REAL ANALYSIS LECTURE NOTES - Atlanta, GA

1 MeasureandpointwiseconvergenceAlthoughco nvergencein measuredoes notimplypointwiseconvergence,we dohave thefollowingweaker (butstillveryuseful) !f, thenthereexistsa subsequenceffnkgk2 Nsuch thatfnk! !f, we can ndn1< n2< such that8n nk; njf fnj>1ko 12k:De neEk=njf fnkj>1koandHm=1Sk=mEk:Thenwe have (Ek)<12kand (Hm) 1Xk=m12k=12m 1:SetZ=1Tm=1Hm:Then (Z) (Hm) 1=2m 1foreverym, so we have (Z) = =2Z, thenx =2 Hmforsomem. Hencex =2 Ekforallk m, which impliesjf(x) fnk(x)j 1k;allk m:Thusfnk(x)!f(x) forallx =2Z. SinceZhasmeasurezero,we thereforehave pointwiseconvergenceoffnktofalmosteverywhere. Asanimportant specialcasewe have !finL1(X), thenthereexistsa subsequenceffnkgk2 Nsuch thatfnk! andexpandonthetext\RealAnalysis:ModernTechniquesandtheirApplications,"2nded.,by MODESOF Cauchycriterionfor convergencein measureAlthoughconvergencein measureis notassociatedwitha particularnorm,thereis stillausefulCauchy criterionforconvergencein , we say thatffngn2 ZisCauchyin measureif8" >0; fjfm fnj "g!

2 0asm; n!1:Precisely, thismeansthat8"; >0;9N >0 such thatm; n > N=) fjfm fnj "g< :Theusefulnessof theCauchy criterionis thatit does notrequireus to know whatthelimitfunctionis | we ,thenexttheoremsays thatforconvergencein measure,Cauchynessis equivalent orderto prove thetheorem,we measurablefunctionsonX.(a)Prove thatE=fx2X: limn!1fn(x) existsgis measurable.(b)Show thatf(x) =8<:limn!1fn(x);if thelimitexists;0;otherwise;is a , thefollowingstatements areequivalent.(a)ffngn2 Nis Cauchy in measure.(b)Thereexistsanfsuch thatfnm! (b))(a).Exercise.(a))(b).Supposethatffng n2 Nis Cauchy in :Show thatthereexistsa subsequencefgjgj2 Nofffngn2 Nsuch that8j2N; njgj gj+1j 12jo 12j:De neEj=njgj gj+1j 12joandHk=1Sj=kEj:Thenwe have (Ej) 12jand (Hk) 1Xj=k12j=12k 1 MODESOF CONVERGENCE3If we setZ=1Tk=1Hk=1Tk=11Sj=kEj=limsupj!

3 1Ej;thenwe haveZ Hkforeveryk, so (Z) = xanyx =2Z. Thenx =2 Hkforsomek, andthereforex =2 Ejforallj k. Hence,forany` j kwe havejgj(x) g`(x)j ` 1Xi=jjgi+1(x) gi(x)j ` 1Xi=j12i 12j 1:Asa consequence,thesequenceofscalarsfgj(x)gj 2 Nis Cauchy. SinceRandCarecomplete,everyCauchy sequenceof ,if we de nef(x) =8<:limj!1gj(x);if thelimitexists;0;otherwise;thenfis measurableby Exercise4, andsincethelimitexistsforeveryx =2 Zwe have thatgj! (if is complete,measurability offis easier,andin factwe couldde neit any way we like onZin thatcase).Now willshow thatgjconvergesin measuretof. Fixanyj. Ifx =2Hj, then,as above,we have thatjgj(x) g`(x)j 12j 1forall` j. Hencejgj(x) f(x)j= gj(x) lim`!1g`(x) =lim`!1jgj(x) g`(x)j 12j 1:Thereforenjgj fj 12j 1o Hk;andso njgj fj 12j 1o (Hj) 12j 1!

4 0asj!1:It followsfromthisthatgjm! ,we have shownthatfgjgj2 Nis a subsequenceofffngn2 Nthatconvergesin measuretof. Exercise:Combinethiswiththefactthatffngn 2 Nis Cauchy in measureto show thatfnm!f. Asanapplication,we canusetheCauchy criterionforconvergencein measureto showthatL1(X) is a Banach space, ,a completenormedspace(seethereviewof metrics,norm,andconvergenceforthede nitionof a completespace).Theorem6(CompletenessofL1 (X)).L1(X) is a Banach alreadyknow thatL1(X) is a normedspace(oncewe ),so we justhave to show thatit is have to showthateveryCauchy sequenceinL1(X) ,supposethatffngn2 Nis a Cauchy sequenceinL1(X). Thismeansthat8" >0;9N >0 such thatm; n > N=) kfm fnk1< " MODESOF CONVERGENCEE xercise:Show thatthisimpliesthatffngn2 Nis Cauchy in measure(verysimilartotheexercisethatshow sthatconvergenceinL1(X) impliesconvergencein measure|useTchebyshev'sinequality).

5 ByTheorem5, thisimpliesthatthereexistsa measurablefunctionfsuch thatfnm! any" >0. Sinceffngn2 Nis Cauchy inL1(X), thereexistsanN >0 such thatm; n > N=) kfm fnk1< ":Fixanyksuch thatnk> N. Thenby Fatou'sLemma,we have thatkf fnkk1=Zjf fnkj=Zlimj!1jfnj fnkj liminfj!1 Zjfnj fnkj=liminfj!1kfnj fnkk1 ":Thusfnk!finL1(X).Exercise:Combinethisw iththefactthatffngn2 Nis Cauchy inL1(X) to show thatfn!finL1(X). 'sTheoremIn general,pointwiseconvergencedoes notimplyconvergencein ,fora nitemeasurespace,thisis true,andin factwe willseein thissectionthatmuch moreis 'stext\RealAnalysis,"hequotesthemathemat icianLittlewood's\ThreePrinciples."Weavi ngquote(romantext)withcomments (italics),thisis:Theextent of knowledgerequiredis nothinglike sogreatasis ,roughlyexpressiblein thefollowingterms:(a)Every(measurable)se tis nearlya niteunionof intervals(becauseit isnearlyan open set);(b)Every(measurable)functionis nearlycontinuous(thisis Lusin's Theo-rem);(c)Everyconvergent sequenceof (measurable)functionsis nearlyuniformlyconvergent(thisis Egoro 's Theorem).

6 SeeRoydenforthefullquote,which is dealwithLittlewood'sthirdprinciple,Egoro 's MODESOF CONVERGENCE5 Theorem7(Egoro 's Theorem).Supposethat is a nitemeasure,andthatfn,f:X! ! ,thenforevery" >0 thereexistsa measurableE Xsuch that(a) (E)< ", and(b)fnconvergesuniformlytofonEC, ,limn!1supx =2 Ejf(x) fn(x)j=0 thesetof measurezerowherefn(x) does notconvergetof(x). Fork,n2N,de nethemeasurablesetsEn(k) =1Sm=nnjf fmj 1koandZk=1Tn=1En(k):Now,ifx2Zk, thenx2En(k) foreveryn. Hence,foreachntheremustexistanm nsuch thatjf(x) fm(x)j>1k. Thereforefn(x) does notconvergetof(x), sox2Z. ThusZk Z;andtherefore (Zk) = 0 by monotonicity. SinceE1(k) E2(k) , we thereforehave bycontinuity fromabove thatlimn!1 (En(k))= (Zk) =0:Choosenow any" >0. Thenforeachk, we can ndannksuch that (Enk(k))<"2k:De neE=1Sk=1 Enk(k);thenwe have by subadditivity that (E) ".

7 Andifx =2E, thenx =2 Enk(k) foreveryk,andthereforejf(x) fm(x)j<1kforallm nk. Thus,we have shownthatforeachk2N,thereexistsannk>0 such thatforallm nkwe havesupx =2 Ejf(x) fm(x)j 1k:Thisimpliesthatfnconvergesuniformlyto fonEC. 'sTheoremLusin'sTheoremis a (Lusin'sTheorem).Givena measurablesetE Rdandgivenf:E!C, thefollowingstatements areequivalent.(a)fis MODESOF convergence (b)For each" >0, thereexistsa closedsetF EwithjEnFj< "such thatfjFiscontinuous, ,8xk; x2F;xk!x=)f(xk)!f(x):Proof.(a))(b).Firstwe prove =PNj=1aj Ejis a simplefunction, " > ,thereexistsa closedFj Ejsuch thatjEjnFjj<"n;j= 1; : : : ; n:ThenF=nSj=1 Ejis closed,andjEnFj< ".IfEis a boundedset,thentheFjarecompact,andhencedist(Fj; Fk)>0ifj6=k. Since is constant oneachFj, it followsthat jFis :Extendto thecasewhereEis notboundedby consideringthesetsEk=fx2E:kxk kg:Now letf:E!

8 Cbe nbe simplefunctionssuchthat n(x)!f(x) foreachx2E. Fix" >0. Bythepreviouscase,foreachnwe can nda closedFn Esuch thatjEnFnj<"2n+1and njFnis Egoro 's Theorem,thereexistsa closedF0 Esuch thatjEnF0j<"2andfnconvergestofuniformlyonF0. De neF=1Tn=0Fn:ThenFis closedsinceeachFnis closed,andjEnFj= 1Sn=0(EnFn) 1Xn=0jEnFnj ":Since njFnis continuous, njFis continuousas nconvergestofuniformlyonF, we have thatfjFis :Extendto thecasewhereEis unboundedby consideringthesetsEk=fx2E:k 1 kxk< MODESOF CONVERGENCE7(b))(a).Supposethatstatement (b) , it su cesto assumethatfis ,foreachn2 Nthereexistsa closedFn Esuch thatjEnFnj<1nandfjFnis :ThenHis anF -set,so is ,foreverynwe have thatjEnHj jEnFnj<1nj;sojEnHj= 0. Thereforewe canwriteE=H[ZwhereZhasmeasurezeroandis we xanya2R, thenwe have thatff > ag=fx2H:f(x)> ag [ fx2Z:f(x)> ag=1Sn=1fx2Fn:f(x)> ag [ fx2Z:f(x)> ag:SinceeachfjFnis continuous,we have thatfx2Fn:f(x)> agis relativelyopenwithrespecttoFn( ,it is theintersectionof anopensetU RdwithFn) complete,we know thatfx2Z:f(x)> concludethatff > agis trueforeveryrealnumbera, we have shownthatfis a measurablefunction.]]]

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