FUNDAMNTALS OF RAL ANALYSIS by Doğn Çömez III. MASURABL FUNCTIONS AND LBSGU INTGRAL III.. Mesurble functions Hving the Lebesgue mesure define, in this chpter, we will identify the collection of functions tht re comptible with Lebesgue mesure nd then define Lebesgue integrl on these functions. Recll: f is continuous if nd only if f (O) is open for ll O open. Also, recll tht O = I k f (O) = f (I k ). Continuous functions re well behved for even Riemnn-integrl; hence we will nturlly be comptible with mesurble sets. We will consider lrger clss of functions: Definition. Let f : R #, where F. f is clled (Lebesgue) mesurble if for ll R, the set {x : f(x) > } F. Fct. Let f : R #, F. The following re equivlent: i) f is mesurble. ii) {x : f(x) } F for ll R. iii) {x : f(x) < } F for ll R. iv) {x : f(x) } F for ll R. Furthermore i)-iv) imply v) {x : f(x) = } F for ll R #. Proof. (i) (ii): First, note tht {x : f(x) } = f ([, ]). Also, since [, ] = n ( n, ], it follows tht (ii) (iii): ( ) {x : f(x) } = f ( n, ] = n= n= f ( ( n, ] ) F. {x : f(x) < } = f ((, )) = f (R \ [, ]) = \ f ([, ]) F. (iii) (iv) nd (iv) (i) re similr nd left s n exercise. For (v), if R, then {f(x) = } = f ({}) = f ((, ]) f ([, ]) F. If =, {f(x) = } = {f(x) > n} F. n=
2 xmples.. Let f : [0, ] R be function such tht { 0 if x Q f(x) = if x Q c. Then, f ( 2, ) = {x [0, ] : f(x) > 2 } = Qc [0, ] F, if =, f (, ) = F, if =, then f (, ) = [0, ] F. It follows tht, for ll R, {x [0, ] : f(x) > } F; hence, f is mesurble. 2. Let f(x) be the piecewise function defined s 0 if x x f(x) = 2 if < x < 2 if x = 2 x if x > If < 0, then f (, ) = {x : f(x) > } = (, 2 ) F. If = 0, then f (0, ) = {x : f(x) > 0} = (, 0) (0, 2) F. If 0 < <, then f (, ) = (, ) (, 2 ) F. If 2, then f (, ) = F. Hence, f is mesurble. Fct.2 f : R, F, is mesurble if nd only if for ll O B, f (O) F. Proof. xercise. Corollry. Continuous functions re mesurble. Proof. xercise. Corollry.2 Let f : R be mesurble function on. i) If D, D F, then f D nd f \D re mesurble on. ii) If f = g lmost everywhere on, then g is mesurble on. Proof. (i) Since f : R be mesurble on, {x D : f(x) > } = (f D ) (, ) = D f (, ) F (ii) Let A = {x : f(x) g(x)}. Note tht m(a) = 0; hence A F. Then, {x : g(x) > } = {x A : g(x) > } {x \ A : g(x) > } = {x A : g(x) > } {x \ A : f(x) > } F. }{{}}{{} F (f \A ) (, ) F Fct.3 Let F nd f, g : R be mesurble functions nd finite lmost everywhere. Then: i) αf + βg is mesurble on for ll α, β R. ii) f 2 is mesurble on (hence, fg is mesurble on ). iii) is mesurble on. f
3 iv) f is mesurble on. Proof. (i) First, we show tht αf is mesurble on for ll α R. If α = 0, then αf = 0. Hence αf is mesurble. If α > 0, then { {x : (αf)(x) > } = x : f(x) > } F. α Similrly for α < 0. Hence αf is mesurble, nd so is βg. Therefore, it is enough to show tht f + g is mesurble on. Notice tht, {x : f(x) + g(x) > } = {x : f(x) > g(x)} Now, there exists r x Q such tht f(x) > r x > g(x). Hence {x : f(x) > g(x)} = r Q({x : f(x) > r} {x : r > g(x)} ) F. }{{}}{{} F F (ii) First, observe tht fg = 2 [(f + g)2 f 2 g 2 ]. Therefore it is enough to show tht f mesurble implies f 2 is mesurble. Now, we hve {x : (f 2 )(x) > } = {x : [f(x)] 2 > } = {x : f(x) > } {x : f(x) < } F. }{{}}{{} F F (iii) If = 0, then {x : (x) > } = {x : f(x) > 0} F. So ssume tht > 0. Then f {x : f } { (x) > = x : f(x) < } F. (iv) Follows from {x : f (x) > } = {x : f(x) > } = {x : f(x) > } {x : f(x) < } F. Definition. For functions f, g : R, define (f g)(x) = mx{f(x), g(x)} (f g)(x) = min{f(x), g(x)} f + (x) = f(x) 0 f (x) = f(x) 0 Notice tht, with the definitions bove, we hve tht f = f + + f nd f = f + f. Corollry. If f, g : R, F is mesurble on, then: i) f +, f re mesurble on. ii) (f g)(x) nd (f g)(x) re mesurble on. Proof. xercise. Convention. For the rest of this chpter f : R. will lwys be mesurble. Definition. Let A R be set, then then the function { if x A χ A (x) = 0 otherwise
4 is clled the chrcteristic function of the set A. sy Fct. F if nd only if χ is mesurble function. For, observe tht, χ { if (, ) = otherwise Hence χ is mesurble (on ). If F, R, then χ is not mesurble. Question: If f nd g re mesurble functions. Wht bout f g? The nswer is No in generl. xmple. Recll tht f : [0, ] [0, 2] defined by f(x) = x + ϕ(x), where φ is the Cntor- Lebesgue function, is strictly incresing. Then, by Fct 9 in Chpter.II, there exists A F such tht f(a) F. xtend f continuously in strictly incresing mnner onto R. Sy g : R R such tht g [0,] = f. Then g is continuous. Next, consider χ A nd define h(x) := (χ A g )(x). Now h ( 2, ) ( ) ( ) = (χ A g ) 2, = (g χ A ) 2, ( ( )) = g χ A 2, = g(a) = f(a) F. Theorem. Let f, g be mesurble functions on their respective domins. If g is continuous, then g f is mesurble. Proof. Define h(x) := (g f)(x). If O R is Borel set, then we hve h (O) = (g f) (O) = f (g (O)). Since g is continuous, g (O) B. Hence f (g (O)) F. Recll tht, for given collection of functions {f n }, f : R, we sy tht f n f pointwise if nd only if for ll x, for ll ɛ > 0, there exists N(ɛ, x) such tht whenever n N we hve f n (x) f(x) < ɛ. Also, f n f uniformly if nd only if for ll ɛ > 0, there exists N(ɛ) such tht whenever n N we hve f n (x) f(x) < ɛ for ll x. Definition. Let {f n } be sequence of rel-vlued functions on mesurble set. We sy tht {f n } converges lmost everywhere (.e.) to f : R if there exists N with m(n) = 0 such tht f n f pointwise on \ N. Remrks. Uniform convergence pointwise convergence lmost everywhere convergence. Note tht the converse is not lwys true (xercise: Give exmples). Observtion. Let f n f pointwise, where ech f n is mesurble (on ). For x, if f(x) >, then m such tht f(x) > +. Thus, since f m n f pointwise, for lrge enough n, we will hve f n (x) > +. This mens tht, m m N such tht, n N, f n (x) > + m. ( )
5 Conversely, if (*) is stisfied, then f(x) = lim n f n (x) +. Hence, we hve m {x : f(x) > } = {x : ( m )( N )( n N)(f n (x) > + /m)} = ( ( { x : f n (x) > + m} )) F. m N n N Theorem.2 Let F nd f n : R be mesurble function for ll n. If f n f lmost everywhere on, then f is mesurble on. Proof. Since the set of x such tht {f n (x)} fils to converge to f(x) hs mesure zero, which is mesurble set, without loss of generlity by restricting f n s to subset of on which f n f pointwise, we cn ssume tht f n f pointwise on. Then by the observtion bove, for ll R {x : f(x) > } = ( ( { x : f n (x) > + m} )) F. m N n N Corollry. Let F, {f n } be fmily of mesurble functions on. If f n f uniformly or pointwise on, then f is mesurble on. Definition. Let f n be sequence of rel-vlued mesurble functions on F. Assume ech f n is bounded (i.e. there exits M > 0 such tht f n (x) < M for ll x ). Fix x nd define: f (x) = sup f n (x) n f (x) = inf n(x) n f(x) = lim sup f n (x) n f(x) = lim inf f n (x) n Then f, f, f, nd f re ll well-defined rel vlued functions on. xercise. Let f n be sequence of rel-vlued mesurble functions on F.. Show tht f, f, f, nd f re ll mesurble on. 2. Show tht f n f lmost everywhere on if nd only if m({x : lim sup f n (x) > lim inf f n (x)}) = 0. Question: If f is mesurble, does it imply tht f is mesurble? The nswer is NO in generl. xercise. Find n exmple of non-mesurble function f for which f is mesurble. Definition. A function f : R which is mesurble nd tkes only finitely mny vlues is clled simple function.
6 Fct.4 A function f : R is simple if nd only if there exists, 2,..., n, 2,... n F such tht n = k nd f(x) = k χ k (x). R nd Proof. ( ) follows from the definition. ( ) In order to show tht f is mesurble, ssume, 2,... n re its vlues. Then we hve f ({ k }) = k F. If i j, then i j nd hence i j =, nd n k =. Let n f(x) = k χ k (x). It is left s n exercise to verify tht both sides gree for ll x. Note tht the formul in the Fct.4 is simple function in cnonicl form. Of course, the sme function cn be expressed in mny different wys involving chrcteristic functions of different sets. Fct.5 Let f nd g be simple functions on F, then i) f + g is simple. ii) fg is simple. iii) f is simple (provided g 0). g Proof. xercise. Fct.6 Let F nd f : R be bounded mesurble function. Then for ll ɛ > 0, there exists simple functions φ ɛ nd ψ ɛ on such tht φ ɛ (x) f(x) ψ ɛ (x) on nd 0 (ψ ɛ φ ɛ )(x) < ɛ on. Proof. Since f is bounded there exists [, b] f(). Prtition [, b] into subintervls of length l < ɛ, sy < y 0 < y < < y n < y n = b. For i = 0,,... n let i = f ((y i, y i+ ]) F. Then Also, n n φ ɛ (x) = y i χ i (x) f(x) y i+ χ i (x) = ψ ɛ (x). i=0 i=0 n ψ ɛ (x) φ ɛ (x) = (y i+ y i )χ i (x) < ɛ. i=0 Theorem.3 (Approximtion by Simple Functions) Let F nd f : R #. Then f is mesurble on if nd only if there exists sequence {φ n } of simple functions on such tht φ n f lmost everywhere nd φ n f on. Furthermore, if f 0, then φ n s cn be chosen s φ n f lmost everywhere on. Proof. Since f = f + f nd f = f + + f without loss of generlity, we cn ssume f is non-negtive (i.e. f 0). Therefore let f 0. Note tht ( ) prt is Theorem.2 bove.
( ) Given n, let n = {x : f(x) n}. Apply the previous fct with ɛ =. So there n exists φ n nd ψ n simple functions on n such tht φ n f ψ n on n nd ψ n φ n < on n n. Now φ n f on n nd f φ n < on n n. xtend φ n to the rest of by letting φ n (x) = n on \ n. Now, we ll show tht φ n f pointwise on. If x is such tht f(x) <, then we hve f(x) < N for some N. Then, for ny n N, since x n for some n, we hve 0 f(x) φ n (x) ; hence, were done. n If x is such tht f(x) =, then tke φ n (x) = n; it follows tht pointwise. Hence φ n f pointwise. φ n (x) f(x) Corollry. Let F, function f : R is bounded nd mesurble if nd only if there exists {φ n }, simple functions on, such tht φ n f uniformly. Proof. xercise. xercises. Let F nd f : R be function. f is mesurble function if nd only if f (O) F, O B. 2. Let {f n } be sequence of R-vlued mesurble functions on F nd f : R be mesurble. f n f lmost everywhere on if nd only if m({x : lim sup f n (x) > lim inf f n (x)}) = 0. n n 3. Find function f : R, F, such tht f is mesurble wheres f is not. 7 III.2 Lebesgue Integrtion of Bounded Mesurble Functions Now tht we sorted out those functions comptible with the (Lebesgue) mesurble sets, we will proceed to define the Lebesgue integrl (of mesurble functions). Throughout we will ssume tht: F with m() <, ll functions f : R re bounded, nd ll simple functions will be in cnonicl form. Definition. Let φ : R be simple function defined s φ(x) = n i= iχ i (x). The (Lebesgue) integrl of φ, denoted by φ(x) dm, is defined s n φ(x) dm = i m( i ). Nottion: i= φ(x) dm = φ(x) dx = φ dm = φ. Remrks.. If we extend φ to R by letting φ(x) = 0 for ll x R \, then φ(x) dm = (φ χ )(x) dm. R 2. The Lebesgue integrl of step function is the sme s the Riemnn integrl of step function.
8 xmple. Let = [0, ] nd let φ(x) = χ Q [0,]. Then φ(x) dx = m(q [0, ]) = 0. Hence φ is Lebesgue integrble. Recll φ is not Riemnn integrble! xercise. Clculte [0,] φ, where φ(x) = χ C Q + 2χ C Q c + 3χ [0,]\C. Fct.7 Let φ, ψ : R be simple functions. Then i) (φ + bψ) = φ + b ψ for ll, b R. ii) φ ψ if φ ψ. iii) φ = φ + 2 φ where, 2 F, = 2, nd 2 =. Proof. i) Let φ(x) = n i= iχ i (x) nd ψ(x) = m j= b jχ Fj (x) be in cnonicl form. Let {A k } N be defined s A k = i F j with n pproprite ordering. Then {A k } N = { i F j } n,m i,j= is disjoint collection with N A k =. Hence N φ + bψ = ( i + bb j )χ Ak where i, j corresponds to the set A k with A k = i F j. Therefore, we hve N N N (φ + bψ) = ( i + bb j )m(a k ) = i m(a k ) + bb j m(a k ) = N i m(a k ) + b = φ + b ψ. N b j m(a k ) = n i m( i ) + b ii) Since φ ψ on, ψ φ 0 on. Also since ψ φ is simple, by i) we hve (ψ φ) 0 ψ φ. Proof of iii) is left s n exercise. i= m b j m(f j ) Recll: Given ny f : R bounded, we cn lwys find (construct) simple functions φ nd ψ such tht φ f ψ on nd 0 (ψ φ)(x) cn be mde rbitrrily smll on. Definition. Let F nd f : R be bounded function. We sy tht the function f is (Lebesgue) integrble over if { } { } sup φ : φ f on = inf ψ : ψ f on, where φ nd ψ re simple functions. The common vlue is clled the (Lebesgue) integrl of f over nd is denoted by f(x) dx = f = f dm. j=
9 For the simplicity of the rguments below, we will denote: inf ψ = U f nd sup φ = L f. f ψ f φ Theorem.4 Let F, m() <, nd f : R be bounded function. mesurble on if nd only f is Lebesgue integrble over Then f is Proof. ( ) Assume f is mesurble nd f M. Prtition the intervl [ M, M] into 2n subintervls of equl length. Let M = y n < y n+ < < y n < y n = M be the endpoints of these subintervls. Define k = f ([y k, y k+ ]). Then, k F for ll k = n,... n with k l = for k l, nd n m( k ) = m(). Now define φ n = n k= n y kχ k k= n where {ψ n } nd {φ n } re integrble. Then ψ Thus, inf f ψ inf f ψ nd ψ n = n k= n y k+χ k. Therefore, on the set, φ n (x) f(x) ψ n (x), ψ n nd sup f φ φ [ n ] ψ sup φ (ψ n φ n ) = (y k+ y k )χ k < M f φ n m(). k= n Since n is rbitrry, we hve L f = U f nd f is Lebesgue integrble. ( ) Given tht f is bounded on nd L f = U f, let 0 < ɛ n = for n. From these n ssumptions nd the definitions of inf / sup, we cn pick sequence of simple functions {ψ n } nd {φ n } on such tht φ n f ψ n on, nd ( ) 0 ψ n φ n < n. Let ψ n = inf n ψ n nd φ n = sup n φ n. Since ech ψ n nd φ n re mesurble, we hve tht both ψ n nd φ n re mesurble on, nd φ n f ψ n on. We need to show ψ n = φ n lmost everywhere on. To show this let A = {x : φ(x) < ψ(x)}. We will show tht m(a) = 0. To do this let A k = {x : ψ(x) φ(x) > /k}, then k A k = A. Now let B n k = {x : ψ n (x) φ n (x) > /k}. Then A k Bk n F nd, on, k(ψ n φ n ) >. So m(bk n ) = k(ψ n φ n ) = k χ B n k It follows tht m(a k ) = 0. Hence, we hve m(a) = 0. φ n. (ψ n φ n ) < k n.
0 Fct.8 Let f : [, b] R be bounded function. If f is Riemnn integrble, then f is Lebesgue integrble with b f dm = f(x) dx. Proof. xercise. [,b] Fct.9 Let F, m() <, f, g : R be bounded functions. Then we hve: ) (αf + βg) = α f + β g for ll α, β R. b) If f = g lmost everywhere on, then f = g. c) If f g lmost everywhere, then f g (Hence f f ). d) If there exists, b R such tht f(x) b on, then m() f bm(). e) If, 2 F, nd = 2 with 2 =, then f = f + f. 2 Proof. xercise. xercise. Let F with m() <, nd {f n } be sequence of rel vlued, bounded mesurble functions on such tht f n f uniformly. Then f n f. The uniform convergence condition in the exercise bove cnnot be relxed s the following exmple shows. xmple. Let f n : [0, ] R, n, be given by n 2 x if x [0, n ] f n (x) = n 2 x + 2n if x [ n, 2 n ] 0 otherwise, nd f 0 on [0, ]. Then, f n = 0 = f! [0,] [0,] Theorem.5 (goroff) Let F with m() <, nd {f n } be sequence of rel vlued mesurble functions on such tht f n f lmost everywhere. Then for ll ɛ > 0, there exists mesurble A ɛ, m( \ A ɛ ) < ɛ, such tht f n f uniformly on A ɛ. Proof. Given ɛ > 0 nd for n, j Z +, define n j = {x : f k (x) f(x) < /j}. k=n
Then n j F for ll n, j. If A such tht f n f pointwise on A (i.e. m( \ A) = 0), then for ll x A, x n j for some n, j. Hence A n. j Therefore A ( n j m n= n= j n ) n= = m(), since f n f.e. (hence m(a) = m()). Observe for fixed j, n j n+. j Hence by continuity of m, m( \ n) j = m() m(n) j nd ( ) lim m( \ j n) = m() lim m( j n) = m() m n j = 0. Consequently, for ll j, we cn find n(j) such tht n n(j) nd m( \ n) j < ɛ. Let 2 j A ɛ = j= j n(j), then A ɛ F. Then ( ) m( \ j n(j) ) = m ( \ j n(j) ) m( \ j n(j) ) < ɛ 2 = ɛ. j j= j= j= j Now, it is left s n exercise to show tht f n f uniformly on A ɛ. Remrk. By the exmple bove, if f n f pointwise on F, in generl, we cnnot expect f n f. However, under somewht mild dditionl conditions we hve positive nswer: n= Theorem.6 (Bounded Convergence Theorem) Let F, m() <, nd {f n } be sequence of uniformly bounded mesurble, rel vlued functions on. If f n f lmost everywhere, then f n f. Proof. Since {f n } is uniformly bounded, we cn find positive rel number M such tht f n (x) M. Then, by goroff s Theorem, given ɛ > 0, there exists A such tht m(\a) < ɛ nd f 4M n f uniformly on A. Then, find N lrge enough tht, for n N, f n (x) f(x) < ɛ for ll x A. Now for such n N, 2m(A) f n f = (f n f) + (f n f), nd hence, f n f A (f n f) + A \A \A (f n f) since f n f 2M. Therefore, we hve f n f < ɛ 2Mɛ m(a) + 2m(A) 4M = ɛ. A f n f + \A 2Mdm, Remrk. The BCT cn be restted s sying tht under the given conditions the lim nd cn be interchnged.
2 xmples.. Given { if x [/2, ] [/4, /3] [/6, /5]... f(x) = 0 otherwise Define f (x) = χ [/2,] f 2 (x) = χ [/4,/3] + χ [/2,]. f n (x) =. n χ [/(2k),/(2k+)] It is strightforwrd to show tht f n f pointwise (hence, is left s n exercise). Then {f n } is uniformly bounded sequence of mesurble functions. By the Bounded Convergence Theorem, [ n ([ ]) ] f = lim f n = lim m [0,] [0,] 2k, 2k [ n ( = lim 2k ) ] [ 2n ] = lim ( ) k+ 2k k ( ) k+ = = ln(2). k 2. Let f(x) = { 2x if x [0, ] Q x if x [0, ] Q c If we define g(x) = x on [0, ], then f = g lmost everywhere nd g is Riemnn Integrble = Lebesgue Integrble. Hence, f(x) dx = f(x) dx + f(x) dx = g(x) dx = [0,] Q c 0 2. [0,] [0,] Q xercises. Using only the definition, evlute the integrls: [0,] 2. Let C denote the Cntor Set. Find f(x) dx if [0,] x 2 if x C Q f(x) = x if x C Q c if x [0, ] \ C. 3. Prove tht f n f uniformly on A ɛ in goroff s Theorem. x dx nd [0,] x/3 dx. III.3. Lebesgue Integrl of Non-negtive Functions In the previous section we defined the Lebesgue integrl for bounded mesurble functions. Now, we will continue with not necessrily bounded functions. We will del with this cse in
two steps: first we will define the Lebesgue integrl of the functions f : R +, nd then extend it to more generl functions f : R. Definition. Let F (m() = is llowed) nd f : R + be mesurble. We define fdm = sup{ hdm : h : R is bounded, mesurble, h f, m({x : h(x) 0}) < }. Remrk. () It is possible tht f = (2) Functions h : R such tht m({x : h(x) 0}) < re clled functions with finite support. The set {x : h(x) 0} is clled the support of h. 3 xercise. Find [0,] x dm. Fct.0 Let F, f, g : + be mesurble functons. Then () (αf + βg) = α f + β g. (b) If f g lmost everywhere on, then f g. (c) f = f + 2 f where = 2 nd 2 =. Proof. The result follows from the the sme fct on bounded functions; hence, it is left s n exercise. Theorem.7 (Ftou s Lemm) Let F nd {f n } be sequence of R + -vlued, mesurble functions on. If f n f lmost everywhere on, then f dm lim inf f n dm. Proof. Let φ be bounded, mesurble, non-negtive function with m({φ 0}) < such tht φ f lmost everywhere on. For ech n, define h n = f n φ = min{f n, φ}. Then: i) h n is bounded nd mesurble for ll n. ii) h n f n. iii) h n s vnish outside A = supp(φ) = {x : φ(x) 0}. iv) h n φ lmost everywhere on A. Then, by the Bounded Convergence Theorem, φ dm = φ dm = lim h n dm lim inf A A (Lst inequlity follows since lim n f n dm my not exist.) Therefore, f dm = sup φ dm lim inf f n dm. φ f f n dm. Remrks.. Strict inequlity in Ftou s Lemm is possible. 2. If non-negtivity of f n s is removed, the conclusion of Ftou s Lemm is not vlid nymore.
4 xercise. Provide exmples for ech of the remrks mde bove. Theorem.8 (Monotone Convergence Theorem) Let F, nd {f n } be sequence of R + -vlued mesurble functions on such tht f n f n+.e. (on ) for ll n. If f n f.e., then lim f n dm = f dm. Proof. From hypothesis, f n f n+ for ll n lmost everywhere on. Hence f n dm f dm for ll n. Therefore by Ftou s Lemm lim sup f n dm f dm lim inf Hence, it follows tht lim f n dm = f dm. Remrks.. MCT Ftou s Lemm. 2. MCT does not hold for Riemnn Integrl. 3. Monotonicity in MCT is essentil. f n dm. xmple. Let = R ([0, ]) nd f n (x) = χ [n, ) (x), n. Then f n 0 pointwise (but not monotoniclly), wheres, = R f n 0 = Corollry. Let F nd {u n } be sequence of positive rel vlued functions on. If f = u n (pointwise), then f dm = u n dm. n= n= R f. Proof. Define f n = n u k, for n. Then f n f.e. on. Apply MCT: n lim f n dm = f dm lim u k dm = u k dm = f dm. Corollry.2 Let F nd { n } F such tht = n n nd i j = if i j. Then for ny f : R + mesurble, f dm = f dm n n= Proof. Let u n = fχ n nd pply Corollry.. Theorem.9 (Chebychev s Inequlity) Let F nd f : R + be mesurble function. Then for ll λ > 0, m({x : f(x) > λ}) f dm. λ
Proof. Let λ = {x : f(x) > λ}. Assume first tht m( λ ) < nd let φ = λχ λ. Then 5 on λ, φ f. Hence φ dm λχ λ dm f dm = λm( λ ) m( λ ) f dm. λ Next we will ssume tht m( λ ) =. If we let λ n = λ [ n, n], then λ n λ nd m(λ n) < for ech n. Now, pplying bove cse to n λ nd φ n = λχ n λ, we obtin λm(λ) n f dm. Then, by continuity of m, λm( λ ) = λ lim m( n λ) f dm. Note tht this gives equlity on both sides since m( λ ) =. Hence, m( λ ) = f dm. λ Theorem.0 Let F nd f : R + be mesurble. Then f dm = 0 f = 0 lmost everywhere on. Proof. ( ) Assume f dm = 0 nd consider, for ny n, the set S = {x : f(x) > /n}. Then by Chebychev s Inequlity, Now, since it follows tht f = 0.e. on. m(s) n f dm = 0. {x : f(x) > 0} = n {x : f(x) > /n}, ( ) Next, ssume tht f = 0 lmost everywhere on. Let, for ll n, n = [ n, n]. On ech n let h be bounded, mesurble, non-negtive function such tht h f, nd φ be ny bounded simple non-negtive function with φ h (i.e. 0 φ h f.e.). Hence φ must be 0 lmost everywhere on n. Then φ = 0 h = 0. n n Since h is rbitrry, f = 0 follows from the definition. Remrk. Theorem.0 implies tht f = g.e. on F f = g. Definition. Let F. A function f : R + is clled (Lebesgue) integrble over if f dm <.
6 xmple. Let s show tht the function f(x) = is Lebesgue-integrble on [, ) nd clculte x 2. For, we will use MCT. First, let, for n, [, ) x 2 f n (x) = x if x n 2 0 if otherwise Hence, ech f n is bounded nd mesurble on [, ). Thus, n f n = f n = x. 2 [, ) [,n] Since is Riemnn-integrble on [, n], it is lso Lebesgue integrble on [, n] (with both x 2 integrls being equl). Hence, f [, ) n = n =. Therefore, by MCT, x 2 n x = lim f 2 n =. n [, ) [, ) Fct. Let F nd f : R + be integrble over. Then f(x) <.e. on. Proof. It is enough to show tht m({x : f(x) = }) = 0. For, first observe tht {x : f(x) = } = n {x : f(x) > n}. Now, by Chebychev s Inequlity, m({x : f(x) = }) m({x : f(x) > n}) f. n Since f <, letting n, we obtin the desired result. Theorem. (Beppo Levi s Lemm) Let {f n } be n incresing sequence of nonnegtive mesurble functions on F. If the sequence { f n} is bounded, then i) f n p f on for some non-negtive mesurble function f : R #, ii) lim n f n = f <, nd iii) f is finite.e. Proof. For ech x, the sequence {f n (x)} is n incresing sequence of rel numbers; hence, it converges to n extended rel number. Therefore, we cn define f pointwise by f(x) = lim n f n (x), for ll x. Since f n f, by MCT, we hve f n f. Therefore, by the fct tht { f n} is bounded, we hve hve f < ; nd by Fct. bove we must lso hve tht f is finite.e. The following fct is nother importnt property of integrble functions. Theorem.2 Let f : R + be n integrble function on F. Then, for ny ɛ > 0 there exists δ > 0 such tht A, A F, with m(a) < δ, we hve A f < ɛ. Proof. If f < M for some M on, tke δ = ɛ. The the ssertion follows. M If f is rbitrry integrble function, let { f(x) if x n f n (x) = n if x > n.
Then {f n } is bounded sequence nd f n p f. on. So, by MCT, f n f. Hence, given ɛ > 0, there exists N such tht if n N, then [ f f n] < ɛ. In prticulr, if n = N, 2 we hve (f f N) < ɛ ɛ. Now, pick δ <, then for ny A with m(a) < δ we hve 2 2N f = (f f N ) + f N (f f N ) + N < ɛ 2 + ɛ 2 = ɛ. A A A A 7 xercise. Let f : R R + be n integrble function. Define set function ν : F [0, ] by ν(a) = fdm, A F. Show tht A i) 0 ν(a) for ll A F. ii) ν(a) ν(b) if A B, A, B F. iii) ν( ) = 0 = ν({}) for ny R. iv) ν( i=a i ) = i= ν(a i) if {A i } F is disjoint fmily of sets. v) ν(a) = 0 if m(a) = 0. Question. For f nd ν s in the exercise bove, wht cn you sy bout i) ν(i) = l(i) for ny intervl I R? ii) ν(a + α) = ν(a) for ny A F nd α R? iii) m(a) = 0 if ν(a) = 0 for A F? III.4. Lebesgue Integrl of Arbitrry Functions In this section, hving defined the (Lebesgue) integrl of non-negtive mesurble function, we will extend it to rbitrry mesurble functions. First, recll tht (i) f = f + f, nd (ii) if f is mesurble, then so re f + nd f. Definition. A function f : R where F, is clled (Lebesgue) integrble over if both f + nd f re integrble over. In tht cse, f = f + f. Remrk. Since f + nd f re non-negtive nd integrble, the set on which they tke s vlue is of mesure zero; hence, the set on which f my tke + or vlues is lso of mesure zero. Fct.2 If F, f : R is integrble nd A, A F with m(a) = 0, then f = f. Proof. xercise. Fct.3 Let F, f : R be mesurble nd g : R + be integrble with f g on. Then f is integrble with f g. In prticulr, f f. \A
8 Proof. xercise. Fct.4 If F nd f, g : R re integrble over, then () αf is integrble over nd αf = α f. (b) f + g is integrble over nd (f + g) = f + g. (c) f g on, then f g. (d) If = 2, where nd 2 re mesurble nd disjoint, then fχ nd fχ 2 re integrble with f = f + f. 2 Proof. The proof of () nd (c) re left s exercises. For (b) without loss of generlity, if necessry, by removing lrger set of mesure zero, we cn ssume tht f + g is finite on. Thus (f + g) + (f + g) = f + g = (f + f ) + (g + g ) nd (f + g) + + f + g = (f + g) + f + + g +. Since both sides re non-negtive, we hve (f + g) + + f + g = (f + g) + f + + g +. If we reorgnize we hve [ (f + g) + (f + g) ] = Hence f + g is integrble. (f + + f ) + (g + g ). For the proof of (d), since f is integrble we know tht fχ f nd fχ 2 f. Now both fχ nd fχ 2 re integrble with χ = χ + χ 2. Theorem.3 (Dominted Convergence Theorem) Let {f n } be sequence of rel vlued mesurble functions on F. Let g : R be n integrble function over such tht f n g for ll n on. If f n f lmost everywhere on, then f n nd f re integrble over nd f n f. Proof. Since g is integrble over nd f n g, then ech f n is integrble over. But then, since f n f lmost everywhere, we lso hve f g on. Hence f is integrble over. Observe tht {g f n } is sequence of non-negtive integrble functions on converging to g f. Similrly {g + f n } is sequence of non-negtive functions converging to g + f. Now by pplying Ftou s Lemm to {g f n } we hve g f = (g f) lim inf (g f n ) = g lim sup f n f lim sup f n. Next, by pplying Ftou s Lemm to {g + f n } we hve f + g = (g + f) lim inf (g + f n ) = f lim inf f n. g + lim inf f n
9 These two inequlities imply tht f n f. xercise. For ech n, let f n (x) = n3/2 x +n 2 x 2, x [0, ]. Show tht () f n 0 lmost everywhere on [0, ]. (b) f n (x) g(x) on [0, ] for ll n where g(x) = (c) lim [0,] f n = 0. Remrk. Converse of the DCT is not vlid! xmples. Let = [0, ]. x.. Let f n (x) = x for ll n nd f(x) = x. Then, f n f, but f n f. 2. Let f n (x) = n n 3. Define x nd f(x) = x. Then f n f, but f n f. f 0 = χ [0,] f = χ [0, 2 ], f 2 = χ [ 2,] f 3 = χ [0, 3 ], f 4 = χ [ 3, 2 3 ], f 5 = χ [ 2 3,] f 6 = χ [0, 4 f 7 = χ [ 4, ], 2 f 8 = χ [ 2, 3 ], 4 f 9 = χ [ 3 4,] f 0 = χ [0, 5.... Then we hve f n 0, but f n 0 lmost everywhere. Theorem.4 (Generlized Dominted Convergence Theorem) Let {f n } be sequence of rel vlued mesurble functions on F such tht f n f lmost everywhere on. Let {g n } be sequence of rel vlued non-negtive functions on such tht g n g lmost everywhere on. Also let f n g n for ll n on. If g n g, then f n f. Proof. xercise. Corollry. Let F nd f : R be integrble over. () If = n (disjoint), n F. Then n= f = n= n f.
20 (b) Given { n } F, n nd n. If A = n n, then f = lim f. A n (c) Given { n } F, n, nd n. If A = n n, then f = lim f. A n Proof. xercise. [Hint: Use Corollry. before Chebychev s Inequlity.] Corollry.2 Let R nd f : R be integrble over. Then given ɛ > 0 there exists δ > 0 such tht for ll A F, A nd m(a) < δ, then f < ɛ. Proof. xercise. A The results stted bove hve some importnt nd interesting rmifictions. Let ν(a) = f A for ll A F, where f : R R is integrble. Then by Corollry.() bove, ν A k = ν(a k ). disjoint Also ν( ) = 0, 0 ν(a), A B ν(a) ν(b). Notice tht ν is not mesure; however, it cts like mesure. Such set functions re clled signed mesures. It turns out tht the converse of Corollry.2 is lso vlid if m() <. Fct.5 Let F, m() <, nd f : R be mesurble function on. If for ll ɛ > 0 there exists δ > 0 such tht for ll A F, A nd m(a) < δ, we hve tht f < ɛ, then A f is integrble over. Proof. WLOG we cn ssume tht f 0. Tke ɛ = nd let δ 0 be the corresponding vlue such tht whenever A, A F, with m(a) < δ 0, we hve A f <. Clim: = n k (disjoint) such tht m( k ) < δ 0 for ll k n. (The proof of this clim is left s n exercise.) Hence we now hve fχ k = f < for ll k n. A k So ech fχ k is integrble. Therefore, f = n fχ k is integrble. xercises. Let f : [, b] R be bounded function. If f is Riemnn-integrble, then f is Lebesgueintegrble nd b f(x)dm = f(x)dx. [,b]
2. Let F with m() < nd {f n } be sequence of R-vlued, bounded mesurble functions on such tht f n f uniformly on. Then f n dm fdm. [Do not use BCT!] 3. The function f(x) = x is mesurble on [0, ]. Clculte [0,] x dm. [Notice: f is not Riemnn-integrble on [0, ].] 4. Find f(x) dm if [0,] where C is the stndrd Cntor Set. x 2 if x C Q f(x) = x if x C Q c if x [0, ] \ C 5. Let {f n } n be sequence of R-vlued functions on [0, ] defined s { 0 if x (0, n+ f n (x) = ) [, ] n x 3 2 if x [, ). n+ n Show tht ) f n 0.e. b) [0,] f n 0. c) There is no integrble function g : [0, ] R such tht f n g.e. for ll n. How would you reconcile this result with the DCT? 2 III.5. Convergence in Mesure In regrds to convergence of sequence of functions, in ddition to uniform, pointwise nd.e. convergence, we cn dd nother mode of convergence s follows. Definition. Let {f n } be sequence of rel vlued mesurble functions on F nd let ech f n be finite lmost everywhere on. For f : R mesurble nd finite lmost everywhere on, we sy tht the sequence {f n } converge in mesure on to f if for ll α > 0 Nottion: f n m f. lim m({x : f n(x) f(x) α}) = 0. Remrk. It is esy to see tht if f n u f, then f n m f. Indeed, with n dditionl condition we cn replce uniform convergence with.e. convergence by the following result. Fct.6 Let m() < nd {f n } be sequence of rel vlued mesurble functions on F, nd f : R be finite lmost everywhere. Then f n.e. f f n m f. Proof. Clerly f is mesurble on (why?). Let α > 0 nd ɛ > 0 be rbitrry. By goroff s Theorem, there exists F, F F, such tht m( \ F ) < ɛ nd f n u f on F. Hence there exists N such tht when n N we hve f n (x) f(x) < α
22 for ll x F. Thus if n N, m({x : f n (x) f(x) α}) m( \ F ) < ɛ. Hence m({x : f n (x) f(x) α}) 0. Remrks.. The condition m() < is essentil. (xercise: Provide counterexmple if m() =.) 2. The converse of Fct.6 is not true in generl, either. (xercise: provide n exmple). However, it s vlid long subsequence: Theorem.5 If f n m f on F, then there exists subsequence {f nk } of {f n } such tht f nk.e. f. Proof. Let ɛ k =. Applying the hypothesis, for ech k there exists N 2 k k such tht when n N k, m({x : f n (x) f(x) }) <. Let k 2 k Then k = {x : f Nk (x) f(x) > k }. m( k ) <. Hence by Borel-Cntelli Lemm, for lmost every x, the point x belongs to t most finitely mny k s. Tht mens for lmost every x, there exists k x such tht x k for k k x. So, for lmost every x, f nk (x) f(x) < k for ll k k x. Hence f nk p f lmost everywhere on. Remrk. In Ftou s Lemm, MCT nd DCT, the condition f n.e. f on cn be replced by f n m f on. III.6. Riemnn Integrl vs Lebesgue Integrl Recll: Let f : [, b] R be bounded, Ṗ n be tgged prt prtition. Then n S(f, Ṗn) = f(t i )(x i+ x i ) S(f, Ṗn) b i= f =: Riemnn Integrl of f on [, b]. Let s look t this process little differently. For f : [, b] R bounded function nd P n prtition of [, b], define step functions {α n } nd {β n } by n n α n (x) = u k χ [xk,x k+ ](x), β n (x) = v k χ [xk,x k+ ](x), respectively, where u k = inf x [xk,x k+ ]{f(x)} nd v k = sup x [xk,x k+ ]{f(x)}. Then, (i) α n nd β n s n pointwise (or s P n 0). (ii) For ll n, α n nd β n re Lebesgue integrble (by construction).
23 (iii) For ll n, α n nd β n re Riemnn-integrble since, b n L n (f) = α n = m k (x k+ x k ) nd U n (f) = b β n = n v k (x k+ x k ), nd tht L n (f) nd U n (f) re specil cses of S(f, Ṗn). We lso hve tht L n (f) nd U n (f). Therefore, f is Riemnn-integrble if nd only if lim n U n = lim n L n (= b f). In generl, since f is bounded, both sequences α n nd β n re uniformly bounded; hence, α n (x) α(x) nd β n (x) β(x) pointwise, for some α : [, b] R nd β : [, b] R. Now, the Dominted Convergence Theorem implies tht L n (f) = α n α, where α = lim α n (x) nd [,b] U n (f) = [,b] [,b] β n [,b] β, where β = lim β n (x). If α = β, then the Lebesgue Integrl of f exists nd is equl to the common vlue. [,b] [,b] Therefore, ny bounded Riemnn-integrble function is Lebesgue integrble. Fct.7 f : [, b] R is continuous t x [, b] if nd only if α(x) = β(x). Proof. xercise. Theorem.6 Let f : [, b] R be bounded. () f is Riemnn Integrble on [, b] if nd only if f is lmost everywhere continuous on [, b]. (b) If f is Riemnn Integrble on [, b], then f is Lebesgue Integrble nd b f = f. Proof. From bove observtion, if () is proven, (b) follows esily. [,b] () ( ) Since f is Riemnn Integrble, then Since α β, we hve [,b] b lim k (β α) = 0 = α k = lim b k b β k b α = b β. (β α) α = β lmost everywhere. Hence by Fct.7, f is continuous lmost everywhere on [, b]. ( ) If f is continuous lmost everywhere, then by Fct.7 for lmost every x [, b] α(x) = β(x)(= f(x)). Since α nd β re Lebesgue Integrble, α = β. [,b] [,b]
24 Hence we must hve tht if L n = b α n = [,b] α n nd U n = b β n = then U n L n = 0. Tht is f is Riemnn Integrble nd we must hve b ) f = lim α n (= lim β n. b b [,b] β n, Importnt Remrk. All the concepts introduced nd developed in this chpter cn be rephrsed in n rbitrry σ-finite mesure spce (X, A, µ) nd for functions f : X R (with pproprite modifictions). We will leve this s n exercise.