Math 301 Integration I

Mth 301 Integrtion I Lecture notes of Prof. Hichm Gebrn hichm.gebrn@yhoo.com Lebnese University, Fnr, Fll 2016-2017 http://fs2.ul.edu.lb/mth.htm http://hichmgebrn.wordpress.com

2 Introduction nd orienttion In previous courses, you studied the Riemnn integrl. With it, you cn integrte mny functions nd you know for exmple tht every piecewise continuous function cn be integrted. There re however functions tht do not possess Riemnn integrl. Consider for instnce the function f : [0, 1] IR defined by f(x) = 1 if x is rtionl nd f(x) = 0 if x is irrtionl. We shll see tht this function is not integrble in the sense of Riemnn. It does hve however n integrl clled the Lebesgue integrl. The Lebesgue integrl is generliztion of the Riemnn integrl: it permits to integrte more functions. One of our trgets in this course is to construct the Lebesgue integrl nd study its properties. There re mny wys to present the subject. We hve chosen the pproch bsed on mesure theory. A mesure is generliztion of the fmilir notions of length, re, volume nd probbility of n event. In this course we shll construct mesure on IR clled the Lebesgue mesure which ssigns length to mny subsets of IR tht re not necessry intervls. Let me mke some remrks bout the subject before we strt. They will become cler s we progress. 1. As we sid, the Lebesgue integrl permits to integrte wider clss of functions thn the Riemnn integrl permits. Also, the sets on which we integrte need not be intervls. 2. The theorems of the Lebesgue theory re stronger nd esier to use thn those of the Riemnn theory. 3. There is n nlogy between the completion of the rtionl numbers by rel numbers nd the completion of Riemnn integrble functions by Lebesgue integrble functions. 4. The Lebesgue theory of mesure nd integrtion is fundmentl to mny fields of Mthemtics like probbility, functionl nlysis, dynmicl systems nd Fourier series. All questions, comments, remrks nd suggestions re welcome. References Pul. R. Hlmos, Mesure theory (Princeton, Vn Nostrnd, 1950). Roger Jen, Mesure et intégrtion (Presses de l Université du Québec, 1982). Wlter Rudin, Rel nd complex nlysis (McGrw-Hill, 1977). Mrc Troynov, Mesure et intégrtion (Lecture notes, PFL, 2005). Thierry Glly, Théorie de l mesure et de l intégrtion (Lecture notes, Université Joseph Fourier, 2009).

Contents 1 Preliminries 5 1.1 Sets nd functions................................... 5 1.2 Topology........................................ 8 1.3 Some fundmentl properties of the rel line nd the extended rel line..... 8 1.3.1 The rel line.................................. 8 1.3.2 The extended rel line............................. 10 1.3.3 Superior limit nd inferior limit........................ 13 2 Axiomtic mesure theory 15 2.1 Mesurble spces................................... 16 2.2 Mesure spces..................................... 18 2.3 Mesurble functions.................................. 23 2.3.1 Mesurble rel vlued functions....................... 25 2.3.2 Simple functions................................ 27 2.4 Outer mesures nd Crtheodory s theorem.................... 29 2.5 Completion of mesure spce............................ 32 3 The Lebesgue mesure on IR 35 3.1 Construction nd properties.............................. 35 3.2 Counterexmples.................................... 41 3.2.1 A Lebesgue non mesurble set........................ 41 3.2.2 The middle third Cntor set.......................... 41 4 The Lebesgue integrl 45 4.1 Construction nd properties.............................. 45 4.1.1 Simple fonctions................................ 45 4.1.2 Nonnegtive mesurble functions...................... 48 4.1.3 Summble functions.............................. 53 4.1.4 Complex vlued functions........................... 54 4.2 The Lebesgue dominted convergence theorem................... 55 4.3 Reltions with the Riemnn integrl......................... 57 4.4 Some pplictions................................... 60 A The Riemnn integrl 63 A.1 Definitions........................................ 63 A.2 Criteri of integrbility................................. 64 A.3 Clsses of integrble functions............................. 67 A.4 Properties of the integrl............................... 69 A.5 Integrtion nd differentition............................. 69 A.6 Limits nd integrtion................................. 71 3

4 CONTNTS

Chpter 1 Preliminries 1.1 Sets nd functions Given set we denote by P() or 2 the set of ll subsets of. The reson for the nottion 2 is tht if is finite nd contins n elements, then there re 2 n subsets of. Let be set nd let A. The chrcteristic function of A is the function χ A defined by { 1 if x A χ A (x) = 0 if x / A. It is lso denoted by 1 A. The complement of A in is the set of points in tht re not in A. It is denoted by one of the following nottions A, A, \A, A, A c. Given fmily of sets (A i ) i I, we set A i = {x j I such tht x A j }, Recll De Morgn s lws i I A i = {x x A i, i I}. i I \ i I \ i I A i = i I(\A i ) A i = i I(\A i ), nd the distributivity lws A A i = A i ) i I i I(A A A i = A i ). i I i I(A Let f : Y be mp between two sets. Let A, the (direct) imge of A under f is f(a) = {y Y x such tht y = f(x)}. Let B Y, the inverse imge of B under f is f 1 (B) = {x f(x) B}. 5

6 CHAPTR 1. PRLIMINARIS Regrdless of wether f is bijection or not, f 1 defines mp This mp stisfies 1. f 1 ( ) = nd f 1 (Y ) =. 2. f 1 (A c ) = (f 1 (A)) c f 1 : P(Y ) P() B f 1 (B). 3. f 1 ( i I B i ) = i I f 1 (B i ) nd f 1 ( i I B i ) = i I f 1 (B i ). We use cpitl letters to denote sets nd clligrphic letters to denote sets of sets. If B is collection of subsets of Y, then we write f 1 (B) = {f 1 (B) B B). This is the direct imge under f 1 of the set B. Note tht if (A i ) i I is fmily of subsets of, then f( i I A i ) = i I f(a i ), but we only hve f( i I A i ) i I f(a i ), the equlity holds if f is one to one (injective). Countble sets A set is clled countble if there is one to one function f : IN where IN is the set of positive integers. This mens tht we cn put the elements of in finite or infinite sequence (x 1, x 2,...). So for exmple the set of ll integers Z nd the set of ll rtionl numbers Q re countble. A countble union of countble sets is countble, tht is, if I is countble nd A i is countble for ll i I, then i I A i is countble. Also finite crtesin product of countble sets is countble, so for exmple IN IN is countble. However P(IN) nd the set of rel numbers IR re uncountble. Limsup nd liminf Let be set nd (A n ) n IN be sequence of subsets of. We set nd lim inf A n = lima n = lim sup A n = lima n = A k, n IN k n A k, n IN k n Therefore, x lim inf A n if strting from s certin rnk, x belongs to ll the A n. In probbility theory, the event lim inf A n is clled the event tht the A n hppen lmost lwys (A n..). On the other hnd, x lim sup A n if x belongs to A n for infinitely mny indices n. In probbility theory, the event lim sup A n is clled the event tht the A n hppen infinitely often (A n i.o.). xmple. Consider the experiment of flipping fir coin infinitely mny times. Let A n be the event: Hed ppers on the n th flip. Then

1.1. STS AND FUNCTIONS 7 1. A n is the event: hed ppers on ll tosses or {(HH... H...)}. 2. lim inf A n is the event: hed ppers strting from certin flip. It is the union of ll events of the form (HH... H...) (HH... H...) (HH... H...)... 3. lim sup A n is the event: hed ppers infinitely mny times. This event is the complement of the event: til ppers strting from certin flip. 4. A n is the event: hed ppers t lest once. Now we clim tht n IN A n lima n lima n Proof. For the first inclusion, let B n = k n A k. Then the first inclusion is equivlent to B 1 n IN B n which is obviously true. For the third inclusion, let C n = k n A k. Then the third inclusion is equivlent to n IN C n C 1 which is obviously true. Now for the second inclusion observe tht x lim inf A n x A k n k n x A k {n IN x / A n } is finite, nd tht x lim sup A n x n IN k n n IN k n n IN A n. A k n k n x A k {n IN x A n } is infinite. Now if {n IN x / A n } is finite then its complement {n IN x A n } is infinite becuse IN is infinite. Hence the inclusion. We sy tht (A n ) converges to A if lima n = lima n = A. The xiom of choice In chpter 3, we shll need the following xiom. Axiom of choice. Let (A i ) i I be fmily of nonempty nd pirwise disjoint sets. Then there exists set C such tht for ll i I, C A i is singleton. Otherwise stted, given fmily of sets s bove, we cn select exctly one element form ech set. This xiom seems trivil (nd this is why we cll it xiom). It hs however some fr reching nd unexpected consequences like the fct tht ny set cn be well ordered (this is known s the well ordering theorem). An ordered set (, ) is sid to be well ordered if every nonempty subset of hs smllest element. For exmple, IN is well ordered in the usul order. However Z nd IR re not well ordered in the usul order. The well ordering theorem implies tht there exists well ordering on IR, fct tht no one cn prove without the xiom of choice. Actully the xiom of choice is equivlent to the well ordering theorem. The well ordering theorem strtled the mthemticl world in the beginning of the twentieth century. A creful nlysis of its proof led to the formultion of the xiom of choice. The xiom of choice sserts the existence of set but gives no procedure to construct it, nd this is why some mthemticins of the beginning of the twentieth century refused it. Nowdys, most mthemticins ccept it nd we shll use it in chpter 3 to construct nonmesurble subset of IR.

8 CHAPTR 1. RVIW AND COMPLMNTS 1.2 Topology Let be set. A fmily T of subsets of (tht is, T P()) is clled topology on provided the following conditions hold. 1., T. 2. If O i T for ll i I, then i I O i T. 3. If O 1, O 2 T, then O 1 O 2 T. The elements of T re clled open sets of. A set is clled closed if its complement is open. If d is distnce on, then d genertes topology on. A function f : Y between two topologicl spces is clled continuous t point x 0 if for every neighborhood V of f(x 0 ), there exists neighborhood of x 0 such tht f(u) V. A function f is clled continuous on if it is continuous t every point of. Then the following conditions re equivlent. (i) f : Y is continuous. (ii) The inverse imge under f of every open set of Y is n open set of. (iii) The inverse imge under f of every closed set of Y is closed set of. A subset A of metric spce is clled dense if the closure of A is equl to. following conditions re equivlent. 1. A is dense in. The 2. very open set of meets A. 3. For every x, there exists sequence in A tht converges to x. Let A be subset of topologicl spce, n open covering of A is fmily of open sets whose union contins A. The subset A is clled compct if ny open covering of A contins finite subcollection tht lso covers A. In IR n, subset is compct if nd only if it is closed nd bounded. A subset A of topologicl spce is connected if the following condition holds. If A B C with B nd C open nd disjoint, then either A B or A C. The union of connected sets hving point in common is connected. The connected subsets of IR re the intervls. A connected component of topologicl spce is mximl (with respect to inclusion) connected subspce, i.e., connected subspce which is not contined properly in bigger connected subspce. 1.3 Some fundmentl properties of the rel line nd the extended rel line 1.3.1 The rel line We strt by formulting two fundmentl properties of the set of rel numbers tht we shll use in the sequel. This set is denoted by IR nd it is nturlly identified to line. An upper bound of subset IR is rel number M such tht x M for ll x. The set is clled bounded from bove if it hs n upper bound. A lower bound of subset IR is rel number m such tht m x for ll x. The set is clled bounded from below if it hs lower bound.

1.3. SOM FUNDAMNTAL PROPRTIS 9 Definition 1.1 Let IR be bounded from bove. The number α is clled the lest upper bound of if the following hold. i) α is n upper bound of. ii) If β < α, then β is not n upper bound of. In this cse, we write α = sup. If is not bounded from bove, we write sup = +. Proposition 1.1 Let IR be bounded from bove. following hold. i) x α for ll x. ii) ε > 0 y such tht α ε < y. Then α = sup if nd only if the Definition 1.2 Let IR be bounded from below. The number α is clled the gretest lower bound of if the following hold. i) α is lower bound of. ii) If β > α, then β is not lower bound of. In this cse, we write α = inf. If is not bounded from below, we write inf =. Proposition 1.2 Let IR be bounded from below. following hold. i) x α for ll x. ii) ε > 0 y such tht α + ε > y. Then α = inf if nd only if the Now we cn stte the two fundmentl properties of IR. Theorem 1.1 Any nonempty subset of IR which is bounded from bove hs lest upper bound. Any nonempty subset of IR which is bounded from below hs gretest lower bound. Any proof of this theorem involves going bck to the construction of the rel numbers from the rtionl numbers. We do not ddress this issue here. Recll now tht subset IR is n intervl if [x, y] whenever x nd y belong to. An intervl hs thus one of the following eleven forms, {}, ], b[, ], b], [, b[, [, b], ], ], ], [, ], + [, [, + [ nd IR. We shll sy tht n intervl is trivil if it is empty or singleton. Theorem 1.2 Any nontrivil intervl of the rel line contins rtionl s well s irrtionl numbers. This theorem is not difficult to prove using the following evident fcts: 1- for ech rel number, there is n integer n bigger thn (this is known s the Archimeden property); 2- every nonempty subset of IN hs smllest element (IN is well ordered). Proposition 1.3 Let nd b be two rel numbers. If b + ε for ll ε > 0 then b. Theorem 1.3 very open set O of the rel line is the union of countble fmily of pirwise disjoint open intervls. Proof. Let (I λ ) λ L be the collection of connected components of O. We know tht ech I λ is n open intervl (recll tht in loclly connected spce, connected component of n open set is open). Since the rtionl numbers re dense in IR, ech I λ contins rtionl number. By the xiom of choice, we cn choose exctly one rtionl number in ech I λ. This defines function between L nd Q which is one to one becuse two distinct components re disjoint. Therefore L is countble. Finlly we hve indeed O = λ L I λ.

10 CHAPTR 1. RVIW AND COMPLMNTS 1.3.2 The extended rel line In mesure n integrtion we will hve to del with sets of infinite mesure nd with unbounded functions. So we need to extend the set of rel numbers. The extended rel line denoted by IR or [, + ] is obtined from IR by dding two objects + nd (which re not rel numbers). Rules in IR I. Order. We extend nturlly the usul order on IR by letting < x < + for every x IR This order is totl, tht is ny two numbers re comprble. Also every nonempty subset of IR hs lest upper bound (sup) nd gretest lower bound (inf). II. Arithmetic opertions. 1. x + = +, for ll x ], ]. 2. x =, for ll x [, [. { + if x ]0, + ] 3. x (+ ) = if x [, 0[. 4. 0 ± = 0. This is convenient convention in mesure theory. The reson is tht we wnt to hve A 0 = 0 even if A hs n infinte mesure. Wrning. The following opertions re not defined:, +, ± ±. III. The topology of IR. We define distnce on IR by setting d(x, y) = rctn x rctn y with the convention tht rctn(+ ) = π 2 nd rctn ( ) = π 2 so tht for exmple d(x, + ) = rctn x π if x IR nd d(+, ) = π. 2 Remrk 1.1 Let (x n ) be sequence of IR. Recll tht (x n ) is sid to tend to + if for every A > 0, we hve x n > A for ll n lrge enough. It is proved in clculus tht x n if nd only if rctn x n π 2. It follows tht x n + if nd only if lim d(x n, + ) = 0 in IR, which mens tht (x n ) converges to the point + in the topology of IR. Similrly, x n if nd only if lim d(x n, ) = 0. Here re some properties of the topology of IR. 1. The restriction of d to IR IR is distnce tht genertes the usul topology of IR. Indeed, let first O be open in (IR, d) nd let x O. Then there is r > 0 such tht B d (x, r) := {y d(x, y) < r} O}. But B(x, r) := {y x y < r} B d (x, r) since d(x, y) = rctn x rctn y x y. Therefore B(x, r) O. Hence O is open for the usul topology which therefore contins the topology generted by d. Conversely, let O be open for the usul topology of IR nd let x O. Then there is ε > 0 such tht B(x, ε) O. Now since the function t tn t is continuous t rctn x, there is δ > 0 such tht tn rctn x tn z < ε whenever z rctn x < δ. We clim tht B d (x, δ) B(x, ε). Indeed, d(x, y) < δ rctn x rctn y < δ x y = tn rctn x tn rctn y < ε. Therefore, O is open for d. 2. IR is homeomorphic to [ π 2, π 2 ] nd therefore to ny compct intervl [, b] ( < b) of IR. Therefore IR is compct nd connected. Indeed, the mp h : IR [ π 2, π 2 ] defined by h(x) = rctn x. First, h is clerly bijection. Next, h is continuous.

1.3. SOM FUNDAMNTAL PROPRTIS 11 Let x IR nd let x n x, then rctn x n rctn x, i.e., h(x n ) h(x). Let x = + nd let x n +, then h(x n ) π 2 = h(+ ). Let x = nd let x n, h(x n ) π 2 = h( ). The inverse mp is given by tn x if x ] π h 1 2, π 2 [ (x) = + if x = π 2 if x = π 2. The proof of the continuity of h 1 is similr to the proof of continuity of h. 3. IR is dense in IR. This follows from the fct tht IR is homeomorphic to ] π 2, π 2 [ which is dense in [ π 2, π 2 ]. Theorem 1.4 very open set O of IR is the union of countble fmily of pirwise disjoint open intervls of IR. Proof. Observe first tht the connected subsets of IR re exctly the intervls of IR. Next, n open intervl of IR contins n open intervl of IR which therefore contins rtionl points. Therefore the proof of Theorem 1.3 cn be repeted. We will hve to del lter with sequences nd series in IR. So it is importnt to formulte some rules for using them. Proposition 1.4 A monotonic sequence (x n ) of IR is convergent. Moreover i) If (x n ) is nondecresing then lim x n = sup{x i i IN }. ii) If (x n ) is nonincresing, then lim x n = inf{x i i IN }. Proof. Let (x n ) be monotonic sequence in IR. Let y n = h(x n ) where h : IR [ π 2, π 2 ] is the homeomorphism defined bove. Then (y n ) is lso monotonic nd therefore convergent (since it is bounded). Continuity of h 1 implies tht (x n ) is convergent. i) Let (x n ) be nondecresing. We distinguish between two cses 1) sup x i < +. We distinguish gin between two cses. ) x n = for ll n IN. Then indeed, lim x n = sup x i =. b) x n0 IR for some n 0. But then x n IR for ll n n 0 becuse the sequence is nondecresing. We modify the sequence by letting x n = x n0 for n n 0. This does not chnge neither the limit nor the sup nd we hve x n IR for ll n. Let ε > 0 be given. By property of the supremum, there exists k IN such tht sup x i ε < x k. Then sup x i ε < x n for ll n k since the sequence is nondecresing. But x n sup x i. It follows tht x n sup x i < ε for ll n k nd this mens tht lim x n = sup x i. 2) sup x i = +. We distinguish gin between two cses.

12 CHAPTR 1. RVIW AND COMPLMNTS ) x n < + for ll n IN. The fct tht sup x i = + mens tht the set {x i i IN } is not bounded from bove. This in turn mens tht for ll M > 0, there exists k such tht x k M. Then x n M for ll n k. This mens tht lim x n = + nd so lim x n = sup x i. b) x n0 = for some n 0. Then x n = for ll n n 0. In this cse, we hve lim x n = + nd so lim x n = sup x i. ii) If (x n ) is nonincresing, then ( x n ) is nondecresing. Therefore lim( x n ) = sup( x i ) = inf(x i ) nd so lim x n = inf x i. Let (x n ) be sequence of [0, + ]. Then the sequence S n = n k=1 x k is monotonic. It is therefore convergent. We set k=1 x k = lim S n. This extends the definition of convergent series of rel numbers. Therefore series of nonnegtive terms re lwys convergent in IR. Note tht the condition k=1 x k < + mens tht x k IR for ll k nd tht the series is convergent in the usul sense. The fmilir rules of convergent series still holds. For exmple, if (x n ) nd (y n ) re sequences in [0, + ] then 1. (x n + y n ) = x n + y n. 2. αx n = α x n for α 0. 3. ( n IN, x n y n ) x n y n. Lemm 1.1 The sum of series of nonnegtive terms does not depend on the order of summtion. Proof. Let (x n ) be sequence of [0, + ] nd let σ : IN IN be bijection. We hve to prove tht x n = x σ(n). Let n be given nd let N = mx{σ(1),, σ(n)}. Then {σ(1),, σ(n)} {1,, N}. It follows tht n k=1 x σ(k) N k=1 x k k=1 x k becuse x k 0. Letting n we get k=1 x σ(k) k=1 x k. By symmetry k=1 x k k=1 x σ(k). Hence the equlity. Remrk 1.2 If the sequence (x n ) is of vrible sign, then the sum of the series x n my depend on the order of integrtion. Here is n exmple. We shll prove lter tht However, 1 1 2 + 1 3 1 4 + + 1 2k 1 1 + = ln 2. 2k 1 1 2 1 4 + 1 3 1 6 1 8 + + 1 2k 1 1 4k 2 1 4k + = 1 ln 2. 2 Indeed, let S n be the n th prtil sum of the first series nd let S n be the n th prtil sum of the second series. Then S 3n = n k=1 ( 1 2k 1 1 4k 2 1 ) = 4k n k=1 ( 1 4k 2 1 ) = 1 4k 2 S 2n. Therefore S 3n 1 2 ln 2. Since S 3n 1 = S 3n + 1 4n nd S 3n 2 = S 3n 1 + 1 4n 2, it follows tht S 3n 1 nd S 3n 2 lso converge to 1 2 ln 2. Consequently, S n converges to 1 2 ln 2.

1.3. SOM FUNDAMNTAL PROPRTIS 13 Let now A be n infinite countble set nd let (x ) A be fmily of [0, + ]. Then there exists bijection ϕ : IN A. We set x = x ϕ(n). A The previous lemm ensures tht this definition mkes sense. Using rguments similr to those used in the proof of Lemm 1.1, one cn prove the following Proposition 1.5 Let (x n,m ) be double sequence of [0, + ]. Then (n,m) IN IN x n,m = m=1 1.3.3 Superior limit nd inferior limit x n,m = m=1 x n,m. Let (x n ) be n rbitrry sequence in IR. Let y n = sup{x k k n}. Then (y n ) is nonincresing nd therefore hs limit in IR. This limit is clled the superior limit of the sequence {x n } nd it is denoted by limx n or lim sup x n. Thus, lim sup x n = inf sup n 1 k n x k = lim sup k n The inferior limit of sequence x n denoted by limx n or lim inf x n is defined by n 1 x k. lim inf x n = sup inf x k = lim inf x k. k n k n xmples. 1. Let x n = ( 1) n. Then y n := inf{x k k n} = 1. Hence lim inf x n = lim y n = 1. On the other hnd, z n := sup{x k k n} = 1. Hence lim sup x n = lim z n = 1. 2. Consider the sequence (1, 2, 1 + 1 2, 2 + 1 2, 1 + 1 3, 2 + 1 3, 1 + 1 4, 2 + 1 4,...). Then lim sup x n = 2 nd lim inf x n = 1. Theorem 1.5 The lim sup nd lim inf stisfy the following properties. (i) x n y n lim inf x n lim inf y n nd lim sup x n lim sup y n. (ii) lim inf x n lim sup x n. (iii) lim inf x n = lim sup x n = l if nd only if lim x n = l. (iv) If x n 0 then lim x n = 0 lim sup x n = 0. (v) lim sup x n is the biggest limit point of (x n ) nd lim inf x n is the smllest limit point of (x n ). Proof. (i). We hve sup{x k k n} sup{y k k n} for ll n. Letting n, we get lim sup x n lim sup y n. The proof for lim inf is similr. (ii) For ll n, we hve inf{x k k n} sup{x k k n}. Letting n, we get the result. (iii) Suppose first tht lim inf x n = lim sup x n = l. If l = + then the sequence y n := inf k n x k tends to. But x n y n nd so x n. If l =, then z n := sup k n x k tends to. But x n z n nd so x n. Now we ssume tht l IR. Let ε > 0. Then sup x k l < ε for k n ll n lrge enough. In prticulr x n sup x k < l + ε. Similrly, inf x k l < ε for n lrge k n k n

14 CHAPTR 1. RVIW AND COMPLMNTS enough. In prticulr x n inf k n x k > l ε. Thus, l ε < x n < l + ε for ll n lrge enough. This proves tht lim x n = l. Conversely, ssume tht lim x n = l. If l = +, then sup k n x k since sup k n x k x n. Therefore lim sup x n = +. Now since x n, we hve for ll A > 0, there is n 0 such tht x n A for ll n n 0 nd so inf k n x k inf k n0 x k A. Letting n we get lim inf x n A. Since A is rbitrry, we conclude tht lim inf x n = +. The cse l = is similr. Now we ssume tht l IR. Let ε > 0 be given. Then there is n 0 such tht l ε < x k < l + ε for ll k n 0. Thus l ε inf x k sup x k l + ε k n for ll n n 0. Letting n, we get l ε lim sup x n l + ε nd l ε lim inf x n l + ε. Since ε is rbitrry, we conclude tht lim sup x n = lim inf x n = l. (iv) If lim x n = 0 it follows from prt (iii) tht lim sup x n = 0. Now conversely, suppose tht lim sup x n = 0. Since x n 0, it follows tht lim inf x n 0. But lim inf x n lim sup x n = 0. Therefore lim sup x n = lim inf x n = 0. Thus, by prt (iii) lim x n = 0. (v) We prove the result for lim sup, the cse of lim inf being similr. We need to show first tht if L := lim sup x n, then there is subsequence (x kp ) tht converges to L. We consider the cse L IR, the other cses being left s n exercise. We hve the following: nd ε > 0 n 0 such tht L sup x k < L + ε n n 0 (by definition of the limsup) k n k n ε > 0 n m n such tht x m > sup x k ε k n (by property of the sup). Thus, we hve the following condition ε > 0 n m n such tht L + ε > x m > L ε. In prticulr, for ε = 1 nd n = 1, k 1 1 such tht L + 1 > x k1 > L 1. Next tke ε = 1/2 nd n = k 1 + 1. Then there exits k 2 k 1 + 1 such tht L + 1 2 > x k 2 > L 1 2. At the pth step, we get n index k p k p 1 + 1 such tht L + 1 p > x k p > L 1 p. This defines subsequence (x k p ) tht converges to L. Let now (x ϕ(n) ) be subsequence tht converges to l. We hve {x ϕ(k) ) k n} {x k k n}. Therefore, sup k n x ϕ(k) sup k n x k. Letting n, we get lim sup x ϕ(n) lim sup x n, tht is, lim x ϕ(n) lim sup x n.

Chpter 2 Axiomtic mesure theory Our trget in this chpter is to develop generl theory of mesure tht includes the notions of length, re, volume nd probbility s specil cses. So let us strt by formulting some nturl requirements tht mesure hs to posess. Consider rndom experiment nd let denote its smple spce, tht is, the set of ll possible outcomes (in probbility theory, is usully denoted by Ω). The subsets of re ususlly clled events. However, we my be interested in ssigning probbility only to some specil subsets of ; we will cll them events. Let A denote the collection of events (to which probbility will be ssigned). First, it is nturl to ssign probbility 1 to the certin event. So should belong to A nd P () = 1. We would like lso to ssign probbility 0 to the impossible event. So should belong to A nd P ( ) = 0. Second, given n event A A, we would like to consider its complement \A s n event nd set P (\A) = 1 P (A). Third, given two events A nd B, we would like to consider their union A B s n event. If moreover A nd B re disjoint we would like to hve P (A B) = P (A) + P (B). It will follow by induction tht if A 1,..., A n re events then their union is n event. Moreover, if these events re disjoint, we would hve P (A 1 A n ) = P (A 1 ) + + P (A n ). Now sometimes we hve to del with n infinite but countble collection of events (A n ) nd we would like to consider their union s n event. If moreover these events re disjoint we would like to hve P (A 1 A n ) = P (A 1 ) + + P (A n ) + Let us summrize our requirements for collection of events A. (i) A. (ii) A. (iii) If A A then \A A. (iv) If (A n ) is sequence of A then A n A. A probbility P on the collection of events A should stisfy 1. P () = 1 2. P ( ) = 0. 3. P (\A) = 1 P (A). 4. P ( A n) = P (A n) if A n re pirwise disjoint. Wht we sid bout events nd probbilities cn be formulted for subsets of the plne nd their re. However, the re of the plne should be considered s infinite, so the requirement 15

16 CHAPTR 2. AIOMATIC MASUR THORY 1. bove is not essentil to our generl theory. Also note tht requirement (ii) is redundnt becuse it cn be deduced from (i) nd (iii). Now we re sufficiently motivted to strt developing our generl theory. A collection of subsets hving properies (i)-(iv) bove is clled σ lgebr. If in requirement (iv) we only consider finite sequences, then the collection is clled n lgebr. 2.1 Mesurble spces Definition 2.1 Let be set. A σ lgebr on is collection A of subsets of (tht is, A P()) stisfying the following properties. (i) A. (ii) A A \A A. We sy tht A is closed or stble under complementtion. (iii) If (A n ) is sequence of A, then countble unions. A n A. We sy tht A is closed or stble under xmples. ) Let be set. Then P() is indeed σ lgebr on. It is the biggest σ lgebr on. b) {, } is σ lgebr on. It is the smllest σ lgebr on. c) Let A. Then {, A, A c, } is σ lgebr on. d) The collection A of ll subsets of IR which re either countble or hve countble complement is σ lgebr on IR. Indeed, (i) A becuse c = is countble. (ii) If is countble or c is countble, then c is countble or = ( c ) c is countble (this condition is symmetric in nd c ). (iii) Let A n be sequence of subsets of IR tht re either countble or hve countble complement. If ll A n re countble then A n is lso countble. If not, some A m is uncountble. But then A c m is therefore countble. Now ( A n) c = Ac n A c m is countble. e) The collection of ll subsets of IN which re either finite or hve finite complement is not σ lgebr on IN. Indeed, tke A n = {2n} then A n is the set of ll even positive integers: it is infinite nd its complement is infinite. Definition 2.2 A mesurble spce is couple (, A) where is set nd A is σ lgebr on. The elements of A re clled the mesurble subsets of. Proposition 2.1 Let A be σ lgebr on. Then 1. A. 2. A n A for n = 1, 2,... intersections. A n A. We sy tht A is closed or stble under countble 3. A n A for n = 1, 2,..., N 4. A n A for n = 1, 2,..., N N A n A. We sy tht A is stble under finite intersections. N A n A. We sy tht A is stble under finite unions.

2.1. MASURABL SPACS 17 Proof. 1. A by the first property of σ lgebr. Therefore, = \ A by the second property. 2. Let A n A for n = 1, 2,.... Then \A n A for ll n by the second property. Therefore \A n A by the third property. But \A n = \ A n. Therefore \ A n A. Thus A n A by the second property. 3. Set A n = for n > N. Then A n A for ll n IN. By property (iii), A n A. But N A n = A n. 4. Set A n = for n > N. Then N A n = A n A by 2. Remrk 2.1 Let (, A) be mesurble spce nd D. Then A D := {D A} is σ lgebr on D. Thus (D, A D ) is mesurble spce tht we my cll subspce of (, A). Note tht A D P(D) A = {F D F A}. If in ddition D A (i.e., D is mesurble), then A D = P(D) A. Lemm 2.1 An intersection of σ lgebrs on set is σ lgebr on. Proof. Let (A i ) i I be fmily of σ lgebrs. We need to show tht A = i I A i is σ lgebr. (i) A since A i for ll i I. (ii) If A A, then A A i for ll i I nd so \A A i for ll i since A i is σ lgebr. Therefore, \A A. (iii) If A n A for ll n = 1, 2..., then A n A i for ll n nd ll i. Therefore A n A i for ll i I since A i is σ lgebr. Thus, A n A. Thnks to this lemm we now cn define the notion of σ lgebr generted by fmily of sets. Definition 2.3 Let be set nd S P() be collection of subsets of. The intersection of ll σ lgebrs contining S is clled the σ lgebr generted by S. It is the smllest (with respect to inclusion) σ lgebr tht contins S. It is denoted by σ(s). xmples. ) If (, T ) is topologicl spce, then the σ lgebr σ(t ) generted by the open sets of is clled the Borel σ lgebr of (, T ). It is lso denoted by B(, T ) or just B() if no confusion rises. An element B B() is clled Borel set or Borel mesurble set. b) Let A be σ lgebr on set nd B be σ lgebr on set Y. We denote by A B the σ lgebr generted by the following fmily {A B Y such tht A A nd B B}. It is clled the product σ lgebr on Y. Remrk 2.2 Let be set. 1. If A is σ lgebr on, then σ(a) = A. Indeed, first, A σ(a). Next, A is σ lgebr contining A, so A contins the intersection σ(a) of ll σ lgebrs contining A. 2. If S T P(), then σ(s) σ(t ).

18 CHAPTR 2. AIOMATIC MASUR THORY Remrk 2.3 (Methodology) 1. To prove tht A = σ(c), we show tht A σ(c) nd tht C A (which implies tht σ(c) σ(a) = A). 2. To prove tht σ(c 1 ) = σ(c 2 ), we show tht C 1 σ(c 2 ) nd tht C 2 σ(c 1 ). The following importnt results will be proved in the exercises. Proposition 2.2 The Borel σ lgebr on IR, tht we denote by B(IR) or B, is generted by nyone of the following collections: (, b IR or, b Q) i. the open intervls. ii. the open intervls of the form ], b[. iii. the closed intervls of the form [, b]. iv. the intervls of the form ], + [. v. the intervls of the form [, + [. vi. the intervls of the form ], b[. vii. the intervls of the form ], b]. viii. ll the intervls. Proposition 2.3 The Borel σ lgebr of IR tht we denote by B(IR) is generted by nyone of the following collections where (where, b Q or, b IR). i. the intervls of the form ], + ]. ii. the intervls of the form [, + ]. iii. the intervls of the form [, b[. iv. the intervls of the form [, b]. 2.2 Mesure spces Let (, A) be mesurble spce (i.e., set equipped with σ lgebr). Definition 2.4 A mesure on (, A) is function µ : A [0, + ] such tht (i) µ( ) = 0. (ii) If {A n } A is sequence of pirwise disjoint elements of A then ( ) µ A n = µ(a n ). Condition (ii) is clled σ dditivity of the mesure. Remrk 2.4 Condition (i) cn be replced by (i ) There exists A A such tht µ(a) <. Indeed, if µ( ) = 0 then µ( ) <. Conversely, suppose tht µ(a) < for some A A. Letting A 1 = A nd A n = for n 2, we get µ(a) = µ(a) + n=2 µ( ). Since µ(a) is finite, subtrcting it form the equlity, we get n=2 µ( ) = 0. This implies tht µ( ) = 0.

2.2. MASUR SPACS 19 Definition 2.5 A mesure spce is triple (, A, µ) such tht A is σ lgebr on nd µ is mesure on (, A). xmples. In xmples (1) to (4), is ny set nd one cn tke A = P(). (1) The trivil mesure. µ(a) = 0 for ll A A. { 0 if A = (2) The infinite mesure. µ(a) = + if A. Proof. (i) It is cler tht µ( ) = 0. (ii) Let now (A n ) be sequence of pirwise disjoint elements of A. If A n = for ll n, then A n = nd so µ( A n) = 0. On the other hnd, µ(a n) = 0 = 0. If A n for some m, A n nd therefore µ( A n) =. On the other hnd the series is equl to since µ(a n) µ(a m ) =. { 1 if x A (3) The Dirc mesure or Dirc mss. Let x. We set δ x (A) = 0 if x / A. Proof. (i) Since x /, it follows tht δ x ( ) = 0. (ii) Let (A n ) be sequence of pirwise disjoint elements of A. We distinguish between two cses. If x belongs to the union, then it belongs to exctly one A m since the A n re pirwise disjoint. Then δ x ( A n) = 1, δ x (A m ) = 1 nd δ x (A k ) = 0 for k m. Therefore δ x(a n ) = 1. Next, if x does not belong to the union, it does not belong to ny of the A n. So both terms re zero. { crd(a) if A is finite (4) The counting mesure. Let µ(a) = + if not. Proof. (i) µ( ) = crd ( ) = 0. (ii) Let (A n ) be sequence of pirwise disjoint elements of A. We distinguish between two cses. Cse 1. A n is finite. Then there is some N IN such tht A n = for ll n N. Otherwise, crd A n 1 for infinitely mny n. So A n would be infinite. Then A n = N A n nd so ( ) ( N ) ( N ) N N µ A n = µ A n = crd A n = crd A n = µ(a n ) = µ(a n ). Cse 2. A n is infinite. Then µ( A n) = +. Here we lso distinguish between two cses. 2) Some A k is infinite. Then µ(a k ) = + nd so µ(a n) = +. 2b) All A n re finite. But then A n for infinitely mny n becuse A n is infinite. But then µ(a n ) = crd (A n ) 1 for infinitely mny n nd so µ(a n) = +. (5) Restriction of mesure. Let (, A, µ) be mesure spce nd let D A. We define new mesure ν on (, A) by ν(a) = µ(a D). This is indeed mesure which is clled the restriction of µ to D. In fct, ν is lso mesure on (D, A D ) where A D = {D A A A}. (6) Positive liner combintion of mesures. Let (µ n ) be sequence of mesures on mesurble spce (, A) nd let (α n ) be sequence of [0, + ]. Then µ := α nµ n is mesure on (, A). Proof. (i) µ( ) = α n µ n ( ) = 0. (ii) Let (A k ) be sequence of pirwise disjoint elements of A. Then µ( k A k ) = α n µ n ( k A k ) = α n µ n (A k ) = α n µ n (A k ) n n k n k = α n µ n (A k ) we cn interchnge the order of summtion by Proposition 1.5 k n = k µ(a k ).

20 CHAPTR 2. AIOMATIC MASUR THORY In prticulr, if (x n ) is sequence of, then µ := n α nδ xn is mesure. A mesure of this form is clled discrete mesure. In prticulr, if = {x 1,..., x N } is finite then µ := 1 N N δ x n is the fmilir probbility mesure µ(a) = crd A crd. Definition 2.6 Let (, A, µ) be mesure spce. We sy tht 1. µ is finite if µ() <. 2. µ is probbility mesure if µ() = 1. 3. µ is σ finite if = A n with A n A nd µ(a n ) < for ll n IN. xmple. The counting mesure is σ finite on IN but not on IR. Proposition 2.4 (lementry properties of mesures) Let (, A, µ) be mesure spce. () If A 1, A 2,..., A k A re pirwise disjoint, then ( k ) µ A n = k µ(a n ). (b) If B A then µ(a\b) + µ(b) = µ(a). In prticulr, if µ(b) < + then µ(a\b) = µ(a) µ(b). (c) If B A then µ(b) µ(a). (d) µ(a B) + µ(a B) = µ(a) + µ(b). In prticulr, if µ(a B) < +, then µ(a B) = µ(a) + µ(b) µ(a B). (e) If {A n } n IN A then ( ) µ A n µ(a n ). This is clled the σ subdditivity (note tht the A n re not necessrily pirwise disjoint). Proof. (). Set A n = for n > k. Then the sequence {A n } n IN is fmily of pirwise disjoint sets. Now, on the one hnd, k A n = A n. On the other hnd, µ(a n ) = 0 for n > k since µ( ) = 0. Therefore µ(a n ) = k µ(a n ) + By the σ dditivity of µ, we hve ( ) µ A n = n=k+1 µ(a n ) = µ(a n ). k µ(a n ). Therefore, ( k ) k µ A n = µ(a n ).

2.2. MASUR SPACS 21 (b) It follows from () tht µ(c D) = µ(c) + µ(d) if C nd D re disjoint. Now note tht A = (A\B) B nd A\B nd B re disjoint. Therefore µ(a) = µ(a\b) + µ(b). (c) follows from (b) nd the fct tht mesure is non-negtive. (d) Observe first tht A B = A B\(A B) nd the two sets re disjoint. It follows from prt () tht µ(a B) = µ(a) + µ(b\(a B)). But µ(b\(a B)) + µ(a B) = µ(b) by prt (b). Hence the conclusion follows. (e) Define sequence {B n } in the following wy: B 1 = A 1, B 2 = A 2 \A 1, B 3 = A 3 \(A 1 A 2 ) nd more generlly, B n = A n \(A 1 A n 1 ). Then the sequence {B n } hs the following properties (see the exercises). 1. The B n re pirwise disjoint. 2. A n = B n. 3. B n A n nd therefore µ(b n ) µ(a n ). It follows from the bove nd the σ dditivity of µ, tht ( ) ( ) µ A n = µ B n = µ(b n ) µ(a n ). Theorem 2.1 (Continuity of mesures) Let (, A, µ) be mesure spce nd let {A n } n IN A. Then the following hold. (i) If {A n } is nondecresing (i.e, A 1 A 2 ) then ( ) µ A n = lim µ(a n). (ii) If the sequence is nonincresing (i.e.,a 1 A 2 ) nd µ(a 1 ) <, then ( ) µ A n = lim µ(a n). Proof. (i). Define sequence {B n } in the following wy: B 1 = A 1, B 2 = A 2 \A 1, B 3 = A 3 \A 2 nd more generlly, B n = A n \A n 1. Then the sequence {B n } hs the following properties 1. The B n re pirwise disjoint. 2. A n = B 1 B 2 B n nd therefore µ(a n ) = n k=1 µ(b k). 3. A n = B n. It follows from the bove nd the σ dditivity of µ, tht ( n ) ( lim µ(a ) n) = lim µ(b k ) = µ(b k ) = µ B k = µ A k. k=1 (ii). Let A = A n nd C n = A 1 \A n. Then C n = A 1 \A nd therefore ( ) µ C n = µ(a 1 \A) = µ(a 1 ) µ(a) k=1 k=1 k=1

22 CHAPTR 2. AIOMATIC MASUR THORY Note tht {C n } is incresing nd therefore by (i) we know tht µ( C n) = lim µ(c n ). From the other hnd, µ(c n ) = µ(a 1 \A n ) = µ(a 1 ) µ(a n ) nd so µ(a 1 ) µ(a) = µ( C n ) = lim µ(c n) = lim (µ(a 1) µ(a n )) = µ(a 1 ) lim µ(a n). Whence µ(a) = lim µ(a n ). Remrk 2.5 The ssumption µ(a 1 ) < is essentil. For exmple, if = IN nd µ is the counting mesure, then the sequence A k = {n IN n k} is decresing but µ(a k ) = nd therefore lim k µ(a k ) =, wheres µ( A k ) = µ( ) = 0. Definition 2.7 Let (, A, µ) be mesure spce. A subset A A is sid to be of full mesure if µ(a c ) = 0. Remrk 2.6 If µ is finite, then A is of full mesure if nd only if µ(a) = µ(). However, if µ is not finite mesure, then subset A stisfying µ(a) = µ() need not be of full mesure. Cn you give n exmple? Definition 2.8 Let (, A, µ) be mesure spce. A subset A is clled µ negligible or simply negligible, if there exists B A such tht A B nd µ(b) = 0. In this definition, the point is tht negligible set need not be mesurble. Here is n rtificil exmple. Let be set, A be proper subset of nd A. Let A = {, A, A c, } nd consider the mesure spce (, A, δ ) where δ is the Dirc mesure. Then ny proper subset of A c is negligible but not mesurble. If every negligible set is mesurble (nd therefore of mesure zero), then the spce (, A, µ) is clled complete. Thus, in complete mesure spce, set is negligible if nd only if it hs mesure zero. xercise. negligible. A subset of negligible set is negligible nd countble union of negligible sets is Definition 2.9 Let P be predicte on mesure spce (, A, µ), tht is, for ech x, there is proposition P (x) which is either true or flse depending on x. We sy tht P holds µ lmost everywhere (µ.e) or just lmost everywhere if the set {x P (x) is flse} is µ negligible. For exmple two functions f, g : IR re equl lmost everywhere if the set {x f(x) g(x)} is negligible. We shll give more exmples in the next chpter. A function from (, A, µ) IR is clled µ negligible if it is equl to 0 lmost everywhere. Remrk 2.7 A predicte P holds µ.e. if nd only if there is A A such tht µ(a c ) = 0 nd P (x) is true x A. Otherwise stted, P holds µ.e. if nd only if it holds on set of full mesure. Indeed, suppose first tht P holds µ.e. This mens tht there exists B A such tht {x P (x) is flse} B nd µ(b) = 0. Let A = B c. Then A A nd A = B c {x P (x) is true}. This mens tht P (x) is true x A with µ(a c ) = 0. Conversely, suppose tht P (x) is true x A. This mens tht A {x P (x) is true}. Therefore {x P (x) is flse} A c with µ(a c ) = 0. This mens precisely tht {x P (x) is flse} is negligible, tht is, P holds µ.e. k=1

2.3. MASURABL FUNCTIONS 23 xercise. Let P 1 nd P 2 be two predictes on mesure spce. If both P 1 nd P 2 hold lmost everywhere then the predicte P 1 P 2 (conjunction) lso holds lmost everywhere. More generlly, if (P n ) is sequence of predictes tht hold lmost everywhere then the predicte P n lso holds lmost everywhere. If P Q nd P holds.e, then Q holds.e. Remrk 2.8 Mny mthemticins sy tht predicte P holds lmost everywhere if {x P (x) is flse} hs mesure 0. Our definition is therefore more generl. However, due to the previous remrk, this mkes little difference in prctise. 2.3 Mesurble functions Definition 2.10 Let (, A) nd (Y, B) be two mesurble spces. A function f : Y is clled (A, B) mesurble if f 1 (B) A, tht is if for ll B B we hve f 1 (B) A. xmples. ) very function f : (, P()) (Y, B) is mesurble. b) A constnt function is mesurble. c) The identity function on is mesurble. The following lemm gives useful criterion for the mesurbility of function. Lemm 2.2 Let (, A) nd (Y, B) be two mesurble spces. Suppose tht B is generted by fmily C P(Y ). Then f : Y is (A, B) mesurble if nd only if f 1 (C) A. Proof. Suppose first tht f is mesurble. Then f 1 (C) f 1 (B) A. Conversely, suppose tht f 1 (C) A nd consider the fmily of subsets of Y F := { Y f 1 () A}. Then one cn check tht F is σ lgebr on Y. But this σ lgebr contins C nd so it contins B since B is the smllest σ lgebr contining C. Now B F mens precisely tht f 1 (B) A for ll B B, tht is, f is (A, B) mesurble. Definition 2.11 Let nd Y be two topologicl spces. A function f : Y is clled Borel-mesurble if it is (B(), B(Y )) mesurble, tht is, if the inverse imge under f of every Borel subset of Y is Borel subset of. xmples. 1) very continuous mp is Borel-mesurble. 2) very monotonic mp f : IR IR is Borel-mesurble. Indeed, we clim first tht the inverse imge under f of n intervl of IR is n intervl of IR nd hence Borel subset of IR. Let I be n intervl of IR nd let x, y f 1 (I) with x y. We need to show tht [x, y] f 1 (I). Let z [x, y]. If f is incresing then f(x) f(z) f(y) nd so f(z) [f(x), f(y)] I, tht is, z f 1 (I). If f is decresing, similr resoning yields z f 1 (I). The clim is proved. Now the Borel σ lgebr of IR is generted by the intervls of IR. Remrk 2.9 Let (, A) nd (Y, B) be two mesurble spce nd A A. We lredy observed tht (A, P(A) A) is mesurble spce. Therefore it mkes sense to sy tht function f : A Y is mesurble. This mens tht f 1 (B) A, whenever B B, becuse f 1 (B) A.

24 CHAPTR 2. AIOMATIC MASUR THORY Lemm 2.3 (The psting lemm) Let (, A) nd (Y, B) be two mesurble spces, nd let A, B A. Let f : A Y nd g : B Y be mesurble functions tht coincide on A B (this condition is stisfied if A B = ). Then the function h : A B Y defined by is mesurble. h(x) = { f(x) g(x) if x A if x B Proof. This follows from the fct tht h 1 () = f 1 () g 1 (). Thus if B, then f 1 () nd g 1 () belong to A (nd they re contined in A B). Corollry 2.1 Let (, A) mesurble spce, A A, nd f : A IR be mesurble. Define f : IR by { f(x) if x A f(x) = 0 if x / A. Then f is mesurble. Proof. A constnt function is mesurble. This corollry mens tht we cn lwys extend mesurble functions to mesurble functions defined on the whole spce. The next corollry shows how one cn modify mesurble function nd still get mesurble function. Corollry 2.2 Let f : IR be mesurble nd let be mesurble set. Then the function h : IR defined by is mesurble. h(x) = { f(x) if x \ 0 if x. Proof. Note tht the restriction of mesurble function to mesurble subset is mesurble. Proposition 2.5 Let f : Y be (A, B) mesurble nd let µ be mesure on (, A). Then the formul ν(b) := µ(f 1 (B)) for ll B B defines mesure on (Y, B). This mesure is denoted by f (µ) nd is clled the imge mesure of µ by f. It is lso clled the pushforwrd mesure of µ by f. Proof. (i). ν( ) = µ(f 1 ( )) = µ( ) = 0. (ii). Let {B n } B be sequence of pirwise disjoint sets, then {f 1 (B n )} A is lso sequence of pirwise disjoint sets nd therefore ( ) ( ( )) ( ) ν B n = µ f 1 B n = µ f 1 (B n ) = µ(f 1 (B n )) = ν(b n ).

2.3. MASURABL FUNCTIONS 25 xmple 2.1 Let nd Y be two rbitrry sets nd let A = P() nd B P(Y ) be n rbitrry σ lgebr on Y. Let µ be the counting mesure on. Then every function f : Y is (A, B) mesurble nd ν = f µ is the mesure tht counts the number of preimges { ν(b) = µ(f 1 crd{x f(x) B} if this set is finite (B)) = + if not. xmple 2.2 Let be set, nd consider the mesure spce (, P(), δ ). Let (Y, B) be n rbitrry mesure spce nd let f : Y be function (it is necessrily mesurble). Then for ny A, we hve { { f (δ (A)) = δ (f 1 1 if f 1 (A) (A)) = 0 if / f 1 (A) = 1 if f() A 0 if f() / A = δ f()(a). Therefore f (δ ) = δ f(). Remrk 2.10 The concept of imge mesure is very importnt in probbility. Let (, A, µ) be probblity spce nd let f : IR is mesurble function (IR is equiped with its Borel σ lgebr). Then f is clled rndom vrible nd f (µ) is clled the probbility lw of f. In order to compute probbilities relted to f, we hve to know wht is f (µ). 2.3.1 Mesurble rel vlued functions In the following two subsections we consider functions tht tke vlues in IR. It is ssumed tht IR is equipped with its Borel σ lgebr B(IR). So given mesurble spce (, A), function f : IR is clled mesurble if it is (A, B(IR))-mesurble. To simplify, we will lso sy in this cse tht f is A mesurble. Remrk 2.11 Let f : IR. To simplify the nottion, we write {f < } insted of f 1 ([, [) = {x f(x) < }. This nottion is used in probbility theory. Most often in probbility theory, probbility spce is denoted by (Ω, F, P ) nd symbols like, Y, Z, W re used to denote rndom vribles, i.e, mesurble functions on Ω. Then P ( < ) denotes the mesure (probbility) of the event { < }. Lemm 2.4 Let (, A) be mesurble spce nd f : IR. Then the following conditions re equivlent. 1. f is mesurble. 2. {f < } is mesurble for every IR. 3. {f } is mesurble for every IR. 4. {f < } is mesurble for every Q. 5. {f } is mesurble for every Q. 6. {f > } is mesurble for every IR. 7. {f } is mesurble for every IR. 8. {f > } is mesurble for every Q. 9. {f } is mesurble for every Q.

26 CHAPTR 2. AIOMATIC MASUR THORY Proof. This follows from Lemm 2.2 nd the fct tht B(IR) is generted by subsets of the form [, [ etc. Theorem 2.2 Let (, A) be mesurble spce. Let f, g : IR be mesurble nd let α be rel constnt. Then the following functions re defined on mesurble subsets nd re mesurble. f + g, αf, fg, f, f g. Proof. f +g is not defined when f(x) = + nd g(x) = or vice vers. Otherwise stted f + g is not defined on the set = f 1 (+ ) g 1 ( ) f 1 ( ) g 1 (+ ). Now observe tht {+ } is closed set in IR. It is therefore Borel subset of IR. It follows tht f 1 ({+ }) is mesurble becuse f is mesurble. Similrly, the sets g 1 ( ), f 1 ( ) g 1 (+ ) re mesurble. Consequently, is mesurble nd so the set \ on which f + g is defined is mesurble. For similr resons, f g is defined on mesurble set. fg is defined everywhere becuse of our convention 0 ± = 0. ) Let h = f + g. Let Q. We clim tht {h < } = {f < p} {g < q} where the union is tken over ll couples (p, q) Q 2 such tht p + q =. Indeed, if f(x) < p nd g(x) < q with p + q =, then h(x) = f(x) + g(x) < p + q =. Conversely suppose tht h(x) <. Note tht this implies tht f(x) < nd g(x) <. Let p be rtionl number such tht f(x) < p < g(x). Let q = p Q. Then g(x) < q. This proves the clim. But the clim mens tht {h < } is countble union of mesurble sets. b) αf is mesurble becuse {f < α } if α > 0 if α = 0 nd 0 {αf < } = if α = 0 nd > 0 {f > α } if α 0. c) f 2 is mesurble becuse d) We cn write {f 2 } = { if < 0 { f } if 0. + if f(x) = g(x) = + or f(x) = g(x) = f(x)g(x) = if (f(x) = + nd g(x) = ) or (f(x) = nd g(x) = + ) ( (f(x) + g(x)) 2 f(x) 2 g(x) 2) if f(x), g(x) IR. 1 2 Since constnt functions re mesurble, it follows from ), b), c) nd the psting lemm tht fg is mesurble. e) f is mesurble becuse { f } = { if < 0 { f } if 0.

2.3. MASURABL FUNCTIONS 27 f) 1 g is mesurble becuse { 1 {g > 1 } {g < 0} if > 0 g < } = {g < 0} if = 0 < g < 0} if < 0. { 1 Theorem 2.3 Let (, A) be mesurble spce nd let (f n ) be sequence of mesurble functions from to IR, then the following functions inf f n, sup f n, lim inf f n, lim sup f n re mesurble. In prticulr, if (f n ) converges pointwise, its limit is mesurble. Proof. ) Let h = sup(f n ). Then h is mesurble becuse {h } = {f n }. b) Let g = inf(f n ). Then g is mesurble becuse {h } = {f n }. c) Consequently, is mesurble. d) Similrly, is mesurble. lim inf f n = sup inf lim sup f n = inf f n n 1 k n sup f n n 1 k n Corollry 2.3 If f nd g re two mesurble functions from to IR, then mx(f, g) nd min(f, g) re mesurble. In prticulr, the functions f + := mx(f, 0) nd f := min(f, 0) = mx( f, 0) re mesurble. 2.3.2 Simple functions Definition 2.12 Let (, A) be mesurble spce nd let f : IR be mesurble function. We sy tht f is simple function if it tkes only finite number of vlues. Note tht we include the condition of mesurbility in our definition of simple functions. Lemm 2.5 Let f : IR be simple function. Then f cn be written in the form f = n i 1 Ai (2.1) i=1 where {A i } re mesurble sets tht form prtition of.

28 CHAPTR 2. AIOMATIC MASUR THORY Proof. Suppose tht f tkes the vlues { 1,..., n } with i j for i j. Set A i = f 1 ( i ) for i = 1,..., n. Then first, ech A i is mesurble s the inverse imge of mesurble set. Second, the sets A i re pirwise disjoint, for if x A i A j, then f(x) = i nd f(x) = j so tht i = j nd this is impossible if i j. Third, if x, then f(x) tkes some vlue k so tht x f 1 ( k ) = A k. This mens tht {A i } i is prtition of. We show now tht f hs the representtion (2.1). Let x. Then s we lredy observed, x belongs to exctly one A k. Now, on the first hnd f(x) = k. On the other hnd, 1 Ak (x) = 1 nd 1 Ai (x) = 0 for i k. Therefore n i=1 i1 Ai (x) = k. Hence the equlity. Remrk. If f : IR is simple function then f could be written in the form (2.1) in severl wys. For exmple consider the function f defined by f(x) = 1 if x < 0 nd f(x) = 1 if x 0. Then f = χ ],0[ + χ [0, [. But lso f = χ ], 2[ χ [ 2,0[ + χ [0, [ + 3χ. Let us sy tht j I b jχ Bj is n dmissible representtion of f if {B j } j J form prtition of. Theorem 2.4 (Approximtion of nonnegtive mesurble functions by simple functions) Let (, A) be mesurble spce nd f : [0, ] be mesurble. Then there exists sequence (h n ) of simple functions such tht ) h n (x) < + for ll n nd ll x. b) 0 h 1 h 2 f. c) h n (x) f(x) s n for ll x. d) If f is bounded then the convergence is uniform. Proof. set For ech n IN divide the intervl [0, n] into n2 n intervls ech of length 2 n nd { {x k2 n f(x) < (k + 1)2 n } if 0 k n2 n 1 n,k = {x f(x) n} if k = n2 n. Then for ech n, the fmily { n,k } k=0,...,n2 n is prtition of. Set { k2 n if x n,k for some 0 k n2 n 1 h n (x) = n if x n,n2 n. Otherwise stted, h n = n2 n k=0 k 2 n 1 n,k is simple function. ) It is cler tht h n (x) < for ll n IN nd ll x. b) It is lso cler tht 0 h n f for ll n. Let us prove tht h n h n+1. Let x. There re 2 cses: (1) k2 n f(x) < (k + 1)2 n for some 0 k n2 n 1. Then h n (x) = k2 n. Also the inequlity of this cse is equivlent to 2k2 n 1 f(x) < (2k + 2)2 n 1. So we distinguish between two cses. (1) 2k2 n 1 f(x) < (2k + 1)2 n 1, tht is, x n+1,2k. In this cse, h n+1 (x) = 2k2 n 1 = k2 n = h n (x). (1b) (2k+1)2 n 1 f(x) < (2k+2)2 n 1, tht is, x n+1,2k+1. In this cse, h n+1 (x) = (2k + 1)2 n 1 > k2 n = h n (x). (2) f(x) n. In this cse, h n (x) = n. We distinguish between two cses.