Measure and Measurable Functions

Transcription

1 3 Measure ad Measurable Fuctios 3.1 Measure o a Arbitrary σ-algebra Recall from Chapter 2 that the set M of all Lebesgue measurable sets has the followig properties: R M, E M implies E c M, E M for N implies E M, ad {E } is a coutable disjoit family i M implies ( ) m E = m(e ). Now, we cosider geeral situatios havig the above properties. Defiitio Let X be a set ad A be a family of subsets of X. The A is called a σ-algebra if (a) X A, (b) A A implies A c A, (c) A A for N implies A M. The pair (X, A) is called a measurable space, ad members of the σ- algebra A are called measurable sets. Defiitio Let (X, A) be a measurable space. A fuctio µ : A [0, ] is called a measure if (a) µ( ) = 0, ad (b) for ay coutable disjoit family {A } i A, ( ) µ A = 31 µ(a ).

2 32 Measure ad Measurable Fuctios The pair (X, A, µ) is called a measure space. For A A, µ(a) is called the measure of A. CONVENTION: We may recall that is a symbol attached to the set R of real umbers such that a + =, a < a R ad + =. We shall attach oe more symbol, to R so as to have the exteded real lie R := R {, } with the followig covetio: a + ( ) =, < a a R, ( ) + ( ) =. However, + ( ) is cosidered to be udefied. As i the case of real umbers, we shall write a for a + ( ). It ca be easily see that the coditio (a) i Defiitio ca be replaced by A 0 A such that µ(a 0 ) <. Remark If (X, A, µ) is a measure space, the we may say that µ is a measure o the measurable space (X, A) or µ is a measure o the σ-algebra A whe the set X is uderstood from the cotext, or µ is a measure o the X whe the σ-algebra A is uderstood from the cotext Theorem Let (X, A, µ) be a measure space ad A, B A. The (i) A B implies µ(a) µ(b). (ii) A B ad µ(a) < imply µ(b \ A) = µ(b) µ(a). Proof. If A B, the B is the disjoit uio of A ad B \ A. Hece, From this (i) ad (ii) follow. µ(b) = µ(a) + µ(b \ A).

3 Measure o a Arbitrary σ-algebra 33 By the above theorem, for ay measure µ o a measurable space (X, A) A A µ(a) µ(x). Defiitio Let (X, A, µ) be a measure space. 1. If µ(x) = 1, the µ is called a probability measure. 2. If µ(x) <, the µ is called a fiite measure, ad (X, A, µ) is called a fiite measure space. 3. If there exists X A, N such that X = =1 X ad µ(x ) < for each N, the µ is called a σ-fiite measure, ad (X, A, µ) is called a σ-fiite measure space. The assertive statemets i the followig examples may be verified by the reader. Example We observe that (R, M, m) is measure space, ad it is a σ-fiite measure space. Example For ay set X, the family {, X} ad the power set 2 X are σ-algebras, ad they are the smallest ad largest (i the sese of set iclusio), respectively, of all σ-algebras o X. Example Give ay ay measurable space (X, A), the fuctio µ defied by µ(a) = 0 for all A A is a measure o (X, A), called the zero measure. Example Observe that, if (X, A, µ) is a measure space, the for every α R, α > 0, E αµ(e), E A, defies a measure o (X, A), ad if 0 < µ(x) <, the E µ(e) µ(x) is a probability measure o (X, A). Example Let X be a o-empty set ad A = {, X}. Let µ be defied by µ( ) = 0 ad µ(x) = 1. The (X, A, µ) is a measure space. Example Let X be ay set, A = 2 X, the power set of X, ad µ be defied by { E µ(e) = # if E is a fiite set if E is a ifiite set. Here, E # deotes the cardiality of E. The µ is a measure space o (X, A). This measure is called the coutig measure o X. Note that a coutig measure o X is 1. fiite if ad oly if X is a fiite set ad

4 34 Measure ad Measurable Fuctios 2. σ-fiite if ad oly if X is a coutable set. Example Let X be ay set, A be the power set of X, x 0 X, ad µ be defied by { 1 if x0 E µ(e) = 0 if x 0 E. The (X, A, µ) is a measure space, ad the measure µ is called the Dirac measure o X cetered at x 0. Let us have a illustratio of the Direc measure: Thik of a thi wire of egligible weight. Suppose a bead of weight 1 uit kept at some poit x 0 o the wire. The weight of a part, say E, of the wire is 1 if x 0 belogs to E, ad the wight is zero if x 0 does ot belog to E. Example Let X be a set, A be the power set of X, x i X ad w i 0 for i {1,..., k}. For E X, let E = {i : x i E}, ad let µ be defied by µ(e) = i E w i. The (X, A, µ) is a measure space. Note hat µ is a probability measure if ad oly if k w i = 1. Aalogous to the illustratio of the Dirac measure we have the followig: Thik of a thi wire of egligible weight. Suppose a fiite umber of beads of weights w 1,..., w k are kept at poits x 1,..., x k respectively o the wire. The the weight of a part, say E, of the wire is i E w i where E = {i : x i E}. We may also have aother illustratio: Let us assume that the wire costs othig, but the beads at x 1,..., x k cost rupees w 1,..., w k respectively. The the cost of a part, say E, of the wire together with the beads o it costs i E w i where E = {i : x i E}. The proof of the followig theorem is left as a exercise. Theorem Suppose (X, A) is a measurable space ad X 0 A. The A 0 = {A X 0 : A A} is a σ-algebra o X 0. Further, if µ is a measure o (X, A), the the fuctio µ 0 : A 0 [0, ] defied by is a measure o (X 0, A 0 ). µ 0 (E) = µ(e), E A 0,

5 Measure o a Arbitrary σ-algebra 35 Proof. Clearly X 0 A 0. Let E A 0. The there exists A A such that E = A X 0, ad X 0 \ E = X 0 \ A X 0 = X 0 (A X 0 ) c. Sice (A X 0 ) c A, we have X 0 \ E A 0. Next, let {E } be a coutably ifiite family i A 0. Let A A such that E = A X 0 for N. The E = =1 ( ) (A X 0 ) = A X 0. =1 Sice =1 A A, we have =1 E A 0. The secod part is obvious from the fact that µ is a measure o A. =1 The σ-algebra A 0 i the above theorem is called the restrictio of A to X 0. If µ is a measure o (X, A), the µ 0 := µ A0 is a measure o the measurable space (X 0, A 0 ). Defiitio Let (X, A) be a measurable space ad X 0 A. The the σ-algebra A 0 give i Theorem is called the restrictio of the σ-algebra A to X 0. If µ is a measure, the the measure µ 0 := µ A0 o (X 0, A 0 ) is called the restrictio of the measure µ to (X 0, A 0 ). NOTATION: I the due course, we shall also deote the restrictio of a σ-algebra to a measurable set X 0 A by A X0, whereas the correspodig restrictio of the measure µ will be deoted by the same otatio µ, but we say that µ is a measure o A X0 or, by abusig the termiology, µ is a measure o X 0. I view of Theorem 3.1.2, we ca talk about Lebesgue measure o ay Lebesgue measurable set X 0. I particular, if I is a iterval, the we have the Lebesgue measure m o the σ-algebra M I or o the iterval I. We kow that a arbitrary family of subsets of a o-empty set X is ot a σ-algebra. However, a arbitrary family of subsets of X is associated with a uique σ-algebra i certai way. Theorem Let X be a set ad S be a family of subsets of X. The the itersectio of all σ-algebras cotaiig S is a σ-algebra o X cotaiig S, ad it is the smallest σ-algebra o X cotaiig S. Proof. Let F be the family of all σ-algebras o X cotaiig S, that is, F = {A : A is a σ-algebra o X ad S A}.

6 36 Measure ad Measurable Fuctios Let A S be the itersectio of all members of F, that is, A S = A F A. First we show that A S is a σ-algebra o X. Sice X A for every A F, we have X A S. Now, let A A S. The A A for every A F. Sice each A is a σ-algebra, A c A for every A F. Hece, A A S. Next, let {A : Λ} be a coutable family i A S. The {A : Λ} A for every A F. Sice each A is a σ-algebra, Λ A A for every A F. Therefore, Λ A A S. Thus, A S is a σ-algebra. Sice S A for every A F, ad A S is the itersectio of members of F, we have S A S, ad A S is the smallest (i terms of set iclusio) σ-algebra cotaiig S. Corollary Itersectio of ay family of σ-algebras o a set X is agai a σ-algebra o X. Proof. If {A α : α Λ} is a family of σ-algebras o a set X, where Λ is some idex set, the takig takig S as either { } or {X} or {, X} i the Theorem 3.1.3, we obtai that α Λ A α is a σ-algebra o X. Defiitio Let X be a set ad S be a family of subsets of X. The A S, the itersectio of all σ-algebras cotaiig S, is called the σ-algebra geerated by S. Note that, for a give family for a give family S of subsets of X, A S is the smallest σ-algebra cotaiig S, i the sese that, if A is ay σ-algebra cotaiig S, the A S A. Defiitio Let Y be a topological space with topology T. The the σ-algebra geerated by T is called the Borel σ-algebra o Y, ad members of the Borel σ-algebra are called Borel sets. We may deote the Borel σ-algebra o a topological Y by B Y. If Y = R, the we shall deote B R by B 1 or B. Note that The m B is a measure o (R, B). Sice M cotais all ope sets ad B is the smallest σ-algebra cotaiig all ope sets, we have B M. We shall see that B a proper subfamily of M. Aother questio of iterest would be the followig:

7 Measure o a Arbitrary σ-algebra 37 Does there exist a σ-algebra A o R cotaiig M such that m is a measure o A? The aswer is egative. We prove this as a corollary to the followig characterizatio of Lebesgue measurability. Theorem A subset E of R is Lebesgue measurable if ad oly if for every ope iterval I. m (I) m (I E) + m (I E c ) Proof. Let E R. Clearly, if Lebesgue measurable, the m (I) m (I E) + m (I E c ) (1) holds for for every ope iterval I. Coversely, suppose (1) holds for for every ope iterval I. We have to show that for every A R. m (A) m (A E) + m (A E c ) (2) Let ε > 0 be give. By the defiitio of m, there exists a coutable family {I } of ope itervals such that A Hece, usig (1) ad (3), I ad l(i ) m (A) + ε. (3) m (A E) + m (A E c ) m ( I E) + m ( I E c ) [m (I E) + m (I E c )] m (I ) Thus, m (A) + ε. m (A E) + m (A E c ) m (A) + ε for every ε > 0. Cosequetly, (2) holds for every A R. Corollary Suppose there is a σ-algebra A o R such that m is a measure o A. The A = M.

8 38 Measure ad Measurable Fuctios Proof. We have to prove that A M. For this, let E A. Sice A cotais itervals, ad sice A is a σ-algebra, we have m (I) = m (I E) + m (I E c ) for every ope iterval I. Therefore, by Theorem 3.1.5, E M. Thus, A M. Defiitio A measure µ o a measurable space (X, A) is called a complete measure if for every A A with µ(a) = 0 ad E A, we have E A, ad i that case the the (X, A, µ) is called a complete measure space. Example Recall that, if A R such that m (A) = 0, the A M. Hece, it follows that the Lebesgue measure o R is complete. Example Let X = {a, b, c} ad A = {, X, {a}, {b, c}}. Let µ o A be defied µ( ) = µ({b, c}) = 0 ad µ(x) = µ({a}) = 1. The µ is a measure o (X, A); it is ot complete, as µ({b, c}) = 0, {b} {b, c} but {b} A. Example The Borel σ-algebra B o R is ot complete. Proof of this fact will be give i the ext chapter after defiig the cocept of measurable fuctios. every measure space ca be com- The followig theorem shows that pleted. Theorem Let (X, A, µ) be a measure space, ad let N := {E X : B A with E B ad µ(b) = 0}. ad à := {A E : A A, E N }. The à is a σ-algebra o X cotaiig A ad µ defied o à by µ(a E) = µ(a) for A A, E N defies a complete measure o (X, Ã) with µ(a) = µ(a) for every A A. Further, (X, Ã, µ) is the smallest complete measure space cotaiig (X, A, µ), i the sese that if (X, Â, ˆµ) is a complete measure space with A  ad µ(a) = µ(a) for every A A, the à Â.

9 Measure o a Arbitrary σ-algebra 39 Proof. First we observe that A à ad N Ã. I particular, X Ã. Suppose A A ad E N. The (A E) c = A c E c. Sice E B for some B A with µ(b) = 0. Hece, E c B c so that E c = B c (E c \ B c ) = B c (E c B). Thus, (A E) c = A c E c = A c [B c (E c B)] = [A c B c ] [A c (E c B)], where A c B c A ad A c (E c B) B with µ(b) = 0. Hece, (A E) c Ã. Next, let A A ad E N for N. We have to show that =1 (A E ) Ã. Note that (A E ) = [ ] [ ] A E. =1 =1 =1 Sice E B for some B A with µ(b ) = 0, we have E B with =1 =1 ( ) B A ad µ B =1 =1 =1 µ(b ) = 0. Therefore, =1 E N so that =1 (A E ) Ã. Thus, we have proved that à is a σ-algebra. Next, let A A ad E N for N be such that {A E } is a disjoit family. The, {A } ad {E } are disjoit families ad hece ( ) ( [ ] [ ] ) µ (A E ) = µ A E =1 Thus, µ is a measure o Ã. =1 ( ) = µ A = = =1 µ(a E ). =1 =1 µ(a ) It remais to show the completeess of µ. For this, let F A E for some A A ad E N with µ(a E) = 0. We have to show that F Ã. Note that F = F (A E) = (F A) (F E), where =1 F A A with µ(a) = µ(a E) = 0 so that F A N. Also, F E E with E N so that F E N. Sice N à ad à is a σ-algebra, F = (F A) (F E) Ã.

10 40 Measure ad Measurable Fuctios Next, suppose that (X, Â, ˆµ) is a complete measure space with A  ad µ(a) = µ(a) for every A A. Sice A  ad ˆµ is a complete measure, we have N Â. Thus, usig the fact that  is a σ-algebra, we also obtai à Â. Defiitio The measure space (X, Ã, µ) defied i Theorem is called the completio of (X, A, µ). We may recall from Corollary that for every E M, there exist a G δ -set G E ad a F σ -set F E such that m(g \ F ) = 0. Hece, Thus E = F (E \ F ), where E \ F G \ F with m(g \ F ) = 0. M is the completio of the Borel σ-algebra B 1. Now some results. Theorem Let (X, A, µ) be a measure space ad A A such that A A +1 for all N. The ( µ(a ) µ A i ) as. Proof. We write A i as a disjoit uio E i by takig E 1 = A 1 ad E i = A i \ A i 1 for i = 2, 3,.... The ( µ ) ( A i = µ E i ) = But, E i = A. Hece, This completes the proof. ( µ meas-moo-ic meas-moo-dec µ(e i ) = lim A i ) = lim µ(a ). ( µ(e i ) = lim µ E i ). Theorem Let (X, A, µ) be a measure space ad A A such that A A +1 for all N ad µ(a 1 ) <. The ( µ(a ) µ A i ) as.

11 Measure o a Arbitrary σ-algebra 41 Proof. Let B = A 1 \ A, N. The B B +1 for all N, so that by the Theorem 3.1.8, ( µ B i ) = lim µ(b ). But, B i = A 1 \ A i. Therefore, sice µ(a 1 ) <, by Theorem (ii), we have ( µ ) ( B i = µ A 1 \ ) ( A i = µ(a 1 ) µ µ(b ) = µ(a 1 \ A ) = µ(a 1 ) µ(a ). Thus, µ(a ) µ ( A i) as. A i ), Remark For a sequece (A ) of sets with A A +1 for every N, sice A i = i=k A i for ay k N, the assumptio µ(a 1 ) < i Theorem ca be replaced by µ(a k ) < for some k N. It is also to be metioed that the coclusio i Theorem eed ot hold if µ(a ) = for every N. To see this cosider the example of (R, M, m) ad A = [, ) for N. I this case, we have A A +1 for every N but m(a ) m( =1 A ), sice m(a ) = for every N ad =1 A =. Now, let us itroduce some set theoretic otios. Defiitio A sequece (A ) of sets is called a (a) a mootoically icreasig if A A +1 for every N, (b) a mootoically decreasig if A A +1 for every N. We may observe that, for ay family {A α : α Λ} of sets, the set α Λ A α is a upper boud of {A α : α Λ} with respect to the partial order of iclusio of sets, ad it is, i fact, the least upper boud. Similarly, the set α Λ A α is a lower boud of {A α : α Λ}, ad it is the greatest lower boud. Thus, for a mootoically icreasig sequece (A ) of sets, the set A i ca be cosidered as the limit of (A ), ad for for a mootoically decreasig sequece (A ) of sets, the set A i ca be cosidered as the limit of (A ). Motivated by these cosideratios, we cosider the followig defiitio.

12 42 Measure ad Measurable Fuctios Defiitio Let (A ) be a sequece of subsets of a set X. The the limit iferior of (A ) ad limit superior of (A ), deoted by lim sup ad lim if A, respectively, are defied by lim sup A = k=1 =k A, lim if A = k=1 =k A. If lim sup A = lim if A, the we say that the limit of (A ) exists, ad the commo set is called the limit of (A ), deoted by lim A, i.e., lim A = lim sup A = lim if A. For a sequece (A ) of subsets of a set X ad x X, the followig results ca be verified easily: 1. x lim sup A x A for ifiitely may N, 2. x lim sup A x A for all but a fiitely may N. 3. If (A ) is mootoically icreasig, the ad hece lim A = lim sup A = lim if A = A. =1 4. If (A ) is mootoically decreasig, the ad hece lim A = lim if A = lim sup A = A. =1 Thus, Theorem ad Theorem ca be restated as follows: =1 =1 A A A If (X, A, µ) is a measure space ad (A ) is a sequece i A which is either mootoically icreasig or mootoically decreasig, the µ( lim A ) = lim µ(a ).

13 Measure o a Arbitrary σ-algebra 43 What ca we say if (A ) is a geeral sequece of sets from A? Theorem Let (X, A, µ) be a measure space ad (A ) be a sequece sets i A. The we have the followig. (i) µ(lim if A ) lim if µ(a ). (ii) If µ( =1 A ) <, the µ(lim sup A ) lim sup µ(a ). (iii) If lim A ad lim µ(a ) exist, the µ( lim A ) lim µ(a ). (iv) If lim A exists ad if µ( =1 A ) <, the lim µ(a ) exists ad µ( lim A ) = lim µ(a ). Proof. For k N, let B k = k A ad C k = k A. The, we have B k B k+1 ad C k C k+1 for all k N. (i) By Theorem 3.1.8, ( µ(lim if A ) = µ But, µ(c k ) µ(a k ). Therefore, k=1 C k ) = lim k µ(c k). lim µ(c k) = lim if µ(c k) lim if µ(a k) k k k so that we obtai the required iequality. (ii) Suppose µ( =1 A ) <, i.e., µ(b 1 ) <. The, by Theorem 3.1.9, ( ) µ(lim sup A ) = µ B k = lim µ(b k). k k=1 But, µ(b k ) µ(a k ). Therefore, lim µ(b k) = lim if µ(b k) lim sup µ(a k ) k k so that we obtai the required iequality. (iii) This follows from (i). (iv) By (ii), lim sup k µ(a ) µ(lim sup A ) = µ( lim A ) = µ(lim if A ) ad by (i), µ(lim if A ) lim if µ(a ). Hece, lim sup µ(a ) µ( lim A ) lim if µ(a ). Thus, lim µ(a ) exists ad µ( lim A ) = lim µ(a ).

14 44 Measure ad Measurable Fuctios 3.2 Measurable Fuctios Recall that a real valued fuctio f defied o a subset E of R is said to be cotiuous if for every ope set G of R, the set f 1 (G) is ope i E. Let A be either B 1 or M, ad A E be the restrictio of A to E. Thus, i case E A, f cotiuous f 1 (G) A E for every ope set G of R. Here, the topology o R is cosidered to be the usual topology. Sice A E cotais sets which are ot ope, the coverse of the above statemet is ot true. I other words, the set F := {f : E R : f 1 (G) A E ope G R} cotais more fuctios tha cotiuous fuctios. fuctio f : R R defied by For example, for the we have f(x) = { 1, x > 0, 0, x 0, (0, ), 1 G, 0 G, f 1 (, 0], 0 G, 1 G, (G) = R, {0, 1} G,, {0, 1} G =. for every ope set G R. Thus, f 1 (G) B 1 for every ope set G R, but the fuctio is ot cotiuous. I view of the above observatios, we have the followig defiitio. Defiitio Let (X, A) be a measurable space. A fuctio f : X R is said to be a real measurable fuctio if f 1 (G) A for every ope set G R. Similarly we ca defie a complex measurable fuctio if the codomai of f is C istead of R More geerally, if f is a fuctio takig values i a topological space, the we ca defie measurability of such fuctios as follows. Defiitio Let (X, A) be a measurable space ad (Y, T ) be a topological space. A fuctio f : X Y is said to be measurable with respect to the pair (A, T ) if f 1 (G) A for every G T. Defiitio For a give topological space Y, a fuctio f : R Y is called

15 Measurable Fuctios 45 (i) Lebesgue measurable if f is measurable with respect to the σ- algebra M o R, ad (ii) Borel measurable if f is measurable with respect to the σ-algebra B 1 o R. We may observe the followig: Let (X, A) be a measurable space. The, a real valued fuctio f : X R is measurable if ad oly if f 1 (I) A for every ope iterval I. Some of the topological spaces that we frequetly use i the study of measure ad itegratio are the followig: R with usual topology, C with usual topology, R k with usual topology. R := [, ], the exteded real lie, with topology geerated by ope subsets of R (uder usual topology) ad itervals of the form (a, ] ad [, b) for a, b R. Note that a subset G of R := [, ] is ope if ad oly if for every x G there exists a iterval of the form (a, b), (a, ] or [, b) cotaiig x ad cotaied i G. I Defiitio 3.2.2, if Y = R, the we say that f is real measurable fuctio, if Y = C, the we say that f is complex measurable fuctio, if Y = R, the we say that f is exteded real valued measurable fuctio. CONVENTION If the σ-algebra, measure ad the topology uder cosideratio is uderstood, the istead of sayig that (X, A) is a measurable space, (X, A, µ) is a measure space, ad (Y, T ) is a topological space, we may simply say that X is a measurable space, X is a measure space, ad Y is a topological space. Also members of a σ-algebra are called measurable sets. A importat class of fuctios i measure ad itegratio is the class of all characteristic fuctios. First, let us recall the defiitio of a characteristic fuctio.

16 46 Measure ad Measurable Fuctios Defiitio Give a set X, the characteristic fuctio of a subset E of X is the fuctio χ E : X R defied by χ E (x) = { 1, x E, 0, x E. With the above defiitio, we have the followig theorem. Theorem Let X be a measurable space ad E X. The χ E is measurable if ad oly if E is measurable. Proof. We observe that for a ope set G, χ 1 E (G) = E, 1 G, 0 G, E c, 1 G, 0 G, X, 1 G, 0 G,, 1 G, 0 G. Thus, χ E is measurable if ad oly if E is measurable. Theorem Let X be a measurable space ad Y be a set ad f : X Y. The the followig results hold. (i) S := {E Y : f 1 (E) A} is a σ-algebra o Y. (ii) If Y is a topological space ad f is measurable, the S cotais the Borel field o Y. (iii) If Y is a topological space, the f is measurable if ad oly if f 1 (B) A for every Borel set B Y. Proof. (i) The fact that S is a σ-algebra o Y follows from the relatios f 1 ( ) =, f 1 (Y ) = X, f 1 (A c ) = [f 1 (A)] c, f 1( ) A = =1 for subsets A, A of Y for N. f 1 (A ) (ii) Suppose that Y is a topological space ad f is measurable. Hece, S cotais all ope sets. Sice Borel σ-algebra B Y o Y is the smallest σ-algebra cotaiig all ope sets, S cotais B Y as well. (iii) This part follows from (ii), as Borel σ-algebra o Y cotais all ope sets i Y. =1

17 Measurable Fuctios 47 Remark Suppose (X 1, A 1 ) ad (X 2, A 2 ) are two measurable spaces. The we may defie a fuctio f : X 1 X 2 to be measurable if f 1 (A) A 1 for every A A 2. The, i view of Theorem 3.2.2, our defiitio of measurability becomes a particular case by takig X 2 a topological space ad A 2 as the σ-algebra of all Borel sets i X 2. Thus, if (X, A) is a measurable space, the a real valued fuctio is measrable if ad oly if f 1 (A) A for every A B. Theorem For a fuctio f : X R := [, ], the followig are equivalet: (i) f is measurable. (ii) {x X : f(x) > a} A a R. (iii) {x X : f(x) a} A a R. (iv) {x X : f(x) < a} A a R. (v) {x X : f(x) a} A a R. Proof. Recall that for ay a R, (a, ] is a ope set i R. Now, for a R, we observe that Note also that {x X : f(x) > a} = f 1 ((a, ]), {x X : f(x) a} = f 1 ([a, ]), {x X : f(x) < a} = f 1 ([, a)), {x X : f(x) a} = f 1 ([, a]). [a, ] = =1 [, a) = R \ [a, ], ad for ay a, b R with a < b, ( a 1, ], [, a] = R \ (a, ], (a, b) = (a, ] [, b]. From these observatios together with the properties of a σ-algebra, the equivalece of (i)-(v) follows. For the ext theorem, ad also to use i the due course, we defie a few otios.

18 48 Measure ad Measurable Fuctios Defiitio Let (f ) be a sequece of fuctios from a set X to [, ]. The supremum, ifimum, limtsuperior, limit iferior of (f), deoted by sup f, if f, lim sup f, lim if f, respectively are defied by (sup f )(x) = sup f (x), (if )(x) = if (x), (lim sup (lim if f )(x) = lim sup f )(x) = lim if f (x), f (x) for x X. If lim sup f = lim if f, the we defie the fuctio lim lim f by lim f := lim sup f = lim if f call it the limit of (f ). f or Theorem Let f : X [, ] be measurable for each N. The sup f, if f, lim sup f, lim if f are measurable fuctios. Proof. For a R, we ote that {x X : sup f > a} = {x X : f > a}, {x X : if f < a} = {x X : f < a}. By Theorem 3.2.3, {x X : f > a} ad {x X : f < a} are measurable fuctios. Hece, by the properties of the σ-algebra, sup f ad if f are also measurable. Sice lim sup f = if sup f, k k lim if f = sup k if k f it also follows that lim sup f ad lim if f are measurable fuctios.

19 Measurable Fuctios 49 Defiitio Suppose (f ) is a sequece of fuctios defied o a measurable set X with values i Y which is either of R, C or R. The we say that (f ) coverges poitwise o a subset E of X if for each x E, the sequece (f (x)) coverges. If f : E Y is defied by f(x) = lim f (x), x E, the we write f f poitwise o E. Remark I the above defiitio, we used a special covetio that if (a ) is a sequece i R which diverges to (respectively, ), the we say that (a ) coverge to (respectively, ) i R := [, ]. Thus, as a corollary to Theorem 3.2.4, we obtai the followig result. Theorem Let (f ) be a sequece of exteded real valued measurable fuctios o X. If (f ) coverges poitwise o X, the the fuctio f defied by f(x) = lim f (x), x X, is measurable. I measure theory, there are other types of covergece. Oe such covergece is the almost everywhere covergece. Defiitio Suppose (f ) is a sequece of measurable fuctios defied o a measure space (X, A, µ) with values i Y which either of R, C or R. The we say that (f ) coverges almost everywhere o a set E A if the set A := {x E : (f (x)) does ot coverge at x} A ad µ(a) = 0, ad we write this fact as (f ) coverges a.e. o E. If (f ) coverges a.e. o E ad if f(x) = lim f (x), x E \ A, where A = {x E : (f (x)) does ot coverge at x}, the the we say (f ) coverges to f almost everywhere ad write this fact as f f a.e. o E. More geerally we have the followig defiitio. Defiitio Let (X, A, µ) be a measure space, E A ad P be a property o the elemets of X. The we say that P holds almost everywhere o E if µ({x E : P does ot hold at x}) = 0, ad we write P holds a.e. o E. The fact P holds a.e. o X is simply writte as P holds a.e.

20 50 Measure ad Measurable Fuctios I particular, if f be a fuctio defied o a measurable space (X, A, µ) with values i Y which is either of R, R or C, the we say that f = 0 a.e. o E if A := {x E : f(x) 0} A ad µ(a) = 0. We have already observed that, give a measurable space X, a fuctio f : X R is measurable if ad oly if f 1 (I) is a measurable set for every ope iterval I. The followig result is aalogous to it for R 2 -valued fuctios. Theorem Let X be a measurable space. The a fuctio f : X R 2 is measurable if ad oly if f 1 (I 1 I 2 ) is a measurable set for every ope rectagle I 1 I 2 i R 2. Proof. Proof follows usig the fact that every ope subset of R 2 is a coutable uio of ope rectagles. I the above we used the followig fact: A set S R 2 is ope if ad oly if for every (x 1, x 2 ) R 2, there exist ope itervals I 1 ad I 2 cotaiig x 1 ad x 2, respectively, such that I 1 I 2 S. Theorem Let X be a measurable space ad u, v are real measurable fuctios o X. The the fuctio f : X R 2 defied by is measurable. f(x) = (u(x), v(x), x X, Proof. Sice every ope subset of R 2 is a coutable uio of ope rectagles, it is sufficiet to show that for ope itervals I 1 ad I 2, f 1 (I 1 I 2 ) is measurable. Note that x f 1 (I 1 I 2 ) f(x) I 1 I 2 u(x) I 1 & v(x) I 2 x u 1 (I 1 ) v 1 (I 2 ). Thus f 1 (I 1 I 2 ) = u 1 (I 1 ) v 1 (I 2 ) which is measurable. Theorem Let X be a measurable space, ad Y ad Z be topological spaces. If f : X Y is measurable ad g : Y Z is cotiuous, the g f : X Z is measurable. Proof. Suppose f : X Y is measurable ad g : Y Z is cotiuous. Let V be a ope set i Z. By cotiuity of g, the set g 1 (V ) is ope i Y,

21 Measurable Fuctios 51 ad by measurability of f, f 1 (g 1 (V )) is measurable i X. Hece, from the idetity (g f) 1 (V ) = f 1 (g 1 (V )) the set (g f) 1 (V ) is measurable i X. As corollaries to the above theorem we have the followig results. Theorem Let X be a measurable space, ad f be a real or complex or exteded real valued measurable fuctio. The the fuctio f defied by f (x) = f(x), x X, is a measurable fuctio. Proof. We ote that f = g f, where g(α) = α for α i the codomai of f. Thus, by Theorem 3.2.8, f is measurable. Theorem Let X be a measurable space. (i) If u, v are real measurable fuctios o X, the the fuctio u + iv defied by (u + i v)(s) := u(x) + iv(x), x X, is a complex measurable fuctio o X. (ii) If f is a complex measurable fuctio o X, the the fuctios Ref, Imf, f, defied by [Ref](x) := Ref(x), [Imf](x) = Imf(x), f (x) := f(x), respectively, for x X, are real measurable fuctios o X. Proof. The results i (i) ad (ii) follow from Theorem ad by Theorem by observig the followig facts: (a) The fuctio x (u(x), v(x)) from X to R 2 is measurable ad the fuctio (α, β) α + iβ from R 2 to C is cotiuous. (b) The fuctio x f(x) is measurable from X to C, ad the fuctios z Rez, z Imz ad z z are cotiuous from C to R. Theorem Let X be a measurable space ad f ad g be real (resp. complex) measurable fuctios o X ad λ be real (resp. complex). The f + g, fg ad λf are real (resp. complex) measurable fuctios o X.

22 52 Measure ad Measurable Fuctios Proof. Let f ad g be real measurable fuctios o X ad λ be a real umber. By Theorem , the fuctio x (f(x), g(x)) is measurable. Also, by the cotiuity of the fuctios (α, β) α+β ad (α, β) αβ from R 2 to R ad α λα from R to R, Theorem implies the measurability of the fuctios f + g, fg ad λf. The case whe f ad g are complex valued will follow by cosiderig their real ad imagiary part ad usig Theorem Now, we itroduce aother defiitio. Defiitio Give a fuctio f : X [, ] defied o a set X, its positive part, egative part ad absolute value or modulus, deoted by f +, f ad f, respectively, are defied by f + (x) = max{f(x), 0}, f (x) = max{ f(x), 0}, f = max{f(x), f(x)} for x X. We may observe that f = f + f, f = f + + f. Theorem Let f : X [, ]. If f is measurable, the f +, f ad f are measurable. I case f is real valued, the f is measurable if ad oly if f + ad f are measurable. Proof. Suppose f is measurable. By Theorem , takig λ := 1, the fuctio f is measurable. Now, measurability of f +, f ad f follows from Theorem 3.2.4, sice f + = sup{f, 0}, f = sup{ f, 0}, f = sup{f, f}. Next, suppose that f is real valued ad f +, f are measurable. The by Theorem , f = f + f is also measurable. 3.3 Simple measurable fuctios I this sectio, we cosider a class of fuctios which are more geeral tha characteristic fuctios. Defiitio Let X be set. A fuctio ϕ : X R is called a simple fuctio if it takes oly a fiite umber of values.

23 Simple measurable fuctios 53 If α 1,..., α are the distict values of a simple fuctio ϕ : X R, the ϕ ca be represeted as ϕ = α i χ Ei, where E i = {x X : ϕ(x) = α i }, i = 1,...,. The above represetatio of ϕ is called its caoical represetatio,. Note that if ϕ = α iχ Ei is the caoical represetatio of a simple fuctio ϕ, the X = E i ad {E 1,..., E } is a disjoit family. I case α 1,..., α are the ozero distict values of ϕ, the also ϕ = α iχ Ei is the caoical represetatio of ϕ, but X = i=0 E i with E 0 = {x X : f(x) = 0}. Suppose a 1,..., a i R ad A i X, i = 1,...,. The we see that the fuctio ϕ : X R defied by ϕ = a i χ Ai takes oly a fiite umber of values, ad hece, it is a simple fuctio. Thus, A fuctio ϕ : X R is simple if ad oly if a 1,..., a i R ad A i X, i = 1,..., such that ϕ = a iχ Ai. Recall that a fuctio f : R R is called a step fuctio if there are disjoit itervals I 1,..., I ad a 1,..., a i R such that f = a i χ Ii. Thus, every real valued step fuctio defied o R is a simple fuctio. Theorem Let (X, A) be a measurable space ad ϕ be a simple fuctio o X with caoical represetatio ϕ = α i χ Ei. The ϕ is measurable if ad oly if E i A for i = 1,...,. Proof. Suppose ϕ is a measurable fuctio. The, each E i is a measurable set as it is the iverse image of the closed set {α i }. Coversely, if each E i is a measurable set, the by Theorem , ϕ is a measurable fuctio.

24 54 Measure ad Measurable Fuctios The ext theorem is oe of the most importat results i the theory of measure ad itegratio, ad it would help us iferrig certai properties of a measurable fuctios from correspodig properties of simple fuctios. Theorem Let (X, A) be a measurable space ad f : X [0, ] be a measurable fuctio.the there exists a sequece (ϕ ) of simple measurable fuctios o X such that (i) 0 ϕ ϕ +1 for every N ad (ii) ϕ f poitwise o X. I fact, ϕ defied by where ϕ = 2 i 1 2 χ E i, + χ F, N, E i, := { x X : i 1 2 f(x) < i } 2, i = 1,..., 2, F := {x X : f(x) }, satisfy the above requiremets. Proof. Note that, for each N, {E i, } {F } is a disjoit family of measurable sets. Let x X. If f(x) =, the ϕ (x) = for all N, so that ϕ (x) f(x) as. I case 0 f(x) <, the there exists k N such that f(x) k. Hece, for every k, x E i, for some i {1, 2,..., 2 }, ad i that case f(x) ϕ (x) 1 2. Therefore, i this case also, ϕ (x) f(x) as. Next suppose that x E i, for some N ad for some i {1, 2,..., 2 }. The ϕ (x) = (i 1)/2, ad { i 1 ϕ +1 (x) 2, i } Thus, ϕ (x) ϕ +1 (x). If x F, the ϕ (x) = ad ϕ +1 (x) {, + 1 }. 2+1 Thus, we get ϕ (x) ϕ +1 (x) for every x X ad for every N.

25 Simple measurable fuctios 55 As a corollary, we obtai the followig theorem. Theorem Let f ad g be exteded real valued o-egative measurable fuctios o a measurable space (X, A). The f + g is measurable. More geerally, if (f ) is a sequece of exteded real valued o-egative measurable fuctios o (X, A), the f := =1 f is measurable. Proof. By Theorem 3.3.2, there exist icreasig sequeces (ϕ ), (ψ ) of o-egative simple measurable fuctios o X which coverge poitwise to f ad g, respectively. Hece, by Theorem 3.2.5, both f ad g are measurable. Secod part of the theorem also follows from Theorem 3.2.5, by cosiderig sequece of partial sums of the series f := =1 f. We ed this chapter by provig oe of the results which we promised earlier. Theorem The Borel σ-algebra B o R is a proper subset of the σ- algebra M of all Lebesgue measurable sets. Proof. We first recall that every x [0, 1] has the biary expasio x = =1 ϕ (x) 2, where ϕ (x) {0, 1}. Now, defie f : [0, 1] R by f(x) = =1 2ϕ (x) 3, x [0, 1]. Note that f is a ijective fuctio ad its rage is cotaied i the Cater set C. We observe that, for each N, ϕ = χ E, where E is a fiite uio of subitervals of [0, 1]. Hece, each ϕ is a Lebesgue measurable fuctio. Therefore, by Theorem 3.3.3, f is also a Lebesgue measurable fuctio. Now, let E 0 [0, 1] be a o-lebesgue measurable set, ad let F = f(e 0 ). Sice F C ad m (C) = 0, we kow that F M. Now, if M = B, the by the Lebesgue measurability of f, E 0 = f 1 (F ) M. Thus, we arrive at a cotradictio. Hece, we ca coclude that B o R is a proper subset of M. Accordig to Theorem 3.3.2, every exteded real valued o-egative measurable fuctio f ca be approximated poitwise by a icreasig sequece (ϕ ) of simple measurable fuctios. Thus, if (X, A) is edowed

26 56 Measure ad Measurable Fuctios with a measure µ, the a atural procedure of defiig the itegral of a exteded real valued o-egative measurable fuctio f would be to defie the itegral X ϕdµ of simple measurable fuctios ϕ, ad the defie X fdµ as the limit (if exists) of the sequece ( X ϕ dµ ) wheever (ϕ ) is a icreasig sequece of o-egative simple measurable fuctios which coverges to f poitwise. But, to have such a itegral to be well-defied, we must also show the followig: If (ϕ ) ad (ψ ) are icreasig sequece of o-egative simple measurable fuctios which coverge to f poitwise, the the limits lim ϕ dµ ad lim ψ dµ exist ad must be equal. X We shall do this i the ext chapter. X