PRAGUE 2005

matfyzpress

Jaroslav LukeJan Mal

Measureand

Integral

Jaroslav Luke, Jan Mal, 2005 MATFYZPRESS by publishing house of the Faculty of Mathematics and Physics Charles University in Prague, 2005

ISBN 80-86732-68-1ISBN 80-85863-06-5 (First edition)

All rights reserved, no part of this publication may be reproduced or transmitted in any form or by any means, electronic, mechanical, photocopying or otherwise, without the prior written permission of the publisher.

Motto: Everybody writes and nobody readsL. Fejer

Preface

This text is based on lectures in measure and integration theory given by theauthors during the past decade at Charles University, and on preliminary lecturenotes published in Czech.

It is impossible to thank individually all colleagues and students who assistedin the preparation of this manuscript, but we will just mention Michal Kubecekwho helped with the translation and TEX processing.

The authors wish to express their deep gratitude to Professor Stylianos Negre-pontis, who was the chief coordinator of TEMPUS project JEP1980. Withoutsupport from him and the Tempus programme the manuscript would never haveappeared.

The preparation of this manuscript was partially supported by the grantNo. 201/93/2174 of the Czech Grant Agency and by the grant No. 354 of theCharles University.

Prague, 1994 Jaroslav Lukes and Jan Maly

Preface to the second edition

We have carried out only minor corrections. We wish to thank all who con-tributed by suggestions and comments.

Prague, 2005 Jaroslav Lukes and Jan Maly

Contents

List of Basic Notations and Frequently Used Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

A. Measures and Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21. The Lebesgue Measure2. Abstract Measures3. Measurable Functions4. Construction of Measures from Outer Measures5. Classes of Sets and Set Functions6. Signed and Complex Measures

B. The Abstract Lebesgue Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277. Integration on R8. The Abstract Lebesgue Integral9. Integrals Depending on a Parameter10. The Lp Spaces11. Product Measures and the Fubini Theorem12. Sequences of Measurable Functions13. The Radon-Nikodym Theorem and the Lebesgue Decomposition

C. Radon Integral and Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5714. Radon Integral15. Radon Measures16. Riesz Representation Theorem17. Sequences of Measures18. Luzins Theorem19. Measures on Topological Groups

D. Integration on R . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8220. Integral and Dierentiation21. Functions of Finite Variation and Absolutely Continuous Functions22. Theorems on Almost Everywhere Dierentiation23. Indenite Lebesgue Integral and Absolute Continuity24. Radon Measures on R and Distribution Functions25. HenstockKurzweil Integral

E. Integration on Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10326. Lebesgue Measure and Integral on Rn

27. Covering Theorems28. Dierentiation of Measures29. Lebesgue Density Theorem and Approximately Continuous Functions30. Lipschitz Functions31. Approximation Theorems32. Distributions33. Fourier Transform

F. Change of Variable and k-dimensional Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14134. Change of Variable Theorem35. The Degree of a Mapping36. Hausdor Measures

G. Surface and Curve Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16337. Integral Calculus in Vector Analysis38. Integration of Dierential Forms39. Integration on Manifolds

40. Measurable Functions41. Vector Measures42. The Bochner Integral43. The Dunford and Pettis Integrals

Appendix on Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203

References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206

A Short Guide to the Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217

Subject Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 219

H. Vector Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191

1List of Basic Notations and Frequently Used Symbols

In this manuscript we use the standard notation.

In all what follows, N, Z, Q, R, C will denote the sets of all natural, integer,rational, real and complex numbers, respectively.

Extended real number set R consists of R together with two symbols and+ equipped with the usual algebraic structure and topology. Remind only that0 and 0 are taken as 0.If X is a set, P(X) denotes the collection of all its subsets.

Rn stands for the Euclidean n-dimensional space under the usual Euclideannorm || and the metric |x y|, where x = [x1, . . . , xn].Remember that when multiplied by a matrix (from the left), the vector x =

[x1, . . . , xn] behaves like a vertical vector, i.e. like a matrix with one column x1...xn

.The horizontal notation is preferred for estetical and typographical reasons.

The standard (or canonical) basis of the space Rn is denoted by {e1, . . . , en},the vector ei = [0, . . . , 0, 1, 0, . . . , 0] with 1 at the ith place. The inner product inRn is denoted by x y.By U(x, r) we denote the open ball in a metric space (P, ) of radius r round the

x. The closed ball is denoted by B(x, r). Thus U(x, r) = {y P : (x, y) < r},B(x, r) = {y P : (x, y) r}. For the diameter of a set we use the symboldiam and for the distance of a pair of sets the symbol dist.

If nothing else is specied, a function on a set X is a mapping of X into R. Ifwe want to emphasise that a function does not attain the values and +,we call it a real function.

Instead of the notation {x X : f(x) > a} we often use the abbreviated version{f > a}.The symbol cA denotes the indicator function of a set A X, i.e. the function

cA(x) =

{1 for x A ,0 for x X \A.

We use fj f to denote the uniform convergence of a sequence of functions.

2 1. The Lebesgue Measure

A. Measures and Measurable Functions

1. The Lebesgue Measure

In the history, people were engaged in the problem of measuring lenghts, areasand volumes. In mathematical formulation the task was, for a given set A, todetermine its size (measure) A. It was required that the volume of a cube orthe area of a rectangle or a circle should agree with the well-known formulae. Itwas also clear by intuition that this measure should be positive and additive, i.e.it should satisfy the equality

Aj =

Aj

provided {Aj} is a nite disjoint collection of sets. For a succesful developmentof the theory a further condition was imposed: The above equality was claimedto hold even for countable disjoint collections of sets. Moreover, the eort waspaid to assign a measure to as many sets as possible.

Now, we are going to show how to proceed on the real line. The same approachwill be used later in the Euclidean space Rn where the proofs will be given.

1.1. Outer Lebesgue Measure. For an arbitrary set A R, dene

A := inf{i=1

(bi ai) :i=1

(ai, bi) A}.

The value A (which can also be +) is called the outer Lebesgue measure of aset A.

1.2. Properties of the Outer Lebesgue Measure. One can see immediatelythat A B if A B and that the measure of a singleton is 0, and withoutmuch eort it becomes clear that I is the length of I in case of I interval of anytype (see Exercise 1.6). Then it is relatively easy to prove that the outer Lebesguemeasure is translation invariant : If A R and x R, then A = (x + A).Another important property is the -subadditivity :

(j=1

Aj) j=1

Aj .

In mathematical terminology, the prex usually relates to countable unions and to countable

intersections.

The question of whether is an additive set function has a negative answer:There are disjoint sets A, B with

(A B) < A+ B(cf. 1.8), and we need to nd a family of sets (as large as possible) on which themeasure is additive. This task will be solved later in Chapter 4 in a much moregeneral case. Now we just briey indicate one of its possible solutions in case ofthe Lebesgue measure.

A. Measures and Measurable Functions 3

1.3. Lebesgue Measurable Sets. Let A be a subset of a bounded interval I.Dening the inner measure A = I (I A), it is natural to investigatethe collection of sets for which A = A (cf. Exercise 1.7). This leads to thefollowing denition.

We say that a set A R is (Lebesgue) measurable if I = (AI)+(I\A)for every bounded interval I R. The collection of all measurable sets onR will be denoted by M. Not every set is measurable as will be seen in 1.8.The set function M M , M M is denoted by and called the Lebesguemeasure. Thus, on measurable sets, the set functions and coincide but fornonmeasurable ones only is dened.Another important property of the measure is contained in the following

theorem which is now presented without proof.

1.4. Theorem. (a) If M1, M2, . . . are elements of M, then also M1 \ M2,Mn and

Mn are elements of M. If, in addition, the sets Mn are pairwise

disjoint, then

(

n

Mn

)=n

Mn.

(b) Intervals of any type are in M.1.5. Remark. The ingenuity of Lebesgues approach to the measure consists in consideringthe countable covers of a set A with intervals. If in the denition of A we consider only nitecovers, we get the notion of so-called Jordan-Peano content. In modern analysis this notion isfar from being as important as the Lebesgue measure.

1.6. Exercise. If I R is an interval (of any type), show that I is its length.Hint . It is sucient to consider the case I = [a, b]. Clearly [a, b] b a (since [a, b] (a , b+ )). Suppose

Si=1(ai, bi) [a, b]. A compactness argument yields the existence of an

index n satisfyingnS

i=1(ai, bi) [a, b]. Using induction (with respect to n) it can be shown that

b a nP

i=1(bi ai).

1.7. Exercise. For every bounded set A R, dene

A := I (I \A)

where I is a bounded interval containing A. Show that:

(a) the value of A does not depend on the choice of I;(b) a bounded set A R is measurable if and only if A = A;(c) a set M R is measurable if and only if its intersection with each bounded interval is

measurable.

In the next part of this chapter we introduce some signicant sets on the realline.

1.8. A Nonmeasurable Set. Now we prove the existence of a nonmeasurable subset of Rand consequently prove that the outer Lebesgue measure cannot be additive.

Set x y if x y is a rational number. It is easy to see that is an equivalence relationon R. Therefore R splits into an uncountable collection of pairwise disjoint classes. A set Vbelongs to this collection if and only if V = x+Q for some x R. By the axiom of choice,

4 1. The Lebesgue Measure

there exists a set E (0, 1) that shares exactly one point with each set V . We show thatE is not in M.Let {qn} be a sequence containing all rational numbers from the interval (1,+1). It is not

very dicult to show that the sets En := qn + E are pairwise disjoint and that

(0, 1) [n

En (1, 2).

Assuming that E M, then also En M and Theorem 1.4 gives SnEn =

Pn

En. Distin-

guishing two cases E = 0 and E > 0 we easily obtain the contradiction.

1.9. Remarks. 1. The proof of the existence of a nonmeasurable set is not a constructiveone (it uses the axiom of choice for an uncountable collection of sets). We return to the topicof nonmeasurable sets in Notes 1.22.2. By a simple argument, an even stronger proposition can be proved: Any measurable setM R of a positive measure contains a nonmeasurable subset. It is sucient to realize thatM =

SqQM (E + q) where E is the nonmeasurable set from 1.8 and that any measurable

subset of E is of zero (Lebesgue) measure.3. Van Vleck [1908] constructed a set E [0, 1] for which E = 1 and E = 0.1.10. Exercise. Show that every countable set S is of measure zero.

Hint . Consider coversSj=1

(rj 2j , rj + 2j) where {rj} is a sequence of all elements of theset S. The assertion also follows from Theorem 1.4 if you realize that singletons have measurezero.

1.11. Examples of Sets of Measure Zero. (a) The set Q of all rational numbers iscountable, thus by Exercise 1.10 it has Lebesgue measure zero.

(b) It can be seen from the hint to the exercise that for every k N there is an open setGk such that Q Gk and Gk 1/k. The set

Tk=1

Gk has also Lebesgue measure zero, it is

dense and uncountable (even residual).

1.12. Cantor Ternary Set. Consider the sequence { n} of nite collections of intervalsdened in the following way: 0 = {[0, 1]}, 1 = {[0, 13 ], [ 23 , 1]}. In each step we construct nfrom n1 as the collection of all closed intervals which are the left or right third of an intervalfrom the collection n1 (the middle thirds are omitted). Then n is a collection of 2n disjointclosed intervals, each of them of length 3n. Let Kn denote the union of the collection n. TheCantor ternary set1 C is dened as

TnKn. It is not dicult to verify that C consists precisely

of points of the formPi=1

ai3i where each ai is 0 or 2. Roughly speaking, in the Cantor set

there are exactly those points of the interval [0, 1] whose ternary expansions do not contain thedigit 1. The Cantor set has the following properties:

(a) C is a compact set without isolated points;

(b) C is a nowhere dense (and totally disconnected) set;

(c) C is an uncountable set;

(d) the Lebesgue measure of C is zero.

1.13 Discontinua of a Positive Measure. If we construct a set D [0, 1] like the Cantorset except that we always omit intervals of length 3n where (0, 1) (note that their centresare not the same as those in the construction of the Cantor set), we get a closed nowhere denseset, for which D = 1 . Sets having this property are called the discontinua of a positivemeasure. Another construction: If G is an open subset of the interval (0, 1), containing allrational points of this interval and G = < 1 then [0, 1]\G is a discontinuum of measure 1.

1sometimes also called the Cantor discontinuum

A. Measures and Measurable Functions 5

1.14. Exercise. Prove that there exists a non-Borel subset of the Cantor set and realize thatthis set is Lebesgue measurable.

Hint . The cardinality argument shows that the set of all Borel subsets of the Cantor set hascardinality of the continuum while the set of all its subsets has greater cardinality.Instead of this, the following idea can be used. Dene

(t) := inf{x [0, 1] : f(x) = t}

where f is the Cantor singular function from 23.1. Show that is increasing on the interval [0, 1],and therefore it is a Borel function. Suppose E is a nonmeasurable subset of [0, 1], B := (E).Then B (as a subset of the Cantor set) is a measurable set. But since 1(B) = E (and is aBorel function), B cannot be a Borel set.

1.15. Lebesgue Measure on Rn. In the same way as for R, we introducethe Lebesgue measure on Rn. Recall that by an interval in Rn we understandan arbitrary Cartesian product of n one-dimensional intervals. If I := (a1, b1) (an, bn) is an open interval, we dene its volume as

vol I = (b1 a1) . . . (bn an).

In the same way we dene vol I for intervals of other types. Given an arbitraryset A Rn, dene the outer Lebesgue measure of A as the quantity

A = inf{

k=1

vol Ik :

k=1

Ik A, Ik is an open interval}.

We say that a set A Rn is measurable if T = (A T ) + (A \ T ) forevery set T Rn. (By analogy with the one-dimensional case we should requirethis equality to hold just for bounded intervals T . We have chosen the presentdenition in order to apply the general approach of Chapter 4. Soon we showthat there is no dierence between these two denitions.) The symbol M againdenotes the collection of all measurable subsets of Rn. For M M we denote byM := M the n-dimensional Lebesgue measure of a set M .

1.16. Theorem. If {Aj} is a sequence of (arbitrary) sets of Rn, then

( j=1

Aj

)

j=1

Aj .

Proof. The assertion follows from Theorem 4.3.

1.17. Theorem. If M1, M2, . . . are elements of M, then also M1 \M2,

Mnand

Mn are elements of M. If, in addition, the sets Mn are pairwise disjoint,

then(

n

Mn

)=n

Mn.

Proof. The assertion follows from general Theorem 4.5.

Compare the following theorem with Exercise 1.6.

6 1. The Lebesgue Measure

1.18. Theorem. If I Rn is a bounded interval, I j

Qj where {Qj} is asequence of open intervals, then

vol I j

volQj .

Thus the n-dimensional Lebesgue measure I is equal to the volume vol I.

Proof. Suppose J is a compact interval contained in I. There exists a p such thatthe intervals {Q1, . . . , Qp} cover J . The interval J can be now divided into anite number of non-overlapping n-dimensional intervals {Ji} (distinct elementsof {Ji} have disjoint interiors) in such a way that the interior of each interval Jiis contained in some of the intervals Qj . Then

volJ =i

volJi p

j=1

volQj j=1

volQj .

Since the dierence vol I volJ can be arbitrarily small, the assertion follows.1.19. Theorem. (a) Any open subset of Rn is measurable.(b) If A = 0, then A is measurable.

Proof. The proof of part (b) is obvious; we will prove (a). First we prove that eachinterval H which is a halfspace (e.g. of the form (, c)Rn1) is measurable.Choose a test set T , T < , and > 0. There exist open intervals {Qj}with

j

Qj T andj

vol Ij < T + .

Now nd open intervals Ij and Jj such that

Ij Jj = Qj , Qj H Ij , Qj \H Jj and Ij + Jj < Qj + 2j .Then

(T I) + (T \ I) j

vol Ij +j

volJj T + .

We proved the measurability of all intervals H of the form of a halfspace. Now,each open set can be expressed as a countable union of intervals and each intervalis a nite intersection of intervals which are halfspaces.

1.20. Theorem. If A Rn, thenA = inf {G : G open, G A}.

Proof. One inequality follows from the monotonicity of . Now if A 0, then there exist open intervals Ij Rn such that

A j

Ij and j

Ij j

vol Ij < A+ .

The reader should compare the following theorem and Exercise 15.19.

A. Measures and Measurable Functions 7

1.21. Theorem. Given a set M Rn, the following are equivalent:(i) M is measurable;(ii) for every bounded interval I, I = (I M) + (I \M);(iii) for every > 0 there exists an open set G M with (G \M) < ;(iv) there exists a G-set D M such that (D \M) = 0;(v) there exist an F-set Bi and a G-set Be such that Bi M Be and

(Be \Bi) = 0.Proof. The implication (i) = (ii) is trivial. Assuming (ii), x > 0 and denoteIk = (k, k)n. By Theorem 1.20 we can nd open sets Gk and Hk such thatIkM Gk, Ik\M Hk, Gk (IkM)+2k and Hk (Ik\M)+2k.We can assume that Gk andHk are subsets of Ik. Then we have Gk\M GkHk.Using (ii) and the measurability of open sets we obtain

Ik+(GkHk) = Gk+Hk (IkM)+(Ik\M)+2k+1 Ik+2k+1.Set G =

k

Gk. Then

(G \M)

k=1

(Gk Hk) 2

so that (iii) holds. That (iii) implies (iv) is evident. It is not very dicult toprove the implication (iv) = (v). If M satises (v), then M = Bi (M \ Bi)where the sets Bi and M \ Bi are measurable by Theorem 1.19 (each one for adierent reason), so that (v) = (i).1.22. Notes. Originally, H. Lebesgue dened the outer measure on the real line usingcountable covers formed by intervals, exactly as explained in the text. He dened measurabilityas in Exercise 1.7.At the end of the last century, various attempts to dene the length or area of geometrical

gures appear; in the works of G. Peano [1887] and C. Jordan [1892] even the measures ofmore complicated sets are considered.The existence of a Lebesgue nonmeasurable set is very closely connected to the axiom of

choice (for uncountable collections of sets) and the assertion that such sets exist was rst provedby G.Vitali [*1905]. Solovays result [1970] says that there exist models of the set theory (ofcourse not satisfying the axiom of choice) in which every subset of real numbers is Lebesgue mea-surable. The existence of a nonmeasurable set can be proved (assuming various set conditions)in other ways as well. Constructions of Bernsteins sets (still assuming the axiom of choice) asexamples of nonmeasurable sets are also interesting. Another construction of a nonmeasurableset (the axiom of choice again) based on results of the graph theory comes from R.Thomas[1985]. Using nonstandard methods, it is possible to prove the existence of a nonmeasurable setassuming the existence of ultralters (a weaker form of the axiom of choice; cf. M.Davis [*1977]).Recently, M.Foreman and F.Wehrung [1991] proved that the existence of a nonmeasurable setfollows from the Hahn-Banach Theorem (which is again a weaker assumption than the axiomof choice).Let us note that the Lebesgue measure can be extended to a translation invariant mea-

sure dened on a wider -algebra than is the collection of all Lebesgue measurable sets. Theconstruction can be found e.g. in S.Kakutani and J.C.Oxtoby [1950]. However, the Lebesguemeasure cannot be extended in a reasonable way to the collection of all subsets of Rn.It is interesting that in R or R2 there exist nitely additive extensions of the Lebesgue

measure to the collection of all subsets which can also be invariant with respect to translations

8 2. Abstract Measures

and rotations. This was rst proved by S.Banach [1923]. However, this result cannot betransferred to spaces of higher dimensions as follows from the famous result of S. Banach andA.Tarski [1924]:If U and V are arbitrary (!) bounded and open sets in the space Rn, n 3, then there exist

sets E1, . . . , Ek and F1, . . . , Fk such that Ei Ej = = Fi Fj for i = j, U =S

Ei, V =S

Fiand Ej are isometric copies of Fj .In this theorem, which is known as the Banach-Tarski paradox, in general all the sets Ei andFi cannot be measurable; realize that U , V can be of dierent measures. More information iscontained in S.Wagon [*1985].

2. Abstract Measures

In this chapter we study an abstract notion of measure which stands as a basisfor modern integration theory. Also in probability theory, the notion of measure(termed a probability there) plays a crucial role. Among many elds of analysiswhich employ measures in an essential way, let us mention e.g. functional analysis,theory of function spaces and theory of distributions, or mathematical modellingof physical quantities.

So far we have the only nontrivial example of the Lebesgue measure. Furtherimportant examples of measures will be introduced later.

Remember that the Lebesgue measure is not dened on the collection of allsubsets of R but on its subcollection which is closed under countable operations.We start with the following denition.

2.1. -algebras. A collection S of subsets of a given set X is called a -algebraif

(a) X S ;(b) if A S , then X \A S ;(c) if An S , then

n=1

An S .The pair (X,S ) is called a measurable space.

Clearly every -algebra is also closed under countable intersections, under dif-ferences and contains the empty set.

Not every collection of sets is a -algebra. However, if T is an arbitrary familyof subsets of X, then there exists the smallest -algebra (T ) which contains T .Such a -algebra is simply the intersection of all -algebras (in X) which containT . It surely exists, since there is at least one such a -algebra (the -algebraP(X) of all subsets of X) and the intersection of any collection of -algebras isagain a -algebra. The collection (T ) is called the -algebra generated by T .2.2. Examples. One of the most important examples is the -algebra M of all Lebesguemeasurable sets on the real line. Another important class of examples yields Borel -algebrasfrom 2.3. For illustration, we add a few simple examples. Suppose X is an arbitrary set. Then

(a) {, X} is a -algebra;(b) the collection (X) of all subsets of X is a -algebra;

(c) {A X : A is countable or X \A is countable} is a -algebra.2.3. Borel Sets. Let P be a topological space. The -algebra B(P ) generatedby the family of all open subsets of P is called the Borel -algebra of P ; its

A. Measures and Measurable Functions 9

elements are called Borel sets. The Borel -algebra B(P ) contains all closedsets, all countable intersections of open sets (these sets are called G sets), allcountable unions of closed sets (F sets), all countable unions of G sets (Gsets), all countable intersections of F sets (F) and so on. Let us note that forthe complete description of all possible types we would have to use (in nontrivialcases) all countable ordinal numbers.

2.4. Measures. Let S be a collection of subsets of a set X. A nonnegative setfunction : S [0,] is called a measure if

(a) S is a -algebra;(b) = 0;(c) for each sequence {An} of pairwise disjoint sets from S , (

An) =

An.

The triplet (X,S , ) is termed a measure space.

From (b) and (c) it immediately follows that each measure is monotone (ifA,B S , A B, then A B); the property (c) is also called the -additivityof a measure.

We say that a measure is nite if X < +, -nite if there exist setsMn Ssuch that Mn < + and X =

n=1

Mn. If X = 1, then we say that is a

probability measure. A measure is said to be complete if whenewer B S is anull set and A B, then also A S (and A = 0).If is a measure on X and E S , we dene E A = (A E) for A S .

A related notion is the restriction |A of a measure to the set A S : Ifwe denote by SA the -algebra {M S : M A} of subsets of A, we dene|A(M) = (M) for M SA. Finally, if T S is a -algebra of subsets of X,then the symbol |T denotes the measure E E, E T .2.5. Examples. (a) The Lebesgue measure on the collection of all Lebesgue measurable setsin Rn. This measure is complete, -nite, but not nite.

(b) Counting measure. Let X be an arbitrary set, = (X) the collection of all subsetsof X and

A =

(the number of elements of A if A is nite,

+ if A is innite.The counting measure is complete; it is -nite if and only if X is countable, and nite justwhen X is nite.

(c) The Dirac measure. Again, let X be an arbitrary set, x X, = (X). We dene

A =

(1, if x A,0, if x X \A.

The measure is called the Dirac measure at x and it is denoted by x. The Dirac measure isa complete probability measure.

(d) Trivial measures. If is an arbitrary -algebra of subsets of X and A = 0 for allA , then is an example of a nite measure which is not complete provided = (X).

2.6. Properties of Measures. Let (X,S , ) be a measure space. Then thethe following propositions hold:

(a) if A1, A2, S , A1 A2 . . . , then (

An) = limAn;

10 2. Abstract Measures

(b) if A1, A2, S , A1 A2 . . . , and A1 < , then (

An) =limAn;

(c) if A1, A2, S , then (

An)

An.

Proof. (a) Since A1, A2 \A1, A3 \A2, . . . are pairwise disjoint, we get

( n=1

An

)=

(A1

n=1

(An+1 \An))= A1 +

n=1

(An+1 \An)

= limk

(A1 +

k1n=1

(An+1 \An))= lim

kAk.

(b) By (a), we obtain

( n=1

An

)= A1

(A1 \

n=1

An

)= A1

( n=1

A1 \An)

= A1 limn(A1 \An) = limnAn.

(c) It is sucient to consider the sequence

B1 = A1, B2 = A2 \A1, B3 = A3 \ (A1 A2), . . . ,

notice that

Bn =

An and that the sequence {Bn} is pairwise disjoint.2.7. Completion of Measures. Now we return to the notion of completenessof a measure. We show that every measure space can be extended to a completemeasure space.

Let (X,S , ) be a measure space and let N denote the collection of all setsA X for which there is a B S such that B = 0 and A B. Further, let Sdenote the collection of all sets of the form M N where M S and N N .We dene a set function on S by

(M N) = M, M S , N N .

Clearly the value of E does not depend on the choice of M and N . The setfunction on S is called the completion of .

2.8. Theorem. S is a -algebra containing S and is a complete measureon S which coincides with on S .

Proof. Obviously, S S . Since both S and N are closed under countableunions, the same is true for S . If E S , there are M S , N N and B Sso that N B, B = 0 and E =MN . Then X\E = (X \ (M B))(B\N) S . It is easy to verify that is a complete measure on S and = on S .

2.9. Remark. Note that there is a variety of extensions of a measure to a complete measure.The completion described above is uniquely determined and in some sense minimal (cf. Exercise2.13).

A. Measures and Measurable Functions 11

2.10. Exercise. Let f be a nonnegative function on a set X and a -algebra on X. ForB we dene

B :=XxB

f(x).

(Recall that by denition

XxB

f(x) = supK

8 0 and foreach set B , B A, either B = 0 or (A \B) = 0.)Hint . Fix 0 < < X and set = {A : A }. There exists a set C such thatA = C for all A , A C. In a similar way nd D , D C, D with thefollowing property: B = D for each set B for which D B C and B . BecauseC \D cannot be an atom, it follows C = D = .

2.16. Notes. In [1895] and [*1898], E. Borel extended the length of intervals to a set function(measure) dened on the collection of all Borel sets. However, H. Lebesgue was the rst whocreated the theory of the integral with the help of this measure. J. Radon in [1913] dened ageneral notion of (Borel) measures in Euclidean spaces. A.N.Kolmogorov introduced in [*1933]the axiomatic theory of probability measures.The Darboux property of real-valued measures is a special version of the general Lyapunov

theorem (A. Lyapunov [1940]) which says that the range of a nite-dimensional nonatomic vectormeasure is a convex compact set.

12 3. Measurable Functions

3. Measurable Functions

As mentioned in the rst chapter, there are serious reasons why the Lebesguemeasure is not dened on the collection all subsets of Rn. Likewise, one cannotexpect a reasonable theory of integration on the class of all functions; it will benecessary to conne to some reasonable family of functions. This class of courseshould contain the indicator functions of all measurable sets and should be closedunder all common (algebraic as well as limit) operations. As we will see, theserequirements are satised by the following natural denition.

In what follows, we assume that S is a -algebra of subsets of a given set X.

3.1. Measurable Functions. Suppose D S . A function f : D R is saidto be S -measurable on D if {x D : f(x) > } S for each R.A complex function on D is S -measurable if its real and imaginary parts are

S -measurable.3.2. Examples. (a) If is a -algebra of all subsets of X, then every function on X is-measurable.

(b) Only the constant functions are -measurable if = {, X}.(c) If is the -algebra of all Borel sets of a topological space X, then -measurable

functions are called shortly Borel functions on X.

3.3. Remark. Since any -algebra is closed for complementation, a function f is -measurable if and only if {f } for each R. Because

{f } =\

n=1

jf > 1

n

,

it is possible to replace the condition {f > } by {f } ) in the denition of-measurability. Other equivalent conditions are stated in Exercise 3.6.

3.4. Theorem. Let f , g be S -measurable functions on X. Then

(a) {x X : f(x) < g(x)} S ;(b) f1(+), f1() S .

Proof. Since

{f < g} =rQ

({f < r} {g > r}),

{x X : f(x) = +} =

n=1

{f > n},

the assertion easily follows.

3.5. Properties of S -measurable Functions. Let f , g, fn be S -measurablefunctions (with possibly dierent domains in S ), R and let be a continuousfunction on an open set G R. Then the following functions are S -measurablewhere dened (and their denition domains are in S ):

(a) f , f + g, max(f, g), min(f, g), |f |, fg, f/g;(b) sup fn, inf fn, lim sup fn, lim inf fn and lim fn;(c) f .

A. Measures and Measurable Functions 13

Proof. We will just indicate the ideas of some of the proofs leaving the others asan exercise for the reader. The assertion (c) is obvious.

(a) Suppose R. Then {f + g > } = {f > g} (and the function gis S -measurable). The functions |f |, f2 and 1/g are S -measurable by (c). Thenwe can use the formulae

max(f, g) =12(f + g + |f g|),

fg =12((f + g)2 f2 g2).

(b) As a hint let us just note that

{sup fn > } =n

{fn > }.

3.6. Exercise. (a) Show that a real-valued function f is -measurable if and only iff1(G) for each open set G R or, if and only if f1(B) for each Borel set B R.Hint . Since each open set G R can be expressed in the form G = S(an, bn) (where (a, b) =(, b) (a,+)), we can see that a function f is -measurable if and only if f1(G) for each open set G R (note that {f > } = f1((,+))!). Then consider the collection

{B R : B Borel, f1(B) },and show that it is a -algebra containing all open sets.

(b) The characterization given in (a) cannot be used for functions having innite valuesunless we introduce the notions of open or Borel subsets of R.

3.7. Exercise. Let {fn} be a sequence of -measurable functions. Show that the sets{x X : lim fn(x) exists } and {x X : lim fn(x) exists and is nite }

are in .

3.8. Simple Functions. By a simple function on X we understand a (nite)linear combination of indicator functions of sets from S . In other words, a real-(or complex-)valued function f is simple if f is S -measurable and f(X) is a nite

set. Thus every simple function is of the formn

i=1icAi where i are numbers and

Ai S . Note that this expression is not uniquely determined!3.9. Theorem. Let f 0 be an S -measurable function on X. Then thereexists a sequence {fn} of nonnegative simple functions on X such that fn f .Proof. For n N and k = 1, 2, . . . , n2n set

Fn,k =

{x X : k 1

2n f(x) < k

2n

}and dene

fn(x) =

{ k12n , if x Fn,k,n if x X \

k

Fn,k.

It is easily seen that fn are simple and fn f .3.10. Exercise. Suppose f , fn have the same meaning as in the previous theorem. Showthat fn f on every set on which f is bounded.

14 3. Measurable Functions

3.11. Exercise. If f is an -measurable function, then there exists a sequence of simplefunctions {fn} such that |fn| |f | and fn f on X.Most frequently we meet the concept of measurability when additionally a

measure is in consideration on a given -algebra. So suppose in what followsthat (X,S , ) is a measure space. The following notion is of great importance inLebesgues integration theory, since usually the null sets are negligible.

3.12. Almost Everywhere. We say that a function h is dened -almosteverywhere (briey almost everywhere) on X if its domain D S satises (X \D) = 0. Suppose f and g are functions dened almost everywhere on X. We saythat f(x) g(x) for -almost all x X, or that f g -almost everywhere ifthere exists a set N such that N = 0 and for all x X \N we have f(x) g(x).Similarly we understand the expressions almost everywhere and almost all inother contexts, e.g. when speaking about the equality of functions or about theconvergence almost everywhere.

3.13. -measurable functions. We say that a function f dened on D Sis -measurable on X if (X \D) = 0 and f is SD-measurable on D.Let us emphasize that we distinguish strictly between -measurable functions

(dened in general only almost everywhere) and S -measurable functions on X(dened everywhere on X).

3.14. Equality almost everywhere. The relation f = g almost everywhereis clearly an equivalence relation on the set of all functions (or all -measurablefunctions) on X. The following observations are quite useful.

(a) Let be a measure on (X,S ) and f be a -measurable function denedon D S . Then there exists an S -measurable function g on X such that f = gon D; in particular, we can take

g =

{f on D,

0 on X \D.

(b) If is a complete measure on (X,S ) and f is a -measurable function,then every function g which is equal to f almost everywhere is -measurable.3.15. Exercise. Find an example of a measure space on which there exists an -measurablefunction f and an -nonmeasurable function g such that f = g almost everywhere.

3.16. Exercises. Suppose (X, , ) is the completion of a measure space (X, , ).

(a) Show that an everywhere dened function f is -measurable if and only if there exist-measurable functions g, h such that g f h on X and g = h -almost everywhere in X.

Hint . The proof of one implication is easy. Let f be -measurable. By Exercise 3.11 nda sequence of simple ( -measurable) functions {fn} such that fn f on X. Then nd -measurable functions gn, hn such that gn fn hn and gn = hn -almost everywhere andset

g := lim sup gn, h := lim inf hn.

(b) If f is an -measurable function on X, then there exists an -measurable function gsuch that f = g -almost everywhere.

Hint . The assertion is obvious for indicator functions of sets from , and therefore also for-measurable simple functions. Then use Exercise 3.11.

A. Measures and Measurable Functions 15

3.17. Exercise. Let {fn} be a sequence of -measurable functions on X which converges toa function f -almost everywhere. Prove that

(a) if the measure is complete, then also f is -measurable;

(b) if f is dened on X and is -measurable, then there exists a sequence {fn} of -measurable functions such that fn = fn -almost everywhere and fn f everywhere on X;(c) if the measure is complete, then a function f is -measurable if and only if there exists

a sequence of simple functions which converges to f -almost everywhere.

3.18. Images and Preimages of Measurable Sets. We know that real-valued -measurable functions are exactly those for which the preimages of Borel sets are in . Toillustrate the situation, let us present the following example for the Lebesgue measure on R.

(a) Let g(x) = 12 (x + f(x)) where f is the Cantor singular function from 23.1. Then g isa continuous and increasing function mapping the interval [0, 1] on [0, 1]. If C is the Cantorset and is the inverse function to g on [0, 1], then 1(C) is a (Lebesgue) measurable set ofpositive measure. As remarked in 1.9.2 there exists a nonmeasurable set E 1(C). Finally,for M := (E) we have M C, so that M is measurable while 1(M) = g(M) = E is anonmeasurable set. Let us note that this M is not even a Borel set (compare also with Exercise1.14.a).

(b) Without presenting a proof, we mention that the continuous image of a Borel set on Ris always measurable but not necessarilly Borel.

4. Construction of Measures from Outer Measures

In this chapter, we will construct measures from the so-called outer measures.This is a very useful method used already before when we introduced the Lebesguemeasure in Rn. We also full our promise to prove propositions from the rstchapter.

4.1. Outer Measure. By an outer measure on a set X we understand a setfunction which assigns to every set A X a nonnegative number A (real or+) such that the following conditions are satised:

(a) = 0;(b) if A B, then A B;(c) (

An)

An.

The property (c) is called the -subadditivity of an outer measure.4.2. Examples of Outer Measures. (a) The outer Lebesgue measure in Rn.

(b) The Hausdor outer measure from Chapter 36.

(c) The counting measure.

(d) If (X, , ) is a measure space, then the set function : A inf{M : M : M , M A} is an outer measure. See also Exercise 4.8.An important example of creating an outer measure is contained in the follow-

ing theorem.

4.3. Theorem. Let G be a collection of subsets of a set X containing , and : G [0,+] be a set function on G with = 0. For A X set

A = inf{

n=1

Gn : Gn G ,

n=1

Gn A}

(note that inf = +). Then is an outer measure.

16 4. Construction of Measures from Outer Measures

Proof. We only have to verify that is -subadditive. So suppose A nAn,

An < +. Fix > 0 and nd Gjn G so thatj

Gjn An andj

Gjn < An +

2n.

Then n,j

Gjn A andn

An A .

4.4. -measurable Sets. A set M X is said to be -measurable (in thesense of Caratheodory) if

T = (T M) + (T \M)

for each test set T X (in other words, if M splits additively each set in X).The collection of all -measurable sets will be denoted by M(). To prove that aset M is -measurable, it is sucient to verify the inequality

T (T M) + (T \M)

for any set T with T < +.4.5. Theorem. M() is a -algebra and is a complete measure on M().

Proof will be divided into a few steps.

(a) It is straightforward to check that X M(), that X \M M() providedM M(), and that A M() whenever A = 0. Suppose M, N M(). Wewould like to show that also M N M(). So choose a test set T X. Then

T = (T M) + (T \M)

and(T M) = (T M N) + ((T M) \N).

Now we use the test set T \ (M N) and -measurability of M to get

(T \ (M N)) = ((T M) \N) + (T \M).

Thanks to last three equalities, it follows that

T = (T (M N)) + (T \ (M N)).

Since M() is closed under complements and nite intersections, it is closed alsounder nite unions.

A. Measures and Measurable Functions 17

(b) In order to show that is -additive onM(), chooseMn M() pairwisedisjoint. Setting T =M1 M2 and using -measurability of M1 we obtain

(M1 M2) = M1 + M2.Thus is nitely additive. Further

n=1

Mn = limk

kn=1

Mn = limk

(k

n=1

Mn

)

( n=1

Mn

),

and since the reverse inequality always holds we reach the conclusion.

(c) Let nowMn M() be pairwise disjoint. Our aim is to show that

n=1Mn

M(). Choosing a test set T X, we have

T =

(T \

kn=1

Mn

)+

(T

kn=1

Mn

)

(T \

n=1

Mn

)+

kn=1

(T Mn)

for each k N. Since is -subadditive, it readily follows that

T (T \

n=1

Mn

)+

(T

n=1

Mn

),

which is what we wanted.

4.6. Exercise. Let be a nonnegative function on (X), = 0. Show that the collectionof all -measurable sets forms a -algebra and that is additive on it.

Hint . It only needs to examine where monotonicity and -subadditivity of the outer measurein the proof of Theorem 4.5 is used.

4.7. Exercise. We say that an outer measure is regular if for each set A X there existsM M() such that A M and A = M .(a) Let be a regular outer measure and A1 A2 A2 . . . an increasing sequence of

sets. Show that (S

An) = lim An.

(b) Show that the outer Lebesgue measure is regular.

(c) There exists a decreasing sequence {Mn} of subsets of [0, 1] such that Mn = 1 andTnMn = (compare with (a)).

4.8. Exercise. Let (X, , ) be a measure space and A X. SetA := inf{M : M , A M}, A := sup{M : M , M A}.

(a) Suppose A

18 5. Classes of Sets and Set Functions

5.1. Systems of sets. A family A of subsets of a set X containing is called:(a) a semiring if

(a1) A B A for each A, B A ,(a2) forA,B A , A B, there exist pairwise disjoint sets C1, . . . , Cn A

such that B \A =n

j=1Cj ;

(b) a ring if given A,B A , then AB, A\B A (so that also AB A );(c) an algebra if it is a ring and X A ;(d) a Dynkin class if

(d1) X A ,(d2) A \B A for each A,B A , B A,(d3) if An A are pairwise disjoint, then

n=1

An A ;

(e) a -system if A B A for each A,B A ;(f) a -ring if A is a ring closed under countable intersections;

(g) a -ring if A is a ring closed under countable unions.

5.2. Premeasure. A nonnegative set function is called a premeasure on X if is dened on a ring A of subsets of X and satises the following conditions:

(a) = 0,(b) if {Aj} A is a sequence of pairwise disjoint sets and

Aj A , then

(

Aj) =

Aj .

A premeasure is in fact a -additive set function dened on a ring of sets. Wesay that a premeasure on X is -nite if there exists a sequence Xj of sets fromA such that Xj

A. Measures and Measurable Functions 19

(f) The collection of all Lebesgue measurable sets of nite measure forms a -ring.

(g) Suppose that is the collection of all unions of a nite number of intervals (includingthe degenerated ones) and = {A Q : A }. Then is an algebra of subsets of Q and : A Q A is a nitely additive set function which is not a premeasure.(h) Let be the algebra of all nite unions of intervals on (0, 1). Dene a set function onby the formula

(A) =

(1 if A contains some interval of the form (0, ), > 0,

0 in other cases.

Then is a nitely additive set function on which is not a premeasure.

5.4. Theorem. Let G be a ring of subsets of a set X and a premeasure onG . If is the outer measure constructed from G and as in Theorem 4.3, then

(a) = on G ;

(b) G M().Proof. Obviously . Suppose G G , G < + and {Gn} G ,

Gn G.

Then

G =

(n

(Gn G))n

(Gn G) n

Gn,

so that G G.Now suppose G G . Choose a test set T X, T < + and > 0. There existGn G such that

T n

Gn andn

Gn T + .

Then

T + n

Gn =n

((Gn G) + (Gn \G)) (T G) + (T \G),

thus G M().5.5. Hopfs Extension Theorem. Let be a premeasure on a ring A ofsubsets of a set X. Then there exists a measure on (A ) which equals on A .This extension of is unique provided is -nite.

Proof. The existence of a measure is an immediate consequence of Theorems 4.3and 5.4 (notice that (A ) M()). To prove uniqueness, under the -nitenessassumptions, let be another measure on (A ), = on A . One can easily ndout (from the construction of the outer measure ) that on (A ). Thenfor Aj A , A =

j

Aj we have

A = limn

nj=1

Aj

= limn

nj=1

Aj

= A.

20 5. Classes of Sets and Set Functions

So if E (A ), E < , then for a given > 0 there exist sets Aj A ,A =

j

Aj such that E A and A < E + . Hence

E A = A = E + (A \ E) E + (A \ E) E + ,

thus E = E. Finally suppose X =j

Xj , Xj < +; one can assume that Xjare pairwise disjoint. If E (A ), then

E =j

(E Xj) =j

(E Xj) = E.

5.6. Exercise. Let be a family of subsets of a set X. Show that there exists the smallestDynkin class ( ) containing .

5.7. Exercise. Show that every -algebra is a Dynkin class. A Dynkin class is a -algebra ifand only if it is a -system.

5.8. Exercise. Prove that if is a -system, then ( ) = ( ).

5.9. Exercise. The assumptions of Hopfs extension theorem can be weakened. Indeed, showthat the assertion of Theorem 5.4 is still true if we only assume that is nitely additive and-subadditive on the semiring and that = 0.Hint . First note that is monotone. Then proceed as in Theorem 5.4 and show that ( ) (). In the essential step use the existence of sets Cjn with the property Gn \ (GGn) =S

jCjn.

5.10. Exercise. Let be a -system on X. Show that if 1, 2 are probability measures on( ) which agree on , then 1 = 2 on ( ).

Hint . Let := {M ( ) : 1(M) = 2(M)}. Show that is a Dynkin class and use Exercise5.7.

5.11. Exercise. Consider

:= {[a, b) \ C : a, b [0, 1]}, ([a, b) \ C) := f(b) f(a),

where C is the Cantor set and f the Cantor function from 23.1.Show that is a semiring and that is nitely additive but not -additive on .

5.12. Exercise. (a) Let X be an arbitrary set and n N. Consider the set function whichis restriction of the counting measure to the collection

n := {A X : A has exactly n elements }

and construct the outer measure as in Theorem 4.3. Investigate the relationship between thecollections n and M() and compare with Theorem 5.4.

(b) Investigate the same problem in case of X = (0, 1), = (X) and

A := inf{kX

i=1

(bi ai) :k[

i=1

(ai, bi) A}

(A is the Jordan-Peano content of a set A).

A. Measures and Measurable Functions 21

5.13. Exercise. Let be an outer measure on X constructed from the premeasure on analgebra by Theorem 4.3 such that X < +. For A X set

A := X (X \A).

Show that A M() if and only if A = A (compare also with Theorem 5.4).5.14. Capacity on Compact Sets. Let X be a locally compact topological space. Areal-valued nonnegative function dened on the collection (X) of all compact subsets of Xis called a Choquet capacity if it satises the following conditions:

(a) if K1 K2, then (K1) (K2);(b) if K1 K2 K3 . . . , then (

TKn) = lim (Kn);

(c) (K1 K2) + (K1 K2) (K1) + (K2) whenever Kn (X) (so-called strongsubadditivity).

An important example is the Newtonian capacity in Rn, n 3. If we dene the Newtonianpotential of a Radon measure on Rn by

(x) :=ZRn

d(t)

|x t|n2 ,

we can introduce the Newtonian capacity capK of a compact set K Rn as

capK := sup{K : supt K, 1 on Rn}.

The proof that this capacity satises the conditions (a), (b), (c) can be found for instance in[KNV].

5.15. Outer Capacity. Let X be a topological space. A mapping c : (X) [0,+] iscalled an outer capacity provided it satises:

(a) if A B, then cA cB;(b) if A1 A2 A3 . . . , then c (

SAn) = sup cAn;

(c) if K1 K2 K3 . . . and Kn are compact, then c(T

Kn) = inf cKn.

A set A X is said to be c-capacitable if

cA = sup{cK : K A, K compact}.

1. If is a Choquet capacity on a locally compact space (Exercise 5.14) and

cA := inf{sup{ (K) : K G, K compact} : G A, G open},

prove that c is an outer capacity.

2. Suppose c is simultaneously an outer capacity and an outer measure. Investigate the relation-ship between the notions of c-capacitability and c-measurability in the sense of Caratheodory.Consider the cases:

(a) X is a two-points set equipped with the discrete topology and cA = 1 if A = , c = 0;(b) X is R with the Euclidean topology and c is the outer Lebesgue measure;

(c) X is R with the discrete topology and c is again the outer Lebesgue measure;

(d) c is the outer capacity derived from the Newtonian capacity. In this case a set A Rnis c-measurable in the sense of Caratheodory if and only if cA = 0 or if c(Rn \ A) = 0 (this isnot quite easy, see M.M.Rao [*1987]). On the other hand, Choquets deep result says that eachBorel set in Rn is c-capacitable.

5.16. Notes. Hopfs extension theorem is usually attributed to H.Hahn or C.Caratheodorybut it is probably originated by M.Frechet [1915]. The proof using Caratheodorys theorem wasgiven independently by H.Hahn [*1924] and A.N.Kolmogorov [*1933].The notions of a -system or of a Dynkin class were investigated by E.B.Dynkin in 1959 as

tools for the probability theory. However, families of similar properties were studied already byW. Sierpinski [1928].

22 6. Signed and Complex Measures

An investigation of the theory of capacities in the classical potential theory during the period19201950 is connected with the names of Ch. de la Vallee Poussin, N.Wiener, O. Frostman orM.Brelot among others. These authors studied mainly the Newtonian capacity and examinedthe role of sets of a small capacity. One could say that the capacity theory is a youngersister of Lebesgue measure theory. In the 50s the general capacity theory was developed andG.Choquet proved his famous capacitability theorem. An interested reader is referred to a nicearticle written by Choquet himself [1986] or [1989].

6. Signed and Complex Measures

In this chapter, we will investigate measures assuming also negative (or evencomplex) values.

6.1. Signed Measures. Let S be a -algebra of subsets of X. A set function : S R is said to be a signed measure on S if

(a) = 0;(b)

( n=1

An

)=

n=1

An whenever An S are pairwise disjoint.6.2. Remarks. 1. A signed measure can assume at most one of the values + and. Indeed, if E = + and F = for E,F , then (E F ) is nite. Therefore(E \ F ) = +, (F \ E) = and the equality (b) does not hold for (disjoint) sets E \ Fand F \ E.2. If (

SAn) is nite, then the series

PAn in the condition (b) converges absolutely (indeed,

this series converges to the same value when its terms are rearranged arbitrarily, which isequivalent to the absolute convergence).

3. Some of basic properties of positive measures hold even for signed measures. Show that thefollowing assertions are true:

(a) if An , An A , then An A,(b) if An , An A, |A1| < + , then An A.

4. A signed measure is not necessarily monotone: If A B, we can no longer assert thatA B.6.3. Hahn Decomposition. We say that a set P S is positive for a signedmeasure if E 0 for each set E S , E P . Analogously we dene negativesets for .

An ordered pair of sets (P,N) is called a Hahn decomposition of X for if

(a) P N = X, P N = ;(b) P is positive and N is negative (for ).

6.4. Examples. (a) The empty set is both positive and negative for any signed measure. Thepair (X, ) is a Hahn decomposition for if and only if is positive.(b) The pairs (R, ) , (R \ Q,Q) , (R \ {5}, {5}) are Hahn decompositions of R for the

Lebesgue measure.

(c) If a measure f has a density f with respect to (i.e. if fA =RA f d where f ()

cf. 8.19), then the set P := {f 0} is positive for f and ({f 0}, {f < 0}) is a Hahndecomposition for f .

6.5. Hahn Decomposition Theorem. For every signed measure thereexists a Hahn decomposition. This decomposition is unique in the following sense:If (P1, N1) and (P2, N2) are two such decompositions, then

(P1 E) = (P2 E), (N1 E) = (N2 E)

A. Measures and Measurable Functions 23

for each set E S .Proof. We start with proof of the uniqueness assertion. Suppose that (P1, N1)and (P2, N2) are Hahn decompositions of X for . Then evidently

(P1 E) = (E (P1 P2)) = (E P2)

and, in the same way, (N1 E) = (N2 E).To prove the existence of the decomposition, assume that S < + for all

S S . We proceed via the following steps.Step 1: If A S , A > and > 0, then there exists a set B A such thatB A and B is positive. The set B will be constructed as the intersection ofa nite or innite nested sequence {An} of subsets of A. We start with A1 = A.In the n-th step, if An is not positive, we can set

kn = min{k N : there exists E An with (E) 1

k

}and nd An+1 An such that (An \ An+1) 1kn . Then we set B =

n=1

An.

The sets An \An+1 are pairwise disjoint and their union is A \B. Thus

+ > (B) = (A)

n=1

(An \An+1) (A) +

n=1

1kn

.

It follows that (B) (A) and kn . If E B is a measurable subset, thenfor each n N we have E An, and thus (E) > 1kn . Hence B is positive.Step 2: To complete the proof, set s = sup{A : A S }. Since S , we haves 0. There exists a sequence {Pn} S such that Pn s. In light of the rststep we can assume P1 P2 . . . (the union of two positive sets is a positiveset!). If P :=

Pn, then P A and P = limPn = s < +. Moreover, P is

positive. Indeed, if E P , then E Pn E and (E Pn) 0. Lastly, to showthat the set N := X \ P is negative we notice that (E P ) = E + P > s forany set E N with E > 0.6.6. Variation of a Measure. Let (P,N) be a Hahn decomposition of X fora signed measure . Dene

+E = (E P ), E = (E N), || (E) = +E + E

for every set E S . By the previous theorem, +E and E do not dependon the choice of the Hahn decomposition (which is not unique!). It is simplychecked that the set functions +, and || are positive measures on S . Theyare called the positive, negative and total variations of a signed measure . Letus summarize our results in the following theorem.

24 6. Signed and Complex Measures

6.7. Theorem (The Jordan decomposition of a signed measure). Let be asigned measure on (X,S ). The functions +, and || are positive measureson S and = + . If also = 1 2 where 1, 2 are positive measures,then 1 +, 2 .Proof. It is enough to prove the last part of the theorem. But for each E S ,one has

+E = (E P ) = 1(E P ) 2(E P ) 1(E P ) 1E.

An important characterization of the total variation of a signed measure, often used as itsdenition, is contained in the following theorem.

6.9. Theorem. Let be a signed measure on (X,S ) and E S . Then

|| (E) = sup{

nk=1

|Ak| : Ak S pairwise disjoint,n

k=1

Ak = E

}.

Proof. It is straightforward to check thatk

|Ak| =k

+Ak Ak k

(+Ak + Ak) =

k

|| (Ak) = || (E).

On the other hand, if (P,N) is a Hahn decomposition of X for , set A1 = EP ,A2 = E N and nd out that the supremum is even attained.6.10. Complex Measures. A complex measure on (X,S ) is a -additiveset function : S C for which = 0. By -additivity we understand that

( k=1

Ak

)=

k=1

Ak whenever {Ak} is a sequence of pairwise disjoint sets fromS . Let us note that, as in Remark 6.2.2, the appearing series must convergeabsolutely or denitely diverge.

Each complex measure can be expressed uniquely in the form = r + iiwhere r and i are (nite!) signed measures on (X,S ). In particular, each nitesigned measure can be understood as a complex measure.

If is a complex measure, we dene its total variation || by

|| (E) = sup{

nk=1

|Ak| : Ak S pairwise disjoint,n

k=1

Ak = E

}.

An appeal to Theorem 6.9 reveals that this denition agrees with the previousone for signed measures.

6.11. Theorem. Let be a complex measure on (X,S ). Then the totalvariation || is a nite positive measure on (X,S ).

A. Measures and Measurable Functions 25

Proof. Apparently, || () = 0 and it is simply to verify that || is nitely additive.If = 1 2 + i(3 4) where 1 2 is the Jordan decomposition of r and3 4 is the Jordan decomposition of i (i.e. j are positive nite measures),then

|| (A) 1A+ 2A+ 3A+ 4A,so that || (A) < +. If An S , An , then limj An = 0 for eachj = 1, 2, 3, 4, and therefore || (An) 0. A routine argument now shows that is -additive.

6.12. Remark. Note that the range of any complex measure is always a bounded subset ofthe complex plane C.

6.13. Exercise. Let be a signed measure on (X, ) and E . Show that +E =sup{B : B , B E}.6.14. Exercise. Let be a complex measure on (X, ) and E . Prove that || (E) =sup

j Pk=1

|Ak| : Ak pairwise disjoint,S

k=1Ak = E

.

6.15. Exercise. Let be a complex measure on (X, ). Prove that || is the smallestpositive measure satisfying A ||A for all A .6.16 Exercise. Investigate whether || = |r|+ |i|, or || =

q|r|2 + |i|2.

6.17. Exercise. We denote by M( ) the set of all nite signed or complex measures on(X, ). If M( ), let = || (X). Show that (M( ), ) is a Banach space (i.e.M( )is a linear space, becomes a norm and M( ) is complete with respect to it).Hint . Only the completeness requires a proof. But if n is a Cauchy sequence inM( ), thenthere exists limnA for each A . Dening A = limnA, it is not hard to show that is-additive and n 0.6.18. Exercise. Suppose that , n M( ), n 0. Prove that nA A forevery set A .6.19. Exercise. Show that the set M( ) of all signed measures on (X, ) equipped withthe order

if A A for each A is a lattice (i.e. for any , M( ) there exist sup(, ) and inf(, ) in M( )). Notice thatsup(, ) does not necessarily mean the set function A sup((A), (A)) such function, ingeneral, is not a measure!.

Hint . Show that 12 (+ + | |) is sup(, ) and 12 (+ | |) is inf(, ).6.20. Finitely Additive Measures Charges. A real-valued set function on an algebraof sets is called a charge if

(a) = 0;(b) (A B) = A+ B whenever A,B , A B = .

If is a charge, dene +A = sup{F : F , F A}, A = inf{F : F , f A},|| (A) = +A+ A.6.21. Exercise. Show that the set functions +, and || are positive nitely additivemeasures on .

6.22. Exercise. Suppose X = [0, 1). Let be the collection of all nite unions of intervals[a, b) [0, 1). Verify that is an algebra. Set

f(x) =

(1x

for x (0, 1),0 for x = 0.

26 6. Signed and Complex Measures

If A =nS

i=1[ai, bi), dene A =

P(f(bi) f(ai)). Show that neither the Jordan decomposition

theorem, nor the Hahn decomposition theorem do hold for . Proceed via the followingsteps:

(a) the denition of A does not depend on the particular representation of A asS[ai, bi);

(b) is a charge on ;

(c) A 0 for A (0, 1);(d) ([0, 1)) = 1;

(e) +([0, 1)) = ([0, 1)) = +.6.23. Exercise. Let be a charge on . Show that = + if and only if is bounded(i.e. if sup{|A| : A } < +).6.24. Notes. Signed measures were investigated by H. Lebesgue already in [1910]; he wasconcerned mostly with measures given by a density (see 8.19).The existence of a Hahn decomposition of the space and the Jordan decomposition of signed

measures were rst established in a full generality by Hahn [*1921].The monograph by K.P.S. Bhaskar Rao and M.Bhaskar Rao [*1983] is devoted to the theory

of charges.

B. The Abstract Lebesgue Integral 27

B. The Abstract Lebesgue Integral

7. Integration on R

7.1. Newton Integral. Let f be a real-valued function on an interval (a, b).A function F on (a, b) is called an antiderivative to f if F (x) = f(x) for allx (a, b). A real number A is called the Newton integral of f over (a, b) if thereis an antiderivative F of f on (a, b) such that A = lim

xbF (x) lim

xa+F (x). The

value of the Newton integral of f on (a, b) does not depend on the choice of Fbecause the dierence between any two antiderivatives is a constant. The Newtonintegral is denoted by N

baf .

7.2. Riemann Integral. Let f be a bounded function on a bounded interval(a, b). If a = 0 < 1 < < m = b is a (nite) sequence of points in [a, b] (socalled partition of the interval [a, b]), and 1, . . . , m a sequence of real numbers,

then the valuem

j=1j(j j1) is called an upper Riemann sum of f if j f

on every interval (j1, j), and a lower Riemann sum of f if j f on every(j1, j). Set

Rf = inf{A : A is an upper Riemann sum of f},Rf = sup{A : A is a lower Riemann sum of f}.

A function f is Riemann integrable on (a, b) provided Rf = Rf ; the commonvalue is then called the Riemann integral of f over (a, b) and it is denoted by

R baf .

7.3. Theorem. Let f be a continuous function on an interval [a, b]. Then boththe Newton integral and the Riemann integral of f on (a, b) exist and are equal.

Proof. Since f is uniformly continuous, we easily obtain that R xaf exists for all

x (a, b). A short reection shows that x R xaf is an antiderivative of f and

this implies the existence of the Newton integral and the required equality.

7.4. Remark. Each Riemann integrable function f is absolutely Riemann integrable, i.e. |f |is Riemann integrable as well. This assertion is no longer true for the Newton integral (considerthe integration of the function x sinx/x over (1,+); the change of variables t = 1/x yieldsan example on a bounded interval).

7.5. Dirichlet Function. The Dirichlet function D is dened as the indicator function ofthe set of all rational numbers. Observe that D = 0 almost everywhere. The Dirichlet functionis nowhere continuous, it has neither an antiderivative (it does not have the Darboux property)nor the Riemann integral over (0, 1) ( D = 1, D = 0).7.6. Riemann function. The Riemann function R is zero on the set of all irrational numbers.If r = p/q is a rational number, p and q are relatively prime and q > 0, then R(r) is dened as1/q (zero is supposed to be 0/1, so R(0) = 1). The Riemann function is continuous at x if andonly if x is irrational. The Riemann function serves as an example of a pathological functionwhich is Riemann but not Newton integrable (it has no antiderivative) on bounded intervals.

7.7. Various Examples. The indicator function of the Cantor ternary set is Riemannintegrable. The indicator function of any discontinuum of a positive measure fails to be Riemannintegrable. An unbounded Newton integrable function cannot be Riemann integrable. Thereare examples of bounded Newton integrable functions which are not Riemann integrable (cf.7.9.f).

28 8. The Abstract Lebesgue Integral

7.8. Chebyshevs Inequality for the Riemann Integral. If f is a nonnegative boundedfunction on (a, b), > f and > 0, then the set {f } can be covered by a union of anite number of intervals, the sum of whose lengths is less than / (so {f } < /).Hint . Find a partition a = 0 < 1 < < m = b and numbers cj such that cj f on(j1, j) and

mPj=1

cj(j j1) < . Then the sum of lengths of those intervals (j1, j) forwhich cj is less than / .7.9. Riemann Integrable Functions. Lebesgue measure theory allows a deeper study ofthe class of Riemann integrable functions.

(a) For any bounded function f on an interval (a, b) we have

f = inf{Z ba

g : g piecewise constant, g f},

f = sup{Z ba

g : g piecewise constant, g f}.

(A function f is said to be piecewise constant on [a, b] if there exist j such that a = 0 < 1 < < m = b and f is constant on each interval (j1, j).)(b) If f is a bounded function on an interval (a, b), then the function

f := inf{g : g f, g continuous}

is called the upper Baire function of f . The denition of the lower Baire function f of f isanalogous.

Show that the function f is upper semi-continuous and that f is continuous at a point x ifand only if f(x) = f(x).(c) Show that

f = inf{RZ ba

g : g f, g continuous }.

(d) Let f be a bounded function on an interval (a, b). Show that the following conditions areequivalent:

(i) f is Riemann integrable;

(ii) for each > 0 there exist continuous functions t, s such that t f s and R R ba (st) 0.(e) Show that any Riemann integrable function is measurable.

Hint . Any semi-continuous function is measurable. Now use (d) and Theorem 3.14.b.

(f) The above condition (v) permits to construct bounded Newton integrable functions whichare not Riemann integrable. An example of Volterra type functions constructed with the aid ofclosed nowhere dense sets of positive measure (Example 1.13) can be found in A.M. Bruckner[*1978].

7.10. Notes. The origins of the integral calculus are connected with the names of I. Newtonand G.W.Leibniz. The modern integration theory has been developed from the beginning ofthe 19th century by A.L.Cauchy, L.Dirichlet, B.Riemann, C. Jordan, E.Borel, H. Lebesgue,G.Vitali and others. Some historical treatments on integration are mentioned in 8.25.

B. The Abstract Lebesgue Integral 29

8. The Abstract Lebesgue Integral

Elementary expositions of the integration theory in usual textbooks of calculusemploy Riemanns or Newtons constructions. However, the examples given inChapter 7 show that these integrals provide rather small classes of integrablefunctions. In deeper applications, we need completeness of normed linear spacesof integrable functions, which is achieved with the aid of Lebesgues integration.A further anvantage of Lebesgues approach consists in the possibility to integrateover more general domains than intervals.

In the sequel, (X,S , ) will be a measure space.The Riemann integral of the indicator function of any interval is its length.

Accordingly, developing the notion of an integral on X, the requirement thatX

cA d = A for any A S is quite natural. Further, it is reasonable toimpose conditions on additivity and monotonicity of the integral and to ask thefamily of all integrable functions to be as large as possible.We introduce the concept of the abstract Lebesgue integral in several steps.

First we deneX

f d in a natural way for nonnegative simple functions and thenwe extend it to nonnegative -measurable functions using their approximation bysimple functions. The general case will be completed using the decomposition ofa function into its positive and negative parts. Building up this theory, we mustbe careful in order to legitimate these steps. For instance, we have to show thatthe denition of

X

sd in the case of simple functions does not depend on theirrepresentation which is not unique. Likewise, we have to deal with problems wheninnite values or indenite expressions appear.

The reader may nd it instructive to remember the most important exampleof the n-dimensional Lebesgue measure in Rn. For integration with respect tothe Lebesgue measure we use a traditional notation

E

f dx :=E

f d , ba

f dx :=(a,b)

f d.

8.1. Simple Functions. Recall that a simple function is a real S -measurablefunction s on X having a nite range. Any simple function s can be expressed as

s =n

j=1

jcBj ,

where B1, . . . , Bn S are pairwise disjoint sets and 1, . . . , n R. Of course,this representation of s is not unique.

Remember again that as a matter of convenience we set 0 = 0 = 0.8.2. Lemma. Let A1, . . . , Am S and B1, . . . , Bn S be pairwise disjointsets and let i and j be nonnegative real numbers. If

mi=1

icAi n

j=1jcBj ,

thenmi=1

iAi n

j=1

jBj .

30 8. The Abstract Lebesgue Integral

Proof. To simplify the proof dene 0 = 0 = 0, A0 = X \mi=1

Ai and B0 =

X \n

j=1Bj (then the collections {Ai}, {Bj} form partitions of X). For i

{0, . . . ,m} and j {0, . . . , n} either Ai Bj = or i j . Thusmi=0

iAi mi=0

nj=0

i(Ai Bj) mi=0

nj=0

j(Ai Bj) n

j=0

jBj .

8.3. Abstract Lebesgue Integral. If D S and s is a nonnegative simplefunction expressed as s =

nj=1

jcBj , where Bj are pairwise disjoint sets and j are

nonnegative coecients, deneD

sd :=n

j=1j(D Bj). The previous lemma

shows that this value does not depend on the particular representation of s. Next,dene

D

f d := sup

{D

sd : 0 s f on D, s simple}

if f 0 is a -measurable function on D S . Lemma 8.2 again ensures that thenew denition and the old one agree in case when f is a simple function.

For a SD-measurable function f on D we deneD

f d :=D

f+ d D

f d provided at least one of these integrals is nite. Remember that f+ :=max(f, 0), f := max(f, 0).If f is a function on X and M S , then apparently

M

f d =X

f cM d

andM

f d =M

f d where = |M is the restriction of to the -algebraSM := {A S : A M} of subsets of M .Therefore, it is no loss of generality to restrict our attention to the integration

over the whole space X.

It is useful to dene the abstract Lebesgue integral even for functions denedonly -almost everywhere. In this case, if f is dened onD S and (X\D) = 0,set

X

f d :=D

f d

if the integral on the right side is dened. It is immediate that this value doesnot depend on the choice of D.The symbol L or L () denotes the family of all -measurable functions

dened -almost everywhere on X for which the abstract Lebesgue integral isdened.

B. The Abstract Lebesgue Integral 31

Further denote

L 1 = L 1() =

{f L () :

X

f d R}

.

If f L 1(), we say that f is -integrable.8.4. Lebesgue Integrable Functions on R. (a) Every bounded -measurable function ona bounded interval (and so every Riemann integrable function) is Lebesgue integrable. Thus,the functions from Examples 7.5 7.6 are Lebesgue integrable as well.

(b) If f is a Riemann integrable function on [a, b], thenR ba f d f = f

R ba f d, so

that the Lebesgue and the Riemann integrals of f coincide.

(c) If a function has both the Newton and the Lebesgue integrals, they are equal. Proof of thisis not easy. We will prove it using the Henstock-Kurzweil integral in Chapter 25.

(d) If f is Newton integrable, then f is -measurable since it can be expressed as the limit of asequence {fn} of continuous functions, in essence, fn(x) = n(F (x+ 1n ) F (x)), where F is anantiderivative to f . It can happen that f is not Lebesgue integrable, but this is only the case iff is not absolutely Newton integrable.

(e) Since nonmeasurable functions are rather rare, the question whether or not f is Lebesgue

integrable can be reduced to the question whether the integralsR ba f

+ dx andR ba f

dx are niteor innite. For instance, the function f : x xp is integrable on (0, 1) if and only if p > 1.The key result is the following monotone convergence theorem (sometimes

called Levis theorem or the Lebesgue monotone convergence theorem) for non-negative functions.

8.5. Theorem. If fn 0 are -measurable functions on X, fn f , thenX

fn dX

f d.

Proof. It is clear that the sequence{

Xfn d

}is nondecreasing, and therefore

there exists := limX

fn d. Since fn f , we have X

f d and theassertion is obvious for = +. Assume < +, x a simple function s,0 s f , and prove that

Xsd . The proof will be given in several steps:

(a) Let (0, 1). Dene En = {x X : fn(x) s(x)}. Then En S , En En+1 and

n=1

En = X (if f(x) = 0, then x E1; if f(x) > 0, then s(x) . ThenX

f d = limX

fn d.

Proof. We have already proved the theorem when fn are nonnegative and fn feverywhere. The general case can easily be reduced to Theorem 8.5. First weredene f and fn on sets of measure zero in such a way that fn f everywhere.It would be clearly sucient to assume that fn L 1 for all n. If gn = fn + f1(notice that f1 L 1), then gn are nonnegative -measurable functions andgn f + f1 . Now, Theorem 8.5 ensures that

X

gn dX(f + f1 ) d. Since

Xfn d =

X

gn dX

f1 d, the assertion easily follows.

8.12. Levis theorem for series. If fn are nonnegative -measurable func-tions on X, then

X

fn d =

X

fn d.

Proof. Use Theorems 8.5 and 8.8.

8.13. Lebesgue dominated convergence theorem. Let {fn} be a sequenceof -measurable functions, fn f almost everywhere. If there exists a functionh L 1 such that |fn| h almost everywhere for all n, then f L 1 andX

f d = limX

fn d.

Proof. The proof may be easily reduced to Levis theorem by considering f =lim sup fn = lim inf fn. Set sn = sup{fn, fn+1, . . . }, tn = inf{fn, fn+1, . . . }.Then h tn fn sn h, sn f , tn f almost everywhere and

0}

and the inequality 0 cAn nf yields (An) = 0.8.17. Theorem. Let f L 1. If

Ef d = 0 for any measurable set E, then

f = 0 almost everywhere.

Proof. If E := {f 0}, then Ef d =

Ef+ d = 0 and using the previous the-

orem we get f+ = 0 almost everywhere. Analogously f = 0 almost everywhere.

8.18. Corollary. Let f, g L 1. IfE

f d E

g d

for every measurable set E, then f g almost everywhere.Proof. Set h = (f g)+. Then h 0 and 0

Ehd =

E{h>0}(f g) d 0.

Thanks to Theorem 8.17, (f g)+ = 0 almost everywhere.8.19. Indenite Lebesgue integral. Let f L . For E S set

f (E) :=E

f d .

The set function f is called the indenite Lebesgue integral of f .

8.20. Theorem. Let f L . Then f is a signed measure on X. Moreover,f (E) = 0 for every set E S with E = 0.Proof. It is enough to prove that f is a measure provided f is nonnegative. Inthis case, notice that fcA =

n

fcAn if A =nAn with Ai Aj = for i = j, and

cite Theorem 8.12.

8.21. Remark. A question arises whether or not for any pair of measures on , say and, there is always a nonnegative -measurable function f on X such that = f , i.e.

E =ZE

f d

for all E . Of course, the answer is negative; if such a function exists, then necessarilyE = 0 whenever E = 0. But if and obey this condition (we say that is absolutelycontinuous with respect to ), then there is a function f with the ascribed property (at leastfor -nite measures). This will be the subject of the chapter concerning the Radon-Nikodymtheorem.

8.22. Exercises. Let f 1. Prove that(a) f is a nite signed measure on ;

(b) for every > 0 there is a > 0 such thatf (E)

< whenever E < .

Hint . (a) To verify the equality f (A) =P

f (An) (An pairwise disjoint) use Theorem8.14 for hn = fcAn .

(b) Suppose there exists an > 0 and a sequence {En} such that En < 2n andREn

|f | .Set E =

Tn=1

Sk=n

Ek. Then E = 0 andRE f d , which is a contradiction.

36 9. Integrals Depending on a Parameter

8.23. Image of a Measure. Let (X, , ) be a measure space, (Y, ) be a measurable spaceand f : X Y a measurable mapping (i.e. f1(E) whenever E ). The set function

E (f1(E)), E ,

is called the image of the measure under the mapping f and it is denoted by f().

(a) Show that f() is a measure on (Y, ).

(b) Let g : Y R be a -measurable function. Show that g is f()-integrable if and onlyif g f 1. In this case Z

Yg df() =

ZX

g f d.

Hint . First consider simple functions and then pass to limits.

8.24. Exercise. Let f be an increasing dierentiable function on R, limx f(x) = . Let

be a measure on (R, (R)) dened by

E =ZE

f dx.

Show that f() = .

8.25. Notes. The modern integration theory starts with Lebesgues doctoral thesis [1902]following a short paper [1901]. His denition uses constructed Lebesgue measure on R. Theidea of using simple functions (not equal to those introduced here) when dening the integral(still on the real line) comes from F.Riesz ([1912], [1920]).Further details can be found in historical notes by T.Hawkins [*1970] or by F.A.Medvedev

[1975], R.Henstock [1988], T.H.Hildebrandt [1953]. It is clear that H. Lebesgue followed investi-gations of his predecessors (B.Riemann, C. Jordan, G.Peano, E.Borel and others). Less knownis the fact that (roughly) at the same time G.Vitali and W.H.Young published similar resultsindependently.Theorem 8.11 (concerning the monotone convergence) was proved by Beppo Levi [1906],

Lemma 8.15 by P.J.L. Fatou [1906] and Theorem 8.13. by H. Lebesgue [1910].One more remark: H. Lebesgue developed his theory of integration mainly for the case of

Lebesgue measures on Rn. Later J. Radon [1913] considered more general measures in Rn

(nowadays called the Radon measures). A general theory of measures on arbitrary -algebras wasgiven by M.Frechet [1915]. Since that time, many results on this subject have been published;the monograph [*1950] by P.R.Halmos is one of the most quoted.

9. Integrals Depending on a Parameter

The Lebesgue dominated convergence theorem has simple but very importantconsequences on continuity and dierentiation of integrals depending on a param-eter.

9.1. Theorem. Let (X,S , ) be a measure space, P a metric space and U aneighbourhood of a point a P . Suppose that a function F : U X R has thefollowing properties:

(a) there exists a set N X of measure zero such that for each x X \ Nthe function F (, x) is continuous at a;

(b) for each t U , the function F (t, ) is -measurable;(c) there exists a function g L 1(X) such that |F (t, )| g almost every-

where for all t U .

B. The Abstract Lebesgue Integral 37

Then for each t U , F (t, ) L 1(X) and the function

f : t X

F (t, ) d

is continuous at a.

Proof. The proof that

limta

X

F (t, ) d =X

F (a, ) d

is achieved by showing that

limj

X

F (tj , ) d =X

F (a, ) d

for each sequence tj a, tj U . To prove the last assertion it suces to use theLebesgue dominated convergence theorem.

9.2. Theorem. Let (X,S , ) be a measure space, N X a set of measurezero and I R an open interval. Suppose that a function F : IX R has thefollowing properties:

(a) For each x X \N , F (, x) is dierentiable on I;(b) for each t I, F (t, ) is -measurable;(c) there exists a function g L 1(X) such that

ddtF (t, x) g(x) for each

x X \N and t I;(d) there is a t0 I such that F (t0, ) L 1(X).

Then F (t, ) L 1(X) for all t I, the function

f : t X

F (t, ) d

is dierentiable on I and

f (t) =X

ddt

F (t, ) d.

Proof. Suppose a, b I, b = a, x X \ N . By the mean value theorem thereexists a between a and b such thatF (b, x) F (a, x)b a

= ddtF (, x) g(x).

For a = t0 it follows that the function

x F (b, x) F (a, x)b a

38 10. The Lp Spaces

is integrable, thus F (b, ) is integrable. Choose a I again. By the Lebesguedominated convergence theorem,

limj

X

F (tj , ) F (a, )tj a d =

X

ddt

F (t, ) d

for each sequence tj a of points of I \ {a}, which yields

ddt

X

F (a, ) d = limta

X

F (t, ) F (a, )t a d =

X

ddt

F (a, ) d.

9.3 Remarks. 1. Notice that in the proofs of the last theorems we made heavy use of theLebesgue dominated convergence theorem. Of course we could state analogous results based onLevis theorem.

2. Since the notions of continuity and derivative are local, it suces to verify the assumption (c)only locally.

3. A number of exercises and clarifying examples can be found in [L-Pr].

10. The Lp Spaces

The Lebesgue spaces Lp are important means linking measure theory and func-tional analysis, and they are essential tools in dierential equations theory, prob-ability theory and other branches of modern analysis.In this chapter, (X,S , ) denotes a xed measure space and p a number from

the interval [1,].10.1. The Set L p. (a) Suppose p

B. The Abstract Lebesgue Integral 39

10.2. Youngs inequality. Let p, q (1,), 1p+ 1q = 1. If a, b are nonnegativenumbers, then

ab ap

p+

bq

q.

Proof. We can assume ab > 0. Making use of concavity of the function ln,

ln

(ap

p+

bq

q

) 1

pln(ap) +

1qln(bq) = ln a+ ln b = ln(ab)

holds, and we easily establish the required inequality.

10.3. Holders inequality. Suppose that f L p and g L q, where p, q (1,), 1p + 1q = 1. Then fg L 1 and

X

fg d

(X

|f |p d)1/p(

X

|g|q d)1/q

.

Proof. Denote

s =

(X

|f |p d)1/p

, t =

(X

|g|q d)1/q

.

We may assume that st > 0. Thanks to Youngs inequality (a = f(x)/s, b =g(x)/t) we have

f(x)g(x)st

|f(x)| |g(x)|st

|f(x)|p

psp+|g(x)|qqtq

for each x X. Thus

1st

X

fg d X|f |p dpsp

+

X|g|q dqtq

1p+1q= 1

which is what we wanted to prove.

10.4. Minkowskis inequality. Let p [1,] and f, g L p. Then f + g L p and

f + gp fp + gp .

Proof. It is not hard to verify that f + g1 f1 + g1.If p = , then |f | s -almost everywhere and |g| t -almost everywhere,which implies |f + g| s+ t -almost everywhere. Therefore f + g s+ t,and consequently

f + g f + g .

40 10. The Lp Spaces

With these trivial cases out of the way, there remains the case 1 < p < .Holders inequality yields

X

|f | |f + g|p1 d

(

X

|f |p d)1/p(

X

|f + g|(p1)q d)1/q

=

(X

|f |p d)1/p(

X

|f + g|p d)11/p

.

AnalogouslyX

|g| |f + g|p1 d (

X

|g|p d)1/p(

X

|f + g|p d)11/p

.

This entailsX

|f + g|p d X

|f | |f + g|p1 d+X

|g| |f + g|p1 d

((

X

|f |p)1/p

+

(X

|g|p)1/p) (

X

|f + g|p d)11/p

.

10.5. The Lp Spaces. The behavior of the function f fp on L p, wherep [1,], resembles axioms of a norm. However, in general, L p is not a linearspace and a nonzero function may have zero norm. To apply the theory of normedlinear spaces, we identify functions which are equal almost everywhere. Formally,we assign to every function f L p the class of functions

[f ] = {g L p : g = f -almost everywhere on X}

and deneLp = Lp(X,S , ) = {[f ] : f L p}.

Then Lp is a linear space equipped with operations

[f ] + [g] := [f + g], [f ] := [f ] ( R),

and with the (true) norm [f ] p := fp .

It can easily be seen that these denitions do not depend on the choice of repre-sentatives.

It is customary not to distinguish between functions and classes, often evenbetween the spaces L p and Lp. For instance, we say that {fj} is a Cauchysequence in Lp while the meaning is that fj are functions and {[fj ]} is a Cauchysequence in Lp.

B. The Abstract Lebesgue Integral 41

10.6. Completeness of Lp. Let {fj} be a Cauchy sequence in Lp. Then {fj}is convergent in Lp, i.e. there exists an f L p such that

X

|f fj |p d 0.

Proof. The easy case p = is left to the reader. Suppose p

42 11. Product Measures and the Fubini Theorem

2. For a counting measure on a set X we write lp(X) := Lp(X, (X), ). In particular, forX = N we get well known spaces of sequences:

the space p, 1 p

B. The Abstract Lebesgue Integral 43

11.1. Product -algebra. First we introduce the notion of the product -algebra. If S and T are some -algebras, the product -algebra S T is the-algebra generated by the collection of all measurable rectangles. Thus, S Tis the smallest -algebra which contains all sets of the form S T where S Sand T T .For M X Y , x X, let

Mx := {y Y : [x, y] M} .The set Mx is called the x-section of M . Analogously, dene the y-section

My = {x X : [x, y] M}for y Y .11.2. Lemma. If M S T and x X, then Mx T (and, of course,similarly My S for y Y ).Proof. Denote A := {E S T : Ex T }. A moments reection showsthat A contains all measurable rectangles. A routine argument yields that(X Y \ E)x = Y \ Ex and (

E)x =

Ex. We see that A is a -algebra,

and therefore A = S T .11.3 Monotone Classes. A familyM of subsets of a set Z is a monotone classif it obeys the following conditions:

(a) if M1 M2 M3 . . . , Mn M , then

Mn M ;(b) if M1 M2 M3 . . . , Mn M , then

Mn M .

Every monotone class which is simultaneously an algebra forms a -algebra.

The following idea, used already in the proof of the previous lemma, will repeatin the proofs of subsequent theorems:If A is the family of sets from S T for which a certain proposition holds

and if A contains all measurable rectangles, then A = S T provided A is a-algebra. It is not always so easy to verify that A a -algebra. Often it is easierto prove that A is only a monotone class and an algebra (mostly using Levistheorem). But even in this case A = S T as follows from the next theorem.11.4. Theorem. If R is an algebra of sets, then the smallest monotone classcontaining R is (R).

Proof. L