Part A: Integration. Alison Etheridge

Transcription

1 Prt A: Integrtion Alison theridge Introduction In Mods you lerned how to integrte step functions nd continuous functions on closed bounded intervls. We begin by reclling some of tht theory. Definition. (Step functions nd their integrls. A function φ defined on the intervl I (with endpoints, b is step function if there is prtition ( i k i of the intervl (tht is < < 2 <... < k b so tht φ is constnt on ech intervl ( j, j, j k. If φ(x c i for x ( i, i, then φ I I φ(xdx k c i ( i i. i Write L step [, b] for the spce of step functions on [, b]. Definition.2 (Lower nd Upper Sums. For rel-vlued function on [, b] define b { b f sup } φ : φ L step [, b], φ f, b The fundmentl result tht you proved ws the following. { b f inf } φ : φ L step [, b], φ f Theorem.3 (Integrting continuous functions over closed bounded intervls. For continuous function f on the closed bounded intervl [, b] b f b f. ( The quntity in eqution ( is then defined to be the integrl of f over [, b]. We write C[, b] for continuous functions on [, b]. Lter in the course you considered sequences of continuous functions: Theorem.4 (xchnging integrls nd limits under uniform convergence. Let {f n } n be sequence of functions in C[, b] for which f n f uniformly (nd so in prticulr f C[, b] then b f n b f s n.

2 You then exploited this to interchnge summtion nd integrtion for power series: Suppose tht the power series f(t k kt k hs rdius of convergence R. Write f n (t n k kt k. Since for x < R, f n (t f(t uniformly on [, x] you deduced tht x f(tdt lim lim x n k k f n (tdt k x k+ k +, k x k+ k + nd this proved to be powerful tool in estblishing deep properties of exponentil nd trigonometric functions. Originlly integrtion ws introduced s the inverse of differentition. The gol of Newton ( nd Leibniz ( ws to solve the problem of primitives: Find the functions F (x which hve derivtive given function f(x. At tht time the notion of function ws not relly well defined, but usully it ment ssociting quntity y to vrible x through n eqution involving rithmetic opertions (ddition, subtrction, multipliction, division, tking roots, trigonometric opertions (sin, cos, tn nd their inverses nd logrithms nd exponentils. With the geometric understnding of integrtion s re under curve cme the generlistion of this ide of function, but still continuous functions were viewed s the rel functions nd the notion of integrtion from Mods sufficed. The version of the Fundmentl Theorem of Clculus tht you proved in Mods showed tht for continuous functions f on closed bounded intervls, the problem of primitives ws solved by the indefinite integrl of f. But then long cme Fourier ( He showed tht trigonometric series which cn be used to represent continuous functions on bounded intervls cn lso be used to represent discontinuous ones. In prticulr, the function { x π, f(x π < x 2π, cn be represented by convergent trigonometric series. The notion of function hd to be extended nd with it the notion of integrl. For finite number of discontinuities, one cn piece together the portions on which the function is continuous nd your methods esily extend - nd in fct you cn even go bit further provided the points of discontinuity re exceptionl in sense tht we ll mke precise lter. But worse ws to come. Dirichlet ( cme cross the following function which now tkes his nme: Definition.5 (Dirichlet function. The function χ(x { if x Q, if x / Q, (2 is known s the Dirichlet function. In fct Dirichlet obtined χ s limit: χ(x lim m lim (cos(m!πx2n. 2

3 The Dirichlet function is nowhere continuous nd yet Dirichlet constructed it s double pointwise limit of sequence of continuous functions. vidently our theory of integrtion from Mods relly cnnot hndle the Dirichlet function. Any step function φ on [, ] stsifying φ χ lso stisfies φ on [, ] nd the constnt step function on [, ] stisfies χ nd so χ(xdx. Likewise, the constnt function step function on [, ] stisfies χ nd ny step function φ on [, ] stisfying χ φ lso stisfies φ on [, ], so χ(xdx. Thus χ(xdx < χ(xdx. Of course there re mny other less perverse wys thn Dirichlet s to obtin the Dirichlet function s pointwise limit of functions tht re integrble in the sense of Mods. xmple.6. Let {r n } n be n enumertion of the rtionls in [, ]. For ech k N define f k : [, ] R by { x rn, n k, f k (x otherwise. Then f k (xdx for ech k, but lim k f k (xdx is not (yet defined. The Lebesgue integrl, introduced by Lebesgue in very short pper of 9 but fully explined in beutiful set of lecture notes published in 94 (from course delivered in 92-3 is n extension of the integrl tht you developed in Mods tht behves well under pssge to the limit. So when will it be the cse tht f n f implies f n f in our new theory? The next exmple shows tht the nswer should certinly be not lwys. xmple.7. Define f k : [, ] R by f k (x { k( kx x /k, otherwise. ch f k is continuous nd f k (x s k for ll x (, ]. So f k f. But f k(xdx /2 for ll k. So lim f k(xdx < lim k k f k (xdx 2. The difficulty from the point of view of the integrtion theory of Mods is tht the convergence is not uniform. It will turn out tht we ll get convergence theorems for our new integrl under other conditions. For exmple, suppose tht we restrict ourselves to functions with f k (x M for ll k nd ll x [, ]. A theorem with this hypothesis is clled bounded convergence theorem. Or, becuse we don t wnt to restrict ourselves to bounded functions on bounded intervls, we could try tking monotone sequence f k f (by which we men tht f k f k+ for ll k nd f k f or f k f. A theorem with this hypothesis is clled monotone convergence theorem. xmple.6 shows tht neither of these conditions is enough to rescue the integrl of Mods, but they will help us out here. The functions of xmple.7 don t stisfy either of our hypotheses nd we greed tht for them we would not expect convergence theorem. It ws nturl tht the integrl of the limit ws strictly less thn the limit of the integrls. We ll see tht for positive functions n inequlity bit like this lwys holds. The limit of the integrls might not exist nd so we replce it with new concept tht we ll define lter clled the lim inf. When the limit does exist then it is equl to the lim inf. Probbly the most 3

4 importnt result of our theory, Ftou s Lemm, will tell us tht for positive functions f k converging to f lim f k(xdx lim inf f k (xdx. k k So wht should the integrl of the Dirichlet function of Definition.5 over [, ] be? The Dirichlet function is just the indictor function of the rtionls. For n intervl I we define the integrl of its indictor function over [, ] to be the length of I [, ]. Similrly for set S which is finite union of disjoint intervls, the integrl over [, ] of the indictor function of S is the totl length of the intervls mking up S [, ]. For n intervl [, b] we ll write m([, b] b for its length. It would mke sense then for the integrl of the Dirichlet function over [, ] to be the length of the set Q [, ], m(q [, ]. So now we hve to work out how to ssign length to this set. For two sets A B our intuition sys tht the length of A should be less thn or equl to tht of B, tht is m(a m(b. Now, given ɛ >, let {r n } n be n enumertion of the rtionls in [, ] nd let I n (r n ɛ 2 n+, r n + ɛ, 2n+ then Q [, ] n I n nd (gin using bit of intuition the totl length of the right hnd side is less thn or equl to n ɛ 2 ɛ. Since ɛ ws rbitrry we conclude tht in resonble world we d n better tke the length of Q [, ] to be zero. We ll mke tht more precise lter, but then with this notion of length, χ(xdx nd our problem with the functions {f k } k of xmple.6 goes wy, χ(xdx lim f k(xdx lim k k f k (xdx. Wht we ve done in this exmple, insted of trying to pproximte χ by step functions bsed on intervls, is we ve sked wht is the length or (s we ll usully sy mesure of the set where χ is equl to one. Lebesgue s key ide ws to extend this. To pproximte function by step functions s you did in Mods, you divide the domin of the function. To pproximte function by simple functions, s we ll do here, we divide the rnge of the function. The simple function in the right hnd picture is defined to be 4 y k Sk (x k where S k {x R : y k f(x < y k } nd S is the indictor function of the set S. Of course in this picture the simple function is step function. But in generl it won t be, s we sw in our Dirichlet function exmple where it ws the indictor function of Q [, ]. (See problem sheet for n exmple of continuous function for which Lebesgue s recipe does not lwys give step function. xercise.8. As nother exmple try pproximting sin x on (, ] by simple functions. Here then is Lebesgue s recipe for integrting bounded function over n intervl [, b]: 4

5 . Subdivide the y-xis by series of points min f(x y < y < < y n > mx f(x. x [,b] x [,b] 2. Form the sum n y k mesure ({x [, b] : y k f(x < y k }. k 3. Let the number of points {y n } go to infinity in such wy tht mx(y k y k. The limit of the sequence of sums in 2 is b f(xdx. In fct we ll see tht for bounded functions on [, b] b f(xdx inf{ b ψ(xdx : ψ simple, ψ f} sup{ b φ(xdx : φ simple, φ f}, so when function is integrble in the sense of Mods we ll get exctly the sme nswer s before. But becuse step functions simple functions we cn define the integrl for wider clss of functions this wy. Moreover, our recipe just needs us to be ble to ssign vlue to mesure ({x : y k f(x < y k } so now we re not restricted to integrting functions f : R R over intervls. We cn integrte functions f : Ω R over ny subset of Ω provided we hve suitble notion of size of sets where f tkes prticulr vlues. In prticulr, Ω could be probbility spce or Z or N or sphere or torus or.... To see the power (nd limittions of the theory though we need to know which sets we cn mesure. So our first tsk is going to be to develop theory of mesure. BUT FIRST: Plese remember tht you cn still integrte everything tht you lredy know how to integrte. Lebesgue integrtion extends the theory of Mods so tht you cn integrte more functions thn before. In prticulr, you cn still integrte continous functions over closed bounded intervls nd technniques tht you justified in Mods like integrtion by prts nd substitution cn still be used for those functions. Thus you my freely use the following results from your Anlysis III notes: Theorem.9 (Integrtion by prts. If u, v re differentible on [, b] nd u, v re in C[, b] then b (u v + b (uv u(bv(b u(v(. Theorem. (Substitution. Suppose tht g C([c, d] is strictly monotone incresing nd continuously differentible (so g C[c, d] nd g > nd suppose tht g(c, g(d b. Then for f C[, b] d c f (g(x g (xdx b f(xdx. 5

6 2 Lebesgue mesurble sets in R In this section we concentrte on the problem tht motivted Lebesgue. Here s the problem tht he proposes: We wish to ttch non-negtive rel number which we shll cll the mesure m( to ech bounded subset of R in such wy tht the following conditions re stisfied:. If is trnsltion of then m( m(. 2. If { i } i is finite or countbly infinite collection of sets with i j whenever i j then ( m i m( i. i 3. m((,. i Let s puse to think bout 2. It seems resonble tht if I split set into two disjoint pieces 2 with 2 sy, then m( m( + m( 2. But we re sking for something bit stronger. If { i } i is countbly infinite fmily of sets with i j for i j, then writing nd i B n \ then if we re going to hve ny hope of developing theory tht behves well under pssge to the limit, it is resonble to sk for kind of continuity property, nmely tht if m( < then m(b n s n. If we ssume tht 2 holds for finite collections of sets then this becomes ( n n m(b n m( m i m( m( i s n. i Tht is m( lim m( n n i lim m( i m( i, i i i which is wht 2 sys for countble disjoint unions. A consequence of Lebesgue s three conditions is (see Problem Sheet tht for ny intervl I [, b] (or (, b], [, b, (, b, the mesure m(i b. So Lebesgue mesure is required to be n extension of our notion of length of n intervl. It turns out (but it is beyond our scope in this course tht if we were to weken 2 by only requiring this property for finite collections, then one cn define notion of length for ll subsets of R. But we re interested in countble opertions (it ws interchnge of limits nd integrtion tht motivted us so we need this stronger version nd then one cn construct sets to which it is not possible to ssign mesure. This ws first chieved by Vitli nd we ll see ( vrint of his fmous set in xmple 2.2. On the other hnd it turns out tht one hs to work hrd to construct non-mesurble set nd so we shll not be gretly inconvenienced by their existence. Lebesgue s pproch to constructing his mesure owes lot to the upper nd lower sums of the integrtion theory from Mods. The key ide is the one tht we used to suggest tht the length of Q [, ] (tht rose in integrting the Dirichlet function over [, ] should be zero. 6 i n i i i

7 Recll tht Lebesgue s conditions lredy tell us tht the mesure of n intervl with endpoints nd b (with < b sy is b. (We lso llow intervls to be empty in which cse they hve length zero. Definition 2. (Outer mesure. For set R, the outer mesure of is { } m ( inf m(i n : I n, I n intervls. n Wht we showed in ws tht m (Q [, ]. Definition 2.2 (Null sets. A set R is null (with respect to Lebesgue mesure if m (. n Lemm 2.3 (Some properties of null sets.. Any subset of null set is null. 2. A countble union of null sets is null. WARNING: NOT ALL NULL STS AR COUNTABL. The proof of Lemm 2.3 nd n exmple of n uncountble null set re both on Problem Sheet. Lemm 2.4 (Some properties of outer mesure.. If is trnsltion of then m ( m (. 2. If { i } i is finite or countbly infinite collection of sets then m ( i i i m ( i. 3. m ([, b] b. The difficulty is 2. It tells us tht m is countbly subdditive, but we cnnot deduce tht it is countbly dditive which ws Lebesgue s condition 2. xercise 2.5. Prove 2. If you get stuck look in, for exmple, Cpinski & Kopp. The outer mesure is relly plying the rôle of the upper sum in our old definition of the integrl if we try to define the integrl of the indictor function of set. To define mesurble sets we need n nlogue of the lower sum. Lebesgue provided this with the inner mesure. Definition 2.6 (Inner mesure. For subset of the intervl [, b], the inner mesure of is given by m ( m ([, b] m ([, b]\ b m ([, b]\. Definition 2.7 (Lebesgue mesurble sets. A subset of R is Lebesgue mesurble if for ech bounded intervl I m ( I + m ( c I m (I. Its Lebesgue mesure is then equl to its outer mesure. denote its Lebesgue mesure by m(. When set is Lebesgue mesurble we 7

8 Notice in prticulr tht ll bounded intervls re Lebesgue mesurble. We cn deduce from this definition tht set is Lebesgue mesurble if nd only if, for every finite intervl I, the inner nd outer mesures of the set I gree. Theorem 2.8. The outer mesure m restricted to the Lebesgue mesurble sets stisfies Lebesgue s conditions, 2 nd 3. This Theorem tells us tht by restricting to the Lebesgue mesurble sets we hve overcome the difficulties of outer mesure nd recovered countble dditivity. This is only going to be good theory if the clss of Lebesgue mesurble sets is sufficiently lrge. So which sets re Lebesgue mesurble? Theorem 2.9. The clss of Lebesgue mesurble sets, denoted M Leb, hs the following properties:. R M Leb. 2. If M Leb then c M Leb. 3. If n M Leb for ll n N then n n M Leb. This Theorem sys tht M Leb forms wht is clled σ-lgebr. It is closed under finite or countble set opertions. In prticulr you cn check, using de Morgn s lws, tht the intersection of countble collection of Lebesgue mesurble sets is lso Lebesgue mesurble. Proofs of Theorems 2.8 nd 2.9 cn be found, for exmple, in Temple (97, Chpter 7. There is lso proof in Cpinski & Kopp, but they tke different definition of Lebesgue mesurble sets. They ssume wht is known s the Crthéodory condition, tht m ( A + m ( c A m (A for ll A R nd not just for bounded intervls. Their definition is equivlent to ours, but tht is not very esy thing to show. As we lredy remrked, ll intervls re Lebesgue mesurble nd so ny set obtined by tking finite or countble unions or intersections of intervls is Lebesgue mesurble. In prticulr, Lemm 2.. All open sets R re Lebesgue mesurble. The clss of Lebesgue mesurble sets is beginning to look rther big. It is ctully slightly lrger still becuse ll null sets re Lebesgue mesurble (nd hve Lebesgue mesure zero. In fct the Lebesgue mesurble sets re precisely those tht cn be obtined from intervls nd null sets by finite or countble sequence of set opertions. There re lot of those sets. Definition 2.. The collection of sets tht cn be obtined by finite or countble sequence of set opertions from the intervls of R is clled the σ-lgebr generted by the intervls, or the Borel σ- lgebr on R. Supplementing these by tking set opertions with sets with m ( is known s tking the completion with respect to the (Lebesgue outer mesure nd yields the Lebesgue mesurble sets. The next (non-exminble exmple illustrtes how hrd one hs to work to construct non-mesurble set. It is vrint of n exmple due to Vitli. It is convenient to work with the unit circle S. The definition of mesurble set trnsltes in n obvious wy since S cn be thought of s [, 2π. xmple 2.2 (A non-mesurble set. Let S {e iθ : θ R}. 8

9 Define n equivlence reltion on S by writing z w if α, β R s.t. z e iα, w e iβ, nd α β Q. Let A be set obtined by choosing exctly one representtive from ech equivlence clss. Define A q e iq A {e iq z : z A}. Then {A q } q Q is collection of pirwise disjoint sets, ech obtined from A by rottion nd S q Q A q. Suppose tht A is mesurble, then so is A q nd m(a q m(a for ll q Q. Then by countble dditivity, 2π m(s { if m(a m(a q if m(a >, q Q which yields contrdiction. Thus we cnnot ssign mesure to A. It turns out tht to construct non-mesurble set one hs to invoke the xiom of choice which is wht llowed us to choose exctly one representtive from ech of the uncountbly infinite number of equivlence clsses. In higher dimensions things re even worse. In three dimensions one cn cut solid bll into finite number of non-overlpping sets which cn then be resssembled to yield two identicl copies of the originl bll. Do web serch on Bnch-Trski prdox to find out more. 3 Other mesure spces In 2 we sw tht we could define notion of length or mesure for wide clss of subsets of R clled the Lebesgue mesurble sets. A fncy wy to sy this is to sy tht (R, M Leb, m is mesure spce. Definition 3. (Mesure spce. We sy tht (Ω, F, µ is mesure spce if Ω is n bstrctly given set, F is collection of subsets of Ω which includes nd is closed under complements nd finite or countble unions (recll tht this sys tht F is σ-lgebr nd µ : F [, ] is countbly dditive set function; tht is if { n } n N is countble collection of sets from F with i j whenever i j then ( µ n µ( n. n There re lots of mesure spces out there, severl of which you re lredy fmilir with. xmple 3.2 (Discrete mesure theory. Let Ω be countble set nd F the power set of Ω (tht is the collection of ll subsets of Ω. A mss function is ny function µ : Ω [, ]. We cn then define mesure on Ω by µ({x} µ(x nd extend to rbitrry subsets of Ω using Property 2. qully given mesure on Ω we cn define mss function. In prticulr,. If x Ω µ(x x Ω µ({x} then µ is probbility mesure nd µ is the corresponding probbility mss function. For A Ω, n µ(a x A µ({x}. 9

10 The null sets correspond in the terminology of Mods probbility to events tht hve probbility zero. 2. If Ω N nd µ(k, k N then we get counting mesure. For A N, In this cse the only null set is the empty set. µ(a {n : n A}. These discrete mesure spces provide toy version of the generl theory where ech of the results we prove for generl mesure spces reduces to some strightforwrd fct bout convergence of series. This is ll we need to do (for exmple discrete probbility nd so tht topic is generlly introduced without discussing mesure theory. But for more generl probbility nd most of nlysis (both pure nd pplied this is not enough. xmple 3.3. Consider the problem from continuous probbility of picking point uniformly from [, ]. We fce mking sense of the probbility tht given number is chosen or tht the number chosen flls within prticulr set. Write U for the (rndom point picked. vidently P[U x] if ll points re eqully likely, for otherwise x [,] P[U x]. This mens tht we won t be ble to distinguish the probbilities of two sets tht differ by just one point, or indeed tht differ by finite number of points. On the other hnd we d expect to be ble to define for n intervl I [, b] [, ] P[U I] b. The Lebesgue mesurble subsets of [, ] re precisely those to which we cn ssign P[U A] in consistent wy. The mesure tht we defined on the Lebesgue mesurble subsets of R ws certinly not the only mesure on (R, M Leb. The form tht it took ws forced upon us by our insistence tht m([, b] b. xmple Define µ([, b] b e x2 /2 dx. 2π Then µ extends to mesure on ll the Lebesgue mesurble subsets of R. This mesure is of centrl importnce in sttistics. 2. For [, b] [, define µ t ([, b] b e tx dx. Then for ech t >, µ t extends to mesure on ll the Lebesgue mesurble subsets of [,. This fmily of mesures is very importnt in the study of differentil equtions. These exmples re ll of either discrete mesures or mesures which re bsolutely continuous with respect to Lebesgue mesure. We ll see the origin of tht term in 7. But of course one cn hve mixtures of the two (see Problem Sheet 2 for n exmple nd in fct in 7 we ll see tht there re more exotic mesures still. Although mostly we ll concentrte on Lebesgue mesure, lwys keep in mind tht our theory is much more generl.

11 4 Mesurble functions We now turn our ttention to the objects t the hert of integrtion theory: the mesurble functions. Lst term in the Anlysis course you sw tht function f : X Y (where X nd Y re subsets of ucliden spce is continuous if nd only if A open in Y f (A open in X. If you re studying topology then you ll know tht this definition pplies more generlly. For the definition of mesurble functions we ll replce open by mesurble. The dvntge of mesurble functions over continuous functions is tht they re much more stble. In prticulr, they behve well under limits: the pointwise limit of sequence of mesurble functions is mesurble - nd this closure of the spce of mesurble functions is t the hert of the convergence theorems. Definition 4. (Mesurble function. Let (Ω, F, µ nd (Λ, G, ν be mesure spces. f : Ω Λ is mesurble (with respect to F, G if nd only if A function G G f (G F. For concreteness, mostly we re going to work in the specil setting of Lebesgue mesurble functions on R. But the rguments tht we use will be insensitive to replcing the Lebesgue mesurble sets, M Leb, by F of Definition 4.. Definition 4.2 (Lebesgue mesurble function. A function f : R R is Lebesgue mesurble if nd only if for every intervl I R, f (I {x R : f(x I} is Lebesgue mesurble set. Similrly, if is mesurble subset of R, then f : R is Lebesgue mesurble if for ech intervl I, f (I is Lebesgue mesurble. In wht follows, if we sy tht f : R R is mesurble, without further qulifiction, then we men Lebesgue mesurble. The definition is equivlent to requiring tht for ll Borel sets A, f (A is Lebesgue mesurble. The point is tht it is enough to check just for intervls nd indeed we don t need to check it for ll intervls I. It is convenient to know the following result. Theorem 4.3. Let be mesurble subset of R nd f : R be function. The following re equivlent:. f is mesurble. 2. For ll, f ((, is mesurble. 3. For ll, f ([, is mesurble. 4. For ll, f (, is mesurble. 5. For ll, f (, ] is mesurble. Proof. Let s just prove one of these. Obviously ( implies ll the others. We prove here tht (2 (. We must show tht for ny intervl I, f (I M Leb. We lredy hve tht f ((, M Leb. Now suppose tht I (, ]. f ((, ] f (R\(, \f ((, M Leb

12 since mesurble sets re closed under tking complements. Now note tht ( f ((, b f (, b n ] n f ((, b n ] M Leb n since countble unions of mesurble sets re mesurble. Tking complements we lso hve f (([b, M Leb. Now let I (, b. f ((, b f ((, b f ((, M Leb. Similrly, f ([, b] f ((, b] f ([, M Leb. The sme resoning gives the hlf open intervls. In prctice, most functions re mesurble. xmple Constnt functions re mesurble. 2. More generlly, continuous functions re mesurble. 3. The indictor function of the set A defined by { if x A A (x otherwise, is mesurble if nd only if A is (Lebesgue mesurble. Proof. We just prove the first two ssertions (the third is on Problem Sheet 2. First suppose tht f(x c. Then { f R if < c ((, otherwise nd in both cses we hve Lebesgue mesurble sets. Now suppose tht f is continuous. Note tht (, is n open set nd, reclling tht for continuous function the inverse imge of n open set is open, f ((, must be n open set. All open sets re mesurble so f is mesurble s required. Theorem 4.5. If f nd g re mesurble functions then so re f + g nd fg. In prticulr, tking g(x c ( constnt cf is mesurble. This result tells us tht the mesurble functions form vector spce which is closed under multipliction. An elegnt proof of Theorem 4.5 uses the following useful Lemm. Lemm 4.6. Suppose tht F : R R R is continuous function. If f nd g re mesurble, then h(x F (f(x, g(x is lso mesurble. 2

13 Theorem 4.5 will follow upon setting F (u, v u + v nd F (u, v uv. Proof of Lemm 4.6. The proof uses the two-dimensionl nlogue of the fct tht open sets in R cn be expressed s finite or countble unions of open intervls. In two dimensions every open set is finite or countble union of open rectngles. So for ny rel, put G F (,, tht is {x : h(x > } {x : (f(x, g(x G }. Then G is open, since F is continuous, nd so cn be written s G n n where the R n re open rectngles, R n ( n, b n (c n, d n sy. So h (, {x : h(x > } R n n {x : f(x ( n, b n } {x : g(x (c n, d n } n which is mesurble by mesurbility of f nd g nd since M Leb is closed under finite or countble set opertions. The following results re on Problem Sheet 2. Proposition 4.7. Let be mesurble subset of R nd f : R function. Let { f + f(x if f(x > (x if f(x, nd Then { f (x if f(x > f(x if f(x,. f is mesurble if nd only if f + nd f re mesurble. 2. If f is mesurble then so is f, but the converse is flse. 3. If f is mesurble then is lso mesurble. f (x { if f(x > f(x if f(x, Before proving our key result bout mesurble functions - stbility under limits - we need to introduce new concept which generlises the ide of limit. Definition 4.8 (limsup nd liminf. For sequence of rel numbers (x n n N we define lim sup x n lim sup x m, m n nd lim inf x n lim inf m n x m. 3

14 To understnd this little better, suppose first tht the sequence {x n } n is bounded. Now notice tht S n sup m n x m decreses s n increses (since we re tking the supremum over smller set. Moreover, since the sequence {x n } n is bounded, {S n } n is lso bounded nd monotone decresing sequence of rel numbers which is bounded below necessrily converges. The limit, lim S n, is denoted lim sup x n. Literlly it is the limit of the suprem. Similrly, I n inf m n x m defines monotone incresing sequence of rel numbers which is bounded bove nd hence convergent. Then lim I n is denoted lim inf x n. Since the infimum of decresing sequence is its limit, one sometimes writes Similrly lim sup x n inf { sup x m }. n m n lim inf x n sup{ inf x m}. m n n We will use this in the proof of Theorem 4.2 below. The lim sup nd lim inf of bounded sequence {x n } n re equl if nd only if the sequence {x n } n converges, then lim inf x n lim sup x n lim x n. In generl the lim inf nd lim sup of sequence will be different. ( xmple Let x n sin n π. Then 2 lim inf x n, lim sup x n. 2. Let x n e sin(nπ/2. Then lim inf x n e, lim sup x n e. We cn use the sme definition for unbounded sequences where we use the convention tht the supremum of set which is not bounded bove is + nd the infimum of set which is not bounded below is. In generl then lim inf nd lim sup cn tke the vlues ±. xmple 4... Let x n ( n n. Then lim inf x n, lim sup x n. 2. Let x n e ( n n 2. Then lim inf x n, For both bounded nd unbounded sequences the ide is tht tht is for sufficiently lrge n. Moreover, tht is given N there is n > N so tht x n > z. lim sup x n. if z > lim sup x n then x n < z eventully, if z < lim sup x n then x n > z infinitely often, 4

15 nd Similrly if v < lim inf x n then x n > v eventully, if v > lim inf x n then x n < v infinitely often. For ny ɛ > the sequence is eventully trpped between lim inf x n ɛ nd lim sup x n + ɛ. In the picture the sequence is not converging, but it is becoming trpped between LI nd LS. xmple Let then. Let x n ( n n2 +2n+6 n then lim inf x n, lim sup x n. ( x n exp + n n + 2 sin(nπ/2, lim inf x n, lim sup x n e 2. Now here is the key result which mkes working with mesurble functions so flexible. Theorem 4.2. Let {f n } n N be sequence of mesurble functions defined on the (mesurble set R. Then the following re lso mesurble: mx f n, n k min f n, n k sup f n, n N inf f n, n N lim sup f n, lim inf f n. Proof. First recll tht if f is mesurble then so is f nd so since min f n mx { f n}, n k n k inf f n sup{ f n } n N n N nd lim inf f n lim sup{ f n } we only hve to consider three of the sttements. Now note tht {x : mx n k f n(x > } k {x : f n (x > } nd ech of the sets on the right hnd side is mesurble (since f n is. Since the union of finite or countble collection of mesurble sets is mesurble we hve shown tht mx n k f n is mesurble. {x : sup f n (x > } n k n {x : f n (x > } so in the sme wy we hve tht sup n k f n is mesurble. In prticulr the cse k gives tht sup n N f n. Finlly, lim sup f n inf { sup f m } n m n nd we hve shown tht h n sup m n f m is mesurble nd so inf n h n is mesurble nd the result is proved. Corollry 4.3. If sequence of mesurble functions {f n } n converges pointwise to limit f then the limit is mesurble function. nk 5

16 Proof. This is immedite since for convergent sequence lim f n lim sup f n ( lim inf f n which Theorem 4.2 tells us is mesurble. The null sets ply very importnt rôle in Lebesgue integrtion. Definition 4.4 (Almost everywhere. We sy tht property holds lmost everywhere, written.e., if it holds except on set of mesure zero. In prticulr, for two functions f nd g defined on set R we sy tht f g.e. if {x : f(x g(x} hs mesure zero. Similrly, if {f n } n is sequence of mesurble functions nd f n (x f(x except on set of mesure zero then we sy tht f n converges to f.e. or for.e. x. Remrk 4.5 (Null sets. Alwys remember tht which sets re of mesure zero is determined by which mesure you re using. For exmple, the null sets for Lebesgue mesure re not the sme s for counting mesure. If it is not completely cler from the context then specify which mesure you re using. xmple 4.6. The Dirichlet function of Definition.5 is equl to.e. with respect to Lebesgue mesure (since the rtionls re Lebesgue null set. Although mesurble sets nd mesurble functions re new concepts, it is worth remembering three principles ptly summrised by Littlewood:. very Lebesgue mesurble set of R of finite mesure is nerly finite union of intervls. 2. very mesurble function on set of finite mesure is nerly continuous. 3. very convergent sequence of mesurble functions defined on set of finite mesure is nerly uniformly convergent. The ctch of course is the word nerly. An exct formultion of 3 is the following importnt result. Theorem 4.7 (gorov s Theorem. Suppose tht {f k } k is sequence of mesurble functions defined on mesurble set with m( < nd ssume tht f k f.e. on. Given ɛ > we cn find closed set A ɛ such tht m(\a ɛ < ɛ nd f k f uniformly on A ɛ. Our proof requires lemm. Lemm 4.8. Let be mesurble set with m( <. Let { i } i N be sequence of mesurble sets with i i+ nd i i. Then given ɛ >, N such tht n N implies m(\ n < ɛ. Proof. This is relly rehsh of n rgument t the beginning of 2 tht we used to motivte Lebesgue s condition of countble dditivity except tht now we re not tking the sets i to be pirwise disjoint nd so we define new sequence of sets F i by F nd F i+ i+ \ i for i. The sets F i re pirwise disjoint nd i F i so by countble dditivity of Lebesgue mesure m( i m(f i. (3 6

17 Now m( < nd so the sum on the right hnd side of eqution (3 converges which implies tht given ɛ >, N so tht n N implies m( i n+ F i in+ m(f i < ɛ which on rewriting in terms of our originl sets i becomes m(\ n < ɛ whenever n N s required. Note. The ssumption tht m( < ws essentil to this proof. Proof of gorov s Theorem. We don t use the extr informtion tht A ɛ is closed in the proofs tht follow nd so we restrict ourselves to showing tht there is mesurble set A ɛ with f k f uniformly on A ɛ nd m(\a ɛ < ɛ. (The fct tht we cn tke A ɛ closed is then consequence of the first of Littlewood s three principles. Without loss of generlity we cn ssume tht f k (x f(x for every x (otherwise remove the set of mesure zero where it fils from. For every pir of non-negtive integers n nd k let k n {x : f j(x f(x < for ll j > k}. n Now fix n. Note tht n k n k+ nd k n k, so by Lemm 4.8, k n such tht m(\ n k n < /2 n. Notice tht by construction Now choose N so tht nn 2 n < ɛ nd let First note tht f j (x f(x < n when j > k n nd x n k n. m(\a ɛ A ɛ n N n k n. m(\k n n < ɛ. nn Next, if δ >, choose n N such tht /n < δ nd note tht x A ɛ implies x k n n so tht f j (x f(x < δ whenever j > k n. Hence f k f uniformly on A ɛ s required. Note tht just s for the proof of Lemm 4.8, it is crucil tht m( <. xmple 4.9. Let R nd let f k (x { if x ( k, k otherwise. Then f k (x s k for ll x, but the convergence is not uniform on sets whose complements hve finite mesure. 5 Integrtion nd the convergence theorems Finlly we re in position to follow Lebesgue s recipe nd define the integrl. We proceed in stges by progressively integrting 7

18 . Simple functions, 2. Bounded functions supported on set of finite mesure, 3. Non-negtive functions, 4. Integrble functions (the generl cse. STP. Simple functions. Definition 5.. A simple function is finite sum φ(x N k k (x (4 where ech k is mesurble set of finite mesure nd the k re constnts. Step functions re simple functions, but there re simple functions tht re not step functions. xmple 5.2. The function is simple function, but it is not step function. k φ(x Q [,] (x A compliction (tht mirrors wht you observed for step functions in Mods is tht ny given simple function cn be written in multitude of different wys s finite liner combintion of indictor functions. As trivil exmple observe tht A A for ny mesurble set A. Fortuntely there is nturl representtion which cn be described unmbiguously. Definition 5.3. The cnonicl form of simple function φ is the unique decomposition s in (4 where the numbers k re distinct nd the sets k re disjoint. Finding the cnonicl form is esy. φ cn tke on only finitely mny vlues c,..., c M sy. Set F k {x : φ(x c k }. vidently these re disjoint so set φ(x M c k Fk (x. k Definition 5.4. If φ is simple function with cnonicl form then we define the Lebesgue integrl of φ by R φ(x M c k Fk (x k φ(xdx M c k m(f k. If is subset of R of finite mesure then φ(x (x is lso simple function nd we define φ(xdx φ(x (xdx. k 8

19 Remrk 5.5. Notice tht there is relly nothing specil bout Lebesgue mesure here. If { k } M k re mesurble with respect to mesure µ then we cn define φ(xµ(dx M c k µ( k. Proposition 5.6. The integrl of simple functions defined in this wy hs the following properties:. Independence of the representtion. If φ(x N k k k (x is ny representtion of φ then φ(xdx k N k m( k. k 2. Linerity. If φ nd ψ re simple functions nd, b R then (φ + bψ(xdx φ(xdx + b ψ(xdx. 3. Additivity. If nd F re disjoint subsets of R with finite mesure then φ(xdx φ(xdx + φ(xdx. F F 4. Monotonicity. If φ ψ re simple functions then φ(xdx ψ(xdx. 5. Tringle inequlity. If φ is simple function then so is φ nd φ(xdx φ(x dx. The only slightly tricky one of these ssertions to check is the first. For tht one, if the k re disjoint then for ech distinct vlue tken by φ, tke the union of the sets k on which φ tkes tht vlue to reduce to cnonicl form nd then use tht for j k, m( j k m( j + m( k. If the k s re not disjoint, then first refine the sets so tht they re disjoint (in much the sme wy s the corresponding proof for step functions in Mods refined the prtition nd thus reduce to the previous cse. Notice tht if f nd g re simple functions tht gree lmost everywhere then f(xdx g(xdx. STP 2. Bounded functions supported on set of finite mesure. Definition 5.7. The support of mesurble function f is defined to be the set of points where f does not vnish supp(f {x : f(x }. We shll sy tht f is supported on the set if f(x whenever x /. Since f is mesurble, so is the set supp(f. The next gol is to integrte bounded functions for which m(supp(f <. The key step is to pproximte such f bove nd below by simple functions in much the sme wy s in Mods we pproximted continuous functions bove nd below by step functions. 9

20 Proposition 5.8. Let f be bounded function on mesurble set with m( <. In order tht inf{ ψ(xdx : ψ simple, ψ f} sup{ φ(xdx : φ simple, φ f} (5 it is necessry nd sufficient tht f be mesurble. We minly cre bout the sufficiency here becuse then we ll define the integrl to be the sup (or equivlently inf in (5. Proof of sufficiency. Suppose tht f is mesurble nd f < M sy. We ll follow Lebesgue s prescription to construct two sequences of simple functions {φ n (x} n nd {ψ n (x} n so tht φ n f, ψ n f nd ψ n φ n s n. The inf is clerly less thn or equl to the sup in (5 nd so this will prove the equlity. Recll then Lebesgue s recipe from (where now [, b] is replced by the mesurble set :. Subdivide the y-xis by series of points min f(x y < y < < y N > mx f(x. x x 2. Let F k {x : y k f(x < y k } nd consider the simple function with integrl N y k Fk (x (6 k N y k m(f k. k 3. Let the number of points {y k } go to infinity in such wy tht mx(y k y k. The sum in (6 gives cndidte for the φ n s. Modifying it to N y k Fk k gives cndidte for the ψ n s. All we need is concrete choice for the points y i R. Since M < f(x < M it is nturl to choose y M nd y N M. We tke N 2n nd subdivide [ M, M] eqully to obtin the other points. It is convenient to write k { (k M x : n k n+ f(x < km n }, n + k n. (This is just F k+n in our previous nottion. Now the sets { k } n k n+ re disjoint nd mesurble (since f is nd n k n+ m( k m(. Define n km n ψ n (x n (k M k (x, φ n (x k (x. n Then k n+ φ n (x f(x ψ n (x. 2

21 Thus nd { inf from which { inf } ψ(xdx : ψ simple, ψ f ψ n (xdx { } sup φ(xdx : φ simple, φ f M n n k n+ n φ n (xdx } { ψ(xdx : ψ simple, ψ f sup { km n k n+ } (k M m( k n m( k M n m(. n k n+ n k n+ km n m( k (k M m( k, n } φ(xdx : φ simple, φ f Since m( <, M < nd n is rbitrry we hve tht { } { inf ψ(xdx : ψ simple, ψ f sup s required. The proof of necessity cn be found, for exmple, in Royden. } φ(xdx : φ simple, φ f Definition 5.9. We define the quntity in (5 to be the Lebesgue integrl of f over nd denote it by f(xdx. If [, b] we ll write this s b f(xdx nd for > b define b b. Notice tht Proposition 5.8 mens tht we cn integrte bounded mesurble functions over sets of finite mesure with no extr conditions. This is lredy big step up from wht we could do with the technology of Mods. But is this theory genuine extension of tht of Mods? Tht is, for functions tht re integrble in the sense of Mods will we get the sme nswer from our modified recipe? Recll the nottion b { b b { b f(xdx sup } φ : φ L step [, b], φ f, f(xdx inf } φ : φ L step [, b], φ f Proposition 5.. Let f be bounded mesurble function on [, b]. If it is integrble in the sense of Mods then it is lso Lebesgue integrble nd the vlue of the two integrls gree. Proof. Since every step function is simple function b f(xdx sup{ φ(xdx : φ simple, φ f} inf{ ψ(xdx : ψ simple, ψ f} b f(xdx nd the result follows. In fct Lebesgue ws ble to chrcterise exctly which functions re integrble in the sense of Mods. 2

22 Theorem 5. (Lebesgue. Let f : [, b] R be bounded. b f(xdx b f(xdx if nd only if the set of points t which f is discontinuous hs Lebesgue mesure zero. xmple The Dirichlet function of Definition.5 is Lebesgue integrble over [, ], but not integrble in the sense of Mods. 2. The popcorn function (lso known s Thome s function or Riemnn s function or... defined by { /q x Q, x p/q, (p, q f(x x / Q, is integrble over [, ] in the sense of Mods nd hence lso in the sense of Lebesgue. Life is often mde esier by the following (intuitively cler result. Lemm 5.3. Let f be bounded mesurble function supported on mesurble set with m( <. If {φ n } n is ny sequence of simple functions, bounded by M, supported by nd such tht φ n (x f(x.e. x then lim φ n(xdx exists nd lim φ n (xdx f(xdx. Proof. Let {ψ n } n be s in the proof of Proposition 5.8. Note tht {φ n ψ n } n is sequence of bounded simple functions on with φ n ψ n.e.. By linerity of the integrl for simple functions it is enough therefore to show tht if {φ n } n is sequence of bounded simple functions with φ n.e. then φ n(xdx. Fix ɛ >, then gorov s Theorem gurntees the existence of (closed mesurble set A ɛ with m(\a ɛ < ɛ nd such tht φ n uniformly on A ɛ. Then φ n (xdx φ n (x dx φ n (x dx + φ n (x dx. A ɛ \A ɛ Now φ n uniformly on A ɛ so N such tht φ n (x < ɛ for ll x A ɛ whenever n N. Then for n N we hve φ n (xdx ɛm(a ɛ + Mm(\A ɛ ɛm( + ɛm. Since ɛ ws rbitrry nd M, m( re finite the result follows. Lemm 5.3 is often combined with the following result. Lemm 5.4 (Approximtion by Step Functions. Suppose tht f is mesurble function supported on set R of bounded Lebesgue mesure with f(x M for ll x. Then there exists sequence of step functions {φ n } n, supported on with φ n (x M for ll n nd ll x nd such tht φ n (x f(x s n for.e. x. 22

23 The proof, which combines Lebesgue s prescription for sequence of simple functions pproximting f with pproximtion of mesurble sets by unions of disjoint intervls (Littlewood s first principle cn be found, for exmple, in Kurtz & Swrtz. Of course in generl mesure spces we don t hve nturl nlogue of step functions nd so this result does not generlise. The following proposition is strightforwrd. Proposition 5.5. Suppose tht f nd g re bounded mesurble functions supported on set of finite mesure. Then the following properties hold:. Linerity. If, b R then (f + bg(xdx f(xdx + b g(xdx. 2. Additivity. If nd F re disjoint subsets of R with finite mesure then f(xdx f(xdx + f(xdx. F F 3. Monotonicity. If f g then f(xdx g(xdx. 4. Tringle inequlity. f(xdx f(x dx. Now we hve our integrl for bounded mesurble functions supported on sets of bounded mesure. Before moving on to Step 3, integrtion of non-negtive mesurble functions, we prove our first convergence theorem. Theorem 5.6 (The Bounded Convergence Theorem. Let {f n } n be sequence of mesurble functions defined on set of finite mesure nd suppose tht there is rel number M such tht f n (x M for ll n nd ll x. If f(x lim f n (x for ech x then f(xdx lim f n (xdx. Proof. Note tht f is mesurble (why? nd so the sttement of the result mkes sense. The conclusion of the Theorem would be trivil if f n were to converge uniformly to f, but we don t hve tht. On the other hnd, gorov s Theorem sys tht it nerly does. So given ɛ >, A ɛ with m(\a ɛ < ɛ nd such tht f n f uniformly on A ɛ. Now choose N lrge enough tht for ll x A ɛ nd n N, f n (x f(x < ɛ. Then for n N f n (xdx f(xdx (f n (x f(xdx f n (x f(x dx f n (x f(x dx + A ɛ f n (x f(x dx \A ɛ < ɛm(a ɛ + 2Mm(\A ɛ ɛ (m( + 2M. 23

24 Since ɛ ws rbitrry we see tht f n (xdx f(xdx s required. s n xercise 5.7. Show tht we could relx the condition f(x lim f n (x in Theorem 5.6 to f(x lim f(x for.e. x. WARNING: Crucil to our proof ws tht {f n } were bounded. The function in xmple.7 shows tht the result cn fil without this condition. We cn lredy mke fir mount of progress with the Bounded Convergence Theorem. xmple 5.8. Find k k(k + 2. Solution. This is n illustrtion of the technique, it is fr from the esiest wy to clculte the sum. We re going to proceed formlly nd then go bck to justify ech step. k k(k ( k k + 2 ( x k x k+ dx k k k k x k ( x 2 dx x k ( x 2 dx ( x 2 ( x dx ( + xdx 3 4. To do this properly we hve to justify the interchnge of integrtion nd summtion in pssing from the third to the fourth line. So set f n (x n x k ( x 2. k This defines sequence of bounded mesurble functions converging to ( + x on [, ] s n so the Bounded Convergence Theorem gives the required justifiction. xmple 5.9 (uler s constnt. Let u n (x x, x [, ]. n(x + n 24

25 By considering N n u n(xdx deduce the existence of ( N The limit is clled uler s constnt. Solution. First note tht N n lim N x N n(x + n n n log(n + n n 2 n., for ll x [, ] n2 N x nd since is monotone incresing s N increses (it is sum of positive terms nd n(x + n n bounded bove it must converge s N to limit. Now the Bounded Convergence Theorem tells us tht N N lim u n (xdx lim u n (xdx. N n N n The left hnd side is finite. We mnipulte the right hnd side: nd so nd N n u n (xdx n u n (x x n(x + n n x + n u n (xdx n [log(x + n] ( n + n log n N ( ( n + n log n N n ( N n log n n + n N n log(n +. n Now pply the Bounded Convergence Theorem to recover the result. Our xmple.7 with n unbounded, lbeit convergent, sequence of mesurble functions on [, ] tells us tht we mustn t chet - we crucilly used boundedness of both the function nd the mesure of its support in our proof. Now we d like to go beyond this to integrte functions where one or both of these ssumptions is relxed. STP 3. Integrtion of non-negtive mesurble functions. We re going to consider non-negtive functions, but we llow them to be extended-rel-vlued, tht is we llow the vlue +, provided tht the set where they re infinite is mesurble. Recll the convention tht the supremum of n unbounded set is +. Definition 5.2 (xtended Lebesgue integrl. For non-negtive mesurble function f on mesurble subset R we define its extended Lebesgue integrl by { } f(xdx sup g(xdx : g bounded, mesurble, m(supp(g <, g f. If the supremum is finite then we sy tht f is Lebesgue integrble over nd we write f L ( or f L (, m. 25

26 Remrk 5.2 (Advnced Wrning. We ll tlk bout L ( in bit more detil in 8. Becuse two integrble functions which differ only on set of mesure zero re indistinguishble from the point of view of integrtion - no mtter which subset of we integrte them over, the integrls will be equl - when we tlk bout points in L ( we don t relly men function, but rther n equivlence clss of functions which differ from one nother only on set of mesure zero. However, we shll loosely use the nottion f L ( to men tht the function f is Lebesgue integrble over. xmple Let Then F L (R if nd only if >. Proof. Consider F (x g n (x + x, x R. + x [ n,n](x. g n is bounded, mesurble nd its support hs bounded mesure. Moreover, g n (x F (x for ll x, so F (xdx sup g n (xdx. n N In prticulr, if the right hnd side is infinite then so is the left. On the other hnd, for ny g with g bounded nd mesurble with support of bounded mesure nd g F, then given ɛ >, since F (x s x, there exists n so tht [ n,n] c g(x < ɛ nd then, for this n, g(xdx gn (xdx + ɛ. Thus rther thn considering ll such g we find tht Suppose then tht. For > note tht n n F (xdx sup n N + x dx 2 2 n g n (xdx. + x dx dx + 2 [ n x dx ] n x n. Tking the supremum over the right hnd side we obtin 2/( < so F is integrble if >. Now suppose tht <. n n + x dx 2 + x dx n dx + 2 [ n + 26 x 2x dx ] n n.

27 This time, since <, the right hnd side diverges to infinity s n so F is not integrble over R for <. For, n n n + x dx 2 + x dx 2 [log( + x] n 2 log(n + s n. So F is lso not integrble over R. Proposition Suppose tht f nd g re non-negtive mesurble functions. Then the integrl hs the following properties:. Linerity. If, b R then (f + bg(xdx f(xdx + b g(xdx. 2. Additivity. If nd F re disjoint subsets of R with finite mesure then f(xdx f(xdx + f(xdx. F F 3. Monotonicity. If f g then f(xdx g(xdx. 4. Comprison. If g is integrble nd f is mesurble with f g then f is integrble. 5. If f is integrble then f(x < for.e. x. 6. If f is non-negtive, mesurble nd f(xdx then f(x for.e. x. Of these 4 is the one you end up using ll the time. To decide whether or not function is integrble we often try to dominte by something tht is integrble or use it to dominte something tht is not. xmple Is the function integrble over [,? x sin x x( + x 2 Solution. Yes, since nd the function on the right is integrble. x sin x x( + x x 2 Remrk As lwys, there ws nothing specil bout Lebesgue mesure in Proposition If insted we integrted ginst the counting mesure, we d recover fmilir results from Mods nlysis bout series of non-negtive terms. 27

28 Another pproch to determining whether or not function is integrble is to pply one of the celebrted convergence theorems nd we now turn to versions of those tht pply to non-negtive mesurble functions. Recll tht in xmple.7 we hd sequence of positive mesurble functions f n on [, ] converging.e. to zero, but for which lim fn (xdx /2 so tht lim f n (xdx < lim fn (xdx. This is specil cse of the cornerstone of the convergence theorems, the result rther modestly known s Ftou s Lemm. verything will follow esily from this. Recll from Definition 4.8 tht for sequence of rel numbers {x n } n N, lim sup x n lim sup x m, m n lim inf x n lim inf m n x m nd tht the ide is tht if z < lim inf x n then eventully x n > z nd lim inf x n is the lrgest possible vlue consistent with this so tht if z > lim inf x n, then for ny N there is n n > N such tht x n > z. Theorem 5.26 (Ftou s Lemm. Suppose tht {f n } n is sequence of mesurble functions with f n. If lim f n (x f(x for.e. x then f(xdx lim inf f n (xdx. Proof. Suppose tht g f where g is bounded nd supported on set of finite mesure. Set g n (x min (g(x, f n (x, then g n is mesurble, supported on nd g n (x g(x.e., so by the Bounded Convergence Theorem g n (xdx g(xdx. By construction we lso hve tht g n f n so tht g n (xdx f n (xdx. (7 Now for convergent sequence the limit nd the liminf re the sme, so g(xdx lim g n (xdx lim inf g n (xdx nd so using eqution (7 we obtin g(xdx lim inf f n (xdx. Now by Definition 5.2 f(xdx is the supremum over g of the left hnd side nd the result follows. We cn immeditely deduce some importnt corollries. NOT. In Corollry 5.27 we llow f(xdx. Corollry Suppose tht f is non-negtive mesurble function nd {f n } n is sequence of non-negtive mesurble functions with f n (x f(x nd f n (x f(x for.e. x. Then f n (xdx f(xdx. lim 28