rudin实分析与复分析答案

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1、
chapterone3.ProvethatiffisarealfunctiononameasurablespaceXsuchthatfx:f(x)rgismeasurableforeveryrationalr,thenfismeasurable.[proof]:Foreachrealnumber,thereexistsandescendingsequencefrngofrationalnumberssuchthatlimrn=.Moreover,wehaven!1[1(;+1)=[rn;+1):n=1Hence

2、,[111f((;+1))=f([rn;+1)):n=1Sincesetsf1([r;+1))aremeasurableforeachn,thesetf1((;+1))isnalsomeasurable.Thenfismeasurable.4.Letfangandfbngbesequencesin[1;+1],andprovethefollowingassertions:(a)limsup(an)=liminfan:n!1n!1(b)limsup(an+bn)limsupan+limsupbn:n!1n!1n!1pr

3、ovidednoneofthesumsisoftheform11.(c)Ifanbnforalln,thenliminfanliminfbn:n!1n!1Showbyanexamplethatstrictinequalitycanholdin(b).[proof]:(a)Sincesup(ak)=infak;n=1;2;


:knknTherefore,letn!1,itobtainslimsup(ak)=liminfak:n!1knn!1kn1Bythedenationsoftheupperandthel

4、owerlimits,thatislimsup(an)=liminfan:n!1n!1(b)Sincesup(ak+bk)supak+supbk;n=1;2;


:knknknHencelimsup(ak+bk)lim[supak+supbk]=limsupak+limsupbk:n!1knn!1knknn!1knn!1knBythedenationsoftheupperandthelowerlimits,thatislimsup(an+bn)limsupan+limsupbn:n!1n!1n!1exa

5、mple:wedenenn+1an=(1);bn=(1);n=1;2;


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


:Butlimsupan=limsupbn=1:n!1n!1(c)Becauseanbnforalln,thenwehaveinf(ak)infbk;n=1;2;


:knknBythedenationsofthelowerlimits,itfollowsliminfanliminfbn:n!1n!15.(a)Suposef:X![1;+1]andg:X![1;+1]aremea

6、surable.Provethatthesetsfx:f(x)

7、x:f(x)

8、easurablereal-valuedfunctionsffn(x)gconverges(toanitelimit).Hence1[111A=fx:jfn(x)fm(x)j