37数学分析原理部分参考答案

chapterone3.ProvethatiffisarealfunctiononameasurablespaceXsuchthatfx:f(x)rgismeasurableforeveryrationalr,thenfismeasurable.[proof]:Foreachrealnumber,thereexistsandescendingsequencefrngofrationalnumberssuchthatlimn!1rn=.Moreover,wehave(;+1)=1[n=1[rn;+1):Hence,f1((;+1))=1[n=1f1([rn;+1)):Sincesetsf1([rn;+1))aremeasurableforeachn,thesetf1((;+1))isalsomeasurable.Thenfismeasurable.4.Letfangandfbngbesequencesin[1;+1],andprovethefollowingassertions:(a)limsupn!1(an)=liminfn!1an:(b)limsupn!1(an+bn)limsupn!1an+limsupn!1bn:providednoneofthesumsisoftheform11.(c)Ifanbnforalln,thenliminfn!1anliminfn!1bn:Showbyanexamplethatstrictinequalitycanholdin(b).[proof]:(a)Sincesupkn(ak)=infknak;n=1;2;


:Therefore,letn!1,itobtainslimn!1supkn(ak)=limn!1infknak:1Bythedenationsoftheupperandthelowerlimits,thatislimsupn!1(an)=liminfn!1an:(b)Sincesupkn(ak+bk)supknak+supknbk;n=1;2;


:Hencelimn!1supkn(ak+bk)limn!1[supknak+supknbk]=limn!1supknak+limn!1supknbk:Bythedenationsoftheupperandthelowerlimits,thatislimsupn!1(an+bn)limsupn!1an+limsupn!1bn:example:wedenean=(1)n;bn=(1)n+1;n=1;2;


:Thenwehavean+bn=0;n=1;2;


:Butlimsupn!1an=limsupn!1bn=1:(c)Becauseanbnforalln,thenwehaveinfkn(ak)infknbk;n=1;2;


:Bythedenationsofthelowerlimits,itfollowsliminfn!1anliminfn!1bn:5.(a)Suposef:X![1;+1]andg:X![1;+1]aremeasurable.Provethatthesetsfx:f(x)g(x)g;ff(x)=g(x)garemeasurable.(b)Provethatthesetofpointsatwhichasequenceofmeasurablereal-valuedfunctionsconverges(toanitelimit)ismeasurable.2[proof]:(a)Sinceforeachrationalr,thesetfx:f(x)rg(x)g=fx:f(x)rg\fx:rg(x)gismeasurable.Andfx:f(x)g(x)g=[r2Qfx:f(x)rg(x)gThereforethesetfx:f(x)g(x)gismeasurable.Alsobeacausejfgjismeasurablefunctionandthesetfx:f(x)=g(x)g=1\n=1fx:jf(x)g(x)j1ngisalsomeasurable.(b)Letffn(x)gbethesequenceofmeasurablereal-valuedfunctions,Abethesetofpointsatwhichthesequenceofmeasurablereal-valuedfunctionsffn(x)gconverges(toanitelimit).HenceA=1\r=11[k=11\n;mkfx:jfn(x)fm(x)j1rg:Therefore,Aismeasurable.7.Supposethatfn:X![0;+1]ismeasurableforn=1;2;3;


;f1f2f3


0;fn(x)!f(x)asn!1,foreveryx2Xandf12L1().Provethatthenlimn!1ZXfnd=ZXfd:andshowthatthisconclusiondoesnotfollowiftheconditionf12L1()isomitted.[proof]:Denegn=f1fn;n=1;2;


,thengnisincreasingmeasurableonXforn=1;2;3;


.Moreover,gn(x)!f1(x)f(x)asn!1,foreveryx2X.BytheLebesgue'smonotoneconvergencetheorem,wehavelimn!1ZXgnd=ZX(f1f)d:Sincef12L1(),itshowslimn!1ZXfnd=ZXfd:3[counterexample]:LetX=(1;+1),andwedenefn=En;En=[n;+1);n=1;2;


:8.SupposethatEismeasurablesubsetofmeasurespace(X;)with(E)0;(XE)0,thenwecanprovethatthestrictinequalityintheFatou'slemmacanhold.10.Supposethat(X)1;ffngisasequenceofboundedcomplexmeasurablefunctionsonX,andfn!funiformlyonX.Provedthatlimn!1ZXfnd=ZXfd;andshowthatthehypothesis(X)1cannotbeomitted.[proof]:ThereexistMn0;n=1;2;


suchthatjfn(x)jMnforeveryx2X;n=1;2;


.Sincefn!funiformlyonX,thereexistsM0suchthatjfn(x)jM;jf(x)jMonX;n=1;2;


:BytheLebesgue'sdominatedconvergencetheorem,wehavelimn!1ZXfnd=ZXfd:[counterexample]:LetX=(1;+1),andwedenefn=1nonX;n=1;2;


:12.Supposef2L1().Provedthattoeach0thereexistsa0suchthatREjfjdwhenever(E).[proof]:SinceZXjfjd=supZXsd;thesupremumbeingtakenoverallsimplemeasurablefunctionsssuchthat0sjfj.Hence,foreach0,thereexistsasimplemeasurablefunctionswith0sjfjsuchthatZXjfjdZXsd+2:SupposethatM0,satisfying0s(x)MonX,andlet=2M,wehaveZEjfjdZEsd+2Mm(E)+2whenm(E).4chaptertwo1.LetffngbeasequenceofrealnonnegativefunctionsonR1,andcon-siderthefollowingfourstatements:(a)Iff1andf2areuppersemicontinuous,thenf1+f2isuppersemicon-tinuous.(b)Iff1andf2arelowersemicontinuous,thenf1+f2islowersemicon-tinuous.(c)Ifeachffngisuppersemicontinuous,then1P1fnisuppersemicontin-uous.(d)Ifeachffngislowersemicontinuous,then1P1fnislowersemicontinu-ous.Showthattreeofthesearetrueandthatoneisfalse.Whathappensifthewordnonnegativeisomitted?IsthetruthofthestatementsaectedifR1isreplacedbyageneraltopologicalspace?[proof]:(a)Becauseforeveryrealnumber,fx:f1(x)+f2(x)g=[r2Q[fx:f1(x)rg\fx:f2(x)rg]andf1andf2areuppersemicontinuous,thereforef1+f2isuppersemicon-tinuous.(b)Becauseforeveryrealnumber,fx:f1(x)+f2(x)g=[r2Q[fx:f1(x)rg\fx:f2(x)rg]andf1andf2areuppersemicontinuous,thereforef1+f2isuppersemicon-tinuous.(c)Thisconclusionisfalse.counterexample:foreachn2N,denefn=(1;x=rn0;others1whichfrngaretheallrationalsonR1.Thenffngisasequenceofrealnon-negativeuppersemicontinuousfunctionsonR1,moreoverwehavef(x)=1X1fn(x)=Q:Butf(x)isnotuppersemicontinuous.(d)Accordingtotheconclusionof(b),itiseasytoobtain.Ifthewordnonnegativeisomitted,(a)and(b)arestilltruebut(c)and(d)isnot.ThetruthofthestatementsisnotaectedifR1isreplacedbyageneraltopologicalspace.2.LetfbeanarbitrarycomplexfunctiononR1,anddene(x;)=supfjf(s)f(t)j:s;t2(x;x+)g(x)=inff(x;):0gProvethatisuppersemicontinuous,thatfiscontinuousatapointxifandonlyif(x)=0,andhencethatthesetofpointsofcontinuityofanarbitrarycomplexfunctionisaG.[proo