A4 Integration (2017 HT)
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1 A4 Integrtion (217 HT) Zhongmin Qin (mil: Mthemticl Institute, Oxford December 3, 216 Contents 1 Introduction 3 2 xtended rel line, upper nd lower limits 5 3 Mesure spces 6 4 The Lebesgue mesure Outer mesures nd Crthéodory s extension theorem The Lebesgue mesure spce The Cntor set nd null subsets An exmple of non mesurble sets Lebesgue sub-spces Mesurble functions Definition nd bsic properties Almost everywhere properties xmples Lebesgue integrtion Lebesgue integrls, nd integrble functions Integrls depending on prmeters xmples Fubini s theorem Mesures on product spces Fubini s theorem nd Tonelli s theorem xmples Chnge of vribles L p -spce nd convergences L p -spces Convergence in mesure Convergence in L p -spce
2 9 Appendix 1 From outer mesures to mesures 58 1 Appendix 2 Further results on Riemnn integrls 64 2
3 These notes re written for the lectures delivered in 216 Hilry term. In writing up these notes, I hve benefited from the notes written by the previous lecturers, in prticulr the notes produced by A. theridge nd C. Btty. I hve dopted mny of their worked exmples, nd in mny plces I hve closely followed their notes. Any errors ppered in these notes however re completely my own responsibility. Plese report errors, typos nd etc. you my discover to qinz@mths.ox.c.uk. 1 Introduction In Prelims Anlysis III Integrtion, the theory of Riemnn integrls for bounded continuous functions on finite intervls were estblished. In ddition to the computtionl techniques such s the methods of chnge of vribles, integrtion by prts nd etc., we hve proved severl fundmentl results bout Riemnn integrls. Let us recll two importnt results. The first is the existence of Riemnn integrls. If f is bounded on [,b] nd is continuous on (,b) (where nd b re two numbers), then f is Riemnn integrble on [,b]. The second is the fundmentl theorem in clculus (FTC): if F hs continuous derivtive on [,b] then b F (x)dx = F(b) F(). These results re importnt nd re still good, nd form the foundtion for further study in your 3rd yer nd 4th yer. Wht you hve lerned in Prelims Clculus nd Anlysis courses re importnt not only for the study of dvnced theories in mthemtics, but lso re essentil in pplying mthemtics to solve prcticl problems in science. The theory of Riemnn integrtion is powerful tool for computtions, it however lcks flexibility in hndling limit procedures such s tking limits under integrtion. Here is simple nd illustrtive exmple. Consider the Dirichlet function f(x) = 1 if x is rtionl nd 1 if x is irrtionl. f is simple function but perhps is not n interesting one. While it ppers s the limit of simple functions. List ll rtionl numbers in [,1] s r 1, r 2,, nd define f n (x) = 1 if x {r 1,,r n } nd f n (x) = 1 otherwise. Clerly f n f on [,1], clerly ech f n is Riemnn 1 integrble on [,1] nd f n(x)dx = 1. Therefore lim n 1 f n(x)dx = 1, but we cn not tke limit lim n f n = f first then tke Riemnn integrl of f, s f is not Riemnn integrble. It ws recognized grdully tht most difficulties one hs with Riemnn integrls come from the limittion of the theory of Riemnn s integrtion. There ws need to extend the theory of integrtion to lrger clss of functions, so tht functions we re interested re integrble. Inordertoexplinthepprochweregoingtodevelopinthiscourse, letusrecllquicklythe min steps in defining Riemnn integrls. The first step is to choose simple clss of functions to which we cn ssign integrls. For the theory of Riemnn integrls we choose the collection L of ll step functions. Recll tht function ϕ is step if ϕ = n i1 Ji, where n is positive integer, J 1,, J n re finite intervls, nd where 1 A denotes the chrcteristic function of A: 1 A (x) = 1 or ccording to x A or not. The integrl of step function ϕ is defined to be I(ϕ) = n i J i, where J denotes the length of n intervl J (which is the mesure of the intervl J). The reson we choose to consider step functions in the Riemnn integrtion lies in the fct tht the mesure of n intervl, nmely its length, mkes sense for intervls. The second step is to define lower nd upper integrls for bounded function f on finite intervl (,b). The lower integrl b f(x)dx is defined to be the supremum of ll I(ϕ) where ϕ is step 3
4 b nd ϕ f1 (,b), nd the upper integrl f(x)dx is the infinimum of ll I(ψ) where ψ is step nd ψ f1 (,b). Finlly, f is Riemnn integrble if the lower nd upper integrls of f over (,b) b coincide, nd its Riemnn integrl f(x)dx is the common upper (nd lower) integrl. By definition of supremum nd infinimum, if we re ble to enlrge the clss L of simple b functions, then the corresponding lower integrl f(x)dx (which is supremum) would become b greter, nd the corresponding upper integrl f(x)dx (n infinimum) become smller, so they hve better chnce to be equl, thus hve better chnce to be integrble. This suggests the following pproch to extend the theory of integrtion to lrger clss of functions. Tke collection of functions ϕ in plce of step functions with the following form n ϕ = i 1 i where i re certin subsets but not necessry intervls. It is required tht we should be ble to define integrl for such ϕ to proceed the definition of integrls for generl functions. Nmely the integrl of ϕ should be defined s I(ϕ) = n i i. The problem to crry out this ide is tht for generl subset,, the length of, is not well-defined. The good news is tht this is the min difficulty we need to overcome in order to estblish new integrtion theory. Therefore our first tsk is to extend the notion of lengths for intervls to lrger clss of subsets thn intervls, the new notion will be clled the Lebesgue mesures. As long s the mesures of certin subsets re estblished, the integrtion theory cn be estblished solely bsed on mesures, no other structures of the rel line will be needed, which is somehow n unexpected rewrd by extending the notion of length to mesures to generl sets beyond intervls. A very short history of Lebesgue s theory of integrtion.. Borel nd H. Lebesgue were interested in the generl construction of functions, nd they wnted to extend the concepts of length, res nd volumes to generl sets. They discovered it ws not lwys possible to do so. It ws H. Lebesgue who recognized the importnce of the countble dditivity in hndling limits under integrtion. His Ph D thesis Intégrle, Longueur, Aire ws finished in 192, nd in book form Lecons sur l intégrtion et l recherche des fonctions primitives which ws published in 194, in Pris, Lebesgue s theory of integrtion ws estblished, nd bsic limit theorems (Monotone Convergence Theorem, Dominted Convergence Theorem), which re the powerful mthemticl tools, were proved. H. Lebesgue pplied his new integrtion theory to the study of trigonometric series, nd published nother monogrph in 196, Lecons sur les séries trigonométriques (Pris). The theory of Lebesgue s integrtion nd its generliztion, clled the theory of mesures ( mesure is generliztion of the concept of length, re, volume), got prominent when, in 1933, A. Kolmogorov firmly estblished the foundtion of probbility theory by interpreting mesure spces s mthemticl models for developing probbility theory nd sttistics. A. Kolmogorov published his finding in the rticle Grundbergriffe der Whrscheinlichkeitsrechnung, rg. Mt. 2, no.3 (1933). The nglish trnsltion in book form, Foundtions of the theory of probbility is still in print, nd is still worthy of reding even tody. A. Kolmogorov not only founded the probbility theory bsed on the theory of mesures, he mde very importnt contributions even for the theory of mesures. It ws him who introduced the concept of conditionl expecttions which plys vitl role in both nlysis nd probbility. He lso developed bsic tools for constructing mesures out of mrginl distributions. Around 194, J. L. Doob systemticlly developed the theory of mrtingles. He turned Kolmogorov s conditionl expecttion into powerful tool, nd identified clss of rndom vribles 4
5 (while, rndom vribles re exctly those mesurble functions ccording to A. Kolmogorov) clled mrtingles which is the most importnt clss of mesurble functions in probbility theory. Doob estblished powerful tools in collection of Doob s inequlities, convergence theorems nd etc. His results on mrtingles were orgnized netly in his clssic Stochstic Processes, published in Menwhile Pul Hlmos wrote book on the subject of mesures, Mesure Theory (195), which ws nd remins stndrd reference on Lebesgue s theory of mesures nd integrtion. 2 xtended rel line, upper nd lower limits We will use the following convention in deling with the symbols nd. Introduce the extended rel line [, ] = { } R { } by dding two symbols nd which re not in R. We mke the following conventions: for every R, < <, + = + =, = + =. = =, nd = =, but nd re not defined. Recll tht if S R is non-empty, nd if S is bounded bove, tht is, there is b R such tht s b for every s S, then the Completeness Axiom sys tht S hs the lest upper bound denoted by sups, which is n upper bound of S, nd if b is n upper bound of S, then sups b. If S is not bounded bove, tht is, for every n = 1,2,, there is s n S, such tht s n > n, then we sy sups =. Clerly, if S R is non-empty, then sups = if nd only if there is sequence {s n }, where ech s n S, such tht s n. Similrly we extend the definition of infs for ny non-empty subset S of R. It is remins true tht infs = sup( S), where ( ) = which is convention, nd S = { x : x S}, for non-empty S R. If { n } is sequence of rel numbers, then, by our extension of the concept for sup nd inf, which is decresing in n, nd which is incresing in n. Define nd sup k = lim mx{ n, n+1,, n+m }, k n m inf k = lim min{ n, n+1,, n+m } k n m lim inf n n = lim n lim m min{ n+1,, n+m } lim sup n = lim lim mx{ n n n+1,, n+m }, m which re clled the lower nd upper limit of { n }, respectively. It follows tht liminf n n = limsup n ( n ) nd limsup n n = liminf n ( n ). It lso follows from definition tht, which re useful to evlute upper nd lower limits, liminf n n is the lest number mong ll possible limits (i.e. numbers or ± which re the 5
6 limits of convergent sub-sequences of { n }), nd similrly, limsup n n is the lrgest number mong ll possible limits (i.e. numbers or ± which re the limits of convergent sub-sequences of { n }). If {b n } is sequence of rel numbers, nd if lim n b n = b exists nd finite (i.e. b or ), then lim inf ( n +b n ) = liminf n + lim b n n n n nd 3 Mesure spces lim sup n ( n +b n ) = limsup n n + lim n b n. Our first tsk is to extend the notion of length to some subsets of R, clled Lebesgue mesures nd denoted by m() (which is the length J if = J is n intervl). Suppose the clss of subsets of R to which we re ble to ssign mesures is denoted by M, then we expect the followings re stisfied: 1. The empty set Ø hs mesure zero, tht is m(ø) =. 2. The mesure of M is non-negtive, i.e. m() (but cn be infinity). 3. m is finite dditive: if 1,, n belong to M, then 1 n M, nd if in ddition i re disjoint, then m( 1 n ) = m( 1 )+ +m( n ). The property of finite dditivity is not enough in order to do integrtion, which should be enhnced. 4. m is countbly dditive: if 1,, n, belong to M, then n=1 n M. If in ddition { n : n = 1,2, } re disjoint, then ( ) m n = m( n ). n=1 We re going to identify crefully the clss M of Lebesgue mesurble subsets of R, nd to define the mesure m s function from M to [, ] which possesses properties 1)-4). The theory of integrtion bsed on the Lebesgue mesure m my be developed in rther generl setting which uses no lgebric or geometric structures of R. Therefore it is beneficil to introduce the concept of mesures, the concept of mesurble spces, nd the concept of mesure spces. Definition 3.1 Let be set [which is clled spce or clled smple spce], nd F be collection of some subsets of. 1) F is clled n lgebr on if ) Ø F nd F, b) if A,B F then A c F nd A B F. n=1 6
7 2) F is clled σ-lgebr (lso clled σ-field) on, if F is n lgebr over, nd if in ddition F is closed under the countble union opertion. Tht is, if A 1,A 2, belong to F, so does n=1a n. 3) A pir (,F), where F is σ-lgebr on, is clled mesurble spce. If F then is clled mesurble subset of (with respect to the σ-lgebr F). Remrk 3.2 (bout nottions) 1) If A is subset of then A c denotes \A if no confusion my rise, so A c = {x : x / A}. 2) Some uthors use AB to denote A B, in prticulr in probbility literture. The nottion is justified since 1 A B = 1 A 1 B, fct which will be used without further comments. 3) According to De Morgn s lw, if F is σ-lgebr nd A n F (n = 1,2, ) then n=1 A n = ( n=1 Ac n) c F. Definition 3.3 Let (,F) be mesurble spce. A function µ : F [, ] is clled mesure on (,F), if 1) µ(ø) = nd µ(a) [, ] (i.e. µ(a) or µ(a) = ) for every A F. 2) [µiscountblydditive]if A n F for n = 1,2,, nda n re disjoint, thenµ( n=1 A n) = n=1 µ(a n). A triple (,F,µ), where F is σ-lgebr nd µ is mesure on the mesurble spce (,F), is clled mesure spce. A mesure µ on mesurble spce (,F) is clled probbility (mesure) if µ() = 1. In this cse is clled smple spce (of fundmentl events), n element A in the σ-lgebr F is clled n event, nd µ(a) is clled the probbility tht the event A hppens. Proposition 3.4 Let (,F,µ) be mesure spce. Then 1) µ(a) µ(b) if A B, 2) if A n F nd A n (tht is A n A n+1 for ll n), then µ( n=1 A n) = lim n µ(a n ), 3) if A n F, A n (tht is A n A n+1 for ll n) nd µ(a 1 ) <, then µ( n=1 A n) = lim n µ(a n ). Proof. 1) If A B then B = A (B A c ), nd A nd B A c re disjoint, by dditivity of the mesure, we hve µ(b) = µ(a)+µ(b A c ) µ(a). 2) Let 1 = A 1 nd n = A n A n 1 for n 2. Then n re disjoint nd n=1 A n = n=1 n. Therefore, by the countble dditivity of µ, [ ] [ ] µ A n = µ n = µ( n ). n=1 n=1 On the other hnd, since A n, n k=1 k = A n, thus which yields 2). k=1 µ( k ) = lim n n k=1 µ( k ) = lim n µ 7 n=1 [ n k=1 k ] = lim n µ(a n )
8 3) Let B n = A 1 \ A n. Then B n. Since A 1 = B n A n, B n nd A n re disjoint, so µ(a 1 ) = µ(b n )+µ(a n ). Hence µ(b n ) = µ(a 1 ) µ(a n ) s µ(a n ) µ(a 1 ) <. According to de Morgn s lw n=1 B n = A 1 \ n=1 A n, thus, by pplying 2) to B n, we obtin tht [ ] [ ] [ ] µ(a 1 ) µ A n = µ A 1 \ A n = µ B n n=1 n=1 n=1 = lim µ(b n ) = lim (µ(a 1 ) µ(a n )) n n = µ(a 1 ) lim µ(a n ) n which implies 3) for µ(a 1 ) <. Interesting exmples of mesures re Lebesgue s mesures on ucliden spces R d which will be constructed in the next section. One of the centrl problems in probbility theory, quntum field theories nd sttisticl mechnics is to construct vrious mesures on some infinite dimensionl spces, which however should be studied in specilized courses. 4 The Lebesgue mesure In this section we construct 1) the σ-lgebr M Leb of Lebesgue mesurble subsets of R, nd 2) the Lebesgue mesure m : M Leb [, ]. Let us first describe cndidte of the Lebesgue mesure m, clled the Lebesgue outer mesure. Let C denote the collection of ll finite intervls with form (,b] where b re two rel numbers. C is π-system over R in the sense tht if A,B C then A B C. If J is n intervl, then J denotes the length of J, so if J = (,b] C then J = b. We build the outer mesure m by { } m (A) = inf J i : where ll J i C such tht J i A (4.1) for A R. Then m possesses the following properties: (1) m (Ø) =, nd m (A) for ny A R. (2) m (A) m (B) if A B, nd (3) m is countbly sub-dditive, s stted in the following Lemm 4.1 [Countbly dditive] If {A n : n = 1,2, } is sequence of subsets, then [ ] m A n n=1 m (A n ). (4.2) Proof. If n=1 m (A n ) = then (4.2) holds. Suppose n=1 m (A n ) <. Then m (A n ) < for every } n. By definition of m (A n ), for every ε > there is countble cover { J (n) i : i = 1,2, of A n, where J (n) i n=1 C, such tht J (n) i m (A n )+ ε 2 n. 8
9 { } While J (n) i : i,n = 1,2, forms countble cover of n=1 A n so tht m A n n=1 = n=1 J (n) i m (A n )+ε. n=1 n=1 ( m (A n )+ ε 2 n ) Since ε > is rbitrry, (4.2) follows immeditely. Although m () is well defined for every subset of R, m is not countbly dditive, thus m is not mesure on P(R) (the σ-lgebr of ll subsets of R). In fct, P(R) is too big for m to be countbly dditive. The min technicl step in the construction of the Lebesgue mesure is to identify the σ-lgebr M Leb on which m is countbly dditive. M Leb should be sufficient lrge nd should include ll intervls. This will be chieved in the celebrted Crthéodory s extension theorem. 4.1 Outer mesures nd Crthéodory s extension theorem This is mjor theorem in the theory of mesures. It is generl theorem so we will formulte it in generl setting. Its proof is not exminble. We will use this theorem to identify some importnt mesurble subsets which we should be fmilir with. Let (,G) be mesurble spce, nd µ : G [, ] be n outer mesure on (,G) in the following sense: 1) µ ( ) =, nd µ (A) for every A G; 2) µ (A) µ (B) if A B, A,B G; 3) µ is countbly sub-dditive: if A n G for n = 1,2,, then [ ] µ A n n=1 µ (A n ). (4.3) Definition 4.2 A subset G is µ -mesurble if stisfies the Crthéodory condition n=1 µ (F) = µ (F )+µ (F c ) for every F G. (4.4) The collection of ll µ -mesurble subsets is denoted by G m. Since F \ = F c, (4.4) my be written s µ (F) = µ (F )+µ (F \) for ny subset F G. (4.5) Since F = (F ) (F c ), by sub-dditivity of µ µ (F) µ (F )+µ (F c ) for ny nd F belonging to G, the Crthéodory condition (4.4) is equivlent to the inequlity µ (F) µ (F )+µ (F c ) (4.6) for ny subset F G. is mesurble if nd only if (4.6) holds for ny subset F G such tht µ (F) <. 9
10 Theorem 4.3 G m is σ-lgebr on, nd µ is mesure on G m. Proof. [The proof is not exminble]. Clerly the empty set G m. Also, ccording to (4.5), G m if nd only if c G m. Let A,B G m. We show tht A B G m so tht G m is n lgebr. Since (A B) c = A c B c, we need to show tht for ny subset F G. Since A is µ -mesurble, so µ (F) = µ (F (A B))+µ (F (A c B c )) µ (F) = µ (F A)+µ (F A c ). (4.7) Since B is lso µ -mesurble, pplying (4.5) to F A (in the plce of F) nd B we obtin Substitute (4.8) into (4.7) to obtin µ (F A) = µ (F A B)+µ (F A B c ). (4.8) µ (F) = µ (F (A B))+µ (F A B c )+µ (F A c ). (4.9) Use gin (4.5) to A (which is mesurble) nd F (A c B c ) to obtin µ (F (A c B c )) = µ (F (A c B c ) A)+µ (F (A c B c ) A c ) = µ (F B c A)+µ (F A c ), here we hve used the elementry equlities (A c B c ) A = B c A nd (A c B c ) A c = A c. Together with (4.9) we deduce tht µ (F) = µ (F (A B))+µ (F (A c B c )) for ny F G, so tht A B G m. By de Morgn lw, it follows lso tht A B = (A c B c ) c G m. Thus G m is n lgebr. We next to show tht G m is σ-lgebr. To this end, consider n G m, n = 1,2,. We show tht = n=1 n G m. Without losing generlity we my ssume tht n re disjoint, otherwise we my consider A n insted, where A 1 = 1 G m nd A n = n \( j<n j ) G m for n 2. A n G m re disjoint, nd = n=1 A n. Since n j G m, by pplying (4.5), we obtin ( n )] ( n ) c ] µ (F) = µ [F j +µ [F j. 1
11 Since n is mesurble, ( n )] µ [F j ( n ) ] ( n ] = µ [F j n +µ [F j ) n c = µ [F n ]+µ [F = = n µ [F j ] ( n 1 j )] here we hve used the following identities: since j re disjoint ( n j ) n = n nd ( n j ) n c = n 1 j. We therefore hve µ (F) = ( n n ) c ] µ [F j ]+µ [F j ( n ) c ] µ (F j )+µ [F j. Letting n to obtin ( ) c ] µ (F) µ (F j )+µ [F j ( )] ( ) c ] µ [F j +µ [F j (4.1) (4.11) which implies tht j G m. Since we must hve ( )] ( ) c ] µ (F) µ [F j +µ [F j for ny F G nd j, the inequlities in (4.11) must be equlities. Hence ( ) c ] µ (F) = µ [F j ]+µ [F j for ny subset F G. In prticulr, by pplying the equlity bove to F = k=1 k we obtin [ ] [( ) ] µ j = µ k j = µ ( j ). k=1 11
12 Tht is, µ is countbly dditive, so tht µ is mesure on G m. It is possible, for some outer mesures µ, the σ-lgebr G m my be trivil, so tht it is useless in this cse. Let us look t some simple exmples of µ -mesurble sets. Definition 4.4 A subset A G is clled µ -null set if it hs zero µ -outer mesure, i.e. µ (A) =. We hve the following simple fct Lemm 4.5 If {A n : n = 1,2, } is sequence of µ -null subsets, then n=1 A n is µ -null s well. This follows immeditely from the fct tht µ is countbly sub-dditive. In fct, if µ (A n ) =, then ( ) µ A n µ (A n ) = n=1 n=1 hence we must hve µ [ n=1 A n] =. n=1 A n is µ -null. Proposition 4.6 If A is µ -null, then A G m. Proof. For every subset F G, we hve µ (F A) µ (A) = so tht µ (F) µ (F A c ) = µ (F A)+µ (F A c ). By definition, A is µ -mesurble. In pplictions, it is importnt to be ble to describe the structure of the σ-lgebr G m of µ -mesurble subsets, which is determined by the outer mesure µ. However, without further informtion bout µ, Prop. 4.6 is the best we cn offer. In prctice, we do hve some priori knowledge, which helps to identify the structure of G m. 4.2 The Lebesgue mesure spce Recll tht C is the π-system of ll intervls (,b] where b re two rel numbers. If J = (,b], then J = b is the length of J. The Lebesgue outer mesure m is defined by { } m () = inf J i : J i C such tht J i (4.12) where issubsetofr, ndtheinf tkesoverll possible countble cover{j i : i = 1,2, } C of. M Leb denotes the collection of ll m -mesurble [from now on, clled Lebesgue mesurble, or simply mesurble if no confusion my rise] subsets. Recll tht is m -mesurble if stifies the Crtheodory condition: m (F) = m (F )+m (F c ) 12
13 for every F R. We hve proved tht M Leb is σ-lgebr, nd m restricted on M Leb is mesure. The mesure spce (R,M Leb,m ) is clled the Lebesgue mesure spce, or simply the Lebesgue spce. We give description of the sets in M Leb in this lecture. The key is the following Lemm 4.7 C M Leb, nd moreover, if J = (,b] C, then m (J) = b. This lemm looks very simple, but its proof is not esy nd is quite technicl. Its proof is not exminble, therefore I don t give the proof in the lecture. By using this lemm, we cn describe the σ-lgebr M Leb. We divide this tsk into severl steps, ll re elementry. 1) Firstly, we clim ny intervl J is Lebesgue mesurble, nd m (J) = J the length of J. If J is unbounded, for exmple J = (, ), then J = n=1 (,n], since ech (,n] is mesurble nd hs mesure n, so tht J M Leb, nd m (J) n, therefore m (J) =. If J = [, ), then J = {} (, ). We only need to show {} = [,] is mesurble. Indeed [,] = n=1 ( 1,], so [,] M n Leb nd m ({}) ( 1 n ) = 1 n for ny n = 1,2,, so m ({}) =. While (,) = [, ) c, nd (,] = (, ) c, so both (,) nd (,] re mesurble. Of course (, ) is mesurble. Now suppose J is bounded intervl with two ends b. By Lemm (,b] is mesurble, nd we hve shown {} is mesurble. Since [,b] = {} (,b], (,b) = n=1 (,b 1 n ], so open / closed intervls re mesurble. Since [,b) = {} (,b) so [,b) is mesurble too. Therefore ny bounded intervl is mesurble. 2) Any open subset of R is mesurble, so is ny closed subset of R. Hence ny closed / open subset is mesurble. According to Question 7, Problem Sheet 1, if G R is open, then G hs decomposition G = i ( i,b i ) (4.13) where ( i,b i ) re disjoint open intervls (bounded or unbounded), t most countbly mny. Since M Leb is σ-lgebr, nd ech intervl ( i,b i ) is mesurble, hence G is lso mesurble. Since m is mesure on M Leb, so it is countbly dditive, m (G) = i (b i i ). Remrk. However we should note tht it is in generl impossible to rrnge intervls ( i,b i ) in the decomposition of G in n order, such tht 1 < b 1 < 2 < b 2 <. The Borel σ-lgebr B(R) over R is defined to be the smllest σ-lgebr contining ll open subsets of R. A subset R is clled Borel mesurble if B(R). Since ny open subset belongs to M Leb nd M Leb is σ-lgebr, therefore M Leb is bigger thn B(R). We thus hve the following importnt conclusion. 13
14 Lemm 4.8 B(R) M Leb, tht is, ny Borel mesurble subset is Lebesgue mesurble. On the other hnd, M Leb is not lot bigger thn B(R). In fct we hve Lemm 4.9 If M Leb, then there re A,B B(R), such tht A B nd m (\A) = m (B \) =. Tht is, M Leb nd B(R) differ by Lebesgue mesure zero subsets. xmple 1. If R, then m ({}) = (see bove), so tht, by the countble dditivity of m on M Leb, ny countble subset is mesurble nd hs Lebesgue mesure zero. In prticulr m (Q) =. xmple 2. There re uncountble Lebesgue null sets. Cntor set (see pges in Lecture Notes) is n exmple. Red the notes crefully for its construction. Also Question 2 prt (), Problem Sheet 2. xmple 3. There re subsets which re not Lebesgue mesurble. See pge 17 Lecture Notes for n exmple. To construct the exmple, we need the following fct: the Lebesgue outer mesure m (nd therefore the Lebesgue mesure m on M Leb ) is trnsltion invrint in the sense tht: m ( +) = m () for every subset, nd for every number R, where + = {x+ : x }. This is prt of Question 1 is Problem Sheet 2. Of course, by symmetry, you only need to show tht m ( +) m (), which you cn prove it by using the definition of m given by (4.12). Let us point out tht exmples of non-lebesgue mesurble subsets re not exminble. Let me finish these notes by sying tht there re Lebesgue mesurble sets which re not Borel mesurble, tht is, the σ-lgebr M Leb is strictly lrger thn the Borel σ-lgebr B(R). 4.3 The Cntor set nd null subsets xmple 4.1 If A is countble subset, then A is Lebesgue null set. There re null sets which re not countble! Here is n exmple. xmple 4.11 The Cntor ternry set. Consider the closed intervl J (1) = [,1]. 1)DivideJ (1) equllyintothreesub-intervls,themiddleopenintervli (1) 1 = ( 1, 2)isremoved 3 3 from J (1), the remining two closed sub-intervls re J (1) 1 = [, 1 ] nd J(1) 3 2 = [ 2,1], ech hs 3 length ) Repet step 1) for ech closed intervl J (1) 1 nd J (1) 2 : divide ech eqully into three subintervls nd remove the middle open ones. From J (1) 1 we remove I (2) 1 = ( 1 remove ( I (2) 2 2 = , ) = 23 ( ) 2 2,. 2 ), 2 (1) 3 2 3, nd from J 2 2 The remining 2 2 closed intervls re denoted by J (2) 1,J (2) 2,J (2) 3,J (2) 4 ech hs length
15 3) Repet the procedure for ech J (2) i, remove the middle open intervls I (2) i (i = 1,,2 2 ) nd the remining 2 3 closed intervls re denoted by J (3) i for i = 1,,2 3, ech hs length Repeting this process, for ech n, we hve 2 n 1 disjoint open intervls I (n) i (i = 1,,2 n 1 ) with equl length 1, nd 2 n disjoint closed intervls J (n) 3 n i (i = 1,2,,2 n ) with length 1 (where 3 n n = 1,2, ). nd For ech n, I (n) i, J (n) k (i = 1,,2 n 1 nd k = 1,,2 n ) re disjoint sub-intervls of [,1], n 2 k 1 k=1 I (k) i ( 2 n k=1 J (n) k ) = [,1]. Cntor s ternry set C is defined to be the subset of [,1] by removing ll middle intervls I (n) i. Tht is C = [,1]\ ( n=1 2 n 1 Lemm 4.12 The Cntor ternry set C is null set. Proof. By the construction, for every n, J (n) i m (C) 2 n I (n) i ). (i = 1,,2 n ) is finite cover of C, so tht J (n) i = 2 n 1 3 n s n. Therefore m (C) =, nd C is null subset by definition. Ternry expnsions For deciml expnsion of x =.x 1 x 2 (,1] we men tht x n x = 1 n. n=1 Similrly x (,1] cn be written in its ternry expnsion n x = 3 =. 1 2 n n=1 where n =,1 or 3, nd we use the convention tht it should not end with k = 2 for ll k N. Then, in terms of ternry expnsion ( 1 3, 2 3 ) = (,1,.2) nd x I (1) 1 = ( 1 3, 2 3 ) then x =. 1 2 with 1 = 1. Let A = {. 1 2 : where i = or 2}. Then by our construction, A C so tht m (A) m (C) =. We leve the reder s n exercise to show tht 1) C is closed, 2) C is uncountble, nd 3) describe C in terms of ternry expnsions for rel numbers. Finlly we should point out tht M Bor is strictly smller thn M Leb. However we hve the following fct Proposition 4.13 If M Leb then there re A,B M Bor such tht A B nd both \A nd B \ re null sets. The proof is outside the syllbus. 15
16 4.4 An exmple of non mesurble sets While not every subset of R is mesurble, though it is not esy to produce n exmple of non mesurble sets, which is good news indeed. In fct we need to invoke the xiom of choice to produce subset which is not in M Leb. If is subset of R, nd is rel number, then +A = {+x : x A} is trnsltion of subset A. It is elementry tht if J is n intervl, so is +J nd +J = J. According to the definition of the outer mesure m we my deduce tht m (+A) = m (A). It follows tht, if M Leb, then + M Leb nd m(+) = m(). Therefore the Lebesgue mesure m is invrint under trnsltions. Consider the unit intervl [,1]. Define the following equivlence reltions: if x,y [,1] nd x y Q, then we sy x y. Divide [,1] into equivlent clsses, nd choose exctly one number from ech equivlent clss to form subset A [,1]. We clim tht A is not mesurble, i.e. A / M Leb. In fct, list the rtionl numbers in [ 1,1] by r 1,r 2,. Then (r i +A) [,1] nd (r i +A) [ 1,2], so tht ( ) 1 m (r i +A) 3. Note tht r i +A re disjoint, nd m (r i +A) = m (A) for every i. If A were mesurble, then r i +A re mesurble for every i, hence, by countble dditivity, we would hve ( ) m (r i +A) = m (r i +A) = m (A) so tht 1 m (A) 3 which is impossible s m (A) = or m (A) = (ccording to m (A) = or not). Thus A is not Lebesgue mesurble. 4.5 Lebesgue sub-spces Suppose R is Lebesgue mesurble, then M Leb () = { A : where A M Leb } = {A : where A nd A M Leb } is σ-lgebr on. On the other hnd M Leb () M Leb so tht the restriction of the Lebesgue mesure m, denoted by m or by m if no confusion is possible, is obviously mesure on (,M Leb ()), clled the Lebesgue mesure on. The triple (,M Leb (),m) is therefore mesure spce, clled Lebesgue subspce. A function f defined on which is M Leb ()- mesurble is clled Lebesgue mesurble on. M Leb () will be simply denoted (by busing nottions) by M Leb if no confusion my rise. 16
17 5 Mesurble functions In this section we identify clss of functions, clled mesurble functions, for which we im to define integrls. The concept of mesurbility cn be developed independent of mesure, we therefore develop this concept for generl mesurble spce (, F). 5.1 Definition nd bsic properties Recll tht B(R) (or denoted by M Bor ) denotes the Borel σ-lgebr on R, which is the smllest σ-lgebr contining ll open sets in R, or equivlently, the smllest σ-lgebr which contins ll intervls. Let us begin with some comments on nottions. If f is function defined on spce nd tkes its vlues in [, ] (i.e. f is mpping from to [, ]), nd if A is subset of [, ], then f 1 (A) is the pre-imge of A under f, tht is, f 1 (A) = {x : f(x) A}. For simplicity, we will lso use {f A} to denote f 1 (A). More generlly, if P is property (or sttement)dependingonx, thenwewilluse{p}todenotethesubset{x : P(x) holds }, if the underlying spce is cler, nd if no confusion my rise. For exmple, f s bove nd R, then {f > } = {x : f(x) > }. Tht is, {f > } is the pre-imge of (, ] under f. Similrly {f = } denotes the set {x : f(x) = } etc. As nother exmple, if f nd g re two mppings from into [, ], then nd nd etc. Let F be σ-lgebr on the spce. {f g} = {x : f(x) = g(x)} {f > g} = {x : f(x) > g(x)} Definition 5.1 Let f : R be function on. Then f is F-mesurble if f 1 (G) F for every G B(R). Tht is, {f G} F for every Borel subset G. In probbility theory, n F-mesurble function on is clled rndom vrible on the mesurble spce (, F). To discuss the mesurbility of functions (in generl mppings, the following fct is very useful. Lemm 5.2 Let X : S be mpping, nd F be σ-lgebr on. Then is σ-lgebr. F X is the push-forwrd σ-lgebr of F by X. The proof is left s n exercise [Problem 4, Sheet 2]. F X = { A S : X 1 (A) F } (5.1) 17
18 Proposition 5.3 Suppose f : R. Then the following sttements re equivlent: 1) f 1 (G) F for every G M Bor, i.e. f is F-mesurble. 2) {f > } F for every R, 3) {f < } F for every R, 4) {f } F for every R, 5) f 1 (J) F for every intervl J. Proof. Let G = F f be defined by (5.1). Then G is σ-lgebr, nd f is F-mesurble, i.e. it stisfies 1), if nd only if B(R) G. Let us for exmple show tht 1) nd 4) re equivlently. Clerly 1) implies 4), so let us prove the other direction tht 4) yields 1). To this end, let A be the fmily of ll intervls (,]. Then 4) sys tht A G. Since G is σ-lgebr, so tht σ{a} G, where σ{a} is the smllest σ-lgebr which contins ll subsets in A. Since σ{a} = B(R), therefore B(R) G. Thus for every G B(R), {f G} F. Definition 5.4 Let be Lebesgue mesurble subset of R, nd M Leb () (or M Leb if no confusion my rise) denote the σ-lgebr of ll Lebesgue mesurble subsets of, tht is, M Leb () = {A M Leb : A }. Then (,M Leb ()) is mesurble spce, nd the Lebesgue mesure restricted on M Leb () is mesure on (,M Leb ). The mesure spce (,M Leb,m) is clled Lebesgue mesure spce. Let f : R be function defined on Lebesgue mesurble subset. If f : R is M Leb -mesurble, then f is clled Lebesgue mesurble on. Definition 5.5 A function f : R R which is B(R)-mesurble, is clled Borel mesurble. Since B(R) M Leb, Borel mesurble function is Lebesgue mesurble. The converse is not true in generl, there re Lebesgue mesurble functions which re not Borel mesurble. xmple. 1) Suppose h : R R is continuous. Then if U is open then h 1 (U) is open, so tht h 1 (U) B(R). Since G = { A R : h 1 (A) B(R) } is σ-lgebr, nd ny open subset belongs to G, by definition, B(R) G. Therefore h 1 (G) B(R) for ny G B(R). Thus continuous function h is B(R)-mesurble. 2) If h : (,b) R is monotone, then h is Borel mesurble. Proposition 5.6 Let (, F) be mesurble spce. 1) Let A. Then 1 A is F-mesurble if nd only if A F. 2) If f : R is F-mesurble, nd h : R R is Borel mesurble, then h f is F-mesurble. In prticulr, if h is continuous, nd f is F-mesurble, then h f is lso F- mesurble. [ While, on the other hnd, there re Lebesgue mesurble functions h nd f such tht h f is not Lebesgue mesurble. ] 3) If R nd f,g : R re F-mesurble, then f, f ± g, fg, f/g (if g ) re F-mesurble. 4) f g = mx{f,g} nd f g = min{f,g} re F-mesurble. Hence f, f + nd f re F-mesurble, where f + = f nd f = ( f). f + (resp. f ) is clled the positive prt (negtive prt) of f. f = f + +f nd f = f + f. 18
19 Proof. 1) For every G, {1 A G} equls A, A c, Ø or, so I A is mesurble if nd only if A F. 2) In fct, for ny G B(R), h 1 (G) B(R), thus (h f) 1 (G) = f 1 (h 1 (G)) F, so h f is F-mesurble. 3) There is nothing to prove if =. If > then for ny b we hve { {f > b} = f > b } F nd if < then {f > b} = { f < b } F so tht f is mesurble. In prticulr g is mesurble if g is mesurble. We show tht f +g is mesurble. For ny b we hve {f +g > b} = q Q{f > q nd g > b q} = q Q{f > q} {g > b q}. Since f,g re mesurble so tht {f > q} F nd {g > b q} F nd therefore {f > q} {g > b q} F q. Since the set Q of ll rtionl numbers is countble, it follows tht {f > q} {g > b q} F q Q nd therefore f +g is mesurble. Since f 2 = h f where h(x) = x 2 is continuous function, so tht f 2 is mesurble if f is mesurble. Now fg = 1 [ (f +g) 2 (f g) 2] 4 is mesurble. Similrly, since h(x) = x is continuous, so tht f = h f is mesurble. Hence f g = 1 [(f +g)+ f g ] 2 nd f g = 1 [(f +g) f g ] 2 re mesurble. In prticulr f + = f nd f = ( f) re mesurble. Definition 5.7 A function f : [, ] is F-mesurble, if {f = }, {f = } re mesurble, nd f 1 (G) F for every Borel subset G. Next we prove tht the clss of F-mesurble functions on mesurble spce (,F) is closed under limiting opertions. 19
20 Proposition 5.8 If {f n : n = 1,2, } is sequence of F-mesurble functions, then sup n 1 f n, inf n 1 f n, limsup n f n nd liminf n f n re F-mesurble. In prticulr, if f = lim n f n exists, then f is F-mesurble. Proof. For ech n consider g n (x) = sup{f n (x),f n+1 (x), } nd h n (x) = inf{f n (x),f n+1 (x), }. Then {g n : n = 1,2, } is [point-wise] decresing sequence nd {h n : n = 1,2, } is n incresing sequence, hence g n g nd h n h, where g nd h my tke vlue ±. g nd h re clledtheupperlimitndlowerlimitof(f n ), denotedbylim n f n (orlimsupf n )nd lim n f n (or by liminff n ) respectively. For ech n, g n tkes vlues in (, ]. For ny rel number we hve nd {g n } = {g n = } = so tht g n is F-mesurble. On the other hnd so tht h n is F-mesurble. Finlly since m=n ( N=1 {f m } {g n N} ) c h n = inf{ f n (x), f n+1 (x), } lim supf n = inf{g n : n 1} n thus limsup n f n is F-mesurble, nd lim n f n = lim n ( f n ) is lso F-mesurble. If (,F) is mesurble spce, then function ϕ : R is clled simple function, or more precisely clled simple F-mesurble function on, if ϕ = k c i 1 i for some positive integer k, some rels c i nd some F-mesurble subsets i. The collection of ll non-negtive, simple functions is denoted by S + (,F) or by S + if no confusion my rise. Theorem 5.9 Suppose tht f is mesurble function tking vlues in [, ]. Then there is n incresing sequence of simple functions (f n ) such tht f n f. 2
21 where Proof. For ech n we define { (n) k = x : f n = 2 2n 1 k= } k k +1 f(x) < 2n 2 n k 2 n1 +2 n 1 (n) An, k nd A n = {x : f(x) 2 n } re mesurble. Then (f n ) is n incresing sequence of non-negtive simple F-mesurble functions. Moreover f f n 1 on {f < 2 n } nd f 2 n n = 2 n on {f 2 n }. Therefore f n f s n. Therefore we hve the following simple structure theorem for mesurble functions in terms of simple mesurble functions. Corollry 5.1 A function f on tking vlues in [, ] is F-mesurble if nd only if there is sequence of simple F-mesurble functions f n : R such tht f n f. Proof. Wenotethtf = f + f ismesurbleifndonlyifbothf + ndf remesurble. Apply the previous theorem to f + nd f. 5.2 Almost everywhere properties Let (,F,µ) be mesurble spce. If P is property depending on x, then {P} denotes the subset {x : P(x) holds }, nd µ[p] deontes the mesure of {P} if {P} is mesurble. Tht is, µ[p] = µ({p}) = µ({x : P(x) holds }). For exmple, if f : [, ] is F-mesurble nd λ is rel number, then If f nd g re two mesurble functions, then µ[f > λ] = µ({x : f(x) > λ}). µ[f g] = µ({x : f(x) g(x)}) etc. We sy the property P holds µ-lmost everywhere (or lmost surely) on, if µ[p doesn t hold ] = µ({x : P(x) doesn t hold }) =. If the underlying spce nd the mesure µ is cler from the context, then we simply sy the property P holds lmost everywhere. For exmple, if f nd g re two functions, then we sy f = g lmost everywhere if µ[f g] =. If (f n ) is sequence of functions, then f n converges to f lmost everywhere, if µ({x : f n (x) does not converge to f(x)}) =. Recll tht mesure spce (,F,µ) is complete, if A F nd µ(a) =, then ny subset of A is F-mesurble, tht is, ny subset of A lso belongs to F. Any Lebesgue mesurble sub-spce (,M Leb,m) (where is Lebesgue mesurble) is complete. However, (R,B(R),m) is not complete. 21
22 Proposition 5.11 Suppose (,F,µ) is complete mesure spce. If f : [, ] is F-mesurble, nd g = f lmost everywhere, then g is lso F-mesurble. This is n exercise in Problem Sheet 2. Proposition 5.12 Suppose (,F,µ) is complete mesure spce. If f n : [, ] re F-mesurble functions, nd f n f lmost everywhere on, then f is F-mesurble. 5.3 xmples We give in this prt further properties bout mesurble functions in terms of exmples. xmple 1. If g : R R is monotone, then g is Borel mesurble. In fct if I is n intervl, then g 1 (I) is union of t most countble mny intervls, so it belongs to B(R). Hence g is Borel mesurble. xmple 2. Let (,F) be mesurble spce. Suppose f n : R re F-mesurble. Recll tht {f n } converges t x if nd only if {f n (x)} is Cuchy sequence. Hence = {x : (f n (x)) converges to number} { f n f m < 1 } k k=1n=1 m,n=n nd = {x : (f n (x)) doesn t converge to number} { f n f m 1 } k k=1n=1 m,n=n re both mesurble. xmple 3. Let (,F) be mesurble spce. If f n,f : R re F-mesurble, then {f n f} = k=1 N=1n=N { f n f < 1 } k nd re mesurble s well. {f n f} = k=1n=1 n=n { f n f 1 } k xmple 4. Let R, nd f,f n : R (n = 1,2, ) be Lebesgue mesurble functions on. Then f n f lmost everywhere on (with respect to the Lebesgue mesure), i.e. m[f n doesn t converge to f] = 22
23 [where {f n doesn t converge to f} = {x : f n (x) f(x)} nd similr nottions pply to other sets], if nd only if for every ε > ( ) m { f n f ε} =. (5.2) N=1 n=n If in ddition m() <, then [Proposition 2.4, item 3)] (5.2) is equivlent to the condition tht ( ) lim m { f n f ε} = (5.3) N for ny ε >. n=n xmple 5. Let = (, ), nd f n (x) = 1 if x (,n] nd f n (x) = if x > n. Then f n f = 1 (, ) everywhere. For every ε (,1) we hve so tht Thus (5.3) doesn t hold. On the other hnd N=1 n=n { f n f ε} = (n, ) m({ f n f ε}) =. { f n f ε} = (N, ) = Ø. [This is simple exmple tht f n f everywhere, but f n does not converge to f in mesure, see Definition 7.5 below]. xmple 6. (gorov s theorem) Let R be mesurble such tht m() <. Suppose f n,f : R re mesurble, nd suppose f n f lmost everywhere on. Then for every δ > there is mesurble δ such tht m( \ δ ) < δ nd f n f uniformly on δ. Proof. Since f n f lmost surely on nd m() <, ccording to (5.3) in xmple 4, for every δ > nd for ny k = 1,2, (pplying (5.2) to ε = 1 k ) there is n k such tht m ( j=n N=1 { f j f 1 } ) δ for n n k 2 k k. Let k = { f j f < 1 } { = f j f < 1 } k k : j n k j=n k nd δ = k=1 k. Then m(k c) δ nd 2 k ( m( \ δ ) = m k=1 c k ) 23 m(k) c k=1 k=1 δ 2 k = δ.
24 We clim tht f n f uniformly on δ. In fct, x δ if nd only if x k for ll k = 1,2,, nd if nd only if f j (x) f(x) < 1 k j n k. Therefore nd f n f uniformly on δ. sup x δ f j (x) f(x) 1 k, j n k 6 Lebesgue integrtion In this section we study Lebesgue s theory of integrtion. Lebesgue s theory of integrtion cn be built on mesure spce, though most of textbooks on integrtion theory del with the cse of Lebesgue mesure first, the process to construct the theory of integrtion however does not become simpler if we restricted to the Lebesgue mesure. First of ll we need to define the clss of simple (mesurble) functions, nd we define integrls for simple nd non-negtive functions, then define integrls for non-negtive mesurble functions. Finlly we define Lebesgue integrls for mesurble functions. In this course, we only del with the theory of Lebesgue s integrtion for the Lebesgue mesure, we will nevertheless develop the theory on mesure spce (,F,µ) if no dditionl effort is required nd if only modifictions of nottions re required. We will however concentrte on the Lebesgue spce (,M Leb,m), where R is Lebesgue mesurble subset, nd m is the Lebesgue mesure s our model of the mesure spce, nd you my red notes s if =, F = M Leb (), µ = m for some Lebesgue mesurble subset R, unless otherwise specified, nd thus F-mesurble mens Lebesgue mesurble on. 6.1 Lebesgue integrls, nd integrble functions Therefore we will develop Lebesgue s theory on mesure spce (, F, µ). The following technicl conditions re enforced in wht follows, unless sid otherwise. 1) (,F,µ) is complete in the sense tht: if A F nd µ(a) =, then ny subset of A belongs to F. 2) (,F,µ) is σ-finite, tht is, there is n incresing sequence G n F (n = 1,2, ), such tht n=1 G n = nd µ(g n ) < for ll n. Our bsic exmple is Lebesgue sub-spce (,M Leb,m), where R is Lebesgue mesurble subset, nd M Leb denotes the σ-lgebr M Leb on for simplicity. S + (,F) denotes the collection of ll non-negtive, simple, F-mesurble functions on. Suppose ϕ = k c i1 i is non-negtive, simple, F-mesurble function on, where c i nd i F, then its Lebesgue integrl is defined by ϕdµ = k c i µ( i ) 24
25 where the convention tht = hs been used, to ensure the finite sum on the right-hnd side is well defined nd is independent of the representtion of ϕ. In fct, if where c i nd j for ll i,j, then k ϕ = c i 1 i = N j 1 Aj k c i µ( i ) = which follows from the dditivity of the mesure µ. N j µ(a j ) Proposition 6.1 Let ϕ,ψ S + (,F), nd λ. Then 1) (ϕ+ψ)dµ = ψdµ. ϕdµ+ 2) λϕdµ = ϕdµ. λ 3) If ϕ ψ then ϕdµ ψdµ. Proof. We my choose mesurble sets i F (i = 1,,k) which re disjoint such tht so tht nd therefore ϕ = (ϕ+ψ)dµ = k c i 1 i, ψ = ϕ+ψ = k d i 1 i k (c i +d i )1 i k (c i +d i )µ( i ) = ϕdµ+ which proves 1). If ϕ ψ then c i d i for ll i for which i Ø, then ϕdµ = k c i µ( i ) k d i µ( i ) = which is 3). Suppose f : [, ] is F-mesurble, then the (Lebesgue) integrl of f is defined by { } fdµ = sup ϕdµ : ϕ S + (,F) s.t. ϕ f. (6.1) A non-negtive mesurble function f is integrble on (with respect to the mesure µ) if fdµ <. By definition, if f, g re non-negtive nd mesurble, f g, nd λ be non-negtive number, then λfdµ = λ fdµ nd fdµ gdµ In prticulr, if g is integrble, then so is f. 25 ψdµ ψdµ,
26 Lemm 6.2 f : [, ] is F-mesurble, then µ[f λ] 1 λ for every λ >, which is clled the Mrkov inequlity. fdµ (6.2) We recll here the convention of our nottion: if P is sttement depending on x, then {P} = {x : P(x)} nd Therefore {f λ} = {x : f(x) λ} nd µ[p] = µ({p}) = µ({x : P(x)}). µ[f λ] = µ({x : f(x) > λ}). Proof. of Lemm 6.2. Since {f λ} is F-mesurble, nd f is non-negtive, the simple function ϕ = λ1 {f λ} f. By definition of integrtion, fdµ ϕdµ = λµ[f λ], which yields (7.1). Proposition 6.3 f : [, ] is integrble, then µ[f = ] =. Tht is, f is finite lmost everywhere on. Proof. Since {f = } is mesurble, so tht, by the Mrkov inequlity µ[f = ] µ[f > λ] 1 fdµ λ for every λ >, where the first inequlity follows from the fct tht {f = } {f > λ} for every λ >. Letting λ in the inequlity bove, we deduce tht nd therefore we must hve µ[f = ] =. µ[f = ] Proposition 6.4 If f : [, ] is F-mesurble, fdµ =, then f = lmost nd everywhere on, i.e. µ[f ] =. Proof. By Mrkov s inequlity µ[f λ] 1 λ fdµ =, so tht µ[f λ] = for every λ >. Since {f > } = { n=1 f 1 n}, we hve [ µ[f > ] µ f 1 ] =. n n=1 26
27 Therefore f = lmost everywhere on. Lebesgue integrble functions on (, F, µ) If f : [, ] is F-mesurble, then its positive prt f + = mx{f,} nd negtive prt f = mx{ f,} re F-mesurble on, f = f + f nd f = f + +f. Thus f+ dµ nd f dµ re both well defined (my be ). If both f+ dµ nd f dµ re finite, then we sy f is (Lebesgue) integrble on, nd define its (Lebesgue) integrl fdµ = f + dµ f dµ. Let L 1 (,F,µ) (L 1 (,µ) or L 1 () if noconfusion is possible from thecontext, for simplicity) denote the spce of ll integrble functions on. If f : [, ] is mesurble, then f L 1 () if nd only if f dµ <. If f L1 (), then µ[ f = ] = [Applying Proposition 6.3 to f ], tht is, f is finite lmost everywhere on. We hve very simple comprison theorem for integrbility. Proposition 6.5 1) Let f nd g be F-mesurble. If g L 1 (,F,µ) nd f g on, then f L 1 (,F,µ) nd f dµ gdµ. 2) If µ() <, nd if f is F-mesurble nd bounded on, then f L 1 (,F,µ). Proof. 1) is obvious by definition, s both f + nd f re dominted by g under ssumption. To show 2), suppose ϕ S + (,F) nd ϕ f. Then ϕ is bounded by C = sup f <, so tht ϕ C1 on. Hence ϕdµ Cµ() <. Hence { } f dµ = sup ϕdµ : ϕ is non-negtive, simple; nd ϕ f Cµ() <. Thus f L 1 (,F,µ). The following theorem, clled Monotone Convergence Theorem (MCT), is one of the most importnt results in Lebesgue s Theory of Integrtion. Theorem 6.6 (MCT, Lebesgue nd B. Levi) Suppose f n : [, ] re F-mesurble nd f n (tht is f n+1 (x) f n (x) for n = 1,2,, nd x ). Let f = lim n f n. Then fdµ = lim f n dµ. (6.3) n Proof. [The proof is not exminble. We use the proof in W. Rudin: Rel nd Complex Anlysis, Third dition, pge 21]. Since f n f, f is mesurble nd tkes vlues in [, ]. Moreover f n f n+1 f so tht f ndµ f n+1dµ fdµ. Hence lim n f ndµ exists (but my be ), nd lim n f ndµ fdµ. We next prove the reversed inequlity tht lim n f ndµ fdµ. Suppose ϕ S + (,F) nd tht ϕ f. Let λ (,1), nd let n = {f n λϕ}. Since f n, n nd n=1 n =. In fct, for ny x, if f(x) = then x 1. If f(x) >, 27
28 then λϕ(x) < f(x) (s ϕ f nd λ < 1), since f n f, there is N such tht f n (x) λϕ(x) for ll n N (though in generl N depends on x) so tht x n for n N. Thus in ny cse x n=1 n, nd therefore n=1 n =. Since it follows tht f n dµ f n f n 1 n λϕ1 n f n 1 n dµ Suppose ϕ = k c i1 Ai where c i nd A i re mesurble, then λϕ1 n dµ = λ ϕ1 n dµ. (6.4) ϕ1 n = k c i 1 Ai 1 n = k c i 1 Ai n so tht ϕ1 n dµ = k c i µ(a i n ). Since A i n A i s n, thus [Proposition 2.4, item 2)] µ(a i n ) µ(a i ) s n for i = 1,,k. Therefore ϕ1 n dµ k c i µ(a i ) = ϕdµ s n. Letting n in (6.4) we obtin lim n f ndµ λ ϕdµ for every ϕ S+ (,F) such tht ϕ f on, which implies tht lim f n dµ λ fdµ. n Since λ (,1) rbitrry, we must hve lim n f ndµ fdµ. Therefore lim n f ndµ = fdµ. Let us drw severl useful consequences which follow directly from MCT. Corollry 6.7 Suppose f,g : [, ] re mesurble. 1) There is sequence of simple non-negtive functions ϕ n S + (,F) such tht ϕ n f on nd fdµ = lim n ϕ ndµ. 2) We hve (f +g)dµ = fdµ+ gdµ nd λfdµ = λ fdµ for every number λ. 28
29 Proof. 1) follows from Theorem 5.9 nd MCT directly. By Theorem 5.9, we my choose ϕ n,ψ n S + (,F) such tht ϕ n f nd ψ n g. Then ϕ n +ψ n f +g, so tht ( ) (f +g)dµ = lim (ϕ n +ψ n )dµ = lim ϕ n dµ+ ψ n dµ n n = lim ϕ n dµ+ lim ψ n dµ = fdµ+ gdµ n n which proved 2). Theorem 6.8 Let (,F,µ) be mesure spce. Then L 1 (,F,µ) is vector spce, nd f fdµ is liner on L1 (,F,µ). Proof. If f,g L 1 (,F,µ), then f+ dµ < nd f dµ <. Since (f +g) + f + +g +, (f +g) f +g so tht (f + g)+ dµ < nd (f + g) dµ <, which shows tht f + g L 1 (,F,µ). Moreover (f +g) + (f +g) = f + f +g + g so tht thus Rerrnge to obtin (f +g) + dµ+ (f +g) + +f + +g + = (f +g) +f +g, f + dµ+ g + dµ = (f +g)dµ = (f +g) dµ+ fdµ+ gdµ. f dµ+ g dµ. Suppose λ is constnt, then λf = λf + λf, thus if f L 1 (,F,µ), then both λf +,λf L 1 (,F,µ), so tht by definition λf is integrble λfdµ = λf + dµ λf dµ = λ fdµ. Since ( f) + = f nd ( f) = f +, so tht f is integrble by definition, nd ( f)dµ = fdµ. Therefore f fdµ is liner, L1 (,F,µ) is vector spce. Theorem 6.9 (MCT for series of non-negtive mesurble functions, B. Levi) Let f n : [, ] be F-mesurble. Then f n dµ = n=1 n=1 Therefore, n=1 f n L 1 (,F,µ) if nd only if n=1 29 f n dµ. f ndµ <.
30 Proof. Apply MCT to the sequence of prtil sum n f i n=1 f n to obtin n n f n dµ = lim f i dµ = lim f i dµ = f n dµ. n n n=1 n=1 Theorem 6.1 Suppose n re F-mesurble (n = 1,2, ) nd re disjoint. Let = n=1 n. Suppose f : [, ] is mesurble. Then fdµ = n=1 n fdµ. Proof. Let f n = n f1 i. Then f n re mesurble nd f n f on. Therefore, ccording to MCT, n fdµ = lim f n dµ = lim f1 i dµ n n n n = lim f1 i dµ = lim fdµ = fdµ. n n i i Corollry 6.11 Suppose h : R [, ] is Lebesgue mesurble. Define µ h () hdm for = ny M Leb. Then µ h is mesure on (R,M Leb ). dµ is denoted by hdm, nd h is clled the density of the mesure µ with respect to the Lebesgue mesure m. The mesure µ h is bsolutely continuous with respect to the Lebesgue mesure in the sense tht if m(a) =, then µ h (A) =. Suppose f : R [, ] is Lebesgue mesurble, then fdµ h = fhdm. R Suppose A is null set, i.e. µ(a) =, then ny function f on A is F-mesurble on the mesure spce (A,A F,µ). If ϕ S + (A,F) is simple function on A, with representtion ϕ = k c i1 Ai where A i A, then A ϕdµ = R k c i µ(a i ) = which yields tht A f+ dµ = A f dµ fdµ = for every null set A. Thus we must hve by = A fdµ = for ny null set A, for ny function f defined on A. definition A If f,g : [, ] be two functions. Suppose f is F-mesurble nd suppose f = g lmost everywhere on. Let A = {f g}. Then µ(a) =. g is F-mesurble too. Thus both A nd \A re F-mesurble. Now g + dµ = g + dµ+ g + dµ = g + dµ = f + dµ = f + dµ \A A nd similrly, f dµ = g dµ. Therefore f L 1 (,F,µ) if nd only if g L 1 (,F,µ), nd in this cse fdµ gdµ. Therefore, the definition of Lebesgue integrls on mesurble = set pplies to mesurble functions which my be well defined on only lmost surely. 3 \A \A
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