4 Sequences of measurable functions

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1 4 Sequences of measurable funcions 1. Le (Ω, A, µ) be a measure space (complee, afer a possible applicaion of he compleion heorem). In his chaper we invesigae relaions beween various (nonequivalen) convergences of sequences of A -measurable funcions f n on Ω. 2. Le us recall he various noions of convergence. Suppose f, f 1, f 2,... are measurable. (a) Convergence µ -almos everywhere (in paricular he poinwise convergence): f n f µ -a.e. in Ω as n iff for some Ω wih µ() = 0, we have f n (x) f(x) for any x Ω \ as n. (b) Convergence uniform on : f n f uniformly on Ω as n iff ( ) lim sup f n (x) f(x) = 0. x (c) Convergence in he Banach space L p, 1 p : f n f in L p as n iff f, f n L p and f n f p 0 as n. (d) Convergence in measure: f n f in measure as n (we wrie f n f in µ as n ) iff > 0 lim µ ( f n f > ) = 0. xcep he convergence in measure, we have encounered hese noions before. An imporan ype of convergence is missing in he above lis laer in he course we will also encouner he weak convergence. All hese convergences are exremely imporan in applicaions such as PDs, applied mahemaics, sochasics, and ohers. 3. In conras wih he basic inegraion heory, i is inconvenien o allow he funcions o be infinie on ses of posiive measure, when dealing wih convergences. For example, if f n f in µ as n, hen f mus be µ -a.e. finie. Indeed, suppose f = wih µ() = m 0 > 0. If all f n are µ -a.e. finie on, hen µ( f n f > ) m 0 for all n and any > 0, which conradics he convergence in µ. The only way o save he convergence is o allow f n be infinie on a se of posiive measure. Bu hen we have o clarify wha do we mean by f n f = on a se of posiive measure. Unforunaely here is no useful universal agreemen for " ", even hough in some paricular problems i is possible o assign a meaning o he equaliy. To make he exposiion concise we make he sanding assumpion 1

2 in all saemens in his chaper all funcions are assumed o be measurable and finie almos everywhere. 4. The a.e. convergence (say, he poinwise convergence) is he mos elemenary noion. I happens o be he mos suble and he mos difficul o deal wih. For example, he heory for λ 1 -a.e. convergence of he e ikx -Fourier series is subsanially more complicaed han, say, he L 2 -convergence. 5. Some relaions beween differen convergences are sraighforward. Prove ha he convergence uniform on Ω implies all oher convergences, assuming ha µ(ω) <. Prove ha he convergence in L p, 1 p implies he convergence in measure (use he Tchebyshev s inequaliy for a finie p ). 6. Le F = F(Ω) be he se of all measurable funcions which are finie µ -a.e. in Ω. Suppose µ(ω) <. For f, g F define f g d(f, g) = d µ (f, g) = 1 + f g dµ. Prove ha: (a) F is a vecor space; (b) (F, d) is a meric space wih he ranslaion invarian meric, d(f, g) = d(f + h, g + h) ; (c) d(, ) merises he convergence in measure, ha is f n f in µ as n if and only if d(f n, f) 0, n ; (d) (F, d) is a complee meric space (his is a more difficul problem, do i las). I is possible o prove (we will no do i) ha here does no exis a norm µ on F such ha he convergence in µ is equivalen o he convergence in he normed space (F, µ ). 7. Firs we invesigae he relaion beween he µ -a.e. convergence and he convergence in µ. The crucial role will be played by he following simple observaions. Recall, ha for a sequence of ses S n lim sup S n = poins which belong o infiniely many S n. For a sequence of measurable funcions ϕ, ϕ 0, lim sup (ϕ ) Ω x Ω: lim sup ϕ (x) Prove his. Can we replace he inclusion by he equaliy here? (Prove or give an example.) In he opposie direcion, prove ha lim sup (ϕ > ) x Ω: lim sup ϕ (x) >. 2.

3 Can we replace he inclusion by he equaliy here? example.) (Prove or give an 8. Theorem 1 Le f be a sequence, such ha f f µ -a.e.,. Suppose ha µ(ω) <. Then f f in µ as. Proof. 1. Se g = f f. Then g is a sequence of measurable funcions, g 0, such ha g 0 µ -a.e. in Ω as. For any > 0 define = x Ω: g (x). (4.1) We mus show ha for any > 0 2. For any > 0 he inclusion lim sup µ( ) 0 as. x Ω: lim sup def g (x) = L holds. A he same ime clearly g 0 µ a.e., µ(l ) = 0 > 0. To use his fac, recall he expression lim sup = k=1 =k. Hence by he coninuiy of µ along he monoone sequences of ses 0 = µ(l ) µ lim sup = lim µ =k lim sup µ(k). 9. Where in he proof of Theorem 1 we used he assumpion µ(ω) <? Prove ha he Theorem 1 does no hold in general if µ(ω) =. Prove ha he Theorem 1 does no hold in general if f is allowed o be infinie on a se of posiive measure. 10. In general, convergence in measure (or convergence in L p for a finie p ) does no imply he a.e. convergence. Show his by example of he sequence of funcions on [0, 1] given by he χ (N) for suiable N and N N. Here (N) = [( 1)2 N, 2 N ) is he dyadic inerval. However, he implicaion holds if we pass o a subsequence. 3

4 11. Theorem 2 Le µ(ω) <, and le f be a sequence, such ha f f in µ as. Then here exiss a subsequence k such ha f k f µ -a.e when k. Proof. 1. Se g = f f. For any ε > 0 define ε = g > ε. (Noice he sric inequaliy here.) Then g is a sequence of measurable funcions, g 0, such ha µ( ε ) 0 as. (4.2) We mus find a subsequence k such ha g k 0 µ -a.e. in Ω as k. 2. Take any vanishing sequence ε k, ε k 0 as k (for example ε k = 1/k ). Observe ha for any subsequence k x Ω: lim sup g k (x) > 0 lim sup ε k k. (4.3) The se in he lef hand side is exacly he se where he sequence g k does no limi o 0 poinwisely. Our goal is o conrol is size. 3. We will choose he desired subsequence k. Tes (4.2) wih ε = 1 o find 1 such ha µ( 1 1 ) < 1/2. Then by inducion es (4.2) wih ε = 1/k o find k such ha k 1 < k, µ( 1/k k ) < 2 k, k = 2, 3,.... Our goal is o esimae he measure of he se on which he convergence of g k o 0 fails. By (4.3) i is enough o show ha ( ) µ lim sup 1/k k = 0. k Bu according o our choice k k=1 µ( 1/k k ) 2 k <. One of he earlier home work problems showed ha for any se sequence F k µ(f k ) < = µ lim sup F k = 0. k k=1 This general saemen is called Borel-Canelli s lemma. heorem. I proves he 4

5 4. In order o make he exposiion independen of he home assignmens, le us prove he Borel-Canelli s lemma. By he definiion lim sup F k = k=1 n=k F n. From he coninuiy of µ along he monoone sequences of ses, derive ha ( ) µ lim sup F k = lim µ F n n=k ( ) = 0, lim n=k µ (F n ) where he las sep holds due o he convergence of he series µ(f n ). 12. Where in he proof did we use µ(ω) <? Prove ha Theorem 2 holds for µ = λ 1 and Ω = R despie λ 1 (R) =. Suppose f n f in L 1 (R) as n, and le f n 0 for all n. Show ha f 0 almos everywhere. 13. The gorov s heorem is imporan in applicaions. I assers ha he µ -a.e. convergence is uniform provided we hrow away a se of arbirary small measure. Theorem 3 Le µ(ω) <, and le f k f µ -a.e.. Then for any ε > 0 here exiss a se F = F ε Ω such ha: (i) µ(f ) < ε ; (ii) f k f uniformly on Ω \ F. Proof. 1. We sar as in he proof of Theorem 1. Se g = f f. Then g is a sequence of measurable funcions, g 0, such ha g 0 µ -a.e. in Ω as. For any > 0 define Then lim sup = x Ω: g (x). x Ω: lim sup g (x). The µ -a.e. convergence o 0 of g implies ha he se in he righ hand side has measure 0. Hence 0 = µ lim sup = µ. k=1 =k 5

6 Consequenly, he coninuiy of µ along he monoone sequences of ses implies lim µ ( F ) k = 0, (4.4) where we defined F k =. =k The ses Fk can be linked wih he uniform convergence of g k. Namely, on one hand sup g k. (4.5) Ω\Fk On he oher hand, we mus find he se F of small measure such ha sup g 0 as. Ω\F 2. Fix any ε > 0. Tes (4.4) wih = 1/m for m = 1, 2,.... For each m uilise (4.4) o find an ineger N m such ha µ F 1/m N m < ε2 m. Then for we derive ha µ(f ) F = A he same ime we claim ha m=1 m=1 F 1/m N m µ F 1/m N m < ε. c sup g 0 as. F c Indeed, by he definiions F c F 1/m N m for any m. Therefore by (4.5) for any m = 1, 2,... we have sup g 1/m N m. Ω\F This implies g 0 uniformly on Ω \ F. 14. Where in he proof did we use ha µ is finie? Prove ha gorov s heorem does no hold in general if µ(ω) =. Prove ha gorov s heorem does no hold in general if f is no finie a.e.. 6

7 15. An applicaion of gorov s heorem is he crierion for he L 1 -convergence of a sequence converging µ -a.e.. Recall, ha for a general µ -a.e converging sequence we canno pass o he limi under he inegral sign. (Why?) To sae he crierion we need he noion of he uniform inegrabiliy. A family of measurable funcions φ α α A, φ α : Ω C, is called uniformly inegrable if for any ε > 0 here is δ > 0 such ha φ α dµ < ε wih µ() < δ α A. (4.6) Here he "uniform" means ha he same δ works simulaneously for all α A. Theorem 4 Le µ(ω) < +, f n L 1, n = 1, 2,..., and le f n f µ -a.e. as n. Then: f n n N is uniformly inegrable f L 1 and lim f n f 1 = 0. Informally, he heorem saes ha he µ -a.e. convergence implies he L 1 convergence if and only if here is no concenraion. 16. Consruc an L 1 -sequence (f ) converging a.e. o some f L 1, for which f f in L 1. Show direcly (wihou using he previous heorem) ha he family f is no uniformly inegrable. How would you describe he concenraion phenomenon in your example? 17. Proof of Theorem Proof of. Fix any ε > 0. By he uniform inegrabiliy of he family f n n N find δ = δ ε such ha (4.6) holds. Since µ is finie he gorov s heorem can be applied. Consequenly for his δ > 0 we find F δ Ω such ha µ(f δ ) < δ and f n f uniformly on Ω \ F δ as n. By he Faou s lemma we deduce ha f dµ = lim n(x) dµ(x) F δ F δ lim inf f n dµ F δ ε. On he se Ω \ F δ (which has a finie measure) he uniform convergence of f n implies he L 1 -convergence. Hence f n f dµ = f n f dµ + f n f dµ Ω F δ Ω\F δ f n dµ + f dµ + f n f dµ F δ F δ Ω\F δ 2ε + f n f L1 (Ω\F δ ). 7

8 Therefore lim sup f n f 1 2ε, which shows ha f n f in L 1 as n. 2. Proof of. Fix any ε > 0. Find N ε such ha f n f 1 < ε n > N ε. Nex, use he absolue coninuiy of he inegral o find δ = δ ε, such ha for any Ω wih µ() < δ he esimae ( f + f f Nε ) dµ < ε (4.7) holds. Then he implicaion µ() < δ = f n dµ < ε holds for n = 1,..., N ε. On he oher hand, for any n > N ε and any wih µ() < δ we use (4.7) o derive f n dµ = f n f + f dµ ( f n f + f ) dµ f n f 1 + f dµ < 2ε. 8