 )dτ2 dξ1 dξ2 ZZZZ = h(τ1 , ξ2 )h(ξ1 , τ2 )dτ1 dτ2 dξ1 dξ2 ZZ ZZ Z ¡ ¢2 = h(τ1 , ξ2 )dτ1 dξ2 · h(ξ1 , τ2 )dτ2 dξ1 = h dλ2 by Fubini’s theorem (252B) used repeatedly, because (by 253D) (τ1 , τ2 , ξ1 , ξ2 ) 7→ h(ξ1 , τ2 )h(τ1 , ξ2 ) is λ4 integrable. (See 251W for a sketch of the manipulations needed to apply 252B, as stated, to the integrals above.) R (b) For the inductive step to k ≥ 3, argue as follows. For y ∈ X k−1 , set g(y) = h(y, t)dt whenever this is defined in R, identifying X k with X k−1 × X, so that g(y) = hk (y, t) whenever either is defined. If 1 ≤ i < k, we can consider Si (y, t) for y ∈ X k−1 and t ∈ X, and we have

R

g(Si (y, t))dt =

RR

h(Si (y, t), u)dudt =

R

hi (y, t)dt

for almost every y ∈ X k−1 . So Z Y Z Z k−1 k Y ( hi )1/(k−1) dλk = ( hi (y, t))1/(k−1) g(y)1/(k−1) dt λk−1 (dy) i=1

Z

i=1 1/(k−1)

=

g(y)

i=1

Z 1/(k−1)

≤

Z k−1 Y ( hi (y, t))1/(k−1) dt λk−1 (dy)

g(y)

k−1 Y

¡

Z

¢1/(k−1) hi (y, t)dt λk−1 (dy)

i=1

(473F) Z g(y)1/(k−1)

=

k−1 Y

gi (y)1/(k−1) λk−1 (dy)

i=1

(where gi is defined from g in the same way as hi is defined from h) Z Z Y ¡ ¢1/(k−1) ¡ k−1 ¢(k−2)/(k−1) ≤ g(y)λk−1 (dy) gi (y)1/(k−2) λk−1 (dy) i=1

1 k−1

k−2 k−1

(by Hölder’s inequality again, this time with + = 1) Z Z ¡ ¢1/(k−1) ≤ g(y)λk−1 (dy) · g(y)λk−1 (dy) (by the inductive hypothesis) =

Z ¡

¢k/(k−1) ¡ g(y)λk−1 (dy) =

Z h(x)λk (dx)

¢k/(k−1)

,

and the induction proceeds. 473H Gagliardo-Nirenberg-SobolevR inequality Suppose that f : Rr → R is a Lipschitz function with compact support. Then kf kr/(r−1) ≤ k grad f kdµ. proof By 473Cc, grad f is measurable and bounded, so k grad f k also is; since it must have compact support, it is integrable. For 1 ≤ i ≤ r, x = (ξ1 , . . . , ξr ) ∈ Rr and t ∈ R write Si (x, t) = (ξ10 , . . . , ξr0 ) where ξi0 = t and ξj0 = ξj for R∞ j 6= i. Set hi (x) = −∞ k grad f (Si (x, t))kdt when this is defined, which will be the case for almost every x. Now, whenever hi (x) is defined,

473I


|f (x)| = |f (Si (x, ξi ))| = |

641

R ξi

∂ f (Si (x, t))dt| −∞ ∂t

≤ hi (x).

(Use 225E and the fact that a Lipschitz function on any bounded interval in R is absolutely continuous.) So |f | ≤a.e. hi for every i ≤ r. Accordingly

R

|f (x)|r/(r−1) µ(dx) ≤

R Qr

i=1

hi (x)1/(r−1) µ(dx) ≤

¡R

k grad f (x)kµ(dx)

¢r/(r−1)

by 473F. Raising both sides to the power (r − 1)/r we have the result. 473I Lemma For any Lipschitz function f : B(0, 1) → R, ¡ r+2 ¢r/(r−1) R √ R r/(r−1) |f | dµ ≤ 2 (1 + . r) k grad f k + |f |dµ B(0,1) B(0,1) proof (a) Set g(x) √ = max(0, 2kxk2 − 1) for x ∈ B(0, 1). Then grad g is defined at every point x such that ∂g kxk < 1 and kxk 6= 2, and at all such points ∂ξ is either 0 or 4ξi for each i, so that k grad g(x)k ≤ 4kxk ≤ 4. i Hence (or otherwise) g is Lipschitz. So f1 = f × g is Lipschitz (473Ca). By Rademacher’s theorem again, grad f1 is defined almost everywhere in B(0, 1). Now Z

Z k grad f1 k =

kf (x) grad g(x) + g(x) grad f (x)kµ(dx)

B(0,1)

B(0,1)

(473Bd)

Z ≤

4|f | + k grad f kdµ. B(0,1)

(b) It will be convenient to have an elementary fact out in the open. Set φ(x) =

x kxk2

for x ∈ Rr \ {0};

note that φ2 (x) = x. Then φ¹ Rr \ B(0, δ) is Lipschitz, for any δ > 0. P P If kxk = α ≥ δ, kyk = β ≥ δ, then we have kφ(x) − φ(y)k2 = = so

1 δ2

1 kxk2 α4

−

1 ¡ kyk2 α2 β 2

2 x.y α2 β 2

+

1 kyk2 β4

¢ 1 − 2x. y + kxk2 ≤ 4 kx − yk2 , δ

is a Lipschitz constant for φ¹ Rr \ B(0, δ). Q Q

(c) Set f2 (x) = f (x) if kxk ≤ 1, f1 φ(x) if kxk ≥ 1. Then f2 is well-defined (because f1 (x) = f (x) if √ 1 kxk = 1), is zero outside B(0, 2) (because g(x) = 0 if kxk ≤ √ ), and is Lipschitz. P P By 473Cb, it will 2

be enough to show that f2 ¹F is Lipschitz, where F = {x : kxk ≥ 1}. But (b) shows that φ¹F is 1-Lipschitz on F , so any Lipschitz constant for f1 is also a Lipschitz constant for f2 ¹F . Q Q If kxk > 1, then, for any i ≤ r, ∂f2 (x) ∂ξi

=

r X ∂f1 j=1

=

∂ξj

(φ(x)) ·

∂ ξ ( j ) ∂ξi kxk2

1 ∂f1 (φ(x)) · ∂ξi kxk2

−

r X ∂f j=1

∂ξj

(φ(x)) ·

ξi ξj kxk4 ∂f

wherever the right-hand side is defined, that is, wherever all the partial derivatives 1 (φ(x)) are defined. ∂ξi √ But H = B(0, 1) \ dom(grad f1 ) is negligible, and does not meet {x : kxk < 2}, so φ¹H is Lipschitz and φ[H] = φ−1 [H] is negligible (262D); while grad f1 (φ(x)) is defined whenever kxk > 1 and x ∈ / φ−1 [H]. So the formula here is valid for almost every x ∈ F , and

642


∂f | 2 (x)| ∂ξi

≤ k grad f1 (φ(x))k · =

473I

v u

1 kxk2

+ k grad f1 (φ(x))k ·

r

X |ξi | u t ξj2 4 kxk j=1

kxk+|ξi | k grad f1 (φ(x))k kxk3

for almost every x ∈ F . But (since we know that grad f2 is defined almost everywhere, by Rademacher’s theorem, as usual) we have √ √ 1+ r k grad f1 (φ(x))k ≤ (1 + r)k grad f1 (φ(x))k k grad f2 (x)k ≤ 2 kxk

for almost every x ∈ F . (c) We are now in a position to estimate Z

Z k grad f2 kdµ =

F

(because grad f1 (x) = 0 if kxk <

√

2)

Z

√ B(0, 2)

Z

√ 2

k grad f2 kdµ −

k grad f2 kdµ B(0,1)

Z

=

k grad f2 (x)kν(dx)dt 1

∂B(0,t)

(265G, as usual) ≤ (1 +

√

√

Z

2

1

(by (b) above) ≤ (1 +

√

Z

r) ∂B(0,t) √

Z

2

Z t2r−2 k grad f1 (y)kν(dy)dt

r) 1

1 t

k grad f1 ( 2 x)kν(dx)dt

∂B(0,1/t)

substituting x = t2 y in the inner integral; the point being that as the function y 7→ t2 y changes all distances by a scalar multiple t2 , it must transform Hausdorff r − 1-dimensional measure by a multiple t2r−2 . But now, substituting s = 1t in the outer integral, we have Z k grad f2 kdµ ≤ (1 +

√

Z r)

F

≤ 2r (1 + = 2r (1 +

√ √

1

1 √ s2r 1/ 2

Z

r)

1

Z k grad f1 (y)kν(dy)ds

Z

√ 1/ 2

Z r)

∂B(0,s)

k grad f1 (y)kν(dy)ds ∂B(0,s)

k grad f1 kdµ B(0,1)

≤ 2r (1 +

√

Z r)

4|f | + k grad f kdµ. B(0,1)

by (a) above. (d) Accordingly Z Rr

Z

Z

k grad f2 kdµ ≤

k grad f kdµ + k grad f2 kdµ F Z √ ≤ 2r+2 (1 + r) |f | + k grad f kdµ. B(0,1)

B(0,1)

But now we can apply 473H to see that

473J


Z

643

Z

Z |f2 |r/r−1 dµ ≤ ( k grad f2 kdµ)r/(r−1) Z ¡ ¢r/(r−1) √ ≤ 2r+2 (1 + r) |f | + k grad f kdµ ,

|f |r/(r−1) dµ ≤ B(0,1)

B(0,1)

as claimed.

473J Lemma Let f : Rr → R be a Lipschitz function. Then

R

B(y,δ)

|f (x) − f (z)|µ(dx) ≤

2r r R δ B(y,δ) k grad f (x)kkx − zk1−r µ(dx) r

whenever y ∈ Rr , δ > 0 and z ∈ B(y, δ). proof (a) To begin with, suppose that f is smooth. In this case, for any x, z ∈ B(y, δ), ¯ |f (x) − f (z)| = ¯ ¯ =¯

Z

1

Z

1

d f (z dt 0

¯ + t(x − z))dt¯

¯ (x − z) . grad f (z + t(x − z))dt¯

0

Z

1

≤ kx − zk

k grad f (z + t(x − z))kdt. 0

So, for η > 0, Z |f (x) − f (z)|ν(dx) B(y,δ)∩∂B(z,η)

Z

1

Z

≤η

k grad f (z + t(x − z))kν(dx)dt 0

B(y,δ)∩∂B(z,η)

(grad f is continuous and bounded, and the subspace measure induced by ν on ∂B(z, η) is a (quasi-)Radon measure (471E, 471Dh), so its product with Lebesgue measure also is (417T), and there is no difficulty with the change in order of integration) Z Z 1 1 k grad f (w)kν(dw)dt =η r−1 t

0

B(y,δ)∩∂B(z,tη)

1

(because if φ(x) = z + t(x − z), then νφ−1 [E] = r−1 νE whenever ν measures E and t > 0) t Z 1Z k grad f (w)kkw − zk1−r ν(dw)dt = ηr 0

Z

= η r−1 0

(substituting s = tη)

B(y,δ)∩∂B(z,tη) ηZ

Z k grad f (w)kkw − zk1−r µ(dw)

= η r−1 B(y,δ)∩B(z,η)

by 265G. By 265G again,

k grad f (w)kkw − zk1−r ν(dw)ds

B(y,δ)∩∂B(z,s)

644


Z

Z

Z

2δ

|f (x) − f (z)|µ(dx) = B(y,δ)

|f (x) − f (z)|ν(dx)dη 0

Z

B(y,δ)∩∂B(z,η)

Z

2δ

η r−1

≤

k grad f (w)kkw − zk1−r µ(dw)dη

0

Z

B(y,δ)∩B(z,η)

Z

2δ

r−1

k grad f (w)kkw − zk1−r µ(dw)dη

η Z

≤ 0 r

=

473J

2 r δ r

B(y,δ)

k grad f (w)kkw − zk1−r µ(dw). B(y,δ)

(b) Now turn to the general case in R which f is not necessarily differentiable everywhere, but is known to be Lipschitz. We need to know that B(y,δ) kx − zk1−r µ(dx) is finite; this is because Z

Z

Z

kx − zk1−r µ(dx) ≤ B(y,δ)

kx − zk1−r µ(dx) = B(z,2δ)

(265G once more)

Z

2δ

=

2δ

t1−r ν(∂B(z, t))dt

0

t1−r rβr tr−1 dt = 2δrβr .

0

˜ n in∈N from 473E. Then hf ∗ h ˜ n in∈N converges uniformly to f (473Ed), while Now take the sequence hh ˜ n )in∈N = hh ˜ n ∗ grad f in∈N (473Dd) is uniformly bounded (473Cc, 473Dg) and converges almost hgrad(f ∗ h ˜ n, everywhere to grad f (473Ee). But this means that, setting fn = f ∗ h Z

Z |f (x) − f (z)|µ(dx) = lim

|fn (x) − fn (z)|µ(dx)

n→∞

B(y,δ)

B(y,δ)

Z

2r r δ k grad fn (x)kkx − zk1−r µ(dx) n→∞ r B(y,δ)

≤ lim (because every fn is differentiable, by 473De) =

Z

2r r δ k grad f (x)kkx − zk1−r µ(dx) r B(y,δ)

by Lebesgue’s Dominated Convergence Theorem. 473K Poincar´ e’s inequality for balls Let B ⊆ Rr be a non-trivial closed ball, and f : B → R a Lipschitz function. Set γ =

1 µB

R

B

¡R where c = 2r+2 (1 +

√

B

f dµ. Then

¢(r−1)/r

|f − γ|r/(r−1) dµ

≤c

R B

k grad f kdµ,

r)(1 + 2r+1 ).

proof (a) To begin with (down to the end of (b)) suppose that B is the unit ball B(0, 1). Then, for any x ∈ B, Z ¯ 1 ¯¯ f (x) − f (z)µ(dz)¯ |f (x) − γ| = µB B Z 1 ≤ |f (x) − f (z)|µ(dz) µB B Z 2r 1 ≤ · k grad f (z)kkz − xk1−r µ(dz), r

µB

B

473K


645

by 473J. Also Z

Z

2

kx − zk1−r µ(dx) = B(z,2)

Z kx − zk1−r ν(dx)dt

0

(265G)

Z

∂B(z,t) 2

=

Z t1−r ν(∂B(z, t))dt =

0

1

t1−r tr−1 ν(∂B(0, 1))dt

0

= 2ν(∂B(0, 1)) = 2rβr by 265H. So Z

2r r

|f (x) − γ|µ(dx) ≤ B

·

1 µB

Z Z k grad f (z)kkz − xk1−r µ(dz)µ(dx)

Z Z

2r rβr

B

B

k grad f (z)kkz − xk1−r µ(dx)µ(dz) ZB B Z 2r kz − xk1−r µ(dx)µ(dz) k grad f (z)k ≤ rβr B B(z,2) Z ≤ 2r+1 k grad f (z)kµ(dz). =

B

(b) Now apply 473I to g = f (x) − γ. We have Z r/(r−1)

|f − γ|

¡

dµ ≤ 2

B

Z

¢r/(r−1) k grad f k + |g|dµ B Z ¢r/(r−1) ¡ r+2 √ k grad f kdµ ≤ 2 (1 + r)(1 + 2r+1 ) r+2

(1 +

√

r)

B

(by (a)) ¡ = c

Z

¢r/(r−1) k grad f kdµ .

B

(c) For the general case, express B as B(y, δ), and set h(x) = f (y +δx) for x ∈ B(0, 1). Then grad h(x) = δ grad f (y + δx) for almost every x ∈ B(0, 1). Now

R B(0,1)

so 1 µB(0,1)

h dµ =

R

hdµ = B(0,1)

1 δr

R B(y,δ)

1 µB(y,δ)

f dµ,

R B(y,δ)

f dµ = γ.

We therefore have Z

Z |f − γ|r/(r−1) dµ = δ r

|h − γ|r/(r−1) dµ

B(y,δ)

B(0,1)

¡ ≤ δr c

Z

¢r/(r−1) k grad hkdµ B(0,1)

(by (a)-(b) above) =δ ¡

¡ r δc δr

Z

Z k grad f kdµ

¢r/(r−1)

B(y,δ)

= c

k grad f kdµ

¢r/(r−1)

.

B(y,δ)

Raising both sides to the power (r − 1)/r we have the result as stated. Remark As will be plain from the way in which the proof here is constructed, there is no suggestion that the formula offered for c gives anything near the best possible value.

646


473L

473L Corollary Let B ⊆ Rr be a non-trivial closed ball, and f : B → [0, 1] a Lipschitz function. Set F0 = {x : x ∈ B, f (x) ≤ 14 }, Then √

F1 = {x : x ∈ B, f (x) ≥ 34 }.

R ¡ ¢(r−1)/r min(µF0 , µF1 ) ≤ 4c B k grad f kdµ,

r)(1 + 2r+1 ). 1 R proof Setting γ = f dµ, µB B Z where c = 2r+2 (1 +

|f − γ|r/(r−1) dµ ≥

1 µF0 4r/(r−1)

if γ ≥ ,

≥

1 µF1 4r/(r−1)

if γ ≤ .

B

So 473K tells us that ¢(r−1)/r 1¡ min(µF0 , µF1 ) 4

≤c

R B

1 2 1 2

k grad f kdµ,

as required. 473M The case r = 1 The general rubric for this section declares that r is taken to be at least 2, which is clearly necessary for the formula in 475J to be appropriate. For the sake of an application in the next section, however, I mention the elementary corresponding result whenR r = 1. In this case, B is just a closed interval, and grad f is the ordinary deriivative of f ; interpreting ( B |f − γ|r/(r−1) )(r−1)/r as kf × χB − γχBkr/(r−1) , it is natural to look at kf × χB − γχBk∞ = supx∈B |f (x) − γ| ≤ supx,y∈B |f (x) − f (y)| ≤

R

B

|f 0 |dµ,

giving a version of 473K for r = 1.R We see that the formula for c remains valid in the case r = 1, with a good deal to spare. As for 473L, if B |f 0 | < 12 then at least one of F0 , F1 must be empty. 473X Basic exercises (a) Set f (x) = max(0, − ln kxk), fk (x) = min(k, f (x)) for x ∈ R2 \ {0}, k ∈ N. Show that limk→∞ kf − fk k2 = limk→∞ k grad f − grad fk k1 = 0, so that all the inequalities 473H-473L are valid for f . (b) Let k ∈ [1, r] be an integer, and set m =

(r−1)! . (k−1)!(r−k)!

Let e1 , . . . , er be the standard orthonormal

basis of Rr and J the family of subsets of {1, . . . , r} with k members. For J ∈ J let VJ be the linear span of {ei : i ∈ J}, πJ : Rr → VJ the orthogonal projection Q and νJ the normalized k-dimensional Hausdorff measure on VJ . Show that if A ⊆ Rr then (µ∗ A)m ≤ J∈J νJ∗ πJ [A]. (Hint: start with A ⊆ [0, 1]r and note Q that ([0, 1]r )m can be identified with J∈J [0, 1]J .) 473Y Further exercises (a) Let D ⊆ Rr be any set and φ : D → Rs any function. Show that D0 = {x : x ∈ D, φ is differentiable at x} is a Borel subset of Rr , and that the derivative of φ is a Borel measurable function. (Compare 225J.) 473 Notes and comments The point of all the inequalities 473H-473L is that they bound some measure of R variance of a function f by the integral of k grad f k. If r = 2, indeed, we are looking at kf k2 (473H) or B |f |2 (473I) or something essentially equal to the variance of probability theory (473K). In higher dimensions we need to look at k kr/(r−1) in place of k k2 , and when r = 1 we can interpret the inequalities in terms of the supremum norm k k∞ (473M). In all cases we want to develop inequalities which will enable us to discuss a function in terms of its first derivative. In one dimension, this is the familiar Fundamental Theorem of Calculus (Chapter 22). We find there a straightforward criterion (‘absolute continuity’) to determine whether a given function of one variable is an indefinite integral, and that if so it is the indefinite integral of its own derivative. Even in two dimensions, this simplicity disappears. The essential problem is that a

474Bc

The distributional perimeter

647

function can be the indefinite integral of an integrable gradient function without being bounded (473Xa). The principal results of this section are stated for Lipschitz functions, but in fact they apply much more widely. The argument suggested in 473Xa involves approximating the unbounded function f by Lipschitz functions fk in a sharp enough sense to make it possible to read off all the inequalities for f from the corresponding inequalities for the fk . This idea leads naturally to the concept of ‘Sobolev space’, which I leave on one side for the moment; see Evans & Gariepy 92, chap. 4, for details.

474 The distributional perimeter The next step is a dramatic excursion, defining (for appropriate sets E) a perimeter measure for which a version of the Divergence Theorem is true (474E). I begin the section with elementary notes on the divergence of a vector field (474B-474C). I then define ‘locally finite perimeter’ (474D), ‘perimeter measure’ and ‘outward normal’ (474F) and ‘reduced boundary’ (474H). The definitions rely on the Riesz representation theorem, and we have to work very hard to relate them to any geometrically natural idea of ‘boundary’. Even half-spaces (474I) demand some attention. From Poincaré’s inequality (473K) we can prove isoperimetric inequalities for perimeter measures (474L). With some effort we can locate the reduced boundary as a subset of the topological boundary (474Xc), and obtain asymptotic inequalities on the perimeter measures of small balls (474N). With much more effort we can find a geometric description of outward normal functions in terms of ‘Federer exterior normals’ (474Q), and get a tight asymptotic description of the perimeter measures of small balls (474R). I end with the Compactness Theorem for sets of bounded perimeter (474S). 474A Notation I had better repeat some of the notation from §473. r ≥ 2 is a fixed integer. µ is Lebesgue measure on Rr , and βr = µB(0, 1) is the volume of the unit ball. Sr−1 = ∂B(0, 1) is the unit sphere. ν is normalized r − 1-dimensional Hausdorff measure on Rr . We shall sometimes need to look at Lebesgue measure on Rr−1 , which I will denote µr−1 . D is the set of smooth functions f : Rr → R with compact support; Dr the set of smooth functions φ : Rr → Rr with compact support, that is, the set of functions φ = (φ1 , . . . , φr ) : Rr → Rr such that φi ∈ D for every i. ˜ n in∈N from 473E; these functions all belong to D, are non-negative I continue to use the sequence hh 1 everywhere and zero outside B(0, n+1 ), are even, and have integral 1. 474B The divergence of a vector field (a) For a function φ from a subset of Rr to Rr , write Pr i div φ = i=1 ∂φ ∂ξi , where φ = (φ1 , . . . , φr ); for definiteness, let us take the domain of div φ to be the set of points at which φ is differentiable (in the strict sense of 262Fa). Note that div φ ∈ D for every φ ∈ Dr . We need the following elementary facts. (b) If f : Rr → R and φ : Rr → Rr are functions, then div(f × φ) = φ. grad f + f × div φ at any point at which f and φ are both differentiable. (Use 473Bc-473Bd.) R (c) If φ : Rr → Rr is a Lipschitz function with compact support, then div φ dµ = 0. P P div φ is defined almost everywhere (by Rademacher’s theorem, 262Q), measurable (473Be), bounded (473Cc) and with compact support, so R Pr R ∂φi dµ div φ dµ = i=1 ∂ξi

is defined in R. For each i ≤ r, Fubini’s theorem tells us that we can replace integration with respect to µ by a repeated integral, in which the inner integral is

R∞

∂φi

−∞ ∂ξi

(ξ1 , . . . , ξr )dξi = 0

because φi (ξ1 , . . . , ξr ) = 0 whenever |ξi | is large enough. So the result. Q Q

R ∂φi ∂ξi

dµ is also zero. Summing over i, we have

648


474Bd

(d) If φ : Rr → Rr and f : Rr → R are Lipschitz functions, one of which has compact support, then

R

R

φ. grad f dµ +

f × div φ dµ = 0.

P P f and φ are both differentiable almost everywhere. So

R

φ. grad f dµ +

R

f × div φ dµ =

R

div(f × φ)dµ = 0

by (c) above. Q Q (e) If f ∈ L∞ (µ), g ∈ L1 (µ) is even (that is, g(−x) is defined and Requal to g(x) for every x ∈ dom g), R and φ : Rr → Rr is a Lipschitz function with compact support, then (f ∗ g) × div φ = f × div(g ∗ φ), where g ∗ φ = (g ∗ φ1 , . . . , g ∗ φr ). P P For each i, Z (f ∗ g) ×

∂φi dµ ∂ξi

ZZ =

(255G/444Od)

f (x)g(y)

∂φi (x + y)µ(dy)µ(dx) ∂ξi

ZZ =

(because g is even)

f (x)g(−y) Z

=

f × (g ∗

∂φi (x + y)µ(dy)µ(dx) ∂ξi

∂φi )dµ ∂ξi

Z =

f×

∂ (g ∂ξi

∗ φi )dµ

by 473Dd. Now take the sum over i of both sides. Q Q (f ) If g ∈ L1 (µ) and φ : Rr → Rr is a bounded Lebesgue measurable function, then k(g ∗ φ)(x)k ≤ kgk1 supy∈R r kφ(y)k r

for every x ∈ R . P P We can find z = (ζ1 , . . . , ζr ) ∈ Sr−1 such that k(g ∗ φ)(x)k = (g ∗ φ)(x) .z =

r Z X

Z ≤

ζi g(x − y)φi (y)µ(dy)

i=1

Q |g(x − y)|kφ(y)kµ(dy) ≤ kgk1 sup kφ(y)k. Q y∈R r

474C Invariance under isometries: Proposition Suppose that T : Rr → Rr is any isometry, and that φ is a function from a subset of Rr to Rr . Then div(T −1 φT ) = (div φ)T . proof Set z = T (0). By 4A4Jb, the isometry x 7→ T x − z is linear and preserves inner products, so there is an orthogonal matrix S such that T (x) = z + Sx for every x ∈ Rr . Now suppose that x ∈ Rr is such that (div φ)(T x) is defined. Then T (y) − T (x) − S(y − x) = 0 for every y, so T is differentiable at x, with derivative S, and φT is differentiable at x, with derivative DS, where D is the derivative of φ at T (x), by 473Bc. Also T −1 (y) = S −1 (y − z) for every y, so T −1 is differentiable at φ(T (x)) with derivative S −1 , and T −1 φT is differentiable at x, with derivative S −1 DS. Now if D is hδij i1≤i,j≤r and S is hσij i1≤i,j≤r and S −1 DS is hτij i1≤i,j≤r , then S −1 is the transpose hσji i1≤i,j≤r of S, because S is orthogonal, so div(T

−1

φT )(x) = =

r X

r X r X

r X

τii = σji δjk σki i=1 i=1 j=1 k=1 r X r r r X X X δjk

j=1 k=1

σji σki =

i=1

δjj = div φ(x)

j=1

Pr because i=1 σji σjk = 1 if j = k and 0 otherwise. If div(T −1 φT )(x) is defined, then (because T −1 is also an isometry)

474E


649

(div φ)(T x) = div(T T −1 φT T −1 )(T x) = div(T −1 φT )(T −1 T x) = div(T −1 φT )(x). So the functions div(T −1 φT ) and (div φ)T are identical. 474D Locally finite perimeter: Definition Let E ⊆ Rr be a Lebesgue measurable set. Its perimeter per E is sup{|

R

E

div φ dµ| : φ : Rr → B(0, 1) is a Lipschitz function with compact support}.

E has locally finite perimeter if

R

sup{|

div φ dµ| : φ : Rr → Rr is a Lipschitz function, kφk ≤ χB(0, n)}

E

is finite for every n ∈ N. 474E Theorem Suppose that E ⊆ Rr has locally finite perimeter. (i) There are a Radon measure λ∂E on Rr and a function ψ : Rr → Sr−1 such that

R

div φ dµ = E

R

φ. ψ dλ∂E

for every Lipschitz function φ : Rr → Rr with compact support. (ii) This formula uniquely determines λ∂E , which can also be defined by saying that λ∂E (G) = sup{|

R

div φ dµ| : φ : Rr → Rr is Lipschitz, kφk ≤ χG}

E

whenever G ⊆ Rr is open. (iii) If ψˆ is another function satisfying the formula in (i), then ψˆ = ψ λ∂E -almost everywhere. proof (a) For each l ∈ N, set γl = sup{|

R E

div φ dµ| : φ : Rr → Rr is Lipschitz, kφk ≤ χB(0, l)}.

Write Ck for the space of continuous functions with compact support from Rr to R. By 473Dc and 473De, ˜ n ∈ D for every f ∈ Ck , n ∈ N. Now the point is that f ∗h Li (f ) = − limn→∞

R

∂

E ∂ξi

˜ n )dµ (f ∗ h

˜ n k∞ → 0 as n → ∞. But this is defined for every f ∈ Ck , i ≤ r. P P Applying 473Ed, we see that kf − f ∗ h (n) ˜ ˜ ˜ n )(x)ei , where ei is the unit means that k(f ∗ hm ) − (f ∗ hn )k∞ → 0 as m, n → ∞. Set φ (x) = (f ∗ h vector (0, . . . , 1, . . . , 0) with a 1 in the ith place. We have some l ∈ N such that |f | ≤ lχB(0, l); in which case φ(n) (x) = 0 whenever kxk > l + 1 and n ∈ N. This means that ¯ ¯

Z

∂ (f ∂ξ i E

Z ˜ m )dµ − ∗h

∂ (f ∂ξ i E

¯ ˜ n )dµ¯ = | ∗h

Z div(φ(m) − φ(n) )dµ|

≤ γl+1 sup kφ(m) (x) − φ(n) (x)k x∈R r

− φ(n) ) ˜ m ) − (f ∗ h ˜ n )k∞ → 0 = γl+1 k(f ∗ h

(applying the definition of γl+1 to a suitable multiple of φ

as m, n → ∞. Thus h

R

∂

E ∂ξi

(m)

˜ n )dµin∈N is a Cauchy sequence and must have a limit. Q (f ∗ h Q

˜ n k∞ = 0 and |f − f ∗ h ˜ n | ≤ kf − f ∗ h ˜ n k∞ χB(0, l + 1) for every If f is Lipschitz, then limn→∞ kf − f ∗ h R ∂ R ∂f ˜ n )dµ = 0, and Li (f ) = − n, so limn→∞ E (f − f ∗ h dµ. E ∂ξi

∂ξi

Note that the argument shows also that |Li (f )| ≤ γl+1 kf k∞ if f (x) = 0 for x ∈ Rr \ B(0, l). R ∂ ˜ n )dµ are linear, Li is linear. Moroever, by the last remark (b) Because all the functionals f 7→ E (f ∗ h ∂ξi

in (a), it is order-bounded when regarded as a linear functional on the Riesz space Ck , so is expressible as − a difference L+ i − Li of positive linear functionals (356B).

650


474E

− + r Representation Theorem (436J), we have Radon λ+ i , λi on R such that Li (f ) = R R By +the Riesz Pr measures − − + − f dλi , Li (f ) = f dλi for every f ∈ Ck . If we set L = i=1 Li + Li , there is another Radon measure R ˜ on Rr such that L(f ) = f dλ ˜ for every f ∈ Ck . Now each λ+ , λ− is an indefinite-integral measure with λ i i ˜ P respect to λ. P Because all these measures are σ-finite and the completions of their restrictions to the Borel − ˜ σ-algebra, 234G tells us that it is enough to check that λ+ i (F ) = λi (F ) = 0 whenever λF = 0. But this is immediate from 416E(a-v). Q Q − ˜ For each i ≤ r, let gi+ , gi− be Radon-Nikod´ ym derivatives of λ+ i , λi with respect to λ. Adjusting them ˜ on a λ-negligible set if necessary, we may suppose that they are all bounded non-negative Borel measurable ˜ must be the completion of its restriction to the Borel functions from Rr to R. (Recall from 256C that λ σ-algebra.) Set gi = gi+ − gi− for each i. Then Z Z Z ∂f − + − dµ = L+ (f ) − L (f ) = f dλ − f dλ− i i i i ∂ξi E Z Z Z ˜ − f × g − dλ ˜ = f × gi dλ ˜ = f × gi+ dλ i

for every Lipschitz function f with compact support (235M). Set g = g (x) − i g(x)

when g(x) 6= 0,

1 √ r

pP r

2 i=1 gi .

For i ≤ r, set ψi (x) =

r

when g(x) = 0, so that ψ = (ψ1 , . . . , ψr ) : R → Sr−1 is Borel measurable. Let

˜ defined by g; then λ∂ is a Radon measure on Rr (256E/416S). λ∂E be the indefinite-integral measure over λ E (c) Now take any Lipschitz function φ : Rr → Rr with compact support. Express it as (φ1 , . . . , φr ) where φi : Rr → R is a Lipschitz function with compact support for each i. Then Z div φ dµ = E

r Z X ∂φi i=1

=−

E

r X

∂ξi

dµ = −

i=1

=−

r X

L− i (φi ) = −

i=1

˜+ φi × gi+ dλ

i=1

(by 235M) =−

r Z X

Li (φi )

i=1

L+ i (φi ) +

r Z X

r X

r Z X

˜+ + φi dλ i

i=1

r Z X

r Z X

˜− φi dλ i

i=1

˜ φi × gi− dλ

i=1

˜= φi × gi dλ

i=1

r Z X

φi × ψi dλ∂E

i=1

(by 235M again, because ψi × g = −gi ) Z = φ.ψ dλ∂E . So we have λ∂E and ψ satisfying (i). (d) Now suppose that G ⊆ Rr is open. If φ : Rr → Rr is a Lipschitz function with compact support and kφk ≤ χG, then |

R

E

div φ dµ| = |

R

φ. ψ dλ∂E | ≤

R

kφkdλ∂E ≤ λ∂E (G).

On the other hand, if γ < λ∂E (G), let G0 ⊆ G be a bounded open set such that γ < λ∂E (G0 ), and set ² = 12 (λ∂E (G0 ) − γ). Let K ⊆ G0 be a compact set such that λ∂E (G0 \ K) ≤ ². Let δ > 0 be such that kx − yk ≥ 2δ whenever y ∈ K and x ∈ Rr \ G0 , and set H = {x : inf y∈K kx − yk < δ}. Now there are f1 , . . . , fr ∈ Ck such that R Pr Pr 2 ∂ i=1 fi ≤ χH, i=1 fi × ψi dλE ≥ γ. R P P For each i ≤ r, we can find a sequence hgmi im∈N in Ck such that |gmi − (ψi × χK)|dλ∂E ≤ 2−m for every m ∈ N (416I); multiplying the gmi by a function which takes the value 1 on K and 0 outside H if necessary, we can suppose that gmi (x) = 0 for x ∈ / H. Set

474F


fmi =

√gmi Pr

max(1,

j=1

2 ) gmj

651

∈ Ck

for each m and i. Now limm→∞ fmi (x) = ψi (x) for everyP i ≤ r whenever limm→∞ gmi (x) = i (x) for every Pψ r r 2 i ≤ r, which is the case for λ∂E -almost every x ∈ K. Also i=1 fmi ≤ χH for every m, so | i=1 fmi × ψi | ≤ χH for every m, while Pr R Pr R limm→∞ i=1 K fmi × ψi dλ∂E = i=1 K ψi2 dλ∂E = λ∂E (K). At the same time, | for every m, so

Pr R i=1 Rr \K

Pr R i=1

fmi × ψi dλ∂E | ≤ λ∂E (H \ K) ≤ ²

fmi × ψi dλ∂E ≥ λ∂E (G0 ) − 2² = γ

for all m large enough, and we may take fi = fmi for such an m. Q Q Now, for n ∈ N, set ˜ n , . . . , fr ∗ h ˜ n ) ∈ Dr . φn = (f1 ∗ h For all n large enough, we shall have kx − yk ≥

1 n+1

for every x ∈ Rr \ G0 and y ∈ H, so that φn (x) = 0 if

x∈ / G0 . By 473Dg, kφn (x)k ≤ supy∈R

pPr i=1

fi (y)2 ≤ 1

for every x and n, so that kφn k ≤ χG0 for all n large enough. Next, φn (x) → (f1 (x), . . . , fr (x)) for every x ∈ Rr (473Ed), so R R R Pr ∂ ∂ div φ dµ = φ . ψ dλ → n n i=1 fi × ψi dλE ≥ γ E E as n → ∞, by Lebesgue’s Dominated Convergence Theorem. As γ is arbitrary,

R

sup{

E

div φ dµ : φ ∈ Dr , kφk ≤ χG} ≥ λ∂E (G),

and the two must be equal. (e) Thus λ∂E must satisfy (ii). By 416Eb, it is uniquely defined. Now suppose that ψˆ is another function from a λ∂E -conegligible set to Rr and satisfies (i). Then

R

φ. ψ dλ∂E =

R

φ . ψˆ dλ∂E

for every Lipschitz function φ : Rr → Rr with compact support. Take any i ≤ r and any compact set K ⊆ Rr . For m ∈ N, set fm (x) = max(0, 1 − 2m inf y∈K ky − xk) for x ∈ Rr , so that hfm im∈N is a sequence of Lipschitz functions with compact support and limm→∞ fm = χK. Set φm = (0, . . . , fm , . . . , 0), where the non-zero term is in the ith position. Then Z K

Z ψi dλ∂E

Z

= lim φm . ψ dλ∂E m→∞ Z Z = lim φm . ψˆ dλ∂E = ψî dλ∂E .

= lim

m→∞

m→∞

fm ×

ψi dλ∂E

K

R R By the Monotone Class Theorem (136B), or otherwise, F ψî dλ∂E = F ψi dλ∂E for every bounded Borel set F , so that ψî = ψi λ∂E -a.e.; as i is arbitrary, ψ = ψˆ λ∂E -a.e. This completes the proof. 474F Definitions In the context of 474E, I will call λ∂E the perimeter measure of E, and if ψ is a function from a λ∂E -conegligible subset of Rr to Sr−1 which has the property in (i) of the theorem, I will call it an outward-normal function for E. The words ‘perimeter’ and ‘outward normal’ are intended to suggest geometric interpretations; much of this section and the next will be devoted to validating this suggestion. Observe that if E has locally finite perimeter, then per E = λ∂E (Rr ).

652


474G

474G The reduced boundary Let E ⊆ Rr be a set with locally finite perimeter; let λ∂E be its perimeter measure and ψ an outward-normal function for E. The reduced boundary ∂ $ E is the set of those y ∈ Rr such that, for some z ∈ Sr−1 , R 1 kψ(x) − zk dλ∂E = 0. limδ↓0 ∂ B(y,δ) λE B(y,δ)

(When requiring that the limit be defined, I mean to insist that λ∂E B(y, δ) should be non-zero for every δ > 0, that is, that y belongs to the support of ∂λ$E . Warning! Some authors use the phrase ‘reduced boundary’ for a slightly larger set.) Note that, writing ψ = (ψ1 , . . . , ψr ) and z = (ζ1 , . . . , ζr ), we must have ζi = limδ↓0

1 λ B(y,δ) ∂ E

R

B(y,δ)

ψi dλ∂E ,

so that z is uniquely defined; call it ψE (y). Of course ∂ $ E and ψE are determined entirely by the set E, because λ∂E is uniquely determined and ψ is determined up to a λ∂E -negligible set (474E). By Besicovitch’s Density Theorem (472D), limδ↓0

1 λ∂E B(x,δ)

R B(x,δ)

|ψi (x) − ψi (y)|λ∂E (dx) = 0

for λ∂E -almost every y ∈ Rr , for every i ≤ r; and for any such y, ψE (y) is defined and equal to ψ(y). Thus ∂ $ E is λ∂E -conegligible and ψE is an outward-normal function for E. I will call ψE : ∂ $ E → Sr−1 the canonical outward-normal function of E. 474H Invariance under isometries: Proposition Let E ⊆ Rr be a set with locally finite perimeter. Let λ∂E be its perimeter measure, and ψ an outward normal function for E. If T : Rr → Rr is any isometry, then T [E] has locally finite perimeter, λ∂T [E] is the image measure λ∂E T −1 , and SψT −1 is an outward normal function for T [E], where S is the derivative of T . proof (a) As noted in 474C, the derivative of T is constant, and is an orthogonal matrix. Suppose that n ∈ N. Let φ : Rr → Rr be a Lipschitz function such that kφk ≤ χB(0, n). Then Z

Z

|

div φ dµ| = | T [E]

(div φ)T dµ| E

(263D, because | det S| = 1)

Z div(T −1 φT )dµ|

=| E

(474C)

Z div(S −1 φT )dµ|

=| E

(because S −1 φT and T −1 φT differ by a constant, and must have the same derivative) ≤ λ∂E (T −1 [B(0, n)]) because S −1 φT is a Lipschitz function and kS −1 φT k = kφT k ≤ χT −1 [B(0, n)]. Since T −1 [B(0, n)] is bounded, λ∂E (T −1 [B(0, n)]) is finite for every n, and T [E] has locally finite perimeter. (b) We can therefore speak of its perimeter measure λ∂T [E] . Let G ⊆ Rr be an open set. If φ : Rr → Rr is a Lipschitz function and kφk ≤ χT [G], then |

R

div φ dµ| = | T [E]

R

E

div(S −1 φT )dµ| ≤ λ∂E (G)

because S −1 φT is a Lipschitz function dominated by χG. As φ is arbitrary, λ∂T [E] (T [G]) ≤ λ∂E (G). Applying the same argument in reverse, with T −1 in the place of T , we see that λ∂E (G) ≤ λ∂T [E] (T [G]), so the two are

474I


653

equal. Now this means that the Radon measures λ∂T [E] and λ∂E T −1 (418I) agree on open sets, and must be identical (416Eb). (c) Now consider SψT −1 . Since ψ is defined λ∂E -almost everywhere and takes values in Sr−1 , ψT and SψT −1 are defined λ∂T [E] -almost everywhere and take values in Sr−1 . If φ : Rr → Rr is a Lipschitz function with compact support, Z

Z T [E]

Z

div(T −1 φT )dµ E E Z Z −1 div(S φT )dµ = (S −1 φT ) .ψ dλ∂E = ZE = (φT ) . (Sψ)dλ∂E

div φ dµ =

(because S is orthogonal)

(div φ)T dµ =

Z φ. (SψT −1 )d(λ∂E T −1 )

= (235I)

Z φ. (SψT −1 )dλ∂T [E]

=

Accordingly SψT −1 is an outward-normal function for T [E]. 474I Half-spaces It will be useful, and perhaps instructive, to check the most elementary special case. Proposition Let H ⊆ Rr be a half-space {x : x . v ≤ α}, where v ∈ S r−1 . Then H has locally finite perimeter; its perimeter measure λ∂H is defined by saying λ∂H (F ) = ν(F ∩ ∂H) whenever F ⊆ Rr is such that ν measures F ∩ ∂H, and the constant function with value v is an outwardnormal function for H. proof (a) Suppose, to begin with, that v is the unit vector (0, . . . , 1) and that α = 0, so that H = {x : ξr ≤ 0}. Let φ : Rr → Rr be a Lipschitz function with compact support. Then for any i < r

R

∂φi H ∂ξi

µ(dx) = 0

because we can regard this as a multiple integral in which the inner integral is with respect to ξi and is therefore always zero. On the other hand, integrating with respect to the rth coordinate first, Z

∂φr µ(dx) ∂ξr H

Z = Rr−1

Z =

Z

0

∂φr (z, t)dt µr−1 (dz) ∂ξr −∞

Z

φr (z, 0)µr−1 (dz) =

Rr−1 r−1

φr (x)ν(dx) ∂H

(identifying ν on Rr−1 × {0} with µr−1 on R ) Z Z = φ. v dν = φ. v dλ ∂H

where λ is the indefinite-integral measure over ν defined by the function χ(∂H). Note that (by 234Da) λ can also be regarded as ν∂H ι−1 , where ν∂H is the subspace measure on ∂H and ι : ∂H → Rr is the identity map. Now ν∂H can be identified with Lebesgue measure on Rr−1 , by 265B or otherwise, so in particular is a Radon measure, and λ also is a Radon measure, by 418I or otherwise. This means that λ and the constant function with value v satisfy the conditions of 474E, and must be the perimeter measure of H and an outward-normal function.

654


474I

(b) For the general case, let S be an orthogonal matrix such that S(0, . . . , 1) = v, and set T (x) = Sx+αv for every x, so that H = T [{x : ξr ≤ 0}]. By 474H, the perimeter measure of H is λT −1 and the constant function with value v is an outward-normal function for H. Now the Radon measure λ∂H = λT −1 is defined by saying that λ∂H F = λT −1 [F ] = ν(T −1 [F ] ∩ {x : ξr = 0}) = ν(F ∩ T [{x : ξr = 0}]) = ν(F ∩ ∂H) whenever ν(F ∩ ∂H) is defined, because ν (being a scalar multiple of a Hausdorff measure) must be invariant under the isometry T . 474J Lemma Let E ⊆ Rr be a set with locally finite perimeter. Let λ∂E be the perimeter measure of E, and ψE its canonical outward-normal function. Then Rr \ E also has locally finite perimeter; its perimeter measure is λ∂E , its reduced boundary is ∂ $ E, and its canonical outward-normal function is −ψE . proof By 474Bc,

R Rr \E

div φ dµ = −

R

div φ dµ = E

R

φ.(−ψE ) dλ∂E

for every Lipschitz function φ : Rr → Rr with compact support. The uniqueness assertions in 474E tell us that Rr \ E has locally finite perimeter, that its perimeter measure is λ∂E , and that −ψE is an outwardnormal function for Rr \ E. Referring to the definition of ‘reduced boundary’ in 474G, we see at once that ∂ $ (Rr \ E) = ∂ $ E and that ψRr \E = −ψE . 474K Lemma Let E ⊆ Rr be a set with locally finite perimeter; let λ∂E be its perimeter measure, and ψ an outward-normal function for E. Let φ : Rr → Rr be a Lipschitz function with compact support, and g ∈ D an even function. Then

R

φ. grad(g ∗ χE)dµ +

R

(g ∗ φ) . ψ dλ∂E = 0.

proof Z

Z φ. grad(g ∗ χE)dµ = −

(474Bd, using 473Df to see that g ∗ χE is Lipschitz)

(g ∗ χE) × div φ dµ Z

=− (474Be)

χE × div(g ∗ φ)dµ Z

=−

(g ∗ φ) . ψ dλ∂E

(because g ∗ φ is smooth and has compact support, so is Lipschitz), as required. 474L Two isoperimetric inequalities: Theorem Let E ⊆ Rr be a set with locally finite perimeter, and λ∂E its perimeter measure. (a) If E is bounded, then (µE)(r−1)/r ≤ per E. (b) If B ⊆ Rr is a closed ball, then where c = 2r+2 (1 +

√

min(µ(B ∩ E), µ(B \ E))(r−1)/r ≤ 2cλ∂E (int B), r)(1 + 2r+1 ).

˜ n . Note proof (a) Let ² > 0. By 473Ef, there is an n ∈ N such that kf − χEkr/(r−1) ≤ ², where f = χE ∗ h that f is smooth (473De) and has compact support, because E is bounded. Let η > 0 be such that

R grad f and set φ = √

η+k grad f k2

k grad f kdµ ≤

R

2

√ k grad f k

η+k grad f k2

dµ + ²,

. Then φ ∈ Dr and kφ(x)k ≤ 1 for every x ∈ Rr . Now we can estimate

474L


Z

655

Z k grad f kdµ ≤

φ. grad f dµ + ² Z ˜ n ∗ φ) . ψ dλ∂ + ² = − (h E

(where ψ is an outward-normal function for E, by 474K) Z ˜ n ∗ φkdλ∂ + ² ≤ per E + ² ≤ kh E ˜ n ∗ φ)(x)k ≤ 1 for every x ∈ Rr , by 473Dg. Accordingly because k(h Z (µE)

(r−1)/r

= kχEkr/(r−1) ≤ kf kr/(r−1) + ² ≤

k grad f kdµ + ²

(473H) ≤ per E + 2². As ² is arbitrary, we have the result. (b)(i) Set α = min(µ(B ∩ E), µ(B \ E)). If α = 0, the result is trivial; so suppose that α > 0. Take any ² ∈ ]0, α]. Let B1 be a closed ball, with the same centre as B and strictly smaller non-zero radius, such that µ(B \ B1 ) ≤ ²; then α − ² ≤ min(µ(B1 ∩ E), µ(B1 \ E)). For f ∈ Lr/(r−1) (µ) set γ0 (f ) =

R

1 µB1

B1

f dµ,

γ1 (f ) = k(f × χB1 ) − γ0 (f )χB1 kr/(r−1) ;

then both γ0 and γ1 are continuous functions on Lr/(r−1) (µ) if we give it its usual pseudometric (f, g) 7→ kf − gkr/(r−1) . Now γ1 (χ(E ∩ B)) ≥ 21 (α − ²)(r−1)/r . P P We have γ0 (χ(E ∩ B)) =

µ(B1 ∩E) , µB1

Z γ1 (χ(E ∩ B))r/(r−1) =

|χ(E ∩ B) − γ0 (χ(E ∩ B))|r/(r−1) B1

¡ µ(B1 ∩E) ¢r/(r−1) ¡ µ(B1 ∩E) ¢r/(r−1) + µ(B1 \ E) = µ(B1 ∩ E) 1 − µB1

µB1

¡ µ(B1 \E) ¢r/(r−1) ¡ µ(B1 ∩E) ¢r/(r−1) = µ(B1 ∩ E) + µ(B1 \ E) . µB1

Either µ(B1 ∩ E) ≥

1 2 µB1

or µ(B1 \ E) ≥

1 2 µB1 ;

γ1 (χ(E ∩ B))r/(r−1) ≥

µB1

suppose the former. Then

1 µ(B1 2r/(r−1)

\ E) ≥

1 (α − ²) 2r/(r−1)

1 2

and γ1 (χ(E ∩ B)) ≥ (α − ²)r/(r−1) . Exchanging B1 ∩ E and B1 \ E we have the same result if µ(B1 ∩ E) ≥ 1 2 µB1 .

Q Q

(ii) Express B as B(y, δ) and B1 as B(y, δ1 ). Take n0 ≥

2 . δ−δ1

Because γ1 is k kr/(r−1) -continuous,

˜ n ∗ χ(E ∩ B) (473Ef); as in part (a) there is an n ≥ n0 such that γ1 (f ) ≥ 21 (α − ²)(r−1)/r − ², where f = h of the proof, f ∈ D. Let η > 0 be such that

R

Let m ≥ n0 be such that

R

B1

√ k grad f k

2

η+k grad f k2

φ. grad f dµ ≥

R

B1

dµ ≥

R

B1

k grad f kdµ − ².

k grad f kdµ − 2², where

˜m ∗ (√ φ=h

grad f η+k grad f k2

× χB1 ).

656


474L

˜ n ∗ φ)(x) = 0 if x ∈ Note that φ(x) = 0 if kx − yk ≥ 21 (δ + δ1 ), so that (h / int B. By 473Dg, kφ(x)k ≤ 1 for ˜ n ∗ φ)(x)k ≤ 1 for every x, so kh ˜ n ∗ φk ≤ χ(int B). every x and k(h Now we have Z

Z ˜ n ∗ χ(E ∩ B)) dµ φ. grad(h Z ˜ n ∗ χ(E ∩ B)) × div φ dµ = − (h

φ. grad f dµ =

(474Bd)

Z ˜ n ∗ φ)dµ div(h

=− E∩B

(474Be)

Z =− E

˜ n ∗ φ)dµ ≤ λ∂ (int B) div(h E

(474E). (iii) Accordingly 1 (α − ²)(r−1)/r 2

Z − ² ≤ γ1 (f ) ≤ c

k grad f kdµ B1

(473K)

Z ¡ ¢ ≤c φ. grad f dµ + 2² ≤ c(λ∂E (int B) + 2²).

As ² is arbitrary, α(r−1)/r ≤ 2cλ∂E (int B), as claimed. 474M Lemma Suppose that E ⊆ Rr has locally finite perimeter, with perimeter measure λ∂E and an outward-normal function ψ. Then for any y ∈ Rr and any Lipschitz function φ : Rr → Rr ,

R

div φ dµ = E∩B(y,δ)

R

φ.ψ dλ∂E + B(y,δ)

R

E∩∂B(y,δ)

1 δ

φ(x) . (x − y) ν(dx)

for almost every δ > 0. proof (a) For t > 0, set w(t) =

R E∩∂B(y,t)

1 t

φ(x) . (x − y) ν(dx)

when this is defined. By 265G, applied to functions of the form ( x−y if x ∈ E and 0 < kx − yk ≤ α, x 7→ φ(x) . kx−yk

0

otherwise,

w is defined almost everywhere in ]0, ∞[ and is measurable (for Lebesgue measure on R). Let δ > 0 be any point in the Lebesgue set of w (223D). Then limt↓0

1 t

R δ+t δ

|w(s) − w(δ)|ds ≤ 2 limt↓0

Let ² > 0. Then there is an η > 0 such that 1 η

R δ+η δ

|w(s) − w(δ)|ds ≤ ²,

R B(y,δ+η)\B(y,δ)

1 2t

R δ+t δ−t

|w(s) − w(δ)| = 0.


k div φkdµ ≤ ².

kφkdλ∂E ≤ ²,

474M


657

(b) Set g(x) = 1 if kx − yk ≤ δ, 1 η

= 1 − (kx − yk − δ) if δ ≤ kx − yk ≤ δ + η, = 0 if kx − yk ≥ δ + η. Then g is continuous, and grad g(x) = 0 if kx − yk < δ or kx − yk > δ + η; while if δ < kx − yk < δ + η, grad g(x) = −

x−y . ηkx−yk

This means that Z φ. grad g dµ = − E

1 − η

= By the choice of η, |

1 η

R

Z Z

Z

δ+η δ

E∩∂B(y,t)

1 (x − y) .φ(x) ν(dx)dt t

δ+η

w(t)dt. δ

φ. grad g dµ + w(δ)| ≤ ².

E

(c) By 474E and 474Bb we have Z

Z (g × φ) .ψ dλ∂E =

div(g × φ)dµ E

(of course g × φ is Lipschitz, by 473Ca and 262Ba) Z Z = φ. grad g dµ + g × div φ dµ. E

Next, by the choice of η,

E

R

| ((g × φ) − φ) . ψdλ∂E | ≤


kφkdλ∂E ≤ ²

and ¯ ¯

Z

Z

¯ 1 φ(x) . (x − y)ν(dx)¯

φ. grad g dµ + E

δ

E∩∂B(y,δ)

Z

¯ = ¯−

δ+η

Z

¯ x−y . φ(x)ν(dx)dt + w(δ)¯ ηkx−yk ∂B(y,t)

δ

(265G) ¯ 1 = ¯− η

Z

δ+η

¯ w(t)dt + w(δ)¯ ≤ ²

δ

and Z Z ¯ ¯ ¯ g × div φ dµ − div φ dµ¯ E E∩B(y,δ) Z ≤ k div φkdµ ≤ ². B(y,δ+η)\B(y,δ)

Putting these together, we have |

R

E∩B(y,δ)

div φ dµ −

R

As ² is arbitrary, this gives the result.

B(y,δ)

φ. ψ dλ∂E −

R E∩∂B(y,δ)

1 δ

φ(x) . (x − y)ν(dx)| ≤ 3².

658


474N

474N Lemma Let E ⊆ Rr be a set with locally finite perimeter, and λ∂E its perimeter measure. Then, for any y ∈ ∂ $ E, µ(B(y,δ)∩E) 1 ; ≥ µB(y,δ) (3r)r βr µ(B(y,δ)\E) 1 (ii) lim inf δ↓0 ≥ ; µB(y,δ) (3r)r βr ∂ λ B(y,δ) 1 (iii) lim inf δ↓0 E r−1 ≥ , δ 2c(3r)r

(i) lim inf δ↓0

where c = 2r+2 (1 + (iv)

√

r)(1 + 2r+1 );

λ∂ B(y,δ) lim supδ↓0 E r−1 δ

≤ 4πβr−2 .

proof (a) Let ψE be the canonical outward-normal function of E (474G). Take y ∈ ∂ $ E. Set Φ = {φ : φ is a Lipschitz function from Rr to B(0, 1)}. Because the space L1 (µ) is separable in its usual (norm) topology (244I), so is {(div φ × χB1 )• : φ ∈ Φ}, (4A2P(a-iv)), and there must be a countable set Φ0 ⊆ Φ such that R ˆ whenever φ ∈ Φ and m ∈ N there is a φˆ ∈ Φ0 such that B(y,1) | div φ − div φ|dµ ≤ 2−m . Now, for each φ ∈ Φ0 , Z Z Z 1 | div φ dµ| = | φ. ψE dλ∂E + φ(x) . (x − y)ν(dx)| E∩B(y,δ)

B(y,δ)

≤

λ∂E B(y, δ)

δ

E∩∂B(y,δ)

+ ν(E ∩ ∂B(y, δ))

for almost every δ > 0, by 474M. But this means that, for almost every δ ∈ ]0, 1], Z per(E ∩ B(y, δ)) = sup | div φ dµ| φ∈Φ E∩B(y,δ) Z = sup | div φ dµ| ≤ λ∂E B(y, δ) + ν(E ∩ ∂B(y, δ)). φ∈Φ0

E∩B(y,δ)

(b) It follows that, for some δ0 > 0, per(E ∩ B(y, δ)) ≤ 3ν(E ∩ ∂B(y, δ)) for almost every δ ∈ ]0, δ0 ]. P P Applying 474M with φ(x) = ψE (y) for every x, we have 0=

R

B(y,δ)

ψE (y) . ψE (x)λ∂E (dx) +

R

E∩∂B(y,δ)

for almost every δ ∈ [0, 1]. But by the definition of ψE (y), limδ↓0

1 λ∂E B(y,δ)

R

B(y,δ)

ψE (y) . ψE dλ∂E = limδ↓0

1 δ

ψE (y) . (x − y)ν(dx)

1 λ∂E B(y,δ)

R B(y,δ)

ψE (y) . ψE (y) dλ∂E = 1.

So there is some δ0 > 0 such that, for almost every δ ∈ ]0, δ0 ], Z λ∂E B(y, δ) ≤ 2

B(y,δ)

ψE (y) . ψE dλ∂E

Z

1 δ

ψE (y) . (x − y)ν(dx) ≤ 2ν(E ∩ ∂B(y, δ)).

= −2 E∩∂B(y,δ)

(1)

But this means that, for almost every such δ, per(E ∩ B(y, δ)) ≤ λ∂E B(y, δ) + ν(E ∩ ∂B(y, δ)) ≤ 3ν(E ∩ ∂B(y, δ)). Q Q Rt (c) Set g(t) = µ(E ∩B(y, t)) for t ≥ 0. By 265G, g(t) = 0 ν(E ∩∂B(y, s))ds for every t, so g is absolutely continuous on [0, 1] and g 0 (t) = ν(E ∩ ∂B(y, t)) for almost every t. Now we can estimate

474P


659

g(t)(r−1)/r = µ(E ∩ B(y, t))(r−1)/r ≤ per(E ∩ B(y, t)) (474La) ≤ 3ν(E ∩ ∂B(y, t)) = 3g 0 (t) for almost every t ∈ [0, δ0 ]. So ¢ d¡ g(t)1/r dt

1 r

= g(t)(1−r)/r g 0 (t) ≥

1 3r

for almost every t ∈ [0, δ0 ]; since t 7→ g(t)1/r is non-decreasing, g(t)1/r ≥

t 3r

(222C) and g(t) ≥ (3r)−r tr for

every t ∈ [0, δ0 ]. (d) Accordingly lim inf δ↓0

µ(B(y,δ)∩E) µB(y,δ)

≥ inf 0<δ≤δ0

µ(B(y,δ)∩E) µB(y,δ)

= inf 0<δ≤δ0

g(δ) βr δ r

≥

1 . (3r)r βr

This proves (i). (e) Because λ∂Rr \E = λ∂E and −ψE is the canonical outward-normal function of Rr \ E (474J), y also belongs to ∂ $ (Rr \ E), so the second formula of this lemma follows from the first. (f ) By 474Lb, λ∂E B(y, δ) ≥

1 2c

¡ ¢(r−1)/r min µ(B(y, δ) ∩ E), µ(B(y, δ) \ E)

for every δ ≥ 0. So lim inf δ↓0

¡ µ(B(y, δ) ∩ E) µ(B(y, δ) \ E) ¢(r−1)/r λ∂E B(y, δ) 1 ≥ lim inf min , 2c δ↓0 δ r−1 δr δr ≥

1 . 2c(3r)r

Thus (iii) is true. (g) Returning to the inequality (1) in the proof of (b) above, we have a δ0 > 0 such that λ∂E B(y, δ) ≤ 2ν(E ∩ ∂B(y, δ)) ≤ 2ν(∂B(y, δ)) = 4πβr−2 δ r−1 for almost every δ ∈ ]0, δ0 ]. But this means that, for any δ ∈ [0, δ0 [, λ∂E B(y, δ) ≤ inf t>δ λ∂E B(y, t) ≤ inf t>δ 4πβr−2 tr−1 = 4πβr−2 δ r−1 , and (iv) is true. 474O Definition Let A ⊆ Rr be any set, and y ∈ Rr . A Federer exterior normal to A at y is a v ∈ Sr−1 such that, limδ↓0

µ∗ ((H4A)∩B(y,δ)) µB(y,δ)

= 0,

where H is the half-space {x : (x − y) . v ≤ 0}. 474P Lemma If A ⊆ Rr and y ∈ Rr , there can be at most one Federer exterior normal to A at y. proof Suppose that v, v 0 ∈ Sr−1 are two Federer exterior normals to E at y. Set H = {x : (x − y) . v ≤ 0},

H 0 = {x : (x − y) .v 0 ≤ 0}.

Then limδ↓0

µ((H4H 0 )∩B(y,δ)) µB(y,δ)

≤ limδ↓0

µ∗ ((H4A)∩B(y,δ)) µB(y,δ)

+ limδ↓0

µ∗ ((H 0 4A)∩B(y,δ)) µB(y,δ)

= 0.

660


474P

But for any δ > 0, (H4H 0 ) ∩ B(y, δ) = y + δ((H0 4H00 ) ∩ B(0, 1)), where H00 = {x : x .v 0 ≤ 0}.

H0 = {x : x . v ≤ 0}, So 0 = limδ↓0

µ((H4H 0 )∩B(y,δ)) µB(y,δ)

µ((H0 4H00 )∩B(0,1)) µB(0,1)

=

=

µ((H0 4H00 )∩B(0,n)) µB(0,n)

for every n ≥ 1, and µ(H0 4H00 ) = 0. Since µ is strictly positive, and H0 and H00 are both the closures of their interiors, they must be identical; and it follows that v = v 0 . 474Q Theorem Let E ⊆ Rr be a set with locally finite perimeter, λ∂E its perimeter measure, ψE its canonical outward-normal function, and y any point of its reduced boundary ∂ $ E. Then ψE (y) is the Federer exterior normal to E at y. proof (a) To begin with (down to the end of (i) below) suppose that y = 0 and that ψE (y) = (0, . . . , 0, 1) = v say. Set √ 2r+2 πβr−1 c0 = , c0 = 2r+1 (1 + r − 1)(1 + 2r ), rβr

(counting β0 as 1, if r = 2),

¢ ¡ √ c1 = 1 + max 4πβr−2 , (3r)r 2r+3 (1 + r)(1 + 2r+1 ) , c2 = 2(r+1)/2 c1 ,

For δ > 0 let Cδ be the solid cylinder

Pr−1

{x :

i=1

c3 =

c2 2

+

8c0 . βr−1

ξi2 ≤ δ 2 , |ξr | ≤ δ}.

For γ ∈ R set ˜ H(γ) = {x : ξr ≤ γ}. For each δ > 0, choose γδ ∈ R such that ˜ δ )) ∩ Cδ ) ≤ δµCδ + inf γ∈R µ((E4H(γ)) ˜ µ((E4H(γ ∩ Cδ ). (b) Take any ² > 0 such that 1 ²<√ ,

1 8

4c0 ² < βr−1 ,

2

Then there is a δ0 ∈ ]0, 1] such that 1 λ∂E B(0,δ)

R B(0,δ)

1 r−1 δ c1

1 4

c2 ²3 < βr−1 .

kψE (x) − vkλ∂E (dx) ≤ ²3 ,

≤ λ∂E B(0, δ) ≤ c1 δ r−1

√ ¤ ¤ for every δ ∈ 0, δ0 2 (using 474N(iii) and 474N(iv) for the inequalities bounding λ∂E B(0, δ)). Then √ B(0, δ) ⊆ Cδ ⊆ B(0, δ 2), so Z

Z kψE (x) −

Cδ

vkλ∂E (dx)

≤

√

B(0,δ 2)

≤

kψE (x) − vkλ∂E (dx)

²3 λ∂E B(0, δ

√

1 2

2) ≤ 2(r−1)/2 c1 ²3 δ r−1 = c2 ²3 δ r−1

474Q


661

whenever 0 < δ ≤ δ0 . ˜ n ∗ χE, where h ˜ n is the smoothing function of 473E, so that fn is smooth and For each n ∈ N, set fn = h Lipschitz (473Df) and grad fn is bounded (473Cc). For n ∈ N, t ∈ R set 3 4

1 4

Fn = {x : x ∈ Rr , fn (x) ≥ },

Fn0 = {x : x ∈ Rr , fn (x) ≤ },

fnt (z) = fn (z, t) for z ∈ Rr−1 , identifying Rr with Rr−1 × R, 3 4

1 4

Fnt = {z : z ∈ Rr−1 , fnt (z) ≥ },

0 Fnt = {z : z ∈ Rr−1 , fnt (z) ≤ }.

Write Vδ ⊆ Rr−1 for the ball {z : kzk ≤ δ}, so that Cδ is Vδ × [−δ, δ]. Set ¡ ¢ 3 γδn = sup {−δ} ∪ {t : t ∈ [−δ, δ], µr−1 (Fnt ∩ Vδ ) ≥ µVδ } , 4

Gδn = {t : t ∈ [−δ, δ],

R Vδ

k grad fnt kdµr−1 > ²2 (δ +

1 r−2 ) }. n+1

The aim of the main part of the proof is to show that, for small δ and large n, Cδ ∩ Fn is well approximated ˜ δn ). by Cδ ∩ H(γ (c) For the time being (down to the end of (f) below) fix δ > 0 and n ∈ N such that δ+

1 n+1

≤ δ0 ,

4c0 ²(δ +

c2 ²3 (δ +

1 r−1 ) n+1

1 r−1 ) n+1

1 8

≤ βr−1 δ r−1 ,

1 4

< βr−1 δ r−1 .

For f : Rr → R set gradr−1 f = grad f − (v . grad f )v = (

∂f ,... ∂ξ1

,

∂f , 0); ∂ξr−1

for definiteness, take the domain of gradr−1 f to be the set of points at which f is differentiable. Then

R

k gradr−1 fn k + max(

Cδ

∂fn , 0)dµ ∂ξr

≤ c2 ²3 (δ +

1 r−1 ) . n+1

P P?? Suppose, if possible, otherwise. Note that because µCδ = 2βr−1 δ r (by Fubini’s theorem), limδ0 ↑δ µCδ0 = µCδ , so there is some δ 0 < δ such that

R

Cδ0



> c2 ²3 (δ +

1 r−1 ) . n+1

For 1 ≤ i < r, set ∂fn ∂ξi (x)

θi (x) =

k gradr−1 fn (x)k = 0 otherwise;

when x ∈ Cδ0 and grad fn (x) 6= 0, r−1

and set θr (x) =

n max( ∂f ∂ξr (x), 0) n | ∂f ∂ξr (x)|

when x ∈ Cδ0 and

∂fn (x) ∂ξr

6= 0,

= 0 otherwise. Then θ : Rr → Rr is Lebesgue measurable and kθk ≤

R

θ . grad fn dµ =

R Cδ 0

√

2χCδ0 , so all the θi are µ-integrable, while



> c2 ²3 (δ +

˜ k k1 ik∈N → 0 for each i; since grad fn is bounded, By 473Ef, hkθi − θi ∗ h

1 r−1 ) . n+1

662


R

˜ k ∗ θ) . grad fn dµ > c2 ²3 (δ + (h

474Q 1 r−1 ) n+1

1 ˜ k ∗ θ, we shall get a smooth function for any k large enough; if we ensure also that ≤ δ − δ 0 , and set φ = h k+1 √ φ, with kφk ≤ 2χCδ (by 473Dc and 473Dg), such that

R

φ. grad fn dµ > c2 ²3 (δ +

1 r−1 ) . n+1

Moreover, ˜ n ∗ φ)(x) . v = (h ˜ n ∗ (h ˜ k ∗ θr ))(x) ≥ 0 (h ˜ n, h ˜ k ant θr are all non-negative. for every x, because h Now φ is smooth (473De) and has compact support, so

c2 ²3 (δ +

1 r−1 ) n+1

Z

Z ˜ n ∗ χE)dµ φ. grad fn dµ = φ. grad(h Z ˜ n ∗ φ) . ψE dλ∂ = − (h E <

(474K)

Z

Z ≤

˜ n ∗ φ)(x)k ≤ (because k(h

√

˜ n ∗ φ) .(v − ψE ) dλ∂ ≤ 2 (h E

Cδ+(1/n+1)

kv − ψE k dλ∂E

˜ n ∗ φ)(x) = 0 if x ∈ 2 for every x, by 473Dg again, and (h / Cδ+(1/n+1) ) ≤ c2 ²3 (δ +

1 r−1 ) ; n+1

which is absurd. X XQ Q (d) Observe that gradr−1 fn (z, t) = (grad fnt (z), 0) for all z and t. So (c) tells us that Z

δ

Z

Z k grad fnt (z)kµr−1 (dz)dt =

−δ

Vδ

k grad fn (x)kµ(dx) Cδ

r−1

≤ c2 ²3 (δ + So, writing µ1 for Lebesgue measure on R, µ1 Gδn ≤ c2 ²(δ +

1 r−1 ) . n+1

1 ). n+1

If t ∈ [−δ, δ] \ Gδn , then 0 min(µr−1 (Fnt ∩ Vδ ), µr−1 (Fnt ∩ Vδ )) ≤ 4c0 ²(δ +

1 r−1 ) . n+1

P P If r > 2, ¡ ¢(r−2)/(r−1) 0 min µr−1 (Fnt ∩ Vδ ), µr−1 (Fnt ∩ Vδ ) ≤ 4c0

Z k grad fnt kdµr−1 Vδ

(473L) ≤ 4c0 ²2 (δ + so that

1 r−2 ) , n+1

474Q


663

0 min(µr−1 (Fnt ∩ Vδ ), µr−1 (Fnt ∩ Vδ )) ≤ (4c0 ²2 )(r−1)/(r−2) (δ +

≤ 4c0 ²(δ + because 4c0 ≥ 1 and

2(r−1) r−2

1 r−1 ) n+1

1 r−1 ) n+1

≥ 1 and ² ≤ 1. If r = 2, then

R Vδ

1 2

k grad fnt kdµr−1 ≤ ²2 < ,

0 0 ∩ Vδ ), µr−1 (Fnt ∩ Vδ )) = 0. ∩ Vδ , Fnt ∩ Vδ is empty, as noted in 473M, and min(µr−1 (Fnt so at least one of Fnt Q Q

3 4

0 (e) The next thing to note is that µr−1 (Fns ∩ Vδ ) < µr−1 Vδ whenever −δ ≤ s < γδn . P P?? Otherwise,

we have −δ ≤ s ≤ t ≤ δ such that 3 4

3 4

0 µr−1 (Fns ∩ Vδ ) ≥ µr−1 Vδ ,

µr−1 (Fnt ∩ Vδ ) ≥ µr−1 Vδ ,

so that 1 2

0 µr−1 (Fns ∩ Fnt ∩ Vδ ) ≥ µr−1 Vδ . 0 ∩ Fnt ∩ Vδ , fn (z, s) ≤ But for any z ∈ Fns

Rδ

max( −δ

1 4

∂fn (z, ξ), 0)dξ ∂ξr

and fn (z, t) ≥ 34 , so ≥

R t ∂fn s ∂ξr

1 2

(z, ξ)dξ = fn (z, t) − fn (z, s) ≥ .

Accordingly Z

Z

∂f max( n (x), 0)µ(dx) ∂ξr Cδ

Z

δ

=

max( Vδ

≥

−δ

1 0 µr−1 (Fns 2

∂fn (z, ξ), 0)dξ µr−1 (dz) ∂ξr 1 4

∩ Fnt ∩ Vδ ) ≥ µr−1 Vδ

1 4

= βr−1 δ r−1 > c2 ²3 (δ +

1 r−1 ) n+1

by the choice of δ and n; but this contradicts (c). X XQ Q (f ) Now ˜ δn )) ∩ Cδ ) ≤ 9µ(Cδ \ (Fn ∪ Fn0 )) + c2 ²(δ + 1 )µr−1 Vδ µ((Fn 4H(γ n+1

+ 16c0 ²δ(δ +

1 r−1 ) . n+1

P P Set 1 0 ˜ = {t : −δ ≤ t ≤ δ, µr−1 (Vδ \ (Fnt ∪ Fnt G )) ≥ µr−1 Vδ }, 8

ˆ = {t : −δ ≤ t ≤ δ, µr−1 (Fnt ∩ Vδ ) ≤ 4c0 ²(δ + 1 )r−1 }, G n+1

1 r−1 0 ˆ 0 = {t : −δ ≤ t ≤ δ, µr−1 (Fnt G ∩ Vδ ) ≤ 4c0 ²(δ + ) }; n+1

ˆ∪G ˆ 0 = [−δ, δ], by (d). Then so that Gδn ∪ G 1 µr−1 Vδ 8

˜ ≤ µ(Cδ \ (Fn ∪ F 0 )), · µ1 G n

664


474Q

ˆ ≤ 8c0 ²δ(δ + 1 )r−1 , µ(Fn ∩ (Vδ × G)) n+1

ˆ 0 )) ≤ 8c0 ²δ(δ + 1 )r−1 . µ(Fn0 ∩ (Vδ × G n+1

So if we set ˜ ∪ Gδn ∪ {γδn })) W = (Cδ \ (Fn ∪ Fn0 )) ∪ (Vδ × (G ˆ ∪ (Fn0 ∩ (Vδ × G ˆ 0 )), ∪ (Fn ∩ (Vδ × G)) we shall have ˜ · µr−1 Vδ + µ1 Gδn · µr−1 Vδ + 16c0 ²δ(δ + 1 )r−1 µW ≤ µ(Cδ \ (Fn ∪ Fn0 )) + µ1 G n+1

≤ 9µ(Cδ \ (Fn ∪ Fn0 )) + c2 ²(δ +

1 )µr−1 Vδ n+1

+ 16c0 ²δ(δ +

1 r−1 ) n+1

(using the first part of (d) to estimate µ1 Gδn ). ˜ δn )) ∩ Cδ \ W . Since t ∈ ?? Suppose, if possible, that there is a point (z, t) ∈ (Fn 4H(γ / Gδn , (d) tells us that 0 min(µr−1 (Fnt ∩ Vδ ), µr−1 (Fnt ∩ Vδ )) ≤ 4c0 ²(δ +

1 r−1 ) n+1

1 8

≤ µr−1 Vδ

7 0 ˆ ∪G ˆ 0 . Also, since t ∈ ˜ µr−1 Fnt + µr−1 Fnt by the choice of δ and n. So t ∈ G / G, ≥ µr−1 Vδ ; so (since t 6= γδn ) 8

either µr−1 Fnt ≥

3 µr−1 Vδ 4

and t < γδn , or

0 µr−1 Fnt

≥

3 µr−1 Vδ 4

and t > γδn (by (e)). Now

3 4

t < γδn =⇒ µr−1 (Fnt ∩ Vδ ) ≥ µr−1 Vδ 0 =⇒ µr−1 (Fnt ∩ Vδ ) ≤ 4c0 ²(δ +

1 r−1 ) n+1

ˆ0 =⇒ t ∈ G =⇒ (z, t) ∈ / Fn0 ˆ 0 )) (because (z, t) ∈ / Fn0 ∩ (Vδ × G =⇒ (z, t) ∈ Fn (because (z, t) ∈ / Cδ \ (Fn ∪

Fn0 )) ˜ δn ) ∩ Cδ , =⇒ (z, t) ∈ Fn ∩ H(γ

which is impossible. And similarly 3 4

0 t > γδn =⇒ µr−1 (Fnt ∩ Vδ ) ≥ µr−1 Vδ

=⇒ µr−1 (Fnt ∩ Vδ ) ≤ 4c0 ²(δ +

1 r−1 ) n+1

ˆ =⇒ t ∈ G =⇒ (z, t) ∈ / Fn ˜ δn ), =⇒ (z, t) ∈ / Fn ∪ H(γ which is equally impossible. X X ˜ δn )) ∩ Cδ ⊆ W has measure at most Thus (Fn 4H(γ 9µ(Cδ \ (Fn ∪ Fn0 )) + c2 ²(δ +

1 )µr−1 Vδ n+1

+ 16c0 ²δ(δ +

1 r−1 ) , n+1

474Q


665

as claimed. Q Q (g) All this has been with fixed δ and n. Now take δ ∈ ]0, δ0 [ and let n → ∞. In this case, fn → χE a.e. (473Ee), so limn→∞ µ((E4Fn ) ∩ Cδ = limn→∞ µ(Cδ \ (Fn ∪ Fn0 )) = 0, and ˜ δn )) ∩ Cδ ) = lim sup µ((Fn 4H(γ ˜ δn )) ∩ Cδ ) lim sup µ((E4H(γ n→∞

n→∞

≤ c2 ²δµr−1 Vδ + 16c0 ²δ r = c3 ²µCδ . Now ˜ δ )) ∩ Cδ ) ≤ δµCδ + inf µ((E4H(γ ˜ δn )) ∩ Cδ ) µ((E4H(γ n∈N

≤ (δ + c3 ²)µCδ by the choice of γδ ((a) above). Since ² is arbitrary, this means that lim supδ↓0

(h) Now limδ↓0

γδ δ

1 ˜ δ )) ∩ Cδ ) µ((E4H(γ µCδ

= 0.

= 0. P P Let ² ∈ ]0, 1]. Then by (g) above and 474N there is a δ0 > 0 such that ˜ δ )) ∩ Cδ ) < µ((E4H(γ µ(B(0, δ) ∩ E) ≥ µB(0, δ) \ E) ≥

²r µCδ , 4βr−1 (3r)r

1 µB(0, δ), 2(3r)r βr 1 µB(0, δ) 2(3r)r βr

whenever 0 < δ ≤ δ0 . ?? Suppose, if possible, that 0 < δ ≤ δ0 and γδ ≤ −²δ. Then ˜ δ )) µ(E ∩ B(0, ²δ)) ≤ µ(E ∩ Cδ \ H(γ <

²r µCδ 4βr−1 (3r)r

=

1 µB(0, ²δ) 2βr (3r)r

which is impossible. X X So −²δ ≤ γδ for all δ ∈ ]0, δ0 ]. In the same way, γδ ≤ ²δ for all δ ∈ ]0, δ0 ]. As ² is arbitrary, we have the result. Q Q (i) This means that limδ↓0

˜ µ((E4H(0))∩C δ) µCδ

≤ limδ↓0

˜ δ ))∩Cδ ) µ((E4H(γ µCδ

+ limδ↓0

γδ δ

= 0,

and limδ↓0

˜ µ((E4H(0))∩B δ) µBδ

≤

2βr−1 βr

limδ↓0

˜ µ((E4H(0))∩C δ) µCδ

= 0.

(j) Thus v is a Federer exterior normal to E at 0 if ψE (0) = v. For the general case, let S be an orthogonal matrix such that SψE (y) = v, and set T (x) = S(x − y) for every x, so that T (y) = 0. The point is of course that 0 ∈ ∂ $ T [E] and ψT [E] (0) = v. P P Because λ∂T [E] = λ∂E T −1 and SψE T −1 is an outward-normal function for T [E] (474H),

666


1 λ∂T [E] B(0,δ)

474Q

Z kSψE T −1 (x) − vkλ∂T [E] (x) B(0,δ) Z 1 = ∂ −1 kSψE T −1 (x) − vk(λ∂E T −1 )(dx) λE T [B(0,δ)] T [B(y,δ)] Z 1 = ∂ kSψE (x) − vkλ∂E (dx) λE B(y,δ)

B(y,δ)

(by 235I, because T −1 is inverse-measure-preserving for λ∂E T −1 and λ∂E ) Z 1 = ∂ kψE (x) − S −1 vkλ∂E (dx) λE B(y,δ) B(y,δ) Z 1 = ∂ kψE (x) − ψE (y)kλ∂E (dx) → 0 λE B(y,δ)

B(y,δ)

as δ ↓ 0. Q Q Now consider ˜ H = {x : (x − y) .ψE (y) ≤ 0} = {x : T (x) . v ≤ 0} = T −1 [H(0)]. Then µ((H4E)∩B(y,δ)) µB(y,δ)

=

˜ µ((H(0)4T [E])∩B(0,δ)) µB(0,δ)

→0

as δ ↓ 0, so ψE (y) is a Federer exterior normal to E at y, as required. 474R Corollary Let E ⊆ Rr be a set with locally finite perimeter, and λ∂E its perimeter measure. Let y be any point of the reduced boundary of E. Then limδ↓0

λ∂E B(y,δ) βr−1 δ r−1

= 1.

proof (a) Set v = ψE (y) and H = {x : (x − y) . v ≤ 0}, as in 474Q. Now

R

H∩∂B(y,δ)

1 δ

v . (x − y)ν(dx) = −βr−1 δ r−1

for almost every δ > 0. P P Set φ(x) = v for every x ∈ Rr . By 474I, φ is an outward-normal function for H, so 474M tells us that, for almost every δ > 0, Z

Z

1 δ

v . (x − y) ν(dx) = H∩∂B(y,δ)

Z div φ dµ −

H∩B(y,δ)

B(y,δ)

v .v dλ∂H

= −λ∂H B(y, δ) = −ν(B(y, δ) ∩ ∂H) (using the identification of λ∂H in 474I) = −βr−1 δ r−1 (identifying ν on the hyperplane ∂H with Lebesgue measure on Rr−1 , as usual). Q Q (b) Now, given ² > 0, there is a δ0 > 0 such that whenever 0 < δ ≤ δ0 there is an η such that δ ≤ η ≤ δ(1 + ²) and |λ∂E B(y, η) − βr−1 η r−1 | ≤ ²δ r−1 . P P Let ζ > 0 be such that ζ(1 +

5π βr−2 )(1 + ²)r r

≤ ²2 .

By 474N(iv) and 474Q and the definition of ψE , there is a δ0 > 0 such that λ∂E B(y, δ) ≤ 5πβr−2 δ r−1 , µ((E4H) ∩ B(y, δ)) ≤ ζδ r ,

474S


R B(y,δ)

667

kψE (x) − vkλ∂E (dx) ≤ ζλ∂E B(y, δ)

whenever 0 < δ ≤ (1 + ²)δ0 . Take 0 < δ ≤ δ0 . Then, for almost every η > 0, we have

R

v . ψE (x) λ∂E (dx) + B(y,η)

R

E∩∂B(y,η)

1 η

v . (x − y) ν(dx) = 0

by 474M, applied with φ ≡ v. Putting this together with (a), we see that, for almost every η ∈ ]0, δ0 ], |λ∂E B(y, η) − βr−1 η r−1 | Z ≤ kψE (x) − vkλ∂E (dx) B(y,η) Z Z ∂ v . ψE (x) λE (dx) + +| B(y,η)

≤

H∩∂B(y,η)

ζλ∂E B(y, η) Z

+|

1 η

v . (x − y) ν(dx)| Z

1 1 v . (x − y) ν(dx) − v . (x − y) ν(dx)| η η H∩∂B(y,η) E∩∂B(y,η)

≤ 5πβr−2 ζη r−1 + ν((E4H) ∩ ∂B(y, η)). Integrating with respect to η, we have Z

δ(1+²) 0

|λ∂E B(y, η) − βr−1 η r−1 |dη ≤

5π βr−2 ζδ r (1 + ²)r r

+ µ((E4H) ∩ B(y, δ(1 + ²)))

5π βr−2 ζδ r (1 + ²)r r

+ ζδ r (1 + ²)r ≤ ²2 δ r

(using 265G, as usual) ≤

by the choice of ζ. But this means that there must be some η ∈ [δ, δ(1 + ²)] such that |λ∂E B(y, η) − βr−1 η r−1 | ≤ ²δ r−1 ≤ ²η r−1 . Q Q (c) Now we see that λ∂E B(y, δ) ≤ λ∂E B(y, η) ≤ (βr−1 + ²)η r−1 ≤ (βr−1 + ²)(1 + ²)r−1 δ r−1 . But by the same argument we have an ηˆ ∈ [

δ , δ] 1+²

such that |λ∂E B(y, ηˆ) − βr−1 ηˆr−1 | ≤ ²ˆ η r−1 , so that

λ∂E B(y, δ) ≥ λ∂E B(y, ηˆ) ≥ (βr−1 − ²)ˆ η r−1 ≥ (βr−1 − ²)(1 + ²)1−r δ r−1 . Thus, for every δ ∈ ]0, δ0 ], (βr−1 − ²)(1 + ²)1−r δ r−1 ≤ λ∂E B(y, δ) ≤ (βr−1 + ²)(1 + ²)r−1 δ r−1 . As ² is arbitrary, limδ↓0

λ∂E B(y,δ) δ r−1

= βr−1 ,

as claimed. 474S The Compactness Theorem Let Σ be the algebra of Lebesgue measurable subsets of Rr , and give it the topology Tm of convergence in measure defined by the pseudometrics ρH (E, F ) = µ((E4F ) ∩ H) for measurable sets H of finite measure (cf. §§245 and 323). Then (a) per : Σ → [0, ∞] is lower semi-continuous. (b) For any γ < ∞, {E : E ∈ Σ, per E ≤ γ} is compact. proof (a) Let hEn in∈N be any Tm -convergent sequence in Σ with limit E R∈ Σ. If φ : Rr → B(0, 1) is a Lipschitz function with compact support, then div φ is integrable, so F 7→ F div φ dµ is truly continuous (225A), and

668


|

R E

div φ dµ| = limn→∞ |

R En

474S

div φ dµ| ≤ supn∈N per En .

As φ is arbitrary, per E ≤ supn∈N per En . This means that {E : per E ≤ γ} is sequentially closed, therefore closed (4A2Ld), for any γ, and per is lower semi-continuous. A = {E : per E ≤ γ}. Let us say that a ‘dyadic cube’ is a set expressible in the form D = Q (b) Set −n −n [2 k (ki + 1)[ where n, k0 , . . . , kr−1 ∈ Z. i, 2 i
2cλ∂E Bi ≥ min(µ(Bi ∩ E), µ(Bi \ E))(r−1)/r ((474Lb)) ≥ min(µ(Di ∩ E), µ(Di \ E))(r−1)/r ≥ (²µDi )(r−1)/r ≥ 2−n(r−1) ². So

µG(E, n, ²) = 2−nr k ≤ ≤

2c 2n ²

k X

λ∂E Bi

i=0

√ 2c(1+ r)r ∂ λE (Rr ) 2n ²

=

c1 γ. 2n ²

Q Q

(ii) Now let hEn in∈N be any sequence in A. Then we can find a subsequence hEn0 in∈N such that whenever n ∈ N, D is a dyadic cube of side length 2−n meeting B(0, n), and i, j ≥ n, then |µ(D ∩ Ei0 ) − µ(D ∩ Ej0 )| ≤ 1 µD. (n+1)r+2

Now µ((Ei0 4Ej0 ) ∩ B(0, n)) ≤

√ 3βr (n+ r)r (n+1)r+2

+ 2−n (n + 1)r+1 c1 γ

−n whenever n ≥ 1 and i, j ≥ n. P P Let E be the√set of dyadic meeting B(0, n); then S cubes of side √ rlength 2 −n every member of E is included in B(0, n+2 r), so µ( E) ≤ βr (n+ r) . Let E1 be the collection of those

dyadic cubes of side length 2−n included in G(Ei0 , n, and µ(Ej0 ∩ D) ≤ µ(D \ Ej0 ) ≤

2 µD (n+1)r+2

1 ). (n+1)r+2

and µ((Ei0 4Ej0 ) ∩ D) ≤

2 µD (n+1)r+2

If D ∈ E \ E1 , either µ(Ei0 ∩ D) ≤

3 µD, (n+1)r+2

or µ(D \ Ei0 ) ≤

3

and µ((Ei0 4Ej0 ) ∩ D) ≤ µD. So (n+1)r+2 X µ((Ei0 4Ej0 ) ∩ B(0, n)) ≤ µ((Ei0 4Ej0 ) ∩ D) D∈E

≤

X

D∈E\E1

≤ ≤

µ((Ei0 4Ej0 ) ∩ D) + µ(

[

E1 )

[ 3 1 µ( E) + µG(Ei0 , n, ) r+2 (n+1) (n+1)r+2 √ 3βr (n+ r)r + 2−n (n + 1)r+2 c1 γ, (n+2)r+1

1 µD (n+1)r+2

1 µD (n+1)r+2

and

474 Notes

The essential boundary

669

as claimed. Q Q S

(iii) means that T This 0 i∈N j≥i Ej , then

P∞ i=0

0 µ((Ei0 4Ei+1 ) ∩ B(0, n)) is finite for each n ∈ N, so that if we set E =

µ((E4Ei0 ) ∩ B(0, n)) ≤

P∞ j=i

0 ) ∩ B(0, n)) → 0 µ((Ej0 4Ej+1

as i → ∞ for every n ∈ N. It follows that limi→∞ ρH (E, Ei0 ) = 0 whenever µH < ∞ (see the proofs of 245Eb and 323Gb). Thus we have a subsequence hEi0 ii∈N of the original sequence hEi ii∈N which is convergent for the topology Tm of convergence in measure. By (a), its limit belongs to A. But since Tm is pseudometrizable (245Eb/323Gb), this is enough to show that A is compact for Tm (4A2Lf). 474X Basic exercises (a) Show directly from 474E that if E ⊆ Rr has locally finite perimeter and F ⊆ Rr is such that µ(E4F ) = 0, then F has locally finite perimeter, λ∂F = λ∂E , and any outward-normal function for E is an outward-normal function for F . (b) Show that for any E ⊆ Rr with locally finite perimeter, its reduced boundary is a Borel set. (c) Show that if E ⊆ Rr has locally finite perimeter, then ∂ $ E ⊆ ∂E. (Hint: 474N(i)-(ii).) (d) Let E ⊆ Rr be a set with locally finite perimeter, and y ∈ ∂ $ E. Show that limδ↓0

µ(E∩B(y,δ)) µB(y,δ)

1 2

= .

(e) In the proof of 474R, use 265E to show that

R H∩∂B(y,δ)

1 δ

v . (x − y)ν(dx) = −βr−1 δ r−1

for every δ > 0. 474Y Further exercises (a) In 474E, explain how to interpret the pair (ψ, λ∂E ) as a vector measure θE : B → Rr , where B is the Borel σ-algebra of Rr , in such a way that we have R(definition: 393O) R div φ dµ = φ dθE for Lipschitz functions φ with compact support. E 474 Notes and comments When we come to the Divergence Theorem itself in the next section, it will be nothing but a repetition of Theorem 474E with the perimeter measure and the outward-normal function properly identified. The idea of the indirect approach here is to start by defining the pair (ψE , λ∂E ) as a kind of ‘distributional derivative’ of the set E. I take the space to match the details with the language of the rest of this treatise, Rbut really 474E amounts to nothing more than the Riesz representation theorem; since the functional φ 7→ E div φ dµ is linear, and we restrict attention to sets E for which it is continuous in an appropriate sense (and can therefore be extended to arbitrary continuous functions φ with compact support), it must be representable by a (vector) measure, as in 474Ya. For the process to be interesting, we have to be able to identify at least some of the appropriate sets E with their perimeter measures and outwardnormal functions. Half-spaces are straightforward enough (474I), and 474Q tells us what the outward-normal functions have to be; but for a proper description of the family of sets with locally finite perimeter we must wait until the next section. I see no quick way to show from the results here that (for instance) the union of two sets with finite perimeter again has finite perimeter. I have tried to find the shortest path to the Divergence Theorem itself, and have not attempted to give ‘best’ results in the intermediate material. In particular, in the isoperimetric inequality 474La, I show only that the measure of a set E is controlled by the magnitude of its perimeter measure. Simple scaling arguments show that if there is any such control, then it must be of the form γ(µE)(r−1)/r ≤ per E; the 1/r identification of the best constant γ as rβr , giving equality for balls, is the real prize, to which I shall come in 476L. I leave the Compactness Theorem (474S) to the end of the section because it is off the line I have chosen to the Divergence Theorem (though it can be used to make the proof of 474Q more transparent; see Evans & Gariepy 92, 5.7.2). I have expressed 474S in terms of the topology of convergence in measure on the algebra of Lebesgue measurable sets. But since the perimeter of a measurable set E is not altered if we

670


474 Notes

change E by a negligible set, ‘perimeter’ can equally well be regarded as a function defined on the measure algebra, in which case 474S becomes a theorem about the usual topology of the measure algebra of Lebesgue measure, as described in §323.

475 The essential boundary The principal aim of this section is to translate Theorem 474E into geometric terms. We have already identified the Federer exterior normal as an outward-normal function (474Q), so we need to find a description of perimeter measures. Most remarkably, it turns out that these are nothing but simple indefinite-integral measures over normalized Hausdorff measures (475G). This description needs the concept of ‘essential boundary’ (475B). In order to complete the programme, we need to be able to determine which sets have ‘locally finite perimeter’; there is a natural criterion in the same language (475L). We now have all the machinery for a direct statement of the Divergence Theorem (for Lipschitz functions) which depends on nothing more advanced than the definition of Hausdorff measure (475N). (The definitions, at least, of ‘Federer exterior normal’ and ‘essential boundary’ are elementary.) This concludes the main work of the middle part of this chapter. But since we are now within reach of a reasonably direct proof of a fundamental fact about the r − 1-dimensional Hausdorff measure of the boundaries of subsets of Rr (475Q), I continue up to Cauchy’s Perimeter Theorem and the Isoperimetric Theorem for convex sets (475S, 475T). 475A Notation As in the last two sections, r will be an integer (greater than or equal to 2, unless explicitly permitted to take the value 1). µ will be Lebesgue measure on Rr ; I will sometimes write µr−1 for Lebesgue measure on Rr−1 and µ1 for Lebesgue measure on R. βr = µB(0, 1) will be the measure of the unit ball in Rr , and Sr−1 = ∂B(0, 1) will be the unit sphere. ν will be normalized r − 1-dimensional Hausdorff measure on Rr (265A), that is, ν = 2−r+1 βr−1 µH,r−1 , where µH,r−1 is r − 1-dimensional Hausdorff measure on Rr . I will take it for granted that x ∈ Rr has coordinates (ξ1 , . . . , ξr ). If E ⊆ Rr has locally finite perimeter (474D), λ∂E will be its perimeter measure (474F), ∂ $ E its reduced boundary (474G) and ψE its canonical outward normal (474G). 475B The essential boundary (In this paragraph I allow r = 1.) Let A ⊆ Rr be any set. The essential closure of A is the set cl*A = {x : lim supδ↓0

µ∗ (B(x,δ)∩A) µB(x,δ)

> 0}.

µ∗ (B(x,δ)∩A) µB(x,δ)

= 1}.

Similarly, the essential interior of A is the set int*A = {x : lim inf δ↓0

(If A is Lebesgue measurable, this is the lower Lebesgue density of A, as defined in 341E.) Finally, the essential boundary ∂*A of A is the difference cl*A \ int*A. 475C Lemma (In this lemma I allow r = 1.) Let A, A0 ⊆ Rr . (a) int A ⊆ int*A ⊆ cl*A ⊆ A, cl*A = Rr \ int*(Rr \ A),

∂*A ⊆ ∂A,

∂*(Rr \ A) = ∂*A.

(b) If A \ A0 is negligible, then cl*A ⊆ cl*A0 and int*A ⊆ int*A0 ; in particular, if A itself is negligible, cl*A, int*A and ∂*A are all empty. (c) int*A, cl*A and ∂*A are Borel sets. (d) cl*(A ∪ A0 ) = cl*A ∪ cl*A0 and int*(A ∩ A0 ) = int*A ∩ int*A0 , so ∂*(A ∪ A0 ), ∂*(A ∩ A0 ) and ∂*(A4A0 ) are all included in ∂*A ∪ ∂*A0 . (e) cl*A ∩ int*A0 ⊆ cl*(A ∩ A0 ), ∂*A ∩ int*A0 ⊆ ∂*(A ∩ A0 ) and ∂*A \ cl*A0 ⊆ ∂*(A ∪ A0 ). (f) ∂*(A ∩ A0 ) ⊆ (cl*A0 ∩ ∂A) ∪ (A ∩ ∂*A0 ). (g) If E ⊆ Rr is Lebesgue measurable, then E4int*E, E4cl*E and ∂*E are Lebesgue negligible.

475C


671

(h) A is Lebesgue measurable iff ∂*A is Lebesgue negligible. proof (a) It is obvious that int A ⊆ int*A ⊆ cl*A ⊆ A, so that ∂*A ⊆ ∂A. Since µ∗ (B(x,δ)∩A) µB(x,δ)

+

µ∗ (B(x,δ)\A) µB(x,δ)

=1

for every x ∈ Rr and every δ > 0 (413Ec), Rr \ int*A = cl*(Rr \ A). It follows that ∂*(Rr \ A) = cl*(Rr \ A)4int*(Rr \ A) = (Rr \ int*A)4(Rr \ cl*A) = int*A4cl*A = ∂*A. (b) If A \ A0 is negligible, then µ∗ (B(x, δ) ∩ A) ≤ µ∗ (B(x, δ) ∩ A0 ),

µ∗ (B(x, δ) ∩ A) ≤ µ∗ (B(x, δ) ∩ A0 )

for all x and δ, so int*A ⊆ int*A0 and cl*A ⊆ cl*A0 . (c) The point is just that (x, δ) 7→ µ∗ (A ∩ B(x, δ)) is continuous. P P For any x, y ∈ Rr and δ, η ≥ 0 we have |µ∗ (A ∩ B(y, η)) − µ∗ (A ∩ B(x, δ))| ≤ µ(B(y, η)4B(x, δ)) = 2µ(B(x, δ) ∪ B(y, η)) − µB(x, δ) − µB(y, η) ¡ ¢ ≤ βr 2(max(δ, η) + kx − yk)r − δ r − η r → 0 as (y, η) → (x, δ). Q Q So x 7→ lim supδ↓0


= inf α∈Q,α>0 supβ∈Q,0<β≤α

1 µ∗ (A ∩ B(x, β)) βr β r

is Borel measurable, and cl*A = {x : lim supδ↓0


= 1}

is a Borel set. Accordingly int*A = Rr \ cl*(Rr \ A) and ∂*A are also Borel sets. (d) For any x ∈ Rr , lim sup δ↓0

µ∗ ((A∪A0 )∩B(x,δ)) µB(x,δ)

≤ lim sup δ↓0

≤ lim sup δ↓0

µ∗ (A∩B(x,δ)) µB(x,δ) µ(A∩B(x,δ)) µB(x,δ)

+

µ∗ (A0 ∩B(x,δ)) µB(x,δ)

+ lim sup δ↓0

µ(A0 ∩B(x,δ)) , µB(x,δ)

so cl*(A ∪ A0 ) ⊆ cl*A ∪ cl*A0 . Of course cl*A ∪ cl*A0 ⊆ cl*(A ∪ A0 ), so we have equality. Accordingly int*(A ∩ A0 ) = Rr \ cl*((Rr \ A) ∪ (Rr \ A0 )) = int*A ∩ int*A0 . Since of course int*(A ∪ A0 ) ⊇ int*A ∪ int*A0 , ∂*(A ∪ A0 ) ⊆ ∂*A ∪ ∂*A0 . Now ∂*(A ∩ A0 ) = ∂*(Rr \ (A ∩ A0 )) ⊆ ∂*(Rr \ A) ∪ ∂*(Rr \ A0 )) = ∂*A ∪ ∂*A0 and ∂*(A4A0 ) ⊆ ∂*(A ∩ (Rr \ A0 )) ∪ ∂*(A0 ∩ (Rr \ A)) ⊆ ∂*A ∪ ∂*A0 . (e) If x ∈ cl*A ∩ int*A0 , then x ∈ / cl*(Rr \ A0 ) so x ∈ / cl*(A \ A0 ). But from (d) we know that x ∈ cl*(A ∩ A0 ) ∪ cl*(A \ A0 ), so x ∈ cl*(A ∩ A0 ). Now

672


475C

∂*A ∩ int*A0 = (cl*A ∩ int*A0 ) \ int*A ⊆ cl*(A ∩ A0 ) \ int*(A ∩ A0 ) = ∂*(A ∩ A0 ), ∂*A \ cl*A0 = ∂*(Rr \ A) ∩ int*(Rr \ A0 ) ⊆ ∂*((Rr \ A) ∩ (Rr \ A0 )) = ∂*(Rr \ (A ∪ A0 )) = ∂*(A ∪ A0 ). (f ) If x ∈ ∂*(A∩A0 )∩∂A, then of course x ∈ cl*(A∩A0 ) ⊆ cl*A, so x ∈ cl*A∩∂A. If x ∈ ∂*(A∩A0 )\∂A, then surely x ∈ A, so x ∈ int A. But this means that µ∗ (B(x, δ) ∩ A0 ) = µ∗ (B(x, δ) ∩ A ∩ A0 ),

µ∗ (B(x, δ) ∩ A0 ) = µ∗ (B(x, δ) ∩ A ∩ A0 )

for all δ small enough, so x ∈ ∂*A0 and x ∈ A ∩ ∂*A0 . (g) Applying 472D to µ and χE, we see that ∂*E ⊆ (E4cl*E) ∪ (E4int*E) ⊆ {x : χE(x) 6= lim δ↓0

µ(E∩B(x,δ)) } µB(x,δ)

are all µ-negligible. (h) If A is Lebesgue measurable then (g) tells us that ∂*A is negligible. If ∂*A is negligible, let E be a measurable envelope of A. Then µ(E ∩ B(x, δ)) = µ∗ (A ∩ B(x, δ)) for all x and δ, so cl*E = cl*A. Similarly, if F is a measurable envelope of Rr \ A, then cl*F = cl*(Rr \ A) = Rr \ int*A (using (a)). Now (g) tells us that µ(E ∩ F ) = µ(cl*E ∩ cl*F ) = µ(cl*A \ int*A) = 0. But now A \ E and E \ A ⊆ (E ∩ F ) ∪ ((Rr \ A) \ F ) are Lebesgue negligible, so A is Lebesgue measurable. 475D Lemma Let E ⊆ Rr be a set with locally finite perimeter, and ∂ $ E its reduced boundary (474G). Then ∂ $ E ⊆ ∂*E and ν(∂*E \ ∂ $ E) = 0. proof (a) By 474N(i), ∂ $ E ⊆ cl*E; by 474N(ii), ∂ $ E ∩ int*E = ∅; so ∂ $ E ⊆ ∂*E. (b) For any y ∈ ∂*E, lim supδ↓0 ¤ ¤ P P We have an ² ∈ 0, 12 such that lim inf δ↓0 Since the function δ 7→



1 λ∂ B(y, δ) δ r−1 E

< 1 − ²,

> 0.

lim supδ↓0


> ².

is continuous, there is a sequence hδn in∈N in ]0, ∞[ such that limn→∞ δn = 0

and ²µB(y, δn ) ≤ µ(E ∩ B(y, δn )) ≤ (1 − ²)µB(y, δn ) for every n. Now from 474Lb we have (²βr )(r−1)/r δnr−1 = (²µB(y, δn ))(r−1)/r ≤ min(µ(B(y, δn ) ∩ E), µ(B(y, δn ) \ E))(r−1)/r ≤ 2cλ∂E B(y, δn ) for every n, where c is the constant there. But this means that lim supδ↓0

1 λ∂ B(y, δ) δ r−1 E

≥ lim supn→∞

1 λ∂E B(y, δn ) r−1 δn

≥

1 (²βr )(r−1)/r 2c

> 0. Q Q

475F


673

(c) Let ² > 0. Set F² = {y : y ∈ Rr \ ∂ $ E, lim supδ↓0

1 λ∂ B(y, δ) δ r−1 E

> ²}.

Because ∂ $ E is λ∂E -conegligible (474G), λ∂E F² = 0. So there is an open set G ⊇ F² such that λ∂E G ≤ ²2 (256Bb). Let δ > 0. Let I be the family of all those non-singleton closed balls B ⊆ G such that diam B ≤ δ and λ∂E B ≥ 2−r+1 ²(diam B)r−1 . Then every point of F² is the centre of arbitrarily small members of I. By Besicovitch’s Covering Lemma (472B), there is a family hIk ik<5r of disjoint countable subsets of I such S S that I ∗ = k<5r Ik covers F² . Now P

r−1 B∈I ∗ (diam B)

≤

P

P

k<5r

B∈Ik

2r−1 ∂ λE B ²

≤

5r 2r−1 ∂ λE G ²

≤ 5r 2r−1 ².

As δ is arbitrary, µ∗H,r−1 F² is at most 5r 2r−1 ² (264Fb/471Dc) and ν ∗ F² ≤ 5r βr−1 ². As ² is arbitrary, ∂*E \ ∂ $ E ⊆ {y : y ∈ Rr \ ∂ $ E, lim supδ↓0

1 λ∂ B(y, δ) δ r−1 E

> 0}

is ν-negligible, as claimed. 475E Lemma Let E ⊆ Rr be a set with locally finite perimeter. (a) If A ⊆ ∂ $ E, then ν ∗ A ≤ (λ∂E )∗ A. (b) If A ⊆ Rr and νA = 0, then λ∂E A = 0. proof (a) Given ², δ > 0 let I be the family of non-trivial closed balls B ⊆ Rr of diameter at most δ such 1

that βr−1 ( diam B)r−1 ≤ (1 + ²)λ∂E B. By 474R, every point of A is the centre of arbitrarily small members 2 S P of I. By 472C, there is a countable family I1 ⊆ I such that A ⊆ I1 and B∈I1 λ∂E B ≤ (λ∂E )∗ A + ². But this means that P 2r−1 r−1 ≤ (1 + ²) ((λ∂E )∗ A + ²). B∈I1 (diam B) βr−1

As δ is arbitrary, ν ∗A =

βr−1 ∗ µ A 2r−1 H,r−1

≤ (1 + ²)((λ∂E )∗ A + ²).

As ² is arbitrary, we have the result. (b) For n ∈ N, set An = {x : x ∈ A, λ∂E B(x, δ) ≤ 2βr−1 δ r−1 whenever 0 < δ ≤ 2−n }. Now, 0, there is a sequence hDi ii∈N of sets covering An such that diam Di ≤ 2−n for every i and P∞ given ² >r−1 ≤ ². Passing over the trivial case An = ∅, we may suppose that for each i ∈ N there is i=0 (diam Di ) an xi ∈ An ∩ Di , so that Di ⊆ B(xi , diam Di ) and (λ∂E )∗ An ≤

∞ ∞ X X (λ∂E )∗ Di ≤ λ∂E B(xi , diam Di ) i=0

≤

∞ X

i=0

2βr−1 (diam Di )r−1 ≤ 2βr−1 ².

i=0

As ² is arbitrary, λ∂E An = 0. And this is true for every n. As A \ 474R, so is A.

S n∈N

An is λ∂E -negligible, by 474G and

475F Lemma Let E ⊆ Rr be a set with locally finite perimeter, and ² > 0. Then λ∂E is inner regular with respect to the family E = {F : F ⊆ Rr is Borel, λ∂E F ≤ (1 + ²)νF }. proof (a) We need some elementary bits of geometry. (i) If x ∈ Rr , δ > 0, α ≥ 0 and v ∈ Sr−1 , then

674


475F

µ{z : z ∈ B(x, δ), |(z − x) . v| ≤ α} ≤ 2αβr−1 δ r−1 . P P Translating and rotating, if necessary, we can reduce to the case x = 0, v = (0, . . . , 1). In this case we are looking at {z : kzk ≤ δ, |ζr | ≤ α} ⊆ {u : u ∈ Rr−1 , kuk ≤ δ} × [−α, α] which has measure 2αβr−1 δ r−1 . Q Q (ii) If x ∈ Rr , δ > 0, 0 < η ≤ 21 , v ∈ Sr−1 , H = {z : z . v ≤ α} and 1 2

|µ(H ∩ B(x, δ)) − µB(x, δ)| < 2−r+1 βr−1 ηδ r , then |x. v − α| ≤ ηδ. P P Again translating and rotating if necessary, we may suppose that x = 0 and v = (0, . . . , 1). Set H0 = {z : ζr ≤ 0}. ?? If α > ηδ, then H ∩ B(0, δ) includes 1 2

H0 ∩ B(0, δ) ∪ {u : u ∈ Rr−1 , kuk ≤ δ} × [0, α], so 1 2

µ(H ∩ B(0, δ)) − µB(0, δ) = µ(H ∩ B(0, δ)) − µ(H0 ∩ B(0, δ)) ≥ 2−r+1 βr−1 δ r−1 α > 2−r+1 βr−1 δ r η, contrary to hypothesis. X X Similarly, if α < −ηδ, then H ∩ B(0, δ) is included in 1 2

H0 ∩ B(0, δ) \ {u : u ∈ Rr−1 , kuk ≤ δ} × ]α, 0], so 1 µB(0, δ) − µ(H 2

∩ B(0, δ)) = µ(H0 ∩ B(0, δ)) − µ(H ∩ B(0, δ)) ≥ 2−r+1 βr−1 δ r−1 |α| > 2−r+1 βr−1 δ r η,

which is equally impossible. So |α| ≤ ηδ. Q Q (b) Let F be such that λ∂E F > 0. Let η, ζ > 0 be such that η < 1,

(1+η)2 (1−η)r−1

≤ 1 + ²,

2(1 + 2r )βr ζ < 2−r βr−1 η.

Because ∂ $ E is λ∂E -conegligible (474G), λ∂E (F ∩ ∂ $ E) > 0. Because λ∂E is a Radon measure (474E) and ψE : ∂ $ E → Sr−1 is dom(λ∂E )-measurable (474G), there is a compact set K1 ⊆ F ∩ ∂ $ E such that λ∂E K1 > 0 and ψE ¹K1 is continuous, by Lusin’s theorem (418J). For x ∈ ∂ $ E, set Hx = {z : (z − x) . ψE (x) ≤ 0}. The function (x, δ) 7→ µ((E4Hx ) ∩ B(x, δ)) : K1 × [0, ∞[ → R is Borel measurable. P P Take a Borel set E 0 such that µ(E4E 0 ) = 0. Then {(x, δ, z) : x ∈ K1 , z ∈ (E 0 4Hx ) ∩ B(x, δ)} is a Borel set in Rr × [0, ∞[ × Rr , so its sectional measure is a Borel measurable function, by 252P. Q Q For each x ∈ K1 , limn→∞

µ((E4Hx )∩B(x,2−n )) µB(x,2−n )

=0

(474P). So there is an n0 ∈ N such that λ∂E F1 > 0, where F1 is the Borel set {x : x ∈ K1 , µ((E4Hx ) ∩ B(x, 2−n )) ≤ ζµB(x, 2−n ) for every n ≥ n0 }. Let K2 ⊆ F1 be a compact set such that λ∂E K2 > 0. For each n ∈ N, the function x 7→ λ∂E B(x, 2−n )) = λ∂E (x + B(0, 2−n )

475F


675

is Borel measurable (444Fe). Let y ∈ K2 be such that λ∂E (K2 ∩ B(y, δ)) > 0 for every δ > 0 (cf. 411Nd). Set v = ψE (y). Let n ≥ n0 be so large that 2βr−1 kψE (x) − vk ≤ βr ζ whenever x ∈ K1 and kx − yk ≤ 2−n . (c) We have |(x − z) . v| ≤ ηkx − zk whenever x, z ∈ K2 ∩ B(y, 2−n−1 ). P P If x = z this is trivial. Otherwise, let k ≥ n be such that 2−k−1 ≤ kx − zk ≤ 2−k , and set δ = 2−k . Set Hx0 = {w : (w − x) . v ≤ 0},

Hz0 = {w : (w − z) . v ≤ 0}.

Since |(w − x) .v − (w − x) . ψE (x)| ≤ 2δkψE (x) − vk whenever w ∈ B(x, 2δ), (Hx 4Hx0 ) ∩ B(x, 2δ) ⊆ {w : w ∈ B(x, 2δ), |(w − x) . v| ≤ 2δkψE (x) − vk} has measure at most 4δkψE (x) − vkβr−1 (2δ)r−1 ≤ 2δβr ζ(2δ)r−1 = ζµB(x, 2δ), by (a-i). So µ((E4Hx0 ) ∩ B(x, 2δ)) ≤ µ((E4Hx ) ∩ B(x, 2δ)) + µ((Hx 4Hx0 ) ∩ B(x, 2δ)) ≤ 2ζµB(x, 2δ) because k ≥ n0 and x ∈ F1 . Similarly, µ((E4Hz0 ) ∩ B(z, δ)) ≤ 2ζµB(z, δ). Now observe that because kx − zk ≤ δ, B(z, δ) ⊆ B(x, 2δ) and µ((E4Hx0 ) ∩ B(z, δ)) ≤ 2ζµB(x, 2δ) = 2r+1 ζµB(z, δ), and µ((Hx0 4Hz0 ) ∩ B(z, δ)) ≤ µ((E4Hx0 ) ∩ B(z, δ)) + µ((E4Hz0 ) ∩ B(z, δ)) ≤ (2 + 2r+1 )ζµB(z, δ). 1 2

Since µ(Hz0 ∩ B(z, δ)) = µB(z, δ), 1 2

|µ(Hx0 ∩ B(z, δ)) − µB(z, δ)| ≤ 2(1 + 2r )ζµB(z, δ) < 2−r βr−1 ηδ r , and (using (a-ii) above) 1 2

Q |(x − z) . v| ≤ ηδ ≤ ηkx − vk. Q (d) Let V be the hyperplane {w : w .v = 0}, and let T : K2 → V be the orthogonal projection, that is, T (x) = x − (x .v)v for every x ∈ K2 . Then (c) tells us that if x, z ∈ F1 , kT (x) − T (z)k ≥ kx − zk − |(x − z). v| ≥ (1 − η)kx − zk. Because η < 1, T is injective. Consider the compact set T [K2 ]. The inverse T −1 of T is and νK2 > 0 (by 475Eb), so νT [K2 ] ≥

1 νF1 (1−η)r−1

1 -Lipschitz, 1−η

> 0 (264G/471J). Let G ⊇ T [K2 ] be an open set such

that ν(G ∩ V ) ≤ (1 + η)νT [K2 ]. (I am using the fact that the subspace measure νV induced by ν on V is a copy of Lebesgue measure on Rr−1 , so is a Radon measure.) Let I be the family of non-trivial closed balls B ⊆ G such that (1 − η)r−1 λ∂E T −1 [B] ≤ (1 + η)ν(B ∩ V ). Then every point of T [K2 ] is the centre of arbitrarily small members of I. P P If x ∈ K2 and δ0 > 0, there is a δ ∈ ]0, δ0 ] such that B(T (x), δ) ⊆ G and λ∂E B(x, δ) ≤ (1 + η)βr−1 δ r−1 (474R). Now consider B = B(T (x), (1 − η)δ). Then T −1 [B] ⊆ B(x, δ), so λ∂E T −1 [B] ≤ λ∂E B(x, δ) ≤ (1 + η)βr−1 δ r−1 =

1+η ν(B (1−η)r−1

∩ V ). Q Q

∼ r−1 S By 261B/472C, applied in V = R , there is a countable disjoint family I0 ⊆ I such that ν(T [K2 ] \ I0 ) = 0. S S Now ν(K2 \ B∈I0 T −1 [B]) = 0, because T −1 is Lipschitz, so λ∂E (K2 \ B∈I0 T −1 [B]) = 0 (475Eb again), and

676


λ∂E K2 ≤

X

λ∂E T −1 [B] ≤

B∈I0

≤

1+η ν(G ∩ V ) (1−η)r−1

1+η (1−η)r−1

≤

X

475F

ν(B ∩ V )

B∈I0

(1+η)2 νT [K2 ] (1−η)r−1

≤

(1+η)2 νK2 (1−η)r−1

(because T is 1-Lipschitz) ≤ (1 + ²)νK2 by the choice of η. Thus K2 ∈ E. (e) This shows that every non-negligible set measured by λ∂E includes a non-negligible member of E. As E is closed under disjoint unions, λ∂E is inner regular with respect to E (412Aa). 475G Theorem Let E ⊆ Rr be a set with locally finite perimeter. Then λ∂E = ν ∂*E, that is, for F ⊆ Rr , λ∂E F = ν(F ∩ ∂*E) whenever either is defined. proof (a) Suppose first that F is a Borel set included in the reduced boundary ∂ $ E of E. Then νF ≤ λ∂E F , by 475Ea. On the other hand, for any ² > 0 and γ < λ∂E F , there is an F1 ⊆ F such that γ ≤ λ∂E F1 ≤ (1 + ²)νF1 ≤ (1 + ²)νF , by 475F; so we must have λ∂E F = νF . (b) Now suppose that F is measured by λ∂E F . Because ∂ $ E is λ∂E -conegligible, and λ∂E is a σ-finite Radon measure, there is a Borel set F 0 ⊆ F ∩ ∂ $ E such that λ∂E (F \ F 0 ) = 0. Now νF 0 = λ∂E F 0 , by (a), and ν(F ∩ ∂ $ E \ F 0 ) = 0, by 475Ea, and ν(∂*E \ ∂ $ E) = 0, by 475D; so ν(F ∩ ∂*E) is defined and equal to λ∂E F . S (c) Let hKn in∈N be a non-decreasing sequence of compact subsets of ∂ $ E such that n∈N Kn is λ∂E S conegligible. By (b), ν(∂*E \ n∈N Kn ) = 0, while νKn = λ∂E Kn is finite for each n. For each n, the subspace measure νKn on Kn is a multiple of Hausdorff r − 1-dimensional measure on Kn (471E), so is a Radon measure (471Dh/471F), and so is (λ∂E )Kn ; since, by (b), νKn and (λ∂E )Kn agree on the Borel subsets of Kn , they are actually identical. So if F ⊆ Rr is such that ν measures F ∩ ∂*E, λ∂E will measure F ∩ Kn for every n, and therefore will measure F ; so that in this case also λ∂E F = ν(F ∩ ∂*E). 475H Proposition Let V ⊆ Rr be a hyperplane, and T : Rr → V the orthogonal projection. Suppose that A ⊆ Rr is such that νA is defined and finite, and for u ∈ V set f (u) = #(A ∩ T −1 [{u}]) if this is finite, = ∞ otherwise. Then

R V

f (u)ν(du) is defined and at most νA.

proof (a) Because ν is invariant under isometries, we can suppose that V = {x : ξr = 0}, so that T (x) = (ξ1 , . . . , ξr−1 , 0) for x = (ξ1 , . . . , ξr ). For m, n ∈ N and u ∈ V set fmn (u) = #({k : k ∈ Z, |k| ≤ 4m , A ∩ ({u} × [2−m k, 2−m (k + 1 − 2−n )[) 6= ∅}); so that f (u) = limm→∞ limn→∞ fmn (u) for every u ∈ V . (b) Suppose for the moment that A is actually a Borel set. Then T [A ∩ (Rr × [α, β[)] is always analytic (423E, 423Bb), therefore measured by ν (432A), and every fmn is measurable. Next, given m, n ∈ N and 0, there is a sequence hFi ii∈N of closed sets of diameter at most 2−m−n , covering A, such that P∞ ² > −r+1 βr−1 (diam Fi )r−1 ≤ νA + ². Now each T [Fi ] is a P compact set of diameter at most diam Fi , so i=0 2 ∞ −r+1 r−1 νT [F ] ≤ 2 β (diam F ) (264H); and if we set g = i r−1 i i=0 χT [Fi ], fmn ≤ g everywhere on V , so R R R fmn dν ≤ g dν ≤ νA + ². As ² is arbitary, fmn dν ≤ νA. Accordingly

R

f dν = limm→∞ limn→∞

R

fmn dν ≤ νA.

475I


677

(c) In general, there are Borel sets E, F such that E \ F ⊆ A ⊆ E and νF = 0, by 264Fc or 471Db. By (b),

R

so

R

R

#(E ∩ T −1 [{u}])ν(du) ≤ νE,

#(A ∩ T −1 [{u}])ν(du) is defined and equal to

R

#(F ∩ T −1 [{u}])ν(du) ≤ νF = 0,

#(E ∩ T −1 [{u}])ν(du) ≤ νA.

475I Lemma (In this lemma I allow r = 1.) Let E ⊆ Rr be a Lebesgue measurable set, and γ < µE. Then there is a compact set K ⊆ E such that µK ≥ γ and K = cl*K. proof (a) Write C ∗ for the set of dyadic (half-open) cubes in Rr , that is, sets expressible in the form Q −m ki , 2−m (ki + 1)[ where m, k0 , . . . , kr−1 ∈ Z. For m ∈ N, x ∈ Rr write C(x, m) for the dyadic i
2

µ (C(x, m) \ A) ≤ βr r

r/2 µ(B(x, 2

−m

√

µB(x, 2−m

r) \ A) √ →0 r)

as m → ∞. Q Q (b) Now, given γ < µE, choose hEn in∈N , hγn in∈N , hmn in∈N and hKn in∈N inductively, as follows. Start with E0 = E and γ0 = γ. Given that γn < µEn , let Kn ⊆ En be a compact set of measure greater than γn . Now, by (a), there is an mn ≥ n such that µ∗ En+1 > γn , where 2 3

En+1 = {x : x ∈ Kn , µ(Kn ∩ C(x, mn )) ≥ µC(x, mn )}; note that En+1 is of the form Kn ∩ measurable and

S

D for some set D of half-open cubes of side 2−mn , so that En+1 is 2 3

µ(En+1 ∩ C(x, mn )) = µ(Kn ∩ C(x, mn )) ≥ µC(x, mn ) for every x ∈ En+1 . Now set γn+1 = max(γn , µEn+1 −

1 3

· 2−mn r ),

and continue. T T At the end of the induction, set K = n∈N Kn = n∈N En . Then K is compact, K ⊆ E and µK = limn→∞ µKn = limn→∞ γn ≥ γ. If x ∈ K and n ∈ N, then √ µ(K ∩ B(x, 2−mn r)) ≥ µ(K ∩ C(x, mn )) ≥ µ(En+1 ∩ C(x, mn )) − µEn+1 + µK 2 3

≥ µC(x, mn ) − µEn+1 + γn+1 ≥

1 3

· 2−mn r ≥

√ 1 µB(x, 2−mn r), r/2 3βr r

so lim supδ↓0

µ(K∩B(x,δ)) µB(x,δ)

≥

1 3βr rr/2

>0

and x ∈ cl*K. (c) Thus K ⊆ cl*K. Since certainly cl*K ⊆ K = K, we have K = cl*K.

678


475J

475J Lemma Let E be a Lebesgue measurable subset of Rr , identified with Rr−1 × R. For u ∈ Rr−1 , set Gu = {t : (u, t) ∈ int*E}, Hu = {t : (u, t) ∈ int*(Rr \ E)} and Du = {t : (u, t) ∈ ∂*E}, so that Gu , Hu and Du are disjoint and cover R. (a) There is a µr−1 -conegligible set Z ⊆ Rr−1 such that whenever u ∈ Z, t < t0 in R, t ∈ Gu and t0 ∈ Hu , there is an s ∈ Du ∩ ]t, t0 [. (b) If E has locally finite perimeter, we can arrange further that, for every u ∈ Z, Du ∩ [−n, n] is finite for every n ∈ N, Gu and Hu are open, and Du = ∂Gu = ∂Hu , so that the constituent intervals of R \ Du lie alternately in Gu and Hu . proof (a)(i) For q ∈ Q, set fq (u) = sup(Gu ∩ ]−∞, q[), taking −∞ for sup ∅. Then f : Rr−1 → [−∞, q] is Lebesgue measurable. P P For α < q, {u : fq (u) > α} = {u : there is some t ∈ ]α, q[ such that (u, t) ∈ int*E}. Because int*E is a Borel set (475Cc), {u : fq (u) > α} is analytic (423E, 423Bc), therefore measurable (432A). Q Q P?? Suppose, if possible, otherwise. (ii) For any q ∈ Q, Wq = {u : fq (u) < q, fq (u) ∈ Gu } is negligible. P Let n ∈ N be such that {u : u ∈ Wq , fq (u) > −n} is not negligible, and set g(u) = max(−n, fq (u)) for u ∈ Rr−1 . Then g is measurable, so by Lusin’s theorem (418J) there is a Borel set F ⊆ Rr−1 such that g¹F is continuous and F1 = {u : u ∈ F ∩ Wq , fq (u) ∈ Gu , fq (u) > −n} is not negligible. Note that F1 is measurable, being the projection of the Borel set {(u, g(u)) : u ∈ F, −n < g(u) < q} ∩ int*E. By 475I, there is a non-negligible compact set K ⊆ F1 such that K ⊆ cl*K, interpreting cl*K in Rr−1 . Because g¹K is continuous, it attains its maximum at u ∈ K say. Set x = (u, fq (u)). Then, whenever 0 < δ ≤ min(n + fq (u), q − fq (u)),

B(x, 2δ) \ int*E ⊇ {(w, t) : w ∈ K ∩ V (u, δ), |t − fq (u)| ≤ δ, fq (w) < t < q} (writing V (u, δ) for {w : w ∈ Rr−1 , kw − uk ≤ δ}) ⊇ {(w, t) : w ∈ K ∩ V (u, δ), fq (u) < t ≤ fq (u) + δ} because fq (w) = g(w) ≤ g(u) = fq (u) for every w ∈ K. So, for such δ, µ(B(x, 2δ) \ int*E) ≥ δµr−1 (K ∩ V (u, δ)) =

βr−1 2r βr

·

µr−1 (K∩V (u,δ)) µr−1 V (u,δ)

· µB(x, 2δ),

and lim sup δ↓0

µ(B(x,δ)\E) µB(x,δ)

= lim sup δ↓0

≥

βr−1 2r βr

µ(B(x,2δ)\int*E) µB(x,2δ)

lim sup δ↓0

µr−1 (K∩V (u,δ)) µr−1 V (u,δ)

>0

because u ∈ cl*K; but x = (u, fq (u)) is supposed to belong to int*E. X XQ Q (iii) Similarly, setting fq0 (u) = inf(Hu ∩ ]u, ∞[), Wq0 = {u : q < fq0 (u) < ∞, fq0 (u) ∈ Hu }, S every Wq0 is negligible. So Z = Rr−1 \ q∈Q (Wq ∪ Wq0 ) is µr−1 -conegligible. Now if u ∈ Z, t ∈ Gu and t0 ∈ Hu , where t < t0 , there is some s ∈ ]t, t0 [ ∩ Du . P P?? Suppose, if possible, otherwise. Since, by hypothesis, neither t nor t0 belongs to Du , Du ∩ [t, t0 ] = ∅. Note that, because u ∈ Z, s = inf(Gu ∩ ]s, ∞[) for every s ∈ Gu ,

s = sup(Hu ∩ ]−∞, s[) for every s ∈ Hu .

Choose hsn in∈N , hs0n in∈N and hδn in∈N inductively, as follows. Set s0 = t, s00 = t0 . and sn ∈ Gu and s0n ∈ Hu , where n is ¤even, set s0n+1 = sup(Gu ∩ [sn , s0n ]). ¤ s0n+1 ∈ Hu , or s0n+1 < s0n and s0n+1 , s0n ∩ Gu = ∅, so s0n+1 ∈ / Gu and again

Given that t ≤ sn < s0n ≤ t0 Then either s0n+1 = s0n , so s0n+1 ∈ Hu . Let δn > 0 be

475J


679

such that δn ≤ 2−n and µ(E ∩ B((u, s0n+1 ), δn )) < 21 βr δnr .£ Because£the function s 7→ µ(E ∩ B((u, s), δn )) is continuous (443C, or otherwise), there is an sn+1 ∈ Gu ∩ sn , s0n+1 such that µ(E ∩ B((u, s), δn )) ≤ 21 βr δnr whenever s ∈ [sn+1 , s0n+1 ]. This is the inductive step from even n. If n is odd and sn ∈ Gu , s0n ∈ Hu and sn < s0n , set sn+1 = inf(Hu ∩ [sn , s0n ]). This time we find that sn+1 must belong to Gu . Let δn ∈ ]0, 2−n ] be such that µ(E ∩ B((u, sn+1 ), δn )) > 21 βr δnr , and let s0n+1 ∈ Hu ∩]sn+1 , s0n ] be such that µ(E ∩B((u, s), δn )) ≥ 12 βr δnr whenever s ∈ [sn+1 , s0n+1 ]. The construction provides us with a non-increasing sequence h[sn , s0n ]in∈N of closed intervals, so there must be some s in their intersection. In this case µ(E ∩ B((u, s), δn ) ≤ 21 βr δnr if n is even, µ(E ∩ B((u, s), δn ) ≥ 12 βr δnr if n is odd. Since limn→∞ δn = 0, lim inf δ↓0

µ(E∩B((u,s),δ)) µB((u,s),δ)

1 2

≤ ,

lim supδ↓0

µ(E∩B((u,s),δ)) µB((u,s),δ)

1 2

≥ ,

and s ∈ Du , while of course t ≤ s ≤ t0 . X XQ Q (iv) Thus the conegligible set Z has the property required by (a). (b) Now suppose that E has locally finite perimeter. Applying (a) to Rr \ E, there is a conegligible set Z1 ⊆ Rr−1 such that if u ∈ Z1 , t ∈ Hu , t0 ∈ Gu and t < t0 , then Du meets ]t, t0 [. Next, we know that ν(∂*E ∩ B(0, n)) = λ∂E B(0,n) is finite for every n ∈ N (475G). By 475H,

R

kuk≤n

#(Du ∩ [−n, n])µr−1 (du) ≤ ν(∂*E ∩ B(0, 2n)) < ∞

for every n ∈ N; but this means that Du ∩ [−n, n] must be finite for almost every u such that kuk ≤ n, for every n, and therefore that Z2 = {u : u ∈ Z ∩ Z1 , Du ∩ [−n, n] is finite for every n} is conegligible. For u ∈ Z2 , R \ Du is an open set, so is made up of a disjoint sequence of intervals with endpoints in Du ∪ {−∞, ∞} (see 2A2I); and because u ∈ Z ∩ Z1 , each of these intervals is included in either Gu or Hu . Now A = {u : there are successive constituent intervals of R \ Du both included in Gu } is negligible. P P?? Otherwise, there are rationals q < q 0 such that F = {u : ]q, q 0 [ \ Gu contains exactly one point} is not negligible. Note that, writing T : Rr → Rr−1 for the orthogonal projection, F = T [(Rr−1 × ]q, q 0 [) \ int*E] [ \ T [(Rr−1 × ]q, q 00 [) \ int*E] ∩ T [(Rr−1 × ]q 00 , q 0 [) \ int*E] q 00 ∈Q,q
is measurable. Take any u ∈ F ∩ int*F , and let t be the unique point in ]q, q 0 [ \ Gu . Then whenever 0 < δ ≤ min(t − q, q 0 − t) we shall have µ(B((u, t), δ) \ E) = µ(B((u, t), δ) \ int*E) ≤ 2δµr−1 (V (u, δ) \ F ), because if w ∈ F then (w, s) ∈ int*E for almost every s ∈ [t − δ, t + δ]. So lim supδ↓0

µ(B((u,t),δ)\E) µB((u,t),δ)

≤

2βr−1 βr

lim supδ↓0

µr−1 (V (u,δ)\F ) µr−1 V (u,δ)

= 0,

and (u, t) ∈ int*E. X XQ Q Similarly, A0 = {u : there are successive constituent intervals of R \ Du both included in Hu } is negligible. So Z2 \ (A ∪ A0 ) is a conegligible set of the kind we need.

680


475K

475K Lemma Suppose that h : Rr → [−1, 1] is a Lipschitz function with compact support; let n ∈ N be such that h(x) = 0 for kxk ≥ n. Suppose that E ⊆ Rr is a Lebesgue measurable set. Then |

R

∂h dµ| E ∂ξr

proof (a) Since (as noted in 473Cc)

¡

¢

≤ 2 βr−1 nr−1 + ν(∂*E ∩ B(0, n)) .

∂h ∂ξr

is defined almost everywhere, is measurable and bounded, and

is zero outside B(0, n), the integral is well-defined. If ν(∂*E ∩ B(0, n)) is infinite, the result is trivial; so henceforth let us suppose that ν(∂*E ∩ B(0, n)) < ∞. Identify Rr with Rr−1 × R. For u ∈ Rr−1 , set f (u) = #({t : (u, t) ∈ ∂*E ∩ B(0, n)}) if this is finite, = ∞ otherwise. R By 475H, f dµr−1 ≤ ν(∂*E ∩ B(0, n)). By 475J, there is a µr−1 -conegligible set Z ⊆ Rr−1 such that whenever u ∈ Z and (u, t) ∈ int*E and (u, t0 ) ∈ Rr \ cl*E, there is an s lying between t and t0 such that (u, s) ∈ ∂*E; we may suppose also that f (u) is finite for every u ∈ Z. For u ∈ Z, set Du0 = {t : (u, t) ∈ B(0, n) \ ∂*E}, and define gu : Du0 → {0, 1} by setting gu (t) = 1 if u ∈ Z and (u, t) ∈ B(0, n) ∩ int*E, = 0 if u ∈ Z and (u, t) ∈ B(0, n) \ cl*E. Now if t, t0 ∈ Du0 and g(t) 6= g(t0 ), there is a point s between t and t0 such that (u, s) ∈ ∂*E; so the variation VarDu gu of gu (224A) is at most f (u). Setting hu (t) = h(u, t) for u ∈ Rr−1 and t ∈ R, we now have Z

Z

∞

∂h (u, t)χ(B(0, n) ∩ int*E)(u, t)dt ∂ξr −∞

= Du

h0u (t)gu (t)dt

/ B(0, n)) (because h0u (t) = 0 if (u, t) ∈ ¡ ≤ 1 + Var gu ) sup | 0 Du

a≤b

Z a

b

h0u (t)|dt

(by 224J, recalling that Du0 is either empty or a bounded interval with finitely many points deleted) ≤ (1 + f (u)) sup |hu (b) − hu (a)| a≤b

(because hu is Lipschitz, therefore absolutely continuous on any bounded interval) ≤ 2(1 + f (u)). Integrating over u, we now have Z |

∂h dµ| ∂ξ r E

Z =|

(475Cg)

∂h dµ| ∂ξ r B(0,n)∩int*E

Z =

Z

∞

∂h (u, t)χ(B(0, n) ∩ int*E)(u, t)dt µr−1 (du) ∂ξ r V (0,n) −∞

(where V (0, n) = {u : u ∈ Rr−1 , kuk ≤ n}) Z ≤ 2(1 + f (u))µr−1 (du) V (0,n)

¡ ¢ ≤ 2 βr−1 nr−1 + ν(∂*E ∩ B(0, n)) .

475M


681

475L Theorem Suppose that E ⊆ Rr . Then E has locally finite perimeter iff ν(∂*E ∩ B(0, n)) is finite for every n ∈ N. proof If E has locally finite perimeter, then ν(∂*E ∩ B(0, n)) = λ∂E B(0, n) is finite for every n, by 475G. So let us suppose that ν(∂*E ∩ B(0, n)) is finite for every n ∈ N. If n ∈ N, then Z sup{| div φ dµ| : φ : Rr → Rr is Lipschitz, kφk ≤ χB(0, n)} E ¡ ¢ ≤ 2r (βr−1 nr−1 + ν(∂*E ∩ B(0, n)) is finite. P P If φ = (φ1 , . . . , φr ) : Rr → Rr is a Lipschitz function and kφk ≤ χB(0, n), then 475K tells us that

R

|

∂φr E ∂ξr

But of course it is equally true that |

R

¡

¢

¡

¢

dµ| ≤ 2 βr−1 nr−1 + ν(∂*E ∩ B(0, n)) .

∂φj dµ| E ∂ξj

≤ 2 βr−1 nr−1 + ν(∂*E ∩ B(0, n))

for every j ≤ r; adding, we have the result. Q Q Since n is arbitrary, E has locally finite perimeter. 475M Corollary (a) The family of sets with locally finite perimeter is a subalgebra of the algebra of Lebesgue measurable subsets of Rr . (b) A set E ⊆ Rr has finite perimeter iff ν(∂*E) < ∞, and in this case ν(∂*E) is the perimeter of E. (c) If E ⊆ Rr has finite measure, then per E = lim inf α→∞ per(E ∩ B(0, α)). proof (a) Recall that the definition in 474D insists that sets with locally finite perimeter should be Lebesgue measurable. If E ⊆ Rr has locally finite perimeter, then so does Rr \ E, by 474J. If E, F ⊆ Rr have locally finite perimeter, then ν(∂*(E ∪ F ) ∩ B(0, n)) ≤ ν(∂*E ∩ B(0, n)) + ν(∂*F ∩ B(0, n)) is finite for every n ∈ N, by 475L and 475Cd; by 475L in the other direction, E ∪ F has locally finite perimeter. (b) If E has finite perimeter (on the definition in 474D), then it is measurable and ν(∂*E) = λ∂E Rr is the perimeter of E, by 475G. If ν(∂*E) < ∞, then µ(∂*E) = 0 (471L) so E is measurable (475Ch); now E has locally finite perimeter, by 475L, and 475G again tells us that ν(∂*E) = λ∂E Rr is the perimeter of E. (c) Now suppose that µE is finite. For any α ≥ 0, ∂*(E ∩ B(0, α)) ⊆ (∂*E ∩ B(0, α)) ∪ ∂B(0, α) by 475Cf; since certainly ∂*(E ∩ B(0, α)) ⊆ cl*(E ∩ B(0, α)) ⊆ cl*E, ∂*(E ∩ B(0, α)) ⊆ ∂*E ∪ (cl*E ∩ ∂B(0, α)). Now we know also that

R∞ 0

ν(cl*E ∩ ∂B(0, α))dα = µE < ∞

(265G), so lim inf α→∞ ν(cl*E ∩ ∂B(0, α)) = 0. This means that lim inf per(E ∩ B(0, α)) = lim inf ν(∂*(E ∩ B(0, α))) α→∞

α→∞

≤ lim inf ν(∂*E) + ν(cl*E ∩ ∂B(0, α)) = ν(∂*E). α→∞

In the other direction, ∂*E ∩ U (0, α) = ∂*E ∩ int*B(0, α) ⊆ ∂*(E ∩ B(0, α))

682


475M

for every α, by 475Cb, so per E = ν(∂*E) = lim ν(∂*E ∩ U (0, α)) α→∞

≤ lim inf ν(∂*(E ∩ B(0, α))) = lim inf per(E ∩ B(0, α)) α→∞

α→∞

and we must have equality. Remark See 475Xj, 475Xk. 475N The Divergence Theorem Let E ⊆ Rr be such that ν(∂*E ∩ B(0, n)) is finite for every n ∈ N. (a) For ν-almost every x ∈ ∂*E, there is a Federer exterior normal vx of E at x. (b) For every Lipschitz function φ : Rr → Rr with compact support,

R

E

div φ dµ =

R

∂*E

φ(x). vx ν(dx).

proof By 475L, E has locally finite perimeter. By 474Q, there is a Federer exterior normal vx = ψE (x) of E at x for every x ∈ ∂ $ E; by 475D, ν-almost every point in ∂*E is of this kind. By 474E-474F,

R

E

div φ dµ =

R

φ. ψE dλ∂E =

R

∂$E

φ. ψE dλ∂E ,

R and this is also equal to ∂*E φ. ψE dλ∂E , because ∂ $ E ⊆ ∂*E. But λ∂E and ν induce the same subspace measures on ∂*E, by 475G, so

R

E

div φ dµ =

R

∂*E

φ. ψE dν =

R

∂*E

φ(x) . vx ν(dx),

as claimed. 475O Lemma Let E ⊆ Rr be a set with locally finite perimeter, and ψE its canonical outward normal. Let v be the unit vector (0, . . . , 0, 1). Identify Rr with Rr−1 × R. Then we have sequences hFn in∈N , hgn in∈N and hgn0 in∈N such that (i) for each n ∈ N, Fn is a Lebesgue measurable subset of Rr−1 , and gn , gn0 : Fn → [−∞, ∞] are Lebesgue measurable functions such that gn (u) < gn0 (u) for every u ∈ Fn ; (ii) ifPm, n R∈ N then gm (u) 6= gn0 (u) for every u ∈ Fm ∩ Fn ; ∞ (iii) n=0 Fn gn0 − gn dµr−1 = µE; (iv) for any continuous function h : Rr → R with compact support, R P∞ R h(x)v . ψE (x) ν(dx) = n=0 F h(u, gn0 (u)) − h(u, gn (u))µr−1 (du), ∂*E n

where we interpret h(u, ∞) and h(u, −∞) as 0 if necessary; (v) for µr−1 -almost every u ∈ Rr−1 , {t : (u, t) ∈ ∂*E} = {gn (u) : n ∈ N, u ∈ Fn , gn (u) 6= −∞} ∪ {gn0 (u) : n ∈ N, u ∈ Fn , gn0 (u) 6= ∞}. proof (a) Take a conegligible set Z ⊆ Rr−1 as in 475Jb. It will be convenient to shrink Z down further, if necessary, to a conegligible Borel set. For u ∈ Z set Du = {t : (u, t) ∈ ∂*E}. (b) For each q ∈ Q, set Fq0 = {u : u ∈ Z, (u, q) ∈ int*E}, so that Fq0 is a Borel set, and for u ∈ Fq0 set fq (u) = sup(Du ∩ ]−∞, q[),

fq0 (u) = inf(Du ∩ ]−∞, q[),

allowing −∞ as sup ∅ and ∞ as inf ∅. Observe that (because Du ∩ [q − 1, q + 1] is finite) fq (u) < q < fq0 (u). Now fq and fq0 are measurable, by the argument of part (a-i) of the proof of 475J. Enumerate Q as hqn in∈N , and set S Fn = Fqn \ m
475O


683

(c) Now let us look at the items (i)-(v) of the statement of this lemma. We have already achieved (i). If u ∈ Fm ∩ Fn , then gm (u) is the infimum of one of the constituent intervals of Gu , and gn0 (u) is either ∞ or the infimum of one of the constituent intervals of Hu ; so gm (u) 6= gn0 (u). Thus (ii) is true. For any u ∈ Z, P 0 n∈N,u∈Fn gn (u) − gn (u) = µ1 {t : (u, t) ∈ int*E}, so

P∞ R n=0 Fn

gn0 − gn dµr−1 =

R Z

µ1 {t : (u, t) ∈ int*E}µr−1 (du) = µ(int*E) = µE.

So (iii) is true. Also, for u ∈ Z, {t : (u, t) ∈ ∂*E} = Du = {gn (u) : n ∈ N, u ∈ Fn , gn (u) 6= −∞} ∪ {gn0 (u) : n ∈ N, u ∈ Fn , gn0 (u) 6= ∞}, so (v) is true. (d) As for (iv), suppose first that h : Rr → R is a Lipschitz function with compact support. Set φ(x) = (0, . . . , h(x)) for x ∈ Rr . Then Z

Z h(x)v .ψE (x) λ∂E (dx)

h(x)v . ψE (x) ν(dx) = ∂*E

(475G)

Z =

(474E)

Z φ.ψE λ∂E (dx)

Z =

∂h ∂ξ r E

Z Z

dµ =

n∈N,u∈Fn

X

=

div φ dµ E

Z

X

=

=

Z

∂h ∂ξ r int*E

dµ

0 gn (u)

∂h (u, t)dt µr−1 (du) ∂ξ r gn (u)

h(u, gn0 (u)) − h(u, gn (u))µr−1 (du)

n∈N,u∈Fn

(with the convention that h(u, ±∞) = limt→±∞ h(u, t) = 0) ∞ Z X = h(u, gn0 (u)) − h(u, gn (u))µr−1 (du) n=0

Fn

because if |h| ≤ mχB(0, m) then Z

X

|h(u, gn0 (u)) − h(u, gn (u))|µr−1 (du)

n∈N,u∈Fn

Z ≤ 2m

#({t : (u, t) ∈ B(0, m) ∩ ∂*E})µr−1 (du)

≤ 2mν(B(0, m) ∩ ∂*E) (475H) is finite. ˜ k for large For a general continuous function h of compact support, consider the convolutions hk = h ∗ h ˜ k, where hk is defined in 473E. If |h| ≤ mχB(0, m) then |hk | ≤ mχB(0, m + 1) for every k, so that P 0 n∈N,u∈Fn |hk (gn (u)) − hk (gn (u))| ≤ 2m#({t : (u, t) ∈ B(0, m + 1) ∩ ∂*E}) for every u ∈ Rr , k ∈ N. Since hk → h uniformly (473Ed),

684


Z

475O

Z h(x)v . ψE (x) ν(dx) = lim

hk (x)v . ψE (x) ν(dx) Z∂*E X = lim hk (u, gn0 (u)) − hk (u, gn (u))µr−1 (du) k→∞

∂*E

k→∞

Z =

n∈N,u∈Fn

X

h(u, gn0 (u)) − h(u, gn (u))µr−1 (du)

n∈N,u∈Fn ∞ XZ

=

n=0

Fn

h(u, gn0 (u)) − h(u, gn (u))µr−1 (du),

as required. 475P Lemma Let v ∈ Sr−1 be any unit vector, and V ⊆ Rr the hyperplane {x : x . v = 0}. Let T : Rr → V be the orthogonal projection. If E ⊆ Rr is any set with locally finite perimeter and canonical outward normal ψE , then

R

∂*E

|v . ψE |dν =

R

#(∂*E ∩ T −1 [{u}])ν(du).

V

proof As usual, we may suppose that the structure R(E, v) is rotated until v = (0, . . . , 1), R so that we can identify T (ξ1 , . . . , ξr ) with (ξ1 , . . . , ξr−1 ) ∈ Rr−1 , and V #(∂*E ∩ T −1 [{u}])ν(du) with Rr−1 #(Du )µr−1 (du), where Du = {t : (u, t) ∈ ∂*E}. (a) Write Φ for the set of continuous functions h : Rr → [−1, 1] with compact support. Then

R

∂*E

P P Of course

|v . ψE |dν = suph∈Φ

R

|v .ψE |dν ≥ ∂*E

for any h ∈ Φ. On the other hand, if γ<

R ∂*E

R

∂*E

R ∂*E

h(x)v . ψE (x) ν(dx).

h(x)v .ψE (x) ν(dx)

|v . ψE |dν =

R

|v . ψE |dλ∂E

(475G), then, because λ∂E is a Radon measure, there is an n ∈ N such that γ<

R

R

|v . ψE |dλ∂E = B(0,n)

h0 (x)v . ψE (x) λ∂E (dx)

where h0 (x) =

v . ψE (x) if x ∈ B(0, n) and v . ψE (x) 6= 0 |v . ψE (x)|

= 0 otherwise. Because λ∂E is a Radon measure, there is a continuous function h1 : Rr → R with compact support such that

R

|h1 − h0 |dλ∂E ≤

R

B(0,n)

|v . ψE |dλ∂E − γ

(416I); of course we may suppose that h1 , like h, takes values in [−1, 1], so that h1 ∈ Φ. Now

R

∂*E

h1 (x)v . ψE (x)ν(dx) =

As γ is arbitrary, suph∈Φ

R

R

h1 (x)v . ψE (x) λ∂E (dx) ≥ γ.

h(x)v . ψE (x) ν(dx) =

R

|v .ψE |dν

as required. Q Q (b) Now take hFn in∈N , hgn in∈N and hgn0 in∈N as in 475O. Then R P∞ R #(Du )µr−1 (du) = suph∈Φ n=0 F h(u, gn0 (u)) − h(u, gn (u))µr−1 (du). n

475P


685

0 P P By 475O(ii), gm (u) 6= gn (u) whenever m, n ∈ N and u ∈ Fm ∩ Fn , while for almost all u ∈ Rr−1

Du = ({gn (u) : n ∈ N, u ∈ Fn } \ {−∞}) ∪ ({gn0 (u) : n ∈ N, u ∈ Fn } \ {∞}). So if h ∈ Φ, X

h(u, gn0 (u)) − h(u, gn (u)) ≤ #({gn (u) : u ∈ Fn , gn (u) 6= −∞})

n∈N,u∈Fn

+ #({gn0 (u) : u ∈ Fn , gn0 (u) 6= ∞}) = #(Du ) r−1

for almost every u ∈ R , and R P∞ R 0 #(Du )µr−1 (du). n=0 Fn h(u, gn (u)) − h(u, gn (u))µr−1 (du) ≤ R R (m) In the other direction, given γ < #(Du )µr−1 (du), there is an m ∈ N such that γ < #(Du )µr−1 (du), R (m) (m) where Du = {t : (u, t) ∈ ∂*E, k(u, t)k < m}. Note that #(Du )µr−1 (du) ≤ ν(∂*E ∩ B(0, m)) is finite. Setting (m)

Hn = {u : u ∈ Fn , gn (u) ∈ Du }, for n ∈ N, we have γ<

P∞ n=0

µr−1 Hn +

(m)

Hn0 = {u : u ∈ Fn , gn0 (u) ∈ Du } P∞ n=0

µr−1 Hn0 < ∞.

By Lusin’s theorem (418J), we can find n ∈ N and compact sets Ki ⊆ Hi , Ki0 ⊆ Hi0 such that gi ¹Ki and gi0 ¹Ki0 are continuous for every i and γ≤

n X

(µr−1 Ki + µr−1 Ki0 ) −

i=0

∞ X

(µr−1 Hi + µr−1 Hi0 )

i=n+1 n X

−

(µr−1 (Hi \ Ki ) + µr−1 (Hi0 \ Ki0 )).

i=0

Set K=

S

i≤n {(u, gi (u))

: u ∈ Ki },

K0 =

S

0 i≤n {(u, gi (u))

: u ∈ Ki0 },

so that K and K 0 are disjoint compact subsets of int B(0, m). Let h : Rr → R be a continuous function such that h(x) = 1 for x ∈ K 0 , = −1 for x ∈ K, = 0 for x ∈ Rr \ int B(0, m) (4A2F(d-viii)); we can suppose that −1 ≤ h(x) ≤ 1 for every x. Then h ∈ Φ, and ∞ Z X i=0

Fi

h(u, gi0 (u)) − h(u, gi (u))µr−1 (du) =

∞ Z X i=0

Hi0

(because h is zero outside int B(0, m)) n Z X ≥ i=0

Ki0

h(u, gi0 (u))µr−1 (du) −

∞ Z X i=0

h(u, gi0 (u))µr−1 (du) −

h(u, gi (u))µr−1 (du)

Hi

n Z X i=0

h(u, gi (u))µr−1 (du) Ki

n ∞ X X 0 0 − (µ(Hi \ Ki ) + µ(Hi \ Ki )) − (µHi0 + µHi ) i=0

≥ γ.

i=n+1

686


This shows that

R

#(Du )µr−1 (du) ≤ suph∈Φ

P∞ R n=0 Fn

475P

h(u, gn0 (u)) − h(u, gn (u))µr−1 (du),

and we have equality. Q Q (c) Putting (a) and (b) together with 475O(iv), we have the result. 475Q Theorem Let E ⊆ Rr be a set with finite perimeter. For v ∈ S r−1 write Vv for {x : x . v = 0}, and let Tv : Rr → Vv be the orthogonal projection. Then ν(∂*E) =

1 2βr−1

R

R

Sr−1

Vv

#(∂*E ∩ Tv−1 [{u}])ν(du)ν(dv).

R proof (a) I start with an elementary fact: there is a constant c such that |w . v|λ∂Sr−1 (dv) = c for every w ∈ Sr−1 ; this is because whenever w, w0 ∈ Sr−1 there is an orthogonal linear transformation taking w to w0 , and this transformation is an automorphism of the structure (Rr , ν, Sr−1 , λ∂Sr−1 ) (474H). (Later in the proof I will come to the calculation of c.) (b) Now, writing ψE for the canonical outward normal of E, we have Z

Z

Z #(∂*E ∩

Sr−1

Vv

Tv−1 [{u}])ν(du)ν(dv)

Z

=

|ψE (x) . v|ν(dx)ν(dv) Sr−1

(475P)

∂*E

ZZ |ψE (x) .v|λ∂E (dx)λ∂Sr−1 (dv)

=

(by 235M, recalling that λ∂Sr−1 and λ∂E are just indefinite-integral measures over ν, while Sr−1 and ∂*E are Borel sets)

ZZ |ψE (x) .v|λ∂Sr−1 (dv)λ∂E (dx)

=

(because . is continuous, so (x, v) 7→ ψE (x) . v is measurable, while λ∂Sr−1 and λ∂E are totally finite, so we can use Fubini’s theorem) = c per E (by (a) above) = cν(∂*E). (c) We have still to identify the constant c. But observe that the argument above applies whenever ν(∂*E) is finite, and in particular applies to E = B(0, 1). In this case, ∂*B(0, 1) = Sr−1 , and for any v ∈ Sr−1 , u ∈ Vv we have #(∂*B(0, 1) ∩ T −1 [{u}]) = 2 if kuk < 1, = 1 if kuk = 1, = 0 if kuk > 1. Since we can identify ν on Vv as a copy of Lebesgue measure µr−1 ,

R

Vv

#(∂*B(0, 1) ∩ T −1 [{u}])ν(du) = 2ν{u : u ∈ Vv , kuk < 1} = 2βr−1 .

This is true for every v ∈ Vu , so from (b) we get 2βr−1 νSr−1 =

R

Sr−1

R

Vv

#(∂*E ∩ Tv−1 [{u}])ν(du)ν(dv) = cνSr−1 ,

and c = 2βr−1 . Substituting this into (b), we have the result.

475S


687

475R Convex sets in Rr For the next theorem it will help to have some elementary facts about convex sets in finite-dimensional spaces out in the open. Lemma (In this lemma I allow r = 1.) Let C ⊆ Rr be a convex set. (a) If x ∈ C and y ∈ int C, then ty + (1 − t)x ∈ int C for every t ∈ ]0, 1]. (b) C and int C are convex. (c) If int C 6= ∅ then C = int C. (d) If int C = ∅ then C lies within some hyperplane. (e) int C = int C. proof (a) Setting φ(z) = x + t(z − x) for z ∈ Rr , φ : Rr → Rr is a homeomorphism and φ[C] ⊆ C, so φ(y) ∈ φ[int C] = int φ[C] ⊆ int C. (b) It follows at once from (a) that int C is convex; C is convex because (x, y) 7→ tx+(1−t)y is continuous for every t ∈ [0, 1]. (c) From (a) we see also that if int C 6= ∅ then C ⊆ int C, so that C ⊆ int C and C = int C. (d) It is enough to consider the case in which 0 ∈ C, since if C = ∅ the result is trivial. ?? If x1 , . . . , xr 1 Pr are linearly independent elements of C, set x = i=1 xi ; then r+1

x+

Pr i=1

αi xi =

Pr

i=1 (αi

+

1 )xi r+1

∈C

Pr 1 whenever i=1 |αi | ≤ . Also, writing e1 , . . . , er for the standard orthonormal basis of Rr , we can express r+1 Pr Pr 1 ei as j=1 αij xj for each j; setting M = (r + 1) maxi≤r j=1 |αij |, we have x ± ei ∈ C for every i ≤ r, so that x + y ∈ C whenever kyk ≤ r

√ r , M

M

and x ∈ int C. X X

So the linear subspace of R spanned by C has dimension at most r − 1. (e) If int C is empty, then (d) shows that C is included in a hyperplane, so that int C is empty. Otherwise, if x ∈ Rr \ int C, there is an e ∈ Rr such that e .y ≤ e . x for every y ∈ int C (4A4Db, or otherwise). Now, by (c), e. y ≤ e. x for every y ∈ C, so x ∈ / int C. This shows that int C ⊆ int C, so that the two are equal. It follows at once that ∂C = ∂C. 475S Corollary: Cauchy’s Perimeter Theorem Let C ⊆ Rr be a bounded convex set with nonempty interior. For v ∈ S r−1 write Vv for {x : x . v = 0}, and let Tv : Rr → Vv be the orthogonal projection. Then ν(∂C) =

R

1 βr−1

Sr−1

ν(Tv [C])ν(dv).

proof (a) The first thing to note is that ∂*C = ∂C. P P Of course ∂*C ⊆ ∂C (475Ca). If x ∈ ∂C, there is a half-space V containing x and disjoint from int C (4A4Db), so that lim supδ↓0

µ(B(x,δ)\C) µB(x,δ)

≥ lim supδ↓0

µ(B(x,δ)\int V ) µB(x,δ)

1 2

= ,

and x ∈ / int*C. On the other hand, if η > 0 is such that B(x0 , η) ⊆ int C, then for any δ > 0 we can set t=

δ , η+kx0 −xk

and then B(x + t(x0 − x), tη) ⊆ B(x, δ) ∩ int C ⊆ B(x, δ) ∩ C,

so that µ(B(x, δ) ∩ C) ≥ βr tr η r =

¡

¢r η µB(x, δ); kx0 −xk+η

as δ is arbitrary, x ∈ cl*C. This shows that ∂C ⊆ ∂*C so that ∂*C = ∂C. Q Q

688


475S

(b) We have a function φ : Rr → C defined by taking φ(x) to be the unique point of C closest to x, for every x ∈ Rr (3A5Ld). This function is 1-Lipschitz. P P Take any x, y ∈ Rr and set e = φ(x) − φ(y). We know that φ(x) − ²e ∈ C, so that kx − φ(x) − ²ek ≥ kx − φ(x)k, for 0 ≤ ² ≤ 1; it follows that (x − φ(x)). e ≥ 0. Similarly, (y − φ(y)). (−e) ≥ 0. Accordingly (x − y) . e ≥ e. e and kx − yk ≥ kek. As x and y are arbitrary, φ is 1-Lipschitz. Q Q Now suppose that that C 0 ⊇ C is a closed bounded convex set. Then ν(∂C 0 ) ≥ ν(∂C). P P Let φ be the function defined just above. By 264G/471J, ν ∗ (φ[∂C 0 ]) ≤ ν(∂C 0 ). But if x ∈ ∂C, there is an e ∈ Rr \ {0} such that x . e ≥ y . e for every y ∈ C. Then φ(x + αe) = x for every α ≥ 0. Because C 0 is closed and bounded, and x ∈ C ⊆ C 0 , there is a greatest α ≥ 0 such that x + αe ∈ C 0 , and in this case x + αe ∈ ∂C 0 ; since φ(x + αe) = x, x ∈ φ[∂C 0 ]. As x is arbitrary, ∂C ⊆ φ[∂C 0 ], and ν(∂C) ≤ ν ∗ (φ[∂C 0 ]) ≤ ν(∂C 0 ). Q Q Since we can certainly find a closed convex set C 0 ⊇ C such that ν(∂C 0 ) is finite (e.g., any sufficiently large ball or cube), ν(∂C) < ∞. It follows at once that µ(∂C) = 0 (471L). (c) The argument so far applies, of course, to every r ≥ 1 and every bounded convex set with non-empty interior in Rr . Moving to the intended case r ≥ 2, and fixing v ∈ Sr−1 for the moment, we see that, because Tv is an open map (if we give Vv its subspace topology), Tv [C] is again a bounded convex set with non-empty (relative) interior. Since the subspace measure induced by ν on V is just a copy of Lebesgue measure, (b) tells us that νTv [C] = ν(int Tv [C]), where here I write int Tv [C] for the interior of Tv [C] in the subspace topology of Vv . Now the point is that int Tv [C] ⊆ Tv [int C]. P P int C is dense in C (475Rc), so W = Tv [int C] is an open convex set which is dense in Tv [C]; now W = int W , taking the interior in V (475Re), so W ⊇ int Tv [C]. Q Q P Tv−1 [{u}] is a straight line meeting int C It follows that #(∂C ∩ Tv−1 [{u}]) = 2 for every u ∈ int Tv [C]. P −1 in y0 say. Because C is a bounded convex set, it meets Tv [{u}] in a bounded convex set, which must be a non-trivial closed line segment with endpoints y1 , y2 say. Now certainly neither y1 nor y2 can be in the interior of C. Moreover, the open line segments between y1 and y0 , and between y2 and y0 , are covered by Q int C, by 475Ra; so Tv−1 [{u}] ∩ ∂C = {y1 , y2 } has just two members. Q (d) This is true for every v ∈ Sr−1 . But this means that we can apply 475Q to see that 1 2βr−1

Z

Z

ν(∂C) = ν(∂*C) = #(∂*E ∩ Tv−1 [{u}])ν(du)ν(dv) Sr−1 Vv Z Z 1 = #(∂*E ∩ Tv−1 [{u}])ν(du)ν(dv) 2βr−1 S Tv [C] r−1 Z Z 1 = #(∂*E ∩ Tv−1 [{u}])ν(du)ν(dv) 2βr−1 S int T [C] v Z r−1 Z 1 1 = ν(int Tv [C])ν(dv) = νTv [C]ν(dv), βr−1

βr−1

Sr−1

Sr−1

as required. 475T Corollary: the Convex Isoperimetric Theorem If C ⊆ Rr is a bounded convex set, then ν(∂C) ≤ rβr ( 12 diam C)r−1 . proof (a) If C is included in some r − 1-dimensional affine subspace, then 1 2

ν(∂C) = νC ≤ βr−1 ( diam C)r−1 by 264H. For completeness, I should check that βr−1 ≤ rβr . P P Comparing 265F with 265H, or working from the formulae in 252Q, we have rβr = 2πβr−2 . On the other hand, by the argument of 252Q, βr−1 = βr−2

R π/2

−π/2

cosr−1 t dt ≤ πβr−2 ,

so (not coincidentally) we have a factor of two to spare. Q Q

475Yb


689

(b) Otherwise, C has non-empty interior (475Rd), and for any orthogonal projection T of Rr onto an 1 2

r − 1-dimensional linear subspace, diam T [C] ≤ diam C, so ν(T [C]) ≤ βr−1 ( diam C)r−1 . Now 475S tells us that 1 2

1 2

ν(∂C) ≤ ( diam C)r−1 νSr−1 = rβr ( diam C)r−1 . 475X Basic exercises (a) Show that if C ⊆ Rr is convex, then either µC = 0 and ∂*C = ∅, or ∂*C = ∂C. (b) Let A, A0 ⊆ Rr be any sets. Show that (∂*A ∩ int*A0 ) ∪ (∂*A0 ∩ int*A) ⊆ ∂*(A ∩ A0 ) ⊆ (∂*A ∩ cl*A0 ) ∪ (∂*A0 ∩ cl*A). (c) Let A ⊆ Rr be any set, and B a non-trivial closed ball. Show that ∂*(A ∩ B)4((B ∩ ∂*A) ∪ (A ∩ ∂B)) ⊆ ((cl*A \ A) ∪ (A \ int*A)) ∩ ∂B. > (d) Let E, F ⊆ Rr be measurable sets, and v a Federer exterior normal to E at x ∈ int*F . Show that v is a Federer exterior normal to E ∩ F at x. (e) Let T be the density topology on Rr (414P) defined from lower Lebesgue density (341E). Show that, for any A ⊆ Rr , A ∪ cl*A is the T-closure of A and int*A is the T-interior of the T-closure of A. (f ) Let A ⊆ Rr be any set. Show that A \ cl*A and int*A \ A are Lebesgue negligible. > (g) Let E ⊆ Rr be such that ν(∂*E) and µE are both finite. Show that, taking vx to be a Federer exterior normal to E at any point x where this is defined,

R

div φ dµ = E

R

∂*E

φ(x) .vx ν(dx).

for every bounded Lipschitz function φ : Rr → Rr . > (h) Let hEn in∈N be a sequence of measurable subsets of Rr such that (i) there is a measurable set E such that limn→∞ µ((En 4E) ∩ B(0, m)) = 0 for every m ∈ NR(ii) supn∈N ν(∂*En ∩ RB(0, m)) is finite for every m ∈ N. Show that E has locally finite perimeter. (Hint: E div φ dµ = limn→∞ En div φ dµ for every Lipschitz function φ with compact support.) (i) Give an example of bounded convex sets E and F such that ∂ $ (E ∪ F ) 6⊆ ∂ $ E ∪ ∂ $ F . (j) (i) Show that if A, B ⊆ Rr then ∂*(A ∩ B) ∩ ∂*(A ∪ B) ⊆ ∂*A ∩ ∂*B. (ii) Show that if E, F ⊆ R have finite perimeter then per(E ∩ F ) + per(E ∪ F ) ≤ per E + per F . (k) Let E ⊆ Rr be a set with finite Lebesgue measure and finite perimeter. (i) Show that if H ⊆ Rr is a half-space, then per(E ∩ H) ≤ per E. (Hint: 475Ja.) (ii) Show that if C ⊆ Rr is convex, then per(E ∩ C) ≤ per E. (Hint: by the Hahn-Banach theorem, C is a limit of polytopes; use 474Sa.) (iii) Show that in 475Mc we have per E = limα→∞ per(E ∩ B(0, α)). 475Y Further exercises (a) Let B be a non-trivial ball in Rr with centre y, and v, v 0 two unit vectors in Rr . Set H = {x : x ∈ Rr , (x − y) . v ≤ 0}, Show that µ((H4H 0 ) ∩ B) =

1 π

H 0 = {x : x ∈ Rr , (x − y) . v 0 ≤ 0}.

arccos(v . v 0 )µB.

(b) Show that µ is inner regular with respect to the family of compact sets K ⊆ Rr such that lim supδ↓0

µ(B(x,δ)∩K) µB(x,δ)

≥

1 2

for every x ∈ K.

690


475Yc

(c) Show that if E ⊆ Rr has finite perimeter then one of E, Rr \ E has finite measure. 475 Notes and comments The successful identification of the distributionally-defined notion of ‘perimeter’, as described in §474, with the geometrically accessible concept of Hausdorff measure of an appropriate boundary, is of course the key to any proper understanding of the results of the last section as well as this one. The very word ‘perimeter’ would be unfair if the perimeter of E ∪ F were unrelated to the perimeters of E and F ; and from this point of view the reduced boundary is less suitable than the essential boundary (475Cd, 475Xh). If we re-examine 474M, we see that it is saying, in effect, that for many balls B the boundary ∂*(E ∩ B) is nearly (B ∩ ∂*E) ∪ (E ∩ ∂B), and that an outward-normal function for E ∩ B can be assembled from outward-normal functions for E and B. But looking at 475Xc-475Xd we see that this is entirely natural; we need only ensure that ν(F ∩ ∂B) = 0 for a µ-negligible set F defined from E; and the ‘almost every δ’ in the statement of 474M is fully enough to arrange this. On the other hand, 475Xg seems to be very hard to prove without using the identification between ν(∂*E) and per E. Concerning 475Q, I ought to emphasize that it is not generally true that νF =

1 2βr−1

R

Sr−1

R

Vv

#(F ∩ Tv−1 [{u}])ν(du)ν(dv)

even for r = 2 and compact sets F with νF < ∞. We are here approaching one of the many fundamental concepts of geometric measure theory which I am ignoring. The key word is ‘rectifiability’; for ‘rectifiable’ sets a wide variety of concepts of k-dimensional measure coincide, including the integral-geometric form above, and ∂*E is rectifiable whenever E has locally finite perimeter (Evans & Gariepy 92, 5.7.3). For the general theory of rectifiable sets, see the last quarter of Mattila 95, or Chapter 3 of Federer 69. I have already noted that the largest volumes for sets of given diameter or perimeter are provided by balls (see 264H and the notes to §474). The isoperimetric theorem for convex sets (475T) is of the same form: once again, the best constant (here, the largest perimeter for a convex set of given diameter, or the smallest diameter for a convex set of given perimeter) is provided by balls.

476 Concentration of measure Among the myriad special properties of Lebesgue measure, a particularly interesting one is ‘concentration of measure’. For a set of given measure in the plane, it is natural to feel that it is most ‘concentrated’ if it is a disk. There are many ways of defining ‘concentration’, and I examine three of them in this section (476J, 476K and 476L); all lead us to Euclidean balls as the ‘most concentrated’ shapes. On the sphere the same criteria lead us to caps (476O, 476Xj). All the main theorems of this section will be based on the fact that semi-continuous functions on compact spaces attain their bounds. The compact spaces in question will be spaces of subsets, and I begin with some very general remarks on topologies on such spaces (476A-476F). The facts to be used here will mostly relate to topologies induced by Hausdorff metrics (476A), but I include a mention of Vietoris topologies (476C) as what might be an appropriate alternative if we had to start from a non-metrizable space. The particular geometric properties of Euclidean space which make all these results possible are described in 476H-476I, where I describe concentrating operators based on reflections. The actual theorems 476J-476L and 476O can now almost be mass-produced. 476A Hausdorff metrics Let (X, ρ) be a metric space, and C the family of closed subsets of X. (a) For a non-empty subset A of X, write ρ(x, A) = inf y∈A ρ(x, y) for every x ∈ X. Note that ρ(x, A) ≤ ρ(x, y) + ρ(y, A) for all x, y ∈ X, so that x 7→ ρ(x, A) : X → R is 1-Lipschitz. For E, F ∈ C \ {∅}, set ρ˜(E, F ) = max(supx∈E ρ(x, F ), supy∈F ρ(y, E)). If E, F ∈ C \ {∅} and z ∈ X, then ρ(z, F ) ≤ ρ(z, E) + supx∈E ρ(x, F ); from this it is easy to see that ρ˜ is a metric on C \ {∅}, the Hausdorff metric. Observe that x 7→ {x} : X → C \ {∅} is an isometry.

476Cc

Concentration of measure

691

(b) If (X, ρ) is complete so is (C \ {∅}, ρ˜). P P Suppose that hFn in∈N is a sequence in C \ {∅} such that ρ˜(Fm , Fn ) ≤ 2−m whenever m ≤ n. Set T S F = lim supn→∞ Fn = n∈N i≥n Fi . Take any n ∈ N and x ∈ Fn . Then we can choose hxi ii∈N inductively by setting x0 = x and, given xi ∈ Fn+i , taking xi+1 ∈ Fn+i+1 such that ρ(xi+1 , xi ) ≤ 2−n−i+1 ; this is possible because ρ(xi , Fn+i+1 ) ≤ ρ˜(Fn+i , Fn+i+1 ) ≤ 2−n−i . Now hxi ii∈N is Cauchy, so has a limit z in X. For every m ∈ N, S z ∈ {xi : i ≥ m} ⊆ i≥m Fi , so z ∈ F ; accordingly ρ(x, F ) ≤ ρ(x, z) ≤

P∞ i=0

ρ(xi , xi+1 ) ≤ 2−n+2 .

Since Fn is not empty, this shows incidentally that F is not empty and belongs to C \ {∅}. Next, {y : ρ(y, Fn ) ≤ 2−n } is a closed set including Fi for every i ≥ n, so includes F . This shows that ρ˜(Fn , F ) ≤ max(2−n+2 , 2−n ) = 2−n+2 . As n is arbitrary, F = limn→∞ Fn in C \{∅}. As hFn in∈N is arbitrary, C \ {∅} is complete under ρ˜. Q Q (c) If (X, ρ) is totally bounded, so is (C \ {∅}, ρ˜). P P Let ² > 0. Then there is a finite set I ⊆ X such that ρ(x, I) ≤ ² for every x ∈ X. Set E = PI \ {∅}. Then E is a finite subset of C \ {∅}. If F ∈ C \ {∅}, set E = {x : x ∈ I, ρ(x, F ) ≤ ²}. Then ρ˜(E, F ) ≤ ². Thus C \ {∅} can be covered by finitely many closed balls of radius ². As ² is arbitrary, (C \ {∅}, ρ˜) is totally bounded. Q Q (d) It follows that if (X, ρ) is compact, so is (C \ {∅}, ρ˜) (4A2Jd). 476B Lemma Let (X, ρ) be a metric space, C1 the set of closed subsets of X, and ρ˜ the Hausdorff metric on C \ {0}. Let σ be the metric on X × X defined by setting σ((x1 , x2 ), (y1 , y2 )) = max(ρ(x1 , y1 ), ρ(x2 , y2 )), C2 the set of closed subsets of X × X and σ ˜ the Hausdorff metric on C2 \ {∅}. Then the map F 7→ F × F : C \ {∅} → C2 \ {∅} is an isometry. proof If E, F ∈ C \ {∅}, then ρ(x, F ) = inf y∈F ρ(x, y) = inf y∈F σ((x, x), (y, y)) ≤ σ ˜ (E × E, F × F ) for any x ∈ E, and similarly ρ(y, E) ≤ σ ˜ (E × E, F × F ) for every y ∈ F ; so ρ˜(E, F ) ≤ σ(E × E, F × F ). In the other direction, σ((x1 , x2 ), F × F ) =

inf σ((x1 , y1 ), (x2 , y2 )) = inf max(ρ(x1 , y1 ), ρ(x2 , y2 )) y1 ,y2 ∈F ¡ ¢ = max inf ρ(x1 , y1 ), inf ρ(x2 , y2 ) ≤ ρ˜(E, F ) y1 ,y2 ∈F

y1 ∈F

y2 ∈F

for all x1 , x2 ∈ E, and similarly σ((y1 , y2 ), E ×E) ≤ ρ˜(E, F ) for all y1 , y2 ∈ F , so σ ˜ (E ×E, F ×F ) ≤ ρ˜(E, F ). 476C Vietoris topologies Let X be a topological space, and write C for the family of closed subsets of X. (a) The Vietoris topology on C is the topology generated by sets of the form {F : F ∈ C, F ⊆ G},

{F : F ∈ C, F ∩ G 6= ∅}

where G ⊆ X is open. (b) If X is regular, then C is Hausdorff in its Vietoris topology. P P Let E1 , E2 ∈ C be distinct. Then at least one of E1 \ E2 , E2 \ E1 is non-empty; suppose the former. Take x ∈ E1 \ E2 and an open set G containing x such that G ∩ E2 = ∅. Then {F : F ∈ C, F ∩ G 6= ∅} and {F : F ∈ C, F ⊆ X \ G} are disjoint open subsets of C containing E1 , E2 respectively. Q Q (c) If X is regular and normal, then C is regular in its Vietoris topology. P P Suppose that F ∈ C and that E is an open subset of C containing F . Then there must be a finite family G of open subsets of X and an open subset H0 of X such that

692


476Cc

F ⊆ {E : E ∈ C, E ∩ G 6= ∅ ∀ G ∈ G, E ⊆ H0 } ⊆ E. Let H1 be an open set such that F ⊆ H1 ⊆ H 1 ⊆ H0 . For each G ∈ G take a point xG ∈ F ∩ G and an open set G0 containing xG such that G0 ⊆ G. Now consider E1 = {E : E ∈ C, E ∩ G0 6= ∅ ∀ G ∈ G, E ⊆ H0 }, E2 =

S

G∈G {E

: E ∈ C, E ⊆ X \ G0 } ∪ {E : E ∈ C, E ∩ (X \ H 1 ) 6= ∅}.

Then E1 and E2 are disjoint open sets in C, F ∈ E1 and E2 ⊇ C \ E. As F and E are arbitrary, C is regular. Q Q (d) A variant of the Vietoris topology will be useful in §495. I will say that the k-Vietoris topology on C is the topology generated by sets of the form {F : F ∈ C, F ∩ G 6= ∅},

{F : F ∈ C, F ∩ K = ∅}

where G ⊆ X is open and K ⊆ X is compact. Now we have the following easy results. T S (i) C is compact in its k-Vietoris topology. P P Let F be an ultrafilter on C. Set C0 = E∈F E. (α) S If G ⊆ X is open and G ∩ C0 6= ∅, set E = {C : C ∈ C, C ∩ G = ∅}. Then E does not meet G, so does not include C0 , and E ∈ / F. Accordingly {C : C ∩ G 6= ∅} = C \ E belongs to F. (β) If K ⊆ X is compact S and C0 ∩ K = ∅, then {K ∩ E : E ∈ F} is a downwards-directed family of relatively closed subsets of K with empty intersection so must contain the empty set, and there is an E ∈ F such that K ∩ C = ∅ for every C ∈ E, that is, {C : C ∩ K = ∅} belongs to F. (γ) By 4A2B(a-v), F → C0 . As F is arbitrary, C is compact. Q Q (ii) If X is Hausdorff, the k-Vietoris topology is coarser than the Vietoris topology. If X is compact and Hausdorff the two topologies agree; in particular, the Vietoris topology on C is compact. (iii) If X is locally compact and Hausdorff, the k-Vietoris topology on X is Hausdorff. P P Let C1 , C2 ∈ C be distinct. Then there is a point belonging to one and not the other; suppose that x ∈ C1 \ C2 . Because X is locally compact and Hausdorff, there is a relatively compact open set G such that x ∈ G and G ∩ C2 = ∅. Now {C : C ∩ G 6= ∅} and {C : C ∩ G = ∅} are disjoint open sets in C containing C1 , C2 respectively. Q Q 476D Proposition Let (X, ρ) be a metric space, and ρ˜ the Hausdorff metric on C \ {∅}, where C is the family of closed subsets of X. Write K ⊆ C for the family of compact subsets of X. (a) If µ is a totally finite topological measure on X, then µ¹C \ {∅} is upper semi-continuous. (b) If µ is a locally finite topological measure on X, then µ¹K \ {∅} is upper semi-continuous for the subspace topology on K \ {∅}. (c)RIf µ is a topological measure on X and f is a non-negative µ-integrable real-valued function, then F 7→ F f dµ : C \ {∅} → R is upper semi-continuous. proof (a) If F ∈ C \ {∅} and µF < α, then for each n ∈ N set Fn = {x : ρ(x, F ) ≤ 2−n }. Since hFn in∈N is a non-decreasing sequence of closed sets with intersection F , and µ is totally finite, there is an n such that µFn < α. If now we take E ∈ C \ {∅} such that ρ˜(E, F ) ≤ 2−n , then E ⊆ Fn so µE < α. As F and α are arbitrary, µ¹C \ {∅} is upper semi-continuous. (b) The argument is the same as in (a), with one refinement. If K ∈ K \ {∅} and we set Fn = {x : ρ(x, K) ≤ 2−n }, then we do not know that every Fn has finite measure. However, since µ is locally finite, every point of K has an open neighbourhood of finite measure; as K is compact, we can cover it by finitely many open sets of finite measure, and K ⊆ G where G ⊆ X is open and µG < ∞. Next, the function x 7→ ρ(x, X \ G) is continuous and strictly positive on K, so has a non-zero lower bound on K, and there is some m ∈ N such that ρ(x, X \ G) > 2−m for every x ∈ K, in which case Fm ⊆ G. It follows now that there is some n ≥ m such that µFn < α, and we can continue as before. (c) Apply (a) to the indefinite-integral measure over µ defined by f .

476Ha


693

476E Proposition Let X be a topological space, and C the family of closed subsets of X, with the Vietoris topology. Write K ⊆ C for the family of closed compact subsets of X. (a) If µ is a totally finite topological measure on X which is inner regular with respect to the closed sets, then µ¹C is upper semi-continuous. (b) If µ is a locally finite topological measure on X which is inner regular with respect to the closed sets, then µ¹K is upper semi-continuous. proof (a) Suppose that F ∈ C and that µF < α. Because µX is finite, there is a closed set F1 ⊆ X \ F such that µF1 > µX − α, so that µ(X \ F1 ) < α. Now E = {E : E ∈ C, E ⊆ X \ F1 } is an open set containing F , and µE < α for every E ∈ E. As F and α are arbitrary, µ¹C is upper semi-continuous. (b) If K ∈ K and µK < α then, because µ is locally finite, there is an open set G ⊇ K such that µG < ∞. Now there is a closed set F1 ⊆ G such that µF1 > µG − α, so that L = {L : L ∈ K, L ⊆ G \ F1 } is an open subset of K and µL < α for every L ∈ L. As K and α are arbitrary, µ¹K is upper semi-continuous. 476F Proposition Let (X, ρ) be a metric space, and ρ˜ the Hausdorff metric on C \ {∅}, where C is the family of closed subsets of X. For non-empty sets A ⊆ X and ² > 0, set U (A, ²) = {x : x ∈ X, ρ(x, A) < ²}. Then for any τ -additive topological measure µ on X, the function (F, ²) 7→ µU (F, ²) : (C \ {∅}) × ]0, ∞[ → [0, ∞] is lower semi-continuous. proof Suppose that α ∈ ]0, ∞[ and that F0 ∈ C \ {∅}, ²0 ∈ ]0, ∞[ are such that µU (F0 , ²0 ) > α. Then {U (I, δ) : I ∈ [F0 ]<ω , δ < ²0 } is an upwards-directed family of open sets with union U (F0 , ²0 ). Because µ is τ -additive, there are δ < ²0 , I ∈ [F0 ]<ω such that µU (I, δ) > α. Now suppose that F ∈ C \ {∅} and ² are such that ρ˜(F, F0 ) < 12 (²0 − δ) and ² > 12 (δ + ²0 ). Then ρ(x, F ) < 21 (²0 − δ) < ² − δ for every x ∈ I, so U (I, δ) ⊆ U (F, ²) and µU (F, ²) > α. As α, F0 and ²0 are arbitrary, (F, ²) 7→ µU (F, ²) is lower semi-continuous. Remark Recall that all ‘ordinary’ topological measures on metric spaces are τ -additive; see 438J. 476G Proposition Let (X, ρ) be a non-empty compact metric space, and suppose that its isometry group G acts transitively on X. Then X has a unique G-invariant Radon probability measure µ, which is strictly positive. proof By 441G, G, with its topology of pointwise convergence, is a compact topological group, and the action of G on X is continuous. So 443Td gives the result. 476H Concentration by partial reflection The following construction will be used repeatedly in the rest of the section. Let X be an inner product space. (In this section, X will be usually be Rr , but in 493J it will be helpful to be able to speak of abstract Hilbert spaces.) For any unit vector e ∈ X and any α ∈ R, write Reα : X → X for the reflection in the hyperplane Veα = {x : x ∈ X, (x|e) = α}, so that Reα (x) = x + 2(α − (x|e))e for every x ∈ X. Next, for any A ⊆ X, we can define a set ψeα (A) by setting ψeα (A) = {x : x ∈ A, (x|e) ≥ α} ∪ {x : x ∈ A, (x|e) < α, Reα (x) ∈ A} ∪ {x : x ∈ Rr \ A, (x|e) ≥ α, Reα (x) ∈ A} = (W ∩ (A ∪ Reα [A])) ∪ (A ∩ Reα [A] \ W ), where W is the half-space {x : (x|e) ≥ α}. Geometrically, we construct ψeα (A) by moving those points of A on the ‘wrong’ side of the hyperplane Veα to their reflections, provided those points are not already occupied. We have the following facts. (a) For non-empty A ⊆ X and ² > 0, set U (A, ²) = {x : ρ(x, A) < ²}, where ρ is the standard metric on X. Now U (ψeα (A), ²) ⊆ ψeα (U (A, ²)). P P Take x ∈ U (ψeα (A), ²). Then there is a y ∈ ψeα (A) such that kx − yk < ².

694


476Ha

case 1 Suppose (x|e) ≥ α. If x ∈ U (A, ²) then certainly x ∈ ψeα (U (A, ²)). Otherwise, because kx − yk < ², y ∈ / A, so Reα (y) ∈ A. But kReα (x) − Reα (y)k = kx − yk < ², so Reα (x) ∈ U (A, ²) and x ∈ ψeα (U (A, ²)). case 2a Suppose (x|e) < α, (y|e) ≥ α. Then kReα (x) − yk = kx − Reα (y)k ≤ kReα (x) − Reα (y)k = kx − yk < ². At least one of y, Reα (y) belongs to A, so both x and Reα (x) belong to U (A, ²) and x ∈ ψeα (U (A, ²)). case 2b Suppose (x|e) < α and (y|e) < α. In this case, both y and Reα (y) belong to A, so both x and Reα (x) belong to U (A, ²) and again x ∈ ψeα (U (A, ²)). Thus x ∈ ψeα (U (A, ²)) in all cases; as x is arbitrary, we have the result. Q Q (b) If F ⊆ X is closed, then ψeα (F ) is closed. P P Use the second formula for ψeα (F ). Q Q 476I Lemma Let X be an inner product space, e ∈ X a unit vector and α ∈ R. Let R = Reα : X → X be the reflection operator, and ψ = ψeα : PX → PX the associated transformation, as described in 476H. For x ∈ A ⊆ X, define φA (x) = x if (x|e) ≥ α, = x if (x|e) < α and R(x) ∈ A, = R(x) if (x|e) < α and R(x) ∈ / A. Let ν be a topological measure on X which is R-invariant, that is, ν is the image measure νR−1 . (a) For any A ⊆ X, φA : A → ψ(A) is a bijection. If α < 0, then kφA (x)k ≤ kxk for every x ∈ A, with kφA (x)k < kxk iff (x|e) < α and R(x) ∈ / A. (b) If E ⊆ X is measured by ν, then ψ(E) is measurable and νψ(E) = νE; moreover, φE is a measure space isomorphism for the subspace measures on RE and ψ(E). R (c) If α < 0 and E ⊆ X is measurable, then E kxkν(dx) ≥ ψ(E) kxkν(dx), with equality iff {x : x ∈ E, (x|e) < α, R(x) ∈ / E} is negligible. (d) Suppose that X is separable. Let λ be the c.l.d. product measure of ν with itself on X × X. If α ≤ 0 and E ⊆ X is measurable, then

R

kx − ykλ(d(x, y)) ≥ E×E

R

ψ(E)×ψ(E)

kx − ykλ(d(x, y)).

(e) Now suppose that X = Rr . Then ν(∂*ψ(A)) ≤ ν(∂*A) for every A ⊆ Rr , where ∂*A is the essential boundary of A (definition: 475B). proof (a) That φA : A → ψ(A) is a bijection is immediate from the definitions of ψ and φA . If α < 0, then for any x ∈ A either φA (x) = x or (x|e) < α and R(x) ∈ / A. In the latter case kφA (x)k2 = kR(x)k2 = kx + 2γek2 (where γ = α − (x|e) > 0) = kxk2 + 4γ(x|e) + 4γ 2 = kxk2 + 4γα < kxk2 , so kφA (x)k < kxk. (b) If we set E1 = {x : x ∈ E, (x|e) ≥ α}, E2 = {x : x ∈ E, (x|e) < α, Reα (x) ∈ E}, E3 = {x : x ∈ E, (x|e) < α, Reα (x) ∈ / E}, E4 = {x : x ∈ Rr \ E, (x|e) > α, Reα (x) ∈ E}, then E1 , E2 , E3 and E4 are disjoint measurable sets, E = E1 ∪ E2 ∪ E3 , ψ(E) = E1 ∪ E2 ∪ E4 and φE ¹E3 = R¹E3 is a measure space isomorphism for the subspace measures on E3 and E4 .

476I


(c) By (a),

R

kxkν(dx) ≥ E

R

kφE (x)kν(dx) = E

695

R ψ(E)

kxkν(dx)

by 235Ic, because φE : E → ψ(E) is inverse-measure-preserving, with equality only when {x : kxk > kφE (x)k} = {x : x ∈ E, (x|e) < α, R(x) ∈ / E} is negligible. (d) Note first that if Λ is the domain of λ then Λ includes the Borel algebra of X × X (because X is second-countable, so this is just the σ-algebra generated by products of Borel sets, by 4A3G); so that (x, y) 7→ kx − yk is Λ-measurable, and the integral is defined in [0, ∞]. Now consider the sets W1 = {(x, y) : x ∈ E, y ∈ E, R(x) ∈ / E, R(y) ∈ E, (x|e) < α, (y|e) < α}, W10 = {(x, y) : x ∈ E, y ∈ E, R(x) ∈ / E, R(y) ∈ E, (x|e) < α, (y|e) > α}, W2 = {(x, y) : x ∈ E, y ∈ E, R(x) ∈ E, R(y) ∈ / E, (x|e) < α, (y|e) < α}, W20 = {(x, y) : x ∈ E, y ∈ E, R(x) ∈ E, R(y) ∈ / E, (x|e) > α, (y|e) < α}. Then (x, y) 7→ (x, R(y)) : W10 → W1 is a measure space isomorphism for the subspace measures induced on W1 and W10 by λ, so Z Z kφE (x) − φE (y)kλ(d(x, y)) = kR(x) − ykλ(d(x, y)) W1 ZW1 kR(x) − R(y)kλ(d(x, y)) = W10

Z =

kx − ykλ(d(x, y)). W10

Similarly, Z

Z W10

kφE (x) − φE (y)kλ(d(x, y)) =

kR(x) − ykλ(d(x, y)) W10

Z =

kR(x) − R(y)kλ(d(x, y)) ZW1

=

kx − ykλ(d(x, y)). W1

So we get

R W1 ∪W10


R W1 ∪W10


In the same way, (x, y) 7→ (R(x), y) is an isomorphism of the subspace measures on W2 and W20 , and we have

R

W2 ∪W20


R

W2 ∪W20


On the other hand, for all (x, y) ∈ (E × E) \ (W1 ∪ W10 ∪ W2 ∪ W20 ), we have kφE (x) − φE (y)k ≤ kx − yk. (Either x and y are both left fixed by φ, or both are moved, or one is on the reflecting hyperplane, or one is moved to the same side of the reflecting hyperplane as the other.) So we get Z Z kx − ykλ(d(x, y)) ≥ kφE (x) − φE (y)kλ(d(x, y)) E×E E×E Z = kx − ykλ(d(x, y)) ψ(E)×ψ(E)

because (x, y) 7→ (φE (x), φE (y)) is an inverse-measure-preserving transformation for the subspace measures on E × E and ψ(E) × ψ(E).

696


476I

(e)(i) Because R is both an isometry and a measure space automorphism, cl*R[A] = R[cl*A] and int*R[A] = R[int*A], where cl*A and int*A are the essential closure and the essential interior of A, as in 475B. Recall that cl*A, int*A and ∂*A are all Borel sets (475Cc), so that ∂*A and ∂*ψ(A) are measurable. (ii) Suppose that x . e = α. Then x ∈ ∂*ψ(A) iff x ∈ ∂*A. P P It is easy to check that B(x, δ) ∩ ψ(A) = ψ(B(x, δ) ∩ A) for any δ > 0, so that ν ∗ (B(x, δ) ∩ ψ(A)) = ν ∗ (ψ(B(x, δ) ∩ A)) for every δ > 0 and x ∈ cl*A iff x ∈ cl*ψ(A). If x = R(x) ∈ int*A, then x ∈ int*R[A] so x ∈ int*(A ∩ R[A]) (475Cd) and x ∈ int*ψ(A). If x ∈ int*ψ(A) then x ∈ int*(R[ψ(A)] ∩ ψ(A)) ⊆ int*A. Q Q (iii) If x ∈ ∂*ψ(A) \ ∂*A then R(x) ∈ ∂*A \ ∂*ψ(A). P P By (ii), x. e 6= α. case 1 Suppose that x .e > α. Setting δ = x . e − α, we see that U (x, δ) ∩ ψ(A) = U (x, δ) ∩ (A ∪ R[A]), while U (R(x), δ) ∩ ψ(A) = U (R(x), δ) ∩ A ∩ R[A]. Since x ∈ / int*ψ(A), x ∈ / int*(A ∪ R[A]) and x ∈ / int*A; since x also does not belong to ∂*A, x ∈ / cl*A. However, x ∈ cl*(A ∪ R[A]) = cl*A ∪ cl*R[A] (475Cd), so x ∈ cl*R[A] and R(x) ∈ cl*A. Next, x ∈ / int*R[A], so R(x) ∈ / int*A and R(x) ∈ ∂*A. Since x ∈ / cl*A, R(x) ∈ / cl*R[A] and R(x) ∈ / cl*ψ(A); so R(x) ∈ ∂*A \ ∂*ψ(A). case 2 Suppose that x . e < α. This time, set δ = α − x . e, so that U (x, δ) ∩ ψ(A) = U (x, δ) ∩ A ∩ R[A] and U (R(x), δ) ∩ ψ(A) = U (R(x), δ) ∩ (A ∪ R[A]). As x ∈ cl*ψ(A), x ∈ cl*(A ∩ R[A]) and R(x) ∈ cl*A. Also x ∈ cl*A; as x ∈ / ∂*A, x ∈ int*A, R(x) ∈ int*R[A] and R(x) ∈ int*ψ(A), so that R(x) ∈ / ∂*ψ(A). Finally, we know that x ∈ int*A but x ∈ / int*(A ∩ R[A]) (because x ∈ / int*ψ(A); it follows that x ∈ / int*R[A] so R(x) ∈ / int*A and R(x) ∈ ∂*A \ ∂*ψ(A). Thus all possibilities are covered and we have the result. Q Q (iv) What this means is that if we set E = ∂*ψ(A) \ ∂*A then R[E] ⊆ ∂*A \ ∂*ψ(A). So ν∂*ψ(A) = νE + ν(∂*ψ(A) ∩ ∂*A) = νR[E] + ν(∂*ψ(A) ∩ ∂*A) ≤ ν∂*A, as required in (e). This ends the proof of the lemma. 476J Theorem Let r ≥ 1 be an integer, and let µ be Lebesgue measure on Rr . For non-empty A ⊆ Rr and ² > 0, write U (A, ²) for {x : ρ(x, A) < ²}, where ρ is the Euclidean metric on Rr . If µ∗ A is finite, then µU (A, ²) ≥ µU (BA , ²), where BA is the closed ball with centre 0 and measure µ∗ A. proof (a) To begin with, suppose that A is bounded. Set γ = µ∗ A, β = µU (A, ²). If γ = 0 then (because A 6= ∅) µU (A, ²) ≥ µU ({0}, ²) = µU (BA , ²), and we can stop. So let us suppose henceforth that γ > 0. Let M ≥ 0 be such that A ⊆ B(0, M ), and consider the family F = {F : F ∈ C \ {∅}, µF ≥ γ, µU (F, ²) ≤ β}, where C is the family of closed subsets of B(0, M ). Because U (A, ²) = U (A, ²), A ∈ F and F is non-empty. By 476D and 476F, F is closed for the Hausdorff metric on C \ {∅}, therefore compact, by 476Ad, because B(0, M ) is compact. Next, the function F 7→

R

F

(M − kxk)µ(dx) : C \ {∅} → [0, ∞[

is upper semi-continuous, by 476Dc. It therefore attains its supremum on F at some F0 ∈ F (4A2Gk). Let F1 ⊆ F0 be a closed self-supporting set of the same measure as F0 ; then U (F1 , ²) ⊆ U (F0 , ²) and µF1 = µF0 , so F1 ∈ F ; also

R

F1

M − kxkµ(dx) =

R

F0

M − kxkµ(dx) ≥

R

F

M − kxkµ(dx)

for every F ∈ F. (b) Now F1 is a ball with centre 0. P P?? Suppose, if possible, otherwise. Then there are x1 ∈ F1 and x0 ∈ Rr \F1 such that kx0 k < kx1 k. Set e = then α=

1 (x0 2kx0 −x1 k

1 (x0 −x1 ), kx0 −x1 k

− x1 ) .(x0 + x1 ) =

1 2

so that e is a unit vector, and α = e . (x0 +x1 );

1 (kx0 k2 2kx0 −x1 k

− kx1 k2 ) < 0.

476K


697

Define R = Reα : Rr → Rr and ψ = ψeα as in 476H. Set F = ψ(F1 ). Then F is closed (476Hb) and µF = µF1 ≥ µ∗ A (476Ib). Also U (F, ²) ⊆ ψ(U (F1 , ²)) (476Ha), so µU (F, ²) ≤ µ(ψ(U (F1 , ²)) = µU (F1 , ²) ≤ β R R R and F ∈ F. It follows that F M −kxkµ(dx) ≤ F1 M −kxkµ(dx); as µF = µF1 , F kxkµ(dx) ≥ F1 kxkµ(dx). By 476Ic, G = {x : x ∈ F1 , x . e < α, R(x) ∈ / F1 } is negligible. But G contains x1 and is relatively open in F1 , and F1 is supposed to be self-supporting; so this is impossible. X XQ Q R

(c) Since µF1 ≥ γ, F1 ⊇ BA , and µU (BA , ²) ≤ µU (F1 , ²) ≤ β = µU (A, ²). So we have the required result for bounded A. In general, given an unbounded set A of finite measure, let δ be the radius of BA ; then µU (BA , ²) = µB(0, δ + ²) = sup µB(0, α + ²) α<δ

≤

sup A0 ⊆A is bounded

µU (BA0 , ²) ≤

sup

µU (A, ²) = µU (A, ²)

A0 ⊆A is bounded

because {U (A0 , ²) : A0 ⊆ A is bounded} is an upwards-directed family of open sets with union U (A, ²), and µ is τ -additive. So the theorem is true for unbounded A as well.

476K Theorem Let r ≥ 1 be an integer, and let µ be Lebesgue measure on Rr ; write λ for the product measure on Rr × Rr . For any measurable set E ⊆ Rr of finite measure, write BE for the closed ball with centre 0 and the same measure as E. Then

R

E×E

kx − ykλ(d(x, y)) ≥

R

BE ×BE


proof (a) Suppose for the moment that E is compact and not empty, and that ² > 0. Let M ≥ 0 be such r that kxk ≤ M for every x ∈ E. For R non-empty sets A ⊆ R set U (A, ²) = {x : ρ(x, A) < ²}, where ρ is r Euclidean distance on R . Set β = U (E,²)×U (E,²) kx−ykλ(d(x, y)). Let F be the family of non-empty closed R subsets F of the ball B(0, M ) = {x : kxk ≤ M } such that µF ≥ µE and U (F,²)×U (F,²) kx−ykλ(d(x, y)) ≤ β. Then F is compact for the Hausdorff metric on the family C \ {∅} of non-empty closed subsets of B(0, M ). P P We know from 476Ad that C \ {∅} is compact, and from 476Da that {F : µF ≥ µE} is closed. To see that F is closed, consider the metric σ on Rr × Rr defined by setting σ((x, y), (x0 , y 0 )) = max(kx − x0 k, ky − y 0 k), and the measure ν which is the indefinite-integral measure over λ defined by the function (x, y) 7→ kx − yk. Writing C2 for the family of non-empty closed subsets of Rr , the map F 7→ F × F : C \ {∅} → C2 \ {∅} is continuous (476B) and the map W 7→ νU (W, ²) : C2 \ {∅} → Rr is lower semi-continuous, where U (W, ²) = {(x, y) : σ((x, y), W ) < ²} (476F). But this means that F 7→ U (F,²)×U (F,²) kx − ykλ(d(x, y)) is lower semiR continuous, and {F : U (F,²)×U (F,²) kx − ykλ(d(x, y)) ≤ β} is closed. Thus F is a closed subset of C \ {∅} and is compact. Q Q R (b) Since E ∈ F, F is not empty. By 476Dc, there is an F0 ∈ F such that F0 M − kxkµ(dx) ≥ R M − kxkµ(dx) for every F ∈ F. Let F1 ⊆ F0 be a closed self-supporting set of the same measure; then F R U (F1 , ²) ⊆ U (F0 , ²), so U (F1 ,²)×U (F1 ,²) kx − ykλ(d(x, y)) ≤ β and F1 ∈ F; also

R

F1

M − kxkµ(dx) =

R

F0

M − kxkµ(dx) ≥

R

F

M − kxkµ(dx)

for every F ∈ F. Now F1 is a ball with centre 0. P P?? Suppose, if possible, otherwise. Then (just as in the proof of 476K) there are x1 ∈ F1 and x0 ∈ Rr \F1 such that kx1 k > kx0 k. Set e =

1 (x0 −x1 ) and α kx0 −x1 k

1 2

= e .(x0 +x1 ) < 0.

Define R = Reα : Rr → Rr and ψ = ψeα as in 476H. Set F = ψ(F1 ). Then F is closed and µF = µF1 ≥ µE and U (F, ²) ⊆ ψ(U (F1 , ²)). So

698


Z

476K

Z kx − ykλ(d(x, y)) ≤

U (F,²)×U (F,²)

kx − ykλ(d(x, y)) ψ(U (F1 ,²))×ψ(U (F1 ,²))

Z ≤

kx − ykλ(d(x, y)) U (F1 ,²))×U (F1 ,²)

(476Id) ≤ β. R R R This means that F ∈ F. Accordingly F1 M −kxkµ(dx) ≥ F M −kxkµ(dx); since µF = µF1 , F1 kxkµ(dx) ≤ R kxkµ(dx). By 476Ic, G = {x : x ∈ F1 , x . e < α, R(x) ∈ / F1 } must be negligible. But G contains x1 and F is relatively open in F1 , and F1 is supposed to be self-supporting; so this is impossible. X XQ Q (c) Since µF1 ≥ µE, F1 ⊇ BE , and Z Z kx − ykλ(d(x, y)) ≤ BE ×BE

kx − ykλ(d(x, y)) U (F1 ,²)×U (F1 ,²)

Z

≤β=

kx − ykλ(d(x, y)). U (E,²)×U (E,²)

At this point, recall that ² was arbitrary. Since E is compact, Z Z kx − ykλ(d(x, y)) = inf kx − ykλ(d(x, y)) ²>0 U (E,²)×U (E,²) E×E Z ≥ kx − ykλ(d(x, y)). BE ×BE

(d) Thus the result is proved for non-empty compact sets E. In general, given a measurable set E of finite measure, then if E is negligible the result is trivial; and otherwise, writing δ for the radius of BE , Z Z kx − ykλ(d(x, y)) = sup kx − ykλ(d(x, y)) α<δ B(0,α)×B(0,α) BE ×BE Z ≤ sup kx − ykλ(d(x, y)) K⊆E is compact BK ×BK Z ≤ sup kx − ykλ(d(x, y)) K⊆E is compact K×K Z = kx − ykλ(d(x, y)), E×E

so the proof is complete. 476L The Isoperimetric Theorem Let r ≥ 1 be an integer, and let µ be Lebesgue measure on Rr . If E ⊆ Rr is a measurable set of finite measure, then per E ≥ per BE , where BE is the closed ball with centre 0 and the same measure as E, while per E is the perimeter of E as defined in 474D. proof (a) Suppose to begin with that E ⊆ B(0, M ), where M ≥ 0, and that per E < ∞. Let F be the family of measurable sets F ⊆ Rr such that F \ B(0, M ) is negligible, µF ≥ µE and per F ≤ per E, with the topology of convergence in measure (474S). Then F = {F : per F ≤ per E, µ(F ∩ B(0, M )) ≥ µE, µ(F ∩ B(0, α)) ≤ µ(F ∩ B(0, M )) for every α ≥ 0} is a closed R subset of {F : per F ≤ per E}, which is compact (474Sb), so F is compact. For F ∈ F set h(F ) = F kxkµ(dx); then |h(F ) − h(F 0 )| ≤ M µ((F 4F 0 ) ∩ B(0, M )) for all F , F 0 ∈ F, so h is continuous.

476N


699

There is therefore an F0 ∈ F such that h(F0 ) ≤ h(F ) for every F ∈ F. Set F1 = cl*F0 , so that F1 4F0 is negligible (475Cg), F1 ∈ F, F1 ⊆ B(0, M ) and h(F1 ) = h(F0 ). (b) Writing δ = supx∈F1 kxk, we have U (0, δ) ⊆ F1 . P P?? Otherwise, there are x0 ∈ Rr \ F1 and x1 ∈ F1 1 (x0 kx0 −x1 k

such that kx0 k < kx1 k. Set e =

1 2

− x1 ), α = e. (x0 + x1 ) < 0, R = Reα : Rr → Rr and ψ = ψeα .

Set F = ψ(F1 ) and let φ = φF1 : F1 → F be the function described in 476I. Then kφ(x)k ≤ kxk for every x ∈ F1 (476Ia). In particular, F = φ[F1 ] ⊆ B(0, M ). Now F is measurable and µF = µF1 ≥ µE, by 476Ib. Also, writing ν for normalized r − 1-dimensional Hausdorff measure on Rr , per F = ν(∂*F ) ≤ ν(∂*F1 ) = per F1 ≤ per E, by 475Mb and 476Ie. So F ∈ F, and

R

F

kxkµ(dx) ≥

R F0

kxkµ(dx) =

R F1

kxkµ(dx).

By 476Ic, G = {x : x ∈ F1 , x . e < α, R(x) ∈ / F1 } is negligible. But now consider G ∩ U (x1 , η) for small η > 0. Since x1 belongs to F1 = cl*F0 = cl*F1 , but x0 does not, lim supη↓0

µ(F1 ∩B(x1 ,η)) µB(x1 ,η)

> 0 = lim supη↓0

µ(F1 ∩B(x0 ,η)) . µB(x0 ,η)

There must therefore be some η > 0 such that η < 12 kx1 −x0 k and µ(F1 ∩B(x0 , η)) < µ(F1 ∩B(x1 , η)). In this case, however, G ⊇ F1 ∩B(x1 , η)\R[F1 ∩B(x0 , η)] has measure at least µ(F1 ∩B(x1 , η))−µ(F1 ∩B(x0 , η)) > 0, which is impossible. X XQ Q (c) Thus U (0, δ) ⊆ F1 ⊆ B(0, δ) and per F1 = per B(0, δ). Since µF1 ≥ µE, the radius of BE is at most δ and per BE ≤ per B(0, δ) = per F1 ≤ per E. (d) Thus the result is proved when E is bounded and has finite perimeter. Of course it is trivial when E has infinite perimeter. Now suppose that E is any measurable set with finite measure and finite perimeter. Set Eα = E ∩ B(0, α) for α ≥ 0; then per E = lim inf α→∞ per Eα (475Mc). By (a)-(c), per Eα ≥ per BEα ; since per BEα → per BE as α → ∞, per E ≥ per BE in this case also. 476M Spheres in inner product spaces For the rest of the section I will use the following notation. Let X be a (real) inner product space. Then SX will be the unit sphere {x : x ∈ X, kxk = 1}. Let HX be the isometry group of SX with its topology of pointwise convergence (441G). A cap in SX will be a set of the form {x : x ∈ SX , (x|e) ≥ α} where e ∈ SX and −1 ≤ α ≤ 1. When X is finite-dimensional, it is isomorphic, as inner product space, to Rr , where r = dim X (4A4Jd). In this case SX is compact, so has a unique HX -invariant Radon probability measure νX , which is strictly positive (476G). If r ≥ 1 is an integer, we know that the r − 1-dimensional Hausdorff measure of the sphere SRr is finite and non-zero (265F). Since Hausdorff measures are invariant under isometries (264G, 471J), and are quasi-Radon measures when totally finite (471Dh), r − 1-dimensional Hausdorff measure on SRr is a multiple of the normalized invariant measure νRr , by 476G. The same is therefore true in any r-dimensional inner product space. 476N Lemma Let X be a real inner product space and f ∈ HX . Then (f (x)|f (y)) = (x|y) for all x, y ∈ SX . Consequently f (αx + βy) = αf (x) + βf (y) whenever x, y ∈ SX and α, β ∈ R are such that αx + βy ∈ SX . proof (a) We have ρ(x, y)2 = (x − y|x − y) = (x|x) − 2(x|y) + (y|y) = 2 − 2(x|y), so 1 2

1 2

(x|y) = 1 − ρ(x, y)2 = 1 − ρ(f (x), f (y))2 = (f (x)|f (y)).

700


476N

(b) kf (αx + βy) − αf (x) − βf (y)k2 = (f (αx + βy) − αf (x) − βf (y)|f (αx + βy) − αf (x) − βf (y)) = 1 + α2 + β 2 − 2α(f (αx + βy)|f (x)) − 2β(f (αx + βy)|f (y)) + 2αβ(f (x)|f (y)) = 1 + α2 + β 2 − 2α(αx + βy|x) − 2β(αx + βy|y) + 2αβ(x|y) = k(αx + βy) − αx − βyk2 = 0.

476O I give a theorem on concentration of measure on the sphere corresponding to 476J. Theorem Let X be a finite-dimensional inner product space of dimension at least 2, SX its unit sphere and νX the invariant Radon probability measure on SX . For a non-empty set A ⊆ SX and ² > 0, write U (A, ²) = {x : ρ(x, A) < ²}, where ρ is the usual metric of X. Then there is a cap C ⊆ SX such that ∗ νX C = νX A, and νX (SX ∩ U (A, ²)) ≥ νX (SX ∩ U (C, ²)) for any such C. proof In order to apply the results of 476H-476I directly, and simplify some of the formulae slightly, it will be helpful to write ν for the Radon measure on X defined by setting νE = νX (E ∩ SX ) whenever this is ∗ defined. By 214Cd, ν ∗ agrees with νX on PSX . (a) The first step is to check that there is a cap C of SX such that νC = ν ∗ A. P P Take any e0 ∈ SX , and set Cα = {x : x ∈ SX , (x|e0 ) ≥ α} for α ∈ [−1, 1]. νCα is defined for every α ∈ R because every Cα is closed and ν is a topological measure. Now examine the formulae of 265F. We can identify X with Rr+1 where r + 1 = dim X; do this in such a way that e0 corresponds to the unit vector (0, . . . , 0, 1). We have r−1 a parametrization φr : Dr → SX , where Dr is a Borel subset of Rr with interior ]−π, π[ × ]0, π[ and φr is differentiable with continuous derivative. Moreover, if x = (ξ1 , . . . , ξr ) ∈ Dr , then φr (x) . e0 = cos ξr , and the Jacobian Jr of φr is bounded R by 1 and never zero on int Dr . Finally, the boundary ∂Dr is negligible. What this means is that νr Cα = Eα Jr dµr , where µr is Lebesgue measure on Rr , νr is normalized Hausdorff r-dimensional measure on Rr+1 , and Eα = {x : x ∈ Dr , cos ξr ≥ α}. So if −1 ≤ α ≤ β ≤ 1 then νr Cα − νr Cβ ≤ µr (Eα \ Eβ ) ≤ 2π r−1 (arccos α − arccos β); because arccos is continuous, so is α 7→ νr Cα . Also, if α < β, then Eα \ Eβ is non-negligible, so R J dµr 6= 0 and νr Cα > νr Cβ . Eα \Eβ r This shows that α 7→ νr Cα is continuous and strictly decreasing; since νr is just a multiple of ν on SX , the same is true of α 7→ νCα . Since νC−1 = νSX = 1 and νC1 = ν{e0 } = 0, the Intermediate Value Theorem tells us that there is a unique α such that νCα = ν ∗ A, and we can set C = Cα . Q Q (b) Now take any non-empty set A ⊆ SX and any ² > 0, and set γ = ν ∗ A, β = νU (A, ²). Let C be a cap such that ν ∗ A = νC; let e0 be the centre of C. Consider the family F = {F : F ∈ C \ {∅}, F ⊆ SX , νF ≥ γ, νU (F, ²) ≤ β}, where C is the family of closed subsets of X. Because U (A, ²) = U (A, ²), A ∈ F and F is non-empty. By 476D and 476F, F is closed for the Hausdorff metric on C \ {∅}, therefore compact, by 475Ad, because SX is compact. Next, the function F 7→

R

F

max(0, 1 + (x|e0 ))ν(dx) : C \ {∅} → [0, ∞[

is upper semi-continuous, by 476Dc. It therefore attains its supremum on F at someR F0 ∈ F . Let F1 ⊆ F0 Rbe a self-supporting closed set with the same measure as F0 ; then F1 ∈ F and F (1 + (x|e0 ))ν(dx) ≥ (1 + (x|e0 ))ν(dx) for every F ∈ F . F1 (c) F1 is a cap with centre e0 . P P?? Otherwise, there are x0 ∈ SX \ F1 and x1 ∈ F1 such that (x0 |e0 ) > (x1 |e0 ). Set e =

1 (x0 kx0 −x1 k

− x1 ). Then e ∈ SX and (e|e0 ) > 0. Set R = Re0 and ψ = ψe0 as defined

in 476H; write F for ψ(F1 ). Note that (x0 + x1 |x0 − x1 ) = kx0 k2 − kx1 k2 = 0, so (x0 + x1 |e) = 0 and

476Xb


701

R(x0 ) = x1 . Also R[SX ] = SX , so ν is R-invariant, because ν is a multiple of Hausdorff r − 1-dimensional measure on SX and must be invariant under isometries of SX . We have νF = νF1 ≥ γ, by 476Ib, and νU (F, ²) ≤ νψ(U (F1 , ²)) = νU (F1 , ²) ≤ νU (F0 , ²) ≤ β by 476Ha. So F ∈ F. But consider the standard bijection φ = φF1 : F1 → F as defined in 476I. We have

R

F1

(1 + (φ(x)|e0 ))ν(dx) =

R

F

(1 + (x|e0 ))ν(dx) ≤

R

F1

(1 + (x|e0 ))ν(dx).

But if we examine the definition of φ, we see that φ(x) 6= x only when (x|e) < 0 and φ(x) = R(x), so that in this case φ(x)−x is a positive multiple of e and (φ(x)|e0 ) > (x|e0 ). So G = {x : x ∈ F1 , (x|e) < 0, R(x) ∈ / F1 } must be ν-negligible. But G includes a relative neighbourhood of x1 in F1 and F1 is supposed to be selfsupporting for ν, so this is impossible. X XQ Q (e) Now ν ∗ A = γ ≤ νF1 , so C ⊆ F1 and νU (A, ²) = β ≥ νU (F1 , ²) ≥ νU (C, ²), as claimed. 476P Corollary For any ² > 0, there is an r0 ≥ 2 such that whenever X is a finite-dimensional inner ∗ ∗ product space of dimension at least r0 , A1 , A2 ⊆ SX and min(νX A1 , νX A2 ) ≥ ², then there are x ∈ A1 , y ∈ A2 such that kx − yk ≤ ². proof Take r0 ≥ 2 such that r0 ²3 > 2. Suppose that dim X = r ≥ r0 . Fix e0 ∈ SX . We need an estimate of νX C²/2 , where C²/2 = {x : x ∈ SX , (x|e0 ) ≥ ²/2} as in 476O. To get this, let e1 , . . . , er−1 be such that e0 , . . . , er−1 is an orthonormal basis of X (4A4Kc). For each i < r, there is an f ∈ HX such that f (ei ) = e0 , so that (x|ei ) = (f (x)|e0 ) for every x (476N), and

R

(x|ei )2 νX (dx) =

R

(f (x)|e0 )2 νX (dx) =

R

(x|e0 )2 νX (dx),

because f : SX → SX is inverse-measure-preserving for νX . Accordingly 1

νX C²/2 = νX {x : x ∈ SX , |(x|e0 )| ≥ ²/2} 2 Z Z 2 ≤ 2 (x|e0 )2 νX (dx) < r² (x|e0 )2 νX (dx) ²

=²

SX

r−1 Z X i=0

SX

SX

Z (x|ei )2 νX (dx) = ²

r−1 X

SX i=0

∗ (x|ei )2 νX (dx) = ² ≤ νX A1 .

So, taking C to be the cap of SX with centre e0 and measure ν ∗ A1 , C = Cα where α < 12 ², and 1 2

νX (SX ∩ U (A1 , 21 ²)) ≥ νX (SX ∩ U (Cα , 12 ²)) ≥ νX Cα−²/2 > . Similarly, νX (SX ∩ U (A2 , 12 ²)) > 12 and there must be some z ∈ SX ∩ U (A1 , 21 ²) ∩ U (A2 , 21 ²). Take x ∈ A1 and y ∈ A2 such that kx − zk < 21 ² and ky − zk < 12 ²; then kx − yk ≤ ², as required. 476X Basic exercises (a) Let (X, ρ) be a metric space and give C \ {∅} its Hausdorff metric, where C is the family of closed subsets of X. (i) Show that x 7→ {x(i) : i < n} : X n → C \ {∅} is continuous for every n ≥ 1. (ii) Show that (E, F ) 7→ E ∪ F : (C \ {∅})2 → C \ {∅} is continuous. (iii) Show that diam : C \ {∅} → [0, ∞] is continuous. (b) Let (X, ρ) be a metric space and K the family of compact subsets of X. (i) Show that the Hausdorff metric topology and the Vietoris topology coincide on K \ {∅}. (ii) Show that if G is the isometry group of X with its topology of pointwise convergence (441G), then (g, K) 7→ g[K] is a continuous action of G on K if K is given the topology induced by the Vietoris topology.

702


476Xc

(c) Let (X, ρ) be a metric space, and C the set of closed subsets of X. Show that the topology induced by the Hausdorff metric on C \ {0} is finer than the subspace topology induced by the k-Vietoris topology on C. (d) Let (X, ρ) be a locally compact metric space, and C the set of closed subsets of X with its k-Vietoris topology. (i) Show that if G is the isometry group of X with its topology of pointwise convergence, then (g, C) 7→ g[C] is a continuous action of G on C. (ii) Show that if X is separable then C is metrizable. (e) Let (X, ρ) be a metric space and give C \ {∅} its Hausdorff metric, where C is the family of closed subsets of X. ShowR that whenever µ is a topological measure on X and f is a µ-integrable real-valued function, then F 7→ F f dµ : C \ {0} → R is Borel measurable. (f ) Let X be a topological space and give C its Vietoris topology, where C is the family of closed subsets of X. (i) Show that if X is T0 then x 7→ {x(i) : i < n} : X n → C} is continuous for every n ≥ 1. (ii) Show that (E, F ) 7→ E ∪ F : C 2 → C is continuous. (g) In the context of 476H, show that diam ψeα (A) ≤ diam A for all A, e and α. > (h) Find an argument along the lines of those in 476J and 476K to prove 264H. (Hint: 476Xg.) > (i) Let X be an inner product space and SX its unit sphere. Show that every isometry f : SX → SX extends uniquely to an isometry Tf : X → X which is a linear operator. (Hint: first check the cases in which dim X ≤ 2.) Show that f is surjective iff Tf is, so that we have a natural isomorphism between the isometry group of SX and the group of invertible isometric linear operators. Show that this isomorphism is a homeomorphism for the topologies of pointwise convergence. (j) Let X be a finite-dimensional inner product space, νX the invariant Radon probability measure on the sphere SX , and E ∈ dom νX ; let C ⊆ SX be a cap R with the same measure Ras E, and let λ be the product measure of νX with itself on SX × SX . Show that C×C kx − ykλ(d(x, y)) ≤ E×E kx − ykλ(d(x, y)). (k) Let X be a finite-dimensional inner product space and νX the invariant Radon probability measure on the sphere SX . (i) Without appealing to the formulae in §265, show that νX (SX ∩ H) = 0 for whenever H ⊆ X is a proper affine subspace. (Hint: induce on dim H.) (ii) Use this to prove that if e ∈ SX then α 7→ νX {x : (x|e) ≥ α} is continuous. 476Y Further exercises (a) Let (X, ρ) and (Y, σ) be metric spaces, and f : X → Y a function. Write CX , CY for the spaces of closed subsets of X, Y respectively, and give CX \ {∅}, CY \ {∅} their Hausdorff metrics. For F ∈ CX \ {∅} set f˜(F ) = f [F ] ∈ CY \ {∅}. Show that the following are equiveridical: (i) f is uniformly continuous (ii); f˜ is continuous; (iii) f˜ is uniformly continuous. (b) Let X and Y be topological spaces, and CX , CY the spaces of closed subsets of X, Y respectively, with their Vietoris topologies. Suppose that f : X → Y is a continuous function and that Y is normal. Show that F 7→ f [F ] : CX → CY is continuous. (c) Let X be a compact metric space and G its isometry group. Suppose that H ⊆ G is a subgroup such that the action of H on X is transitive. Show that X has a unique H-invariant Radon probability measure which is also G-invariant. 476 Notes and comments The main theorems here (476J-476K, 476O), like 264H, are all ‘classical’; they go back to the roots of geometric measure theory, and the contribution of the twentieth century has been to extend the classes of sets for which balls or caps provide the bounding examples. It is very striking that they can all be proved with the same tools (see 476Xh). Of course I should remark that the Compactness Theorem (474S) lies at a much deeper level than the rest of the ideas here. (The proof of 474S relies on the distributional definition of ‘perimeter’ in 474D, while the arguments of 476Ie and 476L work with the Hausdorff measures of essential boundaries; so that we can join these ideas together only after proving all the

476 Notes


703

principal theorems of §§472-475.) So while ‘Steiner symmetrization’ (264H) and ‘concentration by partial reflection’ (476H) are natural companions, 476L is essentially harder than the other results. In all the theorems here, as in 264H, I have been content to show that a ball or a cap is an optimum for whatever inequality is being considered. I have not examined the question of whether, and in what sense, the optimum is unique. It seems that this requires deeper analysis.

704

Gauge integrals

Chapter 48 Gauge integrals For the penultimate chapter of this volume I turn to a completely different approach to integration which has been developed in the last forty years, following Kurzweil 57 and Henstock 63. This depends for its inspiration on a particular formulation of the Riemann integral (see 481Xe), and leads in particular to some remarkable extensions of the Lebesgue integral (§§483-484). While (in my view) it remains peripheral to the most important parts of measure theory, it has deservedly attracted a good deal of interest in recent years, and is entitled to a place here. From the very beginning, in the definitions of §122, I have presented the Lebesgue integral in terms of R almost-everywhere approximations by simple functions. Because the integral lim fn of a limit is not n→∞ R always the limit limn→∞ fn of the integrals, we are forced, from the start, to constrain ourselves by the ordering, and to work with monotone or dominated sequences. This almost automatically leads us to an ‘absolute’ integral, in which |f | is integrable whenever f is, whether we start from measures (as in Chapter 11) or from linear functionals (as in §436). For four volumes now I have been happily developing the concepts and intuitions appropriate to such integrals. But if we return to one of the foundation stones of Lebesgue’s theory, the Fundamental Theorem of Calculus, we find that it is easy to construct a differentiable function f such that the absolute value |f 0 | of its derivative is not integrable (483Xd). It was observed very early (Perron 14) that the Lebesgue integral can be extended to integrate the derivative of any function which is differentiable everywhere. The achievement of Henstock 63 was to find a formulation of this extension which was conceptually transparent enough to lend itself to a general theory, fragments of which I will present here. The first step is to set out the essential structures on which the theory depends (§481), with a first attempt at a classification scheme. (One of the most interesting features of the Kurzweil-Henstock approach is that we have an extraordinary degree of freedom in describing our integrals, and apart from the Henstock integral itself it is not clear that we have yet found the right canonical forms to use.) In §482 I give a handful of general theorems showing what kind of results can be expected and what difficulties arise. In §483, I work through the principal properties of the Henstock integral on the real line, showing, in particular, that it coincides with the Perron and special Denjoy integrals. Finally, in §484, I look at a very striking integral on Rr , due to W.F.Pfeffer.

481 Tagged partitions I devote this section to establishing some terminology (481A-481B, 481E-481G) and describing a variety of examples (481I-481Q), some of which will be elaborated later. The clearest, simplest and most important example is surely Henstock’s integral on a closed bounded interval (481J), so I recommend turning immediately to that paragraph and keeping it in mind while studying the notation here. It may also help you to make sense of the definitions here if you glance at the statements of some of the results in §482; in this section I give only the formula defining gauge integrals (481C), with some elementary examples of its use (481Xb-481Xh). 481A Tagged partitions and Riemann sums The common idea underlying all the constructions of this chapter is the following. We R have a set X and a functional ν defined on some family C of subsets of X. We seek to define an integral f dν, for functions f with domain X, as a limit of finite Riemann sums Pn There is no strict reason, at this stage, to forbid i=0 f (xi )νCi , where xi ∈ X and Ci ∈ C for i ≤ n. repetitions in the string (x0 , C0 ), . . . , (xn , Cn ), but also little to be gained from allowing them, and it will simplify some of the formulae below if I say from the outset that a tagged partition on X will be a finite subset t of X × PX. So one necessary element of the definition will be a declaration of which tagged partitions {(x0 , C0 ), . . . , (xn , Cn )} will be employed, in terms, for instance, of which sets Ci are permitted, whether they are allowed to overlap at their boundaries, whether they are required to cover the space, and whether each tag xi is required to belong to the corresponding Ci . The next element of the definition will be a description of a filter F on the set T of tagged partitions, so that the integral will be the limit (when it exists) of the sums along the filter, as in 481C below.

481Db

Tagged partitions

705

In the formulations studied in this chapter, the Ci will generally be disjoint, but this is not absolutely essential, and it is occasionally convenient to allow them to overlap in ‘small’ sets, as in 481Ya. In some cases, we can restrict attention to families for which the Ci are non-empty and have union X, so that {C0 , . . . , Cn } is a partition of X in the strict sense. 481B Notation Let me immediately introduce notations which will be in general use throughout the chapter. (a) First, a shorthand to describe a particular class of sets of tagged partitions. If X is a set, a straightforward set of tagged partitions on X is a set of the form T = {tt : t ∈ [Q]<ω , C ∩ C 0 = ∅ whenever (x, C), (x0 , C 0 ) are distinct members of t } where Q ⊆ X × PX; S I will say that T is generated by Q. In this case, of course, Q can be recovered from T , since Q = T . Note that no control is imposed on the tags at this point. It remains theoretically possible that a pair (x, ∅) should belong to Q, though in essentially all applications this will be excluded in one way or another. (b) If X is a set and t ⊆ X × PX is a tagged partition, I write S Wt = {C : (x, C) ∈ t }. (c) If X is a set, C is a family of subsets of X, f : X → R and ν : C → R are functions and t ∈ [X × C]<ω is a tagged partition, then P St (f, ν) = (x,C)∈tt f (x)νC. 481C Proposition Let X be a set, C a family of subsets of X, T ⊆ [X × C]<ω a non-empty set of tagged partitions and F a filter on T . For any functions f : X → R and ν : C → R, set Iν (f ) = limt →F St (f, ν) if this is defined in R. (a) Iν is a linear functional defined on a linear subspace of RX . (b) Now suppose that νC ≥ 0 for every C ∈ C. Then (i) Iν is a positive linear functional (definition: 351F); (ii) if f , g : X → R are such that |f | ≤ g and Iν (g) is defined and equal to 0, then Iν (f ) is defined and equal to 0. proof (a) We have only to observe that if f , g ∈ RX and α ∈ R, then St (f + g, ν) = St (f, ν) + St (g, ν),

St (αf, ν) = αSt (f, ν)

for every t ∈ T , and apply 2A3Sf. (b) If g ≥ 0 in RX , that is, g(x) ≥ 0 for every x ∈ X, then St (g, ν) ≥ 0 for every t ∈ T , so the limit Iν (g), if defined, will also be non-negative. Next, if |f | ≤ g, then |St (f, ν)| ≤ St (g, ν) for every t , so if Iν (g) = 0 then Iν (f ) is also zero. 481D Remarks (a) Functionals Iν = limt →F St (., ν), as described above, are called gauge integrals. (b) In fact even greater generality is possible at this point. There is no reason why f and ν should take real values. All we actually need is an interpretation of sums of products f (x) × νC in a space in which we can define limits. So for any linear spaces U , V and W with a bilinear functional φ : U × V → W (253A) and a Hausdorff linear space topology on W , we can set out to construct P an integral of a function f : X → U with respect to a functional ν : C → V as a limit of sums St (f, ν) = (x,C)∈tt φ(f (x), νC) in W . I will not go farther along this path here. But it is worth noting that the constructions of this chapter lead the way to interesting vector integrals of many types.

706

Gauge integrals

481Dc

(c) An extension which is, however, sometimes useful is to allow ν to be undefined (or take values outside R, such as ±∞) on part of C. In this case, set C0 = ν −1 [R]. Provided that T ∩ [X × C0 ]<ω belongs to F, we can still define Iν , and 481C will still be true. 481E Gauges The most useful method (so far) of defining filters on sets of tagged partitions is the following. If X is a set, a gauge on X is a subset δ of X × PX. For a gauge δ, a tagged partition t is δ-fine if t ⊆ δ. Now, for a set ∆ of gauges and a set T of tagged partitions, we can seek to define a filter F on T as the filter generated by sets of the form Tδ = {tt : t ∈ T is δ-fine} as δ runs over ∆. Of course we shall need to establish that T and ∆ are compatible in the sense that {Tδ : δ ∈ ∆} has the finite intersection property; this will ensure that there is indeed a filter containing every Tδ (4A1Ia). In nearly all cases, ∆ will be non-empty and downwards-directed (that is, for any δ1 , δ2 ∈ ∆ there will be a δ ∈ ∆ such that δ ⊆ δ1 ∩ δ2 ); in this case, we shall need only to establish that Tδ is non-empty for every δ ∈ ∆. Note that the filter on T generated by {Tδ : δ ∈ ∆} depends only on T and the filter on X × PX generated by ∆. The most important gauges (so far) are ‘neighbourhood gauges’. If (X, T) is a topological space, a neighbourhood gauge on X is a set expressible in the form δ = {(x, C) : x ∈ X, C ⊆ Gx } where hGx ix∈X is a family of open sets such that x ∈SGx for every x ∈ X. It is useful to note (i) that the family hGx ix∈X can be recovered from δ, since Gx = {A : (x, A) ∈ δ} (ii) that δ1 ∩ δ2 is a neighbourhood gauge whenever δ1 and δ2 are. When (X, ρ) is a metric space, we can define a neighbourhood gauge δh from any function h : X → ]0, ∞[, setting δh = {(x, C) : x ∈ X, C ⊆ X, ρ(y, x) < h(x) for every y ∈ C}. The set of gauges expressible in this form is coinitial with the set of all neighbourhood gauges and therefore defines the same filter on any compatible set T of tagged partitions. Specializing yet further, we can restrict attention to constant functions h, obtaining the uniform metric gauges δη = {(x, C) : x ∈ X, C ⊆ X, ρ(x, y) < η for every y ∈ C} for η > 0, used in the Riemann integral (481I). (The use of the symbol ‘δ’ to represent a gauge has descended from its traditional appearance in the definition of the Riemann integral.) 481F Residual sets The versatility and power of the methods being introduced here derives from the P insistence on taking finite sums (x,C)∈tt f (x)νC, so that all questions about convergence are concentrated in the final limit limt →F St (f, ν). Since in a given Riemann sum we can look at only finitely many sets C of finite measure, we cannot insist, even when ν is Lebesgue measure on R, that Wt should always be X. There are many other cases in which it is impossible or inappropriate to insist that Wt = X for every tagged partition in T . We shall therefore need to add something to the definition of the filter F on T beyond what is possible in the language of 481E. In the examples below, the extra condition will always be of the following form. There will be a collection R of residual families R ⊆ PX. It will help to have a phrase corresponding to the phrase ‘δ-fine’: if R ⊆ PX, and t is a tagged partition on X, I will say that t is R-filling if X \ Wt ∈ R. Now, given a family R of residual sets, and a family ∆ of gauges on X, we can seek to define a filter F(T, ∆, R) on T as that generated by sets of the form TδR = {tt : t ∈ T is δ-fine and R-filling} where δ runs over ∆ and R runs over R. When there is such a filter, that is, the family {TδR : δ ∈ ∆, R ∈ R} has the finite intersection property, I will say that T is compatible with ∆ and R. It is important here to note that we shall not suppose that, for a typical residual family R ∈ R, subsets of members of R again belong to R; there will frequently be a restriction on the ‘shape’ of members of R as well as on their size. On the other hand, it will usually be helpful to arrange that R is a filter base, so that (if ∆ is also downwards-directed) we need only show that every TδR is non-empty, and {TδR : δ ∈ ∆, R ∈ R} will be a base for F(T, ∆, R). If the filter F is defined as in 481E, with no mention of a family R, we can still bring the construction into the framework considered here by setting R = {PX}. If it is convenient to define T in terms which do not impose any requirement on the sets Wt , but nevertheless we wish to restrict attention to sums St (f, ν) for which the tagged partition covers the whole space X, we can do so by setting R = {{∅}}.

481Hf

Tagged partitions

707

481G Subdivisions When we come to analyse the properties of integrals constructed by the method of 481C, there is an important approach which depends on the following combination of features. I will say that (X, T, ∆, R) is a tagged-partition structure allowing subdivisions, witnessed by C, if (i) X is a set. (ii) ∆ is a downwards-directed family of gauges on X. (iii)(α) R is a non-empty downwards-directed collection of families of subsets of X, all containing ∅; (β) for every R ∈ R there is an R0 ∈ R such that A ∪ B ∈ R whenever A, B ∈ R0 are disjoint. (iv) C is a family of subsets of X such that whenever C, C 0 ∈ C then C ∩ C 0 ∈ C and C \ C 0 is expressible as the union of a disjoint finite subset of C. (v) S Whenever C0 ⊆ C is finite and R ∈ R, there is a finite set C1 ⊆ C, including C0 , such that X \ C1 ∈ R. (vi) T ⊆ [X × C]<ω is (in the language of 481Ba) a straightforward set of tagged partitions on X. (vii) Whenever C ∈ C, δ ∈ ∆ and R ∈ R there is a δ-fine tagged partition t ∈ T such that Wt ⊆ C and C \ Wt ∈ R. 481H Remarks(a) Conditions (ii) and (iii-α) of 481G are included primarily for convenience, since starting from any ∆ and R we can find directed sets leading to the same filter F(T, ∆, R). (iii-β), on the other hand, is saying something new. (b) It is important to note, in (vii) of 481G, that the tags of t there are not required to belong to the set C. (c) All the applications below will fall into one of two classes. In one type, the residual families R ∈ R will be families of ‘small’ sets, in some recognisably measure-theoretic sense, and, in particular, we shall have subsets of members of any R belonging to R. In the other type, (vii) of 481G will be true because we can always find t ∈ T such that Wt = C. (d) The following elementary fact got left out of §136 and Chapter 31. Let A be a Boolean algebra and C ⊆ A. Set E = {sup C0 : C0 ⊆ C is finite and disjoint}. If c ∩ c0 and c \ c0 belong to E for all c, c0 ∈ C, then E is a subring of A. P P Write D for the family of finite disjoint subsets of C. (i) If C0 , C1 ∈ D, then for c ∈ C0 , c0 ∈ C1 there is a Dcc0 ∈ D with supremum c ∩ c0 . Now D = supc∈C0 ,c0 ∈C1 Dcc0 belongs to D and has supremum (sup C0 ) ∩ (sup C1 ). Thus e ∩ e0 ∈ E for all e, e0 ∈ E. (ii) Of course e ∪ e0 ∈ E whenever e, e0 ∈ E and e ∩ e0 = 0. (iii) Again suppose that C0 , C1 ∈ D. Then c \ c0 ∈ E for all c ∈ C0 , c0 ∈ C1 . By (i), c \ sup C1 ∈ E for every c ∈ C; by (ii), (sup C0 ) \ (sup C1 ) ∈ E. Thus e \ e0 ∈ E for all e, e0 ∈ E. (iv) Putting (ii) and (iii) together, e 4 e0 ∈ E for all e, e0 ∈ E. (v) As 0 = sup ∅ belongs to E, E is a subring of A. Q Q In particular, if C ⊆ PX has the properties in (iv) of 481G, then S E = { C0 : C0 ⊆ C is finite and disjoint} is a ring of subsets of X. (e) Suppose that X is a set and that R ⊆ PPX S satisfies (iii) of 481G. Then for every R ∈ R there is a non-increasing sequence hRn in∈N in R such that i≤n Ai ∈ R whenever Ai ∈ Ri for i ≤ n and hAi ii≤n is disjoint. P P Take R0 ∈ R such that R0 ⊆ R and A ∪ B ∈ R for all disjoint A, B ∈ R0 ; similarly, for n ∈ N, choose Rn+1 ∈ R such that Rn+1 ⊆ Rn and A ∪ B ∈ Rn for all disjoint A, B ∈ Rn+1 S . Now, given that Ai ∈ Ri for i ≤ n and hAi ii≤n is disjoint, we see by downwards induction on m that m
708

Gauge integrals

481Hf

S finite setSC1 ⊆ C such that X \ C1 ∈ R0 , by 481G(iv). By (d), there is a disjoint family C0 ⊆ C such that S C0 = C1 ; enumerate C0 as hCi ii 0 and C ∈ C, take a disjoint family hCi ii∈I of intervals of length less than 2η covering C, and xi to be the midpoint of Ci for i ∈ I; then t = {(xi , Ci ) : i ∈ I} belongs to T and is δη -fine, in the language of 481E, and Wt = C. 481J The Henstock integral on a bounded interval (Henstock 63) Take X, C, T and R as in 481I. This time, let ∆ be the set of all neighbourhood gauges on [a, b]. Then (X, T, ∆, R) is a tagged-partition structure allowing subdivisions, witnessed by C. proof Again, only 481G(vii) needs more than a moment’s consideration. Take any C ∈ C. Set a0 = inf C, b0 = sup C and let T0 be the family of δ-fine partitions t ∈ T such that Wt is a relatively closed initial subinterval of C, that is, is of the form C ∩ [a0 , yt ] for some yt ∈ [a0 , b0 ]. Set A = {yt : t ∈ T0 }. I have to show that there is a t ∈ T0 such that Wt = C, that is, that b0 ∈ A. Observe that there is an η0 > 0 such that (a0 , A) ∈ δ whenever A ⊆ [a, b] ∩ [a0 − η0 , a0 + η0 ], and now {(a0 , [a0 , a0 +η0 ]∩C)} belongs to T0 , so min(a0 +η0 , b0 ) ∈ A and A is a non-empty subset of [a0 , b0 ]. It follows that c = sup A is defined in [a0 , b0 ]. Let η > 0 be such that (c, A) ∈ δ whenever A ⊆ [a, b] ∩ [c − η, c + η]. There is some t ∈ T0 such that yt ≥ c − η. If yt = b0 , we can stop. Otherwise, set C 0 = C ∩ ]yt , c + η]. Then (c, C 0 ) ∈ δ and c ∈ C 0 and C 0 ∩ Wt is empty, so t 0 = t ∪ {(c, C 0 )} belongs to T0 and yt 0 = min(c + η, b0 ). Since yt 0 ≤ c, this shows that yt 0 = c = b0 and again b0 ∈ A, as required. 481K The Henstock integral on R This time, set X = R and let C be the family of all non-empty intervals in R. Let T be the straightforward set of tagged partitions generated by {(x, C) : C ∈ C, x ∈ C}. Following 481J, let ∆ be the set of all neighbourhood gauges on R. This time, set R = {Rab : a ≤ b ∈ R}, where Rab = {R \ [c, d] : c ≤ a, d ≥ b} ∪ {∅}. Then (X, T, ∆, R) is a tagged-partition structure allowing subdivisions, witnessed by C. proof This time we should perhaps take a moment to look at (iii) of 481G. But all we need to note is that Rab ∩Ra0 b0 = Rmin(a,a0 ),max(b,b0 ) , and that any two members of Rab have non-empty intersection. Conditions (i), (ii), (iv), (v) and (vi) of 481G are again elementary, so once more we are left with (vii). But this can be dealt with by exactly the same argument as in 481J. 481L The symmetric Riemann-complete integral (cf. Carrington 72, chap. 3) Again take X = R, and C the set of all non-empty bounded intervals of X. This time, take T to be the straightforward set of tagged partitions generated by the set of pairs (x, C) where C ∈ C and x is the midpoint of C. As in

481O

Tagged partitions

709

481K, take ∆ to be the set of all neighbourhood gauges on R; but this time take R = {R0a : a ≥ 0}, where R0a = {R \ [−c, c] : c ≥ a} ∪ {∅}. Then T is compatible with ∆ and R. proof Take δ ∈ ∆ and R ∈ R. For x ≥ 0, let θ(x) > 0 be such that (x, D) ∈ δ whenever D ⊆ [x − θ(x), x + θ(x)] and (−x, D) ∈ δ whenever D ⊆ [−x − θ(x), −x + θ(x)]. Write A for the set of those a > 0 such that there is a finite sequence a0 , . . . , an such that 0 < a0 < a1 < . . . < an = a, a0 ≤ θ(0) and ai+1 − ai ≤ 2θ( 12 (ai + ai+1 )) for i < n. ?? Suppose, if possible, that A is bounded above. Then c = inf([0, ∞[ \ A) is defined in [0, ∞[. Observe that if 0 < a ≤ θ(0), then the one-term sequence a witnesses that a ∈ A. So c ≥ θ(0) > 0. Now there must be u, v such that c < u < v < min(c + θ(c), 2c) and ]u, v[ ∩ A = ∅; on the other hand, the interval ]2c − v, 2c − u[ must contain a point x of A. Set y = 2c − x. Then we can find a0 < . . . < an = x such that 0 < a0 ≤ θ(0) and ai+1 − ai ≤ 2θ( 21 (ai + ai+1 )) for i < n; setting an+1 = y, we see that hai ii≤n+1 witnesses that y ∈ A, though y ∈ ]u, v[. X X This contradiction shows that A is unbounded above. So now suppose that R = Ra where a ≥ 0. Take a0 , . . . , an such that 0 < a0 < . . . < an , 0 < a0 ≤ θ(0) and ai+1 − ai ≤ 2θ( 12 (ai + ai+1 )) for i < n, and an ≥ a. For i < n, set xi = 12 (ai + ai+1 ), Ci = ]ai , ai+1 ], x0i = −xi , Ci0 = [−ai+1 , ai [. Then xi , x0i are the midpoints of Ci , Ci0 and (by the choice of the function θ) (xi , Ci ) ∈ δ, (x0i , Ci0 ) ∈ δ for i < n. So if we set t = {(xi , Ci ) : i < n} ∪ {(x0i , Ci0 ) : i < n} ∪ {(0, [−a0 , a0 ])} we shall obtain a δ-fine R-filling member of T . As ∆ and R are both downwards-directed, this is enough to show that T is compatible with ∆ and R. 481M The McShane integral on an interval (McShane 73) As in 481J, take X = [a, b] and let C be the family of non-empty subintervals of [a, b]. This time, take T to be the straightforward set of tagged partitions generated by Q = X × C, so that no condition is imposed relating the tags to their associated intervals. As in 481J, let ∆ be the set of all neighbourhood gauges on X, and R = {{∅}}. Proceed as before. Since the only change is that Q and T have been enlarged, (X, T, ∆, R) is still a tagged-partition structure allowing subdivisions, witnessed by C. 481N The McShane integral on a topological space (Fremlin 95) Now let (X, T, Σ, µ) be any effectively locally finite τ -additive topological measure space, and take C = {E : E ∈ Σ, µE < ∞}, Q = X × C; let T be the straightforward set of tagged partitions generated by Q. Again let ∆ be the set of all neighbourhood gauges on X. This time, define R as follows. For any set E ∈ Σ of finite measure and η > 0, let REη be the set {F : F ∈ Σ, µ(F ∩ E) ≤ η}, and set R = {REη : µE < ∞, η > 0}. Then (X, T, ∆, R) is a tagged-partition structure allowing subdivisions, witnessed by C. proof As usual, everything is elementary except perhaps 481G(vii). But if C ∈ C, δ ∈ ∆, E ∈ Σ, µE < ∞ and η > 0, take for each x ∈ X an open set Gx containing x such that (x, A) ∈ δ whenever A ⊆ Gx . {Gx S : x ∈ X} is an open cover of X, so by S 414Ea there is a finite family hxi ii
710

Gauge integrals

481O

(b) Write D ⊆ C for the family of products of non-empty bounded intervals in R. The next step is to show that if D ∈ D and δ ∈ ∆, then there is a δ-fine tagged partition t ∈ T such that Wt = D and t ⊆ Rr × D. P P Induce on r. For r = 1 this is just 481J again. For the inductive step to r + 1, suppose that D ⊆ Rr+1 is a product of bounded intervals and that δ is a neighbourhood gauge on Rr+1 . Identifying Rr+1 with Rr × R, express D as D0 × L, where D0 ⊆ Rr is a product of bounded intervals and L ⊆ R is a bounded interval. For y ∈ Rr , α ∈ R let G(y, α), H(y, α) be open sets containing y, α respectively such that ((y, α), A) ∈ δ whenever A ⊆ G(y, α) × H(y, α). For y ∈ Rr , set δy = {(α, A) : α ∈ R, A ⊆ H(y, α)}; then δy is a neighbourhood gauge on R. By the one-dimensional case there is a δy -fine tagged partition s y ∈ T1 such that Ws y = L, where I write T1 for the set of tagged partitions used in 481K. Set δ 0 = {(y, A) : y ∈ Rr , A ⊆ G(y, α) for every (α, F ) ∈ s y }. δ 0 is a neighbourhood gauge on Rr . By the inductive hypothesis, there is a δ 0 -fine tagged partition u ∈ Tr such that Wu = D0 , where here Tr is the set of tagged partitions on Rr corresponding to the r-dimensional version of this result. Consider the family t = {((y, α), E × F ) : (y, E) ∈ u , (α, F ) ∈ s y }. For (y, E) ∈ u , (α, F ) ∈ s y , we have y ∈ E,

E ⊆ G(y, α),

α ∈ F,

F ⊆ H(y, α),

so (y, α) ∈ E × F ,

E × F ⊆ G(y, α) × H(y, α),

and ((y, α), E × F ) ∈ δ. If ((y, α), E × F ), ((y 0 , α0 ), E 0 × F 0 ) are distinct members of t , then either (y, E) 6= (y 0 , E 0 ) so E ∩ E 0 = ∅ and (E × F ) ∩ (E 0 × F 0 ) is empty, or y = y 0 and (α, F ), (α0 , F 0 ) are distinct members of s y , so that F ∩ F 0 = ∅ and again E × F , E 0 × F 0 are disjoint. Thus t is a δ-fine member of Tr+1 . Finally, S S S Wt = (y,E)∈uu (α,F )∈ssy E × F = (y,E)∈uu E × L = D0 × L = D. So the induction proceeds. Q Q (c) Now suppose that C0 is an arbitrary member of C and that δ is a neighbourhood gauge on Rr . Set δ 0 = δ ∩ {(x, A) : either x ∈ C 0 or A ∩ C 0 = ∅}. Then δ 0 is a neighbourhood gauge on Rr , being the intersection of δ with the neighbourhood gauge associated with the family hUx ix∈R r , where Ux = Rr if x ∈ C 0 , Rr \ C 0 otherwise. Let D ∈ D be such that C0 ⊆ D. By (b), there is a δ 0 -fine tagged partition t ∈ T such that Wt = D. Set s = {(x, C ∩C0 ) : (x, C) ∈ t , C ∩C0 6= ∅}. Since t ⊆ δ 0 , x ∈ C 0 whenever (x, C) ∈ t and C ∩ C0 6= ∅. By (a), x ∈ C ∩ C0 for all such pairs (x, C); and of course (x, C ∩ C0 ) ∈ δ for every (x, C) ∈ t . So s belongs to T , and Ws = Wt ∩ C0 = C0 . As C0 and δ are arbitrary, 481G(vii) is satisfied. 481P Box products (cf. Muldowney 87, Prop. 1) Let h(Xi , Ti )ii∈I be a non-empty family of nonempty compact metrizable spaces with product (X, T). Set πi (x) = x(i) for x ∈ X, i ∈ I. For each i ∈ I, let Ci ⊆ PXi be such that (α) whenever E, E 0 ∈ Ci then E ∩ E 0 ∈ Ci and E \ E 0 is expressible as the union of a disjoint finite subset of Ci (β) Ci includes a base for Ti . Let C be the set of subsets of X of the form T C = {X ∩ i∈J πi−1 [Ei ] : J ∈ [I]<ω , Ei ∈ Ci for every i ∈ J}, and let T be the straightforward set of tagged partitions generated by {(x, C) : C ∈ C, x ∈ C}. Let ∆ be the set of those neighbourhood gauges δ on X defined by families hGx ix∈X of open sets such that, for some countable J ⊆ I, every Gx is determined by coordinates in J (definition: 254M). Then (X, T, ∆, {{∅}}) is a tagged-partition structure allowing subdivisions, witnessed by C. proof Conditions (i), (iii) and (vi) of 481G are trivial, and (ii), (iv) and (v) are elementary; so we are left with (vii), as usual. ?? Suppose, if possible, that C ∈ C and δ ∈ ∆ are such that there is no δ-fine t ∈ T with Wt = C. Let hGx ix∈X be the family of open sets determining δ, and J ⊆ I a non-empty countable set

481Q

Tagged partitions

711

such that Gx is determined by coordinates in J for every x ∈ X. For i ∈ J, let Ci0 ⊆ Ci be a countable base for Ti (4A2P(a-iii)), and take a sequence h(in , En )in∈N running over {(i, E) : i ∈ J, E ∈ Ci0 }. Write D = {Wt : t ∈ T is δ-fine}. Note that if D1 , D2 ∈ D are disjoint then D1 ∪ D2 ∈ D. So if D ∈ C \ D and C ∈ C, there must be some D0 ∈ C \ D such that either D0 ⊆ D ∩ C or D0 ⊆ D \ C, just because C satisfies 481G(iv). Now choose hCn in∈N inductively so that C0 = C and Cn ∈ C \ D, either Cn+1 ⊆ Cn ∩ πi−1 [En ] or Cn+1 ⊆ Cn \ πi−1 [En ] n n

T for every n ∈ N. Because X, being a product of compact spaces, is compact, there is an x ∈ n∈N Cn . We know that GxQis determined by coordinates in J, so Gx = π ˜ −1 [˜ π [Gx ]], where π ˜ is the canonical map from X onto Y = i∈J Xi . V = π ˜ [Gx ] is open, so there must be a finite set K ⊆ J and a family hVi ii∈K such that x(i) ∈ Vi ∈ Ti for every i ∈ K and {y : y ∈ Y, y(i) ∈ Vi for every i ∈ K} is included in V . This means that {z : z ∈ X, z(i) ⊆ Vi for every i ∈ K} is included in Gx . Now, for each i ∈ K, there is some m ∈ N such that i = im and x(i) ∈ Em ⊆ Vi . Because x ∈ C m+1 , πi−1 [Em ] cannot be disjoint from Cm+1 , and m −1 Cm+1 ⊆ πi−1 [E ] ⊆ π [V ]. m i i m But this means that, for any n large enough, Cn ⊆ Gx and t = {(x, Cn )} is a δ-fine member of T with Wt = Cn ; contradicting the requirement that Cn ∈ / D. X X This contradiction shows that 481G(vii) is also satisfied. 481Q The approximately continuous Henstock integral (Gordon 94, chap. 16) Let µ be Lebesgue measure on R. As in 481K, let C be the family of non-empty bounded intervals in R, and Q the set of pairs (x, C) where C ∈ C and x ∈ C; let T be the straightforward set of tagged partitions defined from Q; and set R = {Rab : a, b ∈ R}, where Rab = {R \ [c, d] : c ≤ a, d ≥ b} ∪ {∅} for a, b ∈ R. This time, define gauges as follows. Let E be the set of families e = hEx ix∈R where every Ex is a measurable set containing x such that x is a density point of Ex (definition: 223B). For e = hEx ix∈R ∈ E, set δe = {(x, C) : C ∈ C, x ∈ C, inf C ∈ Ex and sup C ∈ Ex }. Set ∆ = {δe : e ∈ E}. Then (X, T, ∆, R) is a tagged-partition structure allowing subdivisions, witnessed by C. proof (a) Turning to 481G, we find, as usual, that most of the conditions are satisfied for elementary reasons. Since we have here a new kind of gauge, we had better check 481G(ii); but if e = hEx ix∈R and e 0 = hEx0 ix∈R both belong to E, so does e ∧ e 0 = hEx ∩ Ex0 ix∈R , because lim inf η↓0

1 µ([x − η, x + η] ∩ Ex 2η

≥ lim

1

η↓0 2η

∩ Ex0 )

¡ ¢ µ([x − η, x + η] ∩ Ex ) + µ([x − η, x + η] ∩ Ex0 ) − 2η = 1

for every x; and now δe ∩ δe 0 = δe ∧ee0 belongs to ∆. Everything else we have done before, except, of course, (vii). (b) So take any δ ∈ ∆; express δ as δe , where e = hEx ix∈R ∈ E. For x, y ∈ R, write x _ y if x ≤ y and Ex ∩ Ey ∩ [x, y] 6= ∅; note that we always have x _ x. For x ∈ R, let ηx > 0 be such that µ(Ex ∩ [x − η, x + η]) ≥ 53 η whenever 0 ≤ η ≤ ηx . Fix a < b in R for theSmoment. Say that a finite string (x0 , . . . , xn ) is ‘acceptable’ if a ≤ x0 _ . . . _ xn ≤ b and µ([x0 , xn ] ∩ i x. Now choose sequences hxi ii∈N , hnk ik∈N inductively, as follows. n0 = 0 and x0 = a. Given that xi _ xi+1 for i < nk and that (xnj , . . . , xnk ) is acceptable for

712

Gauge integrals

481Q

every j ≤ k, let nk+1 > nk , (xnk +1 , . . . , xnk+1 ) be such that (xnk , . . . , xnk+1 ) is acceptable and xnk+1 ≥ 1 2 (xnk + h(xnk )); then (xnj , . . . , xnk+1 ) is acceptable for any j ≤ k + 1; continue. At the end of the induction, set c = supi∈N xi = supk∈N xnk . Then there are infinitely many i such that xi _ c. P P For any k ∈ N, µ([xnk , c] ∩

[ i≥nk

¡ Ex+i ) = lim µ [xnk , xnl ] ∩ l→∞

[

Ex+i

¢

nk ≤i
1

≥ lim (xnl − xnk ) l→∞ 2

(because (xnk , . . . , xnl ) is always an acceptable string) 1 2

= (c − xnk ). But this means that if we take k so large that xnk ≥ c − ηc , so that µ(Ec ∩ [xnk , c]) ≥ 32 (c − xnk ), there must S be some z ∈ Ec ∩ i≥nk Ei+ ∩ [xnk , c]; and if i ≥ nk is such that z ∈ Ei+ , then z witnesses that xi _ c. As k is arbitrarily large, we have the result. Q Q ?? If c < b, take any y ∈ Ec such that c < y ≤ min(c + ηc , b). Take k ∈ N such that xnk ≥ c − 31 (y − c), and j ≥ nk such that xj _ c. In this case, (xnk , xnk +1 , . . . , xj , c, y) is an acceptable string, because µEc+ ∩ [xnk , y] ≥ µEc ∩ [c, y] ≥ 32 (y − c) ≥ 12 (y − xnk ). But this means that h(xnk ) ≥ y, so that 1 2

xnk+1 ≤ c < (xnk + h(xnk )), contrary to the choice of xnk +1 , . . . , xnk+1 . X X Thus c = b. We therefore have a j ∈ N such that xj _ b, and a = x0 _ . . . _ xj _ b. (c) Now suppose that C ∈ C. Set a = inf C and b = sup C. If a = b, then t = {(a, C)} ∈ T and Wt = C. Otherwise, (b) tells us that we have x0 , . . . , xn such that a = x0 _ . . . _ xn = b. Choose ai ∈ [xi−1 , xi ] ∩ Exi−1 ∩ Exi for 1 ≤ i ≤ n. Set Ci = [ai , ai+1 [ for 1 ≤ i < n, C0 = [a, a1 [, Cn = [an , b]; set I = {i : i ≤ n, C ∩ Ci 6= ∅}; and check that (xi , C ∩ Ci ) ∈ δ for i ∈ I, so that t = {(xi , C ∩ Ci ) : i ∈ I} is a δ-fine member of T with Wt = C. As C and δ are arbitrary, 481G(vii) is satisfied. 481X Basic exercises (a) Let X, C, T and F be as in 481C. Show that if f : X → R, µ : C → R and ν : C → R are functions, then Iµ+ν (f ) = Iµ (f ) + Iν (f ) whenever the right-hand side is defined. > (b) Let I be any set. Set T = {{(i, {i}) : i ∈ J} : J ∈ [I]<ω }, δ = {(i, {i}) : i ∈ I}, ∆ = {δ}. For J ∈ [I]<ω set RJ = {I \ K : J ⊆ K ∈ [I]<ω } ∪ {∅}; set R = {RJ : J ∈ [I]<ω }. Show that (I, T, ∆, R) is <ω a tagged-partition structure allowing subdivisions, witnessed by [I]<ω . Let → R be any additive P ν : [I] functional. Show that, for a function f : I → R, limt →F (T,∆,R) St (f, ν) = i∈I f (i)ν({i}) if either exists in R. > (c) Set T = {{(n, {n}) : n ∈ I} : I ∈ [Z]<ω }, δ = {(n, {n}) : n ∈ Z}, ∆ = {δ}. For I ∈ [Z]<ω and m, n ∈ N set Rmn = Z \ {i : −m ≤ i ≤ n}, R0n = {Rkl : k, l ≥ n} ∪ {∅}, R00n = {Rkk : k ≥ n} ∪ {∅}, R0 = {R0n : n ∈ N}, R00 = {R00n : n ∈ N}. Show that (Z, T, ∆, R0 ) and (Z, T, ∆, R00 ) are tagged-partition structures allowing subdivisions, witnessed by [Z]<ω . Let µ be P counting measure on Z. Show that, for n a function f : Z → R, (i) limt →F (T,∆,R0 ) St (f, µ) = limm,n→∞ i=−m f (i) if either is defined in R (ii) Pn limt →F (T,∆,R00 ) St (f, µ) = limn→∞ i=−n f (i) if either is defined in R. (d) Set X = N ∪ {∞}, and let T be the straightforward set of tagged partitions generated by {(n, {n}) : n ∈ N} ∪ {(∞, X \ n) : n ∈ N} (interpreting a member of N as the set of its predecessors). For n ∈ N set δn = {(i, {i}) : i ∈ N} ∪ {(∞, A) : A ⊆ X \ n}; set ∆ = {δn : n ∈ N}. Show that (X, T, ∆, {{∅}}) is a taggedpartition structure allowing subdivisions, witnessed by C = [N]<ω ∪ {X \ I : I ∈ [N]<ω }. Let h : N → R be

481Yb

Tagged partitions

713

P any function, and define ν : C → R by setting νI = i∈I h(i), ν(X \ I) = −νI for I ∈ [N]<ω . Let f : X → R Pn be any function such that f (∞) = 0. Show that limt →F (T,∆,{{∅}}) St (f, µ) = limn→∞ i=0 f (i)h(i) if either is defined in R. > (e) Take X, T , ∆ and R as in 481I. Show that if µ is Lebesgue measure on [a, b] then the gauge integral Rb Iµ = limt→F (T,∆,R) St (., µ) is the ordinary Riemann integral Ra as described in 134K. (Hint: show first that they agree on step-functions.) > (f ) Let (X, Σ, µ) be a semi-finite measure space, and let Σf be the family of measurable sets of finite measure. Let T be the straightforward set of tagged partitions generated by {(x, E) : x ∈ E ∈ Σf }. For E ∈ Σf and ² > 0 set RE² = {F : F ∈ Σ, µ(E \ F ) ≤ ²}; set R = {RE²S: E ∈ Σf , ² > 0}. Let E be the family of countable partitions of X into measurable sets, and set δE = E∈E {(x, A) : x ∈ E, A ⊆ E} for E ∈ E, ∆ = {δE : E ∈ E}. Show that (X, T, ∆, R) is aR tagged-partition structure allowing subdivisions, witnessed by Σf . Show that, for a function f : X → R, f dµ = limt →F (T,∆,R) St (f, µ) if either is defined in R. (Hint: when showing that if Iµ (f ) is defined then f is µ-virtually measurable, you will need 413G or something similar; compare 482E.) (g) Let (X, Σ, µ) be a totally finite measure space, and T the straightforward set of tagged partitions generated by S {(x, E) : x ∈ E ∈ Σ}. Let E be the family of finite partitions of X into measurable sets, and set δE = E∈E {(x, A) : x ∈ E, A ⊆ E} for E ∈ E, ∆ = {δE : E ∈ E}. Show that (X, T, ∆, {{∅}}) is a tagged-partition structure allowing subdivisions, witnessed by Σ. Show that, for a function f :RX → R, Iµ (f ) = limt →F (T,∆,R) St (f, µ) is defined iff f ∈ L∞ (µ) (definition: 243A), and that then Iµ (f ) = f dµ. (h) Let X be a zero-dimensional compact Hausdorff space and E the algebra of open-and-closed subsets of X. Let T be the straightforward set of tagged partitions generated by {(x, E) : x ∈ E ∈ E}. Let ∆ be the set of all neighbourhood gauges on X. Show that (X, T, ∆, {{∅}}) is a tagged-partition structure allowing subdivisions, witnessed by E. Now let ν : E → R be an additive functional, and set Iν (f ) = limt →F (T,∆,{{∅}}) St (f, ν) when f : X → R is such that the limit is defined. (i) Show that Iν (χE) = νE for every E ∈ E. (ii) Show that if ν is bounded then Iν (f ) is defined for every f ∈ C(X). (i) Let X be a set, ∆ a set of gauges on X, R a collection of families of subsets of X, and T a set of tagged partitions on X which is compatible with ∆ and R. Let H ⊆ X be such that there is a δ˜ ∈ ∆ such ˜ and set δH = {(x, A ∩ H) : x ∈ H, (x, A) ∈ δ} for δ ∈ ∆, that H ∩ A = ∅ whenever x ∈ X \ H and (x, A) ∈ δ, ∆H = {δH : δ ∈ ∆}, RH = {{R ∩ H : R ∈ R} : R ∈ R}, TH = {{(x, C ∩ H) : (x, C) ∈ t , x ∈ H} : t ∈ T }. Show that TH is compatible with ∆H and RH . 481Y Further exercises (a) Suppose that [a, b], C, T and ∆ are as in 481J. Let T 0 ⊆ [[0, 1] × C]<ω be the set of tagged partitions t = {(xi , [ai , ai+1 ]) : i < n} where a = a0 ≤ x0 ≤ a1 ≤ x2 ≤ a2 ≤ . . . ≤ xn−1 ≤ an = b. Show that T 0 , as well as T , is compatible with ∆ in the sense of 481E; let F 0 , F be the corresponding filters on T 0 and T . Show that if ν : C → R is a functional which is additive in the sense that ν(C ∪ C 0 ) = νC + νC 0 whenever C, C 0 are disjoint members of C with union in C, and if ν{x} = 0 for every x ∈ [a, b] and f : [a, b] → R is any function, then Iν0 (f ) = Iν (f ) if either is defined, where Iν (f ) = limt →F St (f, ν),

Iν0 (f ) = limt →F 0 St (f, ν)

are the gauge integrals associated with (T, F) and (T 0 , F 0 ). (b) Let us say that a family R of residual families is ‘the simple residual structure complementary to H ⊆ PX’ if R = {RH : H ∈ H}, where RH = {X \ H 0 : H ⊆ H 0 ∈ H} ∪ {∅} for each H ∈ H. Suppose that, for each member i of a non-empty finite set I, (Xi , Ti , ∆i , Ri ) is a tagged-partition structure allowing subdivisions, witnessed by an upwards-directed family Ci ⊆ PXi , where Xi is a topological space, ∆i is the set of Q all neighbourhood gauges on Xi , and Ri is the simple residual structure Q complementary to Ci . Set X = i∈I Xi and let ∆ be the set of neighbourhood gauges on X; let C be { i∈I Ci : Ci ∈ Ci for each i ∈ I}, and R the simple residualQstructure based on C; and let T be the straightforward set of tagged partitions generated by {(hxi ii∈I , i∈I Ci ) : {(xi , Ci )} ∈ Ti for every i ∈ I}. Show that (X, T, ∆, R) is a tagged-partition structure allowing subdivisions, witnessed by C.

714

Gauge integrals

481Yc

(c) Give an example to show that, in 481Xi, (X, T, ∆, R) can be a tagged-partition structure allowing subdivisions, while (H, TH , ∆H , RH ) is not. 481 Notes and comments In the examples above I have tried to give an idea of the potential versatility of the ideas here. Further examples may be found in Henstock 91. The goal of 481A-481F is the formula Iν (f ) = limt →F (T,∆,R) St (f, ν); the elaborate notation reflects the variety of the applications. One of these is a one-step definition of the ordinary integral (481Xf). In §483 I will show that the Henstock integral (481J) properly extends the Lebesgue integral on R. In 481Xc I show how adjusting R can change the class of integrable functions; in 481Xd I show how a similar effect can sometimes be achieved by adding a point at infinity and adjusting T and ∆. As will become apparent in later sections, one of the great strengths of gauge integrals is their ability to incorporate special limiting processes. Another is the fact that we don’t need to assume that the functionals ν are countably additive; see 481Xd. In the formulae of this section, I don’t even ask for finite additivity; but of course the functional Iν is likely to have a rather small domain if ν behaves too erratically. ‘Gauges’, as I describe them here, have moved rather briskly forward from the metric gauges δh (481E), which have sufficed for most of the gauge integrals so far described. But the generalization affects only the notation, and makes it clear why so much of the theory of the ordinary Henstock integral applies equally well to the ‘approximately continuous Henstock integral’ (481Q), for instance. You will observe that the sets Q of 481Ba are just ‘gauges’ in the wide sense used here. But (as the examples of this section show clearly) we generally use them in a different way. In ordinary measure theory, we have a fairly straightforward theory of subspaces (§214) and a rather deeper theory of product spaces (chap. 25). For gauge integrals, there are significant difficulties in the theory of subspaces, some of which will appear in the next section (see 482G-482H). For closed subspaces, something can be done, as in 481Xi; but the procedure suggested there may lose some essential element of the original tagged-partition structure (481Yc). For products of gauge integrals, we do have a reasonably satisfying version of Fubini’s theorem (482M); I offer 481O and 481P as alternative approaches. You will note the concentration on ‘neighbourhood gauges’ in the work above. The set ∆ of all neighbourhood gauges on a topological space X has the following special property: if x 7→ δx : X → ∆ is any function, then {(x, A) : x ∈ X, (x, A) ∈ δx } belongs to ∆. By analogy with the notion of ‘full group of automorphisms’ (381Be), we might say that a set ∆ of gauges is ‘full’ if it has this property, and ‘countably full’ if whenever hδn in∈N is a sequence in ∆, and φ : X → N is a function, then {(x, A) : x ∈ X, (x, A) ∈ δφ(x) } belongs to ∆. As will appear repeatedly in the next section, this kind of flexibility in constructing gauges is just what one needs when proving that functions are gauge-integrable. While I have used such phrases as ‘Henstock integral’, ‘symmetric Riemann-complete integral’ above, I have not in fact discussed integrals here, except in the exercises; in most of the examples in 481I-481Q there is no mention of any functional ν from which a gauge integral Iν can be defined. The essence of the method is that we can set up a tagged-partition structure quite independently of any set function, and it turns out that the properties of a gauge integral depend more on this structure than on the measure involved.

482 General theory I turn now to results which can be applied to a wide variety of tagged-partition structures. The first step is a ‘Saks-Henstock’ lemma (482B), a fundamental property of tagged-partition structures allowing subdivisions. In order to relate gauge integrals to the ordinary integrals treated elsewhere in this treatise, we need to know when gauge-integrable functions are measurable (482E) and when integrable functions are gauge-integrable (482F). There are significant difficulties when we come to interpret gauge integrals over subspaces, but I give partial results in 482G and 482L. 482J, 482K and 482M are gauge-integral versions of the fundamental theorem of calculus, B.Levi’s theorem and Fubini’s theorem, while 482H is a new theorem on integrability of a limit function. Henstock’s integral (481J-481K) remains the most important example and the natural test case for the ideas here; I will give the details in the next section, and you may wish to take the two sections in parallel.

482B

General theory

715

482A Lemma Suppose that (X, T, ∆, R) is a tagged-partition structure allowing subdivisions (481G), witnessed by C ⊆ PX. (a) Whenever δ ∈ ∆, R ∈ R and E belongs to the subalgebra of PX generated by C, there is a δ-fine s ∈ T such that Ws ⊆ E and E \ Ws ∈ R. (b) Whenever δ ∈ ∆, R ∈ R and t ∈ T is δ-fine, there is a δ-fine R-filling t 0 ∈ T including t . (c) Suppose that f : X → R, ν : C → R, δ ∈ ∆, R ∈ R and ² ≥ 0 are such that |St (f, ν) − St 0 (f, ν)| ≤ ² whenever t , t 0 ∈ T are δ-fine and R-filling. Then (i) |St (f, ν) − St 0 (f, ν)| ≤ ² whenever t , t 0 ∈ T are δ-fine and Wt = Wt 0 ; (ii) whenever t ∈ T is δ-fine, and δ 0 ∈ ∆, there is a δ 0 -fine s ∈ T such that Ws ⊆ Wt and |Ss (f, ν) − St (f, ν)| ≤ ². (d) Suppose that f : X → [0, ∞[ and ν : C → [0, ∞[ are such that Iν (f ) = limt →F (T,∆,R) St (f, ν) is defined. Then for any ² > 0 there is a δ ∈ ∆ such that St (f, ν) ≤ Iν (f ) + ² for every δ-fine t ∈ T . S proof (a) By 481He, there is a non-increasing sequence hRk ik∈N in R such that i≤k Ai ∈ R whenever Ai ∈ Ri for every i ≤ k and hAi ii≤k is disjoint. Let C0 be a finite subset of C such that E belongs to the subalgebra S of PX generated by C0 , and let C1 ⊇ C0 be a finite subset of C such that X \ W ∈ R0 , where W = C1 (481G(v)). Then eitherSE ⊆ W or E ⊇ X \ W . In either case, E ∩ W belongs to the ring generated by C, so is expressible as i
by the choice of hRk ik∈N . So |St (f, ν) − Ss (f, ν)| = |St 0 (f, ν) − Ss 0 (f, ν)| ≤ ², as required. (d) There are δ ∈ ∆, R ∈ R such that |St (f, ν) − Iν (f )| ≤ ² whenever t ∈ T is δ-fine and R-filling. If t ∈ T is an arbitrary δ-fine tagged partition, there is a δ-fine R-filling t 0 ⊇ t , by (b), so St (f, ν) ≤ St 0 (f, ν) ≤ Iν (f ) + ², as claimed. 482B Saks-Henstock Lemma Let (X, T, ∆, R) be a tagged-partition structure allowing subdivisions, witnessed by C, and f : X → R, ν : C → R functions such that Iν (f ) = limt →F (T,∆,R) St (f, ν) is defined. Let E be the algebra of subsets of X generated by C. Then there is a unique additive functional F : E → R such thatP for every ² > 0 there are δ ∈ ∆ and R ∈ R such that (α) (x,C)∈tt |F (C) − f (x)νC| ≤ ² for every δ-fine t ∈ T , (β) |F (E)| ≤ ² whenever E ∈ E ∩ R. Moreover, F (X) = Iν (f ).

716

Gauge integrals

482B

proof (a) For E ∈ E, write TE for the set of those t ∈ T such that, for every T (x, C) ∈ t , either C ⊆ E or C ∩ E = ∅. For any δ ∈ ∆, R ∈ R and finite D ⊆ E there is a δ-fine t ∈ E∈D TE such that E \ Wt ∈ R for every E ∈SD. P P Let hRn in∈N be a sequence in R such that whenever Ai ∈ Ri for i ≤ n and hAi ii≤n is disjoint then i≤n Ai ∈ R (481He). Let E0 be the subalgebra of E generated by D, and enumerate the atoms of E0 as S hEi ii 0, there are δ ∈ ∆, R ∈ R such that |Iν (f ) − St (f, ν)| ≤ ² for every δ-fine R-filling t ∈ T . Let R0 ∈ R be such that A ∪ B ∈ R for all disjoint A, B ∈ R0 . If t , t 0 belong to TE,δ,R0 = TX\E,δ,R0 , then set s = {(x, C) : (x, C) ∈ t 0 , C ⊆ E} ∪ {(x, C) : (x, C) ∈ t , C ∩ E = ∅}. Then s ∈ TE is δ-fine, and also E \ Ws = E \ Wt 0 , (X \ E) \ Ws = (X \ E) \ Wt both belong to R0 ; so their union X \ Ws belongs to R, and s is R-filling. Accordingly |St E (f, ν) − St 0E (f, ν)| = |St (f, ν) − Ss (f, ν)| ≤ |St (f, ν) − Iν (f )| + |Iν (f ) − Ss (f, ν)| ≤ 2². As ² is arbitrary, this is enough to show that lim inf t →F ∗ St E (f, ν) = lim supt →F ∗ St E (f, ν), so that the limit limt →F ∗ St E (f, ν) is defined (2A3S). Q Q (c) If E, E 0 ∈ E are disjoint, then St E∪E0 (f, ν) = St E (f, ν) + St E0 (f, ν) for any t ∈ TE ∩ TE 0 ; since both TE and TE 0 belong to F ∗ , F (E ∪ E 0 ) = F (E) + F (E 0 ). Thus F is additive. (d) Now suppose that ² > 0. Let δ ∈ ∆, R∗ ∈ R be such that |Iν (f ) − St (f, ν)| ≤ 41 ² for every δ-fine, R -filling t ∈ T . Let R ∈ R be such that A ∪ B ∈ R∗ for all disjoint A, B ∈ R. ∗

P For any η > 0, there is a δ-fine s ∈ T such that (i) If t ∈ T is δ-fine, then |F (Wt ) − St (f, ν)| ≤ 21 ². P |Iν (f ) − Ss (f, ν)| ≤ η, for every (x, C) ∈ s , either C ⊆ Wt or C ∩ Wt = ∅, (X \ Wt ) \P Ws ∈ R, Wt \ Ws ∈ R, |F (Wt ) − (x,C)∈ss,C⊆Wt f (x)νC| ≤ η because the set of s with these properties belongs to F ∗ . Now, setting s 1 = {(x, C) : (x, C) ∈ s , C ⊆ Wt } and t 0 = t ∪ s \ s 1 , t 0 is δ-fine and R∗ -filling, like s , so |F (Wt ) − St (f, ν)| ≤ |F (Wt ) − Ss 1 (f, ν)| + |Ss 1 (f, ν) − St (f, ν)| ≤ η + |Ss (f, ν) − St 0 (f, ν)| 1 2

≤ η + |Ss (f, ν) − Iν (f )| + |Iν (f ) − St 0 (f, ν)| ≤ η + ². As η is arbitrary we have the result. Q Q P (ii) So if t ∈ T is δ-fine, P Set t 0 = {(x, C) : (x, C) ∈ t , F (C) ≤ (x,C)∈tt |F (C) − f (x)νC| ≤ ². P f (x)νC}, t 00 = t \ t 00 . Then both t 0 and t 00 are δ-fine, so X |F (C) − f (x)νC| (x,C)∈tt

=|

X (x,C)∈tt0

F (C) − f (x)νC| + |

X (x,C)∈tt00

F (C) − f (x)νC| ≤ ². Q Q

482D

General theory

717

(iii) Now suppose that E ∈ E ∩ R. Then |F (E)| ≤ ². P P Let R0 ∈ R be such that A ∪ B ∈ R whenever 0 A, B ∈ R are disjoint. Let t be such that t ∈ TE is δ-fine, E \ Wt and (X \ E) \ Wt both belong to R0 , |F (E) − St E (f, ν)| ≤ 21 ²; once again, the set of candidates belongs to F ∗ , so is not empty. Then t and t X\E are both R∗ -filling and δ-fine, so 1 2

|F (E)| ≤ ² + |St E (f, ν)| 1 2

= ² + |St (f, ν) − St X\E (f, ν)| ≤ ². Q Q As ² is arbitrary, this shows that F has all the required properties. (d) I have still to show that F is unique. Suppose that F 0 : E → R is another functional with the same properties, and take E ∈ E and ² > 0. Then there are δ, δ 0 ∈ ∆ and R, R0 ∈ R such that P |F (C) − f (x)νC| ≤ ² for every δ-fine t ∈ T , P(x,C)∈tt 0 0 (x,C)∈tt |F (C) − f (x)νC| ≤ ² for every δ -fine t ∈ T , |F (R)| ≤ ² whenever R ∈ E ∩ R, |F 0 (R)| ≤ ² whenever R ∈ E ∩ R0 . Now taking δ 00 ∈ ∆ such that δ 00 ⊆ δ ∩ δ 0 , and R00 ∈ R such that R00 ⊆ R ∩ R0 , there is a δ 00 -fine t ∈ T such that E 0 = Wt is included in E and E \ E 0 ∈ R00 . In this case X

|F (E) − F 0 (E)| ≤ |F (E \ E 0 )| +

|F (C) − F 0 (C)| + |F 0 (E \ E 0 )|

(x,C)∈tt 0

(because F and F are both additive) ≤ 2² +

X

|F (C) − f (x)νC| +

(x,C)∈tt

X

|F 0 (C) − f (x)νC| ≤ 4².

(x,C)∈tt

As ² and E are arbitrary, F = F 0 , as required.

P (e) Finally, to calculate F (X), take any ² > 0. Let δ ∈ ∆ and R ∈ R be such that (x,C)∈tt |F (C) − f (x)νC| ≤ ² for every δ-fine t ∈ T and |F (E)| ≤ ² whenever E ∈ R. Let t be any δ-fine R-filling member of T such that St (f, ν) − Iν (f )| ≤ ². Then, because F is additive, P |F (X) − Iν (f )| ≤ |F (X) − F (Wt )| + | (x,C)∈tt F (C) − f (x)νC| + |St − Iν (f )| ≤ 3². As ² is arbitrary, F (X) = Iν (f ). 482C Definition In the context of 482B, I will call the function F the Saks-Henstock indefinite integral of f ; of course it depends on the whole structure (X, T, ∆, R, C, f, ν) and not just on (X, f, ν). You should not take it for granted that F (E) = Iν (f × χE); but see 482G below. 482D The Saks-Henstock lemma characterizes the gauge integral, as follows. Theorem Let (X, T, ∆, R) be a tagged-partition structure allowing subdivisions, witnessed by C, and ν : C → R any function. Let E be the algebra of subsets of X generated by C. If f : X → R is any function, then the following are equiveridical: (i) Iν (f ) = limt →F (T,∆,R) St (f, ν) is defined in R; (ii) there is an additive functional F : E → R such Pthat (α) for every ² > 0 there is a δ ∈ ∆ such that (x,C)∈tt |F (C) − f (x)µC| ≤ ² for every δ-fine t ∈ T , (β) for every ² > 0 there is an R ∈ R such that |F (E)| ≤ ² for every E ∈ E ∩ R. proof (i)⇒(ii) is just 482B above; so let us assume (ii) and seek to prove (i). Given ² > 0, take δ ∈ ∆ and R ∈ R such that (α) and (β) are satisfied. Let t ∈ Tα be δ-fine and R-filling. Then

718

Gauge integrals

|F (Rr ) − St (f, µ)| ≤ |F (Rr \ Wt )| +

482D

X

|F (C) − f (x)µC|

(x,C)∈tt

≤ 2². As ² is arbitrary, Iν (f ) is defined and equal to F (Rr ). 482E Theorem Let (X, ρ) be a metric space and µ a complete locally determined measure on X with domain Σ. Let C, Q, T , ∆ and R be such that (i) C ⊆ Σ and µC is finite for every C ∈ C; (ii) Q ⊆ X × C, and for C ∈ C, (x, C) ∈ Q for almost every x ∈ C; (iii) T is the straightforward set of tagged partitions generated by Q; (iv) ∆ is a downwards-directed family of neighbourhood gauges on X containing all the uniform metric gauges; (v) R is a downwards-directed collection of families of subsets of X such that whenever E ∈ Σ, µE < ∞ and ² > 0, there is an R ∈ R such that µ∗ (E ∩ R) ≤ ² for every R ∈ R; (vi) T is compatible with ∆ and R. Let f : X → R be any function such that Iµ (f ) = limt →F (T,∆,R) St (f, µ) is defined. Then f is Σ-measurable. proof ?? Suppose, if possible, otherwise. Because µ is complete and locally determined, there are a measurable set E of non-zero finite measure and α < β in R such that µ∗ {x : x ∈ E, f (x) ≤ α} = µ∗ {x : x ∈ E, f (x) ≥ β} = µE (413G). Let ² > 0 be such that (β − α)(µE − 3²) > 2². Let δ ∈ ∆, R ∈ R be such that |St (f, µ) − Iν (f )| ≤ ² whenever t ∈ T is δ-fine and R-filling. Let hGx ix∈X be a family of open sets such that x ∈ Gx for every x ∈ X and δ = {(x, C) : x ∈ X, C ∈ C, C ⊆ Gx }. For m ≥ 1, set Am = {x : x ∈ E, f (x) ≤ α, U1/m (x) ⊆ Gx }, writing U1/m (x) for {y : ρ(y, x) <

1 m },

Bm = {x : x ∈ E, f (x) ≥ β, U1/m (x) ⊆ Gx }. Then there is some m ≥ 1 such that µ∗ Am ≥ µE − ², µ∗ Bm ≥ µE − ². Let δ 0 ∈ ∆ be such that δ 0 ⊆ δ ∩ {(x, C) : x ∈ C ⊆ U1/3m (x)}. Let R0 ∈ R be such that R0 ⊆ R and µ∗ (R ∩ E) ≤ ² for every R ∈ R0 . Let t be any δ 0 -fine R0 -filling member of T . Enumerate t as h(xi , Ci )ii
J 0 = {i : i < n, Ci ∩ Bm is negligible}.

S

Ci ) ≤ µ∗ (E \ Am ) = µE − µ∗ (E ∩ Am ) ≤ ², S S S 0 and similarly µ(E ∩ i∈JP 0 Ci ) ≤ ². Also, because i 0, while {x : x ∈ Ci , (x, Ci ) ∈ / Q} is negligible, so we can find x0i ∈ Ci ∩ Am 0 00 such that (xi , Ci ) ∈ Q; similarly, there is an xi ∈ Ci ∩ Bm such that (x00i , Ci ) ∈ Q. For other i < n, set x0i = x00i = xi . By conditions (ii)-(iii) of this proposition, s = {(x0i , Ci ) : i < n} and s 0 = {(x00i , Ci ) : i < n} belong to T . Of course they are R0 -filling, therefore R-filling, because t is. We also see that, because 2 for each i < n, so that Ci ⊆ Gx0i and (x0i , Ci ) ∈ δ for each (xi , Ci ) ∈ δ 0 , the diameter of Ci is at most 3m 0 i ∈ K. But since surely (xi , Ci ) ∈ δ ⊆ δ for i ∈ n \ K, this means that s is δ-fine. Similarly, s 0 is δ-fine. We must therefore have µ(E ∩

i∈J

|Ss0 (f, µ) − Ss (f, µ)| ≤ |Ss0 (f, µ) − Iµ (f )| + |Ss (f, µ) − Iµ (f )| ≤ 2². But

482F

General theory

Ss 0 (f, µ) − Ss (f, µ) =

X

719

(f (x00i ) − f (x0i ))µCi

i∈K

≥ (β − α)

X

µCi ≥ (β − α)(µE − 3²) > 2²

i∈K

by the choice of ². X X So we have the result. 482F Proposition Let X, Σ, µ, T, T , ∆ and R be such that (i) (X, Σ, µ) is a measure space; (ii) T is a topology on X such that µ is inner regular with respect to the closed sets and outer regular with respect to the open sets; (iii) T ⊆ [X × Σ]<ω is a set of tagged partitions such that C ∩ C 0 is empty whenever (x, C), 0 (x , C 0 ) are distinct members of any t ∈ T ; (iv) ∆ is a set of gauges on X containing every neighbourhood gauge on X; (v) R is a collection of families of subsets of X such that whenever µE < ∞ and ² > 0 there is an R ∈ R such that µ∗ (E ∩ R) ≤ ² for every R ∈ R; (vi) T is compatible with ∆ and R. R Then Iµ (f ) = limt →F (T,∆,R) St (f, µ) is defined and equal to f dµ for every µ-integrable function f : X → R. ˆ µ proof (a) It is worth noting straight away that we can replace (X, Σ, µ) by its completion (X, Σ, ˆ). P P We need to check that µ ˆ is inner and outer regular. But inner regularity is 412Ha, and outer regularity is equally elementary: if µ Ê < γ, there is an E 0 ∈ Σ such that E ⊆ E 0 and µE 0 = µ Ê (212C), and now there is an open set G ∈ Σ such that E 0 ⊆ G and µG ≤ γ, so thatR E ⊆ G Rand µ ˆG ≤ γ. Since we are not changing T or ∆ or R, Iµˆ (f ) = Iµ (f ) if either is defined; while also f dµ = f dˆ µ if either is defined, by 212Fb. Q Q So let us suppose that µ is actually complete. (b) In this case, f is measurable. Suppose to begin with that it is non-negative. Let ² > 0. For m ∈ Z, set Em = {x : x ∈ X, (1 + ²)m ≤ f (x) < (1 + ²)m+1 }. Then Em is measurable and has finite measure, so there is a measurable open set Gm ⊇ Em such that (1 + R²)m+1 µ(Gm \ Em ) ≤ 2−|m| ². Take a set H0 of finite measure and η0 > 0 such that E f dµ ≤ ² whenever E ∈ Σ and µ(E S ∩ H0 ) ≤ 2η0 (225A); replacing H0 by {x : x ∈ H0 , f (x) > 0} if necessary, we may suppose that H0 ⊆ m∈Z Em . Let F ⊆ H0 be a closed set such that µ(H0 \ F ) ≤ η0 . Define hVx ix∈X by setting Vx = Gm if m ∈ Z and x ∈ Em , Vx = X \ F if f (x) = 0. Let δ ∈ ∆ be the corresponding neighbourhood gauge {(x, C) : x ∈ X, C ⊆ Vx }. Let R ∈ R be such that µ∗ (R ∩ H0 ) ≤ η0 for every R ∈ R. Suppose that t is any δ-fine R-filling member of T . Enumerate t as h(xi , Ci )ii
≤ (1 + ²)

X

m∈Z i∈Jm

(1 + ²) µEm +

m∈Z

Z ≤ (1 + ²)

m

f dµ +

X

X

m∈Z

(1 + ²)m+1 µ(Gm \ Em )

m∈Z −|m|

2

Z

² = (1 + ²)

f dµ + 3².

m∈Z

S Ci . Because i
On the other hand, set F 0 = F ∩

S

i
m∈Z i∈Jm

≥ (1 + ²)−1

X X

(1 + ²)m+1 µ(Ci ∩ F )

m∈Z i∈Jm

Z

f dµ ≥ (1 + ²)−1 (

≥ (1 + ²)−1 F0

Z f dµ − ²).

720

Gauge integrals

What this means is that

482F

R

{tt : (1 + ²)−1 ( f dµ − ²) ≤ St (f, ν) ≤ (1 + ²)

R

f dµ + 3²}

belongs to F(T, ∆, R), for any ² > 0. So limt →F (T,∆,R) St (f, µ) is defined and equal to

R

f dµ.

(c) In general, f is expressible as f + − f − where f + and f − are non-negative integrable functions, so Iν (f ) = Iν (f + ) − Iν (f − ) =

R

f dµ

by 481Ca. 482G Proposition Let X, T, C, ν, T , ∆ and R be such that (i) (X, T, ∆, R) is a tagged-partition structure allowing subdivisions, witnessed by C; (ii) T is a topology on X, and ∆ is the set of neighbourhood gauges on X; (iii) ν : C → R is a functional such that Pn (α) ν is additive in the sense that νC = i=0 νCi whenever C0 , . . . , Cn ∈ C are disjoint S and i≤n Ci = C belongs to C, (β) whenever C ∈ C and ² > 0, there are closed sets F ⊆ C, F 0 ⊆ X \ C such that Pn S sup{ i=0 |νCi | : C0 , . . . , Cn ∈ C are disjoint, i≤n Ci ∩ (F ∪ F 0 ) = ∅} ≤ ²; (iv) whenever E ∈ C there is a δ0 ∈ ∆ such that whenever {(x, C)} ∈ T ∩ δ0 and x ∈ ∂E there is a partition C0 , . . . , Cn of C such that, for each i ≤ n, {(x, Ci )} ∈ T and either Ci ⊆ E or Ci ∩ E = ∅; (v) whenever C ∈ C and R ∈ R, there is an R0 ∈ R such that R ∩ C ∈ R for every R ∈ R0 . Let f : X → R be a function such that Iν (f ) = limt →F (T,∆,R) St (f, ν) is defined. Let E be the algebra of subsets of X generated by C, and F : E → R the Saks-Henstock indefinite integral of f . Then Iν (f × χH) is defined and equal to F (H) for every H ∈ E. proof (a) Because both F and Iν are additive, and F (X) = Iν (f ) (482B), and either H or its complement is a finite disjoint union of members of C (481Hd), P it is enough to consider the case in which H ∈ C. Let ² > 0. Let δ1 ∈ ∆ and R1 ∈ R be such that (x,C)∈tt |F (C) − f (x)νC| ≤ ² for every δ1 -fine t ∈ T , and |F (E)| ≤ ² for every E ∈ E ∩ R1 . Let δ0 ∈ ∆ be such that whenever {(x, C)} ∈ T ∩ δ0 and x ∈ ∂H there is a partition C0 , . . . , Cn of C such that, for each i ≤ n, {(x, Ci )} ∈ T and either Ci ⊆ H or Ci ∩ H = ∅. For each n ∈ N let Fn ⊆ H, Fn0 ⊆ X \ H be such that S Pn 2−n ² sup{ i=0 |νCi | : C0 , . . . , Cn ∈ C are disjoint, i≤n Ci ∩ (F ∪ F 0 ) = ∅} ≤ . n+1

Define hVx ix∈X by setting Vx = int H if x ∈ int H, = X \ H if x ∈ X \ H, = X \ Fn if x ∈ H \ H and n ≤ |f (x)| < n + 1, = X \ Fn0 if x ∈ H \ int H and n ≤ |f (x)| < n + 1, so that each Vx is an open set containing x. Let δ2 be the neighbourhood gauge corresponding to the family hVx ix∈X , and set δ = δ0 ∩ δ1 ∩ δ2 ∈ ∆. Let R ∈ R be such that R ∩ H ∈ R1 whenever R ∈ R. (b) Let T 0 be the set of those t ∈ T such that, for each (x, C) ∈ t, either C ⊆ H or C ∩ H = ∅. The key to the proof is the following fact: if t ∈ T is δ-fine, then there is a δ-fine s ∈ T 0 such that Ws = Wt and Ss (g, ν) = St (g, ν) for every g : X → R. P P Enumerate t as h(xi , Ci )ii
482H

General theory

721

(c) Now |St (f × χH, ν) − F (H)| ≤ 4² for every δ-fine R-filling t ∈ T . P P By (b), there is a δ-fine R-filling s ∈ T 0 such that St (f × χH, ν) = Ss (f × χH, ν). Discarding any terms of the form (x, ∅), we may suppose that C 6= ∅ whenever (x, C) ∈ s . Consider u = {(x, C) : (x, C) ∈ s , C ⊆ H},

v = {(x, C) : (x, C) ∈ s , x ∈ H}.

Then Wu = H ∩ Ws , so we have Ss (f × χH, ν) = Sv (f, ν),

X

|F (H) − Su (f, ν)| ≤ |F (H) − F (Wu )| +

|F (C) − f (x)νC|

u (x,C)∈u

≤ |F (H \ Ws )| + ² (because u is δ-fine) ≤ 2² because X \ Ws ∈ R, so H \ Ws ∈ R1 . Enumerate s as h(xi , Ci )ii
Sv (f × χH, ν) =

P i∈K

f (xi )νCi ,

where J = {i : i < n, Ci ⊆ H}, K = {i : i < n, xi ∈ H}. For m ∈ N, set Jm = {i : i ∈ J \ K, m ≤ |f (x)| < m + 1}, Km = {i : i ∈ K \ J, m ≤ |f (x)| < m + 1}. If i ∈ Jm , then xi ∈ / H but ∅ 6= Ci ⊆ H, so Vxi ∩ H 6= ∅ and xi ∈ H \ H; which implies that Vxi = X \ Fm and Ci ⊆ H \ Fm . Similarly, if i ∈ Km , then xi ∈ H and 0 Ci ∩ H = ∅, so xi ∈ H \ int H and Ci ⊆ (X \ H) \ Fm . Now, for each i, hCi ii∈Jm ∪Km is a disjoint family in 0 C with union included in X ⊆ (Fm ∪ Fm ), so P 2−m ² . i∈Jm ∪Km |νCi | ≤ m+1

So we have |Su (f, ν) − Sv (f, ν)| = | ≤ ≤

X

f (xi )νCi −

i∈J ∞ X

X

f (xi )νCi | ≤

i∈K

¡X

|f (xi )νCi | +

m=0 i∈Jm ∞ X −m

2

X

|f (xi )νCi |

i∈J4K

X

|f (xi )νCi |

¢

i∈Km

² = 2².

m=0

Putting these together, |St (f × χH, ν) − F (H)| ≤ |Sv (f, ν) − Su (f, ν)| + |Su (f, ν) − F (H)| ≤ 2² + 2² = 4². Q Q (d) As ² is arbitrary, Iν (f × χH) is defined and equal to F (H), as claimed. 482H Proposition Suppose that X, T, C, ν, T , ∆ and R satisfy the conditions (i)-(v) of 482G, and that f : X → R, hHn in∈N , H and γ are such that (vi) hHn in∈N is a sequence of open subsets of X with union H, (vii) Iν (f × χHn ) is defined for every n ∈ N, (viii) limt →F (T,∆,R) Iν (f × χWt ¹H ) is defined and equal to γ, where t ¹H = {(x, C) : (x, C) ∈ t , x ∈ H} for t ∈ T . Then Iν (f × χH) is defined and equal to γ. proof Let ² > 0. For each n ∈ N, let Fn be the Saks-Henstock indefinite integral of f × χHn . Let δn ∈ ∆ be such that

722

Gauge integrals

X

2−n ² ≥

482H

|Fn (C) − (f × χHn )(x)νC|

s (x,C)∈s

≥ |Fn (Ws ) − Ss (f × χHn , ν)| whenever s ∈ T is δn -fine. Set δ˜ = {(x, A) : x ∈ X \ H, A ⊆ X} [ [ ∪ {(x, A) : x ∈ Hn \ Hi , A ⊆ Hn , (x, A) ∈ δn }, i
n∈N

˜ then there is some n ∈ N such that x ∈ Hn so that δ˜ ∈ ∆. Note that if x ∈ H and C ∈ C and (x, C) ∈ δ, and C ⊆ Hn , so that Iν (f × χC) = Iν ((f × χHn ) × χC) = Fn (C) ˜ is defined, by 482G; this means that Iν (f × χWt ¹H ) will be defined for every δ-fine t ∈ T . Let δ ∈ ∆, R ∈ R t be such that |γ − Iν (f × χWt ¹H )| ≤ ² whenever ∈ T is δ-fine and R-filling. S ˜ Let tS∈ T be (δ ∩ δ)-fine and R-filling. For n ∈ N, set t n = {(x, C) : (x, C) ∈ t , x ∈ Hn \ i
|γ − St (f × χH, ν)| = |γ −

∞ X

St n (f × χHn , ν)|

n=0

≤ |γ − Iν (f × χWt ¹H )| +

∞ X

|Iν (f × χWt n ) − St n (f × χHn , ν)|

n=0

P∞ (note that t n = ∅ for all but finitely many n, so that Iν (f × χWt ¹H ) = n=0 Iν (f × χWt n )) ∞ X ≤²+ |Iν (f × χHn × χWt n ) − St n (f × χHn , ν)| n=0

(because t is δ-fine and R-filling, while Wt n ⊆ Hn for each n) ∞ X =²+ |Fn (Wt n ) − St n (f × χHn , ν)| n=0

(by 482G) ≤²+

∞ X

2−n ²

n=0

(because every t n is δn -fine) = 3². As ² is arbitrary, γ = Iν (f × χH), as claimed. Remark For applications of this result see 483Bd and 483N. 482I Definitions (a) Let X be a set and ∆ ⊆ P(X × PX) a family of gauges on X. I will say that ∆ is countably full if whenever hδn in∈N is a sequence in ∆, and φ : X → N is any function, then there is a δ ∈ ∆ such that (x, C) ∈ δφ(x) whenever (x, C) ∈ δ. I will say that ∆ is full if whenever hδx ix∈X is a family in ∆, then there is a δ ∈ ∆ such that (x, C) ∈ δx whenever (x, C) ∈ δ. Of course a full set of gauges is countably full. Observe that if (X, T) is any topological space, the set of all neighbourhood gauges on X is full. (b) Let X be a set, C a family of subsets of X, T ⊆ [X ×C]<ω a family of tagged partitions, ν : C → [0, ∞[ a function, and ∆ a family of gauges on X. I will say that ν is moderated (with respect to T and ∆) if there are a δ ∈ ∆ and a function h : X → ]0, ∞[ such that St (h, ν) ≤ 1 for every δ-fine t ∈ T .

482K

General theory

723

482J Integrating a derivative As will appear in the next two sections, the real strength of gauge integrals is in their power to integrate derivatives. I give an elementary general expression of this fact. In the formulae below, we can think of f as a ‘derivative’ of F if ν = θ is strictly positive and additive and we rephrase condition (iii) as ‘limC→Gx

F (C) νC

= f (x)’,

where Gx is the filter on C generated by the sets {C : (x, C) ∈ δ} as δ runs over ∆. Theorem Let X, C ⊆ PX, ∆ ⊆ P(X × PX), R ⊆ PPX, T ⊆ [X × C]<ω , f : X → R, ν : C → R, F : C → R, θ : C → [0, 1] and γ ∈ R be such that (i) T is a straightforward set of tagged partitions which is compatible with ∆ and R, (ii) ∆ is a full set of gauges on X, (iii) for every x ∈ X and ² > 0 there is a δ ∈ ∆ such that |f (x)νC − F (C)| ≤ ²θC whenever (x, C) ∈Pδ, n (iv) i=0 θCi ≤ 1 whenever C0 , . . . , Cn ∈ C are disjoint, P (v) for every ² > 0 there Sis an R ∈ R such that |γ − C∈C0 F (C)| ≤ ² whenever C0 ⊆ C is a finite disjoint set and X \ C0 ∈ R. Then Iν (f ) = limt →F (T,∆,R) St (f, ν) is defined and equal to γ. proof Let ² > 0. For each x ∈ X let δx ∈ ∆ be such that |f (x)νC − F (C)| ≤ ²θC whenever (x, C) ∈ δx . Because ∆ is full, there is a δ ∈ ∆ such that (x, C) ∈ δx whenever (x, C) ∈ δ. Let R ∈ R be as in (v). If t ∈ T is δ-fine and R-filling, then X

|St (f, ν) − γ| ≤

|f (x)νC − F (C)| + |γ −

(x,C)∈tt

X

≤

X

F (C)|

(x,C)∈tt

²θC + ²

(x,C)∈tt

(because (x, C) ∈ δx whenever (x, C) ∈ t , while X \

S (x,C)∈tt

C ∈ R)

≤ 2² by condition (ii). As ² is arbitrary, we have the result. 482K B.Levi’s Theorem Let (X, T, ∆, R) be a tagged-partition structure allowing subdivisions, witnessed by C, such that ∆ is countably full. Let ν : E → R be a non-negative finitely additive functional, where E is a ring of subsets of X including C, which is moderated with respect to T and ∆. Let hfn in∈N be a non-decreasing sequence of functions from X to R with supremum f : X → R. If γ = limn→∞ Iν (fn ) is defined in R, then Iν (f ) is defined and equal to γ. proof As in the proof of 123A, we may, replacing hfn in∈N by hfn − f0 in∈N if necessary, suppose that fn (x) ≥ 0 for every n ∈ N, x ∈ X. (a) Take ² > 0. Then there is a δ ∈ ∆ such that St (f, ν) ≤ γ + 5² for every δ-fine t ∈ T . ˜ P P Fix a strictly positive function h : X → ]0, ∞[ and a δ˜ ∈ ∆ such that St (h, ν) ≤ 1 for every δ-fine t ∈ T. Let ² > 0. For each n ∈ N choose δn ∈ ∆, Rn ∈ R such that |Iν (fn ) − St (f, ν)| ≤ 2−n ² for every δn -fine Rn -filling t ∈ T . For each x ∈ X, take rx ∈ N such that f (x) ≤ frx (x) + ²h(x). Let δ ∈ ∆ be such that (x, C) ∈ δ˜ ∩ δrx for every (x, C) ∈ δ. Suppose that t ∈ T is δ-fine. Enumerate t as h(xi , Ci )ii
fm (xi )νCi ≤

k X m=0

Ss m (fm , ν) + 2−m ² ≤

k X m=0

≤ Ss (fk , ν) + 2² ≤ Iν (fk ) + 3² s (because extends to a δk -fine Rk -filling member of T , by 482Ab)

Ss m (fk , ν) + 2−m ²

724

Gauge integrals

482K

≤ γ + 3². ˜ Now t is δ-fine, so St (h, ν) ≤ 1. Accordingly St (f ) =

X

f (xi )νCi =

f (xi )νCi

m=0 i∈Jm

i
≤

k X X

k X X

(fm (xi ) + ²h(xi ))νCi ≤ γ + 3² + ²St (h, ν) ≤ γ + 4²,

m=0 i∈Jm

as required. Q Q As ² is arbitrary, lim supt →F (T,∆,R) St (f, ν) is at most γ. (b) On the other hand, given ² > 0, there is an n ∈ N such that Iν (f ) ≥ γ − ². So taking δn ∈ ∆, Rn ∈ R as in (a) above, St (f, ν) ≥ St (fn , ν) ≥ Iν (fn ) − ² ≥ γ − 2² for every δn -fine Rn -filling t ∈ T . So lim inf t →F (T,∆,R) St (f, ν) is at least γ, and Iν (f ) = γ. 482L Lemma Let X be a set, C a family of subsets of X, ∆ a countably full downwards-directed set of gauges on X, R ⊆ PPX a downwards-directed collection of residual families, and T ⊆ [X × C]<ω a straightforward set of tagged partitions of X compatible with ∆ and R. Suppose further that whenever δ ∈ ∆, R ∈ R and t ∈ T is δ-fine, there is a δ-fine R-filling t 0 ∈ T including t . If ν : C → [0, ∞[ and f : X → [0, ∞[ are such that Iν (f ) = 0, and g : X → R is such that g(x) = 0 whenever f (x) = 0, then Iν (g) = 0. proof Let ² > 0. For each n ∈ N, let δn ∈ ∆, Rn ∈ R be such that St (f, ν) ≤ 2−n ² for every δn -fine Rn -filling t ∈ T . For x ∈ X, set φ(x) = min{n : |g(x)| ≤ nf (x)}; let δ ∈ ∆ be such that (x, C) ∈ δφ(x) whenever (x, C) ∈ δ. Let t be any δ-fine member of T . Then |St (g, ν)| ≤ 2². P P Enumerate t as h(xi , Ci )ii
P∞ m=0

2−m m² = 2². Q Q

Because T is compatible with ∆ and R, this is enough to show that Iν (g) is defined and equal to 0. 482M Fubini’s theorem Suppose that, for i = 1 and i = 2, we have Xi , Ti , Ti , ∆i , Ci and νi such that (i) (Xi , Ti ) is a topological space; (ii) ∆i is the set of neighbourhood gauges on Xi ; (iii) Ti ⊆ [Xi × Ci ]<ω is a straightforward set of tagged partitions, compatible with ∆i and {{∅}}; (iv) νi : Ci → [0, ∞[ is a function; (v) ν1 is moderated with respect to T1 and ∆1 . Write X for X1 × X2 ; ∆ for the set of neighbourhood gauges on X; C for {C × D : C ∈ C1 , D ∈ C2 }; Q for {((x, y), C × D) : {(x, C)} ∈ T1 , {(y, D)} ∈ T2 }; T for the straightforward set of tagged partitions generated by Q; and set ν(C × D) = ν1 C · ν2 D for C ∈ C1 , D ∈ C2 . (a) T is compatible with ∆ and {{∅}}. (b) Let Iν1 , Iν2 and Iν be the gauge integrals defined by these structures as in 481C-481F. Suppose that f : X → R is such that Iν (f ) is defined. Set fx (y) = f (x, y) for x ∈ X1 , y ∈ X2 . Let g : X1 → R be any function such that g(x) = Iν2 (fx ) whenever this is defined. Then Iν1 (g) is defined and equal to Iν (f ). proof (a) Let δ ∈ ∆; we seek a tagged partition u ∈ T such that Wu = X. Let hVxy i(x,y)∈X be the family of open sets in X defining δ; choose open sets Gxy ⊆ X1 , Hxy ⊆ X2 such that (x, y) ∈ Gxy × Hxy ⊆ Vxy

482M

General theory

725

for all x ∈ X1 , y ∈ X2 . For each x ∈ X1 , let δx be the neighbourhood gauge on X2 defined from the familyThHxy iy∈X2 . Then there is a δx -fine tagged partition t x ∈ T2 such that Wt x = X2 . Set Gx = X1 ∩ (y,D)∈ttx Gxy . The family hGx ix∈X1 defines a neighbourhood gauge δ ∗ on X1 , so there is a δ ∗ -fine s ∈ T1 such that Ws = X1 . Now consider u = {((x, y), C × D) : (x, C) ∈ s , (y, D) ∈ t x }. Then it is easy to check (just as in part (b) of the proof of 481O) that u is a δ-fine member of T with Wu = X. (b)(i) Let A be {x : x ∈ X1 , Iν2 (fx ) is defined}. Let h : X1 → R is any function such that h(x) = 0 for every x ∈ A. For x ∈ X1 , set h0 (x) = inf({1} ∪ {supt ,tt0 ∈F St (fx , ν2 ) − St 0 (fx , ν2 ) : F ∈ F(T2 , ∆2 , {{∅}})}). (Thus Iν2 (fx ) is defined iff h0 (x) = 0.) Then Iν1 (h0 ) = 0. P P Let ² > 0. Then there is a δ ∈ ∆ such that Su (f, ν)−Su 0 (f, ν) ≤ ² whenever u , u 0 ∈ T are δ-fine and Wu = Wu 0 = X. Define hVxy i(x,y)∈X , hGxy i(x,y)∈X , hHxy i(x,y)∈X and hδx ix∈X1 from δ as in (a) above. For each x ∈ X1 , we can find δx -fine partitions t x , t 0x ∈ T2 T such that Wt x = Wt 0x = X2 and St x (fx , ν2 ) − St 0x (fx ,ν1 ≥ 21 h0 (x). Set Gx = X1 ∩ (y,D)∈ttx ∪tt0x Gxy . Let δ ∗ be the neighbourhood gauge on X1 defined from hGx ix∈X1 . Let s be any δ ∗ -fine member of T1 with Ws = X1 . Set u = {((x, y), C × D) : (x, C) ∈ s , (y, D) ∈ t x }, u 0 = {((x, y), C × D) : (x, C) ∈ s , (y, D) ∈ t 0x }, Then u and u 0 are δ-fine members of T with Wu = Wu 0 = X, so X

St (h0 , ν1 ) =

h0 (x)ν1 (C) ≤ 2

(x,C)∈tt

=2

¡ X

X

(St x (fx , ν2 ) − St 0x (fx , ν2 ))ν1 (C)

(x,C)∈tt

X

f (x, y)ν1 (C)ν2 (D) −

t (x,C)∈t tx (y,D)∈t

¡ ¢ = 2 Su (f, ν) − Su 0 (f, ν) ≤ 2².

¢ f (x, y)ν1 (C)ν2 (D)

t (x,C)∈t t 0x (y,D)∈t

As ² is arbitrary, Iν1 (h0 ) = 0. Q Q By 482L, Iν1 (h) is also zero. (ii) Again take any ² > 0, and let δ ∈ ∆ be such that |Su (f, ν) − Iν (f )| ≤ ² for every δ-fine u ∈ T such that Wu = X. Define hVxy i(x,y)∈X , hGxy i(x,y)∈X , hHxy i(x,y)∈X and hδx ix∈X1 from δ as in (a) and (i) above. ˜ : X1 → ]0, ∞[ be such that Ss (h, ˜ ν1 ) ≤ 1 for every δ-fine ˜ s ∈ T1 . Let δ˜ ∈ ∆1 , h ˜ For x ∈ A, let t x ∈ T2 be δx -fine and such that Wt x = X2 and |St x (fx , ν2 ) −TIν2 (fx )| ≤ ²h(x); for x ∈ X1 \ A, let t x be any δx -fine member of T2 such that Wt x = X2 . Set Gx = X1 ∩ (y,D)∈ttx Gxy for every x ∈ X1 . Let δ ∗ be the neighbourhood gauge on X1 defined by hGx ix∈X1 . Set h1 (x) = g(x) − St x (fx , ν2 ) for x ∈ X1 \ A, 0 for x ∈ A. Then Iν1 (|h1 |) = 0, by (a). Let δ 0 ∈ ∆1 be such that δ 0 ⊆ δ ∗ ∩ δ˜ and Ss (|h1 |, ν1 ) ≤ ² for every δ 0 -fine s ∈ T1 such that Ws = X1 . Now suppose that s ∈ T1 is δ 0 -fine and that Ws = X1 . Set u = {((x, y), C × D) : (x, C) ∈ s , (y, D) ∈ t x }, so that u ∈ T is δ-fine and Wu = X. Then

726

Gauge integrals

482M

|Ss (g, ν1 ) − Iν (f )| ≤ |Su (f, ν) − Iν (f )| + |Ss (g, ν1 ) − Su (f, ν)| X ¯ X ¡ ¯ ¢ ≤²+¯ g(x) − f (x, y)ν2 D ν1 C ¯ s (x,C)∈s

(y,D)∈ttx

¯ X ¡ ¯ ¢ =²+¯ g(x) − St x (fx , ν2 ) ν1 C ¯ s (x,C)∈s

X

≤²+

|h1 (x)|ν1 C

s (x,C)∈s

+

X

|g(x) −

s,x∈A (x,C)∈s

≤ 2² +

X

X

f (x, y)ν2 D|ν1 C

(y,D)∈ttx

˜ ˜ ²h(x)ν 1 (C) ≤ 2² + ²Ss (h, ν1 ) ≤ 3².

s,x∈A (x,C)∈s

As ² is arbitrary, Iν (g) is defined and equal to Iν (f ), as claimed. 482X Basic exercises (a) Let (X, T, ∆, R) be a tagged-partition structure allowing subdivisions, witnessed by C, and E the algebra of subsets of X generated by C. Write I for the set of pairs (f, ν) such that f : X → R and ν : C → R are functions and Iν (f ) = limt →F (T,∆,R) St (f, ν) is defined; for (f, ν) ∈ I, let Ff ν : E → R be the corresponding Saks-Henstock indefinite integral. Show that (f, ν) 7→ Ff ν is bilinear in the sense that Ff +g,ν = Ff ν + Fgν ,

Fαf,ν = Ff,αν = αFf ν ,

Ff,µ+ν = Ff µ + Ff ν

whenever (f, ν), (g, ν) and (f, µ) belong to I and α ∈ R. (b) Let (X, T, ∆, R) be a tagged-partition structure allowing subdivisions, witnessed by C, and E the algebra of subsets of X generated by C. Let f : X → R and ν : C → [0, ∞[ be functions such that Iν (f ) = limt →F (T,∆,R) St (f, ν) is defined, and let F : E → R be the corresponding Saks-Henstock indefinite integral. Show that F is non-negative iff Iν (f − ) = 0, where f − (x) = max(0, f (x)) for every x ∈ X. > (c) Let X be a zero-dimensional compact Hausdorff space, E the algebra of open-and-closed subsets of X, Q = {(x, E) : x ∈ E ∈ E}, T the straightforward set of tagged partitions generated by Q, ∆ the set of neighbourhood gauges on X and ν : E → R a non-negative additive functional. Let µ be the corresponding Radon measure on X (416Q) R and Iν the gauge integral defined by (X, T, ∆, {{∅}}) (cf. 481Xh). Show that, for f : X → R, Iν (f ) = f dµ if either is defined in R. (Hint: if f is measurable but not µ-integrable, take x0 such that f is not integrable on any neighbourhood of x0 . Given δ ∈ ∆, fix a δ-fine partition containing (x0 , V0 ) for some V0 ; now replace (x0 , V0 ) by refinements {(x0 , V00 ), (x1 , V1 ), . . . , (xn , Vn )}, where Pn i=1 f (xi )νVi is large, to show that St (f, ν) cannot be controlled by δ.) > (d) Let (X, T, Σ, µ) be an effectively locally finite τ -additive topological measure space in which µ is inner regular with respect to the closed sets and outer regular with respect to the open sets (see 412W). Let T be the straightforward set of tagged partitions generated by X × {E : µE < ∞}, ∆ the set of neighbourhood gauges on X, and R = {REη : µE < ∞, η > 0}, where REη = {F : µ(F ∩ E) ≤ η}, R as in 481N. Show that if Iµ is the associated gauge integral, and f : X → R is a function, then Iµ (f ) = f dµ if either is defined in R. (e) Let C be the set of non-empty subintervals of X = [0, 1], T the straightforward tagged-partition structure generated by [0, 1] × C, and ∆ the set of neighbourhood gauges on [0, 1], as in 481M. Let µ be any Radon measure on [0, 1], and Iµ the gauge integral defined R from µ and the tagged-partition structure ([0, 1], T, ∆, {{∅}}). Show that, for any f : [0, 1] → R, Iµ (f ) = f dµ if either is defined in R. (f ) Let C be the set of non-empty subintervals of [0, 1], T the straightforward tagged-partition structure generated by {(x, C) : C ∈ C, x ∈ C}, and ∆ the set of neighbourhood gauges on X, as in 481J. Let ν : C → R be a bounded additive functional, and Iν the gauge integral defined from ν and (X, T, ∆, {{∅}}). Show that Iν (χ [a, b[) = limx↑a,y↑b ν([x, y]) whenever 0 < a < b ≤ 1.

482 Notes

General theory

727

(g) Let C be the set of non-empty subintervals of X = [0, 1], T the straightforward tagged-partition structure generated by {(x, C) : C ∈ C, x ∈ C}, and ∆ the set of uniform metric gauges on [0, 1], as in 481I. Let µ be the Radon probability measure on [0, 1] such that µ{ 21 } = 1, and let Iµ = limt→F (T,∆,R) St ( , µ) be the corresponding gauge integral. Show that Iµ (χ[0, 1]) is defined but that Iµ (χ[0, 12 ]) is not. > (h) (i) Show that the McShane integral on an interval [a, b] as described in 481M coincides with the Lebesgue integral on [a, b]. (ii) Show that if (X, T, Σ, µ) is a quasi-Radon measure space and µ is outer regular with respect to the open sets then the McShane integral as described in 481N coincides with the usual integral. (i) Explain how the results in 481Xb-481Xc can be regarded as special cases of 482H. (j) Let (X, T) be a topological space, C a ring of subsets of X, T ⊆ [X × C]<ω a straightforward set of tagged partitions, ν : C → [0, ∞[ an additive function, and ∆ the family of neighbourhood gauges on X. Suppose that there is a sequence hGn in∈N of open sets, covering X, such that sup{νC : C ∈ C, C ⊆ Gn } is finite for every n ∈ N. Show that ν is moderated with respect to T and ∆. (k) Let (X, T, Σ, µ) be a quasi-Radon measure space. Let T be the straightforward tagged-partition structure generated by {(x, E) : µE < ∞, x ∈ E} and ∆ the set of all neighbourhood gauges on X. Show that µ is moderated with respect to T and ∆ iff there is a sequence of open sets of finite measure covering X. 482Y Further exercises (a) Set X = R and let C be the family of non-empty bounded intervals in X; set Q = {(x, C) : C ∈ C, x = inf C}, and let T be the straightforward set of tagged partitions generated by Q. Let T be the Sorgenfrey right-facing topology on X, and ∆ the set of neighbourhood gauges for T. Set Rn = {E : E ∈ Σ, µ([−n, n] ∩ E) ≤ 2−n } for n ∈ N, where µ is Lebesgue measure on R and Σ its domain, and write R = {Rn : n ∈ N}. Show that (X, T, ∆, R) is a tagged-partition structure allowing subdivisions. Show that if f : X → R is such that Iµ (f ) is defined, then f is Lebesgue measurable. (b) Give an example of X, T, Σ, µ, T , ∆, C, f and C such that (X, T, Σ, µ) is a compact metrizable Radon probability space, ∆ is the set of all neighbourhood gauges on X, (X, T, ∆, {{∅}}) is a taggedpartition structure allowing subdivisions, witnessed by C, f : X → R is a function such that Iµ (f ) = limt →F (T,∆,{{∅}}) St (f, µ) is defined, C ∈ C is a closed set with negligible boundary, and Iµ (f × χC) is not defined. (c) Suppose that, for i = 1 and i = 2, we have a tagged-partition structure (Xi , Ti , ∆i , Ri ) allowing subdivisions, witnessed by a ring Ci ⊆ PXi , where Xi is a topological space, ∆i is the set of all neighbourhood gauges on Xi , and Ri is the simple residual structure complementary to Ci , as in 481Yb. Set X = X1 × X2 and let ∆ be the set of neighbourhood gauges on X; set C = {C × D : C ∈ C1 , D ∈ C2 }; let R be the simple residual structure on X complementary to C; and let T be the straightforward tagged-partition structure generated by {((x, y), C × D) : {(x, C)} ∈ T1 , {(y, D)} ∈ T2 }. For each i, let νi : Ci → [0, ∞[ be a function moderated with respect to Ti and ∆i , and define ν : C → [0, ∞[ by setting ν(C × D) = ν1 C · ν2 D for C ∈ C1 , D ∈ C2 . Show that T is compatible with ∆ and R. Let Iν1 , Iν2 , Iν be the gauge integrals defined by these structures. Suppose that f : X → R is such that Iν (f ) is defined. Set fx (y) = f (x, y) for x ∈ X1 , y ∈ X2 . Let g : X1 → R be any function such that g(x) = Iν2 (fx ) whenever this is defined. Show that Iν1 (g) is defined and equal to Iν (f ). 482 Notes and comments In 482E, 482F and 482G the long lists of conditions reflect the variety of possible applications of these arguments. The price to be paid for the versatility of the constructions here is a theory which is rather weak in the absence of special hypotheses. As everywhere in this book, I try to set ideas out in maximal convenient generality; you may feel that in this section the generality is becoming inconvenient; but the theory of gauge integrals has not, to my eye, matured to the point that we can classify the systems here even as provisionally as I set out to classify topological measure spaces in Chapters 41 and 43.

728

Gauge integrals

482 Notes

Enthusiasts for gauge integrals offer two substantial arguments for taking them seriously, apart from the universal argument in pure mathematics, that these structures offer new patterns for our delight and new challenges to our ingenuity. First, they say, gauge integrals integrate more functions than Lebesgue-type integrals, and it is the business of a theory of integration to integrate as many functions as possible; and secondly, gauge integrals offer an easier path to the principal theorems. I have to say that I think the first argument is sounder than the second. It is quite true that the Henstock integral on R (481K) can be rigorously defined in fewer words, and with fewer concepts, than the Lebesgue integral. The style of Chapters 11 and 12 is supposed to be better adapted to the novice than the style of this chapter, but you will have no difficulty in putting the ideas of 481A, 481C, 481J and 481K together into an elementary definition of an integral for real functions in which the only non-trivial argument is that establishing the existence of enough tagged partitions (481J), corresponding I suppose to Proposition 114D. But the path I took in defining the integral in §122, though arduous at that point, made (I hope) the convergence theorems of §123 reasonably natural; the proof of 482K, on the other hand, makes significant demands on our technique. Furthermore, the particular clarity of the one-dimensional Henstock integral is not repeated in higher dimensions. Fubini’s theorem, with exact statement and full proof, even for products of Lebesgue measures on Euclidean spaces, is a lot to expect of an undergraduate; but Lebesgue measure on Rr makes sense in a way that it is quite hard to repeat with gauge integrals. (For instance, Lebesgue measure is invariant under isometries; this is not particularly easy to prove – see 263A – but at least it is true; if we want a gauge integral which is invariant under isometries, then we have to use a construction such as 481O, which does not directly match any natural general definition of ‘product gauge integral’ along the lines of 481P, 482M or 482Yc.) In my view, a stronger argument for taking gauge integrals seriously is their ‘power’, that is, their ability to provide us with integrals of many functions in consistent ways. 482E, 482F and 482J give us an idea of what to expect. If we start from a measure space (X, Σ, µ) and build a gauge integral Iµ from a set T ⊆ [X × Σ]<ω of tagged partitions, then we can hope that integrable functions will be gauge-integrable, with the right integrals (482F); while gauge-integrable functions will be measurable (482E). What this means is that forRnon-negative functions, the integrals will coincide. Any ‘new’ gauge-integrable functions f will be R such that f + = f − = ∞; the gauge integral will offer a process for cancelling the divergent parts of these integrals. On the other hand, we can hope for a large class of gauge-integrable derivatives. In the next two sections, I will explain how this works in the Henstock and Pfeffer integrals. For simple examples calling for such procedures, see the formulae of §§282 and 283; for radical applications of the idea, see Muldowney 87. Against this, gauge integrals are not effective in ‘general’ measure spaces, and cannot be, because R + there R is no structure in an abstract measure space (X, Σ, µ) which allows us to cancel an infinite integral f = f F R R against f − = X\F f . Put another way, if a tagged-partition structure is invariant under all automorphisms of the structure (X, Σ, µ), as in 481Xf-481Xg, we cannot expect anything better than the standard integral. In order to get something new, the most important step seems to be the specification of a family C of ‘regular’ sets, preliminary to describing a set T of tagged partitions. To get a ‘powerful’ gauge integral, we want a fine filter on T , corresponding to a small set C and a large set of gauges. The residual families of 481F are generally introduced just to ensure ‘compatibility’ in the sense described there; as a rule, we try to keep them simple. But even if we take the set of all neighbourhood gauges, as in the Henstock integral, this is not enough unless we also sharply restrict both the family C and the permissible tags (482Xc-482Xe). The most successful restrictions, so far, have been ‘geometric’, as in 481J and 481O, and 484F below. Further limitations on admissible pairs (x, C), as in 481L and 481Q, in which C remains the set of intervals, but fewer tags are permitted, also lead to very interesting results. Another limitation in the scope of gauge integrals is the difficulty they have in dealing with spaces of infinite measure. Of course we expect to have to specify a limiting procedure if we are to calculate Iµ (f ) from sums St (f, µ) which necessarily consider only sets of finite measure, and this is one of the functions of the collections R of residual families. But this is not yet enough. In B.Levi’s theorem (482K) we already need to suppose that our set-function ν is ‘moderated’ in order to determine how closely fn (x) needs to approximate each f (x). The condition St (h, ν) ≤ 1 for every δ-fine t of 482Ib is very close to saying that Iµ (h) ≤ 1. But it is not the same as saying that µ is σ-finite; it suggests rather that X should be covered by a sequence of open sets of finite measure (482Xk).

483A

The Henstock integral

729

Because gauge integrals are not absolute – that is,R we can have Iν (f ) defined and finite while Iν (|f |) is not – we are bound to have difficulties with integrals H f , even if we interpret these in the simplest way, as Iν (f × χH), so that we do not need a theory of subspaces as developed in §214. 482G(iii)-(v) are an attempt to find a reasonably versatile sufficient set of conditions. The ‘multiplier problem’, for a given gauge integral Iν , is the problem of characterizing the functions g such that Iν (f × g) is defined whenever Iν (f ) is defined, and even for some intensively studied integrals remains challenging. In 484L I will give an important case which is not covered by 482G. One of the striking features of gauge integrals is that there is no need to assume that the set-function ν is countably additive. We can achieve countable additivity of the integral – in the form of B.Levi’s theorem, for instance – by requiring only that the set of gauges should be ‘countably full’ (482K, 482L; contrast 482Xg). If we watch our definitions carefully, we can make this match the rules for Stieltjes integrals (114Xa, 482Xf). In 481Db I have already remarked on the potential use of gauge integrals in vector integration. It is important to recognise that a value F (E) of a Saks-Henstock indefinite integral (482B-482C) cannot be identified with either Iν (f × χE) or with Iν¹ PE (f ¹E), because in the formula F (E) = limt →F ∗ St E (f, ν) the tags of the partitions t E need not lie in E. The idea of 482G is to impose sufficient conditions to ensure that the contributions of ‘straddling’ elements (x, C), where either x ∈ H and C 6⊆ H or x ∈ / H and C ∩ H 6= ∅, are insignificant. To achieve this we seem to need both a regularity condition on the functional ν (condition (iii-β) of 482G) and a geometric condition on the set C underlying T (482G(iv)). As usual, the regularity condition required is closer to outer than to inner regularity, in contexts in which there is a distinction. I am not sure that I have the ‘right’ version of Proposition 482E. The hypothesis there is that we have a metric space. But in the principal non-metrizable cases the result is still valid (482Xc-482Xe). Proposition 482H is a ‘new’ limit theorem; it has no counterpart in the ordinary theory of integration. The hypotheses seem, from where we are standing at the moment, to be exceedingly restrictive. In the leading examples in §483, however, the central requirement 482H(viii) is satisfied for straightforward reasons. Gauge integrals insist on finite functions defined everywhere. But since we have an effective theory of negligible sets (482L), we can easily get a consistent theory of integration for functions which are defined and real-valued almost everywhere if we say that Iν (f ) = Iν (g) whenever g : X → R extends f ¹f −1 [R] whenever Iν (g) = Iν (g 0 ) for all such extensions. 483 The Henstock integral I come now to the original gauge integral, the ‘Henstock integral’ for real functions. The first step is to check that the results of §482 can be applied to show that this is an extension of both the Lebesgue integral and the improper Riemann integral (483B), coinciding with the Lebesgue integral for non-negative functions (483C). It turns out that any Henstock integrable function can be approximated in a strong sense by a sequence of Lebesgue integrable functions (483G). The Henstock integral can be identified with the Perron and special Denjoy integrals (483J, 483N). Much of the rest of the section is concerned with indefinite Henstock integrals. Some of the results of §482 on tagged-partition structures allowing subdivisions condense into a particularly strong Saks-Henstock lemma (483F). If f is Henstock integrable, it is equal almost everywhere to the derivative of its Henstock integral (483I). Finally, indefinite Henstock integrals can be characterized as continuous ACG∗ functions (483R). 483A Definition The following notation will apply throughout the section. Let C be the family of nonempty bounded intervals in R, and let T ⊆ [R×C]<ω be the straightforward set of tagged partitions generated by {(x, C) : C ∈ C, x ∈ C}. Let ∆ be the set of all neighbourhood gauges on R. Set R = {Rab : a ≤ b ∈ R}, where Rab = {R \ [c, d] : c ≤ a, d ≥ b} ∪ {∅}. Then (R, T, ∆, R) is a tagged-partition structure allowing subdivisions (481K), so T is compatible with ∆ and R (481Hf). The Henstock integral is the gauge integral defined by the process of 481E-481F from (R, T, ∆, R)Rand Lebesgue measure µ. For a function f : R → R I will say that f is Henstock integrable, and that H f = γ, if limt →F (T,∆,R) St (f, µ) is defined Rβ R H f for H f × χ ]α, β[ if this is defined in R. I will use and equal toR γ ∈ R. For α, β ∈ [−∞, ∞] I will write α R the symbol for the ordinary integral, so that f dµ is the Lebesgue integral of f .

730

Gauge integrals

483B

483B Tracing through the theorems of §482, we have the following. Theorem (a) Every Henstock integrable function on R is Lebesgue measurable. (b) Every Lebesgue integrable function f : R → R is Henstock integrable, with the same integral. (c) If f is Henstock integrable so is f × χC for every interval C ⊆ R. (d) Suppose that f : R → R is any function. Then Rb Rβ (α) if β ∈ ]−∞, ∞], H−∞ f = limb↑β H−∞ f if either is defined in R, R∞ R∞ (β) if α ∈ [−∞, ∞[, Hα f = lima↓α Ha f if either is defined in R. proof (a) Apply 482E. (b) Apply 482F, referring to 134F to confirm that condition 482F(ii) is satisfied. (c)-(d)(i) The first thing to check is that the conditions of 482G are satisfied by R, T , ∆, R, C and µ. P P 482G(i) is just 481K, and 482G(ii) is trivial. 482G(iii-β) and 482G(v) are elementary, and so is 482G(iii-β); if you like, it is a special case of 412W(b-iii). As for 482G(iv), if E ∈ C is a singleton {x}, then whenever (x, C) ∈ C we can express C as a union of one or more of the sets C ∩ ]−∞, x[, C ∩ {x} and C ∩ ]x, ∞[, and for any non-empty C 0 of these we have {(x, C 0 )} ∈ T and either C 0 ⊆ E or C 0 ∩ E = ∅. Otherwise, let η > 0 be half the length of E, and let δ be the uniform metric gauge {(x, A) : A ⊆ ]x − η, x + η[. Then if x ∈ ∂E ˜ we can again express C as a union of one or more of the sets C ∩ ]−∞, x[, C ∩ {x} and and (x, C) ∈ T ∩ δ, C ∩ ]x, ∞[, and these will witness that 482G(iv) is satisfied. Q Q Rb (ii) Now suppose that f is Henstock integrable. Then lima→−∞,b→∞ Ha f is defined and equal to Rb R H f. P P Let F be the Saks-Henstock indefinite integral of f (482B-482C). Then 482G tells us that Ha f is defined and equal to F (]a, b[) whenever a < b in R. Now, given ² > 0, there is an R ∈ R such that Rb R |F (R \ C)| ≤ ² whenever C ∈ C and R \ C ∈ R; that is, there are a0 ≤ b0 such that | H f − Ha f | ≤ ² R Rb Q whenever a ≤ a0 ≤ b0 ≤ b. As ² is arbitrary, lima→−∞,b→∞ Ha f is defined and equal to H f . Q Rb (iii) It follows that if f is Henstock integrable and c ∈ R, limb→∞ Hc f is defined. P P Let ² > 0. Then Rb R there are a0 ≤ c, b0 ≥ c such that | Ha f − H f | ≤ ² whenever a ≤ a0 and b ≥ b0 . But this means that if b, b0 ≥ b0 then

Rb

| Hc f −

R b0

Rb

R b0

f | = | Ha f − Ha f | ≤ 2² 0 0 Rb 0 whenever b, b ≥ b0 . As ² is arbitrary, limb→∞ c f is defined. Q Q Rc Similarly, lima→−∞ Ha f is defined. H c

Rc Rb (iv) Moreover, if f is Henstock integrable and a < b in R, then limc↑b Ha f is defined and equal to a f . P P Again let F be the Saks-Henstock indefinite integral of f . Let ² > 0. Then there is a δ ∈ ∆ such that P (x,C)∈tt |F (C) − f (x)µC| ≤ ² whenever t ∈ T is δ-fine. Let η > 0 be such that η|f (b)| ≤ ² and (b, [c, b[) ∈ δ whenever b − η ≤ c < b. Then whenever max(a, b − η) ≤ c < b,

Rb

f− Rc

Rc

f = F ([c, b[) ≤ |F ([c, b[) − f (b)µ [c, b[ | + |f (b)|µ [c, b[ ≤ 2². Rb As ² is arbitrary, limc↑b Ha f = Ha f . Q Q Rb Rb H Similarly, limc↓a c f is defined and equal to Ha f . Rb R∞ (v) Next, if f : R → R and c ∈ R are such that γ = lima↓c,b→∞ Ha f is defined in R, then Hc f is defined and equal to γ. P P I seek to apply 482H, with H = ]c, ∞[. We have already seen that the conditions of 482G are satisfied, and 482H(vi)-(vii) are elementary, if we set Hn = ]c + 2−n , c + 2n [. So we are left with Rb 482H(viii). Given ² > 0, let η ∈ ]0, 1], b0 ≥ c + 1 be such that | Ha f − γ| ≤ ² whenever c < a ≤ c + η and b ≥ b0 . If we set £ ¤ δ0 = {(x, A) : x < c, A ⊆ −∞, 12 (x + c) } H a

H a

∪ {(c, A) : A ⊆ ]c − η, c + η[} ¤ £ ∪ {(x, A) : x > c, A ⊆ 21 (x + c), ∞ },

483C


731

δ0 ∈ ∆. If t ∈ T is δ0 -fine and Rc,b0 -filling, Wt is an interval including [c, b0 ] and Wt ¹H must be an interval R H f × χWt ¹H | ≤ ². As ² is arbitrary, condition with infimum in ]c, c + η[ and supremum at least b . So |γ − 0 R∞ 482H(viii) is satisfied, and Hc f is defined and equal to γ. Q Q Rb Rc Similarly, H∞ f = lima→−∞,b↑c Ha f if the latter is defined. Rb R∞ (vi) We are now in a position to show that if f is Henstock integrable then Hc f = limb→∞ Hc f is R c+1 Rb R c+1 defined for every c ∈ R. P P By (iii) and (iv), lima↓c Ha f = Hc f and limb→∞ Hc+1 are both defined. R Because H f × χ{c} = 0, by (b),

Rb

R c+1

Rb

Rb

lima↓c,b→∞ Ha = Hc f + limb→∞ Hc+1 = limb→∞ Hc f R∞ is defined, and is equal to Hc f , by (v). Q Q Rc Rc Similarly, H−∞ f = lima→−∞ Ha f is defined. Putting these together, we see that if f is Henstock integrable and a < b in R,

Rb

f=

R

× χ ]a, ∞[) × χ ]−∞, b[ R is defined. Once again, because H f × χ{c} is always defined, H f × χC is defined for every interval C, and the proof of (c) is complete. Rβ (vii) Now consider (d). If H−∞ f is defined, then g = f × χ ]−∞, β[ is Henstock integrable. Taking any real c < β, we have R

H a

H (f

Z b Z b Z c Z b Z c Z β lim H f = lim H g = H g + lim H g = H g + H g b↑β

b↑β

−∞

(by (iii) or (iv))

−∞

−∞

b↑β

c

−∞

c

Z β = H f. ∞

In the other direction, suppose that γ = limb↑β

Rb

is defined, and also lima→−∞ Rβ H f , by (v). ∞

lim

a↓0,b→∞

f =γ−

R0

f is defined, where β ∈ R. Then

R β−1

H −∞

f is defined. Then

a

a↓0

a

b→∞

1

Z 1 Z 1 Z 0 = H f +γ− H f =γ− H f −∞

−∞

R

f = γ are defined. R∞ R∞ (viii) Similarly, if α ∈ [−∞, ∞[, Hα f = lima↓α Ha f if either is defined in R. This completes the proof. H 0

H −∞

f and

Rb

Z b Z 1 Z b H f = lim H f + lim H f

0

R∞

H −∞

limb↑β Hβ−1 f = γ − H−∞ f R β−1 Rb R β−1 H f = H−∞ f , by (iii). So lima→−∞,b↑b Ha f = γ, and must be equal to a

For the case β = ∞, suppose that γ = limb→∞

so, by (v),

Rb

H

483C Corollary The and Lebesgue integrals agree on non-negative functions, in the sense that R Henstock R if f : R → [0, ∞[ then H f = f dµ if either is defined in R. proof If f is Lebesgue integrable, it is Henstock integrable,R with the same integral, by 483Bb. If it is Henstock integrable, then it is measurable, by 483Ba, so that f dµ is defined in [0, ∞]; but

732

Gauge integrals

Z

483C

Z f dµ = sup{

(213B)

g dµ : g ≤ f is a non-negative simple function}

Z Z = sup{ H g : g ≤ f is a non-negative simple function} ≤ H f

(481Cb) is finite, so f is Lebesgue integrable. Rα

483D Corollary If f : R → R is Henstock integrable, then α 7→

Rβ

H −∞

f : [∞, ∞] → R and (α, β) 7→

2

f : [−∞, ∞] → R are continuous and bounded. Rx proof Set F (x) = H−∞ f for x ∈ R. Take any x0 ∈ R and ² > 0. By 483B(d-i), there is an η1 > 0 Rx R x0 such that | H−∞ f − H−∞ f | ≤ ² whenever x0 − η1 ≤ x ≤ x0 . By 483B(d-ii), there is an η2 > 0 such R∞ R∞ that | Hx f − Hx0 f | ≤ ² whenever x0 ≤ x ≤ x0 + η2 . But this means that |F (x) − F (x0 )| ≤ ² whenever x0 − η1 ≤ x ≤ x0 + η2 . As ² is arbitrary, R F is continuous at x0 . We know also that limx→∞ F (x) = H f is defined in R; while H α

limx→−∞ F (x) =

R

H

f − limx→−∞

R∞ H x

f =0

is also defined, by 483B(d-ii), as usual. So F is continuous at ±∞. Rβ Now writing G(α, β) = Hα f , we have G(α, β) = F (β) − F (α) if α ≤ β and zero if β ≤ α. So G also is continuous. F and G are bounded because [−∞, ∞] is compact. 483E Definition If f : R → R is Henstock integrable, then its indefinite Henstock integral is the Rx function F : R → R defined by saying that F (x) = H−∞ f for every x ∈ R. 483F In the present context, the Saks-Henstock lemma can be sharpened, as follows. Theorem Let f : R → R and F : R → R be functions. Then the following are equiveridical: (i) f is Henstock integrable and F is its indefinite Henstock integral; (ii)(α) F is continuous, (β) limx→−∞ F (x) = 0 and limx→∞ F (x) is defined in R, (γ) for every ² > 0 there are a gauge δ ∈ ∆ and a non-decreasing function φ : R → [0, ²] such that |f (x)(b − a) − F (b) + F (a)| ≤ φ(b) − φ(a) whenever a ≤ x ≤ b in R and (x, [a, b]) ∈ δ. proof (i)⇒(ii) (α)-(β) are covered Rby 483D. As for (γ), 482G tells us that we can identify the Saks-Henstock H indefinite integral of f with E 7→ R where E is the algebra generated by C. Let ² > 0. P f × χE : E → R, Then there is a δ ∈ ∆ such that (x,C)∈tt |f (x)µC − H f × χC| ≤ ² for every δ-fine t ∈ T . Set R P φ(a) = supt ∈T ∩δ (x,C)∈tt,C⊆]−∞,a] |f (x)µC − H f × χC|, so that φ : R → [0, ²] is a non-decreasing function. Now suppose that a ≤ y ≤ b and that (y, [a, b]) ∈ δ. In this case, whenever t ∈ T ∩ δ, s = {(x, C) : (x, C) ∈ t , C ⊆ ]−∞, a]} ∪ {(y, ]a, b[)} also belongs to T ∩ δ. Rb Now F (b) − F (a) = Ha f , so Z X φ(b) ≥ |f (x)µC − H f × χC| s (x,C)∈s

=

Z |f (x)µC − H f × χC| + |f (y)(b − a) − F (b) + F (a)|.

X (x,C)∈tt,C⊆]−∞,a]

As t is arbitrary, φ(b) ≥ φ(a) + |f (y)(b − a) − F (b) + F (a)|, that is, |f (y)(b − a) − F (b) + F (a)| ≤ φ(b) − φ(a), as called for by (γ).

483G


733

(ii)⇒(i) Assume (ii). Set γ = limx→∞ F (x). Let ² > 0. Let a ≤ b be such that |F (x)| ≤ ² for every x ≤ a and |F (x) − γ| ≤ ² for every x ≥ b. Let δ ∈ ∆, φ : R → [0, ²] be such that φ is non-decreasing and |f (x)(β − α) − F (β) + F (α)| ≤ φ(β) − φ(α) whenever α ≤ x ≤ β and (x, [α, β]) ∈ δ. Let δ 0 ∈ ∆ be such that (x, A) ∈ δ whenever (x, A) ∈ δ 0 . For C ∈ C set λC = F (sup C) − F (inf C), νC = φ(sup C) − φ(inf C); then if (x, C) ∈ δ 0 , (x, [inf C, sup C]) ∈ δ, so |f (x)µC − λC| ≤ νC. Note that λ and ν are both additive in the sense that λ(C ∪ C 0 ) = λC + λC 0 , ν(C ∪ C 0 ) = νC + νC 0 whenever C, C 0 are disjoint members of C such that C ∪ C 0 ∈ C (cf. 482G(iii-α)). Let t ∈ T be δ 0 -fine and Rab -filling. Then Wt is of the form [c, d] where c ≤ a and b ≤ d. So |St (f, µ) − γ| ≤ 2² + |St (f, µ) − F (d) + F (c)| = 2² + |St (f, µ) − λ[c, d]| X X = 2² + | f (x)µC − λC| ≤ 2² + νC (x,C)∈tt

(x,C)∈tt

= 2² + ν[c, d] ≤ 3². As ² is arbitrary, f is Henstock integrable, with integral γ. Rx I still have to check that F is the indefinite integral of f . Set F1 (x) = H−∞ f for x ∈ R, and G = F − F1 . Then (ii) applies equally to the pair (f, F1 ), because (i)⇒(ii). So, given ² > 0, we have δ, δ1 ∈ ∆ and non-decreasing functions φ, φ1 : R → [0, ²] such that |f (x)(b − a) − F (b) + F (a)| ≤ φ(b) − φ(a) whenever a ≤ x ≤ b in R and (x, [a, b]) ∈ δ, |f (x)(b − a) − F1 (b) + F1 (a)| ≤ φ1 (b) − φ1 (a) whenever a ≤ x ≤ b in R and (x, [a, b]) ∈ δ1 . Putting these together, |G(b) − G(a)| ≤ ψ(b) − ψ(a) whenever a ≤ x ≤ b in R and (x, [a, b]) ∈ δ ∩ δ1 , where ψ = φ + φ1 . But if a ≤ b in R, there are a0 ≤ x0 ≤ a1 ≤ x1 ≤ . . . ≤ xn−1 ≤ an such that a = a0 , an = b and (xi , [ai , ai+1 ]) ∈ δ for i < n (481J), so that |G(b) − G(a)| ≤

n−1 X

|G(ai+1 ) − G(ai )|

i=0

≤

n−1 X

ψ(ai+1 ) − ψ(ai ) = ψ(b) − ψ(a) ≤ 2².

i=0

As ² is arbitrary, G is constant. As limx→−∞ F (x) = limx→−∞ F1 (x) = 0, F = F1 , as required. 483G Theorem (see Gordon 94, 9.18) Let f : R → R be a Henstock integrable function. Then there is a countable cover K of R by compact sets such that f × χK is Lebesgue integrable for every K ∈ K. proof (a) For n ∈ N set En = {x : |x| ≤ n, |f (x)| ≤ n}. By 483Ba, f is Lebesgue measurable, so for each n ∈ N we canSfind a compact set Kn ⊆ En such that µ(En \ Kn ) ≤ 2−n ; f is Lebesgue integrable over Kn , and Y = R \ n∈N Kn is Lebesgue negligible. Let F be the indefinite Henstock integral of f , and take a gauge δ0 ∈ ∆ and a non-decreasing function φ : R → [0, 1] such R that |f (x)(b R − a) − F (b) + F (a)| ≤ φ(b) − φ(a) whenever a ≤ x ≤ b and (x, [a, b]) ∈ δ0 (483F). Because H |f |×χY = Y |f |dµ = 0 (483Bb), there is a δ1 ∈ ∆ such that St (|f |×χY, µ) ≤ 1 whenever t ∈ T is δ1 -fine (482Ad). For C ∈ C, set λC = F (sup C) − F (inf C), νC = φ(sup C) − φ(inf C). Set so that

S

Dn = {x : x ∈ En ∩ Y, (x, [a, b]) ∈ δ0 ∩ δ1 whenever x − 2−n ≤ a ≤ x ≤ b ≤ x + 2−n }, S 0 0 0 n∈N Dn = Y ; set Kn = D n , so that Kn is compact and n∈N Kn ⊇ Y .

0 (b) The point is¤ that f × χK P For k ∈ N, let Ak be the set of points ¤ n is 0Lebesgue integrable for each n. P 0 −k x ∈ Kn such that x, x + 2 ∩ K = ∅; then A is finite, because Kn0 ⊆ [−n, n] is bounded. Similarly, if k £ £ n0 S 0 0 −k 0 Ak = {x : x ∈ Kn , x − 2 , x ∩ Kn = ∅}, Ak is finite. Set B = Kn0 \ k∈N (Ak ∪ A0k ), so that Kn0 \ B is countable. Set ¤ £ δ = δ0 ∩ δ1 ∩ {(x, A) : x ∈ R, A ⊆ x − 2−n−1 , x + 2−n−1 },

734

Gauge integrals

483G

so that δ ∈ ∆. Note that if C ∈ C, x ∈ B ∩ C and (x, C) ∈ δ and µC > 0, then int C meets Kn0 (because there are points of Kn0 arbitrarily close to x on both sides) so int C meets Dn ; and if y ∈ Dn ∩ int C then (y, C) ∈ δ0 ∩ δ1 , because diam C ≤ 2−n . This means that if t ∈ T is δ-fine and t ⊆ B × C, then there is a δ0 ∩ δ1 -fine s ∈ T such that s ⊆ Dn × C, Ws ⊆ Wt and whenever (x, C) ∈ t and C is not a singleton, there is a y such that (y, C) ∈ s . Accordingly X X X X |λC| ≤ |λC| ≤ |f (y)µC − λC| + |f (y)|µC (x,C)∈tt

s (y,C)∈s

s (y,C)∈s

X

≤

s (y,C)∈s

νC + Ss (|f | × χY, µ) ≤ 2.

s (y,C)∈s

But this means that if t ∈ T is δ-fine, X

St (|f × χB|, µ) =

|f (x)µC|

(x,C)∈tt¹B

(where t¹B = t ∩ (B × C))

X

≤

|f (x)µC − λC| +

(x,C)∈tt¹B

X

≤

X

|λC|

(x,C)∈tt¹B

νC + 2 ≤ 3.

(x,C)∈tt¹B

It follows that if g is a µ-simple function and 0 ≤ g ≤ |f × χB|, Z

Z g dµ = H g ≤

sup t ∈T is δ-fine

≤

sup t ∈T is δ-fine

St (g, µ)

St (|f × χB|, µ) ≤ 3,

and |f × χB| is µ-integrable, by 213B, so f × χB is µ-integrable, by 122P. As Kn0 \ B is countable, therefore negligible, f × χKn0 is µ-integrable. Q Q (c) So if we set K = {Kn : n ∈ N} ∪ {Kn0 : n ∈ N}, we have a suitable family. 483H Upper and lower derivates: Definition Let F : R → R be any function. For x ∈ R, set D∗ F (x) = lim supy→x

F (y)−F (x) , y−x

D∗ F (x) = lim inf y→x

F (y)−F (x) y−x

in [−∞, ∞]. 483I Theorem Suppose that f : R → R is Henstock integrable, and F is its indefinite Henstock integral. Then F 0 (x) is defined and equal to f (x) for almost every x ∈ R. proof For n ∈ N, set An = {x : |x| ≤ n, D∗ F (x) > f (x) + 2−n }. Then µ∗ An ≤ 2−n+1 . P P Let δ ∈ ∆ and φ : R → [0, 4−n ] be such that φ is non-decreasing and |f (x)(b − a) − F (b) + F (a)| ≤ φ(b) − φ(a) whenever a ≤ x ≤ b and (x, [a, b]) ∈ δ (483F). Let I be the set of non-trivial closed intervals [a, b] ⊆ R such that, for some x ∈ [a, b] ∩ An , (x, [a, b]) ∈ δ and

F (b)−F (a) ≥ b−a S

f (x) + 2−n . By Vitali’s theorem (221A) we can find a

countable disjoint S family I0 ⊆ I such that An \ I0 is negligible; so we have a finite family I1 ⊆ I0 such that µ∗ (An \ I1 ) ≤ 2−n . Enumerate I1 as h[ai , bi ]ii
483J


735

S P P µ( I1 ) = i f (x)} = m∈N n≥m An is negligible. Similarly, or applying the argument to −f , {x : D∗ F (x) < f (x)} is negligible. So D∗ F ≤a.e. f ≤a.e. D∗ F . Since D∗ F ≤ D∗ F everywhere, D∗ F =a.e. D∗ F =a.e. f . But F 0 (x) = f (x) whenever D∗ F (x) = D∗ F (x) = f (x), so we have the result. 483J Theorem Let f : R → R be a function. Then the following are equiveridical: (i) f is Henstock integrable; (ii) for every ² > 0 there are functions F1 , F2 : R → R, with finite limits at both −∞ and ∞, such that D∗ F1 (x) ≤ f (x) ≤ D∗ F2 (x) and 0 ≤ F2 (x) − F1 (x) ≤ ² for every x ∈ R. Rx proof (i)⇒(ii) Suppose that f is Henstock integrable and that ² > 0. Set F (x) = H−∞ f for x ∈ R. Let δ ∈ ∆, φ : R → [0, 12 ²] be such that φ is non-decreasing and |f (x)(b − a) − F (b) + F (a)| ≤ φ(b) − φ(a) whenever a ≤ x ≤ b and (x, [a, b]) ∈ δ (483F). Set F1 = F − φ, F2 = F + φ; then F1 (x) ≤ F2 (x) ≤ F1 (x) + ² for every x ∈ R, and the limits at ±∞ are defined because F and φ both have limits at both ends. If x ∈ R, there is an η > 0 such that (x, A) ∈ δ whenever A ⊆ [x − η, x + η]. So if x − η ≤ a ≤ x ≤ b ≤ x + η and a < b, |

F (b)−F (a) b−a

− f (x)| ≤

φ(b)−φ(a) , b−a

and F1 (b)−F1 (a) b−a

≤ f (x) ≤

F2 (b)−F2 (a) . b−a

In particular, this is true whenever x − η ≤ a < x = b or x = a < b ≤ x + η. So D∗ F1 (x) ≤ f (x) ≤ D∗ F2 (x). As x is arbitrary, we have a suitable pair F1 , F2 . (ii)⇒(i) Suppose that (ii) is true. Take any ² > 0. Let F1 , F2 : R → R be as in the statement of (ii). α) We need to know that F2 − F1 is non-decreasing. P (α P Set G = F2 − F1 . Then lim inf y→x

G(y)−G(x) y−x

= lim inf

F2 (y)−F2 (x) y−x

−

≥ lim inf

F2 (y)−F2 (x) y−x

− lim sup

y→x

y→x

F1 (y)−F1 (x) y−x

y→x

F1 (y)−F1 (x) y−x

(2A3Sf) = D∗ F2 (x) − D∗ F1 (x) ≥ 0. ?? If a < b and G(a) > G(b), set γ =

G(a)−G(b) , 2(b−a)

and choose han in∈N , hbn in∈N inductively as follows.

a0 = a and b0 = b. Given that an < bn and G(an ) − G(bn ) > γ(bn − an ), set c = 21 (an + bn ); then either G(an ) − G(c) > γ(c − an ) or G(c) − G(bn ) ≥ γ(bn − c); in the former case, take an+1 = an and bn+1 = c; in the latter, take an+1 = c and bn+1 = bn . Set x = limn→∞ an = limn→∞ bn . Then for each n, either G(an ) − G(x) > γ(x − an ) or G(x) − G(bn ) > γ(bn − x). In either case, we have a y such that 0 < |y − x| ≤ 2−n (b − a) and

G(y)−G(x) y−x

< −γ. So D∗ G(x) ≤ −γ < 0, which is impossible. X X

Thus G is non-decreasing, as required. Q Q β ) Let a ≤ b be such that |F1 (x) − F1 (a)| ≤ ² whenever x ≤ a and (β R |F1 (x) − F1 (b)| ≤R ² whenever x ≥ b. Let h : R → ]0, ∞[ be a strictly positive integrable function such that h dµ ≤ ². Then H h ≤ ², by 483Bb, so there is a δ0 ∈ ∆ such that St (h, µ) ≤ 2² for every δ0 -fine t ∈ T (482Ad). For x ∈ R let ηx > 0 be such that F1 (y)−F1 (x) y−x

≤ f (x) + h(x),

F2 (y)−F2 (x) y−x

≥ f (x) − h(x)

whenever 0 < |y − x| ≤ 2ηx ; set δ = {(x, A) : (x, A) ∈ δ0 , A ⊆ ]x − ηx , x + ηx [}, so that δ ∈ ∆. Note that if x ∈ R and x − ηx ≤ α ≤ x ≤ β ≤ x + ηx , then

736

Gauge integrals

F1 (β) − F1 (x) ≤ (β − x)(f (x) + h(x)),

483J

F1 (x) − F1 (α) ≤ (x − α)(f (x) + h(x)),

so that F1 (β) − F1 (α) ≤ (β − α)(f (x) + h(x)); and similarly F2 (β) − F2 (α) ≥ (β − α)(f (x) − h(x)). For C ∈ C, set λ1 C = F1 (sup C) − F2 (sup C),

λ2 C = F2 (sup C) − F2 (sup C).

Then if C ∈ C, x ∈ C and (x, C) ∈ δ, λ1 C ≤ (f (x) + h(x))µC,

λ2 C ≥ (f (x) − h(x))µC.

Suppose that t ∈ T is δ-fine and Rab -filling. Then Wt = [α, β] for some α ≤ a, β ≥ b, so that X X St (f, µ) = f (x)µC ≤ λ2 C + h(x)µC = λ2 [α, β] + St (h, µ) (x,C)∈tt

(x,C)∈tt

≤ F2 (β) − F2 (α) + 2² ≤ F1 (β) − F1 (α) + 3² ≤ F1 (b) − F1 (a) + 5². Similarly, X

St (f, µ) =

X

f (x)µC ≥

(x,C)∈tt

λ1 C − h(x)µC

(x,C)∈tt

= λ1 [α, β] − St (h, µ) ≥ F1 (β) − F1 (α) − 2² ≥ F1 (b) − F1 (a) − 4². But this means that if t , t 0 are two δ-fine Rab -filling members of T , |St (f, µ) − St 0 (f, µ)| ≤ 9². As ² is arbitrary, limt →F (T,∆,R) St (f, µ) =

R

H

f

is defined. Remark The formulation (ii) above is a version of the method of integration described by Perron 14. 483K Proposition Let f : R → R be a Henstock integrable function, and F its indefinite Henstock integral. Then F [E] is Lebesgue negligible for every Lebesgue negligible set E ⊆ R. proof Let ² > 0. By 483C and 482Ad, as usual, together with 483F, there are a δ ∈ ∆ and a non-decreasing φ : R → [0, ²] such that St (|f | × χE, µ) ≤ ²,

|f (x)(b − a) − F (b) + F (a)| ≤ φ(b) − φ(a)

whenever t ∈ T is δ-fine, a ≤ x ≤ b and (x, [a, b]) ∈ δ. For n ∈ N and i ∈ Z, set Eni = {x : x ∈ E ∩ [2−n i, 2−n (i + 1)[ , (x, A) ∈ δ whenever A ⊆ [x − 2−n , x + 2−n ]}. Set Jn = {i : i ∈ Z, −4n < i ≤ 4n , Eni 6= ∅}. Observe that S T S E = n∈N m≥n i∈Jm Emi . For i ∈ Jn , take xni , yni ∈ Eni such that xni ≤ yni and min(F (xni ), F (yni )) ≤ inf F [Eni ] + 4−n ², max(F (xni ), F (yni )) ≥ sup F [Eni ] − 4−n ², so that µ∗ F [Eni ] ≤ |F (yni ) − F (xni )| + 2−2n+1 ². Now, for each i ∈ Jn , (xni , [xni , yni ]) ∈ C ∩ δ, while [xni , yni ] ⊆ [2−n i, 2−n (i + 1)[, so t = {(xni , [xni , yni ]) : i ∈ Jn } is a δ-fine member of T , and [ X X µ∗ F [ Eni ] ≤ µ∗ F [Eni ] ≤ 2−2n+1 ² + |F (yni ) − F (xni )| i∈Jn

i∈Jn

≤ 4² +

X

i∈Jn

|f (xni )(yni − xni )| +

i∈Jn

≤ 4² + St (|f | × χE, µ) + ² ≤ 6². Since this is true for every n ∈ N,

X i∈Jn

φ(yni ) − φ(xni )

483N


µ∗ F [E] = µ∗ F [

[ \ [

737

Ei ]

n∈N m≥n i∈Jm

= µ∗ (

[

n∈N

F[

\ [

Ei ]) = sup µ∗ F [ n∈N

m≥n i∈Jm

\ [

Ei ]

m≥n i∈Jm

(132Ae) ≤ 6². As ² is arbitrary, F [E] is negligible, as claimed. Rb 483L Definition If f : R → R is Henstock integrable, I write kf kH for supa≤b | Ha f |. It is elementary to check that this is a seminorm on the linear space of all Henstock integrable functions. (It is finite-valued by 483D.) R 483M Proposition (a) If f : R → R is Henstock integrable, then | H f | ≤ kf kH , and kf kH = 0 iff f = 0 a.e. (b) Write HL1 for the linear space of all Henstock integrable real-valued functions on R, and HL1 for {f • : f ∈ HL1 } ⊆ L0 (µ) (§241). If we write kf • kH = kf kH for every f ∈ HL1 , then HL1 is a normed space. The ordinary space L1 (µ) of equivalence classes of Lebesgue integrable functions is a linear subspace of HL1 , and kukH ≤ kuk1 for every u ∈ L1 (µ). (c) We have a linear operator T : HL1 → Cb (R) defined by saying that T (f • ) is the indefinite Henstock integral of f for every f ∈ HL1 , and kT k = 1. proof (a) Of course

R

| H f | = lima→−∞,b→∞

Rb

f | ≤ kf kH Rb (using 438Bd). Set F (x) = H−∞ f , so that F (b) − F (a) = Ha f whenever a ≤ b. If f = 0 a.e., then Rx F (x) = −∞ f dµ = 0 for every x, by 483B, so kf kH = 0. If kf kH = 0, then F is constant, so f = F 0 = 0 a.e., by 483I. Rx

H a

(b) This follows immediately from (a). (Compare the definitions of the norms k kp on Lp , for 1 ≤ p ≤ ∞, in §§242-244.) By 483Bb, L1 (µ) ⊆ HL1 , and kukH ≤ ku+ kH + ku− kH = ku+ k1 + ku− k1 = kuk1 for every u ∈ L1 (µ), writing u+ and u− for the positive and negative parts of u, as in Chapter 24. (c) If f , g ∈ HL1 and f • = g • , then f and g have the same indefinite integral, by 483Bb or otherwise; so T is defined as a function from HL1 to RR . By 483F, T u is continuous and bounded for every u ∈ HL1 , and by 481C T is linear. If f ∈ HL1 and T f • = F , then kf kH = supx,y∈R |F (y) − F (x)|; since limx→−∞ F (x) = 0, kf kH ≥ kF k∞ ; as f is arbitrary, kT k ≤ 1. On the other hand, for any non-negative Lebesgue integrable function f , kT f • k∞ = kf k1 = kf kH , so kT k = 1. 483N Proposition Suppose that hIk ik∈K is a disjoint family of open intervals in R with union G, and for every k ∈ K. If R P that f : R → R is a function such that fk = f × χIk Ris HenstockPintegrable H f × χG = H fk . k∈K kfk kH < ∞, then f × χG is Henstock integrable, and k∈K proof I seek to apply 482H. We have already seen, in S the proof of 483B, that the conditions of 482G are satisfied by R, T , ∆, R, C, T and µ. Of course G = k∈K Ik is the union of a sequence of open sets over which f is Henstock integrable. So we have only to check 482H(viii). Set S δ0 = k∈K {(x, A) : x ∈ Ik , A ⊆ Ik } ∪ {(x, A) : x ∈ R \ G, A ⊆ R}, so that δ0 ∈ ∆. For each k ∈ K let P Fk be the Saks-Henstock indefinite integral of fk . Let ² > 0. Then there is a finite set K0 ⊆ K such that k∈K\K0 kfk kH ≤ ². Next, there must be δ1 ∈ ∆ and R ∈ R such that

738

Gauge integrals

483N

R

P

| H fk − St (fk , µ)| ≤ ² P for every δ-fine R-filling t ∈ T , and δ2 ∈ ∆ such that k∈K0 |St (fk , µ) − Fk (Wt )| ≤ ² for every δ2 -fine S t ∈ T. t t t t Now let ∈ T be (δ0 ∩ δ1 ∩ δ2 )-fine and R-filling. For each k ∈ K set k = ¹Ik , so that ¹G = k∈K tk . Because Wt is an interval, each Wt k must be an interval, and it is a subinterval of Ik because t is δ0 -fine. So (using 482G) k∈K0

R

|Fk (Wt k )| = | H fk × χWt k | ≤ kfk kH . Also Z X Z X Z X | H fk − H f × χWt k | ≤ | H fk − St (fk , µ)| + |St (fk , µ) − Fk (Wt k | k∈K0

k∈K0

k∈K0

≤ 2². On the other hand,

P

k∈K\K0 |

R H

fk −

R H

P

f × χWtk | ≤ 2

k∈K\K0

kfk kH ≤ 2².

Putting these together, Z X Z X Z X Z | H f × χWt ¹G − H fk | = | H f × χWt k − H fk | k∈K

k∈K

k∈K

P (because t is finite, so all but finitely many terms in the sum k∈K f × χWt k are zero) Z X Z ≤ | H f × χWt k − H fk | ≤ 4². k∈K

As ² is arbitrary, condition 482H(viii) is satisfied, with limt →F (T,∆,R)

R

H

R P f × χWt ¹G = k∈K H fk ,

and 482H gives the result we seek. 483O Definitions (a) For any real-valued function F , write ω(F ) for supx,y∈dom F |F (x) − F (y)|, the oscillation of F . (Interpret sup ∅ as 0, so that ω(∅) = 0.) (b) Let F : R →PR be a function. For A ⊆ R, we say that F is AC∗ on A if for every ² > 0 there is an η P > 0 such that I∈I ω(F ¹I) ≤ ² whenever I is a disjoint family of open intervals with endpoints in A and I∈I µI ≤ η. Note that whether F is AC∗ on A is not determined by F ¹A, since it depends on the behaviour of F on intervals with endpoints in A. (c) Finally, F is ACG∗ if it is continuous and there is a countable family A of sets, covering R, such that F is AC∗ on every member of A. 483P Elementary results (a)(i) If F , G : R → R are functions and A ⊆ B ⊆ R, then ω(F + G¹A) ≤ ω(F ¹A) + ω(G¹A) and ω(F ¹A) ≤ ω(F ¹B). (ii) If F is the indefinite Henstock integral of f : R → R and C ⊆ R is an interval, then kf × χCkH = ω(F ¹C). (iii) If F : R → R is continuous, then (a, b) 7→ ω(F ¹ [a, b]) : R2 → R is continuous, and ω(F ¹A) = ω(F ¹A) for every set A ⊆ R. (b)(i) If F : R → R is AC∗ on A ⊆ R, it is AC∗ on every subset of A. P Let ² > 0. Let η > 0 (ii) If F P : R → R is continuous and is AC∗ on A ⊆ R, it is AC∗ on A. P be such that I∈I ω(F ¹I) ≤ ² whenever I is a disjoint family of open intervals with endpoints in A and P P 1 I∈I µI ≤ η. Let I be a disjoint family of open intervals with endpoints in A and I∈I µI ≤ 2 η. Let

483Q


739

I0 ⊆ I be a non-empty finite set; then we can enumerate I0 as h]ai , bi [ii≤n where a0 , b0 , . . . , an , bn ∈ A and a0 ≤ b0 ≤ a1 ≤ b1 ≤ . . . ≤ an ≤ bn . Because (a, b) 7→ ω(F ¹ [a, b]) is continuous, noted in (a-ii) Pn as 0 0 0 0 0 0 0 0 above, we can find a , . . . , b ∈ A such that a ≤ b ≤ a ≤ . . . ≤ a ≤ b , b − a0i ≤ η, and n n n 0 0 0 1 i=0 i Pn 0 0 i=0 |ω(F ¹ [ai , bi ]) − ω(F ¹ [ai , bi ]) ≤ ²; so that P Pn 0 0 I∈I0 ω(F ¹I) ≤ i=0 ω(F ¹ [ai , bi ]) ≤ 2². P As I0 is arbitrary, I∈I ω(F ¹I) ≤ 2²; as ² is arbitrary, F is AC∗ on A. Q Q 483Q Lemma Let F : R → R be a continuous function, and K ⊆ R a non-empty compact set such that F is AC∗ on K. Write I for the family of non-empty bounded open intervals, disjoint from K, with endpoints in K. P (a) I∈I ω(F ¹I) is finite. (b) Write a∗ for inf K = min K. Then there is a Lebesgue integrable function g : R → R such that Rx P F (x) − F (a∗ ) = a∗ g + J∈I,J⊆[a∗ ,x] F (sup J) − F (inf J) for every x ∈ K. P proof (a) Let η > 0 be such P that I∈J ω(F ¹I) ≤ 1 whenever J is a disjoint family of open intervals with endpoints in K and I∈J µI ≤ η. Let m0 , m1 ∈ Z be such that K ⊆ [m0 η, m1 η]. P For integers m between m and m , let I be the set of intervals in I included in ]jη, (j + 1)η[. Then 0 1 m I∈Im µI ≤ η, S P so I∈Im ω(F ¹I) ≤ 1 for each j. Also every member of J = I \ m0 ≤m
ω(F ¹I) ≤

I∈I

X

ω(F ¹I) +

X

X

ω(F ¹I)

m=m0 I∈Im

I∈J

≤

m 1 −1 X

ω(F ¹I) + m1 − m0 < ∞

I∈J

because F is continuous, therefore bounded on every bounded interval. (b)(i) Set b∗ = sup K = max K. Define G : [a∗ , b∗ ] → R by setting G(x) = F (x) if x ∈ K, =

F (b)(x−a)+F (a)(b−x) b−a

if x ∈ ]a, b[ ∈ I.

Then G is absolutely continuous. P P G is continuous because F is. Let ² > 0. Let η1 > P 0 be such that P ω(F ¹I) ≤ ² whenever J is a disjoint family of open intervals with endpoints in K and I∈J I∈J µI ≤ η1 . P Let I0 ⊆ I be a finite set such that I∈I\I0 ω(F ¹I) ≤ ², and take M > 0 such that |F (b)−F (a)| ≤ M (b−a) whenever ]a, b[ ∈ I0 ; set η = min(η1 ,

² ) M

> 0.

∗

Let J be the set of non-empty open subintervals J of [a∗ ,P b∗ ] such that either J ∩K = ∅ or both endpoints ∗ of J belong to K. Let J ⊆ J be a disjoint family such that I∈J µI ≤ η. Set J 0 = {J : J ∈ J , J ∩K = ∅}. Then X

|G(sup J) − G(inf J)| =

J∈J \J 0

X

≤

X

J∈J \J 0

On the other hand,

|F (sup J) − F (inf J)|

J∈J \J 0

ω(F ¹J) ≤ ².

740

Gauge integrals

X

|G(sup J) − G(inf J)| =

J∈J 0

X

X

|G(sup J) − G(inf J)|

I∈I0 J∈J ,J⊆I

X

+ ≤M

X

483Q

X

|G(sup J) − G(inf J)|

I∈I\I0 J∈J ,J⊆I

X

X

µJ +

I∈I0 J∈J ,J⊆I

|F (sup I) − F (inf I)|

I∈I\I0

(because G is monotonic on I for each I ∈ I) ≤ M η + ² ≤ 2², P so J∈J |G(sup J) − G(inf J)| ≤ 3². Generally, if J is any non-empty open subinterval of [a∗ , b∗ ], we can split it into at most three intervals belonging to J ∗ . So if J is any disjoint family of non-empty open subintervals of [a∗ , b∗ ] with P P P P ∗ ˜ PJ∈J µJ ≤ η, we can find a family J ⊆ J with PJ∈J˜ µJ = J∈J µJ and J∈J˜ |G(sup J) − G(inf J)| ≥ J∈J |G(sup J) − G(inf J)|. But this means that J∈J |G(sup J) − G(inf J)| ≤ 3². As ² is arbitrary, G is Q absolutely continuous. Q Rx (ii) By 225E, G0 is Lebesgue integrable and G(x) = G(a∗ ) + a∗ G0 for every x ∈ [a∗ , b∗ ]. Set g(x) = G0 (x) when x ∈ K and G0 (x) is defined, 0 for other x ∈ R, so that g : R → R is Lebesgue integrable. Now take any x ∈ K. Then Z x F (x) = G(x) = G(a∗ ) + G0 a∗ Z x Z = F (a∗ ) + g+ = F (a∗ ) +

a∗ Z x

G0

[a∗ ,x]\K

X

g+ a∗

I∈I,I⊆[a∗ ,x]

Z G0 I

S (because I is a disjoint countable family of measurable sets, and I∈I,I⊆[a∗ ,x] I = [a∗ , x] \ K) Z x X ∗ = F (a ) + g+ G(sup I) − G(inf I) a∗

I∈I,I⊆[a∗ ,x]

(note that this sum is absolutely summable) Z x = F (a∗ ) + g+ a∗

X

F (sup I) − F (inf I)

I∈I,I⊆[a∗ ,x]

as required. 483R Theorem (see Gordon 94, 11.14) Let F : R → R be a function. Then F is an indefinite Henstock integral iff it is ACG∗ , limx→−∞ F (x) = 0 and limx→∞ F (x) is defined in R. proof (a) Suppose that F is the indefinite Henstock integral of f : R → R. (i) By 483F, F is continuous, with limit zero at −∞ and finite at ∞. So I have just to show that there is a sequence of sets, covering R, on each of which F is AC∗ . Recall that there is a sequence hKm im∈N of compact sets, covering R, such that f × Km is Lebesgue integrable for every m ∈ N (483G). By the arguments of (i)⇒(ii) in the proof of 483J, there is a function F2 ≥ F such that D∗ F2 ≥ f and F2 − F is non-decreasing and takes values between 0 and 1. For n ∈ N, j ∈ Z set Inj = [2−n j, 2−n (j + 1)], Bnj = {x : x ∈ Inj ,

F2 (y)−F2 (x) y−x

≥ −n whenever y ∈ Inj \ {x}}.

483R


741

S Observe that n∈N,j∈Z Bnj = R, so that {Bnj ∩ Km : m, n ∈ N, j ∈ Z} is a countable cover of R. It will therefore be enough to show that F is AC∗ on every Bnj ∩ Km . (ii) Fix m, n ∈ N and j ∈ Z, and set A = Bnj ∩ Km . If A = ∅, then of course F is AC∗ on A; suppose that A is not empty. Set G(x) = F2 (x) + nx for x ∈ Inj , so that F = G − (F2 − F ) − H on Inj , where H(x) = nx. Whenever a, b ∈ Bnj and a ≤ x ≤ b, then G(a) ≤ G(x) ≤ G(b), because x ∈ Inj . So if a0 , b0 , a1 , b1 , . . . , ak , bk ∈ A and a0 ≤ b0 ≤ a1 ≤ b1 ≤ . . . ≤ ak ≤ bk , Pk Pk i=0 ω(G¹ [ai , bi ]) ≤ i=0 G(bi ) − G(ai ) = G(bk ) − G(a0 ) ≤ ω(G¹Inj ), and k X

ω(F ¹ [ai , bi ]) ≤

i=0

k X

ω(G¹ [ai , bi ]) +

i=0

k X

ω(F − F2 ¹ [ai , bi ]) +

i=0

k X

ω(H¹ [ai , bi ])

i=0

≤ ω(G¹Inj ) + ω(F − F2 ¹Inj ) + ω(H¹Inj ) (because F − F2 and H are monotonic) ≤ ω(F ¹Inj ) + 2ω(F − F2 ¹Inj ) + 2ω(H¹Inj ) ≤ ω(F ¹Inj ) + 2(1 + nµInj ) = M say, which is finite, because F is bounded. (iii) By 2A2I (or 4A2Rj), the open set R \ A is expressible as a countable union of disjoint S non-empty open intervals. Two of these are unbounded; let I be the set consisting of the rest, so that A ∪ I = [a∗ , b∗ ] is a closed interval included in Inj . If I, I 0 are distinct members of I and inf I ≤ inf I 0 , then sup I ≤ inf I 0 , 0 because I ∩ I 0 = ∅, and there must be a point P of A in the interval [sup I, inf I ]; so in fact there must be a P If I0 ⊆ I is finite and non-empty, we can point of A in this interval. It follows that I∈I ω(F ¹I) ≤ M . P enumerate it as hIi ii≤k where sup Ii ≤ inf Ii0 whenever i < i0 ≤ k. We can find a0 , . . . , ak+1 ∈ A such that a0 ≤ inf I0 , sup Ii ≤ ai+1 ≤ inf Ii+1 for every i < k, and sup Ik ≤ ak+1 ; so that Pk Pk P i=0 ω(F ¹ [ai , ai+1 ]) ≤ M . i=0 ω(F ¹Ii ) ≤ I∈I0 ω(F ¹I) = As I0 is arbitrary, this gives the result. Q Q (iv) Whenever a∗ ≤ x ≤ y ≤ b∗ , |F (y) − F (x)| ≤

R

|f |dµ + A∩[x,y]

P I∈I

ω(F ¹I ∩ ]x, y[).

P P For each I ∈ I, kf × χ(I ∩ ]x, y[)kH = ω(F ¹I ∩ ]x, y[) ≤ ω(F ¹I) (483P(a-i)). So is finite. Writing H = ]x, y[ ∩ the other hand,

P

S

R

kf × χ(I ∩ ]x, y[)kH ≤

P

ω(F ¹I) R P I, H f × χH is defined and equal to I∈I H f × χ(I ∩ ]x, y[), by 483N. On I∈I

I∈I

]x, y[ \ H ⊆ A ⊆ Km , so f × χ(]x, y[ \ H) is Lebesgue integrable. Accordingly Z |F (y) − F (x)| = | H f × χ ]x, y[ | Z Z ≤ | H f × χH| + | f × χ(]x, y[ \ H)dµ| Z X Z |f |dµ ≤ | H f × χ(I ∩ ]x, y[)| + I∈I

≤

X

I∈I

Z

A∩[x,y]

ω(F ¹I ∩ ]x, y[) +

|f |dµ, A∩[x,y]

742

Gauge integrals

483R

as claimed. Q Q (v) So if a∗ ≤ a ≤ b ≤ b∗ , ω(F ¹ [a, b]) ≤

P I∈I

ω(F ¹I ∩ [a, b]) +

R A∩[a,b]

|f |dµ.

P P We that if a ≤ x ≤ y ≤ b, then ω(F ¹I ∩ ]x, y[) ≤ ω(F ¹I ∩ [a, b]) for every I ∈ I, R have only to observe R and A∩]x,y[ |f |dµ ≤ A∩[a,b] |f |dµ. Q Q Rx (vi) Now let ² > 0. Setting F˜ (x) = a∗ f ×χAdµ for x ∈ [a∗ , b∗ ], F˜ is absolutely continuous (225E), and Pk there is an η0 > 0 such that i=0 |F˜ (bi ) − F˜ (ai )| ≤ ² whenever a∗ ≤ a0 ≤ b0 ≤ a1 ≤ b1 ≤ . . . ≤ ak ≤ bk ≤ b∗ Pk P and i=0 bi − ai ≤ η0 . Take I0 ⊆ I to be a finite set such that I∈I\I0 ω(F ¹I) ≤ ², and let η > 0 be such that η ≤ η0 and η < diam I for every I ∈ I0 . Pk Suppose that a0 , b0 , . . . , ak , bk ∈ A are such that a0 ≤ b0 ≤ . . . ≤ ak ≤ bk and i=0 bi − ai ≤ η. Then Sno member of I0 can be included in any interval [ai , bi ], and therefore, because no ai or bi can belong to I, no member of I0 meets any [ai , bi ]. Also, of course, a∗ ≤ a0 and bk ≤ b∗ . We therefore have k X

ω(F ¹ [ai , bi ]) ≤

i=0

k X X

Z ω(F ¹I ∩ [ai , bi ]) +

=

|f |dµ A∩[ai ,bi ]

i=0 I∈I k X X

ω(F ¹I ∩ [ai , bi ]) +

k X

i=0 I∈I\I0

≤

k X

X

F˜ (bi ) − F˜ (ai )

i=0

ω(F ¹I) + ²

i=0 I∈I\I0 ,I⊆[ai ,bi ]

(because if I ∈ I meets [ai , bi ], it is included in it) X ≤ ω(F ¹I) + ² ≤ 2². I∈I\I0

As ² is arbitrary, F is AC∗ on A. This completes the proof that F is ACG∗ and therefore satisfies the conditions (α) and (β). (b) Now suppose that F satisfies the conditions. Set F (−∞) = 0 and F (∞) = limx→∞ F (x), so that 0 F : [−∞, ∞] → R is continuous. For x ∈ R, set f (x) R = F (x) if this is defined, 0 otherwise. Let J be the family of all non-empty intervals C ⊆ R such that H f × χC is defined and equal to F (sup C) − F (inf C), and let I be the set of non-empty open intervals I such that every non-empty subinterval of I belongs to J . I seek to show that R belongs to I. (i) Of course singleton intervals belong to J . If C1 , C2 ∈ J are disjoint and C = C1 ∪ C2 is an interval, then Z Z Z H f × χC = H f × χC1 + H f × χC2 = F (sup C1 ) − F (inf C1 ) + F (sup C2 ) − F (inf C2 ) = F (sup C) − F (inf C) and C ∈ J . If I1 , I2 ∈ I and I1 ∩ I2 is non-empty, then I1 ∪ I2 is an interval; also any subinterval C of I1 ∪ I2 is either included in one of the Ij or is expressible as a disjoint union C1 ∪ C2 where Cj is a subinterval of Ij for each j; so C ∈ J and I ∈ I. If I1 , I2 ∈ I and sup I1 = inf I2 , then I = I1 ∪ I2 ∪ {sup I1 } ∈ I, because any subinterval of I is expressible as the disjoint union of at most three intervals in J . S (ii) If I0 ⊆ I is non-empty and upwards-directed, then I0 ∈ I. P P This is a consequence of 483Bd. If S we take a non-empty open subinterval J of I0 and express it as ]α, β[, where −∞ R ≤ α < β ≤ ∞, and take any x ∈ J, then for any b ∈ [x, β[ there is some member I of I including [x, b], so H f ×χ ]x, b[ = F (b)−F (x); R now limb↑β F (b) = F (β), so 483Bd tells us that H f × χ ]x, b[ is defined and equal to F (b) − F (x). Similarly,

483X


743

R R using 483B(d-β), H f × χ ]a, x[ = F (x) − F (a), so H f × χ ]a, b[ = F (b)S− F (a). Thus J ∈ J . I wrote this out for open intervals, for convenience; but any non-empty subinterval of I0 is either S a singleton or expressible as an open interval with at most two points added, so belongs to J . Accordingly I0 ∈ I. Q Q (iii) It follows that every member of I is included in a maximal member of I. Let I ∗ be the set of maximal members of I. By (i), these are all disjoint, so no endpoint of any member of I ∗ can belong to S ∗ I . S S S ?? Suppose, if possible, that R ∈ / I. Then I = I ∗ cannot be R, and V = R \ I is a non-empty closed set. By (i), no two distinct members of I ∗ can share a boundary point, so V has no isolated points. We are supposing that F is ACG∗ , so there is a countable family A of sets, covering R, such that F is AC∗ on A for every A ∈ A. By Baire’s theorem (3A3G or 4A2Ma), applied to the locally compact Polish space V , V \ A cannot be dense in V for every A ∈ A, so there are an A ∈ A and a bounded open interval J˜ such that ∅ 6= V ∩ J˜ ⊆ A. Set K = J˜ ∩ A; by 483Pc, F is AC∗ on K, and V ∩ J˜ ⊆ K. Because V has no isolated points, V ∩ J˜ is infinite, so, setting a∗ = min K and b∗ = max K, V ∩ ]a∗ , b∗ [ is non-empty. By 483Q, there is a Lebesgue integrable function g : R → R such that Rx P F (x) = F (a∗ ) + a∗ g + I∈I0 ,I⊆[a∗ ,x] F (sup I) − F (inf I) for every x ∈ K, where P I0 is the family of non-empty bounded open intervals, disjoint from K, with endpoints in K, and I∈I0 ω(F ¹I) is finite. Since every memberR of I0 is disjoint from V , it is included in some member of I ∗ and belongs to J , so F (sup I) − F (inf I) = H f × χI for every I ∈ I0 . If I ∈ I0 , then Z Z kf × χIkH = sup | H f × χI × χC| = sup | H f × χC| C∈C

=

sup

C∈C,C⊆I

F (sup C) − F (inf C) = ω(F ¹I)

C∈C,C⊆I

P because every non-empty subinterval of I belongs to J . So I∈I0 kf × χIkH is finite, and f × χH is S Henstock integrable, where H = I0 , by 483N. Moreover, if x ∈ K, then Z x Z [ H f × χH = H f × χ( {I : I ∈ I0 , I ⊆ [a∗ , x]}) a∗ Z X X = H f × χI = F (sup I) − F (inf I). I∈I0 ,I⊆[a∗ ,x] ∗

I∈I0 ,I⊆[a∗ ,x]

∗

∗

But this means that if y ∈ [a , b ], and x = max(K ∩ [a , y]), so that ]x, y[ ⊆ H, then F (y) = F (x) + F (y) − F (x) Z x X ∗ = F (a ) + g+ a∗

Z = F (a∗ ) +

Z (F (sup I) − F (inf I)) + H f × χ ]x, y[

I∈I0 ,I⊆[a∗ ,x]

Z x Z g × χ(K ∩ [a∗ , y[) + H f × χH + H f × χ ]x, y[

Z y = F (a∗ ) + H h

a∗

−∞

where h = f × χH + g × χK is Henstock integrable because f × χH is Henstock integrable and g × χK is Lebesgue integrable. R Accordingly,R if C is any non-empty subinterval of [a∗ , b∗ ], F (sup C) − F (inf C) = H h × χC. But we also d Hy know that dy h = h(y) for almost every y, by 483I. So F 0 (y) is defined and equal to h(y) for almost −∞ R ∗ ∗ every y ∈ [a , b ], and h = f a.e. on [a∗ , b∗ ]. This means that F (sup C) − F (inf C) = H f × χC for any non-empty subinterval C of [a∗ , b∗ ], and ]a∗ , b∗ [ ∈ I. But ]a∗ , b∗ [ meets V , so this is impossible. X X (iv) This contradiction shows that R ∈ I and that F is an indefinite Henstock integral, as required. 483X Basic exercises >(a) Let [a, b] ⊆ R be a non-empty closed interval, and let Iµ be the gauge integral on [a, b] defined from Lebesgue measure and the tagged-partition structure defined in 481J. Show R that, for f : R → R, Iµ (f ¹ [a, b]) = H f × χ[a, b] if either is defined.

744

Gauge integrals

483Xb

> (b) Extract ideas from the proofs of 482G and 482H to give a direct proof of 483B(c)-(d). f (0) = 0 and f (x) = R ∞(c) Set π H f = . 2 (Hint: 283Da.) 0 (d) Set f (x) =

1 x

cos(

sin x x

for other real x. Show that f is Henstock integrable, and that

1 ) x2

for 0 < x ≤ 1, 0 for other real x. Show that f is Henstock integrable but not R1 d 1 Lebesgue integrable. (Hint: by considering (x2 sin 2 ), show that lima↓0 a f is defined.) dx

x

(e) Let f : R → R be a Henstock integrable function. Show that there is a finitely additive functional λ : PR → R such that for every ² > 0 there are a gauge δ ∈ ∆ and a Radon measure ν on R such that νR ≤ ² and |St (f, µ) − λWt | ≤ νWt for every δ-fine t ∈ T . S (f ) Let f : R → R be a Henstock integrable function. Show that {G : G ⊆ R is open, f is Lebesgue integrable over G} is dense. (Hint: 483G.) > (g) Let f : R → R be a Henstock integrable function and F its indefinite Henstock integral. Show that f is Lebesgue integrable iff F is of bounded variation on R. (Hint: 224I.) > (h) Let F : R → R be a continuous function such that limx→−∞ F (x) = 0, limx→∞ F (x) is defined in R, and F 0 (x) is defined for all but countably many x ∈ R. Show that F is the indefinite Henstock integral of any function f : R → R extending F 0 . (Hint: in 483J, take F1 and F2 differing from F by saltus functions.) (i) Let f : R → R be a Henstock integrable function, and hIn in∈N a disjoint sequence of intervals in R. Show that limn→∞ kf × χIn kH = 0. (j) Show that HL1 is not a Banach space. (Hint: there is a continuous function which is nowhere differentiable.) > (k) Use 483N to replace half of the proof of 483Bd. R (l) Let U be a linear subspace of RR and φ : U → R a linear functional such that (i) f ∈ U and φf = f dµ for every Lebesgue integrable function f : R → R (ii) f × χC ∈ U whenever P f ∈ U and C ∈ C (iii) whenever f ∈ U and ISis a disjoint family of non-empty open intervals such that I∈I supC⊆I,C∈C |φ(f × χC)| < ∞, S P then f × χ( I) ∈ U and φ(f × χ( I)) =R I∈I φ(f × χI). Show that if f : R → R is any Henstock integrable function, then f ∈ U and φ(f ) = H f . (Hint: use the argument of part (b) of the proof of 483R.) Let F : R → R be such that D∗ F and D∗ F are both finite everywhere. Show that F (b) − F (a) = R (m) H D ∗ F × χ[a, b] whenever a ≤ b in R. > (n) Let ν be any Radon measure on R, and let Iν be the gauge integral defined from ν and the tagged-partition structure of 481K and this section. (i) Show that if Iν (f ) is defined, then f Ris dom ν-measurable. R (ii) Show that if f : R → R is such that f dν is defined in R, then Iν (f ) is defined and equal to f dν. (iii) Suppose that f : R → R is a function, and α ∈ ]−∞, ∞]. Show that Iν (f × χ ]−∞, α[) = limβ↑α Iν (f × χ ]−∞, β[) if either is defined in R. (iv) Show that if Iν (f ) is defined, then there is a countable cover of R by compact sets K such that R |f |dν < ∞. K (o) (Bongiorno di Piazza & Preiss 00) Let C be the set of non-empty subintervals of a closed interval [a, b] ⊆ R, and T the straightforward set of tagged partitions generated by [a, b] × C. Let ∆ be the set of neighbourhood gauges on [a, b]. For α ≥ 0 set P Tα = {tt : t ∈ T, (x,C)∈tt ρ(x, C) ≤ α}, writing ρ(x, C) = inf y∈C |x − y| as usual. Show that Tα is compatible with ∆ and R = {{∅}} in the sense of 481F. Show that if Iα is the gauge integral defined from [a, b], Tα , ∆, R and Lebesgue measure, then Iα extends the ordinary Lebesgue integral and Iα (F 0 ) = F (b) − F (a) whenever F : [a, b] → R is differentiable.

483 Notes


745

483Y Further exercises (a) Let us say that a Lebesgue measurable neighbourhood gauge on R is a neighbourhood gauge of the form {(x, A) : x ∈ R, A ⊆ ]x − ηx , x + ηx [} where x 7→ ηx is a Lebesgue ˜ be the set of Lebesgue measurable neighbourhood gauges. measurable function from R to [0, ∞[. Let ∆ ˜ R) and µ is the Henstock Show that the gauge integral defined by the tagged-partition structure (R, T, ∆, integral. (b) Show that if ∆0 ⊆ ∆ is any set of cardinal at most c, then the gauge integral defined by (R, T, ∆0 , R) and µ does not extend the Lebesgue integral, so is not the Henstock integral. (c) Let f : R → R be a Henstock integrable function with indefinite Henstock integral F , and ν a totally finite Radon measure on R. Set G(x) = ν ]−∞, x] for x ∈ R. R Show that f × G is Henstock integrable, with indefinite Henstock integral H, where H(x) = F (x)G(x) − ]−∞,x] F dν for x ∈ R. (d) Let f : R → R be a Henstock integrable function, and G : R → R a function of bounded variation. Show that f × G is Henstock integrable, and

R

H

Rx

f × G ≤ (limx→∞ |G(x)| + VarR G) supx∈R | H−∞ f |.

(Compare 224J.) (e) Let f : R → R be a Henstock integrable function and gR : R → R aR Lebesgue integrable function; let F and G be their indefinite (Henstock) integrals. Show that H f × G + g × F dµ is defined and equal to limx→∞ F (x)G(x). (f ) Let f : R → R be a function. Show that f is Lebesgue integrable iff f × g is Henstock integrable for every bounded continuous function g : R → R. P (g) Let V be a Banach space and f : R → V a function. For t ∈ T , set St (f, µ) = (x,C)∈tt f (x)µC. We R say that f is Henstock integrable, with Henstock integral u = H f ∈ V , if u = limt →F (T,∆,R) St (f, µ). (i) Show that the set HL1V of Henstock integrable functions from RR to V is a linear subspace of V R including the space L1V of Bochner integrable functions (253Yf), and that H : HL1V → V is a linear operator extending the Bochner integral. (ii) RShow that if f : R → V is Henstock integrable, Rso is f × χC for every interval C ⊆ R, and that (a, b) 7→ H f × χ[a, b] is continuous. Set kf kH = supC∈C k H f × χCk. 1 (iii) Show that if inSR, and f : R → V is such that R P f ×RχI ∈ HLV PI is a disjoint family of open intervals for every I ∈ I and I∈I kf × χIkH is finite, then H f × χ( I) is defined and equal to I∈I H f × χI. (iv) Define f : R → `∞ ([0, 1]) by setting f (x) = χ([0, 1] ∩ ]−∞, x]) for x ∈ R. Show that f is Henstock R 1 integrable, but that if F (x) = H f × χ ]−∞, x[ for x ∈ R, then limy→x (F (y) − F (x)) is not defined in `∞ ([0, 1]) for any x ∈ [0, 1].

y−x

483 Notes and comments I hope that the brief account here (largely taken from Gordon 94) will give an idea of the extraordinary power of gauge integrals. While what I am calling the ‘Henstock integral’, regarded as a linear functional on a space of real functions, was constructed long ago by Perron and Denjoy, the gauge integral approach makes it far more accessible, and gives clear pathways to corresponding Stieltjes and vector integrals (483Xn, 483Yg). Starting from our position in the fourth volume of a book on measure theory, it is natural to try to describe the Henstock integral in terms of the Lebesgue integral, as in 483C (they agree on non-negative functions) and 483Xl (offering an extension process to generate the Henstock integral from the Lebesgue integral); on the way, we see that Henstock integrable functions are necessarily Lebesgue integrable over many intervals (483Xf). Alternatively, we can set out to understand indefinite Henstock integrals and their derivatives, just as Lebesgue integrable functions can be characterized as almost everywhere equal to derivatives of absolutely continuous functions (222E, 225E), because if f is Henstock integrable then it is equal almost everywhere to the derivative of its indefinite integral (483I). Any differentiable function (indeed, any continuous function differentiable except on a countable set) is an indefinite Henstock integral (483Xh). Recall that the Cantor

746

Gauge integrals

483 Notes

function (134H) is continuous and differentiable almost everywhere but is not an indefinite integral, so we have to look for a characterization which can exclude such cases. For this we have to work quite hard, but we find that ‘ACG∗ functions’ are the appropriate class (483R). Gauge integrals are good at integrating derivatives (see 483Xh), but bad at integrating over subspaces. Even to show that f × χ [0, ∞[ is Henstock integrable whenever f is (482Bc) involves us in some unexpected manoeuvres. I give an argument which is designed to show off the general theory of §482, and I recommend you to look for short cuts (483Xb), but any method must depend on careful examination of the exact classes C and R chosen for the definition of the integral. We do however have a new kind of convergence theorem in 482H and 483N. One of the incidental strengths of the Henstock integral is that it includes the improper Riemann integral (483Bd, 483Xc); so that, for instance, Carleson’s theorem (286U) can be written in the form

R

∧

1 f (y) = √

2π

H

e−ixy f (x)dx for almost every y

Ra if f : R → C is square-integrable. But to represent the many formulae of the type lima→∞ −a f in §283 (e.g., 283F, 283I and 283L) directly in the form Iµ (f ) we need to change R, as in 481L or 481Xc.

484 The Pfeffer integral I give brief notes on what seems at present to be the most interesting of the multi-dimensional versions of the Henstock integral, leading to Pfeffer’s Divergence Theorem (484N). 484A Notation This section will depend heavily on Chapter 47, and will use much of the same notation. r ≥ 2 will be a fixed integer, and µ will be Lebesgue measure on Rr , while µr−1 is Lebesgue measure on Rr−1 . As in §§473-475, let ν be ‘normalized’ (r − 1)-dimensional Hausdorff outer measure on Rr , as described in §265; that is, ν = 2−r+1 βr−1 µH,r−1 , where µH,r−1 is r −1-dimensional Hausdorff measure on Rr as described in §264, and βr−1 = =

22k k!π k−1 (2k)! πk k!

if r = 2k is even,


is the Lebesgue measure of a ballS of diameter 1 in Rr−1 (264I). For this section only, let us say that a subset of Rr is thin if it is of the form n∈N An where ν ∗ An is finite for every n. Note that by 471L, µA = 0 for every thin set A. For A ⊆ Rr , write ∂A for its ordinary topological boundary. If x ∈ Rr and ² > 0, B(x, ²) will be the closed ball {y : ky − xk ≤ ²}. Q I will use the term dyadic cube for sets of the form i
1 2r−2 ). , rr/2 rβr(r−1)/r

(As will become apparent, the actual value of this constant is of no importance; but the strict logic of the arguments below depends on α∗ being small enough.) As in §475, I write int*A, cl*A and ∂*A for the essential interior, essential closure and essential boundary of a set A ⊆ Rr (475B). Recall that a set A ⊆ Rr has finite perimeter in the sense of 474D iff ν(∂*A) is finite, and then ν(∂*A) = λ∂A (Rr ) = per A is the perimeter of A (475M); we shall also need to remember that A is necessarily Lebesgue measurable. C will be the family of subsets of Rr with locally finite perimeter, and V the family of bounded sets in C, that is, the family of bounded sets with bounded perimeter. Note that C is an algebra of subsets of Rr (475Ma), and that V is an ideal in C.

484B

The Pfeffer integral

747

484B Theorem (Tamanini & Giacomelli 89) Let E ⊆ Rr be a Lebesgue measurable set of finite measure and perimeter, and ² > 0. Then there is a Lebesgue measurable set G ⊆ E such that per G ≤ per E, µ(E \ G) ≤ ² and cl*G = G. proof (Pfeffer 91b) (a) Set α = 1² per E. For measurable sets G ⊆ E set q(G) = per G − αµG. Then there is a self-supporting measurable set G ⊆ E such that q(G) ≤ q(G0 ) whenever G0 ⊆ E is measurable. P P Write Σ for the family of Lebesgue measurable subsets of Rr ; give Σ the topology of convergence in measure defined by the pseudometrics ρH (G, G0 ) = µ((G4G0 ) ∩ H) for measurable sets H of finite measure (cf. 474S). Extend q to Σ by setting q(G) = per(E∩G)−αµ(E∩G) for every G ∈ Σ. Because per : Σ → [0, ∞] is lower semi-continuous for the topology of convergence in measure (474Sa), and G 7→ E ∩ G, G 7→ µ(E ∩ G) are continuous, q : Σ → [0, ∞[ is lower semi-continuous (4A2B(d-ii)). Next, K = {G : G ∈ Σ, per G ≤ per E} is compact (474Sb), while L = {G : µ(G \ E) = 0} is closed, so there is an G0 ∈ L ∩ K such that q(G0 ) = inf G∈L∩K q(G) (4A2Gk). Since G0 ∈ L, per(G0 ∩ E) = per(G0 ) and µ(G0 ∩ E) = µG0 , so we may suppose that G0 ⊆ E. Moreover, there is a self-supporting set G ⊆ G0 such that G0 \ G is negligible (414F), and we still have q(G) = q(G0 ). Of course q(G) ≤ q(E), just because E ∈ L ∩ K. ?? If there is a measurable set G0 ⊆ E such that q(G0 ) < q(G), then per G0 = q(G0 ) + αµG0 ≤ q(E) + αµE = per E, so G0 ∈ K; but this means that G0 ∈ L ∩ K and q(G) = q(G0 ) ≤ q(G0 ). X X So G has the required properties. Q Q (b) Since q(G) ≤ q(E), we must have per G = q(G) + αµG ≤ q(E) + αµE = per E (just as in the last part of the proof of (a)). Also per G + αµ(E \ G) ≤ per E, so µ(E \ G) ≤ ². (c) Next, G ⊆ cl*G. P P Let x ∈ G. For every t > 0, set Ut = {y : ky − xk < t}; then q(G \ Ut ) ≤ q(G). Now

per(G ∩ Ut ) + per(G \ Ut ) = ν(∂*(G ∩ Ut )) + ν(∂*(G \ Ut )) ≤ ν(∂*G ∩ Ut ) + ν(cl*G ∩ ∂Ut ) + ν(∂*G \ Ut ) + ν(cl*G ∩ ∂Ut ) (475Cf, because ∂(Rr \ Ut ) = ∂Ut )) = ν(∂*G) + 2ν(cl*G ∩ ∂Ut ) = per G + 2ν(G ∩ ∂Ut ) for almost every t > 0, because

R∞ 0

ν((G4cl*G) ∩ ∂Ut )dt = µ(G4cl*G) = 0

(265G). So, for almost every t, µ(G ∩ Ut )(r−1)/r ≤ per(G ∩ Ut ) (474La) ≤ per G + 2ν(G ∩ ∂Ut ) − per(G \ Ut ) = q(G) + αµ(G ∩ Ut ) + 2ν(G ∩ ∂Ut ) − q(G \ Ut ) ≤ αµ(G ∩ Ut ) + 2ν(G ∩ ∂Ut ) because q(G) is minimal. 1 r

?? Suppose, if possible, that x ∈ / cl*G. Take β > 0 such that 2β < β (r−1)/r . Then there is some γ > 0 1 r

such that µ(G ∩ Ut ) ≤ βtr for every t ∈ ]0, γ], and we may suppose that 2β < β (r−1)/r − case, for almost every t ∈ ]0, γ],

1 αγ. r+1

In this

748

Gauge integrals

484B

1 2

ν(G ∩ ∂Ut ) ≥ (β (r−1)/r tr−1 − αtr ), so βγ r ≥ µ(G ∩ Uγ ) =

Rγ 0

ν(G ∩ ∂Ut )dt ≥

1 ¡ β (r−1)/r r γ 2 r

−

α r+1 ¢ γ r+1

and 1 r

2β ≥ β (r−1)/r −

1 αγ r+1

> 2β,

which is absurd. X X Thus x ∈ cl*G. As x is arbitrary, G ⊆ cl*G. Q Q Since we certainly have cl*G ⊆ G, this G serves. E is expressible as S 484C Lemma Let E ∈ V and l ∈ N be such that max(per E, diam E) ≤r l. Then r 2 E where hE i is disjoint, per E ≤ 1 for each i < n and n is at most 2 (l + 2) (4r(4l + 1))r/(r−1) + i i i
1 2

diam D}, the family of the 2r dyadic

(a) If l ≤ 1 the result is trivial, so let us suppose that l ≥ 2. Let m ∈ N be minimal subject to 4r(4l2 + 1) ≤ 2m(r−1) , so that 2m ≤ 2(4r(4l2 + 1))1/(r−1) . Then we can cover E by a family L0 of dyadic cubes of side 2−m with #(L0 ) ≤ (2m l + 2)r ≤ 2r (l + 2)r (4r(4l2 + 1))r/(r−1) . (b) Let L1 be the set of those D ∈ D such that D is included in some member of L0 and b2ν(D0 ∩∂*E)c < b2ν(D ∩ ∂*E)c for every D0 ∈ DD . Then #(L1 ) ≤ (2l)2 . P P I show by induction on k that if D0 ∈ D and b2ν(D ∩ ∂*E)c = k then #({D : D ∈ L1 , D ⊆ D0 } ≤ k 2 . For k = 0 this is trivial. For the inductive step to k > 0, given that b2ν(D0 ∩ ∂*E)c = k, let L01 be the set of maximal elements in {D : D ∈ L1 , D ⊆ D0 , b2ν(D ∩ ∂*E)c < k}. Now there can be at most one member D of L1 included in D0 and with b2ν(D ∩ ∂*E)c ≥ k. So

#({D : D ∈ L1 , D ⊆ D0 }) ≤ 1 +

X

#({D : D ∈ L1 , D ⊆ D1 })

D1 ∈L01

≤1+

X

(b2ν(D1 ∩ ∂*E)c)2

D1 ∈L01

(by the inductive hypothesis) ≤ (b2ν(D0 ∩ ∂*E)c)2 (because (bαc)2 + (bβc)2 < (bα + βc)2 whenever α, β ≥ 0 and max(bαc, bβc) < bα + βc)) ≤ k2 . So the induction proceeds. Taking k = 2l, we have the result. Q Q S ˜ = D \ S{D0 : D0 ∈ L1 , D0 ⊆ D}, and consider (c) Set L2 = L0 ∪ {DD : D ∈ L1 }. For D ∈ L2 , set D ˜ : D ∈ L2 }, so that K = {D #(K) ≤ #(L2 ) ≤ #(L0 ) + 2r #(L1 ) ≤ 2r (l + 2)r (4r(4l2 + 1))r/(r−1) + 2r+2 l2 . S (d)(i) K ⊇ E. P P If x ∈ E, there is a smallest member D of L2 containing it, because certainly S ˜ Q x ∈ L0 . But now x cannot belong to any member of L1 included in D, so x ∈ D. Q

484E


749

˜1 ∩ D ˜ 2 = ∅. If D1 ⊂ D2 , then (ii) K is disjoint. P P If D1 , D2 ∈ L2 are disjoint, then of course D ˜ 2 ⊆ D2 \ D is disjoint from D1 . D1 ∈ / L0 , so there is a D ∈ L1 such that D1 ∈ DD ; in this case D ⊆ D2 so D Q Q ˜ 0 ∩∂*E) > 1 , set j = b2ν(D0 ∩∂*E)c ≥ (iii) ν(D ∩∂*E) ≤ 21 for every D ∈ K. P P?? If D0 ∈ L2 and ν(D 2 ˜ 0 ∩ Di 1. Given that Di ∈ D is included in D0 and b2ν(Di ∩ ∂*E)c = j, then ν((D0 \ Di ) ∩ ∂*E) < 21 , so D is non-empty and Di ∈ / L1 . There must therefore be a Di+1 ∈ DDi such that b2ν(Di+1 ∩ ∂*E)c = j. Thus we have a strictly decreasing sequence hDi ii∈N in D such that ν(DiT∩ ∂*E) ≥ j for every i. But (because per E is finite) this means that, writing x for the unique member of i∈N Di , ν{x} ≥ j, which is absurd. X X Q Q (iv) per(D ∩ E) ≤ 1 for every D ∈ K. P P If D ∈ L2 , P 1 0 2 −m(r−1) ˜ ≤ ν(∂D) + ν(∂ D) ≤ D 0 ∈L1 ν(∂D ) ≤ 2r(4l + 1)2 2

by the choice of m. So per(D ∩ E) ≤ ν(∂D) + ν(D ∩ ∂*E) ≤

1 2

+

1 2

=1

by 475Cf. Q Q (e) So if we take hEi ii 0 such that R ∈ Rη whenever µR ≤ ², diam R ≤ γ and per R ≤ γ. (ii) If R ∈ R and γ ≥ 0, there is an ² > 0 such that R ∈ R whenever µR ≤ ² and per R ≤ γ. (c) If R ∈ R there is an R0 ∈ R such that R ∪ R0 ∈ R whenever R, R0 ∈ R0 and R ∩ R0 = ∅. (d)(i) If η ∈ H S and A ⊆ Rr is a thin set, then there S is a set D0 ⊆ D such that every point of A belongs to the interior of D1 for some finite D1 ⊆ D0 , and D1 ∈ Rη for every finite set D1 ⊆ D0 . (ii) If R ∈ RSand A ⊆ Rr is a thin set, then there isSa set D0 ⊆ D such that every point of A belongs to the interior of D1 for some finite set D1 ⊆ D0 , and D1 ∈ R for every finite set D1 ⊆ D0 . (V )

proof (a)(i) Express R as Rη0 where η 0 ∈ H and V ∈ V. Let l ∈ N be such that max(diam V, 1 + per V ) ≤ l, and take n ≥ 2r (l + 2)r (4r(4l2 + 1))r/(r−1) + 2r+2 l2 . Set S η(i) = min{η 0 (j) : ni ≤ j < n(i + 1)} for every i ∈ N, so that η ∈ H. If R ∈ Rη , express it as i∈N Ei where hEiS ii∈N ∈ Mη . For each i, max(diam(Ei ∩ V ), per(Ei ∩ V ) ≤ l, so by 484C we can express Ei ∩ V as ni≤j
(C∩V )

(ii) Express R as Rη , where V ∈ V and η ∈ H. By (i), there is an η 0 ∈ H such that Rη0 ⊆ Rη (V ) Set R0 = Rη0 ∈ R. If R ∈ R0 , then R ∩ V ∈ Rη0 , so C ∩ R ∩ V ∈ Rη and C ∩ R ∈ R.

.

(b)(i) Take l ≥ γ and n ≥ 2r (l + 2)r (4r(4l2 + S 1))r/(r−1) + 2r+2 l2 , and set ² = mini
750

Gauge integrals

484E

(V )

(ii) Express R as Rη . By (i) just above, there is an ² > 0 such that R ∈ Rη whenever diam R ≤ diam V , per R ≤ γ + per V and µR ≤ ²; and this ² serves. (V )

(c) Express R as Rη . Set η 0 (i) = min(η(2i), η(2i + 1)) for every i; then η 0 ∈ H and if hEi ii∈N , hEi0 ii∈N belong to Mη0 and have disjoint unions, then (E0 , E00 , E1 , E10 , . . . ) ∈ Mη ; this is enough to show that (V ) R ∪ R0 ∈ Rη whenever R, R0 ∈ Rη0 are disjoint, so that R ∪ R0 ∈ R whenever R, R0 ∈ Rη0 are disjoint. S α) We can express A as i∈N Ai where µ∗H,r−1 Ai < 1/2r+1 r for every i. P (d)(i)(α P Because A is thin, it is the union of a sequence of sets of finite outer measure for ν, and therefore for µH,r−1 . On each of these the subspace measure is atomless (471E, 471Dg), so that the set can be dissected into finitely many sets of measure less than 1/2r+1 r (215D). Q Q β ) For (β P∞each i ∈ N, we can cover Ai by a sequence hAij ij∈N of sets such that diam Aij < η(i) for every j and j=0 (diam Aij )r−1 < 1/2r+1 r; enlarging the Aij slightly if need be, we can suppose that they are all open. Now we can cover each Aij by 2r cubes Dijk ∈ D in such a way that the side length of each Dijk is at most the diameter of Aij . Setting D0 = {Dijk : i, j ∈ N, k < 2r }, we see that every point of A belongs to an open set Aij which is covered by a finite subsetSof D0 . S If D1 ⊆ D0 is finite, let D10 be the family of maximal elements of D1 , 0 0 so that D1 is disjoint and D1 = D1 . Express D10 as {Dijk : S (i, j, k) ∈ I} where I ⊆ N × N × N and hDijk i(i,j,k)∈I is disjoint. Set Ii = {(j, k) : (i, j, k) ∈ I} and Ei = (j,k)∈Ii Dijk for i ∈ N. Then hEi ii∈N is disjoint, Ei = ∅ for all but finitely many i, and for each i ∈ N r

µEi ≤

∞ 2X −1 X j=0 k=0 ∞ X r

r

µDijk ≤

∞ 2X −1 X

(diam Aij )r

j=0 k=0

η(i)(diam Aij )r−1 ≤ η(i),

≤2

j=0

ν(∂Ei ) ≤

X

r

ν(∂Dijk ) ≤

(j,k)∈Ii

∞ 2X −1 X

r

≤

∞ 2X −1 X j=0 k=0

So

S

D1 =

S i∈N

ν(∂Dijk )

j=0 k=0

2r(diam Aij )r−1 ≤

∞ X

2r+1 r(diam Aij )r−1 ≤ 1.

j=0

Ei belongs to Rη .

(ii) By (a-i), there is an η ∈ HSsuch that Rη ⊆ R. By (i) here, there is a D0 ⊆ D such that every S point of A belongs to the interior of D1 for some finite D1 ⊆ D0 , and D1 ∈ Rη ⊆ R for every finite set D1 ⊆ D 0 . 484F A family of tagged-partition structures For α > 0, let Cα be the family of those C ∈ V such that µC ≥ α(diam C)r and α per C ≤ (diam C)r−1 , and let Tα be the straightforward set of tagged partitions generated by the set {(x, C) : C ∈ Cα , x ∈ cl*C}. 0

Note that if 0 < α ≤ α then Cα ⊆ Cα0 and Tα ⊆ Tα0 . Let Θ be the set of functions θ : Rr → [0, ∞[ such that {x : θ(x) = 0} is thin (definition: 484A), and set ∆ = {δθ : θ ∈ Θ}, where δθ = {(x, A) : x ∈ Rr , θ(x) > 0, A ⊆ B(x, θ(x))}. Then whenever 0 < α < α∗ , (Rr , Tα , ∆, R) is a tagged-partition structure allowing subdivisions, witnessed by C. proof (a) We had better look again at all the conditions in 481G. (i) and (vi) really are trivial. (iv) is true because C is actually an algebra of sets, and (v) is true because V ⊆ C and ∅ ∈ Rη for every η ∈ H. For (ii), we have to observe that the union of two thin sets is thin, so that θ ∧ θ0 ∈ Θ for all θ, θ0 ∈ Θ; since δθ1 ∩ δθ2 = δθ1 ∧θ2 , this is all we need.

484F


751 (V )

(iii)(α) follows from 484Ea: given V , V 0 ∈ V and η, η 0 ∈ H, take η˜, η˜0 ∈ H such that Rη˜ ⊆ Rη (V 0 ) (V ) (V 0 ) (V ∪V 0 )) Rη˜0 ⊆ Rη0 . Then V ∪ V 0 ∈ V (475Cd) and Rη ∩ Rη0 ⊇ Rη˜∧˜η0 . (β) is just 484Ec.

and

(b) Now let us turn to 481G(vii). (i) Fix C ∈ C, δ ∈ ∆ and R ∈ R. Express δ as δθ where θ ∈ Θ. Let R0 ∈ R be such that A ∪ A0 ∈ R (V ) whenever A, A0 ∈ R0 are disjoint. Express R0 as Rη where V ∈ V and η ∈ H. By 475E(b-ii), there is an 0 ² > 0 such that R ∈ R whenever diam R ≤ diam V , per R ≤ 2 per(C ∩ V ) and µR ≤ ². By 484B, there is an E ⊆ C ∩ V such that per E ≤ per(C ∩ V ), µ((C ∩ V ) \ E) ≤ ² and cl*E = E. In this case, diam(C ∩ V \ E) ≤ diam V and per(C ∩ V \ E) ≤ per(C ∩ V ) + per E ≤ 2 per(C ∩ V ), so C ∩ V \ E ∈ R0 , by the choice of ². By 475E(a-ii), there is an R00 ∈ R such that E ∩ R ∈ R0 whenever R ∈ R00 . Now consider A = {x : θ(x) = 0} ∪ ∂*E ∪ {x : lim supζ↓0 supx∈G,0
ν ∗ (G∩∂ *E) (diam G)r−1

> 0}.

Because per E < ∞ and ν ∗ (G∩∂ *E) (diam G)r−1

{x : x ∈ Rr \ ∂*E, lim supζ↓0 supx∈G,0
> 0}

is ν-negligible (471Pc), A is thin. By 484E(d-ii), there is a set D0 ⊆ D such that S S S A ⊆ {int( D1 ) : D1 ∈ [D0 ]<ω }, D1 ∈ R00 for every D1 ∈ [D0 ]ω . (ii) Write T 0 for the set of those δ-fine t ∈ Tα such that every member of t is of the form (x, D ∩ E) for some D ∈ D. If x ∈ int*E \ A, there is an h(x) > 0 such that h(x) ≤ θ(x) and {(x, D ∩ E)} ∈ T 0 whenever D ∈ D, x ∈ D and diam D ≤ h(x). P P Let ²1 > 0 be such that βr rr/2 ²1 ≤ 21 and ¢(r−1)/r 1−rr/2 ²1 1 1 ¡ βr α≤ , ≥ (2r + rr/2 ²1 ) · r−1 ; r/2 r/2 r

α

2

1−r

²1

∗

this is where we need to know that α < α . Because x ∈ / A, θ(x) > 0; let h(x) ≤ θ(x) be such that (α) ν ∗ (D∩∂*E) ≤ ²1 (diam D)r−1 whenever D ∈ D and x ∈ D and diam D ≤ h(x) (β) µ(B(x, t)\E) ≤ ²1 µB(x, t) whenever 0 ≤ t ≤ h(x). Now suppose that D ∈ D and x ∈ D and diam D ≤ h(x). Then, writing γ for the side length of D, µ(D ∩ E) ≥ µD − µ(B(x, diam D) \ E) ≥ γ r − ²1 βr (diam D)r = γ r (1 − rr/2 ²1 ) ≥ αγ r rr/2 = α(diam D)r ≥ α diam(D ∩ E)r . Since also µ(D ∩ E) ≤

1 βr diam(D 2r

∩ E)r (264H), we see that diam(D ∩ E) ≥ 2γ

¡ 1−rr/2 ²1 ¢1/r βr

.

Next,

per(D ∩ E) = ν(∂*(D ∩ E)) ≤ ν(∂D) + ν(D ∩ ∂*E) (475Cf) ≤ 2rγ r−1 + ²1 (diam D)r−1 = γ r−1 (2r + rr/2 ²1 ) ¢(r−1)/r 1 ¡ βr ≤ (2r + rr/2 ²1 ) · r−1 diam(D ∩ E)r−1 r/2 2

≤

1 α

diam(D ∩ E)

r−1

1−²r r

.

752

Gauge integrals

484F

So D ∩ E ∈ Cα . Also, for every s > 0, there is a D0 ∈ D such that D0 ⊆ D and x ∈ D0 and diam D0 ≤ s. In this case 1 2

µ(B(x, diam D0 ) \ E) ≤ ²1 µB(x, diam D0 ) = ²1 βr (diam D0 )r = ²1 βr rr/2 µD0 ≤ µD0 , so µ(D ∩ E ∩ B(x, diam D0 )) ≥ µD0 − µ(B(x, diam D0 ) \ E) 1 2

≥ µD0 =

1 µB(x, diam D0 ). 2βr rr/2

As diam D0 is arbitrarily small, x ∈ cl*(D ∩ E) and t = {(x, D ∩ E)} ∈ Tα . Finally, since diam D ≤ θ(x), (x, D ∩ E) ∈ δ and t ∈ T 0 . Q Q S (iii) Let H be the set of those H ⊆ R such that Wt ⊆ E ∩ H ⊆ Wt ∪ D1 for some δ-fine t ∈ T 0 and finite D1 ⊆ D0 . Then H ∪ H 0 ∈ H whenever H, H 0 ∈ H are disjoint. If hDn iT n∈N is any strictly decreasing sequence in D, then some Dn belongs to H. P P Let x be the unique point of n∈N Dn . case 1 If x ∈ A, then there is a finite subset D1 of D0 whose union is a neighbourhood of x, and therefore includes Dn for some n; so t = ∅ and D1 witness that Dn ∈ H. case 2 If x ∈ E \ A, then x ∈ int*E, so h(x) > 0 and there is some n ∈ N such that diam Dn ≤ h(x). In this case t = {(x, Dn ∩ E)} belongs to T 0 , by the choice of h(x), so that t and ∅ witness that Dn ∈ H. case 3 Finally, if x ∈ / E ∪ A, then x ∈ / cl*E so x ∈ / E and there is some n such that Dn ∩ E = ∅, in which case t = D1 = ∅ witness that Dn ∈ H. Q Q (iv) In fact Rr ∈ H. P P?? Otherwise, because E ⊆ C is bounded, it can be covered by a finite disjoint family in D, and there must be some D0 ∈ D \ H. Now we can find hDn in≥1 in D \ H such that Dn ⊆ Dn−1 XQ Q and diam Dn = 21 diam Dn−1 for every n. But this contradicts (iii).X S (v) We therefore have a δ-fine t ∈ T 0 and aSfinite set D1 ⊆ D0 such that Wt ⊆ E ⊆ Wt ∪ D1 . Now S S 0 0 D1 . D10 = ∅ and Wt 0 ∪ DS we can find t 0 ⊆ t and DS 1 = Wt 0 ∪ 1 ⊆ D1 such that Wt 0 ∩ By the choice of D0 , D1 ∈ R00 ; by the choice of R00 , E \ Wt = E ∩ D1 belongs to R0 . But we know (V ) also that C ∩ V \ E ∈ R0 , that is, C \ E ∈ R0 , because R0 = Rη . By the choice of R0 , C \ Wt ∈ R. 0 And t ∈ T is a δ-fine member of Tα . As C, δ and R are arbitrary, 481G(vii) is satisfied, and the proof is complete. 484G The Pfeffer integral (a) For α ∈ ]0, α∗ [, write Iα for the linear functional defined by setting Iα (f ) = limt →F (Tα ,∆,R) St (f, µ) r

whenever f : R → R is such that the limit is defined. Then if 0 < β ≤ α and Iβ (f ) is defined, so is Iα (f ), and the two are equal. P P All we have to observe is that Cα ⊆ Cβ so that Tα ⊆ Tβ , while F(Tα , ∆, R) is just {A ∩ Tα : A ∈ F(Tβ , ∆, R)}. Q Q R (b) Let f : Rr → R be a function. I will say that it is Pfeffer integrable, with Pfeffer integral Pf f , if

R

Pf

f = limα↓0 Iα (f )

is defined; that is to say, if Iα (f ) is defined whenever 0 < α < α∗ . 484HR The first step is to work through the results of §482 to see which ideas apply directly to the limit integral Pf . R R Proposition (a) The domain of Pf is a linear space of functions, and Pf is a positive linear functional. R R (b) If f , g : Rr → R are such that |f | ≤ g and Pf g = 0, then Pf f is defined and equal to 0. (c) If f : Rr → R is Pfeffer integrable, then there is a unique additive functional F : C → R such that whenever ² > 0 and 0 < α < α∗ there are δ ∈ ∆ and R ∈ R such that P (x,C)∈tt |F (C) − f (x)µC| ≤ ² for every δ-fine t ∈ Tα , |F (E)| ≤ ² whenever E ∈ R.

484K


753

R Moreover, F (Rr ) = Pf f . (d) Every Pfeffer integrable function is Lebesgue measurable. (e) Every Lebesgue integrable function is Pfeffer integrable, with the same integral. (f) A non-negative function is Pfeffer integrable iff it is Lebesgue integrable. proof (a)-(b) Immediate from 481C. (c) For each α ∈ ]0, α∗ [ let Fα be the Saks-Henstock indefinite integral corresponding to the the structure (Rr , Tα , ∆, R, µ). Then all the Fα coincide. P P Suppose that 0 < β ≤ α < α∗ . Then, for any ² > 0, there are δ ∈ ∆, R ∈ R such that P (x,C)∈tt |Fβ (C) − f (x)µC| ≤ ² for every δ-fine t ∈ Tβ , |Fβ (E)| ≤ ² whenever E ∈ R. Since Tα ⊆ Tβ , this means that P (x,C)∈tt

|Fβ (C) − f (x)µC| ≤ ² for every δ-fine t ∈ Tα .

And this works for any ² > 0. By the uniqueness assertion in 482B, Fβ must be exactly the same as Fα . Q Q So we have a single functional F ; and 482B also tells us that F (Rr ) = Iα (f ) =

R

Pf

f

for every α. (d) In fact if there is any α such that Iα (f ) is defined, f must be Lebesgue measurable. P P We have only to check that the conditions of 482E are satisfied by µ, C, Tα , ∆ and R. The only items not covered by the checks above are (ii) and (v) of the list there. 482E(ii) is true because C \ cl*C is negligible for every C ∈ C (475C(f)). 482E(v) P∞is true because if µE < ∞ and ² > 0, there are n ∈ N and η ∈ H such that µ(E \ B(0, n) ≤ 21 ² and i=0 η(i) ≤ 12 ², so that µ(E ∩ R) ≤ µ(E \ B(0, n)) + µ(R ∩ B(0, n)) ≤ ² (B(0,n))

for every R ∈ Rη

.Q Q

(e) This time, we have to check that the conditions of 482F are satisfied by Tα , ∆ and R whenever 0 < α < α∗ . P P Of course µ is inner regular with respect to the closed sets and outer regular with respect to the open sets (134F). Condition 482F(v) just repeats 482E(v), verified in (d) above. Q Q (f ) If f ≥ 0 is integrable in the ordinary R sense, Rthen it Ris Pfeffer integrable, by (e). If it is Pfeffer integrable, then it is measurable; but also g dµ = Pf g ≤ Pf f for every simple function g ≤ f , so f is integrable (213B). 484I Definition If f : Rr → R is Pfeffer integrable, I will call the function F : C → R defined in 484Hc the Saks-Henstock indefinite integral of f . 484J In fact 484Hc characterizes the Pfeffer integral, just as the Saks-Henstock lemma can be used to define general gauge integrals based on tagged-partition structures allowing subdivisions. Proposition Suppose that f : Rr → R and F : C → R are such that (i) F is additive, P (ii) whenever 0 < α < α∗ and ² > 0 there is a δ ∈ ∆ such that (x,C)∈tt |F (C) − f (x)µC| ≤ ² for every δ-fine t ∈ Tα , (iii) for every ² > 0 there is an R ∈ R such that |F (E)| ≤ ² for every E ∈ C ∩ R. Then f is Pfeffer integrable and F is the Saks-Henstock indefinite integral of f . proof By 482D, the gauge integral Iα (f ) is defined and equal to F (Rr ) for every α ∈ ]0, α∗ [. So f is Pfeffer integrable. Now 484Hc tells us that F must be its Saks-Henstock indefinite integral. 484K Lemma Suppose that α > 0 and 0 < α0 < α min( 21 , 2r−1 (

α (r−1)/r ) ). 2βr

If E ∈ C is such that

E ⊆ cl*E, then there is a δ ∈ ∆ such that {(x, C ∩ E)} ∈ Tα0 whenever (x, C) ∈ δ, x ∈ E and {(x, C)} ∈ Tα .

754

Gauge integrals

484K

proof Take ² > 0 such that βr ² ≤ 12 α and 1 α

+ 2r−1 ² ≤

2r−1 ¡ α ¢(r−1)/r . α0 2βr

Set A = ∂*E ∪ {x : limζ↓0 supx∈G,0
ν ∗ (G∩∂ *E) (diam G)r−1

> 0},

so that A is a thin set, as in the proof of 484F. (Of course ∂*E is thin because ν(∂*E ∩ B(0, n)) is finite for every n ∈ N.) For x ∈ E \ A, we have x ∈ int*E (because E ⊆ cl*E), so there is a θ(x) > 0 such that µ(B(x, γ) \ E) ≤ ²µB(x, γ),

ν(∂*E ∩ B(x, γ)) ≤ ²(2γ)r−1

whenever 0 < γ ≤ 2θ(x). If we set θ(x) = 0 for x ∈ E ∩ A and θ(x) = 1 for x ∈ Rr \ E, then θ ∈ Θ and δθ ∈ ∆. Now suppose that x ∈ E, (x, C) ∈ δθ and {(x, C)} ∈ Tα , that is, that C ∈ Cα and x ∈ (E ∩ cl*C) \ A and C ⊆ B(x, θ(x)). Set γ = diam C ≤ 2θ(x). Then µ(C \ E) ≤ µ(B(x, γ) \ E) ≤ ²µB(x, γ) = βr ²γ r , so µ(C ∩ E) ≥ µC − βr ²γ r ≥ (α − βr ²)γ r 1 2

≥ αγ r ≥ α0 γ r ≥ α0 diam(C ∩ E)r . Next, per(C ∩ E) = ν(∂*(C ∩ E)) ≤ ν(∂*C) + ν(cl*C ∩ ∂*E) 1 α

1 α

≤ per C + ν(B(x, γ) ∩ ∂*E) ≤ γ r−1 + ²(2γ)r−1 = ( + 2r−1 ²)γ r−1 . Moreover, 1 2

2−r βr diam(C ∩ E)r ≥ µ(C ∩ E) ≥ αγ r (264H), so diam(C ∩ E) ≥ 2( 1 α

α 1/r ) γ 2βr

and

per(C ∩ E) ≤ ( + 2r−1 ²) ·

1 ¡ 2βr ¢(r−1)/r diam(C 2r−1 α

∩ E)r−1 ≤

1 α0

diam(C ∩ E)r−1 .

Putting these together, we see that C ∈ Cα0 . Finally, because x ∈ cl*C ∩ int*E ⊆ cl*(C ∩ E), {(x, C ∩ E)} ∈ Tα0 . r 484L Proposition Suppose that R f : R → R is Pfeffer integrable, and that F : C → R is its SaksPf Henstock indefinite integral. Then f × χE is defined and equal to F (E) for every E ∈ C.

proof (a) To begin with (down to the end of (d) below), suppose that E ∈ C is such that int*E ⊆ E ⊆ cl*E. For G ∈ C set F1 (G) = F (G ∩ E). I seek to show that F1 satisfies the conditions of 484J. Of course F1 : C → R is additive. If ² > 0, there is an R ∈ R such that |F (G)| ≤ ² whenever G ∈ R, by 484H; now there is an R0 ∈ R such that G ∩ E ∈ R for every G ∈ R0 (484E(a-ii)), so that |F1 (G)| ≤ ² for every G ∈ C ∩ R0 . Thus F1 satisfies (iii) of 484J. (b) Take α ∈ ]0, α∗ [ and ² > 0. Take α0 such that 0 < α0 < α min( 21 , 2r−1 (

α (r−1)/r ) ). 2βr

Applying 484K

to E and its complement, and appealing to the definition of F , we see that there is a δ ∈ ∆ such that (α) {(x, C ∩ E)} ∈ Tα0 whenever (x, C) ∈ δ, x ∈ E and {(x, C)} ∈ Tα , (β) P {(x, C \ E)} ∈ Tα0 whenever (x, C) ∈ δ, x ∈ Rr \ E and {(x, C)} ∈ Tα , (γ) (x,C)∈tt |F (C) − f (x)µC| ≤ ² for every δ-fine t ∈ Tα0 .

484L


755

(For (α), we need to know that E ⊆ cl*E and for (β) we need int*E ⊆ E.) Next, choose for each n ∈ N closed sets Hn ⊆ E, Hn0 ⊆ Rr \ E such that µ(E \ Hn ) ≤ 2−n ² and µ((Rr \ E) \ Hn0 ) ≤ 2−n ². Define θ : Rr → ]0, ∞[ by setting 1 2

θ(x) = min(1, ρ(x, Hn0 )) if x ∈ E and n ≤ |f (x)| < n + 1, 1 2

= min(1, ρ(x, Hn )) if x ∈ Rr \ E and n ≤ |f (x)| < n + 1, writing ρ(x, A) = inf y∈A kx − yk if A ⊆ Rr is non-empty, ∞ if A = ∅. Then δθ ∈ ∆. Let R1 ∈ R be such that A ∩ E ∈ R whenever A ∈ R1 (484E(a-ii)). P (c) Write fE for f × χE. Then (x,C)∈tt |F1 (C) − fE (x)µ(C)| ≤ 11² whenever t ∈ Tα is δ ∩ δθ -fine. P P Set t 0 = {(x, C ∩ E) : (x, C) ∈ t , x ∈ E}. By clause (α) of the choice of δ, t 0 ∈ Tα0 , and of course it is δ-fine. So P (x,C)∈tt,x∈E |F (C ∩ E) − f (x)µ(C ∩ E)| ≤ ² by clause (γ) of the choice of δ. Next, X

|f (x)µ(C \ E)| =

∞ X

X

|f (x)|µ(C \ E)

n=0 (x,C)∈tt,x∈E n≤|f (x)|
(x,C)∈tt,x∈E

≤

∞ X

(n + 1)µ((Rr \ E) \ Hn0 )

n=0

(because diam C ≤ θ(x), so C ∩ Hn0 = ∅ whenever (x, C) ∈ t , x ∈ E and n ≤ |f (x)| < n + 1) ∞ X ≤ 2−n (n + 1)² = 4², n=0

and

P

Similarly,

(x,C)∈tt,x∈E

|F (C ∩ E) − f (x)µC| ≤ 5².

(x,C)∈tt,x∈E /

|F (C \ E) − f (x)µC)| ≤ 5².

P

But as

P (x,C)∈tt,x∈E /

|F (C) − f (x)µC| ≤ ²

(because surely t itself belongs to Tα0 ), we have P (x,C)∈tt,x∈E /

Putting these together, X |F1 (C) − fE (x)µC| = (x,C)∈tt

X

|F (C ∩ E)| ≤ 6².

|F (C ∩ E) − f (x)µC| +

(x,C)∈tt x∈E

X

|F (C ∩ E)|

(x,C)∈tt x∈E /

≤ 5² + 6² = 11². Q Q (d) As α and ² are arbitrary, condition (iii) of 484J is satisfied by F1 and fE , so

R

Pf fE

= F1 (Rr ) = F (E).

(e) This completes the proof when int*E ⊆ E ⊆ cl*E. For a general set E ∈ C, set E1 = (E∪int*E)∩cl*E. Then E4E1 is negligible, so int*E1 = int*E ⊆ E1 ⊆ cl*E = cl*E1 .

756

Also

Gauge integrals

R

Pf f

× χ(E \ E1 ) =

R

R

f × χ(E \ E1 )dµ,

R

Pf

Pf f

R

f × χE =

Pf

484L

R

× χ(E1 \ E) =

f × χ(E \ E1 )dµ are both zero, and

f × χE1 = F (E1 ) = F (E).

(To see that F (E1 ) = F (E), note that E \ E1 and E1 \ E, being negligible sets, have empty essential boundary and zero perimeter, so belong to every member of R, by 484E(b-ii), or otherwise.) 484M Lemma Let G, H ∈ C be disjoint and φ : Rr → Rr a continuous function. If either G ∪ H is bounded or φ has compact support,

R

∂*(G∪H)

φ. ψG∪H dν =

R

∂*G

φ .ψG dν +

R

∂*H

φ. ψG dν,

where ψG , ψH and ψG∪H are the canonical outward-normal functions (474G). proof (a) Suppose first that φ is a Lipschitz function with compact support. Then 475N tells us that Z Z Z Z φ .ψG∪H dν = div φ dµ = div φ dµ + div φ dµ ∂*(G∪H) G∪H G H Z Z = φ.ψG dν + φ. ψG dν. ∂*G

∂*H

(Recall from 474Q that we can identify outward-normal functions with Federer exterior normals, as in the statment of 475N.) ˜ n in∈N be the smoothing (b) Now suppose that φ is a continuous function with compact support. Let hh ˜ n are Lipschitz (473Ca), and hφ ∗ h ˜ n in∈N converges uniformly sequence of 473E. Then all the functions φ ∗ h to φ (473Df, 473Ed). So Z Z ˜ n ) . ψG∪H dν φ. ψG∪H dν = lim (φ ∗ h n→∞ ∂*(G∪H) ∂*(G∪H) Z Z ˜ n ) .ψG dν + lim ˜ n ) . ψH dν = lim (φ ∗ h (φ ∗ h n→∞ ∂*G n→∞ ∂*H Z Z = φ. ψG dν + φ. ψG dν. ∂*G

∂*H

(c) If, on the other hand, φ is continuous and G and H are bounded, then we can find a continuous ˜ function φ˜ with compact support agreeing with φ on G ∪ H (4A2G(e-i), or otherwise); applying (b) to φ, we get the required result for φ. 484N Pfeffer’s Divergence Theorem Let E ⊆ Rr be a set with locally finite perimeter, and φ : Rr → Rr a continuous function with compact support such that {x : x ∈ Rr , φ is not differentiable R at x} is thin. Let vx be the FedererR exterior normal to E at any point x where the normal exists. Then Pf div φ × χE is defined and equal to ∂*E φ(x) .vx ν(dx). R proof (a) Let n be such that φ(x) = 0 for kxk ≥ n. For C ∈ C, set F (C) = ∂*C φ. ψC ν(dx). By 484M, F is additive. (b) If 0 < α < α∗ and ² > 0 and x ∈ Rr is such that φ is differentiable at x, there is a γ > 0 such that |F (C) − div φ(x)µC| ≤ ²µC whenever C ∈ Cα , x ∈ C and diam C ≤ γ. P P Let T be the derivative of φ at x. Let γ > 0 be such that kφ(y) − φ(x) − T (y − x)k ≤ α2 ²ky − xk whenever ky − xk ≤ γ. Let φ˜ : Rr → Rr be ˜ a Lipschitz function with compact support such that φ(y) = φ(x) + (y − x) whenever ky − xk ≤ γ (473Cf). If C ∈ Cα has diameter at most γ and x ∈ C, then Z |F (C) − div φ(x)µC| = |

Z div φ˜ dµ|

φ. ψC ν(dx) − ∂*C

C

˜ (because T is the derivative of φ˜ everywhere on B(x, γ), so div φ(y) = div φ(x) for every y ∈ C)

484N


757

Z ˜ ψC ν(dx)| φ − φ.

=| ∂*C

˜ (applying the Divergence Theorem to φ) ˜ ≤ ν(∂*C) sup kφ(y) − φ(y)k ≤ α2 ² diam C per C y∈C

≤ α²(diam C)r ≤ ²µC because C ∈ Cα . Q Q

P (c) If ² > 0 and α ∈ ]0, α∗ [, there is a δ ∈ ∆ such that (x,C)∈tt |F (C) − div φ(x)µC| ≤ ² whenever t ∈ Tα is δ-fine. P P Let ζ > 0 be such¤ that¤ ζµB(0, n + 2) ≤ ². Set A = {x : x ∈ Rr , φ is not differentiable at x}, and for x ∈ Rr \ A, let θ(x) ∈ 0, 12 be such that |F (C) − div φ(x)µC| ≤ ζµC whenever C ∈ Cα , x ∈ C and diam C ≤ θ(x); for x ∈ A set θ(x) = 0. Now suppose that t ∈ Tα is δθ -fine. Then X

X

|F (C) − div φ(x)µC| =

(x,C)∈tt

|F (C) − div φ(x)µC|

(x,C)∈tt,x∈B(0,n+1)

(because diam C ≤ 1 whenever (x, C) ∈ t, so if kxk > n + 1 then F (C) = div φ(x) = 0) X ≤ ζµC ≤ ζµB(0, n + 2) ≤ ². Q Q (x,C)∈tt,x∈B(0,n+1)

(d) Because φ is a continuous function with compact support, it is uniformly continuous (apply 4A2Je to each of the coordinates of φ). For ζ > 0, let γ(ζ) > 0 be such that kφ(x)−φ(y)k ≤ ζ whenever kx−yk ≤ γ(ζ). If C ∈ C and per C ≤ 1 and µC ≤ ζγ(ζ), where ζ > 0, then |F (C)| ≤ rζ(2kφk∞ + 21 ), writing kφk∞ for supx∈R r kφ(x)k. P P For 1 ≤ i ≤ r, let φi : Rr → R be the ith component of φ, and vi the ith unit vector (0, . . . , 1, . . . , 0); write βi =

Pr

R

∂*C

φi (x)(vi |ψE (x))ν(dx),

so that F (C) = i=1 βi . I start by examining βr . By 475O, we have sequences hHn in∈N , hgn in∈N and hgn0 in∈N such that (i) for each n ∈ N, Hn is a Lebesgue measurable subset of Rr−1 , and gn , gn0 : Hn → [−∞, ∞] are Lebesgue measurable functions such that gn (u) < gn0 (u) for every u ∈ Fn ; (ii) ifPm, n R∈ N then gm (u) 6= gn0 (u) for every u ∈ Hm ∩ Hn ; ∞ (iii) n=0 Hn gn0 − gn dµr−1 = µC; (iv) βr =

P∞ R n=0

Hn

φr (u, gn0 (u)) − φr (u, gn (u))µr−1 (du),

where we interpret φr (u, ∞) and φr (u, −∞) as 0 if necessary; (v) for µr−1 -almost every u ∈ Rr−1 , {t : (u, t) ∈ ∂*C} = {gn (u) : n ∈ N, u ∈ Fn , gn (u) 6= −∞} ∪ {gn0 (u) : n ∈ N, u ∈ Fn , gn0 (u) 6= ∞}. From (iii) we see that gn0 and gn are both finite almost everywhere on Hn , for every n. Consequently, by (v),

P∞

2

n=0

µr−1 Hn =

R

#({t : (u, t) ∈ ∂*C})µr−1 (du) ≤ ν(∂*C) ≤ 1

by 475H. For each n, set Then γ(ζ)

Hn0 = {u : u ∈ Hn , gn0 (u) − gn (u) > γ(ζ)}. P ∞ 0 0 n=0 µHn ≤ µC so n=0 µHn ≤ ζ and

P∞

|

∞ Z X n=0

0 Hn

φr (u, gn0 (u)) − φr (u, gn (u))µr−1 (du)| ≤ 2kφk∞

∞ X n=0

µHn0 ≤ 2ζkφk∞ .

758

Gauge integrals

484N

On the other hand, for n ∈ N and u ∈ Hn \ Hn0 , |φr (u, gn0 (u)) − φr (u, gn (u))| ≤ ζ, so |

∞ Z X n=0

0 Hn \Hn

φr (u, gn0 (u)) − φr (u, gn (u))µr−1 (du)| ≤

∞ X

1 2

ζµ(Hn \ Hn0 ) ≤ ζ.

n=0

Putting these together, 1 2

βr ≤ 2ζkφk∞ + ζ. But of course the same arguments apply to all the βi , so Pr 1 |F (C)| ≤ i=1 |βi | ≤ rζ(2kφk∞ + ), 2

as claimed. Q Q (e) If ² > 0 there is an R ∈ R such that |F (C)| ≤ ² P for every C ∈ C ∩ R. P P Let h²i ii∈N be a sequence of ∞ strictly positive real numbers such that r(2kφk∞ + 21 ) i=0 ²i ≤ ². For each i ∈ N, set η(i) = ²i γ(²i ) > 0. (V ) Set V = B(0, n + 1), and take S any C ∈ C ∩ Rη . Then F (C \ V ) = 0, so F (C) = F (C ∩ V ), while C ∩ V ∈ Rη . Express C ∩ V as i≤n Ei , where hEi ii≤n is disjoint, per Ei ≤ 1 and µEi ≤ η(i) for each i. Then |F (Ei )| ≤ r²i (2kφk∞ + 21 ) for each i, by (d), so Pn |F (C ∩ V )| ≤ i=0 |F (Ei )| ≤ ², as required. Q Q (f ) By 484J, div φ is Pfeffer integrable. Moreover, by the uniqueness assertion in 484Hc, its Saks-Henstock R indefinite integral is just the function F here. By 484L, F (E) = Pf div φ × χE, as required. 484O Differentiating the indefinite integral: Theorem Let f : Rr → R be a Pfeffer integrable function, and F its Saks-Henstock indefinite integral. Then whenever 0 < α < α∗ , f (x) = lim sup{ ζ↓0

= lim inf{ ζ↓0

F (C) µC

F (C) µC

: C ∈ Cα , x ∈ C, 0 < diam C ≤ ζ} : C ∈ Cα , x ∈ C, 0 < diam C ≤ ζ}

for µ-almost every x ∈ Rr . proof (a) It will be useful to know the following: if C ∈ Cα , diam C > 0, x ∈ C and ² > 0, then for any sufficiently small ζ > 0, C ∪ B(x, ζ) ∈ Cα/2 and |F (C ∪ B(x, ζ)) − F (C)| ≤ ². P P Let R ∈ R be such that |F (R)| ≤ ² whenever R ∈ C ∩ R, let R0 ∈ R be such that (Rr \ C) ∩ R ∈ R whenever R0 ∈ R0 , and let η ∈ H be such that Rη ⊆ R0 . Then for all sufficiently small ζ > 0, we shall have per B(x, ζ) ≤ 1 and µB(x, ζ) ≤ η(0), so that B(x, ζ) ∈ Rη and |F (C ∪ B(x, ζ)) − F (C)| = |F (B(x, ζ) \ C| ≤ ². Next, for all sufficiently small ζ > 0, µ(C ∪ B(x, ζ)) ≥ µC ≥ α(diam C)r α 2

≥ (ζ + diam C)r ≥

α 2

diam(C ∪ B(x, ζ))r

and 1 α

per(C ∪ B(x, ζ)) ≤ per C + per B(x, ζ) ≤ (diam C)r−1 + per B(x, ζ) 2 α

≤ (diam C)r−1 ≤ so that C ∪ B(x, ζ) ∈ Cα/2 . Q Q

2 α

diam(C ∪ B(x, ζ))r−1 ,

484P


759

(b) For x ∈ R, set g(x) = limζ↓0 sup{

F (C) µC

: C ∈ Cα , x ∈ C, diam C ≤ δ}.

?? Suppose, if possible, that there are rational numbers q < q 0 and n ∈ N such that A = {x : kxk ≤ n, f (x) ≤ q < q 0 < g(x)} is not µ-negligible. Set (q−q 0 )α ∗ µ A 4·5r βr

²= Let θ ∈ Θ be such that

P

> 0.

|F (C) − f (x)µC| ≤ ²

(x,C)∈tt

for every δθ -fine t ∈ Tα/2 . Let I be the family of all balls B(x, ζ) where x ∈ A, 0 < ζ ≤ θ(x) and there is a F (C) µC

C ∈ Cα such that x ∈ C, diam C = ζ and

> q 0 . Then every member of A1 = A \ θ−1 [{0}] is the centre

of arbitrarily small members of I, so by 472B there is a countable disjoint family J0 ⊆ I such that S µ( J0 ) ≥ 5−r µ∗ A1 = 5−r µ∗ A. S 1 There is therefore a finite family J1 ⊆ J0 such that µ( J1 ) > µ∗ A; enumerate J1 as hB(xi , ζi )ii≤n r 2·5

where, for each i ≤ n, xi ∈ A, 0 < ζi ≤ θi (x) and there is a Ci ∈ Cα such that xi ∈ Ci , diam Ci = ζi and F (Ci ) > q 0 µCi . By (a), we can enlarge Ci by adding a sufficiently small ball around xi to form a Ci0 ∈ Cα/2 such that xi ∈ int Ci0 , Ci0 ⊆ B(xi , ζi ) and F (Ci0 ) ≥ q 0 µCi0 . 0 0 B(xP Consider t = {(xi , Ci0 ) : i ≤ n}. Then, because i , ζi ) are disjoint, and xi ∈ int Ci ⊆ cl*Ci Pn the balls n 0 0 0 0 0 for every i, t is a δθ -fine member of Tα/2 . So i=0 F (Ci ) ≤ ² + i=0 f (xi )µCi . But as F (Ci ) ≥ q µCi and Pn f (xi ) ≤ q for every i, this means that (q 0 − q) i=0 µCi0 ≤ ². But now remember that diam Ci0 ≥ diam Ci = ζi and that Ci0 ∈ Cα/2 for each i. This means that α 2

µCi0 ≥ ζir ≥

α µB(xi , ζi ) 2βr

for each i, and ² ≥ (q − q 0 )

n X

µCi0

i=0

≥ >

n (q−q )α X 0

2βr

µB(xi , ζi )

i=0

(q−q 0 )α ∗ µ A 4·5r βr

=²

which is absurd. X X (c) Since q, q 0 and n are arbitrary, this means that g ≤a.e. f . Similarly (or applying (b) to −f and −F ) f (x) ≤ limζ↓0 inf{

F (C) µC

: C ∈ Cα , x ∈ C, 0 < diam C ≤ ζ}

for almost all x, as required. 484P Lemma Let φ : Rr → Rr be an injective Lipschitz function, and H the set of points at which it is differentiable; for x ∈ H, write T (x) for the derivative of φ at x and J(x) for | det T (x)|. Then, for µ-almost every x ∈ Rr , J(x) = lim sup{ ζ↓0

= lim inf{ ζ↓0

for every α > 0.

µφ[C] µC

µφ[C] µC


760

Gauge integrals

484P

proof By Rademacher’s theorem (262Q), H is conegligible. Let H 0 be the Lebesgue set ofRJ, so that H 0 is also conegligible (261E). Take any x ∈ H 0 and ² > 0. Then there is a ζ0 > 0 such that B(x,ζ) |J(y) − J(x)|µ(dy) ≤ ²µB(x, ζ) for every ζ ∈ [0, ζ0 ]. Now suppose that C ∈ Cα , x ∈ C and 0 < diam C ≤ ζ0 . Then µ(C \ H) = 0 so µφ[C \ H] = 0 (262D), and µφ[C] = µφ[C ∩ H] =

R

C∩H

J dµ

(263D(iv)). So Z Z |µφ[C] − J(x)µC| = | J dµ − J(x)µ(C ∩ H)| ≤ |J(y) − J(x)|dµ C\H C\H Z |J(y) − J(x)|dµ ≤ ²µB(x, diam C) ≤ B(x,diam C)

βr ² µC. α

= βr ²(diam C)r ≤ Thus |

µφ[C] µC

− J(x)| ≤

βr ² α

whenever C ∈ Cα , x ∈ C and 0 < diam C ≤ ζ0 ; as ² is arbitrary,

J(x) = lim sup{ ζ↓0

= lim inf{ ζ↓0

µφ[C] µC

µφ[C] µC

: C ∈ Cα , x ∈ C, 0 < diam C ≤ ζ} : C ∈ Cα , x ∈ C, 0 < diam C ≤ ζ}.

And this is true for µ-almost every x. 484Q Definition If (X, ρ) and (Y, σ) are metric spaces, a function φ : X → Y is a lipeomorphism if it is bijective and both φ and φ−1 are Lipschitz. Of course a lipeomorphism is a homeomorphism. 484R Lemma Let φ : Rr → Rr be a lipeomorphism. (a) For any set A ⊆ Rr , cl*(φ[A]) = φ[cl*A],

int*(φ[A]) = φ[int*A],

∂*(φ[A]) = φ[∂*A].

(b) φ[C] ∈ C for every C ∈ C, and φ[V ] ∈ V for every V ∈ V. (c) For any α > 0 there is an α0 ∈ ]0, α] such that φ[C] ∈ Cα0 for every C ∈ Cα and {(φ(x), φ[C]) : (x, C) ∈ t } belongs to Tα0 for every t ∈ Tα . (d) For any R ∈ R there is an R0 ∈ R such that φ[R] ∈ R for every R ∈ R0 . (e) θφ : Rr → [0, ∞[ belongs to Θ for every θ ∈ Θ. proof Let γ be so large that it is a Lipschitz constant for both φ and φ−1 . Observe that in this case ζ γ

φ−1 [B(φ(x), )] ⊆ B(x, ζ),

ζ γ

φ[B(x, ζ)] ⊇ B(φ(x), )

for every x ∈ Rr , ζ ≥ 0. (a) If A ⊆ Rr and x ∈ cl*A, set ²=

1 2

lim supζ↓0

µ∗ (B(x,ζ)∩A µB(x,ζ)

> 0. ζ γ

ζ γ

Take any ζ0 > 0. Then there is a ζ such that 0 < zeta ≤ ζ0 and µ∗ (B(x, ) ∩ A)] ≥ ²µB(x, ), so that ζ γ

ζ γ

µ∗ (B(φ(x), ζ) ∩ φ[A]) ≥ µ∗ φ[B(x, ) ∩ A] ≥ ²µB(x, ) = As ζ0 is arbitrary, lim supζ↓0 and φ(x) ∈ cl*(φ[A]).

µ∗ (B(φ(x),ζ)∩φ[A] µB(φ(x),ζ)

≥

² γr

> 0,

² µB(φ(x), ζ). γr

484S


761

This shows that φ[cl*A] ⊆ cl*(φ[A]). The same argument applies to φ−1 and φ[A], so that φ[cl*A] must be equal to cl*(φ[A]). Taking complements, φ[int*A] = int*(φ[A]), so that φ[∂*A] = ∂*(φ[A]). (b) Take C ∈ C. Then, for any n ∈ N, φ−1 [B(0, n)] is bounded, so is included in B(0, m) for some m. Now ν(∂*φ[C] ∩ B(0, n)) = ν(φ[∂*C] ∩ B(0, n)) ≤ ν(φ[∂*C ∩ B(0, m)]) ≤ γ r−1 ν(∂*C ∩ B(0, m)) (264G/471J), so is finite. This shows that φ[C] has locally finite perimeter and belongs to C. Since φ[V ] is bounded whenever V is bounded, φ[V ] ∈ V whenever V ∈ V. (c) Set α0 = γ −2r α. Note that as γ 2 is a Lipschitz constant for the identity map, γ ≥ 1, and γ 2−2r α ≤ α ≤ α. If C ∈ Cα , then 0

µφ[C] ≥

1 µC γr

≥

α (diam C)r γr

1 1 α γ

≥ ( diam φ[C])r ≥ α0 (diam φ[C])r , per φ[C] = ν(∂*(φ[C])) = ν(φ[∂*C]) ≤ γ r−1 ν(∂*C) ≤

γ r−1 (diam C)r−1 α

≤

γ r−1 (γ diam φ[C])r−1 α

≤

1 (diam φ[C])r−1 . α0

So C ∈ Cα0 . If now t ∈ Tα , then, for any (x, C) ∈ t, φ[C] ∈ Cα0 and φ(x) ∈ cl*φ[C]; also, because φ is injective, hφ[C]i(x,C)∈tt is disjoint, so {(φ(x), φ[C]) : (x, C) ∈ t} ∈ Tα0 . (V )

(d) Express R as Rη where V ∈ V and η ∈ H,Sso that R ∈ R whenever R ∩ V ∈ R. By 484Ec and 481He, there is a sequence hQi ii∈N in R such that i≤n Ai ∈ R whenever n ∈ N, hAi ii≤n is disjoint and Ai ∈ Qi for every i. By 484E(b-ii), there is an η 0 ∈ H such that R ∈ Qi whenever i ∈ N and R is such (φ−1 [V ])

that µR ≤ γ r η 0 (i) and per R ≤ γ r−1 . Try R0 = Rη0 ∈ R. If R ∈ R0 , we can express R ∩ φ−1 [V ] as S S 0 i≤n φ[Ei ] i≤n Ei where per Ei ≤ 1 and µEi ≤ ηi for each i ≤ n, and hEi ii≤n is disjoint. So φ[R] ∩ V = and hφ[Ei ]ii≤n is disjoint. Now, for each i, µφ[Ei ] ≤ γ r µEi ≤ γ r η 0 (i),

per φ[Ei ] ≤ γ r−1 per Ei ≤ γ r−1 ,

so φ[Ei ] ∈ Qi . By the choice of hQi ii∈N , φ[R] ∩ V ∈ R and φ[R] ∈ R. So R0 has the property we need. (e) θ−1 [{0}],Sthen (θφ)−1 [{0}] = φ−1 [A] is also thin. P P If SWe have only to∗observe that if A is the thin set −1 A = n∈N An where ν An is finite for every n, then φ [A] = n∈N φ−1 [An ], while ν ∗ φ−1 [An ] ≤ γ r−1 ν ∗ An (264G/471J) is finite for every n ∈ N. Q Q 484S Theorem Let φ : Rr → Rr be a lipeomorphism. Let H be the set of points at which φ is differentiable. For x ∈ H, write T (x) for the derivative of φ at x; set J(x) = | det T (x)| for x ∈ H, 0 for x ∈ Rr \ H. Then, for any function f : Rr → Rr ,

R

Pf

f=

R

Pf

J × fφ

if either is defined in R. proof (a) Let H 0 ⊆ H be a conegligible set such that J(x) = lim sup{ ζ↓0

= lim inf{ ζ↓0

µφ[C] µC

µφ[C] µC


762

Gauge integrals

484S

for every α > 0 and every x ∈ H 0 (484P). To begin with (down to the end of (c)), suppose that f is Pfeffer integrable and that f φ(x) = 0 for every x ∈ Rr \H 0 . Let F be the Saks-Henstock indefinite integral of f , and define G : C → R by setting G(C) = F (φ[C]) for every C ∈ C (using 484Rb to see that this is well-defined). (b) G and J × f φ satisfy the conditions of 484J. P P(i) Of course G is additive, because F is. (ii) Suppose that 0 < α < α∗ and ² > 0. Let P α0 ∈ ]0, α∗ [ be such that {(φ(x), φ[C]) : (x, C) ∈ t } ∈ Tα0 whenever t ∈ Tα (484Rc). Let θ1 ∈ Θ be such that (x,C)∈tt |F (C) − f (x)µC| ≤ 21 ² for every δθ1 -fine t ∈ Tα0 . Let θ2 : Rr → ]0, 1] be such that whenever x ∈ H 0 and n ≤ kxk + |f φ(x)| < n + 1 then |µφ[C] − J(x)µC| ≤

µC 2n+3 βr (n+2)r (n+1) 1 γ

whenever C ∈ Cα0 , x ∈ C and diam C ≤ 2θ2 (x). Set θ(x) = min( θ1 φ(x), θ2 (x)) for x ∈ Rr , so that θ ∈ Θ (484Re). If t ∈ Tα is δθ -fine, set t 0 = {(φ(x), φ[C]) : (x, C) ∈ t }. Then t ∈ Tα0 , by the choice of α0 . If (x, C) ∈ t , then θ(x) > 0 so θ1 φ(x) > 0; also C ⊆ B(x, θ(x)) so φ[C] ⊆ B(φ(x), γθ(x)) ⊆ B(φ(x), θ1 φ(x)). This shows that t 0 is δθ1 -fine. We therefore have X

X

|G(C) − J(x)f (φ(x))µC| ≤

(x,C)∈tt

|F (φ[C]) − f (φ(x))µφ[C]|

(x,C)∈tt

+

|f (φ(x))||µφ[C] − J(x)µC|

(x,C)∈tt

X

≤

X

|F (C) − f (x)µC|

(x,C)∈tt0

+

X

|f (φ(x))||µφ[C] − J(x)µC|

t (x,C)∈t x∈H 0

(because if x ∈ / H 0 then f (φ(x)) = 0) 1 2

≤ ²+

1 2

≤ ²+

∞ X

X

n=0

(x,C)∈tt,x∈H 0 n≤kxk+|f (φ(x))|
(n+1)µC 2n+3 βr (n+2)r (n+1)

∞ X (n+1)µB(0,n+2) n=0

2n+3 βr (n+2)r (n+1)

(remembering that θ2 (x) ≤ 1, so C ⊆ B(0, n + 2) whenever (x, C) ∈ t and kxk < n + 1) = ². As t is arbitrary, this shows that G and J × f φ satisfy (ii) of 484J. (iii) Given ² > 0, there is an R ∈ R such that |F (C)| ≤ ² for every C ∈ C ∩ R. Now by 484Rd there is an R0 ∈ R such that φ[C] ∈ R for every R ∈ R0 , so that |G(C)| ≤ ² for every C ∈ C ∩ R0 . Thus G satisfies (iii) of 484J. Q Q (c) This shows that J ×f φ is Pfeffer integrable, with Saks-Henstock indefinite integral G; so, in particular,

R

Pf

f × φ = G(Rr ) = F (Rr ) =

R

Pf

f.

(d) Now suppose that f is an arbitrary Pfeffer integrable function. In this case set f1 = f × χφ[H 0 ]. Because Rr \ H 0 is µ-negligible, so is φ[Rr \ H 0 ], and f1 = f µ-a.e. Also, of course, f φ = f1 φ µ-a.e. Because the Pfeffer integral extends the Lebesgue integral (484He),

484 Notes


R Pf

J × fφ =

R Pf

J × f1 φ =

R Pf

763

f1 =

R Pf

f.

(e) All this has been on the assumption that f is Pfeffer integrable. If g = J × f φ is Pfeffer integrable, ˜ consider J˜ × gφ−1 , where J(x) = | det T˜(x)| whenever the derivative T˜(x) of φ−1 at x is defined, and otherwise is zero. Now −1 ˜ ˜ · J(φ−1 (x))f (x) J(x)g(φ (x)) = J(x) for every x. But, for µ-almost every x, −1 ˜ J(x)J(φ (x)) = | det T˜(x)|| det T (φ−1 (x))| = | det T˜(x)T (φ−1 (x))| = 1 because T˜(x)T (φ−1 (x)) is (whenever it is defined) the derivative at φ−1 (x) of the identity function φ−1 φ, by 473Bc. (I see that we need to know that {x : φ is differentiable at φ−1 (x)} = φ[H] is conegligible.) So J˜ × gφ−1 = f µ-a.e., and f is Pfeffer integrable. This completes the proof. (B(0,n))

484X Basic exercises (a) Show that for every R ∈ R there are η ∈ H and n ∈ N such that Rη

⊆

R. > (b) (Pfeffer 91a) For α > 0 let Cα0 be the family of bounded Lebesgue measurable sets C such that µC ≥ α diam C per C, and Tα0 the straightforward set of tagged partitions generated by {(x, C) : C ∈ Cα0 , 0 0 2 x ∈ cl*C}. (i) Show that if α0 = min(αr , 2−r βr /α) then √ Cα ⊆ Cα ⊆ Cα . (Hint: 474La.) (ii) Show that Tα is compatible with ∆ and R whenever 0 < α < 1/2r r. > (c) For α > 0, let Cα00 be the family of convex sets C such that µC ≥ α(diam C)r . Show that Cα00 ⊆ Cα . (Hint: 475T.) Show that if 0 < α <

2r−1 rβr

and Tα00 = Tα ∩ [Rr × Cα0 ]<ω then Tα00 is compatible with ∆ and R.

(Hint: use the argument of 484F, but in part (b) take C ∈ D.) (d) Describe a suitable filter F to express the Pfeffer integral directly in the form considered in 481C. r R (e) Let f : R → R be a Pfeffer integrable function. Show that there is some n ∈ N such that |f |dµ is finite. Rr \B(0,n)

(f ) Let E ⊆ Rr be a bounded set with finite perimeter, and φ : Rr → Rr a differentiableR function. Let vx be the Federer exterior normal to E at any point x where the normal exists. Show that Pf div φ × χE is R defined and equal to ∂*E φ(x) . vx ν(dx). 484Y Further exercises (a) Let E ⊆ Rr be any Lebesgue measurable set, and ² > 0. Show that there is a Lebesgue measurable set G ⊆ E such that per G ≤ per E, µ(E \ G) ≤ ² and cl*G = G. (b) Give an example of a compact set K ⊆ R2 with zero one-dimensional Hausdorff measure such that whenever θ : K → ]0, ∞[ is a strictly positive function, and γ ∈ S R, there is a disjoint family hB(xi , ζi )ii≤n of balls such that xi ∈ K and ζi ≤ θ(xi ) for every i, while per( i≤n B(xi , ζi )) ≥ γ. 484 Notes and comments Observe that 484B can be thought of as a refinement of 475I. As usual, the elaborate formula in the statement of 484C is there only to emphasize that we have a constant depending only on r. The whole point of ‘gauge integrals’ is that we have an enormous amount of freedom within the framework of §§481-482. There is a corresponding difficulty in making exact choices. The essential ideology of the ‘Pfeffer integral’ is that we take an intersection of a family of gauge integrals, each determined by a family Cα of sets which are ‘Saks regular’ in the sense that their measures, perimeters and diameters are linked. Shrinking Cα and Tα , while leaving ∆ and R unchanged, of course leads to a more ‘powerful’ integral (supposing, at least, that we do not go so far that Tα is no longer compatible with ∆ and R), so that Pfeffer’s Divergence Theorem will remain true. One possibility is to turn to convex sets (484Xc). Similar technical questions arise when choosing the family R of residual classes. Here I use the construc(V ) tion Rη = {R : R ∩ V ∈ Rη } (484D) as a quick method of eliminating any difficulties at infinity (484Xe).

764

Gauge integrals

484 Notes

We do not of course need to look at arbitrary sets V ∈ V here (484Xa). One of the important simplifications in the one-dimensional theory is that we can work with a very much simpler family of residual classes; this is because ‘thin’ sets become countable (see 484Xf), and can be controlled by neighbourhood gauges. I give 484Yb as a pointer to the problems which arise if we do not exclude thin sets by the device of allowing the gauge functions θ to be zero on such sets, thereby removing some potential tags. The many equivalent formulations of the Denjoy-Perron-Henstock integral challenge us to seek corresponding descriptions of integrals on Rr . 484O shows that we can express a Pfeffer integrable function as the derivative almost everywhere (in a strong sense) of its Saks-Henstock indefinite integral. I include 484S as a worked example on the techniques already developed. Note that it depends much more on the fact that the Pfeffer integral can be characterized in the language of 475J, than on the exact choices made in forming R and the Cα . (Of course if we restricted the Cα to the point that they contained only convex sets, we could not expect invariance under lipeomorphisms.) For a discussion of integrals defined by Saks-Henstock lemmas, see Pfeffer 01.

491Ac

Equidistributed sequences

765

Chapter 49 Further topics I conclude the volume with notes on five almost unconnected special topics. In §491 I look at equidistributed sequences and the ideal Z of sets with asymptotic density zero. I give the principal theorems on the existence of equidistributed sequences in abstract topological measure spaces, and examine the way in which an equidistributed sequence induces an embedding of a measure algebra in the quotient algebra PN/Z. In §492 I present some forms of ‘concentration of measure’ which echo ideas from §476 in combinatorial, rather than geometric, contexts, with theorems of Talagrand and Maurey on product measures and the Haar measure of a permutation group. In §493 I show how the ideas of §§449, 476 and 492 can be put together in the theory of ‘extremely amenable’ topological groups. In §494, I give a pair of simple, but surprising, results on subsets of sets of positive measure in product spaces. Finally, in §495, I introduce Poisson point processes, with notes on disintegrations and some special cases in which they can be represented by Radon measures. 491 Equidistributed sequences In many of the most important topological measure spaces, starting with Lebesgue measure (491Xf), there are sequences which are equidistributed in the sense that, in the limit, they spend the right proportion of their time in each part of the space (491Ye). I give the basic results on existence of equidistributed sequences in 491E-491G and 491Q. Investigating such sequences, we are led to some interesting properties of the asymptotic density ideal Z and the quotient algebra Z = PN/Z (491A, 491I-491K). For ‘effectively regular’ measures (491L-491M), equidistributed sequences lead to embeddings of measure algebras in Z (491N). 491A The asymptotic density ideal (a) If I ⊆ N, its upper asymptotic density is d∗ (I) = lim supn→∞ n1 (I ∩ n), and its asymptotic density is d(I) = limn→∞ n1 #(I ∩ n) if this is defined. It is easy to check that d∗ is a submeasure on PN (definition: 392A), so that Z = {I : I ⊆ N, d∗ (I) = 0} = {I : I ⊆ N, d(I) = 0} is an ideal, the asymptotic density ideal. (b) Note that Z = {I : I ⊆ N, limn→∞ 2−n #(I ∩ 2n+1 \ 2n ) = 0}. P P If I ⊆ N and d∗ (I) = 0, then 2−n #(I ∩ 2n+1 \ 2n ) ≤ 2 · 2−n−1 #(I ∩ 2n+1 ) → 0 as n → ∞. In the other direction, if limn→∞ 2−n #(I ∩ 2n+1 \ 2n ) = 0, then for any ² > 0 there is an m ∈ N such that #(I ∩ 2k+1 \ 2k ) ≤ 2k ² for every k ≥ m. In this case, for n ≥ 2m , take kn such that 2kn ≤ n < 2kn +1 , and see that Pkn 1 2k ²) ≤ 2−kn #(I ∩ 2m ) + 2² → 2² #(I ∩ n) ≤ 2−kn (#(I ∩ 2m ) + k=m n

as n → ∞, and d∗ (I) ≤ 2²; as ² is arbitrary, I ∈ Z. Q Q (c) Writing D for the domain of d, 1 n

1

D = {I : I ⊆ N, lim sup #(I ∩ n) = lim inf #(I ∩ n)} n→∞ ∗

n→∞ n

= {I : I ⊆ N, d (I) = 1 − d∗ (N \ I)}, N ∈ D,

if I, J ∈ D and I ⊆ J then J \ I ∈ D,

if I, J ∈ D and I ∩ J = ∅ then I ∪ J ∈ D and d(I ∪ J) = d(I) + d(J). It follows that if I ⊆ D and I ∩ J ∈ I for all I, J ∈ I, then the subalgebra of PN generated by I is included in D (313Ga). But note that D itself is not a subalgebra of PN (491Xa).

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491Ad

(d) The following elementary fact will be useful. If hln in∈N is a strictly increasing sequence in N such that limn→∞ ln+1 /ln = 1, and I ⊆ R, then d∗ (I) ≤ lim supn→∞ P P Set γ = lim supn→∞

1 #(I ln+1 −ln

1 #(I ln+1 −ln

∩ ln+1 \ ln ).

∩ ln+1 \ ln ), and take ² > 0. Let n0 be such that #(I ∩ ln+1 \ ln ) ≤

(γ + ²)(ln+1 − ln ) and ln+1 − ln ≤ ²ln for every n ≥ n0 , and write M for #(I ∩ ln0 ). If m > ln0 , take k such that lk ≤ m < lk+1 ; then #(I ∩ m) ≤ M +

k−1 X

#(I ∩ ln+1 \ ln ) + (m − lk )

n=n0

≤M+

k−1 X

(γ + ²)(ln+1 − ln ) + lk+1 − lk ≤ M + m(γ + ²) + ²m,

n=n0

so 1 #(I m

∩ m) ≤

M m

+ γ + 2².

Accordingly d∗ (I) ≤ γ + 2²; as ² is arbitrary, d∗ (I) ≤ γ. Q Q 491B Equidistributed sequences Let X be a topological space and µ a probability measure on X. I say that a sequence hxn in∈N in X is (asymptotically) equidistributed or uniformly distributed if µF ≥ d∗ ({i : xi ∈ F }) for every measurable closed set F ⊆ X; equivalently, if µG ≤ lim inf n→∞ n1 #({i : i < n, xi ∈ G}) for every measurable open set G ⊆ X. Remark Traditionally, equidistributed sequences have been defined in terms of their action on continuous functions, as in 491Cf. I have adopted the definition here in order to deal both with Radon measures on spaces which are not completely regular (so that we cannot identify the measure with an integral) and with Baire measures (so that there may be closed sets which are not measurable). Note that we cannot demand that the sets {i : xi ∈ F } should have well-defined densities (491Xh). 491C I work through a list of basic facts. The technical details (if we do not specialize immediately to metrizable or compact spaces) are not quite transparent, so I set them out carefully. Proposition Let X be a topological space, µ a probability measure on X and hxn in∈N a sequence in X. Write Cb (X) for the space of bounded on X. R continuous real-valued Pfunctions n 1 (a) hxn in∈N is equidistributed iff f dµ ≤ lim inf n→∞ n+1 f (x ) for every measurable bounded lower i i=0 semi-continuous function f : X → R. R Pn 1 (b) If µ measures every zero set and hxn in∈N is equidistributed, then limn→∞ n+1 f dµ i=0 f (xi ) = for every f ∈ Cb (X). R Pn 1 f dµ for every f ∈ Cb (X), (c) Suppose that µ measures every zero set in X. If limn→∞ n+1 i=0 f (xi ) = then d∗ ({n : xn ∈ F }) ≤ µF for every zero set F ⊆ X. (d) Suppose that X is normal Pn and thatRµ measures every zero set and is inner regular with respect to the 1 closed sets. If limn→∞ n+1 f dµ for every f ∈ Cb (X), then hxn in∈N is equidistributed. i=0 f (xi ) = (e) Suppose that µ is τ -additive and there is a base G for the topology of X, consisting of measurable 1 #({i : i ≤ n, xi ∈ G}) for every G ∈ G. sets and closed under finite unions, such that µG ≤ lim inf n→∞ n+1 Then hxn in∈N is equidistributed. (f) Suppose that X is completely regular Pnand that µRmeasures every zero set and is τ -additive. Then 1 hxn in∈N is equidistributed iff limn→∞ n+1 f dµ for every f ∈ Cb (X). i=0 f (xi ) = (g) Suppose that X is metrizable and that µ is a topological measure. Then hxn in∈N is equidistributed R Pn 1 iff limn→∞ n+1 f (x ) = f dµ for every f ∈ C (X). i b i=0 (h) Suppose that X is compact, Hausdorff and zero-dimensional, and that µ is a Radon measure on X. Then hxn in∈N is equidistributed iff d({n : xn ∈ G}) = µG for every open-and-closed subset G of X.

491C


767

proof (a)(i) Suppose that hxn in∈N is equidistributed. Let f : X → [0, 1] be a measurable lower semicontinuous function and k ≥ 1. For each j < k set Gj = {x : f (x) > kj }. Then lim inf n→∞

1 n+1

Pn j=1

χGj (xi ) = lim inf n→∞

1 #({i n+1

: i ≤ n, xi ∈ Gj }) ≥ µGj

because hxn in∈N is equidistributed and Gj is a measurable open set. Also f − k1 χX ≤ Z f dµ −

1 k

≤

≤

1 k

k X

µGj ≤

j=1

1 1 lim inf k n→∞ n+1

1 k

k X j=1

1 lim inf n→∞ n+1

n X k X

n X

1 k

Pk j=1

χGj ≤ f , so

χGj (xi )

i=0

1 n→∞ n+1

χGj (xi ) ≤ lim inf

i=0 j=1

n X

f (xi ).

i=0

R Pn 1 As k is arbitrary, f dµ ≤ lim inf n→∞ n+1 i=0 f (xi ). The argument above depended on f taking R Pn values in [0, 1]. But multiplying by an appropriate positive 1 scalar we see that f dµ ≤ lim inf n→∞ n+1 i=0 f (xi ) for every bounded measurable lower semi-continuous f : X → [0, ∞[, and adding a multiple of χX we see that the same formula is valid for all bounded measurable lower semi-continuous f : X → R. R Pn 1 (ii) Conversely, if f dµ ≤ lim inf n→∞ n+1 i=0 f (xi ) for every bounded measurable lower semicontinuous f : X → R, and G ⊆ X is a measurable open set, then χG is lower semi-continuous, so 1 µG ≤ lim inf n→∞ n+1 #({i : i ≤ n, xi ∈ G}). As G is arbitrary, hxn in∈N is equidistributed. (b) Apply (a) to the lower semi-continuous functions f and −f . (c) Recall that if µ measures every zero set, then every bounded continuous real-valued function is integrable, by 4A3L. Let F ⊆ X be a zero set, and ² > 0. Then there is a continuous f : X → R such 1 δ

that F = f −1 [{0}]. Let δ > 0 be such that µ{x : 0 < |f (x)| ≤ δ} ≤ ², and set g = (χX − |f |)+ . Then g : X → [0, 1] is continuous and χF ≤ g, so n

1 X g(xi ) n→∞ n+1 i=0

d∗ ({n : xn ∈ F }) ≤ lim sup Z =

g dµ ≤ µ{x : |f (x)| ≤ δ} ≤ µF + ².

As ² and F are arbitrary, we have the result. (d) Let F ⊆ X be a measurable closed set and ² > 0. Because µ is inner regular with respect to the closed sets, there is a measurable closed set F 0 ⊆ X \ F such that µF 0 ≥ µ(X \ F ) − ². Because X is normal, there is a continuous function f : X → [0, 1] such that χF ≤ f ≤ χ(X \ F 0 ). Now R 1 Pn 0 d∗ ({n : xn ∈ F }) ≤ lim supn→∞ i=0 f (xi ) = f dµ ≤ µ(X \ F ) ≤ µF + ². n+1

As F and ² are arbitrary, hxn in∈N is equidistributed. (e) Let F ⊆ X be a measurable closed S set, and ² > 0. Let G1 be the family of members of G disjoint from F . Then G1 is upwards-directed and G1 = X \ F ; because µ is τ -additive, there is a G ∈ G1 such that µG > µ(X \ F ) − ². Now 1 #({i n+1 n→∞

d∗ ({n : xn ∈ F }) ≤ lim sup

= 1 − lim inf

1

n→∞ n+1

: i ≤ n, xi ∈ X \ G})

#({i : i ≤ n, xi ∈ G}) ≤ 1 − µG ≤ µF + ².

As F and ² are arbitrary, hxn in∈N is equidistributed.

R Pn 1 (f ) (i) If hxn in∈N is equidistributed then (b) tells us that f dµ = limn→∞ n+1 i=0 f (xi ) for every R Pn 1 f ∈ Cb (X). (ii) Suppose that f dµ = limn→∞ n+1 f (x ) for every f ∈ C (X). If G ⊆ X is a cozero i b i=0

768

Further topics

491C

1 set, we can apply (c) to its complement to see that µG ≤ lim inf n→∞ n+1 #({i : i ≤ n, xi ∈ G}). So applying (e) with G the family of cozero sets we see that hxn in∈N is equidistributed.

(g) Because every closed set is a zero set, this follows at once from (b) and (c). (h) If hxn in∈N is equidistributed and G ⊆ X is open-and-closed, then d∗ ({n : xn ∈ G}) ≤ µG because G is closed and d∗ ({n : xn ∈ / G}) ≤ 1 − µG because G is open; so d({n : xn ∈ G}) = µG. If the condition is satisfied, then (e) tells us that hxn in∈N is equidistributed. 491D The next lemma provides a useful general criterion for the existence of equidistributed sequences. Lemma Let X be a topological space and µ a probability measure on X. Suppose that there is a sequence hνn in∈N of point-supported probability measures on X such that µF ≥ lim supn→∞ νn F for every measurable closed set F ⊆ X. Then µ has an equidistributed sequence. P proof For each n ∈ N, let P qn : X → [0, 1] be such that νn E = x∈E qn (x) for every E ⊆ X. Let qn0 : X P → [0, 1] be such that x∈X qn0 (x) = 1, Kn = {x : qn0 (x) > 0} is finite, qn0 (x) is rational for every x, and x∈X |qn (x) − qn0 (x)| ≤ 2−n ; then µF ≥ lim supn→∞ νn0 F for every measurable closed F , where νn0 is defined from qn0 . For each n, let sn ≥ 1 be such that rn (x) = qn0 (x)sn is an integer for every x ∈ Kn . Let hxni ii
1 #({i sn

: i < sn , xni ∈ E}) for every E ⊆ X. Pk Let hmk ik∈N be such that sk+1 ≤ 2−k j=0 mj sj for each k. Set l0 = 0. Given ln , take the largest k Pk−1 such that j=0 mj sj ≤ ln ; set ln+1 = ln + sk and xi = xk,i−ln for ln ≤ i < ln+1 ; continue. By the choice of the mk , ln+1 /ln → 1 as n → ∞. For any E ⊆ X, #({i : ln ≤ i < ln+1 , xi ∈ E}) = #({j : j < sk , xkj ∈ E}) Pk Pk−1 whenever j=0 mj sj ≤ ln < j=0 mj sj . So for any measurable closed set F ⊆ X, 1 #({j s k k→∞

d∗ ({i : xi ∈ F }) ≤ lim sup

: j < sk , xkj ∈ F })

(491Ad) = lim sup νk0 F ≤ µF. k→∞

As F is arbitrary, hxn in∈N is an equidistributed sequence for µ. 491E Proposition (a)(i) Suppose that X and Y are topological spaces, µ a probability measure on X and f : X → Y a continuous function. If hxn in∈N is a sequence in X which is equidistributed with respect to µ, then hf (xn )in∈N is equidistributed with respect to the image measure µf −1 . (ii) Suppose that (X, µ) and (Y, ν) are topological probability spaces and f : X → Y is a continuous inverse-measure-preserving function. If hxn in∈N is a sequence in X which is equidistributed with respect to µ, then hf (xn )in∈N is equidistributed with respect to ν. (b) Let X be a topological space and µ a probability measure on X, and suppose that X has a countable network consisting of sets measured by µ. Let λ be the ordinary product measure on X N . Then λ-almost every sequence hxn in∈N in X N is µ-equidistributed. proof (a)(i) Let F ⊆ Y be a closed set which is measured by µf −1 . Then f −1 [F ] is a closed set in X measured by µ. So d∗ ({n : f (xn ) ∈ F }) = d∗ ({n : xn ∈ f −1 [F ]}) ≤ µf −1 [F ]. (ii) Replace ‘µf −1 ’ above by ‘ν’. (b) Let A be a countable network for the given topology S of X consisting of measurable sets, and let E be the countable subalgebra of PX generated by A. Let T ⊇ S be the second-countable topology generated by E; then µ is a τ -additive topological measure with respect to T, and E is a base for T closed under finite unions. If E ∈ E, then d({n : xn ∈ E}) = µE for λ-almost every sequence hxn in∈N in X, by the strong law of large numbers (273J). So

491F


769

d({n : xn ∈ E}) = µE for every E ∈ E for λ-almost every hxn in∈N . Now 491Ce tells us that any such sequence is equidistributed with respect to T and therefore with respect to S. 491F Theorem Let h(Xi , Ti , Σi , µi )ii∈I be a family of τ -additive topological probability spaces,Qeach of which has an equidistributed sequence. If #(I) ≤ c, then the τ -additive product measure λ on X = i∈I Xi (definition: 417G) has an equidistributed sequence. proof (a) For the time being (down to the end of (f)), let us suppose that every µm is inner regular with respect to the Borel sets. (This will make it possible to use the theorems of §417, in particular 417H and 417J.) The formulae of some of the arguments below will be simplified if we immediately re-index the family h(Xi , Ti , Σi , µi )ii∈I as h(XA , TA , ΣA , µA )iA∈A where A ⊆ PN. For each A ∈ A, let htAn in∈N be an equidistributed sequence in XA ; for n ∈ N, let νAn be the point-supported measure on XA defined Q by setting 1 νAn E = n+1 #({i : i ≤ n, tAi ∈ E}) for E ⊆ XA . For each finite set I ⊆ A, set YI = A∈I XA ; set πI (x) = x¹I ∈ YI for x ∈ X. Let λI be the τ -additive product of hµI iA∈I and, for each n, let νˇIn be the product of the measures hνAn iA∈I . (Because I is finite, this is a point-supported probability measure. I do not say ‘τ -additive product’ here because I do not wish to assume that all singleton sets are Borel, so the νAn may not be inner regular with respect to Borel sets.) (b) Suppose that I ⊆ A is finite and that W ⊆ YI is an open set. Then λI W ≤ lim inf n→∞ νˇIn W . P P Induce on #(I). If I = ∅, YI is a singleton and the result is trivial. For the inductive step, if I 6= ∅, take any A ∈ I and set I 0 = I \ {A}. Then we can identify YI with YI 0 × XA , λI with the τ -additive product of λI 0 and µA (417J), and each νˇIn with the product of νˇI 0 n and νAn . Let V be the family of those subsets V of YI which are expressible as a finite union of sets of the form U × H where U ⊆ YI 0 and H ⊆ XA are open. Then V is a base for the topology of YI closed under finite unions. Let ² > 0. Because λI is τ -additive, there is a V ∈ V such that λI V ≥ λW − ². The function t 7→ λI 0 V [{t}] : XA → [0, 1] is lower semi-continuous (417Ba), so 491Ca tells us that Z λI V = λI 0 V [{t}]µA (dt) 1 n→∞ n+1

≤ lim inf

n X

Z λI 0 V [{tAi }] = lim inf

i=0

n→∞

λI 0 V [{t}]νAn (dt).

At the same time, there are only finitely many sets of the form V [{t}], and for each of these we have λI 0 V [{t}] ≤ lim inf n→∞ νˇI 0 n V [{t}], by the inductive hypothesis. So there is an n0 ∈ N such that λI 0 V [{t}] ≤ νˇI 0 n V [{t}] + ² for every n ≥ n0 and every t ∈ XA . We must therefore have Z λI W ≤ λI V + ² ≤ lim inf λI 0 V [{t}]νAn (dt) + ² n→∞ Z ≤ lim inf νˇI 0 n V [{t}]νAn (dt) + 2² n→∞

= lim inf νˇIn V + 2² ≤ lim inf νˇIn W + 2². n→∞

n→∞

As ² and W are arbitrary, the induction proceeds. Q Q

Q (c) For K ⊆ m ∈ N, set AmK = {A : A ∈ A, A ∩ m = K} and ZmK = A∈AmK XA . (If AmK = ∅ then Q ZmK = {∅}.) Then for each m ∈ N we can identify X with the finite product K⊆m ZmK . For K ⊆ m ∈ N and n ∈ N, define zmKn ∈ ZmK by setting zmKn (A) = tAn for A ∈ AmK ; let ν˜mKn be the point-supported 1 #({i : i ≤ n, zmKi ∈ W }) for each W ⊆ ZmK . For measure on ZmK defined by setting ν˜mKn W = n+1 n ∈ N let νñ be the measure on X which is the product of the measures νñKn for K ⊆ n; this too is point-supported. (d) If I ⊆ A is finite, there is an m ∈ N such that νˇIn = νñ πI−1 for every n ≥ m. P P Let m be such that A ∩ m 6= A0 ∩ m for all distinct A,QA0 ∈ I. If n ≥ m, then νñ is the product of the νñKn for K ⊆ n. Now πI , interpreted as a function from K⊆n ZnK onto YI , is of the form πI (hzK iK⊆n ) = hzA∩n (A)iA∈I , so the

770

Further topics

491F

−1 image measure νñ πI−1 is the product of the family h˜ νn,A∩n,n π Â iA∈I , writing π Â (z) = z(A) when A ∩ n = K and z ∈ ZnK (254H, or otherwise). But, looking back at the definitions, −1 νñ,A∩n,n π Â [E] =

1 #({i n+1

−1 : i ≤ n, zn,A∩n,i ∈ π Â [E]})

=

1 #({i n+1

: i ≤ n, zn,A∩n,i (A) ∈ E})

=

1 #({i n+1

: i ≤ n, tAi ∈ E}) = νAn E

for every E ⊆ XA . So νñ πI−1 is the product of hνAn iA∈I , which is νˇIn . Q Q (e) Let W be the family of those open sets W ⊆ X expressible in the form πI−1 [W 0 ] for some finite I ⊆ A and some open W 0 ⊆ YI . If W ∈ W, then λW ≤ lim inf n∈N νñ W . P P Take I ∈ [A]<ω and an open W 0 ⊆ YI −1 0 such that W = πI [W ]. Then λW = λI W 0 (417K) ≤ lim inf νˇIn W 0 n→∞

(by (b) above) = lim inf νñ πI−1 [W 0 ] n→∞

(by (d)) = lim inf νñ W. Q Q n→∞

(f ) If now F ⊆ X is any closed set and ² > 0, then (because W is a base for the topology of X closed under finite unions) there is a W ∈ W such that W ⊆ X \ F and λW ≥ 1 − λF + ². In this case lim supn→∞ νñ F ≤ 1 − lim inf n→∞ νñ W ≤ 1 − λW ≤ λF + ². As ² is arbitrary, lim supn→∞ νñ F ≤ λF ; as F is arbitrary, 491D tells us that there is an equidistributed sequence in X. (g) All this was on the assumption that every µi is inner regular with respect to the Borel sets. For the superficially more general case enunciated, given only that each µi is a τ -additive topological measure with an equidistributed sequence, let µ0i be the restriction of µi to the Borel σ-algebra of Xi for each i ∈ I. Each µ0i is still τ -additive and equidistributed sequences for µi are of course equidistributed for µ0i . If we take λ0 to be the τ -additive product of hµ0i ii∈I , then it must agree with λ on the open sets of X and therefore on the closed sets, and an equidistributed sequence for λ0 will be an equidistributed sequence for λ. This completes the proof. 491G Corollary The usual measure of {0, 1}c has an equidistributed sequence. proof The usual measure of {0, 1} of course has an equidistributed sequence (just set xi = 0 for even i, xi = 1 for odd i), so 491F gives the result at once. 491H Theorem (Veech 71) Any separable compact Hausdorff topological group has an equidistributed sequence for its Haar probability measure. proof Let X be a separable compact Hausdorff topological group. Recall that X has exactly one Haar probability measure µ (442Ic), which is both a left Haar measure and a right Haar measure. (a) We need some elementary facts about convolutions. (i)PIf ν1 and ν2 are point-supported probability measures on X, then ν1 ∗ ν2 is point-supported. P P If P ν1 E = x∈E q1 (x) and ν2 E = x∈E q2 (x) for every E ⊆ X, then

491H


(ν1 ∗ ν2 )(E) = (ν1 × ν2 ){(x, y) : xy ∈ E} = P where q(z) = x∈X q1 (x)q2 (x−1 z) for z ∈ X. Q Q

P xy∈E

771

q1 (x)q2 (y) =

P z∈E

q(z)

(ii) Let ν,R λ be Radon probability measures on X. Suppose that R f ∈ C(X), α ∈ R and ² > 0 are such that | f (yxz)ν(dx) − α| ≤ ² for every y, z ∈ X. Then | f (yxz)(λ ∗ ν)(dx) − α| ≤ ² and R | f (yxz)(λ ∗ ν)(dx) − α| ≤ ² for every y, z ∈ X. P P Z |

f (yxz)(λ ∗ ν)(dx) − α| ZZ =| f (ywxz)ν(dx)λ(dw) − α|

(444C)

Z ≤

Z |

Z f (ywxz)ν(dx) − α|λ(dw) ≤

²λ(dw) = ²,

Z |

f (yxz)(ν ∗ λ)(dx) − α| ZZ =| f (yxwz)ν(dx)λ(dw) − α| Z Z Z ≤ | f (yxwz)ν(dx) − α|λ(dw) ≤ ²λ(dw) = ². Q Q

(b) Let A ⊆ X be a countable dense set. Let N be the set of point-supported probability measures ν on X which are defined by functions q such that {x : q(x) > 0} is a finite subset of A and q(x) is rational for R every x. ThenR N is countable. Now, for every f ∈ C(X) and ² > 0, there is a ν ∈ N such that | f (yxz)ν(dx) − f dµ| ≤ ² for all y, z ∈ X. P P Because X is compact, f is uniformly continuous for the right uniformity of X (4A2Je), so there is a neighbourhood U of the identity e such that |f (x0 ) − f (x)| ≤ 12 ² whenever x0 x−1 ∈ U . Next, again because X is compact, there is a neighbourhood V of e such that yxy −1 ∈ U whenever x ∈ V and y ∈ X (4A5Ej). Because A is dense, xV −1 ∩ A 6= ∅ is not empty for every S x ∈ X, that is, AV = X; once S more because X is compact, there are x0 , . . . , xn ∈ A such that X = i≤n xi V . Set Ei = xi V \ j
Z f (yxz)ν(dx) −

f (x)µ(dx)|

=|

n X

Z αi f (yxi z) −

f (yxz)µ(dx)|

i=0

(because X is unimodular) ≤ ≤

n X i=0 n X

Z |αi f (yxi z) −

f (yxz)µ(dx)| Ei

|αi − µEi ||f (yxi z)| +

i=0

² 2

+

Z |f (yxi z)µEi −

i=0

≤ kf k∞ ≤

n X

n X

i=0 n X i=0

|αi − µEi | +

² µEi 2

n Z X i=0

= ². Q Q

f (yxz)µ(dx)| Ei

|f (yxz) − f (yxi z)|µ(dx) Ei

772

Further topics

491H

(c) Let hνn in∈N be a sequence running over N, and set λn = ν0 ∗ ν1 R∗ . . . ∗ νn for R each n. Then each λn is a point-supported probability measure on X, by (a)(i). Also limn→∞ f dλn = R f dµ for every fR ∈ C(X). P P If f ∈ C(X) and ² > 0, then (b) tells us that there is an m ∈ N such that | f (yxz)νm (dx) − f dµ| ≤ ² 0 00 for R all y, z R∈ X. For any n ≥ m, λn is of the form λ ∗ νm ∗ λ . By (a)(ii), used in both parts successively, | f dλn − f dµ| ≤ ². As ² is arbitrary, we have the result. Q Q (d) If F ⊆ X is closed, then Z µF = inf{

f dµ : χF ≤ f ∈ C(X)} Z = inf lim sup f dλn ≥ lim sup λn F. χF ≤f

n→∞

n→∞

By 491D, µ has an equidistributed sequence. 491I The quotient PN/Z I now return to the asymptotic density ideal Z, moving towards a striking relationship between the corresponding quotient algebra and equidistributed sequences. Since Z C PN, we can form the quotient algebra Z = PN/Z. The functionals d and d∗ descend naturally to Z if we set ¯ • ) = d(I) whenever d(I) is defined. d¯∗ (I • ) = d∗ (I), d(I (a) d¯∗ is a strictly positive submeasure on Z. P P d¯∗ is a submeasure on Z because d∗ is a submeasure on ∗ ∗ PN. d¯ is strictly positive because Z ⊇ {I : d (I) = 0}. Q Q (b) Let ρ¯ be the metric on Z defined by saying that ρ¯(a, b) = d¯∗ (a 4 b) for all a, b ∈ Z. Under ρ, the Boolean operations ∪ , ∩ , 4 and \ and the function d¯∗ : Z → [0, 1] are uniformly continuous (393Bb), and Z is complete. P P Let hcn in∈N be a sequence in Z such that ρ¯(cn+1 , cn ) ≤ 2−n for every n ∈ N; then −i+1 ρ¯(cr , ci ) ≤ 2 for i ≤ r. For each n ∈ N choose Cn ⊆ N such that Cn• = cn ; then d∗ (Cr 4Ci ) ≤ 2−i+1 for i ≤ r. Choose a strictly increasing sequence hkn in∈N in N such that kn+1 ≥ 2kn for every n and, for each n ∈ N, 1 #((Cn 4Ci ) ∩ m) m

≤ 2−i+2 whenever i ≤ n, m ≥ kn .

S Set C = n∈N Cn ∩ kn+1 \ kn , and c = C • ∈ Z. If n ∈ N and m ≥ kn+1 , then take r > n such that kr ≤ m < kr+1 ; in this case ki ≤ 2i−r m for i ≤ r, so

#((C4Cn ) ∩ m) ≤ kn +

r−1 X

#((C4Cn ) ∩ ki+1 \ ki ) + #((C4Cn ) ∩ m \ kr )

i=n

= kn +

r−1 X

#((Ci 4Cn ) ∩ ki+1 \ ki ) + #((Cr 4Cn ) ∩ m \ kr )

i=n

≤ kn +

r−1 X

#((Ci 4Cn ) ∩ ki+1 ) + #((Cr 4Cn ) ∩ m)

i=n+1

≤ kn +

r−1 X

2−n+2 ki+1 + 2−n+2 m

i=n+1

≤ kn +

r−1 X

2−n+2 2i+1−r m + 2−n+2 m

i=n+1

≤ kn + 2−n+3 m + 2−n+2 m. But this means that ρ¯(c, cn ) = d∗ (C4Cn ) ≤ lim

kn

m→∞ m

for every n, and hcn in∈N converges to c in Z. Q Q

+ 2−n+3 + 2−n+2 ≤ 2−n+4

491J


773

*(c) If han in∈N is a non-increasing sequence in Z, there is an a ∈ Z such that a ⊆ an for every T n and d¯∗ (a) = inf n∈N d¯∗ (an ). P P For each n ∈ N, choose In ⊆ N such that In• = an ; replacing In by j≤n Ij if necessary, we can arrange that In+1 ⊆ In for every n. Set γ = inf n∈N d¯∗ (an ) = inf n∈N d∗ (In ). Let hkn in∈N be a strictly increasing sequence in N such that S kn+1 ≥ 2n kn and #(In ∩ kn+1 ) ≥ (γ − 2−n )kn+1 for every n. Set Kn = In ∩ kn+1 \ kn for each n, and I = n∈N Kn , a = I • ∈ Z. Then #(I ∩ kn+1 ) ≥ #(Kn ) ≥ (γ − 2−n+1 )kn+1 for each n, so d¯∗ (a) = d∗ (I) ≥ γ. Also I \ In ⊆ kn is finite, so a ⊆ an for each n. Of course it follows at once that d¯∗ (a) = γ exactly, as required. Q Q *(d) d¯∗ is a Maharam submeasure on Z. (Immediate from (c).) 491J Lemma Let han in∈N be a non-decreasing sequence in Z = PN/Z such that limn→∞ d¯∗ (an ) + ∗ ¯ d (1 \ an ) = 1. Then han in∈N is topologically convergent to a ∈ Z; a = supn∈N an in Z and d∗ (a)+d∗ (1 \ a) = 1. proof (a) The point is that if m ≤ n then d¯∗ (an \ am ) ≤ d¯∗ (an ) + d¯∗ (1 \ am ) − 1. P P Let I, J ⊆ N be such that I • = am and J • = an . For any k ≥ 1, 1 #(k k

1 k

1 k

1 k

∩ J) + #(k \ I) = #(k ∩ J \ I) + (k \ (I \ J)),

so 1 k k→∞

1 k k→∞

d∗ (J) + d∗ (N \ I) = lim sup #(k ∩ J) + lim sup #(k \ I) 1 k

1 k

≥ lim sup #(k ∩ J) + #(k \ I) k→∞

1 k k→∞

1 k

= lim sup #(k ∩ J \ I) + (k \ (I \ J)) 1 k

1

≥ lim sup #(k ∩ J \ I) + lim inf (k \ (I \ J)) = d∗ (J \ I) + 1 k→∞

k→∞ k

because am ⊆ an , so I \ J ∈ Z. But this means that d¯∗ (an \ am ) = d∗ (J \ I) ≤ d∗ (J) + d∗ (N \ I) − 1 = d¯∗ (an ) + d¯∗ (1 \ am ) − 1. Q Q (b) Accordingly lim sup sup ρ¯(am , an ) = lim sup sup d¯∗ (an \ am ) m→∞ n≥m

m→∞ n≥m

≤ lim sup sup d¯∗ (an ) + d¯∗ (1 \ am ) − 1 m→∞ n≥m

= lim sup sup d¯∗ (an ) − d¯∗ (am ) m→∞ n≥m

¯∗

¯∗

(because limm→∞ d (am ) + d (1 \ am ) = 1) = 0, and han in∈N is a Cauchy sequence in Z. (c) Because Z is complete, a = limn→∞ an is defined in Z. For each m ∈ N, am \ a = limn→∞ am \ an = 0 (because \ is continuous), so am ⊆ a; thus a is an upper bound of {an : n ∈ N}. If b is any upper bound of {an : n ∈ N}, then a \ b = limn→∞ an \ b = 0; so a = supn∈N an . Finally, d¯∗ (a) + d¯∗ (1 \ a) = limn→∞ d¯∗ (an ) + d¯∗ (1 \ an ) = 1.

774

Further topics

491K

491K Corollary Set D = {I • : I ⊆ N, d(I) is defined} ⊆ Z = PN/Z. (a) If I ⊆ N and I • ∈ D then d(I) is defined. (b) D = {a : a ∈ Z, d¯∗ (a) + d¯∗ (1 \ a) = 1}; if a ∈ D then 1 \ a ∈ D; if a, b ∈ D and a ∩ b = 0, then a ∪ b ∈ D and d¯∗ (a ∪ b) = d¯∗ (a) + d¯∗ (b); if a, b ∈ D and a ⊆ b then b \ a ∈ D and d¯∗ (b \ a) = d¯∗ (b) − d¯∗ (a). (c) D is a topologically closed subset of Z. (d) If A ⊆ D is upwards-directed, then sup A is defined in Z and belongs to D; moreover there is a sequence in A with the same supremum as A, and sup A belongs to the topological closure of A. (e) Let B ⊆ D be a subalgebra of Z. Then the following are equiveridical: (i) B is topologically closed in Z; (ii) B is order-closed in Z; (iii) setting ν¯ = d¯∗ ¹B, (B, ν¯) is a probability algebra. In this case, B is regularly embedded in Z. (f) If I ⊆ D is closed under either ∩ or ∪ , then the topologically closed subalgebra of Z generated by I, which is also the order-closed subalgebra of Z generated by I, is included in D. proof (a) There is a J ⊆ N such that d(J) is defined and I4J ∈ Z. But in this case d(I4J) = 0, so d(I ∪ J) = d(J) + d(I \ J) is defined; also d(J \ I) = 0, so d(I) = d(I ∪ J) − d(J \ I) is defined. (b) These facts all follow directly from the corresponding results concerning PN and d (491Ac). (c) All we have to know is that a 7→ d¯∗ (a), a 7→ 1 \ a are continuous; so that {a : d¯∗ (a) + d¯∗ (1 \ a) = 1} is closed. (d) Because A is upwards-directed, and d¯∗ is a non-decreasing functional on Z, there is a non-decreasing sequence han in∈N in A such that limn→∞ d¯∗ (an ) = supa∈A d¯∗ (a) = γ say. By 491J, b = limn→∞ an = supn∈N an is defined in Z and belongs to D. If a ∈ A and ² > 0, there is an n ∈ N such that d¯∗ (an ) ≥ γ − ². Let a0 ∈ A be a common upper bound of a and an . Then d¯∗ (a \ b) ≤ d¯∗ (a0 \ an ) = d¯∗ (a0 ) − d¯∗ (an ) ≤ γ − d¯∗ (an ) ≤ ². As ² is arbitrary, a ⊆ b; as a is arbitrary, b is an upper bound of A; as b = supn∈N an , b must be the supremum of A. (e)(i)⇒(ii) Suppose that B is topologically closed. If A ⊆ B is a non-empty upwards-directed subset with supremum b ∈ Z, then (d) tells us that b ∈ A ⊆ B. It follows that B is order-closed in Z (313E). (ii)⇒(iii) Suppose that B is order-closed in Z. If A ⊆ B is non-empty, then A0 = {a0 ∪ . . . ∪ an : a0 , . . . , an ∈ A} is non-empty and upwards-directed, so has a supremum in Z, which must belong to B, and must be the least upper bound of A in B. Thus B is Dedekind (σ-)complete. Now let han in∈N be a disjoint sequence in B and set bn = supi≤n ai for each n. Then hbn in∈N is a non-decreasing sequence in D so has a Pn limit and supremum b ∈ D, and b ∈ B. Also d¯∗ (bn ) = i=0 d¯∗ (ai ) for each n (induce on n), so P∞ ¯∗ P∞ ν¯b = d¯∗ (b) = limn→∞ d¯∗ (bn ) = d (ai ) = ν¯ai . i=0

i=0

Since certainly ν¯0 = 0, ν¯1 = 1 and ν¯b > 0 whenever b ∈ B \ {0}, (B, ν¯) is a probability algebra. (iii)⇒(i) Suppose that (B, ν¯) is a probability algebra. Then it is complete under its measure metric (323Gc), which agrees on B with the metric ρ¯ of Z; so B must be topologically closed in Z. We see also that B is regularly embedded in Z. P P (Compare 323H.) If A ⊆ B is non-empty and downwards-directed and has infimum 0 in B, and b ∈ Z is any lower bound of A in Z, then d¯∗ (b) ≤ inf a∈A d¯∗ (a) = inf a∈A ν¯a = 0 (321F), so b = 0. Thus inf A = 0 in Z. As A is arbitrary, this is enough to show that the identity map from B to Z is order-continuous (313Lb), that is, that B is regularly embedded in Z. Q Q (f ) Suppose that I ⊆ D is closed under ∩ . By 313Gc, D includes the order-closed subalgebra of Z generated by I. If, on the other hand, I is closed under ∪ , then I 0 = {1 \ a : a ∈ I} is a subset of D closed under ∩ , so the order-closed subalgebra generated by I, which is also the order-closed subalgebra generated by I 0 , is again included in D.

491N


775

491L Effectively regular measures The examples 491Xe and 491Yb show that the definition in 491B is drawn a little too wide for comfort, and allows some uninteresting pathologies. These do not arise in the measure spaces we care most about, and the following definitions provide a fire-break. Let (X, Σ, µ) be a measure space, and T a topology on X. (a) I will say that a measurable subset K of X of finite measure is regularly enveloped if for every ² > 0 there are an open measurable set G and a closed measurable set F such that K ⊆ G ⊆ F and µ(F \ K) ≤ ². (b) Note that the family K of regularly enveloped measurable sets of finite measure is closed under finite unions and countable intersections. P P (i) If K1 , K2 ∈ K and ∗ is either ∪ or ∩, let ² > 0. Take measurable open sets G1 , G2 and measurable closed sets F1 , F2 such that Ki ⊆ Gi ⊆ Fi and µ(Fi \ Ki ) ≤ 12 ² for both i. Then G1 ∗ G2 is a measurable open set, F1 ∗ F2 is a measurable closed set, K1 ∗ K2 ⊆ G1 ∗ G2 ⊆ F1 ∗ F2 and µ((F1 ∗ F2 ) \ (K1 ∗ K2 )) ≤ ². As ² is arbitrary, K1 ∗ K2 ∈ K. (ii) If hKn in∈N is a non-increasing sequence in K with intersection K and ² > 0, let n ∈ N be such that µKn < µK + ². Then we can find a measurable open set G and a measurable closed set F such that Kn ⊆ G ⊆ F and µF ≤ µK + ². As ² is arbitrary, K ∈ K. Together with (i), this is enough to show that K is closed under countable intersections. Q Q (c) Now I say that µ is effectively regular if it is inner regular with respect to the regularly enveloped sets of finite measure. 491M Examples (a) Any totally finite Radon measure is effectively regular. P P Let (X, T, Σ, µ) be a totally finite Radon measure space. If K ⊆ X is compact and ² > 0, let L ⊆ X \ K be a compact set such that µL ≥ 1 − µK + ². Let G, H be disjoint open sets including K, L respectively (4A2F(h-i)). Then K ⊆ G ⊆ X \ H, G is open, X \ H is closed, both G and X \ H are measurable, and µ((X \ H) \ K) ≤ ². This shows that every compact set is regularly enveloped, and µ is effectively regular. Q Q (b) Let (X, T, Σ, µ) be a quasi-Radon measure space such that T is a regular topology. Then µ is effectively regular. P P Let E ∈ Σ and take γ < µE. Choose sequences hEn in∈N and hGn in∈N inductively, as follows. E0 ⊆ E is to be any measurable set such that γ < µE0 < ∞. Given that µEn > γ, let G be an open set of finite measure such that µ(En ∩ G) > γ (414Ea), and F ⊆ G \ En a closed set such that µF ≥ µ(G\En )−2−n . Let H be the family of open sets H such that H ⊆ G\F . Then H is upwards-directed and covers En (because T is regular), so there is a Gn ∈ H such that µ(En ∩ Gn ) > γ (414Ea again). Now µ(Gn \ En ) ≤ 2−n . Set En+1 = En ∩ GnT , and continue. At the end of the induction, set K = n∈N En . For each n, K ⊆ Gn ⊆ Gn and limn→∞ µ(Gn \ K) ≤ limn→∞ 2−n + µ(En \ K) = 0, so K is regularly enveloped. At the same time, K ⊆ E and µK ≥ γ. As E and γ are arbitrary, µ is effectively regular. Q Q (c) Any totally finite Baire measure is effectively regular. P P Let µ be a totally finite Baire measure on a topological space X. If F ⊆ X is a zero set, let f : X → R be a continuous function such that F = f −1 [{0}]. For each n ∈ N, set Gn = {x : |f (x)| < 2−n }, Fn = {x : |f (x)| ≤ 2−n }; then Gn is a measurable open set, Fn is a measurable closed set, F ⊆ Gn ⊆ Fn for every n and limn→∞ µFn = µF (because µ is totally finite). This shows that every zero set is regularly enveloped; as µ is inner regular with respect to the zero sets (412D), µ is effectively regular. Q Q (d) A totally finite completion regular topological measure is effectively regular. (As in (c), all zero sets are regularly enveloped.) 491N Theorem Let X be a topological space and µ an effectively regular probability measure on X, with measure algebra (A, µ ¯). Suppose that hxn in∈N is an equidistributed sequence in X. Then we have a unique order-continuous Boolean homomorphism π : A → Z = PN/Z such that πG• ⊆ {n : xn ∈ G}• for every measurable open set G ⊆ X, and d¯∗ (πa) = µ ¯a for every a ∈ A. proof (a) Define θ : PX → Z by setting θA = {n : xn ∈ A}• for A ⊆ X; then θ is a Boolean homomorphism. If F ⊆ X is closed and measurable, then d¯∗ (θF ) ≤ µF , because hxn in∈N is equidistributed. Write K for the family of regularly enveloped measurable sets.

776

Further topics

491N

If K ∈ K, then π0 K = inf{θG : K ⊆ G ∈ Σ ∩ T} is defined in Z, d¯∗ (π0 K) = µK and π0 K ∈ D as defined in 491K. P P For each n ∈ N, let T Gn , Fn ∈ Σ be such that K ⊆ Gn ⊆ Fn , Gn is open, Fn is closed and µ(Fn \ K) ≤ 2−n . Set Hn = X \ i≤n Gi . Then d¯∗ (θHn ) + d¯∗ (1 \ θHn ) ≤ d¯∗ (θHn ) + d¯∗ (θ(

\

Fi )) ≤ µHn + µ(

i≤n

\

Fi )

i≤n

≤ µ(X \ K) + µFn ≤ 1 + 2−n . Also hθHn in∈N is a non-decreasing sequence in Z. By 491J, a = limn→∞ θHn = supn∈N θHn is defined in Z and belongs to D. Set T b = 1 \ a = limn→∞ 1 \ θHn = limn→∞ θ( i≤n Gi ), so that b also belongs to D. If K ⊆ G ∈ Σ ∩ T, then T T b \ θG = limn→∞ θ( i≤n Gi ) \ θG = limn→∞ θ( i≤n Gi \ G), and d¯∗ (b \ θG) = lim d¯∗ (θ( n→∞

Gi \ G))

i≤n

≤ lim d¯∗ (θ( n→∞

\ \

Fi \ G)) ≤ lim µ(

i≤n

n→∞

\

Fi \ G) = 0.

i≤n

This shows that b ⊆ θG wheneverTK ⊆ G ∈ Σ ∩ T. On the other hand, any lower bound of {θG : K ⊆ G ∈ Σ ∩ T} is also a lower bound of {θ( i≤n Gi ) : n ∈ N}, so is included in b. Thus b = inf{θG : K ⊆ G ∈ Σ ∩ T}. To compute d¯∗ (b), observe first that b ⊆ 1 \ θHn ⊆ θFn for every n, so d¯∗ (b) ≤ inf n∈N d¯∗ (θFn ) ≤ inf n∈N µFn = µK. On the other hand, for every n, so

T d¯∗ (θ( i≤n Gi )) ≥ 1 − d¯∗ (θHn ) ≥ 1 − µHn ≥ µK T d¯∗ (b) = limn→∞ d¯∗ (θ( i≤n Gi )) ≥ µK.

Accordingly d¯∗ (b) = µK, and we can take b for π0 K. Q Q (b) If K, L ∈ K, then π0 (K ∩ L) = π0 K ∩ π0 L. P P We know that K ∩ L ∈ K (491Lb). Now

π0 K ∩ π0 L = inf{θG : K ⊆ G ∈ T} ∩ inf{θH : L ⊆ H ∈ T} = inf{θG ∩ θH : K ⊆ G ∈ T, L ⊆ H ∈ T} = inf{θ(G ∩ H) : K ⊆ G ∈ T, L ⊆ H ∈ T} ⊇ π0 (K ∩ L). Now suppose that U ⊇ K ∩ L is a measurable open set and ² > 0. Let G, G0 be measurable open sets and F , F 0 measurable closed sets such that K ⊆ G ⊆ F , L ⊆ G0 ⊆ F 0 , µ(F \ K) ≤ ² and µ(F 0 \ L) ≤ ². Then d¯∗ (π0 K ∩ π0 L \ θU ) ≤ d¯∗ (θG ∩ θG0 \ θU ) = d¯∗ (θ(G ∩ G0 \ U )) ≤ d¯∗ (θ(F ∩ F 0 \ U )) ≤ µ(F ∩ F 0 \ U ) ≤ 2². As ² is arbitrary, π0 K ∩ π0 L ⊆ θU ; as U is arbitrary, π0 K ∩ π0 L ⊆ π0 (K ∩ L). Q Q This means that {π0 K : K ⊆ X is a regularly embedded measurable set} is a subset of D closed under ∩ . By 491Kf, the topologically closed subalgebra B of Z generated by this family is included in D; by 491Ke, B is order-closed and regularly embedded in Z, and (B, d¯∗ ¹ B) is a probability algebra. (c) Now observe that if we set Q = {K • : K ∈ K} ⊆ A, we have a function π : Q → B defined by setting πK • = π0 K whenever K ∈ K. P P Suppose that K, L ∈ K are compact and µ(K4L) = 0. Then

491O


777

d¯∗ (π0 K 4 π0 L) = d¯∗ (π0 K) + d¯∗ (π0 L) − 2d¯∗ (π0 K ∩ π0 L) (because π0 K and π0 L belong to B ⊆ D) = d¯∗ (π0 K) + d¯∗ (π0 L) − 2d¯∗ (π0 (K ∩ L)) = µK + µL − 2µ(K ∩ L) = 0. So π0 K = π0 L and either can be used to define πK • . Q Q Next, the same formulae show that π : Q → B is an isometry when Q is given the measure metric of A, since if K, L ⊆ X are compact ρ¯(πK • , πL• ) = d¯∗ (π0 K 4 π0 L) = µK + µL − 2µ(K ∩ L) = µ(K4L) = µ ¯(K • 4 L• ). As Q is dense in A (412N), there is a unique extension of π to an isometry from A to B. (d) Because π(K • ∩ L• ) = π(K ∩ L)• = π0 (K ∩ L) = π0 K ∩ π0 L = πK • ∩ πL• for all K, L ∈ K, π(a ∩ a0 ) = πa ∩ πa0 for all a, a0 ∈ A. It follows that π is a Boolean homomorphism. P P The point is that d¯∗ (πa) = µ ¯a for every a ∈ Q, and therefore for every a ∈ A. Now if a ∈ A, π(1 \ a) must be disjoint from πa (since certainly π0 = 0), and has the same measure as 1 \ πa (remember that we know that (B, d¯∗ ¹ B) is a measure algebra), so must be equal to 1 \ πa. By 312H, π is a Boolean homomorphism. Q Q By 324G, π is order-continuous when regarded as a function from A to B. Because B is regularly embedded in Z, π is order-continuous when regarded as a function from A to Z. (e) Let G ∈ Σ ∩ T. For any ² > 0, there is a K ∈ K such that K ⊆ G and µ(G \ K) ≤ ². In this case, πK • = π0 K ⊆ θG. So d¯∗ (πG• \ θG) ≤ d¯∗ (πG• \ πK • ) = µ ¯(G• \ K • ) = µ(G \ K) ≤ ². As ² is arbitrary, πG• ⊆ θG. (f ) This shows that we have a homomorphism π with the required properties. To see that π is unique, suppose that π 0 : A → Z is any homomorphism of the same kind. In this case d¯∗ (1 \ π 0 a) = d¯∗ (π 0 (1 \ a)) = µ ¯(1 \ a) = 1 − µ ¯a = 1 − d¯∗ (π 0 a), so π 0 a ∈ D, for every a ∈ A. If K ∈ K, then π 0 K • ⊆ θG whenever K ⊆ G ∈ Σ ∩ T, so π 0 K • ⊆ π0 K = πK • . As both πK • and π 0 K • belong to D, d¯∗ (πK • \ π 0 K • ) = d¯∗ (πK • ) − d¯∗ (π 0 K • ) = µK − µK = 0, and πK • = π 0 K • . As {K • : K ∈ K} is topologically dense in A, and both π and π 0 are continuous, they must be equal. 491O Proposition Let X be a topological space and µ an effectively regular probability measure on X which measures every zero set, and suppose that hxn in∈N is an equidistributed sequence in X. Let A be the measure algebra of µ and π : A → Z = PN/Z the regular embedding described in 491N; let Tπ : L∞ (A) → L∞ (Z) be the corresponding order-continuous Banach algebra embedding (363F). Let S : `∞ (X) → `∞ be the Riesz homomorphism defined by setting (Sf )(n) = f (xn ) for f ∈ `∞ (X) and n ∈ N, and R : `∞ → L∞ (Z) the Riesz homomorphism corresponding to the Boolean homomorphism I 7→ I • : PN → Z. For f ∈ L∞ (µ) let f • be the corresponding member of L∞ (µ) ∼ = L∞ (A) (363I). Then Tπ (f • ) = RSf for every f ∈ Cb (X). proof To begin with, suppose that f : X → [0, 1] is continuous and k ≥ 1. For each i ≤ k set Gi = {x : Pk Pk f (x) > ki }, Fi = {x : f (x) ≥ ki }. Then k1 i=1 χFi ≤ f ≤ k1 i=0 χGi . So 1 k

Pk i=1

1 k

χ(πFi• ) ≤ Tπ f • ≤

1 k

Pk

χ(θFi ) ≤ RSf ≤

1 k

Pk

Pk i=1

i=0

i=0

χ(πG•i ), χ(θGi )

778

Further topics

491O

where θ : PX → Z is the Boolean homomorphism described in 491N, because RS : `∞ → L∞ (Z) is the Riesz homomorphism corresponding to θ (see 363Fa, 363Fg). Now 491N tells us that πG• ⊆ θG for every open G ⊆ X, so 1 k

Tπ f • ≤

1 k

≤

k X

χ(πG•i ) ≤

i=0 k X

1 k

k X

χ(θGi )

i=0 k

χ(θFi ) =

i=0

X 1 χ(θFi ) e+ k

1 k

≤ e + RSf

i=1

where e is the standard order unit of the M -space L∞ (Z). But looking at complements we see that we must have πF • ⊇ θF for every closed set F ⊆ X, so RSf ≤ ≤

1 k 1 k

k X

χ(θGi ) ≤

i=0 k X

1 k 1 k

k X

χ(πFi• ) = e +

i=0

χ(θFi )

i=0 k X

1 k

χ(πFi• ) ≤ e + Tπ f • .

i=1

This means that |Tπ f • − RSf | ≤ k1 e for every k ≥ 1, so that Tπ f • = RSf . This is true whenever f ∈ Cb (X) takes values in [0, 1]; as all the operators here are linear, it is true for every f ∈ Cb (X). 491P Proposition Any probability algebra (A, µ ¯) of cardinal at most c can be regularly embedded as a subalgebra of Z = PN/Z in such a way that µ ¯ is identified with the restriction of the submeasure d¯∗ to the image of A. proof The usual measure of {0, 1}c is a totally finite Radon measure, so is effectively regular (491Ma). It has an equidistributed sequence (491G), so its measure algebra (Bc , ν¯c ) can be regularly embedded in Z in a way which matches ν¯c with d¯∗ (491N). Now if (A, µ ¯) is any probability algebra of cardinal at most c, it can be regularly embedded (by a measure-preserving homomorphism) in (Bc , ν¯c ) (332N), and therefore in (Z, d¯∗ ). This completes the proof. 491Q Corollary Every Radon probability measure on {0, 1}c has an equidistributed sequence. proof Let µ be a Radon probability measure on {0, 1}c , and (A, µ ¯) its measure algebra. By 491P, there is a measure-preserving embedding π : A → Z, and π[A] ⊆ D as defined in 491K. For ξ < c let aξ ∈ A be the equivalence class of {x : x(ξ) = 1}, and let Iξ ⊆ N be such that Iξ• = πaξ in Z. Define xn (ξ), for n ∈ N and ξ < c, by setting xn (ξ) = 1 if n ∈ Iξ , 0 otherwise. Now suppose that E ⊆ {0, 1}c is a basic open set of the form {x : x(ξ) = 1 for ξ ∈ K, 0 for ξ ∈ L}, where K, L ⊆ c are finite. Set b = πE • in Z, T S I = {n : xn ∈ E} = N ∩ ξ∈K Iξ \ ξ∈L Iξ . Then b = πE • = π( inf aξ \ sup aξ ) ξ∈K

ξ∈L

= inf πaξ \ sup πaξ = inf Iξ• \ sup Iξ• = I • . ξ∈K

ξ∈L

ξ∈K

ξ∈L

Since b ∈ D, d(I) is defined and is equal to d¯∗ (b) = µ ¯E • = µE. If we now take E to be an open-and-closed subset of {0, 1}c , it can be expressed as a disjoint union of finitely many basic open sets of the type just considered; because d is additive on disjoint sets, d({n : xn ∈ E}) is defined and equal to µE. But this is enough to ensure that hxn in∈N is equidistributed, by 491Ch. 491X Basic exercises (a) (i) Show that if I, J ∈ D = dom d as defined in 491A, then I ∪ J ∈ D iff I ∩ J ∈ D iff I \ J ∈ D iff I4J ∈ D. (ii) Show that if E ⊆ D is an algebra of sets, then d¹E is additive. (iii) Find I, J ∈ D such that I ∩ J ∈ / D.

491Xm


779

> (b) Suppose that I ⊆ N and that f : N → N is strictly increasing. Show that d∗ (f [I]) ≤ d∗ (I)d∗ (f [N]), with equality if either I or f [N] has asymptotic density. (c) Show that if A ⊆ N and 0 ≤ α ≤ d∗ (A) there is a B ⊆ A such that d∗ (B) = α and d∗ (A \ B) = d (A) − α. ∗

(d) (i) Let I ⊆ N be such that d(J) = d∗ (J ∩ I) + d∗ (J \ I) for every J ⊆ N such that d(J) is defined. Show that either I ∈ Z or N \ IZ. (ii) Show that for every ² > 0 there is an I ⊆ N such that d∗ (I) = ² but d(J) = 1 whenever J ⊇ I and d(J) is defined. (e) Let (X, T, Σ, µ) be a compact Radon probability space. Take any point ∞ not belonging to X, and give X ∪ {∞} the topology generated by {G ∪ {∞} : G ∈ T}. Show that X ∪ {∞} is compact and that the image measure µ∞ of µ under the identity map from X to X ∪ {∞} is a quasi-Radon measure, inner regular with respect to the compact sets. Show that if we set xn = ∞ for every n, then hxn in∈N is equidistributed for µ∞ . > (f ) (i) Show that a sequence htn in∈N in [0, 1] is equidistributed with respect to Lebesgue measure iff 1 limn→∞ n+1 #({i : i ≤ n, ti ≤ β}) = β for every β ∈ [0, 1]. (ii) Show that if α ∈ R is irrational then the sequence hin∈N of fractional parts of multiples of α is equidistributed in [0, 1] with respect to Lebesgue measure. (Hint: 281Yi.) (g) Show that the usual measure on the split interval (419L) has an equidistributed sequence. > (h) Show that if X is a regular Hausdorff space and f : N → X is injective, then there is an open set G ⊆ X such that f −1 [G] does not have asymptotic density. > (i) Let Z, µ, X = Z × {0, 1} and ν be as described in 439K, so that µ is a Radon probability measure on the compact metrizable space Z, X has a compact Hausdorff topology finer than the product topology, and ν is a measure on Z extending µ. (i) Show then {t : t ∈ Z, f (t, 0) 6= f (t, 1)} is R that if f ∈ C(X), R countable. (Hint: 4A2F(h-vii).) (ii) Show that f (t, 0)µ(dt) = f (t, 1)ν(dt) for every f ∈ C(X). (iii) Let λ be the measure νg −1 , where g(t) R Pn = (t, 1) for t ∈ Z. Show that there is a sequence hxn in∈N in Z × {0} 1 such that f dλ = limn→∞ n+1 i=0 f (xi ) for every f ∈ C(X), but that hxn in∈N is not λ-equidistributed. Pn 1 (j) Show that a function f : [0, 1] → R is Riemann integrable iff limn→∞ n+1 i=0 f (xi ) is defined in R for every sequence hxn in∈N in [0, 1] which is equidistributed for Lebesgue measure. (k) Let X be a metrizable space, and µ a quasi-Radon probability measure on X. (i) Show that there is an equidistributed sequence for µ. (ii) Show that if the support of µ is not compact, and hxn in∈N is an equidistributed Pn sequence for µ, then there is a continuous integrable function f : X → R such that 1 limn→∞ n+1 i=0 f (xi ) = ∞. (l) Let φ : c → PN be an injective function. For each n ∈ N let λn be the uniform measure on P(Pn), n giving measure 2−2 to each point. Define ψn : P(Pn) → {0, 1}c by setting ψn (I)(ξ) = 1 if φ(ξ) ∩ n ∈ I, 0 otherwise, and let νn be the image measure λn ψn−1 . Show that νn E is the usual measure of E whenever E ⊆ {0, 1}c is determined by coordinates in a finite set on which the map ξ 7→ φ(ξ) ∩ n is injective. Use this with 491D to prove 491G. > (m) (i) Let Z be the Stone space of the measure algebra of Lebesgue measure on [0, 1], with its usual measure. Show that there is no equidistributed sequence in Z. (Hint: meager sets in Z have negligible closures.) (ii) Show that Dieudonné’s measure on ω1 (411Q) has no equidistributed sequence. (iii) Show that if #(I) > c then the usual measure on {0, 1}I has no equidistributed sequence. (Hint: if hxn in∈N is any sequence in {0, 1}I , there is an infinite J ⊆ I such that hxn (η)in∈N = hxn (ξ)in∈N for all η, ξ ∈ J.) (iv) Show that if X is a topological group with a Haar probability measure µ, and X is not separable, then µ has no equidistributed sequence. (Hint: use 443D to show that every separable subset is negligible.)

780

Further topics

491Xn

(n) Let X be a compact Hausdorff abelian topological group and µP its Haar probability measure. Show n 1 that a sequence hxn in∈N in X is equidistributed for µ iff limn→∞ n+1 i=0 χ(xi ) = 0 for every non-trivial 1 character χ : X → S . (Hint: 281G.) (o) Show that Z is not Dedekind σ-complete. (Hint: 392Hc.) (p) Let han in∈N be a non-decreasing sequence in Z. Show that there is an a ∈ Z such that an ⊆ a for every n ∈ N and d¯∗ (a) = supn∈N d¯∗ (an ). (q) Let Z, d¯∗ and D be as in 491K. Show that if a ∈ D \ {0} and Za is the principal ideal of Z generated by a, then (Za , d¯∗ ¹ Za ) is isomorphic, up to a scalar multiple of the submeasure, to (Z, d¯∗ ). (r) Let (X, Σ, µ) be a semi-finite measure space and T a topology on X. Show that µ is effectively regular iff whenever E ∈ Σ, µE < ∞ and ² > 0 there are a measurable open set G and a measurable closed set F ⊇ G such that µ(F \ E) + µ(E \ G) ≤ ². (s) Let X be a normal topological space and µ a topological measure on X which is inner regular with respect to the closed sets and effectively locally finite. Show that µ is effectively regular. (t) Let X be a topological space and µ an effectively regular measure on X. (i) Show that the completion and c.l.d. version of µ are also effectively regular. (ii) Show that if Y ⊆ X then the subspace measure is again effectively regular. (iii) Show that any totally finite indefinite-integral measure over µ is effectively regular. (u) (i) Let X1 , X2 be topological spaces with effectively regular measures µ1 , µ2 . Show that the c.l.d. product measure on X1 × X2 is effectively regular with respect to the product topology. (Hint: 412R.) (ii) Let hXi ii∈I be a family of topological spaces and µi an Q effectively regular probability measure on Xi for each i. Show that the product probability measure on i∈I Xi is effectively regular. (v) Give [0, 1] the topology T generated by the usual topology and {[0, 1] \ A : A ⊆ Q}. Let µL be Lebesgue measure on [0, 1], and Σ its domain. For E ∈ Σ set µE = µL E + #(E ∩ Q) if E ∩ Q is finite, ∞ otherwise. Show that µ is a σ-finite quasi-Radon measure with respect to the topology T, but is not effectively regular. (w) Let (X, ρ) be a separable metric space and µ a Borel probability measure on X. (i) Show that there is an equidistributed sequence in X. (ii) Show that if hxn in∈N is an equidistributed sequence in X, and hyn in∈N is a sequence in X such that limn→∞ ρ(xn , yn ) = 0, then Pn hyn in∈N is equidistributed. 1 (iii) Show that if f : X → R is a bounded function, then limn→∞ n+1 i=0 f (xi ) − f (yi ) = 0 for all equidistributed sequences hx i , hy i in X iff {x : f is continuous at x} is conegligible, and in this n n∈N n n∈N R Pn 1 f (x ) = f dµ for every equidistributed sequence hx i in X. (Hint: {x : f case limn→∞ n+1 i n n∈N i=0T S is continuous at x} = m∈N Gm , where Gm = {G : G ⊆ X is open, supx,y∈G |f (x) − f (y)| ≤ 2−m }.) Compare 134L. (x) Let (X, T) be a topological space, µ a probability measure on X, and φ : X → X an inverse-measurepreserving function. (i) Suppose that T has a countable network consisting of measurable sets, and that φ is ergodic. Show that hφn (x)in∈N is equidistributed for almost every x ∈ X. (Hint: 372Qb.) (ii) Suppose that µ is either inner regular with respect to the closed sets or effectively regular, and that {x : hφn (x)in∈N is equidistributed} is not negligible. Show that φ is ergodic. (y) Let A be a countable Boolean algebra and ν a finitely additive functional on A such that ν1 = 1. Show that there is a Boolean homomorphism π : A → PN such that d(πa) is defined and equal to νa for every a ∈ A (i) from 491Xc (ii) from 393B, 491P and 341Xf. (z) Let X be a dyadic space. (i) Show that there is a Radon probability measure on X with support X. (ii) Show that the following are equiveridical: (α) w(X) ≤ c; (β) every Radon probability measure on X has an equidistributed sequence; (γ) X is separable. (Hint: 4A2Dd, 418L.)

491Z


781

491Y Further exercises (a) Show that every subset A of N is expressible in the form IA 4JA where d(IA ) = d(JA ) = 21 (i) by a direct construction, with A 7→ IA a continuous function (ii) using 443D. (b) Find a topological space X with a τ -additive probability measure µ on X, a sequence hxn in∈N in X and a base G for the topology of X, consisting of measurable sets and closed under finite intersections, such 1 that µG ≤ lim inf n→∞ n+1 #({i : i ≤ n, xi ∈ G}) for every G ∈ G but hxn in∈N is not equidistributed. (Hint: take #(X) = 4.) (c) Let X be a compact Hausdorff space on which every Radon probability measure has an equidistributed sequence. Show that the cylindrical σ-algebra of C(X) is the σ-algebra generated by sets of the form {f : f ∈ C(X), f (x) > α} where x ∈ X and α ∈ R. (Hint: 436J, 491Cb.) (d) Give ω1 +1 and [0, 1] their usual compact Hausdorff topologies. Let hti ii∈N be a sequence in [0, 1] which is equidistributed for Lebesgue measure µL , and set Q = {ti : i ∈ N}, X = (ω1 × ([0, 1] \ Q)) ∪ ({ω1 } × Q), with the subspace topology inherited from (ω1 + 1) × [0, 1]. (i) Set F = {ω1 } × Q. Show that F is a closed Baire set in the completely regular Hausdorff space X. (ii) Show that if f ∈ Cb (X) then there are a gf ∈ C([0, 1]) and a ζ < ω1 such that such that f (ξ,Rt) = gf (t) R whenever (ξ, t) ∈ X and ζ ≤ ξ ≤ ω1 . (iii) Show that there is a Baire measure µ on X such that f dµ = gf dµL for every f ∈ Cb (X). (iv) Show that R Pn 1 µF = 0. (v) Show that f dµ = limn→∞ n+1 f (ω , t ) for every f ∈ Cb (X), but that h(ω1 , ti )ii∈N is 1 i i=0 not equidistributed with respect to µ. (e) Let (X, T, Σ, µ) be a topological measure space. Let E be the Jordan algebra of X (411Yc). (i) Suppose that µ is a complete probability measure on X and hxn in∈N an equidistributed sequence in X. Show that the asymptotic density d({n : xn ∈ E}) is defined and equal to µE for every E ∈ E. (ii) Suppose that µ is a probability measure on X and that hxn in∈N is a sequence R that d({n : xn ∈ E}) is Pn in X such 1 defined and equal to µE for every E ∈ E. Show that limn→∞ n+1 f (x ) = f dµ for every f ∈ Cb (X). i i=0 (f ) Show that a sequence Pn hxrn in∈N1 in [0, 1] is equidistributed for Lebesgue measure iff there is some r0 ∈ N 1 such that limn→∞ n+1 i=0 xi = r+1 for every r ≥ r0 . (g) Show that a Radon measure on an extremally disconnected compact Hausdorff space has an equidistributed sequence iff it is point-supported. (Hint: see the hint for 326Yo.) (h) Show that there is a countable dense set D ⊆ [0, 1]c such that no sequence in D is equidistributed for the usual measure on [0, 1]c . (i) Let X be a compact Hausdorff topological group.PShow that a sequence hxn in∈N in X is equidistributed n 1 for the Haar probability measure on X iff limn→∞ n+1 i=0 φ(xi ) = 0 for every non-trivial finite-dimensional representation φ of X. (Hint: 446Yb.) (j) Show that Z is weakly σ-distributed. (k) Show that Z ∼ = ZN . (l) Show that Z has the σ-interpolation property (definition: 466G). (m) Let hXξ iξ 1 the sequence h< xn >in∈N of fractional parts of powers of x is equidistributed for Lebesgue measure on [0, 1] (Kuipers & Niederreiter 74, p. 35). But is h< ( 23 )n >in∈N equidistributed?

782

Further topics

491 Notes

491 Notes and comments The notations d∗ , d (491A) are standard, and usefully suggestive. But coming from measure theory we have to remember that d∗ , although a submeasure, is not an outer measure, the domain of d is not an algebra of sets (491Xa), and d and d∗ are related by only one of the Pnformulae we expect to connect a measure with an outer measure (491Ac, 491Xd). The limits limn→∞ n1 i=0 f (xi ) are the Cesaro means of the sequences hf (xn )in∈N . The delicacy of the arguments here arises from the fact that the family of (bounded) sequences with Cesaro means, although a norm-closed linear subspace of `∞ , is neither a sublattice nor a subalgebra. When we turn to the quotient algebra Z = PN/Z, we find ourselves with a natural submeasure to which we can apply results from §393 to good effect (491I). What is striking is that equidistributed sequences induce regular embeddings of measure algebras in Z which can be thought of as measure-preserving (491N). Most authors have been content to define an ‘equidistributed sequence’ to be one such that the integrals of bounded continuous functions are correctly specified (491Cf, 491Cg); that is, that the point-supported Pn 1 δ converge to µ in the vague topology on an appropriate class of measures (437J). measures n+1 x i i=0 I am going outside this territory in order to cover some ideas I find interesting. 491Xi shows that it makes a difference; there are Borel measures on compact Hausdorff spaces which have sequences which give the correct Cesaro means for continuous functions, but lie within negligible closed sets; and the same can happen with Baire measures (491Yd). It seems to be difficult, in general, to determine whether a topological probability space – even a compact Radon probability space – has an equidistributed sequence. In the proofs of 491D-491G I have tried to collect the principal techniques for showing that spaces do have equidistributed sequences. In the other direction, it is obviously impossible for a space to have an equidistributed sequence if every separable subspace is negligible (491Xm). For an example of a separable compact Hausdorff space with a Radon measure which does not have an equidistributed sequence, we seem to have to go deeper (491Yg). 491Z is a famous problem. It is not clear that it is a problem in measure theory, and there is no reason to suppose that any of the ideas of this treatise beyond 491Xf(i) are relevant. I mention it because I think everyone should know that it is there.

492 Combinatorial concentration of measures ‘Concentration of measure’ takes its most dramatic form in the geometrically defined notions of concentration explored in §476. But the phenomenon is observable in many other contexts, if we can devise the right abstract geometries to capture it. In this section I present one of Talagrand’s theorems on the concentration of measure in product spaces, using the Hamming metric (492D), and Maurey’s theorem on concentration of measure in permutation groups (492H). 492A Lemma Let (X, Σ, µ) be a totally finite measure space, α < β in R, φ : [α, β] → R a convex function, and f : X → [α, β] a Σ-measurable function. Then

R

proof If t ∈ [α, β] then t =

φ(f (x))µ(dx) ≤ t−α β β−α

φ(t) ≤

+

β−t α, β−α

φ(β)−φ(α) β−α

R

f dµ +

βφ(α)−αφ(β) µX. β−α

so

t−α β−t φ(β) + φ(α) β−α β−α

=

φ(β)−φ(α) βφ(α)−αφ(β) t+ . β−α β−α

Accordingly φ(f (x)) ≤

φ(β)−φ(α) βφ(α)−αφ(β) f (x) + β−α β−α

for every x ∈ X; integrating with respect to x, we have the result. 492B Corollary Let (X, Σ, µ) be a probability space and f : X → [α, 1] a measurable function, where 0 < α ≤ 1. Then

R

R 1 dµ · f

f dµ ≤

(1+α)2 . 4α

492D

Combinatorial concentration of measures

proof Set γ =

R

783

1 t

f dµ, so that α ≤ γ ≤ 1. By 492A, with φ(t) = ,

R

1 dµ f

≤

γ 1 1 1 (1 − ) + ( 1−α α 1−α α

1+α−γ

Now · γ takes its maximum value R α R for 1/f · f . 1 2

2

492C Lemma (1 + cosh t) ≤ et

/4

(1+α)2 4α

when γ =

− α) =

1+α , 2

1+α−γ . α

so this is also the maximum possible value

for every t ∈ R.

proof For k ≥ 1, 4k k! ≤ 2(2k)! (induce on k), so P∞ t2k P∞ 2t2k 2 1 + cosh t = 2 + k=1 ≤ 2 + k=1 k = 2et /4 . (2k)!

4 k!

492D Theorem (Talagrand 95) Let h(Xi , Σi , µi )ii
ρ(x, y) = #({i : i < n, x(i) 6= y(i)}) for x, y ∈ X. If W ∈ Λ and λW > 0, then

R

etρ(x,W ) λ(dx) ≤

1 t2 /4n e λW

for every t ≥ 0. proof The formulae below will go much more smoothly if we work with the simple Hamming metric σ(x, y) = #({i : x(i) 6= y(i)} instead of ρ. In this case, we can make sense of the case n = 0, and this will be useful. In terms of σ, our target is to prove that if W ∈ Λ and λW > 0, then

R

etσ(x,W ) λ(dx) ≤

1 nt2 /4 e λW

for every t ≥ 0. (a) To begin with, suppose that every Xi = Z = {0, 1}N , every µi is a Borel measure, and W is compact. Note that in this case λ is a Radon measure (because the Xi are compact and metrizable), and S {x : σ(x, W ) ≤ m} = I⊆n,#(I)≤m {x : ∃ y ∈ W, x¹n \ I = y¹n \ I} is compact for every m, so the function x 7→ σ(x, W ) is measurable. Induce on n. If n = 0 we must have W = X = {∅} and σ(x, W ) = Q 0 for every x, so the result is trivial. For the inductive step to n ≥ 1, we have W ⊆ X × Xn , where X = i
Z Z etσ((x,ξ),W ) λ(dx) ≤ min( etσ(x,Vξ ) λ(dx), et etσ(x,V ) λ(dx)) ≤ ent

/4

min(

1 et , λVξ λV

1 et et ) as . 0 λV λV λV max(e−t , ξ ) for ξ ∈ Xn , λV

by the inductive hypothesis, counting min( , It follows that if we set f (ξ) =

2

)

784

Further topics

492D

Z (λ × µn )(W )

etσ((x,ξ),W ) λ(dx)µn (dξ) Z Z 2 1 et ≤ λVξ µn (dξ) · ent /4 min( , )µn (dξ) λVξ λV Z Z 2 λV λVξ µn (dξ) · min(et , )µn (dξ) = ent /4 λV λVξ Z Z 2 1 ≤ ent /4 f (ξ)µn (dξ) · µn (dξ) f (ξ)

2

≤ ent

/4

·

(1+e−t )2 4e−t

(492B) 1 2

2

= ent by 492C, and

R

/4

(1 + cosh t) ≤ e(n+1)t

etσ((x,ξ),W ) λ(dx)µn (dξ) ≤

2

/4

2 1 e(n+1)t /4 , (λ×µn )(W )

so the induction continues.

N (b) Now turn to the general case. If W ∈ Λ, there is a W1 ⊆ W such that W1 ∈ c i 0}, VI0 = {y : y ∈ Y, y¹n \ I ∈ VI }. T Then ν(V \ VI0 ) = 0 for every I ⊆ n, so if we set V 0 = I⊆n VI0 then νV 0 = νV . (Of course V 0 ⊆ V∅0 = V .) Take any γ ∈ ]0, λW [ = ]0, νV 0 [. Let K ⊆ V 0 be a compact set such that νK ≥ γ. Set g(y) = etσ(y,K) for y ∈ Y , where I write σ for the Hamming metric on Y (regarded as a product of n factor spaces). Then g : Y → R is Borel measurable and gh : X → R is Λ-measurable. Also, for any x ∈ X, σ(x, W ) ≤ σ(h(x), K). P P Take y ∈ K such that σ(h(x), y) = σ(h(x), K), and set I = {i : h(x)(i)) 6= y(i)},

u = h(x)¹n \ I = y¹n \ I.

0

Because y ∈ V , u ∈ VIQand νI H > 0, where H = {v : v ∈ Z I , (u, Q v) ∈ V }. But if we write λI for the product measure on i∈I Xi , and hI (z) = hhi (z(i))ii∈I for z ∈ i∈I Xi , then hI is inverse-measure0 preserving for λI and νI ; in particular, h−1 I [H] is non-empty. This means that we can find an x ∈ X such −1 0 0 0 0 that x ¹n \ I = x¹n \ I and x ¹I ∈ hI [H]. In this case, h(x ) ∈ V , so x ∈ W1 ⊆ W , and σ(x, W ) ≤ σ(x, x0 ) ≤ #(I) = σ(h(x), K). Q Q Accordingly etσ(x,W ) ≤ etσ(h(x),K) = g(h(x)) for every x ∈ X, and Z

Z etσ(x,W ) λ(dx) ≤

Z gh dλ =

g dν

(because g is ν-integrable and h is inverse-measure-preserving, see 235I)

492F


≤

785

1 nt2 /4 e νK

(by (a)) 1 γ

2

≤ ent As γ is arbitrary,

R

etσ(x,W ) λ(dx) ≤

/4

.

1 nt2 /4 e , λW

as claimed. 492E Corollary (Talagrand 95) Let h(Xi , Σi , µi )ii 0, then 1 −nγ 2 e λW

λ∗ {x : ρ(x, W ) ≥ γ} ≤

for every γ ≥ 0. 2 (b) If W , W 0 ∈ Λ and γ > 0 are such that e−nγ < λW · λW 0 then there are x ∈ W , x0 ∈ W 0 such that #({i : i < n, x(i) 6= x0 (i)}) < nγ. proof (a) Set t = 2nγ. By 492D, there is a measurable function f : X → R such that f (x) ≥ etρ(x,W ) for R 1 t2 /4n every x ∈ X and f dλ ≤ e . Then λW Z ∗ tγ −tγ λ {x : ρ(x, W ) ≥ γ} ≤ λ{x : f (x) ≥ e } ≤ e f dλ ≤

1 −tγ+t2 /4n e λW

=

1 −nγ 2 e . λW

(b) By (a), λ∗ {x : ρ(x, W 0 ) ≥ γ} < λW , so there must be an x ∈ W such that ρ(x, W 0 ) < γ. 492F The next theorem concerns concentration of measure in permutation groups. I approach this through a general result about slowly-varying martingales (492G). 2

Lemma et ≤ t + et for every t ∈ R. proof If t ≥ 1 then t ≤ t2 so 2

2

et ≤ et ≤ t + et . If 0 ≤ t ≤ 1 then et = 1 + t + ≤1+t+

∞ X tn n=2 ∞ X k=1

n! 2k

t k!

≤1+t+

∞ X

(

k=1

1 (2k)!

+

1 )t2k (2k+1)!

2

= t + et .

If t ≤ 0 then et = 1 + t + ≤1+t+

∞ X tn n=2 ∞ X k=1

n! t2k k!

≤1+t+

∞ X t2k k=1

2

= t + et .

(2k)!

786

Further topics

492G

492G Lemma (Milman & Schechtman 86) Let (X, Σ, µ) be a probability space, and hfn in∈N a martingale on X. Suppose that fn ∈ L∞ (µ) for every n, and set αn = ess sup |fn − fn−1 | for n ≥ 1. Then for any n ≥ 1 and γ ≥ 0, Pn Pr(fn − f0 ≥ γ) ≤ exp(−γ 2 /4 i=1 αi2 ), interpreting exp(−γ 2 /0) as 0 if fn =a.e. f0 . proof Let hΣn in∈N be a non-decreasing sequence of σ-subalgebras of Σ to which hfn in∈N is a adapted. Extending the functions fn if necessary, we may suppose that they are all defined on the whole of X. (a) I show first that E(exp(λ(fn − f0 ))) ≤ exp(λ2

Pn i=1

αi2 )

P0 for any n ≥ 0 and any λ > 0. P P Induce on n. For n = 0, interpreting i=1 as 0, this is trivial. For the inductive step to n + 1, set g = fn − fn−1 and let g1 , g2 be conditional expectations of exp(λg) and exp(λ2 g 2 ) on Σn−1 . Because |g| ≤ αn a.e., exp(λ2 g 2 ) ≤ exp(λ2 αn2 ) a.e. and g2 ≤ exp(λ2 αn2 ) a.e. Because exp(λg) ≤ λg + exp(λ2 g 2 ) wherever g is defined (492F), and 0 is a conditional expectation of g on Σn−1 , g1 ≤ g2 ≤ exp(λ2 αn2 ) a.e. Now observe that fn−1 −f0 is Σn−1 -measurable, so that exp(λ(fn−1 −f0 ))×g1 is a conditional expectation of exp(λ(fn−1 − f0 )) × exp(λg) = exp(λ(fn − f0 )) on Σn−1 (233Eg). Accordingly E(exp(λ(fn − f0 ))) = E(exp(λ(fn−1 − f0 )) × g1 ) ≤ ess sup |g1 | · E(exp(λ(fn−1 − f0 ))) ≤ exp(λ2 αn2 ) exp(λ2

n−1 X

αi2 )

i=1

(by the inductive hypothesis) = exp(λ2

n X

αi2 )

i=1

and the induction continues. Q Q (b) Now Pn take n ≥ 1 and γ ≥ 0. If λ = γ/2 i=1 αi2 . Then

Pn i=1

αi2 = 0 then fn =a.e. f0 and the result is trivial. Otherwise, set

Pr(fn − f0 ≥ γ) = Pr(exp(λ(fn − f0 )) ≥ eλγ ) ≤ e−λγ E(exp(λ(fn − f0 ))) ≤ e−λγ exp(λ2

n X

αi2 )

i=1

(by (a) above) = e−λγ/2 = exp(−γ 2 /4

n X

αi2 )

i=1

as claimed. 492H Theorem (Maurey 79) Let X be a non-empty finite set and G the group of all permutations of X with the discrete topology. For π, φ ∈ G set ρ(π, φ) =

#({x:x∈X, π(x)6=φ(x)}) . #(X)

Then ρ is a metric on G. Give G its Haar probability measure, and let f : G → R be a 1-Lipschitz function. Then Pr(f − E(f ) ≥ γ) ≤ exp(−

γ 2 #(X) ) 16

492I


787

for any γ ≥ 0. proof We may suppose that X = n = {0, . . . , n − 1} where n = #(X). For m ≤ n, p : m → n set Ap = {π : π ∈ G, π¹m = p}, and let Σm be the subalgebra of PG generated by {Ap : p ∈ nm }. Then {∅, G} = Σ0 ⊆ Σ1 ⊆ . . . ⊆ Σn−1 = Σn = PG; for m > n set Σm = PG. For each m let fm be the (unique) conditional expectation of f on Σm , so that P 1 fm (π) = φ∈Ap f (φ) #(Ap )

whenever π ∈ G and p = π¹m. Now we find that |fm (π) − fm−1 (π)| ≤

2 n

for every m ≥ 1 and π ∈ G. P P If

m > n this is trivial. Otherwise, set p = π¹m − 1 and k = π(m − 1). Set J = p[m − 1] = {π(i) : i < m − 1}, and for j ∈ n \ J let pj = pa j be that function from m to n which extends p and takes the value j at m − 1; let αj be the common value of fm (φ) for φ ∈ Apj , so that fm (π) = αk . Now, for each j ∈ n \ (J ∪ {k}), the ← − ← − function φ 7→ (j k)φ is a bijection from Apk to Apj , where (j k) ∈ G is the transposition which exchanges j and k. But this means that ¯ |αj − αk | = ¯ =

1 (n−m)!

1 (n−m)!

X

f (φ) −

φ∈Apj

1 (n−m)!

X

¯ f (φ)¯

φ∈Apk

¯ X ← − ¯ ¯ f (φ) − f ((j k)φ)¯ φ∈Apj

¯ ← − ¯ 2 ≤ sup ¯f (φ) − f ((j k)φ)¯ ≤

n

φ∈Apj

← − because f is 1-Lipschitz and ρ(φ, (j k)φ) = Accordingly ¯ |fm−1 (π) − fm (π)| = ¯ ≤

2 n

for every φ. And this is true for every j ∈ n \ (J ∪ {k}).

1 (n−m+1)!

1 n−m+1

X

¯ ¯ f (φ) − αk ¯ = ¯

φ∈Ap

X

|αj − αk | ≤

j∈n\J

1 n−m+1

1 n−m+1

X

¯ αj − αk ¯

j∈n\J

X 2 j∈n\J

n

2 n

= ,

as claimed. Q Q (b) Now observe that f = fn−1 and that f0 is the constant function with value E(f ), so that

Pr(f − E(f ) ≥ γ) = Pr(fn−1 − f0 ≥ γ) ≤ exp(−γ 2 /4

n−1 X i=1

2 n

( )2 )

(492G) ≤ exp(−

nγ 2 ), 16

which is what we were seeking to prove. 492I Corollary Let X be a non-empty finite set and G the group of all permutations of X; write n = #(X). Let µ be the Haar probability measure of G when given its discrete topology. Suppose that A ⊆ G and µA ≥ 21 . Then µ{π : π ∈ G, ∃ φ ∈ A, #({x : x ∈ X, π(x) 6= φ(x)}) ≤ k} ≥ 1 − exp(− for every k ≤ n.

k2 ) 64n

788

proof If exp(−

Further topics k2 ) 64n

Otherwise, set g(π) =

1 2,

≥ 1 n

492I

this is trivial, since the left-hand-side of the inequality is surely at least

1 2.

minφ∈A #({x : x ∈ X, π(x) 6= φ(x)}) for π ∈ G, so that g is 1-Lipschitz for the

metric ρ of 492H. Applying 492H to f = −g, we see that Pr(E(g) − g ≥

k ) 2n

≤ exp(−

and there must be some π ∈ A such that E(g) − g(π) <

k , 2n

k2 ) 64n

1 2

< ,

so that E(g) <

k . 2n

This means that

µ{π : π ∈ G, ∃ φ ∈ A, #({x : x ∈ X, π(x) 6= φ(x)}) ≤ k} k n

= 1 − µ{π : π ∈ G, g(π) > } ≥ 1 − Pr(g − E(g) ≥

k ) 2n

≥ 1 − exp(−

k2 ), 64n

applying 492H to g itself. 492X Basic exercises (a) Let (X, Σ, µ)p be a probability space, and hfn in∈N a martingale on X. Suppose P∞ 2 that fn ∈ L∞ (µ) for every n, and that σ = n=1 αn is finite and not zero, where αn = ess sup |fn − fn−1 | for n ≥ 1. Show that f = limn→∞ fn is defined a.e., and that Pr(f − f0 ≥ γ) ≤ exp(−γ 2 /4σ 2 ) for every γ ≥ 0. (Hint: show first that kfn k1 ≤ kfn k2 ≤ σ + kf0 k2 for every n, so that we can apply 275G.) (b) Let (X, ρ) be a metric space and µ a topological probability measure on X. Suppose that γ, ² > 0 are such that Pr(f − E(f ) ≥ γ) ≤ ² whenever f : X → [−γ, γ] is 1-Lipschitz. Show that if µF ≥ 21 then µ{x : ρ(x, F ) ≥ 2γ} ≤ ². (c) Let (X, ρ) be a metric space and µ a topological probability measure on X. Suppose that γ, ² > 0 are such that µ{x : ρ(x, F ) > γ} ≤ ² whenever µF ≥ 21 . Show that if f : X → [−1, 1] is a 1-Lipschitz function then Pr(f − E(f ) > 2γ + 2²) ≤ ². (d) Use 492G to show that if h(Xi , Σi , µi )ii
be applied directly to sets W of small measure, at least provided that γ > √ in 492E. We also need a little more measure theory here, since sets which are measured by product measures can be geometrically highly irregular, and our Lipschitz functions x 7→ ρ(x, W ) need not be measurable. In the proof of 492G we have an interesting application of the idea of ‘martingale’. The inequality here is quite different from the standard martingale inequalities like 275D or 275F or 275Yc. It gives a very

493B

Extremely amenable groups

789

strong inequality concerning the difference fn − f0P , at the cost of correspondingly Pstrong hypotheses on the differences fi − fi−1 ; but since we need control of i ess sup |fi − fi−1 |2 , not of i ess sup |fi − fi−1 |, there is scope for applications like 492H. What the inequality tells us is that most of the time the differences fi − fi−1 cancel out, just as in the Central Limit Theorem, and that once again we have a vaguely Gaussian sum fn − f0 . Concentration of measure, in many forms, has been studied intensively in the context of the geometry of normed spaces, as in Milman & Schechtman 86, from which 492F-492I are taken.

493 Extremely amenable groups A natural variation on the idea of ‘amenable group’ (§449) is the concept of ‘extremely amenable’ group (493A). Expectedly, most of the ideas of 449C-449E can be applied to extremely amenable groups (493B493C); unexpectedly, we find not only that there are interesting extremely amenable groups, but that we need some of the central ideas of measure theory to study them. I give a criterion for extreme amenability of a group in terms of the existence of suitably concentrated measures (493D) before turning to four examples: measure algebras under symmetric difference (493E), L0 spaces (493F), automorphism groups of measure algebras (493H) and isometry groups of spheres in infinite-dimensional Hilbert spaces (493J). 493A Definition Let G be a topological group. Then G is extremely amenable or has the fixed point on compacta property if every continuous action of G on a compact Hausdorff space has a fixed point. 493B Proposition (a) Let G and H be topological groups such that there is a continuous surjective homomorphism from G onto H. If G is extremely amenable, so is H. (b) Let G be a topological group and suppose that there is a dense subset A of G such that every finite subset of A is included in an extremely amenable subgroup of G. Then G is extremely amenable. (c) Let G be a topological group with an extremely amenable normal subgroup H such that G/H is extremely amenable. Then G is extremely amenable. (d) The product of any family of extremely amenable topological groups is extremely amenable. proof We can use the same arguments as in 449C, with some simplifications. (a) As in 449Ca, let φ : G → H be a continuous surjective homomorphism, X a non-empty compact Hausdorff space and • : H × X → X a continuous action. Let •1 be the continuous action of G on X defined by the formula a•1 x = φ(a)•x. Then any fixed point for •1 is a fixed point for •. (b) Let X be a non-empty compact Hausdorff space and • a continuous action of G on X. For I ∈ [A]<ω let HI be an extremely amenable subgroup of G including I. The restriction of the action to HI × X is a continuous action of HI on X, so {x : a•x = x for every a ∈ I} ⊇ {x : a•x = x for every a ∈ I} is closed and non-empty. Because X is compact, there is an x ∈ X such that a•x = x for every a ∈ A. Now {a : a•x = x} includes the dense set A, so is the whole of G, and x is fixed under the action of G. As X and • are arbitrary, G is extremely amenable. (c) Let X be a compact Hausdorff space and • a continuous action of G on X. Set Q = {x : x ∈ X, a•x = x for every a ∈ H}; then Q is a closed subset of X and, because H is extremely amenable, is non-empty. Next, b•x ∈ Q for every x ∈ Q and b ∈ G. P P If a ∈ H, then a•(b•x) = (ab)•x = (bb−1 ab)•x = b•((b−1 ab)•x) = b•x, because H is normal, so b−1 ab ∈ H. As a is arbitrary, b•x ∈ Q. Q Q Accordingly we have a continuous action of G on the compact Hausdorff space Q. If b ∈ G, a ∈ H and x ∈ Q, then (ba)•x = b•x. So we have an action of G/H on Q defined by saying that •• b x = b•x for every b ∈ G and x ∈ Q, and this is continuous for the quotient topology on G/H, as in the proof of 449Cc. Because G/H is amenable, there is a point x of Q which is fixed under the action of G/H.

790

Further topics

493B

So b•x = b• •x = x for every b ∈ G, and x is fixed under the action of G. As X and extremely amenable.

•

are arbitrary, G is

(d) By (c), the product of two extremely amenable topological groups is extremely amenable, since each can be regarded as a normal subgroup of the product. It follows that the product of finitely many extremely amenable topological groups is extremely amenable. Now let hGi ii∈I be any family of extremely amenable topological groups with product G. For finite J ⊆ I let HJ be the set of those a ∈QG such that a(i) is the identity in Gi for every i ∈ I \ J. Then HJ is isomorphic (as topological group) to i∈J Gi , so is extremely amenable. Since {HJ : J ∈ [I]<ω } is an upwards-directed family of subgroups of G with dense union, (b) tells us that G is extremely amenable. 493C Proposition Let G be a topological group. Then G is extremely amenable iff there is a point in the greatest ambit Z of G (definition: 449D) which is fixed by the action of G on Z. proof Repeat the arguments of 449E(i) ⇐⇒ (ii), noting that if z0 ∈ Z is a fixed point under the action of G on Z, then its images under the canonical maps φ of 449Dd will be fixed for other actions. 493D Theorem Let G be a topological group, and B its Borel σ-algebra. Suppose that for every open neighbourhood V of the identity of G, ² > 0, finite set I ⊆ G and finite family E of zero sets in G there is a finitely additive functional ν : B → [0, 1] such that νG = 1 and (i) ν(V F ) ≥ 1 − ² whenever F ∈ E and νF ≥ 21 , (ii) |ν(aF ) − νF | ≤ ² for every F ∈ E and a ∈ I. Then G is extremely amenable. proof (a) Write M for the set of finitely additive functionals ν : B → [0, 1] such that νG = 1. If V is an open neighbourhood of the identity e of G, ² > 0, I ∈ [G]<ω and E is a finite family of zero sets in G, let A(V, ², I, E) be the set of those ν ∈ M satisfying (i) and (ii) above. Our hypothesis is that none of these sets A(V, ², I, E) are empty; since A(V, ², I, E) ⊆ A(V 0 , ²0 , I 0 , E 0 ) whenever V ⊆ V 0 , ² ≤ ²0 , I ⊇ I 0 and E ⊇ E 0 , there is an ultrafilter F on M containing all these sets. Let U be the space of bounded real-valued functionals on G which are uniformly continuous for the right uniformity on G. If we identify L∞ (B) with the space of bounded Borel measurable real-valued functions on then U is a Riesz subspace of L∞ (B). ForR each ν ∈ M we have a positive linear functional R G (363Ha), ∞ dν : L (B) → R (363L). For f ∈ U set z(f ) = limν→F f dν. (b) z : U → R is a Riesz homomorphism, and z(χG) = R1. P P Of course z is a positive linear functional taking the value 1 at χG, just because all the integrals dν are. Now suppose that f0 , f1 ∈ U and f0 ∧ f1 = 0. Take any ² > 0. Then there is an open neighbourhood V of e such that |fi (x) − fi (y)| ≤ ² whenever xy −1 ∈ V and i ∈ {0, 1}. Set Fi = {x : fi (x) = 0}, Ei = V Fi for each i. Then F0 ∪ F1 = X, so νF0 + νF1 ≥ 1 for every ν ∈ M , and there is a j ∈ {0, 1} such that A0 = {ν : νFj ≥ 21 } ∈ F. Next, A1 = {ν: if νFj ≥

1 2

then ν(V Fj ) ≥ 1 − ²}

belongs to F. Accordingly limν→F νEj ≥ 1 − ². As fj (x) ≤ ² for every x ∈ Ej , z(fj ) = limν→F

R

fj dν ≤ ²(1 + kfj k∞ ).

This shows that min(z(f0 ), z(f1 )) ≤ ²(1 + kf1 k∞ + kf2 k∞ ). As ² is arbitrary, min(z(f0 ), z(f1 )) = 0; as f0 and f1 are arbitrary, z is a Riesz homomorphism (352G(iv)). Q Q Thus z belongs to the greatest ambit Z of G. (c) a•z = z for every a ∈ G. P P Take any non-negative f ∈ U . For n ∈ N set Fn = {x : x ∈ G, f (x) ≥ n²}. Set m = b 1² kf k∞ c, so that Fn = ∅ for every n > m. Because {ν : |ν(a−1 Fn ) − νFn | ≤ δ} ∈ F for every Pm n ∈ N and every δ > 0, limν→F νFn = limν→F ν(a−1 Fn ) for every n. Next, setting P g = n=1 ²χFn , we m have g ∈ L∞ (B) Since a−1 •g (in the language of 441Ac) is just n=1 ²χ(a−1 Fn ), we R and g ≤ f ≤ gR + ²χG. −1 • have limν→F g dν = limν→F (a g)dν. But also a−1 •g ≤ a−1 •f ≤ a−1 •g + ²χG. So |

R

f dν −

for every ν ∈ M . This means that

R

g dν| ≤ ²,

R

| (a−1 •f )dν −

R

(a−1 •g)dν| ≤ ²

493E


| limν→F

R

f dν − limν→F

As ² is arbitrary, (a•z)(f ) = z(a−1 •f ) = limν→F

R

R

791

(a−1 •f )dν| ≤ 2².

(a−1 •f )dν = limν→F

R

f dν = z(f ).

This is true for every non-negative f ∈ U ; as both z and a•z are linear, (a•z)(f ) = z(f ) for every f ∈ U , and a•z = z. Q Q (d) Thus z ∈ Z is fixed under the action of G on Z; by 493C, this is enough to ensure that G is extremely amenable. 493E I turn now to examples of extremely amenable groups. The first three are groups which we have already studied for other reasons. Theorem Let (A, µ ¯) be an atomless measure algebra. Then A, with the group operation 4 and the measure-algebra topology (definition: 323A), is an extremely amenable group. proof (a) To begin with let us suppose that (A, µ ¯) is a probability algebra; write σ for the measure metric of A, so that σ(a, a0 ) = µ ¯(a 4 a0 ) for a, a0 ∈ A. I seek to apply 493D. (i) Let V be an open neighbourhood of 0 in A, ² > 0, I ∈ [A]<ω and E a finite family of zero sets in A. Let γ > 0 be such that V ⊇ {a : µ ¯a ≤ 2γ}. Let B0 be the finite subalgebra of A generated by I and B0 the set of atoms in B0 . Set t=

1 γ

3 ²

ln ,

n = dmax(t2 , supb∈B0

1 )e. µ ¯b

Because A is atomless, we can split any member of A\{0} into two parts of equal measure (331C); if, starting from the disjoint set B0 , we successively split the largest elements until we have a disjoint set B with just n elements, then we shall have µ ¯b ≤

2 n

for every b ∈ B. We have a natural identification between {0, 1}B

and the subalgebra B of A generated by B, matching x ∈ {0, 1}B with f (x) = sup{b : b ∈ B, x(b) = 1}. Writing ρ for the normalized Hamming metric on {0, 1}B (492D), we have σ(f (x), f (y)) ≤ 2ρ(x, y) for all x, y ∈ {0, 1}B . P P Set J = {b : b ∈ B, x(b) 6= y(b)}, so that P 2 ¯b ≤ #(J) = 2ρ(x, y). Q σ(f (x), f (y)) = µ ¯(f (x) 4 f (y)) = µ ¯(sup J) = b∈J µ Q n

(ii) Let λ be the usual measure on {0, 1}B and set νE = λf −1 [E] for every Borel set E ⊆ A. Then ν is a probability measure. Note that f : {0, 1}B → B is a group isomorphism if we give {0, 1}B the addition +2 corresponding to its identification with ZB 2 , and B the operation 4 . Because λ is translation-invariant for +2 , its copy, the subspace measure νB on the ν-conegligible finite set B, is translation-invariant for 4 . But this means that ν{b 4 d : d ∈ F } = νF whenever b ∈ B and F ⊆ B, and therefore that ν{b 4 d : d ∈ F } = νF whenever b ∈ I and F ∈ E. This shows that ν satisfies condition (ii) of 493D. (iii) Now suppose that F ∈ E and that νF ≥ 21 . Set W = f −1 [F ], so that λW ≥ 21 . By 492D, R tρ(x,W ) 2 e λ(dx) ≤ 2et /4n ≤ 2e1/4 ≤ 3, so λ{x : ρ(x, W ) ≥ γ} = λ{x : tρ(x, W ) ≥ ln 3² } = λ{x : etρ(x,W ) ≥ 3² } ≤ ². Accordingly ν{a4d : a ∈ V, d ∈ F } ≥ ν{a : σ(a, F ) ≤ 2γ} = λ{x : σ(f (x), F ) ≤ 2γ} ≥ λ{x : σ(f (x), f [W ]) ≤ 2γ} ≥ λ{x : ρ(x, W ) ≤ γ} (because f is 2-Lipschitz) ≥ 1 − ². So ν also satisfies (i) of 493D.

792

Further topics

493E

(iv) Since V , ², I and E are arbitrary, 493D tells us that A is an extremely amenable group, at least when (A, µ ¯) is an atomless probability algebra. (b) For the general case, observe first that if (A, µ ¯) is atomless and totally finite then (A, 4 ) is an extremely amenable group; this is trivial if A = {0}, and otherwise there is a probability measure on A which induces the same topology, so we can apply (a). For a general atomless measure algebra (A, µ ¯), set Af = {c : c ∈ A, µ ¯c < ∞} and for c ∈ Af let Ac be the principal ideal generated by c. Then Ac is a subgroup of A and the measure-algebra topology of Ac , regarded as a measure algebra in itself, is the subspace topology induced by the measure-algebra topology of A. So {Ac : c ∈ Af } is an upwards-directed family of extremely amenable subgroups of A with union which is dense in A, so A itself is extremely amenable, by 493Bb. This completes the proof. 493F Theorem (Pestov p01) Let (X, Σ, µ) be an atomless measure space. Then L0 (µ), with the group operation + and the topology of convergence in measure, is an extremely amenable group. proof It will simplify some of the formulae if we move at once to the space L0 (A), where (A, µ ¯) is the measure algebra of (X, Σ, µ); for the identification of L0 (A) with L0 (µ) see 364Jc; for a note on convergence in measure in L0 (A), see 367M; of course A is atomless if (X, Σ, µ) is (322Bg). (a) I seek to prove that S(A), with the group operation of addition and the topology of convergence in measure, is extremely amenable. As in 493E, I start with the case in which (A, µ ¯) is a probability algebra, and use 493D. (i) Take an open neighbourhood V of 0 in S(A), an ² > 0, a finite set I ⊆ S(A) and a finite family E of zero sets in S(A). Let γ > 0 be such that u ∈ V whenever u ∈ S(A) and µ ¯[[u 6= 0]] ≤ 2γ. Let B0 be a finite subalgebra of A such that I is included in the linear subspace of S(A) generated by {χb : b ∈ B0 }, and B0 the set of atoms of B0 . As in the proof of 493E, set t=

1 γ

3 ²

ln ,

n = dmax(t2 , supb∈B0

1 )e, µ ¯b

and let B ⊆ A \ {0} be a partition of unity with n elements, refining B0 , such that µ ¯b ≤ n2 for every b ∈ B. B We have a natural identification of S(A) generated by {χb : b ∈ B}, P between R and the linear subspace B matching x ∈ R with f (x) = b∈B x(b)χb, which is continuous if RB is given its product topology. Writing ρ for the normalized Hamming metric on RB , we have P 2 ¯b ≤ #({b : x(b) 6= y(b)}) = 2ρ(x, y) µ ¯[[f (x) 6= f (y)]] = x(b)6=y(b) µ n

B

for all x, y ∈ R . (ii) Set β = supv∈I kvk∞ (if I = ∅, take β = 0). Let M > 0 be so large that (M + β)n ≤ (1 + 12 ²)M n . On R, write µL for Lebesgue measure and µ0L for the indefinite-integral measure over µL defined by the function

1 χ[−M, M ], 2M

so that µ0L E =

0

1 µL (E ∩ [−M, M ]) 2M

whenever E ⊆ R and E ∩ [−M, M ] is Lebesgue

B

measurable. Let λ, λ be the product measures on R defined from µL and µ0L . Let ν be the Borel probability measure on S(A) defined by setting νF = λ0 f −1 [F ] for every Borel set F ⊆ S(A). Now |ν(v +F ) −νF | ≤ ² for every v ∈ I and Borel set F ⊆ S(A). P P Because B refines B0 , v is expressible as f (y) for some y ∈ RB ; because kvk∞ ≤ β, |y(b)| ≤ β for every b ∈ B. Because f : RB → S(A) is linear, f −1 [v + F ] = y + f −1 [F ]. Now

|ν(v + F ) − νF | = |λ0 f −1 [v + F ] − λ0 f −1 [F ]| ¯ 1 ¯¯ λ(f −1 [v + F ] ∩ [−M, M ]n ) − λ(f −1 [F ] ∩ [−M, M ]n )¯ = n (2M )

(use 253I, or otherwise)

493G


=

1 ¯¯ λ((y (2M )n

=

1 ¯¯ λ(f −1 [F ] ∩ ([−M, M ]n (2M )n

≤

1 λ(([−M, M ]n (2M )n

− y)4[−M, M ]n )

=

2 λ(([−M, M ]n (2M )n

− y) \ [−M, M ]n )

≤

2 λ(([−M (2M )n

=

2 ((M Mn

793

¯ + f −1 [F ]) ∩ [−M, M ]n ) − λ(f −1 [F ] ∩ [−M, M ]n )¯ ¯ − y)) − λ(f −1 [F ] ∩ [−M, M ]n )¯

− β, M + β]n \ [−M, M ]n )

+ β)n − M n ) ≤ ². Q Q

So ν satisfies (ii) of 493D. (iii) Now suppose that F ∈ E and νF ≥ 21 . Set W = f −1 [F ], so that λ0 W ≥ 21 . Just as in the proof of R 2 493E, etρ(x,W ) λ0 (dx) ≤ 2et /4n ≤ 3, so λ0 {x : ρ(x, W ) ≥ γ} = λ0 {x : etρ(x,W ) ≥ 3² } ≤ ², and ν{v + u : v ∈ V, u ∈ F } ≥ ν{w : ∃ u ∈ F, µ ¯[[u 6= w]] ≤ 2γ} ≥ λ0 {x : ρ(x, W ) ≤ γ} ≥ 1 − ². So ν also satisfies (i) of 493D. (iv) Since V , ², I and E are arbitrary, 493D tells us that S(A) is an extremely amenable group, at least when (A, µ ¯) is an atomless probability algebra. (b) The rest of the argument is straightforward, as in 493E. First, S(A) is extremely amenable whenever (A, µ ¯) is an atomless totally finite measure algebra. For a general atomless measure algebra (A, µ ¯), set Af = {c : µ ¯c < ∞}. For each c ∈ Af , let Ac be the corresponding principal ideal of A. Then we can identify S(Ac ), as topological group, with the linear subspace of L0 (A) generated by {χa : a ∈ Ac }, and it is extremely amenable. Since {S(Ac ) : c ∈ Af } is an upwards-directed family of extremely amenable subgroups of L0 (A) with dense union in L0 (A), L0 (A) itself is extremely amenable, by 493Bb, as before. 493G I take the proof of the next theorem in two parts, the first being the leading special case. ¯ its measure algebra with its Lemma Let κ be an infinite cardinal, λ the usual measure on {0, 1}κ and (B, λ) measure metric. Let Autλ¯ B be the group of measure-preserving automorphisms of B with the topology T of pointwise convergence inherited from the isometry group of B (441G). Then Autλ¯ B is extremely amenable. proof (a) For each I ∈ [κ]<ω , let HI be the group of bijections from {0, 1}I to itself. For q ∈ HI let fq : {0, 1}κ → {0, 1}κ be the measure space automorphism defined by saying that fq (x)¹κ \ I = x¹κ \ I,

fq (x)¹I = q(x¹I)

for every x ∈ {0, 1}κ , and ψq the member of Autλ¯ B defined by the formula ψq (E • ) = fq [E]• for every measurable E ⊆ {0, 1}κ . Note that fqq0 = fq fq0 for all q, q 0 ∈ HI , so that q 7→ ψq : HI → Autλ¯ B is a group homomorphism. Set GI = {ψq : q ∈ HI }, so that GI is a subgroup of Autλ¯ B. If I ⊆ J ∈ [κ]<ω then GI ⊆ GJ . P P For any q ∈ HI we can define q 0 ∈ HJ by setting q 0 (z)¹J \ I = z¹J \ I,

q 0 (z)¹I = q(z¹I)

J for every z ∈ fq0 = fq so ψq0 = ψq . As q is arbitrary, GI ⊆ GJ . Q Q S{0, 1} ; then<ω Set G = {GI : I ∈ [κ] }, so that G is a subgroup of Autλ¯ B.

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(b) Let V be a neighbourhood of the identity of G (with the subspace topology inherited from Autλ¯ B), ² > 0 and K ⊆ G a finite set. Then there is a (purely atomic) Borel probability measure ν on G such that (i) ν(πB) = νB for every B ⊆ G and every π ∈ K, (ii) ν(V B ∩ G) ≥ 1 − ² whenever B ⊆ G and νB ≥ 12 . ¯ 4 πa) ≤ δ for every P P There are a finite set A ⊆ B and a δ > 0 such that π ∈ r V whenever π ∈ G and λ(a 1 a ∈ A. Let I ∈ [κ]<ω be such that K ⊆ GI and 2n δ − 1 ≥ 2n+6 ln , where n = #(I). (This is where we ²

need to know that κ is infinite.) Let ν be the Borel probability measure on G such that ν{ψq } = 1/(2n )! for every q ∈ HI . Since ν acts on GI as its Haar probability measure, ν(πB) = ν(GI ∩ πB) = ν(π(GI ∩ B)) = ν(GI ∩ B) = νB for every B ⊆ G and every π ∈ K ⊆ GI . Now suppose that B ⊆ G and νB ≥ 21 . Set C = {q : q ∈ HI , ψq ∈ B}, so that #(C) ≥ 12 #(HI ) = 12 (2n )!. Set r = b2n δc, so that exp(−r2 /2n+6 ) ≤ ². Let C˜ be the set of those q˜ ∈ HI such that there is some q ∈ C with #({z : z ∈ {0, 1}I , q˜(z) 6= q(z)}) ≤ r; then ˜ ≥ (1 − exp(−r2 /2n+6 ))(2n )! ≥ (1 − ²)(2n )! #(C) ˜ νB 0 ≥ 1 − ². Now if π ∈ B 0 , there are q˜ ∈ C, ˜ q ∈ C such that π = ψq˜ by 492I. Writing B 0 = {ψq : q ∈ C}, I and #({z : z ∈ {0, 1} , q˜(z) 6= q(z)}) ≤ r. But this means that ψq ∈ B, while #({z : q −1 (z) 6= q˜−1 (z)}) = #({z : q˜q −1 (z) 6= qq −1 (z)}) = #({z : q˜(z) 6= q(z)}) ≤ r. Setting E0 = {x : x ∈ {0, 1}κ , q˜−1 (x¹I) 6= q −1 (x¹I)}, we see that λE0 ≤ 2−n r ≤ δ and fq−1 (x) = fq˜−1 (x) for every x ∈ {0, 1}κ \ E0 , so that fq [E]4fq˜[E] ⊆ E0 for every E ⊆ {0, 1}κ . Accordingly πa 4 ψq a ⊆ E0• for every a ∈ B, and ¯ 4 πψq−1 a) = λ(π(ψ ¯ ¯ • λ(a q −1 a) 4 ψq (ψq −1 a)) ≤ λE ≤ δ 0

πψq−1

for every a ∈ A. This shows that ∈ V so that π ∈ V B. As π is arbitrary, B 0 ⊆ V B and ν(G ∩ V B) ≥ 0 νB ≥ 1 − ². As B is arbitrary, ν satisfies the condition (ii). Q Q P Take any π ∈ Autλ¯ B and any (c) By 493D, G is extremely amenable. Now G is dense in Autλ¯ B. P neighbourhood V of π in Autλ¯ B. Then there are a0 , . . . , ak ∈ B and ² > 0 such that ¯ V 0 = {ψ : ψ ∈ Aut ¯ B, λ(ψa i 4 πai ) ≤ ² for every i ≤ k} λ

is included in V . Let b0 , . . . , bm be the atoms of the finite subalgebra of B generated by {ai : i ≤ k} ∪ {πai : i ≤ k}. Set δ = ²/12(m+1)(2m+1) > 0. For j ≤ m, let Ej , Fj ⊆ {0, 1}κ be such that Ej• = bj and Fj• = πbj ; let Ej0 , Fj0 be open-and-closed sets such that λ(E4Ej0 ) + λ(Fj 4Fj0 ) ≤ δ. Let I ⊆ κ be a finite set S such that Ej0 and Fj0 are determined by coordinates in I for every j ≤ m. For each j ≤ m, set Ej00 = Ej0 \ i
˜j ) ≤ 6(m + 1)δ λ(Ej 4E

for each j. Similarly λ(Fj 4F˜j ) ≤ 6(m + 1)δ for each j. ˜j is determined by coordinates in I, it is of the form {x : x¹I ∈ Cj } for some Cj ⊆ {0, 1}I . Because E ˜ ˜m are disjoint, so are C0 , . . . , Cm . Similarly, F˜j can be expressed as {x : x¹I ∈ Dj } where Because E0 , . . . , E ˜j = λF˜j , #(Cj ) = #(Dj ) for each j. There is therefore a D0 , . . . , Dm ⊆ {0, 1}I are disjoint. Because λE ˜ • = F˜ • , so q ∈ HI such that q[Cj ] = Dj for each j. Now ψq E j

j

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¯ q bj 4 πbj ) ≤ λ(ψ ¯ q bj 4 F˜ • ) + λ( ¯ F˜ • 4 F • ) λ(ψ j j j • ¯ ˜ ˜ = λ(bj 4 Ej ) + λ(Fj 4Fj ) ˜j ) + 6(2m + 1)δ ≤ 12(2m + 1)δ ≤ λ(Ej 4E for each j. Since every ai is a union of some of the bj , ¯ q ai 4 πai ) ≤ 12(2m + 1)(m + 1)δ = ² λ(ψ for every i ≤ k, and ψq ∈ V 0 ⊆ V . Thus G ∩ V 6= ∅; as V is arbitrary, G is dense in Autλ¯ B. Q Q (d) By 493Bb, with H = {G}, Autλ¯ B is extremely amenable. 493H Theorem (Giordano & Pestov 02) Let (A, µ ¯) be an atomless measure algebra, and set Af = {a : µ ¯a < ∞}. Let Autµ¯ A be the group of measure-preserving automorphisms of A. For π, φ ∈ Autµ¯ A and c ∈ Af , set σc (π, φ) = µ ¯(πc 4 φc). Then {σc : c ∈ Af } is a family of pseudometrics on Autµ¯ A. Under the topology it induces, Autµ¯ A is an extremely amenable topological group. proof (a) It is elementary to check that every σc is a pseudometric. To see that the topology T generated by these pseudometrics is compatible with the group structure, recall that (a, b) 7→ µ ¯(a 4 b) is a metric on Af (323Ae). Now if we set θ(π) = π¹Af , then θ is a group homomorphism from Autµ¯ A to the isometry group Isom(Af ) of Af , and T is just {θ−1 [V ] : V ∈ S}, where S is the topology of pointwise convergence in Isom(Af ). Since (Isom(Af ), S) is a topological group (441Ga), so is (Autµ¯ A, T). (b) For a ∈ A write Aa for the principal ideal of A generated by a and Ga for the subgroup of those π ∈ Autµ¯ A which are supported by a (definition: 381Ba), so that Ga is isomorphic (as topological group) to the group of measure-preserving automorphisms of Aa . (c) Let A be the set of those a ∈ Af such that Aa is homogeneous. Then Ga is extremely amenable for every a ∈ A. P P If a = 0 this is trivial, since Ga has just one element. Otherwise, Aa is isomorphic, up to a scalar multiple of the measure, to the measure algebra B of the usual measure λ on {0, 1}κ where κ is infinite, by Maharam’s theorem (331L). So Ga is isomorphic, as topological group, to Autλ¯ B, which is extremely amenable, by 493G. Q Q (d) Set I = {a0 ∪ . . . ∪ an : a0 , . . . , an ∈ A}. Then Gb is extremely amenable for every b ∈ I. P P Since Aa∪b is homogeneous whenever Aa , Ab are homogeneous and of the same infinite Maharam type, we can express b as a0 ∪ . . . ∪ an where all the Aai are homogeneous and have different Maharam types. (This is really where I use the hypothesis that A is atomless.) In this case, Gb will be isomorphic (as topological Qk group) to i=0 Gai , so is extremely amenable (493Bd). Q Q Note that as πa ∈ A whenever a ∈ A and π ∈ Autµ¯ A, πb ∈ I whenever b ∈ I and π ∈ Autµ¯ A. Also a0 ∈ A whenever a ∈ A and a0 ⊆ a, so I is an ideal of A. S P Suppose that π ∈ Autµ¯ A, a0 , . . . , an ∈ Af and (e) Set G = b∈I Gb . Then G is dense in Autµ¯ A. P ² > 0. Set c = supi≤n ai . By 332A, there is a b0 ∈ I such that b0 ⊆ c and µ ¯(c \ b0 ) ≤ 12 ². Set b = b0 ∪ πb0 ∈ I. Then there is a φ ∈ Gb such that φd = πd for every d ⊆ b0 (332L). Now φai 4 πai ⊆ φ(ai \ b0 ) ∪ π(ai \ b0 ) has measure at most 2¯ µ(c \ b0 ) ≤ ² for every i ≤ n. As π, a0 , . . . , an and ² are arbitrary, G is dense in Autµ¯ A. Q Q (f ) Thus {Gb : b ∈ I} is an upwards-directed family of extremely amenable subgroups of Autµ¯ A with dense union. By 493Bb, Autµ¯ A is extremely amenable. 493I Returning to the ideas of §476, we find another remarkable example of an extremely amenable topological group. I recall the notation of 476M. Let X be a (real) inner product space. SX will be the unit sphere {x : x ∈ X, kxk = 1}. Let HX be the isometry group of SX with its topology of pointwise convergence. When X is finite-dimensional, it is isomorphic, as inner product space, to Rr , where r = dim X. In this case SX is compact, so (if r ≥ 1) has a unique HX -invariant Radon probability measure νX , which is

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strictly positive, and is a multiple of (r − 1)-dimensional Hausdorff measure; also HX is compact (441Gb), so has a unique Haar probability measure λX . Lemma For any m ∈ N and any ² > 0, there is an r(m, ²) ≥ max(2, m) such that whenever X is a finitedimensional inner product space of dimension at least r(m, ²), x0 , . . . , xm−1 ∈ SX , Q1 , Q2 ⊆ HX are closed sets and min(λX Q1 , λX Q2 ) ≥ ², then there are f1 ∈ Q1 , f2 ∈ Q2 such that kf1 (xi ) − f2 (xi )k ≤ ² for every i < m. proof Induce on m. For m = 0, the result is trivial. For the inductive step to m + 1, take r(m + 1, ²) > ∗ ∗ A1 , νX A2 ) ≥ 21 ² then r(m, 31 ²) such that whenever r(m + 1, ²) ≤ dim X < ω and A1 , A2 ⊆ SX and min(νX 1 there are x ∈ A1 , y ∈ A2 such that kx − yk ≤ 3 ²; this is possible by 476P. Now take any inner product space X of finite dimension r ≥ r(m + 1, ²), closed sets Q1 , Q2 ⊆ HX such that min(λX Q1 , λX Q2 ) ≥ ², and orthonormal x0 , . . . , xm ∈ SX . Let Y be the (r − 1)-dimensional subspace {x : x ∈ X, (x|xm ) = 0}, so that dim Y ≥ r(m, 31 ²), and for i < m let yi ∈ Y be a unit vector such that xi is a linear combination of yi and xm . Set HY0 = {f : f ∈ HX , f (xm ) = xm }; then f 7→ f ¹SY is a topological group isomorphism from HY0 to HY . P P (i) If f ∈ HY0 and x ∈ SX , then x ∈ SY ⇐⇒ (x|xm ) = 0 ⇐⇒ (f (x)|f (xm )) = 0 ⇐⇒ f (x) ∈ SY , so f ¹SY is a bijection from SY to itself and belongs to HY . (ii) If g ∈ HY , we can define f ∈ HY0 by setting f (αx + βxm ) = αg(x) + βxm whenever x ∈ SY and α2 + β 2 = 1, and f ¹SY = g. (iii) Note that HY0 is a closed subgroup of HX , so in itself is a compact Hausdorff topological group. Since the map f 7→ f ¹SY : HY0 → HY is a bijective continuous group homomorphism between compact Hausdorff topological groups, it is a topological group isomorphism. Q Q R 0 0 Let λY be the Haar probability measure of HY . Then λX Q1 = λ0Y (HY0 ∩ f −1 Q1 )λX (df ) (443Te), so λ0X Q1 ≥ 21 ², where Q01 = {f : λ0Y (HY0 ∩ f −1 Q1 ) ≥ 12 ²}. Similarly, setting Q02 = {f : λ0Y (HY0 ∩ f −1 Q2 ) ≥ 12 ²}, λX Q02 ≥ 21 ². Next, setting θ(f ) = f (xm ) for f ∈ HX , λX θ−1 is an X-invariant Radon probability measure on SX (443Ta), so must be equal to νX . Accordingly 1 2

νX (θ[Q0j ]) = λX (θ−1 [θ[Q0j ]]) ≥ λX Q0j ≥ ² for both j. We chose r(m + 1, ²) so large that we can be sure that there are z1 ∈ θ[Q01 ], z2 ∈ θ[Q02 ] such that kz1 − z2 k ≤ 31 ². Let h1 ∈ Q01 , h2 ∈ Q02 be such that h1 (xm ) = θ(h1 ) = z1 and h2 (xm ) = z2 . Let h ∈ HX be ˜ 2 = hh1 , so that h ˜ 2 (xm ) = z2 such that h(z1 ) = z2 and kh(x) − xk ≤ 13 ² for every x ∈ SX (4A4Jf). Set h −1 1 ˜ 2 (x)k ≤ ² for every x ∈ SX . Note that h ˜ h2 ∈ H 0 , so that h ˜ 2 and h2 belong to the same and kh1 (x) − h 2 Y 3 left coset of HY0 , and 1 2

˜ −1 Q2 ) = λ0 (H 0 ∩ h−1 Q2 ) ≥ ² λ0Y (HY0 ∩ h 2 2 Y Y by 443Pa. At this point, recall that dim Y ≥ r(m, 31 ²), and that λ0Y is a copy of λY . So we have g1 ∈ HY0 ∩ h−1 1 Q1 , ˜ −1 Q2 such that kg1 (yi ) − g2 (yi )k ≤ 1 ² for every i < m. We have f1 = h1 g1 ∈ Q1 and g2 ∈ HY0 ∩ h 2 3 ˜ 2 g2 ∈ Q2 . For any i < m, f2 = h ˜ 2 g2 (yi )k kf1 (yi ) − f2 (yi )k ≤ kh1 g1 (yi ) − h1 g2 (yi )k + kh1 g2 (yi ) − h 1 3

2 3

≤ kg1 (yi ) − g2 (yi )k + ² ≤ ². Also g1 (xm ) = g2 (xm ) = xm , so ˜ 2 (xm )k ≤ 1 ². kf1 (xm ) − f2 (xm )k = kh1 (xm ) − h 3

If i < m, then xi = (xi |xm )xm + (xi |yi )yi , so fj (xi ) = (xi |xm )fj (xm ) + (xi |yi )fj (yi ) for both j (476N) and

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797

2

kf1 (xi ) − f2 (xi )k ≤ ²|(xi |xm )| + ²|(xi |yi )| 3 3 r p 1 2 ≤ ( ²)2 + ( ²)2 (xi |xm )2 + (xi |yi )2 ≤ ². 3

3

So f1 and f2 witness that the induction proceeds. 493J Theorem Let X be an infinite-dimensional inner product space. Then the isometry group HX of its unit sphere SX , with its topology of pointwise convergence, is extremely amenable. proof I use the same approach as in 493G. (a) Let Y be the family of finite-dimensional subspaces of X. For Y ∈ Y, write Y ⊥ for the orthogonal complement of Y , so that X = Y ⊕ Y ⊥ (4A4Je). For q ∈ HY define θY (q) : SX → SX by saying that θY (q)(αy + βz) = αq(y) + βz whenever y ∈ SY , z ∈ SY ⊥ and α2 + β 2 = 1. Then θY : HY → HX is a injective group homomorphism. Also it is continuous, because q 7→ αq(y) + βz is continuous for all relevant α, β, y and z. If Y , W ∈ Y and Y ⊆ W then θY [HY ] ⊆ θW [HW ]. P P For any q ∈ HY we can define q 0 ∈ HW by saying 0 that q (αy + βx) = αq(y) + βx whenever y ∈ SY , x ∈ SW ∩Y ⊥ and α2 + β 2 = 1. Now θY (q) = θW (q 0 ) ∈ θW [HW ]. Q QS Set G∗ = Y ∈Y θY [HY ], so that G∗ is a subgroup of HX . (b) Let V be an open neighbourhood of the identity in G∗ (with the subspace topology inherited from the topology of pointwise convergence on HX ), ² > 0 and I ⊆ G∗ a finite set. Then there is a Borel probability measure λ on G∗ such that (i) λ(f Q) = λQ for every f ∈ I and every closed set Q ⊆ G∗ , (ii) λ(V Q) ≥ 1 − ² whenever Q ⊆ G∗ is closed and λQ ≥ 12 . P P We may suppose that ² ≤ 21 . Let J ∈ [SX ]<ω and δ > 0 be such that f ∈ V whenever f ∈ G∗ and kf (x) − xk ≤ δ for every x ∈ J. We may suppose that J is non-empty; set m = #(J). Let Y ∈ Y be such that J ⊆ Y and I ⊆ θY [HY ] and dim Y = r ≥ r(m, ²), as chosen in 493I. (This is where we need to know that X is infinite-dimensional.) Set λF = λY θY−1 [F ] for every Borel set F ⊆ G∗ . If f ∈ I and F ⊆ G∗ is closed, then λ(f F ) = λY θY−1 [f F ] = λY (θY−1 (f )θY−1 [F ]) = λY θY−1 [F ] = λF . So λ satisfies condition (i). ?? Suppose, if possible, that Q1 ⊆ G∗ is a closed set such that λQ1 ≥ 12 and λ(V Q1 ) < 1 − ²; set Q2 = G∗ \ V Q1 . Then θY−1 [Q1 ] and θY−1 [Q2 ] are subsets of HY both of measure at least ². Set Rj = {q : q ∈ HY , q −1 ∈ θY−1 [Qj ]} for each j; because HY is compact, therefore unimodular, λY Rj = λY θY−1 [Qj ] = λQj ≥ ² for both j. Because dim Y ≥ r(m, ²), there are q1 ∈ R1 , q2 ∈ R2 such that kq1 (x) − q2 (x)k ≤ ² for x ∈ J. Set f = θY (q2−1 q1 ). If x ∈ J, then kf (x) − xk = kq2−1 q1 (x) − xk = kq1 (x) − q2 (x)k ≤ ². As this is true for every x ∈ J, f ∈ V . On the other hand, θY (q1−1 ) ∈ Q1 and θY (q2−1 ) ∈ Q2 and f θY (q1−1 ) = θY (q2−1 ), so θY (q2−1 ) ∈ V Q1 ∩ Q2 , which is impossible. X X Thus λ satisfies (ii). Q Q (c) By 493D, G∗ is extremely amenable. But G∗ is dense in HX . P P If f ∈ HX and I ⊆ SX is finite and not empty, let Y1 be the linear subspace of X generated by I, and let (y1 , . . . , ym ) be an orthonormal basis of Y1 . Set zj = f (yj ) for each j, so that (z1 , . . . , zm ) is orthonormal (476N); let Y be the linear subspace of X generated by y1 , . . . , ym , z1 , . . . , zm . Set r = dim Y and extend the orthonormal sets (y1 , . . . , ym ) and (z1 , . . . , zm ) to orthonormal bases (y1 , . . . , yr ) and (z1 , . . . , zr ) of Y . Then we have an isometric linear operator T : Y → Y defined by saying that T yi = zi for each i; set q = T ¹SY ∈ HY . By 476N, q(x) = f (x) for every x ∈ I, so θY (q) agrees with f on I, while θY (q) ∈ G∗ . As f and I are arbitrary, G∗ is dense in G. Q Q So 493Bb tells us that HX is extremely amenable, and the proof is complete.

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493K

493K The following result shows why extremely amenable groups did not appear in Chapter 44. Theorem (Veech 77) If G is a locally compact Hausdorff topological group with more than one element, it is not extremely amenable. proof If G is compact, this is trivial, since the left action of G on itself has no fixed point; so let us assume henceforth that G is not compact. (a) Let Z be the greatest ambit of G, a 7→ a ˆ : G → Z the canonical map, and U the space of bounded right-uniformly continuous real-valued functions on G. (I aim to show that the action of G on Z has no fixed point.) Take any z ∗ ∈ Z. Let V0 be a compact neighbourhood of the identity e in G, and let B0 ⊆ G be a maximal set such that V0 b ∩ V0 c = ∅ for all distinct b, c ∈ B0 . Then for any a ∈ G there is a b ∈ B0 such that V0 a ∩ V0 b 6= ∅, that is, a ∈ V0−1 V0 B0 . So if we set Y0 = {ˆb : b ∈ B} ⊆ Z, {a•y : a ∈ V0−1 V0 , y ∈ Y0 } is a compact subset of Z including {ˆ a : a ∈ G}, and is therefore the whole of Z (449Dc). Let a0 ∈ V0−1 V0 , ∗ ˆ y0 ∈ Y0 be such that a0 •y0 = z ∗ , and set B1 = a0 B0 , V1 = a0 V0 a−1 0 ; then z ∈ {b : b ∈ B1 } and V1 b ∩ V1 c = ∅ for all distinct b, c ∈ B1 . (b) Because V1 is compact and G is not compact, there is an a1 ∈ G\V1 . Let V2 ⊆ V1 be a neighbourhood of e such that a1−1 V2 V2−1 a1 ⊆ V1 . Then we can express B1 as D0 ∪ D1 ∪ D2 where a1 Di ∩ V2 Di = ∅ for all i. P P Consider {(b, c) : b, c ∈ B1 , a1 b ∈ V2 c}. Because V2 c ∩ V2 c0 ⊆ V1 c ∩ V1 c0 = ∅ for all distinct c, c0 ∈ B1 , this is the graph of a function h : D → B1 for some D ⊆ B1 . ?? If h is not injective, we have distinct b, c ∈ B1 and d ∈ B1 such that a1 b and a1 c both belong to V2 d. But in this case b and c both belong to a−1 1 V2 d −1 −1 and bc−1 ∈ a−1 V dd V a ⊆ V and b ∈ V c, which is impossible. X X At the same time, if b ∈ B 2 1 1 1 1 , then 1 2 a1 b ∈ / V2 b because a1 ∈ / V2 , so h(b) 6= b for every b ∈ D. Let D0 ⊆ D be a maximal set such that h[D0 ] ∩ D0 = ∅, and set D1 = h[D0 ], D2 = B1 \ (D0 ∪ D1 ). Then h[D0 ] ∩ D0 = ∅ by the choice of D0 ; h[D ∩ D1 ] ∩ D1 = ∅ because h is injective and D1 ⊆ h[D \ D1 ]; and h[D ∩ D2 ] ⊆ D0 because if b ∈ D ∩ D2 there must have been some reason why we did not put b into D0 , and it wasn’t because b ∈ h[D0 ] or because h(b) = b. So h[Di ] ∩ Di = ∅ for all i, which is what was required. Q Q (c) Since z ∗ ∈ {ˆb : b ∈ B1 }, there must be some j ≤ 2 such that z ∗ ∈ {ˆb : b ∈ Dj }. Now recall that the right uniformity on G, like any uniformity, can be defined by some family of pseudometrics (4A2Ja). There is therefore a pseudometric ρ on G such that W² = {(a, b) : a, b ∈ G, ρ(a, b) ≤ ²} is a member of the right uniformity on G for every ² > 0 and W1 ⊆ {(a, b) : ab−1 ⊆ V2 }. If now we set f (a) = min(1, ρ(a, Dj )) = min(1, inf{ρ(a, b) : b ∈ Dj }) for a ∈ G, f : G → R is bounded and uniformly continuous for the right uniformity, so belongs to U . On the other hand, if b, c ∈ Dj , then a1 b ∈ / V2 c, that is, a1 bc−1 ∈ / V2 and ρ(a1 b, c) > 1; as c is arbitrary, f (a1 b) = 1. (d) Now ˆb(f ) = f (b) = 0 for every b ∈ Dj , so z ∗ (f ) = 0. On the other hand, because z 7→ a1 •z is continuous, a1 •z ∗ ∈ {a1 •ˆb : b ∈ Dj } = {ac 1 b : b ∈ Dj }, so (a1 •z ∗ )(f ) ≥ inf b∈Dj ac 1 b(f ) = inf b∈Dj f (a1 b) = 1, so a1 •z ∗ 6= z ∗ . As z ∗ is arbitrary, this shows that the action of G on Z has no fixed point, and G is not extremely amenable. 493X Basic exercises (a) Show that a dense subgroup of an extremely amenable topological group is extremely amenable. b its completion with respect to its bilateral uniformity. (b) Let G be a Hausdorff topological group, and G b is. Show that G is extremely amenable iff G >(c) Let X be a set with more than one member and ρ the metric on X such that ρ(x, y) = 1 for all distinct x, y ∈ X. Let G be the isometry group of X with the topology of pointwise convergence. Show that G is not extremely amenable. (Hint: give X a total ordering ≤, and let x, y be any two points of X.

493Yd


799

For a ∈ G set f (a) = 1 if a−1 (x) < a−1 (y), −1 otherwise. Show that, in the language of 449D, f ∈ U . Show that if (← x− y) is the transposition exchanging x and y then (← x− y)•f = −f , while |z(f )| = 1 for every z in the greatest ambit of G.) (Compare 449Xc.) (d) Show that under the conditions of 493D (i) there is a finitely additive functional ν : B → [0, 1] such that ν(aF ) = νF for every a ∈ G and every zero set F ⊆ G, while ν(V F ) = 1 whenever V is a neighbourhood of the identity, F is a zero set and νF > 21 (ii) there is a positive linear functional p : Cb (G) → R such that p(a•f ) = p(f ) for every f ∈ Cb (G) and a ∈ G. (e) Let (A, µ ¯) be a σ-finite measure algebra of countable Maharam type. Show that Autµ¯ A, with the topology described in 493H, is a Polish group. (Hint: 441Xm.) (f ) Let X be any inner product space. Show that the isometry group of X, with its topology of pointwise convergence, is amenable. (Hint: 449Cd.) (g) Let X be a separable Hilbert space. Show that the isometry group of its unit sphere, with its topology of pointwise convergence, is a Polish group. (h) Let G be a topological group carrying Haar measures. Show that it is extremely amenable iff its topology is the indiscrete topology. (Hint: 443L.) 493Y Further exercises (a) For a Boolean algebra A and a group G with identity e, write S(A; G) for the set of partitions of unity hag ig∈G in A such that {g : ag 6= e} is finite. For hag ig∈G , hbg ig∈G ∈ S(A), write hag ig∈G · hbg ig∈G = hcg ig∈G where cg = sup{ah ∩ bh−1 g : h ∈ G} for g ∈ G. (i) Show that under this operation S(A; G) is a group. (ii) Show that if we write hχa for the member hag ig∈G of S(A; G) such that ah = a and ag = 0 for other g ∈ G, then gχa · hχb = (gh)χ(a ∩ b), and S(A; G) is generated by {gχa : g ∈ G, a ∈ A}. (iii) Show that if A = Σ/I where Σ is an algebra of subsets of a set X and I is an ideal of Σ, then S(A; G) can be identified with a space of equivalence classes in a suitable subgroup of GX . (iv) Devise a universal mapping theorem for the construction S(A; G) to match 361F in the case (G, ·) = (R, +). (v) Now suppose that (A, µ ¯) is a measure algebra and that G is a topological group. Show that we have a topology on S(A, G), making it a topological group, for which basic neighbourhoods of the identity eχ1 are of the form V (c, ², U ) = {hag ig∈G : µ ¯(c ∩ supg∈G\U ag ) ≤ ²} with µ ¯c < ∞, ² > 0 and U a neighbourhood of the identity in G. (vi) Show that if G is an amenable locally compact Hausdorff group and (A, µ ¯) is an atomless measure algebra, then S(A; G) is extremely amenable. (Hint: Pestov p01.) *(vi) Explore possible constructions of spaces L0 (A; G). (See Hartman & Mycielski 58.) (b) Let (A, µ ¯) be an atomless totally finite measure algebra; give A its measure metric. Show that the isometry group of A, with its topology of pointwise convergence, is extremely amenable. (Hint: every isometry of A is of the form a 7→ c 4 πa, where c ∈ A and π ∈ Autµ¯ A; now use 493Bc.) (c) Let (A, µ ¯) be an atomless probability algebra and π : A → A an ergodic measure-preserving automorphism. (i) Show that for any n ≥ 1 there is a periodic measure-preserving automorphism φ of period n and support 1 such that the support of φ−1 π has measure at most

2 . n

(Hint: 386C.) (ii) Show that

−1

{φπφ : φ ∈ Autµ¯ A} is dense in Autµ¯ A with its pointwise topology as described in 493H. (Hint: any two automorphisms with period n and support 1 are conjugate.) (d) Let (A, µ ¯) be the measure algebra of Lebesgue measure on [0, 1], and Autµ¯ A the Polish group of measure-preserving automorphisms of A (493Xe). (i) Show that the entropy function h of 385M is Borel measurable. (Hint: 385Xr.) (ii) Show that the set of ergodic measure-preserving automorphisms is a Gδ set in Autµ¯ A. (Hint: let D ⊆ A be a countable dense set. Show that π ∈ Autµ¯ A is ergodic iff 1 Pn i limn→∞ k ¯d · χdk1 = 0 for every d ∈ D.) i=0 χ(π d) − µ n+1

800

Further topics

493 Notes

493 Notes and comments In writing this section I have relied heavily on Pestov 99 and Pestov p01, where you may find many further examples of extremely amenable groups. It is a striking fact that while the theories of locally compact groups and extremely amenable groups are necessarily almost entirely separate (493K), both are dependent on measure theory. Curiously, what seems to have been the first non-trivial extremely amenable group to be described was found in the course of investigating the Control Measure Problem (Herer & Christensen 75). The theory of locally compact groups has for seventy years now been a focal point for measure theory. Extremely amenable groups have not yet had such an influence. But they encourage us to look again at concentration-of-measure theorems, which are of the highest importance for quite separate reasons. In all four of the principal examples of this section we need concentration of measure in product spaces (493E493F), permutation groups (493H) or on spheres in Euclidean space (493J). 493E and 493F are special cases of a general result in Pestov p01 (493Ya(vi)) which itself extends an idea from Glasner 98. I note that 493E needs only concentration of measure in {0, 1}I , while 493F demands something rather closer to the full strength of Talagrand’s theorem 492D. I have expressed 493J as a theorem about the isometry groups of spheres in infinite-dimensional inner product spaces; of course these are isomorphic to the orthogonal groups of the whole spaces with their strong operator topologies (476Xi). Adapting the basic concentration-of-measure theorem 476O to the required lemma 493I involves an instructive application of ideas from §443.

494 Cubes in product spaces I offer a brief note on a special property of (Radon) product measures. 494A Proposition Let (A, µ ¯) be a semi-finite measure algebra with its measure-algebra topology (323A). Suppose that A ⊆ A is an uncountable analytic set. Then there is a compact set L ⊆ A, homeomorphic to {0, 1}N , such that inf L 6= 0 in A. proof A \ {0} is still an uncountable analytic subset of A. By 423J, it has a subset homeomorphic to ¯) is semi-finite, there {0, 1}N ∼ = PN; let f : PN → A \ {0} be an injective continuous function. Because (A, µ is an a ⊆ f (∅) such that 0 < µ ¯a < ∞; set δ = µ ¯a. Note that (I, J) 7→ µ ¯(a ∩ f (I) \ f (J)) : (PN)2 → R is continuous. Choose a sequence hkn in∈N in N inductively, as follows. Given hki ii
µ ¯(a ∩ inf J⊆Kn f (J)) ≥ δ( + 2−n−1 ) for every n ∈ N. P P Induce on n. If n = 0, then a ∩ inf J⊆Kn f (J) = a ∩ f (∅) has measure δ = δ( 21 + 2−1 ). For the inductive step to n + 1 ≥ 1, observe that

µ ¯(a ∩

inf

J⊆Kn+1

f (J)) = µ ¯(a ∩ inf f (J) ∩ f (J ∪ {kn })) J⊆Kn X ≥µ ¯(a ∩ inf f (J)) − µ ¯(a ∩ f (J) \ f (J ∪ {kn })) J⊆Kn

1 2

≥ δ( + 2−n−1 ) −

J⊆Kn

X

2−2n−2 δ

J⊆Kn

(by the inductive hypothesis and the choice of kn ) 1 2

= δ( + 2−n−2 ). So the induction proceeds. Q Q Set K = {ki : i ∈ N}, c = inf{f (J) : J ⊆ K is finite}. Then

494C

Cubes in product spaces

801 1 2

ν¯(a ∩ c) = inf n∈N µ ¯(a ∩ inf J⊆Kn f (J)) ≥ δ, and c 6= 0. But now observe that L = f [PK] is a subset of A homeomorphic to PK and therefore to {0, 1}N . Also {b : b ⊇ c} is closed (323D(d-i)), so C = {J : f (J) ⊇ c} is closed in PK; as it includes the dense set [K]<ω , C = PK and inf L ⊇ c is non-zero. 494B Proposition (see Brodskiˇı 49, Eggleston 54) Let (X, T, Σ, µ) be an atomless Radon measure ˜ the τ -additive space, (Y, S, T, ν) an effectively locally finite τ -additive topological measure space, and λ ˜ product measure on X × Y (417G). Then if W ⊆ X × Y and λW > 0 there are a non-scattered compact set K ⊆ X and a set F ⊆ Y of positive measure such that K × F ⊆ W . proof (a) To begin with (down to the end of (c)), let us suppose that both µ and ν are totally finite. ˜ is inner regular with respect to the closed sets, we may Let (B, ν¯) be the measure algebra of ν. Because λ suppose that W is closed. In this case, writing λ for the c.l.d. product measure on X × Y , there is a W 0 ⊇ W ˜ (417C(iv)). By 418Tb, there is a µ-conegligible set X0 such that such that λW 0 is defined and equal to λW 0 W [{x}] ∈ T for every x ∈ X0 , B = {W 0 [{x}]• : x ∈ X0 } is separable for the measure-algebra topology of B, and x 7→ W 0 [{x}]• : X0 → B is measurable. Now Fubini’s theorem, applied in the form of 252D to λ ˜ tells us that and in the form of 417Ha to λ, R R ˜ = νW [{x}]µ(dx). νW 0 [{x}]µ(dx) = λW 0 = λW So X1 = {x : x ∈ X0 , W 0 [{x}]• = W [{x}]• } is µ-conegligible. Since the topology of B is metrizable (323Ad), B is separable and metrizable, and x 7→ W [{x}]• : X1 → B is almost continuous (418J, applied to the subspace measure on X1 ). Let K ∗ ⊆ X1 be a compact set of non-zero measure such that x 7→ W [{x}]• : K ∗ → B is continuous. (b) There is a non-zero c ∈ B such that Kc = {x : x ∈ K ∗ , c ⊆ W [{x}]• } is compact and not scattered. P P Because x 7→ W [{x}]• is continuous on S K ∗ , B ∗ = {W [{x}]• : x ∈ K ∗ } is compact and every Kc is ∗ ∗ compact. (i) If B is countable, then K = b∈B ∗ Kb , so there is some c ∈ B ∗ such that µKc > 0. Let E be a non-negligible self-supporting subset of Kc ; then (because µ is atomless, therefore zero on singletons) E has no isolated points. So Kc is not scattered. (ii) If B ∗ is uncountable, then by 494A there is a set D ⊆ B ∗ , homeomorphic to {0, 1}N , with a non-zero lower bound c in B. Now {W [{x}]• : x ∈ Kc } includes D, so {0, 1}N and therefore [0, 1] are continuous images of closed subsets of Kc and Kc is not scattered (4A2G(i-iv)). Q Q (c) Set K = Kc . Then 414Ac tells us that T ( x∈K W [{x}])• = inf x∈K W [{x}]• ⊇ c T is non-zero, so F = x∈K W [{x}] is non-negligible; while K × F ⊆ W . So we have found appropriate sets K and F , at least when µ and ν are totally finite. (d) For the general case, we need observe only that by 417C(iii) there are X 0 ∈ Σ and Y 0 ∈ T, both of ˜ finite measure, such that λ(W ∩ (X 0 × Y 0 )) > 0. Now the subspace measure µX 0 on X 0 is atomless and Radon (214Ja, 416Rb), the subspace measure νY 0 on Y 0 is τ -additive (414K), and the τ -additive product of ˜ (417I). So we can apply (a)-(c) to µX 0 and µX 0 and νY 0 is the subspace measure on X 0 × Y 0 induced by λ νY 0 to see that there are a non-scattered compact set K ⊆ X 0 and a non-negligible measurable set F ⊆ Y 0 such that K × F ⊆ W . 494C Proposition (see Ciesielski & Pawlikowski 03) Let h(Xi , Ti , Σi , µi )ii∈I be a countable family ˜ the product Radon probability measure on X = Q Xi . If of atomless Radon probability spaces, and λ i∈I ˜ > 0, there is a family hKi ii∈I such that Ki ⊆ Xi is a non-scattered compact set for each W ⊆ X and λW Q i ∈ I and i∈I Ki ⊆ W . Q ˜ n be proof (a) To begin with, let us suppose that I = N. For each n ∈ N, set Yn = i≥n Xi and let λ ˜0 = λ ˜ and λ ˜ n can be identified with the product the product Radon probability measure on Yn , so that λ ˜ of µn and λn+1 (417J). Using 494B repeatedly, we can find non-scattered compact sets Kn ⊆ Xn and closed non-negligible sets Wn ⊆ Yn such that W0 ⊆ W and Kn × Wn+1 ⊆ Wn for every n. In this case,

802

Further topics

494C

Q

Q Q Ki × Wn ⊆ W0 for every n. If x ∈ i∈N Ki , then there is for each n ∈ N an xn ∈ i
As x is arbitrary,

x = limn→∞ xn ∈ W0 ⊆ W .

Q i∈N

Ki ⊆ W .

(b) For the general case, we may suppose that I ⊆ N. For i ∈ N \ I, take (Xi , Ti , Σi , µi ) to be [0, 1] with ˜ = {x : x ∈ Q Lebesgue measure. Set W x¹I ∈ W }. By (a), there are non-scattered compact sets i∈N Xi , Q Q ˜ Ki ⊆ Xi such that i∈N Ki ⊆ W , in which case i∈I Ki ⊆ W , as required. 494X Basic exercises (a) Let (X, T, Σ, µ) be a Radon measure space, (Y, S, T, ν) an effectively locally ˜ the τ -additive product measure on X × Y . Show that if finite τ -additive topological measure space, and λ ˜ > 0 there are a compact set K ⊆ X and a set F ⊆ Y of positive measure such that W ⊆ X × Y and λW K × F ⊆ W and K is either non-scattered or non-negligible. (b) Let (X, T, Σ, µ) be an atomless Radon measure space, (Y, T, ν) any measure space, and λ the c.l.d. product measure on X × Y . Show that if W ⊆ X × Y and λW > 0 there are a non-scattered compact set K ⊆ X and a set F ⊆ Y of positive measure such that K × F ⊆ W . (Hint: reduce to the case in which ν is totally finite and T is countably generated, so that the completion of ν is a quasi-Radon measure for an appropriate second-countable topology.) (c) Let h(Xi , Ti , Σi , µiQ )ii∈I be any family of atomless Radon probability spaces, and λ the ordinary product measure on X = i∈I XQ i . Show that if W ⊆ X and λW > 0 then there are non-scattered compact sets Ki ⊆ Xi for i ∈ I such that i∈I Ki ⊆ W . Q (d) Let h(Xi , Ti , Σi , µi )ii∈I be a countable family of atomless Radon probability spaces, and W ⊆ i∈I Xi a set with positive measure for the Radon product of hµi ii∈I . Show that there are atomless Radon probability measures νi on Xi such that W is conegligible for the Radon product of hνi ii∈I . (Hint: 439Xh(vii).) 494Y Further exercises (a) Let h(Xi , Ti , Σ Qi , µi )ii∈I be a family of atomless perfect probability spaces, and λ the ordinary product measure on X = i∈I QXi . Show that if W ⊆ X and λW > 0 then there are sets Ki ⊆ Xi for i ∈ I, all of cardinal c, such that i∈I Ki ⊆ W . 494 Notes and comments I have already noted (325Yc) that a set W of positive measure in a product space need not include the product of two sets of positive measure; this fact is also the basis of 419E. Here, however, we see that if one of the factors is a Radon measure space then W does include the product of a non-trivial compact set and a set of positive measure. There are many possible variations on the result, corresponding to different product measures (494B, 494Xb) and different notions of ‘non-trivial’ (494Xa, 494Ya). The most important of the latter seems to be the idea of a ‘non-scattered’ compact set K; this is a quick way of saying that [0, 1] is a continuous image of K, which is a little stronger than saying that #(K) ≥ c, and arises naturally from the proof of 494B.

495 Poisson point processes A classical challenge in probability theory is to formulate a consistent notion of ‘random set’. Simple geometric considerations lead us to a variety of measures which are both interesting and important. All these are manifestly special constructions. Even in the most concrete structures, we have to make choices which come to seem arbitrary as soon as we are conscious of the many alternatives. There is however one construction which has a claim to pre-eminence because it is both robust under the transformations of abstract measure theory and has striking properties when applied to familiar measures (to the point, indeed, that it is relevant to questions in physics and chemistry). This gives the ‘Poisson point processes’ of 495D-495E. In this section I give a brief introduction to the measure-theoretic aspects of this construction. 495A Poisson distributions We need a little of the elementary theory of Poisson distributions.

495B

Poisson point processes

803

(a) The Poisson distribution with parameter γ > 0 is the point-supported Radon probability meaγn

sure νγ on R such that νγ {n} = e−γ for every n ∈ N. (See 285Q and 285Xo.) Its expectation is n! P∞ γn e−γ = γ. Since νγ N = 1, νγ can be identified with the corresponding subspace measure on N. It n=1 (n−1)!

will be convenient to allow γ = 0, so that the probability measure on R or N with support {0} is a ‘Poisson distribution with expectation 0’. (b) The convolution of two Poisson distributions is a Poisson distribution. P P If α, β > 0 then Z (να ∗ νβ )({n}) =

νβ ({n} − t)να (dt)

(444A) =

n X β n−i i=0

(n−i)!

= e−α−β

1 n!

e−β ·

n X i=0

αi −α e i!

n! αi β n−i i!(n−i)!

=

(α+β)n −α−β e n!

for every n ∈ N, so να ∗ νβ = να+β . Q Q So if f and g are independent random variables with Poisson distributions then f + g has a Poisson distribution (272S). (c) family of random variables with Poisson distributions, and P PIf hfi ii∈I is a countable independent α = i∈I E(fi ) is finite, then f = i∈I fi is defined a.e. and has a Poisson distribution with expectation α. P P For finite I we can induce on #(I), using (b) (and 272L) for the inductive step. For the infinite P∞ case we can suppose that I = N. In this case fi ≥P0 a.e. for each i so f = i=0 fi is defined a.e. and has n expectation α, by B.Levi’s Pn theorem. Setting gn = i=0 fi for each n, so that gn has a Poisson distribution with expectation βn = i=0 αi , we have Pr(f ≤ γ) = lim Pr(gn ≤ γ) = lim n→∞

n→∞

bγc X βi

n

i=0

i!

e

−βn

=

bγc X αi i=0

i!

e−α

for every γ ≥ 0, so f has a Poisson distribution with expectation α. Q Q 495B Theorem Let (X, Σ, µ) be a measure space. Set Σf = {E : E ∈ Σ, µE < ∞}. Then for any γ > 0 there are a probability space (Ω, Λ, λ) and a family hgE iE∈Σf of random variables on Ω such that (i) for every E ∈ Σf , gE has a Poisson distribution with expectation γµE; (ii) whenever hEi ii∈I is a disjoint family in Σf , then hgEi ii∈I is stochastically independent; P∞ (iii) whenever hEi ii∈N is a disjoint sequence in Σf with union E ∈ Σf , then gE =a.e. i=0 gEi . proof (a) Let H ⊆ {H : H ∈ Σ, 0 < µH < ∞} be a maximal family such that H ∩ H 0 is negligible for all distinct H, H 0 ∈ H. For H ∈ H, let µ0H be the normalized subspace measure defined by setting µ0H E = µE/µH for E ∈ Σ ∩ PH, and λH the corresponding product probability measure on H N . Next, for H ∈ H, let νH be the Poisson distribution Q with expectation γµH, regarded as a probability measure on N. Let λ be the product measure on Ω = H∈H (N × H N ), giving each N × H N the product measure νH × λH . For ω ∈ Ω, write mH (ω), xHj (ω) for its coordinates, so that ω = h(mH (ω), hxHj (ω)ij∈N )iH∈H . (b) For H ∈ H and E ∈ Σ, set gHE (ω) = #({j : j < mH (ω), xHj (ω) ∈ E}). Then gHE is measurable and has a Poisson distribution with expectation γµ(H ∩ E); moreover, if E0 , . . . , Er ∈ Σ are disjoint, then gHE0 , . . . , gHEr are independent. P P It is enough to examine the case in which the Ei cover X. Then for any n0 , . . . , nr ∈ N with sum n,

804

Further topics

495B

λ{ω : gHEi (ω) = ni for every i ≤ r} = λ{ω : #({j : j < mH (ω), xHj ∈ Ei }) = ni for every i ≤ r} = λ{ω : mH (ω) = n, #({j : j < n, xHj ∈ Ei }) = ni for every i ≤ r} X = λ{ω : mH (ω) = n, xHj ∈ Ei whenever i ≤ r, j ∈ Ii } J0 ,... ,Jr partition n #(Ji )=ni for each i≤r

X

=

J0 ,... ,Jr partition n #(Ji )=ni for each i≤r

= =

n! 1 −γµH e n0 !...nr ! n! r Y (γµ(H∩Ei ))ni i=0

ni !

(γµH)n −γµH e n!

r Y

r Y ¡ µ(H∩Ei ) ¢ni i=0

µH

(γµ(H ∩ Ei ))ni

i=0

e−γµ(H∩Ei ) ,

which is just what we wanted to know. Q Q P∞ Obviously gHE = i=0 gHEi whenever hEi ii∈N is a sequence in Σ with union E. (c) Suppose that H0 , . . . , Hn ∈ H are distinct and E0 , . . . , Er ∈ Σ are disjoint. Then the random variables gHj Ei are independent. P P For each j ≤ n, gHj Ei is ΛHj -measurable, where ΛHj is the σ-algebra of subsets of Ω which are measured by λ and determined by the single coordinate Hj in the product Q N (N × H ). Now the σ-algebras Λ Hj are independent (272Ma). So if we have any family hnij ii≤r,j≤n H∈H in N, λ{ω : gHj Ei (ω) = nij for every i ≤ r, j ≤ n} n Y = λ{ω : gHj Ei (ω) = nij for every i ≤ r} =

j=0 n Y r Y

λ{ω : gHj Ei (ω) = nij }

j=0 i=0

by (b); and this is what we need to know. Q Q (d) For E ∈ Σf , P set HE = {H : H ∈ H, µ(E ∩ H) > 0}; then HE is countable, because H is almost P disjoint, and µE = H∈HE µ(H ∩ E), because H is maximal. Set gE (ω) = H∈HE gHE (ω) when this is finite. Then gE is defined a.e. and has a Poisson distribution with expectation γµE (495Ac). Also hgEi ii∈I are independent whenever P It is enough to deal S hEi ii∈I is a disjoint family in Σ. P P with the case of finite I (272Bb). Set H∗ = i∈I HEi , so that H∗ is countable, and for i ∈ I set gi0 = H∈H∗ gHEi . Then each gi0 is equal almost everywhere to the corresponding gEi , and hgi0 ii∈I is independent, by 272K. (The point is that each gi0 is Λ∗i -measurable, where Λ∗i is the σ-algebra generated by {gHEi : H ∈ H}, and 272K assures us that the Λ∗i are independent.) It follows at once that hgEi ii∈I is independent (272H). Q Q This proves (i) and (ii). S f f ∗ (e) Similarly, if hE i ii∈N is a disjoint sequence in Σ with union E ∈ Σ , set H = HE ∪ i∈N HEi . For P each i ∈ N, set gi0 = H∈H∗ gHEi ; then gi0 =a.e. gEi . Now ∞ X i=0

as required by (iii).

gEi =a.e.

∞ X i=0

gi0 =

∞ X X H∈H∗ i=0

gHEi =

X H∈H∗

gHE =a.e. gE ,

495D


805

495C Lemma Let X be a set and E a subring of the Boolean algebra PX. Let H be the family of sets of the form {S : S ⊆ X, #(S ∩ Ei ) = ni for every i ∈ I} where hEi ii∈I is a finite disjoint family in E and ni ∈ I for every i ∈ I. Then the Dynkin class T ⊆ P(PX) generated by H is the σ-algebra of subsets of PX generated by H. proof Let Q be the set of functions q from finite subsets of E to N, and for q ∈ Q set Hq = {S : S ⊆ X, #(S ∩ E) = q(E) for every E ∈ dom q}. Our family H is just {Hq : q ∈ Q, dom q is disjoint}. If q ∈ Q and dom q is a subring of E, then Hq ∈ T. P P Being a finite Boolean ring, dom q is a Boolean algebra; let A be the set of its atoms. Then Hq is either empty or equal to Hq¹A ; in either case it belongs to T. Q Q If qSis any member of Q, then Hq ∈ T. P P Let E be the subring of PX generated by dom q. Then Hq = q⊆q0 ∈Q,dom q0 =E Hq0 is the union of a finite disjoint family in T, so belongs to T. Q Q Now observe that H1 = {Hq : q ∈ Q} ∪ {∅} is a subset of T closed under finite intersections, so by the Monotone Class Theorem (136B) T includes the σ-algebra generated by H1 , and must be precisely the σ-algebra generated by H. 495D Theorem Let (X, Σ, µ) be an atomless measure space. Set Σf = {E : E ∈ Σ, µE < ∞}; for E ∈ Σf , set fE (S) = #(S ∩ E) when S ⊆ X meets E in a finite set. Let T be the σ-algebra of subsets of PX generated by sets of the form {S : fE (S) = n} where E ∈ Σf and n ∈ N. Then for any γ > 0 there is a unique probability measure ν with domain T such that (i) for every E ∈ Σf , fE is measurable and has a Poisson distribution with expectation γµE; (ii) whenever hEi ii∈I is a disjoint family in Σf , then hfEi ii∈I is stochastically independent. proof (a) Let H, hνH iH∈H , hµH iH∈H , hµ0H iH∈H , hλH iH∈H , Ω, λ, hHE iE∈Σf and hgE iE∈Σf be as in the proof of 495B. Note that all the µ0H are atomless (234Ff). Define φ : Ω → PX by setting φ(ω) = {xHj (ω) : H ∈ H, j < mH (ω)} for ω ∈ Ω. (b) For E ∈ Σf , let AE be the set of those ω ∈ Ω such that either there are H ∈ H \ HE , j ∈ N such that xHj (ω) ∈ E or there are distinct H, H 0 ∈ HE and j ∈ N such that xHj (ω) ∈ H 0 or there is an H ∈ H such that the xHj (ω), for j ∈ N, are not all distinct. S S Then for any sequence hEi ii∈N in Σf , λ∗ ( i∈N AEi ) = 0. P P Set H∗ = i∈N HEi , so that H∗ is a countable subset of H. For H ∈ H, set S S FH = H \ ( {Ei : i ∈ N, H ∩ Ei is negligible} ∪ {H 0 : H 0 ∈ H∗ , H 0 6= H}), x : x ∈ H N is injective}, WH = {x so that FH is µ0H -conegligible and WH is λH -conegligible (because µ0H is atomless, see 254V). Now S Q N Ω \ k∈N AEk ⊇ H∈H (N × (WH ∩ FH )) has full outer measure in Ω, by 254Lb, and its complement has zero inner measure (413Ec). Q Q ˜ on Ω, extending λ, such that λA ˜ E = 0 for every E ∈ Σf It follows that there is a probability measure λ ˜ −1 . (417A). Let ν0 be the image measure λφ (c) If E ∈ Σf and ω ∈ Ω \ AE , then fE (φ(ω)) = gE (ω) if either is defined. P P If H ∈ H, then all the xHj (ω) are distinct; if H ∈ H \ HE , no xHj (ω) can belong to E; if H, H 0 ∈ HE are distinct, then no xHj (ω) can belong to H 0 . So all the xHj (ω), xH 0 k (ω) for H, H 0 ∈ HE and j, k ∈ N must be distinct, and

806

Further topics

495D

fE (φ(ω)) = #({xHj (ω) : H ∈ H, j < mH (ω), xHj (ω) ∈ E}) = #({(H, j) : H ∈ HE , j < mH (ω), xHj (ω) ∈ E}) X = gHE (ω) = gE (ω) H∈HE

if any of these is finite. Q Q It follows at once that if E0 , . . . , Er ∈ Σf are disjoint, then {ω : fEi (φ(ω)) = gEi (ω) ˜ for every i ≤ r} is λ-conegligible, so that if n0 , . . . , nr ∈ N then ˜ : fE (φ(ω)) = ni for every i ≤ r} ν0 {S : fEi (S) = ni for every i ≤ r} = λ{ω i ˜ : gE (ω) = ni for every i ≤ r} = λ{ω i r Y (γµEi )ni −γµEi = e . i=0

ni !

Thus every fEi is finite ν0 -a.e., belongs to L0 (ν0 ) and has a Poisson distribution with the appropriate expectation, and they are independent. (d) As T is defined to be the σ-algebra generated by the family {fE : E ∈ Σf }, it is included in the domain of ν0 . Set ν = ν0 ¹T; then ν has the properties (i) and (ii). To see that it is unique, observe that if ν 0 also has these properties, then {A : νA = ν 0 A} is a Dynkin class containing every set of the form {S : fEi (S) = ni for i ≤ r} f

where E0 , . . . , Er ∈ Σ are disjoint and n0 , . . . , nr ∈ N. By 495C it contains the σ-algebra generated by this family, which is T. So ν and ν 0 agree on T, and are equal. 495E Definition In the context of 495D, I will call the completion of ν the Poisson point process on X with density γ. Note that the Poisson point process on (X, µ) with density γ > 0 is identical with the Poisson point process on (X, γµ) with density 1. There would therefore be no real loss of generality in the main theorems of this section if I spoke only of point processes with density 1. I retain the extra parameter because applications frequently demand it, and the formulae will be more useful with the γs in their proper places; moreover, there are important ideas associated with variations in γ, as in 495Xd. 495F Proposition Let (X, Σ, µ) be a perfect atomless measure space, and γ > 0. Then the Poisson point process on X with density γ is a perfect probability measure. proof I refer to the construction in 495B-495D. In (b) of the proof of 495D, use the construction set out in ˜ is precisely the family of sets of the form W 4A where W belongs ˜ of λ the proof of 417A, so that the domain Λ to the domain Λ of the product measure λ and A belongs to the σ-ideal A∗ generated by {AE : E ∈ Σf }. ˜ is perfect. P ˜ ˜ a set of non-zero measure. Then λ P Let h : Ω → R be a Λ-measurable function and W ∈ Λ 0 ∗ 0 Then there are a W ∈ Λ and an A S ∈ A such that W 4W ⊆ A and h¹Ω \ A is Λ-measurable; let hEn in∈N f be a sequence in Σ such that A ⊆ n∈N AEn , and h1 : Ω → R a Λ-measurable function agreeing with h on S Ω \ A. Set H∗ = n∈N HEn , so that H∗ is countable. As in the proof of 495D, set S S FH = H \ ( {En : n ∈ N, H ∩ En is negligible} ∪ {H 0 : H 0 ∈ H∗ , H 0 6= H}),

for H ∈ H, so every AEn (as product of the (451Jc again).

x : x ∈ H N is injective}, WH = {x Q 0 N 0 that WH = WH ∩ FH is λH -conegligible. Set Ω0 = H∈H (N × WH ). This is disjoint from 0 in 495D) and therefore from A. The subspace measure λΩ0 on Ω induced by λ is just the 0 measures on N × WH (254La). All of these are perfect (451Jc, 451Dc), so λΩ0 is also perfect Now ˜ > 0. λΩ0 (W ∩ Ω0 ) = λΩ0 (W 0 ∩ Ω0 ) = λW 0 = λW

0 It follows that there is a compact set K ⊆ h1 [W ∩ Ω0 ] such that λΩ0 (h−1 1 [K] ∩ Ω ) > 0. As h and h1 agree 0 on Ω , K ⊆ h[W ], while

495H


807

˜ −1 [K] = λh ˜ −1 [K] λh 1 ˜ = 0) (because λA −1 = λh−1 [K] ∩ Ω0 ) > 0. 1 [K] = λΩ0 (h

˜ is perfect. Q As W and h are arbitrary, λ Q ˜ −1 and its restriction to T are perfect (451Ea); finally, the It follows at once that the image measure λφ completion is perfect, by 451Gc. 495G Proposition Let (X1 , Σ1 , µ1 ) and (X2 , Σ2 , µ2 ) be atomless measure spaces, and f : X1 → X2 an inverse-measure-preserving function. Let γ > 0, and let ν1 , ν2 be the Poisson point processes on X1 , X2 respectively with density γ. Then S 7→ f [S] : PX1 → PX2 is inverse-measure-preserving for ν1 and ν2 ; in particular, PA has full outer measure for ν2 whenever A ⊆ X2 has full outer measure for µ2 . proof Set ψ(S) = f [S] for S ⊆ X1 . (a) If F ∈ Σf2 , then {S : S ⊆ X1 , f ¹f −1 [F ] ∩ S is not injective} is ν1 -negligible. P P Let n ∈ N. Set 1 α = n+1 µ2 F . Because µ2 is atomless, we can find a partition of F into sets F0 , . . . , Fn of measure α. Now S {S : f ¹f −1 [F ] ∩ S is not injective} ⊆ i≤n {S : #(S ∩ f −1 [Fi ]) > 1} has ν1 -outer measure at most (n + 1)(1 − e−γα (1 + γα)) ≤ (n + 1)α2 γ 2 =

1 (γµ2 F )2 . n+1

As n is arbitrary, {S : f ¹f −1 [F ] ∩ S is not injective} is negligible. Q Q (b) It follows that, for any F ∈ Σf2 and n ∈ N, {S : #(f [S] ∩ F ) = n}4{S : #(S ∩ f −1 [F ]) = n} is ν1 -negligible and {S : #(f [S] ∩ F ) = n} is measured by ν1 . So if T2 is the σ-algebra of subsets of PX2 generated by sets of the form {T : #(F ∩ T ) = n} for F ∈ Σf2 and n ∈ N, then ν1 measures ψ −1 [H] for every H ∈ T2 . Next, if hFi ii∈I is a finite disjoint family in Σf2 and ni ∈ N for i ∈ I, ν1 {S : #(f [S] ∩ Fi ) = ni for every i ∈ I} = ν1 {S : #(S ∩ f −1 [Fi ]) = ni for every i ∈ I} Y (γµ1 f −1 [Fi ])n −1 = e−γµ1 f [Fi ] n!

i∈I

(because hf −1 [Fi ]ii∈I is a disjoint family in Σf1 ) Y (γµ2 Fi )n e−γµ2 Fi . = i∈I

n!

So the image measure ν1 ψ −1 satisfies (i) and (ii) of 495D, and must agree with ν2 on T2 ; that is, ψ is inverse-measure-preserving for ν1 and ν2 ¹T2 . As ν1 is complete, ψ is inverse-measure-preserving for ν1 and ν2 (235Hc). (c) If A ⊆ X2 has full outer measure, then we can take µ1 to be the subspace measure on X1 = A and f (x) = x for x ∈ A. In this case, PA = ψ[PA] must have full outer measure for ν2 . ˜ Σ, ˜ µ 495H Lemma Let (X, ˜) be an atomless σ-finite measure space, and γ > 0. Write µL for Lebesgue 0 ˜ × [0, 1], and λ0 for the product measure on measure on [0, 1], µ for the product measure on X 0 = X ˜ ˜ X 0 respectively with density γ. For T ⊆ X ˜ Ω0 = [0, 1]X . Let ν˜, ν 0 be the Poisson point processes on X,

808

Further topics

495H

define ψT : Ω0 → PX 0 by setting ψT (z) = {(t, z(t)) : t ∈ T } for z ∈ Ω0 ; let νT0 be the image measure λ0 ψT−1 on PX 0 . Then hνT0 iT ⊆X˜ is a disintegration of ν 0 over ν˜ (definition: 452E). proof (a) Let E ⊆ X 0 be a measurable set with finite measure, and write HE = {S : S ∩ E 6= ∅}. Then R R 0 ν 0 HE = 1 − e−γµ E ≤ γµ0 E; but also νT0 (HE )˜ ν (dT ) ≤ 2γµ0 E. P P We know that µL E[{t}]˜ µ(dt) = µ0 E ˜ be a conegligible set such that E[{t}] is measurable for every t ∈ Y and t 7→ µL E[{t}] : (252D). Let Y ⊆ X Y → [0, 1] is measurable. Set Fi = {t : t ∈ Y , 2−i−1 < µL E[{t}] ≤ 2−i } for each i ∈ N; let hFi0 ii∈N be ˜ \S ˜ a sequence of sets of finite measure with union X i∈N Fi . Let W be the set of those T ⊆ X such that 0 ˜ T ∩ (X \ Y ) is empty and T ∩ Fi , T ∩ Fi are finite for every i ∈ N; then W is ν˜-conegligible. For any T ∈ W , ψT−1 [HE ] = {z : ψT (z) ∩ E 6= ∅} [ [ [ {z : z(t) ∈ E[{t}]} ∪ = is measured by λ0 and has measure at most if t ∈ T ∩ Fi , and is zero if t ∈ T ∩ Fi0 . So

P∞ i=0

2−i #(T ∩ Fi ), because µL E[{t}] has measure at most 2−i

Z ν (dT ) νT0 (HE )˜

= =

{z : z(t) ∈ E[{t}]}

i∈N t∈T ∩Fi0

i∈N t∈T ∩Fi

Z

[

λ ∞ X

0

ψT−1 [HE ]˜ ν (dT )

2−i #(T ∩ Fi )˜ ν (dT )

i=0

Z 2−i

≤

Z X ∞

∞ X

#(T ∩ Fi )˜ ν (dT ) =

i=0

2−i γ µ ˜ Fi

i=0

(because T 7→ #(T ∩ Fi ) has expectation γ µ ˜F ) Z i ≤ 2γ µL E[{t}]˜ ν (dt) = 2γµ0 E. Q Q (b) Suppose that hFj ij
˜ and z ∈ ψ −1 [H], then for each j < s we must have T ⊆X T #(T ∩ Fj ) = #({t : t ∈ T ∩ Fj , z(t) ∈

[

Cij })

i
=

r−1 X

#({t : t ∈ T ∩ Fj , z(t) ∈ Cij })

i=0

=

r−1 X i=0

#(ψT (z) ∩ Eij ) =

r−1 X

nij = nj .

i=0

Turning this round, we see that if T ∈ / W then ψT−1 [H] = ∅ and νT0 H = 0. If T ∈ W , let Q be the set of all q = hq(i, j)ii
nij !

495H


809

νT0 H = λ0 {z : ψT (z) ∈ H} = λ0 {z : #({t : t ∈ T ∩ Fj , z(t) ∈ Cij }) = nij for all i, j} X = λ{z : z(t) ∈ Cij whenever i < r, j < s, t ∈ q(i, j)} q∈Q

=

X Y q∈Q

Y

X

µL Cij =

(µL Cij )nij =

q∈Q i
i
s−1 Y

Y (µL Cij )nij ¢ ¡ r−1 nj ! . nij ! j=0 i=0

It follows that Z νT0 (H)˜ ν (dT ) = ν˜W ·

s−1 Y

¡

nj !

j=0

=

=

r−1 Y i=0

s−1 Y

s−1 r−1 Y (µL Cij )nij ¢ (γ µ ˜Fj )nj −γ µ˜Fj Y ¡ e · nj ! nj ! nij ! j=0 j=0 i=0

s−1 Y

(γ µ ˜Fj )nj e−γ µ˜Fj

j=0

=

(µL Cij )nij ¢ nij !

s−1 Y r−1 Y j=0 i=0

r−1 Y i=0

0

e−γµ Eij

(µL Cij )nij nij !

(γµ0 Eij )nij = ν 0 H, nij !

as required.

S (ii) For the general case, set Crj = [0, 1] \ i 0. For each i < r we can find an Ei ∈ Σ ⊗ ΣL such that µ (Ei 4Ei0 ) ≤ ² (251Ie). Discarding a ˜ has finite measure. Set negligible set from Ei0 if necessary, we may suppose that the projection of Ei0 on X S 0 0 ˆ ˆ ˜ î Ei = Ei \ k
810

Further topics

495H

ˆ = {S : S ⊆ X 0 , #(S ∩ E î ) = ni for every i < r}. Then (c) tells us that are Set H R 0 still of finite measure. 0 ˆ ν (dT ) = ν H. ˆ νT (H)˜ S î for every i, so Set E = i
Z ˆ ν (dT ) − 2rγ² νT0 (H)˜

= (by (c))

Z ˆ − ν 0 (HE )˜ νT0 (H) ν (dT ) T

≤ (by the other part of (a))

Z

Z (νT0 )∗ (H)˜ ν (dT ) ≤

≤

(νT0 )∗ (H)˜ ν (dT )

Z ≤ As ² is arbitrary,

ˆ + 2rγ² ≤ ν 0 H + 3rγ². ˆ + νT0 (HE )˜ ν (dT ) ≤ ν 0 H νT0 (H)

R R ν 0 H = (νT0 )∗ (H)˜ ν (dT ) = (νT0 )∗ (H)˜ ν (dT ).

In particular, (νT0 )∗ (H) = (νT0 )∗ (H) for almost every T ; since νT0 is the R image of the complete measure λ, it is complete (212Bd), and H is measured by almost every νT0 ; finally, νT0 (H)˜ ν (dT ) is defined and equal to ν 0 H. Q Q R (e) So if we write H for the family of subsets H of PX 0 such that νT0 (H)˜ ν (dT ) is defined and equal to î ) = ni for every i < r} where ν 0 H, and H0 for the family of sets of the form H = {S : S ⊆ X 0 , #(S ∩ E hEi ii 0. Let ν, ν˜ ˜ respectively with density γ. Suppose that f : X → X ˜ is inversebe the Poisson point processes on X, X measure-preserving and that hµt it∈X˜ is a disintegration of µ over µ ˜ consistent with f (definition: 452E) Q ˜ such that every µt is a probability measure. Write λ for the product measure t∈X˜ µt on Ω = X X , and for −1 ˜ define φT : Ω → PX by setting φT (z) = z[T ] for z ∈ Ω; let νT be the image measure λφ on PX. T ⊆X T Then hνT iT ⊆X˜ is a disintegration of ν over ν˜. Moreover ˜ (i) setting f˜(S) = f [S] for S ⊆ X, hνT iT ⊆X˜ is consistent with f˜ : PX → P X; (ii) if µ is strongly consistent with f , then hνT iT ⊆X˜ is strongly consistent with f˜. ˜ let VT be the set of those z ∈ Ω such that f z¹T is injective. We need to know that proof (a) For T ⊆ X ˜ T is countable, VT is λ-conegligible} W = {T : T ⊆ X, ˜ f = {F : F ∈ Σ, ˜ µ is ν˜-conegligible. P P Write Σ ˜F < ∞}. Because µ ˜ is atomless and σ-finite, there is a ˜ such that for every ² > 0 there is a cover of X ˜ by members of E of measure at countable subalgebra E of Σ most ². Set ˜ µt f −1 [F ] = (χF )(t) for every F ∈ E}, Y = {t : t ∈ X,

495I


811

so that Y is µ ˜-conegligible and PY is ν˜-conegligible. For F ∈ E, let WF be the set of those T ⊆ Y such that ˜ f and ² > 0, there is a for every t ∈ T ∩ F there is an F 0 ∈ E such that T ∩ F 0 = {t}. Now, given F ∈ E ∩ Σ partition hFi ii∈I of F into members of E of measure at most ². Then ν˜∗ (PY \ WF ) ≤ ν˜{T : #(T ∩ Fi ) > 1 for some i ∈ I} X X X ≤ 1 − e−γ µ˜Fi (1 + γ µ ˜ Fi ) ≤ (γ µ ˜Fi )2 ≤ ²γ 2 µ ˜Fi = ²γ 2 µ ˜F. i∈I

T

i∈I

i∈I

˜ f } is ν˜-conegligible. As ² is arbitary, WF is ν˜-conegligible; accordingly W = {WF : F ∈ E ∩ Σ 0 f ˜ ˜ Now suppose that T ∈ W . Because X is covered by E ∩ Σ , we see that for every t ∈ T there is an ˜ f containing t, and now there is an F 0 ∈ E such that T ∩ F 0 = {t}. In particular, T is countable, F ∈E ∩Σ so 0

UT = {z : z(t) ∈ f −1 [F ] whenever F ∈ E and t ∈ T ∩ F } is λ-conegligible. Take z ∈ UT . If t, t0 are distinct points of T , there is an F ∈ E containing t but not t0 , and now F contains f (z(t)) but not f (z(t0 )). So z¹T is injective. Thus UT ⊆ VT and VT is λ-conegligible. This is true for every T ∈ W 0 , so W ⊇ W 0 is ν˜-conegligible. Q Q (b) Suppose that hEi ii
R

Ei0 = {(t, α) : t ∈ Y1 , fi (t) ≤ α < fi+1 (t)} fi+1 − fi d˜ ν = µEi for each i, by 252N. H 0 = {S : S ⊆ X 0 , #(S 0 ∩ Ei0 ) = ni for every i < r}, W1 = {T : T ∈ W , T ⊆ Y1 , νT0 H 0 is defined},

so that W1 is ν˜-conegligible. Let T be any member of W1 . Let Q be the set of partitions qS= hq(i)ii≤r of T such that #(q(i)) = ni for every i < r; because T is countable, so is Q. Set Er = X \ i
=

r XY Y

µL Ei0 [{t}]

q∈Q i=0 t∈q(i)

=

r XY Y

µt Ei

q∈Q i=0 t∈q(i)

=

X

λ{z : z ∈ Ω, z(t) ∈ Ei whenever i ≤ r and t ∈ q(i)}

q∈Q

= λ{z : z ∈ Ω, #({t : t ∈ T, z(t) ∈ Ei }) = ni for every i < r} = λ{z : z ∈ VT , #({t : t ∈ T, z(t) ∈ Ei }) = ni for every i < r} = λ{z : z ∈ VT , #(z[T ] ∩ Ei ) = ni for every i < r} = λ{z : z ∈ VT , φT (z) ∈ H} = νT H.

812

Further topics

495I

Since this is true for ν˜-almost every T , Z

Z νT (H)˜ ν (dT ) =

(495H) =

νT0 (H 0 )˜ ν (dT ) = ν 0 H 0 Y (γµ0 E 0 )ni i

i
ni !

0

0

e−γµ Ei =

Y (γµEi )ni ni !

i
Now, just as in part (e) of the proof of 495H, 495C tells us that H, so that hνT iT ⊆X˜ is a disintegration of ν over ν˜.

R

e−γµEi = νH. Q Q

νT (H)˜ ν (dT ) = νH whenever ν measures

˜ f , and take ni ∈ N for i < r. Set (c) Let hFi ii
˜ = {S : S ⊆ X, f [S] ∈ H}. ˜ H = f˜−1 [H]

˜ ) for every T ∈ W2 ∩ H. ˜ P Then νT H = χH(T P For i ≤ r and t ∈ T ∩ Fi , we have λ{z : z(t) ∈ f −1 [Fi ]} = µt f −1 [Fi ] = 1 because T ⊆ Y2 . So V = {z : z ∈ VT , f (z(t)) ∈ Fi whenever i ≤ r and t ∈ T ∩ Fi } is λ-conegligible. But if z ∈ V then f z¹T is injective, so #(f [z[T ]] ∩ Fi ) = #(T ∩ (f z)−1 [Fi ]) = #(T ∩ Fi ) ˜ Thus for every i < r, and z[T ] ∈ H iff T ∈ H. ˜ νT H = λ{z : z[T ] ∈ H} = λV = 1 if T ∈ H, ˜

= λ(X X \ V ) = 0 otherwise. Q Q Setting ˜ :H ˜ ⊆ X, ˜ νT f˜−1 [H] ˜ = χH(T ˜ ) for ν˜-almost every T ∈ H}, ˜ H = {H ˜ #(T ∩ Fi ) = ni for it is easy to check that H is a Dynkin class containing all sets of the form {T : T ⊆ X, f ˜ every i < r} where hFi ii
495K


813

495J Proposition Let (A, µ ¯) be a measure algebra, and γ > 0. Then there are a probability algebra ¯ and a function θ : A → B such that (B, λ) (i) θ(sup A) = sup θ[A] for every non-empty A ⊆ A such that sup A is defined in A; ¯ (ii) λθ(a) = 1 − e−γ µ¯a for every a ∈ A, interpreting e−∞ as 0; (iii) whenever hai ii∈I is a disjoint family in A and Ci is the closed subalgebra of B generated by {θ(a) : a ⊆ ai } for each i, then hCi ii∈I is stochastically independent. proof (a) We may suppose that (A, µ ¯) is the measure algebra of a measure space (X, Σ, µ) (321J). Set Σf = {E : E ∈ Σ, µE < ∞} and Af = {a : a ∈ A, µ ¯a < ∞}. Let (Ω, Λ, λ) and hgE iE∈Σf be as in ¯ to be the measure algebra of (Ω, Λ, λ). Note that if E, F ∈ Σf and µ(E4F ) = 0, 495B, and take (B, λ) then gE\F and gF \E have Poisson distributions with expectation 0, so are zero almost everywhere, while gE =a.e. gE∩F + gE\F and gF =a.e. gE∩F + gF \E ; so that gE =a.e. gF . This means that we can define ¯ θ : Af → B by setting θ(E • ) = {ω : gE (ω) 6= 0}• whenever E ∈ Σf , and we shall have λ(θa) = 1 − e−γ µ¯a f • because gE has a Poisson distribution with expectation µ ¯a whenever E ∈ Σ and E = a. For a ∈ A \ Af set θ(a) = 1B . (b) If a, b ∈ Af are disjoint, they can be represented as E • , F • where E, F ∈ Σf are disjoint. In this case, gE∪F =a.e. gE + gF , so θ(a ∪ b) = θ(a) ∪ θ(b). Of course the same is true if a, b ∈ A are disjoint and either has infinite measure. It follows at once that for any a, b ∈ A, θ(a ∪ b) = θ(a \ b) ∪ θ(a ∩ b) ∪ θ(b \ a) = θ(a) ∪ θ(b). Consequently θ(sup A) = sup θ[A] for any finite set A ⊆ A. If A ⊆ A is an infinite set with supremum a∗ , then A0 = {sup B : B ∈ [A]<ω } is an upwards-directed set with supremum a∗ , so there is a non-decreasing sequence han in∈N in A0 such that limn→∞ µ ¯ an = µ ¯a∗ (321D). In this case, b∗ = supn∈N θ(an ) is defined in B and ¯ ∗ = limn→∞ λθ(a ¯ n ) = limn→∞ 1 − e−γ µ¯an = 1 − e−γ µ¯a∗ = λθ(a ¯ ∗ ). λb So b∗ = θ(a∗ ); since θ(a∗ ) is certainly an upper bound of θ[A0 ], it must actually be the supremum of θ[A0 ] and therefore (because θ preserves finite suprema) of θ[A]. (c) Thus θ satisfies in A, then Q ¯(i) and (ii). As for (iii), note first that if hai ii∈I is a finite disjoint family ¯ P Set J = {i : i ∈ I, µ ¯ai < ∞}. For i ∈ J, represent ai as Ei• where hEi ii∈J λ(inf i∈I θ(ai )) = i∈I λθ(ai ). P is a disjoint family in Σf . Then hgEi ii∈J is independent, so \ ¯ ¯ λ(inf θ(ai )) = λ(inf θ(ai )) = λ(Ω ∩ {ω : gEi (ω) = 0}) i∈I

i∈J

=

Y i∈J

i∈J

λ{ω : gEi (ω) = 0} =

Y i∈J

¯ i) = λθ(a

Y

¯ i ). Q λθ(a Q

i∈I

Now suppose that hai ii∈I is a finite disjoint family in A and that Di is the subalgebra of B generated by ¯ D i Q = ¯{θ(a) : a ⊆ ai } for each i. We know that each Di is closed under ∪ (by (i)) and that λ(inf i∈J di ) = i∈J λdi whenever J ⊆ I and di ∈ Di for each i ∈ J, that is, that hdi ii∈I is stochastically independent whenever di ∈ Di for each i. Setting Di0 = {1 \ d : d ∈ Di } ∪ {0}, we see that Di0 is closed under ∩ and that hdi ii∈I is stochastically independent whenever di ∈ Di0 for each i (as in 272F). An induction on #(J), using 313Ga for i ∈ J, and Q di ∈ Di0 for i ∈ I \ J, then Q the¯inductive step, shows that if J ⊆ I, di ∈ Di for ¯ ¯ ¯ i whenever di ∈ Di λ(inf i∈I di ) = i∈I λdi . At the end of the induction, we see that λ(inf i∈I di ) = i∈I λd for each i, and therefore whenever di belongs to the topological closure of Di for each i, where B is given its measure-algebra topology (§323). Finally, suppose that hai ii∈I is any disjoint family in A, and Ci is the closed subalgebra of B generated by Di = {θ(a) : a ⊆ ai } for each i. Take a finite set J ⊆ I and ci ∈ Ci for each i ∈ J. By 323J, Q C¯i is the ¯ topological closure of the subalgebra Di of B generated by {θ(a) : a ⊆ ai }; so λ(inf i∈J ci ) = i∈J λci . As hci ii∈J is arbitrary, hCi ii∈I is independent. 495K Proposition Let U be any L-space. Then there are a probability space (Ω, Λ, λ) and a positive linear operator T : U → L1 (λ) such that kT uk1 = kuk1 whenever u ∈ L1 (µ)+ and hT ui ii∈I is stochastically independent in L0 (λ) whenever hui ii∈I is a disjoint family in L1 (µ).

814

Further topics

495K

Remarks Recall that a family hui ii∈I in a Riesz space is ‘disjoint’ if |ui | ∧ |uj | = 0 for all distinct i, j ∈ I (352C). A family hvi ii∈I in L0 (λ) is ‘independent’ if hgi ii∈I is an independent family of random variables whenever gi ∈ L0 (λ) represents vi for each i; compare 367W. proof (a) By Kakutani’s theorem, there is a measure algebra (A, µ ¯) such that U is isomorphic, as Banach lattice, to L1 (A, µ ¯); now (A, µ ¯) can be represented as the measure algebra of a measure space (X, Σ, µ), and we can identify U and L1 (A, µ ¯) with L1 (µ) (365B). Set Σf = {E : E ∈ Σ, µE < ∞} and Af = {a : a ∈ A, µ ¯a < ∞} as usual. Take (Ω, Λ, λ) and hgE iE∈Σf from 495B, with γ = 1. As in the proof of 495J, we have gE =a.e. gF whenever E, F ∈ Σf and µ(E4F ) = 0; consequently we can define ψ : Af → L1 (λ) by setting • ψa = gE whenever E ∈ Σf and E • = a. Again as in 495J, gE∪F =a.e. gE + gF whenever E, F ∈ Σf are disjoint, so ψ is additive. Also kψak1 = f

R

gE dλ = µE = µ ¯a

f

whenever E ∈ Σ represents a ∈ A . By 365I, there is a unique bounded linear operator T : L1 (A, µ ¯) → L1 (λ) such that T (χa) = ψa for every a ∈ Af . By 365Ka, T is a positive operator. The set {u : u ∈ L1 (A, µ ¯)+ , kT uk1 = kuk1 } is closed under addition, norm-closed and contains αχa for every a ∈ Af and α ≥ 0, so is the whole of L1 (A, µ ¯)+ , by 365F. (b) Now let hui ii∈I be a disjoint family in L1 (A, µ ¯). Then hT ui ii∈I is independent. P P?? Otherwise, there are a finite set J ⊆ I and a family hVi ii∈J such that Vi is a neighbourhood of T ui in the topology of convergence in measure on L0 (µ) for each i ∈ J, and hvi ii∈J is not independent whenever vi ∈ Vi for each i (367W). Because the embedding L1 (λ) ⊆ L0 (λ) is continuous for the norm topology on L1 (λ) and the topology of convergence in measure (245G), there is a δ > 0 such that T u0i ∈ Vi whenever i ∈ J, 0 f 0 u0i ∈ L1 (A, µ ¯) and ku0i − i k1 ≤ δ. Now we can find such ui ∈ S(A ) with |ui | ≤ |ui | (365F). Pu ni 0 Express each ui as k=0 αik χaik where haik ik≤ni is a disjoint family in Af and no αik is zero (361Eb). In this case, all the aik , P for i ∈ J and k ≤ ni , are disjoint, so all the ψ(aik ) are independent. But this ni αik ψ(aik )ii∈J is independent (272K); which is impossible, because T u0i ∈ Vi means that hT u0i ii∈J = h k=0 for every i ∈ J. X XQ Q So T , regarded as a function from U to L1 (λ), has the required properties. 495L The following is a more concrete expression of the same ideas. Proposition Let (X, Σ, µ) be an atomless measure space, and ν the Poisson point process on X with density γ > 0. R R P (a) If h ∈ L1 (µ), Qh (S) = x∈S∩dom h h(x) is defined for ν-almost every S ⊆ X, and Qh dν = γ h dµ. (b) We have a positive linear operator T : L1 (µ) → L1 (ν) defined by setting T (h• ) = Q•h for every h ∈ L1 (µ). (c) kT uk1 = γkuk1 whenever u ∈ L1 (µ)+ and hT ui ii∈I is stochastically independent in L0 (λ) whenever hui ii∈I is a disjoint family in L1 (µ). R proof (a) In the language of 495D, QχE = fE for every E ∈ Σf . So QχE ∈ L1 (ν) and QχE dν = γµE P P r r 1 for every E ∈ RΣf . If h = i=0 αi χEi is a simple function on X, then Qh =a.e. i=0 αi QχEi ∈ L (ν) R 1 −1 and Qh dν = h dµ. If h ∈ L (µ) is zero a.e., then {S : S ⊆ h [{0}]} is ν-conegligible, so Qh = 0 a.e. It follows that if h ∈ L1 (µ) is non-negative, and hhn in∈N is a non-decreasing sequence of simple functions converging to h almost everywhere, then Qh =a.e. limn→∞ Qhn , while Qhn ≤a.e. Qhn+1 for every n; so Qh is ν-integrable and R R R R Qh dν = limn→∞ Qhn dν = limn→∞ γ hn dµ = γ hdµ. (b) Since Qh =a.e. Qh0 whenever h =a.e. h0 in L1 (µ), we can define T : L1 (µ) → L1 (ν) by setting T (h• ) = (Qh )• for every h ∈ L1 (µ); because Qαh =a.e. αQh and Qh+h0 =a.e. Qh + Qh0 whenever h, h0 ∈ L1 (µ) and α ∈ R, T is linear; because Qh ≥ 0 a.e. whenever h ≥ 0 a.e., T is positive. R R (c) Because Qh dν = γ h dµ for every h ∈ L1 (µ), and T is positive, kT uk1 = γkuk1 for every u ∈ L1 (µ)+ . Finally, if hui ii∈I is a finite disjoint family in L1 (µ), we can find a family hhi ii∈I of measurable functions from X to R such that h•i = ui for each i and the sets Ei = {x : hi (x) 6= 0} are disjoint. For each i ∈ I, let Ti be the σ-algebra of subsets of PX generated by sets of the form {S : #(S ∩ E) = n} where

495M


815

E ⊆ Ei has finite measure and n ∈ N. Then hTi ii∈I is independent (as in part (c) of the proof of 495J), and each Qhi is Ti -measurable, so hQhi ii∈I is independent and hT ui ii∈I is independent. 495M We can identify the characteristic functions of the random variables Qf as defined above. Proposition Let (X, Σ, µ) be an atomless measure space, and ν the Poisson point process on X with density P γ > 0. For f ∈ L1 (µ) set Qf (S) = x∈S∩dom f f (x) when S ⊆ X and the sum is defined in R. Then

¡ R

R

PX

eiyQf dν = exp γ

X

(eiyf − 1)dµ

¢

for any y ∈ R. proof Note that Qf is defined ν-almost everywhere, by 495La.

Pn (a) Consider first the case in which f is a simple function, expressed as j=0 αj χFj where hFj ij≤n is Pn a disjoint family of sets of finite measure and αj ∈ R for each j. Then Qf (S) = j=0 αj #(S ∩ Fj ) for ν-almost every S, so Z eiyQf dν =

Z Y n

eiyαj #(S∩Fj ) ν(dS) =

j=0

n Z Y

eiyαj #(S∩Fj ) ν(dS)

j=0

(because the functions S 7→ #(S ∩ Fj ) are independent) n ∞ n X ∞ Y (γµFj )k −γµFj iyαj k Y −γµFj X (eiyαj γµFj )k e e = e = k! k! j=0 j=0 k=0

k=0

=

n Y

¡ ¢ ¡ exp (eiyαj − 1)γµFj ) = exp γ

j=0

¡ = exp γ

n X

(eiyαj − 1)µFj

¢

j=0

Z (eiyf

¢ − 1)dµ .

(b) Now suppose that f is any integrable function. Then there is a sequence hfn in∈N of simple functions such that |fn | ≤a.e. |f | for every n and limn→∞ fn =a.e. f . Write qn , q for Qfn , Qf . Set D = {x : x ∈ dom f , |fn (x)| ≤ |f (x)| for every n and limn→∞ fn (x) = f (x)}, so that D is µ-conegligible. |f | (S) is defined, then q(S) = limn→∞ qn (S), and this is true for R iyq If S ⊆ D and Q iyqn ν-almost every S; so e dν = limn→∞ e dν, by Lebesgue’s Dominated Convergence Theorem. On the other hand, |eiα − 1| = |

Rα1 0 i

eit dt| ≤ α,

|e−iα − 1| = |

Rα1 0 i

e−it dt| ≤ α

for every α ≥ 0. So if we set g(x) = eiyf (x) − 1, gn (x) = eiyfn (x) − 1 when these are defined, we have |gn | ≤a.e. |yfn | ≤a.e. |yf | for every n. Accordingly

R

(eiyf − 1)dµ =

R

limn→∞ (eiyfn − 1)dµ = limn→∞

R

(eiyfn − 1)dµ

by Lebesgue’s theorem again. It follows that Z

Z e

iyQf

dν = lim

n→∞

e

iyqn

(by (a))

¡ dν = lim exp γ n→∞

Z

¡

= exp γ lim

n→∞

(e

iyfn

¢

Z

¢ (eiyfn − 1)dµ

¡ − 1)dµ = exp γ

Z

¢ (eiyf − 1)dµ ,

as claimed. Remark Recall that a Poisson random variable with expectation γ has characteristic function y 7→ exp(γ(eiy − 1)) (part (a) of the proof of 285Q), corresponding to the case f = χF where µF = 1. The random variables Qf have compound Poisson distributions.

816

Further topics

495N

495N If our underlying measure is a Radon measure, we can look for Radon measures on PX to represent the Poisson point processes on X. There seem to be difficulties in general, but I can offer the following. See also 495Yc. Proposition Let (X, T, Σ, µ) be a σ-compact locally compact Radon measure space, and γ > 0. Give the space C of closed subsets of X its k-Vietoris topology (476Cd). (i) There is a unique Radon measure ν˜ on C such that Qr ν˜{C : #(C ∩ Ei ) 6= ∅ for every i ≤ r} = i=0 1 − e−γµEi whenever E0 , . . . , Er ⊆ X are disjoint sets of finite measure. (ii) If E0 , . . . , Er are disjoint sets of finite measure, none including any singleton set of non-zero measure, and ni ∈ N for i ≤ r, then Qr (γµEi )ni −γµEi e . ν˜{C : #(C ∩ Ei ) = ni for every i ≤ r} = i=0 ni !

(iii) If µ is atomless and ν is the Poisson point process on X with density γ, then C is ν-conegligible and ν˜ extends the subspace measure ν¹C. proof (a) Recall that C is Hausdorff (476Cd). Because X is locally compact and σ-compact, S there is a sequence hHn in∈N of relatively compact open sets covering X. For each n set Hn0 = Hn \ i 0}. Then H is a countable maximal almost-disjoint family of measurable sets of non-zero finite measure. Build Ω, λ and φ : Ω → PX from H as in the proof of 495B. This time, however, observe that all the µ0H are Radon measures (416Rb), while surely the Poisson distributions νH on N also are, so that we have a Radon ˜ on Ω extending λ (417Q). measure λ (b) φ(ω) ∈ C for every ω ∈ Ω. P P For each H ∈ H, H ∩ φ(ω) = {xHn (ω) : n < mH (ω)} is finite. Since each Hn meets only finitely many members of H, φ(ω) ∩ Hn is finite for each n; because {Hn : n ∈ N} is an open cover of the Hausdorff space X, this is enough to ensure that φ(ω) is closed. Q Q (c) φ : Ω → C is continuous. P P (i) If G ⊆ X is open, then S {ω : G ∩ φ(ω) 6= ∅} = H∈H,n∈N {ω : mH (ω) > n, xHn (ω) ∈ G} S is open. (ii) If K ⊆ X is compact, there is some r ∈ N such that K ⊆ i≤r Hi , so that H0 = {H : H ∈ H, H ∩ K 6= ∅} is finite. Now T S T / K} {ω : φ(ω) ∩ K = ∅} = H∈H0 n∈N i
as required by (i). (e) To see that ν˜ is uniquely defined, let ν˜ 0 be another Radon probability measure on C with the same property. Suppose that we have a compact set K ⊆ X and open sets Gi ⊆ X, for i < r, and set V = {C : C ∈ C, C ∩ K = ∅, C ∩ Gi 6= ∅ for every i < r}. S Then ν˜V = ν˜ V . P P For each n ∈ N, set Gni = Gi ∩ j≤n Hj for each i, and set 0

Vn = {C : C ∈ C, C ∩ K = ∅, C ∩ Gni 6= ∅ for every i < r}. Let En be the finite subalgebra of PX generated by {Gni : i < r} ∪ {K}, An the set of atoms of En and Afn the set of members of An of finite measure. For I ⊆ Afn set

495O


Then Vn =

817

VnI = {C : C ∈ C, I = {A : A ∈ Afn , C ∩ A 6= ∅}}.

S I∈In

VnI , where In = {I : I ⊆ Afn , A ∩ K = ∅ for every A ∈ I, for every i < r there is an A ∈ I such that A ⊆ Gni }.

Now the defining formula for ν˜ shows that ν˜VnI = ν˜ 0 VnI for every I ⊆ Afn , so that ν˜Vn = ν˜ 0 Vn ; and now ν˜V = limn→∞ ν˜Vn = limn→∞ ν˜ 0 Vn = ν˜ 0 V . Q Q Since sets V of the type described form a base for the k-Vietoris topology closed under finite intersections, ν˜ = ν˜ 0 (415H(v)). (f ) If E ∈ Σ has finite measure and does not include any non-negligible singleton, then #(E ∩ φ(ω)) = gE (ω) for almost every ω ∈ Ω. P P As in part (c) of the proof of 495D, #(E ∩ φ(ω)) = gE (ω) for every ω ∈ Ω \ AE , where AE = {ω : there are distinct H, H 0 ∈ H and j ∈ N such that xHj (ω) ∈ H 0 } ∪ {ω : there are H ∈ H and distinct i, j ∈ N such that xHi (ω) = xHj (ω) ∈ E}, and AE is negligible, because the subspace measure on E is atomless (414G/416Xa). Q Q So if E0 , . . . , Er are disjoint sets of finite measure, none including any non-negligible singleton, and ni ∈ N for i ≤ r, ˜ : gE (ω) = ni for every i ≤ r} ν˜{C : #(C ∩ Ei ) = ni for every i ≤ r} = λ{ω i r Y (γµEi )ni −γµEi = e , i=0

ni !

as required by (ii). (g) Now suppose that µ is atomless. Let ν be the Poisson point process on X with density γ. For each n, the set {S : S ∩ Hn is finite} is ν-conegligible, so {S : S ∩ Hn is finite for every n} is ν-conegligible; but as hHn in∈N is an open cover of X, this is included in C. So C is ν-conegligible. Let T be the σ-algebra of subsets of PX generated by sets of the form {S : #(S ∩ E) = n} where µE < ∞ and n ∈ N. Then (ii) here and 495C tell us that ν˜(C ∩ S) = νS whenever S belongs to T and therefore whenever ν measures S; so that ν˜ extends the subspace measure νC . 495O Proposition Let (X, T) be a σ-compact locally compact Hausdorff space and MR+ (X) the set of Radon measures on X. Give MR+ (X) the topology generated by sets of the form {µ : µG > α} and {µ : µK < α} where G ⊆ X is open, K ⊆ X is compact and α ∈ R. Let C be the space of closed subsets of X with the k-Vietoris topology, and PR (C) the set of Radon probability measures on C with its narrow topology (definition: 437Jd). Fix γ > 0, and for µ ∈ MR+ (X) let ν˜µ be the Radon measure on C defined in 495N. Then the function µ 7→ ν˜µ : MR+ (X) → PR (C) is continuous. proof (a) Recall that because X is locally compact and Hausdorff, C is compact and Hausdorff (476Cd). Fix an open set W0 ⊆ C, α0 > 0 and µ0 ∈ MR+ (X) such that ν˜µ0 W0 > α0 . Let E be the family of relatively compact Borel subsets E of X such that µ0 (∂E) = 0. Then E is a subring of PX (4A2Bi). Also µ 7→ µE : MR+ (X) → [0, ∞[ is continuous at µ0 for every E ∈ E. P P If E ∈ E and ² > 0, then {µ : µ0 E − η < µE < µ0 E + η} ⊇ {µ : µ(int E) > µ0 (int E) − η, µ(X \ E) > µ0 (X \ E) − η} is a neighbourhood of µ0 . Q Q (b) Next, U = E ∩ T is a base for T (411Gi). It follows that the family V of sets of the form {C : C ∈ C, C ∩ Ui 6= ∅ for i ≤ r, C ∩ U = ∅}, where U , U0 , . . . , Ur ∈ U , is a base for the k-Vietoris topology on C. P P If W ⊆ C is open for the k-Vietoris topology and C0 ∈ W , there are r ∈ N, open sets Gi ⊆ X for i ≤ r and a compact set K ⊆ X such that

818

Further topics

495O

C0 ∈ {C : C ∩ Gi 6= ∅ for each i ≤ r, C ∩ K = ∅} ⊆ W . For each i ≤ r choose xi ∈ C0 ∩ Gi and Ui ∈ U such that xi ∈ Ui ⊆ Gi . Because X is locally compact and Hausdorff, it is regular, so every point of K belongs to a member of U with closure disjoint from C0 ; because U is closed under finite unions, there is a U ∈ U such that K ⊆ U and C0 ∩ U = ∅. Now {C : C ∈ C, C ∩ Ui 6= ∅ for i ≤ r, C ∩ U = ∅} belongs to V, contains C0 and is included in W . As C0 and W are arbitrary, V is a base for the k-Vietoris topology on C. Q Q (c) If V ∈ V, then µ 7→ ν˜µ V : MR+ (X) → [0, 1] is continuous at µ0 . P P Express V as {C : C ∈ C, C ∩ Ui 6= ∅ for i ≤ r, C ∩ U = ∅}, where Ui , U ∈ U. Let A be the set of atoms of the finite subring of E generated by {Ui : i ≤ r} ∪ {U }. For I ⊆ A set VI = {C : C ∈ C, I = {A : A ∈ A, C ∩ A 6= ∅}. Let I be the set of those I ⊆ A such that A ∩ U = ∅ for every A ∈ I and for every i ≤ r there is an A ∈ I such that A ⊆ Ui . Then hVI iI∈I is a partition of V . Moreover, for any µ ∈ MR+ (X) and I ⊆ A, Q Q ν˜µ VI = A∈I e−γµA A∈A\I (1 − e−γµA ). Since each µ 7→ µA is continuous at µ0 , by (a), so are the functionals µ 7→ ν˜µ VI , for I ⊆ A, and µ 7→ ν˜µ V = P ˜µ VI . Q Q I∈I ν (d) Let V ∗ be the family of Borel subsets V of C such that µ 7→ ν˜µ V : MR+ (X) → [0, ∞[ is continuous at µ0 . Then V ⊆ V ∗ (by (c)), C ∈ V ∗ and V \ V 0 ∈ V ∗ whenever V , V 0 ∈ V ∗ and V 0 ⊆ V . Because V is closed under finite intersections, it follows that V ∗ includes the algebra of subsets of C generated by V (313Ga); in particular, any finite union of members of V belongs to V ∗ . (e) Let us return to the open set W0 ⊆ C and the α0 ∈ R of part (a). Because ν˜µ0 is τS-additive and V is a base for the topology of C ((b) above), there is a finite family V0 ⊆ V such that V0 = V0 is included in W0 and ν˜µ0 V0 > α0 . But this means that {µ : µ ∈ MR+ (X), ν˜µ W0 > α0 } ⊇ {µ : µ ∈ MR+ (X), ν˜µ V0 > α0 } is a neighbourhood of µ0 . As µ0 is arbitrary, {µ : ν˜µ W0 > α0 } is open; as W0 and α0 are arbitrary, µ 7→ ν˜µ is continuous. 495P There are many constructions which, in particular cases, can be used as an alternative to the method of 495B-495D in setting up Poisson point processes. I give one which applies to the half-line [0, ∞[ with Lebesgue measure. Theorem Let γ > 0, and let ν be the Poisson point process on [0, ∞[, with Lebesgue measure, with density γ. Let λ0 be the exponential distribution with expectation 1/γ, regarded as a Radon probability measure N N on ]0, ∞[, and λ the corresponding product measure on ]0, ∞[ . Define φ : ]0, ∞[ → P([0, ∞[) by setting Pn N N φ(x) = { i=0 x(i) : n ∈ N} for x ∈ ]0, ∞[ . Then φ is a measure space isomorphism between ]0, ∞[ and a ν-conegligible subset of P([0, ∞[). Remark As I seem not to have mentioned exponential distributions earlier in this treatise, I remark now that the exponential distribution with parameter γ has distribution function F (t) = 0 if t < 0, 1 − e−γt if t ≥ 0, and probability density function f (t) = 0 if t ≤ 0, γe−γt if t > 0; its expectation is

R∞ 0

γte−γt dt = −

R∞ 0

d γt+1 −γt e dt dt γ

1 γ

= .

Because (when regarded as a Radon probability measure on R, following my ordinary rule set out in §271) it gives measure zero to ]−∞, 0], it can be identified with the subspace measure on ]0, ∞[, as here.

495P


819

proof (a) For each n ∈ N, #(S ∩ [0, n]) is finite for ν-almost every S; so the set Q0 = {S : S ⊆ [0, ∞[, #(S ∩ [0, n]) is finite for every n} is ν-conegligible. Next, the sets {S : S ∩ [n, n + 1[ 6= ∅} are ν-independent and have measure 1 − e−γ > 0, so {S : S ∩ [n, n + 1[ 6= ∅ for infinitely many n} is ν-conegligible (273K). Finally, ν{S : 0 ∈ S} = 0, so Q = {S : S ∈ Q0 , 0 ∈ / S, S is infinite} is ν-conegligible. For S ∈ Q, let hgn (S)in∈N be the increasing enumeration of S. Let T be the σ-algebra of subsets of P([0, ∞[) generated by sets of the form {S : #(S ∩ E) = n} where E ⊆ [0, ∞[ has finite measure and n ∈ N. Then, for n ∈ N and α ≥ 0, {S : gn (S) ≤ α} = {S : #(S ∩ [0, α]) ≥ n + 1} belongs to T, so gn is T-measurable. N Set h0 (S) = g0 (S), hn (S) = gn (S) − gn−1 (S) for n ≥ 1, and h(S) = hhn (S)in∈N ; then h : Q → ]0, ∞[ is a bijection, and its inverse is φ. £ £ (b) For each k ∈ N, IS = {i : i ∈ N, S ∩ 2−k i, 2−k (i + 1) 6= ∅} is infinite for every S ∈ Q. So we can define gkn : Q → ]0, ∞[, for each n, by taking gkn (S) = 2−k (j + 1) if j ∈ IS and #(IS ∩ j) = n. Because all the sets {S : j ∈ IS } belong to T, each gkn is T-measurable, and hgkn ik∈N is a non-increasing sequence with limit gn . Set hk0 (S) = gk0 (S) and hkn (S) = gk,n+1 (S) − gkn (S) for n ≥ 1. Then hn = limk→∞ hkn . Pr (d) For any n ∈ N, j0 , . . . , jn ∈ N, k ∈ N, set jr0 = i=0 ji for r ≤ n. Then ν{S : S ∈ Q, hki (S) = 2−k (ji + 1) for every i ≤ n} = ν{S : S ∈ Q, gkr (S) = 2−k (r + 1 + jr0 ) for every r ≤ n} £ £ = ν{S : S ∩ 2−k (r + jr0 ), 2−k (r + 1 + jr0 ) 6= ∅ for every r ≤ n, £ £ S ∩ 0, 2−k j0 = ∅, £ £ 0 S ∩ 2−k (r + 1 + jr0 ), 2−k (r + 1 + jr+1 ) = ∅ for every r < n} Y 0 = (1 − exp(−2−k γ))n+1 exp(−2−k γj0 ) exp(−2−k γ(jr+1 − jr0 )) r
=

n Y

(1 − exp(−2−k γ)) exp(−2−k γji )

i=0

This means that the hki , for i ∈ N are independent, with Pr(hki = 2−k (j + 1)) = (1 − e−2

−k

γ

)e−2

−k

γj

for each j. Since hki → hi ν-a.e. for each i, hhi ii∈N is also independent (367W). Now, for any α > 0, Pr(hki ≤ α) =

X

(1 − exp(−2−k γ)) exp(−2−k γj)

2−k (j+1)≤α

= 1 − exp(−2−k γb2k αc) → 1 − e−γα as k → ∞. So Pr(hi ≤ α) = inf lim inf Pr(hki ≤ β) β>α i→∞

(271L) = inf 1 − e−γβ = 1 − e−γα β>α

for every α ≥ 0 and every i ∈ N. (e) Accordingly hhi ii∈N is an independent sequence of random variables, each exponentially distributed N with expectation 1/γ. It follows that h : Q → ]0, ∞[ is inverse-measure-preserving for the subspace measure νQ and λ (254G).

820

Further topics

495P

Observe next that if E ⊆ [0, ∞[ is Lebesgue measurable and n ∈ N, then S Pj N {x : x ∈ ]0, ∞[ , #(φ(x) ∩ E) = n} = I∈[N]n {x : i=0 x(i) ∈ E ⇐⇒ j ∈ I} ∈ Λ, writing Λ for the domain of λ. So φ is (Λ, T)-measurable. Now, for any W ∈ T, λφ−1 [W ] = ν(h−1 [φ−1 [W ]]) = ν(W ∩ Q) = νW . So φ is inverse-measure-preserving for λ and for ν¹T. Since λ is complete and ν is defined as the completion of its restriction to T, φ is inverse-measure-preserving for λ and ν. Thus φ and h are the two halves of an N isomorphism between (]0, ∞[ , λ) and the subspace (Q, νQ ), as claimed. 495X Basic exercises (a) Let (X, Σ, µ) be an atomless measure space, and ν a Poisson point process on X. (i) Show that [X]≤ω has full outer measure for ν. (ii) Show that if µ is semi-finite then [X]≤ω is conegligible iff µ is σ-finite. (iii) Show that if µ is semi-finite, then [X]<ω is non-negligible iff [X]<ω is conegligible iff µ is totally finite. (b) Let (X, Σ, µ) be an atomless measure space and E a countable partition of X into measurable sets; fix γ > 0. Let ν be the Poisson point process on X with density γ, and for each E ∈ E let νE be the Poisson point process on E with density γ corresponding to the subspace measure µE on E. Let λ be the Q product of the family hνE iE∈E . Show that the map S 7→ hS ∩ EiE∈E : PX → E∈E PE is a measure space isomorphism for ν and λ. > (c) Let (X, Σ, µ) be an atomless measure space and T a topology on X such that X is covered by a sequence of open sets of finite outer measure. Let ν be a Poisson point process on X. Show that ν-almost every set S ⊆ X is locally finite in the sense that X is covered by the open sets meeting S in finite sets; in particular, if X is Hausdorff, then ν-almost every subset of X is closed. > (d) Let (X, Σ, µ) be an atomless measure space, and for γ > 0 let νγ be the Poisson point process on X with density γ. Show that for any γ, δ > 0 the map (S, T ) 7→ S ∪ T : PX × PX → PX is inverse-measurepreserving for the product measure νγ × νδ and νγ+δ . (e) Let (X, Σ, µ) be an atomless semi-finite measure space, and ν the Poisson point process on X with density 1. Show that ν is perfect iff µ is. f

(f ) Let (A, µ ¯) be a measure algebra. Show that there is a probability measure λ on RA such that (i) for f every a ∈ A the corresponding marginal measure on R is the Poisson distribution with expectation µ ¯a (ii) f Af whenever a0 , . . . , an ∈ A are disjoint, the functions z 7→ z(ai ) : R → R are stochastically independent with respect to λ. (Hint: prove the result for finite A and use 454D.) Use this to prove 495J. ¯ and a linear operator (g) Let U be a Hilbert space. Show that there are a probability algebra (B, λ) 0 T : U → L (B) such that (i) for every u ∈ U , T u has a normal distribution with expectation 0 and variance ¯ kuk22 (ii) if hui ii∈I is an orthogonal family in U then hT ui ii∈I is λ-independent. (Hint: see the proof of 456G.) ¯ and a linear operator T : (h) Let U be an L-space. Show that there are a probability algebra (B, λ) 1 ¯ U → L (B, λ) such that (i) for every u ∈ U , T u has a standard gamma distribution (definition: 455Xe) ¯ with expectation kuk (ii) if hui ii∈I is a disjoint family in U then hT ui ii∈I is λ-independent. (i) Let ν be the Poisson point process with density 1 on [0, ∞[ with Lebesgue measure. Set Q0 = {S : S ⊆ [0, ∞[, S ∩ [0, n] is finite for every n} and for S ∈ Q0 set ψ(S)(t) = #(S ∩ [0, t]) for t ∈ [0, ∞[. Show that ψ is inverse-measure-preserving for the subspace measure νQ0 and the Baire measure on R[0,∞[ described in 455Xc. (j) Let (Y, T, ν) be a probability space, and λ0 the exponential distribution with expectation 1, regarded N N N as a Radon measure on ]0, ∞[. Let λ be the product measure λN 0 × ν on ]0, ∞[ × Y . Set φ(x, y) = Pn N N N N {( i=0 x(i), y(n)) : n ∈ N} for x ∈ ]0, ∞[ and y ∈ Y . Show that φ : ]0, ∞[ × Y → P(]0, ∞[ × Y ) is a N measure space isomorphism between (]0, ∞[ × Y N , λ) and a conegligible set for the Poisson point process on ]0, ∞[ × Y with density 1 for the c.l.d. product measure µL × ν, where µL is Lebesgue measure.

495Yc


821

(k) Let C be the family of closed subsets of [0, ∞[. Let ρ be the usual metric on [0, ∞[ and ρ˜ the corresponding Hausdorff metric on C \ {∅}. Let ν the Poisson point process on [0, ∞[ with density 1 over Lebesgue measure. Show that any ρ˜-compact subset of C \ {∅} is ν-negligible. (l) Show that the topology on MR+ (X) described in 495O is just the topology induced by the natural embedding of MR (X) into Ck (X)∼ (436J) and the weak topology Ts (Ck (X)∼ , Ck (X)), where Ck (X) is the Riesz space of continuous real-valued functions on X with compact support. (m) Let X be a σ-compact locally compact Hausdorff space, C the family of closed subsets of X with its k-Vietoris topology, and PR (C) the set of Radon probability measures on C with its narrow topology. Let µ be a Radon measure on X and γ > 0, and ν˜ the Radon measure on C defined from µ and γ as in 495N. Let Q be the set of families q = h(xqi , Eqi )ii∈Iq such that Iq is a countable set, hEqi ii∈Iq is a disjoint family of measurable sets of finite measure, and hxqi ii∈Iq is a family of distinct points in X; for q ∈ Q set Aq = {xqi : i ∈ Iq }. (i) Show that for each q ∈ Q we have a Radon measure ν˜q on C defined by saying that ν˜q (PAq ) = 1 and P ν˜q {C : xqi ∈ / C for i ∈ J} = exp(−γ i∈J µEqi ) for every J ⊆ Iq . (ii) Show that for every neighbourhood V of ν˜ for the narrow topology on PR (C) there are anSopen cover H of X, a compact set K ⊆ X and an ² > 0 such that ν˜q ∈ V whenever q ∈ Q, µ(K \ i∈Iq Eqi ) ≤ ² and for every i ∈ Iq there is an H ∈ H including Eqi ∪ {xqi }. (n) Let C be the set of closed subsets of [0, ∞[ with its k-Vietoris topology. For δ ∈ ]0, 1] let λδ be the measure on {0, 1}N which is the product of copies of the measure on {0, 1} in which {1} is given measure δ. Define φδ : {0, 1}N → C by setting φδ (x) = {nδ : n ∈ N, x(n) = 1}, and let ν˜δ be the Radon measure λδ φ−1 δ on C. Show that the Radon measure on C representing the Poisson point process on [0, ∞[ with density 1 over Lebesgue measure is the limit limδ↓0 ν˜δ for the narrow topology on the space of Radon probability measures on C. (o) Show that the standard gamma distribution with expectation 1 is the exponential distribution with expectation 1. (p) Let r ≥ 1 be an integer; let µ be Lebesgue measure on Rr and βr the volume of the unit ball in Rr . Set ψ(t) = (t/βr )1/r for t ≥ 0, so that the volume of a ball of radius ψ(t) is t. Let Sr−1 be the unit sphere in Rr and θ the invariant Radon probability measure on Sr−1 , so that θ is a multiple of r − 1-dimensional Hausdorff measure (476M). Let λ0 be the exponential distribution with expectation 1, regarded as a Radon N N N probability measure on ]0, ∞[, and λ the product measure λN 0 × θ on ]0, ∞[ × Sr−1 . Set Pn φ(x, z) = {ψ( i=0 x(i))z(n) : n ∈ N} N

N

N N → P(Rr ) is a measure space isomorphism for x ∈ ]0, ∞[ and z ∈ Sr−1 . Show that φ : ]0, ∞[ × Sr−1 N N between ]0, ∞[ × Sr−1 and a conegligible set for the Poisson point process on Rr with density 1.

¯ 495Y Further exercises (a) Let U be an L-space. Show that there are a probability algebra (B, λ) 0 1 and a linear operator T : U → L (B) such that (i) for every u ∈ L (A, µ ¯), T u has a Cauchy distribution ¯ with centre 0 and scale parameter kuk (ii) if hui ii∈I is a disjoint family in U then hT ui ii∈I is λ-independent. ¯ y : L0 (A) → L0 (B) (b) Let (A, µ ¯) be a measure algebra. For α, y ∈ R set hy (α) = eiyα , and let h C (definition: 364Yn) be the corresponding operator (to be defined, following the ideas of 364I or otherwise). ¯ and a positive linear operator T : L1 (A, µ ¯ such Show that there are a probability algebra (B, λ) ¯) → L1 (B, λ) 1 + 0 ¯ that (i) kT uk1 = kuk1 whenever uR ∈ L (A, µ ¯) (ii) hT whenever hui ii∈I is R ui ii∈I is λ-independent in L (B) ¯ y (T u)dλ ¯ = exp( (h ¯ y (u) − χ1)d¯ a disjoint family in L1 (A, µ ¯) (iii) h µ) for every u ∈ L1 (A, µ ¯) and y ∈ R. (c) Let (X, ρ) be a totally bounded metric space, µ a non-totally-finite Radon measure on X, and γ > 0. Let C be the set of closed subsets of X, and ρ˜ the Hausdorff metric on C \ {∅}. (i) Show that there is a unique Radon measure ν˜ on C \ {∅} such that Qr ν˜{C : #(C ∩ Ei ) 6= ∅ for every i ≤ r} = i=0 1 − e−γµEi

822

Further topics

495Yc

whenever E0 , . . . , Er ⊆ X are disjoint sets of finite measure. (ii) Show that if E0 , . . . , Er are disjoint sets of finite measure, none including any singleton set of non-zero measure, and ni ∈ N for i ≤ r, then Qr (γµEi )ni −γµEi ν˜{C : #(C ∩ Ei ) = ni for every i ≤ r} = i=0 e . ni !

(iii) Show that if µ is atomless and ν is the Poisson point process on X with density γ, then C \ {∅} is ν-conegligible and ν˜ extends the subspace measure ν¹C \ {∅}. 495 Notes and comments The underlying fact on which this section relies is that the Poisson distributions form a one-parameter semigroup of infinitely divisible distributions, with να ∗ νβ = να+β for all α, β > 0. Other well-known families with this property are normal distributions, Cauchy distributions and gamma distributions; for each of these we have results corresponding to 495B and 495K (495Xg, 495Ya, 495Xh). The same distributions appeared, for the same reason, in the evolving processes of 455C and 455Xc-455Xe. Observe that the version for the normal distribution is related to the Gaussian processes of §456. The ‘compound Poisson’ distributions of 495M provide further examples, which approach the general form of `ve 77, §23, or Fristedt & Gray 97, §16.3). infinitely divisible distributions (Loe The special feature of the Poisson point process, in this context, is the fact that (for atomless measure spaces (X, µ)) it can be represented by a measure on PX rather than on some abstract auxiliary space (495D); so that we have a notion of ‘random subset’, and can discuss the expected topological properties of subsets of X (495Xa, 495Xc). In Euclidean spaces the geometric properties of these random subsets are also of great interest; see Meester & Roy 96. Here I look at the relations between this construction and others which have been prominent in this book, such as inverse-measure-preserving functions (495G) and disintegrations (495H-495I). In the latter we find ourselves in an interesting difficulty. If, as in 495H, we have a measure ˜ × [0, 1], where X ˜ is an atomless measure space, then it is natural to suppose that our Poisson space X = X ˜ and then, for each t ∈ T , a random process on X can be represented by picking a random subset T of X ˜ × [0, 1]X˜ → PX. (t, α) ∈ X. The obvious model for this idea is the map (T, z) 7→ {(t, z(t)) : t ∈ T } : P X ˜ The problem with this model is that the map is simply not measurable for the standard σ-algebras on P X, ˜ X ˜ × [0, 1] and PX. When we have a canonical ordering in order type ω of almost every subset of X ˜ PX ˜ of course), as in 495Xp, there can be a way (‘almost every’ with respect to the Poisson point process on X, ˜

around this, cutting [0, 1]X down to a countable product and re-inventing the representation of pairs (T, z) as subsets of X. But in the general case it seems that we have to set up a disintegration of the Poisson point ˜ which does not correspond to any measure on a product process on X over the Poisson point process on X ˜ P X × Ω. Following my usual custom, I have expressed the theorems of this section in terms of arbitrary (atomless) measure spaces. The results are not quite without interest when applied to totally finite measures, but their natural domain is the class of non-totally-finite σ-finite measures, as in 495N-495P. There is an unavoidable obstacle if we wish to extend the ideas to measure spaces which are not atomless. The functions S 7→ #(S∩E) can no longer have Poisson distributions, since if E is a singleton of positive measure then we shall have a non-trivial two-valued random variable. In 495N-495O I take one of the possible resolutions of this, with measures ν˜ on spaces of subsets for which at least the functions S 7→ #(S ∩ E), for disjoint E, are independent (495Xm). An alternative which is sometimes appropriate is to work with functions h : X → N P and x∈E h(x) in place of subsets S of X and #(S ∩ E); see Fristedt & Gray 97, §29. In 495J-495L we have a little cluster of results which are relevant to rather different questions, to which I expect to return in Chapter 52 of Volume 5. The objective here is to connect the structure of a measure algebra or Banach lattice of arbitrarily large cellularity with something which can be realized in a probability space. In each case, disjointness is transformed into stochastic independence. Once again, the special feature of the Poisson point process is that we have a concrete representation of a linear operator which can also be described in a more abstract way (495L). The construction of 495B-495D seems to be the most straightforward way to generate Poisson point processes. It fails however to give a direct interpretation of one of the most important approaches to these processes, as limits of purely atomic processes in which sets are chosen by including or excluding individual points independently (495Xm-495Xn). In order to make sense of the limit here it seems that we need to put some further structure onto the underlying measure space, and ‘σ-finite locally compact Radon measure space’ is sufficient to give an interesting pair of positive results (495N-495O).

4A1Bc

Set theory

823

Appendix to Volume 4 Useful facts As is to be expected, we are coming in this volume to depend on a wide variety of more or less recondite information, and only an exceptionally broad mathematical education will have covered it all. While all the principal ideas are fully expressed in standard textbooks, there are many minor points where I need to develop variations on the familiar formulations. A little under half the material, by word-count, is in general topology (§4A2), where I begin with some pages of definitions. I follow this with a section on Borel and Baire σ-algebras, Baire property algebras and cylindrical algebras (§4A3), worked out a little more thoroughly than the rest of the material. The other sections are on set theory (§4A1), linear analysis (§4A4), topological groups (§4A5) and Banach algebras (§4A6). 4A1 Set theory For this volume, we need fragments from four topics in set theory and one in Boolean algebra. The most important are the theory of closed cofinal sets and stationary sets (4A1B-4A1C) and infinitary combinatorics (4A1D-4A1H). Rather more specialized, we have the theory of normal (ultra)filters (4A1J-4A1L) and a mention of Ostaszewski’s ♣ (4A1M-4A1N), used for an example in §439. I conclude with a simple result on the cardinality of σ-algebras (4A1O). 4A1A Cardinals again (a) An infinite cardinal κ is regular if it is not the supremum of fewer than κ smaller cardinals, that is, if cf κ = κ. Any infinite successor cardinal is regular. (Kunen 80, I.10.37; Jech 78, p. 40; Just & Weese 96, 11.18; Levy 79, IV.3.11.) In particular, ω1 = ω + is regular. (b) If ζ is an ordinal, X a set then I say that a family hxξ iξ<ζ in X runs over X with cofinal repetitions if {ξ : ξ < ζ, xξ = x} is cofinal with ζ for every x ∈ X. Now if X is any non-empty set and κ is a cardinal greater than or equal to max(ω, #(X)), there is a family hxξ iξ<κ running over X with cofinal repetitions. P P By 3A1Ca, there is a surjection ξ 7→ (xξ , αξ ) : κ → X × κ. Q Q (c) The cardinal c (i) Every non-trivial interval in R has cardinal c. (Enderton 77, p. 131.) (ii) cf(2κ ) > κ for every infinite cardinal κ; in particular, cf c > ω. (Kunen 80, I.10.40; Jech 78, p. 46; Just & Weese 96, 11.24; Levy 79, V.5.2.) (d) The Continuum Hypothesis This is the statement ‘c = ω1 ’; it is neither provable nor disprovable from the ordinary axioms of mathematics, including the Axiom of Choice. As such, it belongs to Volume 5 rather than to the present volume. But I do at one point refer to one of its immediate consequences. If the continuum hypothesis is true, then there is a well-ordering 4 of [0, 1] such that ([0, 1], 4) has order type ω1 (because there is a bijection f : [0, 1] → ω1 , and we can set s 4 t if f (s) ≤ f (t)). 4A1B Closed cofinal sets Let κ be a cardinal. (a) Note that a subset F of κ is closed in the order topology iff sup A ∈ F whenever A ⊆ F is non-empty and sup A < κ. (4A2S(a-ii).) (b) If κ has uncountable cofinality, and A ⊆ κ has supremum κ, then A0 = {ξ : 0 < ξ < κ, ξ = sup(A∩ξ)} is a closed cofinal set in κ. P P (i) For any η < κ we can choose inductively a strictly increasing sequence hξn in∈N in A starting from ξ0 ≥ η; now ξ = supn∈N ξn ∈ A0 and ξ ≥ η; this shows that A0 is cofinal with κ. (ii) If ∅ 6= B ⊆ A0 and sup B = ξ < κ, then A ∩ ξ ⊇ A ∩ η for every η ∈ B, so S ξ = sup B = supη∈B sup(A ∩ η) = sup( η∈B A ∩ η) ≤ sup A ∩ ξ ≤ ξ and ξ ∈ A0 . Q Q In particular, taking A = κ = ω1 , the set of non-zero countable limit ordinals is a closed cofinal set in ω1 . T (c) If F is a non-empty family of closed cofinal sets in κ and #(F) < cf κ, then F is a closed cofinal set in κ. (Kunen 80, II.6.8; Just & Weese 97, Theorem 21.3; Levy 79, IV.4.15.) In particular, the intersection of any sequence of closed cofinal sets in ω1 is again a closed cofinal set in ω1 .

824

Appendix

4A1Bd

(d) If hFξ iξ<κ is a family of subsets of κ, the diagonal intersection of hFξ iξ<κ is {ξ : ξ < κ, ξ ∈ Fη for every η < ξ}. If κ is a regular uncountable cardinal and hFξ iξ<κ is any family of closed cofinal sets in κ, its diagonal intersection is again a closed cofinal set in κ. (Kunen 80, II.6.14; Jech 78, p. 58; Just & Weese 97, Theorem 21.7; Levy 79, IV.4.20.) In particular, if f : κ → κ is any function, then {ξ : ξ < κ, f (η) < ξ for every η < ξ} is a closed cofinal set in κ, being the diagonal intersection of hκ \ (f (ξ) + 1)iξ<κ . 4A1C Stationary sets (a) Let κ be a cardinal. A subset of κ is stationary in κ if it meets every closed cofinal set in κ; otherwise it is non-stationary. (b) If κ is a cardinal of uncountable cofinality, the intersection of any stationary subset of κ with a closed cofinal set in κ is again a stationary set (because the intersection of two closed cofinal sets is a closed cofinal set); the family of non-stationary subsets of κ is a σ-ideal. (Kunen 80, II.6.9; Just & Weese 97, Lemma 21.11; Levy 79, IV.4.35.) (c) Pressing-Down Lemma (Fodor’s theorem) If κ is a regular uncountable cardinal, A ⊆ κ is stationary and f : A → κ is such that f (ξ) < ξ for every ξ ∈ A, then there is a stationary set B ⊆ A such that f is constant on B. (Kunen 80, II.6.15; Jech 78, Theorem 22; Just & Weese 97, Theorem 21.2; Levy 79, IV.4.40.) (d) There are disjoint stationary sets A, B ⊆ ω1 . (This is easily deduced from 419G or 438Cd, and is also a special case of very much stronger results. See Kunen 80, II.6.12; Jech 78, p. 59; Just & Weese 97, Corollary 23.4; Levy 79, IV.4.48.) 4A1D ∆-systems (a) A family hIξ iξ∈A of sets is a ∆-system with root I if Iξ ∩ Iη = I for all distinct ξ, η ∈ A. (b) ∆-system Lemma If #(A) is a regular uncountable cardinal and hIξ iξ∈A is any family of finite sets, there is a set D ⊆ A such that #(D) = #(A) and hIξ iξ∈D is a ∆-system. (Kunen 80, II.1.6; Just & Weese 97, Theorem 16.3. For the present volume we need only the case #(A) = ω1 , which is treated in Jech 78, p. 225.) 4A1E Free sets (a) Let A be a set of cardinal at least ω2 , and hJξ iξ∈A a family of countable sets. Then there are S distinct ξ, η ∈ A such that ξ ∈ / Jη and η ∈ / Jξ . P P Let K ⊆ A be a set of cardinal ω1 , and set L = K ∪ i∈K Ji . Then L has cardinal ω1 , so there is a ξ ∈ A \ L. Now there is an η ∈ K \ Jξ , and this pair (ξ, η) serves. Q Q (b) If hKξ iξ∈A is a disjoint family of sets indexed by an uncountable subset A of ω1 , and hJη iη<ω1 is a family of countable sets, there is an uncountable B ⊆ A such that Kξ ∩ Jη = ∅ whenever η, ξ ∈ B and η < ξ. P P Choose hξζ iζ<ω1 inductively in such a way that ξζ ∈ A and Kξζ ∩ Jξη = ∅, ξζ > ξη for every η < ξ. Set B = {ξζ : ζ < ω1 }. Q Q T 4A1F Selecting subsequences (a) Let hKi ii∈I be a countable family of sets such that i∈J Ki is infinite for every finite subset J of I. Then there is an infinite set K such that K \ Ki is finite and Ki \ K P We can suppose that I ⊆ N. Choose hkn in∈N inductively such that T is infinite for every i ∈ I. P kn ∈ i∈I,i≤n Ki \ {ki : i < n} for every n ∈ N, and set K = {k2n : n ∈ N}. Q Q Consequently there is a family hKξ iξ<ω1 of infinite subsets of N such that Kξ \ Kη is finite if η ≤ ξ, infinite if ξ < η. (Choose the Kξ inductively.) (b) Let hJi ii∈I be a countable family of subsets of [N]ω such that Ji ∩ PK 6= ∅ for every K ∈ [N]ω and i ∈ I. Then there is an infinite K ⊆ N such that for every i ∈ I there is a J ∈ Ji such that K \ J is finite. P P The case I = ∅ is trivial; suppose that hin in∈N runs over I. Choose Kn , kn inductively, for n ∈ N, by taking K0 = N, kn ∈ Kn , Kn+1 ⊆ Kn \ {kn }, for every n; set K = {kn : n ∈ N}. Q Q

Kn+1 ∈ Jin

4A1K

Set theory

825

4A1G Ramsey’s Theorem If n ∈ N, K is finite and h : [N]n → K is any function, there is an infinite ´ s 79, p. 105, Theorem 3; Jech 78, 29.1; Just & Weese I ⊆ N such that h is constant on [I]n . (Bolloba 97, 15.3; Levy 79, IX.3.7. For the present volume we need only the case n = #(K) = 2.) 4A1H The Marriage Lemma again In 449J it will be useful to have an infinitary version of the Marriage Lemma available. Proposition Let X and Y be sets, and R ⊆ X × Y a set such that R[{x}] is finite for every x ∈ X and #(R[I]) ≥ #(I) for every finite set I ⊆ X. Then there is an injective function f : X → Y such that (x, f (x)) ∈ R for every x ∈ X. proof For each finite J ⊆ X there is an injective function fJ : J → Y such that fJ ⊆ R (identifying fJ with its graph), by the ordinary Marriage Lemma (3A1K) applied to R ∩ (J × R[J]). Let F be any ultrafilter on [X]<ω containing {J : I ⊆ J ∈ [X]<ω } for every I ∈ [X]<ω ; then for each x ∈ X there must be an f (x) ∈ R[{x}] such that {J : fJ (x) = f (x)} ∈ F , because R[{x}] is finite. Now f ⊆ R is a function from X to Y and must be injective, because for any x, x0 ∈ X there is a J ∈ [X]<ω such that f and fJ agree on {x, x0 }. 4A1I Filters (a) Let X be a non-empty set. If E ⊆ PX is non-empty and has the finite intersection property, T F = {A : A ⊆ X, A ⊇ E 0 for some non-empty finite E 0 ⊆ E} is the smallest filter on X including E, the filter generated by E. If E ⊆ PX is non-empty and downwards-directed, then it has the finite intersection property iff it does not contain ∅; in this case we say that E is a filter base; F = {A : A ⊆ X, A ⊇ E for some E ∈ E}, and E is a base for the filter F. In general, if E is a family of subsets of a non-empty set X, then there is a filter on X including E iff E has the finite intersection property; in this case, there is an ultrafilter on X including E (2A1O). (b) If κ is a cardinal and F is a filter then F is κ-complete if Every filter is ω-complete.

T

E ∈ F whenever E ⊆ F and 0 < #(E) < κ.

(c) A filter F on a regular uncountable cardinal κ is normal if (i) κ \ ξ ∈ F for every ξ < κ (ii) whenever hFξ iξ<κ is a family in F, its diagonal intersection belongs to F. 4A1J Lemma A normal filter F on a regular uncountable cardinal κ is κ-complete. proof If λ < κ and T hFξ iξ<λ is a family in F, set Fξ = κ for λ ≤ ξ < κ, and let F be the diagonal intersection of hFξ iξ<κ ; then ξ<λ Fξ ⊇ F \ λ belongs to F. 4A1K Theorem Let X be a set and F a non-principal ω1 -complete ultrafilter on X. Let κ be the least T cardinal of any non-empty set E ⊆ F such that E ∈ / F. Then κ is a regular uncountable cardinal, F is κ-complete, and there is a function g : X → κ such that g[[F]] is a normal ultrafilter on κ. proof (a) By the definition of κ, F is κ-complete. Because F is ω1 -complete, κ > ω. Let H be the set of −1 all functions h : X → κ such that P Let hEξ iξ<κ be T h [κ \ ξ] ∈ F for every ξ < κ. Then H is not empty. P a family in F such that E = ξ<κ Eξ ∈ / F. Because F is an ultrafilter, X \ E ∈ F. Set h(x) = 0 if x ∈ E, h(x) = min{ξ : x ∈ / Eξ } if x ∈ X \ E; then T h−1 [κ \ ξ] ⊇ (X \ E) ∩ η<ξ Eη ∈ F for every ξ < κ, because F is κ-complete, so h ∈ H. Q Q (b) For h, h0 ∈ H, say that h ≺ h0 if {x : h(x) < h0 (x)} ∈ F. Then there is a g ∈ H such that h 6≺ g for any h ∈ H. P P?? Otherwise, there is a sequence hhn in∈N in H such that hn+1 ≺ hn for every n ∈ N. Then T En = {x : hn+1 (x) < hn (x)} ∈ F for every n. Because F is ω1 -complete, there is an x ∈ n∈N En ; but now hhn (x)in∈N is a strictly decreasing sequence of ordinals, which is impossible. X XQ Q

826

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(c) I should check that κ is regular. P P If hαξ iξ<λ is any family in κ with λ < κ, then g −1 [κ \ αξ ] ∈ F for every ξ, so (because F is κ-complete) T g −1 [κ \ supξ<λ αξ ] = ξ<λ g −1 [κ \ αξ ] ∈ F, Q and supξ<λ αξ 6= κ. Q (d) The image filter g[[F]] is an ultrafilter on κ, by 2A1N. Because g ∈ H, g −1 [κ\ξ] ∈ F and κ\ξ ∈ g[[F]] for any ξ < κ. ?? Suppose, if possible, that g[[F]] is not normal. Then there is a family hAξ iξ<κ in g[[F]] such that its diagonal intersection A does not belong to g[[F]], that is, g −1 [A] ∈ / F and X \ g −1 [A] ∈ F . Define h : X → κ by setting h(x) = 0 if g(x) ∈ A = min{η : η < g(x), g(x) ∈ / Aη } if g(x) ∈ / A. Then h−1 [κ \ ξ] ⊇ (X \ g −1 [A]) ∩

T η<ξ

g −1 [Aη ] ∈ F

for every ξ < κ. Thus h ∈ H. But also h(x) < g(x) for every x ∈ X \ g −1 [A], so h ≺ g, contrary to the choice of g. X X Thus g[[F]] is a normal filter, and the theorem is proved. 4A1L Theorem Let κ be a regular uncountable cardinal, and F a normal ultrafilter on κ. If S ⊆ [κ]<ω , there is a set F ∈ F such that, for each n ∈ N, [F ]n is either a subset of S or disjoint from S. (Jech 78, Theorem 70; Just & Weese 97, Corollary 23.18; Levy 79, IX.4.30. Compare 4A1G.) 4A1M Ostaszewski’s ♣ This is the statement Let Ω be the family of non-zero countable limit ordinals. Then there is a family hθξ (n)iξ∈Ω,n∈N such that (i) for each ξ ∈ Ω, hθξ (n)in∈N is a strictly increasing sequence with supremum ξ (ii) for any uncountable A ⊆ ω1 there is a ξ ∈ Ω such that θξ (n) ∈ A for every n ∈ N. This is an immediate consequence of Jensen’s ♦ (Just & Weese 97, Exercise 22.9), which is itself a consequence of Gödel’s Axiom of Constructibility (Kunen 80, §II.7; Jech 78, §22; Just & Weese 97, §22). 4A1N Lemma Assume ♣. Then there is a family hCξ iξ<ω1 of sets such that (i) Cξ ⊆ ξ for every ξ < ω1 (ii) Cξ ∩ η is finite whenever η < ξ < ω1 (iii) for any uncountable sets A, B ⊆ ω1 there is a ξ < ω1 such that A ∩ Cξ and B ∩ Cξ are both infinite. proof (a) Let hθξ (n)iξ∈Ω,n∈N be a family as in 4A1M. Let f : ω1 → [ω1 ]2 be a surjection (3A1Cd). For ξ ∈ Ω, set S Cξ = i∈N f (θξ (i + 1)) ∩ ξ \ θξ (i). Then Cξ ⊆ ξ, and if η < ξ there is some n ∈ N such that θξ (n) ≥ η, so that S Cξ ∩ η ⊆ i≤n f (θξ (i)) is finite. For ξ ∈ ω1 \ Ω set Cξ = ∅. Then hCξ iξ<ω1 satisfies (i) and (ii) above. (b) Now suppose that A, B ⊆ ω1 are uncountable. Choose hαξ iξ<ω S 1 , hβξ iξ<ω1 , hIξ iξ<ω1 inductively, as follows. βξ is to be the smallest ordinal such that {αη : η < ξ} ∪ η<ξ Iη ⊆ βξ ; Iξ is to be a doubleton S subset of ω1 \ (βξ ∪ η≤βξ f (η)) meeting both A and B; and αξ < ω1 is to be such that f (αξ ) = Iξ . Set D = {αξ : ξ < ω1 }. This construction ensures that hαξ iξ<ω1 and hβξ iξ<ω1 are strictly increasing, with βξ < αξ < βξ+1 for every ξ, so that f (δ) meets both A and B for every δ ∈ D, while f (δ) ⊆ δ 0 and f (δ 0 ) ∩ f (δ) = ∅ whenever δ < δ 0 in D. By the choice of hθξ (n)iξ∈Ω,n∈N , there is a ξ ∈ Ω such that θξ (n) ∈ D for every n ∈ N. But this means that for every i ∈ N, so Cξ =

S

f (θξ (i)) ⊆ θξ (i + 1) ⊆ ξ, i∈N

f (θξ (i + 1)) ∩ f (θξ (i)) = ∅

f (θξ (i)) meets both A and B in infinite sets.

§4A2 intro.

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4A1O The size of σ-algebras: Proposition Let A be a Boolean algebra, B a subset of A, and B the σ-subalgebra of A generated by B (331E). Then #(B) ≤ max(4, #(B N )). In particular, if #(B) ≤ c then #(B) ≤ c. proof (a) If #(B) ≤ 1, this is trivial, since then #(B) ≤ 4. So we need consider only the case #(B) ≥ 2. (b) Set κ = #(B N ); then whenever #(A) ≤ κ, that is, there is an injection from A to B N , then #(AN ) ≤ #((B N )N ) = #(B N×N ) = #(B N ) = κ. As we are supposing that B has more than one element, κ ≥ #({0, 1}N ) = c ≥ ω1 . (c)SDefine hBξ iξ<ω1 inductively, as follows. B0 = B ∪ {0}. Given hBη iη<ξ , where 0 < ξ < ω1 , set Bξ0 = η<ξ Bη and Bξ = {1 \ b : b ∈ Bξ0 } ∪ {sup bn : hbn in∈N is a sequence in Bξ0 with a supremum in A}; n∈N

continue. An easy S induction on ξ (relying on 3A1Cc and (b) above) shows that every Bξ has cardinal at most κ. So C = ξ<ω1 Bξ has cardinal at most κ. (d) Now hBξ iξ<ω1 is a non-decreasing family, so if hcn in∈N is any sequence in C there is some ξ < ω1 0 such that every cn belongs to Bξ ⊆ Bξ+1 . But this means that if supn∈N cn is defined in A, it belongs to Bξ+1 ⊆ C. At the same time, 1 \ c0 ∈ Bξ+1 ⊆ C. This shows that C is closed under complementation and countable suprema; since it contains 0, it is a σ-subalgebra of A; since it includes B, it includes B, and #(B) ≤ #(C) ≤ κ, as claimed. (d) Finally, if #(B) ≤ c = #({0, 1}N ), then #(B) ≤ κ ≤ #({0, 1}N×N ) = c. 4A1P An incidental ii∈I of strictly positive P fact If I is a countable set and ² > 0, there is a family h²i−f P Let f : I → N be an injection and set ²i = 2 (i)−1 ². Q Q real numbers such that i∈I ²i ≤ ². P

4A2 General topology Even more than in previous volumes, naturally enough, the work of this volume depends on results from general topology. We have now reached the point where some of the facts I rely on are becoming hard to find as explicitly stated theorems in standard textbooks. I find myself therefore writing out rather a lot of proofs. You should not suppose that the results to which I attach proofs, rather than references, are particularly deep; on the contrary, in many cases I am merely spelling out solutions to classic exercises. The style of ‘general’ topology, as it has evolved over the last ninety years, is to develop a language capable of squeezing the utmost from every step of argument. While this does sometimes lead to absurdly obscure formulations, it remains a natural, and often profitable, response to the remarkably dense network of related ideas in this area. I therefore follow the spirit of the subject in giving the results I need in the full generality achievable within the terminology I use. For the convenience of anyone coming to the theory for the first time, I repeat some of them in the forms in which they are actually applied. I should remark, however, that in some cases materially stronger results can be proved with little extra effort; as always, this appendix is to be thought of not as a substitute for a thorough study of the subject, but as a guide connecting standard approaches to the general theory with the special needs of this volume.

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4A2A Definitions I begin the section with a glossary of terms not defined elsewhere. Base of neighbourhoods If X is a topological space and x ∈ X, a base of neighbourhoods of x is a family V of neighbourhoods of x such that every neighbourhood of x includes some member of V. boundary If X is a topological space and A ⊆ X, the boundary of A is ∂A = A \ int A. ˇ ˇ Cech-complete A completely regular Hausdorff topological space X is Cech-complete if it is homeomorphic to a Gδ subset of a compact Hausdorff space. closed interval Let X be a totally ordered set. A closed interval in X is an interval of one of the forms [x, y], ]−∞, y], [x, ∞[ or X = ]−∞, ∞[ where x, y ∈ X (see the definition of ‘interval’ below); for definiteness, let me say that I count ∅ as a ‘closed interval’. coarser topology If S and T are two topologies on a set X, we say that S is coarser than T if S ⊆ T. compact support Let X be a topological space and f : X → R a function. I say that f has compact support if {x : x ∈ X, f (x) 6= 0} is compact in X. countably compact A topological space X is countably compact if every countable open cover of X has a finite subcover. (Warning! some authors reserve the term for Hausdorff spaces.) A subset of a topological space is countably compact if it is countably compact in its subspace topology. countably paracompact A topological space X is countably paracompact if given any countable open cover G of X there is a locally finite family H of open sets which refines G and covers X. countably tight A topological space X is countably tight (or has countable tightness) if whenever A ⊆ X and x ∈ A there is a countable set B ⊆ A such that x ∈ B. direct sum, disjoint union Let h(Xi , Ti )ii∈I be a family of topological spaces, and set X = {(x, i) : i ∈ I, x ∈ Xi }. The disjoint union topology on X is T = {G : G ⊆ X, {x : (x, i) ∈ G} ∈ Ti for every i ∈ I}; (X, T) is the (direct) sum of h(Xi , Ti )ii∈I . If X is a set, hXi ii∈I a partition of X, and Ti a topology on Xi for every i ∈ I, then the disjoint union topology or sum topology on X is {G : G ⊆ X, G ∩ Xi ∈ Ti for every i ∈ I}. dyadic A Hausdorff space is dyadic if it is a continuous image of {0, 1}I for some set I. finer topology If S and T are two topologies on a set X, we say that S is finer than T if S ⊇ T. first-countable A topological space X is first-countable if every point has a countable base of neighbourhoods. half-open Let X be a totally ordered set. A half-open interval in X is a set of one of the forms [x, y[, ]x, y] where x, y ∈ X and x < y (see the definition of ‘interval’ below). hereditarily Lindel¨ of A topological space is hereditarily Lindel¨ of if every subspace is Lindelöf. hereditarily metacompact A topological space is hereditarily metacompact if every subspace is metacompact. indiscrete If X is any set, the indiscrete topology on X is the topology {∅, X}. interval Let (X, ≤) be a totally ordered set. Adjoin points ∞, −∞ to X and extend ≤ to a total order on X ∪ {−∞, ∞} by declaring that −∞ < x < ∞ for every x ∈ X. For u, v ∈ X ∪ {−∞, ∞}, set ]u, v[ = {x : u < x < v}, [u, v] = {x : u ≤ x ≤ v}, ]u, v] = {x : u < x ≤ v}, [u, v[ = {x : u ≤ x < v}; call sets of these forms intervals. Note that if X is not Dedekind complete then not every set which is order-convex in the sense of 351Xb will be an ‘interval’ in this sense. S isolated If X is a topological space, a family A of subsets of X is isolated if A ∩ (A S \ {A}) is empty for every A ∈ A; that is, if A is disjoint and every member of A is a relatively open set in A. Lindel¨ of A topological space is Lindel¨ of if every open cover has a countable subcover. (Warning! some authors reserve the term for regular spaces.) locally finite If X is a topological space, a family A of subsets of X is locally finite if for every x ∈ X there is an open set which contains x and meets only finitely many members of A. lower semi-continuous If X is a topological space and Y a totally ordered set, a function f : X → Y is lower semi-continuous if {x : f (x) > y} is open for every y ∈ Y . (Cf. 225H, 3A3Cf.) metacompact A topological space is metacompact if every open cover has a point-finite refinement which is an open cover. (Warning! some authors reserve the term for Hausdorff spaces.) neighbourhood If X is a topological space and x ∈ X, a neighbourhood of x is any subset of X including an open set which contains x. network Let (X, T) be a topological space. A network for T is a family E ⊆ PX such that whenever x ∈ G ∈ T there is an E ∈ E such that x ∈ E ⊆ G.

4A2A

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normal A topological space X is normal if for any disjoint closed sets E, F ⊆ X there are disjoint open sets G, H such that E ⊆ G and F ⊆ H. (Warning! some authors reserve the term for Hausdorff spaces.) open interval Let X be a totally ordered set. An open interval in X is a set of the form ]u, v[ where u, v ∈ X ∪ {−∞, ∞} (see the definition of ‘interval’ above). open map If (X, T) and (Y, S) are topological spaces, a function f : X → Y is open if f [G] ∈ S for every G ∈ T. order topology Let (X, ≤) be a totally ordered set. Its order topology is that generated by sets of the form ]x, ∞[ = {y : y > x}, ]−∞, x[ = {y : y < x} as x runs over X. paracompact A topological space is paracompact if every open cover has a locally finite refinement which is an open cover. (Warning! some authors reserve the term for Hausdorff spaces.) perfectly normal A topological space is perfectly normal if it is normal and every closed set is a Gδ set. (Warning! remember that some authors reserve the term ‘normal’ for Hausdorff spaces.) point-countable, point-finite A family A of sets is point-countable if no point belongs to more than countably many members of A. Similarly, an indexed family hAi ii∈I of sets is point-finite if {i : x ∈ Ai } is finite for every x. Polish A topological space X is Polish if it is separable and its topology can be defined from a metric under which X is complete. pseudometrizable A topological space (X, T) is pseudometrizable if T is defined by a single pseudometric (2A3F). refine(ment) If A is a family of sets, a refinement of A is a family B of sets such that every member of B is in some member of A; in this case I say that B refines A. (Warning! I do not suppose that S included S B = A.) relatively countably compact If X is a topological space, a subset A of X is relatively countably compact if every sequence in A has a cluster point in X. (Warning! This is not the same as supposing that A is included in a countably compact subset of X.) scattered A topological space X is scattered if every non-empty subset of X has an isolated point (in its subspace topology). second-countable A topological space is second-countable if the topology has a countable base, that is, if its weight is at most ω. semi-continuous see lower semi-continuous, upper semi-continuous. separately continuous Let hXi ii∈I be a family of topological spaces with product X, and Y another topological space. Then a function f : X → Y is separately continuous if for every j ∈ I and z ∈ Q X the function t 7→ f (z a t) : Xj → Y is continuous, where z a t is the member of X extending z i i∈I\{j} and such that (z a t)(j) = t. sequential A topological space is sequential if every sequentially closed set in X is closed. sequentially closed If X is a topological space, a subset A of X is sequentially closed if x ∈ A whenever hxn in∈N is a sequence in A converging to x ∈ X. sequentially compact A topological space is sequentially compact if every sequence has a convergent sequence. A subset of a topological space is sequentially compact if it is sequentially compact in its subspace topology. (Warning! some authors reserve the term for Hausdorff spaces.) sequentially continuous If X and Y are topological spaces, a function f : X → Y is sequentially continuous if hf (xn )in∈N → f (x) in Y whenever hxn in∈N → x in X. subbase If (X, T) is a topological space, a subbase for T is a family U ⊆ T which generates T, in the sense that T is the coarsest topology on X including U. (Warning! most authors reserve the term for families U with union X.) totally bounded If (X, W) is a uniform space, a subset A of X is totally bounded if for every W ∈ W there is a finite set I ⊆ X such that A ⊆ W [I]. If (X, ρ) is a metric space, a subset of X is totally bounded if it is totally bounded for the associated uniformity (3A4B). uniform convergence If X is a set and A is a family of subsets of X then the topology of uniform convergence on members of A is the topology on RX generated by the pseudometrics (f, g) 7→ min(1, supx∈A |f (x) − g(x)|) as A runs through A \ {∅}. (It is elementary to verify that the formula here defines a pseudometric.) upper semi-continuous If X is a topological space, a function f : X → R is upper semi-continuous if {x : f (x) < α} is open for every α ∈ R.

830

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A topological space (X, T) is weakly α-favourable if there is a function σ : S weakly α-favourable n+1 (T \ {∅}) → T \ {∅} such that (i) σ(G0 , . . . , Gn ) ⊆ Gn whenever G0 , . . . , Gn are non-empty open n∈N sets (ii) whenever hG i is a sequence in T \ {∅} such that Gn+1 ⊆ σ(G0 , . . . , Gn ) for every n, then n n∈N T G is non-empty. n n∈N weight If X is a topological space, its weight w(X) is the smallest cardinal of any base for the topology. Fσ If X is a topological space, an Fσ set in X is one expressible as the union of a sequence of closed sets. Gδ If X is a topological space, a Gδ set in X is one expressible as the intersection of a sequence of open sets. Kσ If X is a topological space, a Kσ set in X is one expressible as the union of a sequence of compact sets. lim sup If hAi ii∈I is a non-empty family of sets in a topological space and F is a filter on I, I write T S lim supi→F Ai for J∈F i∈J Ai . PX If X is any set, the usual topology on PX is that generated by the sets {A : A ∩ J = K} where J ⊆ X is finite and K ⊆ J. T0 If (X, T) is a topological space, we say that it is T0 if for any two distinct points of X there is an open set containing one but not the other. T1 If (X, T) is a topological space, we say that it is T1 if {x} is a closed set for every x ∈ X. π-base If (X, T) is a topological space, a π-base for T is a set U ⊆ T such that every non-empty open set includes a non-empty member of U. σ-compact A topological space X is σ-compact if there is a sequence of compact sets covering X. S σ-disjoint A family A of sets is σ-disjoint if it is expressible as n∈N An where every An is disjoint. S σ-isolated If X is a topological space, a family of subsets of X is σ-isolated if it is expressible as n∈N An where every An is an isolated family. σ-metrically-discrete S If (X, ρ) is a metric space, a family A of subsets of X is σ-metrically-discrete if it is expressible as n∈N An where ρ(x, y) ≥ 2−n whenever n ∈ N, A and B are distinct members of An , x ∈ A and y ∈ B. 4A2B Elementary facts about general topological spaces (a) Bases and networks (i) Let (X, T) be a topological space and U a subbase for T. Then {X} ∪ {U0 ∩ U1 ∩ . . . ∩ Un : U0 , . . . , Un ∈ U} is a base for T. (For this is a base for a topology, by 3A3Mc.) (ii) Let X and Y be topological spaces, and V a subbase for the topology of Y . Then a function f : X → Y is continuous iff f −1 [V ] is open for every V ∈ V. (Engelking 89, 1.4.1(ii)). (iii) If X and Y are topological spaces, E is a network for the topology of Y , and f : X → Y is a function such that f −1 [E] is open for every E ∈ E, then f is continuous. (The topology generated by E includes the given topology on Y .) (iv) If U is a (sub-)base for a topology on X, and Y ⊆ X, then {Y ∩ U : U ∈ U} is a (sub-)base for the ´sza ´r 78, 2.3.13(e)-(f).) subspace topology of Y . (Csa (v) If X is a topological space and U is a subbase for the topology of X, then a filter F on X converges to x ∈ X iff {U : x ∈ U ∈ U} ⊆ F. (If the condition is satisfied, F ∪ {A : A ⊆ X, x ∈ / A} is a topology on X including U.) (vi) If X and Y are topological spaces with subbases U, V respectively, then {U ×Y : U ∈ U}∪{X ×V : V ∈ V} is a subbase for the product topology of X × Y . (Kuratowski 66, §15.I.) (vii) If X is a topological space, E is a network for the topology of X, and Y is a subset of X, then {E ∩ Y : E ∈ E} is a network for the topology of Y . (viii) If X is a topological space and A is a (σ-)isolated family of subsets of X, then {A ∩ Y : A ∈ A0 } is (σ-)isolated whenever Y ⊆ X and A0 ⊆ A. (ix) If a topological space X has a σ-isolated network, so does every subspace of X. (b) If hHi ii∈I is a partition ofSa topological space X into open S sets and Fi ⊆ Hi is closed (either in X or in Hi ) for each i ∈ I, then F = i∈I Fi is closed in X. (X \ F = i∈I (Hi \ Fi ).) (c) If X is a topological space, A ⊆ X and x ∈ X, then x ∈ A iff there is an ultrafilter on X, containing A, which converges to x. ({A}∪ {G : x ∈ G ⊆ X, G is open} has the finite intersection property; use 4A1Ia.)

4A2Bi

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(d) Semi-continuity Let X be a topological space. ˇ (i) A function f : X → R is lower semi-continuous iff −f is upper semi-continuous. (Cech 66, 18D.8.) (ii) If f : X → R is lower semi-continuous, Y is another topological space, and g : Y → X is continuous, then f g : Y → R is lower semi-continuous. ({y : (f g)(y) > α} = g −1 [{x : f (x) > α}].) Similarly, if f : X → R is upper semi-continuous and g : Y → X is continuous, then f g : Y → R is upper semi-continuous. ˇ (iii) If f , g : X → ]−∞, ∞] are lower semi-continuous so is f + g : X → ]−∞, ∞]. (Cech 66, 18D.8.) ˇ (iv) If f , g : X → [0, ∞] are lower semi-continuous so is f × g : X → [0, ∞]. (Cech 66, 18D.8.) (v) If Φ is any non-empty set of lower semi-continuous functions from X to [−∞, ∞], then x 7→ supf ∈Φ f (x) : X → [−∞, ∞] is lower semi-continuous. (vi) f : X → R is continuous iff f is both upper semi-continuous and lower semi-continuous iff f and −f are both lower semi-continuous. (e) Separable spaces (i) If hAi ii∈I is a countable family of separable subsets of a topological space X S S then i∈I Ai is separable. (If Di ⊆ Ai is countable and dense for each i, i∈I Di is countable and dense in S A .) i∈I i Q (ii) If hXi ii∈I is a family of separable topological spaces and #(I) ≤ c, then i∈I Xi is separable. (Engelking 89, 2.3.16.) (iii) A continuous image of a separable topological space is separable. (Engelking 89, 1.4.11.) (f )(i)QLet hXi ii∈I be any family of topological spaces, with product X. If J ⊆ I is any set, and we write XJ for i∈J Xi , then the canonical map x 7→ x¹J : X → XJ is open. (Engelking 89, p. 79.) (ii) Let X and Y be topological spaces and f : X → Y a continuous open map. Then int f −1 [B] = f −1 [int B] and f −1 [B] = f −1 [B] for every B ⊆ Y . P P Because f is continuous, f −1 [int B] is an open set −1 −1 included in f [B], so is included in int f [B]. Because f is open, f [int f −1 [B]] is an open set included in f [f −1 [B]] ⊆ B, so f [int f −1 [B]] ⊆ int B, that is, int f −1 [B] ⊆ f −1 [int B]. Now apply this to Y \ B and take complements. Q Q (g) Let hXi ii∈I be a family of topological spaces with product X. (i) If A ⊆ X is determined by coordinates inQJ ⊆ I in the sense of 254M, then A and int A are also determined by coordinates in J. P P Let π : X → i∈J Xi be the canonical map. Then A = π −1 [π[A]], so −1 (f) tells us that int A = π [int π[A]] and A = π −1 [π[A]]; but these are both determined by coordinates in J. Q Q (ii) If F ⊆ X is closed, there is a smallest set J ∗ ⊆ I such that F is determined by coordinates in J ∗ . P P Let J be the family of all those sets J ⊆TI such that F is determined by coordinates in J . If J1 , J2 ∈ J , then J1 ∩ J2 ∈ J (254Ta). Set J ∗ = J . ?? Suppose, if possible, that F is not determined by coordinates in J ∗ . Then there are x ∈ F , y ∈ X \ F such that x¹J ∗ = y¹J ∗ . Because X \ F is open, there is a finite set K ⊆ I such that z ∈ / F whenever z ∈ X and z¹K = y¹K. Because J is closed under finite intersections, there is a J ∈ J such that K ∩ J = K ∩ J ∗ . Define z ∈ X by setting z(i) = x(i) for i ∈ J, z(i) = y(i) for i ∈ I \ J. Then z¹J = x¹J, so z ∈ F , but z¹K = y¹K, so z ∈ / F. X X Thus J ∗ ∈ J and is the required smallest member of J . Q Q (h) Let X be a topological space. S (i) If E is a locally finite family of closed subsets of X, then E 0 is closed for every E 0 ⊆ E. (Engelking 89, 1.1.11.) (ii) If hfi ii∈I is a family in C(X) such that h{x : fi (x) P6= 0}ii∈I is locally finite, then we have a continuous function f : X → R defined by setting f (x) = P For any x, i∈I fi (x) for every x ∈ X. P {i : fi (x) 6= 0} is finite, so f is well-defined. If x0 ∈ X and ² > 0, there is a neighbourhood V of x0 such that J = {i : iP∈ I, fi (x) 6= 0Pfor some x ∈ V } is finite; now there is a neighbourhood W of x0 , included in V , such that i∈J |fi (x) − i∈J fi (x0 )| < ² for every x ∈ W , so that |f (x) − f (x0 )| < ² for every x ∈ W . As x0 and ² are arbitrary, f is continuous. Q Q (i) Let X be a topological space and A, B two subsets of X. Then ∂(A ∗ B) ⊆ ∂A ∪ ∂B, where ∗ is any of ∪, ∩, \, 4. (Generally, if F ⊆ X, {A : ∂A ⊆ F } = {A : A \ F ⊆ int A} is a subalgebra of PX.)

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4A2C

4A2C Gδ , Fσ , zero and cozero sets Let X be a topological space. (a) (i) The union of two Gδ sets in X is a Gδ set. (Engelking 89, p. 26; Kuratowski 66, §5.V.) (ii) The intersection of countably many Gδ sets is a Gδ set. (Engelking 89, p. 26; Kuratowski 66, §5.V.) (iii) If Y isTanother topological space, f : X → Y is continuous and E ⊆ Y is Gδ in Y , then f −1 [E] is T Gδ in X. (f −1 [ n∈N Hn ] = n∈N f −1 [Hn ].) (iv) If Y is a Gδ set in X and Z ⊆ Y is a Gδ set for the subspace topology of Y , then Z is a Gδ set in X. (Kuratowski 66, §5.V.) (v) A set E ⊆ X is an Fσ set iff X \ E is a Gδ set. (Kuratowski 66, §5.V.) (b) (i) A zero set is closed. A cozero set is open. ´ sza ´ r 78, 4.2.36.) The intersection of two cozero sets (ii) The union of two zero sets is a zero set. (Csa is a cozero set. (iii) The intersection of a sequence of zero sets is a zero set. (If fn : X → R is continuous for each n, P ∞ x 7→ n=0 min(2−n , |fn (x)|) is continuous.) The union of a sequence of cozero sets is a cozero set. (iv) If Y is another topological space, f : X → Y is continuous and L ⊆ Y is a zero set, then f −1 [L] ˇ is a zero set. (Cech 66, 28B.3.) If f : X → Y is continuous and L ⊆ Y is a cozero set, then f −1 [L] ˇ is a cozero set. (Cech 66, 28B.3.) If K ⊆ X, L ⊆ Y are zero sets then K × L is a zero set in X × Y . (K × L = π1−1 [K] ∩ π2−1 [L].) (v) If H ⊆ X is a (co-)zero set and Y ⊆ X, then H ∩ Y is a (co-)zero set in Y . (Use (iv).) (vi) A cozero S set is the union of a non-decreasing sequence of zero sets. (If f : X → R is continuous, X \ f −1 [{0}]) = n∈N gn−1 [{0}], where gn (x) = max(0, 2−n − |f (x)|).) In particular, a cozero set is an Fσ set; taking complements, a zero set is a Gδ set. (vii) If G is a partition of X into open sets, and H ⊆ X is such that H ∩ G is a cozero set in G for every G ∈ G, then H is a cozero set in X. (If fG : G → R is continuous for every G ∈ G, then f : X → R is continuous, where f (x) = fG (x) for x ∈ G ∈ G.) Similarly, if F ⊆ X is such that F ∩ G is a zero set in G for every G ∈ G, then F is a zero set in X. 4A2D Weight Let X be a topological space. (a) w(Y ) ≤ w(X) for every subspace Y of X (4A2B(a-iv)). (b) A disjoint family G of non-empty open sets in X has cardinal at most w(X). (If U is a base for the topology of X, then every non-empty member of G includes a non-empty member of U, so we have an injective function from G to U.) (c) A point-countable family G of open sets in X has cardinal at most max(ω, w(X)). P P If X = ∅, this is trivial. Otherwise, let U be a base for the topology of X with #(U) = w(X) > 0. Choose a function f : G → U such that ∅ 6= f (G) ⊆ G whenever G ∈ G \ {∅}. Then GU = {G : f (G) = U } is countable for every U ∈ U , so there is an injection hU : GU → N; now G 7→ (f (G), hf (G) (G)) : G → U × N is injective, so #(G) ≤ #(U × N) = max(ω, w(X)). Q Q (d) If X is a dyadic space then X is a continuous image of {0, 1}w(X) . P P By definition, X is Hausdorff and there are a set I and a continuous surjection f : {0, 1}I → X; because any power of {0, 1} is compact, so is X. If w(X) is finite, #(X) = w(X) ≤ #({0, 1}w(X) ) and the result is trivial; so we may suppose that w(X) is infinite. Let U be a base for the topology of X with cardinality w(X). Set Z = {0, 1}I and let E be the algebra of subsets of Z determined by coordinates in finite sets, so that E is an algebra of subsets of Z and is a base for the topology of Z. For each pair U , V of members of U such that U ⊆ V , f −1 [V ] ⊆ Z is open; the set {E : E ∈ E, E ⊆ f −1 [V ]} is upwards-directed and covers the compact set f −1 ][U ], so there is an EU V ∈ E such that f −1 [U ] ⊆ EU V ⊆ f −1 [V ]. Let J ⊆ I be a set of cardinal at most max(ω, w(X)) such that every EU V is determined by coordinates in J. Fix any w ∈ {0, 1}I\J and define g : {0, 1}J → X by setting g(z) = f (z, w) for every z ∈ {0, 1}J , identifying Z with {0, 1}J × {0, 1}I\J . Then g is continuous. ?? If g is not surjective, set H = X \ g[{0, 1}J ]. Take x ∈ H; take V ∈ U such that x ∈ V ⊆ H; take an open set G such that x ∈ G ⊆ G ⊆ V (this must be possible because X, being

4A2Fd

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compact and Hausdorff, is regular); take U ∈ U such that x ∈ U ⊆ G, so that x ∈ U ⊆ U ⊆ V . Because f is surjective, there is a (u, v) ∈ {0, 1}J × {0, 1}I\J such that f (u, v) = x. Now (u, v) ∈ f −1 [U ] ⊆ EU V ; as EU V is determined by coordinates in J, (u, w) ∈ EU V ⊆ f −1 [V ] and g(u) = f (u, w) ∈ V ; but V is supposed to be disjoint from g[{0, 1}J ]. X X So g is surjective, and X is a continuous image of {0, 1}J . Since J #(J) ≤ max(ω, w(X)) = w(X), {0, 1} and X are continuous images of {0, 1}w(X) . Q Q (e) If X is a dyadic space then X is separable iff it is a continuous image of {0, 1}c . P P {0, 1}c is separable (4A2B(e-ii)), so any continuous image of it is separable. If X is a separable dyadic space, let A ⊆ X be a countable dense set. If G, G0 ⊆ X are distinct regular open sets, then G ∩ A 6= G ∩ A0 . Thus X has at most c regular open sets; since X is compact and Hausdorff, therefore regular, its regular open sets form a base, and w(X) ≤ c. By (d), X is a continuous image of {0, 1}max(ω,w(X)) which is in turn a continuous image of {0, 1}c . Q Q Q 4A2E The countable chain condition (a)(i) Q Let hXi ii∈I be a family of topological spaces. If XJ = i∈J Xi is ccc for every finite J ⊆ I, then X = i∈I Xi is ccc. (Kunen 80, II.1.9; Fremlin 84, 12I.) (ii) A separable topological space is ccc. (If D is a countable dense set and G is a disjoint family of non-empty open sets, we have a surjection from a subset of D onto G.) (iii) The product of any family of separable topological spaces is ccc. P P By 4A2B(e-ii) and (ii) here, the product of finitely many separable spaces is separable, therefore ccc; so we can apply (i). Q Q Q (b) Let hXi ii∈I Q be a family of topological spaces, and suppose that X = i∈I Xi is ccc. For J ⊆ I and x ∈ X set XJ = i∈J Xi , πJ (x) = x¹J. (i) If G ⊆ X is open, there is an open set W ⊆ G determined by coordinates in a countable subset of I such that G ⊆ W . P P Let W be the family of subsets of X determined by coordinates in countable sets. Then W is a σ-algebra (254Mb) including the standard base U for the topology of X. Let U0 be a maximal S disjoint family in {U : U ∈ U, U ⊆ G}. Then U0 is countable, so W = U0 belongs to W. No member of U can be included in G \ W , so G \ W must be empty, and we have a suitable set. Q Q So G and int G are determined by coordinates in a countable set (4A2B(g-ii)); in particular, if G is a regular open set, then it is determined by coordinates in a countable set. (ii) If f : X → R is continuous, there are a countable set J ⊆ I and a continuous function g : XJ → R such that f = gπJ . P P For each q ∈ Q, set Fq = {x : f (x) < q}. By (i), Fq is determined by coordinates in a countable set. Because Q is countable,Sthere is a countable J ⊆ I such that every Fq is determined by coordinates in J. Also {x : f (x) < α} = q∈Q,q<α Fq is determined by coordinates in J for every α ∈ R, so f (x) = f (y) whenever x¹J = y¹J, and there is a g : XJ → R such that f = gπJ . Now if H ⊆ R is open, g −1 [H] = πJ [f −1 [H]] is open, so g is continuous. Q Q ˇ 4A2F Separation axioms (a) Hausdorff spaces (i) A Hausdorff space is T1 . (Cech 66, 27A.1.) ˇ Any subspace of a Hausdorff space is Hausdorff. (Engelking 89, 2.1.6; Cech 66, 27A.3; Kuratowski ´ sza ´ r 78, 2.5.21.) 66, §5.VIII; Csa (ii) If X is a Hausdorff space and hxn in∈N is a sequence in X, then a point x of X is a cluster point of hxn in∈N iff there is a non-principal ultrafilter F on N such that x = limn→F xn . (If x is a cluster point of hxn in∈N , apply 4A1Ia to {{n : n ≥ n0 , xn ∈ G} : n0 ∈ N, G ⊆ X is open, x ∈ G}.) ˇ (b) Regular spaces A regular T1 space is Hausdorff. (Cech 66, 27B.7; Gaal 64, p. 81.) Any subspace of a regular space is regular. (Engelking 89, 2.1.6; Kuratowski 66, §14.I.) (c) Completely regular spaces In a completely regular space, the cozero sets form a base for the ˇ topology. (Cech 66, 28B.5.) (d) Normal spaces (i) Urysohn’s Lemma If X is normal and E, F are disjoint closed subsets of X, then there is a continuous function f : X → [0, 1] such that f (x) = 0 for x ∈ E and f (x) = 1 for x ∈ F . (Engelking 89, 1.5.11; Kuratowski 66, §14.IV.) (ii) A regular normal space is completely regular. ´ sza ´ r 78, 4.2.5; (iii) A normal T1 space is Hausdorff (Gaal 64, p. 86) and completely regular (Csa Gaal 64, p. 110).

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(iv) If X is normal and E, F are disjoint closed sets in X there is a zero set including E and disjoint from F . (Take a continuous function f which is zero on E and 1 on F , and set Z = {x : max(0, f (x) − 12 ) = 0}.) (v) In a normal space a closed Gδ set is a zero set. (Engelking 89, 1.5.12.) (vi) If X is a normal space and hGi ii∈I is a point-finite cover of X by open sets, there is a family ˇ 66, hHi ii∈I of open sets, still covering X, such that H i ⊆ Gi for every i. (Engelking 89, 1.5.18; Cech 29C.1; Gaal 64, p. 89.) (vii) If X is a normal space and hGi ii∈I is a point-finite cover of X by open sets, there is a family hHi0 ii∈I of cozero sets, still covering X, such that Hi0 ⊆ Gi for every i. (Take hHi ii∈I from (vi), and apply (iv) to the disjoint closed sets H i , X \ Gi to find a suitable cozero set Hi0 for each i.) (viii) Tietze’s theorem Let X be a normal space, F a closed subset of X and f : F → R a continuous function. Then there is a continuous function g : X → R extending f . (Engelking 89, 2.1.8; Kuratowski 66, §14.IV; Gaal 64, p. 203.) It follows that if F ⊆ X is closed and f : F → [0, 1]I is a continuous function from F to any power of the unit interval, there is a continuous function from X to [0, 1]I extending f . (Extend each of the functionals x 7→ f (x)(i) for i ∈ I.) (ix) If X is a normal space and hGi ii∈I is a locally finite cover of X by open sets,Pthere is a family hgi ii∈I of continuous functions from X to [0, 1] such that gi ≤ χGi for every i ∈ I and i∈I gi (x) = 1 for every x ∈ X. (Engelking 89, proof of 5.1.9.) (e) Paracompact spaces A Hausdorff paracompact space is regular. (Engelking 89, 5.1.5.) A regular paracompact space is normal. (Engelking 89, 5.1.5; Gaal 64, p. 160.) (f ) Countably paracompact spaces A normal space X is countably paracompact iff whenever hFn in∈N is a sequence of closed subsets of X with empty intersection, there is a sequence hGn in∈N of open sets, also ´ sza ´ r 78, 8.3.56(f).) with empty intersection, such that Fn ⊆ Gn for every n ∈ N. (Engelking 89, 5.2.2; Csa (g) Metacompact spaces (i) A paracompact space is metacompact. (ii) A closed subspace of a metacompact space is metacompact. ´ sza ´ r 78, (iii) A normal metacompact space is countably paracompact. (Engelking 89, 5.2.6; Csa 8.3.56(c).) (h) Separating compact sets (i) If X is a Hausdorff space and K and L are disjoint compact subsets ´sza ´r 78, 5.3.18.) of X, there are disjoint open sets G, H ⊆ X such that K ⊆ G and L ⊆ H. (Csa (ii) If X is a regular space, F ⊆ X is closed, and K ⊆ X \ F is compact, there are disjoint open sets G, H ⊆ X such that K ⊆ G and F ⊆ H. (Engelking 89, 3.1.6.) (iii) If X is a completely regular space, G ⊆ X is open and K ⊆ G is compact, there is a continuous function f : X → [0, 1] such that f (x) = 1 for x ∈ K and f (x) = 0 for x ∈ X \ G. P P For each x ∈ K there is a continuous functionSfx : X → [0, 1] such that fx (x) = 1 and fx (y) = 0 for y ∈SX \ G. Set Hx = {y : fx (y)P> 12 }. Then x∈K Hx ⊇ K, so there is a finite set I ⊆ K such that K ⊆ x∈I Hx . Set f (y) = min(1, 2 x∈I fx (y)) for y ∈ X. Q Q (iv) If X is a completely regular Hausdorff space and K and L are disjoint compact subsets of X, there are disjoint cozero sets G, H ⊆ X such that K ⊆ G and L ⊆ H. P P By (i), there are disjoint open sets G0 , H 0 such that K ⊆ G0 and L ⊆ H 0 . By (iii), there is a continuous function f : X → [0, 1] such that f (x) = 1 for x ∈ K and f (x) = 0 for x ∈ X \ G0 ; set G = {x : f (x) 6= 0}, so that G is a cozero set and K ⊆ G ⊆ G0 . Similarly there is a cozero set H including L and included in H 0 . Q Q (v) If X is a completely regular space and K ⊆ X is a compact Gδ set, then K is a zero set. P P Let hGn in∈N be a sequence of open sets with intersection K. For each n ∈ N there is a continuousTfunction fn : X → [0, 1] such that fn (x) = 1 for x ∈ K, fn (x) = 0 for x ∈ X \ Gn , by (iii). Now K = n∈N {x : 1 − fn (x) = 0} is a zero set, by 4A2C(b-iii). Q Q (vi) If hXn in∈N is a sequence of topological spaces with product X, K ⊆ X is compact, F ⊆ X is closed and K ∩ F = ∅, there is some n ∈ N such that x¹n 6= y¹n for any x ∈ F , y ∈ K. P P?? Otherwise, we can find for each n ∈ N points xn ∈ F , yn ∈ K such that xn ¹n = yn ¹n. Let y ∈ K be a cluster point of hyn in∈N (2A3Ob). Then y ∈ / F , so there must be an m ∈ N and open sets Gi ⊆ Xi , for i ≤ m, such that y ∈ G ⊆ X \ F , where G = {z : z ∈ X, z(i) ∈ Gi for every i ≤ m}. Now there is an n > m such that yn ∈ G; but in this case xn ∈ G and xn ∈ / F , which is impossible. X XQ Q

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(vii) If X is a compact Hausdorff space, f : X → R is continuous, and U is a subbase for T, then there is a countable set U0 ⊆ U such that f (x) = f (y) whenever {U : x ∈ U ∈ U} = {U : y ∈ U ∈ U}. (Apply (i) to sets of the form K = {x : f (x) ≤ α}, L = {x : f (x) ≥ β}.) (i) Perfectly normal spaces A topological space X is perfectly normal iff every closed set is a zero set. (Engelking 89, 1.4.9.) Consequently, every open set in a perfectly normal space is a cozero set (and, of course, an Fσ set). 4A2G Compact and locally compact spaces (a) In any topological space, the union of two compact subsets is compact. ´ sza ´ r 78, 5.3.23; Gaal 64, p. (b) A compact Hausdorff space is normal. (Engelking 89, 3.1.9; Csa 139.) (c) If X is a compact Hausdorff space, Y ⊆ X is a zero set and Z ⊆ Y is a zero set in Y , then Z is a zero set in X. (By 4A2C(b-vi) and 4A2C(a-iv), Z is a Gδ set in X; now use 4A2F(d-v).) (d) If X T is a Hausdorff space, V is a downwards-directed family of compact neighbourhoods of a point x of X and V = {x}, then V is a base of neighbourhoods of {x}. P P Let G be any open set containing x. Fix any V0 ∈ V. Note that because X is Hausdorff, every member of V is closed (3A3Dc). So {V0 ∩V \G : V ∈ V} is a family of (relatively) closed subsets of V0 with empty intersection, cannot have the finite intersection property (3A3Da), and there is a V ∈ V such that V0 ∩ V \ G = ∅. Now there is a V 0 ∈ V such that V 0 ⊆ V0 ∩ V and V 0 ⊆ G. As G is arbitrary, V is a base of neighbourhoods of x. Q Q (e) Let (X, T) be a locally compact Hausdorff space. (i) If K ⊆ X is a compact set and G ⊇ K is open, then there is a continuous f : X → [0, 1] with compact support such that χK ≤ f ≤ χG. (Let V be the family of relatively compact open subsets of X. Then V is upwards-directed and covers X, so there is a V ∈ V including K. By 3A3Bb, T is completely regular; now 4A2F(h-iii) tells us that there is a continuous f : X → [0, 1] such that χK ≤ f ≤ χ(G ∩ V ), so that f has compact support.) (ii) T is the coarsest topology on X such that every T-continuous real-valued function with compact support is continuous. P P Let Φ be the set of continuous functions of compact support for T. If S is a topology on X such that every member of Φ is continuous, and x ∈ G ∈ T, then there is an f ∈ Φ such that f (x) = 1 and f (y) = 0 for y ∈ X \ G, by (i). Now f is S-continuous, by hypothesis, so H = {y : f (y) > 21 } belongs to S and x ∈ H ⊆ G. As x is arbitrary, G = intS G belongs to S; as G is arbitrary, T ⊆ S. Q Q (f ) (i) A topological space X is countably compact iff every sequence in X has a cluster point in X, that ´ sza ´ r 78, 5.3.31(e); Gaal 64, is, X is relatively countably compact in itself. (Engelking 89, 3.10.3; Csa p. 129.) T topological space and hFn in∈N is a sequence of closed sets such that T (ii) If X is a countably compact ´ ´ F = 6 ∅ for every n ∈ N, then i≤n i n∈N Fn 6= ∅. (Engelking 89, 3.10.3; Csaszar 78, 5.3.31(c).) (iii) In any topological space, a relatively compact set is relatively countably compact (2A3Ob). (iv) Let X and Y be topological spaces and f : X → Y a continuous function. If A ⊆ X is relatively countably compact in X, then f [A] is relatively countably compact in Y . P P Let hyn in∈N be a sequence in f [A]. Then there is a sequence hxn in∈N in A such that f (xn ) = yn for every n ∈ N. Because A is relatively countably compact, hxn in∈N has a cluster point x ∈ X. If n0 ∈ N and H is an open set containing f (x), there is an n ≥ n0 such that xn ∈ f −1 [H], so that yn ∈ H. Thus f (x) is a cluster point of hyn in∈N ; as hyn in∈N is arbitrary, f [A] is relatively countably compact. Q Q (v) A relatively countably compact set in R must be bounded. (If A ⊆ R is unbounded there is a sequence hxn in∈N in A such that |xn | ≥ n for every n.) So if X is a topological space, A ⊆ X is relatively countably compact and f : X → R is continuous, then f [A] is bounded. (vi) If X and Y are topological spaces and f : X → Y is continuous, then f [A] is countably compact whenever A ⊆ X is countably compact. (Engelking 89, 3.10.5.) (g) (i) Let X and Y be topological spaces and φ : X ×Y → R a continuous function. Define θ : X → C(Y ) by setting θ(x)(y) = φ(x, y) for x ∈ X, y ∈ Y . Then θ is continuous if we give C(Y ) the topology of uniform

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Appendix

4A2Gg

convergence on compact subsets of Y . (As noted in Engelking 89, pp. 157-158, the topology of uniform convergence on compact sets is the ‘compact-open’ topology of C(Y ), so the result here is covered by Engelking 89, 3.4.1.) (ii) In particular, if Y is compact then θ is continuous if we give C(Y ) its usual norm topology. (h) (i) Suppose that X is a compact space such that there are no non-trivial convergent sequences in X, that is, no convergent sequences which are T not eventually constant. If hFn in∈N is a non-increasing sequence of infinite closed subsets of X, then F = n∈N Fn is infinite. P P Because X is compact, F cannot be empty (3A3Da). T Choose a sequence hxn in∈N such that xn ∈ Fn \ {xi : i < n} for each n ∈ N. If G ⊇ F is an open set, then n∈N Fn \ G = ∅, so there must be some n ∈ N such that Fn ⊆ G, and xi ∈ G for i ≥ n. ?? If F = {y0 , . . . , yk }, let l ≤ k be the first point such that whenever G ⊇ {y0 , . . . , yl } is open, then {i : xi ∈ / G} is finite. Then there is an open set G0 ⊇ {yj : j < l} such that I = {i : xi ∈ / G0 } is infinite. But if H is any open set containing yl , then {i : xi ∈ / G0 ∪ H} is finite, so {i : i ∈ I, xi ∈ / H} is finite. Thus if we 0 0 re-enumerate hxi ii∈I as hxn in∈N , hxn in∈N converges to yl and is a non-trivial convergent sequence. X X Thus F is infinite, as claimed. Q Q (ii) If X is an infinite scattered compact Hausdorff space it has a non-trivial convergent sequence. P P T Let hxn in∈N be any sequence of distinct points in X. Set Fn = {xi : i ≥ n} for each n, so that F = n∈N Fn is a non-empty set. Because X is scattered, F has an isolated point z say; let G be an open set such that F ∩ G = {z}, and H an open set such that z ∈ H ⊆ H ⊆ G (3A3Bc). In this case, I = {i : xi ∈ H} must be infinite; re-enumerate hxi ii∈I as hx0n in∈N . ?? If hx0n in∈N does not converge to z, there is an open set H 0 containing z such that {n : x0n ∈ / H 0 } is infinite, that is, {i : i ∈ I, xi ∈ / H 0 } is infinite. In this case, 0 0 X Thus hx0n in∈N is a Fn ∩ H \ H is non-empty for every n ∈ N, but F ∩ H \ H = ∅, which is impossible. X non-trivial convergent sequence in X. Q Q (iii) If X is an extremally disconnected Hausdorff space (definition: 3A3Ae), it has no non-trivial convergent sequence. P P?? Suppose, if possible, that there is a sequence hxn in∈N converging to x ∈ X such that {n : xn 6= x} is infinite. Choose hni ii∈N and hGi ii∈N inductively, as follows. Given that x ∈ / Gj for j < i, there is an ni such that xni 6= x and xni ∈ / Gj for every j < i; now let Gi be an open set such that xni ∈ Gi and x ∈ / Gi , and continue. Since all the ni must be distinct, hxni ii∈N → x. But consider S S S S G = i∈N G2i \ j<2i Gj , H = i∈N G2i+1 \ j≤2i Gj . Then G and H are disjoint open sets and xn2i ∈ G, xn2i+1 ∈ H for every i. So x ∈ G ∩ H. But G is open (because X is extremally disconnected), and is disjoint from H, and now H is disjoint from G; so they cannot both contain x. X XQ Q (i) (i) Let X be a non-empty compact Hausdorff space without isolated points. Then there are a closed S set F ⊆ X and a continuous surjection f : F → {0, 1}N . P P For σ ∈ n∈N {0, 1}n choose closed sets Vσ ⊆ X inductively, as follows. V∅ = X. Given that Vσ is a closed set with non-empty interior, there are distinct points x, y ∈ int Vσ (because X has no isolated points); let G, H be disjoint open subsets of X such that x ∈ G and y ∈ H; and let Vσa 0 , Vσa 1 be closed sets such that x ∈ int Vσa 0 ⊆ G ∩ int Vσ ,

y ∈ int Vσa 1 ⊆ H ∩ int Vσ .

(This is possible because X is regular.) The construction ensures that Vτ ⊆ Vσ whenever τS∈ {0, 1}n extends T m n σ ∈ {0, 1} , and that Vτ ∩ Vσ = ∅ whenever τ , σ ∈ {0, 1} are different. Set F = n∈N σ∈{0,1}n Vσ ; then F is a closed subset of X and we have a continuous function f : F → {0, 1}N defined by saying that f (x)(i) = σ(i) whenever n ∈ N, σ ∈ {0, 1}n , i < n and x ∈ Vσ . Finally, f is surjective, because if z ∈ {0, 1}N then hVz¹n in∈N is a non-increasing sequence of closed sets in the compact space X, so has non-empty intersection V say, and f (x) = z for any x ∈ V . Q Q (ii) If X is a non-empty compact Hausdorff space without isolated points, then #(X) ≥ c. (Use (i).) (iii) If X is a compact Hausdorff space which is not scattered, it has an infinite closed subset without isolated points with a countable π-base. P P Because X is not scattered, it has a non-empty subset A without isolated points. Then A is compact and has no isolated points; by (i), there are a closed set F0 ⊆ A and a N continuous surjection f : F0 → {0, 1}N . Let F be the family of closed T 0 sets F ⊆ F0 such that f [F ] = {0, 1}N . 0 0 If F ⊆ F is non-empty and downwards-directed, then F = F is closed; also, for every z ∈ {0, 1} ,

4A2Hc

General topology

837

{F ∩ f −1 [{z}] : F ∈ F 0 } is a downwards-directed family of closed sets in the compact space X, so has non-empty intersection, and F 0 ∩ f −1 [{z}] 6= ∅; as z is arbitrary, f [F 0 ] = {0, 1}N , and F 0 ∈ F. Thus every totally ordered subset of F has a lower bound in F, and F has a minimal element F1 say. Let U be a countable base for the topology of {0, 1}N , and set V = {F1 ∩ f −1 [U ] : U ∈ U}, so that V is a countable family of relatively open subsets of F1 . If H ⊆ F1 is relatively open and not empty, then F1 \ H is a proper closed subset of F1 ; because F1 is minimal in F, F1 \ H ∈ / F and f [F1 \ H] 6= {0, 1}N . But F1 \ H N is compact, so f [F1 \ H] is compact and its complement in {0, 1} is open; there is therefore a non-empty U ∈ U such that U ∩ f [F1 \ H] = ∅ and f −1 [U ] ∩ F1 ⊆ H. But as f [F1 ] = {0, 1}N , f −1 [U ] ∩ F1 is not empty, and is a non-empty member of V included in H. As H is arbitrary, V is a π-base for the topology of F1 , and F1 is a closed subset of X with a countable π-base. Q Q (iv) Let X be a compact Hausdorff space. Then there is a continuous surjection from X onto [0, 1] iff X is not scattered. P P (α) Suppose that f : X → [0, 1] is a continuous surjection. Let E be the family of closed sets F ⊆ X such that f [F ] = [0, 1]. If F ⊆ E is non-empty and downwards-directed, then for any t ∈ [0, 1] the family {F ∩ f −1 [{t}] : F ∈ F} is a downwards-directedTfamily of non-empty closed sets, so (because X is compact) has non-empty intersection; this shows that F ∈ E. By Zorn’s Lemma, E has a minimal element K say. ?? If x ∈ K is an isolated point of K, then K \ {x} is closed, therefore compact, and f [K \ {x}] ⊇ [0, 1] \ {f (x)} is compact and dense in [0, 1]. But this means that K \ {x} ∈ E and K is not minimal. X X Thus K ⊆ X is a non-empty set with no isolated points and X is not scattered. (β) If X is not scattered, let A ⊆ X be a non-empty set with no isolated points. Then A is a non-empty compact subset of X with no isolated points, so there is a continuous surjection → {0, 1}N ((i) above). Now P∞g : A−n−1 N y(n) for y ∈ {0, 1}N ), so there is a continuous surjection h : {0, 1} → [0, 1] (e.g., set h(y) = n=0 2 we have a continuous surjection hg : A → [0, 1]. By Tietze’s theorem (4A2F(d-viii)), there is a continuous function f0 : X → R extending hg; setting f (x) = max(0, min(f0 (x), 1)) for x ∈ X, we have a continuous surjection f : X → [0, 1]. Q Q (v) A Hausdorff continuous image of a scattered compact Hausdorff space is scattered. (Immediate from (iv).) (vi) If X is an uncountable first-countable compact Hausdorff space, it is not scattered. P P Let G be the family of countable open subsets of X, and G∗ its union. No finite subset of G can cover X, so X \ G∗ is non-empty. ?? If x is an isolated point of X \ G∗ , then {x} ∪ G∗ is a neighbourhood of x; let hUn in∈N run over a base of open neighbourhoods of x with U 0 ⊆ {x} ∪ G∗ . For each n ∈ N, Fn = U 0 \ Un is a compact set included Sin G∗ , so is covered by finitely many members of G, and is countable. But this means that X Thus X \ G∗ is a non-empty set with no isolated U0 = {x} ∪ n∈N U0 \ Un is countable, and x ∈ G∗ . X points, and X is not scattered. Q Q It follows that there is a continuous surjection from X onto [0, 1], by (iv). ˇ (j) A locally compact Hausdorff space is Cech-complete. (Engelking 89, p. 196.) (k) If X is a topological space, f : X → R is lower semi-continuous, and K ⊆ X is compact and not empty, then there is an x0 ∈ K such that f (x0 ) = inf x∈K f (x). Similarly, if g : X → R is upper semi-continuous, there is an x1 ∈ K such that g(x1 ) = supx∈K g(x). 4A2H Lindel¨ of spaces (a) If X is a topological space, then a subset Y of X is Lindelöf (in its subspace topology) iff for every family G of open subsets of X covering Y there is a countable subfamily of G still covering Y . (b) (i) A regular Lindelöf space X is paracompact, therefore normal and completely regular. (Engelking 89, 3.8.11 & 5.1.2.) (ii) If X is a Lindelöf space and A is a locally finite family of subsets of X then A is countable. P P The family G of open sets meeting only finitely many members of A is an open cover of X. If G0 ⊆ G is a countable cover of X then {A : A ∈ A, A meets some member of G0 } = A \ {∅} is countable. Q Q (c) (i) A topological space X is hereditarily S S Lindelöf iff for any family G of open subsets of X there is a countable family G0 ⊆ G such that G0 = SG. P P (α) If X is hereditarily Lindelöf and G is a family of open subsets of X, then G is an open cover of G, so has a countable subcover. (β) If X is not hereditarily Lindelöf, let Y ⊆ X be a non-Lindelöf subspace, and H a cover of Y by relatively open sets which has no

838

Appendix

4A2Hc

countable S subcover; setting G = {G : G ⊆ X is open, G ∩ Y ∈ H}, there can be no countable G0 ⊆ G with union G. Q Q (ii) Let X be a regular hereditarily Lindelöf space. Then every closed subset of X is a zero set, so X is perfectly normal. P PSLet F ⊆ X be closed. Let G be the family of open sets G ⊆ X such that G ∩ F = ∅; because X is regular, G = X \ F ; because X is hereditarily Lindelöf, there is a sequence hGn in∈N in G S T such that X \ F = n∈N Gn . This means that F = n∈N X \ Gn is a Gδ set. But X is normal ((b) above), so F is a zero set (4A2F(d-v)). By 4A2Fi, X is perfectly normal. Q Q (d) Any σ-compact topological space is Lindelöf. (Engelking 89, 3.8.5.)

ˇ 4A2I Stone-Cech compactifications (a) Let X be a completely regular Hausdorff space. Then there ˇ is a compact Hausdorff space βX, the Stone-Cech compactification of X, in which X can be embedded as a dense subspace. If Y is another compact Hausdorff space, then every continuous function from X to Y ´ sza ´r has a unique continuous extension to a continuous function from βX to Y . (Engelking 89, 3.6.1; Csa ˇ 78, 6.4d; Cech 66, 41D.5.) ˇ (b) Let I be any set, and write βI for its Stone-Cech compactification when I is given its discrete topology. (i) Let Z be the Stone space of the Boolean algebra PI. Then there is a canonical homeomorphism φ : βI → Z defined by saying that φ(i)(a) = χa(i) for every i ∈ I, a ⊆ Z. P P Recall that Z is the set of ring homomorphisms from PI onto Z2 (311E). If i ∈ I, let î be the corresponding member of Z defined by setting î(a) = χa(i) for every a ⊆ I. Then i 7→ î : I → Z is continuous, while Z is compact and Hausdorff (311I), so has a unique extension to a continuous function φ : βI → Z. If G ⊆ Z is open and not empty, it includes a set of the form b a = {θ : θ ∈ Z, θ(a) = 1} where a ⊆ I is not empty; if i is any member of a, î ∈ G so G ∩ φ[βI] 6= ∅. This shows that φ[βI] is dense in Z; as βI is compact, φ[βI] is compact, therefore closed, and is equal to Z. Thus φ is surjective. If t, u are distinct points of βI, there is an open subset H of βI such that t ∈ H and u ∈ / H. Set a = H ∩I. Then t ∈ a, the closure of a regarded as a subset of βI, so φ(t) ∈ φ[a] (3A3Cd). But φ[a] = {î : i ∈ a} ⊆ b a, which is open-and-closed, so φ(t) ∈ b a. Similarly, setting b = I \ H, φ(u) ∈ bb; since b a ∩ bb = a[ ∩ b is empty, φ(t) 6= φ(u). This shows that φ is injective, therefore a homeomorphism between βI and Z (3A3Dd). Q Q (ii) C(βI) is isomorphic, as Banach lattice, to `∞ (I). P P Let Z be the Stone space of PI. By 363Ha, we can identify `∞ (I), as Banach lattice, with L∞ (PI) = C(Z). But (i) tells us that we have a canonical identification between C(Z) and C(βI). Q Q (iii) We have a one-to-one correspondence between filters F on I and non-empty closed sets F ⊆ βI, T got by matching F with {b a : a ∈ F}, or F with {a : a ⊆ I, F ⊆ b a}, where b a ⊆ βI is the open-and-closed set corresponding to a ⊆ I. P P The identification of βI with the Stone space Z of PI ((i) above) means that we can regard the map a 7→ b a as a Boolean isomorphism T between PI and the algebra of open-and-closed subsets of βI (311I). For any filter F on I, set H(F) = {b a : a ∈ F}; because {b a : a ∈ F} is a downwardsdirected family of non-empty closed sets in the compact Hausdorff space βI, H(F) is a non-empty closed set. If F ⊆ βI is a non-empty closed set, then it is elementary to check that H(F ) = {a : F ⊆ b a} is a filter on I, and evidently H(H(F )) ⊇ F . But if t ∈ βI \ F , then (because {b a : a ⊆ I} is a base for the topology of βI, see 311I again) there is an a ⊆ I such that t ∈ b a and F ∩ b a = ∅, that is, F ⊆ Id \ a; so Id \ a ∈ H(F ) d and H(H(F )) ⊆ I \ a and t ∈ / H(H(F )). Thus H(H(F )) = F for every non-empty closed set F ⊆ βI. If F1 and F2 are filters on I and a ∈ F1 \ F2 , then {bd \ a : b ∈ F2 } is a downwards-directed family of non-empty closed sets in βI, so has non-empty intersection; if t ∈ bd \ a = bb \ b a for every b ∈ F2 , then t ∈ H(F2 ) \ H(F1 ). This shows that F 7→ H(F) is injective. It follows that F 7→ H(F), F 7→ H(F ) are the two halves of a bijection, as claimed. Q Q (iv) There are no non-trivial convergent sequences in βI. P P Because PI is Dedekind complete, its Stone space is extremally disconnected (314S), so, by the identification in (i), βI is extremally disconnected, and has no non-trivial convergent sequences (4A2G(h-iii)). Q Q (Compare Engelking 89, 3.6.15.)

4A2Jg

General topology

839

4A2J Uniform spaces (See §3A4.) Let (X, W) be a uniform space; give X the associated topology T (3A4Ab). (a) W is generated by a family of pseudometrics. (Engelking 89, 8.1.11; Bourbaki 66, IX.1.4; ´sza ´ r 78, 4.2.32.) More precisely: if hWn in∈N is any sequence in W, there is a pseudometric ρ on X such Csa that (α) {(x, y) : ρ(x, y) ≤ ²} ∈ W for every ² > 0 (β) whenever n ∈ N and ρ(x, y) < 2−n then (x, y) ∈ Wn (Engelking 89, 8.1.10). It follows that T is completely regular, therefore regular (3A3Be). T is defined by the bounded uniformly continuous functions, in the sense that it is the coarsest topology S on X such that these are all continuous. P P Let P be the family of pseudometrics compatible with W in the sense of (α) just above. If x ∈ G ∈ T, there is a ρ ∈ P such that {y : ρ(x, y) < 1} ⊆ G; setting f (y) = ρ(x, y), we see that f is uniformly continuous, therefore S-continuous, and that x ∈ intS G. As x is arbitrary, G ∈ S; as G is arbitrary, T ⊆ S; but of course S ⊆ T just because uniformly continuous functions are continuous. Q Q (b) If W ∈ W and x ∈ X then x ∈ int W [{x}]. (Engelking 89, 8.1.3.) If A ⊆ X then A = (Engelking 89, 8.1.4.)

T W ∈W

W [A].

´ sza ´ r 78, (c) Any subset of a totally bounded set in X is totally bounded. (Engelking 89, 8.3.2; Csa 3.2.70.) The closure of a totally bounded set is totally bounded. P P If A is totally bounded and W ∈ W, take W 0 ∈ W such that W 0 ◦ W 0 ⊆ W . Then there is a finite set I ⊆ X such that A ⊆ W 0 [I]. In this case A ⊆ W 0 [A] ⊆ W 0 [W 0 [I]] = (W 0 ◦ W 0 )[I] ⊆ W [I] by (b). As W is arbitrary, A is totally bounded. Q Q (d) A subset of X is compact iff it is complete (definition: 3A4F) (for its subspace uniformity) and totally ˇ ´ sza ´ r 78, 5.2.22; Gaal 64, pp. 278-279.) So if X bounded. (Engelking 89, 8.3.16; Cech 66, 41A.8; Csa is complete, every closed totally bounded subset of X is compact, and the totally bounded sets are just the relatively compact sets. (A closed subspace of a complete space is complete.) (e) If f : X → R is a continuous function with compact support, it is uniformly continuous. P P Set 1 K = {x : f (x) 6= 0}. Let ² > 0. For each x ∈ X, there is a Wx ∈ W such that |f (y) − f (x)| ≤ 2 ² whenever 0 y ∈ Wx [{x}]. Let Wx0 ∈ W be such that Wx0 ◦ Wx0 ⊆ Wx . Set Gx , by (b). Because S Gx = int Wx [{x}]; then x ∈T K is compact, there is a finite set I ⊆ K such that K ⊆ x∈I Gx . Set W = (X × X) ∩ x∈I Wx0 ∈ W. Take any (y, z) ∈ W ∩ W −1 . If neither y nor z belongs to K, then of course |f (y) − f (z)| ≤ ². If y ∈ K, let x ∈ I be such that y ∈ Gx . Then y ∈ Wx0 [{x}] ⊆ Wx [{x}],

z ∈ W [Wx0 [{x}]] ⊆ Wx0 [Wx0 [{x}]] ⊆ Wx [{x}],

so |f (y) − f (z)| ≤ |f (y) − f (x)| + |f (z) − f (x)| ≤ ². The same idea works if z ∈ K. So |f (y) − f (z)| ≤ ² for all y, z ∈ W ∩ W −1 ; as ² is arbitrary, f is uniformly continuous. Q Q (f ) (i) If (Y, S) is a completely regular space, there is a uniformity on Y compatible with S. (Engelking 89, 8.1.20.) (ii) If (Y, S) is a compact Hausdorff space, there is exactly one uniformity on Y compatible with S; it is induced by the set of all those pseudometrics on Y which are continuous as functions from Y × Y to R. (Engelking 89, 8.3.13.) (iii) If (Y, S) is a compact Hausdorff space and V is the uniformity on Y compatible with S, then any continuous function from Y to X is uniformly continuous. (g) The set U of uniformly continuous real-valued functions on X is a Riesz subspace of RX containing the constant functions. If a sequence in U converges uniformly, the limit function again belongs to U .

840

Appendix

4A2K

4A2K First-countable, sequential and countably tight spaces (a) Let X be a countably tight topological space. If hFξ iξ<ζ is a non-decreasing family of closed subsets of X indexed by an ordinal ζ, then S E = ξ<ζ Fξ is closed unless cf ζ = ω, in which case E is an Fσ set. P P If cf ζ = 0, that is, ζ = 0, then E = ∅ is closed. If cf ζ = 1, that is, ζ = ξ + 1 for some ordinal ξ, then E = Fξ is closed. If cf ζ = ω, there is S a sequence hξn in∈N in ζ with supremum ζ, so that E = n∈N Fξn is Fσ . If cf ζ > ω, take x ∈ E. Then there is a sequence hxn in∈N in E such that x ∈ {xn : n ∈ N}. For each n there is a ξn < ζ such that xn ∈ Fξn , and now ξ = supn∈N ξn < ζ and x ∈ F ξ = Fξ ⊆ E. As x is arbitrary, E is closed. Q Q (b) If X is countably tight, any subspace of X is countably tight, just because if A ⊆ Y ⊆ X then the closure of A in Y is the intersection of Y with the closure of A in X. If X is compact and countably tight, then any Hausdorff continuous image of X is countably tight. P P Let f : X → Y be a continuous surjection, where Y is Hausdorff, B a subset of Y and y ∈ B. Set A = f −1 [B]. Then A is compact, so f [A] is compact, therefore closed; because f is surjective, y ∈ f [A] ⊆ f [A], and there is an x ∈ A such that f (x) = y. Now there is a countable set A0 ⊆ A such that x ∈ A0 , in which case y = f (x) ∈ f [A0 ] ⊆ f [A0 ], while f [A0 ] is a countable subset of B. Q Q S (c) If X is a sequential space, it is countably tight. P P Suppose that A ⊆ X and x ∈ A. Set B = {C : C ⊆ A is countable}. If hxn in∈N is a sequence in B convergingSto x ∈ X, then for each n ∈ N we can find a countable set Cn ⊆ A such that xn ∈ C n , and now C = n∈N Cn is a countable subset of A such that x ∈ C ⊆ B. So B is sequentially closed, therefore closed, and x ∈ B. As A and x are arbitrary, X is countably tight. Q Q (d) If X is a sequential space, Y is a topological space and f : X → Y is sequentially continuous, then it is continuous. (Engelking 89, 1.6.15.) (e) First-countable spaces are sequential. (Engelking 89, 1.6.14.) (f ) Let X be a locally T compact Hausdorff space in which every singleton set is Gδ . Then X is firstcountable. P P If {x} = n∈N Gn where each T Gn is open, then for each n ∈ N we can find a compact set Fn Q such that x ∈ int Fn ⊆ Gn . By 4A2Gd, { i≤n Fi : n ∈ N} is a base of neighbourhoods of x. Q 4A2L (Pseudo-)metrizable spaces ‘Pseudometrizable’ spaces, as such, hardly appear in this volume, for the usual reasons; they surface briefly in §463. It is perhaps worth noting, however, that all the ideas, and very nearly all the results, in this paragraph apply equally well to pseudometrics and pseudometrizable topologies. (a) Any subspace of a (pseudo-)metrizable space is (pseudo-)metrizable (2A3J). A topological space is metrizable iff it is pseudometrizable and Hausdorff (2A3L). ˇ ´sza ´r 78, 8.3.16; Cech (b) Metrizable spaces are paracompact (Engelking 89, 5.1.3; Csa 66, 30C.2; Gaal 64, p. 155), therefore hereditarily metacompact ((a) above and 4A2F(g-i)). ´ sza ´ r 78, 8.4.5.), so every closed (c) A metrizable space is perfectly normal (Engelking 89, 4.1.13; Csa set is a zero set and every open set is a cozero set (in particular, is Fσ ). (d) If X is a pseudometrizable space, it is first-countable. (If ρ is a pseudometric defining the topology of X, and x is any point of X, then {{y : ρ(y, x) < 2−n } : n ∈ N} is a base of neighbourhoods of x.) So X is sequential and countably tight (4A2Ke, 4A2Kc), and if Y is another topological space and f : X → Y is sequentially continuous, then f is continuous (4A2Kd). (e) Relative compactness Let X be a pseudometrizable space and A a subset of X. Then the following are equiveridical: (α) A is relatively compact; (β) A is relatively countably compact; (γ) every sequence in A has a subsequence with a limit in X. P P Fix a pseudometric ρ defining the topology of X. (α)⇒(β) by 4A2G(f-iii). If hxn in∈N is a sequence in A with a cluster point x ∈ X, then we can choose hni ii∈N inductively

4A2N

General topology

841

such that ρ(xni , x) ≤ 2−i and ni+1 > ni for every i; now hxni ii∈N → x; it follows that (β)⇒(γ). Now assume that (α) is false. Then there is an ultrafilter F on X containing A which has no limit in X (3A3Be, 3A3De). T If F is a Cauchy filter, choose Fn ∈ F such that ρ(x, y) ≤ 2−n whenever x, y ∈ Fn , and xn ∈ A ∩ i≤n Fi for each n; then it is easy to see that hxn in∈N is a sequence in A with no convergent subsequence. If F is not a Cauchy filter, let ² > 0 be such that there is no F ∈ F such that ρ(x, y) ≤ ² for every x, S y ∈ F . Then X \ U (x, 21 ²) ∈ F for ever x ∈ X, so we can choose hxn in∈N inductively such that xn ∈ A \ i qn0 } (interpreting inf ∅ as ∞). Then U = hGnξ iξ<κ,n∈N is a σ-metrically-discrete family of open sets. If G ⊆ X is open and x ∈ G, let ² > 0 be such that U (x, 2²) ⊆ G. Let ξ < κ be minimal such that ρ(x, xξ ) < ², and let n ∈ N be such that ρ(x, xξ ) < qn < qn0 < ²; then x ∈ Gnξ ⊆ G. As x and G are arbitrary, U is a base for the topology of X. Q Q (ii) Consequently, any metrizable space has a σ-disjoint base. (Compare Engelking 89, 4.4.3; ´sza ´ r 78, 8.4.5; Kuratowski 66, §21.XVII.) Csa ´ sza ´r (h) The product of a countable family of metrizable spaces is metrizable. (Engelking 89, 4.2.2; Csa 78, 7.3.27.) (i) Let X be a metrizable space and κ ≥ ω a cardinal. Then w(X) ≤ κ iff X has a dense subset of cardinal at most κ. (Engelking 89, 4.1.15.) (j) If (X, ρ) is any metric space, then the balls B(x, δ) = {y : ρ(y, x) ≤ δ} are all closed sets (cf. 1A2G). In particular, in a normed space (X, k k), the balls B(x, δ) = {y : ky − xk ≤ δ} are closed. 4A2M Complete metric spaces (a) Baire’s theorem for complete metric spaces In a complete metric space, the intersection of a sequence of dense Gδ sets is dense. (Engelking 89, 4.3.36 & 3.9.4; ´sza ´ r 78, 9.2.1 & 9.2.8; Gaal 64, p. 287; Kechris 95, 8.4.) So a non-empty complete metric space is Csa not meager (cf. 3A3Ha). (b)QLet h(Xi , ρi )ii∈I be a countable family of complete metric spaces. Then there is a complete metric on X = i∈I Xi which defines the product topology on X. (Engelking 89, 4.3.12; Kuratowski 66, §33.III.) (c) Let (X, ρ) be a complete metric space, and E ⊆ X a Gδ set. Then there is a complete metric on E which defines the subspace topology of E. (Engelking 89, 4.3.23; Kuratowski 66, §33.VI; Kechris 95, 3.11.) ˇ (d) Let (X, ρ) be a complete metric space. Then it is Cech-complete. (Engelking 89, 4.3.26.) (e) A non-empty complete metric space without isolated points is uncountable. (If x is not isolated {x} is nowhere dense.) 4A2N Countable networks: Proposition (a) If X is a topological space with a countable network, any subspace of X has a countable network. (b) Let X be a space with a countable network. Then X is hereditarily Lindelöf. If it is regular, every closed set is a zero set. (c) If X is a topological space, and hAn in∈NSis a sequence of subsets of X each of which has a countable network (for its subspace topology), then A = n∈N An has a countable network. (d) A continuous image of a space with a countable network has a countable network.

842

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4A2N

(e) Let hXi ii∈I be a countable family of topological spaces with countable networks, with product X. Then X has a countable network. (f) If X is a Hausdorff space with a countable network, there is a countable family G of open sets such that whenever x, y are distinct points in X there are disjoint G, H ∈ G such that x ∈ G and y ∈ H. (g) If X is a regular topological space with a countable network, it has a countable network consisting of closed sets. (h) A compact Hausdorff space with a countable network is metrizable. (i) If a topological space X has a countable network, then any dense set in X includes a countable dense set; in particular, X is separable. (j) If a topological space X has a countable network, then C(X), with the topology of pointwise convergence inherited from the product topology of RX , has a countable network. proof (a) If E is a countable network for the topology of X, and Y ⊆ X, then {Y ∩ E : E ∈ E} is a countable network for the topology of Y . (b) By Engelking 89, 3.8.12 X is Lindelöf. Since any subspace of X has a countable network ((a) above), it also is Lindelöf, and X is hereditarily Lindelöf. By 4A2H(c-ii), if X is regular, closed sets are zero sets. S (c) If En is a countable network for the topology of An for each n, then n∈N En is a countable network for the topology of A. (d) Let X be a topological space with a countable network E, and Y a continuous image of X. Let f : X → Y be a continuous surjection. Then {f [E] : E ∈ E} is a network for the topology of Y . P P If H ⊆ Y is open and y ∈ H, then f −1 [H] is an open subset of X and there is an x ∈ X such that f (x) = y. Now there must be an E ∈ E such that x ∈ E ⊆ f −1 [H], so that y ∈ f [E] ⊆ H. Q Q But {f [E] : E ∈ E} is countable, so Y has a countable network. (e) For each i ∈ I let Ei be a Q countable network for the topology of Xi . For each finite J ⊆ I, let CJ be the family of sets expressible as i∈I Ei where Ei ∈ Ei for each i ∈ J and Ei = Xi for i ∈ I \ J; then CJ is countable because for each i ∈ J. Because the family [I]<ω of finite subsets of I is countable S Ei is countable <ω (3A1Cd), E = {CJ : J ∈ [I] } is countable. But E is a network for the topology of X. P P If G ⊆ X is open andQx ∈ G, then there is a family hGi ii∈I such that every Gi ⊆ Xi is open, J = {i : Gi 6= Xi } is finite, and x ∈ i∈I Gi . For i ∈ J, there is an Ei ∈ Ei such that x(i) ∈ Ei ⊆ Gi ; set Ei = Xi for i ∈ I \ J. Then Q E = i∈I Ei ∈ CJ ⊆ E and x ∈ E ⊆ G. Q Q So E is a countable network for the topology of X. (f ) By (b) and (e), X × X is hereditarily Lindelöf. In particular, W = {(x, y) : x 6= y} is Lindelöf. Set V = {G × H : G, H ⊆ X are open, G ∩ H = ∅}. Because X is Hausdorff, V is a cover of W . So there is a countable V0 ⊆ V covering W . Set G = {G : G × H ∈ V0 } ∪ {H : G × H ∈ V0 }. Then G is a countable family of open sets separating the points of X. (g) Let X be a regular space and E a countable network for the topology of X. Set E 0 = {E : E ∈ E}. If G ⊆ X is open and x ∈ G, then (because the topology is regular) there is an open set H such that x ∈ H ⊆ H ⊆ G. Now there is an E ∈ E such that x ∈ E ⊆ H, in which case E ∈ E 0 and x ∈ E ⊆ G. So E 0 is a countable network for X consisting of closed sets. (h) Engelking 89, 3.1.19. (i) Let D ⊆ X be dense, and E a countable network for the topology of X. Let D0 ⊆ D be a countable set such that D0 ∩ E 6= ∅ whenever E ∈ E and D ∩ E 6= ∅. If G ⊆ X is open and not empty, there is an x ∈ D ∩ G; now there is an E ∈ E such that x ∈ E ⊆ G, and as x ∈ D ∩ E there must be an x0 ∈ D0 ∩ E, so that x0 ∈ D0 ∩ G. As G is arbitrary, D0 is dense in X. Taking D = X, we see that X has a countable dense subset.

4A2Q

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(j) Let E be a countable network for the topology of X and U a countable base for the topology of R (4A2Ua). For E ∈ E, U ∈ U set H(E, U ) = {f : f ∈ C(X), E ⊆ f −1 [U ]}. Then the set of finite intersections of sets of the form H(E, U ) is a countable network for the topology of pointwise convergence on C(X). (Compare 4A2Oe.) 4A2O Second-countable spaces (a) Let (X, T) be a topological space and U a countable subbase for T. Then T is second-countable. ({X} ∪ {U0 ∩ U1 ∩ . . . ∩ Un : U0 , . . . , Un ∈ U } is countable and is a base for T, by 4A2B(a-i).) ´ sza ´ r 78, 2.4.17.) (b) Any base of a second-countable space includes a countable base. (Csa (c) Let (X, T) be a second-countable space. T has a countable network (because a base is also a network), so is separable and hereditarily Lindelöf (Engelking 89, 1.3.8 & 3.8.1, 4A2Nb, 4A2Ni). (d) The product of a countable family of second-countable spaces is second-countable. (Engelking 89, 2.3.14.) (e) If X is a second-countable space then C(X), with the topology of uniform convergence on compact sets, has a countable network. P P (See Engelking 89, Ex. 3.4H.) Let U be a countable base for the topology of X and V a countable base for the topology of R (4A2Ua). For U ∈ U , V ∈ V set H(U, V ) = {f : f ∈ C(X), U ⊆ f −1 [V ]}. Then the set of finite intersections of sets of the form H(U, V ) is a countable network for the topology of uniform convergence on compact subsets of X. Q Q 4A2P Separable metrizable spaces (a) (i) A metrizable space is second-countable iff it is separable. ´ sza ´ r 78, 2.4.16; Gall p. 120.) (Engelking 89, 4.1.16; Csa ´ sza ´ r 78, 5.3.35; Kura(ii) A compact metrizable space is separable (Engelking 89, 4.1.18; Csa towski 66, §21.IX), so is second-countable and has a countable network. (iii) Any base of a separable metrizable space includes a countable base (4A2Ob), which is also a countable network, so the space is hereditarily Lindelöf (4A2Nb). (iv) Any subspace of a separable metrizable space is separable and metrizable (4A2La, 4A2Na, 4A2Ni). (v) A countable product of separable metrizable spaces is separable and metrizable (4A2B(e-ii), 4A2Lh). (b) A topological space is separable and metrizable iff it is second-countable, regular and Hausdorff. ´ sza ´ r 78, 7.1.57; Kuratowski 66, §22.II.) (Engelking 89, 4.2.9; Csa (c) Let X be a compact metrizable space, Y a Hausdorff space and f : X → Y a continuous surjection. Then Y is metrizable. (It is a compact Hausdorff space, by 2A3N(b-ii), with a countable network, by 4A2Nd, so is metrizable, by 4A2Nh.) (d) A metrizable space is separable iff it is ccc iff it is Lindelöf. (Engelking 89, 4.1.16.) (e) If X is a compact metrizable space, then C(X) is separable under its usual norm topology defined from the norm k k∞ . (Engelking, 3.4.16.) 4A2Q Polish spaces: Proposition (a) A countable discrete space is Polish. (b) A compact metrizable space is Polish. (c) The product of a countable family of Polish spaces is Polish. (d) A Gδ subset of a Polish space is Polish in its subspace topology; in particular, a set which is either open or closed is Polish. (e) The disjoint union of countably many Polish spaces is Polish. (f) If X is any set and hTn in∈N is a sequence of Polish S topologies on X such that Tm ∩ Tn is Hausdorff for all m, n ∈ N, then the topology T∞ generated by n∈N Tn is Polish. (g) If X is a Polish space, it is homeomorphic to a Gδ set in a compact metric space. proof (a) Any set X is complete under the discrete metric ρ defined by setting ρ(x, y) = 1 whenever x, y ∈ X are distinct. This defines the discrete topology, and if X is countable it is separable, therefore Polish.

844

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4A2Q

(b) By 4A2P(a-ii), it is separable; by 4A2Jd, any metric defining the topology is complete. (c) If hXi ii∈I is a countable family of Polish spaces with product X, then surely X is separable (4A2B(eii)); and 4A2Mb tells us that its topology is defined by a complete metric. (d) If X is Polish and E is a Gδ set in X, then E is separable, by 4A2P(a-iv), and its topology is defined by a complete metric, by 4A2Mc. So E is Polish. Any open set is of course a Gδ set, and any closed set is a Gδ set by 4A2Lc. S (e) Let hXi ii∈I be a countable disjoint family of Polish spaces, and X = i∈I Xi . For each i ∈ I let ρi be a complete metric on Xi defining the topology of Xi . Define ρ : X × X → [0, ∞[ by setting ρ(x, y) = min(1, ρi (x, y)) if i ∈ I and x, y ∈ Xi , ρ(x, y) = 1 otherwise. It is easy to check that ρ is a complete metric on X defining the disjoint union topology on X. X is separable, by 4A2B(e-i), therefore Polish. (f ) This result is in Kechris 95, 13.3; but I spell out the proof because it is an essential element of some measure-theoretic arguments. On X N take the product topology T of the topologies Tn . This is Polish, by (c). Consider the diagonal ∆ = {x : x ∈ X N , x(m) = x(n) for all m, n ∈ N}. This is closed in X N . P P If x ∈ X N \ ∆, let m, n ∈ N be such that x(m) 6= x(n). Because Tm ∩ Tn is Hausdorff, there are disjoint G, H ∈ Tm ∩ Tn such that x(m) ∈ G and x(n) ∈ H. Now {y : y ∈ X N , y(m) ∈ G, y(n) ∈ H} is an open set in X N containing x and disjoint from ∆. As x is arbitrary, ∆ is closed. Q Q By (d), ∆, with its subspace topology, is a Polish space. Let f : X → ∆ be the natural bijection, setting f (t) = x if x(n) = t for every n, and let S be the topology on X which makes f a homeomorphism. The topology on ∆ is generatedSby {{x : x ∈ ∆, x(n) ∈ G} : n ∈ N, G ∈ Tn }, so S is generated by {{t : t ∈ X, t ∈ G} : n ∈ N, G ∈ Tn } = n∈N Tn . Thus S = T∞ and T∞ is Polish. (g) Kechris 95, 4.14. 4A2R Order topologies Let (X, ≤) be a totally ordered set and T its order topology. (a) The set U of open intervals in X (definition: 4A2A) is a base for T. (b) [x, y], [x, ∞[ and ]−∞, x] are closed sets for all x, y ∈ X. (c) T is Hausdorff and normal. (d) If A ⊆ X then A is the set of elements of X expressible as either suprema or infima of non-empty subsets of A. (e) F ⊆ X is closed iff it is order-closed. (f) If hxn in∈N is a non-decreasing sequence in X with supremum x, then x = limn→∞ xn . (g) A set K ⊆ X is compact iff sup A and inf A are defined in X and belong to K for every non-empty A ⊆ K. (h) X is Dedekind complete iff [x, y] is compact for all x, y ∈ X. (i) X is compact iff it is either empty or Dedekind complete with greatest and least elements. (j) If X is Dedekind complete, any open set G ⊆ X is expressible as a union of disjoint open intervals. (k) If X is well-ordered it is locally compact. (l) In X × X, {(x, y) : x < y} is open and {(x, y) : x ≤ y} is closed. (m) If F ⊆ X is either compact or an interval, then the subspace topology on F is induced by the inherited order of F . (n) If X is ccc it is hereditarily Lindelöf, therefore perfectly normal. (o) If Y is another totally ordered set with its order topology, an order-preserving function from X to Y is continuous iff it is order-continuous. proof (a) Put the definition of ‘order topology’ (4A2A) together with 4A2B(a-i). (b) Their complements are either X, or members of U, or unions of two members of U. (c) Fix a well-ordering 4 of X. If x < y ∈ X, define Uxy , Uyx as follows: if ]x, y[ is empty, Uxy = ]−∞, x] = ]−∞, y[ and Uyx = [y, ∞[ = ]x, ∞[; otherwise, let z be the 4-least member of ]x, y[ and set Uxy = ]−∞, z[, Uyx = ]z, ∞[. This construction ensures that if x, y are any distinct points of X, Uxy and Uyx are disjoint open sets containing x, y respectively, so T is Hausdorff.

4A2R

General topology

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T Now suppose that F ⊆ X is closed and that x ∈ X \ F . Then VxF = int(X ∩ y∈F Uxy ) contains x. P P There are u, v ∈ X ∪ {−∞, ∞} such that x ∈ ]u, v[ ⊆ X \ F . If ]u, x[ = ∅, set u0 = u; otherwise, let u0 be the 4-least member of ]u, x[. Similarly, if ]x, v[ = ∅, set v 0 = v; otherwise, let v 0 be the 4-least member of ]x, v[. Then u0 < x < v 0 . Now suppose that y ∈ F and y > x. If ]x, v[ = ∅, then Uxy ⊇ ]−∞, x] = ]−∞, v 0 [. Otherwise, v 0 ∈ ]x, v[ ⊆ ]x, y[, so Uxy = ]−∞, z[ where z is the 4-least member of ]x, y[. But this means that z 4 v 0 and either z = v 0 or z ∈ / ]x, v[; in either case, v 0 ≤ z and ]−∞, v 0 [ ⊆ Uxy . 0 0 Similarly, ]u , ∞[ ⊆ Uxy whenever y ∈ F and y < x. So x ∈ ]u0 , vS [ ⊆ VxF . Q Q S Finally, suppose that E and F are disjoint closed sets. Set G = x∈E VxF , H = y∈F VyE . Then G and H are open sets including E, F respectively. If x ∈ E and y ∈ F , then VxF ∩ VyE ⊆ Uxy ∩ Uyx = ∅, so G ∩ H = ∅. As E and F are arbitrary, T is normal. (d) Let B be the set of such suprema and infima. For x ∈ X set Ax = A ∩ ]−∞, x], A0x = A ∩ [x, ∞[. Then x ∈ B iff either x = sup Ax or x = inf A0x , so x∈ / B ⇐⇒ x 6= sup Ax and x 6= inf A0x ⇐⇒ there are u < x, v > x such that Ax ⊆ ]−∞, u] and A0x ⊆ [u, ∞[ ⇐⇒ there are u, v such that x ∈ ]u, v[ ⊆ X \ A ⇐⇒ x ∈ / A. Thus B = A, as claimed. (e) Because X is totally ordered, all its subsets are both upwards-directed and downwards-directed; so we have only to join the definition in 313Da to (d) above. (f ) If x ∈ ]u, v[ then there is some n ∈ N such that xn ≥ u, and now xi ∈ ]u, v[ for every i ≥ n. (g)(α) If K is compact and A ⊆ K is non-empty, let B be the set of upper bounds for A in X ∪ {−∞, ∞}, S and set G = {]−∞, a[ : a ∈ A}∪{]b, ∞[ : b ∈ B}. Then no finite subfamily of G can cover K; and if c ∈ K\ G then c = sup A. Similarly, any non-empty subset of K has an infimum in X which belongs to K. (β) Now suppose that K satisfies the condition. By (d) above, it is closed. If it is empty it is certainly compact. Otherwise, a0 = inf K and b0 = sup K are defined in X and belong to K. Let G be an open cover of K. Set A = {x : x ∈ X, K ∩ [a0 , x] is not covered by any finite G0 ⊆ G}. Note that A is bounded below by a0 . ?? If b0 ∈ A, then c = inf A is defined and belongs to [a0 , b0 ], because X is Dedekind complete. If c ∈ / K then there are u, v such that c ∈ ]u, v[ ⊆ X \ K; if c ∈ K then there are u, v such that c ∈ ]u, v[ ⊆ G for some G ∈ G. In either case, u ∈ / A, so that K ∩ [a0 , v[ ⊆ (K ∩ [a0 , u]) ∪ ]u, v[ is covered by a finite subset of G, and A does not meet [a0 , v[, that is, A ⊆ [v, ∞[ and v is a lower bound of A. X X Thus b0 ∈ / A, and K = K ∩ [a0 , b0 ] is covered by a finite subset of G. As G is arbitrary, K is compact. (h) Use (g). (i) Use (h). (j) Gaal 64, p. 36. (Compare 2A2I.) (k) Use (h). (l) Write W for {(x, y) : x < y}. If x < y, then either there is a z such that x < z < y, in which case ]−∞, z[ × ]z, ∞[ is an open set containing (x, y) and included in W , or ]x, y[ = ∅, so ]−∞, y[ × ]x, ∞[ is an open set containing (x, y) and included in W . Thus W is open. Now {(x, y) : x ≤ y} = (X × X) \ {(x, y) : y < x} is closed. (m) The subspace topology TF on F is generated by sets of the form F ∩ ]−∞, x[, F ∩ ]x, ∞[ where x ∈ X (4A2B(a-iv)), while the order topology S on F is generated by sets of the form F ∩ ]−∞, x[, F ∩ ]x, ∞[ where x ∈ F . So S ⊆ TF . Now suppose that F is either compact or an interval, and that x ∈ X. If x ∈ F , or F ∩ ]−∞, x[ is either F or ∅, then of course F ∩ ]−∞, x[ ∈ S. Otherwise, F meets both ]−∞, x[ and [x, ∞[ and does not contain

846

Appendix

4A2R

x, so is not an interval, and is compact. Accordingly x0 = inf(F ∩ [x, ∞[) is defined and belongs to F , by (g), and F ∩ ]−∞, x[ = F ∩ ]−∞, x0 [ ∈ S. Similarly, F ∩ ]x, ∞[ ∈ S for every x ∈ X. But this means that TF ⊆ S, so the two topologies are equal, as stated. (n)(α) Let G be a family of open subsets of X with union H. Set S A = {A : A ⊆ G0 for some countable G0 ⊆ G}. S (I seek to show that H ∈ A.) Of course A0 ∈ A for every countable subset A0 of A. (β) Let C be the family of members C of A which are order-convex (that is, [x, y] ⊆ C for all x, y ∈ C), and C ∗ the family of maximal members of C. If C ∈ C is not included in any member of C ∗ , then there is a C 0 ∈ C such that C ⊆ C 0 and int(C 0 \ C) 6= ∅. P P Since no member of C including C can be maximal, we can find C 0 ∈ C such that C ⊆ C 0 and #(C 0 \ C) ≥ 5. Because C is order-convex, every point of X \ C is either a lower bound or an upper bound of C, and there must be three points x < y < z of C 0 \ C on the same side of C. In this case, y ∈ ]x, z[ ⊆ int(C 0 \ C), so we have an appropriate C 0 . Q Q (γ) In fact, every member of C is included in a member of C ∗ . P P?? Suppose, if possible, otherwise. Then we can choose a strictly increasing family hCξ iξ<ω1 in C inductively, as follows. Start from any non-empty C0 ∈ C not included in any member of C ∗ . Given that C0 ⊆ Cξ ∈ C, then Cξ cannot be included in any member of C ∗ , so by (β) above there is a Cξ+1 ∈ C such that int(C S ξ+1 \ Cξ ) is non-empty. Given hCη iη<ξ where ξ < ω1 is a non-zero countable limit ordinal, set Cξ = η<ξ Cη ; then Cξ is order-convex, because {Cη : η < ξ} is upwards-directed, and belongs to A, because A is closed under countable unions, so Cξ ∈ C and the induction proceeds. Now, however, hint(Cξ+1 \ Cξ )iξ<ω1 is an uncountable disjoint family of non-empty open sets, and X is not ccc. X XQ Q (δ) Since C ∪ C 0 is order-convex whenever C, C 0 ∈ C and C ∩ C 0 6= ∅, C ∗ is a disjoint family. Moreover, if x ∈ H, there is some open interval containing x and belonging to C, so x ∈ int S C for some C ∈ C ∗ ; this ∗ ∗ shows that C is an open cover of H. Because X is ccc, C is countable, so H = C ∗ ∈ A. Thus there is some countable G0 ⊆ G with union H; as G is arbitrary, X is hereditarily Lindelöf, by 4A2H(c-i). (²) By 4A2H(c-ii), X is perfectly normal. (o)(α) Suppose that f is continuous. If A ⊆ X is a non-empty set with supremum x in X, then x ∈ A, by (d), so f (x) ∈ f [A] (3A3Cc) and f (x) is less than or equal to any upper bound of f [A]; but f (x) is an upper bound of f [A], because f is order-preserving, so f (x) = sup f [A]. Similarly, f (inf A) = inf f [A] whenever A is non-empty and has an infimum, so f is order-continuous. (β) Now suppose that f is order-continuous. Take any y ∈ Y and consider A = f −1 [ ]−∞, y[ ], B = X \ A. If x ∈ A then f (x) cannot be inf f [B] so x cannot be inf B and there is an x0 ∈ X such that x < x0 ≤ z for every z ∈ B; in which case x ∈ ]−∞, x0 [ ⊆ A. So A is open. Similarly, f −1 [ ]y, ∞[ ] is open. By 4A2B(a-ii), f is continuous. 4A2S Order topologies on ordinals (a) Let ζ be an ordinal with its order topology. (i) ζ is locally compact (4A2Rk); all the sets [0, η] = ]−∞, η + 1[, for η < ζ, are open and compact (4A2Rh). If ζ is a successor ordinal, it is compact, being of the form [0, η] where ζ = η + 1. (ii) For any A ⊆ ζ, A = {sup B : ∅ 6= B ⊆ A, sup B < ζ}. (4A2Rd, because inf B ∈ B ⊆ A for every non-empty B ⊆ A.) (iii) If ξ ≤ ζ, then the subspace topology on ξ induced by the order topology of ζ is the order topology of ξ. (4A2Rm.) (b) Give ω1 its order topology. (i) ω1 is first-countable. P P If ξ < ω1 is either zero or a successor ordinal, then {ξ} is open so {{ξ}} is a base of neighbourhoods of ξ. If ξ is a non-zero limit ordinal, there is a sequence hξn in∈N in ξ with supremum ξ, and {]ξn , ξ] : n ∈ N} is a base of neighbourhoods of ξ. Q Q (ii) Singleton subsets of ω1 are zero sets. (Assemble 4A2F(d-v), 4A2Rc and (i) above.)

4A2Ud

General topology

847

(iii) If f : ω1 → R is continuous, there is a ξ < ω1 such that f (η) = f (ξ) for every η ≥ ξ. P P?? Otherwise, we may define a strictly increasing family hζξ iξ<ω1 in ω1 by saying that ζ0 = 0, ζξ+1 = min{η : η > ζξ , f (η) 6= f (ζξ )} for every ξ < ω1 , ζξ = sup{ζη : η < ξ} for non-zero countable limit ordinals ξ. Now S ω1 = k∈N {ξ : |f (ζξ+1 ) − f (ζξ )| ≥ 2−k }, so there is a k ∈ N such that A = {ξ : |f (ζξ+1 ) − f (ζξ )| ≥ 2−k } is infinite. Take a strictly increasing sequence hξn in∈N in A and set ξ = supn∈N ξn = supn∈N (ξn + 1). Then hζξn in∈N and hζξn +1 in∈N are strictly increasing sequences with supremum ζξ , so both converge to ζξ in the order topology of ω1 (4A2Rf), and f (ζξ ) = limn→∞ f (ζξn ) = limn→∞ f (ζξn +1 ). But this means that limn→∞ f (ζξn ) − f (ζξn +1 ) = 0, which is impossible, because |f (ζξn ) − f (ζξn +1 )| ≥ 2−k for every n. X XQ Q 4A2T Upper limits of families of sets For the sake of §446 I outline some elementary ideas. Let X be a topological space, hAi ii∈I a family of subsets of X, and F a filter on I. (a) lim supi→F Ai is closed. If F ⊆ X is closed and Ai ⊆ F for every i, then lim supi→F Ai ⊆ F . If hBi ii∈I is another family of sets, and Bi ⊆ Ai for every i, then lim supi→F Bi ⊆ lim supi→F Ai . If x ∈ Ai for every i then x ∈ lim supi→F Ai . (Immediate from the definition in 4A2A.) If X is Hausdorff, xi ∈ Ai for every i and x = limi→F xi is defined in X, then x ∈ lim supi→F Ai . (lim supi→F {xi } = {x}.) If f : X → X is a homeomorphism then f [lim supi→F Ai ] = lim supi→F f [Ai ]. S (b) If K ⊆ X is compact and K ∩ lim supi→F Ai = ∅, then {i : Ai ∩ K = ∅} ∈ F. ({K ∩ i∈J Ai : J ∈ F} is a downwards-directed family of relatively closed subsets of K with empty intersection, so contains ∅.) (c) Suppose that F is an ultrafilter. If G ⊆ X is open and G∩lim supi→F Ai 6= ∅ then {i : G∩Ai 6= ∅} ∈ F. (Otherwise, J = {i : G ∩ Ai = ∅} ∈ F and S lim supi→F Ai ⊆ i∈J Ai ⊆ X \ G.) 4A2U Old friends (a) R, with its usual topology, is metrizable (2A3Ff) and separable (the countable set Q is dense), so is second-countable (4A2P(a-i)). Every subset of R is separable (4A2P(a-iv)); in particular, every dense subset of R has a countable subset which is still dense. (b) N N is Polish in its usual topology (4A2Qc), so has a countable network (4A2P(a-iii), or 4A2Ne), and is hereditarily Lindelöf (4A2Nb or 4A2Pd). Moreover, it is homeomorphic to [0, 1] \ Q(Kuratowski 66, §36.II; Kechris 95, 7.7). P∞ (c) The map x 7→ 23 j=0 3−j x(j) (cf. 134Gb) is a homeomorphism between {0, 1}N and the Cantor set C ⊆ [0, 1]. (It is a continuous bijection.) (d) If I is any set, then the map A 7→ χA : PI → {0, 1}I is a homeomorphism (for the usual topologies on PI and {0, 1}I , as described in 4A2A and 3A3K). So PI is zero-dimensional, compact (3A3K) and Hausdorff. If I is countable, then PI is metrizable, therefore Polish (4A2Qb).

848

Appendix

§4A3 intro.

4A3 Topological σ-algebras I devote a section to some σ-algebras which can be defined on topological spaces. While ‘measures’ will not be mentioned here, the manipulation of these σ-algebras is an essential part of the technique of measure theory, and I will give proofs and exercises as if this were part of the main work. I look at Borel σ-algebras (4A3A-4A3J), Baire σ-algebras (4A3K-4A3P), Baire property algebras (4A3Q, 4A3R) and cylindrical σalgebras on linear spaces (4A3T-4A3V). 4A3A Borel sets If (X, T) is a topological space, the Borel σ-algebra of X is the σ-algebra B(X) of subsets of X generated by T. Its elements are the Borel sets of X. If (Y, S) is another topological space with Borel σ-algebra B(Y ), a function f : X → Y is Borel measurable if f −1 [H] ∈ B(X) for every H ∈ S, and is a Borel isomorphism if it is a bijection and B(Y ) = {F : F ⊆ Y, f −1 [F ] ∈ B(X)}, that is, f is an isomorphism between the structures (X, B(X)) and (Y, B(Y )). 4A3B (Σ, T)-measurable functions It is time I put the following idea into bold type. (a) Let X and Y be sets, with σ-algebras Σ ⊆ PX and T ⊆ PY . A function f : X → Y is (Σ, T)measurable if f −1 [F ] ∈ Σ for every F ∈ T. (b) If Σ, T and Υ are σ-algebras of subsets of X, Y and Z respectively, and f : X → Y is (Σ, T)measurable while g : Y → Z is (T, Υ)-measurable, then gf : X → Z is (Σ, Υ)-measurable. (If H ∈ Υ, g −1 [H] ∈ T so (gf )−1 [H] = f −1 [g −1 [H]] ∈ Σ.) (c) Let hXi ii∈I be a family of sets with product X, Y another set, and f : X → Y a function. If N T ⊆ PY , Σi ⊆ PXi are σ-algebras for each i, then f is (T, c i∈I Σi )-measurable iff πi f : Y → Xi is (T, Σi )N measurable for every i, where πi : X → Xi is the coordinate map. P P πi is ( c j∈I Σj , Σi )-measurable, so N if f is (T, c j∈I Σj )-measurable then πi f must be (T, Σi )-measurable. In the other direction, if every πi f is measurable, then {H : H ⊆ X, f −1 [E] ∈ T} is a σ-algebra of subsets of X containing πi−1 [E] whenever N Q i ∈ I and E ∈ Σi , so includes c i∈I Σi , and f is measurable. Q 4A3C Elementary facts (a) If X is a topological space and Y is a subspace of X, then B(Y ) is just the subspace σ-algebra {E ∩ Y : E ∈ B(X)}. P P {E : E ⊆ X, E ∩ Y ∈ B(Y )} and {E ∩ Y : E ∈ B(X)} are σ-algebras containing all open sets, so include B(X), B(Y ) respectively. Q Q (b) If X is a set, Σ is a σ-algebra of subsets of X, (Y, S) is a topological space and f : X → Y is a function, then f is (Σ, B(Y ))-measurable iff f −1 [H] ∈ Σ for every H ∈ S. P P If f is (Σ, B(Y ))-measurable then f −1 [H] ∈ Σ for every H ∈ S just because S ⊆ B(Y ). If f −1 [H] ∈ Σ for every H ∈ S, then {F : F ⊆ Y, f −1 [F ] ∈ Σ} is a σ-algebra of subsets of Y (cf. 112E) including S, so includes B, and f is (Σ, B)-measurable. Q Q (c) If X and Y are topological spaces, and f : X → Y is a function, then f is Borel measurable iff it is (B(X), B(Y ))-measurable. (Apply (b) with Σ = B(X).) So if X, Y and Z are topological spaces and f : X → Y , g : Y → Z are Borel measurable functions, then gf : X → Z is Borel measurable. (Use 4A3Bb.) (d) If X and Y are topological spaces and f : X → Y is continuous, it is Borel measurable. (Immediate from the definitions in 4A3A.) (e) If X is a topological space and f : X → [−∞, ∞] is lower semi-continuous, then it is Borel measurable. (The inverse image of a half-open interval ]α, β] is a difference of open sets, so is a Borel set, and every open subset of [−∞, ∞] is a countable union of such half-open intervals.) N (f ) If hXi ii∈I is a family of topological spaces with product X, then B(X) ⊇ c i∈I B(Xi ). (Put (d) and 4A3Bc together.)

4A3F

Topological σ-algebras

849

(g) Let X be a topological space. (i) The algebra A of subsets generated by the open sets is precisely the family of sets expressible as a S disjoint union i≤n Gi ∩ Fi where every Gi is open and every Fi is closed. P P Write A for the family of sets expressible in this form. Of course A ⊆ A. In the other direction, observe that X ∈ A, if E, E 0 ∈ A then E ∩ E 0 ∈ A, if E ∈ A then X \ E ∈ A because if Gi is open and Fi is closed for i ≤ n, then (identifying {0, . . . , n} with n + 1) [ \ X\ (Gi ∩ Fi ) = (X \ Gi ) ∪ (Gi \ Fi ) i≤n

i≤n

=

[ ¡\

I⊆n+1 i∈I

(Gi \ Fi ) ∩

\

(X \ Gi )

¢

i∈(n+1)\I

belongs to A. So A is an algebra of sets and must be equal to A. Q Q S (ii) B(X) is the T smallest family E ⊇ A such that n∈N En ∈ E for every non-decreasing sequence hEn in∈N in E and n∈N En ∈ E for every non-increasing sequence hEn in∈N in E. (136G.) 4A3D Hereditarily Lindel¨ of spaces (a) Suppose that X is a hereditarily Lindelöf space and U is a subbase for the topology of X. Then B(X) is the σ-algebra of subsets of X generated by U. P P Write Σ for the σ-algebra generated by U . Of course Σ ⊆ B(X) just because every member of U is open. In the other direction, set V = {X} ∪ {U0 ∩ U1 ∩ . . . ∩ Un : U0 , . . . , Un ∈ U}; then V ⊆ Σ and V isSa base for the topology of X (4A2B(a-i)). If G ⊆ X is open, set V1 = {V : V ∈ V, V ⊆S G}; then G = V1 . Because X is hereditarily Lindelöf, there is a countable set V0 ⊆ V1 such that G = V0 (4A2H(c-i)), so that G ∈ Σ. Thus every open set belongs to Σ and B(X) ⊆ Σ. Q Q (b) Let X be a set, Σ a σ-algebra of subsets of X, Y a hereditarily Lindelöf space, U a subbase for the topology of Y , and f : X → Y a function. If f −1 [U ] ∈ Σ for every U ∈ U, then f is (Σ, B(Y ))-measurable. P P {F : F ⊆ Y , f −1 [F ] ∈ Σ} is a σ-algebra of subsets of Y including U, so contains every open set, by (a). Q Q (c) Let hXi ii∈I be a family of topological spaces with product X. Suppose that X is hereditarily Lindelöf. N N N P By 4A3Cf, B(X) ⊇ c i∈I B(Xi ). On the other hand, c i∈I B(Xi ) is a (i) B(X) = c i∈I B(Xi ). P σ-algebra including U = {πi−1 [G] : i ∈ I, G ⊆ Xi is open}, where πi (x) = x(i) for i ∈ I and x ∈ X; since U is a subbase for the topology of X, (a) tells us that N c B(Xi ) includes B(X). Q Q i∈I (ii) If Y is another topological space, then a function f : Y → X is Borel measurable iff πi f : Y → Xi is Borel measurable for every i ∈ I, where πi : X → Xi is the canonical map. (Use 4A3Bc.) 4A3E Applications Recall that any topological space with a countable network (in particular, any second-countable space, any separable metrizable space) is hereditarily Lindelöf (4A2Nb, 4A2Oc, 4A2P(aiii)); and so is any ccc totally ordered space (4A2Rn). So 4A3Da-4A3Db will be applicable to these. As for product spaces, the product of a countable family of spaces with countable networks again has a countable network (4A2Ne), so for such spaces we shall be able to use 4A3Dc. For instance, B({0, 1}N ) is the σ-algebra of subsets of {0, 1}N generated by the sets {x : x(n) = 1} for n ∈ N. 4A3F Spaces with countable networks (a) Let X be a topological space with a countable network. Then #(B(X)) ≤ c. P P Let E be a countable network for the topology of X and Σ the σ-algebra of S subsets of X generated by E. Then #(Σ) ≤ c (4A1O). If G ⊆ X is open, there is a subset E 0 of E such that E 0 = G; but E 0 is necessarily countable, so G ∈ Σ. It follows that B(X) ⊆ Σ and #(B(X)) ≤ c. Q Q

850

Appendix

4A3Fb

(b) #(B(N N )) = c. P P N N has a countable network (4A2Ub), so #(B(N N )) ≤ c. On the other hand, N B(N ) contains all singletons, so #(B(N N )) ≥ #(N N ) ≥ #({0, 1}N ) = c. Q Q 4A3G Second-countable spaces Suppose that X is a second-countable space and Y is any topological b b space. Then B(X × Y ) = B(X)⊗B(Y ). P P By 4A3Cf, B(X × Y ) ⊇ B(X)⊗B(Y ). On the other hand, let U be a countable base for the topology of X. If W ⊆ X × Y is open, set S VU = {H : H ⊆ Y is open, U × H ⊆ W } S b b for U ∈ U . Then W = U ∈U U × VU belongs to B(X)⊗B(Y ). As W is arbitrary, B(X × Y ) ⊆ B(X)⊗B(Y ). Q Q 4A3H Borel sets in Polish spaces: Proposition Let (X, T) be a Polish space and E ⊆ X a Borel set. Then there is a Polish topology S on X, including T, for which E is open. proof Let E be the union of all the Polish topologies on X including T. Of course X ∈ E. If E ∈ E then X \ E ∈ E. P P There is a Polish topology S ⊇ T such that E ∈ S. Now both E and X \ E are Polish in the subspace topologies SE , SX\E induced by S (4A2Qd), so the disjoint union topology S0 of SE and SX\E is also Polish (4A2Qe). Now S0 ⊇ S ⊇ T and X \ E ∈ S0 , so X \ E ∈ E. Q Q Moreover, the union of any sequence hEn in∈N in E belongs to E. P P For each n ∈ N let Sn ⊇ T be a Polish topology containing S En . If m, n ∈ N then Sm ∩ Sn includes T, so is Hausdorff. By 4A2Qf, the topology S generated by n∈N Sn is S S Q Polish. Of course S ⊇ T, and n∈N En ∈ S, so n∈N En ∈ E. Q Thus E is a σ-algebra. Since it surely includes T, it includes B(X, T), as claimed. 4A3I Corollary If (X, T) is a Polish space and hEn in∈N is a sequence of Borel subsets of X, then there is a Polish topology T0 on X, including T, for which every En is open-and-closed. proof For each n ∈ N we can find a Polish topology Tn ⊇S T containing Ek (if n = 2k is even) or X \ Ek (if n = 2k + 1 is odd); now the topology T0 generated by n∈N Tn is Polish, by 4A2Qf, and every En is open-and-closed for T0 . 4A3J Borel sets in ω1 : Proposition A set E ⊆ ω1 is a Borel set iff either E or its complement includes a closed cofinal set. proof (a) Let Σ be the family of all those sets E ⊆ ω1 such that either E or ω1 \ E includes a closed cofinal set. Of course Σ is closed under complements. Because the intersection of a sequence of closed cofinal sets is a closed cofinal set (4A1Bc), the union of any sequence in Σ belongs to Σ; so Σ is a σ-algebra. If E is closed, then either it is cofinal with ω1 , and is a closed cofinal set, or there is a ξ < ω1 such that E ⊆ ξ, in which case ω1 \ E includes the closed cofinal set ω1 \ ξ; in either case, E ∈ Σ. Thus every open set belongs to Σ, and Σ includes B(ω1 ). (b) Now suppose that E ⊆ ω1 is such that there is a closed cofinal set F ⊆ ω1 \ E. For each ξ < ω1 let fξ : ξ → N be an injective function. Define g : E → N by setting g(η) = fα(η) (η), where α(η) = min(F \ η) S for η ∈ E. Set An = g −1 [{n}] for n ∈ N, so that E = n∈N An . If ξ ∈ An \ An and ξ 0 < ξ, there must be η, η 0 ∈ An such that ξ 0 ≤ η < η 0 < ξ. But now, because fα(η0 ) is injective, while g(η) = g(η 0 ) = n, α(η) < α(η 0 ), so α(η) ∈ F ∩ ]ξ 0 , ξ[. As ξ 0 is arbitrary, ξ ∈ F =SF . This shows that An ⊆ An ∪ F and An = An \ F is a Borel set. This is true for every n ∈ N, so E = n∈N An is a Borel set. (c) If E ⊆ ω1 includes a closed cofinal set, then (b) tells us that ω1 \ E and E belong to Σ. Thus Σ ⊆ B(ω1 ) and Σ = B(ω1 ), as claimed. 4A3K Baire sets When we come to study measures in terms of the integrals of continuous functions (§436), we find that it is sometimes inconvenient or even impossible to apply them to arbitrary Borel sets, and we need to use a smaller σ-algebra, as follows.

4A3N


851

(a) Definition Let X be a topological space. The Baire σ-algebra Ba(X) of X is the σ-algebra generated by the zero sets. Members of Ba(X) are called Baire sets. (Warning! Do not confuse ‘Baire sets’ in this sense with ‘sets with the Baire property’ in the sense of 4A3Q, nor with ‘sets which are Baire spaces in their subspace topologies’.) (b) For any topological space X, Ba(X) ⊆ B(X) (because every zero set is closed, therefore Borel). If T is perfectly normal – for instance, if it is metrizable (4A2Lc), or is regular and hereditarily Lindelöf (4A2H(c-iii)) – then Ba(X) = B(X) (because every closed set is a zero set, by 4A2Fi, so every open set belongs to Ba(X)). (c) Let X and Y be topological spaces, with Baire σ-algebras Ba(X), Ba(Y ) respectively. If f : X → Y is continuous, it is (Ba(X), Ba(Y ))-measurable. P P Let T be the σ-algebra {F : F ⊆ Y, f −1 [F ] ∈ Ba(X)}. If g : Y → R is continuous, then gf : X → R is continuous, so f −1 [{y : g(y) = 0}] = {x : gf (x) = 0} ∈ Ba(X), and {y : g(y) = 0} ∈ T. Thus every zero set belongs to T, and T ⊇ Ba(Y ). Q Q (d) In particular, if X is a subspace of Y , then E∩X ∈ Ba(X) whenever E ∈ Ba(Y ). More fundamentally, F ∩ X is a zero set in X for every zero set F ⊆ Y , just because g¹ X is continuous for any continuous g : Y → R. (e) If X is a topological space and Y is a separable metrizable space, a function f : X → Y is Baire measurable if f −1 [H] ∈ Ba(X) for every open H ⊆ Y . Observe that f is Baire measurable in this sense iff it is (Ba(X), B(Y ))-measurable iff it is (Ba(X), Ba(Y ))-measurable. 4A3L Lemma Let (X, T) be a topological space. Then Ba(X) is just the smallest σ-algebra of subsets of X with respect to which every continuous real-valued function on X is measurable. proof (a) Let f : X → R be a continuous function and α ∈ R. Set g(x) = max(0, f (x) − α) for x ∈ X; then g is continuous, so {x : f (x) ≤ α} = {x : g(x) = 0} is a zero set and belongs to Ba(X). As α is arbitrary, f is Ba(X)-measurable. (b) On the other hand, if Σ is any σ-algebra of subsets of X such that every continuous real-valued function on X is Σ-measurable, and F ⊆ X is a zero set, then there is a continuous g such that F = g −1 [{0}], so that F ∈ Σ; as F is arbitrary, Σ ⊇ Ba(X). 4A3M Product spaces Let hXi ii∈I be a family of topological spaces with product X. N (a) Ba(X) ⊇ c i∈I Ba(Xi ). (Apply 4A3Bc to the identity map from X to itself; compare 4A3Cf.) (b) Suppose that X is ccc. Then every Baire subset of X is determined by coordinates in a countable set. P P By 254Mb, the family W of sets determined by coordinates in countable sets is a σ-algebra of subsets of X. By 4A2E(b-ii), every continuous real-valued function is W-measurable, so W contains every zero set and every Baire set. Q Q 4A3N Products of separable metrizable spaces: Proposition Let hXi ii∈I be a family of separable metrizable spaces, with product X. N N (a) Ba(X) = c i∈I Ba(Xi ) = c i∈I B(Xi ). (b) Ba(X) is the family of Borel subsets of X which are determined by coordinates in countable sets. (c) A set Z ⊆ X is a zero set iff it is closed and determined by coordinates in a countable set. (d) If Y is a dense subset of X, then the Baire σ-algebra Ba(Y ) of Y is just the subspace σ-algebra Ba(X)Y induced by Ba(X). (e) If Y is a set, T is a σ-algebra of subsets of Y , and f : Y → X is a function, then f is (T, Ba(X))measurable iff πi f : Y → Xi is (T, B(Xi ))-measurable for every i ∈ I, where πi (x) = x(i) for x ∈ X, i ∈ I.

852

Appendix

4A3N

proof (a) X is ccc (4A2E(a-iii)), so if f : X → R is continuous, thereQare a countable set J ⊆ I and a continuous function g : XJ → R such that f = g˜ πJ , where XJ = i∈J XJ and π ˜J : X → XJ is the canonical map (4A2E(b-ii)). Now XJ is separable and metrizable (4A2P(a-v)), therefore hereditarily N N N Lindelöf (4A2P(a-iii)), so B(XJ ) = c i∈J B(Xi ), by 4A3D(c-i). By 4A3Bc, π ˜J is ( c i∈I B(Xi ), c i∈J B(Xj ))N N measurable, so f is c i∈J B(Xj )-measurable. As f is arbitrary, Ba(X) ⊆ c i∈I B(Xi ) (4A3L). Also B(Xi ) = N N Ba(Xi ) for every i (4A3Kb), so c i∈I B(Xi ) = c i∈I Ba(Xi ). With 4A3Ma, this proves the result. (b)(i) If W ∈ Ba(X) it is certainly a Borel set (4A3Kb), and by 4A3Mb it is determined by coordinates in a countable set. (ii) Q If W is a Borel subset Q of X determined by coordinates in a countable subset J of X, then write XJ = i∈J Xi and XI\J = i∈I\J Xi ; let π ˜J : X → XJ be the canonical map. We can identify W with W 0 × XI\J , where W 0 is some subset of XJ . Now if z ∈ XI\J , W 0 = {w : w ∈ XJ , (w, z) ∈ W } is a Borel subset of XJ , because w 7→ (w, z) : XJ → X is continuous. (I am passing over the trivial case X = ∅.) Since XJ is metrizable (4A2P(a-v)), W 0 ∈ Ba(XJ ) (4A3Kb) and W = π ˜J−1 [W 0 ] is a Baire set (4A3Kc). (c)(i) If Z is a zero set, it is surely closed; and it is determined by coordinates in a countable set by (b) above, or directly from 4A2E(b-ii). (ii) If Z is closed and determined by coordinates in a countable set J, then (in the language of (b) above) it can be identified with Z 0 × XI\J for some Z 0 ⊆ XJ . As in the proof of (b), Z 0 is closed (at least, if XI\J 6= ∅), so is a zero set (4A2Lc), and Z = π ˜J−1 [Z 0 ] is a zero set (4A2C(b-iv)). (d)(i) Ba(Y ) ⊇ Ba(X)Y by 4A3Kd. (ii) Let f : Y → R be any continuous function. For each n ∈ N, there is an open set Gn ⊆ X such that Gn ∩ Y = {y : f (y) > 2−n }. Now Gn is determined by coordinates in a countable set (4A2E(b-i)), so is a zero set, by (c)Shere. Because Y is dense in X, Gn = Gn ∩ Y does not meet {y : f (y) = 0}, and {y : f (y) > 0} = Y ∩ n∈N Gn belongs to Ba(X)Y . Thus Ba(X)Y contains every cozero subset of Y and includes Ba(Y ). (e) Put (a) and 4A3Bc together. 4A3O Compact spaces (a) Let (X, T) be a topological space, U a subbase for T, and A the algebra of subsets of X generated by U . If H ⊆ X is open and K ⊆ H is compact, there is an E ∈ A such that K ⊆ E ⊆ H. P P Set V = {X} ∪ {U0 ∩ U1 ∩ . . . ∩ Un : U0 , . . . , Un ∈ U}, so that V is a base for T and V ⊆ A. {U : U ∈ V, U ⊆ H} is an open cover of the compact set K, so there is a finite set U0 ⊆ V such S that E = U0 includes K and is included in H; now E ∈ A. Q Q (b) Let (X, T) be a compact space and U a subbase for T. Then every open-and-closed subset of X belongs to the algebra of subsets of X generated by U. (If F ⊆ X is open-and-closed, it is also compact; apply (a) here with K = H = F .) (c) Let (X, T) be a compact space and U a subbase for T. Then Ba(X) is included in the σ-algebra of subsets of X generated by U. P P Let Σ be the σ-algebra generated by U. If Z ⊆ X is a zero set, there is a sequence hHn in∈N of open sets with intersection Z (4A2C(b-vi)); now we can find a sequence hEn in∈N in T Σ such that Z ⊆ En ⊆ Hn for every n, by (a), so that Z = n∈N En ∈ Σ. This shows that every zero set belongs to Σ, so Σ must include Ba(X). Q Q N (d) If hXi ii∈I is a family of compact Hausdorff spaces with product X, then Ba(X) = c i∈I Ba(Xi ). P P N By 4A3Ma, Ba(X) ⊇ c i∈I Ba(Xi ). On the other hand, let Ui be the family of cozero sets in Xi for each i. Because Xi is completely regular (3A3Bb), Ui is a base for its topology (4A2Fc). Set Q W = { i∈I Ui : Ui ∈ Ui for every i ∈ I, {i : Ui 6= Xi } is finite}, N so that W ⊆ c i∈I Ba(Xi ) is a base for the topology of X. By (c) above, Ba(X) is the σ-algebra generated N by W, and Ba(X) ⊆ c Ba(Xi ). Q Q i∈I

4A3R


853

(e) In a compact Hausdorff zero-dimensional space the Baire σ-algebra is the σ-algebra generated by the open-and-closed sets. (Apply (c) with U the family of open-and-closed sets.) (f ) In particular, for any set I, Ba({0, 1}I ) is the σ-algebra generated by sets of the form {x : x(i) = 1} as i runs over I. 4A3P Proposition The Baire σ-algebra Ba(ω1 ) of ω1 is just the countable-cocountable algebra (211R). proof We see from 4A2S(b-iii) that every continuous function is measurable with respect to the countablecocountable algebra, so Ba(ω1 ) is included in the countable-cocountable algebra. On the other hand, [0, ξ] = {η : η ≤ ξ} = [0, ξ + 1[ = ω1 \ ]ξ, ω1 [ is an open-and-closed set (4A2S(a-i)), therefore a zero set, therefore belongs to Ba(ω1 ), for every ξ < ω1 . Now if ξ < ω1 , it is itself a countable set, so S [0, ξ[ = η<ξ [0, η] ∈ Ba(ω1 ), {ξ} = [0, ξ] \ [0, ξ[ ∈ Ba(ω1 ). It follows that every countable set belongs to Ba(ω1 ) and the countable-cocountable algebra is included in Ba(ω1 ). 4A3Q Baire property If X is a topological space, a subset X has the Baire property if it is expressible in the form G4M where G ⊆ X is open and M ⊆ X is meager; that is, A ⊆ X has the Baire property if b there is an open set G ⊆ X such that G4A is meager. (For A = G4M iff M = G4A.) The family B(X) of all such sets is the Baire property algebra of X. (See 4A3R.) (Warning! do not confuse the ‘Baire property algebra’ Bb with the ‘Baire σ-algebra’ Ba as defined in 4A3K.) 4A3R Proposition Let X be a topological space. (a) Let A ⊆ X be any set. (i) There is a unique regular open set H ⊆ X such that A \ H is meager and H ∩ G is empty whenever G ⊆ X is open and A ∩ G is meager. (ii) H is in itself a Baire space. (iii) If A has the Baire property, H4A is meager. (iv) If A = X, then H is the smallest comeager regular open set in X. b (b) B(X) is a σ-algebra of subsets of X including B(X). b If X has a countable (c) Write M for the family of meager subsets of X, so that M is a σ-ideal of B. b network, then the quotient algebra B/M has a countable order-dense set (definition: 313J). proof (a) (See Kechris 95, 8.29.) S (i) S Set G = {G : G ⊆ X is open, A ∩ G is meager}. Let G0 ⊆ G be a maximal disjoint set, and G0 = G0 . Then A ∩ G0 is meager.SP P For each G ∈ G0 , let hFGn in∈N be a sequence of nowhere dense closed sets covering A ∩ G. Set An = G∈G0 G ∩ FGn . If U ⊆ X is any non-empty open set, either U ∩ An is empty or there is a G ∈ G0 such that U ∩ G 6= ∅, in which case U ∩ G \ FGn is a non-empty open subset of S U not meeting An . Thus An is nowhere dense. This is true for every n, so A ∩ G0 ⊆ n∈N An is meager. Q Q Set H = int(X \ G0 ), so that H is a regular open set (definition: 314O). A \ H ⊆ (G0 \ G0 ) ∪ (A ∩ G0 ) is meager. If G ∈ G then G ∩ H = G \ G0 belongs to G and is disjoint from every member of G0 ; by the maximality of G0 , it is empty. If H 0 6= H is another regular open set, either H \ H 0 is not empty and A \ H 0 ⊇ A ∩ H \ H 0 is non-meager, or H 0 \ H is not empty and meets A in a meager set; so H is the only regular open set with the declared properties. 0 (ii) If now hHn in∈N is a sequence of open subsets of H which T are dense in H, and H ⊆ H is any 0 0 non-empty open set, H \ Hn is nowhere dense for every T n, so H \ n∈N Hn is meager.TOn the other hand, A ∩ H 0 is non-meager so H 0 also is, and H 0 must meet n∈N Hn . As H 0 is arbitrary, n∈N Hn is dense in H; as hHn in∈N is arbitrary, H is a Baire space in its subspace topology.

854

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(iii) If A has the Baire property, there is an open set G such that A4G is meager. In this case, A ∩ H \ G must be meager, so H ⊆ G and H \ A ⊆ (G \ A) ∪ (G \ G) is meager and H4A = (A \ H) ∪ (H \ A) is meager. (iv) If A = X, then H is comeager and is disjoint from every meager open set. So any comeager regular open set has complement disjoint from H and includes H; H is the smallest such set. ˇ (b) (See Cech 66, §22C; Kuratowski 66, §11.III; Kechris 95, 8.22.) Of course X = X4∅ belongs to b b B(X). If E ∈ B(X), let G ⊆ X be an open set such that E4G is meager. Then (X \ G)4(X \ E) ⊆ (G \ G) ∪ (G4E) b b is meager, so X \ E ∈ B(X). If hEn in∈N is a sequence in B(X), then for each n ∈ N we can find an open set Gn such that Gn 4En is meager, and now S S S ( n∈N Gn )4( n∈N En ) ⊆ n∈N Gn 4En S b is meager, so n∈N En ∈ B(X). b This shows that B(X) is a σ-algebra of subsets of X. Since it contains every open set, it must include the Borel σ-algebra. •

(c) Suppose that X has a countable network A. For A ∈ A, set dA = A , the equivalence class of b b A ∈ Bb in B/M. If b ∈ B/M is non-zero, take E ∈ Bb such that E • = b, and an open set G ⊆ X suchSthat E4G ∈ M. Then G is not meager. Set A1 = {A : A ∈ A, A ⊆ G}; then A1 is countable and G = A1 , so there is a non-meager A ∈ A1 . In this case A is not meager, so dA 6= 0, while A \ G ⊆ G \ G is nowhere b dense, so dA ⊆ b. As b is arbitrary, {dA : A ∈ A} is order-dense in B/M, and is countable because A is. *4A3S The following result will be useful in §424 and in Volume 5. Lemma Let X and Y be sets, Σ a σ-algebra of subsets of X, T a σ-algebra of subsets of Y and J a σ-ideal of T. Suppose that the quotient Boolean algebra T/J has a countable order-dense set. b and A ⊆ Y . (a) {x : x ∈ X, W [{x}] ∩ A ∈ J } belongs to Σ for any W ∈ Σ⊗T b (b) For every W ∈ Σ⊗T there S are sequences hEn in∈N in Σ, hVn in∈N in T such that (W 4W1 )[{x}] ∈ J for every x ∈ X, where W1 = n∈N En × Vn . proof (Compare Kechris 95, 16.1.) (a) Let D ⊆ T/J be a countable order-dense set. Let V ⊆ T be a countable set such that D = {V • : V ∈ V}. Let Λ be the family of subsets W of X × Y such that W [{x}] ∈ T for every x ∈ X, {x : x ∈ X, W [{x}] ∩ A ∈ J } ∈ Σ for every A ⊆ Y . (i) Of course ∅ ∈ Λ, because if W = ∅ then W [{x}] = ∅ for every x ∈ X, and {x : W [{x}]∩A ∈ J } = X for every A ⊆ Y . (ii) Suppose that W ∈ Λ, and set W 0 = (X × Y ) \ W . Then W 0 [{x}] = Y \ W [{x}] ∈ T for every x ∈ X. Now suppose that A ⊆ Y . Set V ∗ = {V : V ∈ V, A ∩ V ∈ / J }, E = {x : W 0 [{x}] ∩ A ∈ J }, E 0 = {x : V ∩ W [{x}] ∈ / J for every V ∈ V ∗ }. Then E = E 0 . P P (α) If x ∈ E and V ∈ V ∗ , then A∩W 0 [{x}] = A\W [{x}] belongs to J , so V ∩A\W [{x}] ∈ J and V ∩ W [{x}] ⊇ V ∩ A ∩ W [{x}] is not in J . As V is arbitrary, x ∈ E 0 ; as x is arbitrary, E ⊆ E 0 . (β) If x ∈ / E, then W 0 [{x}] ∩ A ∈ / J . Set V1 = {V : V ∈ V, V \ W 0 [{x}] ∈ J }, D1 = {V • : V ∈ V1 }. Then 0 D1 = {d : d ∈ D, d ⊆ W [{x}]• }, so W 0 [{x}]• = sup D1 in T/J (313K). S Because J is a σ-ideal, the map F 7→ F • : T → T/J is sequentially order-continuous (313Qb), and ( V1 )• = sup D1 (313Lc), that is,

4A3U


855

S W 0 [{x}]4 V1 ∈ J . There must therefore be a V ∈ V1 such that V ∩ A ∈ / J , that is, V ∈ V ∗ . At the same 0 time, V ∩ W [{x}] = V \ W [{x}] belongs to J , so V witnesses that x ∈ / E 0 . As x is arbitrary, E 0 ⊆ E. Q Q Since W ∈ Λ, S E = E 0 = X \ V ∈V ∗ {x : V ∩ W [{x}] ∈ J } belongs to T. As A is arbitrary, W 0 ∈ Λ. Thus the complement of any member of Λ belongs to Λ. (iii) If hWn in∈N is any sequence in Λ with union W , then S W [{x}] = n∈N Wn [{x}] ∈ T for every x, while {x : W [{x}] ∩ A ∈ J } =

T

n∈N {x

: Wn [{x}] ∩ A ∈ J } ∈ Σ

for every A ⊆ Y . So W ∈ Λ. (iv) What this shows is that Λ is a σ-algebra of subsets of X × Y . But if W = E × F , where E ∈ Σ and F ∈ T, then W [{x}] ∈ {∅, F } ⊆ T for every x ∈ X, {x : W [{x}] ∩ A ∈ J } ∈ {X \ E, X} ⊆ Σ b as claimed. for every A ⊆ Y ; so W ∈ Λ. Accordingly Λ must include Σ⊗T, (b) Let hVn in∈N be a sequence running over V ∪ {∅}. For each n ∈ N, set En = {x : Vn \ W [{x}] ∈ J };

S by (a), En ∈ Σ. Set W1 = n∈N En × Vn . Take any x ∈ X. Then W1 [{x}] = n∈I Vn where I = {n : x ∈ En }. Since Vn \ W [{x}] ∈ J for every n ∈ I, W1 [{x}] \ W [{x}] ∈ J . ?? If W [{x}] \ W1 [{x}] ∈ / J , there is a d ∈ D such that 0 6= d ⊆ W [{x}]• \ W1 [{x}]• . Let n ∈ N be such that Vn• = d; then Vn \ W [{x}] ∈ J , so n ∈ I and Vn ⊆ W1 [{x}] and d ⊆ W1 [{x}]• , which is impossible. X X As x is arbitrary, W1 has the required properties. S

4A3T Cylindrical σ-algebras I end the section with a note on a particular type of Baire σ-algebra. Definition Let X be a linear topological space. Then the cylindrical σ-algebra of X is the smallest σ-algebra Σ of subsets of X such that every continuous linear functional on X is Σ-measurable. 4A3U Proposition Let X be a linear topological space and Ts = Ts (X, X ∗ ) its weak topology. Then the cylindrical σ-algebra of X is just the Baire σ-algebra of (X, Ts ). proof (a) Let hfi ii∈I be a Hamel basis for X ∗ (4A4Ab). For x ∈ X, set T x = hfi (x)ii∈I ; then T : X → RI is a linear operator. Now Y = T [X] is dense in RI . P P Y is a linear subspace of RI , so its closure Y also I ∗ is (2A5Ec). If φ ∈ (R ) is such Pthat φ(T x) = 0 for every x ∈ X, there are a finitePset J ⊆ I and a family hαi ii∈J ∈ RJ such that φ(y) = i∈J αi y(i) for every y ∈ RI (4A4Be). In this case, i∈J αi fi (x) = φ(T x) = 0 for every x ∈ X; but hfi ii∈I is linearly independent, so αi = 0 for every i ∈ J and φ = 0. By 4A4Eb, Y must be the whole of RI . Q Q (b)(i) Set T0s = {T −1 [H] : H ⊆ Y is open}. Then T0s = Ts . P P Because every fi is continuous, T is continuous, so T0s ⊆ Ts . On the other hand, any f ∈ X ∗ is a linear combination of the fi , so is T0s -continuous, and Ts ⊆ T0s . Q Q (ii) If g : X → R is Ts -continuous, there is a continuous g1 : Y → R such that g = g1 T . P P If x, x0 ∈ X 0 are such that T x = T x , then every Ts -open set containing one must contain the other, so g(x) = g(x0 ). This means that there is a function g1 : Y → R such that g = g1 T . Next, if U ⊆ R is open, T −1 [g1−1 [U ]] = g −1 [U ] belongs to Ts = T0s , so g1−1 [U ] must be open in Y ; as U is arbitrary, g1 is continuous. Q Q (c) Now let Σ be the cylindrical σ-algebra of X. Then every fi : X → R is Σ-measurable, so T : X → RI is (Σ, Ba(RI ))-measurable, by 4A3Ne. Because Y is dense in RI , Ba(Y ) is the subspace σ-algebra induced by Ba(RI ) (4A3Nd). So T : X → Y is (Σ, Ba(Y ))-measurable. Now if g : X → R is a continuous function,

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there is a continuous function g1 : Y → R such that g = g1 T ; g1 is Ba(Y )-measurable, so g is Σ-measurable. As this is true for every Ts -continuous g : X → R, Ba(X) ⊆ Σ. (d) On the other hand, Σ ⊆ Ba(X) just because every member of X ∗ is Ts -continuous. So Σ = Ba(X), as claimed. 4A3V Proposition Let (X, T) be a separable metrizable locally convex linear topological space, and Ts = Ts (X, X ∗ ) its weak topology. Then the cylindrical σ-algebra of X is also both the Baire σ-algebra and the Borel σ-algebra for both T and Ts . proof (a) For any linear topological space, we have Σ ⊆ Ba(X, Ts ) ⊆ Ba(X, T) ⊆ B(X, T),

Ba(X, Ts ) ⊆ B(X, Ts ) ⊆ B(X, T),

writing Σ for the cylindrical σ-algebra and Ba(X, Ts ), Ba(X, T), B(X, Ts ) and B(X, T) for the Baire and Borel σ-algebras of the two topologies. So all I have to do is to show that B(X, T) ⊆ Σ. Let hτn in∈N be a sequence of seminorms defining the topology of X (4A4Cf), and D ⊆ X a countable dense set; for n ∈ N, δ > 0 and x ∈ X, set Un (x, δ) = {y : τi (y − x) < δ for every i ≤ n}. Then every Un (x, δ) is a convex open set. Set U = {Un (z, 2−m ) : z ∈ D, m, n ∈ N}. P Set UG = {U : U ∈ U , U ∩ G = ∅}. For each U ∈ UG , (b) If G ⊆ X is any convex open set, G ∈ Σ. P there are fU ∈ X ∗ , αU ∈ R such that fU (x) < αU for every x ∈ G and fU (x) > αU for every x ∈ U (4A4Db). So F = {x : fU (x) ≤ αU for every U ∈ UG } belongs to Σ. Of course F ⊇ G. On the other hand, if x ∈ / G, there are m, n ∈ N such that G ∩ Un (x, 2−m ) = ∅; if we take z ∈ D ∩ Un (x, 2−m−1 ), U = Un (z, 2−m−1 ) then Q x ∈ U ∈ UG , so fU (x) > αU and x ∈ / F . Thus G = F belongs to Σ. Q (c) In particular, V = {U : U ∈ U } is included in Σ. But V is a countable network for T. So every member of T is a union of countably many members of Σ and belongs to Σ. It follows at once that B(X, T) ⊆ Σ, as required. 4A3X Basic exercises (a) (i) Let X be a regular space with a countable network and Y any topological b space. Show that B(X × Y ) = B(X)⊗B(Y ). (Hint: 4A2Ng.) (ii) Set X = [0, 1] with the topology generated by {X \ {x} : x ∈ X} and Y = [0, 1] with the discrete topology. Show that X has a countable network and that {(x, x) : x ∈ [0, 1]} is a closed subset of X × Y not belonging to B(X) × B(Y ). (b) Let X be a topological space, and E ⊆ X. Show that the following are equiveridical: (i) E ∈ Ba(X); (ii) there are a continuous function f : X → [0, 1]N and F ∈ B([0, 1]N ) such that E = f −1 [F ]; (iii) there are a metric space Z, a continuous function f : X → Z and an F ∈ B(Z) such that E = f −1 [F ]. > (c) Let X be a topological space, and K ⊆ X a compact set such that K ∈ Ba(X). Show that K is a zero set. (Hint: 4A3Xb.) (d) Let X be a compact Hausdorff space such that Ba(X) = B(X). Show that X is perfectly normal. (e) Let S be the topology on R generated by the usual topology and {{x} : x ∈ R \ Q}. Show that S is completely regular and Hausdorff and that Q is a closed Baire set which is not a zero set. N c

(f ) Let hXi ii∈I be a family of spaces with countable networks, and X their product. Show that Ba(X) = i∈I Ba(Xi ).

> (g) (i) Let I be an uncountable set with its discrete topology, and X the one-point compactification of I. Show that Ba(I) is not the subspace σ-algebra generated by Ba(X). (ii) Show that ω1 has a subset I such that Ba(I) is not the subspace σ-algebra induced by Ba(ω1 ). ˇ 4A3Y Further exercises (a) Let X be a Cech-complete space and hEn in∈N a sequence of Borel sets in ˇ X. Show that there is a Cech-complete topology on X, finer than the given topology, with the same weight, and containing every En .

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(b) Let hXi ii∈I be a family of regular spaces with countable networks, with product X. Show that (i) Ba(X) is the family of Borel subsets of X which are determined by coordinates in countable sets; (ii) a set Z ⊆ X is a zero set iff it is closed and determined by coordinates in a countable set; (iii) if Y ⊆ X is dense, then Ba(Y ) is just the subspace σ-algebra Ba(X)Y induced by Ba(X). 4A3 Notes and comments Much of this section consists of easy technicalities. It is however not always easy to guess at the exact results obtainable by these methods. It is important to notice that Baire σalgebras on subspaces can give difficulties which do not arise with Borel σ-algebras (4A3Xg, 4A3Ca). I have expressed 4A3D in terms of ‘hereditarily Lindelöf’ spaces. Of course the separable metrizable spaces form by far the most important class of these, but there are others (the split interval, for instance) which are of great interest in measure theory. Similarly, there are important products of non-metrizable spaces which are ccc (e.g., 417Xu(vii)), so that 4A3Mb has something to say. 4A3H-4A3I are a most useful tool in studying Borel subsets of Polish spaces, especially in conjunction with the First Separation Theorem (422I); see 424G and 424H. I include 4A3Ya and 4A3Yb to show that some more of the arguments here can be adapted to non-separable or non-metrizable spaces. You will note my caution in the definition of ‘Baire measurable’ function (4A3Ke). This is supposed to cover the case of functions taking values in [−∞, ∞] without taking a position on functions between general topological spaces. 4A4 Locally convex spaces As in §3A5, all the ideas, and nearly all the results as stated below, are applicable to complex linear spaces; but for the purposes of this volume the real case will almost always be sufficient, and for definiteness you may take it that the scalar field is R, except in 4A4J-4A4K. (Complex Hilbert spaces arise naturally in §445.) 4A4A Linear spaces (a) If U is a linear space, a Hamel basis for U isPa maximal linearly independent family hui ii∈I in U , so that every member of U is uniquely expressible as i∈J αi ui for some finite J ⊆ I, hαi ii∈J ∈ (R \ {0})J . ¨ the 69, §7.3.) (b) Every linear space has a Hamel basis. (Schaefer 71, p. 10; Ko (c) If U is a linear space, I write U 0 for the algebraic dual of U , the linear space of all linear functionals from U to R. 4A4B Linear topological spaces (see §2A5) (a) If U is a linear topological space, and V is a linear subspace of U , then V , with the linear structure and topology induced by those of U , is again a linear ¨ the 69, §15.2.) topological space. (Bourbaki 87, I.1.3; Schaefer 71, §I.2; Ko Q (b) If hUi ii∈I is any family of linear topological spaces, then U = i∈I Ui , with the product linear space structure and topology, is again a linear topological space. (Bourbaki 87, I.1.3; Schaefer 71, §I.2; ¨ the 69, §15.4.) In particular, RX , with its usual linear and topological structures, is a linear topological Ko space, for any set X. (c) If U and V are linear topological spaces, the set of continuous linear operators from U to V is a linear subspace of the space L(U ; V ) of all linear operators from U to V . If U , V and W are linear topological spaces, and T : U → V and S : V → W are continuous linear operators, then ST : U → W is a continuous linear operator. (d) If U is a linear topological space, I will write U ∗ for the dual of U , the space of all continuous linear functionals from U to R (compare 2A4H). U ∗ is a linear subspace of U 0 as defined in 4A4Ac. The weak topology on U , Ts (U, U ∗ ), is that defined by the method of 2A5B from the seminorms u 7→ |f (u)| as f runs through U ∗ (compare 2A5Ia). The weak* topology on U ∗ , Ts (U ∗ , U ), is that defined by the method of 2A5B from the seminorms f 7→ |f (u)| as u runs through U (compare 2A5Ig). By 2A5B, both are linear space topologies. If U and V are linear topological spaces, T : U → V is a continuous linear operator, and g ∈ V ∗ , then gT ∈ U ∗ ((c) above).

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Q (e)PIf U = i∈I Ui is a product of linear topological spaces, then every element of U ∗ is of the form u 7→ i∈J fi (u(i)) where J ⊆ I is finite and fi ∈ Ui∗ for every i ∈ J. (Bourbaki 87, II.6.6; Schaefer 71, ¨ the 69, §22.5.) IV.4.3; Ko (f ) If U is a linear topological space, a set A ⊆ U is bounded if for every neighbourhood G of 0 there is an n ∈ N such that A ⊆ nG. The closure of a bounded set is bounded. (Schaefer 71, I.5.1; Rudin 91, 1.13.) (g) If U and V are linear topological spaces, T : U → V is a continuous linear operator and A ⊆ U is a ¨ the 69, §15.6; Rudin 91, 1.32.) bounded set, then T [A] ⊆ V is bounded. (Schaefer 71, I.5.4; Ko (h) Let U be a linear topological space. If G is a neighbourhood of 0 in U , then its polar set in U ∗ , G = {f : f ∈ U ∗ , f (x) ≤ 1 for every x ∈ G}, is a Ts (U ∗ , U )-compact subset of U ∗ (compare 3A5F). ¨ the 69, §20.9; Rudin 91, 3.15.) (Schaefer 71, III.4.3; Ko ◦

(i) Let U be a linear topological space. If D ⊆ U is non-empty and closed under addition and multiplication by rationals, D is a linear subspace of U . P P The linear span Pn V = { i=0 αi ui : u0 , . . . , un ∈ D, α0 , . . . , αn ∈ R} of

Pn

D={

i=0

αi ui : u0 , . . . , un ∈ D, α0 , . . . , αn ∈ Q}

is included in D, because addition and scalar multiplication are continuous; so D = V is a linear subspace. Q Q If A ⊆ U is separable, then the closed linear subspace generated by A is separable. P P let D0 ⊆ A be a countable dense subset; then Pn D = { i=0 αi ui : u0 , . . . , un ∈ D0 , α0 , . . . , αn ∈ Q} is countable, and D is separable; but D is the closed linear subspace generated by A. Q Q (j) P u ∈ U , we say that PIf hui ii∈I is an indexed family in a Hausdorff linear topological space U and u = i∈I ui if for every neighbourhood G of u there is a finite set J ⊆ I such that i∈K ui ∈ G whenever K ⊆ I is finite and J ⊆ K (compare 226Ad). another Hausdorff linear topological space and T : UP→ V is a continuous linear operator, P If now V is P P Set u = i∈I ui . If H is an open set containing T u = T ( i i∈I ui ) if the right-hand-side is defined. P i∈I P T u, then T −1 [H] is an open set containing u, so there is a J ∈ [I]<ω (notation: 3A1J) such that i∈K ui ∈ P Q T −1 [H] and i∈K T ui ∈ H whenever J ⊆ K ∈ [I]<ω . Q (k) If U is a Hausdorff linear topological space, then any finite-dimensional linear subspace of U is closed. (Schaefer 71I.3.3; Taylor 64, 3.12-C; Rudin 91, 1.2.1.) 4A4C Locally convex spaces (a) A linear topological space U (2A5A) is locally convex if the convex open sets form a base for the topology. (b) A linear topological space is locally convex iff its topology can be defined by a family of seminorms ¨ the 69, §18.1.) (2A5B, 2A5D). (Bourbaki 87, II.4.1; Schaefer 71, §II.4; Ko (c) Let U be a linear space and τ a seminorm on U . Then Nτ = {u : τ (u) = 0} is a linear subspace of X. On the quotient space U/Nτ we have a norm defined by setting ku• k = τ (u) for every u ∈ U . (Bourbaki 87, II.1.3; Schaefer 71, II.5.4; Rudin 91, 1.43.) (d) Let U be a locally convex linear topological space, and T the family of continuous seminorms on U . For each τ ∈ T, write Nτ = {u : τ (u) = 0}, as in (c) above, and πτ for the canonical map from U to S Uτ = U/Nτ . Give each Uτ its norm, and set Gτ = {πτ−1 [H] : H ⊆ Uτ is open}. Then τ ∈T Gτ is a base for the topology of X closed under finite unions. (Schaefer 71, II.5.4.)

4A4Ef


859

(e) A linear subspace of a locally convex linear topological space is locally convex. (Bourbaki 87, II.4.3; ¨ the 69, §18.3.) The product of any family of locally convex linear topological spaces (4A4Bb) is locally Ko ¨ the 69, §18.3.) convex. (Bourbaki 87, II.4.3; Ko (f ) If U is a metrizable locally convex linear topological space, its topology can be defined by a sequence ¨ the 69, §18.2.) of seminorms. (Bourbaki 87, II.4.1; Ko (g) Let U be a linear space and V a linear subspace of the space U 0 of all linear functionals on U . Let Ts (V, U ) be the topology on V generated by the seminorms f 7→ |f (u)| as u runs over U (compare 4A4Bd), and let φ : V → R be a Ts (V, U )-continuous linear functional. Then there is a u ∈ U such that ¨ the 69, §20.2; Rudin 91, φ(f ) = f (u) for every f ∈ V . (Bourbaki 87, IV.1.1; Schaefer 71, IV.1.2; Ko 3.10; Dunford & Schwartz 57, II.3.9; Taylor 64, 3.81-A.) (h) Grothendieck’s theorem If U is a complete locally convex Hausdorff linear topological space, and φ is a linear functional on the dual U ∗ such that φ¹G◦ is Ts (U ∗ , U )-continuous for every neighbourhood G of 0 in U (defining G◦ as in 4A4Bh), then φ is of the form f 7→ f (u) for some u ∈ U . (Bourbaki 87, ¨ the 69, §21.9.) III.3.6; Schaefer 71, IV.6.2; Ko 4A4D Hahn-Banach theorem (a) Let U be a linear space and θ : U → [0, ∞[ a seminorm. (i) If V ⊆ U is a linear subspace and g : V → R is a linear functional such that |g(v)| ≤ θ(v) for every v ∈ V , then there is a linear functional f : U → R, extending g, such that |f (u)| ≤ θ(u) for every u ∈ U . (ii) If u0 ∈ U then there is a linear functional f : U → R such that f (u0 ) = θ(u0 ) and |f (u)| ≤ θ(u) for every u ∈ U . (Bourbaki 87, II.3.2; Rudin 91, 3.3; Dunford & Schwartz 57, II.3.11; Taylor 64, 3.7-C.) (b) Let U be a linear topological space and G, H two disjoint convex sets in U , of which one has nonempty interior. Then there are an f ∈ U ∗ and an α ∈ R such that f (u) ≤ α ≤ f (v) for every u ∈ G, v ∈ H. If G and H are both open, then f (u) < α < f (v) for every u ∈ G, v ∈ H. (Bourbaki 87, II.5.2; Schaefer ¨ the 69, §17.1.) 71, II.9.1; Ko 4A4E The Hahn-Banach theorem in locally convex spaces Let U be a locally convex linear topological space. (a) If V ⊆ U is a linear subspace, then every member of V ∗ extends to a member of U ∗ (compare 3A5Ab). ¨ the 69, §20.1; Rudin 91, 3.6; Taylor 64, 3.8-D.) (Bourbaki 87, II.4.1; Schaefer 71, II.4.2; Ko ∗ Consequently Ts (V, V ) is just the subspace topology on V induced by Ts (U, U ∗ ). (b) Let C ⊆ U be a non-empty closed convex set. If u ∈ U then u ∈ C iff f (u) ≤ supv∈C f (v) for every ¨ the 69, f ∈ U ∗ iff f (u) ≥ inf v∈C f (v) for every f ∈ U ∗ . (Bourbaki 87, II.5.3; Schaefer 71, II.9.2; Ko §20.7; Dunford & Schwartz 57, V.2.12.) If V ⊆ U is a closed linear subspace and u ∈ U \ V there is an f ∈ U ∗ such that f (u) 6= 0 and f (v) = 0 ¨ the 69, §20.1; Rudin 91, 3.5; Taylor 64, 3.8-E.) for every v ∈ V . (Bourbaki 87, II.5.3; Ko (c) If U is Hausdorff, U ∗ separates its points (compare 3A5Ae). (Bourbaki 87, II.4.1; Rudin 91, 3.4.) (d) If u ∈ U belongs to the Ts (U, U ∗ )-closure of a convex set C ⊆ U , it belongs to the closure of C ¨ the 69, §20.7; Rudin 91, 3.12.) In particular, if C is closed, (compare 3A5Ee). (Schaefer 71, II.9.2; Ko it is Ts (U, U ∗ )-closed. (Dunford & Schwartz 57, V.2.13.) (e) If C, C 0 ⊆ U are disjoint non-empty closed convex sets, of which one is compact, there is an f ∈ U ∗ such that supu∈C f (u) < inf u∈C 0 f (u). (Apply (b) to C − C 0 . See Bourbaki 87, II.5.3; Schaefer 71, ¨ the 69, §20.7; Rudin 91, 3.4; Dunford & Schwartz 57, V.3.13.) II.9.2; Ko (f ) Let V be a linear subspace of U 0 . Let K ⊆ U be a non-empty Ts (U, V )-compact convex set, and φ0 : V → R a linear functional such that φ0 (f ) ≤ supu∈K f (u) for every f ∈ V . Then there is a u0 ∈ K such that φ0 (f ) = f (u0 ) for every f ∈ V . P P Give U the topology Ts (U, V ) and V 0 the topology Ts (V 0 , V )

860

Appendix

4A4Ef

(4A4Cg). For u ∈ U , f ∈ V set u ˆ(f ) = f (u); then u 7→ u ˆ is a continuous linear operator from U to V 0 (use ˆ = {ˆ 2A3H), so K u : u ∈ K} is a compact convex subset of V 0 . ˆ By (b), there is a continuous linear functional θ : V 0 → R such ?? Suppose, if possible, that φ0 ∈ / K. that θ (φ) > supu∈K θ (ˆ u). But there is an f ∈ V such that θ (φ) = φ(f ) for every φ ∈ V 0 (4A4Cg), so that φ0 (f ) > supu∈K f (u), contrary to hypothesis. X X So there is a u0 ∈ K such that φ0 = u ˆ0 , as claimed. Q Q (g) The Bipolar Theorem Let A ⊆ U 0 be a non-empty set. Set A◦ = {u : u ∈ U, f (u) ≤ 1 for every f ∈ A} (compare 4A4Bh). If g ∈ U 0 is such that g(u) ≤ 1 for every u ∈ A◦ , then g belongs to the Ts (U 0 , U )-closed convex hull of A ∪ {0}. P P Put 4A4Cg and (b) above together, as in (f). See Bourbaki ¨ the 69, 20.8. Q 87, II.6.3; Schaefer 71, IV.1.5; Ko Q 4A4F The Mackey topology Let U be a linear space and V a linear subspace of U 0 . The Mackey topology Tk (V, U ) on V is the topology of uniform convergence on convex Ts (U, V )-compact subsets of U . Every Tk (V, U )-continuous linear functional on V is of the form f 7→ f (u) for some u ∈ U (use 4A4Ef). So every Tk (V, U )-closed convex set is Ts (V, U )-closed, by 4A4Ed. (See Bourbaki 87, IV.1.1; Schaefer 71, ¨ the 69, 21.4.) IV.3.2; Ko 4A4G Extreme points (a) Let X be a real linear space, and C ⊆ X a convex set. An element of C is an extreme point of C if it is not expressible as a convex combination of two other members of C; equivalently, if it is not expressible as 12 (x + y) where x, y ∈ C are distinct. (b) The Kreˇın-Mil’man theorem Let U be a Hausdorff locally convex linear topological space and K ⊆ U a compact convex set. Then K is the closed convex hull of the set of its extreme points. (Bourbaki ¨ the 69, §25.1; Rudin 91, 3.22.) 87, II.7.1; Schaefer 71, II.10.4; Ko (c) Let U and V be Hausdorff locally convex linear topological spaces, T : U → V a continuous linear operator, K ⊆ X a compact convex set and v any extreme point of T [K] ⊆ V . Then there is an extreme point u0 of K such that T u0 = v. P P Set K1 = {u : u ∈ K, T u = v}. Then K1 is a compact convex set so has an extreme point u0 . ?? If u0 is not an extreme point of K, it is expressible as αu1 + (1 − α)u2 where u1 , u2 are distinct points of K and α ∈ ]0, 1[. So v = αT u1 + (1 − α)T u2 , and we must have T u1 = T v = T u2 , because v is an extreme point of T [K]; but this means that u1 , u2 ∈ K1 and u0 is not an extreme point of K1 . X X So u0 has the required properties. Q Q 4A4H Proposition Let I be a set, W a closed linear subspace of RI , U a linear topological space and V a Hausdorff linear topological space. Let K ⊆ U be a compact set and T : U × RI → V a continuous linear operator. Then T [K × W ] is closed. proof Take v0 ∈ T [K × W ]. Let J ⊆ I be a maximal set such that whenever L ⊆ J is finite and H ⊆ V is an open set containing v0 , there are a u ∈ K and an x ∈ W such that x(i) = 0 for every i ∈ L and T (u, x) ∈ H. Let F be the filter on U × RI generated by the closed set K × W , the sets {(u, x) : x(i) = 0} for i ∈ J, and the sets T −1 [H] for open sets H containing v0 . Then for any j ∈ I there is an F ∈ F such that {x(j) : (u, x) ∈ F } is bounded. P P?? Suppose, if possible, otherwise. Then, in particular, j ∈ / J. So there must be a finite set L ⊆ J and an open set H containing v0 such that x(j) 6= 0 whenever u ∈ K, x ∈ W , x(i) = 0 for every i ∈ L and T (u, x) ∈ H. By 2A5C, or otherwise, there is a neighbourhood G0 of 0 in V such that v0 + G0 − G0 ⊆ H and αv ∈ G0 whenever v ∈ G0 and |α| ≤ 1. Fix u∗ ∈ K, x∗ ∈ W such that x∗ (i) = 0 for every i ∈ L and T (u∗ , x∗ ) ∈ v0 + G0 . If x ∈ W and x(i) = 0 for every i ∈ L and x(j) = x∗ (j), then T (u∗ , x∗ − x) ∈ / v0 + G0 − G0 so T (0, x) ∈ / G0 . It follows that T (0, x) ∈ / G0 whenever x ∈ W and x(i) = 0 for every i ∈ L and |x(j)| ≥ |x∗ (j)|. Let G be a neighbourhood of 0 in V such that G + G − G − G ⊆ G0 . We are supposing that {x(j) : (u, x) ∈ F } is unbounded for every F ∈ F. So for every n ∈ N there are un ∈ K and xn ∈ W such that xn (i) = 0 for i ∈ L, T (un , xn ) ∈ v0 + G and |xn (j)| ≥ n. Let u ∈ K be a cluster point of hun in∈N . Then T (u, 0) is a cluster point of hT (un , 0)in∈N , so M = {n : T (un , 0) ∈ T (u, 0) + G} is infinite. For n ∈ M , T (0, xn ) = T (un , xn )−T (un , 0) ∈ v0 −T (u, 0)+G−G; so if m, n ∈ M , T (0, xm −xn ) ∈ G−G−(G−G) ⊆ G0 .

4A4Jb


861

But note now that (xm − xn )(i) = 0 for all m, n ∈ N and i ∈ L, and that because M is infinite there are certainly m, n ∈ M such that |xm (j) − xn (j)| ≥ |x∗ (j)|; which contradicts the last paragraph. X XQ Q Now let G be any ultrafilter on U × RI including F. Then for every i ∈ I there is a γi < ∞ such that G contains {(u, x) : |x(i)| ≤ γi }. It follows that G has a limit (ˆ u, x ˆ) in K × W . Now the image filter T [[G]] (2A1Ib) converges to T (ˆ u, x ˆ); since T [[F]] → v0 , and the topology of V is Hausdorff, v0 = T (ˆ u, x ˆ) ∈ T [K × W ]. As v0 is arbitrary, T [K × W ] is closed. 4A4I Normed spaces (a) Two norms k k, k k0 on a linear space U give rise to the same topology iff they are equivalent in the sense that, for some M ≥ 0, kxk ≤ M kxk0 ,

kxk0 ≤ M kxk

¨ the 69, §14.2; Taylor 64, 3.1-D; Jameson 74, 2.8.) for every x ∈ U . (Ko (b) If U and V are normed spaces, T : U → V is a linear operator and gT : X → R is continuous for every g ∈ V ∗ , then T is a bounded operator. (Jameson 74, 27.6.) (c) If U is any normed space, its dual U ∗ , under its usual norm (2A4H), is a Banach space. (Rudin 91, ¨ the 69, §14.5.) 4.1; Dunford & Schwartz 57, II.3.9; Ko (d) If U is a separable normed space, its dual U ∗ (regarded as a normed space, as in 2A4H) is isometrically isomorphic to a closed linear subspace of `∞ . P P Let hxn in∈N run over a dense subset of the unit ball of U (4A2P(a-iv)); define T : U ∗ → `∞ by setting (T f )(n) = f (xn ) for every n. T is a linear isometry between U ∗ and T [U ∗ ], which is closed because U ∗ is complete (4A4Ic, 3A4Fd). Q Q P (e) Let U be a Banach space. Suppose that hui ii∈I is a family in U such that γ = i∈I kui k < ∞. P P P P For J ∈ [I]<ω , set vJ = i∈J ui . For each n ∈ N, there (i) i∈I ui is defined in the P sense of 4A4Bj. −n <ω whenever Jn ⊆ K ∈ [I]<ω . Now is a Jn ∈ [I] such that γ − i∈K kui k ≤ 2 P kvJm − vJn k ≤ i∈Jm 4Jn kui k ≤ 2−m + 2−n for all m, n ∈ N, so hvJn in∈N is a Cauchy sequence and has a limit v say. If now n ∈ N and Jn ⊆ K ∈ [I]<ω , kv −

X i∈K

ui k = lim kvJm − m→∞

X

ui k ≤ lim sup m→∞

i∈K

X

kui k

i∈Jm 4K

≤ lim 2−m + 2−n = 2−n , m→∞

P

Q so v = i∈I ui . Q P P (ii) Now if hIj ij∈J is any partition of I, wj = i∈Ij ui is defined for every j, and j∈J wj is defined P and equal to i∈I ui . (f ) Let U be a normed space. For u ∈ U , define u ˆ ∈ U ∗∗ = (U ∗ )∗ by setting u ˆ(f ) = f (u) for every ∗ f ∈ U . Then {ˆ u : u ∈ U, kuk ≤ 1} is weak*-dense in {φ : φ ∈ U ∗∗ , kφk = 1}. (Apply 4A4Eg with A = {ˆ u : kuk ≤ 1}.) 4A4J Inner product spaces (a) Let U be an inner product space over R C (3A5L). An orthonormal basis in U is a family hei ii∈I in U such that (α) (ei |ej ) = 0 if i 6= j, 1 if i = j (β) the closed linear subspace of U generated by {ei : i ∈ I} is U itself. (b) If U , V are inner product spaces over R and T : U → V is an isometry such that T (0) = 0, then (T u|T v) = (u|v) for all u, v ∈ U and T is linear. P P (α) 1 2

1 2

(T u|T v) = (kT uk2 + kT vk2 − kT u − T vk2 ) = (kuk2 + kvk2 − ku − vk2 ) = (u|v). (β) For any u, v ∈ U ,

862

Appendix

4A4Jb

kT (u + v) − T u − T vk2 = kT (u + v)k2 + kT uk2 + kT vk2 − 2(T (u + v)|T u) − 2(T (u + v)|T v) + 2(T u|T v) 2

= ku + vk + kuk2 + kvk2 − 2(u + v|u) − 2(u + v|v) + 2(u|v) = 0. So T is additive. (γ) Consequently T (qu) = qT u for every u ∈ U and q ∈ Q; as T is continuous, it is linear. Q Q (c) If U is an inner product space over R C , a linear operator T : U → U is self-adjoint if (T u|v) = (u|T v) for all u, v ∈ U . (d) If U is a finite-dimensional inner product space over R, it is isomorphic to Euclidean space Rr , where r = dim U . (Taylor 64, 3.21-A.) In particular, any finite-dimensional inner product space is a Hilbert space. ⊥ (e) If U is an inner product space over R = {x : x ∈ C and V ⊆ U is a linear subspace of U , then V 2 2 2 U, (x|y) = 0 for every y ∈ V } is a linear subspace of U , and kx + yk = kxk + kyk for x ∈ V , y ∈ V ⊥ . If V is complete (in particular, if V is finite-dimensional), then U = V ⊕ V ⊥ . (Rudin 91, 12.4; Bourbaki 87, V.1.6; Taylor 64, 4.82-A.)

(f ) If U is an inner product space over R and v1 , v2 ∈ U are such that kv1 k = kv2 k = 1, there is a linear operator T : U → U such that T v1 = v2 and kT uk = kuk and kT u − uk ≤ kv1 − v2 kkuk for every u ∈ U . P P If v2 is a multiple of v1 , say v2 = αv1 , take T u = αu for every u. Otherwise, set w = v2 − (v2 |v1 )v1 and w1 =

1 w, kwk

so that v2 = cos θv1 + sin θw1 , where θ = arccos(v2 |v1 ). Let V be the two-dimensional linear

subspace of U generated by v1 and w1 , so that U = V ⊕ V ⊥ . Define a linear operator T : U → U by saying that T v1 = v2 , T w1 = cos θw1 − sin θv1 and T u = u for u ∈ V ⊥ . Then T acts on V as a simple rotation through an angle θ, so kT vk = 1 and kT v − vk = kv2 − v1 k whenever v ∈ V and kvk = 1; generally, if u ∈ U , then kT uk = kuk and kT u − uk = kT (P u) − P uk = kv2 − v1 kkP uk ≤ kv2 − v1 kkuk, where P is the orthogonal projection of U onto V . Q Q (g) Let U be an inner product space over R C , and hui ii∈I a countable family in U . Then there is a countable orthonormal family hvj ij∈J in U such that {vj : j ∈ J} and {ui : i ∈ I} span the same linear subspace of U . P P We can suppose that I ⊆ N; set ui = 0 for i ∈ N \ I. Define hvn in∈N inductively by setting P 1 0 vn = un − i
4A4K Hilbert spaces (a) If U is a real or complex Hilbert space, its unit ball is compact in the weak topology Ts (U, U ∗ ); any bounded set is relatively compact for Ts (U, U ∗ ). (Bourbaki 87, V.1.7; Dunford & Schwartz 57, IV.4.6.) (b) If U is a real or complex Hilbert space, any norm-bounded sequence in U has a weakly convergent subsequence. (462D; Dunford & Schwartz 57, IV.4.7.) (c) If U is a real or complex Hilbert space and hui ii∈I is any orthonormal family in U , then it can be extended to an orthonormal basis. (Dunford & Schwartz 57, IV.4.10; Taylor 64, 3.2-I.) In particular, U has an orthonormal basis. 4A4L Compact operators (see 3A5Ka) (a) Let U , V and W be Banach spaces. If T ∈ B(U ; V ) and S ∈ B(V ; W ) and either S or T is a compact operator, then ST is compact. (Dunford & Schwartz 57, VI.5.4; Jameson 74, 34.2.)

4A5Bc

Topological groups

863

(b) If U is a Banach space, T ∈ B(U ; U ) is a compact linear operator and γ 6= 0 then {u : T u = γu} is finite-dimensional. (Rudin 91, 4.18; Taylor 64, 5.5-C; Dunford & Schwartz 57, VII.4.5; Jameson 74, 34.8.) 4A4M Self-adjoint compact operators If U is a Hilbert space and T : U → U is a self-adjoint compact linear operator, then T [U ] is the closed linear span of {T v : v is an eigenvector of T }. (Taylor 64, 6.4-B.) 4A4N Max-flow Min-cut Theorem (Ford & Fulkerson 56) Let (V, E, γ) be a (finite) transportation network, that is, V is a finite set of ‘vertices’, E ⊆ {(v, v 0 ) : v, v 0 ∈ V, v 6= v 0 } is a set of (directed) ‘edges’, γ : E → [0, ∞[ is a function; we regard a member e = (v, v 0 ) of E as ‘starting’ at v and ‘ending’ at v 0 , and γ(e) is the ‘capacity’ of the edge e. Suppose that v0 , v1 ∈ V are distinct vertices such that no edge ends at v0 and no edge starts at v1 . Then we have a ‘flow’ φ : E → [0, ∞[ and a ‘cut’ X ⊆ E such that (i) for every v ∈ V \ {v0 , v1 }, P P e∈E,e starts at v φ(e) = e∈E,e ends at v φ(e), (ii) φ(e) ≤ γ(e) for every e ∈ E, (iii) P there is no path from v0 P to v1 using only edges inPE \ X, (iv) e∈E,e starts at v0 φ(e) = e∈E,e ends at v1 φ(e) = e∈X γ(e). ´ s 79, §III.1; Anderson 87, 12.3.1.) (Bolloba

4A5 Topological groups For Chapter 44 we need a variety of facts about topological groups. Most are essentially elementary, ´ sza ´ r 78 and Hewitt & Ross 63. In and all the non-trivial ideas are covered by at least one of Csa 4A5A-4A5C and 4A5I I give some simple definitions concerning groups and group actions. Topological groups, properly speaking, appear in 4A5D. Their simplest properties are in 4A5E-4A5G. I introduce ‘right’ and ‘bilateral’ uniformities in 4A5H; the latter are the more interesting (4A5M-4A5O), but the former are also important (4A5P). 4A5J-4A5L deal with quotient spaces, including spaces of cosets of non-normal subgroups. I conclude with notes on metrizable groups (4A5Q-4A5S). 4A5A Notation If X is a group, x0 ∈ X, and A, B ⊆ X I write x0 A = {x0 x : x ∈ A}, AB = {xy : x ∈ A, y ∈ B},

Ax0 = {xx0 : x ∈ A}, A−1 = {x−1 : x ∈ A}.

A is symmetric if A = A−1 . Observe that (AB)C = A(BC), (AB)−1 = B −1 A−1 for any A, B, C ⊆ X. 4A5B Group actions (a) If X is a group and Z is a set, an action of X on Z is a function (x, z) 7→ x•z : X × Z → Z such that (xy)•z = x•(y •z) for all x, y ∈ X and z ∈ Z, e•z = z for every z ∈ Z where e is the identity of X. In this context I may say that ‘X acts on Z’, taking the operation (b) An action x•w = z. (c) If

•

•

•

for granted.

of a group X on a set Z is transitive if for every w, z ∈ Z there is an x ∈ X such that

is an action of a group X on a set Z, I write x•A = {x•z : z ∈ A} whenever x ∈ X and A ⊆ Z.

864

Appendix

4A5Bd

(d) If • is an action of a group X on a set Z, then z 7→ x•z : Z → Z is a bijection for every x ∈ X. (For it has an inverse z 7→ x−1 •z.) So if Z is a topological space and z 7→ x•z is continuous for every x, it is a homeomorphism for every x. (e) An action • of a group X on a set Z is faithful if whenever x, y ∈ X are distinct there is a z ∈ Z such that x•z 6= y •z; that is, the natural homomorphism from X to the group of bijections from Z to itself is injective. An action of X on Z is faithful iff for any x ∈ X which is not the identity there is a z ∈ Z such that x•z 6= z. (f ) If • is an action of a group X on a set Z, then Yz = {x : x ∈ X, x•z = z} is a subgroup of X (the stabilizer of z) for every z ∈ Z. If • is transitive, then Yw and Yz are conjugate subgroups for all w, z ∈ Z. (If x•w = z, then Yz = xYw x−1 .) 4A5C Examples Let X be any group. (a) Write x•l y = xy,

x•r y = yx−1 ,

x•c y = xyx−1

for x, y ∈ X. These are all actions of X on itself, the left, right and conjugacy actions. (b) If A ⊆ X, we have an action of X on the set {xA : x ∈ X} defined by setting x•(yA) = xyA for x, y ∈ X. 4A5D Definitions (a) A topological group is a group X endowed with a topology such that the operations (x, y) 7→ xy : X × X → X and x 7→ x−1 : X → X are continuous. (b) A Polish group is a topological group in which the topology is Polish (definition: 4A2A). 4A5E Elementary facts Let X be any topological group. (a) For any x ∈ X, the functions y 7→ xy, y 7→ yx and y 7→ y −1 are all homeomorphisms from X to itself. (Hewitt & Ross 63, 4.2; Folland 95, 2.1.) (b) The maps (x, y) 7→ x−1 y, (x, y) 7→ xy −1 and (x, y) 7→ xyx−1 from X × X to X are continuous. (c) {G : G is open, e ∈ G, G−1 = G} is a base of neighbourhoods of the identity e of X. (Hewitt & Ross 63, 4.6; Folland 95, 2.1.) (d) If G ⊆ X is an open set, then AG and GA are open for any set A ⊆ X. (Hewitt & Ross 63, 4.4.) (e) If F ⊆ X is closed and x ∈ X \ F , there is a neighbourhood U of e such that U xU U ∩ F U U = ∅. P P Set U1 = X \ x−1 F . Let U2 be a neighbourhood of e such that U2 U2 U2 U2−1 U2−1 ⊆ U1 . Let U be a neighbourhood of e such that U ⊆ U2 ∩ xU2 x−1 ; this works. Q Q (f ) If K ⊆ X is compact and F ⊆ X is closed then KF and F K are closed. If K, L ⊆ X are compact so is KL. (Hewitt & Ross 63, 4.4.) (g) If there is any compact set K ⊆ X such that int K is non-empty, then X is locally compact. (h) If K ⊆ X T is compact and F is a downwards-directed family of closed T T subsets of X with intersection F0 , then KF0 = F ∈F KF and F0 K = F ∈F F K. P P Of course KF0 ⊆ F ∈F KF . If x ∈ X \ KF0 , then K −1 x ∩ F0 is empty; because K −1 xTis compact, there is some F T ∈ F such that K −1 x ∩ F = ∅ (3A3Db), so that x ∈ / KF . Accordingly KF0 = F ∈F KF . Similarly, F0 K = F ∈F F K. Q Q (i) If K ⊆ X is compact and G ⊆ X is open, then {x : xK ⊆ G} and {x : Kx ⊆ G} are open. P P {x : xK 6⊆ G} = (X \ G)K −1 }, which by (e) above is closed. Similarly, {x : Kx 6⊆ G} = K −1 (X \ G) is closed. Q Q

4A5J

Topological groups

865

(j) If X is Hausdorff, K ⊆ X is compact and U is a neighbourhood of e, there is a neighbourhood V of e such that xy ∈ U whenever x, y ∈ K and yx ∈ V ; that is, y −1 zy ∈ U whenever z ∈ V and y ∈ K. P P If U is open, then {xy : x, y ∈ K, yx ∈ / U } is a compact set not containing e. Compare 4A5Oc below. Q Q ´ sza ´ r 78, 11.2.12; Hewitt & Ross 63, 5.5; Folland (k) Any open subgroup of X is also closed. (Csa 95, 2.1.) (l) If X is locally compact, it has an open subgroup which is σ-compact. (Hewitt & Ross 63, 5.14; Folland 95, 2.3.) 4A5F Proposition (a) Let (X, T) and (Y, S) be topological groups. If φ : X → Y is a group homomor´ sza ´ r 78, 11.2.17; Hewitt & Ross phism which is continuous at the identity of X, it is continuous. (Csa 63, 5.40.) (b) Let X be a group and S, T two topologies on X both making X a topological group. If every Sneighbourhood of the identity is a T-neighbourhood of the identity, then S ⊆ T. (Apply (a) to the identity map from (X, T) to (X, S).) Q 4A5G Proposition If hXi ii∈I is any family of topological groups, then i∈I Xi , with the product topology and the product group structure, is again a topological group. (Hewitt & Ross 63, 6.2.) 4A5H The uniformities of a topological group Let (X, T) be a topological group. Write U for the set of open neighbourhoods of the identity e of X. (a) For U ∈ U, set WU = {(x, y) : xy −1 ∈ U } ⊆ X × X. The family {WU : U ∈ U} is a filter base, and the filter on X × X which it generates is a uniformity on X, the right uniformity of X. (Warning!! ´ sza ´ r 78, 11.2.7). Some authors call this the ‘left uniformity’.) This uniformity induces the topology T (Csa It follows that T is completely regular, therefore regular (4A2Ja, or Hewitt & Ross 63, 8.4). ˜ U = {(x, y) : xy −1 ∈ U, x−1 y ∈ U } ⊆ X × X. The family {W ˜ U : U ∈ U} is a filter (b) For U ∈ U , set W base, and the filter on X × X which it generates is a uniformity on X, the bilateral uniformity of X. ´ sza ´ r 78, 11.3.c.) This uniformity induces the topology T. (Csa 4A5I Definitions If X is a topological group and Z a topological space, an action of X on Z is ‘continuous’ or ‘Borel measurable’ if it is continuous, or Borel measurable, when regarded as a function from X × Z to Z. Of course the left, right and conjugacy actions of a topological group on itself are all continuous. 4A5J Quotient groups: Proposition Let X be a topological group, Y a subgroup of X, and Z the set of left cosets of Y in X. Set πx = xY for x ∈ X. (a) We have a topology on Z defined by saying that V ⊆ Z is open iff π −1 [V ] is open in X. The canonical map π : X → Z is continuous and open. (b) Z is Hausdorff iff Y is closed. If X is locally compact, so is Z. (c) We have a continuous action of X on Z defined by saying that x•πx0 = π(xx0 ) for any x, x0 ∈ X. (d) If Y is a normal subgroup of X, then the group operation on Z renders it a topological group. proof (a) By Hewitt & Ross 63, 5.15-5.16 the given formula defines a topology on Z for which π is continuous and open. (b) By Hewitt & Ross 63, 5.21, Z is Hausdorff iff Y is closed. By Hewitt & Ross 63, 5.22 or Folland 95, 2.2, Z is locally compact if X is. (c) I have noted in 4A5Cb that the formula given defines an action. If V ⊆ Z is open and x0 ∈ X, z0 ∈ Z are such that x0 •z0 ∈ V , take x00 ∈ X such that πx00 = z0 , and observe that x0 x00 ∈ π −1 [V ], which is open. So there are open neighbourhoods V0 , V00 of x0 , x00 respectively such that V0 V00 ⊆ π −1 [V ], and x•z ∈ V whenever x ∈ V0 and z ∈ π[V00 ]. Since π[V00 ] is an open neighbourhood of z0 , this is enough to show that • is continuous at (x0 , z0 ). ´ sza ´ r 78, 11.2.15; Hewitt & Ross 63, 5.26; Folland 95, 2.2. (d) Csa

866

Appendix

4A5K

4A5K Proposition Let X be a topological group with identity e. (a) Y = {e} is a closed normal subgroup of X. (b) Writing π : X → X/Y for the canonical map, (i) a subset of X is open iff it is the inverse image of an open subset of X/Y , (ii) a subset of X is closed iff it is the inverse image of a closed subset of X/Y , (iii) π[G] is a regular open set in X/Y for every regular open set G ⊆ X, (iv) π[F ] is nowhere dense in X/Y for every nowhere dense set F ⊆ X, (v) π −1 [V ] is nowhere dense in X for every nowhere dense V ⊆ X/Y . ´ sza ´ r 78, 11.2.13; Hewitt & Ross 63, 5.4; Folland 95, 2.3. proof (a) Csa (b)(i)-(ii) If G ⊆ X is open and x ∈ G, then x{e} = {x} ⊆ G, because X is regular (4A5Ha). So G = GY = π −1 [π[G]]. Since π is an open map (4A5Ja), π[G] is open and G is the inverse image of an open set. If F ⊆ X is closed, F = X \ π −1 [π[X \ F ]] = π −1 [(X/Y ) \ π[X \ F ]] is the inverse image of a closed set, and π[F ] = (X/Y ) \ π[X \ F ] is closed. Conversely, because π is continuous, the inverse image of an open or closed set is again open or closed. (iii) If A ⊆ X, then π[A] is a closed set included in π[A] (because π is continuous), so π[A] = π[A]. If G ⊆ X is a regular open set, then π −1 [int π[G]] is an open subset of π −1 [π[G]] = G, so is included in int G = G. But this means that the open set π[G] includes int π[G] = int π[G], and π[G] = int π[G] is a regular open set. (iv) If F ⊆ X is nowhere dense, then its closure is of the form π −1 [V ] for some closed set V ⊆ X/Y . Now if H ⊆ X/Y is a non-empty open set, π −1 [H] is a non-empty open subset of X, so is not included in F , and H cannot be included in V . Thus V is nowhere dense; but V ⊇ π[F ], so π[F ] is nowhere dense. (v) If V ⊆ X/Y is nowhere dense, and G ⊆ X is open and not empty, then G = π −1 [H] for some non-empty open H ⊆ X/Y . In this case, H \ V is non-empty, so π −1 [H \ V ] is a non-empty open subset of G disjoint from π −1 [V ]. As G is arbitrary, π −1 [V ] is nowhere dense. 4A5L Theorem Let X be a topological group and Y a normal subgroup of X. Let π : X → X/Y be the canonical homomorphism. (a) If X 0 is another topological group and φ : X → X 0 a continuous homomorphism with kernel included in Y , then we have a continuous homomorphism ψ : X/Y → X 0 defined by the formula ψπ = φ; ψ is injective iff Y is the kernel of φ. (b) Suppose that K1 , K2 are two subgroups of X/Y such that K2 C K1 . Set Y1 = π −1 [K1 ] and Y2 = π −1 [K2 ]. Then Y2 C Y1 and Y1 /Y2 and K1 /K2 are isomorphic as topological groups. proof (a) This is elementary group theory, except for the claim that ψ is continuous. But if H ⊆ X 0 is open, then ψ −1 [H] = π[φ−1 [H]] is open because φ is continuous and π is open (4A5Ja); so ψ is continuous. (b) See Hewitt & Ross 63, 5.35. 4A5M Proposition Let X be a topological group. (a) Let Y be any subgroup of X. If X is given its bilateral uniformity, then the subspace uniformity on ´ sza ´ r 78, 11.3.13.) Y is the bilateral uniformity of Y . (Csa ´ sza ´ r 78, 11.3.21.) If X is (b) If X is locally compact it is complete under its right uniformity. (Csa ´ sza ´ r 78, 11.3.10.) complete under its right uniformity it is complete under its bilateral uniformity. (Csa (c) Suppose that X is Hausdorff and that Y is a subgroup of X which is locally compact in its subspace topology. Then Y is closed in X. P P Putting (a) and (b) together, we see that Y is complete in its subspace uniformity; by 3A4Fd, it is closed. Q Q b under its bilateral 4A5N Theorem Let X be a Hausdorff topological group. Then its completion X uniformity can be endowed (in exactly one way) with a group structure rendering it a Hausdorff topological b represents X as a dense subgroup of X. b (Csa ´sza ´r 78, group in which the natural embedding of X in X 11.3.15.) If X has a neighbourhood of the identity which is totally bounded for the bilateral uniformity, b is locally compact. (Csa ´ sza ´ r 78, 11.3.24.) then X

4A5O

Topological groups

867

4A5O Proposition Let X be a topological group. (a) If A ⊆ X, then the following are equiveridical: (i) A is totally bounded for the bilateral uniformity of X; (ii) for every neighbourhood U of the identity there is a finite set I ⊆ X such that A ⊆ IU ∩ U I. (b) If A,SB ⊆ X are totally bounded for the bilateral uniformity of X, so are A ∪ B, A−1 and AB. In particular, i≤n xi B is totally bounded for any x0 , . . . , xn ∈ X. (c) If A ⊆ X is totally bounded for the bilateral uniformity, and U is any neighbourhood of the identity, then {y : xyx−1 ∈ U for every x ∈ A} is a neighbourhood of the identity. (d) If X is the product of a family hXi ii∈I of topological Q groups, a subset A of X is totally bounded for the bilateral uniformity of X iff it is included in a product i∈K Ai where Ai ⊆ Xi is totally bounded for the bilateral uniformity of Xi for every i ∈ I. (e) If X is locally compact, a subset of X is totally bounded for the bilateral uniformity iff it is relatively compact. proof (a)(i)⇒(ii) Suppose that A is totally bounded, and that U is a neighbourhood of the identity e of X. Set W = {(x, y) : xy −1 ∈ U −1 , x−1 y ∈ U } = {(x, y) : y ∈ U x ∩ xU }; then W belongs to the uniformity, so there is a finite set I ⊆ X such that A ⊆ W [I]. But W [I] ⊆ U I ∩ IU , so A ⊆ U I ∩ IU . (ii)⇒(i) Now suppose that A satisfies the condition, and that W belongs to the uniformity. Then there is a neighbourhood U of e such that {(x, y) : xy −1 ∈ U, x−1 y ∈ U } ⊆ W . Let V be a neighbourhood of e such that V V −1 ⊆ U and V −1 V ⊆ U . Let I ⊆ X be a finite set such that A ⊆ V I ∩ IV . For w, z ∈ I set Awz = A ∩ V w ∩ zV . If x, y ∈ Awz , xy −1 ∈ V ww−1 V −1 ⊆ U and x−1 y ∈ V −1 z −1 zV ⊆ U . But this means that Awz × Awz ⊆ W . So if we take a finite set J which meets every non-empty Awz , A ⊆ W [J]. As W is arbitrary, A is totally bounded. (b) Of course A ∪ B is totally bounded; this is immediate from the definition of ‘totally bounded’. If U is a neighbourhood of e, so is U −1 , so there is a finite set I ⊆ X such that A ⊆ IU −1 ∩ U −1 I and A−1 ⊆ U I −1 ∩ I −1 U ; as U is arbitrary, A−1 is totally bounded. To see that AB is also totally bounded, let U be a neighbourhood of e, and take a neighbourhood V of e such that V V ⊆ U . Then there is a finite set I ⊆ X such that A ⊆ V I and B ⊆ IV . Let W be a neighbourhood of e such that zW z −1 ∪ z −1 W z ⊆ V for every z ∈ I, and J a finite set such that B ⊆ W J and A ⊆ JW . Then zW ⊆ V z,

W z ⊆ zV

IW ⊆ V I,

W I ⊆ IV

for every z ∈ I, so and AB ⊆ V IW J ⊆ V V IJ ⊆ U K,

AB ⊆ JW IV ⊆ JIV V ⊆ KU

where K = IJ ∪ JI is finite. As U is arbitrary, AB is totally bounded. (c) Let V be a neighbourhood of e such that V V V −1 ⊆ U . Let I be a finite set such that A ⊆ V I. Let W be a neighbourhood of e such that zW z −1 ⊆ V for every z ∈ I. If now y ∈ W and x ∈ A, there is a z ∈ I such that x ∈ V z, so that xyx−1 ∈ V zW z −1 V −1 ⊆ V V V −1 ⊆ U . Turning this round, {y : xyx−1 ∈ U for every x ∈ A} includes W and is a neighbourhood of e. (d)(i) Suppose that A is totally bounded. Set Ak = πk [A] for each k ∈ K, where πk (x) = x(k) for x ∈ X. If U is a neighbourhood of the identity in Xk , then V = πk−1 [U ] is a neighbourhood of the identity in X, so there is a finite set I ⊆ X such that A ⊆ IV ∩ V I; now Ak ⊆ JU ∩QU J, where J = πk [I] is finite. As U is arbitrary, Ak is totally bounded. This is true for every k, and A ⊆ k∈K Ak . Q (ii) Suppose that A ⊆ k∈K Ak where Ak ⊆ Xk is totally bounded for each k ∈ K. If A is empty, of course it is totally bounded; assume that A 6= ∅. If K = ∅, then X = {∅} is the trivial group, and again

868

Appendix

4A5O

A is totally bounded; so assume that K is non-empty. Let V be a neighbourhood of the identity in X. Then there are a non-empty finite set L ⊆ T K and a family hUk ik∈L such that Uk is a neighbourhood of the identity in Xk for each k ∈ L, and V ⊇ k∈L πk−1 Uk . For each k ∈ L, let Ik be a finite subset of Xk such that Ak ⊆ Ik Uk ∩ Uk Ik . Set I = {x : x ∈ X, x(k) is the identity for k ∈ K \ L, x(k) ∈ Ik for k ∈ L}. Then I is finite and A ⊆ IV ∩ V I. As V is arbitrary, A is totally bounded. (e) By 4A5Mb and 4A2Jd, the totally bounded sets are the relatively compact sets. 4A5P Lemma Let X be a locally compact Hausdorff topological group. Take f ∈ Ck (X), the space of continuous real-valued functions on X with compact supports. (a) Let K ⊆ X be a compact set. Then for any ² > 0 there is a neighbourhood W of the identity E of X such that |f (xay) − f (xby)| ≤ ² whenever x ∈ K, y ∈ X and ab−1 ∈ W . (b) For any x0 ∈ X, there is a non-negative f ∗ ∈ Ck (X) such that for every ² > 0 there is an open set G containing x0 such that |f (xy) − f (x0 y)| ≤ ²f ∗ (y) for every x ∈ G and y ∈ X. proof (a) By 4A2Je and 4A5Ha, f is uniformly continuous for the right uniformity of X. There is therefore a symmetric neighbourhood U of e, included in V , such that |f (y1 ) − f (y2 )| ≤ ² whenever y1 , y2 ∈ X and y1 y2−1 ∈ U . By 4A5Oc, there is a symmetric neighbourhood W of e such that xzx−1 ∈ U whenever x ∈ K and z ∈ W . Now suppose that x ∈ K, y ∈ X and ab−1 ∈ W . Then (xay)(xby)−1 = xab−1 x−1 ∈ U , so |f (xay) − f (xby)| ≤ ², as required. (b) We need a trifling refinement of the ideas above. (i) Suppose for the moment that x0 = e. Set L = {x : f (x) 6= 0} and let V be a compact symmetric neighbourhood of the identity e, so that L and V L are compact. Let f ∗ ∈ Ck (X) be such that f ∗ ≥ χ(V L) (4A2G(e-i)). Given ² > 0, take U as in (a), so that U is a symmetric neighbourhood of e and |f (y1 )−f (y2 )| ≤ ² whenever y1 y2−1 ∈ U ; this time arrange further that U ⊆ V . Then if x ∈ U and y ∈ X, either y and xy belong to V L, while (xy)y −1 ∈ U , so |f (xy) − f (y)| ≤ ² ≤ ²f ∗ (y) or neither y nor xy belongs to L, so |f (xy) − f (y)| = 0 ≤ ²f ∗ (y). (ii) For the general case, set f0 (x) = f (x0 x) for x ∈ X. Because x 7→ x0 x is a homeomorphism, f0 ∈ Ck (X). By (i), we have a non-negative f ∗ ∈ Ck (X) such that for every ² > 0 there is a neighbourhood G² of e such that |f0 (xy) − f0 (y)| ≤ ²f ∗ (y) whenever x ∈ G² and y ∈ X. Now, given ² > 0, G0 = x0 G² is a neighbourhood of x0 and |f (xy) − f (x0 y)| ≤ ²f ∗ (y) whenever x ∈ G0 and y ∈ X. So f ∗ witnesses that the result is true. 4A5Q Metrizable groups: Proposition Let (X, T) be a topological group. Then the following are equiveridical: (i) X is metrizable; (ii) the identity e of X has a countable neighbourhood base; (iii) there is a metric ρ on X, inducing the topology T, which is right-translation-invariant, that is, ρ(x1 , x2 ) = ρ(x1 y, x2 y) for all x1 , x2 , y ∈ X; (iv) there is a right-translation-invariant metric on X which induces the right uniformity of X. ´sza ´r 78, 11.2.10; Hewitt & Ross 63, 8.3. proof Csa Warning! A Polish group (4A5Db) is of course metrizable, so has a right-translation-invariant metric inducing its topology. At the same time, it has a complete metric inducing its topology. But there is no suggestion that these two metrics should be the same, or even induce the same uniformity. 4A5R Corollary If X is a locally compact topological group and {e} is a Gδ set in X, then X is metrizable. (Put 4A5Q and 4A2Kf together, or see Hewitt & Ross 63, 8.5.)

4A6Ab

Banach algebras

869

4A5S Lemma Let X be a σ-compact locally compact Hausdorff topological group and hUn iT n∈N any sequence of neighbourhoods of the identity in X. Then X has a compact normal subgroup Y ⊆ n∈N Un such that X/Y is metrizable. proof Let hKn in∈N be a sequence of compact sets covering X. Choose inductively a sequence hVn in∈N of compact neighbourhoods of e such that, for each n ∈ N, (α) Vn+1 ⊆ Vn−1 , Vn+1 Vn+1 ⊆ Vn , Vn ⊆ S Un , (β) xyx−1 ∈ Vn whenever y ∈ Vn+1 and x ∈ i≤n Ki . (When we come to choose Vn+1 , we can achieve (α) because inversion and multiplication are continuous, and (β) by 4A5Oc; and we can then shrink Vn+1 to a closed neighbourhood of e because X is regular, byT4A5Ha.) T Set Y = n∈N Vn . Then (α) is enough to ensure that Y is a compact subgroup of X included in n∈N Un , while (β) ensures that Y is normal, because for any x ∈ X there is an n ∈ N such that xVm+1 x−1 ⊆ Vm for every m ≥ n. T Let π : X → X/Y be the canonical map. Then C = n∈N π[int Vn ] is a Gδ subset of X/Y , because π is open. But \ \ \ π −1 [C] ⊆ π −1 [π[Vn+1 ]] = Vn+1 Y ⊆ Vn = Y, n∈N

n∈N

n∈N

so C = {eX/Y }, writing eX/Y for the identity of X/Y . Thus {eX/Y } is a Gδ set; as X/Y is locally compact and Hausdorff (4A5Jb), it is metrizable (4A5R). *4A5T I shall not rely on the following fact, but it will help you to make sense of some of the results of this volume. Theorem A compact Hausdorff topological group is dyadic. proof Uspenskiˇı 88. 4A6 Banach algebras I give results which are needed for Chapter 44. Those down to 4A6K should be in any introductory text on normed algebras; 4A6L-4A6N, as expressed here, are a little more specialized. As with normed spaces or linear topological spaces, Banach algebras may be defined over either R or C. In §445 we need complex Banach algebras, but in §446 I think the ideas are clearer in the context of real Banach algebras. Accordingly, as in §2A4, I express as much as possible of the theory in terms applicable equally to either, speaking of ‘normed algebras’ or ‘Banach algebras’ without qualification, and using the symbol R C to represent the field of scalars. Since (at least, if you keep to the path indicated here) the ideas are independent of which field we work with, you will have no difficulty in applying the arguments given in Folland 95 or Hewitt & Ross 63 for the complex case to the real case. In 4A6B and 4A6I-4A6K, however, we have results which apply only to ‘complex’ Banach algebras, in which the underlying field is taken to be C. 4A6A Definition (a) I repeat a definition from §2A4. A normed algebra is a normed space U together with a multiplication, a binary operator × on U , such that u × (v × w) = (u × v) × w, u × (v + w) = (u × v) + (u × w),

(u + v) × w = (u × w) + (v × w),

(αu) × v = u × (αv) = α(u × v), ku × vk ≤ kukkvk for all u, v, w ∈ U and α ∈ R C . A normed algebra is commutative if its multiplication is commutative. (b) A Banach algebra is a normed algebra which is a Banach space. A unital Banach algebra is a Banach algebra with a multiplicative identity e such that kek = 1. (Warning: some authors reserve the term ‘Banach algebra’ for what I call a ‘unital Banach algebra’.) In a unital Banach algebra I will always use the letter e for the identity.

870

Appendix

4A6B

4A6B Stone-Weierstrass Theorem: fourth form Let X be a locally compact Hausdorff space, and C0 = C0 (X; C) the complex Banach algebra of continuous functions f : X → C such that {x : |f (x)| ≥ ²} is compact for every ² > 0. Let A ⊆ C0 be such that A is a linear subspace of C0 ; f × g ∈ A for every f , g ∈ A; the complex conjugate f¯ of f belongs to A for every f ∈ A; for every x ∈ X there is an f ∈ A such that f (x) 6= 0: whenever x, y are distinct points of X there is an f ∈ A such that f (x) 6= f (y). Then A is k k∞ -dense in C0 . proof Let X∞ = X ∪ {∞} be the one-point compactification of X (3A3O). For f ∈ C0 write f # for the extension of f to X ∪ {∞} with f # (∞) = 0, so that f # ∈ Cb (X∞ ; C). Let B ⊆ Cb (X∞ ; C) be the set of all functions of the form f # + αχX∞ where f ∈ A and α ∈ C. Then B is a subalgebra of Cb (X ∪ {∞}) which contains complex conjugates of its members and constant functions and separates the points of X∞ . Take any h ∈ C0 and ² > 0. By the ‘third form’ of the Stone-Weierstrass theorem (281G), there is a g ∈ B such that kg − h# k∞ ≤ 21 ². Express g as f # + αχX∞ where f ∈ A and α ∈ C. Then |α| = |g(∞)| = |g(∞) − h# (∞)| ≤ 21 ², so kh − f k∞ = kh# − f # k∞ ≤ kh# − gk∞ + kg − f # k∞ ≤ 21 ² + |α| ≤ ². As h and ² are arbitrary, A is dense in C0 . 4A6C Proposition If U is any Banach space other than {0}, then the space B(U ; U ) of bounded linear ¨ the 69, 14.6.) operators from U to itself is a unital Banach algebra. (Ko 4A6D Proposition Any normed algebra U can be embedded as a subalgebra of a unital Banach algebra V , in such a way that if U is commutative so is V . (Folland 95, §1.3; Hewitt & Ross 63, C.3.) 4A6E Proposition Let U be a unital Banach algebra and W ⊆ U a closed proper ideal. Then U/W , with the quotient linear structure, ring structure and norm, is a unital Banach algebra. (Hewitt & Ross 63, C.2.) 4A6F Proposition If U is a Banach algebra and φ : U → R C is a multiplicative linear functional, then |φ(u)| ≤ kuk for every u ∈ U . 1

proof ?? Otherwise, there is a u such that |φ(u)| > kuk; set v = u, so that φ(v) = 1 and kvk < 1. Since φ(u) P∞ n n n kv k ≤ kvk for every n ≥ 1, w = n=1 v is defined in U (4A4Ie), and w = vw + v. But this means that φ(w) = φ(v)φ(w) + φ(v) = φ(w) + 1, which is impossible. X X 4A6G Definition Let U be a normed algebra and u ∈ U . (a) For any u ∈ U , limn→∞ kun k1/n is defined and equal to inf n≥1 kun k1/n . (Hewitt & Ross 63, C.4.) (b) This common value is called the spectral radius of u. 4A6H Theorem If U is a unital Banach algebra, then the set R of invertible elements is open, and u 7→ u−1 is a continuous function from R to itself. Writing e for the identity of U , v ∈ R and kv −1 − ek ≤ kv−ek 1−kv−ek

10.12.)

whenever kv − ek < 1. (Folland 95, 1.4; Hewitt & Ross 63, C.8 & C.10; Rudin 91, 10.7 &

4A6M

Banach algebras

871

4A6I Theorem Let U be a complex unital Banach algebra and u ∈ U . Write r for the spectral radius of u. (a) If ζ ∈ C and |ζ| > r then ζe − u is invertible. (b) There is a ζ such that |ζ| = r and ζe − u is not invertible. proof Folland 95, 1.8; Hewitt & Ross 63, C.24; Rudin 91, 1.13. 4A6J Theorem Let U be a commutative complex unital Banach algebra, and u ∈ U . Then for any ζ ∈ C the following are equiveridical: (i) there is a non-zero multiplicative linear functional φ : U → C such that φ(u) = ζ; (ii) ζe − u is not invertible. proof Folland 95, 1.13; Hewitt & Ross 63, C.20; Rudin 91, 11.5. 4A6K Corollary Let U be a commutative complex Banach algebra and u ∈ U . Then its spectral radius r(u) is max{|φ(u)| : φ is a multiplicative linear functional on U }. (Folland 95, 1.13; Hewitt & Ross 63, C.20; Rudin 91, 11.9.) 4A6L Exponentiation Let U be a unital Banach algebra. For any u ∈ U ,

P∞ k=0

1 k!

k uk k ≤

P∞

1

k

k=0 k! kuk

is finite, so P∞

1 k k=0 k! u

exp(u) =

is defined in U (4A4Ie). (In this formula, interpret u0 as e for every u.) 4A6M Lemma Let U be a unital Banach algebra. (a) If u, v ∈ U and max(kuk, kvk) ≤ γ then k exp(u) − exp(v) − u + vk ≤ ku − vk(exp(γ) − 1). So if max(kuk, kvk) ≤ 12 and exp(u) = exp(v) then u = v. (b) If ku − ek ≤ 16 then there is a v such that exp(v) = u and kvk ≤ 2ku − ek. (c) If u, v ∈ U and uv = vu then exp(u + v) = exp(u) exp(v). proof (a) Note first that if k ≥ 1 then

kuk − v k k = k

k−1 X

uk−i v i − uk−i−1 v i+1 k = k

i=0

≤

k−1 X

k−1 X

uk−i−1 (u − v)v i k

i=0

kukk−i−1 ku − vkkvki ≤

i=0

k−1 X

γ k−1 ku − vk = kγ k−1 ku − vk.

i=0

So k exp(u) − exp(v) − u + vk = k

∞ X 1 k=2

≤

∞ X k k=2

Now if exp(u) = exp(v) and γ ≤

1 2,

k!

k!

(uk − v k )k ≤

∞ X 1 k=2

k!

kuk − v k k

γ k−1 ku − vk = ku − vk(exp(γ) − 1).

exp(γ) − 1 < 1 so ku − vk = 0 and u = v.

(b) Set γ = ku − ek. Define hvn in∈N in U by setting v0 = 0, vn+1 = vn + u − exp(vn ) for n ∈ N. Then kvn+1 − vn k = ku − exp(vn )k ≤ 2−n γ,

kvn k ≤ 2(1 − 2−n )γ

for every n ∈ N. P P Induce on n. The induction starts with kv0 k = 0 and ku − exp(v0 )k = ku − ek = γ. Given that kvn k ≤ 2(1 − 2−n )γ and ku − exp(vn )k ≤ 2−n γ, then

872

Appendix

4A6M

kvn+1 k ≤ kvn k + kvn+1 − vn k = kvn k + ku − exp(vn )k ≤ 2(1 − 2−n )γ + 2−n γ = 2(1 − 2−n−1 )γ. Now max(kvn+1 k, kvn k) ≤ 2γ ≤ 31 , so 1 3

ku − exp(vn+1 )k = kvn+1 − vn + exp(vn ) − exp(vn+1 )k ≤ kvn+1 − vn k(exp( ) − 1) 1 2

≤ kvn+1 − vn k ≤ 2−n−1 γ, and the induction continues. Q Q P∞ Since kv − v k is finite, v = limn→∞ vn is defined in U , and kvk = limn→∞ kvn k ≤ 2γ. n+1 n n=0 Accordingly k exp(v) − exp(vn )k ≤ kv − vn k + k exp(v) − exp(vn ) − v + vn k 1 3

≤ kv − vn k(1 + exp( ) − 1) → 0 as n → ∞, and exp(v) = limn→∞ exp(vn ) = u. Pm m! (c) Because uv = vu, (u + v)m = j=0 uj v m−j for every m ∈ N (induce on m; the point is that j!(m−j)! P 1 uv j = v j u for every j ∈ N). Next, j,k∈N kukj kvkk is finite. So j!k!

exp(u + v) = =

∞ X m=0 ∞ X

1 (u + v)m m!

¡ X

m=0 j+k=m

(using 4A4I(e-ii)) =

X¡X 1 j∈N k∈N

(4A4I(e-ii) again) =

X¡ 1 j∈N

(by 4A4Bj, because w 7→

1 j u w j!

j!

uj

j!k!

1 j k¢ u v j!k!

X

=

(j,k)∈N×N

1 j k u v j!k!

uj v k )

X1 k∈N

k!

vk

¢

is a continuous linear operator for each j) =

X1 j∈N

j!

uj exp(v) =

¡X 1 j∈N

j!

¢ uj exp(v)

(4A4Bj again) = exp(u) exp(v), as claimed. 4A6N Lemma If U is any unital Banach algebra, u ∈ U and u 6= e, there is some n ∈ N such that kun − ek > 16 . n

proof ?? Otherwise, for every n ∈ N take vn such that exp(vn ) = u2 and kvn k ≤ 31 (4A6Mb). Then exp(vn+1 ) = exp(vn )2 = exp(2vn ) (4A6Mc) and kvn+1 k ≤ 13 , k2vn k ≤ 13 so vn+1 = 2vn for every n (4A6Ma). Inducing on n, vn = 2n v0 for every n, so that kv0 k ≤ 2−n kvn k → 0 as n → ∞, and u = exp(v0 ) = e. X X

452I

Concordance

873

Concordance I list here the section and paragraph numbers which have (to my knowledge) appeared in print in references to this volume, and which have since been changed. 411I Completion regular measures The definition of ‘completion regular’ measure, referred to in Fremlin 00, has been moved to 411J. 414N Density topologies The note on ‘density topologies’, referred to in the 2001 edition of Volume 2, has been moved to 414P. 415Yd Sorgenfrey topology This exercise, referred to in the 2001 edition of Volume 2, has been moved to 415Ye. 416P The algebra of open-and-closed subsets This paragraph, referred to in the 2002 edition of Volume 3, has been moved to 416Q. 416T Kakutani’s theorem The description of the usual measure on {0, 1}κ , referred to in Fremlin & Plebanek 03, has been moved to 416U. 417E τ -additive product measures The reference in Fremlin 00 to Kakutani’s theorem that the product measure on {0, 1}I is completion regular should be directed to 415E or 416U rather than 417E. §445 Convolutions The material mentioned in the 2001 edition of Volume 2 has been moved to §444. The particular result referred to as ‘445K’ is now 444R. 445Xq The exercise 445Xq, referred to in the 2002 edition of Volume 3, has been moved to 445Xp. 452I In Fremlin 00 I quote Pachl’s result that if (X, Σ, µ) is countably compact, (Y, T, ν) is strictly localizable and f : X → Y is inverse-measure-preserving, then ν is countably compact; this is now in 452R. References Fremlin D.H. [00] ‘Weakly α-favourable measure spaces’, Fundamenta Math. 165 (2000) 67-94. Fremlin D.H. & Plebanek G. [03] ‘Large families of mutually singular Radon measures’, Bull. Acad. Pol. Sci. (Math.) 51 (2003) 169-174.

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Shelah S. & Fremlin D.H. [93] ‘Pointwise compact and stable sets of measurable functions’, J. Symbolic Logic 58 (1993) 435-455. [§465 notes.] Sierpi´ nski W. [45] ‘Sur une suite infinie de fonctions de classe 1 dont toute fonction d’accumulation est non mesurable’, Fundamenta Math. 33 (1945) 104-105. [464C.] Solecki S. [01] ‘Haar null and non-dominating sets’, Fundamenta Math. 170 (2001) 197-217. [444Ye.] Solovay R.M. [71] ‘Real-valued measurable cardinals’, pp. 397-428 in Scott 71. [438C.] Souslin M. [17] ‘Sur une définition des ensembles mesurables B sans nombres infinis’, C.R.Acad.Sci. (Paris) 164 (1917) 88-91. [421D, §421 notes.] Steen L.A. & Seebach J.A. [78] Counterexamples in Topology. Springer, 1978. [434Ya, §439 notes.] Steinlage R.C. [75] ‘On Haar measure in locally compact T2 spaces’, Amer. J. Math. 97 (1975) 291-307. [441C.] Strassen V. [65] ‘The existence of probability measures with given marginals’, Ann. Math. Statist. 36 (1965) 423-439. [457D.] Sullivan J.M. [94] ‘Sphere packings give an explicit bound for the Besicovitch covering theorem’, J. Geometric Analysis 4 (1994) 219-231. [§472 notes.] Swierczkowski S. [58] ‘On a free group of rotations of the Euclidean space’, Indagationes Math. 20 (1958) 376-378. [449Yf.] Talagrand M. [75] ‘Sur une conjecture de H.H.Corson’, Bull. Sci. Math. 99 (1975) 211-212. [467M.] Talagrand M. [78a] ‘Sur un théorème de L.Schwartz’, C.R.Acad. Sci. Paris 286 (1978) 265-267. [466Ia.] Talagrand M. [78b] ‘Comparaison des boreliens pour les topologies fortes et faibles’, Indiana Univ. Math. J. 21 (1978) 1001-1004. [466Z.] Talagrand M. [80] ‘Compacts de fonctions mesurables et filtres non-mesurables’, Studia Math. 67 (1980) 13-43. [463O, 464C, 464D.] Talagrand M. [81] ‘La τ -régularité de mesures gaussiennes’, Z. Wahrscheinlichkeitstheorie und verw. Gebiete 57 (1981) 213-221. [§456 intro., 456K.] Talagrand M. [82] ‘Closed convex hull of set of measurable functions, Riemann-measurable functions and measurability of translations’, Ann. Insitut Fourier (Grenoble) 32 (1982) 39-69. [465M.] Talagrand M. [84] Pettis integral and measure theory. Mem. Amer. Math. Soc. 307 (1984). [463Ya, 463Yd, 463Za, §465 intro., 465B, 465N, 465R-465T, 466Ia, §466 notes.] Talagrand M. [87] ‘The Glivenko-Cantelli problem’, Ann. of Probability 15 (1987) 837-870. [§465 intro., 465L, 465M, 465Yb, §465 notes.] Talagrand M. [88] ‘On liftings and the regularization of stochastic processes’, Prob. Theory and Related Fields 78 (1988) 127-134. [465U.] Talagrand M. [89] ‘A “regular oscillation” property of stable sets of measurable functions’, Contemporary Math. 94 (1989) 309-313. [§465 notes.] Talagrand M. [95] ‘Concentration of measure and isoperimetric inequalities in product spaces’, Publ. ´ Math. Inst. Hautes Etudes Scientifiques 81 (1995) 73-205. [492D, 492E.] Talagrand M. [96] ‘The Glivenko-Cantelli problem, ten years later’, J. Theoretical Probability 9 (1996) 371-384. [§465 notes.] Tamanini I. & Giacomelli C. [89] ‘Approximation of Caccioppoli sets, with applications to problems in image segmentation’, Ann. Univ. Ferrara, Sez. VII (N.S.) 35 (1989) 187-213. [484B.] Taylor A.E. [64] Introduction to Functional Analysis. Wiley, 1964. [§4A4.] Ulam S. [30] ‘Zur Masstheorie in der allgemeinen Mengenlehre’, Fund. Math. 16 (1930) 140-150. [419G, 438C.] Uspenskiˇı V.V. [88] ‘Why compact groups are dyadic’, pp. 601-610 in Frol´ık 88. [4A5T.] Veech W.A. [71] ‘Some questions of uniform distribution’, Annals of Math. (2) 94 (1971) 125-138. [491H.] Veech W.A. [77] ‘Topological dynamics’, Bull. Amer. Math. Soc. 83 (1977) 775-830. [493K.] Vinokurov V.G. & Mahkamov B.M. [73] ‘O prostranstvah s soverxenno$ i, no ne kompaktno$ i mero$ i’, Nauqye zapiski Taxkentskogo instituta narodnogo hoz$ istva 71 (1973) 97-103. [451T.] Wage M.L. [80] ‘Products of Radon spaces’, Russian Math. Surveys 35:3 (1980) 185-187. [§434 notes.] Weil A. [40] L’intégration dans les groupes topologiques et ses applications, Hermann, 1940 (Actualités Scientifiques et Industrielles 869). [445N.]

880

Concordance

Wheeler R.F. [83] ‘A survey of Baire measures and strict topologies’, Expositiones Math. 1 (1983) 97-190. [§435 notes, §437 notes.]

Bernoulli

Principal topics and results

881

Index to volumes 1-4 Principal topics and results The general index below is intended to be comprehensive. Inevitably the entries are voluminous to the point that they are often unhelpful. I am therefore preparing a shorter, better-annotated, index which will, I hope, help readers to focus on particular areas. It does not mention definitions, as the bold-type entries in the main index are supposed to lead efficiently to these; and if you draw blank here you should always, of course, try again in the main index. Entries in the form of mathematical assertions frequently omit essential hypotheses and should be checked against the formal statements in the body of the work. absolutely continuous real functions §225 —– as indefinite integrals 225E absolutely continuous additive functionals §232 —– characterization 232B —– on measure algebras 327B actions of groups on topological spaces —– and invariant measures 441C, 443P, 443Q, 443T, 449A —– inducing actions on measure algebras and function spaces 443C, 443G, 444F additive functionals on Boolean algebras §326, §327, §362 —– dominating or subordinate to a given functional 391E, 391F —– on PI §464 almost continuous functions §418 —– and measurable functions 418E, 418J, 451S —– and image measures of Radon measures 418I amenable group §449 analytic spaces §423, §433 —– K-analytic spaces with countable networks 423C —– and Souslin’s operation 423E angelic spaces §462 Archimedean Riesz spaces §353 associate norm (on a dual function space) §369 atomless measure spaces —– have elements of all possible measures 215D automorphisms and automorphism groups —– of Boolean algebras §381, §382, §384 —– —– automorphisms as products of involutions 382M —– —– normal subgroups of automorphism groups 382R, 383I —– —– isomorphisms of automorphism groups 384D, 384E, 384J, 384M —– of measure algebras §383, §385, §386, §387, §388, 493H Baire measures §435 Banach lattices §354 —– order-bounded linear operators are continuous 355C —– duals 356D Banach-Ulam problem 363S band algebra in a Riesz space 352Q barycenter of a measure on a locally convex space §461 —– and convex functions 461C —– existence of barycenters of given measures 461D, 461E, 461G —– existence of measures with given barycenters 461H, 461K, 461M, 461N Becker-Kechris theorem (on Borel measurable actions of Polish groups) 424H Bernoulli shifts §387 —– calculation of entropy 385R —– as factors of given homomorphisms (Sinaˇı’s theorem) 387E, 387L —– specified by entropy (Ornstein’s theorem) 387I, 387K

882

Index

Besicovitch

Besicovitch’s Covering Lemma (for balls in Rr ) 472B, 472C Besicovitch’s Density Theorem (for Radon measures on Rr ) 472D biduals of Riesz spaces 356F, 356I Bochner’s theorem (positive definite functions are Fourier-Stieltjes transforms/characteristic functions) 445N Boolean algebras Chap. 31 Boolean homomorphisms §312 —– between measure algebras §324 —– induced by measurable functions 324A —– continuity and uniform continuity 324F —– when measure-preserving 324K —– represented by measurable functions 343B, 344B, 344E Boolean rings §311 Borel measures §434 Borel sets in Rr 111G —– and Lebesgue measure 114G, 115G, 134F bounded variation, real functions of §224 —– as differences of monotonic functions 224D —– integrals of their derivatives 224I —– Lebesgue decomposition 226C Brownian motion 455C —– as a Radon measure on C([0, ∞[) 455D Cantor set and function 134G, 134H capacity 432I, 432J Carathéodory’s construction of measures from outer measures 113C Carleson’s theorem (Fourier series of square-integrable functions converge a.e.) §286 Cauchy’s Perimeter Theorem 475S Central Limit Theorem (sum of independent random variables approximately normal) §274 —– Lindeberg’s condition 274F, 274G change of measure in the integral 235M change R of variable in R the integral §235 —– J × gφ dµ = g dν 235A, 235E, 235L —– finding J 235O; —– —– J = | det T | for linear operators T 263A; J = | det φ0 | for differentiable operators φ 263D —– —– —– when the measures are Hausdorff measures 265B, 265E —– when φ is inverse-measure-preserving 235I R R —– gφ dµ = J × g dν 235T characteristic function of a probability distribution §285 —– sequences of distributions converge in vague topology iff characteristic functions converge pointwise 285L Choquet’s theorem (analytic sets are capacitable) 432J Choquet’s theorem (on measures on sets of extreme points) 461K closed subalgebra of a measure algebra 323H, 323J —– classification theorem 333H, 333K —– defined by a group of automorphisms 333R compact measure space §342, §343, §451 —– and representation of homomorphisms between measure algebras 343B Compactness Theorem (for sets with bounded perimeter) 474S complete measure space §212 completion of a measure 212C et seq. concentration of measure 264H, §476, §492 conditional expectation —– of a function §233 —– as operator on L1 (µ) 242J

direct


construction of measures —– image measures 112E —– from outer measures (Carathéodory’s method) 113C —– subspace measures 131A, 214A —– product measures 251C, 251F, 251W, 254C —– from inner measures 413C —– extending given functionals or measures 413I, 413J, 413O —– —– yielding a Borel measure 435C —– —– yielding a quasi-Radon measure 415K —– —– yielding a Radon measure 416J, 416M, 432D, 435B —– dominating given functionals 413S —– τ -additive product measures 417C, 417E —– as pull-backs 132G, 418L —– as projective limits of Radon measures 418M —– from integrals §436 —– —– from sequentially smooth functionals 436D —– —– from smooth functionals 436H —– —– yielding Radon measures (Riesz’ theorem) 436J, 436K —– of invariant measures 441C, 441E, 441H, 448P continued fractions 372M et seq. Control Measure Problem §393 convergence theorems (B.Levi, Fatou, Lebesgue) §123 convergence in measure (linear space topology on L0 ) —– on L0 (µ) §245 —– on L0 (A) §367 —– when Hausdorff/complete/metrizable 245E, 367N —– and positive linear operators 367P —– and pointwise compact sets of measurable functions §463, 465G convex functions 233G et seq., §461 Convex Isoperimetric Theorem 475T convolution of functions —– (on Rr ) §255, 473D, 473E —– (on groups) §444 R general topological R —– h × (f ∗ g) = h(x + y)f (x)g(y)dxdy 255G, 444O —– f ∗ (g ∗ h) = (f ∗ g) ∗ h 255J, 444O convolution of measures —– (on Rr ) §257 —– (on groups) §444 R general topological R —– h d(ν1 ∗ ν2 ) = h(x + y)ν1 (dx)ν2 (dy) 257B, 444C —– of absolutely continuous measures 257F, 444Q convolution of measures and functions 444H, 444J countable sets 111F, 1A1C et seq. countable-cocountable measure 211R countably compact measures §451, §452 —– and disintegrations 452I, 452M —– and inverse-measure-preserving functions 452R counting measure 112Bd Dedekind (σ)-complete Boolean algebras §314 —– Riesz spaces §353 Dedekind completion of a Boolean algebra 314T differentiable functions (from Rr to Rs ) §262, §263 direct sum of measure spaces 214K

883

884

Index

disintegration

disintegration of measures §452 —– of one Radon measure over another, using a strong lifting to concentrate on fibers 453K —– of Poisson point processes 495I —– of symmetric measures on product space 459E, 459G distribution of a finite family of random variables §271 —– as a Radon measure 271B X ) 271J —– of φ(X —– of an independent family 272G —– determined by characteristic functions 285M —– of an infinite family of random variables 454J distributive laws in Boolean algebras 313B —– in Riesz spaces 352E divergence of a vector field §474 Divergence Theorem 475N, 484N Doob’s Martingale Convergence Theorem 275G duals —– of topological groups §445 —– of Riesz spaces §356, §362, 369C, 369J —– of Banach lattices 356D Duality Theorem (for abelian locally compact Hausdorff groups) 445U Dye’s theorem (on the full subgroup generated by a measure-preserving automorphism) 388K, 388L effectively regular measure 491N entropy (of a partition or a function) §385, §386 —– calculation (Kolmogorov-Sinaˇı theorem) 385P; (Bernoulli shifts) 385R equidistributed sequences §491 Ergodic Theorem §372 —– (Pointwise Ergodic Theorem) 372D, 372E —– (Maximal Ergodic Theorem) 372C essential boundary (of a set in Rr ) §475 exchangeable random variables 459B, 459C exhaustion, principle of 215A extended real line §135 extension of measures 413K, 413O, 433J, 435C, 442Yc, 464D —– yielding quasi-Radon measures 415L, 415M, 415N —– yielding Radon measures 416F, 416M, 416N, 416O, 416P, 416Q, 432D, 432F, 433I, 435B extension of order-continuous positive linear operators 355F, 368B, 368M extremely amenable topological group §493 R R Fatou’s Lemma ( lim inf ≤ lim inf for sequences of non-negative functions) 123B Federer exterior normal 474Q Féjer sums (running averages of Fourier sums) converge to local averages of f 282H —– uniformly if f is continuous 282G de Finetti’s theorem 459C Fourier series §282 —– norm-converge in L2 282J —– converge at points of differentiability 282L —– converge to midpoints of jumps, if f of bounded variation 282O —– and convolutions 282Q —– converge a.e. for square-integrable function 286V Fourier transforms —– on R §283, §284 —– on abelian topological groups §445 Rd ∧ —– formula for c f in terms of f 283F —– and convolutions 283M, 445G

indefinite


885

—– as a multiplicative operator 445K —– in terms of action on test functions 284H et seq. —– of square-integrable functions 284O, 286U, 445R —– inversion formulae for differentiable functions 283I; for functions of bounded variation 283L; for positive definite functions 445P ∧ R∞ 2 —– f (y) = lim²↓0 √12π −∞ e−iyx e−²x f (x)dx a.e. 284M free products of Boolean algebras §315 —– universal mapping theorem 315I free products of measure algebras §325 —– universal mapping theorems 325D, 325J Fremlin’s Alternative (for sequences of measurable functions) 463H Fubini’s R theorem §252 RR —– f d(µ × ν) = f (x, y)dxdy 252B —– when both factors σ-finite 252C, 252H —– for characteristic functions 252D, 252F —– for τ -additive product measures 417H function spaces §369, §374, §376 Rx Rb d Fundamental Theorem of the Calculus ( dx f = f (x) a.e.) §222, 483I; ( a F 0 (x)dx = F (b) − F (a)) a 225E, 483R Gagliardo-Nirenberg-Sobolev inequality 473H gauge, gauge integral §481, §482 Gaussian distributions §456 —– are τ -additive 456K Haar measure (on topological groups) Chap. 44 —– construction 441E —– uniqueness 442B —– action of the group on the measure algebra 443C —– —– and on L0 443G —– derivable from Haar measure on locally compact groups 443L —– completion regular 443M —– has a translation-invariant lifting 447J —– —– which is a strong lifting 453B Hahn decomposition of a countably additive functional 231E —– on a Boolean algebra 326I Hahn-Banach property see Nachbin-Kelley theorem (363R) Hardy-Littlewood Maximal Theorem 286A Hausdorff measures (on Rr ) §264 —– are topological measures 264E —– r-dimensional Hausdorff measure on Rr a multiple of Lebesgue measure 264I —– on general metric spaces §471 —– Increasing Sets Lemma 471G —– inner regularity properties 471D, 471I, 471S —– density theorem 471P —– need not be semi-finite 439H —– (r − 1) dimensional measure on Rr 265F-265H, Chap. 47 Henry’s theorem (on extending measures to Radon measures) 416M Henstock integral §483 image measures 112E —– of Radon measures 418I —– of countably compact measures 452R indefinite integrals —– differentiate back to original function 222E, 222H, 483I

886

Index

indefinite

—– to construct measures §234 —– Saks-Henstock indefinite integrals 482B-482D, 484J independent random variables §272 —– joint distributions are product measures 272G —– limit theorems §273, §274 inner measures 413A et seq. inner regularity of measures §412 —– (with respect to closed sets) Baire measures 412D; Borel measures on metrizable spaces 412E —– (with respect to compact sets) Lebesgue measure 134F; Radon measures §256, §416 —– in complete locally determined spaces 412J —– in product spaces 412S, 412U, 412V —– and Souslin’s operation 431D integral operators §376 —– represented by functions on a product space 376E —– characterized by action on weakly convergent sequences 376H —– into M -spaces or out of L-spaces 376M —– and weak compactness 376P —– disintegrated 376S integration of real-valued functions, construction §122 —– as a positive linear functional 122O —– —– acting on L1 (µ) 242B —– by parts 225F —– characterization of integrable functions 122P, 122R —– over subsets 131D, 214E —– functions and integrals with values in [−∞, ∞] §133 —– measures derived from integrals §436 integration of families of measures §452 invariant means on topological groups 449E, 449I invariant measures §441, §448 —– on measure algebras 394P —– on locally compact spaces, under continuous group actions 441C, 443P, 443Q, 443T —– on locally compact groups (Haar measures) 441E —– on locally compact metric spaces, under isometries 441H —– on Polish groups 448P isoperimetric theorems 474L, 475T Jensen’s inequality 233I-233J —– expressed in L1 (µ) 242K Kakutani’s theorem (on the representation of L-spaces as L1 -spaces) 369E Kolmogorov’s extension theorem (for a consistent family of probability measures on finite subproducts) 454D Kolmogorov-Sinaˇı theorem (on the entropy of a measure-preserving homomorphism) 385P Komlós’ subsequence theorem 276H Kwapien’s theorem (on maps between L0 spaces) 375I Lebesgue’s Density Theorem (in R) §223 R x+h —– limh↓0 h1 x f = f (x) a.e. 223A R x+h 1 —– limh↓0 2h |f (x − y)|dy = 0 a.e. 223D x−h —– (in Rr ) 261C, 261E —– (for Hausdorff measure) 471P —– (in topological groups with B-sequences) 447D Lebesgue measure, construction of §114, §115 —– further properties §134 —– characterized as measure space 344I

measurable


887

R R Lebesgue’s Dominated Convergence Theorem ( lim = lim for dominated sequences of functions) 123C, 367J R R R P PR B.Levi’s theorem ( lim = lim for monotonic sequences of functions) 123A; ( = ) 226E lifting (of a measure) —– complete strictly localizable spaces have liftings 341K —– translation-invariant liftings on Rr and {0, 1}I §345 —– respecting product structures §346 —– strong liftings §453 —– Haar measures have translation-invariant liftings 447J Lipschitz functions §262 —– differentiable a.e. 262Q localizable measure space —– assembling partial measurable functions 213N —– which is not strictly localizable 216E locally finite perimeter (for sets in Rr ) §474, §475 locally uniformly rotund norm §467 Loomis-Sikorski representation of a Dedekind σ-complete Boolean algebra 314M lower density (on a measure space) —– strictly localizable spaces have lower densities 341H —– respecting a product structure 346G Lusin’s theorem (measurable functions are almost continuous), 418J —– (on Rr ) 256F —– (on Radon measure spaces) 451S Lusin’s theorem (injective continuous images of Polish spaces are Borel) 423I Maharam’s theorem —– (homogeneous probability algebras of Maharam type κ are isomorphic) 331I —– —– to measure algebra of {0, 1}κ ) 331L —– classification of localizable measure algebras 332J Maharam type —– (calculation of) 332S —– —– for measure algebras of qproduct spaces 334A, 334C —– relative to a subalgebra 333E Markov processes §455 Q —– as Baire measures on t∈T Xt defined from conditional distributions 455A martingales §275 —– L1 -bounded martingales converge a.e. 275G, 367K —– when of form E(X|Σn ) 275H, 275I measurable algebra §391 —– characterized by weak (σ, ∞)-distributivity and existence of strictly positive additive functional 391D; by existence of uniformly exhaustive strictly positive Maharam submeasure 392J —– Control Measure Problem §393 measurable envelopes —– elementary properties 132E —– existence 213L measurable functions —– (real-valued) §121 —– —– sums, products and other operations on finitely many functions 121E —– —– limits, infima, suprema 121F —– into general topological spaces §418 —– and almost continuous functions 418E, 418J, 451S —– as selectors 423M, 433F-433G measurable sets —– and Souslin’s operation 431A

888

Index

measure

measure algebras —– properties related to those of measure spaces 322B —– relationships between elementary properties 322C —– satisfying the countable chain condition 322G —– topologies and uniformities on §323, §324; uniqueness of topology and uniformity 324H measure-free cardinals §438 measure-preserving Boolean homomorphisms —– extension from a closed subalgebra 333C modular function (of a topological group carrying Haar measures) 442I —– and calculation of integrals 442K —– of a subgroup 443Q —– of a quotient group 443S Monotone Class Theorem 136B monotonic functions —– are differentiable a.e. 222A Nadkarni-Becker-Kechris theorem (on measures invariant under Polish group actions) 448P narrow topologies on spaces of signed measures 437J et seq. von Neumann-Jankow selection theorem 423N non-measurable set (for Lebesgue measure) 134B non-paradoxical group (of automorphisms of a Boolean algebra) §394, §448 —– characterizations 394D, 448D —– invariant functionals 394O, 394P, 395B, 448P order-bounded linear operators §355 order-closed ideals and order-continuous Boolean homomorphisms 313P order-continuous linear operators §355 order*-convergence of a sequence in a lattice §367 order-dense sets in Boolean algebras 313K —– in Riesz spaces 352N Ornstein’s theorem (on isomorphism of Bernoulli shifts with the same entropy) 387I, 387K outer measures constructed from measures §132 —– elementary properties 132A outer regularity of Lebesgue measure 134F partially ordered linear spaces §351 perfect measure space 342K et seq., §451 —– and pointwise compact sets of measurable functions 463I perimeter measure (for sets in Rr ) §474, 475G Pfeffer integral §484 Plancherel Theorem (on Fourier series and transforms of square-integrable functions) 282K, 284O, 445R Poincaré’s inequality 473K pointwise convergence (in a space of functions) §462, §463, 465D, 465G —– (in an isometry group) 441G Poisson point processes §495 —– and disintegrations 495I Poisson’s theorem (a count of independent rare events has an approximately Poisson distribution) 285Q product of two measure spaces §251 —– basic properties of c.l.d. product measure 251I —– Lebesgue measure on Rr+s as a product measure 251M —– more than two spaces 251W —– Fubini’s theorem 252B, 417H —– Tonelli’s theorem 252G, 417H —– and L1 spaces §253 —– continuous bilinear maps on L1 (µ) × L1 (ν) 253F

representation


889

—– conditional expectation on a factor 253H —– and measure algebras 325A et seq. —– of τ -additive measures 417C; of quasi-Radon measures 417N; of Radon measures 417P —– of compact, countably compact and perfect measures 451I product of any number of probability spaces §254 —– basic properties of the (completed) product 254F —– characterization with inverse-measure-preserving functions 254G —– products of subspaces 254L —– products of products 254N —– determination by countable subproducts 254O —– subproducts and conditional expectations 254R —– and measure algebras 325I et seq. —– having homogeneous measure algebras 334E —– of separable metrizable spaces 415E; of compact metrizable spaces 416U —– of τ -additive measures 417E; associative law 417J; of quasi-Radon measures 415E, 417O; of Radon measures 416U, 417Q —– of compact, countably compact and perfect measures 451J Prokhorov’s theorem (on projective limits of Radon measures) 418M quasi-Radon measures §415 —– are strictly localizable 415A —– subspaces are quasi-Radon 415B —– specification and uniqueness 415H —– construction 415K —– —– from smooth linear functionals 436H —– products 415E, 417N, 417O —– on locally convex spaces §466 Rademacher’s theorem (Lipschitz functions are differentiable a.e.) 262Q Radon measures, §416 —– on Rr §256 —– as completions of Borel measures 256C, 433C —– indefinite-integral measures 256E, 416S —– image measures 256G, 418I —– product measures 256K, 416U, 417P, 417Q —– extending given functionals or measures 416J, 416M-416Q, 432D —– subspaces 416R —– and Stone spaces of measure algebras 416V —– and measurable functions 451S —– as pull-backs 418L, 433D —– and disintegrations 453K —– Haar measures 441E; other invariant measures 441H, 448P —– on locally convex spaces §466 Radon-Nikod´ ym theorem (truly continuous additive set-functions have densities) 232E —– in terms of L1 (µ) 242I —– on a measure algebra 327D, 365E rearrangements (monotonic rearrangements of measurable functions) §373, §374 regular open algebras 314P, §513 representation —– of Boolean algebras see Stone’s theorem (311E) —– of Dedekind σ-complete Boolean algebras 314M —– of measure algebras 321J —– of homomorphisms between measure algebras §343, §344 —– of linear functionals as integrals §436 —– of partially ordered linear spaces 351Q

890

Index

representation

—– of Riesz spaces 352L, 353M, 368E et seq., 369A —– of Riesz space duals 369C —– of L-spaces 369E —– of M -spaces 354L representations of topological groups 446B Riesz representation theorem (construction of a Radon measure from an integral) 436J, 436K Riesz spaces (vector lattices) §352 —– identities 352D, 352F, 352M Saks-Henstock lemma 482B —– indefinite integrals 482D, 484J Shannon-McMillan-Breiman theorem (entropy functions of partitions generated by a measure-preserving homomorphism converge a.e.) 386E signed measures 437C, 437H simple product of Boolean algebras —– universal mapping theorem 315B —– of measure algebras 322K Sinaˇı’s theorem (on existence of Bernoulli partitions for a measure-preserving automorphism) 387E, 387L Souslin’s operation §421 —– on measurable sets §431 split interval 343J, 419L stable set of functions §465 —– is relatively pointwise compact 465D —– pointwise convergence and convergence in measure 465G —– uniform strong law of large numbers 465M —– in L0 465P —– in L1 465R —– R-stable set 465S Stone space of a Boolean ring or algebra 311E et seq. —– as topological space 311I —– and Boolean homomorphisms 312P, 312Q —– of a Dedekind complete algebra 314S —– and the countable chain condition 316B —– and weak (σ, ∞)-distributivity 316I —– of a measure algebra 321J, 322N, 322Q —– of a Radon measure space 416V Stone-Weierstrass theorem §281 —– for Riesz subspaces of Cb (X) 281A —– for subalgebras of Cb (X) 281E —– for *-subalgebras of Cb (X; C) 281G —– for *-subalgebras of C0 (X; C) 4A6B strictly localizable measures —– sufficient condition for strict localizability 213O —– complete locally determined effectively locally finite τ -additive topological measures are strictly localizable 414J Pn 1 strong law of large numbers (limn→∞ n+1 i=0 (Xi − E(Xi )) = 0 a.e.) §273 P∞ 1 —– when n=0 Var(Xn ) < ∞ 273D 2 (n+1)

—– when supn∈N E(|Xn |1+δ ) < ∞ 273H —– for identically distributed Xn 273I —– for martingale difference sequences 276C, 276F —– convergence of averages for k k1 , k kp 273N —– and functions on product spaces 465H —– nearly uniformly on stable sets 465M strong liftings §453

L


submeasure §392 subspace measures —– for measurable subspaces §131 —– for arbitrary subspaces §214 surface measure in Rr §265 symmetric group 449Xc, 492H, 493Xc symmetric measures on product spaces 459E, 459G tagged-partition structures §481, §482 Talagrand’s measure §464 Tamanini-Giacomelli theorem 484B tensor products of L1 spaces §253, 376C RR Tonelli’s theorem (f is integrable if |f (x, y)|dxdy < ∞) 252G, 417H topological groups chap. 44 —– amenable groups §449 —– duality theorem §445 —– extremely amenable groups §493 —– Haar measure §441, §442, §443 —– structure theory §446 uniformly exhaustive submeasures —– and additive functionals 392F uniformly integrable sets in L1 §246 —– criteria for uniform integrability 246G —– in L-spaces 354P et seq. —– and convergence in measure 246J —– and weak compactness 247C, 356Q —– in dual spaces 356O, 362E universally measurable sets and functions §434 vague topologies on spaces of signed measures 437J et seq. Vitali’s theorem (for coverings by intervals in R) 221A —– (for coverings by balls in Rr ) 261B —– (in general metric spaces) 471O —– (in topological groups with B-sequences) 447C weak compactness in L1 (µ) §247 —– in L-spaces 356Q —– in C0 (X) 462E weak topologies of locally convex spaces §466 weakly compactly generated normed spaces 467L weakly (σ, ∞)-distributive Boolean algebras §316 —– and measure algebras 322F, 391D —– Riesz spaces 368Q Weyl’s Equidistribution Theorem 281N c.l.d. version 213D et seq. C(X) §436, §462 Cb (X) §436, §437 K-analytic spaces §422, §432 —– basic properties 422G —– and Souslin-F sets 422H —– are capacitable 432B, 432J K-countably determined spaces 467H et seq L-space 354N et seq., §371 L0 (µ) (space of equivalence classes of measurable functions) §241 —– as Riesz space 241E

891

892

Index

L0 (A) (based on a Boolean algebra) §364, §368, §369, §375 —– as a quotient of a space of functions 364D —– algebraic operations 364E —– calculation of suprema and infima 364M —– linear operators induced by Boolean homomorphisms 364R —– *when A is a regular open algebra 364U —– —– *of a compact extremally disconnected space 364W —– as the domain of a linear operator 375A, 375C —– positive linear operators between L0 spaces can be derived from Riesz homomorphisms 375I L1 (µ) (space of equivalence classes of integrable functions) §242 —– norm-completeness 242F —– density of simple functions 242M —– (for Lebesgue measure) density of continuous functions and step functions 242O L1 (A, µ ¯) (based on a measure algebra) §365 —– related to L1 (µ) 365B —– as L-space 365C, 369E —– linear operators induced by Boolean homomorphisms 365H, 365O, 365P —– duality with L∞ 365M —– universal mapping theorems 365I, 365K —– stable sets 465R Lp (µ) (space of equivalence classes of pth-power-integrable functions, where 1 < p < ∞) §244 —– is a Banach lattice 244G —– has dual Lq (µ), where p1 + 1q = 1 244K —– and conditional expectations 244M Lp (A) (based on a Boolean algebra) §366 —– related to Lp (µ) 366B —– as Banach lattice 366C —– linear operators induced by Boolean homomorphisms 366H L∞ (µ) (space of equivalence classes of bounded measurable functions) §243 —– duality with L1 243F-243G —– norm-completeness 243E —– order-completeness 243H L∞ (A) (where A is a Boolean algebra) §363 —– universal mapping theorems 363E —– linear operators induced by Boolean homomorphisms 363F —– order-completeness 363M —– Nachbin-Kelley theorem (generalized Hahn-Banach theorem) 363R `∞ and its dual §464 L∼ (U ; V ) (space of order-bounded linear operators) §355, 371B, 371C L× (U ; V ) (space of differences of order-continuous linear operators) §355, 376E M -space 354G et seq. S(A) (space of ‘simple’ functions corresponding to a Boolean ring A) §361 —– universal mapping theorems 361F-361I —– linear operators induced by ring homomorphisms 361J —– dual spaces §362 T , T (0) , T × (classes of operators which are norm-reducing for every k kp ) 371F-371G, §372, §373, §374 usco-compact relations §422 σ-algebras of sets §111 —– generated by given families 136B, 136G σ-finite measures 215B τ -additive measures §414, §417 —– and semi-continuous functions 414A, 414B —– products

L

almost

General index

893

—– —– of two τ -additive measures 417C —– —– of many τ -additive probability measures 417E —– on quasi-dyadic spaces 434Q {0, 1}κ —– usual measure is (Maharam) homogeneous with Maharam type κ 331K; is homogeneous in strict sense 344L; is a completion regular Radon measure 416U

General index References in bold type refer to definitions; references in italics are passing references. Definitions marked with > are those in which my usage is dangerously at variance with that of some other author. Abel’s theorem 224Yi Abel-Poisson kernel 282Yg abelian topological group 441Ia, 444D, 444Og, 444Sb, 444Xd, 444Xk, §445, 446Xb, 449Cf, 491Xn absolute summability 226Ac absolutely continuous additive functional 232Aa, 232B, 232D, 232F-232I, 232Xa, 232Xb, 232Xd, 232Xf, 232Xh, 234Ce, 256J, 257Xf, 327A, 327B, 327C, 362C, 362Xh, 362Xi, 363S, 414D, 414Xd, 438Xb absolutely continuous function 225B, 225C-225G, 225K-225O, 225Xa-225Xh, 225Xn, 225Xo, 225Ya, 225Yc, 232Xb, 233Xc, 244Yh, 252Yj, 256Xg, 262Bc, 263I, 264Yp, 265Ya, 282R, 282Yf, 283Ci absolutely continuous submeasure 393D, 393E, 393H, 393Yg Absoluteness Theorem see Shoenfield’s Absoluteness Theorem abstract integral operator 376D, 376E, 376F, 376H, 376J, 376K, 376M, 376P, 376Xc, 376Xk, 376Yh376Yj, 376Yl, 376Yn accumulation point see cluster point (3A3La) action of a group on a set 441A, 441C, 441Ya, 442Xe, 444Xe, 448Xi, 449J, 449L, 449Xr, 4A5B, 4A5C —– Borel measurable action 424H, 448P, 448Xf, 4A5I —– continuous action 424H, 441B, 441Ga, 441Xa, 441Yc, 441Yj, 443C, 443G, 443O-443Q, 443T, 443Xd, 443Yb, 444F, 449A, 449B, 449D, 461Yc, 471Xc, 471Xj, 4A5I, 4A5Jc —– of a semigroup 449Ya —– see also conjugacy action (4A5Ca), left action (4A5Ca), right action (4A5Ca), transitive action (4A5Bb) additive functional on an algebra of sets see finitely additive (136Xg, 231A), countably additive (231C) additive function(al) on a Boolean algebra see finitely additive (326A, 361B), countably additive (326E), completely additive (326J) additivesee also τ -additive (411C) adherence point see cluster point (3A3La) adjoint operator 243Fc, 243 notes, 371Xe, 371Yd, 373S-373U, 373Xu, 373Yg, 3A5Ed, 437Xd affine subspace 476Xk algebra (over R) 361Xb, 361Xf; see also algebra of sets (231A), Baire property algebra (314Yd), Banach algebra (2A4Jb, 4A6Ab), Boolean algebra (311A), category algebra (314Yd), Jordan algebra (491Ye), normed algebra (2A4J, 4A6Aa), f -algebra (352W), regular open algebra (314Q) algebra of sets 113Yi, 136E, 136F, 136G, 136Xg, 136Xh, 136Xk, 136Ya, 136Yb, 231A, 231B, 231Xa, 311Bb, 311Xb, 311Xh, 312B, 315G, 315L, 362Xg, 363Yf, 381Xa, 475M, 4A3Cg; see also Boolean algebra (311A), σ-algebra (111A) algebraic dual 4A4Ac, 4A4Cg Alexandroff compactification see one-point compactification (3A3O) almost continuous function 256F, 411M, 411Ya, 418D-418K, 418Nb, 418P, 418Q, 418Sb, 418Tb, 418Xc418Xn, 418Xp, 418Xq, 418Xs-418Xu, 418Ye-418Yh, 419Xe, 419Xh, 432Xe, 433E, 434Yb, 438F, 438G, 443Pf, 451S, 452Xe, 453K, 454N almost every, almost everywhere 112Dd; see also Haar almost everywhere (443Ae) almost isomorphic (inverse-measure-preserving functions) 385U, 385V, 385Xn-385Xp, 385Yf almost Lindelöf see measure-compact (435D)

894

Index

almost

almost strong lifting 453A, 453D-453G, 453K, 453Xf, 453Xi, 453Ya, 453Zb; see also strong lifting (453A) almost surely 112De ambit see greatest ambit (449D) amenable group 394 notes, > 449A, 449C, 449E-449G, 449I, 449K, 449L 449Xa-449Xe, 449Xg-449Xl, 449Xn-449Xt, 449Yb, 449Yf, 461Yc, 493Ya; see also extremely amenable (493A) amenable semigroup 449Ya, 449Yh analytic (complex) function 133Xc analytic (topological) space 423A, 423B-423O, 423Xa-423Xh, 423Yb-423Yd, 424Xb, 424Xd, 433C-433I, 433Xb, 433Xc, 433Yb, 433Yd, 434B, 434Dc, 434Ka, 434Xd, 434Xi, 434Ya, 471F, 471I, 471S, 471Xc, 471Xh, 471Yc, 439Yc, 452N, 452P, 453G, 454O, 454P, 494A; see also K-analytic (422F) angelic topological space 2A5J, 462Aa, 462B-462D, 462Xa, 462Yb antichain 316 notes aperiodic Boolean homomorphism 381Ad, 381H, 381Nf, 381P, 381Xc, 381Xj, 381Xl, 382J, 386C, 386D, 386Ya, 388F, 388H, 388J, 388K, 388Yb; see also nowhere aperiodic (381Xk) approximately continuous Henstock integral 481Q Archimedean partially ordered linear space 351R, 351Xe, 361Gb Archimedean Riesz space 241F, 241Yb, 242Xc, 353A-353F, 353Ha, 353L-353P, 353Xa, 353Xb, 353Xd, 353Ya, 353Yd-353Yh, 354Ba, 354F, 354I-354K, 354Yi, 355Xd, 356G-356I, 361Ee, 367C, 367F, 367Xh, 367Ya, 367Yf, 368B, 368E-368G, 368I, 368J, 368M, 368O, 368P, 368R, 368Ya, 368Yb area see surface measure arrow (‘double arrow space’, ‘two arrows space’) see split interval (343J) associate extended Fatou norm 369H, 369I-369L, 369O, 369R, 369Xd, 374B, 374C, 374Xh, 376S asymptotic density asymptotic density 273G, 464Jb, 491A, 491I, 491Ka, 491Xa, 491Xb, 491Xd, 491Xh, 491Xy, 491Ya, 491Ye; see also upper asymptotic density (491A) asymptotic density ideal 491A, 491I, 491J; see also Z asymptotically equidistributed see equidistributed (281Yi, 491B) atom (in a Boolean algebra) 316L, 316M, 316Xr-316Xt, 316Xy, 316Yq, 322Bf, 324Ye, 331H, 333Xb, 343Xe, 362Xe; see also relative atom (331A) atom (in a measure space) 211I, 211Xb, 246G, 322Bf, 414G, 415Xa, 416Xa atom (in a Riesz space) 362Xe atomic see purely atomic (211K, 316Lc) atomless Boolean algebra 316Lb, 316Mb, 316Xr, 316Xs, 316Xu-316Xw, 316Yl-316Yn, 316Yp, 316Yq, 322Bg, 322Kc, 324Kf, 324Ye, 331Xk, 332I, 332P, 332Ya, 375B, 375Xb, 375Yb, 381P, 382Q, 384E, 384F, 384Xb, 386D, 386M, 387Cd, 387C-387H, 387L, 392Xg; see also relatively atomless (331A) atomless additive functional 326Ya, 326Yb-326Yd, 326Yf, 362Xf, 362Xg, 362Yi atomless measure algebra 331C, 369Xh, 374Xl, 383G, 383H, 383J, 383Xg, 383Xh, 383Ya, 384Ld, 384M, 384O, 384P, 384Xd, 493E, 493H, 493Ya, 493Yb atomless measure (space) 211J, 211Md, 211Q, 211Xb, 211Yd, 211Ye, 212G, 213He, 214Xe, 215D, 215Xe, 215Xf, 216A, 216Ya, 234Ff, 234Yd, 251T, 251Wo, 251Xq, 251Xs, 251Yc, 252Yp, 252Yr, 252Yt, 254V, 254Yf, 256Xd, 264Yg, 322Bg, 325Ye, 342Xc, 343Cb, 343Xi, 344I, 415Xa, 417Xn, 419Xc, 433Xf, 433Yb, 439D, 439Xh, 451Xb, 471Dg, 493F, 494B, 494C, 494Xb-494Xd, 494Ya, 495D, 495F-495I, 495L-495N, 495Xa-495Xd atomless vector measure 326Ye automorphism see Banach lattice automorphism, Boolean automorphism, Borel automorphism,cyclic automorphism,ergodic automorphism, group automorphism (3A6B), induced automorphism (381M), inner automorphism (3A6B), measure-preserving Boolean automorphism, measure-preserving ring automorphism, measure space automorphism, outer automorphism (387B) automorphism group of a Boolean algebra 381A, 381B, 381Xa, 382O-382S, 382Xg, 382Yb, 383Xh, 383Ya, 383Yb, §384, §394; of a measure algebra 366Xh, 374J, 381Xj, §§383-384, 394F, 394R, 493G, 493H, 493Xe, 493Yc, 493Yd axiom see Banach-Ulam problem, choice (2A1J), constructibility, continuum hypothesis (4A1Ad), countable choice, Jensen’s ♦, Ostaszewski’s ♣ (4A1M) axiom of countability (for topological spaces) see first-countable (4A2A), second-countable (4A2A)

bilateral

General index

895

Baire measurable function 417Be, 437Yg, 4A3Ke Baire measure 411K, 411Xf, 411Xh, 412D, 415N, 416Xk, 432F, §435, 436E, 436F, 436Xj, 436Yc, 437E, 452Xc, 454B, 454M, 454P, 455Xa, 461M, 491Mc, 491Yd; see also signed Baire measure (437G) Baire property (for subsets of a topological space) 314Yd, 4A3Q, 4A3Ra Baire property algebra 314Yd, 316Yi, 341Yb, 367Yj, 414Yd, 431F, 466Xh, 4A3Q, 4A3R Baire property envelope 431Fa Baire set 412Yd, 417Be, 417Xu, 421L, 422Xd, 422Xe, 434Hc, 434Pd, 435H, 4A3K Baire space 314Yd, 341Yb, 364Yi, 364Yl, 367Yh, 367Yj, 414Yd, 4A3Ra Baire’s theorem 3A3G, 4A2Ma Baire σ-algebra 254Xs, 341Yc, 341Zb, 343Xc, 344E, 344Ya, 344Yc-344Ye, 411R, 415Xp, 417U, 417V, 421Xh, 421Yb, 423Db, 434Yf, 435Xa, 435Xb, 437E, 439A, 443Yg, 452N, 461Xg, 4A3K, 4A3L-4A3P, 4A3U, 4A3V, 4A3Xb-4A3Xd, 4A3Xf, 4A3Xg, 4A3Yb ball (in Rr ) 252Q, 252Xh, 474L; (in other metric spaces) 466 notes, 4A2Lj; see also sphere Banach algebra > 2A4Jb, 363Xa, 444E, 444S, 444Xu, 444Yb, 444Yc, 445H, 446A, 4A6Aa, 4A6F; see also commutative Banach algebra (4A6Aa), unital Banach algebra (4A6Ab) Banach function space §369, §374 Banach lattice 242G, 242Xc, 242Yc, 242Ye, 243E, 243Xb, 326Yj, 354A, 354C, 354Ee, 354L, 354Xa, 354Xb, 354Xe-354Xj, 354Xn-354Xp, 354Yd, 354Yi-354Yl, 355C, 355K, 355Xb, 355Ya, 356D, 356M, 356Xc, 356Xk, 356Yg, 356Yh, 356Yi, 363E, 365K, 366C, 366Xd, 367D, 367P, 369B, 369G, 369Xe, 371B-371E, 371Xa, 371Xc, 375Xb, 376C, 376M, 376Xg, 376Yj, 467N; see also extended Fatou norm, L-space (354M), Lp , M -space (354Gb), Orlicz space Banach lattice automorphism 366Xh Banach-Mazur game 451U Banach space 231Yh, 262Yi, 2A4D, 2A4E, 2A4I, 326Yk, 354Yk, 3A5H, 3A5I, 466B, 466F, 466L, 466M, 467F, 467G, 467Ic, 467M, 467Xg, 467Xh, 483Yg, 4A4I; see also Banach algebra (2A4Jb), Banach lattice (242G), separable Banach space Banach-Tarski paradox 449Yg Banach-Ulam problem 232Hc, 326 notes, 363S, 376Yg, 439J, 439L; see also measure-free cardinal (438A) band in a Riesz space 352O, 352P-352V, 352Xe, 352Xf, 352Xj, 353B, 353C, 353E, 353H, 353I, 354Bd, 354O, 355H, 355I, 356B, 356L, 362B, 362C, 362Xf-362Xi, 364Xn, 364Xo, 414Yb, 436L, 436Xr, 437A, 437F, 437Xq; see also complemented band, complement of a band, principal band, projection band band algebra (of an Archimedean Riesz space) 353B, 353D, 356Yc, 356Yd, 361Yc, 362Ya, 362Yb, 365S, 365Xm, 366Xb, 368E, 368R; see also complemented band algebra (352Q), projection band algebra (352S) band projection 352R, 352Sb, 352Xm, 355Yh, 356C, 356Xe, 356Xf, 356Yb, 362B, 362D, 362Ye, 362Yi barycenter 461Aa, 461B-461E, 461G, 461H, 461J, 461K, 461M, 461N, 461Xb-461Xd, 461Xj, 461Yb, 464Xb base for a filter 4A1Ia base for a topology 3A3M, 4A2Ba, 4A2Fc, 4A2Lg, 4A2O, 4A2Pa —– see also π-base (4A2A) base of neighbourhoods 4A2A, 4A2Gd basically disconnected topological space 314Yf basis see Hamel basis (4A4Aa), orthonormal basis (4A4Ja) Becker-Kechris theorem 424H; see also Nadkarni-Becker-Kechris theorem A.Bellow’s problem 463Za Bernoulli partition 387A, 387B, 387D-387H, 387J —– shift 385Q, 385R, 385S, 385Xi-385Xm, 386Xc, 387B, 387I, 387K, 387L, 387Xa, 387Xc, 387Ya Berry-Esséen theorem 274Hc, 285 notes Besicovitch’s Covering Lemma 472A-472C, 472Yb Besicovitch’s Density Theorem 472D, 472Ye bidual of a normed space (U ∗∗ ) 342Yc, 356Xh, 3A5Ga, 437Ib, 4A4If; see also order-continuous bidual bidual of a topological group 445E, 445O, 445U Bienaymé’s equality 272R bilateral uniformity on a topological group 443H, 443I, 443K, 443Xk, 443Yd, 444Xs, 445Ab, 445E, 445Ya, 449Xh, 493Xb, 4A5Hb, 4A5M-4A5O

896

Index

bilinear

bilinear map §253 (253A), 255Xb, 363L, 376B, 376C, 376Ya-376Yc, 436Yf, 437M, 437Xn, 437Ym, 444Sa, 482Xa Bipolar Theorem 4A4Eg Bochner integral 253Yf, 253Yg, 253Yi, 354Yl, 451Yf, 463Yb, 463Yc, 483Yg Bochner’s theorem 285Xr, 445N, 445Xh, 445Xo Boolean algebra Chap. 31 (311Ab), 363Xf; see also algebra of sets, complemented band algebra (352Q), Dedekind (σ-)complete Boolean algebra, measure algebra (321A), projection band algebra (352S), regular open algebra (314Q) Boolean automorphism 363Ye, 366Yh, 374Yd, 381C-381E, 381G, 381I, 381J, 381L, 381Q, 381Xc, 381Xf, 381Xg, 381Xn, 381Xo, 381Yb, 382A, 382B, 382D, 382E, 382G-382M, 382Xa, 382Xc, 382Xd, 384A, 388D, 388Xa, 388Ya, 394Ge, 394Ya, 395A, 424Xk, 442H, 443Af; see also Borel automorphism, cyclic automorphism (381R), induced automorphism (381M), involution, measure-preserving Boolean automorphism, periodic automorphism (381Bc), von Neumann automorphism (388D), weakly von Neumann automorphism (388D) Boolean homomorphism 312F, 312G-312K, 312N, 312P-312T, 312Xe, 312Xg, 312Xj, 312Ya, 313L-313N, 313P-313R, 313Xp, 313Yb, 314I, 314K, 314Xh, 316Yo, §324, 326Be, 332Yc, 343A, 352Ta, 361Xd, 363F, 363Xb, 363Xc, 381B, 381H, 381Xj, 385K; see also aperiodic Boolean homomorphism (381Bd), Boolean isomorphism, measure-preserving Boolean homomorphism, (sequentially) order-continuous Boolean homomorphism, periodic Boolean homomorphism (381Bc) Boolean-independent subalgebras 315Xn Boolean isomorphism 312L, 314I, 332Xj, 344Yb, 366Yh Boolean ring §311 (311Aa), §361 Boolean subalgebra see subalgebra of a Boolean algebra (312A) ‘Boolean value’ (of a proposition) 364Bb Borel algebra see Borel σ-algebra Borel automorphism 419Xh Borel-Cantelli lemma 273K Borel constituent see constituent (423P) Borel isomorphism 254Xs, 419Xh, 423Yc, 424C, 424D, 4A3A Borel lifting 345F, 345Xg, 345Yc Borel measurable function 121C, 121D, 121Eg, 121H, 121K, 121Yf, 134Fd, 134Xd, 134Yt, 135Ef, 135Xc, 135Xe, 225H, 225J, 225Yf, 233Hc, 241Be, 241I, 241Xd, 256M, 364I, 364J, 364Xg, 364Yd, 367Yo, 417Bb, 418A, 423G, 423Ib, 423O, 423Xi, 423Yc, 423Ye, 433D, 434Xi, 437Jd, 437Yg, 471Xf, 443Jb, 444F, 444G, 444Xg, 466Xh, 476Xe, 493Yd, 4A3A, 4A3C, 4A3Dc; see also Borel isomorphism Borel measure >411K, 411Xf-411Xh, 412E, 412Xf, 414L, 414Xj, 415C, 415D, 416F, 416H, 417Xc, 417Xd, 432Xc, 433C, §434, 435C, 435Xa-435Xd, 435Xf, 435Xg, 451Xi, 452Xa, 471S, 438I, 438S, 438Yc, 438Yj, 444E, 466H, 491Xw; (on R) 211P, 216A, 341Lg, 343Xd, 345F; see also signed Borel measure (437G) Borel-measure-compact (topological) space 434Ga, 434H, 434I, 434Na, 434Xj, 434Xk-434Xn, 434Xu, 434Yg, 434Yj, 435Fd, 435Xk, 438J, 438Xk, 438Xo, 438Yd, 439P Borel-measure-complete (topological) space 434Gb, 434I, 434Ka, 434N, 434Xj, 434Xk-434Xn, 434Xr, 434Xs, 434Xu, 438Jc, 438M, 438Xk, 438Xo, 438Ye, 439P, 439Xh, 466H Borel sets in R, Rr 111G, 111Yd, 114G, 114Yd, 115G, 115Yb, 115Yd, 121Ef, 121K, 134F, 134Xc, 135C, 136D, 136Xj, 212Xc, 212Xd, 254Xs, 225J, 264E, 264F, 264Xb, 342Xi, 364H, 364Yc, 474Xb, 475Cc; —– in other topological spaces 411A, 414Ye, 418A, 421Xj, 421Xk, 422J, 422Xf, 422Yc, 422Yd, 423E, 423I, 423K, 423L, 423Xe, 424F, 434Fc, 443Jb, 443Xn, 444F, 444Xg, 451N, 4A3A, 4A3H-4A3J, 4A3Ya Borel space see standard Borel space (424A) Borel σ-algebra (of subsets of Rr ) 111Gd, 114Yg, 114Yh, 114Yi, 121J, 121Xd, 121Xe, 121Yb, 216A, 251L, 364G, 382Yb, 439A; —– (of other spaces) 135C, 135Xb, 256Ye, 271Ya, 314Ye, 333Ya, 411K, 421H, 421Xe-421Xg, 421Xk, 421Xm, 423N, 423Xd, 423Yc, 424Xb, 424Xd-424Xf, 424Yb, 431Xa, 433I, 434Eb, 435Xb, 435Xd, 437H, 443Jb, 443Yh, 452N, 461Xi, 466E, 466Yb, 4A3A, 4A3C-4A3G, 4A3Kb, 4A3N, 4A3R, 4A3V, 4A3Xa, 4A3Xd; see also standard Borel space (424A), Effros Borel structure (424Ya) boundary of a set in a topological space 411Gi, 411Yc, 474Xc, 475C, 475J, 475S, 475T, 475Xa, 4A2A, 4A2Bi; see also essential boundary (475B), reduced boundary (474G)

Chacon

General index

897

bounded bilinear map 253Ab bounded linear operator 253Ab, 253F, 253Gc, 253L, 253Xc, 253Yf, 253Yj, 253Yk, 2A4F, 2A4G-2A4I, 2A5If, 355C, 3A5Ed, 466L, 466M, 466Yc, 4A4Ib, 4A4La; see also (weakly) compact linear operator (3A5K), order-bounded linear operator (355A), B(U ; V ) (2A4F) bounded set (in Rr ) 134E; (in a normed space) 2A4Bc, 3A5H, 4A4Ka; (in a linear topological space) 245Yf, 367Xw, 4A4B; see also order-bounded (2A1Ab) bounded support, function with see compact support (4A2A) bounded variation, function of §224 (224A, 224K), 225Cb, 225M, 225Oc, 225Xh, 225Xn, 225Yc, 226Bc, 226C, 264Yp, 264Yb, 282M, 282O, 283L, 283Xj, 283Xk, 283Xm, 283Xn, 283Xq, 284Xk, 284Yd, 343Yc, 354Xt, 438Xp, 438Yi, 483Xg, 483Yd bounding see ω ω -bounding (316Yg) box product of tagged-partition structures 481P Breiman see Shannon-McMillan-Breiman theorem (386E) Brownian motion 455C, 455D, 455Yc, 456Xg; see also fractional Brownian motion (456Ye) canonical outward-normal function 474G, 474J, 474Q, 475O, 475P; see also Federer exterior normal (474O) Cantor function 134H, 134I, 222Xd, 225N, 226Cc, 262K, 264Xf, 264Yn Cantor measure 256Hc, 256Ia, 256Xk, 264Ym Cantor set 134G, 134H, 134I, 134Xe, 256Hc, 256Xk, 264J, 264Ym, 264Yn, 412Ya, 471Xg, 4A2Uc; see also {0, 1}I cap (in a sphere) 476M capacity 432I, 432J, 432Xf-432Xh; see also Hausdorff capacity (471H) Carathéodory complete measure space see complete (211A) Carathéodory’s method (of constructing measures) 113C, 113D, 113Xa, 113Xd, 113Xg, 113Yc, 114E, 114Xa, 121Yc, 132Xc, 136Ya, 212A, 212Xg, 213C, 213Xa, 213Xb, 213Xd, 213Xf, 213Xg, 213Xj, 213Ya, 214H, 214Xc, 216Xb, 251C, 251Wa, 251Xd, 264C, 264K, 413Xd, 413Xl, 431C, 436Xc, 436Xk, 471C, 471Ya, 452Xi cardinal 2A1Kb, 2A1L, 3A1B-3A1E, 4A1A, 4A1B; see also measure-free cardinal (438A), regular cardinal (4A1Aa) cardinal function (of a Boolean algebra) see cellularity (332D, Maharam type (331F) —– (of a partially ordered set) see cofinality (3A1F) —– (of a topological space) see cellularity (332Xd, density (331Yf), weight (4A2A) Carleson’s theorem 282K remarks, 282 notes, 284Yg, 286U, 286V carry Haar measures (in ‘topological group carrying Haar measures’) 442D, 442Xb, 442Xc, 443A, 443C443F, 443H, 443J, 443K, 443N, 443Xa, 443Xd-443Xf, 443Xi-443Xk, 443Xn, 443Ya, 443Ye, 444J, 444L, 444Xi-444Xk, 444Xm, 444Yi, 445J, 445O, 445Xc, 445Yi, 445Yj, 447A, 447B, 447J, 449Yd, 493Xh; see also Haar measure category algebra (of a topological space) 314Yd, 4A3Rc Cauchy distribution 285Xm, 367Xw, 455Xd, 455Yf, 495Ya Cauchy filter 2A5F, 2A5G, 3A4F Cauchy’s inequality 244E Cauchy’s Perimeter Theorem 475S Cauchy sequence 242Yc, 2A4D, 354Ed, 356Yf, 356Yg, 3A4Fe ˇ Cech-complete space 434Jg, 437Wf, 437Yl, 4A2A, 4A2Gj, 4A2Md, 4A3Ya ˇ ˇ Cech-Stone compactification see Stone-Cech compactification (4A2I) cellularity(of a topological space) 332Xd, 332Xe, 438Ye, 438Yj, ; —– (of a Boolean algebra) §332 (332D), 365Ya, 438Xc, 443Ya; —– see also ccc (316A), magnitude (332Ga) centered Gaussian distribution see Gaussian distribution (456Ab) Central Limit Theorem 274G, 274I-274K, 274Xc-274Xg, 285N, 285Xn, 285Ym Cesaro sums 273Ca, 282Ad, 282N, 282Xn Chacon-Krengel theorem 371 notes, 465 notes

898

Index

chain

chain condition (in Boolean algebras or topological spaces) see ccc (316A), σ-linked (391L), σ-m-linked (391Yh) chain rule for differentiation 473Bc chain rule for Radon-Nikod´ ym derivatives 235Xi change of variable in integration §235, 263A, 263D, 263F, 263G, 263I, 263Xc, 263Xe, 263Yc character (on a topological group) 445Aa, 491Xn characteristic function (of a set) 122Aa —– (of a probability distribution) §285 (285Aa), 445Xh, 454M, 454Xk, 466J, 466K; see also FourierStieltjes transform (445C) —– (of a random variable) §285 (285Ab), 495M charge 326A; see finitely additive functional chargeable Boolean algebra 391D, 391J, 391N, 391X, 391Xa-391Xf, 391Yb, 391Yc, 391Yf, 392F, 394Xd; topological space 417Xu Chebyshev’s inequality 271Xa choice, axiom of 134C, 1A1G, 254Da, 2A1J, 2A1K-2A1M, 361 notes, 441 notes; see also countable choice choice function 2A1J Choquet capacity see capacity Choquet’s theorem (analytic sets are capacitable) 432J Choquet’s theorem (on measures on sets of extreme points) 461K circle group 255M, 255Yh, 345Xb, 345Xg, 443Xt, 445B clopen algebra 311I, 311J, 311Xh, 437Xl, 481Xh closed convex hull 253G, 2A5E, 461H, 461I, 461Xf, 461Xj, 462J, 463Ye, 4A4Eg, 4A4Gb Closed Graph Theorem 466 notes closed interval (in R or Rr ) 114G, 115G, 1A1A; (in general partially ordered spaces) 2A1Ab; (in totally ordered spaces) 4A2A, 4A2Rb closed set (in a topological space) 134Fb, 134Xc, 1A2E, 1A2F, 1A2G, 2A2A, 2A3A, 2A3D, 2A3Nb; (in an ordered space) see order-closed (313D); (spaces of closed sets) see Hausdorff metric (476A), Vietoris topology (476C) closed subalgebra of a measure algebra 323H-323K (323I), 323Xb, 323Xd, 323Yd, 325Xa, 327F, 331B, 331D, 332P, 332T, §333, 334Xb, 364Xe, 366J, 372G, 372J, 385D, 385G, 385Xq, 386L-386N, 386Xf, 386Yb, 386Yc, 387Be, 388I; see also order-closed subalgebra closed subgroup of a topological group 443O-443S, 443Xt, 443Xy, 443Yj, 443Yk, 4A5Jb, 4A5Mc; see also compact subgroup closure of a set 2A2A, 2A2B, 2A3D, 2A3Kb, 2A5Ec, 3A5Eb, 475Ca, 475R, 4A2Bg, 4A2Rd, 4A2Sa, 4A4Bi; see also essential closure (475B) club see Ostaszewski’s ♣ cluster point (of a filter) 3A3L; (of a sequence) 2A3O, 4A2Fa, 4A2Gf coanalytic set 423P, 423Q, 423Yc Coarea Theorem 265 notes coarser topology 4A2A cofinal repetitions 4A1Ab cofinal set (in a partially ordered set) 3A1F, 4A1B, 4A1Ca cofinality (of a partially ordered set) 3A1F, 424Yc; (of a cardinal) 4A1Ac; (of the Lebesgue null ideal) 521B, 521E, 521F, 521P, 521T-521V, 521Yb, 522O, 523Oc, 523Qf, 523S, 526H, 527Bd, 528Pc, 529Xb, 529Xc; see also uncountable cofinality cofinite see finite-cofinite algebra (231Xa, 316Yk) comeager set 3A3Fb, 3A3G commutative (ring, algebra) 2A4Jb, 363Xf, 3A2A, 3A2Ff, 3A2Hc, 4A6Aa commutative Banach algebra 224Yb, 243D, 255Xc, 257Ya, 363B, 444Ed, 444Sb, 444Xd, 4A6B, 4A6D, 4A6J, 4A6K commutative group see abelian group commutator in a group 382Xe, 382Xf, 382Yc commuting Boolean homomorphisms 381E, 381Sc

complete

General index

899

compact class of sets 342Ab, 342D, 342Yc, 356Yi, 451Aa, 451Ha, 451Xh compact Hausdorff space 364V, 364W, 364Yj, 364Yl, 364Ym, 3A3D, 3A3Ha, 3A3K, 422Gc, 434B, 437O, 437Wa, 452N, 462Z, 463J, 463Zc, 476Cd, 481Xh, 4A2Fh, 4A2G, 4A2Nh, 4A2Ri, 4A2Ud, 4A3Xd compact linear operator 371Ya, 371Yb, 376Yn, 3A5Ka, 444V, 444Xt, 4A4L, 4A4M; see also weakly compact operator (3A5Kb) compact measure 342A, 342E-342H, 342J, 342M, 342N, 342Xd-342Xh, 342Xl, 342Xm, 342Xo, 342Ya, 342Yb, 343J, 343K, 343Xa, 343Xb, 343Ye, 415Ym, 416W, 416Xx, 433Xe, 439Xa, 451Ab, 451Da, 451F, 451Ga, 451Ia, 451Ja, 451L, 451P-451R, 451Xi, 451Xj, 451Xm, 451Ya, 451Yd, 451Ye, 454Ab, 454B, 454F; see also locally compact measure (342Ad) compact metrizable space 423Dc, 424Xf, 434Kb, 437R, 441Xl, 441Ya, 441Yg, 441Gb, 441Xm, 449Xu, 461Xf, 461Xi, 462H, 463C-463E, 463I, 463Xb, 463Xd, 463Xk, 463Za, 465Xm, 476Ad, 476G, 476Yc, 481P, 4A2Nh, 4A2P, 4A2Q, 4A2Ud compact-open topology 441Ye, 441Yf; see also uniform convergence on compact sets compact set (in a topological space) 247Xc, 2A2D, 2A2E-2A2G, > 2A3N, 2A3R, 3A3D, 421M, 431D, 421Xo, 461H, 461I, 461Xe, 461Xj, §463, 4A2Fh, 4A2G, 4A2Jd, 4A2Rh, 4A2Sa, 4A2Tb, 4A3Xc, 4A4Bh, 4A4H, 4A4Ka, 4A5E; (spaces of compact subsets) 476Db, 476Eb;see also convex compact, relatively compact (2A3N), relatively weakly compact (2A5Id), weakly compact (2A5Ic) compact subgroup of a topological group 443R, 443Xw, 443Yl compact support, function with 242O, 242Pd, 242O, 242Xh, 244H, 244Yi, 256Be, 256D, 256Xh, 262Yd262Yg, 416I, 416Xj, 436J, 436Xo, 443O, 444Xr, 473Be, 473Db, 473Ed, 473H, 475K, 475N, 495Xl, 4A2A, 4A2Ge, 4A5P compact support, measure with 284Yi compact topological group 441Xd, 441Xk, 443Ag, 444Xf, 445A, 445Xi, 445Xm, 446B, 446C, 446Yb, 447Ya, 491H, 491Xn, 491Yi, 4A5T compact topological space > 2A3N, 3A3Dd, 3A3J, 415Yb, 437Md, 462Yd, 4A2G, 4A2Jf, 4A2Kb, 4A2Lf, 4A3O; see also compact Hausdorff space, countably compact (4A2A), Eberlein compactum 467O, locally compact (3A3Ag), metacompact (4A2A), paracompact (4A2A) ˇ compactification see one-point compactification (3A3O), Stone-Cech compactification (4A2I) compactly generated see weakly compactly generated (467L) Compactness Theorem (for sets with bounded perimeter) 474S, 476 notes compatible set of tagged partitions 481F, 481Hf, 481Xi, 481Ya complement of a band (in a Riesz space) 352P, 352Xh complementary Young’s functions 369Xc, 369Xd, 369Xr complemented band (in a Riesz space) 352P, 352Xf, 352Xh, 352Xm, 353B, 353Xa, 464M, 464O, 464R; see also complemented band algebra (352Q) complemented band algebra 352Q, 352S, 352Xg, 353B, 356Yc complete generation of a Boolean algebra see Maharam type (331F) complete lattice 314Bf complete linear topological space 245Ec, 2A5F, 2A5H, 367Nc, 461E, 461I, 461Xe, 461Xj, 466A, 4A4Ch complete locally determined measure (space) 213C, 213D, 213H, 213J, 213L, 213O, 213Xg, 213Xi, 213Xl, 213Yd, 213Ye, 214I, 214Xc, 216D, 216E, 234Fb, 251Yb, 252B, 252D, 252E, 252N, 252Xb, 252Yh, 252Yj, 252Yr-252Yt, 253Yj, 253Yk, 325B, 341M, 341Pb, 342N, 364Xm, 376S, 376Xj, 411Xd, 412J-412L, 412Xh412Xk, 413F, 413G, 413I-413K, 413M, 413O, 413Xh, 413Xj, 414I, 414J, 414P, 414R, 414Xf, 414Xh, 414Xr, 416Dd, 416R, 417C, 417D, 417H, 418E, 431A, 431B, 431D, 431Xa, 431Xc, 432A, 434Db, 434S, 438E, 438Xf, 438Xi, 439Fb, 439Xc, 451Rb, 451Yq, 463Xc, 465D, 482E; see also c.l.d. version (213E) complete measure (space) 112Df, 113Xa, 122Ya, 211A, 211M, 211N, 211R, 211Xc, 211Xd, §212, 214I, 214J, 216A, 216C, 216Ya, 234A, 234Dd, 254Fd, 254G, 254J, 264Dc, 321K, 341J, 341K, 341Mb, 341Xc, 341Yd, 343B, 343Xi, 344C, 344I, 344Xb-344Xd, 385V, 412M, 413C, 413S, 431Ya, 415Xd, 417A, 418T, 451N, 463Xf; see also complete locally determined measure complete metric space 224Ye, 323Yd, 393Xb, 3A4Fe, 3A4Hc, 434Jg, 438H, 438Xf, 441Ya, 476Ab, 4A2L; see also Banach space (2A4D), Polish space (4A2A) complete normed space see Banach space (2A4D), Banach lattice (242G, 354Ab) complete Riesz space see Dedekind complete (241Fc) complete uniform space 323G, 3A4F, 3A4G, 3A4H, 4A2Jd, 4A5Mb; see also completion (3A4H)

900

Index

complete

complete (in ‘κ-complete filter’) 368F, 438Yb, 4A1Ib, 4A1J, 4A1K complete see also Dedekind complete (314Aa), Dedekind σ-complete (314Ab), uniformly complete (354Yi) completed indefinite-integral measure see indefinite-integral measure (234B) completely additive functional on a Boolean algebra 326J, 326K-326P, 326Xe, 326Xf, 326Xh, 326Xi, 326Yp-326Ys, 327B-327E, 327Ya, 332Xo, 362Ad, 362B, 362D, 362Xe-362Xg, 363K, 363S, 365Ea, 394Xe, 438Xc; see also Mτ —– (on PI) 464Fb completely additive part (of an additive functional on a Boolean algebra) 326Yq, 362Bd; (of a linear functional on a Riesz space) 355Yh completely continuous linear operator see compact linear operator (3A5Ka) completely generate see τ -generate (331E) completely regular topological space 353Yb, 367Yh, > 3A3Ac, 3A3B, 411Gi, 414Yf, 415I, 415P, 415Xp, 415Yk, 416T, 416Xi, 416Xk, 416Xv, 422Ya, 434Jg, 434Yc, 434Yd, 436Xl, 437Kc, 437L, 453Cb, 459F, 4A2F, 4A2Hb, 4A2J completion (of a Boolean algebra) see Dedekind completion (314U) —– (of a measure (space)) §212 (212C), 213Fa, 213Xa, 213Xb, 213Xk, 214Xb, 214Xj, 232Xe, 234Db, 235D, 235Hc, 241Xb, 242Xb, 243Xa, 244Xa, 245Xb, 251S, 251Wn, 251Xp, 252Ya, 254I, 256C, 322Da, 342Gb, 342Ib, 342Xn, 343A, 411Xc, 411Xd, 412H, 412Xh, 412Xi, 413E, 416Xm, 433Cb, 451G, 451Ym, 452A-452C, 452Xg, 456Xg, 465Cj, 458Xl, 491Xt —– (of a metric space) 393B —– (of a normed space) 3A5Ib —– (of a normed Riesz space) 354Xh, 354Yg —– (of a topological group) 443K, 445Ya, 449Xh, 493Xb, 4A5N —– (of a uniform space) 325Ea, 325Yb, 3A4H completion regular measure 411Jb, 411Pb, 411Yb, 415D, 415E, 415Xj, 416U, 417Xv, 417Xw, 419C, 419D, 419E, 434A, 434Q, 434Xa, 434Xq, 434Ya, 435A, 435Xb, 439J, 443M, 451T, 453N, 491Md; see also inner regular with respect to zero sets complex measure 437Yb complexification of a Riesz space 354Yk, 355Yk, 356Yh, 362Yj, 363Xj, 364Yn, 437Yb, 444Yc complex-valued function §133 component (in a Riesz space) 352Rb, 352Sb, 356Yb, 368Yj, 437Ya; (in a topological space) 111Ye; see also Maharam-type-κ component (332Gb) composition of relations 422Df compound Poisson distribution 495M concave function 385A concentration of measure 476J-476L, 476O, 476P, 476Xh, 476Xj, 492D, 492E, 492H, 492I, 492Xb-492Xd, 493I concentration by partial reflection 476H, 476I, 476Xg conditional distributions 454H, 455A, 455B conditional entropy 385D, 385E, 385G, 385N, 385Xb-385Xd, 385Xq, 385Ya, 385Yb, 386L conditional expectation 233D, 233E, 233J, 233K, 233Xg, 233Yc, 235Yc, 242J, 246Ea, 253H, 253Le, 275Ba, 275H, 275I, 275K, 275Ne, 275Xi, 275Ya, 275Yk, 275Yl, 365R, 369Xq, 372K, 452Qb, 452Xr, 458A, 458Xt, 459A, 459D, 465M, 465Xq conditional expectation operator 242Jf, 242K, 242L, 242Xe, 242Yk, 243J, 244M, 244Yk, 246D, 254R, 254Xp, 275Xd, 275Xe, 327Xb, 365R, 365Xl, 366J, 372G, 385B, 385D, 385E, 465Xr conditional probability see regular conditional probability (452E) conditionally (order)-complete see Dedekind complete (314A) cone 3A5B, 3A5D; see also positive cone (351C) conegligible set 112Dc, 214Cc connected set 222Yb, 434Ya conjugacy action of a group 441Xd, 4A5Ca, 4A5I —– class in a group 441Xd

convolution

General index

901

conjugate Boolean homomorphisms 381Gd, 381Sb, 385T conjugate operator see adjoint operator (3A5Ed) consistent disintegration 452E, 452G, 452I, 452K, 452O, 452Q, 452Xl, 452Xm, 452Xq, 452Yb, 452Ye, 454Xf, 495I; see also strongly consistent (452E) consistent lifting 346I, 346J, 346L, 346Xe, 346Yd constituent (of a coanalytic set) 421Ng, 421O, 421Yd, 422K, 423P, 423Q, 423Ye, 439Xe, 439Yg constructibility, axiom of 4A1M continued fractions 372M-372O; (continued-fraction approximations) 372M, 372Xj, 372Yf; (continuedfraction coefficients) 372M, 372O, 372Xh-372Xj, 372Xt, 372Ye continuity, points of 224H, 224Ye, 225J continuous function 121D, 121Yf, 262Ia, 2A2C, 2A2G, 2A3B, 2A3H, 2A3Nb, 2A3Qb, 3A3C, 3A3Eb, 3A3Ib, 3A3Nb, 418D, 418L, 422Dc, 434Xe, 434Xi, 473Db, 4A2B, 4A2C, 4A2F, 4A2G, 4A2Ia, 4A2Kd, 4A2Ld, 4A2Ro, 4A2Sb, 4A3Cd, 4A3Kc, 4A3L, 4A5Fa continuous at a point 3A3C continuous image 493Ba, 4A2Gi, 4A2Kb, 4A2Nd, 4A2Pc continuous linear functional 284Yj; see also dual linear space (2A4H, 4A4Bd) continuous linear operator 2A4Fc, 456B, 456Eb, 456G, 461B, 461L, 4A4B, 4A4Gc, 4A4H; see also bounded linear operator (2A4F) continuous outer measure see Maharam submeasure (392G) continuous see also almost continuous (411M), order-continuous (313H), semi-continuous (225H, 4A2A), uniformly continuous (3A4C) continuum see c (2A1L) continuum hypothesis 463Yd, 463Ye, 463Za, 4A1Ad; see also generalized continuum hypothesis control measure 393O, 393P, 393S, 393Xg, 393Yh, 393Yi Control Measure Problem §393, 493 notes convergence in mean (in L1 (µ) or L1 (µ)) 245Ib >245A), 246J, 246Yc, 247Ya, 416Xj, 418R, 418S, 418Yj, 444F, convergence in measure (in L0 (µ)) §245 (> 445I, 465Xi, 465Xp, 493F >245A), 271Yd, 274Yd, 285Xs, §463, 464E, 464Yb, 465G, 465Xi, 465Xj —– (in L0 (µ)) §245 (> —– (in the algebra of measurable sets) 232Ya, 474S —– (in L0 (A, µ ¯)) 367M, 367N-367W, 367Xo, 367Xp, 367Xr, 367Xt, 367Xv-367Xx, 367Yo, 367Yr, 369M, 369Xf, 369Ye, 393M, 443Ad, 443G —– (of sequences) §245, 246J, 246Xh, 246Xi, 253Xe, 255Yf, 271L, 273Ba, 275Xk, 275Yp, 276Yf, 367Na, 367Q, 369Yf, 376Xf convergent almost everywhere 245C, 245K, 273Ba, 276G, 276H, 463G-463I, 463Xb, 463Xj, 473Ee; see also ergodic theorem, strong law convergent filter 2A3Q, 2A3S, 2A5Ib, 3A3Ic, 3A3Lb, 4A2B; —– sequence 135D, 245Yi, 2A3M, 2A3Sg, 3A3Lc, 4A2Gh, 4A2Ib, 4A2Le, 4A2Rf; see also convergence in measure (245Ad), order*-convergent (367A), pointwise convergence convex function 233G, 233H-233J, 233Xb-233Xf (233Xd), 233Xh, 233Ya, 233Yc, 233Yd, 242K, 242Yi, 242Yj, 242Yk, 244Xm, 244Yg, 255Yk, 275Yg, 365Rb, 367Xx, 369Xb, 369Xc, 369Xd, 373Xb, 461A, 461C, 461Xa; see also mid-convex (233Ya) convex hull 2A5E, 462Xd, 465N, 465Xo; see also closed convex hull (2A5E) Convex Isoperimetric Theorem 475T convex polytope 481O convex set 233Xd, 244Yj, 262Xh, 2A5E, 326Ye, 351Ce, 3A5C, 3A5Ee, 3A5Ld, 461Ac, 461C, 465Xb, 475R-475T, 475Xa, 475Xk, 484Xc, 4A4Db, 4A4E, 4A4F, 4A4Ga —– compact set 437S, 438Xl, 461D, 461I-461N, 461Xh, 461Xi, 461Yb, 463D, 463Xd, 4A4E-4A4G —– see also cone (3A5B), order-convex (351Xb) convex structure 373Xv, 373Yf, 437P, 459Xd, 459Xe convolution in L0 255Fc, 255Xc, 255Yf, 255Yk, 444S, 444V, 444Xt, 444Xu, 444Yk convolution of functions 255E, 255F-255K, 255O, 255Xa-255Xc, 255Xf-255Xj, 255Ya, 255Yb, 255Yd, 255Ye, 255Yi, 255Yl, 255Ym, 255Yn, 262Xj, 262Yd, 262Ye, 262Yh, 263Ya, 282Q, 282Xt, 283M, 283Wd,

902

Index

convolution

283Wf, 283Wj, 283Xl, 284J, 284K, 284O, 284Wf, 284Wi, 284Xb, 284Xd, 373Xg, 376Xa, 376Xd, 444O, 444P-444S, 444Xq-444Xs, 444Yi, 444Yl, 445G, 445Sc, 445Xn, 449H; see also smoothing by convolution convolution of measures §257 (257A), 272S, 285R, 285Yn, 364Xf, 444A, 444B-444E, 444Q, 444Xa-444Xc, 444Xe, 444Xf, 444Xp, 444Yb, 444Yi, 445D, 455Yd, 458Yc, 495Ab convolution of measures and functions 257Xe, 284Xo, 284Yi, 444H, 444I, 444J, 444K, 444M, 444P, 444T, 444Xh-444Xk, 444Xn, 444Xp, 444Yi, 449Yc convolution of sequences 255Xe, 255Yo, 282Xq, 352Xl, 376Ym countable (set) 111F, 114G, 115G, §1A1, 226Yc, 4A1P countable antichain condition 316 notes (see ccc (316A)) countable chain condition see ccc (316A) countable choice (axiom of) 134C countable-cocountable algebra 211R, 211Ya, 232Hb, 316Yk, 368Xa, 454Xg, 4A3P countable-cocountable measure 211R, 232Hb, 252L, 326Xf, 342M, 356Xc, 411R, 435Xb, 451B, 463Xh countable network 415Xj, 417T, 417Yh, 417Yi, 418Yb, 418Yc, 418Yg, 418Yh, 419Xe, 423Be, 423C, 423Ib, 423Ya, 423Yb, 423Yd, 433A, 433B, 433Xa, 433Ya, 434Pc, 434Xr, 434Xt, 434Yb, 451M, 451N, 453Fa, 453I, 453J, 491Eb, 491Xx, 491Ym, 4A2N-4A2P, 4A3E, 4A3F, 4A3Rc, 4A3Xa, 4A3Xf, 4A3Yb countable separation property see σ-interpolation property (466G) countable sup property (in a Riesz space) 241Yd, 242Yd, 242Ye, 244Yb, 353Ye, 354Yc, 355Yi, 356Ya, 363Yb, 364Yb, 366Yb, 368Yh, 376Ye, 376Yf countably additive functional (on a σ-algebra of sets) §231 (231C), §232, 246Yg, 246Yi, 327C, 362Xg, 363S, 416Ya; (on a Boolean algebra) > 326E, 326F-326I, 326L, 326Xb-326Xi, 327B, 327C, 327F, 327G, 327Xa, 327Xd, 327Xe, 362Ac, 362B, 363S, 363Yh, 392Xf, 392Yb, 438Xc; see also completely additive (326J), countably subadditive (392H), Mσ countably additive part (of an additive functional on a Boolean algebra) 326Yn, 362Bc, 362Ye, 393Yg countably compact class of sets 413L, 413M, 413O, 413R, 413S, 413Ye, 451Ac, 451B, 451Hb, 451Xd, 451Xh, 452H, 452I, 452M, 457E countably compact measure (space) 451B, 451C, 451Db, 451Gb, 451Ib, 451Jb, 451T, 451Xc-451Xe, 451Yb, 451Yj, 452H, 452I, 452M, 452R, 452S, 452Xk, 452Xl, 452Xn, 452Xr, 452Yb, 452Ye, 454Ab, 454Xe, 459E countably compact measure property 454Xe, 454Xf, 454Xg, 454Xi countably compact set, topological space 413Ye, 421Ya, 422Yc, 434M, 434N, 436Yc, 462B, 462C, 462F462J, 462Xb, 462Ya, 463I, 463Xb, > 4A2A, 4A2Gf, 4A2Lf countably full family of gauges 482Ia, 482K, 482L countably full group (of automorphisms of a Boolean algebra) 381Af, 381I, 381Nh, 381Yb, 382H-382M countably full local semigroup (of partial automorphisms of a Boolean algebra) §448 (448A) countably generated σ-algebra 415Xi, 419Yb, 423Xd, 424Ba, 424Xb-424Xd, 424Xh-424Xj, 433I, 433J, 451F, 452Xl, 452Xn countably K-determined see K-countably determined (467H) countably metacompact (topological) space 434Yf, 435Ya countably paracompact (topological) space 435C, 435Xg, 435Xh, 435Ya, 4A2A, 4A2Ff countably separated (measure space, σ-algebra of sets) 343D, 343E-343H, 343K, 343L, 343Xe, 343Xh, 343Xi, 343Yb, 343Ye, 343Yf, 344B, 344C, 344I, 344Xb-344Xd, 381Xm, 382Xd, 383Xa, 385V, 388A, 388B, 424E, 433B, 433Xe, 451Ad, 452I, 453Xf countably subadditive functional 392H countably tight (topological) space 423Ya, 434N, 434Yg, 463E, 463Xd, 4A2A, 4A2K, 4A2Ld counting measure 112Bd, 122Xd, 122 notes, 211N, 211Xa, 213Xb, 226A, 241Xa, 242Xa, 243Xl, 244Xi, 244Xn, 245Xa, 246Xc, 251Xb, 251Xh, 252K, 255Yo, 264Db, 324Xe, 416Xb, 417Xi, 442Ie, 445Xi, 445Xj, 481Xc covariance matrix (of a Gaussian distribution) 456Ac, 456B, 456F, 456Xa, 456Xg covariant measure 441 notes cover see measurable envelope (132D) covering theorem 221A, 261B, 261F, 261Xc, 261Ya, 261Yi, 261Yk, 472A-472C, 472Xb, 472Ya-472Yc cozero set 3A3Pa, 412Xg, 416Xi, 421Xh, 443N, 443Yi, 4A2Cb, 4A2F, 4A2Lc

density

General index

903

cycle notation (for automorphisms of Boolean algebras) 381R, 381S, 381T, 382Fb; see also pseudo-cycle (388Xc) cyclic automorphism 381R; see also exchanging involution (381R) cylinder (in ‘measurable cylinder’) 254Aa, 254F, 254G, 254Q, 254Xa cylindrical σ-algebra (in a locally convex space) 461D, 461Xg, 461Xi, 464R, 466J, 466Xc, 466Xd, 466Xf, 491Yc, 4A3T, 4A3U, 4A3V Daniell integral 436Ya, 436Yb Davies R.O. 325 notes decimal expansions 273Xf decomposable distribution see infinitely divisible decomposable measure (space) see strictly localizable (211E) decomposition (of a measure space) 211E, 211Ye, 213O, 213Xh, 214Ia, 214K, 214M, 214Xi, 322L, 412I, 416Xe, 417Xe; see also disintegration (452E), Hahn decomposition of an additive functional (231F), Jordan decomposition of an additive functional (231F), Lebesgue decomposition of a countably additive functional (232I), Lebesgue decomposition of a function of bounded variation (§226) decreasing rearrangement (of a function) 252Yp; (of an element of M 0,∞ (A)) §373 (373C), 374B-374D, 374J, 374L, 374Xa, 374Xj, 374Ya-374Yd Dedekind complete Boolean algebra 314B, 314Ea, 314Fa, 314G, 314Ha, 314I, 314Ja, 314K, 314P, 314S, 314T, 314Va, 314Xa, 314Xb, 314Xf, 314Xg, 314Xh, 314Yd, 314Yg, 314Yh, 315F, 315Yf, 316Fa, 316K, 316Xt, 316Yi, 316Yn, 331Yd, 332Xa, 333Bc, 352Q, 363Mb, 363P-363S, 364O, 368A-368D, 368K, 368M, 368Xe, 375C, 375D, 375Xa, 375Xc, 375Ya, 381C, 381D, 381F, 381Ib, 381Xb, 381Xf, 381Xk, 381Yc, 382E, 382F, 382N, 382Q-382S, 382Xe-382Xj, 384D, 384J, 393A-393C, 393J, 393Xc-393Xf, 393Ya, 393Yc-393Yf 438Xc, 448Ya Dedekind complete partially ordered set 135Ba, 314Aa, 314B, 314Ya, 315De, 343Yc, 3A6C, 419L, 419Xf, 4A2Rsee also Dedekind complete Boolean algebra, Dedekind complete Riesz space Dedekind complete Riesz space 241Fc, 241G, 241Xf, 242H, 242Yc, 243H, 243Xj, 244L, 353G, 353I, 353Jb, 353K, 353Yb, 354Ee, §355, 356B, 356K, 356Xj, 356Ye, 361H, 363Mb, 363P, 363S, 364O, 366C, 366G, 368H-368J, 371B-371D, 371Xd, 371Xe, 375J, 375Ya, 437Yj, 438Xd Dedekind completion of a Boolean algebra 314T, 314U, 314Xe, 314Yh, 315Yd, 322O, 361Yc, 365Xm, 366Xk, 368Xb, 384Lf, 384Xc, 391Xc; see also localization (322P) Dedekind completion of a Riesz space 368I, 368J, 368Xb, 368Yb, 369Xp Dedekind σ-complete Boolean algebra 314C-314F, 314Hb, 314Jb, 314M, 314N, 314Xa, 314Xc, 314Ye, 314Yf, 315O, 316C, 316Fa, 316Hb, 316J, 316Xg, 316Yi, 316Yk, 321A, 324Yb, 324Yc, 326Fg, 326I, 326O, 326P, 326Xc, 326Xi, 326Yf, 326Yo, 344Yb, 362Xa, 363Ma, 363P, 363Xh, 363Yh, §364, 367Xk, 368Yd, 375Yb, 381C, 381D, 381Ia, 381K-381N, 381Xb, 381Xk, 381Xp, 381Yb, 382D, 382E, 382G-382M, 382S, 382Xk, 382Xl, 382Ya, 382Yc, 392H, 392Xd, 392Xf, 392Xg, 393A, 393E, 393M, 393O-393Q, 393S, 393Xb, 393Yg-393Yi, 424Xd, 491Xo Dedekind σ-complete partially ordered set 314Ab, 315De, 367Bf, 367Xb, 466G; see also Dedekind σcomplete Boolean algebra, Dedekind σ-complete Riesz space Dedekind σ-complete Riesz space 241Fb, 241G, 241Xe, 241Yb, 241Yh, 242Yg, 243H, 243Xb, 353G, 353H, 353J, 353Xb, 353Yc, 353Yd, 354Xn, 354Yi, 354Yl, 356Xc, 356Xd, 363M-363P, 364C, 364E delta function see Dirac’s delta function (284R) delta system see ∆-system (4A1D) De Morgan’s laws 311Xf; see also distributive laws dense set in a topological space 242Mb, 242Ob, 242Pd, 242Xi, 243Ib, 244H, 244Ob, 244Yi, 254Xo, 281Yc, 2A3U, 2A4I, 313Xj, 323Dc, 367O, 3A3E, 3A3G, 3A3Ie, 3A4Ff, 412N, 412Yc, 417Xu, 449Xg, 493Xa, 4A2Ma, 4A2Ni, 4A2Ua; see also nowhere dense (3A3Fa), order-dense (313J, 352Na), quasi-order-dense (352Na) density (of a topological space) 331Ye, 331Yf, 365Ya, 393Yf density function (of a random variable) 271H, 271I-271K, 271Xc-271Xe, 272T, 272Xd, 272Xj, 495P; see also Radon-Nikod´ ym derivative (232If, 234B) density point 223B, 223Xi, 223Yb, 481Q density topology 223Yb, 223Yc, 223Yd, 261Yf, 414P, 414R, 414Xk-414Xo, 414Xq, 414Yd-414Yh, 418Xl, 451Yq, 453Xg, 475Xe

904

Index

density

density see also asymptotic density (273G, 491A), Besicovitch’s Density Theorem (472D), Lebesgue’s Density Theorem (223A), lower density (341C) derivate see lower derivate (483H), upper derivate (483H) derivative of a function (of one variable) 222C, 222E, 222F, 222G, 222H, 222I, 222Yd, 225J, 225L, 225Of, 225Xc, 226Be, 282R; (of many variables) 262F, 262G, 262P, 473B, 473Ya;see also gradient, partial derivative derived set (of a set of ordinals) 4A1Bb; derived tree 421N, 421O determinant of a matrix 2A6A determined see locally determined measure space (211H), locally determined negligible sets (213I) determined by coordinates (in ‘W is determined by coordinates in J’) 254M, 254O, 254R-254T, 254Xp, 254Xr, 311Xh, 325N, 325Xh, 415E, 417M, 417Xu, 456H, 481P, 4A2Bg, 4A2Eb, 4A3Mb, 4A3N, 4A3Yb Devil’s Staircase see Cantor function (134H) diagonal intersection (of a family of sets) 4A1Bd, 4A1Ic diamond see Jensen’s ♦ Dieudonné’s measure 411Q, 411Xj, 434Xf, 434Xx, 491Xm differentiability, points of 222H, 225J differentiable function (of one variable) 123D, 222A, 224I, 224Kg, 224Yc, 225L, 225Of, 225Xc, 225Xn, 233Xc, 252Yj, 255Xg, 255Xh, 262Xk, 265Xd, 274E, 282L, 282Xs, 283I-283K, 283Xm, 284Xc, 284Xk, 483Xh; (of many variables) 262Fa, 262Gb, 262I, 262Xg, 262Xi, 262Xj, 473B, 473Cd, 484N, 484Xf; see also derivative ‘differentiable relative to its domain’ 262Fb, 262I, 262M-262Q, 262Xd-262Xf, 262Yc, 263D, 263Xc, 263Xd, 263Yc, 265E, 282Xk diffused measure see atomless measure (211J) dilation 286C dimension see Hausdorff dimension (264 notes, 471Xe) Dini’s theorem 436Ic Dirac’s delta function 257Xa, 284R, 284Xn, 284Xo, 285H, 285Xp direct image (of a set under a function or relation) 1A1B direct sum of measure spaces 214K, 214L, 214Xi-214Xl, 241Xg, 242Xd, 243Xe, 244Xg, 245Yh, 251Xh, 251Xi, 322Kb, 332C, 342Gd, 342Ic, 342Xn, 343Yb, 411Xh, 412Xl, 414Xb, 415Xb, 416Xc, 451Xe, 451Yh direct sum of σ-algebras 424Xa direct sum of topological spaces 4A2A, 4A2Qe; see also disjoint union topology (4A2A) directed set see downwards-directed (2A1Ab), upwards-directed (2A1Ab) Dirichlet kernel 282D; see also modified Dirichlet kernel (282Xc) disconnected see extremally disconnected (3A3Ae), basically disconnected discrete see σ-metrically-discrete (4A2A) discrete topology (3A3Ah), 416Xb, 417Xi, 436Xg, 439Cd, 439Xh, 442Ie, 445A, 445Xj, 449G, 449K, 449Xj, 449Xr, 449Xs, 449Yb, 4A2Ib, 4A2Qa disintegration of a measure 452E, 452F-452K, 452M-452Q, 452Xf-452Xh, 452Xk-452Xn, 453K, 453Xi, 453Ya, 455A, 455Ya, 459E-459G, 459Xc, 495H; see also consistent disintegration (452E) disjoint family (of sets) 112Bb; (in a Boolean ring) 311Gb; (in a Riesz space) 352C, 352Fb disjoint sequence theorem 246G, 246Ha, 246Yd, 246Ye, 246Yf, 246Yj, 354Rb, 356O, 356Xm disjoint set (in a Boolean ring) 311Gb; (in a Riesz space) 352C disjoint union topology 315Xe, 411Xh, 412Xl, 414Xb, 415Xb, 416Xc, 4A2A, 4A2Bb, 4A2Cb; see also direct sum of topological spaces distribution see Schwartzian distribution, tempered distribution distribution of a random variable 241Xc, 271E, 271F, 271Ga, 272G, 272S, 272Xe, 272Yc, 272Yf, 272Yg, 285H, 285Xg, 415Fb, 415Xq, 454L; (of an element of L0 (A, µ ¯)) 364Xd, 364Xf, 367Xv, 373Xh; see also Cauchy distribution (285Xm), empirical distribution (273 notes), exponential distribution (495P), gamma distribution (455Xe), Poisson distribution (285Xo), relative distribution (458C) —– of a finite family of random variables 271B, 271C, 271D-271G, 272G, 272Ye, 272Yf, 285Ab, 285C, 285Mb, 454J, 454K, 456M, 459C —– of an infinite family of random variables 454J-454M (454K), 454Xj; see also Gaussian distribution (456Ab distribution function of a random variable >271G, 271L, 271Xb, 271Yb, 271Yc, 271Yd, 272Xe, 272Yc, 273Xg, 273Xh, 274F-274L, 274Xd, 274Xg, 274Xh, 274Ya, 274Yc, 285P, 495P

equidecomposable

General index

905

distributive lattice 367Yc, 382Xh, 3A1Ic; see also (2, ∞)-distributive (367Yd) distributive laws in Boolean algebras 313B; see also De Morgan’s laws, weakly σ-distributive (316Yg), weakly (σ, ∞)-distributive (316G) distributive laws in Riesz spaces 352E; see also weakly σ-distributive, weakly (σ, ∞)-distributive (368N) divergence (of a vector field) 474B, 474C-474E, 475N, 475Xg, 484N Divergence Theorem 475N, 475Xg, 484N, 484Xf divisible distribution see infinitely divisible Dominated Convergence Theorem see Lebesgue’s Dominated Convergence Theorem (123C) Doob’s Martingale Convergence Theorem 275G double arrow space see split interval (343J) doubly recurrent Boolean automorphism 381Ag, 381M, 381N, 381Xk, 381Xp, 381Xo, 382J doubly stochastic matrix 373Xe Dowker space 439O downwards-directed partially ordered set 2A1Ab dual group (of a topological group) §445 (445A) dual linear operator see adjoint operator (3A5Ed) dual linear space (of a linear topological space) 461Aa, 461F, 461G, 4A4B, 4A4C-4A4F dual normed space 243G, 244K, 2A4H, 356D, 356N, 356O, 356P, 356Xg, 356Yh, 365L, 365M, 366C, 366D, 369K, 3A5A, 3A5C, 3A5E-3A5H, 436Ib, 437I, 4A4I —– see also algebraic dual, bidual, order-bounded dual (356A), order-continuous bidual, order-continuous dual (356A), sequentially order-continuous dual (356A), sequentially smooth dual (437Aa), smooth dual (437Ab) Duality Theorem (Pontryagin-van Kampen) 445U Dunford’s theorem 376N dyadic cube 484A dyadic cycle system (for a Boolean automorphism) 388D, 388H, 388J dyadic numbers 445Xp dyadic (topological) space 434 notes, 491Xz, 4A2A, 4A2D, 4A5T; see also quasi-dyadic (434O) Dye’s theorem 388L dynamical system 437S, 461P Dynkin class 136A, 136B, 136Xb, 452A, 495C Eberlein compactum 467O, 467P Eberlein’s theorem 2A5J, 356 notes, 462D effectively locally finite measure 411Fb, 411G, 411P, 411Xd, 411Xh, 411Xi, 412F, 412G, 412Se, 412Xe, 412Xf, 414A-414M, 414P, 414R, 414Xd-414Xh, 414Xj, 415C, 415D, 415M-415O, 416H, 417B-417D, 417H, 417I, 417S, 417T, 417V, 417Xc, 417Xe, 417Xh, 417Xj, 417Yb, 417Yc, 418Ye, 419A, 433Xb, 434Ib, 434Ja, 434R, 435Xe, 481N, 482Xd; see also locally finite (411Fa) effectively regular measure 491L, 491M-491O, 491Xr-491Xv, 491Xx Effros Borel structure 424Ya Egorov’s theorem 131Ya, 215Yb, 412Xr, 418Xc Egorov property (of a Boolean algebra) 316Yg, 316Yh eigenvector 4A4Lb, 4A4M embedding see regular embedding (313N) empirical distribution 273Xh, 273 notes, 465H, 465I, 465M, 465Xn, 465Ya entropy §385; (of a partition of unity) 385C, 386E, 386I, 386L, 386O, 387B-387D, 387F-387H; (of a measure-preserving Boolean homomorphism) 385M, 386L, 386Xe, 386Yb, 386Yc, 387C-387E, 387I-387L, 388Yb, 493Yd; see also conditional entropy (385D), entropy metric (385Xq) entropy metric 385Xq, 385Yg enumerate, enumeration 3A1B envelope see Baire property envelope, measurable envelope (132D), upr(a, C) (314V) enveloped see regularly enveloped (491L) equidecomposable (in ‘G-equidecomposable’) 395Ya; (in ‘G-τ -equidecomposable’) §394 (394A); (in ‘Gσ-equidecomposable’) §448 (448A)

906

Index

equidistributed

equidistributed sequence (in a topological probability space) 281N, 281Yi, 281Yj, 281Yk, 491B, 491C491H, 491N, 491O, 491Q, 491Xe-491Xg, 491Xi-491Xn, 491Xw, 491Xx, 491Xz, 491Yb-491Yi, 491Ym, 491Z Equidistribution Theorem see Weyl’s Equidistribution Theorem (281N) equivalent norms 355Xb, 369Xd, 4A4Ia equiveridical 121B, 212B, 312B, 411P ergodic automorphism (of a Boolean algebra) 363F, 381Xp, 382Xk, 385Se, 385Ta, 387C, 387D, 387E, 387J, 387Xb, 388Xg, 388Yc, 394Ge ergodic Boolean homomorphism 372Pa, 381P, 386D, 386F ergodic group of automorphisms (of a Boolean algebra) 394Ge, 394Q, 394R, 394Xa, 394Xg, 394Yd ergodic inverse-measure-preserving function 372P, 372Q, 372Xo, 372Xs, 372Yi, 372Yj, 372Yl, 388Yd, 461O, 461P, 461Yd, 491Xx ergodic measure-preserving Boolean homomorphism (of a measure algebra) 372Q, 372Xk, 372Xl, 372Yl, 385Se, 386D, 386F, 387C-387E, 387J, 387Xb, 388Xg, 388Yc, 395Xc, 493Yc, 493Yd Ergodic Theorem 372D, 372E, 372Ya, 386Xc, 449Xu; see also Maximal Ergodic Theorem (372C), Mean Ergodic Theorem (372Xa), Wiener’s Dominated Ergodic Theorem (372Yb) essential boundary 475B, 475C, 475D, 475G, 475J-475L, 475N-475Q, 475Xb, 475Xc, 475Xg, 475Xh, 475Xj, 476Ie, 484Ra essential closure 475B, 475C, 475Xb, 475Xc, 475Xe, 475Xf, 484B, 484K, 484Ra, 484Ya essential interior 475B, 475C, 475J, 475Xb-475Xf essential supremum of a family of measurable sets 211G, 213K, 215B, 215C; of a real-valued function 243D, 243I, 376S, 376Xn, 443Gb, 444R, 492G, 492Xa, 492Xd essentially bounded function 243A Etemadi’s lemma 272U Euclidean metric (on Rr ) 2A3Fb, 443Xx Euclidean topology §1A2, §2A2, 2A3Ff, 2A3Tc even function 255Xb, 283Yb, 283Yc exchangeable family of random variables 276Xe, 459B, 459C, 459Xa, 459Xb, 459Ya exchanging involution 381Ra, 381S, 382C, 382E, 382H, 382K, 382L, 382M, 382P, 382Ya exhaustion, principle of 215A, 215C, 215Xa, 215Xb, 232E, 246Hc, 342B, 365 notes exhaustive submeasure 392Bb, 392C, 392Hc, 392Xb-392Xd, 393A-393C, 393H; see also uniformly exhaustive (392Bc) expectation of a random variable 271Ab, 271E, 271F, 271I, 271Xa, 272Q, 272Xb, 272Xi, 285Ga, 285Xo; see also conditional expectation (233D) exponential distribution 495P, 495Xj, 495Xo, 495Xp exponentially bounded group 449Xp exponentiation (in a Banach algebra) 4A6L, 4A6M extended Fatou norm §369 (369F), 376S, 376Xn, 376Yl, 415Yl; see also T -invariant extended Fatou norm (374Ab), rearrangement-invariant extended Fatou norm (374Eb) extended real line 121C, §135 exterior see Federer exterior normal (474O) extension of finitely additive functionals 391G, 413K, 413N, 413Q, 413S, 413Yd, 449L, 449Xt, 457A, 457C, 457D, 457Hc, 457Xb-457Xe, 457Ya extension of measures 132Yd, 212Xk, 327Xf, 413O, 415L-415N, 415Yi, 416F, 416M, 416O, 416P, 416Q, 416Yb, 416Yc, 417A, 417C, 417E, 417Xa, 417Ya, 417Ye-417Yg, 432D, 432F, 432Ya, 433I, 433J, 433Xf, 434A, 434Ha, 434Ib, 434R, 435B, 435C, 435Xa-435Xd, 435Xg, 435Xj, 439A, 439M, 439O, 449M, 454C454F, 454Xa-454Xc, 457E, 457G, 457Hc, 457J, 457Xa, 457Xf, 457Xg, 457Xk, 457Yc, 458Xr, 464D, 466H, 466Xd, 466Za; see also completion (212C), c.l.d. version (213E) extremally disconnected topological space 314S, 314Xj, 353Yb, 363Yc, 364W, 364Yk-364Ym, 368G, 368Yc, 3A3Ae, 3A3Bd, 4A2Gh, 491Yg extreme point of a convex set 353Yf, 437P, 437Xp, 438Xl, 459Xd, 459Xe, 461O, 461P, 461Xb, 461Xe, 461Xk, 461Xl, 461Yd, 461Ye, 4A4G; (of a compact set in a Hausdorff linear topological space) 461Xj —– set of extreme points 461K-461N, 461Xh-461Xi, 461Yb extremely amenable topological group §493 (493A)

Fourier

General index

907

factor (of an automorphism of a Boolean algebra) 387Ac, 387L, 387Xb fair-coin probability 254J faithful action 449De, 4A5Be faithful representation 446A, 446N Fatou’s Lemma 123B, 133K, 135G, 135Hb, 365Xc, 365Xd, 367Xf Fatou norm on a Riesz space 244Yf, 354Da, 354Eb, 354J, 354Xk, 354Xn, 354Xo, 354Ya, 354Ye, 356Da, 367Xg, 369G, 369Xe, 371B-371D, 371Xc, 371Xd; see also extended Fatou norm (369F) favourable see weakly α-favourable (451U) Federer exterior normal 474O, 474P, 474Q, 475N, 475Xd, 475Xg, 484M, 484N, 484Xf; see also canonical outward-normal function (474G) Féjer integral 283Xf, 283Xh-283Xj Féjer kernel 282D Féjer sums 282Ad, 282B-282D, 282G-282I, 282Yc Feller, W. Chap. 27 intro. fiber product 458La, 458Xf field (of sets) see algebra (136E) filter 2A1I, 2A1N, 2A1O, 2A5F, 464C, 464D, 464Fc, 4A1I, 4A2Ib; see also complete filter (4A1Ib), convergent filter (2A3Q), normal filter (4A1Ic), ultrafilter (2A1N) filter base 4A1Ia fine (in ‘δ-fine tagged partition’) 481E, 481G finer topology 4A2A de Finetti’s theorem 459C finite-cofinite algebra 231Xa, 231Xc, 316Yk, 392Xc finite-dimensional linear subspace 4A4Bk, 4A4Je finite-dimensional representation 446A, 446B, 446C, 446N, 446Xa, 446Xb, 446Ya, 446Yb, 491Yi finite group 445Ye finite intersection property 3A3D, 434M, 4A1Ia finite perimeter (for sets in Rr ) 474D, 475Mb, 475Q, 484B, 484C, 484F, 484Xf finite rank operators see abstract integral operators (376D) finitely additive functional on an algebra of sets 136Xg, 136Ya, 136Yb, 231A, 231B, 231Xb-231Xe, 231Ya231Yh, 232A, 232B, 232E, 232G, 232Ya, 232Ye, 232Yg, 243Xk, 327C, 413H, 413K, 413N, 413P, 413Q, 413S, 413Xl, 413Xm, 413Xp, 416N-416Q, 416Xl, 416Xm, 416Xr, 471Q, 449J, 449Xr, 457A-457D, 482G, 482H, 493D; see also countably additive, completely additive finitely additive function(al) on a Boolean algebra 326A, 326B-326D, 326Fa, 326G, 326Kg, 326Xa-326Xc, 326Xe, 326Xg, 326Xi, 326Yt-326Yd, 326Yg-326Yj, 326Yk, 326Yl-326Yn, 326Yq, 326Yr, 327B, 327C, 331B, 361Xa-361Xc, 391D-391G, 391I, 391J, 391N, 393Be, 393C, 393Xi, 394N, 394O, 413Xo, 416Q, 481Xh, 491Xy; see also chargeable Boolean algebra (391X), completely additive functional (326J), countably additive functional (326E) finitely additive function(al) on a Boolean ring 361B, 361C, 361E-361I, 365Eb finitely additive measure 326A finitely full subgroup 381Xg first category see meager (3A3Fb) first-countable topological space 434R, 434Xx, 434Yk, 438S, 439K, 439Xf, 452Xa, 462Xa, 462Xc, 4A2A, 4A2K, 4A2Ld, 4A2Sb, 4A5Q First Separation Theorem (of descriptive set theory) 422I, 422Xd, 422Yc fixed-point subalgebra (in a Boolean algebra) 312T, 313Xq, 333Q, 333R, 361Xk, 364Xv, 372Pa, 386B, 386Xa, 381P, 381Xj, 381Xp, 388L, 394G, 394H, 394I, 394K-394N, 394P, 394Xb-394Xf, 394Yb, 394Yc fixed point on compacta property see extremely amenable (493A) Fodor’s theorem see Pressing-Down Lemma (4A1Cc) Fourier coefficients 282Aa, 282B, 282Cb, 282F, 282I, 282J, 282M, 282Q, 282R, 282Xa, 282Xg, 282Xq, 282Xt, 282Ya, 283Xu, 284Ya, 284Yg Fourier’s integral formula 283Xm Fourier series 121G, §282 (282Ac)

908

Index

Fourier

Fourier sums 282Ab, 282B-282D, 282J, 282L, 282O, 282P, 282Xi-282Xk, 282Xp, 282Xt, 282Yd, 286V, 286Xb Fourier transform 133Xd, 133Yc, §283 (283A, 283Wa), §284 (284H, 284Wd), 285Ba, 285D, 285Xd, 285Ya, 286U, 286Ya, §445 (445F) Fourier-Stieltjes transform 445C, 445D, 445Ec, 445Xc, 445Xd, 445Xo, 445Yf; see also characteristic function (285A) fractional Brownian motion 456Ye Fréchet filter 2A3Sg Fréchet-Urysohn topological space 462Aa free group 449G, 449Yf, 449Yg free product of Boolean algebras §315 (315H), 316Yl, §325, 326Q, 326Yt, 326Yu, 361Yf, 384Le, 391Xb, 391Ye; see also localizable measure algebra free product (325Ea), probability algebra free product (325K), relative free product (458I) free set 4A1E Fremlin’s Alternative 463H Frol´ık’s theorem 382E Frostman’s lemma 471Xb Fubini’s theorem 252B, 252C, 252H, 252R, 393Td, 417H, 434R, 434Xv, 452E, 444N, 444Xo, 482M, 482Yc full family of gauges 482Ia, 482J; see also countably full (482Ia) full local semigroup (of partial automorphisms of a Boolean algebra) 394A, 394B, 394C, 394Gc, 394Xc, 394Ya, 394Yd, 448Ya; see also countably full (448A) full outer measure 132F, 132G, 132Xk, 132Yd, 133Yf, 134D, 134Yt, 214F, 254Yf, 322Jb, 324Cb, 343Xa, 415J, 415Qc, 419Xd, 432G, 433Xf, 443Yj, 456D, 495G; full outer Haar measure 443Ac, 443K full subgroup of Aut A 381Ae, 381Ib, 381J, 381Qc, 381Xe, 381Xf, 381Xn, 381Yb, 381Yc, 382N, 382Q, 382Xe, 382Xh, 382Xi, 382Xj, 383C, 388A-388C, 388G, 388H, 388J-388L, 388Xa, 388Ya; see also countably full (381Bf), finitely full (381Xg) —– subsemigroup of Aut A 381Yd fully non-paradoxical group of automorphisms of a Boolean algebra 394E, 394F, 394H, 394I, 394K-394R, 394Xd-394Xf, 394Xh, 394Ya-394Yc, 395B, 395Ya fully symmetric extended Fatou norm see T -invariant (374Ab) function 1A1B function space see Banach function space (§369) functionally closed see zero set (3A3Pa) functionally open see cozero set (3A3Pa) Fundamental Theorem of Calculus 222E, 222H, 222I, 225E, 482J, 483I, 484O; see also Divergence Theorem Fundamental Theorem of Statistics 273Xh, 273 notes Gagliardo-Nirenberg-Sobolev inequality 473H Gaifman’s example 391N, 391Yf-391Yh, game see infinite game, Banach-Mazur game gamma distribution see standard gamma distribution (455Xe) Gamma function see Γ-function (225Xj) gauge 481E, 481G, 481Q, 481Xf, 481Xg; see also neighbourhood gauge (481E), uniform metric gauge (481E) gauge integral 481D, 481Xa-481Xh, 481Ya, 482B, 482D-482H, 482J-482M, 482Xa-482Xg, 482Ya-482Yc, 483Xo, 483Yb, 484G, 484Xc; see also Henstock integral (483A), Pfeffer integral (484G) Gauss C.F. 372Yg Gauss-Green theorem see Divergence Theorem (475N) Gaussian distribution §456 (456A), 495 notes; see also standard normal distribution (274Aa) Gaussian process 456M, 456N, 456Xg, 456Xh, 456Ya, 456Ye Gaussian quasi-Radon measure 456Xf Gaussian random variable see normal random variable (274Ad) generated filter 4A1Ia generated set of tagged partitions 481Ba

Henry

General index

909

generated topology 3A3Ma generated (σ-)algebra of sets 111Gb, 111Xe, 111Xf, 121J, 121Xd, 136B, 136C, 136G, 136Xc, 136X, 136Xl, 136Yb, 421F, 421Xi generated (σ)-subalgebra of a Boolean algebra 313F, 313M, 314G, 314Ye, 331E, 4A1O generating Bernoulli partition 387Ab, 387Bg generating set in a Boolean algebra 331E; see also σ-generating (331E), τ -generating (331E) Giacomelli see Tamanini-Giacomelli theorem (484B) Glivenko-Cantelli class 465 notes Glivenko-Cantelli theorem 273 notes Gödel K. 4A1M gradient (of a scalar field) 473B, 473C, 473Dd, 473H-473L, 474Bb graph of a function 264Xf, 265Yb greatest ambit (of a topological group) 449D, 449E, 493C Green’s theorem see Divergence Theorem (475N) Grothendieck’s theorem 4A4Ch group 255Yn, 255Yo; see also abelian group, amenable group, automorphism group, circle group, ergodic group, finite group, free group, isometry group, locally compact group, orthogonal group, Polish group (4A5Db), simple group, topological group (4A5Da) group automorphism 3A6B; see also inner automorphism (3A6B), outer automorphism (3A6B) group homomorphism 442Xi, 445Xe, 4A5Fa, 4A5L ‘Haar almost everywhere’ 443Ae Haar measurable envelope 443Ab, 443Xg, 443Xn Haar measurable function 443Ae, 443G, 444Xo Haar measurable set 442H, 442Xd, 443A, 443D, 443F, 443J, 443Pc, 447A, 447B, 447D-447H Haar measure Chap. 44 (441D), 452Xj, 491H, 491Xm, 491Xn, 491Yg, 492H, 492I Haar measure algebra 442H, 442Xe, 443A, 443C, 443Jb, 443Xd, 443Xe, 443Xj, 447A, 447Xa Haar negligible set 442H, 443A, 443F, 443Jb, 443Pc, 444L, 444Xm, 444Ye Haar null set 444Ye Hahn-Banach theorem 363R, 373Yd, 3A5A, 3A5C, 4A4D, 4A4E Hahn decomposition of an additive functional 231Eb, 231F, 326I, 326O —– see also Vitali-Hahn-Saks theorem (246Yg) Hajian-Ito theorem 395B half-open interval (in R or Rr ) 114Aa, 114G, 114Xe, 114Yj, 115Ab, 115Xa, 115Xc, 115Yd; (in general totally ordered spaces) 4A2A Hall’s Marriage Lemma 3A1K, 4A1H Halmos-Rokhlin-Kakutani lemma 386C Hamel basis 4A4A Hamming metric on a product space 492D, 492E, 492Xd Hardy-Littlewood Maximal Theorem 286A Hausdorff capacity 471H, 471Xb Hausdorff dimension 264 notes, 471Xe Hausdorff measure §264 (264C, 264Db, 264K, 264Yo), 265Yb, 343Ye, 345Xb, 345Xg, 411Yb, 439H, 439Xk, 441Yd, 441Yh, 441Yi, 442Ya, 443Ym, 456Xb, §471 (471A, 471Ya), 476M; see also normalized Hausdorff measure (265A) Hausdorff metric (on a space of closed subsets) 246Yb, 441Ya, 441Yb, 476A, 476B, 476D, 476F, 476Xa, 476Xb, 476Xc, 476Xe, 476Ya, 495Xk, 495Yc Hausdorff outer measure §264 (264A, 264K, 264Yo), 471A, 471Dc, 471J Hausdorff topology 2A3E, 2A3L, 2A3Mb, 2A3S, 3A3Aa, 3A3D, 3A3Id, 3A3K, 3A4Ac, 3A4Fd, 476Cb, 4A2F, 4A2La, 4A2Nf, 4A2Qf, 4A2Rc, 4A4Ec Hausdorff uniformity 3A4Ac Helly space 438Qb, 438R, 438Xl Henry’s theorem 416M, 432 notes

910

Index

Henstock

Henstock integrable function §483 (483A, 483Yg) —– integral 481J, 481K, §483 (483A, 483Yg); see also approximately continuous Henstock integral, indefinite Henstock integral, Saks-Henstock lemma Henstock-Stieltjes integral 483Xn hereditarily Lindelöf (topological space) 411Xd, 414O, 418Ya, 419L, 419Xf, 422Xe, 423Da, 433Xb, 434Ie, 434Ke, 438Lc, 439D, 4A2A, 4A2Hc, 4A2N-4A2P, 4A2Rn, 4A3D, 4A3E, 4A3Kb hereditarily metacompact 438Lb, 4A2A, 4A2Lb hereditarily separable 423Ya hereditarily weakly submetacompact see hereditarily weakly θ-refinable (438K) hereditarily weakly θ-refinable 438K, 438L-438N, 438P, 438Q, 438Xm, 466Eb, 467Pb hereditarily σ-relatively metacompact see hereditarily weakly θ-refinable (438K) Hilbert space 244N, 244Yj, 3A5L, 456F, 456Xa, 456Xd, 456Yb, 466Xk, 493Xg, 495Xg, 4A4K, 4A4M; see also inner product space (3A5L) Hilbert transform 376Ym Hölder’s inequality 244Eb homeomorphism 3A3Ce homogeneous Boolean algebra 331M, 331N, 331Xj, 331Xk, 331Yg-331Yj, 381D, 382Q, 382S, 384E-384G, 384La, 384Xb, 424Xk; see also Maharam-type-homogeneous (331Fc), Martin-number-homogeneous, relatively Maharam-type-homogeneous (333Ac) —– measure algebra 331N, 331Xk, 332M, 373Yb, 374H, 374Yc, 375Kb, 383E, 383F, 383I, 394R; see also quasi-homogeneous (374G) —– measure space 344L —– probability algebra 333P, 333Yc, 334E, 372Xk homomorphic image (of a Boolean algebra) 314M; (of a group) 449Ca, 493Ba homomorphism see Boolean homomorphism (312F), group homomorphism, lattice homomorphism (3A1I), ring homomorphism (3A2D) —– of topological groups 446I; see also finite-dimensional representation hull see convex hull (2A5E), closed convex hull (2A5E) Humpty Dumpty’s privilege 372 notes Huntingdon’s metric 437Yi hypo-radonian space see pre-Radon (434Gc) ideal in an algebra of sets 232Xc, 312C, 363Yf; see also σ-ideal (112Db) —– in a Banach algebra 4A6E —– in a Boolean ring or algebra 311D, 312C, 312K, 312O, 312Xd, 312Xi, 312Xk, 382R, 392Xd; see also principal ideal (312D), σ-ideal (313E) —– in a Riesz space see solid linear subspace (352J) —– in a ring 352Xm, 3A2E, 3A2F, 3A2G —– see also asymptotic density idempotent 363Xf, 363Xg identically distributed random variables 273I, 273Xh, 274 notes, 276Yg, 285Xn, 285Yc, 372Xg; see also exchangeable family (276Xe, 459C) image filter 2A1Ib, 2A3Qb, 2A3S, 4A1K image measure 112E, 112F, 112Xd, 112Xg, 123Ya, 132G, 132Xk, 132Yb, 132Yf, 211Xd, 212Bd, 212Xg, 235L, 254Oa, 256G, 342Xh, 342Xj, 343Xb, 343Yf, 387Xd, 413Xc, 415Rb, 418I, 418L, 418Xa, 418Xd, 418Xo, 418Xp, 418Yf, 432G, 433D, 437Q, 443Pd, 444Xa, 451N, 451R, 451Xg, 454J, 456B, 456Xb, 461B, 474H, 491Ea image measure catastrophe 235J, 343 notes, 418 notes Increasing Sets Lemma 471G indecomposable transformation see ergodic inverse-measure-preserving function (372Pb) indefinite Henstock integral 483D, 483E, 483F, 483I, 483K, 483Mc, 483Pa, 483R, 483Xg, 483Xh, 483Yc indefinite integral 131Xa, 222D-222F, 222H, 222I, 222Xa-222Xc, 222Yc, 224Xg, 225E, 225Od, 225Xh, 232D, 232E, 232Yf, 232Yi; see also indefinite Henstock integral, indefinite-integral measure, Pettis integral (463Ya), Saks-Henstock indefinite integral (482C, 484I)

inner regular

General index

911

indefinite-integral measure §234 (234A), 235M, 235P, 235Xi, 253I, 256E, 256J, 256L, 256Xe, 256Yd, 257F, 257Xe, 263Ya, 275Yi, 275Yj, 285Dd, 285Xe, 285Ya, 322Xf, 342Xd, 342Xn, 342Yd, 365T, 412Q, 414H, 414Xe, 415O, 416S, 416Xt, 416Yf, 471F, 442L, 444K, 444P, 444Q, 445F, 445Q, 451Xc, 451Yi, 452Xf, 452Xh, 453Xe, 465Ck, 475G, 491Xt; see also uncompleted indefinite-integral measure independence §272 (272A) independent family in a probability algebra 325Xe, 325Yf, 371Yc, 495J independent family in L0 (A) 364Xe, 364Xf, 364Xt, 367W, 367Xw, 367Xy, 495K, 495L, 495Xg, 495Xh, 495Ya, 495Yb independent increments 455C, 455Xc-455Xe, 455Yd independent random variables 272Ac), 272D-272I, 272L, 272M, 272P-272U, 272Xb, 272Xd, 272Xh-272Xj, 272Ya-272Yd, 272Yf, 272Yg, 273B, 273D, 273E, 273H, 273I, 273L-273N, 273Xh, 273Xi, 273Xk, 274B-274D, 274F-274K, 274Xc, 274Xd, 274Xg, 275B, 275Yh, 276Af, 285I, 285Xf, 285Xg, 285Xm-285Xo, 285Xs, 285Yc, 285Yk, 285Yl, 454Xj, 456Xe, 456Xh, 495Ab, 495B, 495D, 495Xf; see also independent increments, relatively independent (458A) independent sets 272Aa, 272Bb, 272F, 272N, 273F, 273K, 464Qb independent subalgebras of a probability algebra 325L, 325Xe, 325Xf, 325Xg, 327Xe, 364Xe, 385Q, 385Sf, 385Xd-385Xf, 387B; see also Boolean-independent (315Xn) independent σ-algebras 272Ab, 272B, 272D, 272F, 272J, 272K, 272M, 272O, 275Ym indiscrete topology 4A2A Individual Ergodic Theorem see Ergodic Theorem (372D, 372E) induced automorphism (on a principal ideal of a Boolean algebra) 381M, 381N, 381Qe, 381Xk, 381Xp, 381Xo, 382J, 382K, 388F, 388G, 388Xb induced topology see subspace topology (2A3C) inductive definitions 2A1B inequality see Cauchy, Chebyshev, Gagliardo-Nirenberg-Sobolev, Hölder, isodiametric, Jensen, Poincaré, Wirtinger, Young infimum in a Boolean algebra 313A-313C infinite game 451U infinitely decomposable see infinitely divisible infinitely divisible distribution 455Xc-455Xe, 455Yc, 455 notes, 495 notes infinity 112B, 133A, §135 information 385Cb initial ordinal 2A1E, 2A1F, 2A1K injective ring homomorphism 3A2Da, 3A2G inner automorphism of a group 3A6B inner measure 113Yh, 212Yc, 213Xe, 213Yc, 324Cb, 413A, 413C, 413Xa-413Xc, 413Ya —– —– defined by a measure 413D, 413E, 413F, 413Xd-413Xg, 413Yb, 418Xt, 431Xe, 443Xa, 451Ob, 463F inner product space 244N, 253Xe, 3A5L, 456Xd, 476H, 476I, 476M-476P, 476Xi-476Xk, 493I, 493J, 493Xf, 4A4J; see also Hilbert space (3A5L) inner regular finitely additive functional 413N, 413Q inner regular measure 256A, 342Aa, 342C, 342Xa, 411B, 412A, 412C, 412H, 412I, 412J, 412K, 412L, 412N, 412O, 412Q, 412R, 412T, 412Xc, 412Xd, 412Xh, 412Xk, 412Xr, 412Ya, 413E-413G, 413I-413K, 413M, 413O, 413S, 413Xh, 413Xj, 414M, 414Xg, 416Yb, 416Yc, 417Ye-417Yg, 418Yi, 418Yj, 431Xb, 434L, 443L, 443Xr, 451K, 452I, 452M, 452Xd, 454A, 454N, 454Yd, 462H, 462Yc, 463Yd, 465Xf, 475F, 475Yb —– —– with respect to Borel sets 412B, 412Pb, 412Sd, 412Uc, 412Xi, 412Xl, 412Xn, 412Yb, §417, 419H, 432Ya, 433A, 433Xa, 433Xf, 434Xl, 459F —– —– with respect to closed sets 256Bb, 256Ya, 342F, 412B, 412E, 412G, 412Pc, 412S, 412Ua, 412Wb, 412Xa, 412Xf, 412Xg, 412Xl, 412Xn, 412Xo, 412Yb, 414M, 414N, 414Xc, 414Ya, 415Xe, 416P, 416Xi, 417C, 417E, 418J, 418Xr, 418Yg-418Yi, 419A, 431Xb, 432D, 432G, 432Xc, 432Xf, 433E, 434A, 434Ga, 434Ha, 434M, 434Yb, 434Yf, 435C, 471De, 471Xh, 471Yb, 438F, 454N, 454Yd, 466H, 482F, 482Xd, 491Xx; see also quasi-Radon measure (411Ha) —– —– with respect to compact sets 256Bc; see also tight (411Ja)

912

Index

inner regular

—– —– with respect to zero sets 412B, 412D, 412Pd, 412Sc, 412Ub, 412Xl, 412Xn, 412Xs, 413Yc, 454J; see also completion regular (411Jb) integrable function §122 (122M), 123Ya, 133B, 133Db, 133Dc, 133F, 133J, 133Xa, 135Fa, 212B, 212F, 213B, 213G, 412Wa, 436Yb, 482F, 482Xc-482Xe; see also Bochner integrable function (253Yf), Henstock integrable function (483A), Pfeffer integrable function (484G), L1 (µ) (242A) integral §122 (122E, 122K, 122M), 363L, 436A; (on an L-space) 356P; see also Henstock integral (483A), integrable function, Lebesgue integral (122Nb), lower integral (133I), McShane integral, Perron integral, Pettis integral (463Ya), Pfeffer integral (484G), Riemann integral (134K), symmetric Riemanncomplete integral, upper integral (133I) integral-geometric measure 475Q integral operator see kernel operator, abstract integral operator (§376) integration by parts 225F, 225Oe, 252Xi, 483Yc, 483Ye integration by substitution see change of variable in integration integration of measures 452A-452D, 452Xa-452Xe, 452Ya, 453Xk interior of a set 2A3D, 2A3Ka, 475Ca, 475R, 4A2Bg; see also essential interior (475B) interpolation see σ-interpolation property (466G), Riesz Convexity Theorem intersection number (of a family in a Boolean algebra) 391H, 391I-391K, 391Xh, 391Xi, 392E interval (in a totally ordered set) 4A2A; see also closed interval (4A2A), half-open interval (114Aa, 115Ab, 4A2A), open interval (111Xb, 4A2A), split interval (343J) invariant function (in ‘G-invariant function’) 394A, 394N-394P, 394Xd, 394Xe, 394Z, 395B, 395Xd —– additive functional 449J, 449Xq-449Xs —– ideal 382R —– integral 441J, 441L —– lower density 346Xd —– mean 449E, 449H, 449I, 449Xi, 449Xo, 449Yc —– measure 436Xs, 441Ad, 441B, 441C, 441H-441L, 441Xa-441Xd, 441Xk, 441Xp, 441Yj, 441Yk, 441Yb, 441Yc, 442Ya, 442Z, 443P, 443Q, 443T, 443Xs, 443Xu, 443Xv, 443Xz, 447E, §448, 449E, 461O, 461P, 461Xl, 461Xm, 461Ye, 461Ye, 476G, 476I, 476M, 476Yc; see also Haar measure (441D), symmetric measure —– see also rearrangement-invariant (374E), T -invariant (374A), translation-invariant inverse (of an element in an f -algebra) 353Pc, 353Yh inverse Fourier transform 283Ab, 283B, 283Wa, 283Xb, 284I, 445P, 445Yg; see also Fourier transform inverse image (of a set under a function or relation) 1A1B inverse-measure-preserving function 132G, 134Yl, 134Ym, 134Yn, 235G, 235H, 235I, 235Xe, 235Ya, 235Yb, 241Xh, 242Xf, 243Xn, 244Xo, 246Xf, 254G, 254H, 254Ka, 254O, 254Xc-254Xf, 254Xh, 254Yb, 324M, 324N, 341P, 341Xd, 341Yc, 341Yd, 343C, 343J, 343Xd, 343Xh, 343Yd, 365Xi, 372H-372L, 372Xe, 372Xf, 372Yh, 372Ym, 386Xb, 385S-385V, 411Ne, 412K, 412M, 413Eh, 413Yc, 414Xa, 414Xq, 414Xs, 416W, 417Xg, 418H, 418M, 418P, 418Q, 418Xf, 418Xg, 418Xn, 418Xs, 418Xt, 418Yl, 419L, 419Xf, 432G, 433K, 433Xg, 433Yd, 436Xs, 437S, 441Xa, 442Xa, 443L, 451E, 452E, 452I, 452O-452R, 452Xl, 452Xm, 452Yb, 453K, 454G, 454J, 456Ba, 456Eb, 456G, 456Yb, 457F, 457Xf-457Xj, 457Yd, 457Za, 457Zb, 458Xo, 459A, 461O, 461P, 461Xl-461Xm, 461Yd, 461Ye, 464B, 465Xe, 491Ea, 495G, 495I, 495Xd, 495Xi; isomorphic inverse-measure-preserving functions 385Tb; see also almost isomorphic (385U); see also ergodic inversemeasure-preserving function (372P), image measure (112E), mixing inverse-measure-preserving function (372P) Inversion Theorem (for Fourier series and transforms) 282G-282I, 282L, 282O, 282P, 283I, 283L, 284C, 284M, 445P; see also Carleson’s theorem invertible element (in a Banach algebra) 4A6H-4A6J;(in an f -algebra) 353Pc, 353Yh involution in a group 3A6A involution in Aut A 382C, 382N, 382Q, 382Xc-382Xe, 382Xg, 382Xi, 382Ya, 382Yc, 383D, 383G, 383Xa, 385Xl, 388Ye; see also exchanging involution (381R), ‘many involutions’ (382O) irreducible continuous function 313Ye isodiametric inequality 264H isolated family of sets 438Nd, 466D, 4A2A, 4A2Ba isolated point 316M, 316Yj, 316Yq, 3A3Pb

Lebesgue

General index

913

isometry 324Yh, 3A4Ff, 3A4G, 441Xl, 448Xf, 448Xi, 474C, 474H, 476N, 476Xi, 4A4Jb —– group (of a metric space) 441F, 441G, 441H, 441Xm-441Xp, 441Ya, 441Yb, 441Ye, 441Yg, 441Yj, 443Xx, 443Xz, 443Ym, 448Xk, 449Xc, 449Xd, 476G, 476M, 476Xb, 476Xd, 476Xi, 476Yc, 493J, 493Xc, 493Xf, 493Xg, 493Yb isomorphic see almost isomorphic (385U) isomorphism see Boolean isomorphism, Borel isomorphism (4A3A), measure space isomorphism, orderisomorphism Isoperimetric Theorem 475 notes, 476L; see also Convex Isoperimetric Theorem Jacobian 263Ea, 484P, 484S Jensen’s inequality 233I, 233J, 242Yi, 365Rb Jensen’s ♦ 4A1M joint distribution see distribution (271C) Jordan algebra > 411Yc, 491Ye Jordan decomposition of an additive functional 231F, 231Ya, 232C, 326D, 326H, 326M Kadec norm 466C, 466D-466F, 467B Kadec-Klee norm see Kadec norm (466C) Kakutani’s theorem 369E, 369Yb; see also Halmos-Rokhlin-Kakutani lemma (386C) Kalton-Roberts theorem 392F, 392J Kampen, E.R.van see Duality Theorem (445U) Kawada’s theorem 394Q Kechris A.S. see Becker-Kechris theorem, Nadkarni-Becker-Kechris theorem Kelley’s criterion 391L; see also Nachbin-Kelley theorem kernel (of a ring homomorphism) 3A2D, 3A2Eb, 3A2G; (of a Souslin scheme) 421B, 421C, 421I, 421Ng, 421O, 421Q; see also Abel-Poisson kernel (282Yg), Dirichlet kernel (282D), Féjer kernel (282D), modified Dirichlet kernel (282Xc) kernel operator 373Xf, §376 Kirzbraun’s theorem 262C Kolmogorov’s extension theorem 454D-454G Kolmogorov’s Strong Law of Large Numbers 273I, 275Yn, 372Xg Kolmogorov-Sinaˇı theorem 385P Komlós’ theorem 276H, 276Yh, 367Xn, 463Xe Kreˇın’s theorem 461I Kreˇın-Mil’man theorem 4A4Gb Kronecker’s lemma 273Cb Kwapien’s theorem 375I Lacey-Thiele Lemma 286M Laplace’s central limit theorem 274Xe Laplace transform 123Xc, 123Yb, 133Xc, 225Xe laterally complete Riesz space 368L, 368M lattice 2A1Ad, 311L, 311Xi, 311Yb, 315Xb, 315Yg, 367A, 367B, 367Xa, 3A1I; see also Riesz space (=vector lattice) (352A), Banach lattice (242G, 354Ab) —– homomorphism 3A1Ia —– norm see Riesz norm (242Xg, 354Aa) —– of normal subgroups 382Xh, 382Xj, 383H, 383J, 383Xg, 383Xh, 383Ya, 3A6C law of a random variable see distribution (271C) law of large numbers see strong law (§273) law of rare events 285Q Lebesgue, H. Vol. 1 intro., Chap. 27 intro. Lebesgue Covering Lemma 2A2Ed Lebesgue decomposition of a countably additive functional 232I, 232Yb, 232Yg, 256Ib Lebesgue decomposition of a function of bounded variation 226C, 226Dc, 226Ya, 232Yb Lebesgue density see lower Lebesgue density (341E)

914

Index

Lebesgue

Lebesgue’s Density Theorem §223, 261C, 275Xg, 471P, 447D, 447Xc, 475Yb; see also Besicovitch’s Density Theorem (472D) Lebesgue’s Dominated Convergence Theorem 123C, 133G, 363Yh, 367J, 367Xf, 436Yb Lebesgue extension see completion (212C) Lebesgue integrable function 122Nb, 122Yb, 122Ye, 122Yf, 483Bb, 483C, 483G, 483Q, 483Xf, 483Xg, 483Xl, 483Yf, 484H Lebesgue integral 122Nb Lebesgue measurable function 121C, 121D, 134Xd, 225H, 233Yd, 262K, 262P, 262Yc, 466Xh, 483Ba, 484Hd Lebesgue measurable set 114E, 114F, 114G, 114Xe, 114Ye, 115E, 115F, 115G, 115Yc Lebesgue measure (on R) §114 (114E), 131Xb, 133Xc, 133Xd, 134G-134L, 212Xc, 216A, Chap. 22, 242Xi, 246Yd, 246Ye, 252N, 252O, 252Xf, 252Xg, 252Yj, 252Yp, §255, 342Xo, 343H, 344K, 345Xc, 346Xe, 372Xd, 383Xf, 384Q, 412Ya, 413Xe, 413Xk, 415Xc, 419J, 419Xf, 453Xb, 439E, 439F, 445Xk, 454Yb, 481Q, 482Ya, 483A —– —– (on Rr ) §115 (115E), 132C, 132Ef, 133Yc, §134, 211M, 212Xd, 245Yj, 251M, 251Wi, 252Q, 252Xh, 252Yu, 254Xk, 255A, 255K, 255L, 255Xd, 255Yc, 255Yd, 256Ha, 256J-256L, 264H, 264I, 342Ja, 344K, 345B, 345D, 345Xf, 411O, 411Ya, 415Ye, 419Xd, 441Ia, 441Xg, 442Xf, 443Yj, 453B, 449M, 471Yi, 476J-476L, 495Xp —– —– (on [0, 1], [0, 1[) 211Q, 216A, 252Yq, 254K, 254Xh, 254Xj-254Xl, 324Yg, 343Cb, 343J, 343Xd, 344K, 385Xj, 385Xk, 385Yd, 415F, 417Xi, 417Xq, 418Xd, 419L, 419Xg, 436Xm, 438Ce, 439Xa, 448Xk, 451Ad, 454Xa, 457G-457I, 457Xg-457Xj, 457Yd, 457Za, 457Zb, 463Xj, 463Yc, 463Ye, 491Xf, 491Yf, 491Z —– —– (on other subsets of Rr ) 242O, 244Hb, 244I, 244Yh, 246Yf, 246Yl, 251Q, 252Ym, 255M, 255N, 255O, 255Ye, 255Yh, 343Cc, 343H, 343M, 344J, 345Xd, 346Xb, 372Yj, 372Yk, 388Yd, 433Xc, 433Yb, 442Xc, 463Zb, 495P, 495Xi, 495Xk, 495Xn —– —– algebra 331Xd, 331Xe, 373C, 374C, 374D, 374Xa, 374Xf, 374Xh, 374Ya, 374Yd, 375Xe, 493Yd Lebesgue negligible set 114E, 115E, 134Yk Lebesgue outer measure 114C, 114D, 114Xc, 114Yd, 115C, 115D, 115Xb, 115Xd, 115Yb, 132C, 134A, 134D, 134Fa Lebesgue set of a function 223D, 223Xf, 223Xg, 223Xh, 223Yg, 261E, 261Ye, 282Yg Lebesgue-Stieltjes measure 114Xa, 114Xb, 114Yb, 114Yc, 114Yf, 131Xc, 132Xg, 134Xc, 211Xb, 212Xd, 212Xi, 225Xf, 232Xb, 232Yb, 235Xb, 235Xg, 235Xh, 252Xi, 256Xg, 271Xb, 224Yh, 342Xe left action (of a group on itself) 4A5Ca, 4A5I left Haar measure 441D; see Haar measure left modular function (of a topological group) see modular function (442I) left-translation-invariant see translation-invariant length of a curve 264Yl, 265Xd, 265Ya length of a finite function 421A length of an interval 114Ab B.Levi’s theorem 123A, 123Xa, 133K, 135G, 135Hb, 226E, 242E, 365Df, 365Dh, 482K Levi property of a normed Riesz space 242Yb, 244Ye, 354Db, 354J, 354N, 354Xi, 354Xn, 354Xo, 354Yi, 356Da, 356L, 356Xk, 356Yi, 365C, 366D, 369G, 369Xe, 369Yg, 371C, 371D, 371Xc, 371Xd Lévy’s martingale convergence theorem 275I, 367Kb Lévy’s metric 274Ya, 285Yd, 437Yi lexicographic ordering 351Xa Liapounoff’s central limit theorem 274Xg Liapounoff’s vector measure theorem 326Ye lifting of a measure (algebra) Chap. 34 (341A, 341Ya), 363Xe, 363Yg, 414Xp; see also almost strong lifting (453A), lifting topology (414Q), linear lifting (363Xe), strong lifting (453A), translation-invariant lifting (345A, 447A) Lifting Theorem 341K, 363Yg lifting topology 414Q, 414R, 414Xo-414Xt, 414Yh, 418Xm, 453Xd limit of a filter 2A3Q, 2A3R, 2A3S limit of a sequence 2A3M, 2A3Sg limit ordinal 2A1Dd

locally

General index

915

Lindeberg’s central limit theorem 274F-274H, 285Ym Lindeberg’s condition 274H Lindel¨ of space 411Ge, 412Xo, 422De, 422Gg, 422Xb, 434Hb, 435Fb, 435Xd, 435Xj, 439Xm, 439Yg, 462Yd, 466Xd, 467Xa, > 4A2A, 4A2H, 4A2Ub; see also hereditarily Lindelöf (4A2A) Lindel¨ of-Σ space see K-countably determined (467H) linear functional 3A5D, 4A4Ac; see also positive linear functional, tight linear functional (436Xn) linear lifting 341Xe, 341Ye, 345Yb, 363Xe, 363Yg linear operator 262Gc, 263A, 265B, 265C, §2A6, 355D, 3A5E, 456Xd, 466Xi, 4A4J; see also bounded linear operator (2A4F), (weakly) compact linear operator (3A5K), continuous linear operator, order-bounded linear operator (355A), positive linear operator (351F), self-adjoint linear operator (4A4Jc) linear order see totally ordered set (2A1Ac) linear space 4A4A linear space topology see linear topological space (2A5A), weak topology (2A5I), weak* topology (2A5I), Ts (U, V ) (3A5E) linear subspace (of a normed space) 2A4C; (of a partially ordered linear space) 351E, 351Rc; (of a linear topological space) 2A5Ec, 4A4B-4A4E linear topological space 245D, 284Ye, 2A5A, 2A5B, 2A5C, 2A5Eb, 2A5F, 2A5G, 2A5H, 2A5I, 367N, 367Ym, 367Yn, 393L, 393M, 393O-393Q, 393Yh, 393Yi, 3A4Ad, 3A4Bd, 3A4Fg, 418Xj, 445Yc, 463A, 4A3T, 4A3U, 4A4A, 4A4B, 4A4H; see also locally convex space (4A4Ca) linked set(in a Boolean algebra) 391L; (in κ-linked set, <κ-linked set) 391Yh; see also σ-linked (391L), σ-m-linked (391Yh) lipeomorphism 484Q, 484S Lipschitz constant 262A, 262C, 262Yi, 471J Lipschitz function 225Yc, 262A, 262B-262E, 262N, 262Q, 262Xa-262Xc, 262Xh, 262Yi, 263F, 264G, 264Yj, 282Yb, 437Yi, 471J, 471Xc, 473C-473E, 473H-473J, 474B, 474D, 474E, 474K, 474M, 475K, 475N, 475Xg, 492H, 492Xb-492Xd; see also lipeomorphism (484Q) local convergence in measure see convergence in measure (245A) local semigroup see full local semigroup (394A), countably full local semigroup (448A) local type (lifting) see strong lifting (453A) localizable measure algebra 322Ae, 322Be, 322C, 322K-322O, 323Gc, 323H, 323J, 323K, 323Xc, 323Yb, 323Z, 324Yf, 325C, 325D, 332B, 332C, 332J, 332O, 332S, 332T, 332Xo, 332Ya, 332Yc, 333H, 333Ia, 364Xk, 365M, 365N, 365Sa, 365T, 366Xb, 366Xf, 367Nc, 367V, 369A, 369B, 369D, 369E, 369R, 373L-373N, 373R, 373T, 373U, 375Yd, 381Xn, 383B-383H, 383Xb, 383Xc, 383Xe, 383Xh, 384M, 384N, 384Pa, 384Ya-384Yd, 391Cb, 394Xg, 394Xh, 394Yd, 395A, 411Pd, 434T; see also localization (322P) localizable measure algebra free product 325D, 325Ea, 325H, 325Xa, 325Xb, 325Xc, 327Ya, 333E-333H, 333K, 333Q, 333R, 334B, 334Xb, 376B, 376E, 376F, 376Xb, 376Yb, 438S, 465O; see also probability algebra free product (325K) localizable measure (space) 211G, 211L, 211Ya, 211Yb, 212G, 213Hb, 213L-213N, 213Xl, 213Xm, 214Id, 214J, 214Xa, 214Xd, 214Xf, 216C, 216E, 216Ya, 216Yb, 234Fc, 234G, 234Ye, 241G, 241Ya, 243G, 243H, 245Ec, 245Yf, 252Yq, 252Yr, 252Yt, 254U, 322Be, 322N, 325B, 342N, 342Yb, 363S, 364Xm, 419C, 419D, 433A, 438Xb, 463Xm; see also strictly localizable (211E) localization of a semi-finite measure algebra 322P, 322Yb, 332Yb, 365Xm, 366Xe, 366Xf, 369Xp locally compact Hausdorff space 416G-416I, 416L, 416T, 416Xj, 416Xv, 416Yd, 419Xf, 434Jg, 435Xe, 436J, 436K, 436Xo, 436Xq, 437I, 441C, 441H, 441Xp, 462E, 476Cd, 476Xd, 495N, 495O, 495Xm, 4A2Ge, 4A2Kf, 4A2Rk, 4A2Sa locally compact measure (space) 342Ad, 342H, 342I, 342L-342N, 342Xc, 342Xd, 343B, 343Xa, 343Xg, 343Ya, 344H locally compact topological group 434 notes, 441E, 441Xh, 441Xn, §442, 443E, 443K, 443L, 443O-443S, 443Xs, 443Xt, 443Xw, 443Xx, 443Yb, 443Yc, 443Yk, 443Yl, 445Ac, 445J, 445U, §446, 447E-447I, 447Xa, 449H, 449I, 449Xo, 449Xq, 493K, 4A5E, 4A5Jb, 4A5M-4A5P, 4A5R, 4A5S locally compact topological space 3A3Ag, 3A3B, 3A3G, 3A3Hb, 414Xj, 415Ym, 422Xb, 441Yc; see also locally compact Hausdorff space, locally compact topological group locally convex linear topological space 393Yi, §461, 462Yb, 463A, 466A, 466K, 466Xa, 466Xc, 466Xd, 466Xj, 466Ya, 4A3V, 4A4C, 4A4D-4A4G

916

Index

locally

locally determined measure (space) 211H, 211L, 211Ya, 216Xb, 216Ya, 216Yb, 251Xc, 252Ya, 322N, 465Xi; see also complete locally determined measure locally determined negligible sets 213I, 213J-213L, 213Xj-213Xl, 214Ib, 214Xg, 214Xh, 216Yb, 234Yb, 252Yb, 376Yk, 431Xd locally finite family of sets 4A2A, 4A2Bh, 4A2Fd, 4A2Hb locally finite measure 256A, 256C, 256G, 256Xa, 256Ya, 411Fa, 411G, 411Pe, 411Xc, 411Xg-411Xi, 412Xo, 414Xj, 415G, 416C, 416Dd, 416F-416H, 416M, 416P, 416S, 416T, 416Xi, 417Xb, 419A, 432C-432G, 432Xf, 433Xf, 434J, 435B, 441Xj, 442Aa, 451N, 451Rb, 466A; see also effectively locally finite (411Fb) locally finite perimeter 474D, 474E-474N, 474Q, 474R, 474Xa-474Xd, 474Ya, 475D-475G, 475J, 475L475P, 475Xh, 484L-484N, 484Rb locally finite set 495Xc locally integrable function 242Xi, 255Xh, 255Xi, 256E, 261Xa, 262Yg, 411Fc, 411G, 411Xb, 411Xe, 414Xf, 415Pb, 415Xo, 447Xc, 472D, 473D, 473Ee locally uniformly convex see locally uniformly rotund (467A) locally uniformly rotund norm 467A, 467B-467G, 467K, 467N, 467Yb Loeb measure 413Xn, 451Xd Loève, M. Chap. 27 intro. logistic map see quadratic map (372Xn) Loomis-Sikorski theorem 314M Lorentz norm 374Xj, 374Yb Losert’s example 453N, 453Xk lower density on a measure space §341 (341C, 341Ya), 343Yc, §345, 346F, 346G, 346Xa, 346Xc, 346Xd, 346Ye, 346Yf, 346Zb, 363Xe, 414Yf, 447Yb, 453Xh, 472Xc; see also density topology (414P), lower Lebesgue density (341E), translation-invariant lower density (447A) lower derivate 483H, 483J, 483Xm lower integral 133I, 133J, 133Xe, 135H, 364Xm lower Lebesgue density 223Yf, 341E, 414Xm, 414Ye, 414Yg, 414Yh, 475B, 475Xe lower limit topology see Sorgenfrey topology (415Xc) lower Riemann integral 134Ka lower semi-continuous function 225H, 225I, 225Xl, 225Xm, 225Yd, 225Ye, 323Cb, 367Xx, 412W, 412Xq, 414A, 414B, 417B, 437Jd, 443Xc, 444Fa, 452C, 452D, 452Xa, 461C, 461Xa, 472F, 476F, 4A2A, 4A2Bd, 4A2Gk, 4A3Ce Lusin measurable function see almost continuous (411M) Lusin’s theorem 134Yc, 256F, 451S Luxemburg, W.A.J. 363 notes Mackey topology (on a linear space) 466Ya, 4A4F magnitude of (an element in) a measure algebra 332Ga, 332J, 332O, 383Xd, 383Xe, 383Xh magnitude of a measure space 332Ga, 343Yb, 344I, 438Xb Maharam algebra see localizable measure algebra (322Ae) Maharam measure (space) see localizable (211G) Maharam submeasure 392G, 392H-392J, 392Xc, 392Xd, 392Xf, 392Xg, 392Ya, 392Yb, 393A-393C, 393E, 393M, 393Xc-393Xf, 393Ya, 393Yf, 393Yg, 438Ya, 491Id Maharam’s theorem 331I, 332B Maharam type of a Boolean algebra 331F, 331H, 331Xc, 331Xg, 331Xh, 331Xj, 331Yd, 332H, 365Ya, 373Yb, 382Xl, 393Xe, 393Xf, 393Ya, 393Yf, 448L; see also relative Maharam type (333Aa) —– —– of a measure algebra 331I-331K, 331Xd-331Xf, 331Xi, 332M, 332N, 332R-332T, 333D, §334, 369Xg, 375Kb, 387K, 387L, 387Xc, 388K, 388L, 438S, 438Xh, 448Xj, 493Xe —– —– of a measure (space) 331Fc, 365Xp, 367Xr, 416Yg, 433A, 433Xa, 433Ya, 434Yl, 438Yc, 438Yj, 439A, 453Zb, 456Yd, 464Qb Maharam-type-homogeneous Boolean algebra 331Fb, 331H, 331Xh, 331Xj, 332A, 332H, 332Xa; see also homogeneous (331M) —– —– measure algebra 331I, 331K, 331L, 331N, 331Xd, 331Xe, 331Yj, 344Xe; see also relatively Maharam-type-homogeneous (333Ac)

measurable

General index

917

—– —– measure (space) 331Fc, 334Xe, 334Xf, 334Ya, 341Yc, 341Yd, 346E Maharam-type-κ component in a Boolean algebra 332Gb, 332H, 332J, 332O, 332P, 332Xj, 332Ya, 384N, 394Xh many involutions (group with many involutions) 382O, 382P-382R, 382Xe, 382Xg-382Xj, 383G, 384A, 384C, 384D, 384I, 394Xc Marczewski functional 343E marginal measure 452L, 452M, 452N, 459G Maˇr´ık’s theorem 435C, 439O Markov process 455B Markov time see stopping time (275L) Marriage Lemma see Hall’s Marriage Lemma (3A1K) martingale §275 (275A, 275Cc, 275Cd, 275Ce); see also reverse martingale martingale convergence theorems 275G-275I, 275K, 275Xf, 367K, 369Xq martingale difference sequence 276A, 276B, 276C, 276E, 276Xd, 276Ya, 276Yb, 276Ye, 276Yg martingale inequalities 275D, 275F, 275Xb, 275Yc-275Ye, 276Xb, 372 notes, 492G, 492Xa Max-flow Min-cut Theorem 332Xk, 4A4N maximal element in a partially ordered set 2A1Ab Maximal Ergodic Theorem 372C maximal ideal space of a Banach algebra 445H maximal theorems 275D, 275Yc, 275Yd, 276Xb, 286A, 286T, 372C, 372Yb, 472E, 472F, 472Yd Mazur S. see Banach-Mazur game McMillan see Shannon-McMillan-Breiman theorem (386E) McShane integral 481M, 481N, 482Xh meager set 314L, 314M, 314Yd, 316I, 316Yf, 316Yi, 341Yb, 3A3F, 3A3Ha, 433Yc, 434Ye, 4A2Ma mean (of a random variable) see expectation (271Ab) Mean Ergodic Theorem 372Xa —– see also convergence in mean (245Ib) measurable additive functional (on PI) 464I, 464J, 464K, 464M, 464Q measurable algebra 391B, 391C, 391D, 391K, 391Xf, 391Z, 392J, 393A, 393J, 393Ye, 394P, 394Q, 394Xf, 395B, 395Xd measurable cover see measurable envelope (132D) measurable envelope 132D, 132E, 132F, 132Xf, 132Xg, 134Fc, 134Xc, 213K-213M, 214G, 216Yc, 322I, 322J, 413Ei, 419J, 414Xl, 414Xo, 414Xp; see also Haar measurable envelope (443Ab), full outer measure (132F) measurable envelope property 213Xl, 214Xl, 364Xm, 431Ya, 433Xa measurable function (taking values in R) §121 (121C), 122Ya, 212B, 212F, 213Yd, 214La, 214Ma, 235C, 235K, 252O, 252P, 256F, 256Yb, 256Yc, 316Yi, 322Yf, 413G, 414Xf, 414Xk, 414Xr, 418Xl, 418Xm, 418Xr, 463K, 482E, 482Ya —– —– (taking values in Rr ) 121Yf), 256G, 411Ya —– —– (taking values in other spaces) 133Da, 133E, 133Yb, 135E, 135Xd, 135Yf, 411L, 418A-418C, 418E, 418J, 418K, 418R-418T, 418Xa-418Xd, 418Xx, 418Ya-418Yc, 418Yf-418Yh, 419Xe, 419Xg, 431Xa, 433E, 438D-438G, 438Xf-438Xi, 451N, 451O, 451Q-451S, 451Xn, 451Xp, 451Yf, 452Xb, 452Xc, 4A3Cb, 4A3Db, 4A3Ke —– —– ((Σ, T)-measurable function) 121Yb, 235Xc, 251Ya, 251Yc, 343A, 411L, 423O, 424B, 424Xh, 434De, 443Yg, 451Yl, 459B, 4A3B, 4A3C, 4A3Db, 4A3K, 4A3Ne —– —– see also Borel measurable, Haar measurable (443Ae), Lebesgue measurable, measurable additive functional (464I), measurable selector, scalarly measurable (463Ya), universally measurable (434Dd), universally Radon-measurable (434Ec) measurable neighbourhood gauge 483Ya measurable selector 423M-423O, 423Xf, 423Xg, 424Xg, 424Xh, 433F, 433G measurable set 112A; µ-measurable set 212Cd; see also Haar measurable (442H), relatively measurable (121A), universally measurable (434D), universally Radon-measurable (434E) measurable space 111Bc measurable space with negligibles 431Yb, 431Yc

918

Index

measurable

measurable transformation §235, 365H; see also inverse-measure-preserving function measure 112A; see also control measure (393O), vector measure (393O) —– (in ‘µ measures E’, ‘E is measured by µ’) > 112Be measure algebra 211Yb, 211Yc, vol. 3 (321A), 448Xj, 448Xk, 493H, 495J, 495Xf, 495Xh; isomorphic measure algebras 331I, 331L, 332B, 332C, 332J, 332K, 332P, 332Q, 332Ya, 332Yb measure algebra of a measure space 321H-321K (321I), 322B, 322N, 327C, 412N, 414A, 415L, 416C, 416W, 418T, 441Ka, 441Yk, 474 notes, 493G; see also Haar measure algebra (442H), usual topology on a measure algebra (323A) measure-algebra topology §323 (323A), 324F-324H, 324K, 324Xb, 324Xc, 327B, 327C, 365Ea, 367R, 367Xs, 367Yo, 411Yc, 412N, 443Aa, 443C, 444Fc, 444Yd, 448Xj, 493E, 494A; see also strong measurealgebra topology (323Ae) measure-algebra uniformity 323A, 323B-323D, 323G, 324F, 324H, 327B, 443Aa measure-compact (topological) space 435D, 463F-463H, 435Xb, 435Xh, 435Xi, 435Xk, 435Yb, 436Xf, 436Xg, 438Jb, 438Xk, 439P, 439Q, 454Pa, 454Xm; see also strongly measure-compact (435Xk), Borelmeasure-compact (434Gb) measure-free cardinal 438A, 438B, 438C, 438I, 438R, 438S, 438Xa-438Xe, 438Xh, 438Xj, 438Xn-438Xp, 438Yb-438Yf, 438Yi, 439Cd, 439J, 439L, 452Yb, 454Yb, 466F; see also Banach-Ulam problem measure-free weight 438D-438H, 438J, 438M, 438S, 438Xf, 438Xg, 438Xi, 438Xk, 448Xi, 467Pb measure metric (on a measure algebra) 323A, 323Xg, 324Yh, 448Xk, 493G, 493Yb measure-preserving Boolean automorphism 332L, 333Gb, 333P-333R, 333Yc, 343Jc, 344C, 344E-344G, 344Xf, 364Xt, 372Xk, 372Xl, 372Yi, 373Yb, 374E, 374G, 374L, 374Yc, 383D, 383Yc, 385P, 385Sb, 385Sf, 385Xg, 385Yd, 386N, 387B-387H, 387J, 387L, 388A, 388C, 388E-388H, 388J-388L, 388Xa-388Xe, 388Yd, 395Xc; see also automorphism group of a measure algebra, two-sided Bernoulli shift (385Qb) measure-preserving Boolean homomorphism 324I, 324J-324P, 324Xd, 324Xe, 324Yf, 324Yh, 325C, 325D, 325I, 325J, 325Xd, 327Xa, 331D, 332L-332O, 332Q, 332Xm, 332Xn, 333C, 333D, 333F, 333Gb, 333Xc, 333Yd, 343B, 366Xf, 369Xm, 369Xn, 372G, 372H, 372Xc, 373Bd, 373U, 373Xi, 383Yc, 385L-385Q, 385T, 385V, 385Xf-385Xi, 385Xm, 385Xp, 385Yb-385Ye, 386A-386F, 386L, 386Xe, 386Xf, 386Ya-386Yc, 387A, 387B, 387F, 458K, 491N, 491Xy; isomorphic measure-preserving Boolean homomorphisms 385Ta; see also automorphism of a measure algebra, Bernoulli shift (385Q), measure-preserving ring homomorphism measure-preserving function see inverse-measure-preserving function (235G), measure space automorphism, measure space isomorphism measure-preserving ring automorphism 366L, 366Xh measure-preserving ring homomorphism 361Ad, 365O, 365Q, 366H, 366K, 366L, 366Xg, 366Xh, 366Yg, 369Xl, 372F; see also measure-preserving Boolean homomorphism measure space §112 (112A), 113C, 113Yi measure space automorphism 255A, 255Ca, 255N, 255Ya, 255 notes, 344C, 344E-344G, 344Xf, 345Ab, 345Xa, 346Xd, 385Sb, 419Xh, 443Xb, 482 notes measure space isomorphism 254K, 254Xj-254Xl, 255Ca, 255Mb, 344I, 344J, 344K, 344L, 344Xa-344Xc, 344Xe-344Xg, 344Yc, 433Xf, 433Yb measure zero, cardinal of see measure-free cardinal (438A) metacompact (topological) space 435Xj, 438J, 438Yd-438Yf, 451Ye, > 4A2A, 4A2Fg; see also countably metacompact (434Yf), hereditarily metacompact (4A2A) metric 2A3F, 2A4Fb; see also entropy metric (385Xq), Euclidean metric (2A3Fb), Hamming metric (492D), Hausdorff metric (246Yb, 476A), Huntingdon’s metric 437Yi, Lévy’s metric (274Ya), measure metric (323Ad), pseudometric (2A3F), right-translation-invariant metric (4A5Q) metric gauge see uniform metric gauge (481E) metric outer measure 264Xb, 264Yc, 471B, 471C, 471Ya metric space 224Ye, 261Yi, 316Yj, 3A4Fe, 3A4Ff, 3A4Hc, 434L, §471, 448Xi, 4A2Lg, 482E, 484Q; see also complete metric space, isometry group (441F), metrizable space (2A3Ff) metrically discrete see σ-metrically-discrete (4A2A) metrically transitive transformation see ergodic inverse-measure-preserving function (372Pb) metrizable (topological) space 2A3Ff, 2A3L, 412E, 418B, 418F, 418G, 418Xc, 421Yb, 423D, 434Ye, 437Yj, 437Yn, 438D-438G, 438I, 438Jc, 438Xk, 438Yc, 439Yd, 441Xm, 451Q-451S, 451Xm, 461K, 461N,

norm

General index

919

462Xa, 491Cg, 491Xk, 4A2L, 4A2Nh, 4A3Kb, 4A4Cf; see also compact metrizable space, metric space, separable metrizable space, Polish space (4A2A) —– linear topological space 462Yb, 466A, 466Xa, 466Xc, 466Xd, 4A3V; see also normed space —– topological group 446Ec, 4A5Q-4A5S —– uniformity 3A4Bc, 445Ya mid-convex function 233Ya, 233Yd minimal element in a partially ordered set 2A1Ab mixing inverse-measure-preserving function 372P, 372Q, 372R, 372Xk, 372Xm-372Xo, 372Xq, 372Xr, 372Xu, 372Xv, 372Yh, 372Yl mixing measure-preserving homomorphism 333P, 333Yc, 372P, 372Q, 372R, 372Xk, 372Xl, 372Xp, 372Yl, 372Yn, 385Se, 387Xb, 388Yc mixture of distributions 459C; see also disintegration (452E) moderated set function 482Ib, 482K, 482M, 482Xj, 482Xk, 482Yc modified Dirichlet kernel 282Xc modular function (of a group carrying Haar measures) 442I, 442J, 442K, 442Xf, 442Xh, 443G, 443Q-443S, 443Xo, 443Yo, 444J, 444M, 444O, 444Xn, 447E, 447F modular functional on a lattice 413Xp; see also supermodular (413P) modulation 286C monocompact class of sets 413Yd Monotone Class Theorem 136B, 312Xc, 313G, 313Xd Monotone Convergence Theorem see B.Levi’s theorem (133A) monotonic function 121D, 222A, 222C, 222Yb, 224D, 313Xh; (space of monotonic functions) 463Xg, 463Xh Monte Carlo integration 273J, 273Ya, 465H, 465M multilinear map 253Xc multiplicative identity (in a Banach algebra) 4A6Ab; (in an f -algebra) 353P, 353Yg multiplicative linear functional on a Banach algebra 445H, 4A6F, 4A6J, 4A6K multiplicative Riesz homomorphism 352Xm, 353Pd, 361J, 363F, 364R Nachbin-Kelley theorem 363R, 363Yj Nadkarni-Becker-Kechris theorem 448P narrow topology (on a space of measures) > 437J, 437K-437M, 437O, 437Q-437S, 437Ub, 437V, 437Xi437Xk, 437Xo, 437Xs, 437Yk, 437Yl, 437Yn, 438Xl, 444Xc, 449Yc, 452Xb, 452Xe, 458Yf, 459F, 459G, 459Xc-459Xe, 495O, 495Xm, 495Xn natural extension (of an inverse-measure-preserving function) 383Yc negligible set 112D, 131Ca, 214Cb, 344H, 344Yf, 412Jb, 412Xi, 413Xh, 436Yb, 482L; see also Haar negligible (442H), Lebesgue negligible (114E, 115E), null ideal (112Db), universally negligible (439B) neighbourhood (of a point in a topological space) 4A2A; see also base of neighbourhoods (4A2A) neighbourhood gauge 481E, 481J-481P, 481Xh, 482E-482I, 482Xc-482Xf, 482Xj, 482Xk, 482Ya-482Yc, 483A, 483Ya; see also uniform metric gauge (481E) network in a topological space 438Ld, 466D, 467Pb, 4A2A, 4A2Ba; see also countable network Neumann, J. von 341 notes; von Neumann-Jankow selection theorem 423O, 423Xg, 423Yd, 424Xh, 433F, 433G Nikod´ ym see Radon-Nikod´ ym Nirenberg see Gagliardo-Nirenberg-Sobolev inequality (473H) no small subgroups (in ‘topological group with no small subgroups’) 446E, 446H-446K, 446M, 446O non-decreasing sequence of sets 112Ce non-increasing sequence of sets 112Cf non-measurable set 134B, 134D, 134Xg, 419J non-principal ultrafilter 464Ca, 464Jc non-scattered compact set 494B, 494C, 494Xa-494Xc non-stationary set 4A1C norm 2A4B, 4A4Cc, 4A4Ia; (of a linear operator) 2A4F, 2A4G, 2A4I; (of a matrix) 262H, 262Ya; (norm topology) 242Xg, 2A4Bb; see also Fatou norm (354Da), Kadec norm (466C), Riesz norm (354Aa)

920

Index

normal

normal density function 274A, 283N, 283We, 283Wf normal distribution 274Ad, 495Xg; see also standard normal (274A) normal distribution function 274Aa, 274F-274K, 274M, 274Xe, 274Xg normal filter 4A1Ic, 4A1J-4A1L normal random variable 274A, 274B, 285E, 285Xm, 285Xn, 455C, 456Ab, 456M normal subgroup 382R, 382Xg, 382Xi, 382Yb, 383Ic, 383Xe, 383Yb; see also lattice of normal subgroups —– —– of a topological group 443R, 443S, 443Xv-443Xx, 443Yc, 443Yk, 449C, 493Bc, 4A5J-4A5L, 4A5S normal subspace (of a Riesz space) see band (352O) normal topological space 411Yc, 412Xg, 412Yd, 414Yg, 415Yj, 418Xr, 435C, 435Xg, 435Xh, 438Jb, 438Yf, 476Cc, 476Yb, 491Cd, 491Xs, > 4A2A, 4A2F-4A2H, 4A2Rc; see also perfectly normal (4A2A) —– see also canonical outward-normal function (474G), Federer exterior normal (474O), outward-normal function (474F) normalized Haar measure 442Ie normalized Hamming metric 492D normalized Hausdorff measure 264 notes, §265 (265A), Chap. 47 normalizer of a subgroup 443Oc normed algebra 2A4J, 3A5J, 4A6Aa, 4A6D normed space 224Yf, §2A4 (2A4Ba), 451Xp, 461F, 462D, 466C-466E, §467, 4A4I; see also Banach space (2A4D) Novikov’s Separation Theorem see Second Separation Theorem nowhere aperiodic Boolean homomorphism 381Xk nowhere dense set 313R, 313Yb, 315Yb, 316I, 316Yc, 316Yf, 316Yh, 3A3F, 443N, 4A5Kb nowhere rigid Boolean algebra 384H, 384I-384L, 384Xa-384Xc null ideal 112Db; see also Lebesgue null ideal null set see negligible (112Da), Haar null (444Ye) odd function 255Xb, 283Yd odometer transformation 388E, 388Ye, 445Xp Ogasawara’s representation theorem for Riesz spaces 368 notes one-parameter subgroup (in a topological group) 446I one-point compactification 311Ya, 3A3O, 423Xi, 435Xb, 462Xa, 467Xi open interval 111Xb, 114G, 115G, 1A1A, 2A2I, 4A2A, 4A2Ra open map 314Xj, 4A2A, 4A2Bf, 4A5Ja open set (in Rr ) 111Gc, 111Yc, 114Yd, 115G, 115Yb, 133Xb, 134Fa, 134Yj, 135Xa, 1A2A, 1A2B, 1A2D, 256Ye, 2A3A, 2A3G; (in R) 111Gc, 111Ye, 114G, 134Xc, 2A2I; see also topology (2A3A) operation A see Souslin’s operation (421B) operator topology see strong operator topology (3A5M), very weak operator topology (373K) optional time see stopping time (275L) orbit of a bijection 382Xa, 388A, 388Xa, 388Xb, 381Qc, 388Ya —– of a group action 441C, 441Xc, 441Yg order (in a Boolean ring) 311H order-bounded dual Riesz space (U ∼ ) §356 (356Aa), 362A, 363K, 365L, 365M, 366D, 366Ya, 368P, 371Xe, 375Yb, 436I, 437A order-bounded set (in a partially ordered space) 2A1Ab, 354Xd, 368Xe order-bounded linear operator (between Riesz spaces) §355 (355A); see also order-bounded dual (356A), L∼ (355A) order-closed set in a partially ordered space 313Da, 313Fa, 313Id, 313Xb, 313Xc, 314Ya, 316Fb, 323D, 352Oa, 354Xp, 4A2Re; (ideal in a Boolean algebra) 313Eb, 313Pa, 313Qa, 313Xo, 314Xb, 314Yh, 316Xh, 316Xi, 316Xn, 316Xu; (Riesz subspace of a Riesz space) 353J, 354Xp, 375C; (subalgebra of a Boolean algebra) 313E-313G, 314E-314H, 314Ja, 314Xg, 314Xh, 314Yc, 315Xm, 322Md, 322Xh, 323H-323J, 323Yc, 331E, 331G, 331Yb, 331Yc, 333Bc, 393Xd; see also sequentially order-closed (313Db) order-complete see Dedekind complete (241Ec, 314Aa) order-continuous bidual Riesz space (U ×× ) 356I, 356J, 369D

orthonormal

General index

921

order-continuous Boolean homomorphism 313L-313N, 313Xl-313Xn, 313Xp, 313Xq, 313Yf, 314Fa, 314G, 314R, 314T, 314Xj, 315Jc, 315P, 315Ya, 316Fd, 316Xy, 322Yd, 324E-324G, 324K, 324Ya, 324Ye, 324Yf, 325Aa, 325C, 325D, 325H, 326Kf, 331H, 331J, 332Xm, 332Xn, 334Xb, 343B, 344A, 344C, 344E, 344F, 352Xi, 363Ff, 364Rc, 364Yh, 366Xf, 369Xm, 369Xn, 373Bd, 373U, 375Ya, 381Ec, 381F, 395Xa, 416W, 491Nsee also regular embedding (313N) order-continuous dual Riesz space (U × ) §356 (356Ac), 362A, 363K, 363S, 365L, 365M, 366D, 366Ya, 366Yc, 367Xf, 368Pc, 368Yj, 369A, 369C, 369D, 369K, 369Q, 369Xa, 369Xe, 369Yh, 371Xe, 375B, 375J, 375Yd, 437A, 437C, 438Xd, 464R order-continuous norm (on a Riesz space) 242Yc, 242Ye, 244Yd, 313Yd, 326Yj, 354Dc, 354E, 354N, 354Xi, 354Xj, 354Xl, 354Xm, 354Xo, 354Xp, 354Yc-354Yh, 355K, 355Yg, 356D, 356M, 356Yg, 356Yi, 365C, 366D, 367E, 367Xu, 367Ym, 367Yq, 369B, 369Xf, 369Xg, 369Yc-369Yf, 371Xa-371Xc, 371Ya, 374Xe, 374Xg, 374Yb, 376L, 376M, 415Yl, 444Yj, 467N order-continuous order-preserving function 313Ha, 313I, 313Xi, 313Yc, 313Yd, 315D, 315Yg, 316Fc, 326Kc, 361Cf, 361Gb, 363Eb, 363Ff, 366Yd, 367Xb, 367Yb, 385Ya, 386Yb, 386Yc, 392I, 392Xb, 394N, 4A2Ro; see also sequentially order-continuous order-continuous positive linear operator 351Ga, 355G, 355H, 355K, 364Rc, 365P, 366H, 368I, 375Xa, 375Xb, 375Yd; see also order-continuous dual (356A), order-continuous Riesz homomorphism, L× (355G) order-continuous Riesz homomorphism 327Yb, 351Xc, 352N, 352Oe, 352Rb, 352Ub, 352Xe, 353O, 353Xc, 356I, 361Je, 355F, 365O, 366H, 367Xh, 368B, 375C, 375Xc, 375Ya order-continuous ring homomorphism (between Boolean rings) 361A, 361Je, 365P order*-convergent sequence in a partially ordered set 245Xc, 356Xd, 367A, 367B-367K, 367Xa-367Xn, 367Xs-367Xu, 367Ya-367Yk, 367Yq, 368Yg, 368Yh, 369Xq, 376G, 376H, 376Xe, 376Xg, 376Ye, 376Yh, 393Xa, 393Yd, 393Ye; (in C(X)) 367L, 367Yh-367Yj; (in L0 (µ)) 245C, 245K, 245L, 245Xc, 245Xd, 376J; (in L0 (A)) 367Na, 367Q, 367Xq, 367Xt, 367Yg, 367Yp, 368Yi, 372D-372G, 372Xb, 386E, 386F order-convex set in a partially ordered set 351Xb; see also interval (4A2A) order-dense set in a Boolean algebra 313J, 313K, 313Xm, 314Yg, 332A, 412N, 4A3Rc, 4A3S order-dense Riesz subspace of a Riesz space 352N, 353A, 353D, 354Ef, 354I, 354Ya, 355F, 355J, 356I, 363C, 363Xd, 364L, 365F, 365G, 367Ob, 367Ye, 368B-368E, 368G-368I, 368M, 368P, 368S, 368Ya, 368Ye, 369A, 369B, 369C, 369D, 369G, 375D, 375Xc; see also quasi-order-dense (352Na) order-dense subalgebra of a Boolean algebra 313O, 313Xj, 313Xn, 313Ye, 314I, 314T, 314Xf, 316Xf, 316Xo, 316Xr, 316Yk, 316Yn, 323Dc, 363Xd, 391Xj order-isomorphism 312L order*-limit see order*-convergent (367A) order-preserving function 313H, 313I, 313La, 315D, 324Yg, 326Bf, 361Ce order topology (on a partially ordered set) 313Xb, 313Xj, 313Yc, 326Yr, 367Yk, 393Xa, 393Yd; (on a totally ordered set) 434Yl, 438Yj, 4A2A, 4A2R, 4A2S order unit (in a Riesz space) 243C, 353L, 353M, 363N, 368Ya; see also order-unit norm (354Ga), standard order unit (354Gc), weak order unit (353L) order-unit norm 354F, 354G (354Ga), 354I-354K, 354Yi, 353Yf, 355Xc, 356N, 356O; see also M -space (354Gb) ordered set see partially ordered set (2A1Aa), totally ordered set (2A1Ac), well-ordered set (2A1Ae) ordinal 2A1C, 2A1D-2A1F, 2A1K ordinate set 252N, 252Yg, 252Yh Orlicz norm 369Xd, 369Xi, 369Xj, 369Xn, 369Yc, 369Yd, 369Ye, 369Yg, 374Xc Orlicz space 369Xd, 369Xi, 369Xj, 373Xm Ornstein’s theorem 387I, 387K orthogonal complement 4A4Je orthogonal group on Rr 441Xk, 441Yi, 445Yd orthogonal matrix 2A6B, 2A6C orthogonal projection in Hilbert space 244Nb, 244Yj, 244Yk, 366K —– in Rr 475H, 475P, 475Q, 475S orthonormal basis 416Yg, 444Yk, 456Yb, 4A4Ja orthonormal vectors 2A6B, 4A4Jg

922

Index

oscillation

oscillation of a function 483O, 483P, 483Q Ostaszewski’s ♣ 439O, 4A1M, 4A1N outer automorphism of a group 384G, 384O, 384Pb, 384Q, 384Yc, 384Yd, 3A6B outer measure §113 (113A), 114Xd, 132B, 132Xg, 136Ya, 212Ea, 212Xa, 212Xb, 212Xg, 213C, 213Xa, 213Xg, 213Xk, 213Ya, 251B, 251Wa, 251Xd, 254B, 264B, 264Xa, 264Ya, 264Yo, 393T, 393Xh, 393Yh, 413Xd, 452Xi, 463F, 463Xa; see also Lebesgue outer measure (114C, 115C), metric outer measure (264Yc, 471B), regular outer measure (132Xa), submeasure (392A) —– —– defined from a measure 113Ya, §132 (132B), 213C, 213F, 213Xa, 213Xg-213Xj, 213Xk, 213Yd, 214Cd, 215Yc, 235Ya, 251O, 251R, 251Wk, 251Wm, 251Xm, 251Xo, 252Yh, 254G, 254L, 254S, 254Xb, 254Xq, 254Yd, 264Fb, 264Ye, 412Ib, 412Jc, 413E-413G, 413Xd, 417Xj, 431E, 431Xd, 432Xf, 471Dc, 471Yb, 451Oc, 451Xk, 464C, 464D outer regular measure 256Xi, 411D, 411O, 411Pe, 411Xa, 411Yb, 412W, 412Xo-412Xq, 412Yf, 415Xh, 415Yc, 416Yd, 419C, 432Xb, 482F, 482Xd, 482Xh outward-normal function 474E, 474F, 474H, 474I, 474K, 474M, 474Ya; see also canonical outward-normal function (474G) paracompact (topological) space 415Xo, > 4A2A, 4A2F, 4A2Hb, 4A2Lb; see also countably paracompact (4A2A) Parseval’s identity 284Qd partial derivative 123D, 252Yj, 262I, 262J, 262Xh, 262Yb, 262Yc partial lower density on a measure space 341Dc, 341N, 447Ab, 447B, 447F-447H partial order 441Yj; see also partially ordered set (2A1Aa) partially ordered linear space 241E, 241Yg, 326C, §351 (351A), 355Xa, 361C, 361G, 362Aa; see also Riesz space (352A) partially ordered set 2A1Aa, 313D, 313F, 313H, 313I, 313Xb, 313Xg, 313Xh, 313Yc, 315C, 315D, 315Xb, 315Yg, 466G partition 1A1J partition of unity in a Boolean ring or algebra 311G, 313K, 313L, 313Xk, 315E, 315F, 315Xk, 315Xq, 316Xp, 322E, 332E, 332I, 332Xi, 332Yb, 352T, 375H, 375I, 381C, 381D, 381H, 381Ia, 381N, 381Xf, 381Xg, 382Fb, 383F, 385C, 385D, 385G-385P, 385R, 385Xa-385Xe, 385Xi, 385Xq, 385Xr, 385Ya-385Yc, 385Yg, 386E, 386I, 386J, 386L, 386M, 386O, 386Xd, 386Xe, 388Ib, 392Xg; see also Bernoulli partition (387A) pathological submeasure 393Tc Peano curve 134Yl-134Yo perfect measure (space) 342K, 342L, 342M, 342Xh-342Xo, 343K-343M, 343Xg, 343Xh, 344C, 344I, 344Xb-344Xd, 344Yf, 385V, 416W, 451Ad, 451C-451G, 451I-451O, 451Xa, 451Xc, 451Xe, 451Xf, 451Xi, 451Xl, 451Yj, 454Ab, 454C-454E, 454G, 454J, 457F, 463F-463I, 463Xl, 494Ya, 495F, 495Xe; see also perfect measure property (454Xd) perfect measure property 454I, 454Xd, 454Xe, 454Xh, 454Yb, 454Yc, 455Ba perfect Riesz space 356J, 356K-356M, 356P, 356Xg, 356Xi-356Xk, 356Ye, 356Yg, 365C, 365N, 366D, 369D, 369K, 369Q, 369Ya, 374M, 374Xk, 374Xl, 374Ya, 376P, 376Xl, 437Yj perfectly normal topological space 412E, 422Xe, 434Xj, 434Yj, 438Jc, > 4A2A, 4A2Fi, 4A2Hc, 4A2Lc, 4A2Rn, 4A3Kb, 4A3Xd perimeter 474D, 475Mb, 475Xj, 475Xk, 475Yc, 484B, 484C, 484Ya, 484Yb; see also Cauchy’s Perimeter Theorem (475S), locally finite perimeter (474D) perimeter measure 474E-474R (474F), 474Xa, 474Ya, 475E-475G periodic Boolean automorphism 381Ac, 381H, 381R, 381Xj, 381Xl, 381Xm, 382F, 493Yc periodic extension of a function on ]−π, π] 282Ae permutation group see symmetric group Perron integral 483J Pettis integrable function 463Ya, 463Yb, 463Yc Pettis integral 463Ya, 464Yb Pfeffer integrable function 484G, 484H-484J, 484L, 484O, 484Xe Pfeffer integral 484G, 484H, 484N, 484S, 484Xd Pfeffer’s Divergence Theorem 484N

principal

General index

923

Plancherel Theorem (on Fourier series and transforms of square-integrable functions) 282K, 284O, 284Qd, 445R Poincaré’s inequality 473K point process see Poisson point process (495E) point-countable family of sets 4A2A, 4A2Dc point-finite family of sets 438B, 438Xj, 438Ya, 451Yd, 4A2A, 4A2Fd point-supported measure 112Bd, 112Xg, 211K, 211O, 211Qb, 211Rc, 211Xb, 211Xf, 213Xo, 234Xc, 215Xr, 256Hb, 343Xi, 416Xb, 419G, 465Yd, 491D, 491Yg pointwise compact set of functions 462E, 462F, 462J, 462Xd, 462Ya, 463C, 463D, 463F, 463G, 463K, 463N, 463Xb-463Xe, 463Xg, 463Xi, 463Xj, 463Xl-463Xn, 463Ya, 463Yc-463Ye, 463Za, 463Zb, 464E, 464Yb, 465Db, 465V, 465Xg, 465Xj, 465Xm see also Eberlein compactum (467O) pointwise convergence (topology on a space of functions) 281Yf, 437Xl, 438P, 438Xp, 438Yh, 438Yi, 443Yn, 454Pa, 462Ab, 462B, 462E-462H, 462J, 462Xd, 462Ya, 462Yd, 462Z, §463, 464E, 464Yb, 465Cb, 465G, 465Xi, 465Xj, 467Pa, 4A2Nj; see also pointwise compact —- (on an isometry group) 441G, 441Xm-441Xo, 441Ye, 441Yj, 443Xx, 443Ym, 448Xk, 449Xc, 449Xd, 476Xi, 493G, 493J, 493Xc, 493Xf, 493Xg, 493Yb pointwise convergent (sequence or filter) 462Ab, 462Xb; see also order*-convergent (245Cb, 367A) pointwise topology see pointwise convergence Poisson distribution 285Q, 285Xo, 455Xc, 455Ye, 495A, 495B, 495D, 495M, 495Xf; see also compound Poisson distribution Poisson point process 495E, 495F-495I, 495L-495N, 495P, 495Xa-495Xe, 495Xi-495Xk, 495Xn, 495Xp, 495Yc Poisson’s theorem 285Q —– see also Abel-Poisson kernel (282Yg) Pol, R. 463 notes polar coordinates 263G, 263Xf polar set 4A4Bh, 4A4Ch, 4A4Eg Polish group 424H, 441Xm, 441Xn, 441Yf, 444Ye, 448P, 448Xg, 448Xj, 448Xk, 449G, 493Xe, 493Xg, 4A5Db, 4A5Q Polish (topological) space 284Ye, 418C, 423Ba, 423I, 423Xa, 424F, 424H, 424Ya, 424Yb, 433Yc, 434Kb, 437Rb, 437Wg, 437Yo, 438Yh, 441Xm, 454B, 454O, 454Xl, 454Xm, 461Xi, 4A2A, 4A2Q, 4A2Ud, 4A3H, 4A3I; see also standard Borel space (424A) Pólya’s urn scheme 275Xc polynomial (on Rr ) 252Yu polytope see convex polytope (481O) Pontryagin Duality Theorem see Duality Theorem (445U) porous set 223Ye, 261Yg, 262L positive cone 253G, 253Xi, 253Yd, 351C positive definite function 283Xt, 285Xr, 445L, 445M, 445N, 445P, 445Xe-445Xh, 445Xo positive linear functional (on a Riesz space) 436D, 436H, 436J, 436K, 436Xe, 436Xp, 436Yc positive linear operator (between partially ordered linear spaces) 351F, 355B, 355E, 355K, 355Xa, 361G, 367P, 375I-375K, 375Xa, 375Xe, 375Xg, 376Cc, 452H, 481C, 495K, 495Lb, 495Xh, 495Yb; see also (sequentially) order-continuous positive linear operator, Riesz homomorphism (351H) positive measure everywhere see self-supporting (411Na) predictable sequence 276Ec pre-Radon (topological) space 434Gc, 434J, 434Ka, 434Xn, 434Xo, 434Xs, 434Yd, 435Xk, 439Xl, 462Z, 466B, 466Xf; see also Radon space (434C) presque partout 112De Pressing-Down Lemma 4A1Cc primitive product measure 251C, 251E, 251F, 251H, 251K, 251Wa, 251Xa-251Xc, 251Xe, 251Xf, 251Xj, 251Xl-251Xp, 252Yc, 252Yd, 252Yg, 253Ya-253Yc, 253Yg, 325Ya, 412Ye principal band (in a Riesz space) 352V, 353C, 353H, 353Xb, 362Yd, 437Yj

924

Index

principal

principal ideal in a Boolean ring or algebra 312D, 312E, 312J, 312S, 312Xf, 312Yf, 314E, 314Xb, 314Xi, 315E, 315Xl, 316Xb, 316Xk, 316Xv, 321Xa, 322H-322J, 322Xf, 323Xe, 325Xb, 331Fb, 331H, 331M, 332A, 332L, 332P, 332Xh, 352Sc, 352Xg, 364Xo, 384Lb, 391Xe, 391Xj, 491Xq principal projection property (of a Riesz space) 353Xb, 353Yd, 354Yi principal ultrafilter 2A1N probability algebra 322Aa, 322Ba, 322Ca, 322G, 322Mb, §385, 386E, 386H-386J, 386L-386O, 386Xe, 386Yb, 386Yc, §387, 391B, 491Ke, 491P probability algebra free product 325F, 325G, 325I-325M (325K) 325Xd, 334D, 334Xc, 372Yl, 385Xm, 388Yc, 458Xe probability density function see density (271H) probability space 211B, 211L, 211Q, 211Xb, 211Xc, 211Xd, 212G, 213Ha, 215B, 243Xi, 253H, 253Xh, §254, Chap. 27, 322Ba, 441Xe, 444Xu process see Gaussian process (456N), Markov process, Poisson point process (495E) product Boolean algebra see simple product (315A), free product (315H) product f -algebra 352Wc, 364S product linear topological space 467Xc, 4A4B product measure Chap. 25, 434R, 434Xv, 434Xw, 434Xx, 434Yk, 436F, 436Xh, 436Yd, 436Ye, 438S, 439L, 439R; see also c.l.d. product measure (251F, 251W), primitive product measure (251C), product probability measure (254C), quasi-Radon product measure (417R), Radon product measure (417R), τ additive product measure (417G) product partial order 315C, 315D, 315Xb product partially ordered linear space 351L, 351Rd, 351Xc, 351Xd; see also product Riesz space product probability measure §254 (254C), 272G, 272J, 272M, 275J, 275Yi, 275Yj, 281Yk, 325I, 334C, 334E, 334Xf, 334Xg, 334Ya, 342Gf, 342Xn, 343H, §346, 372Xf, 385S, 412T-412V 413Yb, 415E, 415F, 415Xj, 416U, 417E, 417Xt, 417Xw, 417Yf, 417Yg, 433H, 443Xq, 451J, 451K, 451Yo, 453I, 453J, 454Xj, 456Aa, 456B, 456Xb, 458Xk, 459E, 465H-465M, 491Eb, 491Xu, 494Xc, 494Ya, 495P, 495Xb, 495Xj, 495Xp; see also quasi-Radon product measure (417R), Radon product measure (417R), relative product measure (458Lb), τ -additive product measure (417G) product Riesz space 352K, 352T, 354Xb product ring 3A2H product tagged-partition structure 481P, 481Yb product topological group 442Xb, 442Xh, 443Xb, 443Xd, 443Xq, 443Xt, 445Bd, 445Xl, 445Yb, 449Ce, 493Bd, 4A5G, 4A5Od product topology 281Yc, 2A3T, 315Xf, 315Yb, 315Yd, 323L, 391Yc, 3A3I, 3A3J, 3A3K, 417Xu, 418B, 418Dd, 418Xb, 418Xd, 418Yb, 419Xg, 422Dd, 422Ge, 423Bc, 434Jh, 434Pb, 434Xc, 434Xe, 434Xr-434Xu, 435Xk, 436Xg, 437Wd, 437Yo, 438E, 438Xm, 438Xn, 439Xg, 439Yg, 463J, 463K, 467Xa, 4A2B, 4A2Cb, 4A2E, 4A2Fh, 4A2H, 4A2L, 4A2O-4A2R, 4A3C-4A3E, 4A3G, 4A3M-4A3O, 4A3Xa, 4A3Yb product uniformity 3A4E —– see also inner product space projection (on a subalgebra of a Boolean algebra) see upper envelope (314Va); (on a Riesz space) see band projection (352Rb) projection band (in a Riesz space) 352R, 352S, 352T, 352Xh, 353E, 353H, 353I, 353Xb, 361Xh, 361Ye, 362B projection band algebra 352S, 352T, 361K, 363J, 363N, 363O, 364Q projection pair (of linear subspaces) 467Hc, 467J projectional resolution of the identity (on a Banach space) 467G, 467K projective system of probability spaces 418M-418Q, 418Xu, 418Xv, 418Yk, 418Yl, 433K, 433Yd, 451Yb, 451Yc Prokhorov space 437V, 437W, 437Xs, 437Xt, 437Yn, 437Yo, 439S, 439Yi, 452Xe; see also strongly Prokhorov (437Yo) Prokhorov’s theorem 418M, 418N, 418P, 418Q, 433K, 433Yd pseudo-cycle (for a Boolean automorphism) 388Xc pseudometric 2A3F, 2A3G, 2A3H, 2A3I, 2A3J, 2A3K, 2A3L, 2A3Mc, 2A3S, 2A3T, 2A3Ub, 2A5B, 323A, 3A3Be, 3A4B, 3A4D-3A4F, 458Yf, 4A2Ja

Radon

General index

925

pseudometrizable topology 463A, 4A2A, 4A2L pseudo-simple function 122Ye, 133Ye pull-back measures 132G, 418L, 432G, 433D purely atomic Boolean algebra 316Lc, 316M, 316Xr-316Xx, 316Yk, 316Yl, 316Yq, 322Bh, 322Kc, 324Kg, 331Yd, 332Xb, 332Xc, 362Xf, 385J, 385Xq, 394Xf purely atomic measure (space) 211K, 211N, 211R, 211Xb, 211Xc, 211Xd, 212G, 213He, 214Xe, 234Xb, 251Xr, 322Bh, 342Xn, 343Xi, 375Ya, 416Xb, 438Ce, 451Xa, 451Xo, 451Yg purely infinite measurable set 213 notes purely non-measurable additive functional 464I, 464Jc, 464L, 464M, 464P, 464Xa push-forward measure see image measure (112F) quadratic map 372Xn quasi-dyadic space 434O, 434P, 434Q, 434Yh, 434Yi, 434Yj; see also dyadic (434 notes) quasi-homogeneous measure algebra 374G, 374H-374M, 374Xl, 374Ye, 394Xg quasi-order-dense Riesz subspace (of a Riesz space) 352N, 353A, 353K, 353Ya quasi-Radon measure (space) 256Ya, 263Ya, 411Ha, 411Pd, 411Yc, 414Xj, §415, 416A, 416C, 416G, 416R, 416T, 416Xu, 418Hb, 418K, 418Xi, 418Xn, 432E, 434A, 434H-434J, 434L, 435D, 436H, 436Xl, 437N, 471Dh, 438G, 439Xh, 441B, 441Xc, §444, 445D, 445Ec, 445M, 445Xc, 445Yf, 451T, 453Ga, 453J, 453Xd, 453Xg, 453Xh, 454Xj, 454Xk, 456Aa, 456L, 456Xf, 459E, 466A, 466B, 466K, 466Xa-466Xd, 466Xj, 482Xk, 491Mb, 491Xk, 491Xv; see also Haar measure (441D), Radon measure (411Hb) —– product measure (of finitely many spaces) 417N, 417R, 417Xv, 418Xf, 418Xh, 436Xh, 441Xi, 443Xb, 443Xp, 444A, 444N, 444O, 444Ya, 444Yl; (of infinitely many probability spaces) 417O, 417R, 417Xq, 417Xw, 417Yh, 417Yi, 433H, 441Xi, 443Xq, 453J, 456Xf, 458Xk, 459G; see also Radon product measure (417R), τ -additive product measure (417G) quasi-simple function 122Yd, 133Yd quasi-Stonian topological space 314Yf quotient Banach algebra 4A6E quotient Boolean algebra (or ring) 312K, 312Xk, 313Q, 313Xo, 314C, 314D, 314M, 314N, 314Yd, 316C, 316D, 316Xh, 316Xi, 316Xn, 316Xu, 316Ye, 316Yp, 341Xf, 361M, 363Gb, 391Xe, 391Yb, 392Xd, 424Xk, 491I quotient partially ordered linear space 241Yg, 351J, 351K, 351Xb; see also quotient Riesz space quotient Riesz space 241Yg, 241Yh, 242Yg, 352J, 352U, 354Yi, 361M, 364C, 364U quotient ring 3A2F, 3A2G quotient topological group 443S, 443Xw-443Xy, 443Yk, 446Ab, 446O, 446P, 447E, 447F, 449Cc, 493Bc, 4A5J-4A5L, 4A5S quotient topology 245Ba, 443O-443R, 443Xs, 443Yl Rademacher’s theorem 262Q Radon, J. 416 notes Radon measure > 411Hb, 414Xj, §416, 418I, 418L, 418M, 418P, 418Q, 418Xo, 418Xp, 418Xs, 432C432H, 432Xe, 433Cb, 433D, 433Ha, 433I, 434Ea, 434Fa, 434J, 434Ya, 435B, 435Xa-435Xd, 435Xk, 435Yb, 436J, 436K, 436Xn, 436Xp, 437N-437P, 437S, 471F, 471Q, 439Xh, 441Bb, 441E, 441H, 441L, 444Xa, 444Xl, 444Ye, 445N, 445Xb, 451Ab, 451N, 451Rb, 451S, 451Xk, 451Xn-451Xp, 453Gb, 453K, 453Xf, 453Za, 453Zb, 454J, 454K, 454O, 454Pb, 455D, 455Xe, 455Ye, 456Aa, 457Xf, 458O, 458Xr, 458Yf, 459Xc-459Xe, 462G462I, 463J, 463K, 463Xk, 466A, 466Xk, 466Xl, 466Ya, 482Xc, 482Xe, 482Xg, 491Ma, 491Xz, 491Yg, 494B, 494C, 494Xa-494Xc, 495N(on R or Rr ) §256 (256A), 284R, 284Yi, 342Jb, 342Xj, 411O, 419C, 419E, 452D, 452O, 452P, 452Xe, 472C-472F, 483Yc; (on ]−π, π]) 257Yb; see also complex Radon measure, quasi-Radon measure (411Ha), tight Borel measure Radon-Nikod´ ym derivative 232Hf, 232Yj, 234B, 234Ca, 234Yd, 234Yf, 235Xi, 256J, 257F, 257Xe, 257Xf, 272Xe, 272Yc, 275Ya, 275Yi, 285Dd, 285Xe, 285Ya, 363S, 452Qa, 458Lb, 458P, 472Ye Radon-Nikod´ ym theorem 232E-232G, 234G, 235Xk, 242I, 244Yk, 327D, 365E, 365Xf Radon probability measure (on R or Rr ) 271B, 271C, 271Xb, 285Aa, 285M, 364Xd, 372Xf, 415Fb; (on other spaces) 256Ye, 271Ya, 416U, 417Q, 418M, 461H, 461J, 461K, 461N, 461Xc, 461Xj, 461Yb

926

Index

Radon

Radon product measure (of finitely many spaces) 256K, 417P, 417R, 417Xi, 417Xo, 419E, 419Yb, 436Xo, 436Yf, 453Za; (of infinitely many probability spaces) 416U, 417Q, 417R, 417Xp, 417Xr, 418Na, 433Ha, 453J, 459Xc-459Xe, 461Ye, 494C, 494Xd Radon (topological) space 434C, 434F, 434K, 434Nd, 434Xh, 434Xm, 434Xp-434Xu, 434Yc, 434Za, 434Zb, 438H, 438R, 438Xn, 438Xp, 438Yi, 439Ca, 439K, 439P, 439Xb, 454Pa, 454Xh, 454Xm, 455Ye, 466F, 466H, 466Xe, 467Pb; see also pre-Radon (434Gc) Ramsey’s theorem 4A1G, 4A1L random variable 271Aa rank (of a tree) 421N, 421O, 421Q, 423Yh rapidly decreasing test function §284 (284A, 284Wa), 285Dc, 285Xd, 285Ya rare see nowhere dense (3A3Fa) realcompact (topological) space 436Xg, 439Xm rearrangement 373Ya; see also decreasing rearrangement (252Yp, 373C) rearrangement-invariant extended Fatou norm 374Eb, 374F, 374K, 441Yk rearrangement-invariant set 374Ea, 374F, 374M, 374Xk, 374Xl, 374Ye rectifiable 475 notes Recurrence Theorem (for a measure-preserving Boolean homomorphism) 386Xa, 386Xb, 386A recurrent Boolean homomorphism 381Ag, 381L, 381O, 381P, 381Q, 381Xh, 381Xi recursion 2A1B reduced boundary 474G, 474N, 474Q, 474R, 474Xb, 474Xc, 474Xd, 475D, 475E, 475Xi; see also essential boundary (475B) reduced power of R 351M, 351Q, 351Yd, 352L, 368F reduced product 413Xn refinable see hereditarily weakly θ-refinable (438K) refine 311Gd, > 4A2A refinement > 4A2A refinement property see σ-refinement property (448K) reflexive Banach space 372A, 3A5G, 461G, 467Xf regular cardinal 4A1Aa, 4A1Bd, 4A1Cc, 4A1I-4A1L regular conditional probability see disintegration (452E) regular embedding of Boolean algebras 313N regular measure see completion regular (411Jb), effectively regular (491L), inner regular (256Ac, 411B), outer regular (411D) regular open algebra (of a topological space) 314P-314S (314Q), 314Xi, 314Xj, 314Yd, 315Xe, 315Yd, 316Xe, 316Yb, 316Yd, 316Yf, 316Yj, 316Yq, 332Xe, 363Yc, 364U-364W, 364Yh-364Yl, 368Ya, 391Yc, 391Yf, 417Xu; (of R) 316K, 316Yb, 316Yn, 316Yo, 331Xh, 331Yi, 367Yp, 375Xe, 375Yf, 375Z, 376Yi, 391Xd, 392Yb regular open set 314O, 314P, 314Q, 417Xu, 443N, 443Yi, 4A2Eb, 4A3Ra, 4A5Kb regular operator (between Riesz spaces) 355 notes regular outer measure 132C, 132Xa, 213C, 214Hb, 251Xm, 254Xb, 264Fb, 471Dc, 471Yb regular Souslin scheme 421Xn, 431Xc regular topological space 2A5J, 316Yf, > 3A3Ab, 3A3Ba, 3A3De, 414M, 414Xj, 414Ya, 415C, 415D, 415J, 415Qf, 415Rc, 416K, 418H, 418Xn, 422E, 422K, 422Xe, 434Ib, 434Jc, 462Aa, 462C, 476C, 4A2F, 4A2H, 4A2Ja, 4A2N, 4A2Pb; see also completely regular (3A3Ac) regularly embedded (subalgebra of a Boolean algebra) 313N, 313O, 313P, 316Xm, 316Xs, 326Kf, 367Xi, 491Ke, 491P; (Riesz subspace) 352Ne, 352Xe, 354Xk, 354Xm, 367F, 368Pa, 368S regularly enveloped set 491L relation 1A1B relative atom in a Boolean algebra 331A relative distribution 458C, 458D, 458E, 459C, 458Xb, 458Ya, 458Yc, 459C relative free product of probability algebras 458I, 458J, 458K, 458M, 458Xe relative Maharam type of a Boolean algebra over a subalgebra 333A, 333B, 333C, 333E, 333F, 333Yb relative product measure >458Lb, 458M-458P, 458Xg-458Xr, 458Ye

ring

General index

927

relatively atomless (Boolean algebra) 331A, 381P, 386C, 388I, 394Xc relatively compact set 2A3Na, 2A3Ob, 3A3De, 3A5M, 4A2Le relatively countably compact set (in a topological space) 462Aa, 462Yc, 463I, 463Xf, 4A2A, 4A2Gf, 4A2Le relatively independent family (in a measure algebra) 458H; (of measurable sets) 458Aa relatively independent closed subalgebras 458H, 458Xd relatively independent random variables 458A, 458E, 459C, 458Xt, 458Ya, 458Yc, 458Yd, 459C relatively independent σ-algebras 458A, 458B, 458F, 458G, 458Xa, 458Xc, 458Xj, 458Yb, 459B, 459D relatively invariant measure 441 notes relatively Maharam-type-homogeneous 333Ac, 333Bb relatively measurable set 121A relatively von Neumann transformation 388Da, 388K, 388Xg, 388Yb, 388Yc, 388Yf relatively weakly compact set (in a normed space) 247C, 2A5I, 356Q, 356Xl, 3A5Gb, 3A5Hb, 3A5Kb; (in other linear spaces) 376O, 376P, 376Xl repeated integral §252 (252A), 434R, 436F, 436Xo; see also Fubini’s theorem, Tonelli’s theorem representation (of a group) see finite-dimensional representation (446A), action (441A) representation of homomorphisms (between Boolean algebras) 344Ya, 344Yd, 364Xr, 424Xk; (between measure algebras) 324A, 324B, 343A, 343B, 343G, 343J, 343M, 343Xc, 343Xf, 343Yd, 344A-344C, 344E344G, 344Xf, 344Yc, 416W, 451Ab residual family 481F, 481G, 481H, 481K, 481L, 481N, 481O, 481Q, 481Xb, 481Xc, 481Xf, 481Yb, 484D resolution of the identity 4A2Fd; see also projectional resolution of the identity (467G) ‘respects coordinates’ (said of a lifting) 346A, 346C, 346E, 346Xg, 346Yc, 346Za reverse martingale 275K Riemann-complete integral see Henstock integral, symmetric Riemann-complete integral Riemann integrable function 134K, 134L, 281Yh, 281Yi, 491Xw Riemann integral 134K, 242 notes, 363Yi, 436Xm, 481E, 481I, 481Xe, 491Xj Riemann-Lebesgue lemma 282E, 445Ka Riemann sum 481A Riesz Convexity Theorem 244 notes Riesz homomorphism (between partially ordered linear spaces) 351H, 351J, 351L, 351Q, 351Xc, 351Ya, 352G; (between Riesz spaces) 352G-352J, 352U, 352W, 352Xb, 352Xe, 353Pd, 353Yf, 355Xe, 356Xh, 361Gc, 361J, 361Xg, 362Xe, 363Ec, 363F, 363Xb, 363Xc, 364R, 365K, 375I, 375J, 375Xg, 375Ya, 376Cc, 436Xg; see also order-continuous Riesz homomorphism Riesz norm 242Xg, 354A, 354B, 354D, 354F, 354M, 354Xc-354Xf, 354Xh, 354Yb, 354Yf, 354Yk, 355Xc, 356D, 356Xg, 356Xh, 438Xi, 466H, 467Yb; see also Fatou norm (354Da), order-continuous norm (354Dc), order-unit norm (354Ga) Riesz Representation Theorem (for positive linear functionals) 436J, 436K Riesz space (= vector lattice) 231Yc, 241Ed, 241F, 241Yc, 241Yg, chap. 35 (352A), 361Gc, 367C, 367E, 367Xc, 367Yn; see also Archimedean Riesz space (§352), Banach lattice (354Ab), Riesz norm (354A) Riesz subspace (of a partially ordered linear space) 352I; (of a Riesz space) 352I, 352J, 352L, 353A, 354O, 354Rc, 4A2Jg; see also band (352O), (quasi-)order-dense Riesz subspace (352N), solid linear subspace (351I) right action (of a group on itself) 4A5Ca, 4A5I right Haar measure 441D, 441Xe, 442C, 442E, 442F, 442H, 442I, 442L, 442Xd, 442Xf, 442Xg, 443Xh, 443Xi, 444Yn right modular function (on a topological group) 442Ib right-translation-invariant lifting 447Xa, 447Ya right-translation-invariant metric 441Xn, 4A5Q right uniformity (of a topological group) 449B, 449D, 449E, 449H, 449Yc, > 4A5Ha, 4A5Mb, 4A5Q rigid Boolean algebra 384Ha, 384L; see also nowhere rigid (384Hb) ring §3A2 (3A2A); see also Boolean ring (311Aa) ring homomorphism 3A2D, 3A2F-3A2H ring homomorphism between Boolean rings 311D, 312Xf, 312Xg, 312Xh, 312Yc, 312Yd, 312Ye, 312Yf, 361A, 361Cc, 361J, 361Xe, 361Xg, 375H

928

Index

Rokhlin

Rokhlin see Halmos-Rokhlin-Kakutani lemma (386C) root (of a ∆-system) 4A1Da root algebra (of a Bernoulli shift) 385Q, 385R, 385S, 387B, 387Ya rotund see locally uniformly rotund (467A) Saks see Vitali-Hahn-Saks theorem (246Yg) Saks-Henstock indefinite integral 482B, 482C, 482D, 482G, 482Xa, 482Xb, 484I, 484J, 484L, 484O Saks-Henstock lemma 482A, 482B, 483F, 483Xe, 484Hc saltus function 226B, 226Db, 226Xa scalarly measurable 463Ya scattered topological space 439Ca, 439Xf, 439Xh, 4A2A, 4A2G; see also non-scattered Schröder-Bernstein theorem 2A1G, 332 notes, 344D, 344Xa Schwartz function see rapidly decreasing test function (284A) Schwartzian distribution 284R, 284 notes; see also tempered distribution (284 notes) second-countable topological space 434Ya, 454Yd, 4A2A, 4A2O, 4A2P, 4A2Ua, 4A3G Second Separation Theorem (of descriptive set theory) 422Yd selection theorem 423M-423O, 423Xf, 423Xg, 424Xg, 424Xh, 433F, 433G selector see measurable selector self-adjoint linear operator 444V, 4A4Jc, 4A4M self-supporting set (in a topological measure space) 256Xf, 411N, 414F, 415E, 415Xk, 416Dc, 416Xe, 417Ma, 443Xa, 443Xl, 443Yf, 456D semicompact paving see countably compact class (413L) semi-continuous function see lower semi-continuous (225H, 4A2A), upper semi-continuous (4A2A) semi-finite measure algebra §322 (322Ad), 323Dd, 323Ga, 323Xa, 324K, 324Xb, 325Ae, 325D, 327B, 331C, 331Xl, 332E, 332F, 332I, 332R, 332Xi, 332Yb, 364L, 365E, 365G, 365M, 365P, 365Sb, 366E, 366Xe, 366Xf, 366Xk, 367Nb, 368S, 369H, 369Xa, 371Xc, 373R, 375I, 383E, 383F, 383Ga, 383I, 384Ld, 384P, 384Xe, 391Ca semi-finite measure (space) 211F, 211L, 211Xf, 211Ya, 212G, 213A, 213B, 213Hc, 213Xc, 213Xd, 213Xj, 213Xl, 213Xm, 213Ya-213Yc 214Xe, 214Xh, 215B, 216Xa, 216Yb, 234Fa, 235O, 235Xd, 235Xe, 241G, 241Ya, 241Yd, 243G, 245Ea, 245J, 245Xd, 245Xj, 245Xl, 246J, 246Xh, 251J, 251Xc, 252P, 252Yf, 253Xf, 253Xg, 322Bd, 322Yd, 327C, 327D, 342L, 342Xa, 342Xc, 342Xn, 343B, 344H, 365Xp, 367Xr, 411Gd, 412Xb, 413E, 413Xf, 414N, 418F, 431Xb, 431Xe, 471S, 438B, 438Ce, 438I, 463Cb, 481Xf semi-finite version of a measure 213Xc, 213Xd, 322Xb, 413Xf semigroup see amenable semigroup (449Ya), countably full local semigroup (448A), full local semigroup (394A), topological semigroup (444Yb) semi-martingale see submartingale (275Yf) seminorm 2A5D, 4A4C, 4A4Da semi-radonian see pre-Radon (434Gc) semi-ring of sets 115Ye separable (topological) space 2A3Ud, 316Xd, 316Yb, 316Yj, 367Xr, 391Yc, 3A3E, 417Xu, 491H, 491Xz, 4A2Be, 4A2De, 4A2Ea, 4A2Ni, 4A2Oc, 4A4Bi; see also hereditarily separable (423Ya) separable Banach space 244I, 254Yc, 365Xp, 366Xc, 369Xg, 424Xe, 456Yd, 466M, 466Xe, 493Xg separable metric space 476Xd, 491Xw separable metrizable space 245Yj, 264Yb, 284Ye, 415E, 415Xk, 417T, 418B, 418G, 418J, 418K, 421Xg, 423Yc, 434Ie, 434O, 471Df, 471Xf, 438D, 439Xm, 441Xm, 451O, 454N-454P, 454Yd, 4A2P, 4A2Ua, 4A3E, 4A3N, 4A3V, 4A4Id; see also Polish space (4A2A) separable normed space 467E, 467Xe separate zero sets 473J separatedsee countably separated (343D) separately continuous function 436Ye, 437Mc, 462I, 463J, 463K, 463Zc, 4A2A separation see First Separation Theorem, Second Separation Theorem; (in ‘countable separation property’) see σ-interpolation property (466G) separator (for a Boolean automorphism) 382Aa, 382B-382E, 382I, 382J, 382L, 382M, 382Xa, 382Xd, 382Xl

solid

General index

929

sequential (topological) space 436F, 436Yd, 436Ye, 4A2A, 4A2K, 4A2Ld; see also Fréchet-Urysohn (462Aa) sequentially closed set 4A2A sequentially compact set, topological space 413Ye, 434Za, 462C, 463Cd, > 4A2A, 4A2Lf sequentially continuous function 463B, 4A2A, 4A2Kd, 4A2Ld sequentially order-closed set in a partially ordered space 313Db, 313Xg, 313Yc, 316Fb, 353Ja, 364Xm, 367Yb; see also σ-ideal (313Ec), σ-subalgebra (313Ec) sequentially order-continuous additive function (on a Boolean algebra) 326Gc, 363Eb —– —– Boolean homomorphism 313Lc, 313Pb, 313Qb, 313Xq, 313Yb, 314Fb, 314Hb, 314Xc, 314Ye, 315Ya, 316Fd, 324A, 324B, 324Kd, 324Xa, 324Xe, 324Yc, 326Ff, 343Ab, 363Ff, 364G, 364H, 364R, 364Xp, 364Xv, 364Yc, 364Yf, 364Yn, 365H, 365Xg, 375Ya, 375Ye, 381K, 424Xk —– —– dual (of a Riesz space)(Uc∼ ) 356Ab, 356B, 356D, 356L, 356Xa, 356Xb, 356Xc, 356Xd, 356Xf, 356Ya, 362Ac, 363K, 363S, 437Aa, 437B, 438Xd —– —– function 313Hb, 313Ic, 313Xg, 313Yc, 315D, 316Fc, 361Cf, 361Gb, 367Xb, 367Yb, 375Xd, 392H —– —– positive linear operator or functional 351Gb, 355G, 355I, 361Gb, 363Eb, 363Ff, 364R, 375A, 436A, 436Xi; see also sequentially order-continuous dual (356A), L∼ c (355G) —– —– Riesz homomorphism 361Jf, 375Ya, 437C-437E, 437Hb, 437Xc, 437Xe, 437Xn, 437Yd —– —– ring homomorphism between Boolean rings 361Ac, 361Jf, 375Ka sequentially smooth dual (of a Riesz subspace of RX , Uσ∼ ) 437Aa, 437B-437E, 437Xa, 437Xc, 437Ya, 437Yd, 437Yi —– —– linear operator 437D —– —– positive linear functional 436A, 436C-436E, 436G, 436Xc, 436Xe, 436Xf, 436Ya, 436Yc; see also sequentially smooth dual (437Aa) Shannon-McMillan-Breiman theorem 386E, 386Xc shift operators (on function spaces based on topological groups) 286C, 443G, 444F, 444Of, 444Xg, 444Xp, 444Yk; see also Bernoulli shift (385Q) Shoenfield’s Absoluteness Theorem §393 notes Sierpi´ nski Class Theorem see Monotone Class Theorem (136B) signed measure 416Ya, 437B, 437E, 437G, 437I, 444E, 444Sb, 444Xd, 444Yb, 445Yi, 445Yj; see also countably additive functional (231C), Mσ signed Baire measure 437E, 437G signed Borel measure 437F, 437G signed tight Borel measure 437G; see Mt signed τ -additive Borel measure 437G; see Mτ >122A), 242M, 361D simple function §122 (> simple group 382S, 383I, 383Xb-383Xd simple product of Boolean algebras §315 (315A), 316Xc, 316Xl, 316Xw, 332Xa, 332Xg, 364S, 384Lc, 391Xb, 391Xj —– of measure algebras 322K, 323L, 325Xc, 332B, 332Xm, 332Xn, 333H, 333Ia, 333K, 333R, 366Xi simple product property 417Yi, 419Yc Sinaˇı’s theorem 387E, 387L singular additive functional 232Ac, 232I, 232Yg singular measures 231Yf, 232Yg, 362Xa, 461Xl small subgroups see ‘no small subgroups’ (446E) smooth dual (of a Riesz subspace of RX )(Uτ∼ ) 437Ab, 437Xa, 437Xc, 437Xd, 437Xq, 437Ya; see also Mτ smooth function (on R or Rr ) 242Xi, 255Xi, 262Yd-262Yg, 284A, 284Wa, 473Bf, 473D, 473E smooth positive linear functional 436G, 436H, 436Xf, 436Xi-436Xl, 439I; see also sequentially smooth (436A), smooth dual (437Ab) smoothing by convolution 261Ye, 473D, 473E Sobolev space 282Yf, 473 notes; see also Gagliardo-Nirenberg-Sobolev inequality (473H) solid hull (of a subset of a Riesz space) 247Xa, 352Ja solid set (in a partially ordered linear space) 351I; (in a Riesz space) 354Xg —– linear subspace (of a partially ordered linear space) 351J, 351K, 351Yb; (of a Riesz space) 352J, 353J, 353K, 353N, 355F, 355J, 355Yj, 368Ye, 383J, 411Xe, 464K, 464L; see also band (352O)

930

Index

Sorgenfrey

Sorgenfrey topology on R 415Xc, 415Yd, 415Ye, 416Xx, 417Yi, 418Yf, 419Xf, 423Xe, 434Xm, 434Yi, 438Xm, 438Xo, 439Q, 453Xb, 462Xa, 482Ya Souslin scheme 421B —– set see Souslin-F set (421K) —– space see analytic space (423A) Souslin-F set 421J, 421K, 421L, 421Xe, 421Xf, 421Xj, 421Xl, 421Xm, 421Yd, 422H, 422K, 422Ya, 422Yc, 423E, 423O, 423Xf, 423Ye, 431B, 431E, 431Xb, 434Dc, 434Hc, 435G, 435Xk, 436Xg, 466M, 467Xa; σ-algebra generated by Souslin-F sets 423O, 423Xg, 423Xi Souslin property see ccc (316Ab) >421B), 422Hc, 422Xc, 423E, 423M, 423N, 423Yb, 424Xg, 424Xh, 424Yc, Souslin’s operation §421 (> §431, 434Dc, 434Eb, 434Fd, 434Xn, 471Dd, 454Xh space-filling curve 134Yl spectral radius (of an element in a Banach algebra) 445Kd, 445Yk, 4A6G, 4A6I, 4A6K spectrum (of an M -space) 354L sphere, surface measure on 265F-265H, 265Xa-265Xc, 265Xe, 456Xb, 457Xi —– isometry group of 441Xo spherical polar coordinates 263Xf, 265F split interval (= ‘double arrow space’) 343J, 343Xf, 343Yc, 344Xf, 412Yb, 419L, 419Xg, 419Xh, 419Yc, 424Yd, 433Xg, 434Ke, 434Yi, 438Qa, 438R, 438Xl, 438Xn, 453Xc, 463Xg, 491Xg —– line 419Xf square-integrable function 244Na; see also L2 stabilizer subgroup 4A5Bf stable set of functions §465 (465B); (in L0 ) 465O, 465P-465R; see also R-stable (465S) >424A), 433J, 433K, 433Yd, 451L, 452N, 452Xm, 454F, 454H, 454Xg, 455A standard Borel space §424 (> standard extension of a countably additive functional 327F, 327G, 327Xa, 327Xd, 327Xe, 327Yc standard gamma distribution 455Xe, 495Xh, 495Xo standard normal distribution, standard normal random variable 274A, 456Aa, 456M, 456Xe standard order unit (in an M -space) 354Gc, 354H, 354L, 356N, 356P, 363Ba, 363Ye stationary set 4A1C Steiner symmetrization 264H, 476 notes step-function 226Xa Stieltjes measure see Henstock-Stieltjes integral, Lebesgue-Stieltjes measure (114Xa) Stirling’s formula 252Yn stochastic see doubly stochastic matrix stochastically independent see independent (272A, 325L, 325Xe) ˇ Stone-Cech compactification 418Yk, 422Ya, 434Yc, 434Yd, 435Yb, 464P, 4A2I; (of N) 416Yd, 434Yg, 463M Stone’s condition see truncated Riesz space (436B) Stone representation of a Boolean ring or algebra 311E, 352 notes; see also Stone space Stone space of a Boolean ring or algebra 311E, 311F, 311I-311K, 311Xg, 311Ya, 312O-312S, 312Xi312Xk, 312Yc-312Yf, 313C, 313R, 313Xp, 313Yb, 314M, 314S, 314T, 314Yd, 315H, 315Xc, 316B, 316I, 316Yb, 316Yc, 316Yi, 363A, 363Yf, 381Q, 381Xl, 382Xa, 413Yc, 416Q, 436Xp, 4A2Ib Stone space of a measure algebra 321J, 321K, 322N, 322Q, 322Xi, 322Yb, 322Yf, 341O, 341P, 342Jc, 343B, 344A, 344Xe, 346K, 346L, 346Xf, 411P, 414Xs, 415Q, 415R, 416V, 416Xw, 419E, 453M, 453Xa, 453Za, 463Ye, 491Xm Stone-Weierstrass theorem 281A, 281E, 281G, 281Ya, 281Yg, 4A6B stopping time 275L, 275M-275O, 275Xi, 275Xj straightforward set of tagged partitions 481Ba, 481G Strassen’s theorem 457D strategy (in an infinite game) 451U strictly localizable measure (space) 211E, 211L, 211N, 211Xf, 211Ye, 212G, 213Ha, 213J, 213L, 213O, 213Xa, 213Xh, 213Xn, 213Ye, 214Ia, 214J, 215Xf, 216E, 234Fd, 235P, 251N, 251P, 251Xn, 252B, 252D, 252E, 252Ys, 252Yt, 322Kd, 322N, 322Qb, 322Xi, 325He, 341H, 341K, 342Hb, 343B, 343Xi, 344C, 344I,

successor

General index

931

344Xb, 344Xc, 346Xd, 385V, 412I, 412Yf, 414J, 414Xt, 415A, 416B, 416W, 417C, 417Ya, 451Xa, 452O, 452R, 452Yb, 452Yc, 453E, 453F, 453H, 465Pb, 465Xp strictly positive additive functional (on a Boolean algebra) 391D, 391J, 391N, 395Xd, 417Xu; see also chargeable Boolean algebra (391X) —– —– measure (on a topological space) 411N, 411O, 411Pc, 411Xh, 415E, 415Fb, 415Xj, 416U, 417M, 417Sc, 417Yd, 417Yh, 418Xg, 433Hb, 435Xj, 441Xc, 441Xh, 441Yj, 442Aa, 443Td, 443Yi, 444Xm, 453D, 453I, 453J, 463E, 463J, 463Xd, 463Xk —– —– submeasure (on a Boolean algebra) 392B, 392F, 392I, 392J, 392Xc, 392Xg, 392Ya, 393Xa, 393Xb, 393Yd, 491Ia strong law of large numbers 273D, 273H, 273I, 273Xh, 275Yn, 276C, 276F, 276Ye, 276Yg, 458Yd, 459Xa, 465H, 465M strong lifting 453A, 453B-453D, 453H-453J, 453M, 453N, 453Xa-453Xe, 453Za; see also almost strong lifting (453A) strong measure-algebra topology 323Ae, 323Xg, 366Yi, 412Yc strong operator topology 366Yi, 372Yn, 388Yf, 3A5M strongly consistent disintegration 452E, 452G, 452P, 458N, 458Xp, 458Xq, 495I strongly measure-compact (topological) space 435Xk strongly mixing see mixing (372P) strongly Prokhorov topological space 437Yo subalgebra of a Boolean algebra 312A, 312B, 312M, 312N, 312Xb, 312Xc, 312Xj, 313Fc, 313G, 313Xd, 313Xe, 315Xn, 315Xp, 316Xa, 331E, 331G, 332Xf, 363G, 391Xb, 391Xf, 391Xj, 457A, 457C, 457D; see also Boolean-independent subalgebras (315Xn), closed subalgebra (323I), fixed-point subalgebra (394Ga), independent subalgebras (325L), order-closed subalgebra, order-dense subalgebra, regularly embedded subalgebra (313N),σ-subalgebra (233A, 313E) subbase (of a topology) > 4A2A, 4A2Ba, 4A2Fh, 4A2Hc, 4A2Oa, 4A3D, 4A3O subdivision see tagged-partition structure allowing subdivisions (481G) subgroup of a topological group 443Xg, 4A5E, 4A5J; see also closed subgroup, compact subgroup, normal subgroup, one-parameter subgroup subhomomorphism see σ-subhomomorphism (375E) sublattice 3A1Ib submartingale 275Yf, 275Yg submeasure (on a Boolean algebra) §392 (392A), 393B, 393T, 393U, 393Xi, 393Yb, 491Aa; see also Maharam submeasure (392G), outer measure (113A), pathological submeasure (393Tc), strictly positive submeasure submodular functional 413Xq subring 3A2C —– of a Boolean ring 311Xd, 312Xa, 312Xg, 481Hd subspace measure 113Yb, 214A, 214B, 214C, 214H, 214I, 214Xb-214Xh, 216Xa, 216Xb, 241Ye, 242Yf, 243Ya, 244Yc, 245Yb, 251P, 251Q, 251Wl, 251Xn, 251Yb, 254La, 254Ye, 264Yf, 322I, 322J, 322Xg, 322Yd, 343H, 343M, 343Xa, 411Xg, 412O, 412P, 412Xp, 413Xc, 414K, 414Xn, 414Xp, 415B, 415J, 415Yb, 416R, 416T, 417I, 417Xf, 435Xb, 471E, 443K, 443Xg, 451Xf, 451Ya, 453E; (on a measurable subset) 131A, 131B, 131C, 132Xb, 214J, 214K, 214Xa, 214Xi, 241Yf, 247A, 342Ga, 342Ia, 342Xn, 343L, 344J, 344Xa, 344Xe, 344Xf, 414Yc, 415Xd, 415Yh, 416Xo, 416Xu, 416Xv, 419A, 443F, 451D, 451Yk, 453Dd, 454Pb, 491Xt, 495Xb; (integration with respect to a subspace measure) 131D, 131E-131H, 131Xa-131Xc, 133Dc, 133Xa, 214D, 214E-214G, 214M, 214Xm, 412J subspace of a normed space 2A4C subspace topology 2A3C, 2A3J, 3A4Db, 422Gf, 434F, 434Id, 434J, 438La, 462C, 4A2B-4A2D, 4A2F, 4A2Ha, 4A2Kb, 4A2L, 4A2N, 4A2P-4A2S, 4A2Ua, 4A3Ca, 4A3Kd subspace uniformity 3A4D, 4A5Ma subspace σ-algebra 121A, 214Ce, 418A, 424Bd, 424E, 424G, 454Xd, 454Xe, 461Xg, 461Xi, 4A3Ca, 4A3Kd, 4A3Nd, 4A3Xg, 4A3Yb substitution see change of variable in integration successor cardinal 2A1Fc, 4A1Aa —– ordinal 2A1Dd

932

Index

sum

sum over arbitrary index set 112Bd, 226A sum of measures 112Xe, 112Ya, 212Xe, 212Xh, 212Xi, 212Xj, 212Yd, 212Ye, 334Xb, 334Xd, 415Ya, 416Xd, 437Yj sum topology see disjoint union topology (4A2A) summable family of real numbers 226A, 226Xf supermodular functional on a lattice 413P, 413Xo support of an additive functional on a Boolean algebra 326Xi support of a Boolean homomorphism 381B, 381G, 381H, 381Xm, 382E, 382I, 382K, 382N, 382P-382R, 384B, 493Yc support of a topological measure 256Xf, 257Xd, > 411N, 415Qb, 417C, 417E, 417Xq, 434Ha, 434Ia, 444Xb, 456D, 456L, 456Xf, 464Xa, 491Xz; (of a Gaussian distribution) 456D, 456E, 456L support of a submeasure on a Boolean algebra 392Xe support see also bounded support, compact support (4A2A) supported see point-supported (112Bd) supporting (element in a Boolean algebra supporting an automorphism) 381Ba, 381E, 381Jb, 381Qb, 381Sa, 381Xk, 382D, 382N, 382O, 384B; see also self-supporting set (256Xf, 411Na), support supremum 2A1Ab —– in a Boolean algebra 313A-313C surface measure see normalized Hausdorff measure (265A) symmetric distribution 272Ye symmetric difference (in a Boolean algebra) 311G, 448Xj, 493E symmetric group 449Xc, 492H, 492I, 493Xc symmetric measure on a product space 459D, 459E, 459G, 459H, 459Xc, 459Xd, 459Xe, 459Ya symmetric Riemann-complete integral 481L symmetric set (in a group) 4A5A; (in X n ) 465J symmetrization see Steiner symmetrization tag 481A tagged partition §481 (481A) tagged-partition structure allowing subdivisions 481G, 481H-481K, 481M-481Q, 481Xb-481Xd, 481Xf, 481Xf, 481Yb, 481Yc, 482A, 482B, 482E, 482G, 482H, 482K, 484F Talagrand’s measure 464D, 464E, 464N, 464R, 464Xb, 464Z, 466Xm Tamanini-Giacomelli theorem 484B, 484Ya Tarski’s theorem §394 notes, 449J, 449Ye; see also Banach-Tarski paradox tempered distribution 284 notes tempered function §284 (284D), 286D tempered measure 284Yi tensor product of linear spaces 253 notes, 376Ya-376Yc tent map 372Xm, 385Xk, 461Xn test function 242Xi, 284 notes; see also rapidly decreasing test function (284A) thick set see full outer measure (132F) thin set 484A, 484E, 484N Tietze’s theorem 4A2Fd tight functional 436Xn, 437Tb, 437Xq —– measure 342Xh, 343Yc, 411E, 411Gf, 411Ja, 411Yb, 412B, 412Sb, 412V, 412Xa, 412Xe, 412Xl, 412Xn, 412Ye, 414Xj, 416D, 416M, 417C, 417E, 419D, 432B, 432Xc, 433Ca, 434A, 434Gc, 434Ja, 434Xb, 435Xb-435Xe, 451M, 451Ra, 471I, 471S, 471Yc, 471Yc; see also Radon measure (411Hb), signed tight Borel measure (437G) —– see also countably tight (4A2A), uniformly tight (285Xj, 437T) Tonelli’s theorem 252G, 252H, 252R, 417Hc topological group chap. 44, §4A5 (4A5Da); see also abelian topological group, compact Hausdorff group, locally compact group, Polish group (4A5Db), unimodular group (442I) topological measure (space) 256A, >411A, 411G, 411Jb, 411N, 411Xg, 411Xh, 411Xj, 411Yb, 411Yc, 414B, 414C, 414E-414K, 414P, 414R, 414Xh, 414Ya, 419H, 431E, 432B, 433A, 433B, 434D, 461C, 463E,

uniform

General index

933

463Xd, 464Z, 471C, 471Da, 471Xf, 476D, 476E, 476I, 476Xe, 481N; see also Borel measure (411K), quasiRadon measure space (411Ha), Radon measure space (411Hb) topological semigroup 444Yb, 449Ya topological space §2A3 (2A3A), §3A3, §4A2 topological vector space see linear topological space (2A5A) topology §2A2, §2A3 (2A3A); see also convergence in measure (245A, 367M), linear space topology (2A5A), measure-algebra topology (323Ab), order topology (4A2A), pointwise convergence (462Ab), uniform convergence (4A2A) total order 418Xw; see also totally ordered set (2A1Ac) total variation (of an additive functional) 231Yh; (of a function) see variation (224A) totally bounded set (in a metric space) 434L, 441Ya, 448Xi, 476Ac, 495Yc, 4A2A; (in a uniform space) 443H, 443I, 443Xk, 443Yd, 463Xb, 4A2A, 4A2J, 4A5O totally finite measure algebra 322Ab, 322Bb, 322C, 322Mb, 322Qc, 323Ad, 323Ca, 324Kb, 324P, 327B, 331B, 331D, 332M, 332P, 332Q, 333C-333G, 333J-333N, 333Q, 333R, 366Yh, 373Xj, 373Xn, 373Yb, 375Fb, 375H, 383Fb, 383I, 383J, 383Xg, 383Ya, 386Xa, 384O, 384Xd, 386A-386D, 386K, 386Xd, 386Xf, 391B, 394R, 395Xa totally finite measure (space) 211C, 211L, 211Xb, 211Xc, 211Xd, 212G, 213Ha, 214Ia, 214Ja, 215Yc, 232Bd, 232G, 243I, 243Xk, 245Fd, 245Xe, 245Ye, 246Xi, 246Ya, 322Bb, 386Xb, 412Xg, 441Xe, 441Xh, 442Ie, 471Dh, 481Xg, 495Xa totally ordered set 135Ba, 2A1Ac, 3A1F tower (in PN) 4A1Fb trace (of a σ-algebra) see subspace σ-algebra (121A) transfinite recursion 2A1B transitive action 442Z, 443T, 443Xz, 448Xi, 476G, 476Yc, 4A5B translation-invariant inverse-measure-preserving function 387Xd translation-invariant lifting 345A, 345B-345D, 345Ya-345Yc, 346C, 447Aa, 447I, 447J, 447Xa, 447Ya, 453B translation-invariant lower density 345Xf, 345Ya, 447Ab, 447B, 447F-447H translation-invariant measure 114Xf, 115Xd, 134A, 134Ye, 134Yf, 255A, 255Ba, 255Yn, 345A, 345Xb, 345Ya; see also Haar measure (441D) translation-invariant σ-ideal 444Ye transportation network 4A4N transversal (for a Boolean automorphism) 382Ab, 382B, 382C, 382G-382I, 382L, 382Xa, 382Xk tree (of sequences) > 421N truly continuous additive functional 232Ab, 232B-232E, 232H, 232I, 232Xa, 232Xb, 232Xf, 232Xh, 232Ya, 232Ye, 234Ce, 327C, 362Xh, 363S, 414D, 438Xb, 444Xl truncated Riesz subspace of RX 436B, 436C, 436D, 436H two-sided invariant mean 449Xs Tychonoff space see completely regular (3A3Ac) Tychonoff’s theorem 3A3J type see Maharam type (331F) Ulam S. see Banach-Ulam problem ultrafilter 254Yd, 2A1N, 2A1O, 2A3R, 2A3Se, 351Yd, 3A3De, 3A3Lc, 4A1Ia, 4A1K, 4A1L, 4A2Bc, 4A2Tc; see also non-principal ultrafilter, principal ultrafilter (2A1N) Ultrafilter Theorem 2A1O uncompleted indefinite-integral measure 234Cc, 465Ck uncountable cofinality, cardinal of 4A1A-4A1C uniferent homomorphism 312F Uniform Boundedness Theorem 3A5H uniform convergence, topology of 4A2A uniform convergence on compact sets, topology of 454N-454P, 454Xl, 454Xm, 455D, 456Ye, 4A2Gg, 4A2Oe; see also Mackey topology (4A4F) uniform metric gauge 481E, 481I, 482E, 482Xg uniform space 2A5F, §3A4 (3A4A), 441Yc, 4A2J

934

Index

uniformity

uniformity §3A4 (3A4A), 3A5I, 4A2J; (of a linear topological space) 463Xb; see also bilateral uniformity (4A5Hb), measure-algebra uniformity (323Ab), right uniformity (4A5Ha), uniform space (3A4A) uniformly complete Riesz space 354Yi, 354Yk uniformly continuous function 224Xa, 255K, 3A4C, 3A4Ec, 3A4G, 444Xs, 449B, 473Da, 473Ed, 476Ya, 4A2J uniformly convergent (sequence of functions) 3A3N, 4A2Jg uniformly distributed sequence see equidistributed (281Yi, 491B) uniformly exhaustive submeasure 392Bc, 392C, 392E, 392F, 392J, 392Xc, 393A, 438Ya >246A), 252Yp, 272Yd, 273Na, 274J, 275H, 275Xi, 275Yl, 276Xd, uniformly integrable set (in L1 ) §246 (> 276Yb; (in L1 (µ)) §246 (246A), 247C, 247D, 247Xe, 253Xd, 354Q; (in an L-space) 354P, 354Q, 354R, 356O, 356Q, 356Xm, 362E, 362Yf-362Yh, 371Xf; (in L1 (A, µ ¯)) 367Xm, 373Xj, 373Xn uniformly rotund see locally uniformly rotund (467A) uniformly smooth (set of linear functionals) 437Ye uniformly tight (set of linear functionals) 437Ta, 437Ua, 437Yo;(set of measures) 285Xj, 285Xk, 285Ye, 285Yf, 437Tb, 437U, 437V, 437Xr-437Xt, 437Yf, 437Ym, 452D, 452Xd unimodular topological group 442I, 442Xf, 442Xg, 442Yb, 443Ag, 443Xw-443Xy, 443Yk, 443Yn, 444R, 444Yh, 449Xe, 449Xl, 449Xo unit ball in Rr 252Q unital (Banach) algebra 4A6Ab, 4A6C-4A6E, 4A6I, 4A6J, 4A6L-4A6N universal Gaussian distribution 456F, 456G, 456H, 456Yb, 456Yc universal mapping theorems 253F, 254G, 315B, 315I, 315 notes, 325C, 325H, 325J, 369Xk universally complete Riesz space see laterally complete (368L) universally measurable set 432A, > 434D, 434Eb, 434F, 434S, 434T, 434Xc, 434Xd, 434Xf, 434Xg, 434Xy, 434Yc, 434Yb, 438Xj, 439Xd, 439Xl, 444Ye, 451Xi, 454Xi; see also Σum (434Dc) —– —– function > 434D, 434S, 434T, 437Xe, 437Yd, 463Zc, 466Xi —– —– see also universally Radon-measurable (434E) universally negligible topological space 439B, 439C, 439F, 439G, 439Xb-439Xh, 439Ya-439Yd; see also universally τ -negligible (439Xh) universally Radon-measurable set > 434E, 434Fc, 434J, 434Xe-434Xg, 434Xi, 434Yd, 437Ib; see also ΣuRm (434Eb) —– —– function 434Ec, 437Ib, 466L universally τ -negligible topological space 439Xh, 439Yd up-crossing 275E, 275F upper asymptotic density 491A, 491I, 491Xb, 491Xd upper derivate of a real function 483H, 483J, 483Xm upper envelope see upr(a, C) (314V) upper integral 133I, 133J, 133K, 133Xe, 133Yf, 135H, 214Xm, 252Ye, 252Yh, 252Yi, 253J, 253K, 463Xj, 464Ab, 464H, 464Kb, 465M, 465Xn upper Riemann integral 134Ka upper semi-continuous function 414A, 471Xf, 476D, 476E, 4A2A, 4A2Bd —– —– relation 422A upwards-directed partially ordered set 2A1Ab Urysohn’s lemma 4A2Fd —– see also Fréchet-Urysohn (462Aa) usual measure on {0, 1}I 254J; see under {0, 1}I usual measure on [0, 1]I 416U; see under [0, 1]I usual measure on PX 254J; see under PX usual topology on PX 4A2A; see under PX vague topology (on a space of signed measures) 274Ld, 274Xh, 274Ya-274Yd, 275Yp, 285K, 285L, 285S, 285U, 285Xk, 285Xq, 285Xs, 285Yd, 285Yg-285Yi, 285Yn, 367Xv, >437J, 437L, 437M, 437Xf-437Xh, 437Xl, 437Xo, 437Xt, 437Ye-437Yh, 437Ym, 437Yo, 444Xf, 445Yh, 452Xc, 491 notes, 495Xl; see also narrow topology (437Jd)

well-ordered

General index

935

variance of a random variable 271Ac, 271Xa, 272R, 272Xf, 285Gb, 285Xo variation of a function §224 (224A, 224K, 224Yd, 224Ye), 226B, 226Db, 226Xc, 226Xd, 226Yb, 264Xf, 265Yb, 463Xi, 463Xj, 463Yc, 465Xc; see also bounded variation (224A) —– of a measure see total variation (231Yh) vector integration see Bochner integral (253Yf), Henstock integral (483Yg), Pettis integral (463Ya) vector lattice see Riesz space (241E, 352A) vector measure 326Ye, 361Gb, 393O, 393P, 393Q, 393S, 393Xg, 393Yh, 393Yi, 474Ya very weak operator topology 373K, 373L, 373Xp, 373Xv, 373Yg Vietoris topology 476C, 476E, 476Xb, 476Xf, 476Yb; see also k-Vietoris (476Cd) virtually measurable function 122Q, 122Xe, 122Xf, 212Bb, 212F, 241A, 252E Vitali cover 261Ya Vitali’s theorem 221A, 221Ya, 221Yc-221Ye, 261B, 261Yk, 447C, 471N, 471O; see also Besicovitch’s Covering Lemma (472A-472C) Vitali-Hahn-Saks theorem 246Yg, 362Yh volume 115Ac —– of a ball in Rr 252Q, 252Xh von Neumann transformation 388D, 388E, 388Xd, 388Xd, 388Xg see also relatively von Neumann (388D), weakly von Neumann (388D) Wald’s equation 272Xh wandering see weakly wandering (395C) weak operator topology see very weak operator topology (373K) weak order unit (in a Riesz space) 353L, 353P, 353Yg, 368Yd weak topology (of a linear topological space) 461I, 462Yb, 466A, 466Xa, 466Xb, 466Xd, 466Ya, 466Yb, 4A3U, 4A3V, 4A4Bd, 4A4Cg, 4A4E weak topology (of a normed space) 247Ya, 2A5I, 356Yf, 356Yg, 3A5Eb, 436Xq, 462D, 462E, 462Xb, 464Z, 465E, 466B-466F, 466H, 466Xe, 466Xf, 466Xm, 466Yb, 467Xg, 467Xh, 4A4K —– see also very weak operator topology (373K), (relatively) weakly compact, weakly compactly generated, weakly convergent, weakly K-countably determined weak* topology on a dual space 253Yd, 285Yg, 2A5Ig, 3A5E, 3A5F, 437K, 4A4Bd, 4A4If; see also vague topology (274Ld, 437J) weakly compact linear operator 371Xb, 376Q, 3A5Kb; see also compact linear operator (3A5Ka) weakly compact set (in a linear topological space) 247C, 247Xa, 247Xc, 247Xd, 2A5I, 376Yj, 461I, 462E, 462F, 4A4F, 4A4Ka; see also Eberlein compactum (467O), relatively weakly compact (2A5Id) weakly compactly generated normed space 467L, 467M, 467Xd-467Xg weakly convergent sequence in a normed space 247Yb, 367Xy, 4A4Kb weakly K-countably determined normed space 467Hb, 467I-467K, 467M, 467Xc, 467Xh, 467Ya weakly measurable function see scalarly measurable (463Ya) weakly von Neumann transformation 388D, 388F, 388H, 388Xf, 388Yc weakly wandering element 395C weakly α-favourable measure space 451U, 451Yg-451Yq, 454Ya —– —– topological space 451Yq, 4A2A weakly θ-refinable see hereditarily weakly θ-refinable (438K) weakly σ-distributive Boolean algebra 316Yg, 316Yh, 362Ye, 375Yc, 391Ya, 392Ya weakly (σ, ∞)-distributive Boolean algebra 316G, 316H-316K, 316Xk-316Xq, 316Xx, 316Yf, 316Yg, 316Yi, 316Yj, 316Yp, 322F, 325Yd, 362D, 367Yk, 368Q, 368R, 368Xe, 368Yg, 368Yi, 376Yf, 391D, 391K, 391Ya, 391Yg, 391Z, 392I, 393Yd, 394Yc, 395Ya weakly (σ, ∞)-distributive Riesz space 368N, 368O-368S, 368Xd, 368Xe, 368Ye, 368Yf, 368Yh, 368Yj 376H Weierstrass’ approximation theorem 281F; see also Stone-Weierstrass theorem weight of a topological space 437Yk, 491Xz, 4A2A, 4A2D, 4A2Li; see also measure-free weight well-distributed sequence 281Xh well-ordered set 2A1Ae, 2A1B, 2A1Dg, 2A1Ka, 4A1Ad, 4A2Rk; see also ordinal (2A1C)

936

Index

well-ordering

Well-ordering Theorem 2A1Ka Weyl’s Equidistribution Theorem 281M, 281N, 281Xh, 372Xo, 491Xf; Wiener’s Dominated Ergodic Theorem 372Yb Wiener measure see Brownian motion (455C) Wirtinger’s inequality 282Yf Young’s function > 369Xc, 369Xd, 369Xr, 369Yc, 369Yd, 373Xm Young’s inequality 255Ym, 444Yh Zermelo’s Well-ordering Theorem 2A1Ka Zermelo-Fraenkel set theory 3A1A zero-dimensional topological space 311I-311K, 315Xf, 316Xo, 316Yd, 353Yc, 3A3Ad, 3A3Bd, 414R, 416Q, 419Xa, 437Xl, 481Xh, 4A2Ud, 4A3Oe, 482Xc, 491Ch zero-one law 254S, 272O, 272Xf, 272Xg, 325Xg, 417Xx, 458Yb, 464Ac zero set in a topological space 313Yb, 316Yh, 324Yb, 3A3Pa, 411Jb, 412Xg, 416Xi, 421Xh, 423Db, 443N, 443Yf, 491C, 4A2Cb, 4A2F, 4A2Gc, 4A2Hc, 4A2Lc, 4A2Nb, 4A2Sb, 4A3Kd, 4A3Nc, 4A3Xc, 4A3Yb Zorn’s lemma 2A1M, 3A1G A-operation see Souslin’s operation (421B) AC∗ function 483O, 483Pb, 483Q ACG∗ function 483Oc, 483R a.e. (‘almost everywhere’) 112Dd a.s. (‘almost surely’) 112De Aut (in Aut A) see automorphism group of a Boolean algebra (381A); (in Autµ A) see automorphism group of a measure algebra (383A) AL (Lebesgue measure algebra) 373C B (in B(x, δ), closed ball) 261A, 2A2B B (in B(U ; V ), space of bounded linear operators) 253Xb, 253Yj, 253Yk, 2A4F, 2A4G, 2A4H, 371B-371D, 371G, 371Xd, 371Yc, 376M, 3A5H, 444Yg; (B(U ; U )) 395Xb, 4A6C; (B(Rr , Rr )) 446A B (in B(X)) see Borel σ-algebra (4A3A) b Bb (in B(X)) see Baire property algebra (4A3Q) B-sequence (in a topological group) 446L, 446M, 446N, 446P, 446Xc, 447C, 447D, 447F, 447Xb, 447Xc Ba (in Ba(X)) see Baire σ-algebra (4A3K) c (in c(A), where A is a Boolean algebra) see cellularity (332D) c (in c(X), where X is a topological space) see cellularity (332Xd) c (the cardinal of R or PN) 2A1H, 2A1L, 343I, 343Yb, 344H, 344Yf, 383Xd, 391Yd, 391Ye, 416Yg, 419I, 419Xd, 421Xd, 421Xi, 421Yc, 423K, 423Xh, 423Ye, 424Db, 434Xg, 436Xg, 438C, 438R, 438Xn-438Xp, 438Yb, 438Yc, 438Yi, 439P, 439Xm, 454Yb, 491F, 491G, 491P, 491Q, 491Xz, 4A1A, 4A1O, 4A2Be, 4A2De, 4A2Gi, 4A3F + c+ (the successor of c) 523P, 531Xa; see also 2c C \ {0} (the multiplicative group) 441Xf, 441Yh C (in C(X), where X is a topological space) 243Xo, 281Yc, 281Ye, 281Yf, 352Xk, 353M, 353Xd, 354L, 354Yf, 353Yc, 363A, 367L, 367Yh-367Yj, 368Ya, 416Xi, 416Yg, 424Xf, 436Xe, 436Xg, 437Yk, 454N-454P, 454Xl, 454Xm, 455D, 456Ye, 462B, 462G, 462H, 462J, 462Xd, 462Ya, 462Yd, 462Z, 463Xd, 466Xf, 481Xh, 4A2Ib, 4A2N-4A2P; (in C(X; C)) 437Yb C([0, 1]) 242 notes, 352Xh, 356Xb, 368Yf, 436Xi, 436Xm, 437Yc Cb (in Cb (X), where X is a topological space) 281A, 281E, 281G, 281Ya, 281Yd, 281Yg, 285Yg, 352Xk, 354Hb, 436E, 436Ia, 436L, 436Xf, 436Xl, 436Yc, 437J, 437Xd, 437Yi, 462Xb, 483Mc, 491C Ck (in Ck (X), where X is a topological space) see compact support (4A2A) C0 (in C0 (X), where X is a topological space) 436I, 436K, 436Xq, 437I, 443Yb, 445K, 449Ba, 462E, 4A6B C ∞ (in C ∞ (X), for extremally disconnected X) 364W, 364Yl, 368G c 354Xq, 354Xs, 355Ye c0 354Xa, 354Xd, 354Xi, 371Yc, 461Xc; see also `∞ /cc0

General index

H

937

cac (‘countable antichain condition’) 316 notes ccc Boolean algebra 316Aa, 316C-316F, 316Xa-316Xj, 316Yc-316Ye, 316Yg, 316Yp, 322G, 324Yd, 325Yd, 326L, 326Xi, 331Ge, 332D, 332H, 363Yb, 364Yb, 367Yk, 368Yg, 368Yi, 381Yc, 391M, 391Xa, 392Ca, 392I, 393J, 393Yd, 393Ye, 394Xf, 394Yb, 448Ya ccc topological space 316Ab, 316B, 316Xd, 316Xe, 316Ya, 316Yd, 411Ng, 418Ye, 443Xm, 444Xm, 4A2E, 4A2Pd, 4A2Rn, 4A3E, 4A3Mb cf (in cf P ) see cofinality (3A1Fb) c.l.d. product measure §§251-253 (251F, 251W), 254Db, 254U, 254Ye, 256K, 256L, 325A, 325B, 325C, 325H, 334A, 334Xa, 342Ge, 342Id, 342Xn, 343H, 354Yl, 376J, 376R, 376S, 376Yc, 411Xi, 412R, 412S, 413Xg, 417C, 417D, 417S, 417T, 417V, 417Xb, 417Xj, 417Xs, 417Xv, 417Ye, 417Yi, 418S, 418T, 419E, 419Ya, 434R, 436Yd, 438Xg, 438Xh, 439Xj, 443Xp, 451I, 451Yn, 464A, 464B, 464Hd, 464Qc, 464Yb, 471Yi, 491Xu, 494Xb, 495Xj c.l.d. version of a measure (space) 213E, 213F-213H, 213M, 213Xb-213Xe, 213Xg, 213Xj, 213Xk, 213Xn, 213Xo, 213Yb, 214Xf, 214Xj, 232Ye, 234Yf, 241Ya, 242Yh, 244Ya, 245Yc, 251Ic, 251S, 251Wn, 251Xd, 251Xj, 251Xk, 252Ya, 322D, 322Qb, 322Xc, 322Xi, 322Yb, 324Xc, 324Xe, 342Gb, 342Ib, 342Xn, 343H, 343Ye, 411Xc, 411Xd, 412H, 413Eg, 413Xi, 413Xl, 414Xj, 415Xn, 416F, 416H, 436Xa, 436Xc, 436Xk, 436Yd, 471Xh, 451G, 451Ym, 465Cj, 471Yi, 491Xt cl* see essential closure (475B) CM∗ (formulations of the Control Measure Problem) 393A, 393H, 393J, 393L, 393P, 393Xe, 393Xg d (in d(X)) see density (331Yf) D (in Dn (A, π), where A is a subset of a Boolean algebra, and π is a homomorphism) 385K, 385L, 385M; (in Dn (A), where A is a subset of a topological group) 446D; (in Dk (A, E, α, β), where E is a set and A is a set of functions) 465Ae D∗ (in D∗ F ) see upper derivate (483H) D∗ (in D∗ F ) see lower derivate (483H) D (smooth functions with compact support) 473Be, 473C-473E, 474Ba, 474K Dr (smooth functions with compact support) 474A, 474Ba ∂ (in ∂A, where A is a subset of a topological space) see boundary (4A2A); (in ∂T , ∂ ξ T , where T is a tree) see derived tree (421N) ∂* see essential boundary (475B) ∂ $ see reduced boundary (474G) diam (in diam A) = diameter div (in div φ) see divergence (474B) dom (in dom f ): the domain of a function f ess sup see essential supremum (243Da) exp see exponentiation (4A6L) E (in E(X), expectation of a random variable) 271Ab; (in E(X|T), conditional expectation) 465M f

(in Af ) 361Ad Fσ set 412Xf, 414Ye, 443Jb, 466Yc, 4A2A, 4A2C, 4A2Fi, 4A2Ka, 4A2Lc F (in F(B ↑), F(B↓)) 323D f -algebra 241H, 241 notes, 352W, 352Xk-352Xn, 353O, 353P, 353Xd, 353Yg, 353Yh, 361Eh, 363B, 364C-364E, 367Yf Gδ set 264Xd, 419B, 419Xb, 434Xb, 437We, 437Yo, 443Jb, 461Xi, 461Xj, 471Db, 493Yd, 4A2A, 4A2C, 4A2Fd, 4A2Kf, 4A2L, 4A2Q, 4A5R GL(r, R) (the general linear group) 446A grad f see gradient (473B), 473C, 473Dd, 473H, 473I, 473J, 473K, 473L h (in h(π)) see entropy (385M); (in h(π, A)) 385M, 385N-385P, 385Xq, 385Yb, 387C H (in H(A)) see entropy of a partition (385C); (in H(A|B)) see conditional entropy (385D) HL1 483M HL1V 483Yg HL1 483M, 483Xj

938

Index

I

I (in Iσ , where σ ∈ N k ) 421A; (in Iν (f ) see gauge integral (481C) I k see split interval (343J) int* see essential interior (474B) JKR-space 439K K-analytic set, topological space 422F, 422G-422K, 422Xb, 422Xd, 422Xf, 422Ya, 422Yb, 422Yd-422Yf, 423C-423E, §432, 434B, 434Dc, 434Hb, 434Jf, 434Ke, 434Xp, 434Xq, 435Fb, 438P, 438Q, 455Ye, 467Xg K-countably determined topological space 467H, 467Xa, 467Xb, 467Xh; see also weakly K-countably determined (467Hb) k-space 462J k-Vietoris topology 476Cd, 476Xc, 476Xd, 495N, 495O, 495Xm, 495Xn Kσ set 451Xl, 4A2A Kσδ set 422Yb `1 (in `1 (X)) 242Xa, 243Xl, 246Xd, 247Xc, 247Xd, 354Xa, 356Xc, 464R `1 (= `1 (N)) 246Xc, 354M, 354Xd, 356Xl `2 244Xn, 282K, 282Xg, 355Yb, 371Ye, 376Yh, 376Ym, 376Yn `p (in `p (X)) 244Xn, 354Xa, 466C `∞ (in `∞ (X)) 243Xl, 281B, 281D, 354Ha, 354Xa, 361D, 361L, 461Xd, 464F, 464R, 464Xb, 464Z, 466I, 466Xl, 466Xm, 483Yg, 4A2Ib `∞ (= `∞ (N)) 243Xl, 354Xj, 356Xa, 371Yd, 383J, 4A4Id `∞ -complemented subspace 363Yd `∞ /cc0 466Ib L-space 354M, 354N-354P, 354R, 354Xt, 354Yj, 356N, 356P, 356Q, 356Xm, 356Yf, 362A, 362B, 362Yj, 365C, 365Xc, 365Xd, 367Xn, 369E, 371A-371E, 371Xa, 371Xb, 371Xf, 371Ya, 376M, 376P, 376Yi, 376Yj, 436Ib, 436Yb, 437B, 437C, 437E, 437F, 437H, 437I, 437Yc, 437Yj, 444E, 461Xl, 467Yb, 495K, 495Xh, 495Ya L0 (in L0 (µ)) 121Xb, 121Ye, §241 (241A), §245, 253C, 253Ya, 443Ae, 443G; (in L0 (Σ)) 345Yb, 364C, 364D, 364E, 364J, 364Yi, 463A-463D, 463I, 463Xa-463Xc, 463Xa, 463Xb, 463Xc, 463Xf, 463Xl, 463Xm, 463Yd, 463Za; see also L0 (241C), L0strict (241Yh), L0C (241J) L0strict 241Yh L0C (in L0C (µ)) 241J, 253L L0 (in L0 (µ)) §241 (241A), 242B, 242J, 243A, 243B, 243D, 243Xe, 243Xj, §245, 253Xe, 253Xf, 253Xg, 271De, 272H, 323Xf, 345Yb, 352Xk, 364Jc, 376Yc, 416Xj, 418R, 418S, 418Xx, 438Xg, 441Kb, 442Xg, 466Xl, 493F; (in L0C (µ)) 241J; (in L0 (A)) §364 (364A), 368A-368E, 368H, 368K, 368M, 368Qb, 368R, 368S, 368Xa, 368Xe, 368Ya, 368Yd, 368Yi, §369, 372H, §375, 376B, 376Yb, 393M, 394I, 443A, 443G, 443Jb, 443Xi; (in L0C (A)) 364Yn, 495Yb; see also L0 (241A, 364C) L1 (in L1 (µ)) 122Xc, 242A, 242Da, 242Pa, 242Xb, 443P, 444P-444R; (in L1strict (µ) 242Yg, 341Ye; (in 1 LC (µ) 242P, 255Yn; (in L1V (µ)) 253Yf; see also L1 , k k1 L1 (in L1 (µ)) §242 (242A), 243De, 243F, 243G, 243J, 243Xf, 243Xg, 243Xh, 245H, 245J, 245Xh, 245Xi, §246, §247, §253, 254R, 254Xp, 254Ya, 254Yc, 255Xc, 257Ya, 282Bd, 327D, 341Ye, 354M, 354Q, 354Xa, 365B, 376N, 376Q, 376S, 376Yl, 443Pf, 444S, 467Yb, 483Mb, 495L; (in L1V (µ)) 253Yf, 253Yi, 354Yl; (in L1 (A, µ ¯) or L1µ¯ ) §365 (365A), 366Yc, 367J, 367K, 367U, 367Ys, 369E, 369N, 369O, 369P, 371Xc, 371Yb, 371Yc, 371Yd, 372B, 372C, 372E, 372G, 372Xc, 376C, 386E, 386F, 386H, 465R, 495Yb; see also L1 , L1C , k k1 L1C (µ) 242P, 243K, 246K, 246Yl, 247E, 255Xc; (as Banach algebra, when µ is a Haar measure) 445H, 445I, 445K, 445Yk; see also convolution of functions L2 (in L2 (µ)) 244Ob, 253Yj, §286, 465E, 465F; (in L2C (µ)) 284N, 284O, 284Wh, 284Wi, 284Xi, 284Xk284Xm, 284Yg; see also L2 , Lp , k k2 L2 (in L2 (µ)) 244N, 244Yk, 247Xe, 253Xe, 355Ye, 372 notes, 416Yg, 444V, 444Xt, 444Xu, 444Yk, 445R, 445S, 445Xm, 445Xn, 456J, 456Yd, 465E; (in L2C (µ)) 282K, 282Xg, 284P; (in L2 (A, µ ¯)) 366K, 366L, 366Xh, 395A, 395Xb; see also L2 , Lp , k k2 Lp (in Lp (µ)) §244 (244A), 246Xg, 252Ym, 253Xh, 255K, 255Og, 255Yc, 255Yd, 255Yl, 255Ym, 261Xa, 263Xa, 273M, 273Nb, 281Xd, 282Yc, 284Xj, 286A, 411Gh, 412Xc, 415Pa, 415Yj, 415Yk, 416I, 443G, 444R444U, 444Xs, 444Yh, 444Ym, 472F, 473Ef; see also Lp , L2 , k kp

M

General index

939

Lp (in Lp (µ), 1 < p < ∞) §244 (244A), 245G, 245Xj, 245Xk, 245Yg, 246Xh, 247Ya, 253Xe, 253Xi, 253Yk, 255Yf, 354Xa, 354Yk, 366B, 376N, 411Xe, 418Yj, 441Kc, 442Xg, 443A, 443G, 443Xi, 444M; (in Lp (A, µ ¯) = Lpµ¯ , 1 < p < ∞) §366 (366A), 369L, 371Gd, 372Xp, 372Xq, 372Yb, 373Bb, 373F, 376Xb;(in p LC (µ), 1 < p < ∞) 354Yk;(in Lp (A, µ ¯), 0 < p < 1) 366Ya, 366Yg, 366Yi;see also Lp , k kp ∞ ∞ L (in L (µ)) 243A, 243D, 243I, 243Xa, 243Xl, 243Xn, 481Xg; (in L∞ (Σ)) 341Xe, 363H, 414Xt, 437B-437E, 437H, 437Ib, 437Xe, 437Yd;see also L∞ L∞ C 243K, 437Yb L∞ strict 243Xb, 363I L∞ (in L∞ (µ)) §243 (243A), 253Yd, 341Xe, 352Xk, 354Hc, 354Xa, 363I, 376Xn, 418Yi, 442Xg, 463Yc; (in L∞ (A)) §363 (363A), 364K, 364Xh, 365L, 365M, 365N, 365Xk, 367Oc, 368Q, 394N, 436Xp, 437B, 437J, 443Ad, 443Jb, 447Yb, 457A; see also L∞ , L∞ C , k k∞ L∞ 243K, 243Xm C Lτ (where τ is an extended Fatou norm) 369G, 369J, 369K, 369M, 369O, 369R, 369Xi, 374Xd, 374Xi; see also Orlicz space (369Xd), Lp , M 1,∞ (369N), M ∞,1 (369N) L (in L(U ; V ), space of linear operators) 253A, 253Xa, 351F, 351Xd, 351Xe, 4A4Bc L∼ (in L∼ (U ; V ), space of order-bounded linear operators) §355 (355A), 356Xi, 361H, 361Xc, 361Yb, 363Q, 365K, 371B-371E, 371Gb, 371Xb-371Xe, 371Ya, 371Yc-371Ye, 375Kb, 376J, 376Xe, 376Ym; see also order-bounded dual (356A) ∼ L∼ c (in Lc (U ; V )) 355G, 355I, 355Yi, 376Yf; see also sequentially order-continuous dual (356A) × × L (in L (U ; V )) 355G, 355H, 355J, 355K, 355Yg, 355Yi, 355Yj, 371B-371D, 371Gb, 376D, 376E, 376H, 376K, 376Xj, 376Yf; see also order-continuous dual (356A) lim (in lim F) 2A3S; (in limx→F ) 2A3S lim inf (in lim inf n→∞ ) §1A3 (1A3Aa), 2A3Sg; (in lim inf δ↓0 ) 2A2H; (in lim inf x→F ) 2A3S lim sup (in lim supn→∞ ) §1A3 (1A3Aa), 2A3Sg; (in lim supδ↓0 ) 2A2H, 2A3Sg; (in lim supx→F f (x)) 2A3S; (in lim supi→F Ai ) 446G, 4A2A, 4A2T ln+ 275Yd M (in M (A), space of bounded finitely additive functionals) 362B, 362E, 363K, 436Xr, 437J; (when A = PX) 464G-464M, 464O-464Q M -space 354Gb, 354H, 354L, 354Xq, 354Xr, 356P, 356Xj, 363B, 363O, 371Xd, 376M, 449D; see also order-unit norm (354Ga) M 0 (in M 0 (A, µ ¯) = Mµ¯0 ) 366F, 366G, 366H, 366Yb, 366Yd, 366Yg, 373D, 373P, 373Xk M0,∞ 252Yp M 0,∞ (in M 0,∞ (A, µ ¯) = Mµ¯0,∞ ) 373C, 373D, 373E, 373F, 373I, 373Q, 373Xo, 374B, 374J, 374L M 1,0 (in M 1,0 (A, µ ¯) = Mµ¯1,0 ) 366F, 366G, 366H, 366Ye, 369P, 369Q, 369Yh, 371F, 371G, 372D, 372Ya, 373G, 373H, 373J, 373S, 373Xp, 373Xr, 374Xe M 1,∞ (in M 1,∞ (µ)) 234Yd, 244Xl, 244Xm, 244Xo, 244Yc; (in M 1,∞ (A, µ ¯) = Mµ¯1,∞ ) 369N, 369O-369Q, 369Xi-369Xk, 369Xm, 369Xq, §373, 374A, 374B, 374M M ∞,0 (in M ∞,0 (A, µ ¯)) 366Xd, 366Yc M ∞,1 (in M ∞,1 (A, µ ¯) = Mµ¯∞,1 ) 369N, 369O, 369P, 369Q, 369Xi, 369Xj, 369Xk, 369Xl, 369Yh, 373K, 373M, 374B, 374M, 374Xa, 374Ya Mm see measurable additive functional (464I) Mpnm see purely non-measurable additive functional (464I) Mr (space of r × r matrices) 446A Mt (space of signed tight Borel measures) 437Fb, 437G, 437Ia, 437Xh, 437Xl, 437Xt, 437Yb, 437Yi, 437Ym, 437Yo Mσ (in Mσ (A), space of countably additive functionals) 362B, 362Xd, 362Xh, 362Xi, 362Ya, 362Yb, 363K, 437C-437F, 437Xe, 438Xa Mτ (in Mτ (A), space of completely additive functionals) 326Yp, 327D, 362B, 362D, 362Xd, 362Xg, 362Xi, 362Ya, 362Yb, 363K, 438Xa, 464H, 464Ja, 464R; (in Mτ (X), space of signed τ -additive measures) 437F, 437G, 437H, 437L, 437M, 437Xf, 437Xh, 437Yg-437Yi, 444E, 444S, 444Xd-444Xf, 444Yb, 444Yc, 444Yg, 445Yi, 445Yj ˜ + 437Jd, 437K M med (in med(u, v, w)) 352Xd

940

Index

N

N 3A1H; see also PN N × N 111Fb N N 372Xi, 421A, 424Cb, 4A2Ub —– as topological space 421A, 421H-421K, 421M, 421Xf, 421Xo, 431D, 4A3Fb On (the class of ordinals) 3A1E p (in p(t)) 386G, 386H P (in PX) 311Ba, 311Xe, 312B, 312C, 313Ec, 313Xf, 363S, 382Xc, 383Yb, 438A, 438Xa;(usual measure on PX) 254J, 254Xf, 254Xq, 254Yd, 441Xg, 464A-464D, 464H-464L, 464O-464Q, 464Ya; (usual topology on PX) 423Ye, 463A, 463F, 463G, 4A2A, 4A2Ud PN 1A1Hb, 2A1Ha, 2A1Lb, 315O, 316Yo, 324Yg, 326Yg, 374Xk; (usual measure on) 273G, 273Xd, 273Xe PN/[N]<ω 517Yd P(N N ) 316Yg Pω1 419F, 419G, 419Yb p.p. (‘presque partout’) 112De per (in per E, the distributional perimeter of E) 474D X ∈ E) etc. 271Ad, 434Xy Pr(X > a), Pr(X Q (the set of rational numbers) 111Eb, 1A1Ef, 364Yg, 439S, 442Xc; (as topological group) 445Xa q (in q(t)) 385A, 386M; (in qp (x, y), qk k (x, y)) 467C R (the set of real numbers) 111Fe, 1A1Ha, 2A1Ha, 2A1Lb, 352M, 4A1Ac, 4A2Gf, 4A2Ua; (as topological group) 442Xc, 445Ba, 445Xa, 445Xk RX 245Xa, 256Ye, 352Xk, 375Ya, 3A3K, 435Xc, 4A4Bb; (as linear topological space) 4A4H;see also Euclidean metric, Euclidean topology, pointwise convergence RX |F see reduced power (351M) R C 2A4A R see extended real line (§135) R \ {0} (the multiplicative group) 441Xf r (in r(u), where u is in a Banach algebra) see spectral radius (4A6G); (in r(T ), where T is a tree) see rank (421N) RO (in RO(X)) see regular open algebra (314Q) R-stable set of functions 465S, 465T-465V, 465Yc, 465Yd S (in S(A)) 243I, §361 (361D), 363C, 363Xg, 364K, 364Xh, 365F, 367Oc, 368Q, 369O; (in S f ∼ = S(Af )) ∼ ∼ × 242M, 244H, 365F, 365Gb, 369O, 369P; (in S(A) ) 362A; (in S(A)c ) 362Ac; (in S(A) ) 362Ad; (in SC (A)) 361Xj, 361Yd; (in S(A; G) 493Ya; (in St (f, ν) 481Bc S (in S(E)) see Souslin’s operation (421B) S see rapidly decreasing test function (284A) S 1 (the unit circle, as topological group) see circle group S r−1 (the unit sphere in Rr ) see sphere S6 (the group of permutations of six elements) 384 notes ∗ sf (in µsf ) see semi-finite version of a measure (213Xc); (in µsf ) 213Xf, 213Xg, 213Xk T0 topology 437Xi, 451M, 476Xf, 4A2A T1 topology 4A2A, 4A2F T2 topology see Hausdorff (2A3E, 3A3Aa) (0) T (0) (in Tµ¯,¯ν ) 371F, 371G, 371H, 372D, 372Xb, 372Yb, 372Yc, 373B, 373G, 373J, 373R, 373S, 373Xp, 373Xq, 373Xr, 373Xu, 373Xv T (in Tµ¯,¯ν ) 244Xm, 244Xo, 244Yc, 246Yc, §373 (373A), 444Ym; see also T -invariant (374A) T × (in Tµ¯×,¯ν ) §373 (373Ab), 376Xa, 376Xh T -invariant extended Fatou norm 374Ab, 374B-374D, 374Fa, 374Xb, 374Xd-374Xj, 374Yb, 444Yf, 444Yj T -invariant set 374Aa, 374M, 374Xa, 374Xi, 374Xk, 374Xl, 374Ya, 374Ye Tm see convergence in measure (245A) Tp see pointwise convergence (462Ab)

General index

σ

941

Ts (in Ts (U, V )) 373M, 373Xq, 376O, 3A5E, 465E, 465F; see also weak topology (2A5I), weak* topology (2A5Ig) U (in U (x, δ)) 1A2A; (in U (A, δ)) 476F upr (in upr(a, C)) 314V, 314Xg, 333Xa, 365Rc, 386B, 394G, 394I, 394K-349N, 458Hc usco-compact relation 422A, 422B, 422C, 422D, 422E, 422F, 422G, 422Xa, 432Xh, 443Yl, 467Ha Var (in Var(X)) see variance (271Ac); (in VarD f , Var f ) see variation (224A) w (in w(X)) see weight (4A2A) w∗ -topology see weak* topology 2A5Ig W (in Wt ) 481Bb Z (the set of integers) 111Eb, 1A1Ee; (as topological group) 255Xe, 441Xa, 445B Z2 (the group {0, 1}) 311Bc, 311D Z see asymptotic density zero ideal (491A) Z = PN/Z 491I, 491J, 491K, 491N-491P, 491Xo-491Xq, 491Yj-491Yl ZFC see Zermelo-Fraenkel set theory α-favourable see weakly α-favourable (451U, 4A2A) βr (volume of unit ball in Rr ) 252Q, 252Xh, 265F, 265H, 265Xa, 265Xb, 265Xe, 474N, 474R, 474Xe ˇ βX see Stone-Cech compactification (4A2I) Γ-function 225Xj, 225Xk, 252Xh, 252Yk, 252Yn, 255Xj ∆ (the modular function of a topological group) 442I; (in ∆(θ)) 464G ∆-system 2A1Pa, 4A1D θ-refinable see hereditarily weakly θ-refinable λ∂ (in λ∂E ) see perimeter measure (474F) µG (standard normal distribution) 274Aa µ ¯L (in §373) 373C νX see distribution of a random variable (271C) π-base for a topology 411Ng, 4A2A, 4A2Gi π-λ Theorem see Monotone Class Theorem (136B) σ-additive see countably additive (231C, 326E) σ-algebra of sets §111 (111A), 136Xb, 136Xi, 212Xk, 314D, 314M, 314N, 314Yd, 316D, 322Ya, 326Ys, 343D, 344D, 362Xg, 363H, 382Xd; see also Baire property algebra (4A3Q), Baire σ-algebra (4A3K), Borel σ-algebra (111G, 4A3A), cylindrical σ-algebra (4A3T), standard Borel space (424A) σ-algebra defined by a random variable 272C, 272D σ-compact topological space 422Xb, 441Xh, 467Xb, 495N, 495O, 495Xm, 4A2A —– locally compact group 443P, 443Xm, 443Yg, 447E-447G, 4A5El, 4A5S σ-complete see Dedekind σ-complete (241Fb, 314Ab) σ-discrete see σ-metrically-discrete (4A2A) σ-disjoint family of sets 4A2A, 4A2Lg σ-field see σ-algebra (111A) σ-finite measure algebra 322Ac, 322Bc, 322C, 322G, 322M, 323Gb, 323Ya, 324K, 325Eb, 327B, 331N, 331Xl, 362Xd, 367Nd, 367Q, 367Xq, 367Xs, 369Xg, 383E, 448Xj, 493Xe σ-finite measure (space) 211D, 211L, 211M, 211Xe, 212G, 213Ha, 213Ma, 214Ia, 214Ja, 215B, 215C, 215Xe, 215Ya, 215Yb, 216A, 232B, 232F, 234Fe, 235O, 235R, 235Xe, 235Xk, 241Yd, 243Xi, 245Eb, 245K, 245L, 245Xe, 251K, 251Wg, 252B-252E, 252H, 252P, 252R, 252Xc, 252Yb, 252Yl, 322Bc, 342Xi, 362Xh, 365Xp, 367Xr, 376I, 376J, 376N, 376S, 411Ge, 411Ng, 411Xd, 412Xh, 412Xi, 412Xo, 414D, 415D, 415Xg, 415Xh, 415Xi, 415Xo, 415Xp, 416Yd, 417Xh, 417Xu, 418G, 418R-418T, 418Xh, 418Ye, 433Xb, 434R, 436Yd, 438Bc, 438S, 438Yc, 452I, 441Xe, 441Xh, 443Xm, 451Oc, 451Xl, 463C-463E, 463H, 463I, 463L, 463Xb, 463Xd, 463Xf, 463Xk, 463Yd, 495H, 495I, 495Xa σ-generating set in a Boolean algebra 331E

942

Index

σ-ideal

σ-ideal (in a Boolean algebra) 313E, 313Pb, 313Qb, 314C, 314D, 314L, 314N, 314Yd, 316C, 316D, 316Xi, 316Ye, 321Ya, 322Ya, 392Xd —– (of sets) 112Db, 211Xc, 212Xf, 212Xk, 313Ec, 322Ya, 363H, 464Pa, 4A1Cb; see also translationinvariant σ-ideal σ-interpolation property (in a partially ordered set) 466G, 466H, 491Yl σ-isolated family of sets 438K, 438Ld, 438N, 466D, 466Eb, 467Pb, 4A2A vthreeσ-linked Boolean algebra 391L, 391M, 391N, 391Xj, 391Yb, 391Yd, 391Ye, 391Z, 393Xf; σ-mlinked Boolean algebra 391Yh, 393Ya σ-metrically-discrete family of sets 4A2A, 4A2Lg σ-order complete see Dedekind σ-complete (314Ab) σ-order-continuous see sequentially order-continuous (313H) σ-refinement property (for subgroups of Aut A) 448K, 448L-448O σ-subalgebra of a Boolean algebra 313E, 313F, 313G, 313Xd, 313Xe, 313Xq, 314Eb, 314Fb, 314Hb, 314Jb, 314Xc, 315Yc, 321G, 321Xb, 322M, 323Z, 324Xb, 326Fg, 331E, 331G, 364Xc, 364Xu, 366I, 4A1O; see also order-closed subalgebra σ-subalgebra of sets §233 (233A), 321Xb, 323Xb, 412Ab, 465Xg σ-subhomomorphism between Boolean algebras 375E, 375F-375H, 375Xd, 375Yc, 375Ye (σ, ∞)-distributive see weakly (σ, ∞)-distributive (316G) Σum (algebra of universally measurable sets) 434D, 434T Σ PuRm (algebra of universally Radon-measurable sets) 434E, 437Ib i∈I ai 112Bd, 222Ba, 226A, 4A4Bj, 4A4Ie τ (in τ (A)) see Maharam type (331Fa); (in τC (A)) see relative Maharam type (333Aa) τ -additive functional on a Boolean algebra see completely additive (326J) τ -additive measure 256M, 256Xb, 256Xc, 411C, 411E, §414, 415C, 415L-415M, 415Xn, §417, 418Ha, 418Xi, 418Ye, 419A, 419D, 419H, 432D, 432Xc, 434G, 434Ha, 434Ja, 434Q, 434R, 434Xa, 434Yl, 435D, 435E, 435Xa, 435Xc-435Xe, 436Xg, 436Xj, 437Kc, 439Xh, 444Yb, 451Xm, 452C, 453Dc, 453H, 454Pb, 456J, 456K, 459F, 461E, 462Yc, 465S, 465T, 466H, 466Xc, 466Xm, 476F, 481N, 482Xd, 491Ce; see also quasi-Radon measure (411Ha), signed τ -additive measure (437G) —– positive linear operator 437Xc —– product measure §417 (417G), 418Xg, 434R, 437L, 453I, 459G, 465S, 465T, 491F, 491Ym, 494B, 494Xa; see also quasi-Radon product measure (417R), Radon product measure (417R) τ -generating set in a Boolean algebra 313Fb, 313M, 331E, 331F, 331G, 331Yb, 331Yc τ -negligible see universally τ -negligible (439Xh) τ -regular see τ -additive (411C) Φ see normal distribution function (274Aa) χ (in χA, where A is a set) 122Aa; (in χa, where a belongs to a Boolean ring) 361D, 361Ef, 361L, 361M, 364K; (the function χ : A → L0 (A)) 364Kc, 367R ψ (in ψE ) see canonical outward-normal function (474G) ω (the first infinite ordinal) 2A1Fa, 3A1H; (in [X]<ω ) 3A1Cd, 3A1J ω ω -bounding Boolean algebra see weakly σ-distributive (316Yg) ω1 (the first uncountable ordinal) 2A1Fc, 419F, 419G, 419Yb, 421P, 435Xb, 435Xi, 435Xk, 438C, 463Xh, 463Yd, 463Ye, 4A1A, 4A1Bb, 4A1Eb, 4A1M, 4A1N; see also Pω1 —– (with its order topology) 411Q, 411R, 412Yb, 434Kc, 434Yi, 435Xi, 437Xt, 439N, 439Yf, 439Yh, 454Yc, 4A2Sb, 4A3J, 4A3P, 4A3Xg ω1 -saturated ideal in a Boolean algebra 316C, 316D, 341Lh, 344Yd, 344Ye, 431Yc ω1 +1 (with its order topology) 434Kc, 434Xb, 434Xf, 434Xk, 434Xl, 434Yi, 435Xa, 435Xc, 435Xi, 437Xk, 463D —– see also Pω1 ω2 2A1Fc, 4A1Ea ωξ (the ξth uncountable initial ordinal) 3A1E, 438Xe ω (in ω(F ¹A)) see oscillation (483Oa)

special symbols

General index

943

see closure (2A2A, 2A3Db) ¯ ¯ (in h(u), where h is a Borel function and u ∈ L0 ) 241I, 241Xd, 241Xi, 245Dd, 364I, 364J, 364Xg, 364Xq, 364Yd, 364Ye, 367I, 367S, 367Xl, 367Yr, 495Yb; (when h is universally measurable) 434T =a.e. 112Dg, 112Xh, 222E, 241C, 444Ob ≤a.e. 112Dg, 112Xh, 212B ≥a.e. 112Dg, 112Xh, 233I 0 (in U 0 ) see algebraic dual; (in T 0 ) see adjoint operator (3A5Ed) ∗ (in f ∗ g, u ∗ v, λ ∗ ν, ν ∗ f , f ∗ ν) see convolution (255E, 255O, 255Xe, 255Yn, 444A, 444H, 444K, 444O, 444S) * (in weak*) see weak* topology (2A5Ig); (in U ∗ = B(U ; R), linear topological space dual) see dual (2A4H, 4A4Bd); (in u∗ ) see decreasing rearrangement (373C); (in µ∗ ) see outer measure defined by a measure (132B) ∗ (in µ∗ ) see inner measure defined by a measure (113Yh, 413D) ∼ (in U ∼ ) see order-bounded dual (356A); (local usage in §471) 471M ∼ ∼ c (in Uc ) see sequentially order-continuous dual (356A) ∼ ∼ σ (in Uσ ) see sequentially smooth dual (437Aa) ∼ ∼ τ (in Uτ ) see smooth dual (437Ab) ˜ (in a ˜, f˜, u ˜) 443Af, 443Xe, 443Xi, 444R, 444V, 444Xs, 444Xt × (in U × ) see order-continuous dual (356A); (in U ×× ) see order-continuous bidual \ (in E \ F , ‘set difference’) 111C 4 (in E4F , ‘symmetric difference’) 111C, 311Ba ∪ , ∩ (in a Boolean ring or algebra) 311Ga, 313Xi, 323B \ , 4 (in a Boolean ring or algebra) 311Ga, 323B ⊆ , ⊇ (in a Boolean ring or algebra) 311H, 323Xa S S S T (in Tn∈N En ) 111C; (in T A) 1A1F (in n∈N En ) 111C; (in E) 1A2F ∨, ∧ (in a lattice) 121Xa, 2A1Ad; (in A ∨ B, where A, B are partitions of unity in a Boolean algebra) 385F R R R R R (in f , f dµ, f (x)µ(dx)) 122E, 122K, 122M, 122Nb, > 363L; (in A f ) 131D, 214D, 235Xf, R 482G, 482H, 482Yb; R R see also subspace measure; (in u) 242Ab, 242B, 242D, 363L, 365D, 365Xa; (in u) 242Ac; (in u) 365D, 365Xb; see also upper integral, lower integral (133I) AR a R see upper integral (133I) see lower integral (133I) R Rβ H H R , α see Henstock integral (483A) Pf see Pfeffer integral (484G) R R see Riemann integral (134K) ¹ (in f ¹A, the restriction of a function to a set) 121Eh | | (in a Riesz space) 241Ee, 242G, 352C, 352D, 352Fb, 352K, 354Aa, 354Bb k ke see order-unit norm (354Ga) k k1 (on L1 (µ)) §242 (242D), 246F, 253E, 275Xd, 282Ye, 483Mb; (on L1 (µ)) 242D, 242Yg, 273Na, 273Xi, 415Pb, 473Da; (on L1 (A, µ ¯)) 365A, 365B, 365C, 386E, 386F; (on the `1 -sum of Banach lattices) 354Xb, 354Xo k k2 244Da, 273Xj, 282Yf, 366Yh, 473Xa; see also L2 , k kp k kp (for 1 < p < ∞) §244 (244Da), 245Xm, 246Xb, 246Xh, 246Xi, 252Ym, 252Yp, 253Xe, 253Xh, 273M, 273Nb, 275Xe, 275Xf, 275Xh, 276Ya, 366A, 366C, 366D, 366H, 366J, 366Xa, 366Xi, 366Yf, 367Xo, 369Oe, 372Xb, 372Yb, 374Xb, 415Pa, 415Yj, 415Yk, 416I, 443G, 444M, 444R, 444T, 444U, 444Ym, 473H, 473I, 473K; see also Lp , Lp , k kp,q k kp,q (the Lorentz norm) 374Yb k k∞ 243D, 243Xb, 243Xo, 244Xh, 273Xk, 281B, 354Xb, 354Xo, 356Xc, 361D, 361Ee, 361I, 361J, 361L, 361M, 363A, 364Xh, 436Ic, 463Xi, 473Da, 4A6B; see also essential supremum (243D), L∞ , L∞ , `∞ k k1,∞ 369O, 369P, 369Xh-369Xj, 371Gc, 372D, 372F, 373F, 373Xl; see also M 1,∞ , M 1,0 k k∞,1 369N, 369O, 369Xi, 369Xj, 369Xl; see also M ∞,1 k kH 483L, 483M, 483N, 483Xi, 483Yg

944

Index

special symbols

⊗ (in f ⊗ g) 253B, 253C, 253J, 253L, 253Ya, 253Yb; (in u ⊗ v) §253 (253E); (in A ⊗ B, a ⊗ b) see free product (in Σ ⊗ T) 457Fa N (315M); N N N (in i∈I Ai ) see free product (315H); (in i∈I Σi ) 457Fb; (in I Σ) 465Ad b (in Σ⊗T) b ⊗ 251D, 251K, 251L, 251Xk, 251Ya, 252P, 252Xd, 252Xe, 253C, 418R, 418T, 419F, 421H, 424Yd, 443Yg, 452Bb, 452M, 452N, 454C, 4A3S, 4A3Xa N c (in N c Σi ) 251Wb, 251Wf, 254E, 254F, 254Mc, 254Xc, 254Xi, 254Xs, 343Xb, 424Bb, 454A, 454D, i∈I N c 454Xd, 4A3Cf, 4A3Dc, Q 454Xe, Q Q4A3M-4A3O, 4A3Xf; (in I Σ) 459D, 459E, 465Ad, 465I, 465K (in i∈I αi ) 254F; (in i∈I Xi ) 254Aa # (in #(X), the cardinal of X) 2A1Kb C (in I C R) see ideal (3A2Ea) −− ←−−−− −) (in (← (← a π b), (a π b φ c) etc.) see cycle notation (381R), cyclic automorphism, exchanging involution (381R) + (in κ+ , successor cardinal) 2A1Fc, 438Cd; (in f + , where f is a function) 121Xa, 241Ef; (in u+ , where u belongs to a Riesz space) 241Ef, 352C, 352D, 352Fc; (in U + , where U is a partially ordered linear space) 351C; (in F (x+ ), where F is a real function) 226Bb − (in f − , where f is a function) 121Xa, 241Ef; (in u− , in a Riesz space) 241Ef, 352C, 352D; (in F (x− ), where F is a real function) 226Bb 2 (in 2κ ) 3A1D, 438Cf, 4A1Ac ⊥ (in A⊥ , in a Boolean algebra) 313Xo; (in A⊥ , in a Riesz space) 352O, 352P-352R, 352Xh; (in V ⊥ , in a Hilbert space) see orthogonal complement; see also complement of a band (352P) ◦ (in A◦ ) see polar set (4A4Bh) a (in z a i) 3A1H, 421A • (in a• f ) 441Ac; (in x• a) 443C; (in a• u) 443G c c c c • (in a• f ) 441Ac, 444F; (in x• a) 443C; (in a• u) 443G l l l l • • f ) 441Ac; (in x• a) 443C; (in a• u) 443G (in a r r r r ∧

∨

, ∨ (in f , f ) see Fourier transform, inverse Fourier transform (283A) {0, 1}I (usual measure on) 254J, 254Xd, 254Xe, 254Yc, 272N, 273Xb, 331J-331L, 332B, 332C, 332N, 332Xm, 332Xn, 341Yc, 341Yd, 341Zb, 342Jd, 343Ca, 343I, 343Xc, 343Yd, 344G, 344L, 344Xg, 345Ab, 345C-345E, 345Xa, 346C, 383Xc, 416U, 441Xg, 453B, 491G, 491Xm, 493G; —– —– (when I = N) 254K, 254Xj, 254Xq, 256Xk, 261Yd, 341Xb, 343Cb, 343H, 343M, 345Yc, 346Zb, 388E, 471Xa; see also PX —– (usual topology of) 311Xh, 3A3K, 434Kd, 434Xb, 491G, 491Q, 4A2Ud, 4A3Of; (when I = N) 314Ye, 423J, 4A2Gi, 4A2U, 4A3E —– (open-and-closed algebra of) 311Xh, 315Xh, 316Xq, 316Yj, 316Ym, 331Yg, 391Xd, 393A, 393F —– (regular open algebra of) 316Yj —– (and Hausdorff measures) 471Xa, 471Yh [0, 1]I (usual measure on) 416U, 419B, 491Yh —– (usual topology of) 412Yb, 434Kd I ]0, 1[ (usual measure on) 415F —– (usual topology of) 434Xo (2, ∞)-distributive lattice 367Yd ¿ (in ν ¿ µ) see absolutely continuous (232Aa) 4σG (in a 4σG b) §448 (448A) 4τG (in a 4τG b) 394A, 394G, 394I, 394K, 394Ma, 394Xb ∞ see infinity [ ] (in [a, b]) see closed interval (115G, 1A1A, 2A1Ab, 4A2A); (in f [A], f −1 [B], R[A], R−1 [B]) 1A1B; (in [X]κ , [X]<κ , [X]≤κ ) 3A1J; (in [X]<ω ) 3A1Cd, 3A1J [[ ]] (in f [[F]]) see image filter (2A1Ib) [[ ]] (in [[u > α]], [[u ≥ α]], [[u ∈ E]] etc.) 363Xh, 364A, 364B, 364H, 364Xa, 364Xc, 364Yc, 434T; (in [[µ > ν]]) 326O, 326P [ [ (in [a, b[) see half-open interval (115Ab, 1A1A, 4A2A) ] ] (in ]a, b]) see half-open interval (1A1A) ] [ (in ]a, b[) see open interval (115G, 1A1A, 4A2A) ∧

special symbols

General index

d e (in db : ae) 394I, 394J, 394K-394M, 394Xa, 448I, 448J b c (in bb : ac) 394I, 394J, 394K-394M, 394Xa, 448I, 448J (in µ E) 234E, 235Xf, 475G ♦ see Jensen’s ♦ ♣ see Ostaszewski’s ♣ (4A1M)

945

Measure Theory Fremlin Vol4

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