Measure Theory Notes (2010)

MEASURE THEORY NOTES (2010)

Mr. Andrew L Pinchuck

Department of Mathematics (Pure & Applied)

Rhodes University Contents

Introduction 1

1 Riemann integration 2 1.1 Introduction...... 2 1.2 Basicdeﬁnitionsandtheorems ...... 2 1.3 Fundamentaltheoremofcalculus...... 11

2 Measure spaces 14 2.1 Introduction...... 14 2.2 Notationandpreliminaries ...... 15 2.3 Measureonaset ...... 17 2.4 Outermeasure...... 20 2.5 Lebesgue measure on R ...... 25

3 The Lebesgue integral 31 3.1 Introduction...... 31 3.2 Measurablefunctions ...... 31 3.2.1 Basicnotions...... 31 3.2.2 Convergence of sequences of measurable functions ...... 36 3.3 Deﬁnitionoftheintegral ...... 41 3.3.1 Integralofanonnegativesimplefunction ...... 41 3.3.2 Integralofa nonnegativemeasurable function ...... 45 3.3.3 Integrablefunctions ...... 54 3.4 LebesgueandRiemannintegrals ...... 59 3.5 Lp spaces...... 60 3.5.1 H¨olderandMinkowskiinequalities ...... 61

4 Decomposition of measures 66 4.1 Signedmeasures ...... 66

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Introduction

In mathematics, more speciﬁcally in measure theory, a measure on a set is a systematic way to assign to each suitable subset a number, intuitively interpreted as the size of the subset. In this sense, a measure is a generalization of the concepts of length, area, volume, etc. A particularly important example is the Lebesgue measure on a Euclidean space, which assigns the conventional length, area and volume of Eu- clidean geometry to suitable subsets of Rn. Most measures met in practice in analysis (and in many cases also in probability theory) are Radon measures. Radon measures have an alternative deﬁnition in terms of linear functionals on the locally convex space of continuous functions with compact support. In this course, however, we will mainly concentrate on the Lebesgue measure.

Measure theory was developed in successive stages during the late 19th and early 20th century by Emile Borel, Henri Lebesgue, Johann Radon and Maurice Frchet, among others. The main applications of measures are in the foundations of the Lebesgue integral, in Andrey Kolmogorov’s axiomatisation of probability theory and in ergodic theory. In integration theory, specifying a measure allows one to define integrals on spaces more general than subsets of Euclidean space; moreover, the integral with respect to the Lebesgue measure on Euclidean spaces is more general and has a richer theory than its predecessor, the Riemann integral. Probability theory considers measures that assign to the whole set, the size 1, and considers measurable subsets to be events whose probability is given by the measure. Ergodic theory considers measures that are invariant under, or arise naturally from, a dynamical system. The importance of integration and how measure theory puts integration and probability theory on an axiomatic foundation is a principle motivation for the development of this theory. Measure theory is useful in functional analysis in defining Lp function spaces, which cannot otherwise be defined properly.

These notes are intended to be an introduction to measure theory and integration. I make no claims of originality with regards to this material, and I have used a number of different sources as references in the compilation of these notes. Chapter 1 deals with the theory of Riemann integration and highlights some of its shortcomings. In Chapter 2, we define the fundamental concepts of measure theory and present some well-known,significant results. In Chapter 3, we define the Lebesgue integral and consider some results that follow from this and lastly, in Chapter 4, we provide a brief introduction to the study of signed measures.

Throughout the course notes you will ﬁnd exercises that need to be attempted for complete under- standing of the text. Once again, I should mention that I have found wikipedia to be a useful resource and recommend its use to students. All the course information and material can be found on the Rhodes mathematics web site: www.ru.ac.za/mathematics by following the appropriate links.

1 Chapter 1

Riemann integration

1.1 Introduction

In the branch of mathematics known as real analysis, the Riemann integral, created by Bernhard Riemann (1826-1866), was the first rigorous definition of the integral of a function on an interval. The Riemann integral was clearly an improvement upon the way in which integration was initially conceived by Leibnitz and Newton, i.e., simply as the anti-derivative. While the Riemann integral is one of the easiest integrals to define, and has obvious uses in numerical analysis, it is unsuitable for many theoretical purposes. Some of these technical deficiencies can be remedied by the RiemannStieltjes integral, and most of them disappear in the Lebesgue integral. The idea behind the Riemann integral is to use very simple approximationsfor the area of S. By taking better and better approximations, we can say that ‘in the limit’, we get exactly the area of S under the curve.

1.2 Basic deﬁnitions and theorems

In this section we discuss the construction of the Riemann integral. This discussion reveals some of the shortcomings of the Riemann integral and we present these as a motivation for the construction of the Lebesgue integral. 1.2.1 Deﬁnition Let Œa; b be a closed interval in R. A partition of Œa; b is a set P x0; x1; x2;:::; xn of points in R such that D f g a x0 < x1 < x2 < < xn b: D D Let f be a real-valued function which is bounded on Œa; b and let P x0; x1; x2;:::; xn be a partition of Œa; b. Denote by D f g M sup f .x/ and m inf f .x/: D aÄxÄb D aÄxÄb

Since f is bounded on Œa; b, it is bounded on each subinterval Œxi1; xi, for each i 1; 2;:::; n. Let D Mi sup f .x/ xi1 x xi and mi inf f .x/ xi1 x xi ; D f W Ä Ä g D f W Ä Ä g for each i 1; 2;:::; n. Clearly, D m mi Mi M; for each i 1; 2;:::; n: Ä Ä Ä D We can now form the sums n n U.f; P/ Mi .xi xi1/ and L.f; P/ mi .xi xi1/: D D XiD1 XiD1

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1.2.2 Deﬁnition The sums U.f; P/ and L.f; P/ are called, respectively, the upper and the lower sum of f relative to the partition P. 1.2.3 Remark [1] It is important to note that U.f; P/ and L.f; P/ depend on the partition P. If f is nonnegative on Œa; b, then the upper sum U.f; P/ is the sum of the areas of the rectangles whose heights are Mi and whose bases are Œxi1; xi. Similarly, L.f; P/ is the sum of the areas of the rectangles whose heights are mi , and whose bases are Œxi1; xi.

[2] It also follows that U.f; P/ L.f; P/. 1.2.4 Theorem Let f be a real-valued function which is bounded on Œa; b and let P x0; x1; x2;:::; xn be a partition of Œa; b. Then D f g m.b a/ L.f; P/ U.f; P/ M.b a/: Ä Ä Ä PROOF.

Since Mi M , for each i 1; 2;:::; n, it follows that Ä D n n n U.f; P/ Mi .xi xi1/ M.xi xi1/ M .xi xi1/ M.b a/: D Ä D D XiD1 XiD1 XiD1

Similarly, since m mi , for each i 1; 2;:::; n, it follows that Ä D n n n L.f; P/ mi .xi xi1/ m.xi xi1/ m .xi xi1/ m.b a/: D D D XiD1 XiD1 XiD1 2 This theorem says that the set A U.f; P/ P is a partition of Œa; b is bounded below by m.b a/. Hence, A has an inﬁmum, †.f /, say.Df That is, W g

†.f / inf U.f; P/; D P where the inﬁmum is taken over all possible partitions P of Œa; b. This theorem also shows that the set B L.f; P/ P is a partition of Œa; b is bounded above by M.b a/ and hence B has a supremum, .fD /, f say. ThatW is, g .f / sup L.f; P/; D P where the supremum is taken over all possible partitions P of Œa; b. It follows that

m.b a/ †.f / M.b a/; and Ä Ä m.b a/ .f / M.b a/: Ä Ä 1.2.5 Deﬁnition Let f be a real-valued function which is bounded on Œa; b. The upper integral of f on Œa; b is deﬁned by

b f .x/dx inf U.f; P/; D P Za

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and the lower integral of f on Œa; b is deﬁned by

b f .x/dx sup L.f; P/; D P Za

where, once again, the inﬁmum and supremum are taken over all possible partitions P of Œa; b.

It is intuitively clear that bf .x/dx bf .x/dx. Of course, this is not enough and we will therefore a Ä a prove this fact shortly. R R 1.2.6 Definition Let P x0; x1; x2;:::; xn be a partition of Œa; b. A partition P of Œa; b is called a refinement of D f g P, denoted by P P , if xi P , for each i 0; 1; 2;:::; n. A partition P is called a common Â 2 D refinement of the partitions P1 and P2 of Œa; b if P is a refinement of both P1 and P2.

The next theorem states that refining a partition decreases the upper sum and increases the lower sum. 1.2.7 Lemma (Refinement Lemma) Let f be a real-valued function which is bounded on Œa; b. If P is a refinement of a partition P D x0; x1; x2;:::; xn of Œa; b, then f g L.f; P/ L.f; P / U.f; P / U.f; P/: Ä Ä Ä PROOF. Suppose that P has one more point that P, say a point x which lies in the subinterval Œxr1; xr . Let

L1 sup f .x/ xr1 x x ; L2 sup f .x/ x x xr and D f W Ä Ä g D f W Ä Ä g l1 inf f .x/ xr1 x x ; l2 inf f .x/ x x xr : D f W Ä Ä g D f W Ä Ä g Recalling that

Mr sup f .x/ xr1 x xr ; and mr inf f .x/ xr1 x xr ; D f W Ä Ä g D f W Ä Ä g we observe that mr l1; mr l2; L1 Mr ; and L2 Mr : Ä Ä Ä Ä It follows that

mr .xr xr1/ mr .xr x / mr .x xr1/ l2.xr x / l1.x xr1/: D C Ä C Hence,

r1 n L.f; P / mj .xj xj1/ l1.x xr1/ l2.xr x / mj .xj xj1/ D C C C jXD1 jDXrC1 r1 n

mj .xj xj1/ mr .xr xr1/ mj .xj xj1/ C C jXD1 jDXrC1 n mj .xj xj1/ L.f; P/: D D jXD1 Similarly,

Mr .xr xr1/ Mr .xr x / Mr .x xr1/ L2.xr x / L1.x xr1/; D C C

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and so

r1 n U.f; P/ Mj .xj xj1/ L1.x xr1/ L2.xr x / Mj .xj xj1/ D C C C jXD1 jDXrC1 r1 n Mj .xj xj1/ Mr .xr xr1/ Mj .xj xj1/ Ä C C jXD1 jDXrC1 n

Mj .xj xj1/ U.f; P/: D D jXD1 the case where P contains k 2 more points than P can be proved by simply repeating the above argument k times. 2 1.2.8 Theorem Let f be a real-valued function which is bounded on Œa; b. Then,

b b f .x/dx f .x/dx: Ä Za Za

PROOF. Let P1 and P2 be any two partitions of Œa; b and let P be their common reﬁnement. Then by Lemma 1.2.7, L.f; P1/ L.f; P / U.f; P / U.f; P2/: Ä Ä Ä Since P1 is any partition of Œa; b, it follows that

sup L.f; P/ U.f; P2/; P Ä

and since P2 is any partition of Œa; b, we have that sup L.f; P/ inf U.f; P/; P Ä P where the inﬁmum and the supremum are taken over all possible partitions P of Œa; b. Thus,

b b f .x/dx f .x/dx: Ä Za Za

2 1.2.9 Remark Implicit in the proof of Theorem 1.2.8, is the fact that no lower sum can exceed an upper sum. That is, every lower sum is less than or equal to every upper sum. 1.2.10 Deﬁnition Let f be a real-valued function on Œa; b. We say that f is Riemann-integrable on Œa; b if f is bounded on Œa; b and b b f .x/dx f .x/dx: D Za Za

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If f is Riemann-integrable on Œa; b, we deﬁne the integral of f on Œa; b to be the common value of the upper and lower integrals, i.e,

b b b f .x/dx f .x/dx f .x/dx: D D Za Za Za

We shall denote by RŒa; b the collection of all functions that are Riemann-integrableon Œa; b. 1.2.11 Remark In the definition of the integral of f on Œa; b, we have tacitly assumed that a < b. If a b, then D b f .x/dx a f .x/dx 0. Also if b < a, then define a D a D R R a b f .x/dx f .x/dx: D Zb Za 1.2.12 Examples [1] Show that if f is a constant function on Œa; b, then f RŒa; b and find it’s integral. 2

Solution: Let f .x/ k, for all x Œa; b and let P x0; x1; x2;:::; xn by any partition of Œa; b. Then D 2 Df g

Mi sup f .x/ xi1 x xi k and mi inf f .x/ xi1 x xi k; D f W Ä Ä gD D f W Ä Ä gD for each i 1; 2;:::; n. Therefore, D n n U.f; P/ Mi .xi xi1/ k .xi xi1/ k.b a/ and D D D iD1 iD1 Xn Xn L.f; P/ mi .xi xi1/ k .xi xi1/ k.b a/ D D D XiD1 XiD1 Since P is any partition of Œa; b, it follws that U.f; P/ L.f; P/ k.b a/, for all partitions P of Œa; b. Therefore, D D b b f .x/dx k.b a/ f .x/dx: D D Za Za That is, f is integrable on Œa; b and

b f .x/dx k.b a/: D Za

[2] Let f be a function deﬁned by [3]

1 if x Q Œ0; 1 f .x/ 2 \ D 0 if x Q Œ0; 1: 62 \ Show that f is not Riemann-integrable on Œ0; 1.

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Solution: Firstly we observe that f is bounded on Œ0; 1. Let P x0; x1; x2;:::; xn be any D f g partition of Œ0; 1. Since for each i 1; 2;:::; n, the subinterval Œxi1; xi contains both rational and irrational numbers, we have that D

Mi sup f .x/ xi1 x xi sup 0; 1 1; and D f W Ä Ä gD f gD mi inf f .x/ xi1 x xi inf 0; 1 0; D f W Ä Ä gD f gD for each i 1; 2;:::; n. Thus, D n n

U.f; P/ Mi .xi xi1/ .1/ .xi xi1/ 1 0 1, and D D D D XiD1 XiD1 n n L.f; P/ mi .xi xi1/ .0/ .xi xi1/ 0: D D D XiD1 XiD1 Since P is any partition of Œ0; 1, it follows that U.f; P/ 1 0 1 and L.f; P/ 0, for all partitions of P of Œ0; 1. Therefore, D D D

1 1 f .x/dx 0 and f .x/dx 1; D D Z0 Z0

and hence f is not Riemann-integrable on Œ0; 1.

Notice that in the last example, the given function f is discontinuous at every point. 1.2.13 Theorem (Darboux’s integrability condition) Let f be a real-valued function which is bounded on Œa; b. Then f is integrable on Œa; b if and only if, for any > 0, there exists a partition P of Œa; b such that

U.f; P / L.f; P /<: PROOF. Assume that f is integrable on Œa; b and let > 0. Since

b b f .x/dx f .x/dx sup L.f; P/; D D P Za Za

there is a partition P1 of Œa; b such that

b f .x/dx < L.f; P1/: 2 Za Again, since b b f .x/dx f .x/dx inf U.f; P/; D D P Za Za

there exists a partition P2 of Œa; b such that

b U.f; P2/ < f .x/dx : C 2 Za

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Let P be a common reﬁnement of P1 and P2. Then

b b f .x/dx < L.f; P1/ L.f; P / U.f; P / U.f; P2/ < f .x/dx : 2 Ä Ä Ä C 2 Za Za It now follows that U.f; P / L.f; P /<: For the converse, assume that given > 0, there is a partition P of Œa; b such that

U.f; P / L.f; P /<: Then, b b f .x/dx inf U.f; P/ U.f; P /; and f .x/dx sup L.f; P/ L.f; P /: D P Ä D P Za Za Thus, b b 0 f .x/dx f .x/dx U.f; P / L.f; P /<: Ä Ä Za Za Since > 0 is arbitrary, we have that

b b f .x/dx f .x/dx: D Za Za

That is, f RŒa; b. 2 2 We now highlight the following important fact which is contained in the ﬁrst part of the proof of Dar- boux’s integrability condition: 1.2.14 Theorem If f is integrable on Œa; b, then for each > 0, there exists a partition P of Œa; b such that

b b f .x/dx < L.f; P/ U.f; P/ < f .x/dx : Ä C Za Za 1.2.15 Theorem If f is continuous on Œa; b, then it is integrable there.

PROOF. Since f is continuous on Œa; b, we have that f is bounded on Œa; b. Furthermore, f is uniformly continuous on Œa; b. Hence, given > 0, there is a ı > 0 such that f .x/ f .y/ < , whenever x; y Œa; b and x y < ı: j j b a 2 j j Let P x0; x1; x2;:::; xn be any partition of Œa; b such that xi xi1 < ı, for each i 1; 2;:::; n. D f g D By the Extreme-Value Theorem (applied to f on Œxi1; xi, for each i 1; 2;:::; n), there exist points ti D and si in Œxi1; xi, for each i 1; 2;:::; n such that D f .ti / sup f .x/ xi1 x xi Mi and f .si / inf f .x/ xi1 x xi mi : D f W Ä Ä gD D f W Ä Ä gD

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Since xi xi1 <ı, it follows that ti si <ı, and so j j Mi mi f .ti / f .si / f .ti / f .si / < ; for all i 1; 2;:::; n: D Dj j b a D Thus,

n n n

U.f; P/ L.f; P/ Mi .xi xi1/ mi .xi xi1/ .Mi mi /.xi xi1/ D D iD1 iD1 iD1 Xn X n X f .ti / f .si / .xi xi1/ < .xi xi1/ Ä j j b a iD1 iD1 Â Ã X X .b a/ : D b a D It now follows from Theorem 1.2.13, that f is integrable on Œa; b. 2 1.2.16 Theorem If f is monotone on Œa; b, then f is integrable there.

PROOF. Assume that f is monotone increasing on Œa; b and f .a/ Ä0, there isÄ a partition 2 P of Œa; b such that U.f; P/ L.f; P/ < . Let > 0 be given an let P x0; x1; x2;:::; xn be any D f g partition of Œa; b such that xi xi1 < f.b/f.a/ , for each i 1; 2;:::; n. Since f is increasing on Œa; b, we have that D

Mi sup f .x/ xi1 x xi f .xi / and mi inf f .x/ xi1 x xi f .xi1/; D f W Ä Ä gD D f W Ä Ä gD for each i 1; 2;:::; n. Hence, D n n n

U.f; P/ L.f; P/ Mi .xi xi1/ mi .xi xi1/ .Mi mi /.xi xi1/ D D iD1 iD1 iD1 Xn X X n Œf .xi / f .xi1/.xi xi1/ < Œf .xi / f .xi1/ Ä f .b/ f .a/ iD1 iD1 X X Œf .b/ f .a/ : D f .b/ f .a/ D It then followsfrom Theorem 1.2.13, that f is Riemann-integrable on Œa; b. The case where f is monotone decreasing can be proved in exactly the same way. 2 1.2.17 Theorem If f is integrable on Œa; b and a c < d b, then f is integrable on Œc; d. Ä Ä PROOF. Since f is integrable on Œa; b, it is bounded there. Hence f is bounded on Œc; d. Furthermore, given any > 0, there is a partition P of Œa; b, such that

U.f; P/ L.f; P/<: Let P P c; d . Then P is a reﬁnement of P, and hence D [f g U.f; P / L.f; P / U.f; P/ L.f; P/<: Ä

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Let Q1 P Œa; c; Q2 P Œc; d; Q3 P Œd; b. Then P Q1 Q2 Q3, and so, D \ D \ D \ D [ [ U.f; P / U.f; Q1/ U.f; Q2/ U.f; Q3/; and D C C L.f; P / L.f; Q1/ L.f; Q2/ L.f; Q3/: D C C Hence,

ŒU.f; Q1/ L.f; Q1/ ŒU.f; Q2/ L.f; Q2/ ŒU.f; Q3/ L.f; Q3/ U.f; P / L.f; P /<: C C D

Note that all terms on the left are nonnegative. Therefore Q2 is a partition of Œc; d with the property that

U.f; Q2/ L.f; Q2/<: This implies that f is integrable on Œc; d. 2 1.2.18 Corollary If f is integrable on Œa; b and a < c < b, then f is integrable on both Œa; c and Œc; b.

The following theorem states that the converse of Corollary 1.2.18 also holds. 1.2.19 Theorem If a < c < b and f is integrable on both Œa; c and Œc; b, then f is integrable on Œa; b and

b c b f .x/dx f .x/dx f .x/dx: D C Za Za Zc

PROOF.

Let > 0 be given. Since f is integrable on Œa; c and on Œc; b, there are partitions P1 and P2 of Œa; c and Œc; b respectively, such that U.f; P1/ L.f; P1/ < ; and 2 U.f; P2/ L.f; P2/ < : 2

Let P P1 P2. Then P is a partition of Œa; b and D [

U.f; P/ L.f; P/ U.f; P1/ U.f; P2/ L.f; P1/ L.f; P2/ D C U.f; P1/ L.f; P1/ U.f; P2/ L.f; P2/ < : D C 2 C 2 D Therefore f is integrable on Œa; b. Furthermore,

f .x/dx U.f; P/ U.f; P1/ U.f; P2/ Ä D C Za < L.f; P1/ L.f; P2/ C 2 C C 2 Ä Ä L.f; P1/ L.f; P2/ D C C c b f .x/dx f .x/dx : Ä C C Za Zc

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Since is arbitrary, it follows that

b c b f .x/dx f .x/dx f .x/dx (1.1) Ä C Za Za Zc Also, b

f .x/dx L.f; P/ L.f; P1/ L.f; P2/ D C Za > U.f; P1/ U.f; P2/ 2 C 2 Ä Ä U.f; P1/ U.f; P2/ D C Since is arbitrary, it follows that

b c b f .x/dx f .x/dx f .x/dx: (1.2) C Za Za Zc Combining (1.1) and (1.2), we have that

b c b f .x/dx f .x/dx f .x/dx: D C Za Za Zc

2 1.2.20 Corollary Let f be deﬁned on Œa; b and suppose that a c0 < c1 < < cn1 < cn b. Then f is integrable on D D Œa; b if and only if f is integrable on Œck1; ck , for each k 1; 2;:::; n. In this case, D c b n k f .x/dx f .x/dx: D kD1 Za XckZ1 1.2.21 Corollary If f is continuousat all but a ﬁnite set of points in Œa; b, then f is Riemann-integrable on Œa; b.

1.3 Fundamental theorem of calculus 1.3.1 Theorem (Fundamental theorem of calculus) Let f be a bounded real-valued integrable function on Œa; b. Suppose that F Œa; b R is continuous on Œa; b and differentiable on .a; b/ and that W !

F 0.x/ f .x/; for each x .a; b/: D 2 Then, b f .x/dx F.b/ F.a/: D Za

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PROOF.

Let P x0; x1; x2;:::; xn be a partition of Œa; b. Df g

Mi sup f .x/ xi1 x xi and mi inf f .x/ xi1 x xi ; D f W Ä Ä g D f W Ä Ä g

for each i 1; 2;:::; n. For each i 1; 2;:::; n, F is continuous on Œxi1; xi and differentiable on D D .xi1; xi/. By the Mean Value Theorem, there is a i .xi1; xi/ such that 2 0 F.xi / F.xi1/ F .i /.xi1 xi / f .i /.xi1 xi /: D D

Since, for each i 1; 2;:::; n mi f .i / Mi , it followsthat D Ä Ä

mi .xi xi1/ f .i /.xi1 xi / Mi .xi xi1/ mi .xi xi1/ F.xi / F.xi1/ Mi .xi xi1/ Ä Ä , Ä Ä n n n mi .xi xi1/ ŒF.xi / F.xi1/ Mi .xi xi1/ ) Ä Ä XiD1 XiD1 XiD1 L.f; P/ F.b/ F.a/ U.f; P/: , Ä Ä Since P is an arbitrary partitionof Œa; b, it followsthat F.b/ F.a/ isan upper boundfor theset L.f; P/ P is a partition of Œa; b and a lower bound for the set U.f;P/ P is a partition of Œa; b . Therefore,f W g f W g b b f .x/dx sup L.f; P/ F.b/ F.a/ inf U.f; P/ f .x/dx: D P Ä Ä P D Za Za

Since f is integrable on Œa; b, we have that

b b b f .x/dx f .x/dx f .x/dx; D D Za Za Za

b and consequently f .x/dx F.b/ F.a/ D Za 2 1.3.2 Exercise [1] Let f be the function on Œ0; 1 given by

x if x is rational f .x/ D 0 if x is irrational: Show that f is not Riemann-integrable on Œ0; 1. [2] Let f be the function on Œ0; 1 given by

1 if 0 x < 1 f .x/ Ä 2 D x 1 if 1 x < 1: 2 2 Ä (a) Show, from ﬁrst principles, that f is Riemann-integrable on Œ0; 1.

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(b) Quote a result that assures us that f is Riemann-integrable. 1 (c) Find f .x/dx:

(d) Let r1; r2; r3;::: be an enumeration of rationals in the interval Œ0; 1. For each n N, deﬁne f g 2

1 if x r1; r2; r3;:::; rn fn.x/ 2f g D 0 otherwise:

Show that .fn/ is a nondecreasing sequence of functions that are Riemann-integrable on Œ0; 1. Show also that the sequence .fn/ converges pointwise to the function

1 if x Q Œ0; 1 f .x/ 2 \ D 0 otherwise; and that f is not Riemann-integrable.

13 Chapter 2

Measure spaces

2.1 Introduction

We now concern ourselves withtheidea ofa defininga measure on the real numbers, which putsthe intuitive idea of length on an axiomatic foundation. This turns out to be more difficult than you might expect and we begin with the motivation for developing measure theory. If we are to properly define the definite integral of a real-valued function, we need a meaningful notion of the length of a set A R.

If we consider the set of real numbers, we would like to be able to deﬁne a function that measures the ‘length’ of a set. That is a function m P.R/ Œ0; such that: W ! 1 [1] .Œa; b/ b a, for a b. D Ä [2] is translation invariant, i.e., .A/ m.x A/, for A R and x R. D C 2 [3] . / 0. ; D [4] If .Ai / is a sequence of mutually disjoint sets in .R/ (i.e., Ai Aj , for every i j ), then \ D; 6D 1 1 Ai Ai : D Â i[D1 Ã XiD1

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This turns out to be impossible. That is, you cannot deﬁne such a function whose domain consists of the whole of P.R/. To show this we present the following example by Giuseppe Vitali (1905):

2.1.1 Example A Vitali set V , is a subset of Œ0; 1, which for each real number r, V contains exactly one number v such that v r is rational (This implies that V is uncountable, and also, that v u is irrational, for any u; v V , u v). Such sets can be shown to exist provided we assume the axiom of choice. 2 6D To construct the Vitali set V , consider the additive quotient group R=Q. Each element of this group is a ‘shifted copy’ of the rational numbers. That is, a set of the form Q r, for some r R. Therefore, the elements of this group are subsets of R and partition R. There are uncountablyC many elements.2 Since each element intersects Œ0; 1, we can use the axiom of choice to choose a set V Œ0; 1 containing exactly one representative out of each element R=Q.

A Vitali set is non-measurable. To show this, we argue by contradiction and assume that V is measurable. Let q1; q2;::: be an enumeration of the rational numbers in Œ 1; 1. From the construction of V , note that the translated sets Vk V qk v qk v V , k 1; 2;::: are pairwise disjoint, and further D C D f C W 2 g D note that Œ0; 1 k Vk Œ 1; 2 (to see the ﬁrst inclusion, consider any real number r in Œ0; 1 and let v be the representativeÂ in V Âfor the equivalence class Œr, the r v q for some rational number q in Œ 1; 1). Now assumeS that there exists a that satisﬁes the four conditionsD listed above. Apply to these inclusions using condition [4]: 1

1 .Vk / 3: Ä Ä kXD1 Since is translation invariant, .Vk / .V / and D 1 1 .V / 3: Ä Ä kXD1 This is impossible. Summing infinitely many copies of the constant .V / yields either zero or infinity, according to whether the constant is zero or positive. In neither case is the sum in Œ1; 3. So, V cannot have been measurable after all, i.e., must not define any value for .V /.

It is important to stress that this whole argument depends on the assumption of the axiom of choice.

2.2 Notation and preliminaries

Let X be a set.

We will use the standard notation for the compliment of A: Ac x X x A Df 2 W 62 g We denote the complement of B with respect to A by A B A Bc . n D \ We denote by P.X /, the powerset of X , i.e., the set of all subsets of X . Let A and B be in P.X /. The symmetric difference of A and B, denoted by A B, is given by 4 A B .A B/ .B A/: 4 D n [ n A subset C of P.X / is said to be:

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[1] closed under finite intersections if, whenever A1; A2;:::; An is a finite collection of elements n f g of C, then Aj C. 2 j\D1 [2] closed under finite unions if, whenever A1; A2;:::; An is a finite collection of elements of C, n f g then Aj C. 2 j[D1 [3] closed under countable intersections if, whenever .An/ is a sequence of elements of C, then 1

Aj C. 2 j\D1 1 [4] closed under countable unions if, whenever .An/ is a sequence of elements of C, then Aj 2 jD1 C. [

Let .An/ be a sequence in P.X /. We denote by 1 1 1 1

limsup An Am, and liminfAn Am: n!1 n!1 D mDn D mDn n\D1 [ n[D1 \ 2.2.1 Exercise Let X be a nonempty set. Prove that if a collection C of P.X / is closed under countable intersections (resp., unions), then it is closed under finite intersections (resp., unions). 2.2.2 Definition An algebra (in X ) is a collection † of subsets of X such that: (a) ; X †. ; 2 (b) If A †, then Ac †. 2 2 n n (c) If A1;:::; An †, then Ai and Ai are in †. 2 [iD1 \iD1 † is a -algebra (or -field) if in addition:

1 1 (d) If A1; A2;::: are in †, then Ai and Ai are in †. [iD1 \iD1 2.2.3 Examples [1] Let X be any nonempty set. Then there are two special -algebras on X : †1 ; X and †2 P.X / (the power set of X ). D f; g D [2] Let X Œ0; 1. Then † ; X; Œ0; 1 ; . 1 ; 1 is a -algebra on X . D D f; 2 2 g [3] Let X R and let D † A R A is countable or Ac is countable : Df W g † is a -algebra on X . Parts (a) and (b) of the deﬁnition are easy to show. Suppose A1; A2;::: are c all in †. If each of the Ai ’s are countable, then i Ai is countable,and so in †. If A is countable for [ i0 some i0, then c c c . i Ai / i A A [ D \ i i0 c c is countable, and again i Ai is in †. Since i Ai . i Ai / , then the countable intersection of sets in † is again in †. [ \ D [ 2.2.4 Remark Let X R and let † be a -algebra on X . If .An/ is a sequence of sets in †, then liminfn!1 An †. D 2

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If S is any collection of subsets of a set X , then by takingthe intersectionof all -algebras containing S, we obtain the smallest -algebra containing S. Such a -algebra, denoted by .S/ is said to be generated by S. It is clear that if S is a -algebra, then .S/ S. D 2.2.5 Deﬁnition The smallest -algebra conataining all open subsets of R is called the Borel -algebra of R and is denoted by B.R/. Elements of this -algebra are called Borel sets. 2.2.6 Proposition Let I Œa; b/ a; b R; a b . Then .I/ B.R/. Df W 2 Ä g D PROOF. If a; b R and a b, then since 2 Ä 1 1 Œa; b/ a ; b ; D n n\D1 Á it follows that Œa; b/ B.R/ and consequently, I B.R/. It now follows that .I/ B.R/. 2 Â Â Conversely, let A be an open subset of R. Then A is a countable union of open intervals. Since any open interval .a; b/ can be expressed in the form

1 1 .a; b/ a ; b ; n D nDn C [0 h Á 1 I I B where n < b a, it follows that A . /. Thus . / .R/. 0 2 D 2 2.2.7 Exercise Let X be a nonempty set and S the smallest -algebra in X which contains all singletons x . Show that f g S A P.X / either A or Ac is countable : Df 2 W g 2.3 Measureonaset 2.3.1 Definition A measurable space is a pair .X; †/, where X is a set and † is a -algebra on X . The elements of the -algebra † are called measurable sets. 2.3.2 Definition Let .X; †/ be a measurable space. A measure on .X; †/ is a function † Œ0; such that: W ! 1 (a) . / 0, and ; D (b) if .Ai / is a sequence of mutually disjoint sets in † (i.e., Ai †, for each i 1; 2; 3;:::, and 2 D Ai Aj for every i j ), then \ D; 6D 1 1 Ai Ai : D Â i[D1 Ã XiD1 Property (b) in the above definition is known as the countable additivity property of the measure . If n n (b) is replaced by the condition that . iD1 Ai / iD1 Ai for every finite sequence A1; A2;:::; An of pairwise disjoint sets in †, then we say that is finitelyD additive. It follows immediately, that if a measure S P

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is countably additive, then it must also be finitely additive. If you have a finite sequence A1; A2;:::; An in †, then you can consider it to be an infinite sequence .Ai / with AnC1 AnC2 : : : . Then D D ; n 1 1 n Ai Ai Ai Ai : D D D Â i[D1 Ã Â i[D1 Ã XiD1 XiD1 2.3.3 Definition A measure space is a triple .X;†;/, where X is a set, † a -algebra in X , and a measure on †.

2.3.4 Remark Let .X;†;/ be a measure space. If .X / 1, then .X;†;/ is called a probability space and is called a probability measure. D 2.3.5 Examples [1] Let X be a nonempty set and † P.X /. Deﬁne † Œ0; by D W ! 1 n if A has n elements .A/ D otherwise: 1 Then is a measure on X called the counting measure.

[2] Let .X; †/ be a measurable space. Choose and ﬁx x X . Deﬁne x † Œ0; by 2 W ! 1 1 if x A x.A/ 2 D 0 otherwise:

Then x is a measure on X called the unit point mass at x. 2.3.6 Theorem Let .X;†;/ be a measure space. Then the following statements hold: [1] If A; B † and A B, then .A/ .B/. 2 Â Ä [2] If .An/ is a sequence in †, then

1 1

An .An/: Ä Â n[D1 Ã nXD1

[3] If .An/ is a sequence in † such that An AnC1, for each n N, then Â 2 1

An lim .An/: D n!1 Â n[D1 Ã

[4] If .An/ is a sequence in † such that An AnC1, for each n N and .A1/ < , then Ã 2 1 1 An lim .An/: D n!1 Â n\D1 Ã

PROOF.

[1] Write B A .B A/, a disjointunion. Since is ﬁnitely additive, we have that D [ n .B/ .A/ .B A/ .A/: D C n

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n1 [2] Set B1 A1, and for each natural number n > 1, let Bn An Ai . Then D D n iD1 n1 n1 S c Bn .An Ai / An A : D n D i i\D1 i\D1

Therefore, for each n 1; 2; 3;:::, we have that Bn †. Clearly, Bn An, for each n D 2 Â D 1; 2; 3;:::. Let l and k be integers such that l < k. Then Bl Al , and so k1 c Bl Bk Al Bk Al Ak A \ \ D \ i iD1 \ c Al Ak Al D \ c \\ \ Al A Ak Ak D \ l \ \\D;\ \\ : D;

That is, Bi Bj : i6Dj D; 1 1 Claim: nTD1 An nD1 Bn: D 1 1 Since BnS An, for eachS n 1; 2; 3;:::; we have that nD1 Bn nD1 An. Â1 D Â Let x nD1 An: Then x An for some n. Let k beS the smallestS index such that x An. Then 2 2 c 2 x Ak and x Aj , for each j 1; 2;:::; k 1. Therefore, x A , for each j 1; 2;:::; k 1. 2 62 D 2 j D S k1 c 1 That is, x Bk Ak A , whence x Bn, which proves the claim. 2 D jD1 j 2 nD1 Therefore, T 1 1 S 1 1 An Bn .Bn/ .An/; D D Ä Â n[D1 Ã Â n[D1 Ã nXD1 nXD1 by part [1].

k [3] Let B1 A1 and Bn An An1 for all n > 1. Then Bi i6Dj Bj ; Ak nD1 Bn and 1 D 1 D n D; D An Bn. Therefore, nD1 D nD1 T S S S1 1 1 k An Bn .Bn/ lim .Bn/ lim .Ak /: D D D k!1 D k!1 Â n[D1 Ã Â n[D1 Ã nXD1 nXD1 1 [4] Let A nD1 An. Then D 1 1

T A1 A A1 An .A1 An/; n D n D n n\D1 n[D1 and A1 An A1 AnC1 , for each natural number n. Therefore, by [3] above, n n 1 .A1 A/ .A1 An/ lim .A1 An/: (2.1) n D n D n!1 n Â n[D1 Ã

Since A A1 and .A1/ < , we have that .A/ < . Also, since An A1, for each natural 1 1 number n, it follows that .A/ < , for each n. Writing A1 as a disjoint union A1 .A1 A/ A, 1 D n [ we have that .A1/ .A1 A/ .A/. It now follows that D n C

.A1 A/ .A1/ .A/: (2.2) n D

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Similarly, by expressing A1 as A1 .A1 An/ An, we get that D n [

.A1 An/ .A1/ .An/: (2.3) n D From equations 2.1, 2.2 and 2.3, we have that

.A1/ .A/ lim Œ.A1/ .An/ .A1/ lim .An/: D n!1 D n!1

Since .A1/ < , we have that 1 1

.A/ lim .An/; that is An lim .An/: D n!1 D n!1 Â n\D1 Ã 2 2.3.7 Theorem (Borel-Cantelli Lemma) 1 Let .X;†;/ be a measure space. If .An/ is a sequence of measurable sets such that .An/ < , nD1 1 then limsup An 0. P n!1 D PROOFÂ. Ã 1 1 Since limsup An Ak , it follows that for each n N, n!1 D nD1 kDn 2 T S 1 1 limsup An Ak .Ak / 0 n!1 Ä Ä ! Â Ã Â n[D1 Ã nXD1 as n . ! 1 2 2.3.8 Definition Let .X;†;/ be a measure space. The measure is said to be: [1] finite if .X / < , and 1 1 [2] -finite if there is a sequence .Xn/ of sets in † with .Xn/ < , for each n such that X Xn. 1 D nD1 S 2.4 Outermeasure

Constructing a measure is a nontrivial exercise as illustrated in the introduction to this chapter. The basic idea that we will use is as follows: We will begin by defining an ‘outer measure’ on all subsets of a set. Then we will restrict the outer measure to a reasonable class of subsets and in so doing we will define a measure. This is indeed how Lebesgue solved the problems associated with measure and integration. Both were published as part of his dissertation in 1902. 2.4.1 Definition Let X be a set. An outer measure on X is a set function P.X / Œ0: satisfying the following properties: W ! 1 (a) . / 0: ; D (b) If A B, then .A/ .B/. Â Ä 1 1 (c) For any sequence .An / of subsets of X , An .An /: nD1 Ä nD1 Â Ã S P Property (b) above is referred to as monotonicity, and property (c) is countable subadditivity

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2.4.2 Examples [1] Let X be a set. Deﬁne on P.X / by:

1 if A .A/ 6D ; D 0 if A : D; Then is an outer measure on X . [2] Let X be an uncountable set. Deﬁne on P.X / by:

0 if A is countable .A/ D 1 if A is uncountable: Then is an outer measure on X .

We can now use the concept of outer measure to construct a measure. 2.4.3 Deﬁnition Let X be a set and an outer measure on X . A subset E of X is said to be -measurable if, for every subset A of X , we have .A/ .A E/ .A Ec/: D \ C \ 2.4.4 Remark [1] Since A .A E/ .A Ec / and is subadditive, we always have that D \ [ \ .A/ .A E/ .A Ec /: Ä \ C \ Therefore, a subset E of X is -measurable if, for every subset A of X ,

.A/ .A E/ .A Ec /: \ C \ [2] Replacing E by Ec in the deﬁnition of -measurability, we have that

.A/ .A Ec / ŒA .Ec /c .A Ec/ .A E/: D \ C \ D \ C \ Thus, E is -measurable if and only if Ec is -measurable. 2.4.5 Lemma Let X be a set, an outer measure on X and E a subset of X . If .E/ 0, then E is -measurable. D PROOF. Let A be any subset of X . Then A E E. Since is monotone, .A E/ .E/ 0. Also, since A Ec A, it follows that .A\ EÂc / .A/. Therefore, \ Ä D \ Â \ Ä .A/ .A E/ .A Ec /; \ C \ and therefore, E is -measurable. 2 2.4.6 Corollary The empty set is -measurable. ; 2.4.7 Lemma Let X be a set, an outer measure on X , and E and F -measurable subsets of X . Then E F is -measurable. [

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PROOF. Let A be any susbset of X . Then since E is -measurable,

.A/ .A E/ .A Ec/: D \ C \ Also, since F is -measurable,

.A Ec / Œ.A Ec / F Œ.A Ec / F c : \ D \ \ C \ \ From the two equations above we get:

.A/ .A E/ Œ.A Ec/ F Œ.A Ec/ F c : D \ C \ \ C \ \ Therefore, .A/ .A E/ Œ.A Ec/ F Œ.A Ec/ F c D \ C \ \ C \ \ Œ.A E/ ..A Ec / F/ Œ.A Ec/ F c \ [ \ \ C \ \ ŒA .E F/ ŒA .E F/c ; D \ [ C \ [ whence E F is measurable. [ 2 2.4.8 Theorem Let X be a set and an outer measure on X . Denote by M, the collection of all -measurable subsets of X . Then M is an algebra on X .

PROOF. This follows from Remark 2.4.4 [2], Corollary 2.4.6 and Lemma 2.4.7 2 2.4.9 Proposition Let X be a set and an outer measure on X . If E1; E2 M and E1 E2 , then 2 \ D; ŒA .E1 E2/ .A E1/ .A E2/: \ [ D \ C \ PROOF. Since E1 is -measurable,

c .A/ .A E1/ .A E /; D \ C \ 1

for each subset A of X . Replacing A by A .E1 E2/, we have \ [ c ŒA .E1 E2/ Œ.A .E1 E2// E1 Œ.A .E1 E2// E1 \ [ D \ [ \ C \ [ \ c Œ..A E1/ .A E2// E1 Œ..A E1/ .A E2// E1 D \ [ \ \ C \ [ \c \ c Œ.A E1 E1/ .A E2 E1/ Œ.A E1 E1/ .A E2 E1/ D \ \ [ \ \ C \ \ [ \ \ .A E1/ .A E2/: D \ C \ 2 2.4.10 Theorem Let X be a set and an outer measure on X . Suppose that .En/ be a sequence of mutually disjointsets in M. Then for any n 1 and any subset A of X , n n A Ej .A Ej /: (2.4) \ D \ Â j[D1 Ã jXD1

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PROOF. We prove this theorem by induction. The case n 1 is obviously true. The case n 2 has been proved in Proposition 2.4.9. Assume that D D

n1 n1 A Ej .A Ej /: (2.5) \ D \ Â j[D1 Ã jXD1 Firstly, we note that n A Ej En A En, and \ \ D \ Âj[D1 Ã n n1 c A Ej E A Ej : \ \ n D \ Â j[D1 Ã Â j[D1 Ã Since En is -measurable,

n n n c A Ej A Ej En A Ej E \ D \ \ C \ \ n Ä Â j[D1 Ã Ä Â j[D1 Ã Ä Â j[D1 Ã n1 .A En/ A Ej D \ C \ Ä Â j[D1 Ã n1 .A En/ .A Ej / (by 2.5) D \ C \ jXD1 n .A Ej /: D \ jXD1 2 By taking A X in (2.4), we obtain D n n Ej .Ej /: D Â j[D1 Ã jXD1 i.e., is ﬁnitely additive on M. The following theorem is the key to using an outer measure to construct a measure. 2.4.11 Theorem Let X be a set and an outer measure on X . Denote by M, the collection of all -measurable subsets of X . Then [1] M is a -algebra, and

[2] the restriction of M of to M is a measure on .X; M/. j PROOF.

[1] We have already shown in Theorem 2.4.8 that M is an algebra. It remains to show that M is closed under countable unions. We will show that M is in fact closed under pairwise disjoint countable

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n 1 unions. Let .En/ be a pairwise disjoint sequence of sets in M, Sn Ej and S Ej . D jD1 D \jD1 Then, since Sn M, we have that for any A X , 2 Â S c .A/ .A Sn/ .A Sn / D \ C \ c c c .A Sn/ .A S / (since S S and is monotone) \ C \ n n c .A Ej / .A S / (by Theorem 2.4.10). D \ C \ jXD1 Since this is true for every natural number n, it follows that

1 c .A/ .A Ej / .A S / \ C \ jXD1 .A S/ .A S c / (since is countably subadditive). \ C \ 1 Therefore, Ej M. jD1 2 [2] By deﬁnitionS of the outer measure, . / 0. We have to show that is countably additive. We ; D do this as follows, let .En/ be a pairwise disjoint sequence of sets in M. As shown in [1], for any A X , Â 1 1 c .A/ .A Ej / A Ej : \ C \ jXD1 Ä Âj[D1 Ã 1 If A is replaced by Ej , we have that [jD1 1 1 Ej .Ej /: (2.6) Â j[D1 Ã jXD1 Since is, by deﬁnition, countably subadditive, we have that

1 1 Ej .Ej /: (2.7) Ä Â j[D1 Ã jXD1 From (2.6) and (2.7), we conclude that

1 1 Ej .Ej /: D Â j[D1 Ã jXD1 Thus, is a measure on M. 2 The measure M on X is often referred to as the measure on X induced by the outer measure . D j 2.4.12 Exercise [1] In provingTheorem 2.4.11 [1], we showed that M is closed under pairwise disjoint countable unions. Deduce from this that M is closed under countable unions. [2] Let be an outer measure on a set X such that .A/ 0, for some subset A of X . Show that if B is a subset of X , then .A B/ .B/. D [ D

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2.5 Lebesgue measure on R

We now use the approach discussed in the previous section to construct the Lebesgue measure on R. The Lebesgue measure will lead to the deﬁnition of the Lebesgue integral in the next chapter. Let I be a collection of open intervals in R. For I I, let l.I/ denote the length of I. 2 2.5.1 Theorem For any subset A of R, deﬁne m P.R/ Œ0; by W ! 1

m .A/ inf l.In/ In I, for each n and A In : D n W 2 Â n X [ Then m is an outer measure on R.

PROOF.

(i) Since .a; a/, for any real number a, it follows that 0 m. / a a 0. Thus, m. / 0. ; Â Ä ; Ä D ; D

(ii) Let A and B be subsets of R such that A B. If B n In, where .In/ is a sequence of open Â Â1 intervals in R, then A In. Therefore, m .A/ l.In/. It followsthat m .A/ m .B/. Â n Ä nDS1 Ä (iii) Let .An/ be a sequence ofS subsets of R. If m .An / P for some n, then D 1 1 1 m An m .An /: Ä Â n[D1 Ã nXD1 Suppose that m .An / < , for each n. Then given > 0, there is a sequence .Ink /k of open 1 1 intervals in R such that An Ink , and Â kD1 S 1 l.Ink / < m .An/ : C 2n kXD1 1 1 1 Now, since An Ink , it follows that nD1 Â nD1 kD1 S 1 S 1S 1 1 1 m An l.Ink / < m .An/ m .An/ : Ä C 2n D C Â n[D1 Ã nXD1 kXD1 nXD1 Á nXD1

1 1 Since is arbitrary, we have that m An m .An /. nD1 Ä nD1 Â Ã S P 2 The outer measure deﬁned in Theorem 2.5.1 is called the Lebesgue outer measure. 2.5.2 Deﬁnition A set E R is said to be Lebesgue-measurable (or simply measurable) if Â m.A/ m.A E/ m.A Ec/; D \ C \ for every set A R. Â Denote by M, the set of all Lebesgue-measurable subsets of R. We have already shown in Theorem 2.4.11, that M is a -algebra and that m m M is a measure on M. This measure is called the Lebesgue measure on R. D j

The Lebesgue outer measure has other important properties that are listed in the following propositions.

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2.5.3 Proposition The outer measure of an interval I is its length. That is, m.I/ l.I/: D PROOF. Case 1: Let I be a closed and bounded interval, say I Œa; b. Then Œa; b .a ; b /, for each > 0. Thus, D C m.Œa; b/ l.a ; b / b a 2: Ä C D C Since is arbitrary, m.I/ m.Œa; b/ b a. It remains to show that m.I/ b a. Suppose that 1 D Ä m I kD1 Ik , where Ik is open for each k N. Since I is compact, there is a ﬁnite subcollection Ik kD1 m 2 mf g of intervals such that I . Since a I, there is an interval .a1; b1/ in the collection Ik such S kD1 2 f gkD1 that a1 < a < b1. If b b1, then we are done. Otherwise, a < b1 < b and so b1 I. Therefore, there Ä S m 2 is an interval .a2; b2/ in the collection Ik kD1 such that a2 < b1 < b2. If b < b2, then we are done. f g m Otherweise, a < b2 < b. That is, b2 I. So there is an interval .a3; b3/ in the collection Ik such that 2 f gkD1 a3 < b2 < b3. Continuingin this fashion, we obtain a sequence of intervals .a1; b1/; .a2; b2/;:::;.ar ; br / m m from the collection Ik such that ai < bi1 < bi for i 2; 3;:::; r. Since Ik is a ﬁnite f gkD1 D f gkD1 collection, this process must terminate after s m steps with b < bs. Now, Ä

1 m s l.Ij / l.Ij / l.aj ; bj / jXD1 jXD1 jXD1 .bs as/ .bs1 as1/ .bs2 as2/ .b1 a1/ D C C CC bs .as bs1/ .as1 bs2/ .a2 b1/ a1 D > bs a1 > b a: It follows from this that m.I/ b a l.I/. D Case 2: Let I be an interval of the form .a; b/, .a; b or Œa; b/. Then for each 0 <<.b a/ with Œa ; b I Œa ; b : C 2 2 2 C 2 Hence, mŒa ; b m.I/ mŒa ; b C 2 2 Ä Ä 2 C 2 l Œa ; b m.I/ Œa ; b , C 2 2 Ä Ä 2 C 2 b a m.I/ b a ( by case 1). , Ä Ä C Since is arbitrary, it follows that m.I/ b a l.I/: D D Case 3: Suppose that I is an unbounded interval. Then for any real number k, there is a closed interval J I such that l.J/ k. Hence, D m.I/ m.J/ l.J/ k: D D Thus, m.I/ . D 1 2 2.5.4 Proposition For each x R, m. x / 0: 2 f g D PROOF.

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Given any > 0, x .x ; x /. Thus f g 2 C 2 0 m. x / m.x ; x / l.x ; x / : f g Ä 2 C 2 D 2 C 2 D Since is arbitrary, we have that m. x / 0. f g D 2 2.5.5 Corollary Every countable subset of R has outer measure zero.

PROOF.

Let A be a countable subset of R. Then A an . Therefore, D nf g S 0 m .A/ m an m . an / 0: Ä D n f g Ä n f g D Â [ Ã X Thus, m.A/ 0. D 2 2.5.6 Corollary The interval Œ0; 1 is uncountable.

2.5.7 Proposition The Lebesgue outer measure is translation-invariant. That is, m.A x/ m.A/ for all A R and x R. C D 2 PROOF.

Let .In/ be a sequence of open intervals such that A In. Then A x .In x/. Thus, n C n C m .A x/ l.In xS/ l.In/: S C Ä n C D n X X That is, m .A x/ is a lower bound for the sums l.In/, with A In. Therefore, C n n m.A Px/ m.A/: S (2.8) C Ä To prove the reverse inequality, we only need to notice that m.A/ m.A x . x// m.A x/; D C C Ä C where the inequality follows from (4.1). Thus, m.A x/ m.A/. C D 2 2.5.8 Proposition For any set A R and any > 0, there is an open set V such that A V and m.V / < m.A/ . C PROOF. By deﬁnition of m .A/, there is a sequence .In/ of open intervals such that A In and n S l.In/ < m .A/ : 2 n C X I .a ; b / J .a ; b / A J V J V R If n n n , let n n 2nC1 n . Then n n. Let n n. Then isan open set in and D D D m.V / l.J / lS.I / < m.AS/ : n n 2 Ä n D n C C X X 2

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2.5.9 Proposition For any a R, the set .a; / is measurable. 2 1 PROOF. Let A be any subset of R. We must show that m.A/ m.A .a; // m.A .a; /c / m.A .a; // m.A . ; a/: \ 1 C \ 1 D \ 1 C \ 1 Let A1 A .a; / and A2 A . ; a. Then we need to show that D \ 1 D \ 1 m .A/ m .A1/ m .A2/: C If m.A/ , then there is nothing to prove. Assume that m.A/ < . From the deﬁnition of m, given D 1 1 > 0, there is a sequence .In/ of open intervals such that A In and n m .A/ > l.InS/: C n X 0 0 Let Jn In .a; / and Jn In . ; a. Then Jn and Jn are disjoint (possibly empty) intervals and D \0 1 D \ 1 In Jn J . Therefore, D [ n 0 0 l.In/ l.Jn/ l.J / m .Jn/ m .J /: D C n D C n 0 Since A1 Jn and A2 J , it followsthat n n n S S m .A1/ m Jn m .Jn/ l.Jn/; and Ä n Ä n D n Â [ Ã X X 0 0 0 m .A2/ m Jn m .Jn/ l.Jn/: Ä n Ä n D n Â [ Ã X X Therefore, 0 m .A1/ m .A2/ Œl.Jn/ l.Jn/ l.In/ < m .A/ : C Ä n C D n C X X Since is arbitrary, m .A1/ m .A2/ m .A/. C Ä 2 2.5.10 Corollary [1] Every interval in R is measurable. [2] Every open set in R is measurable. [3] Every closed set in R is measurable. [4] Every set that is a countable intersection of open sets in R is measurable. [5] Every set that is a countable union of closed sets in R is measurable.

PROOF.

[1] We have already shown that for each a R, the interval .a; / is measurable. It follows from the fact that M is an algebra that . ; a 2.a; /c is measurable1 for each a R. 1 D 1 2 Since Œa; / a .a; / and m. a / 0, it follows from Lemma 2.4.5 that a is measurable. Therefore,1 D fŒa;g [/ is measurable.1 f Sinceg DM is an algebra, we have that . ; a/f g Œa; /c is measurable for each1a R. 1 D 1 2 If a and b are real numbers such that a < b, then .a; b/ . ; b/ .a; /. Since . ; b/ and .a; / are both measurable, it follows thta .a; b/ is also measurable.D 1 \ 1 1 1 If a and b are real numbers such that a < b, then Œa; b/ a .a; b/, .a; b .a; b/ b , and Œa; b a .a; b/ b . Therefore, the sets Œa; b/, .a; bD, andf g [Œa; b are all measurable.D [f g Df g [ [f g

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[2] Every open set in R is a countable union of open intervals. Since each open interval is measurable and M is a -algebra, it follows that the union of open intervals is also measurable. [3] A closed set F in R is a complement of an open set V in R. That is F V c for some open set V in R. Since V is measurable, V c F is also measurable. D D 1 [4] Let U iD1 Vi , where Vi is open in R for each i N. Then by [2], Vi is measurable, for each i N, andD so is V c for each i N. Since M is a -algebra,2 we have that 2 T i 2 1 1 c c U Vi V D D i i\D1 Â i[D1 Ã is measurable.

1 c [5] Let F iD1 Fi , where Fi is closed in R, for each i N. Then Fi is open in R, for each D 1 c 2 M i N. By [4], we have that iD1 Fi is measurable. Since is a -algebra, it follows that 2 Sc 1 c 1 F Fi F Tis measurable. iD1 i D iD1 D Â Ã T S 2 2.5.11 Deﬁnition (a) A set that is a union of a countable collection of closed sets is called a F -set.

(b) A set that is an intersection of a countable collection of open sets is called a Gı-set. 2.5.12 Remark [1] The previous corollary says, amongst other things, that every F -set and every Gı-set is measurable. [2] The previous corollary also states that all ‘nice’ sets in R are measurable. There are however sets that are not measurable. The point being that these sets are certainly not ‘natural’ but somewhat strange. [3] The previous corollary shows that every open set in R is measurable. The smallest -algebra containing all the open sets is called the Borel -algebra.

The following result states that any measurable subset of R can be approximated by the ‘nice’ ones. 2.5.13 Proposition Let E R and m the Lebesgue outer measure on R. The following statements are equivalent: [1] E is measurable. [2] For each > 0, there is an open set V such that E V and m.V E/ < . n [3] There is a Gı-set G such that E G and m .G E/ 0. n D [4] For each > 0, there is a closed set F such that F E and m.E F/ < . n [5] There is an F -set H such that H E and m .E H/ 0. n D PROOF.

[1] [2]: Suppose that E is measurable and let > 0. If m.E/ < , then by Proposition 2.5.8, there is an ) open set V such that E V and 1 m.V / < m.E/ : C Since V and E are measurable, we can rewrite the equation above as

m.V / < m.E/ : C

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Hence, m.V E/ m.V / m.E/ < . Suppose that m.E/ . Write R In, a union n D D 1 D n2N of disjoint ﬁnite intervals. For each n N, let En E In. Then E En, and for each 2 D \ D n2NS n N, m.En/ < , Therefore, by the ﬁrst part, there is an open set Vn such that En Vn and 2 1 S m.Vn En/ < n . Write V Vn. Then V is an open set and n 2 D n2N S V E Vn En .Vn En/: n D n D n n2N n2N n2N [ [ [ Hence m.V E/ m.Vn En/ < : n Ä n 2n D n2N n2N X X 1 [2] [3]: For each n N, let Vn be an open set such that E Vn and m .Vn E/ < . Let G Vn. ) 2 n n D n2N Then G is a Gı-set, E G, and for each n N, 2 T

1 m .G E/ m .Vn E/ m .Vn E/ : n D n Ä n Ä n n2N Â \ Ã Thus, m.G E/ 0: n D [3] [1]: Since G is a Gı-set, it is measurable. Also G E is measurable since m .G E/ 0. Therefore, the ) set E G .G E/ is also measurable. n n D D n n [1] [4]: Suppose that E is measurable and let > 0. Then Ec is also measurable. Since [1] implies [2], ) we have, by [2], that there is an open set V such that Ec V and m.V Ec / < . Note that V Ec E V E V c . Take F V c . Then F is a closed set, F E, andn m.E F/ < . n D \ D n D n 1 [4] [5]: For each n N, let Fn be a closed set such that Fn E and m .E Fn/ < . Let H Fn. ) 2 n n D n2N Then H is an F -set, H E, and for each n N, 2 S

1 m .E H/ m E Fn .E Fn/ m .E Fn/ < : n D n D n Ä n n n2N n2N Â [ Ã \ Thus, m.E H/ 0. n D [5] [1]: Since H is an F -set, it is measurable. Also E H is measurable since m .E H/ 0. Therefore, ) the set E .E H/ H is also measurable. n n D D n [ 2

30 Chapter 3

The Lebesgue integral

3.1 Introduction

We now turn our attention to the construction of the Lebesgue integral of general functions, which, as already discussed, is necessary to avoid the technical deﬁciencies associated with the Riemann integral.

3.2 Measurable functions 3.2.1 Basic notions The extended real number system, R, is the set of real numbers together with two symbols and . That is, R R ; Œ ; . The algebraic operations for these two infinities1 are: C1 D [ f1 C1g D 1 1 [1] For r R, r . 2 ˙1C D ˙1 [2] For r R, r. / , if r > 0 and r. / if r < 0. 2 ˙1 D ˙1 ˙1 D 1 [3] . / and . / . C1C C1 D C1 1 C 1 D 1 [4] . / is undefined. 1C 1 [5] 0 . / 0. ˙1 D 3.2.1 Definition Let .X; †/ be a measurable space and E †. A function f E R is said to be measurable if for each ˛ R, the set x E f .x/>˛ is measurable.2 W ! 2 f 2 W g 3.2.2 Proposition Let .X; †/ be a measurable space and E †, and f R. Then the following are equivalent: 2 ! [1] f is measurable. [2] For each ˛ R, the set x E f .x/ ˛ is measurable. 2 f 2 W g [3] For each ˛ R, the set x E f .x/<˛ is measurable. 2 f 2 W g [4] For each ˛ R, the set x E f .x/ ˛ is measurable. 2 f 2 W Ä g PROOF.

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[1] [2]: If f is measurable, then the set ) 1 x E f .x/ ˛ x E f .x/>˛ ; f 2 W gD 2 W n n2N \ which is an intersection of measurable sets, is measurable. [2] [3]: If the set x E f .x/ ˛ is measurable, then so is the set ) f 2 W g E x E f .x/ ˛ x E f .x/<˛ : nf 2 W gDf 2 W g [3] [4]: If the set x E f .x/<˛ is measurable, then so is the set ) f 2 W g 1 x E f .x/ ˛ x E f .x/<˛ f 2 W Ä gD 2 W C n n2N \ since it is an intersection of measurable sets. [4] [1]: If the set x E f .x/ ˛ , is measurable, so is its complement in E. Hence, ) f 2 W Ä g x E f .x/>˛ E x E f .x/ ˛ f 2 W gD nf 2 W Ä g is measurable. It follows from this that f is measurable. 2 3.2.3 Examples [1] The constant function is measurable. That is, if .X; †/ is a measurable space, c R, and E †, then the function f E R given by f .x/ c, for each x E, is measurable. 2 2 W ! D 2 Let ˛ R. If ˛ c, then the set x E f .x/>˛ , and is thus measurable. 2 f 2 W gD; If ˛ < c, then the set x E f .x/>˛ E, and is therefore measurable. Hence, f is measurable. f 2 W g D [2] Let .X; †/ be a measurable space and let A †. The characteristic function, , is deﬁned by 2 A

1 if x A .x/ 2 A D 0 if x A: 62

The characteristic function A is measurable. This follows immediately from the observation that, for each ˛ R and E †, the set x E >˛ is either E, A or . 2 2 f 2 W A g ; [3] Let B be the Borel -algebra in R and E B. Then, any continuous function f E R is measurable. This is an immediate consequence2 of the fact that, if f E R is continuousW ! and ˛ R, then the set x E f .x/>˛ is open and hence belongs to BW. ! 2 f 2 W g 3.2.4 Proposition [1] If f and g are measurable real-valued functions deﬁned on a common domain E † and c R, then the functions 2 2 (a) f c, C (b) cf , (c) f g, ˙ (d) f 2,

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(e) f g, (f) f , j j (g) f g, _ (h) f g are also measurable. ^ [2] If .fn/ is a sequence of measurable functions deﬁned on a common domain E †, then the functions 2 (a) supn fn,

(b) infn fn,

PROOF.

[1] (a) For any real number ˛, x E f .x/ c >˛ x E f .x/>˛ c : f 2 W C gf 2 W g Since the set on the right-hand side is measurable, we have that f c is measurable. C (b) If c 0, then cf is obviously measurable. Assume that c < 0. Then, for each real number ˛, D ˛ x E cf .x/>˛ x E f .x/ < : f 2 W gD 2 W c Since the set x E f .x/ < ˛ is measurable, it follows that cf is also measurable. f 2 W c g (c) Let ˛ be a real number. Since the rationals are dense in the reals, there is a rational number r such that f .x/ < r <˛ g.x/; whenever f .x/ g.x/<˛. Therefore, C x E f .x/ g.x/<˛ . x E f .x/ < r x E g.x/<˛ r /: f 2 W C gD f 2 W g\f 2 W g r2Q [ Since the sets x E f .x/ < r and x E g.x/<˛ r are measurable, so is the set x E f .x/f <2r W x g.x/<˛g rf , and2 consequently,W theg set x E f .x/ g.x/ < ˛f ,2 beingW a countableg\f unionW of measurable g sets, is also measurable. f 2 W C g If g is measurable, it followsfrom (b) that . 1/g is also measurable. Hence, so is f . 1/g f g. C D (d) Let ˛ R and E †. If ˛ < 0, then x E f 2.x/>˛ E, which is measurable. 2 2 f 2 W gD If ˛ 0, then x E f 2.x/>˛ x E f .x/ > p˛ x E f .x/ < p˛ : f 2 W gDf 2 W g[f 2 W g Since the two sets on the right hand side are measurable, it follows that the set x E f 2.x/>˛ is also measurable. Hence f 2 is measurable. f 2 W g (e) Since f g 1 Œ.f g/2 .f g/2, it followsfrom (b), (c), and (d) that f g is measurable. D 4 C (f) Let ˛ R and E †. If ˛ < 0, then x E f .x/ >˛ E, which is measurable. 2 2 f 2 Wj j gD If ˛ 0, then x E f .x/ >˛ x E f .x/>˛ x E f .x/ < ˛ : f 2 Wj j gDf 2 W g[f 2 W g Since the two sets on the right hand side are measurable, it follows that the set x E f .x/2 >˛ is also measurable. Thus, f is measurable. f 2 W g j j

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1 (g) It is sufﬁcient to observe that f g 2 f g f g . It now follows from (b), (c), and (d) that f g is measurable. _ D f C Cj jg _ 1 (h) It is sufﬁcient to observe that f g 2 f g f g . It now follows from (b), (c), and (d) that f g is measurable. ^ D f C j jg ^ [2] (a) Let ˛ R. Then 2 1 x E sup fn >˛ x E fn.x/>˛ : f 2 W n gD f 2 W g n[D1 Since for each n N, fn is measurable, it follows that the set x E fn.x/>˛ is 2 f 2 W g measurable for each n N. Therefore, the set x E supn fn.x/>˛ is measurable as it is a countable union of measurable2 sets. f 2 W g

(b) It is sufﬁcient to note that infn fn sup . fn/. D n

(d) Notice that liminfn fn sup infkn fk and use (a) and (b) above. D n1 Â Ã 2 3.2.5 Corollary Let .X; †/ be a measurable space. If f is a pointwise limit of a sequence .fn/ of measurable functions deﬁned on a common domain E †, then f is measurable. 2 PROOF.

If the sequence .fn/ converges pointwise to f , then, for each x E, 2

f .x/ lim fn.x/ limsup fn.x/: D n!1 D n Now, by Proposition 3.2.4 [2], f is measurable. 2 3.2.6 Definition Let f be a real-valued function defined on a set X . The positive part of f , denoted by f C, is the function f C max f; 0 f 0 and the negativepart of f , denoted by f , is thefunction f max f; 0 . fD / 0. f gD _ D f gD _ Immediately we have that if f is a real-valued function defined on X , then

f f C f and f f C f : D j jD C Note that 1 1 f C Œ f f and f Œ f f : D 2 j jC D 2 j j Let .X; †/ be a measurable space and f X R. It is trivial to deduce from Proposition 3.2.4 that f is measurable if and only if f C and f areW measurable.! 3.2.7 Deﬁnition n Let .X; †/ be a measurable space. A simple function on X is a function of the form cj , D jD1 Ej where, for each j 1; 2;:::n, cj is an extended real number and Ej †. D 2 P Since is measurable for each j 1; 2;:::; n, it follows from Proposition 3.2.4 that is also Ej D measurable.

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The representation of inDeﬁnition3.2.7 is notunique. If assumes distinct nonzero values a1; a2;:::; an, then n aj ; D Aj jXD1 where Aj x X .x/ aj ; j 1; 2;:::; n . This is called the cannonical representation of . D f 2 W D D g n Note that the sets Aj are pairwise disjoint and X iD1 Aj . What Corollary 3.2.5 is saying is that the pointwiseD limit of a sequence of simple functions is a measurable function. The converse also holds. That is,S every measurable function is a pointwise limit of a sequence of simple functions which we state as a theorem. 3.2.8 Theorem Let .X; †/ be a measurable space and f a nonnegative measurable function. Then there is a monotonic increasing sequence .n/ of nonnegative simple functions which converge pointwise to f .

PROOF. For each n N and for k 1; 2;:::; n2n, let 2 D k 1 k Enk x X f .x/ < ; and D 2 W 2n Ä 2n Fn x X f .x/ n : Df 2 W g n Then Enk k 1; 2;::: n2 Fn n N is a collection of pairwise disjoint measurable sets whose unionf is X . W D g[f W 2 g For each n N, deﬁne n by 2 n n2 k 1 n n : D 2n Enk C Fn kXD1 Then the n are nonnegative simple functions, n f , and n nC1, for each n N. If x X is such that f .x/ < , then, if f .x/ < n, Ä Ä 2 2 1 k .k 1/ 1 0 f .x/ n.x/ < : Ä 2n 2n D 2n 1 Since limn!1 2n 0, we have that limn!1 n.x/ n, and so limn!1 n.x/ f .x/: D D D 2 3.2.9 Corollary Let .X; †/ be a measurable space and f a measurable function. Then there is a sequence .n/ of simple functions which converge pointwise to f .

PROOF. The functions f C and f are both nonnegative and measurable. By Theorem 3.2.8, there exist two C sequences .n/ and . n/ of nonnegative simple functions which converge pointwise to f and f respectively. The sequence .n/, where, for each n N, n n n, is a sequence of simple functions which converges pointwise to f C f f . 2 D D 2 3.2.10 Deﬁnition Let .X;†;/ be a measure space and E †. A property P.x/, where x E, is said to hold almost everywhere (abbreviated, a.e.) on E if the2 set N x E P.x/ fails belongs2 to † and .N / 0. Df 2 W g D 3.2.11 Examples [1] f g a.e. if . x X f .x/ g.x/ / 0. D f 2 W 6D g D

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[2] f g a.e. if . x X f .x/ > g.x/ / 0. Ä f 2 W g D 3.2.12 Definition Let .X;†;/ be a measure space. A function f X R is said to be almost everywhere real-valued, denoted by a.e. real-valued, if W ! . x X f .x/ / 0: f 2 Wj jD1g D We call a set of measure zero a null set. 3.2.13 Definition A measure space .X;†;/ is said to be complete if † contains all subsets of sets of measure zero. That is, if E †, .E/ 0 and A E, then A †. 2 D 2 It follows from Theorem 2.3.6 [1] that if a measure space .X;†;/ is complete, E †, .E/ 0 and A E, then .A/ 0. 2 D D 3.2.14 Proposition Let .X;†;/ be a complete measure space and f g a.e. If f is measurable on E †, then so is g. D 2 PROOF. Let ˛ R and N x E g.x/ f .x/ . Then N † and .N / 0. Now, 2 Df 2 W 6D g 2 D x E g.x/>˛ x E N g.x/>˛ x N g.x/>˛ f 2 W gDf 2 n W g[f 2 W g x E N f .x/>˛ x N g.x/>˛ : Df 2 n W g[f 2 W g The first set on the right hand side is measurable since f is measurable. The second set is measurable since it is a subset of the null set N and the measure is complete. 2

3.2.2 Convergence of sequences of measurable functions We now consider various notions of convergence of a sequence of measurable functions examine the rela- tionships that exist between them. 3.2.15 Deﬁnition Let .X †;/ be a measure space. A sequence .fn/ of a.e. real-valued measurable functions on X is said to a.e. [1] converge almost everywhere to an a.e. real-valued measurable function f . denoted by fn f , if for each > 0 and each x X , there is a set E † and a natural number N N./ such! that .E/ < and 2 2 D fn.x/ f .x/ < , for each x X E and each n N: j j 2 n a.u. [2] converge almost uniformly to an a.e. real-valued measurable function f , denoted by fn f , if for each > 0, there is a set E † and a natural number N N./ such that .E/ < !and 2 D fn f 1 sup fn.x/ f .x/ < , for each n N: k k D x2X nE j j

[3] converge in measure to an a.e. real-valued measurable function f , denoted by fn f , if for every > 0, ! lim . x X fn.x/ f .x/ / 0: n!1 f 2 Wj j g D (if is a probability measure, then this mode of convergence is called convergence in probability). 3.2.16 Proposition Let .X;†;/ be a complete measure space and .fn/ a sequence of measurable functions on E † which converges to f a.e. Then f is measurable on E. 2

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PROOF.

Let ˛ R and N x E limn!1 fn.x/ f .x/ . Then N †;.N / 0 and 2 Df 2 W 6D g 2 D limn!1 fn.x/ f .x/, for each x E N . Since E and N are measurable, so is E N . For each D 2 n n n N, deﬁne gn and g by 2

0 if x N gn.x/ 2 D fn.x/ if x E N : 2 n and

0 if x N g.x/ 2 D f .x/ if x E N : 2 n Then gn fn a.e. and f g a.e. It follows from Proposition3.2.4, that for each n N, gn is measurable. If x ND, then D 2 2 lim gn.x/ lim 0 0 g.x/; (3.1) n!1 D n!1 D D and if x E N , then 2 n lim gn.x/ lim fn 0 f .x/ g.x/ (3.2) n!1 D n!1 D D D It follows from (3.1) and (3.2) that the sequence .gn/ converges pointwise to g everywhere on E. ByCorol- lary 3.2.5, that g is measurable. Since g f a.e., we have, by Proposition 3.2.14, that f is measurable. D 2 3.2.17 Theorem Let .X;†;/ be a measure space and .fn/ be a sequence of real-valued measurable functions on X . If the sequence .fn/ converges almost uniformly to f , then it converges in the measure to f . That is, almost uniform convergence implies convergence in the measure.

PROOF.

Let > 0 be given. Since the sequence .fn/ converges almost uniformly to f , there is a measurable set E and a natural number N such that .E/ < and

fn.x/ f .x/ < , for all x X E and all n N: j j 2 n It now followsthat, for all n N; x X fn.x/ f .x/ E. Therefore, for all n N, f 2 Wj j g 2 . x X fn.x/ f .x/ /<: f 2 Wj j g

2 3.2.18 Theorem Let .X;†;/ be a measure space and .fn/ be a sequence of a.e. real-valued measurable functions on X . If the sequence .fn/ converges almost uniformly to f , then it converges almost everywhere to f . That is, almost uniform convergence implies convergence almost everywhere.

PROOF.

Suppose that the sequence .fn/ converges almost uniformly to f . Then, for each n N, there is a 1 2 1 measurable set En, with .En/ < n such that fn f uniformly on X En. Let A nD1.X En/. Then ! n D n 1 1 S .X A/ En .En/ 0: n D Ä D n ! Â n\D1 Ã

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a.e. That is, .X A/ 0. Furthermore, for each x A, fn.x/ f .x/ as n . Hence, fn f . n D 2 ! ! 1 ! 2 In general, the converses of the previous two theorems do not hold. The following is an example of a sequence that converges almost everywhere and in measure, but does not converge almost uniformly. 3.2.19 Example

Let X Œ0; / with the Lebesgue measure . For each n N, let fn 1 , i.e., D 1 2 D Œn;nC n

1 1 if x Œn; n n fn.x/ 2 C D 0 otherwise: n!1 Then, for each x X , fn.x/ 0. That is, the sequence .fn/ converges pointwise on X to the zero 2 ! function. Since x X fn.x/ 0 , it follows that . x X fn.x/ 0 / 0, Hence, a.e. f 2 W 6! gD; f 2 W 6! g D fn 0. ! The sequence .fn/ converges in measure to the zero function. Indeed, for each > 0,

1 n!1 . x X fn.x/ 0 / 0: f 2 Wj j g D n !

We show that the sequence .fn/ does not converge almost uniformly to the zero function. Assume, on the contrary that .fn/ converges almost uniformly to the zero function. Then, there is a Lebesgue measurable c set E with .E/ 0 such that fn 0 such that fn 0 uniformly on E . Therefore, with 1, there is an N N suchD that for all x E!c and all n N , ! D 2 2

fn 0 1 fn 1 < 1 fn.x/ fn.x/ < 1: k k Dk k ,j jD 1 c For each n N, let An Œn; n . Then, E An , and so An E. Thus, 2 D C n \ nN D; nN S 1 1 S 1 0 .E/ An .An/ ; D D D n D 1 Â n[N Ã nXDN nXDN

which is absurd. Therefore .fn/ does not converge almost uniformly to the zero function. 3.2.20 Theorem (Egoroff’s Theorem) Let .X;†;/ be a ﬁnite measure space and .fn/ be a sequence of a.e. real-valued measurable functions on X . If the sequence .fn/ converges almost everywhere to f , then it converges almost uniformly to f .

PROOF.

Without loss of generality, we assume that .fn/ converges to f everywhere on X . For m; n N , let 2 1 m 1 E x X fk .x/ f .x/ : n D f 2 Wj j m g k[Dn m m m m Then, for m; n N, En is a measurable set and EnC1 En . That is, .En /n is a decreasing sequence of 2 1 n!1 m measurable subsets of X . Since fn.x/ f .x/ for each x X , we have that E . Also, ! 2 n D; nD1 since .X / < , we have that .Em/ < . By Theorem 2.3.6 [4], it followsthat \ 1 1 1 1 0 . / Em lim .Em/: D ; D n D n!1 n Â n\D1 Ã

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1 m ı m For ı > 0, let km N be such that .E / < m . Let Eı E . Then, Eı is a measurable set and 2 km 2 D km m[D1 1 m .Eı/ .E /<ı: Ä km mXD1 Notice that 1 1 m m X Eı X E .X E / n D n km D n km m[D1 m\D1 1 1 1 X x X fk .x/ f .x/ D n 2 Wj j m m\D1 Ä kD[km 1 1 1 x X fk .x/ f .x/ < : D 2 Wj j m m\D1 k\Dkm 1 Let > 0. Choose m N such that < . Then for any x X Eı and any k km, 2 m 2 n 1 fk .x/ f .x/ < < : j j m

That is, .fn/ converges uniformly to f on X Eı. n 2 3.2.21 Corollary Let .X;†;/ be a ﬁnite measure space and .fn/ be a sequence of a.e. real-valued measurable functions on X . If the sequence .fn/ converges almost everywhere to f , then it converges in measure to f .

PROOF. This follows from Theorem 3.2.17 and Theorem 3.2.20. 2 The hypothesis that .X / < in Egoroff’s Theorem is essential. Let X R with the Lebesgue 1 D measure . For each n N, let fn , i.e., 2 D Œn;1/

1 if x Œn; / fn.x/ 2 1 D 0 otherwise: n!1 Then, for each x R, fn.x/ 0. That is, the sequence .fn/ converges pointwise on R to the zero 2 ! a.e. function. Since x R fn.x/ 0 , it follows that . x R fn.x/ 0 / 0. Thus, fn 0. f 2 W 6! gD; f 2 W 6! g D ! We show that the sequence .fn/ does not converge almost uniformly to the zero function. Assume, on the contrary, that .fn/ converges almost uniformly to the zero function. Then, there is a Lebesgue measurable c set E with .E/ 0 such that fn 0 uniformly on E . Therefore, with 1, there is an N N such that for all x EcDand all n N , ! D 2 2

fn 0 1 fn 1 < 1 fn.x/ fn.x/ < 1: k k Dk k ,j jD c c For each n N, let An Œn; /. Then, E An E AN , and so AN E. 2 D 1 \ D \ D; n[N Hence, 0 .E/ .AN / , which is a contradiction. Therefore .fn/ does not converge almost uniformly to theD zero function. D 1

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3.2.22 Theorem (F. Riesz) Let .X;†;/ be a measurable space and .fn/ be a sequence of a.e. real-valued measurable functions on X .

If the sequence .fn/ converges in measure to f , then there is a subsequence .fnk / of .fn/ which converges almost everywhere to f .

PROOF.

Let n1 N such that for all n n1, 2 1 . x X fn.x/ f .x/ 1 / < : f 2 Wj j g 2

Now, choose n2 N such that n2 > n1 and for all n n2, 2 1 1 x X fn.x/ f .x/ < : f 2 Wj j 2g 22 Â Ã

Next, choose n3 N such that n3 > n2 and for all n n3, 2 1 1 x X fn.x/ f .x/ < : f 2 Wj j 3g 23 Â Ã

Continue this process obtaining an increasing sequence .nk / of natural numbers with 1 1 x X fn .x/ f .x/ < f 2 Wj k j k g 2k Â Ã For each k N, let 2 1 Ak x X fn .x/ f .x/ : Df 2 Wj k j k g 1 1 Since, for each k N, .Ak / < , we have that the series .Ak / converges. By the Borel-Cantelli 2 2k kXD1 Lemma (Theorem 2.3.7), .limsup Ak / 0. Let D 1 1

A limsup Ak Aj : D D k\D1 j[Dk

Choose x X A. Then there is an index jx N such that x X Aj , Note that 2 n 2 2 n x 1 1 X Aj X x X fn .x/ f .x/ n x D n f 2 Wj k j k g k[Djx 1 1 X x X fn .x/ f .x/ D nf 2 Wj k j k g k\Djx Ä 1 1 x X fn .x/ f .x/ : D f 2 Wj k j k g k\Djx 1 1 Hence, for all k jx; fn .x/ f .x/ < . Let > 0 be given. Choose k0 jx such that < . Then, j k j k k0 for all k k0, 1 fn .x/ f .x/ < < : j k j k a.e. Therefore, for each x X A, fnk .x/ f .x/, and hence, fnk f . 2 n ! ! 2

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3.3 Deﬁnition of the integral

In this section, we define a Lebesgue integral. This is done in three stages: firstly, we define the integral of a nonnegative simple function; then, using the integral of a nonnegative simple function, we define the integral or any measurable function. We also prove three key results: Monotone Convergence Theorem, Dominated Convergence Theorem and Fatou’s Lemma. Unless otherwise specified, we will work in the measure space .X;†;/.

3.3.1 Integral of a nonnegative simple function

3.3.1 Deﬁnition n Let be a nonnegative simple function with the canonical representation ai . The Lebesgue D Ei XiD1 integral of with respect to , denoted by X d, is the extended real number

R n d ai .Ei /: D XZ XiD1 If A †, we deﬁne 2 d d: D A AZ XZ

The function is integrable if X d is ﬁnite. R We also write d for X d. It is straightforward to show that if A is a measurable set and is as above, then R R n d ai .A Ei/: D \ AZ XiD1 Furthermore, d d d .A/: D A D A D AZ AZ XZ It is necessary to show that the above deﬁnition of the Lebesgue integral is unambiguous, i.e, the integral is independent of the representation of . Assume that

n m ai bj ; D Ei D Fi XiD1 jXD1 n

where Ei Ek , for all 1 i k n, X Ei; Fk Fl , for all 1 j l m, and \ D; Ä 6D Ä D \ D; Ä 6D Ä iD1 m [

Fj . Then for each i 1; 2;:::; n and each j 1; 2;:::; m, D D j[D1 m m Ei Ei X Ei Fj .Ei Fj /, a disjoint union, and D \ D \ D \ Âj[D1 Ã j[D1 n n

Fj Fj X Ei .Ei Fj /, a disjoint union. D \ D D \ Â i[D1 Ã i[D1

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If Ei Fj , then, for x Ei Fj , ai .x/ bj . That is, ai bj in this case. If Ei Fj , \ 6D ; 2 \ D D D \ D; then .Ei Fj / 0. Thus, ai .Ei Fj / bj .Ei Fj /, for all 1 i n and for all 1 j m. Therefore \ D \ D \ Ä Ä Ä Ä n n m n m ai .Ei / ai .Ei Fj / ai .Ei Fj / D \ D \ XiD1 XiD1 Âj[D1 Ã XiD1 jXD1 n m n m

ai .Ei Fj / bj .Ei Fj / D \ D \ XiD1 jXD1 XiD1 jXD1 m n m n bj .Ei Fj / bj .Ei Fj / D \ D \ jXD1 XiD1 jXD1 Â i[D1 Ã m

bj .Fj /: jXD1 Hence the deﬁnition of the integral of is unambiguous. 3.3.2 Proposition Let .X;†;/ be a measure space, and nonnegative simple functions, and c a nonnegative real number. Then [1] . /d d d. X C D X C X [2] R cd c Rd: R X D X [3] IfR , thenR d d. Ä X Ä X [4] If A and B are disjointR measurableR sets, then

d d d: D C A[ZB AZ BZ

[5] The set function † Œ0; deﬁned by W ! 1 .A/ d; D AZ for A †, is a measure on X . 2 [6] If A † and .A/ 0, then d 0. 2 D A D PROOF. R n m Let ai , and bj be canonical representations of and respectively. iD1 Ei jD1 Fj D D n Then Ei Ek , for all i i k n, Fj Fl , for all 1 j l m, X iD1Ei, and m \ PD; Ä 6D PÄ \ D; Ä 6D Ä D [ X Fj . D [jD1 [1] For 1 i n and 1 j m, let Gij Ei Fj . Then, for each x Gij , .x/ .x/ ai bj , Ä Ä Ä Ä D \ 2 C D C i.e., the function takes the values ai bj on Ei Fj . For all 1 i; k n and 1 j ; l m, C C \ Ä Ä Ä Ä

Gij Gkl .Ei Fj / .Ek Fl / \ D \ \ \ .Ei Ek / .Fj Fl / D \ \ \ : D;\;D;

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That is, Gij 1 i n and 1 j m is a collection of nm pairwise disjoint sets. Furthermore, f W Ä Ä Ä Ä g m m m Ei Ei X Ei Fj .Ei Fj / Gij ; D \ D \ D \ D Âj[D1 Ã j[D1 j[D1 n n n Fj Fj X Fj Ei .Ei Fj / Gij , and D \ D \ D \ D Â i[D1 Ã i[D1 i[D1 n m n m n m

X X X Ei Fj .Ei Fj / Gij D \ D \ D \ D Â i[D1 Ã Â j[D1 Ã i[D1 j[D1 i[D1 j[D1 Therefore, n m

.ai bj / : C D C Gij XiD1 jXD1 With this representation of , the numbers ai bj are not necessarily distinct. Let c1; c2;:::; ck be the distinct values assumedC by . Then C C k . /d cr . x .x/ .x/ cr / C D f W C D g XZ rXD1 k cr Ei Fj D \ ! rXD1 ai C[bj Dcr k cr .Ei Fj / D \ rXD1 ai CXbj Dcr n m .ai bj /.Gij / D C XiD1 jXD1 n m n m ai .Gij / bj .Gij / D C XiD1 jXD1 XiD1 jXD1 n m n m

ai .Gij / bj .Gij / D C XiD1 jXD1 XiD1 jXD1 n m ai .Ei / bj .Fj / D C XiD1 jXD1 d d: D C XZ XZ

[2] If c 0, then c vanishes identically and hence, X cd c X d. Assume that c > 0. D n D n If ai is the canonical representation ofR, then c R cai is the canonical D Ei D iD1 Ei iD1 representationX of c. Therefore, P

n n cd cai .Ei / c ai .Ei / c d: D D D XZ XiD1 XiD1 XZ

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[3] As shown above,

n n m n m ai .Ei / ai Ei Fj ai .Ei Fj /, and D D \ D \ XZ XiD1 XiD1 Âj[D1 Ã XiD1 jXD1 m m n m n bj .Fj / bj Fj Ei bj .Ei Fj /: D D \ D \ XZ jXD1 jXD1 Â i[D1 Ã jXD1 XiD1

Now suppose that . Then, for each x Ei Fj , ai .x/ .x/ bj . i.e., ai bj , for Ä 2 \ D Ä D Ä all i; j such that Ei Fj . It follows that d d. \ 6D ; X Ä X [4] Let be as above. Then R R

n d ai ..A B/ Ei/ D [ \ A[ZB XiD1 n ai Œ.A Ei/ .B Ei / D \ [ \ iD1 Xn ai Œ.A Ei / .B Ei / D \ C \ iD1 Xn n ai .A Ei / .B Ei/ D \ C \ XiD1 XiD1 d d: D C AZ BZ

[5] We have that, for A †, 2 n .A/ d ai .A Ei /: D D \ AZ XiD1 If A , then A Ei , for each i 1; 2;:::; n. Hence, .A Ei/ 0, for i 1; 2;:::; n and soD;. / 0. \ D; D \ D D ; D 1

Let .Ak / be a sequence of pairwise disjoint measurable sets and A Ak : Then, for each i D D kD1 1; 2;:::; n, [ 1 1

Ei A Ei Ak .Ei Ak /, a disjointunion. \ D \ D \ Â k[D1 Ã k[D1

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Thus, n n

.A/ ai .A Ei/ D \ XiD1 XiD1 n 1

ai Ei Ak D \ XiD1 Â k[D1 Ã n 1

ai .Ei Ak / D \ XiD1 kXD1 1 n

ai .Ei Ak / D \ kXD1 XiD1 1

.Ak /: D kXD1 1 1

That is, Ak .Ak /: D Â k[D1 Ã kXD1 [6] We have by [5], that

n n 0 d ai .A Ei / ai.A/ 0: Ä D \ Ä D AZ XiD1 XiD1 Thus, d 0. A D R 2 The next result asserts that changing a simple function on a null set does not change the integral. 3.3.3 Corollary Let .X;†;/ be a measure space and a nonnegative simple function. If A and B are measurable sets such that A B and .B A/ 0, then Â n D d d: D AZ BZ

PROOF. Noting that B A .B A/, a disjoint union, we have by Proposition3.3.2 [4] and [6], that D [ n d d d d d 0 d: D D C D C D BZ A[.BZnA/ AZ BZnA AZ AZ

3.3.2 Integral of a nonnegative measurable function We showed in Theorem 3.2.8 that every nonnegative measurable function is a pointwiselimit of an increasing sequence of of nonnegative simple functions. Using this fact, we are able to deﬁne the integral of a nonnegative measurable function using the integrals of nonnegative simple functions.

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3.3.4 Deﬁnition Let f be a nonnegative measurable function. The Lebesgue integral of f with respect to , denoted by

X f d, is defined as R f d sup d 0 f; is a simple function : D W Ä Ä XZ XZ If E †, then we define the Lebesgue integral of f over E with respect to as 2 f d f d: D E EZ Z Notice that if f is a nonnegative simple function, then Definitions 3.3.1 and 3.3.4 coincide. 3.3.5 Proposition Let f and g be nonnegative measurable functions and c a nonnegative real number. [1] cf d c f d: X D X [2] IfR f g, thenR f d gd: Ä X Ä X [3] If A and B areR measurableR sets such that A B, then Â f d f d: Ä AZ BZ

PROOF.

[1] If c 0, then the equality holds trivially. Assume that c > 0. Then D cf d sup d 0 cf; a simple function D W Ä Ä XZ XZ sup c d 0 f; a simple function D c W Ä c Ä XZ c sup d 0 f; a simple function D c W Ä c Ä XZ c f ı: D XZ

[2] Since f g, it followsthat 0 f; a simple function 0 g; a simple function . Thus, Ä f W Ä Ä gf W Ä Ä g f d sup d sup d gd: D 0ÄÄf Ä 0ÄÄg D XZ XZ XZ XZ

[3] If A B, then A B . Therefore, for any nonnegative measurable function f , we have that f Â f . By [2],Ä it follows that A Ä B

f d f d f d B f d: D A Ä B D2 AZ XZ XZ

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2 3.3.6 Proposition (Chebyshev’s inequality) Let .X;†;/ be a measure space, f a nonnegative measurable function and c a positive real number. Then

1 . x X f .x/ c / f d: f 2 W g Ä c XZ

PROOF. Let A x X f .x/ c . Then, since c f on A and A X , Df 2 W g Ä f f f: A Ä A Ä X D By Proposition 3.3.5, we have that cd f d A Ä XZ XZ c d f d ) A Ä XZ XZ c d f d ) Ä AZ XZ c.A/ f d ) Ä XZ 1 .A/ f d ) Ä c XZ 1 i.e., . x X f .x/ c / c X f d: f 2 W g Ä 2 R 3.3.7 Proposition

Let .X;†;/ be a measure space and f a nonnegative measurable function on X . Then, X f d 0 if and only if f 0 a.e. on X . D D R PROOF.

1 Assume that f d 0. For each n N, let En x X f .x/ : Then by Chebyshev’s X D 2 D 2 W n inequality (PropositionR 3.3.6),

0 .En/ n f d 0: Ä Ä D XZ

Hence, .En/ 0 for each n N. Since En EnC1, for each n N and D 2 2 1

E x X f .x/ > 0 En; Df 2 W gD n[D1 it follows, by Theorem 2.3.6, that

.E/ En lim .En/ 0: D D n!1 D Â n[D1 Ã That is, f 0 a.e. on X . D

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n Conversely, assume that f 0 a.e. on X . Let ai be a nonnegative simple function, D D Ei XiD1 written in canonical form, such that f . Then 0 a.e. on X . Since Ei x X .x/ ai; i n Ä D Df 2 W D D 1; 2;:::; n ; we have that Ei 0. But since g D Â i[D1 Ã n n Ei .Ei /; D Â i[D1 Ã XiD1 we have that .Ei / 0, for each i 1; 2;:::; n. Therefore, D D n d ai .Ei / 0: D D XZ XiD1 It then follows that f d sup d 0: D 0ÄÄf D XZ XZ 2 3.3.8 Proposition Let .X;†;/ be a measure space and f a nonnegative measurable function. If A † and .A/ 0, then 2 D f d 0: D AZ

PROOF. Let be a nonnegative simple function such that f . Then, by Proposition 3.3.2 [6], we have that d 0. Therefore Ä A D R f d sup d sup 0 0: D 0ÄÄf D 0ÄÄf f gD AZ AZ 2 3.3.9 Theorem (Monotone convergence theorem) Let .X;†;/ be a measure space, .fn/ a sequence of measurable functionson X such that 0 fn fnC1, Ä Ä for every n N and limn!1 fn.x/ f .x/, for each x X . Then f is measurable and 2 D 2

f d lim fn lim fnd: D n!1 D n!1 Z Z Z X X X PROOF. We have already shown in Corollary 3.2.5 that f is measurable. Since 0 fn fnC1 f , for each n N, it follows, by Proposition 3.3.5, that Ä Ä Ä 2

fnd fnC1d f d; Ä Ä XZ XZ XZ for each n N. Hence, 2 lim fnd f d n!1 Ä (3.3) XZ XZ

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We now prove the reverse inequality. Let be a simple function such that 0 f . Choose and ﬁx such that 0 < < 1. For each n N, let Ä Ä 2 An x X fn.x/ .1 /.x/ : Df 2 W g Now, for each n N, An is measurable, and since .fn/ is an increasing sequence, we have that An AnC1. 2 1 1 Â Furthermore X An. Since for each n N, An X , it follows that An X . For the reverse D 2 Â n[D1 n[D1 containment, let x X . If f .x/ 0, then .x/ 0, and therefore x A1. If on the other hand, 2 D D 2 f .x/ > 0, then .1 /.x/

We have already shown in Proposition 3.3.2 [5], that is a measure on †. Since .An/ is an increasing 1 sequence of measurable sets such that X An, we have, by Theorem 2.3.6 [3], that D n[D1 1

.X / An lim .An /: D D n!1 Â n[D1 Ã That is, d limn!1 d. Letting n in (3.4), we have that X D An ! 1 R R lim fnd .1 / d: n!1 XZ XZ Since is arbitrary, it follows that

lim fnd d; n!1 XZ XZ for any nonnegative simple function such that f . Then, Ä

lim fnd sup d is simple and 0 f f d: n!1 W Ä Ä D (3.5) XZ XZ XZ From (3.3) and (3.5), we have that

lim fnd f d: n!1 D XZ XZ 2 3.3.10 Corollary Let f and g be nonnegative measurable functions. Then

.f g/d f d gd: C D C XZ XZ XZ

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PROOF.

By Theorem 3.2.8, there are monotonic increasing sequences .n/ and . n/ of nonnegative simple functions such that n f and n g, as n . Then .n n/ is a monotonic increasing sequence which converges to f !g. By the Monotone! Convergence! 1 TheoremC (Theorem 3.3.9), we have that C

.f g/d lim .n n/d C D n!1 C XZ XZ

lim nd nd D n!1 C ÄXZ XZ

lim nd lim nd D n!1 C n!1 XZ XZ f d gd (by the Monotone Convergence Theorem) D C XZ XZ

2 3.3.11 Corollary Let .X;†;/ be a measure space and .fn/ a sequence of nonnegative measurable functions on X . Then, for each n N, 2 n n fk d fk d: D XZ kXD1 kXD1XZ PROOF. (Induction on n). If n 1, then the result is obviously true. If n 2, then, by Corollary 3.3.10, the result holds. Assume that D D n1 n1

fk d fk d: D XZ kXD1 kXD1XZ then n n1

fk d fk fn d D C XZ kXD1 XZ Â kXD1 Ã n1

fk fnd (by Corollary 3.3.10) D C XZ kXD1 XZ n1

fk d fnd (by the induction hypothesis) D C kXD1XZ XZ n

fk d: D kXD1XZ 2 3.3.12 Corollary (Beppo Levi) Let .X;†;/ be a measure space .fn/ a sequence of nonnegative measurable functions on X and f D

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1 nD1 fn. Then f is measurable and P 1 1 f d fn d fnd: D D XZ XZ Â nXD1 Ã nXD1XZ PROOF. m For each m N, let gm fn: Then, for each m N, gm is measurable, 0 gm gmC1 and 2 D nD1 2 Ä Ä gm f as m . By the Monotone Convergence Theorem (Theorem 3.3.9), we have that ! ! 1 P

f d lim gmd D m!1 XZ XZ m

lim fnd D m!1 XZ nXD1 m

lim fnd (by Corollary 3.3.11) D m!1 nXD1XZ 1

fnd D nXD1XZ 2 3.3.13 Corollary Let .X;†;/ be a measure space and f a nonnegative measurable function on X . If N is a null set in †, then f d f d: D XZ NZc

PROOF. By deﬁnition,

f .x/ if x N f .x/ 2 c N D 0 if x N : 2 That is, f 0 a.e. By Proposition 3.3.7 and Corollary 3.3.10, we have that N D f d .f f / f d f d: D N C N c D N C N c XZ XZ XZ XZ 2 In the hypothesis if the Monotone Convergence Theorem (Theorem 3.3.9), we required that the monotone sequence of nonnegative measurable functions converge pointwise to the function f . We can weaken the ‘pointwise convergence’ to a.e. convergence and require that the limit be a measurable function. 3.3.14 Theorem (Monotone Convergence Theorem) Let .X;†;/ be a measure space, .fn/ a sequence of measurable functionson X such that 0 fn fnC1, a.e. Ä Ä for every n N and fn f , where f is measurable. Then 2 !

lim fnd f d: n!1 D XZ XZ

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PROOF. a.e. c Since fn , there is a set N † such that .N / 0 and fn f pointwise on N . By Theorem 3.3.9, ! 2 D !

lim fnd f d: n!1 D NZc NZc From Corollary 3.3.13, it follows that

f d f d lim fnd lim fnd: D D n!1 D n!1 XZ NZc NZc XZ

2 3.3.15 Proposition Let .X;†;/ be a measure space and f a nonnegative measurable function on X . Then the set function † Œ0; deﬁned by W ! 1 .A/ f d; A † D 2 AZ is a measure on X . Furthermore, for any nonnegative measurable function g,

gd gf d: D XZ XZ

PROOF. We note that .A/ f d f d: D D A AZ XZ

Since f 0, it followsthat .A/ 0, for each A †. If A , then A 0 and so f 0. Therefore 2 D; D A D . / .A/ f d 0d 0: ; D D A D D XZ XZ 1 Let .An/ be a sequence of pairwise disjoint measurable sets and A An. Then D nD1 1 S

A 1 A An : D nD1 n D nD1 S X Therefore, f 1 f , and so A D nD1 An P 1 .A/ f d f d D A D An XZ XZ nXD1 1 f d (by Corollary 3.3.12) D An nXD1XZ 1 f d D nD1 XAZn 1 .An/: D nXD1

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1 1 That is, An .An/. nD1 D nD1 If g Â , for someÃ A †, i.e., g is a characteristic function, then D SA P2 gd d d .A/ D A D D XZ XZ AZ and gf d f d f d .A/: D A D D XZ XZ AZ n Thus, gd gf d. Assume that g , a nonnegative simple function and let ai X X iD1 Ei be the canonicalD representation of . Then, D D R R P n gd d ai d D D Ei XZ XZ XZ Â XiD1 Ã n ai d (by Corollary 3.3.11 and Proposition 3.3.5) D Ei XiD1 XZ n ai d D iD1 Z X Ei n ai .Ei /; D XiD1 and n gf d f d ai f d D D Ei XZ XZ XZ Â XiD1 Ã n

ai f d (by Corollary 3.3.11 and Proposition 3.3.5) D Ei XiD1 XZ n

ai f d D iD1 Z X Ei n

ai .Ei /: D XiD1 Therefore, if g is a nonnegative simple function, then X gd X gf d. Let g be a nonnegative measurable function. Then, by TheoremD 3.2.8, there is an increasing sequence nR!1 R n!1 .n/ of nonnegative simple functions such that n g. Then nf gf . Clearly the nf are nonnegative and measurable functions for each n !N (see Proposition 3.2.4).! By the Monotone Conver- gence Theorem (Theorem 3.3.9), we have that 2

gd lim nd D n!1 XZ XZ

lim nf d D n!1 XZ gf d: D XZ

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2 3.3.16 Theorem (Fatou’s Lemma) Let .X;†;/ be a measure space and .fn/ a sequence of nonnegative measurable functions on X . Then

liminffnd liminf fnd: n!1 Ä n!1 XZ XZ

PROOF.

For each k N, let gk infnk fn. Then each gk is a nonnegative measurable function. Furthermore, 2 D for each k N, gk gkC1 and limk!1 gk liminfn!1 fn. By the Monotone Convergence Theorem (Theorem 3.3.9),2 weÄ have that D

liminffnd lim gk d: n!1 D k!1 (3.6) XZ XZ

Since gk fn for all n k, it follows, by Proposition 3.3.5, that Ä

gk d fnd; (for all n k/: Ä XZ XZ

That is, the number X gk d is a lower bound for the sequence of numbers X fnd . Hence, Â Ãnk R R

gk d inf fnd; Ä nk XZ XZ and consequently

lim gk d liminf fnd: k!1 Ä n!1 (3.7) XZ XZ From (3.6) and (3.7), we have that

liminffnd liminf fnd: n!1 Ä n!1 XZ XZ 2

3.3.3 Integrable functions Let f be a measurable function on X . Recall that the positivepart of f , denoted by f C, is the nonnegative measurable function f C max f; 0 ; D f g and the negative part of f , denoted f , is the nonnegative measurable function

f max f; 0 : D f g We have already seen that f f C f and f f C f : D j jD C Suppose now that f g h, where g and h are nonnegative measurable functions. Then f C g and f h. Indeed, if f D 0, then f C f . Hence, for any x X , Ä Ä D 2 f C.x/ f .x/ g.x/ h.x/ g.x/: D D Ä

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If f < 0, then f C 0 and so, for any x X , D 2 f C.x/ 0 g.x/: D Ä A similar argument shows that f h. Ä 3.3.17 Deﬁnition C Let .X;†;/ be a measure space and f a measurable function on X . If X f d < or X f d < , then we deﬁne the integral of f , denoted by f d, as the extended real number 1 1 X R R R f d f Cd f d: D XZ XZ XZ

If f Cd < and f d < , then f is said to be (Lebesgue) integrable on X . The set of all X 1 X 1 integrable functions is denoted by L1.X; /. R R It is clear that f is integrable if and only if f d < . In this case, X j j 1 R f d f Cd f d: j j D C XZ XZ XZ 3.3.18 Remark Let .X;†;/ be a measure space and f a measurable function on X . If f is integrableon X , then f < a.e. on X . Indeed, if A † with .A/ > 0 and f on A, then, for each n N, f > n .j Hence,j 1 2 j j D 1 2 j j A

f d n d n d n.A/: j j A D D XZ XZ AZ

Letting n tend to inﬁnity, we have that f d . X j j D 1 3.3.19 Theorem R Let .X;†;/ be a measure space, f; g L1.X; / and let c R. Then 2 2 [1] cf L1.X; / and .cf /d c f d. 2 X D X [2] f g L1.X; / andR .f g/dR f d gd. C 2 X C D X C X PROOF. R R R

[1] Assume that c 0. Then, .cf /C max cf; 0 c max f; 0 c.f C/ and D f gD f gD .cf / max .cf /; 0 c max f; 0 c.f / D f gD f gD Therefore, .cf /Cd cf Cd c f Cd < and D D 1 XZ XZ XZ .cf /d cf d c f d < : D D 1 XZ XZ XZ

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Hence, cf is integrable and

.cf /d .cf /Cd .cf /d D XZ XZ XZ c f Cd c f d D XZ XZ

c f Cd f d D ÂXZ XZ Ã c f d: D XZ

If c < 0, then c k for some k > 0. Therefore D .cf /C max cf; 0 max kf; 0 k max f; 0 kf cf and D f gD f gD f gD D .cf / max .cf /; 0 max cf; 0 max kf; 0 k max f; 0 kf C cf C: D f gD f gD f gD f gD D From this it follows that

.cf /Cd . cf /d c f d < and D D 1 XZ XZ XZ .cf /d . cf C/d c f Cd < : D D 1 XZ XZ XZ

Therefore, cf is integrable and

.cf /d .cf /Cd .cf /d D XZ XZ XZ c f d . c/ f Cd D XZ XZ c f d c f Cd D C XZ XZ

c f Cd f d D ÂXZ XZ Ã c f d: D XZ

[2] Since f g .f g/C .f g/ and C D C C f g .f C f / .gC g/ .f CgC/ .f g/; C D C D C we have that .f g/C f CgC and .f g/ f g: C Ä C Ä C

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By Proposition3.3.5 [2], it follows that

.f g/C d .f C gC/d f Cd gCd < and C Ä C Ä C 1 XZ XZ XZ XZ .f g/d .f g/d f d gd < : C Ä C Ä C 1 XZ XZ XZ XZ

Thus, f g is integrable. Since .f g/C .f g/ f g f C f gC g, we have that C C C D C D C .f g/C f g .f g/ f C gC: C C C D C C C By Corollary 3.3.10, it follows that

Œ.f g/C f f d Œ.f g/ f C gCd C C C D C C C XZ XZ .f g/Cd f d gd .f g/d f Cd gCd , C C C D C C C XZ XZ XZ XZ XZ XZ .f g/Cd .f g/d f Cd f d gCd gd , C C D C C XZ XZ XZ XZ XZ XZ .f g/d f d gd: , C D C XZ XZ XZ 2 3.3.20 Corollary Let .X;†;/ be a measure space. Then L1.X; / is a vector space with respect to the usual operations of addition and scalar multiplication. 3.3.21 Proposition Let .X;†;/ be a measure space, f; g L1.X; /. If f g a.e. on X , then 2 Ä

f d gd: Ä XZ XZ

PROOF. Since g f 0 a.e. on X , we have that gd f d .g f /d 0: D XZ XZ XZ

It now follows that X f d X gd. Ä 2 R R 3.3.22 Lemma Let .X;†;/ be a measure space. If g is a Lebesgue integrable function on X and f is a measurable function such that f g a.e., then f is Lebesgue integrable on X . j j Ä PROOF.

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Since f f C f , it follows that f C g and f g. Hence, j jD C Ä Ä 0 f Cd gd and 0 f d gd: Ä Ä Ä Ä XZ XZ XZ XZ

Therefore f C and f are both integrable. 2 The following theorem, known as Lebesgue’s Dominated Convergence Theorem, provides a useful criterion for the interchange of limits and integrals and is of fundamental importance in measure theory. As a motivation for the measure theoretic approach to integration, think about how careful one must be when interchanging the limits and integrals in Riemann integration. 3.3.23 Theorem (Lebesgue’s Dominated Convergence Theorem) n!1 Let .X;†;/ be a measure space, .fn/ a sequence of measurable functions on X such that fn f a.e., for f a measurable function. If there is a Lebesgue integrable function g such that for each! n N, 2 fn g a.e., then f is Lebesgue integrable and j j Ä

lim fnd f d: n!1 D XZ XZ

PROOF.

Since fn g a.e., for each n N, it follows that f g a.e. By Lemma 3.3.22, we have that f is Lebesguej integrable.j Ä 2 j j Ä Since g fn 0, for each n N, we have by Fatou’s Lemma (Theorem 3.3.16), that ˙ 2

gd f d .g f /d liminf .g fn/d, and C D C Ä n C (3.8) XZ XZ XZ XZ

gd f d .g f /d liminf .g fn/d: D Ä n (3.9) XZ XZ XZ XZ Now,

liminf .g fn/d gd liminf fnd, and n C D C n (3.10) XZ XZ XZ

liminf .g fn/d gd limsup fnd: n D n (3.11) XZ XZ XZ Since g is Lebesgue integrable, we have, from (3.8) and (3.10), that

f d liminf fnd: Ä n (3.12) XZ XZ Similarly, from (3.9) and (3.11), we have that

limsup fnd f d n Ä (3.13) XZ XZ From (3.12) and (3.13), it follows that

limsup fnd f d liminf fnd; n Ä Ä n XZ XZ XZ

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whence, lim fnd f d. n!1 D Z Z X X 2 Note that in the previous theorem, if we replace the almost everywhere convergence with pointwise convergence, then we can remove the assumption that f is measurable since f would be guaranteed to be measurable in that case. Furthermore, if we have a complete measure space .X;†;/, then the almost everywhere limit of .fn/ is deﬁnitely measurable. 3.3.24 Remark The existence of a Lebesgue-integrable dominating function g in Lebesgue’s Dominated Convergence The- orem (Theorem 3.3.23) is essential. Let be the Lebesgue measure on Œ0; 1 and, for each n N, let a.e. 1 2 fn n 1 . Then fn 0 and fn 0, but 0 fnd 1 0. D Œ0; n ! ! D 6D R 1 For another example, let again be the Lebesgue measure on R. For each n N, let fn . 2 D n Œ0;n Then fn 0 uniformly on R, but fnd 1 0. ! R D 6D R 3.4 Lebesgue and Riemann integrals

In this section, we show that the Lebesgue integral is a generalization of the Riemann integral. the phrase ‘f is Lebesgue-integrable on Œa; b’ in this section means that f is integrable with respect to the Lebesgue measure on Œa; b. 3.4.1 Theorem Let f be a bounded real-valued function on Œa; b. If f is Riemann-integrable on Œa; b then f is Lebesgue- integrable on Œa; b and the two integrals coincide.

PROOF. Let Œa; b be a closed and bounded interval of real numbers, f a bounded real-valued function on Œa; b and P x0; x1;:::; xn a partition of Œa; b. Let Df g

Mi sup f .x/ xi1 x xi , and mi inf f .x/ xi1 x xi ; D f W Ä Ä g D f W Ä Ä g for each i 1; 2;:::; n. Recall that the upper sum of f relative to the partition P is D n

U.f; P/ Mi .xi xi1/ D XiD1 and the lower sum of f relative to the partition P is

n u.f; P/ mi .xi xi1/: D XiD1 Deﬁne u.f; P/ Œa; b R and l.f; p/ Œa; b R by W ! W !

u.f; P/ Mi if xi1 x xi ; i 1; 2;:::; n D Ä Ä D l.f; P/ mi if xi1 x xi ; i 1; 2;:::; n: D Ä Ä D Note that u.f; P/ and l.f; P/ are simple functions since they can be expressed in the form:

n n u.f; P/ Mi and l.f; P/ mi : D Œxi1 ;xi D Œxi1 ;xi XiD1 XiD1

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Let m be the Lebesgue measure on Œa; b. It is clear that the upper sum of f relative to the partition P, U.f; P/, is just the Lebesgue integral of the simple function u.f; P/ and the lower sum of f relative to the partition P, L.f; P/, is the Lebesgue integral of the function simple l.f; P/. Let .Pn/ be a sequence of partitions of Œa; b such that, for each n N, PnC1 is a reﬁnement of Pn and 2 maxi xj maxj .xi xi1/ tends to zero as n tends to inﬁnity. For each n N, let n u.f; Pn/ and 4 D 2 D n l.F; Pn /. Then, D 1 2 f 2 1: Ä ÄÄ ÄÄ Ä That is, the sequence .n/ is decreasing and bounded below, while the sequence . n/ is increasing and bounded above. Hence, in the limit, , where limn!1 n infn n and limn!1 n Ä D D D sup n. Note that and are measurable functions. If f is bounded by M , then so are the functions n j j n and n , for each n N. Since a constant function M is Lebesgue-integrable on Œa; b, it follows, by jLebesgue’sj j Dominatedj Convergence2 Theorem (Theorem 3.3.23), that

d lim nd lim U.f; Pn/, and D n!1 D n!1 ŒaZ;b ŒaZ;b

d lim nd lim L.f; Pn/ D n!1 D n!1 ŒaZ;b ŒaZ;b

b Since f is Riemann-integrable on Œa; b, limn!1 U.f; Pn/ a f .x/dx limn!1 L.f; Pn/. This implies that D D R d d d . /d 0: D D , D ŒaZ;b ŒaZ;b ŒaZ;b ŒaZ;b By Proposition 3.3.7, we have that a.e. with respect to the Lebesgue measure on Œa; b. Since f , it followsthat f Da.e. []. It follows that f is Lebesgue integrable and Ä Ä D D b d f d f .x/dx: D D ŒaZ;b ŒaZ;b Za 2

3.5 Lp spaces

Let .X;†;/ be a measure space and 1 p < . Denote by Lp .X; / the set of all measurable functions f X R such that f p is integrable;Ä i.e., 1 W ! j j Lp .X; / f X R f is measurable and f pd < : Df W ! W j j 1g XZ

If ˛ R and f Lp.X; /, then ˛f is measurable and 2 2 ˛f p d ˛ p f pd < ; j j Dj j j j 1 XZ XZ i.e., ˛f Lp.X; /. If f;2g Lp.X; /, then f g is measurable and, since 2 C f g p 2p1. f p g p /; j C j Ä j j j j

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it follows that

f g p d 2p1. f p g p /d 2p1 f p d g p d < ; j C j Ä j j Cj j D j j C j j 1 XZ XZ ÂXZ XZ Ã and so f g Lp.X; /. C 2 3.5.1 Deﬁnition Let .X;†;/ be a measure space. A measurable function f X R is said to be essentially bounded (with respect to ) if there is a constant M > 0 such that W ! f .x/ M a.e. on X (3.14) j j Ä The smallest constant M such that (3.14) holds is called the essential supremum of f and is denoted by j j f 1; i.e., k k f 1 ess sup f inf M R f M a.e. on X : k k D X j j WD f 2 Wj j Ä g 3.5.2 Remark Note that f is essentially bounded if there is a constant M > 0 such that . x X f .x/ > M / 0 f 2 Wj j g D I i.e., the sets of points in X where f fails to be bounded is a null set.

Denote by L1.X; / the set of all essentially bounded functions on a measure space .X:†; /. 3.5.3 Proposition Let .X;†;/ be a measure space and 1 p . Then Lp.X; / is a linear space. Ä Ä 1 PROOF. Exercise. 2

3.5.1 Holder¨ and Minkowski inequalities 3.5.4 Deﬁnition 1 1 Let p and q be positive real numbers. If 1 < p < and 1,orif p 1 and q ,orif p 1 p C q D D D 1 D 1 and q 1, then we say that p and q are conjugate exponents. D

3.5.5 Lemma (Young’s inequality) Let p and q be conjugate exponents, with 1 < p; q < , ˛ 0 and ˇ 0. Then 1 ˛p ˇq ˛ˇ : Ä p C q If p 2 q, then the inequality follows from the fact that .˛ ˇ/2 0. ConsiderD theD function f Œ0; / R given by W 1 ! ˛p ˇq f .˛/ ˛ˇ; for ﬁxed ˇ > 0: D p C q ˇq Then f .0/ > 0; f .ˇq=p / 0; and lim f .˛/ . Hence f has a global minimum on D q D ˛!C1 D C1 0 Œ0; /, i.e., there is an ˛0 Œ0; / such that f .˛0/ f .˛/ for all ˛ Œ0; /: Now, since f .˛0/ 0, it follows1 that 2 1 Ä 2 1 D

0 p1 p1 1 q=p 0 f .˛0/ ˛ ˇ ˇ ˛ ˛0 ˇ p1 ˇ : D D 0 ” D 0 ” D D

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Hence, for all ˛ Œ0; /, 2 1 ˛p ˇq ˛p ˇq 0 f .ˇq=p / f .˛/ ˛ˇ ˛ˇ : D Ä D p C q ” Ä p C q 3.5.6 Theorem (H¨older’s inequality) Let 1 p; q be conjugate exponents, .X;†;/ a measure space, f Lp.X; / and g Lq .X; /. Ä Ä 1 2 2 Then fg L1.X; / and 2 fg d f p g q: j j Äk k k k XZ

PROOF.

Firstly, we consider the case that 1 < p < . If f p 0 or g q 0, then the result follows 1 k k D k k D trivially. We therefore assume that f > 0 and g > 0. Let F f and G g . Then, p q kf kp kgkq p q k k k k D D F L .X; / and G L .X; / and F p 1 G q . By Young’s inequality (Lemma 3.5.5), we have that2 2 k k D k k 1 1 FG F p G q : j j Ä p j j C q j j It follows that 1 1 1 1 FG d F p d G q d 1; j j Ä p j j C q j j D p C q D Zx XZ XZ and so,

fg d f p g q: j j Äk k k k XZ

Now assume that p 1 and q . Let M g 1. Then fg M f a.e. on X and so D D 1 Dk k j j Ä j j

fg d M f d M f d g 1 f 1: j j Ä j j Ä j j Dk k k k XZ XZ XZ

2 3.5.7 Theorem Let .X;†;/ be a ﬁnite measure space and 1 r < p . Then Lp.X; / Lr .X; /. Ä Ä 1 Â PROOF. p p 1 1 Lp Let t r and s pr . Then t s 1; i.e., s and t are conjugate exponents. Let f .X; /. Then, by H¨older’sD inequalityD (TheoremC 3.5.6),D 2

1 1 t s f r f r d . f r /t d 1sd k kr D j j Ä j j XZ ÂXZ Ã ÂXZ Ã r 1 p p s . f r / r d 1s d D j j ÂXZ Ã ÂXZ Ã 1 f r Œ.X / s : Dk kp Hence, 1 1 1 f r f p Œ.X / f pŒ.X / r p < ; k k Äk k sr Dk k 1 and so f Lr .X; /. 2 2

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3.5.8 Theorem (Minkowski’s inequality) Let 1 0 . If f; g Lp.X; /, then f g Lp .X; / and Ä Ä 1 2 C 2

f g p f p g p: k C k Äk k Ck k PROOF. Lp p p We have shown that f g .X; /. Let q p1 . If X f g d 0, then the result follows C 2 p D j C j D trivially. Assume then that X f g d 0. Then 2 j C j 6D R f g p d f g f g p1d j C j D j C jj C j XZ XZ f f g p1s g f g p1d Ä j jj C j C j jj C j XZ XZ 1 1 1 q p p f g .p1/q f p d g p d Ä j C j j j C j j ÂXZ Ã ÄÂXZ Ã ÂXZ Ã 1 q p f g d Œ f p g p: D j C j k k Ck k ÂXZ Ã

1 q Dividing both sides by f g p d , gives the desired result, that is, X j C j Â Ã R 1 1 p 1 q p p f g d f g d f p g q f g p f p g p : j C j D j C j Äk k Ck k ,k C k Äk k Ck k ÂXZ Ã ÂXZ Ã If p , then for each x X , D 1 2

f .x/ g.x/ f .x/ g.x/ f 1 g 1: j C jÄj jCj jÄk k Ck k Thus, f g L1.X; / and C 2 f g 1 f 1 g 1: k C k Äk k Ck k

2 p Let .X;†;/ be a measure space and 1 p < . Deﬁne a real-valued function p on L .X; / by Ä 1 kk

p f p f d ; k k WD j j ÂXZ Ã

for f Lp .X; /. Note2 that if f Lp.X; /, then 2

p f p 0 f d 0 f 0 a.e. on X f 0 a.e. on X: k k D , j j D ,j jD , D XZ

p This shows that p does not deﬁne a norm on L .X; /. The way thatkk we deal with this,is we deﬁne a relation on Lp.X; / as follows: For f; g Lp .X; /, 2 f g f g a.e. on X: , D

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It is easy to show that deﬁnes an equivalence relation on Lp.X; /. The relation identiﬁes elements of Lp.X; / that only differ on a null set. We denote by Lp .X; / the quotient spaceLp.X;/= . That is Lp.X; / Lp.X;/=N ; D where N f Lp.X; / f 0 a.e. on X . Note that Lp.X; / is a set of equivalence classes of functions.D f For 2f Lp .X; /W , denoteD by Œf theg equivalence class of f . That is, Œf is the set of all functions in Lp.X; /2that differ from f on a null set.

Similarly, denote L1.X; /, the quotient space obtained by identifying functions in L1.X; / that coincide almost everywhere. 3.5.9 Remark We can consider the elements of Lp.X; / as functions. This is technically incorrect since Lp.X; / is really the set of equivalence classes of functions. Never-the-less, we will continue to abuse the language by not distinguishing between elements Œf of Lp.X; / and the function f representing the equivalence class Œf . 3.5.10 Theorem Let .X;†;/ be a measure space and 1 p . Deﬁne addition and scalar multiplication in Lp.X; / as follows: Ä Ä 1

Œf Œg Œf g and ˛Œf Œ˛f ; where f; g Lp .X; / and ˛ R: C D C D 2 2 Then Lp.X; / is a real vector space.

PROOF. Exercise. 2 Now that we know that Lp .X; / is a linear space, we want to define a norm on it. If 1 p < and p p Ä 1 f L .X; /, define p L .X; / Œ0; / by 2 kk W ! 1 1 p p Œf p f p f d ; k k Dk k D j j ÂXZ Ã and for p , define D 1 Œf p f 1 esssup f : k k Dk k D X j j Note that it is important that the definition of p (or 1) is independent of the choice of a representative of the equivalence class Œf , for f kkLp.X; /kk(or f Lp.X; /). Indeed, let f; g Lp.X; / such that g Œf . Then Œf Œg and f 2 g a.e. on X . Therefore,2 2 2 D D f d gd: D XZ XZ 3.5.11 Theorem p 1 Let .X;†;/ be a measure space. Then .L .X;/; p / and .L .X;/; 1/ are normed linear spaces. kk kk 3.5.12 Example Let X N; † P.N/, and be the counting measure on .N; P.N//. Then the functions on N are sequences,D integrationD is summation, and integrable functions correspond to absolutely summable sequences.

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p p p For 1 p < , then L .N; / l .N/, the spaces of real sequences .xn/ such that n2N xn < . The normÄ on l p .N1/ is D j j 1 1 P p p .xn/ p xn : k k D j j n2N Â X Ã 1 1 If p , then L .N/ l .N/, then the spaces of bounded real sequences .xn/ and D 1 D

.xn/ 1 sup xn : k k D n2N j j

65 Chapter 4

Decomposition of measures

4.1 Signed measures

In this chapter, we discuss the Hahn decomposition, Jordan decomposition, Radon-Nikodym Theorem and the Lebesgue decomposition Theorem, which are amongst the most important results in measure theory. 4.1.1 Deﬁnition Let .X; †/ be a measurable space. A signed measure on † is an extended real-valued set function such that (a) assumes at most one of the values and , 1 1 (b) . / 0, ; D (c) for any sequence .An/ of disjoint measurable sets,

An .An /: D n2N n2N Â [ Ã X It follows from the above definition that every measure space is a signed measure, but not conversely. 4.1.2 Example Let .X;†;/ be a measure space and f a measurable function on X such that f C or f is integrable. Define on † by .A/ f d; D AZ for A †. Then is a signed measure on †. 2 4.1.3 Definition Let be a signed measure on a measurable space .X; †/. A set A † is said tobe a 2 [1] positive set with respect to if .A E/ 0, for any E †. \ 2 [2] negative set with respect to if .A E/ 0, for any E †. \ Ä 2 4.1.4 Remarks [1] It can beshown thta set A † is positive (resp., negative) with respect to if and only if .E/ 0 (resp., .E/ 0) for every2 measurable subset E of A. Ä [2] One can also show that A † is a null set if and onlyif .E/ 0, for all E † such that E A. 2 D 2

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[3] Any measurable subset of a positive (resp., negative) set is positive (resp., negative). Also, a countable union of positive (resp., negative) sets is positive (resp., negative).

We show the latter for positive sets: Let .An/ be a sequence of positive sets in †. then there is a sequence .Bn/ in † such that Bn An, for each n N, Bi Bj if i j and An Bn. 2 \ D; 6D D n2N n2N [ [ Let E † such that E An. Then, since 2 n2N [

E E An E Bn .E Bn/ (a disjoint set); D \ D \ D \ n2N n2N n2N Â [ Ã Â [ Ã [ it follows that 1

.E/ .E Bn/: D \ nXD1 Since An is positive for each n N and E Bn An, it follows that .E Bn/ 0, for each 2 \ \ n N. Hence, .E/ 0. That is, An is a positive set. 2 n2N [ [4] If A is a positive for each set with respect to , then .A/ 0. The converse is not true in general. D 4.1.5 Theorem Let .X; †/ be a measurable space and a signed measure on †. If E † and 0 < .E/ < , then E contains a positive set A with .A/ > 0. 2 1

PROOF. If E is positive with respect to , then take A E. Assume that E is not positive. Then there is a set D B † with B E and .B/ < 0. Let k1 be the smallest positive integer such that there is an E1 † 2 1 2 with E1 E and .E1/ < . Since k1 1 0 < .E/ < .E/ .E1/ .E E1/ < ; C k1 D n 1

we can re[eat the above argument used for E in the case of E E1. Let k1 be the smallest positive integer n 1 such that there is an E2 † with E2 E E1 and .E2/ < . Then 2 n k2 1 0 < .E E1/ < .E E1/ .E2/ .E .E1 E2// < : n C k2 n D n [ 1

Continuing in this fashion, we obtain a sequence .kn/ or positive integers such that for each n N, kn is 2 the smallest positive integer for which there is a measurable set En such that n1 En E Ej n j[D1 and .E / E n1 E and .E / < 1 . If the process stops at n , then let A E s E . Then n jD1 j n kn s jD1 j n s D n S S A is a positive set and .A/ > 0 (if .A/ 0, then .E/ .Ej / < 0, contradicting the fact that D D jD1 1 X 1 .E/ > 0). Otherwise, let A E Ej . Then E A .Ej /, as disjoint union. Hence, D n D [ jD1 jD1 [ S 1

.E/ .A/ .Ej /: D C jXD1

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1 If .A/ 0, then .E/ jD1 .Ej / < 0, contradicting the hypothesis that .E/ > 0. Similarly, if D D 1 .A/ < 0, then .E/ < 0.P Thus .A/ > 0. Since 0 < .E/ < , the series .Ej / converges. 1 jXD1 1 1 1 Therefore < .Ej /, whence < . In particular, we have that kj as j . We 1 kj 1 ! 1 ! 1 jXD1 jXD1 show that A is a positive set. Since limn!1 kj , there is a j0 N such that kj > 1, for all j j0. D 1 j 2

Let B be a measurable subset of A and j > j0. Then B E Ei and so n i[D1 1 .B/ : kj 1 This is true for all j j0. Letting j , we get that .B/ 0. Hence A is a positive set. ! 1 2 4.1.6 Theorem (Hahn decomposition) Let be a signed measure on a measurable space .X; †/. Then thereisa pair .P; N / of measurable subsets of X such that [1] P is a positive set and N a negative set. [2] P N . \ D; [3] X P N . D [ PROOF. Since assumes at most one of the values of or , we may assume that is the value that is never attained. That is, .E/ < , for all E †.1 Let } be1 the collection of all sets1 in † that are positive with respect to . Then } is nonempty1 since 2 }. Let ; 2 ˛ sup .A/ A } : D f W 2 g

Then ˛ 0 since }. Let .An/ be a sequence in } such that ˛ limn!1 .An / and let P n2N An. Since a countable; union 2 of positive sets is positive, we have thatDP is positive with respectD to . By the S definition of ˛, we have that ˛ .P/. Since P An P, for each n N and P is positive with respect n 2 to , it follows that .P An / 0, for each n N. Therefore, n 2 .P/ .An / .P An / .An /; D C n for each n N. Thus, ˛ .P/. Hence, .P/ ˛. 2 Ä D Let N X P. We show that N is negative with respect to . Let E be a measurable subset of N . If .E/ > 0,D then,n by Theorem 4.1.5, E contains a subset B such that .B/ > 0. But then P B (a disjoint union) would be a positive set with .P B/ .P/ .B/>˛, contradicting the definition[ of ˛. [ D C It is clear from the above that P N and X P N . \ D; D [ 2 4.1.7 Definition The pair of sets .P; N / as discussed in Theorem 4.1.6, is called a Hahn decomposition of X with respect to the signed measure .

The Hahn decomposition .P; N / is not unique. Indeed, if M is a null set with respect to , then the pairs .P M; N M / and .P M; N M / are also Hahn decompositions of X with respect to . Despite this lack[ of uniqueness,n we haven the following[ proposition.

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4.1.8 Proposition Let .X; †/ be a measurable space and a signed measure on X . If .P1; N1/ and .P2; N2/ are Hahn decompositions of X with respect to , then, for all E †, 2

.E P1/ .E P2/ and .E N1/ .E N2/: \ D \ \ D \ PROOF. We ﬁrstly note that since

c P1 P2 P1 and P1 P2 P1 P P1 N2; n n D \ 2 D \ we have that, for any E †, 2

E .P1 P2/ E P1 P1 and E .P1 P2/ E .P1 N2/ N2: \ n \ \ \ n D \ \ Therefore, .E .P1 P2// 0 and .E .P1 P2// 0: \ n \ n Ä It now follows that .E .P1 P2// 0. A similar argument shows that .E .P2 P1// 0. \ n D \ n D Now, E P1 E Œ.P1 P2/ .P1 P2/ \ D \ n [ \ .E .P1 P2// .E .P1 P2// (a disjoint union): D \ n [ \ \ Thus, .E P1/ .E .P1 P2// .E .P1 P2// .E .P1 P2//: \ D \ n C \ \ D \ \ Similarly, E P2 E Œ.P2 P1/ .P1 P2/ \ D \ n [ \ .E .P2 P1// .E .P1 P2// (a disjoint union): D \ n [ \ \ Therefore .E P2/ .E .P2 P1// .E .P1 P2// .E .P1 P2//: \ D \ n C \ \ D \ \ It follows from this that .E P1/ .E P2/: \ D \ The proof that .E N1/ .E N2/ is left as an exercise to the reader. \ D \ 2 4.1.9 Definition Let be a signed measure on a measurable space .X; †/ and let P; N be a Hahn decomposition of X with respect to . The positive variation C, negative variation f, andgtotal variation of are finite measures defined on † by j j

C.E/ .E P/ D \ .E/ .E N / D \ .E/ .E P/ .E N / C .E/ .E/: j j D \ \ D C It is clear from Proposition 4.1.8, that the deﬁnitions of C , , and are independent of any Hahn decomposition of X with respect to the signed measure . j j

Let .P; N / be a Hahn decomposition of X with respect to a signed measure . Then X P N and P N . Let E †, then D [ \ D; 2 E E X E .P N / .E P/ .E N /, a disjoint union. D \ D \ [ D \ [ \

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Therefore, .E/ .E P/ .E N / C.E/ .E/: D \ C \ D Hence, C : (This is called the Jordan decompostion of .) D Since assumes at moest one of the values and , either C or must be finite. If both C and are finite, then is called a finite signed measure.1 1 4.1.10 Example Let .X;†;/ be a measure space and f L1./. Let be a set function defined on † by 2

.E/ f d; E †: D 2 EZ Show that is a signed measure on X and

C.E/ f Cd; .E/ f d; and .E/ f d: D D j j D j j EZ EZ EZ Solution: That is a signed measure on X follows from Proposition 3.3.15. Let P x X f .x/ > 0 and P x X f .x/ 0 . Then X P N and P N . Let E †. ThenDf 2 W g Df 2 W Ä g D [ \ D; 2 .E P/ f d f d \ D D EZ\P fx2EWf.Z x/>0g

f Cd 0; D EZ and .E N / f d f d \ D D E\ZN fx2X Wf.Z x/Ä0g

f d 0: D Ä EZ Thus, P is a positive set with respect to and N is a negative set with respect to . that is, .P; N / is a Hahn decomposition of X with respect to . By deﬁnition of C; , and we have that, for each E †, j j 2 C.E/ .E P/ f Cd; D \ D EZ .E/ .E N / f d; D \ D EZ .E/ C.E/ .E/ f Cd f d .f C f /d f d: j j D C D C D C D j j EZ EZ EZ EZ 4.1.11 Deﬁnition Two measures and on a measurable space .X; †/ are said to be mutually singular, denoted by , if there is an A † such that .A/ 0 .Ac /. ? 2 D D If and are signed measures, then if . ? j j?j j

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4.1.12 Example Let be a signed measure on a measurable space .X; †/. Then the measures C and are mutually singular. Indeed, let .P; N / be a Hahn decomposition of X with respect to . Then X P N and P N . Therefore, D [ \ D; C.N / .N P/ . / 0 and D \ D ; D .P/ .N P/ . / 0: D \ D ; D Thus C . ? 4.1.13 Deﬁnition Let and be measures on a measurable space .X; /. We say that is absolutely continuous with respect to , denoted by , if .E/ 0, for every E † for which .E/ 0. D 2 D If and are signed measures, then if . j jj j 4.1.14 Example Let .X;†;/ be a measure space and f a nonnegative measurable function on X . We have shown in Proposition 3.3.15 that the set function

.E/ f d; E † D 2 (4.1) EZ deﬁnes a measure on X . The measure will be ﬁnite if and only if f is integrable. Since the integral over a set E †, with .E/ 0 is zero, it follows that is absolutely continuous with respect to . 2 D 4.1.15 Proposition Let ;, and be measures on a measurable space .X; †/ and c1; c2 numbers. Then

[1] If and , then .c1 c2/ . C [2] If and , then .c1 c2/ . ? ? C ? [3] If and , then 0. ? ? D PROOF. We prove [3] and leave [1] and [2] as exercises.

Since , there is a set E † such that .E/ 0 .Ec /. Using the fact that , it follows that .Ac / ?0. Therefore, 2 D D D .X / .A Ac / .A/ .Ac / 0: D [ D C D Hence, 0. D 2 The following theorem, known as the Radon-Nikodym Theorem shows that if .X;†;/ is a -ﬁnite measure space, then every measure on X that is absolutely continuouswith respect to arises in the manner indicated by (4.1). 4.1.16 Theorem (Radon-Nikodym Theorem) Let and be -ﬁnite measures on a measurable space .X; †/. Suppose that is absolutely continuous with respect to . Then there is a nonnegative measurable function f on X such that, for each E †, 2 .E/ f d: D EZ Moreover, the function f is uniquely determined almost everywhere Œ.

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The function f as described in the Radon-Nikodym Theorem is also called the Radon-Nikodym derivative of with respect to . We write f d or d f d. D d D 4.1.17 Theorem (Lebesgue Decomposition Theorem) Let .X;†;/ be a -finite measure space and a -finite measure on †. Then there is a unique pair of measures a and s , with a s such that a and s . D C ? PROOF. Let . Then is a -finite measure with and . By the Radon-NikodymTheorem, there are nonnegativeD C measurable functions f and g onX such that

.E/ f d and .E/ gd; E †: D D 2 EZ EZ

Let A x X f .x/ > 0 and B x X f .x/ 0 . Then X A B, A B , and D f 2 W g D f 2 W D g D [ \ D; .B/ f d 0. Deﬁne a and s by D B D

R a.E/ .E A/ and s .E/ .E B/: D \ D \ Let E †. Then 2 E E X E .A B/ .E A/ .E B/, a disjoint union. D \ D \ [ D \ [ \ Therefore, .E/ .E A/ .E B/ a.E/ s .E/ .a s/.E/: D \ C \ D C D C Hence, a s . D C Since, s .A/ .A B/ . / 0, we have that s .A/ 0 .B/. That is, s . We now D \ D ; D D D ? show that a . Let E † such that .E/ 0. Then 2 D 0 .E/ f d: D D EZ

Therefore, f 0 a.e. Œ. Now, D 0 f d f d 0: Ä Ä D A\ZE EZ Hence, by Proposition 3.3.7, f d 0. Since f > 0 on A E, we must have that .A E/ 0. A\E D \ \ D Since , it follows that .A E/ 0 and therefore a.E/ 0. That is, a . R \ D D

For uniqueness, ;et !a and !s be another pair of measures with !a !s such that !a and D C !s . Then ? a s !a !s a !a !s s: C D C , D Hence, .a !a/ and .!s s/ . But since a !a !s s , it follows that .a !a/ . That is, ? D ? .a !a/ ; .a !a/ and ? .!s s/ .!s s/ : ? By Proposition 4.1.15, we have that a !s 0 and !s s 0. Hence, a !a and s !s . D D D D 2