数学分析原理部分参考答案.pdf
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数学分析原理部分参考答案.pdf

数学分析原理部分参考答案.pdf

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¢¡¤£¦¥¢§©¨chapterone3.ProvethatiffisarealfunctiononameasurablespaceXsuchthatfx:f(x)≥rgismeasurableforeveryrationalr,thenfismeasurable.[proof]:Foreachrealnumberα,thereexistsandescendingsequencefrngofrationalnumberssuchthatlimrn=α.Moreover,wehaven!11(α;+1)=[rn;+1):n[=1Hence,1−1−1f((α;+1))=f([rn;+1)):n[=1−1−1Sincesetsf([rn;+1))aremeasurableforeachn,thesetf((α;+1))isalsomeasurable.Thenfismeasurable.4.Letfangandfbngbesequencesin[−∞;+1],andprovethefollowingassertions:(a)limsup(−an)=−liminfan:n!1n!1(b)limsup(an+bn)≤limsupan+limsupbn:n!1n!1n!1providednoneofthesumsisoftheform1−1.(c)Ifan≤bnforalln,thenliminfan≤liminfbn:n!1n!1Showbyanexamplethatstrictinequalitycanholdin(b).[proof]:(a)Sincesup(−ak)=−infak;n=1;2;···:k≥nk≥nTherefore,letn!1,itobtainslimsup(−ak)=−liminfak:n!1k≥nn!1k≥n1Bythedefinationsoftheupperandthelowerlimits,thatislimsup(−an)=−liminfan:n!1n!1(b)Sincesup(ak+bk)≤supak+supbk;n=1;2;···:k≥nk≥nk≥nHencelimsup(ak+bk)≤lim[supak+supbk]=limsupak+limsupbk:n!1k≥nn!1k≥nk≥nn!1k≥nn!1k≥nBythedefinationsoftheupperandthelowerlimits,thatislimsup(an+bn)≤limsupan+limsupbn:n!1n!1n!1example:wedefinenn+1an=(−1);bn=(−1);n=1;2;···:Thenwehavean+bn=0;n=1;2;···:Butlimsupan=limsupbn=1:n!1n!1(c)Becausean≤bnforalln,thenwehaveinf(ak)≤infbk;n=1;2;···:k≥nk≥nBythedefinationsofthelowerlimits,itfollowsliminfan≤liminfbn:n!1n!15.(a)Suposef:X![−∞;+1]andg:X![−∞;+1]aremeasurable.Provethatthesetsfx:f(x)<g(x)g;ff(x)=g(x)garemeasurable.(b)Provethatthesetofpointsatwhichasequenceofmeasurablereal-valuedfunctionsconverges(toafinitelimit)ismeasurable.2[proof]:(a)Sinceforeachrationalr,thesetfx:f(x)<r<g(x)g=fx:f(x)<rg\fx:r<g(x)gismeasurable.Andfx:f(x)<g(x)g=fx:f(x)<r<g(x)gr[2QThereforethesetfx:f(x)<g(x)gismeasurable.Alsobeacausejf−gjismeasurablefunctionandtheset11fx:f(x)=g(x)g=fx:jf(x)−g(x)j<gnn\=1isalsomeasurable.(b)Letffn(x)gbethesequenceofmeasurablereal-valuedfunctions,Abethesetofpointsatwhichtheseque