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首页 Rudin 实分析前三章.pdf

Rudin 实分析前三章.pdf

Rudin 实分析前三章.pdf

上传者: 盛涛 2012-09-15 评分1 评论0 下载151 收藏0 阅读量395 暂无简介 简介 举报

简介:本文档为《Rudin 实分析前三章pdf》,可适用于高等教育领域,主题内容包含REALANALYSISZuoshunhuaShiNovember,SchoolofMathematicalSciences,GraduateUni符等。

REALANALYSISZuoshunhuaShiNovember,SchoolofMathematicalSciences,GraduateUniversityoftheChineseAcademyofSciences,Beijing,PRChinaEmailaddress:shizuoshunhuabmailsgucasaccnContentsAbstractIntegrationPositiveBorelMeasureLpspacesChapterAbstractIntegrationExercisesDoesthereexistaninfiniteσalgebrawhichhasonlycountablymanymembersSolutionITheanswerisnegativeWeshallprovethatthecardinalnumberofanyinfiniteσalgebraSisnotlessthanthatofthecontinuumWefirstintroduceausefullemmaLemmaIfB={B,B,,Bn,}isasequenceofstrictlyincreasingsets,thenanyσalgebraMcontainingBisuncountableThesamestatementholdsforstrictlydecreasingsequencesProofPutG=B,Gn=BnBnfornThenG={G,G,,Gn,}isasubclassofMandanysubclassofGisalsoasubclassofMItiseasytoseethatfordifferentnonemptysubclassofGhavedifferentunionsInotherwords,supposeHandHaresubclassofGwemusthaveH=HButthecollectionofallsubclassofGhascontinuumcardinalitythereforethecardinalnumberoftheσalgebraMcontainingGisnotlessthancontinuumForthecaseinwhichsequencesarestrictlydecreasing,theproofissimilarThiscompletestheproofWeshallprovethefollowinginterestingtheoremwhichimpliesourassertionTheorem:IfBisaninfiniteσalgebra,thenthereexistsasequenceofdisjointandnonemptymeasurablesetsinBConstructionProofSupposeB={,X}{Bα:αΣ}isaninfiniteσalgebrainXInthiscontext,each=Bα=XandBα=Bβforα=βWewanttofindasequence{B,B,,Bn,}from{Bα:αΣ}TakeαΣarbitrarilyChooseα{α:α=αandBαBα}if{α:α=αandBαBα}isnotemptyGenerally,if{Bα,Bα,Bαn}hasbeenchosen,wetakeαn{α:α=αi,i=,,,n,BαnBα}Ifthisprocesscontinues,twocasesmayoccur:(a)Wefindasequence{Bα,Bα,Bαn,}(b)ThereexistsanαNΣsuchthat{αΣ:α=NandBαNBα}=Case(a)isjustrequiredIfweareinthecase(b),thenweclaimthatB=BBαN={BBαN:BB}isaninfiniteσalgebrainBαNThesimplebutkeyobservationisthateitherBαBαNorBcαNBαforanyαΣOtherwise,ifBαisnotcontainedinneitherBαNnorBcαNforsomeαΣ,thenBαBcαNisanonemptyandpropersubsetofBcαNwhichimpliesthatBNisapropersubsetof(BαBcαN)BNIfBcontainsfinitelymanymembers,then{αΣ:BαBαN}isfiniteRecallthatwehaveobservedthateitherBαBαNorBcαNBαforanyαΣItfollowsfromthefactΣisinfinitethatthereexistβΣandβΣsuchthatBcαNBβandBcαNBβandBβBN=BβBNItfollowsimmediatelyBβ=BβwhichcontradictsourassumptionWehaveprovedthatBisaninfiniteσalgebrainBαNAgain,werepeataboveprocesstofindastrictlyincreasingmeasurablesetsinBinplaceofBIfwecanfindsuchasequence,thenthetheoremfollowsfromLemmaIfnot,wealsoobtainanotherσalgebraBwhichcontainsinfinitelymanymembersIfthisprocesscontinues,twocasesmayoccur:(a)Wefindasequence{Bα,Bα,Bαn,}ofstrictlyincreasingmeasurablesets(b)Thereexistsasequence{Bn}n=ofstrictlydecreasingσalgebrawhichcontaininfinitelymanymembersIfweareincase(a),thetheoremisprovedIfweareinthecase(b),itisclearthatwehaveobtainedasequenceofstrictlydecreasingmeasurablesetsthisfactimpliesthetheoremTheproofiscompleteItfollowsfromLemmathatthecardinalnumberofanyinfiniteσalgebraSisnotlessthanthatofthecontinuumArgumentbycontradictionSupposeMisaninfiniteσalgebrawhichhascountablymanymembersAssumeM={Ei:iN}ForeveryxX,letEx={Ei:xEi,iN}WeclaimthatEx=EyorExEy=φforanypairx,yXInfact,ifExEy=andxisnotcontainedinEy,thenxEcyandEx={Ei:xEi}EcywhichcontradictstheassumptionExEy=ConsiderS={Ex:xX}MNotethatEi=xEiExThereforeScontainsinfinitelymanymembersTheassertionfollowsProveananalogueofTheoremfornfunctionsProofThisstatementfollowsfromthefactthatanygivenopensetinRkisacountableunionofopenrectanglesProvethatiffisarealfunctiononameasurablespaceXsuchthat{x:f(x)r}ismeasurableforeveryrationalr,thenfismeasurableProof{x:f(x)>a}isameasurablesetforeveryrealnumberasince{x:f(x)a}canbewrittenasacountableunionofthesets{x:f(x)r},whererisrationalSimilarly,wecanshowthat{x:f(x)<b}isalsomeasurableforeveryrealnumberbsince{x:f(x)<r}ismeasurableforeveryrationalnumberrHence{x:a<f(x)<b}ismeasurableThefactthateveryopensetonthereallineistheunionofcountablymanyopenintervalsimpliesthat{x:f(x)V}isameasurablesetforeveryopensetVThiscompletestheproofLetanandbnbesequencesin,,andprovethefollowingassertions:(a)limnsup(an)=limninfan,(b)limnsup(anbn)limnsupanlimnsupbn,providednoneofthesumsisoftheform(c)If{an}{bn}foralln,then(c)limninfanlimninfbn,Showbyanexamplethatstrictinequalityholdin(b)ProofForanyfixedpositiveintegern,wehave(a)supkn(ak)=infkn(ak),hence(a)holdsSimilarly,wehave(b)supkn(akbk)supkn(ak)supkn(bk),foranygivenn,providedtherightsideoftheinequalityisnotoftheform(c)holdsbyapplying(a)and(b)Putan=()nandbn=()n,then(b)isastrictinequalityProof(a)Suppose{r,r,,rn,}isanenumerationoftherationalsonthereallineThenitiseasytoverifythat{x:f(x)<g(x)}=n={x:f(x)<rn}{x:g(x)>rn}Itiseasytoseethat{x:f(x)>g(x)}isalsomeasurablebythesameapproach{x:f(x)=g(x)}whichisthecomplementoftheunionofthetwosetsdescribedabovethereforeismeasurable(b)Suppose{fn(x)}isasequenceofmeasurablefunctionsdefinedonXBothg(x)=limninffn(x)andh(x)=limnsupfn(x)aremeasurable,soisS={x:g(x)=h(x)}Sisjustthesetofpointsatwhich{fn(x)}convergesForthecaseinwhichthelimitisfinite,wecanwritetherequiredsetask=N=mNnN{xX:|fm(x)fn(x)|<k}ProofFromtheconstructionofM,bothXandΦbelongtoM,sodoesthecomplementofGifGMIf{An}isasequenceinM,eitherAniscountableforeverynorAnhasacountablecomplementforsomenn=AnismeasurableinbothcasesThisprovesthatMisaσalgebraIf{An}isasequenceofdisjointcountablesetsinM,weclaimthatAnisuncountableforatmostonenThisobservationassuresthecountableadditivityofµThisshowsthatµisameasurein(M,X)Todescribemeasurablefunctionson(M,X),weconsiderfirstrealmeasurablefunctionsPutα=sup{c:f(c,)isuncountable},thenαiswelldefinedsinceXisuncountable(Note:X=n={x:f(x)>n},thereissomemsuchthat{x:f(x)>m}isauncountablesetwhosecomplementiscountablesinceXisuncountableand{x:f(x)>m}ismeasurable)Ifαisfinite,wehavef=αaeµandhencetheintegraloffonXisαIfαisinfinite,wehavef=aeµandthereforetheintegraloffdoesnotexistForcomplexvaluedfunctionsf,itiseasytoseethatfisconstantalmosteverywheref=CaeµimpliesthatXfdµ=CProofPutgn=ffn,thengggnffApplyingLebesgue’smonotoneconvergencetheorem,wehavelimnX(ffn)=X(ff)andlimnXfn=XfsincefL(µ)IffL(µ),thisequalitymayfailLetfnbethecharacteristicfunctionof(n,)andµbetheLebesguemeasure,itiseasytoseethattheequalitydoesnotholdQEDProofThisexampleshowsthatthestrictinequalityoccursinFatou’sLemmawheneverbothEandEcarenotofmeasureProofSinceXfdµ=c>,S={x:f(x)>a}hasfiniteandpositivemeasureforsomea>When<α<,wehavenlog(f(x)n)α>nlog(an)αforeveryxSThefactthatnlog(an)αtendstoasnimpliesthatlimnXfdµ=Whenα=,nlog(f(x)n)αf(x)foreverypointxXandthelimitoftheleftsideequalstof(x)ApplyingLebesgue’sdominatedconvergencetheoremyieldstherequiredequalityWhenα>,itiseasytoverifythatnlog(f(x)n)ααf(x)byasimpleinequalityapbp<(ab)pwherea,b>,p>Ontheotherhand,nlog(f(x)n)αtendstoforalmosteveryxXThestatementcanbeprovedbyapplyingLebesgue′sconvergencetheoremTheproofiscompleteProofWithoutoflossgenerality,itisenoughtoprovethecaseinwhichfispositivefunctionSuppose|fn(x)|MnforsomeMn<ThefactthatfnfuniformlyonXimpliesthat|f|MforsomepositiveMSinceµispositivefinitemeasure,thestatementfollowsfromLebesgue’sdominatedconvergencetheoremWhentheconditionµ(X)<isomitted,thestatementmayfailAnexampleisgivenbyputtingfn=nmultipleofthecharacteristicfunctionof(n,n)QEDProofWeclaimthatxAifonlyandonlyifxbelongstoinfinitelymanymembersof{En}TheproofisomittedhereForanygivenpositiveintegern,itiseasytoseethatµ(A)k=n(Ek)Sincen=µ(En)<,thenk=n(Ek)tendstoasnThisyieldsµ(A)=ThiscompletestheproofProofWithoutlossofgenerality,wecansupposethatfispositiveInthisspecialcase,thereexistsanondecreasingsequenceofsimplefunctions{sn}convergestofforeverypointinXSincefL(µ),forgivenpositivenumberthereexistssomeNsuchthatX|fsN|dµ<byLebesgue’smonotoneconvergencetheoremOntheotherhand,theimageofsNconsistsoffinitelymanynonnegativenumbers,hencesNisboundedAssumingthatMisaupperboundofsN,thenEsNdµ<wheneverµ(E)<M=δTherefore,E|f|dµ<wheneverµ(E)<M=δForthecaseinwhichfiscomplexvalued,wecanwritef=ffififThestatementfollowsbyconsideringthesefourpositivefunctionsseparatelyProofBothsidesareeitheroratthesametimeChapterPositiveBorelMeasureExercisesProofWeclaimthat(a),(b)and(d)aretrueand(c)mayfailundertheassumptionthat{fn}arenonnegativeThisconclusionstillholdsifRisreplacedbyageneraltopologicalspaceButthestatement(c)mayfailiftheword”nonnegative”isomittedthestatement(a)and(b)stillholdinthissituationThedetailsofproofsisgivenbelow(a)Foranyrealnumberr,weobservethat{x:f(x)f(x)<r}=sR{x:f(x)<s}{x:f(x)<rs}(b)Theproofissimilartothatof(a)(d)Since{fn}arenonnegative,then,foranyrealnumberr,{x:n=fn(x)>r}=n={x:nk=fk(x)>r}Consequently,thestatement(d)followsfromthefactthat(b)isstilltrueforfinitesums(c)Thereexistsasequenceofclosedsets{Fn}onthereallinewhoseunionisneitherclosednoropenDefinef(x)=n=χFn(x),whereχFnisthecharacteristicfunctionofFnItiseasytoseethatfisneitheruppersemicontinuousnorlowersemicontinuousForinstance,suppose{rn}n=isthesetofallrationalnumbersdefinefnasthecharacteristicfunctionof{rn}Observethatf=n=fnconvergesabsolutelyand{x:f(x)<c}isnotanopensetinRfor<cHencefisnotuppercontinuouseventhougheach{fn}isanonnegativesequenceofuppercontinuousfunctionsIftheword”nonnegative”isomitted,(a)and(b)arestilltruebut(c)and(d)mayfailProofWeonlyprovethatϕisuppersemicontinuousThisfollowsfromtheobservationthatify{x:ϕ(x)<r},thereexistsapositiveintegernsuchthatϕ(y,n)<r,hence(yn,yn){x:ϕ(x)<r}ProofWeonlyprovethefirstpartofthestatementForanypairx,xX,wehaveρ(x,y)ρ(x,x)ρ(x,y)foreveryyEsoforρE(x)ρ(x,x)ρ(x,y)foreveryyEsoforρE(x)ρ(x,x)ρE(x)Forsymmetry,wealsohaveρE(x)ρ(x,x)ρE(x)Consequently,wehave|ρE(x)ρE(x)|ρ(x,x)whichimpliesthatρE(x)isuniformlycontinuousinXProof(a)Recallthatµ(E)=inf{µ(G):Gisopen,EG}Itfollowsimmediatelythatµ(EE)µ(E)µ(E)Ontheotherhand,foreveryopensetGwhichcontainsEE,GVandGVcontainEandErespectivelySinceVandVaredisjointopensets,wehaveµ(G)µ(GV)µ(GV)andµ(G)µ(E)µ(E)TakingtheinfimumoverallopenGcontainingEE,weobtainthatµ(EE)µ(E)µ(E)Collectingtheresultsobtained,wehaveprovedtheassertion(b)ForeveryEMF,wecanchooseopensets{Gi}i=andcompactsets{Ki}i=suchthatKiEGiµ(Gi)µ(Ki)<iIfwesetN=Ei=Ki,thenµ(N)=ProofOnecanprovem(E)=withoutdifficultyWeonlygiveanimportantobservationherewhichimpliesthatthecardinalnumberof”middlethirds”setisthatofcontinuumThekeyobservationisthatxEifonlyandonlyifxcanberepresentedastheformx=n=pnn,wherepn{,}Thisexpansionisuniqueexceptforthecasethatx=amwhere<a<manddoesnotdivideaInthiscasexhasafiniteexpansionoftheformx=ppmmwherepm=orpm=Ifpm=weusethisfiniteexpansionforx,butifpm=,weprefertheexpansionx=ppmmk=mkThuswehaveassignedtoauniqueternaryexpansiontoeachx,Itisclearthatthemappingn=pkk{pk}isaonetoonecorrespondencebetweenthemiddlethirdssetand{,}NThereforeEandRhavethesamecardinalityProofSuppose{r,r,,rn,}istherationalnumbersinRLetI=n=(rnn,rnn)Onewillfindthat<m(I)<PutK=,I,thenKisdisconnectedandwithpositivemeasureIfvislowersemicontinuousandvχK,itiseasytoseethat{x:v(x)>c}KforeverypositivenumbercBut{x:v(x)>c}isopenwhichimpliesthat{x:v(x)>c}isemptysinceKdoesnothaveaconnectedsubset(iesomeinterval)Therefore,vholdsProofSuppose{r,r,,rn,}arethesetofallrationalnumberin,Withoutlossofgenerality,weassumethatr=PutEa=n=(rnan,rnan)foreverya,Defineafunctionon,byf(a)=m(Ea,)Thenfiswelldefinedandf()=f()=Wegiveanimportantobservationthatfisalsocontinuouson,Toprovethisstatement,onecanverifythatf(a)f(b)(ab)if<b<aandf(b)f(a)(ba)if<a<bThen|f(a)f(b)||ab|holdsThereforefiscontinuousLetβ=inf{a>:f(a)>}Itfollowsfromthefactthatfiscontinuousthatf(β)=PutE=Eβ(,)andhenceEisanopensetoftheclosedunitintervalwithm(E)=ProofWewilluseavariantofCantor′sconstructionofthefamiliar”middlethirds”setItwillbeconvenienttostatethisasalemmaLemmaSupposea,bisagivenclosedintervalForevery<δ<,thereexistsanopensubsetGsuchthat(a)Gisadensesubsetofa,b(b)GhasLebesguemeasureδ(ba)(c)foranyopenintervalIG,|I|<(ba)Theproofofthislemmaisessentiallythesameasthatoftheconstructionof”middlethirds”setIndeed,withoutlossofgenerality,wemayassumethata,b=,Ifwedeletensegmentsoflength()nfromnremainedintervalsrespectivelyinthenstepoftheCantor′sconstruction,weobtainadenseandopensubsetGof,suchthat(i)Gistheunionofthosedeletedintervals(ii)|G|=n=n()n=,<<(iii)foranyopenintervalIG,|I|<Ifwetakesuchthatδ=,theproofofthelemmaiscompleteSuppose>δ>δ>>δn>andδnItfollowsfromthelemmathatthereexistsanopensetGhasthepropertiesdescribedinthelemmawith,andδinplaceofa,bandδ,respectivelyWriteG=i=I()iastheunionofdisjointopenintervalsForeachsuchintervalI()i,weusethelemmatoobtainanopensetG()isuchthat(i)|G()i|=δδ|I()i|(ii)foranyopenintervalIG()i,|I|<|I()i|(iii)G()iisdenseinI()iLetGbetheunionofallsuchopensetsG()iGenerally,foreachn,wehaveanopensubsetGnofGnsuchthat(i)GnisdenseinGn(ii)foranyopenintervalIGn,|I|<n(iii)Gn=i=I(n)iand|I(n)iGn|=δnδn|I(n)i|LetE=GnSupposeIisanyintervalcontainedin,WeclaimthatforallsufficientlylargeNwith()N<|I|thereexistssomejsuchthatI(N)jISinceGnisdensein,foreachn,foranysegmentJwiththesamecenterwithI,theintersectionGNJisnotemptyItfollowsimmediatelythatI(N)jITheassertionisprovedNoticethat|GNiI(N)j|=δNiδN|I(N)j|foreachiTherefore|EI(N)j|=δN|I(N)j|HencetheproofiscompleteintheunitintervalifwerecallthatδN>PutE=n(nE)ThenEsatisfiedthedesiredinequalityEmayhavefiniteLebesguemeasureIndeed,supposeGisadenseandopensubsetofRwithfinitemeasureThenEGisrequiredSolutionDefineafunctionfa,b(x)asfollows:fa,b(x)=(xa)baifaxab(bx)baifabxbotherwiseItiseasytoseethatfa,biscontinuousandsupportedina,bAsequenceofcontinuousfunctionsisconstructedasfollows:m=,f=f,f=f,f=f,m=,f=f,,f=f,,f=f,,f=f,,f=f,m=k,,fkk=fk,k,,fkk=fk,kforkItisclearthat{fn}hasthefollowingtwoproperties:(a)Toeveryx,therecorrespondssomenwithkknkksuchthatfn(x)(b)fn(x)dxkforallnwithkknkkConsequently,{fn(x)}hasdesiredpropertiesProof(TheoriginalproofisduetoWFEberlin)WefirstgiveausefullemmaLemmaIffandfnarecontinuousandnonnegativefunctionson,suchthatf(x)n=fn(x)foreveryx,,thenf(x)dxn=fn(x)dxTheproofoftheabovelemma(duetoMHStone)iseasyBythecompactnessoftheunitinterval,forevery>,wecanfindsomeN()suchthat|f(x)|N()n=|fn(x)|,foreveryx,Also,f(x)dxN()n=fn(x)dxSinceisarbitrary,thelemmafollowsAsimplebutkeyobservationisthattheconditionsandassertioncanbereducedasfollows:If(a)fnlieinC,foralln,(b)supn‖f‖,(c)limnfn=pointwise,(d)fnforalln,(e)limfn(x)dx=L,thenL=LetKnbetheconvexhulloftheset{fm:mn}Clearlyanysequence{gn}n=withgnKnsatisfies(a)(e)Setdn=inf{‖gn‖:gnKn}SinceKnKn,itfollows

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