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LECTURE NOTES IN MEASURE THEORY

lecture notes measure theory and probability and lecture notes on measure theory and integration pdf free downlaod
1 LECTURENOTES INMEASURETHEORY Christer Borell Matematik Chalmers och Göteborgs universitet 412 96 Göteborg (Version: January 12)2 PREFACE These are lecture notes on integration theory for a eight-week course at the Chalmers University of Technology and the Göteborg University. The parts de…ning the course essentially lead to the same results as the …rst three chaptersintheFollandbook F; whichisusedasatextbookonthecourse. Theproofsinthelecturenotessometimesdi¤erfromthosegiveninF:Here is a brief description of the di¤erences to simplify for the reader. In Chapter 1 we introduce so called -systems and -additive classes, which are substitutes for monotone classes of sets F. Besides we prefer to emphasizemetricoutermeasuresinsteadofsocalledpremeasures. Through- out the course, a variety of important measures are obtained as image mea- sures of the linear measure on the real line. In Section 1.6 positive measures inRinducedbyincreasingrightcontinuousmappingsareconstructedinthis way. Chapter 2 deals with integration and is very similar to F and most other texts. Chapter3startswithsomestandardfactsaboutmetricspacesandrelates the concepts to measure theory. For example Ulam’s Theorem is included. The existence of product measures is based on properties of -systems and -additive classes. Chapter 4 deals with di¤erent modes of convergence and is mostly close to F: Here we include a section about orthogonality since many students have seen parts of this theory before. The Lebesgue Decomposition Theorem and Radon-Nikodym Theorem 2 in Chapter 5 are proved using the von Neumann beautiful L -proof. Toillustratethepowerof abstractintegrationthesenotescontainseveral sections, which do not belong to the course but may help the student to a better understanding of measure theory. The corresponding parts are set between the symbols and """ respectively.3 Finally I would like to express my deep gratitude to the students in my classes for suggesting a variety of improvements and a special thank to Jonatan Vasilis who has provided numerous comments and corrections in my original text. Göteborg 2006 Christer Borell4 CONTENT 1 Measures 1.1-Algebras and Measures 1.2 Measure Determining Classes 1.3 Lebesgue Measure 1.4 Carathéodory’s Theorem 1.5 Existence of Linear Measure 2 Integration 2.1 Integration of Functions with Values in 0;1 2.2 Integration of Functions with Arbitrary Sign 2.3 Comparison of Riemann and Lebesgue Integrals 3 Further Construction Methods of Measures 3.1 Metric Spaces 3.2 Linear Functionals and Measures 3.3 q-Adic Expansions of Numbers in the Unit Interval 3.4 Product Measures 3.5 Change of Variables in Volume Integrals 3.6 Independence in Probability 4 Modes of Convergence 1 2 4.1 Convergence in Measure, inL (); and inL () 4.2 Orthogonality 4.3 The Haar Basis and Wiener Measure 5 Decomposition of Measures 5.1 Complex Measures 5.2 The Lebesgue Decomposition and the Radon-Nikodym Theorem 5.3TheWienerMaximalTheoremandLebesgueDi¤erentiationTheorem5 5.4AbsolutelyContinuousFunctionsandFunctionsofBoundedVariation 5.5 Conditional Expectation 6 Complex Integration 6.1 Complex Integrand 6.2 The Fourier Transform 6.3 Fourier Inversion 6.4 Non-Di¤erentiability of Brownian Paths References6 CHAPTER1 MEASURES Introduction The Riemann integral, dealt with in calculus courses, is well suited for com- putations but less suited for dealing with limit processes. In this course we will introduce the so called Lebesgue integral, which keeps the advantages of theRiemannintegralandeliminatesitsdrawbacks. Atthesametimewewill develop a general measure theory which serves as the basis of contemporary analysis and probability. In this introductory chapter we set forth some basic concepts of measure theory, which will open for abstract Lebesgue integration. 1.1. -Algebras and Measures Throughout this course N =f0;1;2;:::g (the set of natural numbers) Z =f:::;2;1;0;1;;2;:::g (the set of integers) Q = the set of rational numbers R = the set of real numbers C = the set of complex numbers. If AR; A is the set of all strictly positive elements inA: + Iff isafunctionfromasetAintoasetB;thismeansthattoeveryx2A there corresponds a point f(x)2B and we write f : A B: A function is often called a map or a mapping. The functionf is injective if (x6=y)) (f(x)6=f(y))7 and surjective if to each y 2 B; there exists an x 2 A such that f(x) = y: An injective and surjective function is said to be bijective. A set A is …nite if either A is empty or there exist an n 2 N and a + bijection f :f1;:::;ngA: The empty set is denoted by : A set A is said to be denumerable if there exists a bijection f : N A: A subset of a + denumerable set is said to be at most denumerable. Let X be a set. For any AX; the indicator function  of A relative A toX is de…ned by the equation  1 if x2A  (x) = A c 0 if x2A : The indicator function  is sometimes written 1 : We have the following A A relations:  c = 1 A A  = min( ; ) =  A\B A B A B and  = max( ; ) = +   AB A B A B A B: De…nition 1.1.1. LetX be a set. a) A collectionA of subsets of X is said to be an algebra in X ifA has the following properties: (i)X 2A: c c (ii)A2A)A 2A; whereA is the complement of A relative toX: (iii) If A;B2A thenAB2A: (b) A collectionM of subsets of X is said to be a -algebra in X if M is an algebra with the following property: 1 If A 2M for all n2N , then A 2M: n + n n=18 If M is a -algebra in X; (X;M) is called a measurable space and the members of M are called measurable sets. The so called power set P(X), that is the collection of all subsets of X, is a -algebra in X: It is simple to provethattheintersectionofanyfamilyof-algebrasinX isa-algebra. It follows that ifE is any subset ofP(X); there is a unique smallest -algebra (E) containingE; namely the intersection of all -algebras containingE: The-algebra(E) iscalledthe-algebrageneratedbyE: The-algebra generated by all open intervals in R is denoted byR. It is readily seen that the -algebra R contains every subinterval of R. Before we proceed, recall thatasubsetE ofRisopeniftoeachx2E thereexistsanopensubinterval ofR contained inE and containingx; the complement of an open set is said to be closed. We claim thatR contains every open subset U of R: To see this suppose x 2 U and let x 2 a;b  U; where 1 a b 1: Now pickr;s2Q such thatarxsb: Thenx2 r;sU and it follows that U is the union of all bounded open intervals with rational boundary points contained inU: Since this family of intervals is at most denumberable we conclude that U 2R: In addition, any closed set belongs to R since its complementsisopen. Itisbynomeanssimpletograspthede…nitionofRat this stage but the reader will successively see that the-algebraR has very nice properties. At the very end of Section 1.3, using the so called Axiom of Choice, we will exemplify a subset of the real line which does not belong to R. In fact, an example of this type can be constructed without the Axiom of Choice (see Dudley’s book D). In measure theory, inevitably one encounters 1: For example the real line has in…nite length. Below 0;1 = 0;1f1g: The inequalitiesxy and x y have their usual meanings if x;y 2 0;1. Furthermore, x1 if x 2 0;1 and x 1 if x 2 0;1: We de…ne x +1 = 1 +x = 1 if x;y2 0;1; and  0 if x = 0 x1 =1x = 1 if 0x1: Sums and multiplications of real numbers are de…ned in the usual way. IfA X;n2N , andA \A = ifk = 6 n, the sequence (A ) is n + k n n n2N + calledadisjointdenumerablecollection. If(X;M)isameasurablespace,the 1 collection is called a denumerable measurable partition of A if A = A n n=1 and A 2M for every n2N : Some authors call a denumerable collection n + of sets a countable collection of sets.9 De…nition 1.1.2. (a) Let A be an algebra of subsets of X: A function  :A 0;1 is called a content if (i)() = 0 (ii)(AB) =(A)+(B) if A;B2A andA\B =: (b) If (X;M) is a measurable space a content de…ned on the-algebraM is called a positive measure if it has the following property: For any disjoint denumerable collection (A ) of members ofM n n2N + 1 1 ( A ) =  (A ): n n n=1 n=1 If (X;M) is a measurable space and the function  : M 0;1 is a positivemeasure, (X;M;) iscalledapositivemeasurespace. Thequantity (A) is called the -measure of A or simply the measure of A if there is no ambiguity. Here (X;M;) is called a probability space if (X) = 1; a …nite positive measure space if (X) 1; and a -…nite positive measure space ifX is a denumerable union of measurable sets with …nite-measure. The measure  is called a probability measure, …nite measure, and -…nite measure, if (X;M;) is a probability space, a …nite positive measure space, and a -…nite positive measure space, respectively. A probability space is often denoted by ( ;F;P): A memberA ofF is called an event. As soon as we have a positive measure space (X;M;), it turns out to be a fairly simple task to de…ne a so called-integral Z f(x)d(x) X as will be seen in Chapter 2.10 The class of all …nite unions of subintervals of R is an algebra which is denoted byR : If A2R we denote by l(A) the Riemann integral 0 0 Z 1  (x)dx A 1 and it follows from courses in calculus that the function l :R 0;1 is a 0 content. The algebra R is called the Riemann algebra and l the Riemann 0 content. If I is a subinterval of R, l(I) is called the length of I: Below we follow the convention that the empty set is an interval. If A 2 P(X), c (A) equals the number of elements in A, when A is a X …nite set, andc (A) =1 otherwise. Clearly, c is a positive measure. The X X measurec is called the counting measure onX: X Givena2X;theprobabilitymeasure de…nedbytheequation (A) = a a  (a); if A2P(X); is called the Dirac measure at the point a: Sometimes A we write =  to emphasize the setX: a X;a If  and  are positive measures de…ned on the same -algebra M, the sum + isapositivemeasureonM:Moregenerally,  +  isapositive measure for all real ;  0: Furthermore, if E 2M; the function (A) = (A\E); A 2 M; is a positive measure. Below this measure  will be E E denotedby andwesaythat isconcentratedonE: IfE2M; theclass M = fA2M; AEg is a -algebra of subsets of E and the function E (A) =(A), A2M ; is a positive measure. Below this measure  will be E denoted by and is called the restriction of  toM : E jE LetI ;:::;I be subintervals of the real line. The set 1 n n I :::I =f(x ;:::;x )2R ; x 2I ; k = 1;:::;ng 1 n 1 n k k n is called an n-cell in R ; its volume vol(I :::I ) is, by de…nition, equal 1 n to n vol(I :::I ) =  l(I ): 1 n k k=1 IfI ;:::;I are open subintervals of the real line, then-cellI :::I is 1 n 1 n n called an open n-cell. The -algebra generated by all open n-cells in R is denoted byR : In particular, R =R. A basic theorem in measure theory n 1 statesthatthereexistsauniquepositivemeasurev de…nedonR suchthat n n themeasureofanyn-cellisequaltoitsvolume. Themeasurev iscalledthe n n volume measure onR or the volume measure onR : Clearly,v is-…nite. n n 2 The measure v is called the area measure on R and v the linear measure 2 1 on R:11 n Theorem 1.1.1. The volume measure on R exists. Theorem1.1.1willbeprovedinSection1.5inthespecialcasen = 1. The general case then follows from the existence of product measures in Section 3.4. An alternative proof of Theorem 1.1.1 will be given in Section 3.2. As soonastheexistenceofvolumemeasureisestablishedavarietyofinteresting measures can be introduced. Next we prove some results of general interest for positive measures. Theorem 1.1.2. Let A be an algebra of subsets of X and  a content de…ned on A. Then, (a)  is …nitely additive, that is (A :::A ) =(A )+:::+(A ) 1 n 1 n if A ;:::;A are pairwise disjoint members of A: 1 n (b) if A;B2A; (A) =(AnB)+(A\B): Moreover, if (A\B) 1; then (AB) =(A)+(B)(A\B) (c) AB implies (A)(B) if A;B2A: (d) …nitely sub-additive, that is (A :::A )(A )+:::+(A ) 1 n 1 n if A ;:::;A are members of A: 1 n If (X;M;) is a positive measure space12 (e) (A )(A) if A = A ; A 2M; and n n2N n n + A A A ::: : 1 2 3 (f) (A )(A) if A =\ A ; A 2M; n n2N n n + A A A ::: 1 2 3 and (A )1: 1 (g) is sub-additive, that is for any denumerable collection (A ) of n n2N + members of M, 1 1 ( A )  (A ): n n n=1 n=1 PROOF (a) If A ;:::;A are pairwise disjoint members ofA; 1 n n n ( A ) =(A ( A )) k 1 k k=1 k=2 n =(A )+( A ) 1 k k=2 and, by induction, we conclude that is …nitely additive. (b) Recall that c AnB =A\B : NowA = (AnB)(A\B) and we get (A) =(AnB)+(A\B): Moreover, sinceAB = (AnB)B; (AB) =(AnB)+(B) and, if (A\B)1; we have (AB) =(A)+(B)(A\B). (c) Part (b) yields (B) =(BnA)+(A\B) =(BnA)+(A); where the last member does not fall below(A):13 n (d) If (A ) is a sequence of members of A de…ne the so called disjunction i i=1 n n (B ) of the sequence (A ) as k i k=1 i=1 k1 B =A andB =A n A for 2kn: 1 1 k k i i=1 k k ThenB A ; A = B;k = 1;::;n;andB \B =ifi6=j:Hence, k k i i i j i=1 i=1 by Parts (a) and (c), n n n ( A ) =  (B )  (A ): k k k k=1 k=1 k=1 (e) Set B = A and B = A nA for n 2: Then A = B ::::B ; 1 1 n n n1 n 1 n 1 B \B = if i6=j andA = B : Hence i j k k=1 n (A ) =  (B ) n k k=1 and 1 (A) =  (B ): k k=1 Now e) follows, by the de…nition of the sum of an in…nite series. (f) PutC =A nA ; n 1: ThenC C C :::; n 1 n 1 2 3 1 A nA = C 1 n n=1 and(A)(A )(A )1: Thus n 1 (C ) =(A )(A ) n 1 n and Part (e) shows that (A )(A) =(A nA) = lim (C ) =(A ) lim (A ): 1 1 n 1 n n1 n1 This proves (f). (g) The result follows from Parts d) and e). This completes the proof of Theorem 1.1.2.14 Thehypothesis”(A )1”inTheorem1.1.2(f)isnotsuper‡uous. If 1 c isthecountingmeasureonN andA =fn;n+1;:::g;thenc (A ) = N + n N n + + 1 1 1 for all n butA A :::: andc (\ A ) = 0 since\ A =: 1 2 N n n + n=1 n=1 If A;BX; the symmetric di¤erenceAB is de…ned by the equation AB = (AnB)(BnA): def Note that  =j  j: AB A B Moreover, we have c c AB =A B and 1 1 1 ( A )( B ) (A B ): i i i i i=1 i=1 i=1 Example 1.1.1. Let  be a …nite positive measure on R: We claim that to each set E 2 R and " 0; there exists a set A; which is …nite union of intervals (that is, A belongs to the Riemann algebraR ), such that 0 (EA)": To see this let S be the class of all sets E 2 R for which the conclusion c is true. Clearly  2 S and, moreover, R  S: If A 2 R , A 2 R and 0 0 0 c thereforeE 2S ifE2S: NowsupposeE 2S;i2N : Thentoeach" 0 i + i andi there is a setA 2R such that(E A ) 2 ": If we set i 0 i i 1 E = E i i=1 then 1 1 (E( A ))  (E A )": i i i i=1 i=1 Here 1 1 c c 1 E( A ) =fE\(\ A )gfE \( A )g i i i=1 i=1 i i=1 and Theorem 1.1.2 (f) gives that n c c 1 (fE\(\ A )gf(E \( A )g)" i i=1 i i=1 if n is large enough (hint: \ (D F) = (\ D )F): But then i2I i i2I i n n c c n (E A ) =(fE\(\ A )gfE \( A )g)" i i i=1 i=1 i i=115 if n is large enough we conclude that the set E 2 S: Thus S is a -algebra and sinceR S R it follows thatS =R: 0 Exercises 1. Prove that the sets NN =f(i;j); i;j2Ng and Q are denumerable. 2. SupposeA is an algebra of subsets of X and  and  two contents onA such that and(X) =(X)1: Prove that =: 3. Suppose A is an algebra of subsets of X and  a content on A with (X)1: Show that (ABC) =(A)+(B)+(C) (A\B)(A\C)(B\C)+(A\B\C): 4. (a)AcollectionC ofsubsetsofX isanalgebrawiththefollowingproperty: 1 If A 2C; n2N andA \A = if k6=n, then A 2C. n + k n n n=1 Prove thatC is a-algebra. (b)AcollectionCofsubsetsofX isanalgebrawiththefollowingproperty: 1 If E 2C andE E ; n2N ; then E 2C . n n n+1 + n 1 Prove thatC is a-algebra. 1 5. Let (X;M) be a measurable space and ( ) a sequence of positive k k=1 measures onM such that   ::: . Prove that the set function 1 2 3 (A) = lim  (A); A2M k k1 is a positive measure.16 6. Let (X;M;) be a positive measure space. Show that q n n n (\ A )  (A ) k k k=1 k=1 for all A ;:::;A 2M: 1 n 7. Let (X;M;) be a-…nite positive measure space with(X) =1: Show that for anyr2 0;1 there is someA2M withr(A)1: 8. Show that the symmetric di¤erence of sets is associative: A(BC) = (AB)C: 9. (X;M;) is a …nite positive measure space. Prove that j(A)(B)j(AB): 10. LetE = 2N: Prove that c (EA) =1 N if A is a …nite union of intervals. 11. Suppose(X;P(X);)isa…nitepositivemeasurespacesuchthat(fxg) 0 for everyx2X: Set d(A;B) =(AB); A;B2P(X): Prove that d(A;B) = 0 ,A =B; d(A;B) =d(B;A)17 and d(A;B)d(A;C)+d(C;B): 12. Let (X;M;) be a …nite positive measure space. Prove that n n ( A )  (A ) (A \A ) i i 1ijn i j i=1 i=1 for all A ;:::;A 2M and integers n 2: 1 n 13. Let(X;M;)beaprobabilityspaceandsupposethesetsA ;:::;A 2M 1 n P n n satisfy the inequality (A )n1: Show that(\ A ) 0: i i 1 1 1.2. Measure Determining Classes Suppose and areprobabilitymeasuresde…nedonthesame-algebraM, which is generated by a classE: If and agree onE; is it then true that and agree onM? The answer is in general no. To show this, let X =f1;2;3;4g and E =ff1;2g;f1;3gg: 1 Then(E) =P(X): If  = c and X 4 1 1 1 1  =  +  +  +  X;1 X;2 X;3 X;4 6 3 3 6 then = onE and = 6 : Inthissectionwewillproveabasicresultonmeasuredeterminingclasses for -…nite measures. In this context we will introduce so called -systems and -additive classes, which will also be of great value later in connection with the construction of so called product measures in Chapter 3.18 De…nition 1.2.1. A class G of subsets of X is a -system if A\B 2 G for all A;B2G: n The class of all openn-cells inR is a-system. De…nition 1.2.2. A class D of subsets of X is called a -additive class if the following properties hold: (a)X 2D: (b) If A;B2D andAB; thenBnA2D: (c) If (A ) is a disjoint denumerable collection of members of the n n2N + 1 classD; then A 2D: n n=1 Theorem 1.2.1. If a -additive class M is a -system, then M is a - algebra. c PROOF. If A 2 M; then A = XnA 2 M since X 2 M and M is a - additiveclass. Moreover,if (A ) isadenumerablecollectionofmembers n n2N + ofM; c c c A :::A = (A \:::\A ) 2M 1 n 1 n 1 for eachn; sinceM is a-additive class and a-system. Let (B ) be the n n=1 1 disjunctionof (A ) :Then(B ) isadisjointdenumerablecollectionof n n n2N n=1 + 1 1 members ofM and De…nition 1.2.2(c) implies that A = B 2M: n n n=1 n=1 Theorem 1.2.2. Let G be a -system and D a -additive class such that G  D: Then (G)D: PROOF. Let M be the intersection of all -additive classes containing G: TheclassMisa-additiveclassandGMD. InviewofTheorem1.2.1 M is a-algebra, ifM is a-system and in that case(G)M: Thus the theorem follows if we show thatM is a-system. GivenC X; denotebyD betheclassofallDX suchthatD\C 2 C M.19 CLAIM 1. If C 2M; thenD is a-additive class. C PROOF OF CLAIM 1. First X 2D since X\C =C 2M: Moreover, if C A;B2D andAB; thenA\C;B\C 2M and C (BnA)\C = (B\C)n(A\C)2M: Accordinglyfromthis,BnA2D :Finally,if(A ) isadisjointdenumer- C n n2N + ablecollectionofmembersofD , then (A \C) isdisjointdenumerable C n n2N + collection of members ofM and ( A )\C = (A \C)2M: n2N n n2N n + + Thus A 2D : n2N n C + CLAIM 2. If A2 G; thenMD : A PROOF OF CLAIM 2. If B 2 G; A\B 2 GM: Thus B 2 D : We A have proved thatGD and remembering thatM is the intersection of all A -additiveclassescontainingG Claim2followssinceD isa-additiveclass. A TocompletetheproofofTheorem1.2.2, observethatB2D ifandonly A if A2D : By Claim 2, ifA2G andB2M; thenB2D that isA2D : B A B ThusG D if B 2M. Now the de…nition ofM implies thatMD if B B B 2 M: The proof is almost …nished. In fact, if A;B 2 M then A 2 D B that isA\B2M: Theorem 1.2.2 now follows from Theorem 1.2.1. Theorem 1.2.3. Let  and  be positive measures on M = (G), where G is a -system, and suppose (A) = (A) for every A2G: (a) If  and  are probability measures, then  =: 1 (b) Suppose there exist E 2G; n2N ; such that X = E ; n + n n=120 E E :::; and 1 2 (E ) =(E )1; all n2N : n n + Then  =: PROOF. (a) Let D =fA2M; (A) = (A)g: It is immediate thatD is a -additive class and Theorem 1.2.2 implies that M =(G)D sinceGD andG is a-system. (b) If (E ) =(E ) = 0 for all all n2N , then n n + (X) = lim (E ) = 0 n n1 and, in a similar way, (X) = 0: Thus =: If (E ) =(E ) 0; set n n 1 1  (A) = (A\E ) and (A) = (A\E ) n n n n (E ) (E ) n n for eachA2M: By Part (a) = and we get n n (A\E ) =(A\E ) n n for eachA2M: Theorem 1.1.2(e) now proves that  =: Theorem 1.2.3 implies that there is at most one positive measure de…ned n onR such that the measure of any openn-cell inR equals its volume. n Next suppose f : X Y and let A  X and B  Y: The image of A and the inverse image of B are f(A) =fy; y =f(x) for somex2Ag and 1 f (B) =fx; f(x)2Bg21 respectively. Note that 1 f (Y) =X and 1 1 f (Y nB) =Xnf (B): Moreover, if (A ) is a collection of subsets ofX and (B ) is a collection i i2I i i2I of subsets of Y f( A ) = f(A ) i2I i i2I i and 1 1 f ( B ) = f (B ): i2I i i2I i Given a classE of subsets of Y; set  1 1 f (E) = f (B); B2E : 1 If (Y;N)isameasurablespace,itfollowsthattheclassf (N)isa-algebra inX: If (X;M) is a measurable space  1 B2P(Y); f (B)2M is a-algebra inY. Thus, given a classE of subsets of Y; 1 1 (f (E)) =f ((E)): De…nition 1.2.3. Let (X;M) and (Y;N) be measurable spaces. The func- 1 tionf :X Y is said to be (M;N)-measurable iff (N)M. If we say that f : (X;M) (Y;N) is measurable this means that f : X Y is an (M;N)-measurable function. Theorem 1.2.4. Let (X;M) and (Y;N) be measurable spaces and suppose E generates N: The function f :X Y is (M;N)-measurable if 1 f (E)M: PROOF. The assumptions yield 1 (f (E))M:22 Since 1 1 1 (f (E)) =f ((E)) =f (N) we are done. Corollary 1.2.1. A function f :X R is (M;R)-measurable if and only 1 if the set f ( ;1)2M for all 2R: If f :X Y is (M;N)-measurable and  is a positive measure onM, the equation 1 (B) =(f (B)), B2N 1 de…nes a positive measure  on N: We will write  = f ;  = f() or  =  : The measure  is called the image measure of  under f and f is f said to transport to: Two (M;N)-measurable functionsf :X Y and g :X Y are said to be-equimeasurable if f() =g(): n n n As an example, leta2R and de…nef(x) =x+a ifx2R : IfBR ; 1 f (B) =fx; x+a2Bg =Ba: 1 Thus f (B) is an open n-cell if B is, and Theorem 1.2.4 proves that f is (R ;R )-measurable. Now, granted the existence of volume measurev ; for n n n everyB2R de…ne n (B) =f(v )(B) =v (Ba): n n Then (B) = v (B) if B is an open n-cell and Theorem 1.2.3 implies that n  =v : We have thus proved the following n n Theorem 1.2.5. For any A2R and x2R n A+x2R n and v (A+x) =v (A): n n23 Suppose( ;F;P)isaprobabilityspace. Ameasurablefunction de…ned on is called a random variable and the image measure P is called the  probability law of : We sometimes write L() =P :  Here are two simple examples. If the range of a random variable  consists of n points S = fs ;:::;s g 1 n 1 (n  1) and P = c ;  is said to have a uniform distribution in S. Note  S n that 1 n P =   :  s k=1 k n Suppose  0 is a constant. If a random variable  has its range in N and n  1  P =  e   n n=0 n then is said to have a Poisson distribution with parameter: Exercises 1. Letf :X Y, AX; andBY: Show that 1 1 f(f (B))B andf (f(A))A: 2. Let (X;M) be a measurable space and suppose A  X: Show that the function is (M;R)-measurable if and only if A2M: A 3. Suppose (X;M) is a measurable space and f : X R; n 2 N; a n sequence of (M;R)-measurable functions such that lim f (x) exists and =f(x)2R n n1 for eachx2X: Prove thatf is (M;R)-measurable.24 4. Suppose f : (X;M) (Y;N) and g : (Y;N) (Z;S) are measurable. Prove thatgf is (M;S)-measurable. n 5. Grantedtheexistenceofvolumemeasurev , showthatv (rA) =r v (A) n n n if r 0 andA2R : n 2 2 6. Let  be the counting measure on Z and f(x;y) = x; (x;y) 2 Z : The positive measure  is -…nite. Prove that the image measure f() is not a -…nite positive measure. 7. Let; :R0;1betwopositivemeasuressuchthat(I) =(I)1 for each open subinterval of R: Prove that =: n k 8. Letf :R R be continuous. Prove thatf is (R ;R )-measurable. n k 9. SupposehasaPoissondistributionwithparameter:ShowthatP 2N =   e cosh: 9. Find a-additive class which is not a-algebra. 1.3. Lebesgue Measure Once the problem about the existence of volume measure is solved the exis- tence of the so called Lebesgue measure is simple to establish as will be seen in this section. We start with some concepts of general interest. If (X;M;) is a positive measure space, the zero set Z of  is, by  de…nition, the set at all A 2 M such that (A) = 0: An element of Z is  called a null set or-null set. If (A2Z andBA))B2M 25 themeasurespace (X;M;)issaidtobecomplete. Inthiscasethemeasure  is also said to be complete. The positive measure space (X;f;Xg;); whereX =f0;1gand = 0;isnotcompletesinceX 2Z andf0g2=f;Xg:  1 Theorem 1.3.1 If (E ) is a denumerable collection of members of Z n  n=1 1 then E 2Z : n  n=1 PROOF We have 1 1 0( E )  (E ) = 0 n n n=1 n=1 which proves the result. Granted the existence of linear measurev it follows from Theorem 1.3.1 1 that Q2Z since Q is countable andfag2Z for each real numbera. v v 1 1 Suppose (X;M;) is an arbitrary positive measure space. It turns out that  is the restriction to M of a complete measure. To see this suppose M istheclassofallEX issuchthatthereexistsetsA;B2Msuchthat AEB andBnA2Z : It is obvious thatX 2M sinceMM : If  E 2M ; choose A;B 2M such that A E  B and BnA2Z : Then  c c c c c c B E A andA nB =BnA2Z and we conclude thatE 2M : If  1 (E ) is a denumerable collection of members ofM ; for eachi there exist i i=1 setsA;B 2M such thatA EB andB nA 2Z : But then i i i i i i  1 1 1 A  E  B i i i i=1 i=1 i=1 1 1 1 1 where A; B 2M. Moreover, ( B )n( A )2Z since i i i i  i=1 i=1 i=1 i=1 1 1 1 ( B )n( A ) (B nA ): i i i i i=1 i=1 i=1 1 Thus E 2M andM is a-algebra. i i=1 IfE2M; supposeA;B 2M are such thatA EB andB nA 2 i i i i i i Z fori = 1;2: Then for eachi; (B \B )nA 2Z and  1 2 i  (B \B ) =((B \B )nA )+(A ) =(A ): 1 2 1 2 i i i Thus the real numbers(A ) and(A ) are the same and we de…ne (E) to 1 2 be equal to this common number. Note also that (B ) =  (E): It is plain 126 1 that  () = 0: If (E ) is a disjoint denumerable collection of members i i=1 of M; for each i there exist sets A;B 2 M such that A  E  B and i i i i i B nA 2Z : From the above it follows that i i  1 1 1 1  ( E ) =( A ) =  (A ) =   (E ): i i i i i=1 i=1 n=1 n=1 We have proved that   is a positive measure on M . If E 2 Z the   de…nition of   shows that any set A  E belongs to the -algebra M : It follows that the measure  is complete and its restriction toM equals: The measure   is called the completion of  andM is called the com- pletion ofM with respect to: n De…nition 1.3.1 The completion of volume measure v on R is called n n Lebesgue measure onR and is denoted bym : The completion ofR with n n n respect to v is called the Lebesgue -algebra in R and is denoted byR : n n n A member of the class R is called a Lebesgue measurable set in R or a n n n LebesguesetinR :Afunctionf :R RissaidtobeLebesguemeasurable if it is (R ;R)-measurable. Below, m is writtenm if this notation will not 1 n lead to misunderstanding. Furthermore,R is writtenR . 1 n Theorem 1.3.2. Suppose E 2 R and x 2R : Then E +x 2 R and n n m (E +x) =m (E): n n PROOF. Choose A;B 2R such that AE B and BnA2Z : Then, n v n by Theorem 1.2.5, A +x;B +x 2 R ; v (A +x) = v (A) = m (E); and n n n n (B +x)n(A+x) = (BnA)+x2Z : Since A+x E +x B +x the v n theorem is proved. n The Lebesgue -algebra in R is very large and contains each set of interestinanalysisandprobability. Infact,inmostcases,the-algebraR is n n n su¢ cientlylargebuttherearesomeexceptions. Forexample,iff :R R is continuous and A2 R , the image set f(A) need not belong to the class n R (see e.g. the Dudley book D). To prove the existence of a subset of the n real line, which is not Lebesgue measurable we will use the so called Axiom of Choice.27 Axiom of Choice. If (A ) is a non-empty collection of non-empty sets, i i2I there exists a functionf :I A such thatf(i)2A for everyi2I: i2I i i Let X and Y be sets. The set of all ordered pairs (x;y); where x 2 X andy2Y is denoted byXY: An arbitrary subsetR ofXY is called a relation. If(x;y)2R,wewritexsy:Arelationissaidtobeanequivalence relation onX if X =Y and (i)xsx (re‡exivity) (ii)xsy )ysx (symmetry) (iii) (xsy andysz))xsz (transitivity) The equivalence classR(x) = fy; ysxg: The de…nition of the equiv- def alence relations implies the following: (a)x2 R(x) (b)R(x)\R(y) = 6 )R(x) =R(y) (c) R(x) =X: x2X AnequivalencerelationleadstoapartitionofX intoadisjointcollection of subsets of X:   1 1 LetX = ; andde…neanequivalencerelationfornumbersx;y inX 2 2 by stating thatxsy ifxy is a rational number. By the Axiom of Choice it is possible to pick exactly one element from each equivalence class. Thus there exists a subsetNL ofX which contains exactly one element from each equivalence class. 1 IfweassumethatNL2R wegetacontradictionasfollows. Let (r ) i i=1 be an enumeration of the rational numbers in 1;1. Then 1 X  (r +NL) i i=1 and it follows from Theorem 1.3.1 that r +NL2= Z for some i: Thus, by i m Theorem 1.3.2, NL2=Z : m28 0 00 Now assume (r +NL)\ (r +NL) = 6 : Then there exist a;a 2 NL i j 0 00 0 00 0 00 such that r +a =r +a or a a =r r: Hence a sa and it follows i j j i 0 00 0 00 thata anda belong to the same equivalence class. But thena =a : Thus r = r and we conclude that (r +NL) is a disjoint enumeration of i j i i2N + Lebesgue sets. Now, since   3 3 1 (r +NL) ; i i=1 2 2 it follows that 1 1 3m( (r +NL)) =  m(NL): i i=1 n=1 But thenNL2Z ; which is a contradiction. Thus NL2=R : m In the early 1970’Solovay S proved that it is consistent with the usual axioms of Set Theory, excluding the Axiom of Choice, that every subset of R is Lebesgue measurable. From the above we conclude that the Axiom of Choice implies the exis- tenceofasubsetofthesetofrealnumberswhichdoesnotbelongtotheclass R: Interestingly enough, such an example can be given without any use of the Axiom of Choice and follows naturally from the theory of analytic sets. The interested reader may consult the Dudley book D: Exercises 1. (X;M;) is a positive measure space. Prove or disprove: If AE B and(A) =(B) thenE belongs to the domain of the completion:  2. Prove or disprove: If A and B are not Lebesgue measurable subsets of R; thenAB is not Lebesgue measurable. 3. Let (X;M;) be a complete positive measure space and suppose A;B2 M, where BnA is a -null set. Prove that E 2M if A E  B (stated otherwiseM =M).29 4. SupposeER andE2=R . Show there is an" 0 such that m(BnA)" for all A;B2R such thatAEB: 5. Suppose (X;M;) is a positive measure space and (Y;N) a measurable space. Furthermore, suppose f : X Y is (M;N)-measurable and let 1 1  = f ; that is (B) = (f (B)); B 2 N: Show that f is (M ;N )- measurable, whereM denotes the completion ofM with respect to and N the completion ofN with respect to: 1.4. Carathéodory’s Theorem In these notes we exhibit two famous approaches to Lebesgue measure: One isbasedontheCarathéodoryTheorem,whichwepresentinthissection,and the other one, due to F. Riesz, is a representation theorem of positive linear functionals on spaces of continuous functions in terms of positive measures. The latter approach, is presented in Chapter 3. Both methods depend on topological concepts such as compactness. De…nition 1.4.1. A function  : P(X) 0;1 is said to be an outer measure if the following properties are satis…ed: (i)() = 0: (ii)(A)(B) if AB: 1 (iii) for any denumerable collection (A ) of subsets of X n n=1 1 1 ( A )  (A ): n n n=1 n=130 Since c E = (E\A)(E\A ) an outer measure satis…es the inequality c (E)(E\A)+(E\A ): If  is an outer measure on X we de…ne M() as the set of all A  X such that c (E) =(E\A)+(E\A ) for all EX or, what amounts to the same thing, c (E)(E\A)+(E\A ) for all EX: The next theorem is one of the most important in measure theory. Theorem 1.4.1. (Carathéodory’s Theorem) Suppose  is an outer measure. The class M() is a -algebra and the restriction of  to M() is a complete measure. c PROOF. Clearly,  2 M() and A 2 M() if A 2 M(): Moreover, if A;B2M() andEX; c (E) =(E\A)+(E\A ) c =(E\A\B)+(E\A\B ) c c c +(E\A \B)+(E\A \B ): But c c AB = (A\B)(A\B )(A \B) and c c c A \B = (AB) and we get c (E)(E\(AB))+(E\(AB) ): It follows thatAB2M() and we have proved that the classM() is an algebra. Now if A;B2M() are disjoint c (AB) =((AB)\A)+((AB)\A ) =(A)+(B)31 and therefore the restriction of  toM() is a content. 1 Next we prove thatM() is a -algebra. Let (A ) be a disjoint denu- i i=1 merable collection of members ofM() and set for eachn2N 1 B = A andB = A n 1in i i i=1 (hereB =). Then for anyEX; 0 c (E\B ) =(E\B \A )+(E\B \A ) n n n n n =(E\A )+(E\B ) n n1 and, by induction, n (E\B ) =  (E\A ): n i i=1 But then c (E) =(E\B )+(E\B ) n n n c   (E\A )+(E\B ) i i=1 and lettingn1; 1 c (E)  (E\A )+(E\B ) i i=1 1 c ( (E\A ))+(E\B ) i i=1 c =(E\B)+(E\B )(E): Alltheinequalitiesinthelastcalculationmustbeequalitiesandweconclude thatB2M() and, choosingE =B; results in 1 (B) =  (A ): i i=1 Thus M() is a -algebra and the restriction of  to M() is a positive measure. Finally we prove that the the restriction of  to M() is a complete measure. SupposeBA2M() and(A) = 0: If EX; c c (E)(E\B)+(E\B )(E\B )(E) and soB2M(): The theorem is proved.32 Exercises 1. Suppose  : P(X) 0;1; i = 1;2; are outer measures. Prove that i  = max( ; ) is an outer measure. 1 2 2. Supposea;b2R anda = 6 b: Set = max( ; ): Prove that a b fag;fbg2=M(): 1.5. Existence of Linear Measure The purpose of this section is to show the existence of linear measure on R using the Carathéodory Theorem and a minimum of topology. First let us recall the de…nition of in…mum and supremum of a non- empty subset of the extended real line. Suppose A is a non-empty subset of 1;1 = Rf1;1g: We de…ne 1  x and x  1 for all x 2 1;1: An elementb2 1;1 is called a majorant ofA ifxb for all x2 A and a minorant if x b for all x2 A: The Supremum Axiom states that A possesses a least majorant, which is denoted by supA. From this follows that if A is non-empty, then A possesses a greatest minorant, which is denotedby infA. (Actually, theSupremumAxiomisatheoremincourses where time is spent on the de…nition of real numbers.) Theorem 1.5.1. (The Heine-Borel Theorem; weak form) Let a;b be a closed bounded interval and (U ) a collection of open sets such that i i2I U  a;b: i2I i Then U  a;b i2J i for some …nite subset J of I:33 PROOF. LetA be the set of all x2 a;b such that U  a;x i2J i for some …nite subset J of I: Clearly, a 2 A since a 2 U for some i: Let i c = supA: There exists an i such that c2U : Let c2 a ;b U ; where 0 i 0 0 i 0 0 a b : Furthermore, by the very de…nition of least upper bound, there 0 0 exists a …nite setJ such that U  a;(a +c)=2: i2J i 0 Hence U  a;(c+b )=2 i2Jfi g k 0 0 and it follows thatc2A andc =b. The lemma is proved. A subsetK ofR is called compact if for every family of open subsetsU; i i2 I; with U  K we have U  K for some …nite subset J of I: i2I i i2J i The Heine-Borel Theorem shows that a closed bounded interval is compact. If x;y2R andE; F R; let d(x;y) =jxyj be the distance betweenx andy; let d(x;E) = inf d(x;u) u2E be the distance fromx toE; and let d(E;F) = inf d(u;v) u2E;v2F be the distance between E and F (here the in…mum of the emty set equals 1): Note that for anyu2E; d(x;u)d(x;y)+d(y;u) and, hence d(x;E)d(x;y)+d(y;u)34 and d(x;E)d(x;y)+d(y;E): By interchanging the roles of x andy and assuming thatE = 6 ; we get jd(x;E)d(y;E)jd(x;y): Note that if F R is closed andx2= F; thend(x;F) 0: An outer measure :P(R)0;1 is called a metric outer measure if (AB) =(A)+(B) for all A;B2P(R) such thatd(A;B) 0: Theorem 1.5.2. If  : P(R)0;1 is a metric outer measure, then RM(): PROOF. Let F 2P(R) be closed. It is enough to show that F 2M(): To this end we chooseEX with(E)1 and prove that c (E)(E\F)+(E\F ): Letn 1 be an integer and de…ne   1 c A = x2E\F ; d(x;F) : n n Note thatA A and n n+1 c 1 E\F = A : n n=1 Moreover, since is a metric outer measure (E)((E\F)A ) =(E\F)+(A ) n n and, hence, proving c (E\F ) = lim (A ) n n1 we are done.35 c LetB =A \A : It is readily seen that n n+1 n 1 d(B ;A ) n+1 n n(n+1) since if x2B and n+1 1 d(x;y) n(n+1) then 1 1 1 d(y;F)d(y;x)+d(x;F) + = : n(n+1) n+1 n Now (A )(B A ) =(B )+(A ) 2k+1 2k 2k1 2k 2k1 k :::  (B ) 2i i=1 and in a similar way k (A )  (B ): 2k 2i1 i=1 But(A )(E)1 and we conclude that n 1  (B )1: i i=1 We now use that c 1 E\F =A ( B ) n i i=n to obtain c 1 (E\F )(A )+ (B ): n i i=n c Now, since(E\F )(A ), n c (E\F ) = lim (A ) n n1 and the theorem is proved. PROOF OF THEOREM 1.1.1 IN ONE DIMENSION. Suppose  0: If AR; de…ne 1  (A) = inf l(I )  k k=1 the in…mum being taken over all open intervalsI with l(I ) such that k k 1 A I : k k=136 1 Obviously, () = 0 and (A) (B) ifAB: Suppose (A ) is a    n n=1 denumerable collection of subsets of R and let " 0: For each n there exist open intervalsI ;k2N ; such that l(I ); kn + kn 1 A  I n kn k=1 and 1 n  l(I ) (A )+"2 : kn  n k=1 Then 1 1 A = A  I def n kn n=1 k;n=1 and 1 1  l(I )   (A )+": kn  n k;n=1 n=1 Thus 1  (A)   (A )+"   n n=1 and, since" 0 is arbitrary, 1  (A)   (A ):   n n=1 It follows that is an outer measure.  If I is an open interval it is simple to see that  (I) l(I):  Toprovethereverseinequality,chooseaclosedboundedintervalJ I:Now, if 1 I  I k k=1 whereeachI isanopenintervalof l(I );itfollowsfromtheHeine-Borel k k Theorem that n J  I k k=1 for somen: Hence n 1 l(J)  l(I )  l(I ) k k k=1 k=1 and it follows that l(J) (I)  and, accordingly from this, l(I) (I): 37 Thus, if I is an open interval, then  (I) = l(I).  Note that  if 0  : We de…ne   1 2 1 2  (A) = lim (A) if AR: 0  0 It obvious that is an outer measure such that (I) =l(I); if I is an open 0 0 interval. Tocompletetheproofweshowthat isametricoutermeasure. Tothis 0 end letA;BR andd(A;B) 0: Suppose 0d(A;B) and 1 AB I k k=1 where eachI is an open interval with l(I ): Let k k =fk; I \A6=g k and =fk; I \B6=g: k Then \ =; A I k2 k and B I k2 k and it follows that 1  l(I )  l(I )+ l(I ) k k2 k k2 k k=1  (A)+ (B):   Thus  (AB) (A)+ (B)    and by letting 0 we have  (AB) (A)+ (B) 0 0 0 and  (AB) = (A)+ (B): 0 0 038 Finally by applying the Carathéodory Theorem and Theorem 1.5.2 it follows that the restriction of  toR equalsv . 0 1 We end this section with some additional results of great interest. Theorem 1.5.3. For any  0;  = : Moreover, if AR  0 1  (A) = inf l(I ) 0 k k=1 the in…mum being taken over all open intervals I ; k 2 N ; such that k + 1 I A: k k=1 PROOF. It follows from the de…nition of  that    : To prove the 0  0 reverseinequalityletARandchooseopenintervalsI ;k2N ;suchthat k + 1 I A: Then k k=1 1 1  (A) ( I )   (I ) 0 0 k 0 k k=1 k=1 1 =  l(I ): k k=1 Hence 1  (A) inf l(I ) 0 k k=1 the in…mum being taken over all open intervals I ; k 2 N ; such that k + 1 I  A: Thus  (A)   (A); which completes the proof of Theorem k 0  k=1 1.5.3. Theorem 1.5.4. If AR;  (A) = inf  (U): 0 0 UA U open Moreover, if A2M( ); 0  (A) = sup  (K): 0 0 KA K closed bounded39 PROOF. If AU,  (A) (U): Hence 0 0  (A) inf  (U): 0 0 UA U open Next let " 0 be …xed and choose open intervals I ; k 2 N ; such that k + 1 I A and k k=1 1  l(I ) (A)+" k 0 k=1 (here observe that it may happen that  (A) = 1). Then the set U = 0 def 1 I is open and k k=1 1 1  (U)   (I ) =  l(I ) (A)+": 0 0 k k 0 k=1 k=1 Thus inf  (U) (A) 0 0 UA U open and we have proved that  (A) = inf  (U): 0 0 UA U open If K A;  (K) (A) and, accordingly from this, 0 0 sup  (K) (A): 0 0 KA K closed bounded To prove the reverse inequality we …rst assume thatA2M( ) is bounded. 0 Let" 0 be …xed and supposeJ is a closed bounded interval containingA: Thenweknowfromthe…rstpartofTheorem1.5.4alreadyprovedthatthere exists an open setU JrA such that  (U) (JrA)+": 0 0 But then  (J) (JrU)+ (U) (JrU)+ (JrA)+" 0 0 0 0 0 and it follows that  (A)" (JnU): 0 040 SinceJrU is a closed bounded set contained inA we conclude that  (A) sup  (K): 0 0 KA K closed bounded If A2M( ) let A =A\n;n; n2N : Then given " 0 and n2 0 n + N ; letK beaclosedboundedsubsetofA suchthat (K ) (A )": + n n 0 n 0 n Clearly, there is no loss of generality to assume that K  K  K  ::: 1 2 3 and by lettingn tend to plus in…nity we get lim  (K ) (A)": 0 n 0 n1 Hence  (A) = sup  (K): 0 0 KA K compact and Theorem 1.5.4 is completely proved. Theorem1.5.5. Lebesguemeasure m equalstherestrictionof  toM( ): 1 0 0 PROOF. Recall that linear measurev equals the restriction of toR and 1 0 m =v : First supposeE2R and chooseA;B2R such thatAEB 1 1 andBrA2Z :Butthen (ErA) = 0andE =A(ErA)2M( )since v 0 0 1 the Carathéodory Theorem gives us a complete measure. Hence m (E) = 1 v (A) = (E). 1 0 ConverselysupposeE2M( ):WewillprovethatE2R andm (E) = 0 1  (E). FirstassumethatE isbounded. Thenforeachpositiveintegernthere 0 exist openU E and closed boundedK E such that n n n  (U ) (E)+2 0 n 0 and n  (K ) (E)2 : 0 n 0 1 1 The de…nitions yieldA = K ; B =\ U 2R and n n 1 1  (E) = (A) = (B) =v (A) =v (B) =m (E): 0 0 0 1 1 141 It follows thatE2R and (E) =m (E): 0 1 In the general case set E =E\n;n; n2N : Then from the above n + E 2 R and  (E ) = m (E ) for each n and Theorem 1.5.5 follows by n 0 n 1 n lettingn go to in…nity. The Carathéodory Theorem can be used to show the existence of volume n measure onR but we do not go into this here since its existence follows by several other means below. By passing, let us note that the Carathéodory Theorem is very e¢ cient to prove the existence of so called Haussdor¤mea- sures(seee.g. F);whichareofgreatinterestinGeometricMeasureTheory. Exercises 1. Prove that a subset K of R is compact if and only if K is closed and bounded. 2. Suppose A 2 R and m(A) 1: Set f(x) = m(A\ 1;x); x 2 R: Prove thatf is continuous. 3 3. SupposeA2Z andB =fx ;x2Ag: Prove thatB2Z : m m 4. LetA be the set of all real numbersx such that p 1 jx j 3 q q for in…nitely many pairs of positive integersp andq: Prove thatA2Z : m 5. LetI ;:::;I be open subintervals of R such that 1 n n Q\0;1 I : k k=1 n Prove that  m(I ) 1: k k=142 6. If E 2R and m(E) 0; for every 2 0;1 there is an interval I such 1 that m(E\I) m(I). (Hint: m(E) = inf m(I ), where the in…mum k k=1 1 is taken over all intervals such that I E:) k k=1 7. IfE2R andm(E) 0;thenthesetEE =fxy;x;y2Egcontains an open non-empty interval centred at 0:(Hint: Take an interval I with 3 1 m(E\I) m(I):Set" = m(I):Ifjxj";then(E\I)\(x+(E\I))6=:) 4 2 1 8. Letbetherestrictionofthepositivemeasure   1 toR:Provethat k=1 R; k inf (U)(A) U A U open if A =f0g: 1.6. Positive Measures Induced by Increasing Right Continuous Functions SupposeF :R 0;1 is a right continuous increasing function such that lim F(x) = 0: x1 Set L = lim F(x): x1 Wewillprovethatthereexistsauniquepositivemeasure :R 0;Lsuch that (1;x) =F(x); x2R: This measure will often be denoted by : F The special caseL = 0 is trivial so let us assumeL 0 and introduce H(y) = inffx2R;F(x)yg; 0yL: The de…nition implies that the functionH increases.43 Supposea is a …xed real number. We claim that fy2 0;L; H(y)ag = 0;F(a)\0;L: To prove this …rst suppose that y 2 0;L and H(y)  a: Then to each n positive integern; there is anx 2 H(y);H(y)+2 such thatF(x )y: n n Thenx H(y)asn1andweobtainthatF(H(y))y sinceF isright n continuous. Thus, remembering that F increases, F(a)  y: On the other hand, if 0y L and 0yF(a); then, by the very de…nition of H(y); H(y)a: We now de…ne  =H(v ) 1j0;L and get (1;x) =F(x); x2R: The uniqueness follows at once from Theorem 1.2.3. Note that the measure  is a probability measure if L = 1: Example 1.1.1. If  0 if x 0 F(x) = 1 if x 0 then is the Dirac measure at the point 0 restricted toR. F Example 1.1.2. If Z x 2 t dt 2 F(x) = e p (a Riemann integral) 2 1 then is called the standard Gaussian measure on R: F Exercises 1. SupposeF :RR is a right continuous increasing function. Prove that there is a unique positive measure onR such that (a;x) =F(x)F(a); if a;x2R andax:44 2. Suppose F : R R is an increasing function. Prove that the set of all discontinuity points of F is at most denumerable. (Hint: Assume …rst that F is bounded and prove that the set of all points x 2 R such that F(x+)F(x)" is …nite for every" 0.) 3. Suppose  is a -…nite positive measure on R: Prove that the set of all x2R such that(fxg) 0 is at most denumerable. 4. Suppose is a-…nite positive measure onR : Prove that there is an at n most denumerable set of hyperplanes of the type x =c (k = 1;:::;n; c2R) k with positive-measure. 5. Construct an increasing function f : RR such that the set of discon- tinuity points of f equalsQ.45 CHAPTER2 INTEGRATION Introduction In this chapter Lebesgue integration in abstract positive measure spaces is introduced. A series of famous theorems and lemmas will be proved. 2.1. Integration of Functions with Values in 0;1 Recall that 0;1 = 0;1f1g: A subinterval of 0;1 is de…ned in the natural way. We denote byR the-algebra generated by all subintervals 0;1 of 0;1: The class of all intervals of the type ;1; 0  1; (or of the type ;1; 0 1) generates the -algebra R and we get the 0;1 following Theorem 2.1.1. Let (X;M) be a measurable space and suppose f : X 0;1: 1 (a) The function f is (M;R )-measurable if f ( ;1) 2 M for 0;1 every 0 1: 1 (b) The function f is (M;R )-measurable if f ( ;1) 2 M for 0;1 every 0 1: Notethattheset ff g2Mforallreal iff is(M;R )-measurable. 0;1 Iff;g :X 0;1are(M;R )-measurable,thenmin(f;g); max(f;g), 0;1 andf +g are (M;R )-measurable, since, for each 2 0;1; 0;1 min(f;g) , (f  andg ) max(f;g) , (f  org )46 and ff +g g = (ff qg\fgqg): q2Q Given functions f : X 0;1; n = 1;2;:::; f = sup f is de…ned n n n1 by the equation f(x) = supff (x); n = 1;2;:::g: n Note that 1 1 1 f ( ;1) = f ( ;1) n=1 n for every real  0 and, accordingly from this, the function sup f is n1 n (M;R )-measurable if each f is (M;R )-measurable. Moreover, f = 0;1 n 0;1 inf f is given by n1 n f(x) = infff (x); n = 1;2;:::g: n Since 1 1 1 f (0; ) = f (0; ) n=1 n foreveryreal  0weconcludethatthefunctionf = inf f is(M;R )- n1 n 0;1 measurable if eachf is (M;R )-measurable. n 0;1 Below we write f "f n iff ;n = 1;2;:::; andf arefunctionsfromX into 0;1 suchthatf f n n n+1 for eachn andf (x)f(x) for eachx2X as n1: n An (M;R )-measurable function ' : X 0;1 is called a simple 0;1 measurable function if'(X) is a …nite subset of 0;1: If it is neccessary to be more precise, we say that' is a simpleM-measurable function. Theorem 2.1.2. Let f :X 0;1 be (M;R )-measurable. There exist 0;1 simple measurable functions ' ; n2N ; on X such that ' "f : + n n PROOF. Givenn2N , set +   i1 i 1 E =f ( ; ); i2N in + n n 2 247 and 1 X i1  =  +1 1 : n E f (f1g) n in 2 i=1 It is obvious that  f and that  : Now set' = min(n; ) and n n n+1 n n we are done. Let (X;M;) be a positive measure space and ' :X 0;1 a simple measurablefunction:If ;:::; arethedistinctvaluesofthesimplefunction 1 n 1 ', and if E =' (f g); i = 1;:::;n; then i i n ' =   : i i=1 E i Furthermore, if A2M we de…ne Z n n E i (A) = 'd =  (E \A) =   (A): i i i i=1 k=1 A n Note that this formula still holds if (E ) is a measurable partition ofX and i 1 ' = onE for eachi = 1;:::;n: Clearly, is a positive measure since each i i term in the right side is a positive measure as a function of A. Note that Z Z 'd = 'd if 0 1 A A and Z 'd =a(A) A if a2 0;1 and' is a simple measurable function such that' =a onA: If is another simple measurable function and' ; Z Z 'd d: A A  1 To see this, let ;:::; be the distinct values of and F = ( ); j 1 p j j = 1;:::;p: Now, puttingB =E \F ; ij i j Z 'd =( (A\B )) ij ij A Z Z =  (A\B ) =  'd =  d ij ij ij ij i A\B A\B ij ij48 Z Z   d = d: ij j A\B A ij In a similar way one proves that Z Z Z ('+ )d = 'd+ d: A A A From the above it follows that Z Z n ' d =   d i A i=1 E \A i A A Z n n =   d =  (E \A) i i i i=1 E \A i=1 i A and Z Z ' d = 'd: A A A If f :X 0;1 is an (M;R )-measurable function and A2M, we 0;1 de…ne Z Z  fd = sup 'd; 0'f; ' simple measurable A A   Z c = sup 'd; 0'f; ' simple measurable and' = 0 onA : A The left member in this equation is called the Lebesgue integral off overA withrespecttothemeasure:Sometimeswealsospeekofthe-integraloff overA: The two de…nitions of the-integral of a simple measurable function ' :X 0;1 overA agree. Fromnowon in this section, an (M;R )-measurable functionf :X 0;1 0;1 is simply called measurable. The following properties are immediate consequences of the de…nitions. The functions and sets occurring in the equations are assumed to be mea- surable. R R (a) If f; g 0 andf g onA; then fd gd: A A49 R R (b) fd =  fd: A A X R R (c) If f  0 and 2 0;1, then fd = fd: A A R (d) fd = 0 if f = 0 and(A) =1: A R (e) fd = 0 if f =1 and(A) = 0: A If f :X 0;1 is measurable and 0 1; thenf   1 = f ( ;1)  and ff g Z Z Z fd  d =  d: ff g ff g X X X This proves the so called Markov Inequality Z 1 (f  ) fd X where we write(f  ) instead of the more precise expression(ff  g): Example 2.1.1. Supposef :X 0;1 is measurable and Z fd1: X We claim that 1 ff =1g =f (f1g)2Z :  To prove this we use the Markov Inequality and have Z 1 (f =1)(f  ) fd X50 for each 2 0;1: Thus(f =1) = 0: Example 2.1.2. Supposef :X 0;1 is measurable and Z fd = 0: X We claim that 1 ff 0g =f (0;1)2Z :  To see this, note that   1 1 1 1 f (0;1) = f ( ;1 ) n=1 n Furthermore, for every …xedn2N ; the Markov Inequality yields + Z 1 (f )n fd = 0 n X and we get ff 0g2Z since a countable union of null sets is a null set.  We now come to one of the most important results in the theory. Theorem 2.1.3. (Monotone Convergence Theorem) Let f : X n 0;1 , n = 1;2;3;::::; be a sequence of measurable functions and suppose that f "f; that is 0f f ::: and n 1 2 f (x)f(x) as n1, for everyx2X: n Then f is measurable and Z Z f d fd as n1: n X X PROOF. The functionf is measurable sincef = sup f : n n151 R R R The inequalitiesf f f yield f d f d fd and n n+1 n n+1 X X X we conclude that there exists an 2 0;1 such that Z f d as n1 n X and Z  fd: X To prove the reverse inequality, let ' be any simple measurable function such that 0  '  f, let 0  1 be a constant, and de…ne, for …xed n2N ; + A =fx2X; f (x)'(x)g: n n If ;:::; are the distinct values of '; 1 p p A = (fx2X; f (x) g\f' = g) n n k k k=1 and it follows that A is measurable. Clearly, A  A  ::: . Moreover, if n 1 2 f(x) = 0; then x2 A and if f(x) 0; then '(x) f(x) and x2 A for 1 n 1 all su¢ ciently largen. Thus A =X: Now n n=1 Z Z  f d 'd n A A n n and we get Z  'd X R since the mapA 'd is a positive measure onM: By letting" 1, A Z  'd X and, hence Z  fd: X The theorem follows. Theorem 2.1.4. (a) Let f;g :X 0;1 be measurable functions. Then Z Z Z (f +g)d = fd+ gd: X X X52 (b) (Beppo Levi’s Theorem) If f : X 0;1 , k = 1;2;::: are mea- k surable, Z Z 1 1  f d =  f d k k k=1 k=1 X X 1 1 PROOF.(a)Let (' ) and ( ) besequencesofsimpleandmeasurable n n=1 n n=1 functions such that 0' "f and 0 "g: We proved above that n n Z Z Z (' + )d = ' d+ d n n n n X X X and, by letting n 1; Part (a) follows from the Monotone Convergence Theorem. (b) Part (a) and induction imply that Z Z n n  f d =  f d k k k=1 k=1 X X and the result follows from monotone convergence. Theorem 2.1.5. Suppose w : X 0;1 is a measurable function and de…ne Z (A) = wd; A2M: A Then  is a positive measure and Z Z fd = fwd; A2M A A for every measurable function f :X 0;1: 1 PROOF. Clearly, () = 0. Suppose (E ) is a disjoint denumerable col- k k=1 1 lection of members ofM and setE = E : Then k k=1 Z Z Z 1 1 ( E ) = wd =  wd =   wd k k=1 E k=1 E k E X X53 where, by the Beppo Levi Theorem, the right member equals Z Z 1 1 1   wd =  wd =  (E ): k k=1 E k=1 k=1 k X E k This proves that is a positive measure. Let A 2 M. To prove the last part in Theorem 2.1.5 we introduce the classC of all measurable functions f :X 0;1 such that Z Z fd = fwd: A A The indicator function of a measurable set belongs to C and from this we concludethateverysimplemeasurablefunctionbelongstoC:Furthermore,if f 2C;n2N;andf "f ;theMonotoneConvergenceTheoremprovesthat n n f 2C: Thus in view of Theorem 2.1.2 the classC contains every measurable functionf :X 0;1: This completes the proof of Theorem 2.1.5. The measure in Theorem 2.1.5 is written  =w or d =wd: 1 Let( ) beasequencein1;1:Firstput = inff ; ; ;:::g n k k+1 k+2 n=1 k 1 and = supf ; ; ;::g = lim :Wecall thelowerlimitof ( ) n1 n 1 2 3 n n=1 and write = liminf : n n1 Note that = lim n n1 ifthelimitexists. Nowput = supf ; ; ;:::gand = inff ; ; ;::g = k k+1 k+2 k 1 2 3 1 lim : We call the upper limit of ( ) and write n1 n n n=1 = limsup : n n1 Note that = lim n n154 if the limit exists. Given measurable functions f : X 0;1; n = 1;2;:::; the function n liminf f is measurable. In particular, if n1 n f(x) = lim f (x) n n1 exists for everyx2X; thenf is measurable. Theorem 2.1.6. (Fatou’s Lemma) If f :X 0;1; n = 1;2;:::; are n measurable Z Z liminff d liminf f d: n n n1 n1 X X PROOF. Introduce g = inf f : k n nk The de…nition gives thatg " liminf f and, moreover, k n1 n Z Z g d f d; nk k n X X and Z Z g d inf f d: k n nk X X The Fatou Lemma now follows by monotone convergence. Below we often write Z f(x)d(x) E instead of Z fd: E Example 2.1.3. Suppose a 2 R and f : (R;R ) (0;1;R ) is 0;1 measurable. We claim that Z Z f(x+a)dm(x) = f(x)dm(x): R R55 First if f = ; whereA2R , A Z Z f(x+a)dm(x) =  (x)dm(x) =m(Aa) = Aa R R Z m(A) = f(x)dm(x): R Next it is clear that the relation we want to prove is true for simple mea- surable functions and …nally, we use the Monotone Convergence Theorem to deduce the general case. 1 Example 2.1.3, Suppose  a is a positive convergent series and letE be n 1 the set of all x2 0;1 such that p a n min jx j p2f0;:::;ng n n for in…nitely manyn2N : We claim thatE is a Lebesgue null set. + To prove this claim for …xed n 2 N ; let E be the set of all x 2 0;1 + n such that p a n min jx j : p2N n n + Then if B(x;r) = xr;x+r; x2 0;1; r 0; we have n p a n E  B( ; ) n n n p=0 and 2a n m(E ) (n+1)  4a : n n n Hence 1 X m(E )1 n 1 and by the Beppo Levi theorem Z 1 1 X  dm1: E n 0 156 Accordingly from this the set ( ) 1 X F = x2 0;1;  (x)1 E n 1 is of Lebesgue measure 1. SinceE 0;1nF we havem(E) = 0: Exercises 1. Supposef :X 0;1; n = 1;2;:::; are measurable and n 1  (f 1)1: n n=1 Prove that   limsupf 1 2Z . n  n1 2 2. Set f =n  ; n2N : Prove that 1 n + 0; n Z Z liminff dm = 01 = liminf f dm n n n1 n1 R R (the inequality in the Fatou Lemma may be strict). 3. Supposef : (R;R ) (0;1;R ) is measurable and set 0;1 1 g(x) =  f(x+k); x2R: k=1 Show that Z gdm1 if and only if ff 0g2Z : m R 4. Let (X;M;) be a positive measure space and f : X 0;1 an (M;R )-measurable function such that 0;1 f(X)N57 and Z fd1: X For everyt 0, set F(t) =(f t) andG(t) =(f t): Prove that Z 1 1 fd =  F(n) =  G(n): n=0 n=1 X 2.2. Integration of Functions with Arbitrary Sign Asusualsuppose (X;M;)isapositivemeasurespace. Inthissectionwhen we speak of a measurable function f :X R it is understood that f is an (M;R)-measurable function, if not otherwise stated. If f;g : X R are measurable, the sumf +g is measurable since ff +g g = (ff qg\fgqg) q2Q for each real : Besides the function f and the di¤erence f g are mea- surable. It follows that a function f : X R is measurable if and only if + the functions f = max(0;f) and f = max(0;f) are measurable since + f =f f : 1 We writef 2L () if f :X R is measurable and Z jf jd1 X and in this case we de…ne Z Z Z + fd = f d f d: X X X Note that Z Z j fdj jf jd X X58 + sincejf j=f +f : Moreover, if E2M we de…ne Z Z Z + fd = f d f d E E E and it follows that Z Z fd =  fd: E E X Note that Z fd = 0 if (E) = 0: E Sometimes we write Z f(x)d(x) E instead of Z fd: E 1 If f;g2L (); settingh =f +g; Z Z Z jhjd jf jd+ jgjd1 X X X 1 and it follows thath+g2L (): Moreover, + + + h h =f f +g g and the equation + + + h +f +g =f +g +h gives Z Z Z Z Z Z + + + h d+ f d+ g d = f d+ g d+ h d: X X X X X X Thus Z Z Z hd = fd+ gd: X X X Moreover, Z Z fd = fd X X59 for each real : The case  0 follows from (c) in Section 2.1. The case + + =1 is also simple since (f) =f and (f) =f : Theorem 2.2.1. (Lebesgue’s Dominated Convergence Theorem) Suppose f :X R; n = 1;2;:::; are measurable and n f(x) = lim f (x) n n1 1 exists for every x2X: Moreover, suppose there exists a function g2L () such that jf (x)jg(x); all x2X andn2N : n + 1 Then f 2L (), Z lim jf f jd = 0 n n1 X and Z Z lim f d = fd n n1 X X Proof. Since j f j g, the function f is real-valued and measurable since + f andf are measurable. Note here that   f (x) = lim f (x);all x2X: n n1 We now apply the Fatous Lemma to the functions 2g j f f j;n = n 1;2;:::; and have Z Z 2gd liminf (2gjf f j)d n n1 X X Z Z = 2gdlimsup jf f jd: n n1 X X R But 2gd is …nite and we get X Z lim jf f jd = 0: n n1 X Since Z Z Z Z j f d fdj=j (ff )dj jff jd n n n X X X X60 the last part in Theorem 2.2.1 follows from the …rst part. The theorem is proved. Example2.2.1. Supposef : a;bX R isafunctionsuchthatf(t;)2 f 1 L () for eacht2 a;b and, moreover, assume exists and t f j (t;x)jg(x) for all (t;x)2 a;bX t 1 whereg2L (). Set Z F(t) = f(t;x)d(x) if t2 a;b: X We claim thatF is di¤erentiable and Z f 0 F (t) = (t;x)d(x): t X 1 To see this lett 2 a;b be …xed and choose a sequence (t ) in a;bn  n n=1 ft g which converges tot : De…ne   f(t ;x)f(t ;x) n  h (x) = if x2X: n t t n  Here eachh is measurable and n f lim h (x) = (t ;x) for all x2X: n  n1 t Furthermore,foreach…xednandxthereisa 2 t ;t suchthath (x) = n;x n  n f ( ;x) and we conclude thatjh (x)jg(x) for everyx2X: Since n;x n t Z F(t )F(t ) n  = h (x)d(x) n t t n  X theclaimabovenowfollowsfromtheLebesgueDominatedConvergenceThe- orem. Suppose S(x) is a statement, which depends on x2X: We will say that S(x) holds almost (or -almost) everywhere if there exists an N 2Z such 61 that S(x) holds at every point of XnN: In this case we write ”S holds a.e. ”or ”S holds a.e. ”. Sometimes we prefer to write ”S(x) holds a.e.” or ”S(x) holds a.e. ”: If the underlying measure space is a probability space, we often say ”almost surely”instead of almost everywhere. The term ”almost surely”is abbreviated a.s. Supposef :X R; is an(M;R)-measurablefunctionsandg :X R: If f = g a.e.  there exists an N 2 Z such that f(x) = g(x) for every  x2 XnN: We claim that g is (M ;R)-measurable. To see this let 2 R and use that fg g = ff g\(XnN)fg g\N: Now if we de…ne A =ff g\(XnN) the setA2M and Afg gAN: Accordingly from thisfg g2M andg is (M ;R)-measurable since is an arbitrary real number. Next supposef :X R;n2N ; is a sequence of (M;R)-measurable n + functions andf :X R a function. Recall if lim f (x) =f(x); all x2X n n1 thenf is (M;R)-measurable since  1 ff g = \ f +l ; all 2R: k;l2N nk n + If we only assume that lim f (x) =f(x); a.e.  n n1 then f need not be (M;R)-measurable but f is (M ;R)-measurable. To see this supposeN 2Z and  lim f (x) =f(x); all x2XnN: n n1 Then lim  (x)f (x) = (x)f(x) n XnN XnN n162 and it follows that the function  f is (M;R)-measurable. Since f = XnN  f a.e.  it follows that f is (M ;R)-measurable. The next example XnN shows thatf need not be (M;R)-measurable. Example 2.2.2. Let X =f0;1;2g;M =f;f0g;f1;2g;Xg; and (A) =  (0); A 2 M: Set f =  ; n 2 N ; and f(x) = x; x 2 X: Then each n + A f1;2g f is (M;R)-measurable and n lim f (x) =f(x) a.e.  n n1 since n o x2X; lim f (x) =f(x) =f0;1g n n1 andN =f1;2g is a-null set. The functionf is not (M;R)-measurable. 1 Supposef;g2L ():Thefunctionsf andg areequalalmosteverywhere with respect to  if and only if ff = 6 gg 2 Z : This is easily seen to be an  1 equivalencerelationandthesetofallequivalenceclassesisdenotedbyL (): Moreover, if f =g a.e. ; then Z Z fd = gd X X since Z Z Z Z Z fd = fd+ fd = fd = gd X ff=gg ff= 6 gg ff=gg ff=gg and, in a similar way, Z Z gd = gd: X ff=gg 1 1 BelowweconsidertheelementsofL ()membersofL ()andtwomembers 1 of L () are identi…ed if they are equal a.e. : From this convention it is 1 straight-forward to de…ne f +g and f for all f;g 2 L () and 2 R: Moreover, we get Z Z Z 1 (f +g)d = fd+ gd if f;g2L () X X X63 and Z Z 1 fd = fd if f 2L () and 2R: X X Next we give two theorems where exceptional null sets enter. The …rst one is a mild variant of Theorem 2.2.1 and needs no proof. Theorem 2.2.2. Suppose (X;M;) is a positive complete measure space and let f :X R; n2N ; be measurable functions such that n + sup jf (x)jg(x) a.e.  n n2N + 1 where g2L (): Moreover, suppose f :X R is a function and f(x) = lim f (x) a.e. : n n1 1 Then, f 2L (), Z lim jf f jd = 0 n n1 X and Z Z lim f d = fd: n n1 X X Theorem 2.2.3. Suppose (X;M;) is a positive measure space. (a) If f : (X;M ) (0;1;R ) is measurable there exists a measur- 0;1 able function g : (X;M) (0;1;R ) such that f =g a.e. : 0;1 (b) If f : (X;M ) (R;R) is measurable there exists a measurable function g : (X;M) (R;R) such that f =g a.e. : + PROOF.Sincef =f f itisenoughtoprovePart(a):Thereexistsimple M -measurable functions' ;n2N ; such that 0' "f: For each …xed + n n n suppose ;:::; arethedistinctvaluesof' andchooseforeach…xed 1n k n n n 1 1 i = 1;:::;k a set A  ' (f g) such that A 2M and ' ( )nA n in in in in in n n 2Z . Set   k n =   : in n A i=1 in64 k 1 n Clearly (x) " f(x) if x 2 E = \ ( A ) and (XnE) = 0: We def in n n=1 i=1 now de…ne g(x) =f(x); if x2E; and g(x) = 0 if x2XnE: The theorem is proved. Exercises 2 1. Supposef andg are real-valued measurable functions. Prove thatf and fg are measurable functions. 1 2. Supposef 2L (): Prove that Z lim jf jd = 0: 1 jfj R R (Here means .) jfj fjfj g 1 3. Suppose f 2 L (): Prove that to each " 0 there exists a  0 such that Z jf jd" E whenever(E): 1 4. Let (f ) be a sequence of (M;R)-measurable functions. Prove that n n=1 1 the set of all x 2 R such that the sequence (f (x)) converges to a real n n=1 limit belongs toM: 5. Let (X;M;R) be a positive measure space such that (A) = 0 or1 for 1 everyA2M: Show thatf 2L () if and only if f(x) = 0 a.e. : 6. Let (X;M;) be a positive measure space and suppose f and g are non-negative measurable functions such that Z Z fd = gd; all A2M: A A65 (a) Prove thatf =g a.e.  if  is-…nite. (b) Prove that the conclusion in Part (a) may fail if  is not-…nite. 7. Let(X;M;)bea…nitepositivemeasurespaceandsupposethefunctions f : X R; n = 1;2;:::; are measurable. Show that there is a sequence n 1 ( ) of positive real numbers such that n n=1 lim f = 0 a.e. : n n n1 8. Let(X;M;)beapositivemeasurespaceandletf :X R;n = 1;2;:::; n 1 be a sequence in L () which converges to f a.e.  as n 1: Suppose 1 f 2L () and Z Z lim jf jd = jf jd: n n1 X X Show that Z lim jf f jd = 0: n n1 X 1 9. Let (X;M;) be a …nite positive measure space and suppose f 2 L () is a bounded function such that Z Z Z 2 3 4 f d = f d = f d: X X X Prove thatf = for an appropriateA2M: A 10. Let (X;M;) be a …nite positive measure space and f : X R a 1 measurable function. Prove thatf 2L () if and only if 1  (jf jk)1: k=1 1 1 11. Suppose f 2L (m): Prove that the series  f(x+k) converges for k=1 m-almost all x:66 R 12. a) Suppose f : R0;1 is Lebesgue measurable and fdm 1: R Prove that lim m(f  ) = 0: 1 b) Find a Lebesgue measurable function f : R0;1 such that f 2= 1 L (m); m(f 0)1; and lim m(f  ) = 0: 1 13. (a) SupposeM is an-algebra of subsets ofX and a positive measure onM with(X)1: LetA ;:::;A 2M: Show that 1 n  = 1(1 ):::(1 ) A A :::A A A 1 2 n 1 n and conclude that X X (A A :::A ) = (A ) (A \A ) 1 2 n i i i 1 2 1in 1i i n 1 2 X n+1 + (A \A \A ):::+(1) (A \:::\A ): i i i 1 n 1 2 3 1i i i n 1 2 3 (b) Let X be the set of all permutations (bijections) x : f1;2;:::;ng 1 f1;2;:::;ng and let = c . A random variable : X has the uniform X n distribution inX or, stated otherwise, the image measureP equals: Find  the probability that has a …xed point, that is …nd P (i) =i for somei2f1;2;:::;ng: (Hint: SetA =fx2X; x(i) =ig;i = 1;:::;n;andnotethattheprobability i in question equals (A A :::A ).) 1 2 n 14. Let (X;M;) be a positive measure space and f : X R an (M,R)- measurable function. Moreover, for eacht 1; let 1 X n n n+1 a(t) = t (t jf jt ): n=167 Show that Z lim a(t) = jf jd: + t1 X 15. Let (X;M;) be a positive measure space and f :X R; n 2 N ; a n + sequence of measurable functions such that 2 2 limsupn (jf jn )1: n n1 P 1 Prove that the series f (x) converges for-almost all x2X: n n=1 16. Let (X;M;) be a positive measure space and f:X R a measurable function. Furthermore, suppose there are strictly positive constants B and C such that Z 2 a C af 2 e dBe if a2R: X Prove that 2 t 2C (jf jt) 2Be if t 0: 2.3 Comparison of Riemann and Lebesgue Integrals In this section we will show that the Lebesgue integral is a natural general- ization of the Riemann integral. For short, the discussion is restricted to a closed and bounded interval. Let a;b be a closed and bounded interval and suppose f : a;bR is a bounded function. For any partition  :a =x x :::x =b 0 1 n of a;b de…ne n S f =  ( sup f)(x x )  i i1 i=1 x ;x i1 i and n s f =  ( inf f)(x x ):  i i1 i=1 x ;x i1 i68 The functionf is Riemann integrable if infS f = sups f     R b and the Riemann integral f(x)dx is, by de…nition, equal to this common a value. Below an ((R ) ;R)-measurable function is simply called Lebesgue a;b measurable. Furthermore, we writem instead of m : ja;b Theorem 2.3.1. A bounded function f : a;b R is Riemann integrable if and only if the set of discontinuity points of f is a Lebesgue null set. Moreover, if the set of discontinuity points of f is a Lebesgue null set, then f is Lebesgue measurable and Z Z b f(x)dx = fdm: a a;b 0 0 0 0 PROOF. A partition  : a = x x ::: x = b is a re…nement of a 0 0 1 n 0 partition  :a =x x :::x =b if eachx is equal to somex and in 0 1 n k l 0 this case we write  : The de…nitions giveS f S 0f ands f s 0f     0 if  : We de…ne, mesh() = max (x x ): 1in i i1 First supposef is Riemann integrable. For each partition  let n G =f(a) + ( sup f)  fag i=1 x ;x i1 i x ;x i1 i and n g =f(a) + ( inf f)  fag i=1 x ;x i1 i x ;x i1 i and note that Z G dm =S f   a;b and Z g dm =s f:   a;b69 Suppose  ;k = 1;2;:::;isasequenceofpartitionssuchthat    , k k k+1 Z b S f f(x)dx  k a and Z b s f " f(x)dx  k a as k 1: Let G = lim G and g = lim g : Then G and g are k1  k1  k k (R ;R)-measurable, gf G; and by dominated convergence a;b Z Z Z b Gdm = gdm = f(x)dx: a a;b a;b But then Z (Gg)dm = 0 a;b so that G = g a.e. m and therefore G = f a.e. m: In particular, f is Lebesgue measurable and Z Z b f(x)dx = fdm: a a;b Set N =fx; g(x)f(x) orf(x)G(x)g: Weprovedabovethatm(N) = 0:LetM betheunionofallthosepointswhich belong to some partition  : Clearly, m(M) = 0 since M is denumerable. k We claim that f is continuous o¤N M: If f is not continuous at a point 1 c 2= N M, there is an " 0 and a sequence (c ) converging to c such n n=1 that jf(c )f(c)j" all n: n Since c2= M, c is an interior point to exactly one interval of each partition  and we get k G (c)g (c)"   k k and in the limit G(c)g(c)": But thenc2N which is a contradiction.70 Conversely, supposethesetofdiscontinuitypointsoff isaLebesguenull 1 set and let ( ) is an arbitrary sequence of partitions of a;b such that k k=1    and mesh( ) 0 as k1: By assumption, k k+1 k lim G (x) = lim g (x) =f(x)   k k k1 k1 at each point x of continuity of f. Therefore f is Lebesgue measurable and dominated convergence yields Z Z lim G dm = fdm  k k1 a;b a;b and Z Z lim g dm = fdm:  k k1 a;b a;b Thusf is Riemann integrable and Z Z b f(x)dx = fdm: a a;b In the following we sometimes write Z f(x)dx (A2R ) A instead of Z fdm (A2R ): A In a similar way we often prefer to write Z f(x)dx (A2R ) n A instead of Z fdm (A2R ): n n A R R b Furthermore, fdm means fdm: Here, however, a warning is moti- a a;b vated. Itissimpleto…ndareal-valuedfunctionf on0;1,whichisbounded71 on each bounded subinterval of 0;1; such that the generalized Riemann integral Z 1 f(x)dx 0 is convergent, that is Z b lim f(x)dx b1 0 exists and the limit is a real number, while the Riemann integral Z 1 jf(x)jdx 0 sinx is divergent (take e.g. f(x) = ): In this case the function f does not x 1 belong toL with respect to Lebesgue measure on 0;1 since Z Z b jf jdm = lim jf(x)jdx =1: b1 0;1 0 Example 2.3.1. To compute Z n x n (1 ) n lim p dx n1 x 0 t supposen2N and use the inequality 1+te; t2R; to get + x n x  (x)(1 ) e if x 0: 0;n n From this x n x (1 ) e n f (x) =  (x) p  p ; x 0 n def 0;n x x and, in addition, x e lim f (x) = p : n n1 x x x e 1 e p p Here 2L (m on 0;1) since  0 and 1 x x Z Z 1 1 x p e 2 x p dx = 2 e dx = : x 0 072 Moreover f  0 for every n2N and by using dominated convergence we n + get Z Z x n n 1 (1 ) n p lim dx = lim f (x)dx = n n1 n1 x 0 0 Z Z 1 1 x p e lim f (x)dx = p dx = : n n1 x 0 0 Exercises 1. Let f : 0;1 0;1, n 2 N; be a sequence of Riemann integrable n functions such that lim f (x) exists =f(x) all x2 0;1: n n1 Show by giving an example thatf need not be Riemann integrable. 2 njxj 2. Suppose f (x) =n jxje ; x2R; n2N : Compute lim f and n + n1 n R lim f dm. n1 n R 3. Compute the following limits and justify the calculations: (a) Z 1 x sin(e ) lim dx: 2 n1 1+nx 0 (b) Z n x n lim (1+ ) cosxdx: n1 n 0 (c) Z n x n 2x lim (1+ ) e dx: n1 n 0 (d) Z 1 x x n n lim (1+ ) exp((1+ ) )dx: n1 n n 0 (e) Z 1 x sin( ) n lim n dx: 2 n1 x(1+x ) 073 (f) Z n x 1+nx n lim (1 ) cosxdx n1 n n+x 0 (g) Z 1 x 2 n nx lim (1+ ) e dx: n1 n 0 (h) Z 1 2 1+nx lim dx: 2 n n1 (1+x ) 0 (i) Z 1 p p 2 n lim n (1t ) (1+ nj sintj)dt: n1 1 1 4. Let (r ) be an enumeration of Q and de…ne n n=1 1 n f(x) =  2 '(xr ) n n=1 1 2 where'(x) = x if 0x 1 and'(x) = 0 if x 0 orx 1: Show that a) Z 1 f(x)dx = 2: 1 b) Z b 2 f (x)dx =1 if ab: a c) f 1 a.s. m: d) sup f(x) = +1 if ab: axb 2 x x n 5. Letn2N and de…nef (x) =e (1 ) ; x2R: Compute + n 2n p Z 2n lim f (x)dx: n p n1 2n74 p p1 n 6. Suppose p2N and de…ne f (x) =n x (1x) ; 0x 1; for every + n n2N : Show that + Z 1 lim f (x)dx = (p1): n n1 0 7. Supposef:0;1R is a continuous function. Find Z 1 2 (nmin(x;1x)) lim n f(x)e dx: n1 0 2.4. Expectation Suppose ( ;F;P) is a probability space and  : ( ;F) (S;S) a random variable. Recallthattheprobabilitylawof isgivenbytheimagemeasure P : By de…nition,  Z Z  d =  ()dP B B S for everyB2S; and, hence Z Z 'd = '()dP S for each simpleS-measurable function ' on S (we sometimes write fg = f(g)): By monotone convergence, we get Z Z fd = f()dP S for every measurable f : S 0;1: Thus if f : S R is measurable, 1 1 f 2L () if and only if f()2L (P) and in this case Z Z fd = f()dP: S 1 In the special case when is real-valued and2L (P); Z Z xd(x) = dP: R The integral in the right-hand side is called the expectation of  and is denoted byE:75 CHAPTER3 Further Construction Methods of Measures Introduction In the …rst section of this chapter we collect some basic results on metric spaces, which every mathematician must know about. Section 3.2 gives a version of the Riesz Representation Theorem, which leads to another and perhaps simpler approach to Lebesgue measure than the Carathéodory The- orem. A reader can skip Section 3.2 without losing the continuity in this paper. The chapter also treats so called product measures and Stieltjes in- tegrals. 3.1. Metric Spaces The construction of our most important measures requires topological con- cepts. For our purpose it will be enough to restrict ourselves to so called metric spaces. A metricd on a setX is a mappingd :XX 0;1 such that (a)d(x;y) = 0 if and only if x =y (b)d(x;y) =d(y;x) (symmetry) (c)d(x;y)d(x;z)+d(z;y) (triangle inequality). Here recall, if A ;:::;A are sets, 1 n A :::A =f(x ;:::;x ); x 2A for all i = 1;:::;ng 1 n 1 n i i AsetX equippedwithametricd iscalledametricspace. Sometimeswe write X = (X;d) to emphasize the metric d: If E is a subset of the metric76 space (X;d); the function d (x;y) = d(x;y); if x;y 2 E; is a metric on jEE E: Thus (E;d ) is a metric space. jEE The function'(t) = min(1;t); t 0; satis…es the inequality '(s+t)'(s)+'(t): Therefore, if d is a metric on X, min(1;d) is a metric on X: The metric min(1;d) is a bounded metric. The setR equipped with the metricd (x;y) =jxyj is a metric space. 1 n More generally, R equipped with the metric d (x;y) =d ((x ;:::;x );(y ;:::;y )) = max jx y j n n 1 n 1 n k k 1kn n is a metric space. If not otherwise stated, it will always be assumed thatR is equipped with this metric. LetC0;Tdenotethevectorspaceofallreal-valuedcontinuousfunctions on the interval 0;T; whereT 0: Then d (x;y) = max jx(t)y(t)j 1 0tT is a metric onC0;T: If (X ;e ); k = 1;:::;n, are metric spaces, k k d(x;y) = max e (x ;y ); x = (x ;:::;x ) ;y = (y ;:::;y ) k k k 1 n 1 n 1kn is a metric on X :::X : The metric d is called the product metric on 1 n X :::X : 1 n If X = (X;d) is a metric space andx2X andr 0; the open ball with centre atx and radiusr is the setB(x;r) =fy2X;d(y;x)rg: IfEX andE iscontainedinanappropriateopenballinX itissaidtobebounded. The diameter of E is, by de…nition, diamE = sup d(x;y) x;y2E and it follows that E is bounded if and only if diamE 1. A subset of X which is a union of open balls in X is called open. In particular, an open ballisanopenset. Theemptysetisopensincetheunionofanemptyfamily of sets is empty. An arbitrary union of open sets is open. The class of all77 open subsets of X is called the topology of X: The metrics d and min(1;d) determine the same topology. A subset E of X is said to be closed if its c complement E relative to X is open. An intersection of closed subsets of  X is closed. If E X, E denotes the largest open set contained in E and   E (or E) the smallest closed set containing E: E is the interior of E and E its closure. The -algebra generated by the open sets in X is called the Borel -algebra inX and is denoted byB(X). A positive measure onB(X) is called a positive Borel measure. 1 A sequence (x ) inX converges tox2X if n n=1 lim d(x ;x) = 0: n n1 1 If, in addition, the sequence (x ) converges toy2X; the inequalities n n=1 0d(x;y)d(x ;x)+d(x ;y) n n imply thaty =x and the limit pointx is unique. If EX andx2X; the following properties are equivalent: (i)x2 E : (ii)B(x;r)\E = 6 ; all r 0: 1 (iii) There is a sequence (x ) inE which converges tox: n n=1 c If B(x;r)\E =, then B(x;r) is a closed set containing E but not x: c  Thus x2= E : This proves that (i))(ii). Conversely, if x2= E ; since E is c c  openthereexistsanopenballB(y;s)suchthatx2B(y;s)E E :Now choose r =sd(x;y) 0 so that B(x;r)B(y;s): Then B(x;r)\E =: This proves (ii))(i). 1 If (ii) holds choose for each n 2 N a point x 2 E with d(x ;x) + n n n and (iii) follows. If there exists an r 0 such that B(x;r)\E = ; then (iii) cannot hold. Thus (iii))(ii).  If EX, the setE nE is called the boundary ofE and is denoted by E: A set AX is said to be dense inX if A =X: The metric space X is called separable if there is an at most denumerable dense subset of X: For n n n example,Q is a dense subset of R : The space R is separable.78 n Theorem 3.1.1. B(R ) =R : n n PROOF. The -algebra R is generated by the open n-cells in R and an n n n open n-cell is an open subset of R : HenceR B(R ): Let U be an open n n n n subset inR and note that an open ball inR = (R ;d ) is an openn-cell. n n If x 2 U there exist an a 2Q \U and a rational number r 0 such that x 2 B(a;r)  U: Thus U is an at most denumerable union of open n-cells n and it follows thatU 2R : ThusB(R )R and the theoremis proved. n n Let X = (X;d) and Y = (Y;e) be two metric spaces. A mapping f : X Y (orf : (X;d) (Y;e) to emphasize the underlying metrics) is said to be continuous at the point a2 X if for every " 0 there exists a  0 such that x2B(a;))f(x)2B(f(a);"): 1 Equivalently this means that for any sequence (x ) inX which converges n n=1 1 toa inX; the sequence (f(x )) converges tof(a) inY: Iff is continuous n n=1 at each point of X, the mapping f is called continuous. Stated otherwise this means that 1 f (V) is open if V is open or 1 f (F) is closed if F is closed. The mappingf is said to be Borel measurable if 1 f (B)2B(X) if B2B(Y) or, what amounts to the same thing, 1 f (V) 2B(X) if V is open. A Borel measurable function is sometimes called a Borel function. A continuous function is a Borel function. Example 3.1.1. Letf : (R;d ) (R;d ) be a continuous strictly increasing 1 1 function and set(x;y) =jf(x)f(y)j;x;y2R: Then is a metric onR.79 De…ne j(x) =x; x2R: The mapping j : (R;d ) (R;) is continuous. We 1 claim that the map j : (R;) (R;d ) is continuous: To see this, let a2R 1 1 and suppose the sequence (x ) converges to a in the metric space (R;); n n=1 that isjf(x )f(a)j 0 as n1: Let" 0: Then n f(x )f(a)f(a+")f(a) 0 if x a+" n n and f(a)f(x )f(a)f(a") 0 if x a": n n Thus x 2 a";a+" if n is su¢ ciently large. This proves that he map n j : (R;) (R;d ) is continuous. 1 The metrics d and  determine the same topology and Borel subsets of 1 R: A mapping f : (X;d) (Y;e) is said to be uniformly continuous if for each " 0 there exists a  0 such that e(f(x);f(y)) " as soon as d(x;y): If x2X andE; F X; let d(x;E) = inf d(x;u) u2E be the distance fromx toE and let d(E;F) = inf d(u;v) u2E;v2F  bethedistancebetweenE andF:Notethatd(x;E) = 0ifandonlyifx2E: If x;y2X andu2E; d(x;u)d(x;y)+d(y;u) and, hence d(x;E)d(x;y)+d(y;u) and d(x;E)d(x;y)+d(y;E): Next supposeE = 6 : Then by interchanging the roles ofx andy; we get jd(x;E)d(y;E)jd(x;y)80 and conclude that the distance function d(x;E); x 2 X; is continuous. In fact, it is uniformly continuous. Ifx2X andr 0; the so called closed ball  B(x;r) = fy2X; d(y;x)rg is a closed set since the map y d(y;x); y2X; is continuous. If F X is closed and" 0, the continuous function 1 X  = max(0;1 d(;F)) F;" " X X X ful…ls 0   1 and  = 1 onF: Furthermore,  (a) 0 if and only F;" F;" F;" if a2F = fx2X; d(x;F)"g: Thus " def X     : F F;" F " 1 Let X = (X;d) be a metric space. A sequence (x ) in X is called n n=1 a Cauchy sequence if to each " 0 there exists a positive integer p such 1 thatd(x ;x )" for alln;mp: If a Cauchy sequence (x ) contains a n m n n=1 1 convergent subsequence (x ) it must be convergent. To prove this claim, n k k=1 1 suppose the subsequence (x ) converges to a pointx2X: Then n k k=1 d(x ;x)d(x ;x )+d(x ;x) m m n n k k can be made arbitrarily small for all su¢ ciently largem by choosingk su¢ - 1 ciently large. Thus (x ) converges tox: n n=1 A subset E of X is said to be complete if every Cauchy sequence in E converges to a point in E: If E  X is closed and X is complete it is clear that E is complete. Conversely, if X is a metric space and a subset E of X is complete, thenE is closed: It is important to know that R is complete equipped with its standard 1 metric. Toseethislet (x ) beaCauchysequence. Thereexistsapositive n n=1 integer such thatjx x j 1 if n;mp: Therefore n m jx jjx x j +jx j 1+jx j n n p p p 1 for all n  p: We have proved that the sequence (x ) is bounded (the n n=1 reader can check that every Cauchy sequence in a metric space has this property). Now de…ne a = supfx2R; there are only …nitely manyn withx xg: n 1 The de…nition implies that there exists a subsequence (x ) ; which con- n k k=1 verges to a (since for any r 0; x 2 B(a;r) for in…nitely many n). The n81 originalsequenceisthereforeconvergentandweconcludethatRiscomplete (equippedwithitsstandardmetricd ):Itissimpletoprovethattheproduct 1 n of n complete spaces is complete and we conclude that R is complete. Let E  X: A family (V ) of subsets of X is said to be a cover of E i i2I 0 if V  E and E is said to be covered by the V s: The cover (V ) is i2I i i i2I i said to be an open cover if each member V is open. The set E is said to be i totally bounded if, for every " 0; E can be covered by …nitely many open balls of radius ": A subset of a totally bounded set is totally bounded. The following de…nition is especially important. De…nition 3.1.1. A subsetE of a metric spaceX is said to be compact if toeveryopencover (V ) ofE,thereisa…nitesubcoverofE,whichmeans i i2I there is a …nite subsetJ of I such that (V ) is a cover of E: i i2J IfK isclosed,K E;andE iscompact,thenK iscompact. Toseethis, let (V ) be an open cover of K: This cover, augmented by the set XnK i i2I is an open cover of E and has a …nite subcover since E is compact. Noting thatK\(XnK) =; the assertion follows. Theorem 3.1.2. The following conditions are equivalent: (a) E is complete and totally bounded. (b) Every sequence in E contains a subsequence which converges to a point of E: (c) E is compact. 1 PROOF. (a))(b). Suppose (x ) is a sequence in E: The set E can be n n=1 1 covered by …nitely many open balls of radius 2 and at least one of them 1 must contain x for in…nitely many n 2 N : Suppose x 2 B(a ;2 ) if n + n 1 1 n 2 N  N = N ; where N is in…nite. Next E\B(a ;2 ) can be 1 0 def + 1 1 2 covered by …nitely many balls of radius 2 and at least one of them must 1 contain x for in…nitely many n 2 N : Suppose x 2 B(a ;2 ) if n 2 N ; n 1 n 2 2 j where N  N is in…nite. By induction, we get open balls B(a ;2 ) and 2 1 j j in…nite sets N  N such that x 2 B(a ;2 ) for all n2 N and j  1: j j1 n j j82 1 Let n n :::, where n 2 N ; k = 1;2;::: . The sequence (x ) is a 1 2 k k n k k=1 Cauchy sequence, and sinceE is complete it converges to a point of E . (b))(a). If E is not complete there is a Cauchy sequence in E with no limit in E: Therefore no subsequence can converge in E; which contradicts (b). On the other hand if E is not totally bounded, there is an " 0 such that E cannot be covered by …nitely many balls of radius ": Let x 2 E 1 n1 be arbitrary. Having chosen x ;:::;x ; pick x 2 En B(x;"); and 1 n1 n i i=1 1 so on. The sequence (x ) cannot contain any convergent subsequence as n n=1 d(x ;x )" if n = 6 m; which contradicts (b). n m f(a) and (b)g )(c). Let (V ) be an open cover of E: Since E is totally i i2I bounded it is enough to show that there is an " 0 such that any open ball of radius " which intersects E is contained in some V: Suppose on the i n contrary that for every n 2 N there is an open ball B of radius  2 + n which intersects E and is contained in no V: Choose x 2 B \ E and i n n 1 assume without loss of generality that (x ) converges to some point x in n n=1 E by eventually going to a subsequence. Suppose x2V and choose r 0 i 0 such that B(x;r)  V : But then B  B(x;r)  V for large n, which i n i 0 0 contradicts the assumption onB : n 1 (c))(b). If (x ) is a sequence in E with no convergent subsequence in n n=1 E, then for everyx2E there is an open ballB(x;r ) which containsx for x n only …nitely many n: Then (B(x;r )) is an open cover of E without a x x2E …nite subcover. n Corollary 3.1.1. A subset of R is compact if and only if it is closed and bounded. PROOF. Suppose K is compact. If x 2 K and x 2= B(0;n) for every n n 1 n 2 N ; the sequence (x ) cannot contain a convergent subsequence. + n n=1 ThusK is bounded. SinceK is complete it is closed.83 n Conversely, supposeK is closed and bounded. SinceR is complete and K is closed, K is complete. We next prove that a bounded set is totally n bounded. It is enough to prove that any n-cell in R is a union of …nitely manyn-cellsI :::I whereeachintervalI ;:::;I hasaprescribedpositive 1 n 1 n length. This is clear and the theorem is proved. Corollary 3.1.2. Suppose f :X R is continuous and X compact: (a) There exists an a2X such that max f =f(a) and a b2X X such that min f =f(b): X (b) The function f is uniformly continuous: PROOF. (a) For eacha2X; letV =fx2X :f(x) 1+f(a)g: The open a cover(V ) ofX hasa…nitesubcoveranditfollowsthatf isbounded. Let a a2K 1 (x ) be a sequence inX such thatf(x ) sup f asn1: SinceX is n n n=1 K 1 compact there is a subsequence (x ) which converges to a point a2X: n k k=1 Thus, by the continuity of f; f(x )f(a) as k1: n k The existence of a minimum is proved in a similar way. (b) If f is not uniformly continuous there exist " 0 and sequences 1 1 n (x ) and (y ) such that j f(x )f(y ) j " and j x y j 2 n n n n n n n=1 n=1 1 for every n 1: Since X is compact there exists a subsequence (x ) of n k k=1 1 1 (x ) which converges to a point a 2 X: Clearly the sequence (y ) n n n=1 k k=1 converges toa and therefore jf(x )f(y )jjf(x )f(a)j +jf(a)f(y )j 0 n n n n k k k k ask1 sincef is continuous. Butjf(x )f(y )j" and we have got n n k k a contradiction. The corollary is proved. Example 3.1.2. Suppose X = 0;1 and de…ne  (x;y) = d (x;y) and 1 1 1 1  (x;y) =j j;x;y2X:AsinExample3.1.1weconcludethatthemetrics 2 x y  and  determine the same topology of subsets of X: The space (X; ) 1 2 184 totally bounded but not complete. However, the space (X; ) is not totally 2 1 bounded but it is complete. To see this, let (x ) be a Cauchy sequence in n n=1 (X; ): AsaCauchysequenceitmustbeboundedandthereforethereexists 2 an " 2 0;1 such that x 2 ";1 for all n: But then, by Corollary 3.1.1, n 1 (x ) contains a convergent subsequence in (X; ) and, accordingly from n n=1 1 this, the same property holds in (X; ): The space (X; ) is not compact, 2 2 since (X; ) is not compact, and we conclude from Theorem 3.1.2 that the 1 space (X; ) cannot be totally bounded. 2 ˆ Example 3.1.3. SetR=Rf1;1g and d(x;y) =j arctanxarctanyj ˆ if x;y2 R: Here   arctan1 = and arctan1 = : 2 2 Example 3.1.1 shows that the standard metric d and the metric d 1 jRR determine the same topology. ˆ We next prove that the metric spaceRis compact. To this end, consider 1 ˆ asequence(x ) inR. IfthereexistsarealnumberM suchthatjx jM n n n=1 1 forin…nitelymanyn;thesequence(x ) containsaconvergentsubsequence n n=1 sincetheinterval M;Miscompact. Intheoppositecase,foreachpositive real number M, either x  M for in…nitely many n or x  M for n n in…nitely many n: Suppose x  M for in…nitely many n for every M 2 n  N : Then d(x ;1) =j arctanx j 0 as k 1 for an appropriate + n n k k 2 1 subsequence (x ) : n k k=1 ˆ ˆ The spaceR= (R,d) is called a two-point compacti…cation of R. It is an immediate consequence of Theorem 3.1.2 that the product of n …nitely many compact metric spaces is compact. Thus R equipped with the product metric is compact. We will …nish this section with several useful approximation theorems. Theorem 3.1.3. Suppose X is a metric space and  positive Borel measure 1 in X: Moreover, suppose there is a sequence (U ) of open subsets of X n n=1 such that 1 X = U n n=185 and (U )1; all n2N : n + Then for each A2B(X) and " 0; there are a closed set F  A and an open set V A such that (V nF)": In particular, for every A2B(X); (A) = inf (V) VA V open and (A) = sup (F) FA F closed 1 If X =R and (A) =   1(A) ; A2R; then(f0g) = 0 and (V) = n=1 n 1 for every open set containingf0g: The hypothesis that the sets U ; n2 n N ;areopen(andnotmerelyBorelsets)isveryimportantinTheorem3.1.3. + PROOF. First suppose that is a …nite positive measure. Let A be the class of all Borel sets A in X such that for every " 0 there exist a closedF A and an openV A such that(V nF)": IfF  1 is a closed subset of X and V = x;d(x;F) ; then V is open and, by n n n Theorem 1.1.2 (f), (V ) (F) as n1. Thus F 2A and we conclude n thatA contains all closed subsets of X: c Now suppose A2A: We will prove that A 2A: To this end, we choose " 0 andaclosedsetF AandanopensetV Asuchthat(VnF)": c c c c c ThenV A F and, moreover, (F nV )" since c c V nF =F nV : c c c If we note thatV is closed andF open it follows thatA 2A: 1 Next let (A ) be a denumerable collection of members of A. Choose i i=1 " 0: By de…nition, for each i 2 N there exist a closed F  A and an + i i i openV A such that(V nF ) 2 ": Set i i i i 1 V = V: i i=186 Then 1 1 (V n( F ))( (V nF )) i i i i=1 i=1 1   (V nF )": i i i=1 But 1 1 n V n( F ) =\ fV n( F )g i i i=1 n=1 i=1 and since is a …nite positive measure 1 n (V n( F )) = lim (V n( F )): i i i=1 i=1 n1 Accordingly, from these equations n (V n( F ))" i i=1 if n is large enough. Since a union of open sets is open and a …nite union of 1 closed sets is closed, we conclude that A 2A: This proves that A is a i i=1 -algebra. SinceA contains each closed subset of X;A =B(X): U n We now prove the general case. Suppose A2B(X): Since  is a …nite positivemeasuretheprevioustheoremgivesusanopensetV A\U such n n U n n that (V n(A\U ))"2 :ByeventuallyreplacingV byV \U wecan n n n n n U n n assumethatV U :Butthen(V n(A\U )) = (V n(A\U ))"2 : n n n n n n 1 SetV = V and note thatV is open. Moreover, n n=1 1 V nA (V n(A\U )) n n n=1 and we get 1 (V nA)  (V n(A\U ))": n n n=1 c By applying the result already proved to the complement A we conclude c there exists an open setW A such that c c (AnW ) =(W nA )": c Thus if F = W it follows that F  A  V and (V nF) 2": The def theorem is proved. If X is a metric space C(X) denotes the vector space of all real-valued continuous functions f : X R: If f 2 C(X); the closure of the set of87 all x where f(x) = 6 0 is called the support of f and is denoted by suppf: The vector space of all all real-valued continuous functions f :X R with compact support is denoted byC (X): c n Corollary 3.1.3. Suppose  and  are positive Borel measures in R such that (K)1 and(K)1 n for every compact subset K of R : If Z Z n f(x)d(x) = f(x)d(x); all f 2C (R ) c n n R R then  =: PROOF. Let F be closed. Clearly (B(0;i)) 1 and (B(0;i)) 1 for every positive integer i: Hence, by Theorem 3.1.3 it is enough to show that  (F) = (F): Now …x a positive integer i and set K = B(0;i)\F: It is enough to show that(K) =(K): But Z Z n n R R  (x)d(x) =  (x)d(x) j j K;2 K;2 n n R R for each positive integerj and letting j1 we are done. A metric space X is called a standard space if it is separable and com- plete. Standardspaceshaveaseriesofverynicepropertiesrelatedtomeasure theory; an example is furnished by the following Theorem 3.1.4. (Ulam’s Theorem) Let X be a standard space and suppose  is a …nite positive Borel measure on X: Then to each A2B(X) and " 0 there exist a compact K  A and an open V  A such that (V nK)":88 PROOF. Let " 0: We …rst prove that there is a compact subset K of X suchthat(K)(X)":Tothisend,letAbeadensedenumerablesubset 1 of X and let (a ) be an enumeration of A: Now for each positive integer i i=1 1 j j; B(a;2 ") =X; and therefore there is a positive integern such that i j i=1 n j j j ( B(a;2 "))(X)2 ": i i=1 Set n j j  F = B(a;2 ") j i i=1 and 1 L =\ F : j j=1 ThesetListotallybounded. SinceX iscompleteandLclosed,Liscomplete. Therefore, the setL is compact and, moreover c 1 c (K) =(X)(L ) =(X)( F ) j=1 j 1 c 1 (X) (F ) =(X) ((X)(F )) j j=1 j j=1 1 j (X) 2 " =(X)": j=1 Depending on Theorem 3.1.3 to each A 2 B(X) there exists a closed F A and an openV A such that(V nF)": But V n(F \L) = (V nF)(F nL) and we get (V n(F \L))(V nF)+(XnK) 2": Since the setF \L is compact Theorem 3.1.4 is proved. n Two Borel sets in R are said to be almost disjoint if their intersection has volume measure zero. n Theorem 3.1.5. Every open set U in R is the union of an at most denu- merable collection of mutually almost disjoint cubes.89 n Before the proof observe that a cube in R is the same as a closed ball n inR equipped with the metricd . n k PROOF. For each,k2N ; letQ be the class of all cubes of side length 2 + k k whose vertices have coordinates of the formi2 ;i2Z: LetF be the union 1 of those cubes in Q which are contained in U: Inductively, for k  1; let 1 F be the union of those cubes in Q which are contained in U and whose k k k1 n interiors are disjoint from F : Since d(x;R nU) 0 for every x2U it j j=1 1 follows thatU = F : j j=1 Exercises d n 1. Suppose f : (X;M) (R ;R ) andg : (X;M) (R ;R ) are measur- d n d+n able. Set h(x) = (f(x);g(x)) 2 R if x 2 X. Prove that h : (X;M) d+n (R ;R ) is measurable. d+n 2. Suppose f : (X;M) (R;R) and g : (X;M) (R;R) are measurable. Prove thatfg is (M;R)-measurable. 3. The function f :RR is a Borel function. Set g(x;y) =f(x); (x;y)2 2 2 R : Prove thatg :R R is a Borel function. 4. Suppose f : 0;1R is a continuous function and g : 0;10;1 a Borel function. Compute the limit Z 1 n lim f(g(x) )dx: n1 0 5. SupposeX andY aremetricspacesandf :X Y acontinuousmapping. Show thatf(E) is compact if E is a compact subset of X.90 6. SupposeX andY aremetricspacesandf :X Y acontinuousbijection. 1 Show that the inverse mappingf is continuous if X is compact. 7. Construct an open bounded subsetV of R such thatm(V) 0: 8. The function f : 0;1 R has a continuous derivative. Prove that the 0 1 setf(K)2Z if K = (f ) (f0g): m 9. Let P denote the class of all Borel probability measures on 0;1 and L the class of all functions f : 0;1 1;1 such that jf(x)f(y)jjxyj; x;y2 0;1: For any;2P; de…ne Z Z (;) = supj fd fdj: f2L 0;1 0;1 (a) Show that (P;) is a metric space. (b) Compute (;) if  is linear n1 1 measure on 0;1 and  =   k; where n2N (linear measure on 0;1 + n k=0 n is Lebesgue measure on 0;1 restricted to the Borel sets in 0;1). n 10. Suppose  is a …nite positive Borel measure on R : (a) Let (V ) be a i i2I n family of open subsets of R andV = V . Prove that i2I i (V) = sup (V :::V ): i i 1 k i ;:::;i 2I 1 k k2N + n (b) Let (F ) be a family of closed subsets of R and F = \ F . Prove i i2I i2I i that (F) = inf (F \:::\F ): i i 1 k i ;:::;i 2I 1 k k2N +91 3.2. Linear Functionals and Measures Let X be a metric space. A mapping T :C (X)R is said to be a linear c functional onC (X) if c T(f +g) =Tf +Tg; all f;g2C (X) c and T( f) = Tf; all 2R; f 2C (X): c If in addition Tf  0 for all f  0; T is called a positive linear functional on C (X): In this case Tf  Tg if f  g since gf  0 and TgTf = c T(gf) 0: Note thatC (X) =C(X) if X is compact. c The main result in this section is the following Theorem 3.2.1. (The Riesz Representation Theorem) Suppose X is a compact metric space and let T be a positive linear functional on C(X): Then there exists a unique …nite positive Borel measure  in X with the following properties: (a) Z Tf = fd; f 2C(X): X (b) For every E2B(X) (E) = sup (K): KE K compact (c) For every E2B(X) (E) = inf (V): VE V open92 The property (c) is a consequence of (b), since for each E 2 B(X) and " 0 there is a compactK XnE such that (XnE)(K)+": But then (XnK)(E)+" and X nK is open and contains E: In a similar way, (b) follows from (c) sinceX is compact. The proof of the Riesz Representation Theoremdepends on properties of continuous functions of independent interest. Suppose K  X is compact andV X is open. If f :X 0;1 is a continuous function such that f  and suppf V V we write f V and if  f  and suppf V K V we write K f V: Theorem 3.2.2. Let K be compact subset X. (a) Suppose K  V where V is open. There exists a function f on X such that K f V: (b) Suppose X is compact and K V :::V ; where K is compact and 1 n V ;:::;V are open. There exist functions h ;:::;h on X such that 1 n 1 n h V; i = 1;:::;n i i and h +:::+h = 1 onK: 1 n93 1 c PROOF. (a) Suppose " = min d(;V ): By Corollary 3.1.2, " 0: The K 2 X continuous function f =  satis…es  f  ; that is K f K : " K K K;" " Part (a) follows if we note that the closure (K ) of K is contained inV: " " (b) For each x2K there exists an r 0 such that B(x;r )V for some x x i 1 i. Let U = B(x; r ): It is important to note that (U )  V and (U ) x x x i x 2 is compact since X is compact. There exist points x ;:::;x 2K such that 1 m m U K: If 1in; letF denote the union of those (U ) which are x i x j=1 i j contained in V: By Part (a), there exist continuous functions f such that i i F f V; i = 1;:::;n: De…ne i i i h = f 1 1 h = (1f )f 2 1 2 :::: h = (1f ):::(1f )f : n 1 n1 n Clearly, h V; i = 1;:::;n: Moreover, by induction, we get i i h +:::+h = 1(1f ):::(1f )(1f ): 1 n 1 n1 n n Since F K we are done. i i=1 The uniqueness in Theorem 3.2.1 is simple to prove. Suppose  and 1  are two measures for which the theorem holds. Fix " 0 and compact 2 K X and choose an open setV so that (V) (K)+": IfK f V; 2 2 Z Z  (K) =  d  fd =Tf 1 K 1 1 X X Z Z = fd   d = (V) (K)+": 2 V 2 2 2 X X Thus  (K)   (K): If we interchange the roles of the two measures, the 1 2 opposite inequality is obtained, and the uniqueness of  follows. ToprovetheexistenceofthemeasureinTheorem3.2.1;de…neforevery openV inX, (V) = supTf: fV94 Here() = 0 sincethesupremumovertheemptyset, byconvention, equals 0: Note also that(X) =T1: Moreover,(V )(V ) ifV andV are open 1 2 1 2 andV V : Now set 1 2 (E) = inf (V) if E2B(X): VE V open Clearly, (E )  (E ); if E  E and E E 2 B(X): We therefore say 1 2 1 2 1; 2 that is increasing. Lemma 3.2.1. (a) If V ;:::;V are open, 1 n n n ( V )  (V ): i i i=1 i=1 (b) If E ;E ;:::2B(X); 1 2 1 1 ( E )  (E ): i i i=1 i=1 (c) If K ;:::;K are compact and pairwise disjoint, 1 n n n ( K ) =  (K ): i i i=1 i=1 PROOF. (a) It is enough to prove (a) for n = 2: To this end …rst choose gV V andthenh V ,i = 1;2; suchthath +h = 1 onsuppg: Then 1 2 i i 1 2 g =h g +h g 1 2 and it follows that Tg =T(h g)+T(h g)(V )+(V ): 1 2 1 2 Thus (V V )(V )+(V ): 1 2 1 295 (b) Choose" 0 and for eachi2N , choose an openV E such(V ) + i i i i 1 (E ) + 2 ": Set V = V and choose f  V: Since suppf is compact, i i i=1 f V :::V for somen: Thus, by Part (a), 1 n n 1 Tf (V :::V )  (V )  (E )+" 1 n i i i=1 i=1 and we get 1 (V)  (E ) i i=1 1 since" 0 is arbitrary. But E V and it follows that i i=1 1 1 ( E )  (E ): i i i=1 i=1 (c) It is enough to treat the special case n = 2: Choose " 0: Set  = d(K ;K ) and V = (K ) and V = (K ) : There is an open set U  1 2 1 1 =2 2 2 =2 K K suchthat(U)(K K )+"andtherearefunctionsf U\V 1 2 1 2 i i such thatTf (U\V )" fori = 1;2: Now, using that increases i i (K )+(K )(U\V )+(U\V ) 1 2 1 2 Tf +Tf +2" =T(f +f )+2": 1 2 1 2 Sincef +f U; 1 2 (K )+(K )(U)+2"(K K )+3" 1 2 1 2 and, by letting" 0; (K )+(K )(K K ): 1 2 1 2 The reverse inequality follows from Part (b). The lemma is proved. Next we introduce the class 8 9 = M = E2B(X); (E) = sup (K) : ; KE K compact96 Since is increasingM contains every compact set. Recall that a closed setinX iscompact,sinceX iscompact. Especially,notethatandX 2M. COMPLETION OF THE PROOF OF THEOREM 3.2.1: CLAIM 1. M contains every open set. PROOF OF CLAIM 1. Let V be open and suppose (V): There exists anf V suchthat Tf:IfU isopenandU K = suppf;thenf U; def and hence Tf  (U): But then Tf  (K): Thus (K) and Claim 1 follows sinceK is compact andK V: 1 CLAIM 2. Let (E ) be a disjoint denumerable collection of members of i i=1 1 M and putE = E: Then i i=1 1 (E) =  (E ) i i=1 andE2M: PROOFOFCLAIM2. Choose" 0andforeachi2N ,chooseacompact + i K E such that (K )(E )2 ": Set H =K :::K : Then, by i i i i n 1 n Lemma 3.2.1 (c), n n (E)(H ) =  (K )  (E )" n i i i=1 i=1 and we get 1 (E)  (E ): i i=1 1 Thus, by Lemma 3.2.1 (b), (E) =  (E ). To prove that E2M; let " i i=1 be as in the very …rst part of the proof and choosen such that n (E)  (E )+": i i=197 Then (E)(H )+2" n and this shows thatE2M: CLAIM 3. Suppose E 2M and " 0: Then there exist a compact K and an openV such thatK EV and(V nK)": PROOF OF CLAIM 3. The de…nitions show that there exist a compact K and an openV such that " " (V) (E)(K)+ : 2 2 The set V nK is open and V nK 2 M by Claim 1. Thus Claim 2 implies that (K)+(V nK) =(V)(K)+" and we get(V nK)": CLAIM 4. If A2M; thenXnA2M: PROOF OF CLAIM 4. Choose" 0: Furthermore, choose compactK A and openV A such that(V nK)": Then XnA (V nK)(XnV): Now, by Lemma 3.2.1 (b), (XnA)"+(XnV): SinceXnV is a compact subset of XnA; we conclude thatXnA2M: Claims 1, 2 and 4 prove that M is a -algebra which contains all Borel sets. ThusM =B(X):98 We …nally prove (a). It is enough to show that Z Tf  fd X for eachf 2C(X): For once this is known Z Z Tf =T(f) fd fd X X and (a) follows. Choose " 0: Set f(X) = a;b and choose y y :::y such that 0 1 n y =a, y =b; andy y ": The sets 1 n1 i i1 1 E =f (y ;y ); i = 1;:::;n i i1 i constituteadisjointcollectionofBorelsetswiththeunionX:Now,foreachi; " 1 pickanopensetV E suchthat(V )(E )+ andV f (1;y ): i i i i i i n n ByTheorem3.2.2therearefunctionsh V;i = 1;:::;n;suchthat  h = i i i i=1 1 on suppf andhf yh for all i: From this we get i i i n n n Tf =  T(hf)  yTh   y(V ) i i i i i i=1 i=1 i=1 " n n   y(E )+ y i i i i=1 i=1 n n   (y ")(E )+"(X)+(b+")" i i i=1 Z n   fd+"(X)+(b+")" i=1 E i Z = fd+"(X)+(b+")": X Since" 0 is arbitrary, we get Z Tf  fd: X This proves Theorem 3.2.1. n It is now simple to show the existence of volume measure in R : For pedagogical reasons we …rst discuss the so called volume measure in the unit n n cubeQ = 0;1 inR :99 The Riemann integral Z f(x)dx; Q is a positive linear functional as a function of f 2 C(Q): Moreover, T1 = 1 and the Riesz Representation Theorem gives us a Borel probability measure  inQ such that Z Z f(x)dx = fd: Q Q SupposeAQ is a closedn-cell andi2N : Then + Z Q vol(A)  (x)dxvol(A i) i 2 A;2 Q and Q  (x) (x) as i1 i A A;2 n for everyx2R : Thus (A) =vol(A): The measure is called the volume measure in the unit cube. In the special case n = 2 it is called the area measure in the unit square and if n = 1 it is called the linear measure in the unit interval. ˆ PROOFOFTHEOREM1.1.1. LetR=Rf1;1g bethetwo-pointcom- n pacti…cationofRintroducedinExample3.1.3andletR denotetheproduct of n copies of the metric space R: Clearly, n o n n n B(R ) = A\R ; A2B(R ) : n Moreover, letw :R 0;1 be a continuous map such that Z w(x)dx = 1: n R Now we de…ne Z n Tf = f(x)w(x)dx; f 2C(R ): n R100 n Note that T1 = 1. The function T is a positive linear functional on C(R ) and the Riesz Representation Theorem gives us a Borel probability measure n  on R such that Z Z n f(x)w(x)dx = fd; f 2C(R ): n n R R As above we get Z w(x)dx =(A) A n for each compactn-cell in R : Thus Z n (R ) = lim w(x)dx = 1 i1 n i;i n and we conclude that  is concentrated on R : Set  (A) = (A); A 2 0 n B(R ); and 1 dm = d : n 0 w n Then, if f 2C (R ); c Z Z f(x)w(x)dx = fd 0 n n R R and by replacingf byf=w; Z Z f(x)dx = fdm : n n n R R From thism (A) =vol(A) for every compactn-cellA and it follows thatm n n n is the volume measure onR . Theorem 1.1.1 is proved. """ 3.3 q-Adic Expansions of Numbers in the Unit Interval To begin with in this section we will discuss so called q-adic expansions of real numbers and give some interesting consequences. As an example of an101 application, we construct a one-to-one real-valued Borel map f de…ned on a proper interval such that the range of f is a Lebesgue null set. Another example exhibits an increasing continuous function G on the unit interval with the range equal to the unit interval such that the derivative of G is equal to zero almost everywhere with respect to Lebesgue measure. In the nextsectionwewillgivemoreapplicationsofq-adicexpansionsinconnection with in…nite product measures. Tosimplifynotationlet( ;P;F) = (0;1;B(0;1);v ). Furthermore, 1j0;1 let q  2 be an integer and de…ne a function h : Rf0;1;2;:::;q1g of period one such that k k +1 h(x) =k; x ; k = 0;:::;q1: q q Furthermore, set for eachn2N ; + n1  () =h(q ); 0 1: n Then 1 P  =k = ; k = 0;:::;q1: n q Moreover, if k ;:::;k 2f0;1;2;:::;q1g; it becomes obvious on drawing a 1 n …gure that     q1 P  =k ;:::; =k =  P  =k ;:::; =k ; =i 1 n1 1 n1 1 n1 i=0 1 n1 n whereeachterminthesumintheright-handside hasthesamevalue. Thus     P  =k ;:::; =k =qP  =k ;:::; =k ; =k 1 n1 1 n1 n 1 n1 1 n1 n and     P  =k ;:::; =k ; =k =P  =k ;:::; =k P  =k : 1 n1 n 1 n1 n 1 n1 n 1 n1 n By repetition,   n P  =k ;::: =k ; =k =  P  =k : 1 n1 n i 1 n1 n i=1 i From this we get   n P  2A ;::: 2A ; 2A =  P  2A 1 n1 n i 1 n1 n i i=1102 for all A ;:::;A f0;1;2;:::;q1g: 1 n Note that each2 0;1 has a so calledq-adic expansion  () 1 i =  : i=1 i q (q) If necessary, we write = to indicateq explicitly. n n Letk 2f0;1;2;:::;q1g be …xed and consider the eventA that a num- 0 ber in 0;1 does not have k in its q-adic expansion. The probability of A 0 equals P A =P  = 6 k ; i = 1;2;::: = lim P  6=k ; i = 1;2;:::;n 0 0 i i n1 q1 n n = lim  P  6=k = lim( ) = 0: 0 i=1 i n1 n1 q In particular, if n o (3) D = 2 0;1;  6= 1; i = 1;:::;n : n i 1 then, D =\ D is aP-zero set. n n=1 Set (2) 2 () 1 i f() =  ; 0 1: i=1 i 3 0 0 We claim that f is one-to-one. If 0  ; 1 and = 6 let n be the (2) (2) (2) 0 least i such that  () 6=  ( ); we may assume that  () = 0 and i i n (2) 0  ( ) = 1: Then n (2) (2) 0 0 2 ( ) 2 ( ) 2 0 n i n1 i f( )  =  + i=1 i=1 i i n 3 3 3 (2) (2) 2 () 4 2 () n1 i 1 1 i =  +  =f(): i=1 i=n+1 i=1 i i i 3 3 3 Thus f is one-to-one. We next prove that f( ) = D: To this end choose (3) y 2D: If  (y) = 2 for all i2 N ; then y = 1 which is a contradiction. If + i (3) (3) k  1 is …xed and  (y) = 0 and  (y) = 2; i  k + 1; then it is readily k i (3) seen that (y) = 1 which is a contradiction. Now de…ne k (3) 1  (y) 1 i 2 =  i=1 i 2103 and we havef() =y: 1 Let C = D ; n 2 N : The set C = \ C ; is called the Cantor set. n + n n n=1 The Cantor set is a compact Lebesgue zero set. The construction of the Cantor set may alternatively be described as follows. FirstC = 0;1. Then 0     1 2 1 2 trisectC and remove the middle interval ; to obtainC =C n ; = 0 1 0 3 3 3 3     1 2 0; ;1 : At the second stage subdivide each of the closed intervals 3 3 of C into thirds and remove from each one the middle open thirds. Then 1     1 2 7 8 C =C n( ; ; ): What is left fromC isC de…ned above. The 2 1 n1 n 9 9 9 9 n n n set 0;1nC is the union of 2 1 intervals numbered I ; k = 1;:::;2 1; n k n n where the interval I is situated to the left of the interval I if kl: k l Supposen is …xed and letG : 0;1 0;1 be the unique monotone in- n creasing continuous function, which satis…es G (0) = 0;G (1) = 1;G (x) = n n n n n k2 for x 2 I and which is a¢ ne on each interval of C It is clear that n: k n n G =G on each interval I , k = 1;:::;2 1: Moreover,jG G j n n+1 n n+1 k n1 2 and thus n+k n jG G j  jG G j 2 : n n+k k k+1 k=n Let G(x) = lim G (x); 0 x 1: The continuous and increasing func- n1 n 0 tion G is constant on each removed interval and it follows that G = 0 a.e. with respect to linear measure in the unit interval.The function G is called the Cantor function or Cantor-Lebesgue function. Nextweintroducethefollowingconvention,whichisstandardinLebesgue integration. Let (X;M;) be a positive measure space and supposeA2M c 1 and(A ) = 0: If two functionsg;h2L () agree onA; Z Z gd = hd: X X 1 If a function f : A R is the restriction to A of a function g 2 L () we de…ne Z Z fd = gd: X X Now suppose F : R R is a right continuous increasing function and let be the unique positive Borel such that (a;x) =F(x)F(a) if a;x2R andax: 1 If h2L () andE2R; the so called Stieltjes integral Z h(x)dF(x) E104 is by de…nition equal to Z hd: E If a;b2R, ab; andF is continuous at the pointsa andb; we de…ne Z Z b h(x)dF(x) = hd a I whereI is any interval with boundary points a andb: The reader should note that the integral Z h(x)dF(x) R in general is di¤erent from the integral Z 0 h(x)F (x)dx: R Forexample, ifG is the CantorfunctionandG is extendedsothatG(x) = 0 for negativex andG(x) = 1 forx larger than 1, clearly Z 0 h(x)G(x)dx = 0 R 0 sinceG(x) = 0 a.e. m: On the other hand, if we chooseh = ; 0;1 Z h(x)dG(x) = 1: R 3.4. Product Measures Suppose (X;M) and (Y;N) are two measurable spaces. If A 2 M and B2N;thesetAB iscalledameasurablerectangleinXY:Theproduct -algebraM N is,byde…nition,the-algebrageneratedbyallmeasurable rectangles inXY: If we introduce the projections  (x;y) =x; (x;y)2XY X and  (x;y) =y; (x;y)2XY; Y105 the product-algebraM N is the least-algebraS of subsets of XY, which makes the maps  : (XY;S) (X;M) and  : (XY;S) X Y 1 1 (Y;N) measurable, that isM N =( (M) (N)):. X Y SupposeE generatesM;whereX 2E;andF generatesN;whereY 2F. We claim that the class EF =fEF;E2E andF 2Fg generates the-algebraM N: First it is clear that (EF)M N: Moreover, the class  1 fE2M;EY 2(EF)g =M\ EX;  (E)2(EF) X is a -algebra, which contains E and therefore equals M. Thus AY 2 (EF) for all A 2 M and, in a similar way, XB 2 (EF) for all B 2N and we conclude that AB = (AY)\(XB)2 (EF) for all A2M and all B2N: This proves that M N (EF) and it follows that (EF) =M N: Thus (EF) =(E) (F) if X 2E andY 2F: n Sincethe-algebraR isgeneratedbyallopenn-cellsinR , weconclude n that R =R R : k+n k n GivenEXY; de…ne E =fy; (x;y)2Eg if x2X x and y E =fx; (x;y)2Eg if y2Y: If f :XY Z is a function andx2X; y2Y, let f (y) =f(x;y); if y2Y x106 and y f (x) =f(x;y); if x2X: y Theorem 3.4.1 (a) If E 2M N; then E 2N and E 2M for every x x2X and y2Y: (b) If f : (XY;M N) (Z;O) is measurable, then f is (N;O)- x y measurable for each x2X and f is (M;O)-measurable for each y2Y: Proof. (a) Choosey2Y andde…ne' :X XY by'(x) = (x;y): Then 1 1 1 M =(' (MN)) =' ((MN)) =' (M N) y and it follows thatE 2M: In a similar wayE 2N for everyx2X: x (b) For any setV 2O; 1 1 (f (V)) = (f ) (V) x x and 1 y y 1 (f (V)) = (f ) (V): Part (b) now follows from (a). Belowan(M;R )-measurableor(M;R)-measurablefunctionissimply 0;1 calledM-measurable. Theorem 3.4.2. Suppose (X;M;) and (Y;N;) are positive -…nite measurable spaces and suppose E2M N. If y f(x) =(E ) andg(y) =(E ) x for every x 2 X and y 2 Y; then f is M-measurable, g is N-measurable, and Z Z fd = gd: X Y Proof. We …rst assume that (X;M;) and (Y;N;) are …nite positive measure spaces.107 Let D be the class of all sets E 2 M N for which the conclusion of the theorem holds. It is clear that the class G of all measurable rectangles in X Y is a subset of D and G is a -system. Furthermore, the Beppo Levi Theoremshows thatD is a-additive class. Therefore, using Theorem 1.2.2,M N =(G)D and it follows thatD =M N: 1 In the general case, choose a denumerable disjoint collection (X ) of k k=1 1 members ofM and a denumerable disjoint collection (Y ) of members of n n=1 N such that 1 1 X =X and Y =Y: k n k=1 n=1 Set  = , k = 1;2;::: k X k and  = , n = 1;2;::: . n Y n Then, by the Beppo Levi Theorem, the function Z 1 f(x) =   (x;y) (y)d(y) n=1 E Y n Y Z 1 1 =   (x;y) (y)d(y) =   (E ) n x n=1 E Y n=1 n Y isM-measurable. Again, by the Beppo Levi Theorem, Z Z 1 fd =  fd k=1 k X X and Z Z Z 1 1 1 fd =  (  (E )d (x)) =   (E )d (x): n x n x k k k=1 n=1 k;n=1 X X X In a similar way, the functiong isN-measurable and Z Z Z 1 1 y 1 y gd =  (  (E )d (y)) =   (E )d (y): n n n=1 k=1 k k;n=1 k Y Y Y Since the theorem is true for …nite positive measure spaces, the general case follows.108 De…nition 3.4.1. If (X;M;) and (Y;N;) are positive-…nite measur- able spaces andE2M N, de…ne Z Z y ()(E) = (E )d(x) = (E )d(y): x X Y The function is called the product of the measures  and: NotethatBeppoLevi’sTheoremensuresthat isapositivemeasure. Beforethenexttheoremwerecallthefollowingconvention. Let(X;M;) c be a positive measure space and suppose A 2 M and (A ) = 0: If two 1 functionsg;h2L () agree onA; Z Z gd = hd: X X 1 If a function f : A R is the restriction to A of a function g 2 L () we de…ne Z Z fd = gd: X X Theorem 3.4.3. Let (X;M;) and (Y;N;) be positive -…nite measur- able spaces. (a) (Tonelli’s Theorem) If h : XY 0;1 is (M N)-measurable and Z Z f(x) = h(x;y)d(y) andg(y) = h(x;y)d(x) Y X for every x 2 X and y 2 Y; then f is M-measurable, g is N-measurable, and Z Z Z fd = hd() = gd X XY Y (b)(Fubini’s Theorem) (i) If h :XY R is (M N)-measurable and Z Z ( jh(x;y)jd(y))d(x)1 X Y109 1 then h2L (): Moreover, Z Z Z Z Z ( h(x;y)d(y))d(x) = hd() = ( h(x;y)d(x))d(y) X Y XY Y X 1 1 (ii) If h2L (() ); then h 2L () for -almost all x and x Z Z Z hd() = ( h(x;y)d(y))d(x) XY X Y 1 y 1 (iii) If h2L (() ); then h 2L () for -almost all y and Z Z Z hd() = ( h(x;y)d(x))d(y) XY Y X PROOF.(a)Thespecialcasewhenhisanon-negative(M N)-measurable simple function follows from Theorem 3.4.2. Remembering that any non- negativemeasurablefunctionisthepointwiselimitofanincreasingsequence of simple measurable functions, the Lebesgue Monotone Convergence Theo- rem implies the Tonelli Theorem. (b) PART (i) : By Part (a) Z Z Z + + 1 ( h (x;y)d(y))d(x) = h d() X Y XY Z Z + = ( h (x;y)d(x))d(y) Y X and Z Z Z 1 ( h (x;y)d(y))d(x) = h d() X Y XY Z Z = ( h (x;y)d(x))d(y): Y X + 1 In particular, h =h h 2L (): Let  + 1 A = x2X; (h ) ;(h ) 2L () : x x110 c ThenA is a-null set and we get Z Z Z + + ( h (x;y)d(y))d(x) = h d() A Y XY and Z Z Z ( h (x;y)d(y))d(x) = h d(): A Y XY Thus Z Z Z ( h(x;y)d(y))d(x) = hd() A Y XY and, hence, Z Z Z ( h(x;y)d(y))d(x) = hd(): X Y XY The other case can be treated in a similar way. The theorem is proved. PART (ii) : We …rst use Theorem 2.2.3 and write h = ' + where ' 2 1 L (); is (M N) -measurable and = 0 a.e. : Set  + 1 A = x2X; (' ) ;(' ) 2L () : x x Furthermore, supposeEf(x;y); (x;y) = 6 0g; E2M N and ()(E) = 0: Then, by Tonelli’s Theorem Z 0 = (E )d(x): x X Let B = fx2X; (E ) = 6 0g and note that B 2 M: Moreover (B) = 0 x and ifx2= B, then = 0 a.e.  that ish =' a.e. : Now, by Part (i) x x x Z Z Z Z hd() = 'd() = ( '(x;y)d(y))d(x) XY XY A Y Z Z Z Z = ( '(x;y)d(y))d(x) = ( h(x;y)d(y))d(x) c c A\B Y A\B Y111 Z Z = ( h(x;y)d(y))d(x): X Y Part (iii) is proved in the same manner as Part (ii): This concludes the proof of the theorem. If (X;M ); i = 1;:::;n; are measurable spaces, the product -algebra i i M ::: M is, by de…nition, the -algebra generated by all sets of the 1 n form A :::A 1 n whereA 2M;i = 1;:::;n:Nowassume(X;M; );i = 1;:::;n;are-…nite i i i i i positivemeasurespaces. Byinduction,wede…ne = and =  ; 1 k k1 1 k k = 1;2;:::;n:Themeasure, iscalledtheproductofthemeasures ;:::; n 1 n and is denoted by ::: : It is readily seen that 1 n R =R ::: R (n factors) n 1 1 and v =v :::v (n factors): n 1 1 Moreover, n R  (R ) = R ::: R (n factors): def n 1 1 1 If A 2 P(R)nR ; by the Tonelli Theorem, the set Af0;:::;0g (n 1 1 zeros) is an m -null set, which, in view of Theorem 3.4.1, cannot belong to n n the-algebra (R ) : Thus the Axiom of Choice implies that 1 n R = 6 (R ) : n 1 Clearly, the completion of the measure m :::m (n factors) equals 1 1 m : n Sometimes we prefer to write Z f(x ;:::;x )dx :::dx 1 n 1 n A :::A 1 n instead of Z f(x)dm (x) n A :::An 1112 or Z f(x)dx: A :::A 1 n Moreover, the integral Z Z ::: f(x ;:::;x )dx :::dx 1 n 1 n A A n 1 is the same as Z f(x ;:::;x )dx :::dx : 1 n 1 n A :::A 1 n De…nition 3.4.2. (a) The measure Z 2 x dx 2 (A) = e p ; A2R 1 2 A is called the standard Gauss measure in R: (b) The measure = ::: (n factors) n 1 1 n is called the standard Gauss measure in R : Thus, if q 2 n 2 jxj= x +:::+x ; x = (x ;:::;x )2R 1 n 1 n we have Z 2 jxj dx 2 (A) = e p ; A2R : n n n 2 A (c) A Borel measure  in R is said to be a centred Gaussian measure if  =f( ) for some linear mapf :RR: 1 (d) A real-valued random variable  is said to be a centred Gaussian random variable if its probability law is a centred Gaussian measure in R: Statedotherwise,isareal-valuedcentredGaussianrandomvariableifeither L() = (abbreviated2N(0;0)) 0113 or there exists a 0 such that  L( ) = (abbreviated2N(0;)). 1  (e)Afamily( ) ofreal-valuedrandomvariablesissaidtobeacentred t2T t real-valued Gaussian process if for all t ;:::;t 2T; ;:::; 2R and every 1 n 1 n n2N ; the sum + n  =   k k=1 t k is a centred Gaussian random variable: p 2 n 2 Example 3.4.1 Supposejxj = x +:::+x if x = (x ;:::;x )2 R : We 1 n 1 n claim that Z n 2 Y p x +x n n i i k 16 lim (1+ ) d (x) = 2 e n k1 4k n R i=1 t To prove this claim we …rst use that e  1+t for every real t and have for each …xedi2f1;:::;ng; 2 2 x +x x +x i i i i 4k 1+ e : 4k Moreover, if k2N ; then + 2 x +x 1 1 1 i i 2 1+ = ((x + ) +4k ) 0 i 4k 4k 2 4 and we conclude that 2 2 x +x x +x i i i i k 4 (1+ ) e : 4k Thus, for anyk2N ; + n n 2 2 Y Y x +x x +x i i i i k 4 0f (x) = (1+ )  e = g(x) k def def 4k i=1 i=1 1 whereg2L ( ) since n Z Z n 2 Y x x i dx i 4 p g(x)d (x) = e =fTonelli’s Theoremg = n n n n 2 R R i=1114 Z n 2 Y p x x n i i dx n i 4 16 p e = 2 e : 2 R i=1 Moreover, lim f (x) =g(x) k k1 and by dominated convergence we get Z Z n 2 Y p x +x n n i i k 16 lim (1+ ) d (x) = g(x)d (x) = 2 e : n n k1 4k n n R R i=1 Exercises 1. Let (X;M;) and (Y;N;) be two -…nite positive measure spaces. Let 1 1 f 2 L () and g 2 L () and de…ne h(x;y) = f(x)g(y); (x;y) 2 X Y: 1 Prove thath2L () and Z Z Z hd() = fd gd: XY X Y 2. Let (X;M;) be a -…nite positive measure space and f :X 0;1 a measurable function. Prove that Z fd = (m)(f(x;y); 0yf(x); x2Xg): X 3. Let (X;M;) be a -…nite positive measure space and f : X R a measurable function. Prove that (m)(f(x;f(x)); x2X g) = 0: 1 4. Let E 2R and E  0;10;1: Suppose m(E ) for m-almost all x 2 2 x2 0;1: Show that 1 y m(fy2 0;1;m(E ) = 1g) : 2115 5. Letc be the counting measure onR restricted toR and D =f(x;x); x2Rg: De…ne for everyA2 (RR)fDg; Z Z (A) = (  (x;y)dv (x))dc(y) 1 A R R and Z Z (A) = (  (x;y)dc(y))dv (x): 1 A R R (a) Prove that and agree onRR: (b) Prove that (D)6=(D): 6. LetI = 0;1 and 2 2 x y h(x;y) = ; (x;y)2II: 2 2 2 (x +y ) Prove that Z Z  ( h(x;y)dy)dx = ; 4 I I Z Z  ( h(x;y)dx)dy = 4 I I and Z jh(x;y)jdxdy =1: II 7. Fort 0 andx2R let 2 1 x 2t g(t;x) = p e 2t and g h(t;x) = : t Givena 0; prove that Z Z 1 1 ( h(t;x)dt)dx =1 1 a116 and Z Z 1 1 ( h(t;x)dx)dt = 0 a 1 and conclude that Z jh(t;x)jdtdx =1: a;1R (Hint: First prove that Z 1 g(t;x)dx = 1 1 and 2 g 1 g = :) 2 t 2x 1 8. Givenf 2L (m); let Z x+1 1 g(x) = f(t)dt; x2R: 2 x1 Prove that Z Z jg(x)jdx jf(x)jdx: R R 9. Let I = 0;1 and suppose f : I R is a Lebesgue measurable function such that Z jf(x)f(y)jdxdy1: II Prove that Z jf(x)jdx1: I 1 10. SupposeA2R andf 2L (m): Set Z d(y;A)f(y) g(x) = dy; x2R: 2 jxyj R Prove that Z jg(x)jdx1: A117 11. Suppose that the functions f;g : R0;1 are Lebesgue measurable and introduce  = fm and  = gm: Prove that the measures  and  are -…nite and Z ()(E) = f(x)g(y)dxdy if E2R R : E n n 12. Suppose  is a …nite positive Borel measure on R and f : R R a n Borel measurable function. Set g(x;y) =f(x)f(y); x;y2R : Prove that 1 1 f 2L () if and only if g2L (): 13. Arandom variable is non-negative and possesses the distribution func- R 1 tionF(x) =P x: Prove thatE = (1F(x))dx: 0 14. Let (X;d) be a metric space and suppose Y 2B(X): Then Y equipped with the metricd is a metric space. Prove that jYY B(Y) =fA\Y; A2B(X)g: 15. The continuous bijection f : (X;d) (Y;e) has a continuous inverse. Prove thatf(A)2B(Y) if A2B(X) 16. Areal-valuedfunctionf(x;y);x;y2R; isaBorelfunctionofx forevery …xed y and a continuous function of y for every …xed x: Prove that f is a Borel function. Is the same conclusion true if we only assume thatf(x;y) is a real-valued Borel function in each variable separately? 17. Supposea 0 and 1 n X a a  =e  n a n n=0118 where (A) = (n) if n2N =f0;1;2;:::g andA N: Prove that n A 1 (  )s = a b a+b for all a;b 0; if s(x;y) =x+y; x;y2N: 18. Suppose Z 2 1 x xe f(t) = dx; t 0: 2 2 x +t 0 Compute Z 1 lim f(t) and f(t)dt: t0+ 0 Finally, prove thatf is di¤erentiable. 19. Suppose Z 1 ln(1+x) tx f(t) = e dx; t 0: 1+x 0 R 1 a) Show that f(t)dt1: 0 b) Show thatf is in…nitely many times di¤erentiable. 20. Suppose f is Lebesgue integrabel on 0;1: (a) Show that the function R R 1 1 1 g(x) = t f(t)dt; 0 x 1; is continuous. (b) Prove that g(x)dx = x 0 R 1 f(x)dx: 0 3.5 Change of Variables in Volume Integrals If T is a non-singularn byn matrix with real entries, we claim that 1 T(v ) = v n n j detT j119 n n (here T is viewed as a linear map of R into R ). Remembering Corollary 3.1.3 this means that the following linear change of variables formula holds, viz. Z Z 1 n f(Tx)dx = f(x)dx all f 2C (R ): c j detT j n n R R The casen = 1 is obvious. Moreover, by Fubini’s Theorem the linear change of variables formula is true for arbitraryn in the following cases: (a)Tx = (x ;:::;x );whereisapermutationofthenumbers1;:::;n: (1) (n) (b)Tx = ( x ;x ;:::;x ); where is a non-zero real number. 1 2 n (c)Tx = (x +x ;x ;:::;x ): 1 2 2 n Recallfromlinearalgebrathateverynon-singularnbynmatrixT canbe row-reduced to the identity matrix, that is T can by written as the product of …nitely many transformations of the types in (a),(b), and (c). This proves the above linear change of variables formula. Our main objective in this section is to prove a more general change n of variable formula. To this end let and be open subsets of R and 1 G : a C di¤eomorphism, that is G = (g ;:::;g ) is a bijective 1 n g 0 i continuously di¤erentiable map such that the matrixG(x) = ( (x)) 1i;jn x j is non-singular for each x 2 : The inverse function theorem implies that 1 1 G : is aC di¤eomorphism DI: Theorem 3.5.1. If f is a non-negative Borel function in ; then Z Z 0 f(x)dx = f(G(x))j detG(x)jdx: The proof of Theorem 3.5.1 is based on several lemmas. n Throughout, R is equipped with the metric d (x;y) = max jx y j: n k k 1kn LetK be a compact convex subset of : Then if x;y2K and 1in; Z 1 d g (x)g (y) = g (y +t(xy))dt i i i dt 0 Z 1 g i n =  (y +t(xy))(x y )dt k k k=1 x k 0120 and we get d (G(x);G(y))M(G;K)d (x;y) n n where g i n M(G;K) = max  maxj (z)j: k=1 1in z2K x k  Thus if B(a;r) is a closed ball contained inK;   G(B(a;r))B(G(a);M(G;K)r): 1 Lemma3.5.1. Let (Q ) be a sequence of closed balls contained in such k k=1 that Q Q k+1 k and diamQ 0 as k1: k Then, there is a unique point a belonging to each Q and k v (G(Q )) n k 0 lim sup j detG(a)j: v (Q ) n1 n k PROOF. The existence of a point a belonging to each Q is an immediate k consequence of Theorem 3.1.2. The uniqueness is also obvious since the 0 1 diameter of Q converges to 0 as k 1: Set T = G(a) and F = T G: k  Then, if Q =B(x ;r ); k k k 1 1  v (G(Q )) =v (T(T G(Q ))) =j detT j v (T G(B(x ;r ))) n k n k n k k 1 1 1 n  j detT jv (B(T G(x );M(T G;Q )r ) =j detT j M(T G;Q ) v (Q ): n k k k k n k Since 1 lim M(T G;Q ) = 1 k k1 the lemma follows at once. Lemma 3.5.2. Let Q be a closed ball contained in : Then Z 0 v (G(Q)) j detG(x)jdx: n Q121 PROOF. Suppose there is a closed ball Q contained in such that Z 0 v (G(Q)) j detG(x)jdx: n Q This will lead us to a contradiction as follows. Choose" 0 such that Z 0 v (G(Q)) (1+") j detG(x)jdx: n Q n 2 n Let Q = B where B ;:::;B are mutually almost disjoint closed balls k 1 2 1 with the same volume. If Z 0 n v (G(B )) (1+") j detG(x)jdx; k = 1;:::;2 n k B k we get n 2 v (G(Q))  v (G(B )) n n k k=1 Z Z n 2 0 0  (1+") j detG(x)jdx = (1+") j detG(x)jdx k=1 B Q k which is a contradiction. Thus Z 0 v (G(B )) (1+") j detG(x)jdx n k B k 1 for some k: By induction we obtain a sequence (Q ) of closed balls con- k k=1 tained in such that Q Q ; k+1 k diamQ 0 as k1 k and Z 0 v (G(Q )) (1+") j detG(x)jdx: n k Q k But applying Lemma 3.5.1 we get a contradiction.122 1 PROOF OF THEOREM 3.5.1. Let U  be open and write U = Q i i=1 0 where theQs are almost disjoint cubes as in Theorem 3.1.5. Then i Z 1 1 0 v (G(U))  v (G(Q ))  j detG(x)jdx n n i i=1 i=1 Q i Z 0 = j detG(x)jdx: U Using Theorem 3.1.3 we now have that Z 0 v (G(E)) j detG(x)jdx n E for each Borel setE : But then Z Z 0 f(x)dx f(G(x))j detG(x)jdx for each simple Borel measurable function f  0 and, accordingly from this and monotone convergence, the same inequality holds for each non-negative Borel function f: But the same line of reasoning applies to G replaced by 1 0 G andf replaced byf(G)j detG j, so that Z Z 0 0 1 1 0 f(G(x))j detG(x)jdx f(x)j detG(G (x))jj det(G )(x)jdx Z = f(x)dx: This proves the theorem. 2 Example 3.5.1. If f : R 0;1 is (R ;R )-measurable and 0 " 2 0;1 R1, the substitution G(r;) = (rcos;rsin) yields Z Z Z R 2 f(x ;x )dx dx = lim f(rcos;rsin)rddr 1 2 1 2 p + 2 2 0 " x +x R "  1 2123 Z Z R 2 = f(rcos;rsin)rddr " 0 and by letting" 0 andR1; we have Z Z Z 1 2 f(x ;x )dx dx = f(rcos;rsin)rddr: 1 2 1 2 2 R 0 0 The purpose of the example is to show an analogue formula for volume n measure in R : n1 n n LetS =fx2R ;jxj= 1g be the unit sphere inR : We will de…ne a n1 so called surface area Borel measure onS such that n1 Z Z Z 1 n1 f(x)dx = f(r)r d ()dr n1 n n1 R 0 S n for any (R ;R )-measurable functionf :R 0;1: To this end de…ne n 0;1 n n1 G :R nf0g 0;1S by settingG(x) = (r;); where x r =jxj and = : jxj 1 n1 n Note thatG : 0;1S R nf0g is given by the equation 1 G (r;) =r: Moreover, 1 1 n1 G (0;aE) =aG (0;1E) if a 0 andES : n1 If E2B(S ) we therefore have that 1 n 1 v (G (0;aE)) =a v (G (0;1E)): n n We now de…ne 1 n1  (E) =nv (G (0;1E)) if E2B(S ) n1 n and Z n1 (A) = r dr if A2B(0;1): A124 Below, by abuse of language, we write v n = v : Then, if 0 a  njR nf0g n n1 b1 andE2B(S ); G(v )(0;aE) =(0;a) (E) n n1 and G(v )(a;bE) =(a;b) (E): n n1 Thus, by Theorem 1.2.3, G(v ) = n n1 and the claim above is immediate. To check the normalization constant in the de…nition of  ; …rst note n1 that Z Z R n R n1 n1 v (jxjR) = r d()dr =  (S ) n n1 n1 n 0 S and we get d n1 n1 v (jxjR) =R  (S ): n n1 dR Exercises 1. Extend Theorem 3.5.1 to Lebesgue measurable functions. R 2. The function f :R0;1 is Lebesgue measurable and fdm = 1: R R Determine all non-zero real numbers such that hdm1; where R 1 n h(x) =  f( x+n); x2R: n=0 3. Suppose Z n n jxj n (A) = jxj e dx; A2B(R ); A p n 2 2 where j x j = x +:::+x if x = (x ;:::;x ) 2 R : Compute the 1 n 1 n limit n n lim  ln(fx2R ;jxjg): 1125 4. Compute then-dimensional Lebesgue integral Z ln(1jxj)dx jxj1 n where j x j denotes the Euclidean norm of the vector x 2 R : (Hint: n=2 n1 2 (S ) = :) (n=2) 1 5. Suppose f 2 L (m ): Show that lim f(nx) = 0 for m -almost all 2 n1 2 2 x2R : 3.6. Independence in Probability Suppose( ;F;P)isaprobabilityspace. Therandomvariables :( ;P) k (S ;S ); k = 1;:::;n are said to be independent if k k n P = P : ( ;:::; )  k=1 1 n k Afamily ( ) ofrandomvariablesissaidtobeindependentif ;:::; i2I i i i 1 n are independent for any i ;:::i 2 I with i = 6 i if k 6= l: A family of 1 n k l events (A ) is said to be independent if ( ) is a family of independent i i2I i2I A i random variables. Finally a family (A ) of sub--algebras ofF is said to i i2I be independent if, for any A 2 A; i 2 I; the family (A ) is a family of i i i i2I independent events. Example 3.6.1. Let q  2 be an integer. A real number 2 0;1 has a q-adic expansion (q)  1 k =  : k=1 k q (q) 1 The construction of the Cantor set shows that ( ) is a sequence of k=1 k independent random variables based on the probability space (0;1;v ;B(0;1)): 1j0;1126 Example3.6.2. Let(X;M;)beapositivemeasurespaceandletA 2M; i i2N ; be such that + 1  (A )1: i i=1 The …rst Borel-Cantelli Lemma asserts that -almost all x 2 X lie in A i for at most …nitely many i: This result is an immediate consequence of the Beppo Levi Theorem since Z Z 1 1   d =   d1 i=1 A i=1 A i i X X implies that 1   1 a.e. : i=1 A i 1 Suppose ( ;F;P) is a probability space and let (A ) be independent i i=1 events such that 1  P A =1: i i=1 The second Borel-Cantelli Lemma asserts that almost surelyA happens for i in…nitely manyi: To prove this, we use the inequality x 1+xe ; x2R to obtain   k+n c k+n c P \ A =  P A i i i=k i=k k+n k+n k+n PA  PA i i i=k =  (1P A )  e =e : i i=k i=k By lettingn1; 1 c P \ A = 0 i=k i or 1 P A = 1: i i=k But then 1 1 P \ A = 1 i k=1 i=k and the second Borel-Cantelli Lemma is proved.127 Theorem 3.6.1. Suppose  ;:::; are independent random variables and 1 n  2N(0;1); k = 1;:::;n: If ;:::; 2R; then 1 n k n n 2   2N(0; ) k k=1 k k=1 k PROOF. The case ;:::; = 0 is trivial so assume 6= 0 for somek: We 1 n k have for each open interval A; Z n P   2A = d (x ):::d (x ) k 1 n k=1 k 1 1 n  x 2A k k k=1 Z 1 1 2 2 (x +:::+x ) n 2 1 p e dx :::dx : n 1 n n 2  x 2A k k k=1 p 2 2 Set  = +:::+ and let y = Gx be an orthogonal transformation 1 n such that 1 y = ( x +:::+ x ): 1 1 1 n n  Then, since detG = 1; Z 1 1 2 2 1 (y +:::+y ) n 2 1 P   2A = p e dy :::dy k n 1 n k=1 k 2 y 2A 1 Z 1 1 2 y 2 1 = p e dy 1 2 y 2A 1 where we used Fubini’s theorem in the last step. The theorem is proved. Finally, in this section, we prove a basic result about the existence of in…nite product measures. Let  ; k 2 N be Borel probability measures + k N 1 + in R. The space R is, by de…nition, the set of all sequences x = (x ) k k=1 N + of real numbers. For each k 2 N ; set  (x) = x : The -algebra R + k k N + is the least -algebra S of subsets of R which makes all the projections N +  : (R ;S) (R;R); k 2 N ; measurable. Below, ( ;:::; ) denotes k + 1 n N n + the mapping of R into R de…ned by the equation ( ;:::; )(x) = ( (x);:::; (x)): 1 n 1 n128 N + Theorem3.6.1. There is a unique probability measure  on R such that  = ::: ( ;:::; ) 1 n 1 n for every n2N : + The measure  in Theorem 3.6.1 is called the product of the measures  ; k2N ; and is often denoted by + k 1   : k=1 k PROOF OF THEOREM 3.6.1. Let ( ;P;F) = (0;1;v ;B(0;1) and 1j0;1 set (2)  () 1 k () =  ; 2 : k=1 k 2 1 We already know that P = P: Now suppose (k ) is a strictly increasing  i i=1 sequence of positive integers and introduce (2)  () k 0 1 i  =  ; 2 : i=1 i 2 (2) (2) n Note that for each …xed positive integern; theR -valued maps ( ;:::; ) 1 n (2) (2) and ( ;:::; ) areP-equimeasurable. Thus, if f : R is continuous, k k 1 n Z Z (2)  () n k f()dP = lim f( )dP() k=1 k n1 2 Z Z (2)  () k n i 0 = lim f( )dP() = f( )dP i=1 i n1 2 0 and it follows thatP =P =P:   By induction, we de…ne for each k2N an in…nite subset N of the set + k k1 k N n N suchthatthesetN n N containsin…nitelymanyelements + i + i i=1 i=1 and de…ne (2)  () n 1 ik  =  k i=1 i 2 1 where (n ) is an enumeration of N : The map ik k i=1 1 () = ( ()) k k=1129 N N + + is a measurable map of ( ;F) into (R ;R ) and 1 P =  i k=1 where =P for eachi2N : i + For each i2N there exists a measurable map ' of ( ;F) into (R;R) + i such thatP = (see Section 1.6). The map ' i i 1 (x) = (' (x )) i i i=1 N N + + is a measurable map of (R ;R ) into itself and we get  = (P ) . This completes the proof of Theorem 3.6.1. """130 CHAPTER4 MODES OFCONVERGENCE Introduction Inthischapterwewilltreatavarietyofdi¤erentsortsofconvergencenotions 2 in measure theory. So calledL -convergence is of particular importance. 1 2 4.1. Convergence in Measure,in L (); and inL () Let (X;M;) be a positive measure space and denote byF(X) the class of measurable functionsf : (X;M) (R;R). For anyf 2F(X); set Z kf k = jf(x)jd(x) 1 X and s Z 2 kf k = f (x)d(x): 2 X The Cauchy-Schwarz inequality states that Z jfgjdkf k kgk if f;g2F(X): 2 2 X To prove this, without loss of generality, it can be assumed that 0kf k 1 and 0kgk 1: 2 2 We now use the inequality 1 2 2  ( + ); ; 2R 2131 to obtain Z Z 2 2 jf j jgj 1 f g d ( + )d = 1 2 2 kf k kgk 2 kf k kgk X 2 2 2 2 and the Cauchy-Schwarz inequality is immediate. If not otherwise stated, in this sectionp is a number equal to 1 or 2: If it is important to emphasize the underlying measurekf k is writtenkf k : p p; We now de…ne p L () =ff 2F(X); kf k 1g: p The special case p = 1 has been introduced earlier. We claim that the following so called triangle inequality holds, viz. p kf +gk kf k +kgk if f;g2L (): p p p The casep = 1; follows by-integration of the relation jf +gjjf j +jgj: To prove the casep = 2; we use the Cauchy-Schwarz inequality and have 2 2 kf +gk kjf j +jgjk 2 2 Z 2 2 =kf k +2 jfgjd+kgk 2 2 X 2 2 2 kf k +2kf k kgk +kgk = (kf k +kgk ) 2 2 2 2 2 2 and the triangle inequality is immediate. p Supposef;g2L ():Thefunctionsf andg areequalalmosteverywhere with respect to  if ff = 6 gg 2 Z : This is easily seen to be an equivalence  p relation and the set of all equivalence classes is denoted by L (): Below p p we consider the elements of L () as members of L () and two members p of L () are identi…ed if they are equal a.e. : From this convention it is p straight-forward to de…ne f +g and f for all f;g 2L () and 2 R and (p) p the function d (f;g) =kfgk is a metric on L (): Convergence in the p p p (p) p metric spaceL () = (L (),d ) is called convergence inL (): A sequence 1 (f ) inF(X) converges in measure to a functionf 2F(X) if k k=1 lim (jf f j") = 0 all " 0: k k1132 1 If the sequence (f ) in F(X) converges in measure to a function f k k=1 2F(X) as well as to a functiong 2F(X); thenf =g a.e.  since n o n o " " fjfgj"g jff j jf gj k k 2 2 and " " (jfgj")(jff j )+(jf gj ) k k 2 2 1 for every " 0 and positive integer k: A sequence (f ) inF(X) is said k k=1 to be Cauchy in measure if for every" 0; (jf f j") 0 as k;n1: k n p BytheMarkovinequality,aCauchysequenceinL ()isCauchyinmeasure. p Example 4.1.1. (a) If f = k 1 ; k2N ; then k + 0; k 1 kf k = 1 and kf k = p : k 2;m k 1;m k 1 2 Thusf 0 inL (m) as k1 butf 9 0 inL (m) as k1: k k 1 2 (b)L (m)L (m) since 1 2 1  (x) 2L (m)nL (m) 1;1 jxj 2 1 andL (m)L (m) since 1 1 2 p  (x) 2L (m)nL (m): 0;1 jxj Theorem 4.1.1. Suppose p = 1 or 2: p (a) Convergence in L () implies convergence in measure:133 2 1 2 (b) If (X)1; then L ()L () and convergence in L () implies 1 convergence in L (): 1 p Proof. (a) Suppose the sequence (f ) converges to f in L () and let n n=1 " 0: Then, by the Markov inequality, Z 1 1 p p (jf f j") jf f j d = kf f k n n n p p p " " X and (a) follows at once. (b) The Cauchy-Schwarz inequality gives for anyf 2F(X); Z Z Z 2 2 ( jf j1d)  f d 1d X X X or p kf k kf k (X) 1 2 and Part (b) is immediate. Theorem 4.1.2. Suppose f 2F(X); n2N : n + 1 (a) If (f ) is Cauchy in measure, there is a measurable function f : n n=1 X R such that f f in measure as n 1 and a strictly increasing n 1 sequence of positive integers (n ) such that f f a.e.  as j1. j n j=1 j (b) If  is a …nite positive measure and f f 2 F(X) a.e.  as n n1; then f f in measure. n (c) (Egoro¤’s Theorem) If  is a …nite positive measure and f n f 2F(X) a.e.  as n1; then for every " 0 there exists E2M such that (E)" and sup jf (x)f(x)j 0 as n1: k kn c x2E PROOF. (a) For each positive integer j; there is a positive integer n such j that j j (jf f j 2 ) 2 ; all k;ln : k l j134 There is no loss of generality to assume thatn n ::: : Set 1 2  j E = jf f j 2 j n n j j+1 and 1 F = E : k j j=k c If x2F andijk k X jf (x)f (x)j jf (x)f (x)j n n n n i j l+1 l jli X l j+1  2 2 jli 1 c and we conclude that (f (x)) is a Cauchy sequence for everyx2F : Let n j j=1 k 1 c G = F and note that for every …xed positive integerk; k=1 k 1 X c j k+1 (G )(F ) 2 = 2 : k j=k c Thus G is a -null set. We now de…ne f(x) = lim f (x) if x2 G and j1 n j f(x) = 0 if x2= G: 1 We next prove that the sequence (f ) converges to f in measure. If n n=1 c x2F andjk we get k j+1 jf(x)f (x)j 2 : n j Thus, if jk j+1 k+1 (jff j 2 )(F ) 2 : n k j Since " " (jf f j")(jf f j )+(jf f j ) n n n n j j 2 2 if " 0; Part (a) follows at once. (b) For each" 0; Z (jf f j") =  (jf f j)d n n ";1 X135 and Part (c) follows from the Lebesgue Dominated Convergence Theorem. (c) Set for …xedk;n2N ; +   1 1 E = jf f j : kn j j=n k We have 1 \ E 2Z kn  n=1 and since is a …nite measure (E ) 0 as n1: kn k 1 Given" 0pickn 2N suchthat(E )"2 :Then,ifE = E , k + kn kn k k=1 k (E)". Moreover, if x2= E andjn k 1 jf (x)f(x)j : j k The theorem is proved. 1 2 Corollary 4.1.1. The spaces L () and L () are complete. 1 p PROOF.Suppose p = 1or 2andlet (f ) beaCauchysequenceinL (): n n=1 1 Weknowfromtheprevioustheoremthatthereexistsasubsequence (f ) n j j=1 whichconvergespointwisetoafunctionf 2F(X)a.e. :Thus, byFatou’s Lemma, Z Z p p jff j d liminf jf f j d k n k j j1 X X p p and it follows that f f 2 L () and, hence f = (f f ) +f 2 L (): k k k Moreover, we have thatkff k 0 ask1: This concludes the proof k p of the theorem. 2 2 Corollary 4.1.2. Suppose  2N(0; ); n2N ; and   in L (P) as + n n n n1: Then  is a centred Gaussian random variable.136 q   2 PROOF. We have that k  k = E  =  and k  k k  k =  2 n 2 2 def n n n asn1: Suppose f is a bounded continuous function on R. Then, by dominated convergence, Z Z Ef( ) = f( x)d (x) f(x)d (x) n n 1 1 R R 1 as n 1. Moreover, there exists a subsequence ( ) which converges n k=1 k to a.s. Hence, by dominated convergence   E f( ) Ef() n k as k1 and it follows that Z Ef() = f(x)d (x): 1 R By using Corollary 3.1.3 the theorem follows at once. Theorem 4.1.3. Suppose X is a standard space and  a positive -…nite 1 2 Borel measure on X. Then the spaces L () and L () are separable. 1 PROOF. Let (E ) be a denumerable collection of Borel sets with …nite k k=1 1 -measures and such that E E and E = X: Set  =  and k k+1 k k=1 k E k p …rst suppose that the set D is at most denumerable and dense in L ( ) k k for every k 2 N : Without loss of generality it can be assumed that each + member of D vanishes o¤E : By monotone convergence k k Z Z fd = lim fd , f  0 measurable, k k1 X X 1 p anditfollowsthattheset D isatmostdenumerableanddenseinL (): k k=1 From now on we can assume that  is a …nite positive measure. Let A be an at most denumerable dense subset of X and and suppose the subset fr ; n2N ;g of 0;1 is dense in 0;1: Furthermore, denote byU the n +137 classofallopensetswhichare…niteunionsofopenballsofthetypeB(a;r ); n a2A; n2N . If U is any open subset of X + U =V :V U andV 2U and, hence, by the Ulam Theorem (U) = supf(V); V 2U andV Ug: Let K be the class of all functions which are …nite sums of functions of the type  ; where  is a positive rational number and U 2U. It follows U thatK is at most denumerable. p Suppose " 0 and that f 2 L () is non-negative. There exists a 1 sequence of simple measurable functions (' ) such that i i=1 0' "f a.e. : i p p Sincejf' j f ; the Lebesgue Dominated Convergence Theorem shows i " that k f ' k for an appropriate k: Let ;:::; be the distinct p 1 l k 2 positive values of ' and set k l C = 1+ : k k=1 Now for each …xed j 2 f1;:::;lg we use Theorem 3.1.3 to get an open 1 " U ' (f g) such thatk  1 k and from the above we j j p k U ' (f g) j j 4C k " get aV 2U such thatV U andk  k : Thus j j j p U V j j 4C " k  1 k p V j ' (f g) j k 2C and l kf  k " j p k=1 V j Now it is simple to …nd a 2 K such that k f k ": From this we p deduce that the set KK =fgh; g;h2Kg p is at most denumerable and dense inL ():138 Thesetofallreal-valuedandin…nitelymanytimesdi¤erentiablefunctions n (1) n de…ned on R is denoted byC (R ) and  (1) n (1) n C (R ) = f 2C (R ); suppf compact : c Recall that the support suppf of areal-valuedcontinuous functionf de…ned n on R is the closure of the set of all x wheref(x) = 6 0: If n Y n f(x) = f'(1+x )'(1x )g; x = (x ;:::;x )2R k k 1 n k=1 1 1 n where'(t) = exp(t ); if t 0; and'(t) = 0; if t 0; thenf 2C (R ) . c The proof of the previous theorem also gives Part (a) of the following n Theorem 4.1.4. Suppose  is a positive Borel measure in R such that n (K)1 for every compact subset K of R : The following sets are dense 1 2 in L (); and L () : (a) the linear span of the functions n  ; I open boundedn-cell in R ; I (1) n (b)C (R ): c PROOF. a) The proof is almost the same as the proof of Theorem 4.1.3. First theE :s can be chosen to be open balls with their centres at the origin k n sinceeachboundedsetinR has…nite-measure. Moreover, asintheproof of Theorem 4.1.3 we can assume that is a …nite measure. Now letA be an n at most denumerable dense subset of R and for eacha2A let R(a) =fr 0; (fx2X; jx a j=rg) 0 for somek = 1;:::;ng: k k Then R(a) is at most denumerable and there is a subsetfr ; n2N g a2A n + of 0;1n R(a) which is dense in 0;1: Finally, letU denote the class a2A of all open sets which are …nite unions of open balls of the type B(a;r ); n a 2 A; n 2 N ; and proceed as in the proof of Theorem 4.1.3. The result + follows by observing that the characteristic function of any member of U equals a …nite sum of characteristic functions of open bounded n-cells a.e. :139 Part (b) in Theorem 4.1.4 follows from Part (a) and the following n Lemma 4.1.1. Suppose K U R , where K is compact and U is open. (1) n Then there exists a function f 2C (R ) such that c K f U that is  f  and suppf U: K U 1 n PROOF. Suppose2C (R ) is non-negative, suppB(0;1); and c Z dm = 1: n n R 1 Moreover, let" 0 be …xed. For anyg2L (v ) we de…ne n Z n 1 f (x) =" g(y)(" (xy))dy: " n R Since k +:::+k 1 n  1 jgj maxj j2L (v ); all k ;:::;k 2N n 1 n k n 1 k n R x :::x 1 n 1 n the Lebesgue Dominated Convergent Theorem shows that f 2 C (R ): " 1 n n Heref 2C (R ) if g vanishes o¤a bounded subset of R : In fact, " c suppf  (suppg) : " " 1 c Now choose a positive number" d(K;U ) and de…neg = : Since K 2 " Z f (x) = g(x"y)(y)dy " n R we also have thatf (x) = 1 if x2K: The lemma is proved. "140 1 n Example 4.1.2. Suppose f 2 L (m ) and let g : R R be a bounded n Lebesgue measurable function. Set Z n h(x) = f(xy)g(y)dy; x2R : n R We claim thath is continuous. To see this …rst note that Z h(x+x)h(x) = (f(x+xy)f(xy))g(y)dy n R and Z jh(x+x)h(x)jK jf(x+xy)f(xy)jdy n R Z =K jf(x+y)f(y)jdy n R n n ifjg(x)jK foreveryx2R : Now…rstchoose" 0 andthen'2C (R ) c such that kf'k ": 1 Using the triangle inequality, we get Z jh(x+x)h(x)jK(2kf'k + j'(x+y)'(y)jdy) 1 n R Z K(2"+ j'(x+y)'(y)jdy) n R where the right hand side is smaller than 3K" ifj xj is su¢ ciently small. This proves thath is continuous. Example 4.1.3. SupposeA2R andm (A) 0: We claim that the set n n AA =fxx;x2Ag contains a neighbourhood of the origin. To show this there is no loss of generality to assume that m (A) 1: n Set n f(x) =m (A\(A+x)); x2R : n141 Note that Z f(x) =  (y) (yx)dy A A n R and Example 4.1.2 proves thatf is continuous. Sincef(0) 0 there exists a  0 suchthatf(x) 0 ifjxj: Inparticular,A\(A+x) = 6  ifjxj; which proves that B(0;)AA: The following three examples are based on the Axiom of Choice. Example 4.1.4. Let NL be the non-Lebesgue measurable set constructed in Section 1.3. Furthermore, assume A  R is Lebesgue measurable and A  NL: We claim that m(A) = 0: If not, there exists a  0 such that B(0;)AANLNL: If 0r andr2Q, there exista;b2NL such that a =b+r: But thena = 6 b and at the same timea andb belong to the same equivalence class, which is a contradiction. Accordingly from this, m(A) = 0:   1 1 Example4.1.5. SupposeA ; is Lebesguemeasurableandm(A) 2 2 0: We claim there exists a non-Lebesgue measurable subset ofA: To see this note that 1 A = ((r +NL)\A) i i=1 1 where(r ) isanenumerationoftherationalnumbersintheinterval1;1: i i=1 If each set (r +NL)\A; is Lebesgue measurable i 1 m(A) =  m((r +NL)\A) i i=1 and we conclude that m((r +NL)\A) 0 for an appropriate i: But then i m(NL\(Ar )) 0 andNL\(Ar )NL; which contradicts Example i i 4.1.4. Hence (r +NL)\A is non-Lebesgue measurable for an appropriatei: i If AisaLebesguemeasurablesubsetofthereallineofpositiveLebesgue measure, we conclude thatA contains a non-Lebesgue measurable subset.142 Example 4.1.6. Set I = 0;1: We claim there exist a continuous function 1 f : I I and a Lebesgue measurable set L  I such that f (L) is not Lebesgue measurable. FirstrecallfromSection3.3theconstructionoftheCantorsetC andthe CantorfunctionG. FirstC = 0;1. ThentrisectC andremove the middle 0 0         1 2 1 2 1 2 interval ; to obtain C = C n ; = 0; ;1 : At the second 1 0 3 3 3 3 3 3 stage subdivide each of the closed intervals of C into thirds and remove 1     1 2 7 8 from each one the middle open thirds. Then C =C n( ; ; ): We 2 1 9 9 9 9 repeat the process and what is left fromC isC . The set 0;1nC is the n1 n n n n n n unionof 2 1 intervalsnumberedI ;k = 1;:::;2 1; wheretheintervalI k k n 1 is situatedtotheleft of theintervalI ifkl: TheCantorsetC =\ C : n l n=1 Supposen is …xed and letG : 0;1 0;1 be the unique the monotone n increasingcontinuousfunction,whichsatis…esG (0) = 0;G (1) = 1;G (x) = n n n n n k2 for x 2 I and which is linear on each interval of C It is clear that n: k n n G = G on each interval I , k = 1;:::;2 1: The Cantor function is n n+1 k de…ned by the limitG(x) = lim G (x); 0x 1: n1 n Now de…ne 1 h(x) = (x+G(x)); x2I 2 where G is the Cantor function. Since h :I I is a strictly increasing and 1 continuous bijection, the inverse function f = h is a continuous bijection fromI ontoI: Set A =h(InC) and B =h(C): Recall from the de…nition of G that G is constant on each removed interval n I and thath takes each removed interval onto an interval of half its length. k 1 1 Thusm(A) = andm(B) = 1m(A) = : 2 2 By the previous example there exists a non-Lebesgue measurable subset 1 M of B: Put L = h (M): The set L is Lebesgue measurable since L  C 1 andC is a Lebesgue null set. However, the setM =f (L) is not Lebesgue measurable. Exercises143 1. Let (X;M;) be a …nite positive measure space and suppose '(t) = min(t;1);t 0:Provethatf f inmeasureifandonlyif'(jf f j) 0 n n 1 inL (): 2. Let  = m : Find measurable functions f : 0;1 0;1; n 2 N ; j0;1 n + 2 such thatf 0 inL () as n1; n liminff (x) = 0 all x2 0;1 n n1 and limsupf (x) = 1 all x2 0;1: n n1 3. If f 2F(X) set kf k = inff 2 0;1; (jf j ) g: 0 Let 0 L () =ff 2F(X); kf k 1g 0 0 and identify functions inL () which agree a.e. : (0) 0 (a) Prove that d =kfg k is a metric on L () and that the corre- 0 sponding metric space is complete. 0 (b) Show thatF(X) =L () if  is a …nite positive measure. p p 4. SupposeL (X;M;)isseparable,wherep = 1or2:ShowthatL (X;M ; ) is separable. 5. Suppose g is a real-valued, Lebesgue measurable, and bounded function of period one. Prove that Z Z Z 1 1 1 lim f(x)g(nx)dx = f(x)dx g(x)dx n1 1 1 0144 1 for everyf 2L (m): 6. Leth (t) = 2+sinnt; 0t 1; andn2N : Find real constants and n + such that Z Z 1 1 lim f(t)h (t)dt = f(t)dt n n1 0 0 and Z Z 1 1 f(t) lim dt = f(t)dt n1 h (t) 0 n 0 for every real-valued Lebesgue integrable functionf on 0;1: n n n 7. If k = (k ;:::;k )2 N ; set e (x) =  sinkx; x = (x ;:::;x )2 R ; 1 n k i i 1 n + i=1 1 n 2 2 andjkj= ( k ) : Prove that i=1 i Z lim fe dm = 0 k n jkj1 n R 1 for everyf 2L (m ): n 1 n 8. Suppose f 2L (m ); where m denotes Lebesgue measure on R . Com- n n pute the following limit and justify the calculations: Z lim jf(x+h)f(x)jdx: jhj1 n R 9. The setAR has positive Lebesgue measure and A+Q =fx+y; x2A andy2Qg where Q stands for the set of all rational numbers. Show that the set Rn(A+Q) is a Lebesgue null set. (Hint: The functionf(x) =m(A(Ax));x2R; is continuous.)145 4.2 Orthogonality 2 Suppose (X;M;) is a positive measure space. If f;g2L (); let Z hf;gi = fgd def X be the so called scalar product of f andg: The Cauchy-Schwarz inequality jhf;gijkf k kgk 2 2 2 shows that the mapf hf;gi of L () intoR is continuous. Observe that 2 2 2 kf +gk =kf k +2hf;gi+kgk 2 2 2 and from this we get the so called Parallelogram Law 2 2 2 2 kf +gk +kfgk = 2(kf k +kgk ): 2 2 2 2 We will say thatf andg are orthogonal (abbr. f ?g) ifhf;gi = 0: Note that 2 2 2 kf +gk =kf k +kgk if and only if f ?g: 2 2 2 Since f ? g implies g ? f; the relation ? is symmetric. Moreover, if ? f ? h and g ? h then ( f + g) ? h for all ; 2R. Thus h = def 2 2 ff 2L (); f ?hg is a subspace of L (); which is closed since the map 2 2 f hf;hi; f 2L () is continuous. If M is a subspace of L (); the set ? ? M = \ h def h2M 2 is a closed subspace of L (): The functionf = 0 if and only if f ?f: 2 2 IfM is a subspace ofL () andf 2L () there exists at most one point ? g 2 M such that f g 2 M : To see this, let g ;g 2 M be such that 0 1 ? ? fg 2M ; k = 0;1: Theng g = (fg )(fg )2M and hence k 1 0 0 1 g g ?g g that isg =g : 1 0 1 0 0 1 2 Theorem 4.2.1. Let M be a closed subspace in L () and suppose f 2 2 L (): Then there exists a unique point g2M such that kfgk kfhk all h2M: 2 2146 Moreover, ? fg2M : The function g in Theorem 4.2.1 is called the projection of f on M and is denoted by Proj f: M PROOF OF THEOREM 4.2.1. Set (2) d = d (f;M) = inf kfgk : def 2 g2M 1 and let (g ) be a sequence inM such that n n=1 d = lim kfg k : n 2 n1 Then, by the Parallelogram Law 2 2 2 2 k (fg )+(fg )k +k (fg )(fg )k = 2(kfg k +kfg k ) k n k n k n 2 2 2 2 that is 1 2 2 2 2 4kf (g +g )k +kg g k = 2(kfg k +kfg k ) k n n k k n 2 2 2 2 2 1 and, since (g +g )2M; we get k n 2 2 2 2 2 4d +kg g k  2(kfg k +kfg k ): n k k n 2 2 2 2 Heretherighthandconvergesto4d askandngotoin…nityandweconclude 1 2 that (g ) is a Cauchy sequence. Since L () is complete and M closed n n=1 there exists ag2M such thatg g as n1: Moreover, n d =kfgk : 2 ? We claim that f g 2 M : To prove this choose h 2 M and 0 arbitrarily and use the inequality 2 2 k (fg)+ hk kfgk 2 2 to obtain 2 2 2 2 kfgk +2 hfg;hi+ khk kfgk 2 2 2147 and 2 2hfg;hi+ khk  0: 2 By letting 0;hfg;hi 0 and replacingh byh;hfg;hi 0: Thus ? ? fg2h and it follows thatfg2M : The uniqueness in Theorem 4.2.1 follows from the remark just before the formulation of Theorem 4.2.1. The theorem is proved. 2 2 A linear mapping T : L () R is called a linear functional on L (): 2 2 If h 2 L (); the map h hf;hi of L () into R is a continuous linear 2 functional onL (): It is a very important fact that every continuous linear 2 functional onL () is of this type. 2 Theorem 4.2.2. Suppose T is a continuous linear functional on L (): 2 Then there exists a unique w2L () such that 2 Tf =hf;wi all f 2L (): 0 2 0 2 PROOF.Uniqueness: If w;w 2L ()andhf;wi=hf;wiforallf 2L (); 0 2 0 thenhf;wwi = 0 for allf 2L (): By choosingf =ww we getf ?f 0 that isw =w: 1 Existence: The set M = T (f0g) is closed since T is continuous and def 2 2 M is a linear subspace of L () since T is linear. If M = L () we choose 2 w = 0: Otherwise, pick ag2L ()nM: Without loss of generality it can be assumed that Tg = 1 by eventually multiplying g by a scalar. The previous ? theorem gives us a vector h2 M such that u = gh2 M : Note that def 2 0kuk =hu;ghi =hu;gi: 2 2 Toconclude the proof, let …xedf 2L () be…xed; andusethat (Tf)g f 2M to obtain h(Tf)gf;ui = 0 or (Tf)hg;ui =hf;ui:148 By setting 1 w = u 2 kuk 2 we are done. 4.3. The Haar Basis and Wiener Measure In this section we will show the existence of Brownian motion with continu- ous paths as a consequence of the existence of linear measure  in the unit interval. The so called Wiener measure is the probability law on C0;1 of real-valued Brownian motion in the time interval 0;1: The Brownian mo- tion process is named after the British botanist Robert Brown (1773-1858). It was suggested by Lous Bachelier in 1900 as a model of stock price ‡uctua- tionsandlaterbyAlbertEinsteinin1905asamodelofthephysicalphenom- enonBrownianmotion. TheexistenceofthemathematicalBrownianmotion process was …rst established by Norbert Wiener in the twenties. Wiener also provedthatthemodelcanbechosensuchthatthepathtW(t); 0t 1; is continuous a.s. Today Brownian motion is a very important concept in probability,…nancialmathematics,partialdi¤erentialequationsandinmany other …elds in pure and applied mathematics. Suppose n is a non-negative integer and set I =f0;:::;ng: A sequence n 2 (e ) in L () is said to be orthonormal if e ? e for all i 6= j; i;j 2 I i i2I i j n n 2 andke k= 1 for eachi2I : If (e ) is orthonormal andf 2L (); i n i i2I n f hf;eie ?e all j2I i2I i i j n and Theorem 4.2.1 shows that kf hf;eie k kf e k all real ;:::; : i2I i i 2 i2I i i 2 1 n n n Moreover 2 2 2 kf k =kf hf;eie k +k  hf;eie k i2I i i i2I i i 2 n 2 n 2 and we get 2 2  hf;ei kf k : i2I i n 2149 2 Wesaythat (e ) isanorthonormal basisinL () if itisorthonormal n n2I n and 2 f =  hf;eie all f 2L (): i2I i i n 1 2 n A sequence (e ) in L () is said to be orthonormal if (e ) is ortho- i i i=0 i=0 2 normal for each non-negative integern: In this case, for eachf 2L (); 1 2 2  hf;ei kf k i i=0 2 and the series 1  hf;eie i i i=0 converges since the sequence n 1 ( hf;eie ) i i i=0 n=0 2 1 of partial sums is a Cauchy sequence in L (): We say that (e ) is an i i=0 2 orthonormal basis inL () if it is orthonormal and 1 2 f =  hf;eie for all f 2L (): i i i=0 1 2 Theorem 4.3.1. An orthonormal sequence (e ) in L () is a basis of i i=0 2 L () if (hf;ei = 0 all i2N))f = 0 i 2 Proof. Letf 2L () and set 1 g =f hf;eie: i i i=0 Then, for anyj2N; 1 hg;e i =hf hf;eie;e i j i i j i=0 1 =hf;e i hf;eihe;e i =hf;e ihf;e i = 0: j i i j j j i=0 Thusg = 0 or 1 f =  hf;eie: i i i=0 The theorem is proved.150 As an example of an application of Theorem 4.3.1 we next construct an 2 orthonormal basis of L (), where  is linear measure in the unit interval. Set H(t) = (t) (t); t2R 1 1 0; ;1 2 2 n1 Moreover,de…neh (t) = 1; 0t 1;andforeachn 1andj = 1;:::;2 , 00 n1 n1 2 h (t) = 2 H(2 tj +1); 0t 1: jn n1 Stated otherwise, we have for eachn 1 andj = 1;:::;2 8 1 n1 j j1 2 2 2 ; t ; n1 n1 2 2 1 n1 j j h (t) = 2 jn 2 2 ; t ; n1 n1 2 2 : 0; elsewhere in 0;1: n1 It is simple to show that the sequence h h ;j = 1;:::;2 ; n  1; is 00; jn 2 orthonormal in L (). We will prove that the same sequence constitute an 2 2 orthonormal basis of L (): Therefore, suppose f 2 L () is orthogonal to n1 each of the functionsh h ;j = 1;:::;2 ; n 1: Then for eachn 1 and 00; jn n1 j = 1;:::;2 1 j Z Z j 2 n1 n1 2 2 fd = fd 1 j j1 2 n1 n1 2 2 and, hence, Z j Z 1 n1 2 1 fd = fd = 0 n1 j1 2 0 n1 2 since Z Z 1 1 fd = fh d = 0: 00 0 0 Thus Z k n1 2 n1 fd = 0; 1jk 2 j n1 2 and we conclude that Z Z 1 b 1 fd = fd = 0; 0ab 1: a;b 0 a151 Accordingly from this, f = 0 and we are done. 1 2 The above basis (h ) = (h h ;h ;h ;h ;h ;h ;h ;:::) of L () k 00; 11 12 22 13 23 33 43 k=0 is called the Haar basis. Let 0t 1 and de…ne for …xedk2N Z Z 1 t a (t) =  (x)h (x)dx = h d k k k 0;t 0 0 so that 1 2  =  a (t)h inL (): k k 0;t k=0 Then, if 0s;t 1; Z 1 1 min(s;t) =  (x) (x)dx =h a (s)h ; i k k 0;s 0;t 0;t k=1 0 1 1 =  a (s)hh ; i =  a (s)a (t): k k k k k=0 0;t k=0 Note that 1 2 t =  a (t): k=0 k 1 If (G ) is a sequence ofN(0;1) distributed randomvariables based on k k=0 a probability space ( ;F;P) the series 1  a (t)G k k k=0 2 converges inL (P) andde…nes aGaussianrandomvariable whichwe denote byW(t): From the above it follows that (W(t)) is a real-valued centred 0t1 Gaussian stochastic process with the covariance EW(s)W(t) = min(s;t): Such a process is called a real-valued Brownian motion in the time interval 0;1: Recall that 1 (h h ;h ;h ;h ;h ;h ;h ;:::) = (h ) : 00; 11 12 22 13 23 33 43 k k=0 We de…ne 1 (a a ;a ;a ;a ;a ;a ;a ;:::) = (a ) 00; 11 12 22 13 23 33 43 k k=0 and 1 (G G ;G ;G ;G ;G ;G ;G ;:::) = (G ) : 00; 11 12 22 13 23 33 43 k k=0152 It is important to note that for …xedn; Z t a (t) =  (x)h (x)dx6= 0 for at most onej: jn jn 0;t 0 Set U (t) =a (t)G 0 00 00 and n1 2 U (t) =  a (t)G ; n2N : n jn jn + j=1 We know that 1 2 W(t) =  U (t) inL (P) n n=0 for …xedt: The spaceC0;1 will from now on be equipped with the metric d(x;y) =kxyk 1 wherekxk = max jx(t)j: Recall that everyx2 C0;1 is uniformly 1 0t1 continuous. From this, remembering that R is separable, it follows that the spaceC0;1 is separable. SinceR is complete it is also simple to show that the metric spaceC0;1 is complete. Finally, if x 2C0;1; n2N; and n 1  kx k 1 n 1 n=0 the series 1  x n n=0 converges since the partial sums n s =  x ; k2N n k k=0 forms a Cauchy sequence. We now de…ne 1  =f2 ; kU k 1g: n 1 n=0 Here 2F since kU k = sup jU (t)j n 1 n 0t1 t2Q for eachn: Next we prove that n is a null set.153 To this end letn 1 and note that     n n 4 4 P kU k 2 P max (ka k jG j) 2 : n 1 jn 1 jn n1 1j2 But 1 ka k = jn 1 n+1 2 2 and, hence, h i   n n 1 n1 + 4 4 2 P kU k 2  2 P jG j 2 : n 1 00 Since Z 1 dy 2 2 y =2 x =2 x 1)P jG jx 2 ye p e 00 x 2 x we get   n n=2 n 2 4 P kU k 2  2 e n 1 and conclude that " 1 1 X X   n 4 n E 1 = P kU k 2 1: n 1 4 kU k 2 n 1 n=0 n=0 From this and the Beppo Levi Theorem (or the …rst Borel-Cantelli Lemma) P  = 1: The trajectory t W(t;); 0  t  1; is continuous for every 2 : Without loss of generality, from now on we can therefore assume that all trajectories of Brownian motion are continuous (by eventually replacing by ): Suppose 0t :::t  1 1 n and letI ;:::;I be open subintervals of the real line. The set 1 n S(t ;:::;t ;I ;:::;I ) =fx2C0;1; x(t )2I ; k = 1;:::;ng 1 n 1 n k k is called an open n-cell in C0;1: A set in C0;T is called an open cell if thereexistsann2N suchthatitisanopenn-cell. The-algebragenerated + by all open cells in C0;1 is denoted by C: The construction above shows that the map W : C0;1154 which maps to the trajectory tW(t;); 0t 1 is (F;C)-measurable. The image measure P is called Wiener measure in W C0;1: The Wiener measure is a Borel measure on the metric spaceC0;1: We leave it as an excersice to prove that C =B(C0;1): """155 CHAPTER5 DECOMPOSITIONOFMEASURES Introduction InthissectionaversionofthefundamentaltheoremofcalculusforLebesgue integrals will be proved. Moreover, the concept of di¤erentiating a measure withrespecttoanothermeasurewillbedevelopped. Averyimportantresult in this chapter is the so called Radon-Nikodym Theorem. 5:1: Complex Measures Let (X;M) be a measurable space. Recall that if A  X; n 2 N , and n + A \A = ifi6=j, the sequence (A ) is called a disjoint denumerable i j n n2N + 1 collection. ThecollectioniscalledameasurablepartitionofAifA = A n n=1 andA 2M for everyn2N : n + A complex function onM is called a complex measure if 1 (A) =  (A ) n n=1 1 foreveryA2M andmeasurablepartition (A ) ofA: Notethat() = 0 n n=1 if  is a complex measure. A complex measure is said to be a real measure if it is a real function. The reader should note that a positive measure need not be a real measure since in…nity is not a real number. If  is a complex measure = +i ,where =Reand =Imarerealmeasures. Re Im Re Im 1 If (X;M;) is a positive measure andf 2L () it follows that Z (A) = fd; A2M A is a real measure and we writed =fd.156 A function :M 1;1 is called a signed measure measure if (a) :M 1;1 or :M 1;1 (b)() = 0 and 1 (c) for everyA2M and measurable partition (A ) of A; n n=1 1 (A) =  (A ) n n=1 where the latter sum converges absolutely if (A)2R: Here 11 = 1 and 1 +x = 1 if x 2 R: The sum of a positivemeasureandarealmeasureandthedi¤erenceofarealmeasureand apositivemeasureareexamplesofsignedmeasuresanditcanbeprovedthat there are no other signed measures (see Folland F). Below we concentrate on positive, real, and complex measures and will not say more about signed measures here. Suppose is a complex measure onM and de…ne for everyA2M 1 jj (A) = sup j(A )j; n n=1 1 where the supremum is taken over all measurable partitions (A ) of A: n n=1 Note thatjj () = 0 and jj (A)j(B)j if A;B2M andAB: Thesetfunctionjjiscalledthetotalvariationoforthetotalvariation measure of : It turns out that j  j is a positive measure. In fact, as will shortly be seen,jj is a …nite positive measure. Theorem5.1.1. The total variation jj of a complex measure is a positive measure. 1 PROOF. Let (A ) be a measurable partition of A: n n=1157 1 For each n; suppose a j  j (A ) and let (E ) be a measurable n n kn k=1 partition of A such that n 1 a  j(E )j: n kn k=1 1 Since (E ) is a partition of A it follows that kn k;n=1 1 1  a  j(E )jjj (A): n kn n=1 k;n=1 Thus 1  jj (A )jj (A): n n=1 1 Toprovetheoppositeinequality, let (E ) beameasurablepartitionof k k=1 1 1 A:Then,since (A \E ) isameasurablepartitionofE and (A \E ) n k k n k n=1 k=1 a measurable partition of A ; n 1 1 1  j(E )j=  j  (A \E )j k n k k=1 k=1 n=1 1 1   j(A \E )j  jj (A ) n k n k;n=1 n=1 and we get 1 jj (A)  jj (A ): n n=1 Thus 1 jj (A) =  jj (A ): n n=1 Sincejj () = 0, the theorem is proved. Theorem5.1.2. The total variation jj of a complex measure  is a …nite positive measure. PROOF. Since jjj j +j j Re Im there is no loss of generality to assume that is a real measure. Supposejj (E) =1 for someE2M: We …rst prove that there exist disjoint sets A;B2M such that AB =E158 and j(A)j 1 and jj (B) =1: 1 To this endletc = 2(1+j(E)j) and let (E ) be ameasurable partition k k=1 of E such that n  j(E )jc k k=1 for some su¢ ciently largen: There exists a subsetN off1;:::;ng such that c j  (E )j : k2N k 2 c SetA = E andB =EnA: Thenj(A)j  1 and k2N k 2 j(B)j=j(E)(A)j c j(A)jj(E)j j(E)j= 1: 2 Since 1 =j  j (E) =j  j (A)+ j  j (B) we have j  j (A) = 1 or j  j (B) = 1: If j  j (B) 1 we interchange A and B and have j(A)j 1 andjj (B) =1: Supposejj (X) =1: SetE =X and choose disjoint setsA ;B 2M 0 0 0 such that A B =E 0 0 0 and j(A )j 1 and jj (B ) =1: 0 0 SetE =B and choose disjoint setsA ;B 2M such that 1 0 1 1 A B =E 1 1 1 and j(A )j 1 and jj (B ) =1: 1 1 1 By induction, we …nd a measurable partition (A ) of the set A = n def n=0 1 A such that j (A ) j 1 for every n: Now, since  is a complex n n n=0 measure, 1 (A) =  (A ): n n=0 But this series cannot converge, since the general term does not tend to zero asn1: This contradiction shows thatjj is a …nite positive measure.159 If  is a real measure we de…ne 1 +  = (jj +) 2 and 1  = (jj): 2 + The measures  and  are …nite positive measures and are called the positive and negative variations of ; respectively . The representation +  =  is called the Jordan decomposition of : Exercises 1. Suppose (X;M;) is a positive measure space and d = fd; where 1 f 2L (): Prove thatdjj=jf jd: 2. Suppose;; and are real measures de…ned on the same-algebra and  and: Prove that  min(;) where 1 min(;) = (+jj): 2 3. Suppose  :MC is a complex measure and f;g :X R measurable functions. Show that j(f 2A)(g2A)jjj (f = 6 g)160 for everyA2R: 5.2. The Lebesque Decomposition and the Radon-Nikodym Theo- rem Let  be a positive measure on M and  a positive or complex measure onM: The measure  is said to be absolutely continuous with respect to  (abbreviated  ) if (A) = 0 for every A2M for which (A) = 0: If we de…ne Z =fA2M; (A) = 0g  it follows that if and only if Z Z :   For example, v andv : n n n n E The measure  is said to be concentrated on E 2 M if  =  , where E  (A) = (E\A) for every A 2 M: This is equivalent to the hypoth- def esis that A 2 Z if A 2 M and A\E = : Thus if E ;E 2 M, where  1 2 E E ; and  is concentrated on E ; then  is concentrated on E : More- 1 2 1 2 over, if E ;E 2 M and  is concentrated on both E and E ; then  is 1 2 1 2 concentrated on E \E : Two measures  and  are said to be mutually 1 2 1 2 singular (abbreviated ? ) if there exist disjoint measurable setsE and 1 2 1 E such that is concentrated onE and is concentrated onE : 2 1 1 2 2 Theorem 5.2.1. Let  be a positive measure and ; ; and  complex 1 2 measures. (i) If   and  ; then (  +  )  for all complex 1 2 1 1 2 2 numbers and : 1 2 (ii) If  ?  and  ? ; then (  +  ) ?  for all complex 1 2 1 1 2 2 numbers and : 1 2 (iii) If  and ?; then  = 0: (iv) If ; then jj: PROOF. The properties (i) and (ii) are simple to prove and are left as exer- cises.161 E Toprove(iii)supposeE2M isa-nullsetand = : IfA2M, then (A) = (A\E) and A\E is a -null set. Since   it follows that A\E 2Z and, hence, (A) =(A\E) = 0: This proves (iii)  1 To prove (iv) suppose A 2 M and (A) = 0: If (A ) is measurable n n=1 partition of A; then (A ) = 0 for every n: Since  ; (A ) = 0 for n n everyn and we conclude thatjj (A) = 0: This proves (vi). Theorem5.2.2. Let beapositivemeasureonMandacomplexmeasure on M: Then the following conditions are equivalent: (a): (b) To every " 0 there corresponds a  0 such that j (E)j " for all E2M with (E): If is a positive measure, the implication (a)) (b) in Theorem 5.2.2 is, in general, wrong. To see this take  = and  =v : Then  and if 1 1 we chooseA = n;1;n2N ; then(A ) 0 asn1 but(A ) =1 n + n n for eachn: PROOF. (a))(b). If (b) is wrong there exist an " 0 and sets E 2M, n n n2N ; such thatj(E )j" and(E ) 2 : Set + n n 1 1 A = E andA =\ A : n k n k=n n=1 n+1 Since A  A  A and (A ) 2 , it follows that (A) = 0 and n n+1 n using thatjj (A )j(E )j; Theorem 1.1.2 (f) implies that n n jj (A) = lim jj (A )": n n1 This contradicts thatjj: (b))(a). If E 2M and (E) = 0 then to each " 0; j (E)j "; and we conclude that(E) = 0: The theorem is proved:162 Theorem 5.2.3. Let  be a -…nite positive measure and  a real measure on M. (a) (The Lebesgue Decomposition of ) There exists a unique pair of real measures  and  onM such that a s  = + ;  ; and ?: a s a s If  is a …nite positive measure,  and  are …nite positive measures. a s 1 (b)(TheRadon-NikodymTheorem) There exits a unique g2L () such that d =gd: a If  is a …nite positive measure, g 0 a.e. : The proof of Theorem 5.2.3 is based on the following Lemma5.2.1. Let (X;M;) be a …nite positive measure space and suppose 1 f 2L (): (a) If a2R and Z fda(E); all E2M E then f a a.e. . (b) If b2R and Z fdb(E); all E2M E then f b a.e. . PROOF. (a) Setg =fa so that Z gd 0; all E2M: E Now chooseE =fg 0g to obtain Z Z 0 gd =  gd 0 E E X163 as  g 0 a.e. : But then Example 2.1.2 yields  g = 0 a.e.  and we E E getE2Z : Thusg 0 a.e.  orf a a.e. :  Part (b) follows in a similar way as Part (a) and the proof is omitted here. (k) (k) PROOF.Uniqueness: (a)Suppose and arerealmeasuresonMsuch a s that (k) (k) (k) (k)  = + ;  ; and ? a s a s fork = 1;2: Then (1) (2) (2) (1)   =  a a s s and (1) (2) (1) (2)    and  ?: a a a a (1) (2) (1) (2) Thus by applying Theorem 5.2.1,   = 0 and  =  : From this a a a a (1) (2) we conclude that = . s s 1 (b) Supposeg 2L ();k = 1;2; and k d =g d =g d: a 1 2 Thenhd = 0 whereh =g g : But then 1 2 Z hd = 0 fh0g and it follows that h 0 a.e. : In a similar way we prove that h 0 a.e. 1 1 . Thush = 0 inL (); that isg =g inL (): 1 2 Existence: The beautiful proof that follows is due to von Neumann. Firstsupposethat and are…nitepositivemeasuresandset =+: 1 1 2 Clearly, L ()L ()L (): Moreover, if f :X R is measurable s Z Z Z p 2 jf jd jf jd f d (X) X X X and from this we conclude that the map Z f fd X164 2 is a continuous linear functional on L (): Therefore, in view of Theorem 2 4.2.2, there exists ag2 L () such that Z Z 2 fd = fgd all f 2L (): X X SupposeE2M and putf = to obtain E Z 0(E) = gd E and, since; Z 0 gd(E): E But then Lemma 5.2.1 implies that 0  g  1 a.e. : Therefore, without loss of generality we can assume that 0  g(x)  1 for all x 2 X and, in addition, as above Z Z 2 fd = fgd all f 2L () X X that is Z Z 2 f(1g)d = fgd all f 2L (): X X A S Put A = f0g 1g, S = fg = 1g;  =  ; and  =  : Note that a s A S  = + : Thechoicef = gives(S) = 0 andhence ?: Moreover, s S the choice n f = (1+:::+g ) E whereE2M; gives Z Z n+1 n (1g )d = (1+:::+g )gd: E E By lettingn1 and using monotone convergence Z (E\A) = hd: E where n h = lim(1+:::+g )g: n1165 Sinceh is non-negative and Z (A) = hd X 1 it follows that h2L (): Moreover, the construction above shows that  =  + : a s In the next step we assume that  is a -…nite positive measure and  1 a …nite positive measure. Let (X ) be a measurable partition of X such n n=1 that (X ) 1 for every n: Let n be …xed and apply Part (a) to the pair n X X X X n n n n  and to obtain …nite positive measures ( ) and ( ) such that a s X X X X X X X n n n n n n n  = ( ) +( ) ; ( )  ; and ( ) ? a s a s and X X X X n n n n d( ) =h d (or ( ) =h  ) a n a n 1 X n where 0  h 2 L ( ): Without loss of generality we can assume that n X n h = 0 o¤X and that ( ) is concentrated on A X where A 2Z : n n s n n n  X n In particular, ( ) =h : Now a n 1 X n  =h+ ( ) s n=1 where 1 h =  h n n=1 and Z hd(X)1: X X 1 1 n 1 Thush2L (): Moreover, =  ( ) is concentratedon A 2 s def s n n=1 n=1 Z : Hence ?:  s Finally if  is a real measure we apply what we have already proved to the positive and negative variations of  and we are done. Example 5.2.1. Let  be Lebesgue measure in the unit interval and  the counting measure in the unit interval restricted to the class of all Lebesgue measurablesubsetsoftheunitinterval. Clearly,:Supposethereisan166 1 h2L () such that d =hd. We can assume that h 0 and the Markov inequality implies that the setfh"g is …nite for every" 0: But then n (h2 0;1) = lim (h 2 ) = 0 n1 R and it follows that 1 =(h = 0) = hd = 0; which is a contradiction. fh=0g Corollary 5.2.1. Suppose  is a real measure. Then there exists 1 h2L (jj) such that jh(x)j= 1 for all x2X and d =hdjj: PROOF. Since j (A) jj  j (A) for every A 2 M, the Radon-Nikodym 1 Theorem implies that d = hd j  j for an appropriate h 2 L (j  j): But thendjj=jhjdjj (see Exercise 1 in Chapter 5.1): Thus Z jj (E) = jhjdjj; all E2M E and Lemma 5.2.1 yields h = 1 a.e. jj: From this the theorem follows at once. Theorem 5.2.4. (Hahn’s Decomposition Theorem) Suppose  is a real measure. There exists an A2M such that c + A A  = and = : PROOF. Let d = hd j  j where j h j= 1: Note that hd = d j  j : Set A =fh = 1g: Then 1 1 + d = (djj +d) = (h+1)d = d A 2 2167 and + d =d d = ( 1)d = d: c A A The theorem is proved. If a real measure  is absolutely continuous with respect to a -…nite positivemeasure, theRadon-NikodymTheoremsaysthatd =fd foran 1 appropritef 2L (): We sometimes write d f = d and call f the Radon-Nikodym derivate of  with respect to: Exercises 1. Let be a-…nite positive measure on (X;M) and (f ) a sequence of n n2N measurablefunctionswhichconvergesin-measuretoameasurablefunction f: Moreover, suppose  is a …nite positive measure on (X;M) such that  : Prove thatf f in-measure. n 2. Suppose  and  ;n 2N, are positive measures de…ned on the same n 1 -algebra and set =   . Prove that n n=0 a) ?  if  ?; all n2N: n b)  if  ; all n2N: n 3. Suppose  is a real measure and  =  ; where  and  are …nite 1 2 1 2 + positive measures. Prove that  and  : 1 2 4. Let  and  be mutually singular complex measures on the same - 1 2 algebra: Show thatj j?j j: 1 2168 5. Let (X;M;) be a -…nite positive measure space and suppose  and  are two probability measures de…ned on the-algebraM such that and : Prove that Z 1 d d sup j(A)(A)j= j jd: 2 d d A2M X 6. Let (X;M) be a measurable space and suppose;:MR and are real measures. Prove that + + + (+)  + : 5.3. The Wiener Maximal Theorem and the Lebesgue Di¤erentia- tion Theorem n We say that a Lebesgue measurable functionf inR is locally Lebesgue in- 1 1 tegrableandbelongstotheclassL (m )iff 2L (m )foreachcompact n n loc K n 1 subsetK of R : In a similar wayf 2L (v ) if f is a Borel function such n loc 1 n 1 that f 2 L (v ) for each compact subset K of R : If f 2 L (m ); we n n K loc de…ne the averageA f(x) of f on the open ball B(x;r) as r Z 1 A f(x) = f(y)dy: r m (B(x;r)) n B(x;r) It follows from dominated convergence that the map (x;r) A f(x) of r n R 0;1 into R is continuous. The Hardy-Littlewood maximal function   f is, by de…nition, f = sup A jf j or, stated more explicitly, r r0 Z 1  n f (x) = sup jf(y)jdy; x2R : m (B(x;r)) r0 n B(x;r)  n n The functionf : (R ;B(R )) (0;1;R ) is measurable since 0;1  f = supA jf j: r r0 r2Q169 Theorem 5.3.1. (Wiener’s Maximal Theorem) There exists a positive 1 constant C =C 1 such that for all f 2L (m ); n n C  m (f ) kf k if 0: n 1 The proof of the Wiener Maximal Theorem is based on the following remarkable result. n Lemma5.3.1. Let C be a collectionof openballs in R and set V = B: B2C If cm (V) there exist pairwise disjoint B ;:::;B 2C such that n 1 k k n  m (B ) 3 c: n i i=1 PROOF.LetKV becompactwithm (K)c;andsupposeA ;:::;A 2C n 1 p 0 cover K: Let B be the largest of the As (that is, B has maximal radius), 1 1 i 0 let B be the largest of the As which are disjoint from B ; let B be the 2 1 3 i 0 largestoftheAswhicharedisjointfromB B ;andsoonuntiltheprocess 1 2 i  k  stops afterk steps. IfB =B(x;r ) putB =B(x;3r ): Then B K i i i i i i i=1 i and k  n k c  m (B ) = 3  m (B ): n n i i=1 i i=1 The lemma is proved. PROOF OF THEOREM 5.3.1. Set  E =ff g: Foreachx2E chooseanr 0suchthatA jf j (x) :Ifcm (E ); x r n x byLemma5.3.1thereexistx ;:::;x 2E suchthattheballsB =B(x;r ); 1 k i i x i i = 1;:::;k; are mutually disjoint and k n  m (B ) 3 c: n i i=1 But then Z Z n n 3 3 n k k c 3  m (B )  jf(y)jdy jf(y)jdy: n i i=1 i=1 n B R i170 The theorem is proved. 1 Theorem 5.3.2. If f 2L (m ); n loc Z 1 lim f(y)dy =f(x) a.e. m : n r0 m (B(x;r)) n B(x;r) 1 PROOF. Clearly, there is no loss of generality to assume that f 2 L (m ): n n n Supposeg2C (R ) = ff 2C(R ); f(x) = 0 if jxj large enoughg. Then c def n limA g(x) =g(x) all x2R : r r0 SinceA ff =A (fg)(fg)+A gg; r r r  limjA ff j (fg) +jfgj: r r0 Now, for …xed 0; m (limjA ff j ) n r r0  m ((fg) )+m (jfgj ) n n 2 2 and the Wiener Maximal Theorem and the Markov Inequality give m (limjA ff j ) n r r0 2C 2  ( + )kfgk : 1 n 1 Remembering thatC (R ) is dense inL (m ); the theorem follows at once. c n 1 If f 2L (m ) we de…ne the so called Lebesgue setL to be n f loc  Z  1 L = x; lim jf(y)f(x)jdy = 0 : f r0 m (B(x;r)) n B(x;r) Note that if q is real and   Z 1 E = x; lim jf(y)qjdy =jf(x)qj q r0 m (B(x;r)) n B(x;r)171 c thenm ( E ) = 0: If x2\ E ; n q2Q q2Q q q Z 1 lim jf(y)f(x)jdy 2jf(x)qj r0 m (B(x;r)) n B(x;r) c for all rational numbers q and it follows thatm (L ) = 0: n f n A family E = (E ) of Borel sets in R is said to shrink nicely to a x x;r r0 n pointx inR ifE B(x;r) for eachr and there is a positive constant ; x;r independent of r; such thatm (E ) m (B(x;r)): n x;r n Theorem 5.3.3. (The Lebesgue Di¤erentiation Theorem) Suppose 1 f 2L (m ) and x2L : Then n f loc Z 1 lim jf(y)f(x)jdy = 0 r0 m (E ) n x;r E x;r and Z 1 lim f(y)dy =f(x): r0 m (E ) n x;r E x;r PROOF. The result follows from the inequality Z Z 1 1 jf(y)f(x)jdy jf(y)f(x)jdy: m (E ) m (B(x;r)) n x;r n E B(x;r) x;r Theorem5.3.4. Suppose  is a real or positive measure on R and suppose n ? v : If  is a positive measure it is assumed that (K) 1 for every n n compact subset of R . Then (E ) x;r lim = 0 a.e. v n r0 v (E ) n x;r IfE =B(x;r) and isthecountingmeasurec n restrictedtoR then x;r Q n n ?v butthelimitinTheorem5.3.4equalsplusin…nityforallx2R :The n n hypothesis "(K)1 for every compact subset ofR " in Theorem 5.3.4 is not super‡ous.172 PROOF.Sincej(E)jjj (E)ifE2R ;thereisnorestrictiontoassume n that is a positive measure (cf. Theorem 3.1.4). Moreover, since (E ) (B(x;r)) x;r  v (E ) v (B(x;r)) n x;r n it can be assumed that E = B(x;r): Note that the function (B(;r)) x;r is Borel measurable for …xed r 0 and (B(x;)) left continuous for …xed n x2R : A SupposeA2Z andv = (v ) : Given 0; it is enough to prove that  n n F 2Z where v n   (B(x;r)) F = x2A; lim  r0 m (B(x;r)) n To this end let" 0 and use Theorem3.1.3 to get an openU A such that (U)": For eachx2F there is an open ball B U such that x (B )v (B ): x n x If V = B and cv (V) we use Lemma 5.3.1 to obtain x ;:::;x such x2F x n 1 k thatB ;:::;B are pairwise disjoint and x x 1 k n k n 1 k c 3  v (B ) 3   (B ) n x x i=1 i i=1 i n 1 n 1  3  (U) 3  ": 1 n Thus v (V) 3  ": Since V F 2R and " 0 is arbitrary, v (F) = 0 n n n and the theorem is proved. 0 Corollary5.3.1. Suppose F :RRis an increasing function. Then F (x) exists for almost all x with respect to linear measure. PROOF. LetD be the set of all points of discontinuity ofF: Suppose1 ab1 and" 0: If ax :::x b; wherex ;:::;x 2D and 1 n 1 n F(x +)F(x )"; k = 1;:::;n k k173 then n n"  (F(x +)F(x ))F(b)F(a): k k k=1 Thus D\ a;b is at most denumerable and it follows that D is at most N denumerable. Set H(x) = F(x+)F(x); x 2 R; and let (x ) be an j j=0 enumeration of the members of the setfH 0g: Moreover, for anya 0; X X H(x ) (F(x +)F(x )) j j j jx ja jx ja j j F(a)F(a)1: Now, if we introduce N (A) =  H(x ) (A); A2R j x j=0 j then  is a positive measure such that (K) 1 for each compact subset K of R. Furthermore, if h is a non-zero real number; 1 1 1 j (H(x+h)H(x)j (H(x+h)+H(x)) 4 (B(x;2jhj) h jhj 4jhj 0 and Theorem 5.3.4 implies that H (x) = 0 a.e. v . Therefore, without loss 1 of generality it may be assumed thatF is right continuous and, in addition, there is no restriction to assume thatF(+1)F(1)1: By Section 1.6F induces a …nite positive Borel measure such that (x;y) =F(y)F(x) if xy: Now consider the Lebesgue decomposition d =fdv +d 1 1 wheref 2L (v ) and?v : If xy; 1 1 Z y F(y)F(x) = f(t)dt+(x;y) x and the previous two theorems imply that F(y)F(x) lim =f(x) a.e. v 1 yx yx174 If yx; Z x F(x)F(y) = f(t)dt+(y;x) y and we get F(y)F(x) lim =f(x) a.e. v : 1 y"x yx The theorem is proved. Exercises 1 1. Suppose F : RR is increasing and let f 2 L (v ) be such that 1 loc 0 F (x) =f(x) a.e. v : Prove that 1 Z y f(t)dtF(y)F(x) if1xy1: x 5.4. Absolutely Continuous Functions and Functions of Bounded Variation Throughoutthissectionaandbarerealswithabandtosimplifynotation 1 we setm =m : Iff 2L (m ) we know from the previous section that a;b ja;b a;b the function Z x (If)(x) = f(t)dt; axb def a has the derivativef(x) a.e. m ; that is a;b Z x d f(t)dt =f(x) a.e. m : a;b dx a Our next main task will be to describe the range of the linear mapI: A function F : a;b R is said to be absolutely continuous if to every " 0 there exists a 0 such that n n  jb a j implies  jF(b )F(a )j" i i i i i=1 i=1175 whenever a ;b ;:::;a ;b are disjoint open subintervals of a;b. It is ob- 1 1 n n vious that an absolutely continuous function is continuous. It can be proved that the Cantor function is not absolutely continuous.1. 1 Theorem 5.4.1. If f 2L (m ); then If is absolutely continuous. a;b PROOF. There is no restriction to assumef  0: Set d =fdm : a;b By Theorem 5.2.2, to every " 0 there exists a  0 such that (A) " for each Lebesgue setA in a;b such thatm (A): Now restrictingA to a;b be a …nite disjoint union of open intervals, the theorem follows. Suppose1 1 and F : ; R: For every x2 ; we de…ne n T (x) = sup jF(x )F(x )j F i i1 i=1 where the supremum is taken over all positive integers n and all choices n (x ) such that i i=0 x x :::x =x : 0 1 n The functionT : ; 0;1 is called the total variation ofF: Note that F T isincreasing. IfT isaboundedfunction,F issaidtobeofboundedvaria- F F tion. AboundedincreasingfunctiononRisofboundedvariation. Therefore the di¤erence of two bounded increasing functions on R is of bounded vari- ation. Interestingly enough, the converse is true. In the special case ; = R we writeF 2BV if F is of bounded variation. Example 5.4.1. Letf:RR be a Lebesgue integrable function and de…ne Z 1 jyj g(x) = e f(xy)dy if x2R: 1176 We claim thatg is a continuous function of bounded variation. jxj To prov this claim puth(x) =e if x2R so that Z 1 g(x) = h(xy)f(y)dy. 1 We …rst prove that the function h is continuous. To this end suppose (a ) is a sequence of real numbers which converges to a real number a: n n2N + Then jh(a y)f(y))j2jf(y))j if n2N andy2R n + 1 and sincef 2L (m) by dominated convergence, Z 1 lim g(a ) = lim h(a y)f(y)dy = n n n1 n1 1 Z 1 h(ay)f(y)dy =g(a) 1 and it follows thatg is continuous. We next prove that the function h is of bounded variation. Recall that the total variation functionT (x) ofh at the pointx is the supremum of all h sums of the type n X jh(x )h(x )j i i1 i=1 where 1x x :::x =x1: 0 1 n Weclaimthathisthedi¤erenceoftwoboundedincreasingfunctions. Setting min(0;x) (x) =e and observing that h(x) = (x)+ (x)1 the claim above is obvious and C = supT 1: def h Moreover, if1x x :::x 1; 0 1 n n X jg(x )g(x )j= i i1 i=1177 Z Z n 1 1 X j h(x y)f(y)dy h(x y)f(y)dy j i i1 1 1 i=1 Z n 1 X  jh(x y)h(x y)jjf(y)jdy i i1 1 i=1 Z n 1 X jh(x y)h(x y)j jf(y)jdy i i1 1 i=1 Z Z 1 1  C jf(y)jdy =C jf(y)jdy1: 1 1 Henceg is of bounded variation. Theorem 5.4.2. Suppose F 2BV: (a) The functions T +F and T F are increasing and F F 1 1 F = (T +F) (T F): F F 2 2 In particular, F is di¤erentiable almost everywhere with respect to linear measure. (b) If F is right continuous, then so is T : F PROOF. (a) Letxy and" 0: Choosex x :::x =x such that 0 1 n n  jF(x )F(x )jT (x)": i i1 f i=1 Then T (y)+F(y) F n   jF(x )F(x )j +jF(y)F(x)j +(F(y)F(x))+F(x) i i1 i=1 T (x)"+F(x) F and, since" 0 is arbitrary,T (y)+F(y)T (x)+F(x): HenceT +F is F F F increasing. Finally, replacing F by F it follows that the function T F F is increasing.178 (b) If c2R andxc; n T (x) =T (c)+sup jF(x )F(x )j f F i i1 i=1 where the supremum is taken over all positive integers n and all choices n (x ) such that i i=0 c =x x :::x =x: 0 1 n SupposeT (c+)T (c) wherec2R: Then there is an" 0 such that F F T (x)T (c)" F F for allxc: Now, sinceF is right continuous at the pointc, for …xedxc there exists a partition cx :::x =x 11 1n 1 such that n 1  jF(x )F(x )j": 1i 1i1 i=2 But T (x )T (c)" F 11 F and we get a partition cx :::x =x 21 2n 11 2 such that n 2  jF(x )F(x )j": 2i 2i1 i=2 Summing up we have got a partition of the interval x ;x with 21 n n 2 1  jF(x )F(x )j + jF(x )F(x )j 2": 2i 2i1 1i 1i1 i=2 i=2 By repeating the process the total variation of F becomes in…nite, which is a contradiction. The theorem is proved. Theorem 5.4.3. Suppose F : a;b R is absolutely continuous. Then 1 there exists a unique f 2L (m ) such that a;b Z x F(x) =F(a)+ f(t)dt; axb: a179 Inparticular, therangeofthemapI equalsthesetofallreal-valuedabsolutely continuous maps on a;b which vanish at the point a: PROOF. SetF(x) =F(a) ifxa andF(x) =F(b) ifxb: There exists a  0 such that n n  jb a j implies  jF(b )F(a )j 1 i i i i i=1 i=1 whenever a ;b ;:::;a ;b are disjoint subintervals of a;b: Let p be the 1 1 n n least positive integer such that a+p  b: Then T  p and F 2 BV: Let F 1 1 F = GH; where G = (T +F) and H = (T F): There exist …nite F F 2 2 positive Borel measures and such that G H  (x;y) =G(y)G(x); xy G and  (x;y) =H(y)H(x); xy: H If we de…ne =  ; G H (x;y) =F(y)F(x); xy: Clearly, (x;y) =F(y)F(x); xy sinceF is continuous. Ournexttaskwillbetoprovethatv . Tothisend, supposeA2R 1 andv (A) = 0: Now choose" 0 and let 0 be as in the de…nition of the 1 absolute continuity of F on a;b: For each k2N there exists an open set + V A such thatv (V ) and lim (V ) =(A): But each …xedV is k 1 k k1 k k 1 a disjoint union of open intervals (a;b ) and hence i i i=1 n  jb a j i i i=1 for everyn and, accordingly from this, 1  jF(b )F(a )j" i i i=1 and 1 j(V )j  j(a;b )j": k i i i=1180 Thusj(A)j" and since " 0 is arbitrary, (A) = 0: From this v 1 and the theorem follows at once. Suppose (X;M;) is a positive measure space. From now on we write 1 1 c f 2L () if there exist a g2L () and an A2M such that A 2Z and  f(x) =g(x) for all x2A: Furthermore, we de…ne Z Z fd = gd X X (cfthediscussioninSection2). Notethatf(x) neednotbede…nedforevery x2X: Corollary 5.4.1. A function f : a;bR is absolutely continuous if and only if the following conditions are true: 0 (i) f (x) exists for m -almost all x2 a;b a;b 0 1 (ii) f 2L (m ) a;b R x 0 (iii) f(x) =f(a)+ f (t)dt; all x2 a;b: a Exercises 1. Supposef : 0;1R satis…es f(0) = 0 and 1 2 f(x) =x sin if 0x 1: 2 x Prove thatf is di¤erentiable everywhere butf is not absolutely continuous. 2. Suppose is a positive real number and f a function on 0;1 such that 1 f(0) = 0 and f(x) = x sin ; 0 x  1. Prove that f is absolutely x continuous if and only if 1:181 3. Suppose f(x) = xcos(=x) if 0 x 2 and f(x) = 0 if x 2 Rn0;2: Prove thatf is not of bounded variation on R. 4 A function f : a;b R is a Lipschitz function, that is there exists a positive real numberC such that jf(x)f(y)jC jxyj 0 for all x;y 2 a;b: Show that f is absolutely continuous and j f (x) j C a.e. m : a;b 5. Supposef : a;bR is absolutely continuous. Prove that Z x 0 T (x) = jf (t)jdt; axb g a if g is the restriction of f to the open interval a;b: 6. Suppose f and g are real-valued absolutely continuous functions on the compact interval a;b. Show that the function h = max(f;g) is absolutely 0 0 0 continuous andh  max(f ;g ) a.e. m . a;b 1 7. Suppose(X;M;)isa…nitepositivemeasurespaceandf 2L ():De…ne Z g(t) = jf(x)tjd(x); t2R: X Prove thatg is absolutely continuous and Z t g(t) =g(a)+ ((f s)(f s))ds if a;t2R: a 8. Letand beprobabilitymeasureson(X;M)suchthatjj (X) = 2: Show that?: 182 5.5. Conditional Expectation 1 Let ( ;F;P) be a probability space and suppose  2 L (P): Moreover, supposeG F is a-algebra and set (A) =P A; A2G and Z (A) = dP; A2G: A Itistrivial thatZ =Z \GZ andtheRadon-NikodymTheoremshows  P  1 there exists a unique2L () such that Z (A) = d all A2G A or, what amounts to the same thing, Z Z dP = dP all A2G: A A Note that  is (G;R)-measurable. The random variable  is called the con- ditional expectation of  givenG and it is standard to write =EjG: 1 A sequence of -algebras (F ) is called a …ltration if n n=1 F F F: n n+1 1 1 If (F ) is a …ltration and ( ) is a sequence of real valued random n n=1 n n=1 variables such that for eachn; 1 (a) 2 L (P) n (b) is (F ;R)-measurable n n   (c)E  jF = n n+1 n 1 then ( ;F ) is called a martingale. There are very nice connections n n n=1 between martingales and the theory of di¤erentiation (see e.g Billingsley B and Malliavin M): """183 CHAPTER6 COMPLEXINTEGRATION Introduction In this section, in order to illustrate the power of Lebesgue integration, we collect a few results, which often appear with uncomplete proofs at the un- dergraduate level. 6.1. Complex Integrand Sofarwe have onlytreatedintegrationof functionswiththeirvalues inRor 0;1 and it is the purpose of this section to discuss integration of complex valued functions. 1 Suppose (X;M;) is a positive measure. Letf;g2L (): We de…ne Z Z Z (f +ig)d = fd+i gd: X X X If and are real numbers, Z Z ( +i )(f +ig)d = (( f g)+i( g + f))d X X Z Z = ( f g)d+i ( g + f)d X X Z Z Z Z = fd gd+i gd+i fd X X X X Z Z = ( +i )( fd+i gd) X X Z = ( +i ) (f +ig)d: X184 1 1 1 Wewritef 2L (;C)ifRef;Imf 2L ()andhave,foreveryf 2L (;C) and complex ; Z Z fd = fd: X X 1 Clearly, if f;g 2L (;C); then Z Z Z (f +g)d = fd+ gd: X X X Now suppose is a complex measure on M: If 1 1 1 f 2L (;C) = L ( ;C)\L ( ;C) Re Im def we de…ne Z Z Z fd = fd +i fd : Re Im X X X 1 It follows for everyf;g2L (;C) and 2C that Z Z fd = fd: X X and Z Z Z (f +g)d = fd+ gd: X X X 6.2. The Fourier Transform n Below, if x = (x ;:::;x ) andy = (y ;:::;y )2R ; we let 1 n 1 n n hx;yi =  x y : k k k=1 and p jxj= hx;yi: If is a complex measure onR (orR ) the Fourier transform of is n n de…ned by Z ihx;yi n  (y) = e d(x); y2R : n R185 Note that n  (0) =(R ): 1 The Fourier transform of a functionf 2L (m ;C) is de…ned by n f(y) = (y) whered =fdm : n n Theorem6.2.1. The canonical Gaussian measure in R has the Fourier n transform 2 jyj 2 (y) =e : n PROOF. Since = ::: (n factors) n 1 1 it is enough to consider the special casen = 1: Set Z 2 1 x 2 g(y) = (y) = p e cosxydx: 1 2 R Note thatg(0) = 1: Since cosx(y +h)cosxy j jjxj h the Lebesgue Dominated Convergence Theorem yields Z 2 1 x 0 2 g (y) = p xe sinxydx 2 R (Exercise: Prove this by using Example 2.2.1). Now, by partial integration, Z h i 2 x=1 2 1 x y x 0 2 2 g (y) = p e sinxy p e cosxydx x=1 2 2 R that is 0 g (y)+yg(y) = 0 and we get 2 y 2 g(y) =e :186 n 1 If  = ( ;:::; ) is an R -valued random variable with  2 L (P); 1 n k k = 1;:::;n; the characteristic functionc of  is de…ned by    ih;yi n c (y) =E e =P (y); y2R :   For example, if 2N(0;); then =G; whereG2N(0;1); and we get   ihG;yi c (y) =E e = (y)  1 2 2  y 2 =e : Choosingy = 1 results in   1 2 i E  2 E e =e if 2N(0;): n Thus if ( ) is a centred real-valued Gaussian process k k=1     n 1 i y  n 2 k k k=1 E e = exp( E ( y  ) k k k=1 2     1 n 2 2 = exp(  y E   y y E   ): 1jkn j k k=1 k k j k 2 In particular, if   E   = 0; j = 6 k j k we see that 2 y   2 n k i y  n E  k k k k=1 2 E e =  e k=1 or     n i y  n iy  k k k k k=1 E e =  E e : k=1 n Statedotherwise,theFouriertranformsofthemeasuresP and P ( ;:::; )  k=1 1 n k n agree. Below we will show that complex measures in R with the same Fourier transforms are equal and we get the following n Theorem 6.2.2. Let ( ) be a centred real-valued Gaussian process with k k=1 uncorrelated components, that is   E   = 0; j = 6 k: j k187 Then the random variables  ;:::; are independent. 1 n 6.3 Fourier Inversion 1 1 Theorem 6.3.1. Suppose f 2 L (m ): If f 2 L (m ) and f is bounded n n and continuous Z dy ihy;xi n f(x) = e f(y) ; x2R : n d (2) R PROOF. Choose" 0: We have Z Z Z  2 2 " " 2 dy 2 dy ihy;xi jyj ihy;xui jyj 2 2 e e f(y) = f(u) e e du n n n (2) n n (2) R R R where the right side equals   Z Z Z 1 xu 1 2 dv du 2 du ihv; i jvj juxj 2 " 2 2" f(u) e e p p = f(u)e p n n n n n n n n 2 2 " 2 " R R R Z 1 dz 2 jzj 2 = f(x+"z)e p : n n 2 R Thus Z Z 2 " dy 1 dz 2 2 ihy;xi jyj jzj 2 2 e e f(y) = f(x+"z)e p : n n n (2) n 2 R R By letting" 0 and using the Lebesgue Dominated Convergence Theorem, Theorem 6.3.1 follows at once. 1 n n Recall that C (R ) denotes the class of all functions f : R R c with compact support which are in…nitely many times di¤erentiable. If f 2 1 n 1 C (R ) then f 2 L (m ). To see this, suppose y 6= 0 and use partial n k c integration to obtain Z Z 1 ihx;yi ihx;yi 0 f(y) = e f(x)dx = e f (x)dx x k d iy d R k R and Z 1 ihx;yi (l) f(y) = e f (x)dx; l2N: x l k (iy ) d k R188 Thus Z l (l) jy jjf(y)j jf (x)jdx; l2N k x k n R and we conclude that n+1 sup(1+jyj) jf(y)j1: n y2R 1 and, hence, f 2L (m ): n 1 n 1 Corollary 6.3.1. If f 2 C (R ); then f 2L (m ) and n c Z dy ihy;xi n f(x) = e f(y) ; x2R : n (2) n R n Corollary 6.3.2 If  is a complex Borel measure in R and  = 0; then  = 0: f(y) 1 n PROOF.Choosef 2C (R ). Wemultiplytheequation (y) = 0by n c (2) n and integrate over R with respect to Lebesgue measure to obtain Z f(x)d(x) = 0: n R 1 n Sincef 2C (R ) is arbitrary it follows that = 0: The theorem is proved. c 6.4. Non-Di¤erentiability of Brownian Paths Let ND denote the set of all real-valued continuous function de…ned on the unit interval which are not di¤erentiable at any point. It is well known that ND is non-empty. In fact, if  is Wiener measure on C0;1, x 2 ND a.e. : The purpose of this section is to prove this important property of Brownian motion.189 Let W = (W(t)) be a real-valued Brownian motion in the time 0t1 interval 0;1 suchthateverypathtW(t); 0t 1 iscontinuous. Recall that EW(t) = 0 and EW(s)W(t) = min(s;t): If 0t :::t  1 0 n and 1j kn E(W(t )W(t ))(W(t )W(t ) k k1 j j1 =E(W(t )W(t )EW(t )W(t )EW(t )W(t )+EW(t )W(t ) k j k j1 k1 j k1 j1 =t t t +t = 0: j j1 j j1 From the previous section we now infer that the random variables W(t )W(t );:::;W(t )W(t ) 1 0 n n1 are independent. Theorem 7. The function t W(t); 0  t  1 is not di¤erentiable at any point t2 0;1 a.s. P: PROOF. Without loss of generality we assume the underlying probability space is complete. Let c;" 0 and denote by B(c;") the set of all 2 such that jW(t)W(s)jcjtsjif t2 s";s+"\0;1 for somes2 0;1: It is enough to prove that the set 1 1 1 B(j; ): k j=1 k=1 is of probability zero. From now on let c;" 0 be …xed. It is enough to proveP B(c;") = 0 :190 Set j +1 j X = max jW( )W( )j n;k kjk+3 n n for each integern 3 andk2f0;:::;n3g: Letn 3 be so large that 3 ": n We claim that   6c B(c;") min X  : n;k 0kn3 n If 2B(c;") there exists ans2 0;1 such that jW(t)W(s)jcjtsjif t2 s";s+"\0;1: Now choosek2f0;:::;n3g such that   k k 3 s2 ; + : n n n If kj k +3; j +1 j j +1 j jW( )W( )jjW( )W(s)j +jW(s)W( )j n n n n 6c  n 6c and, hence, X  : Now n;k n   6c B(c;") min X  n;k 0kn3 n and it is enough to prove that   6c lim P min X  = 0: n;k n1 0kn3 n But     n3 X 6c 6c P min X   P X  n;k n;k 0kn3 n n k=0191     6c 6c = (n2)P X  nP X  n;0 n;0 n n   1 6c 6c 3 3 p =n(P jW( )j ) =n(P(jW(1)j ) n n n 12c 3 n(p ) : 2n where the right side converges to zero as n1. The theorem is proved. Recall that a function of bounded variation possesses a derivative a.e. withrespecttoLebesguemeasure. Therefore,withprobabilityone,aBrown- ian path is not of bounded variation. In view of this an integral of the type Z 1 f(t)dW(t) 0 cannot be interpreted as an ordinary Stieltjes integral. Nevertheless, such an integral can be de…ned by completely di¤erent means and is basic in, for example, …nancial mathematics. """192 CHAPTER6 COMPLEXINTEGRATION Introduction In this section, in order to illustrate the power of Lebesgue integration, we collect a few results, which often appear with uncomplete proofs at the un- dergraduate level. 6.1. Complex Integrand Sofarwe have onlytreatedintegrationof functionswiththeirvalues inRor 0;1 and it is the purpose of this section to discuss integration of complex valued functions. 1 Suppose (X;M;) is a positive measure. Letf;g2L (): We de…ne Z Z Z (f +ig)d = fd+i gd: X X X If and are real numbers, Z Z ( +i )(f +ig)d = (( f g)+i( g + f))d X X Z Z = ( f g)d+i ( g + f)d X X Z Z Z Z = fd gd+i gd+i fd X X X X Z Z = ( +i )( fd+i gd) X X Z = ( +i ) (f +ig)d: X193 1 1 1 Wewritef 2L (;C)ifRef;Imf 2L ()andhave,foreveryf 2L (;C) and complex ; Z Z fd = fd: X X 1 Clearly, if f;g 2L (;C); then Z Z Z (f +g)d = fd+ gd: X X X Now suppose is a complex measure on M: If 1 1 1 f 2L (;C) = L ( ;C)\L ( ;C) Re Im def we de…ne Z Z Z fd = fd +i fd : Re Im X X X 1 It follows for everyf;g2L (;C) and 2C that Z Z fd = fd: X X and Z Z Z (f +g)d = fd+ gd: X X X 6.2. The Fourier Transform n Below, if x = (x ;:::;x ) andy = (y ;:::;y )2R ; we let 1 n 1 n n hx;yi =  x y : k k k=1 and p jxj= hx;yi: If is a complex measure onR (orR ) the Fourier transform of is n n de…ned by Z ihx;yi n  (y) = e d(x); y2R : n R194 Note that n  (0) =(R ): 1 The Fourier transform of a functionf 2L (m ;C) is de…ned by n f(y) = (y) whered =fdm : n n Theorem6.2.1. The canonical Gaussian measure in R has the Fourier n transform 2 jyj 2 (y) =e : n PROOF. Since = ::: (n factors) n 1 1 it is enough to consider the special casen = 1: Set Z 2 1 x 2 g(y) = (y) = p e cosxydx: 1 2 R Note thatg(0) = 1: Since cosx(y +h)cosxy j jjxj h the Lebesgue Dominated Convergence Theorem yields Z 2 1 x 0 2 g (y) = p xe sinxydx 2 R (Exercise: Prove this by using Example 2.2.1). Now, by partial integration, Z h i 2 x=1 2 1 x y x 0 2 2 g (y) = p e sinxy p e cosxydx x=1 2 2 R that is 0 g (y)+yg(y) = 0 and we get 2 y 2 g(y) =e :195 n 1 If  = ( ;:::; ) is an R -valued random variable with  2 L (P); 1 n k k = 1;:::;n; the characteristic functionc of  is de…ned by    ih;yi n c (y) =E e =P (y); y2R :   For example, if 2N(0;); then =G; whereG2N(0;1); and we get   ihG;yi c (y) =E e = (y)  1 2 2  y 2 =e : Choosingy = 1 results in   1 2 i E  2 E e =e if 2N(0;): n Thus if ( ) is a centred real-valued Gaussian process k k=1     n 1 i y  n 2 k k k=1 E e = exp( E ( y  ) k k k=1 2     1 n 2 2 = exp(  y E   y y E   ): 1jkn j k k=1 k k j k 2 In particular, if   E   = 0; j = 6 k j k we see that 2 y   2 n k i y  n E  k k k k=1 2 E e =  e k=1 or     n i y  n iy  k k k k k=1 E e =  E e : k=1 n Statedotherwise,theFouriertranformsofthemeasuresP and P ( ;:::; )  k=1 1 n k n agree. Below we will show that complex measures in R with the same Fourier transforms are equal and we get the following n Theorem 6.2.2. Let ( ) be a centred real-valued Gaussian process with k k=1 uncorrelated components, that is   E   = 0; j = 6 k: j k196 Then the random variables  ;:::; are independent. 1 n 6.3 Fourier Inversion 1 1 Theorem 6.3.1. Suppose f 2 L (m ): If f 2 L (m ) and f is bounded n n and continuous Z dy ihy;xi n f(x) = e f(y) ; x2R : n d (2) R PROOF. Choose" 0: We have Z Z Z  2 2 " " 2 dy 2 dy ihy;xi jyj ihy;xui jyj 2 2 e e f(y) = f(u) e e du n n n (2) n n (2) R R R where the right side equals   Z Z Z 1 xu 1 2 dv du 2 du ihv; i jvj juxj 2 " 2 2" f(u) e e p p = f(u)e p n n n n n n n n 2 2 " 2 " R R R Z 1 dz 2 jzj 2 = f(x+"z)e p : n n 2 R Thus Z Z 2 " dy 1 dz 2 2 ihy;xi jyj jzj 2 2 e e f(y) = f(x+"z)e p : n n n (2) n 2 R R By letting" 0 and using the Lebesgue Dominated Convergence Theorem, Theorem 6.3.1 follows at once. 1 n n Recall that C (R ) denotes the class of all functions f : R R c with compact support which are in…nitely many times di¤erentiable. If f 2 1 n 1 C (R ) then f 2 L (m ). To see this, suppose y 6= 0 and use partial n k c integration to obtain Z Z 1 ihx;yi ihx;yi 0 f(y) = e f(x)dx = e f (x)dx x k d iy d R k R and Z 1 ihx;yi (l) f(y) = e f (x)dx; l2N: x l k (iy ) d k R197 Thus Z l (l) jy jjf(y)j jf (x)jdx; l2N k x k n R and we conclude that n+1 sup(1+jyj) jf(y)j1: n y2R 1 and, hence, f 2L (m ): n 1 n 1 Corollary 6.3.1. If f 2 C (R ); then f 2L (m ) and n c Z dy ihy;xi n f(x) = e f(y) ; x2R : n (2) n R n Corollary 6.3.2 If  is a complex Borel measure in R and  = 0; then  = 0: f(y) 1 n PROOF.Choosef 2C (R ). Wemultiplytheequation (y) = 0by n c (2) n and integrate over R with respect to Lebesgue measure to obtain Z f(x)d(x) = 0: n R 1 n Sincef 2C (R ) is arbitrary it follows that = 0: The theorem is proved. c 6.4. Non-Di¤erentiability of Brownian Paths Let ND denote the set of all real-valued continuous function de…ned on the unit interval which are not di¤erentiable at any point. It is well known that ND is non-empty. In fact, if  is Wiener measure on C0;1, x 2 ND a.e. : The purpose of this section is to prove this important property of Brownian motion.198 Let W = (W(t)) be a real-valued Brownian motion in the time 0t1 interval 0;1 suchthateverypathtW(t); 0t 1 iscontinuous. Recall that EW(t) = 0 and EW(s)W(t) = min(s;t): If 0t :::t  1 0 n and 1j kn E(W(t )W(t ))(W(t )W(t ) k k1 j j1 =E(W(t )W(t )EW(t )W(t )EW(t )W(t )+EW(t )W(t ) k j k j1 k1 j k1 j1 =t t t +t = 0: j j1 j j1 From the previous section we now infer that the random variables W(t )W(t );:::;W(t )W(t ) 1 0 n n1 are independent. Theorem 7. The function t W(t); 0  t  1 is not di¤erentiable at any point t2 0;1 a.s. P: PROOF. Without loss of generality we assume the underlying probability space is complete. Let c;" 0 and denote by B(c;") the set of all 2 such that jW(t)W(s)jcjtsjif t2 s";s+"\0;1 for somes2 0;1: It is enough to prove that the set 1 1 1 B(j; ): k j=1 k=1 is of probability zero. From now on let c;" 0 be …xed. It is enough to proveP B(c;") = 0 :199 Set j +1 j X = max jW( )W( )j n;k kjk+3 n n for each integern 3 andk2f0;:::;n3g: Letn 3 be so large that 3 ": n We claim that   6c B(c;") min X  : n;k 0kn3 n If 2B(c;") there exists ans2 0;1 such that jW(t)W(s)jcjtsjif t2 s";s+"\0;1: Now choosek2f0;:::;n3g such that   k k 3 s2 ; + : n n n If kj k +3; j +1 j j +1 j jW( )W( )jjW( )W(s)j +jW(s)W( )j n n n n 6c  n 6c and, hence, X  : Now n;k n   6c B(c;") min X  n;k 0kn3 n and it is enough to prove that   6c lim P min X  = 0: n;k n1 0kn3 n But     n3 X 6c 6c P min X   P X  n;k n;k 0kn3 n n k=0200     6c 6c = (n2)P X  nP X  n;0 n;0 n n   1 6c 6c 3 3 p =n(P jW( )j ) =n(P(jW(1)j ) n n n 12c 3 n(p ) : 2n where the right side converges to zero as n1. The theorem is proved. Recall that a function of bounded variation possesses a derivative a.e. withrespecttoLebesguemeasure. Therefore,withprobabilityone,aBrown- ian path is not of bounded variation. In view of this an integral of the type Z 1 f(t)dW(t) 0 cannot be interpreted as an ordinary Stieltjes integral. Nevertheless, such an integral can be de…ned by completely di¤erent means and is basic in, for example, …nancial mathematics. """201 REFERENCES B Billingsley, P. (1995) Probability and Measure. Wiley&Sons. BO Bogachev, V. I. (2007) Measure Theory, Springer. DDudley,R.M.(1989)RealAnalysisandProbability. Wadsworth&Brooks. DI Diedonné, J. (1960) Foundations of Modern Analysis. Academic Press. F Folland, G. B. (1999) Real Analysis; Modern Techniques and Their Ap- plications. Second Edition. Wiley&Sons. M Malliavin, P. (1995) Integration and Probability. Springer. R Rudin, W. (1966) Real and Complex Analysis. McGraw-Hill. S Solovay R. M. (1970) A model of set theory in which every set of reals is Lebesgue measurable. Annals of Mathematics92, 1-56.
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