Folland Real Analysis Solutions.pdf

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1、MATH605,HW1SOLUTIONSFolland’sRealAnalysis;Chapter1:4.)Thisfollowssinceanycountableunioncanbewrittenasanincreasingcountableunion:hi∞∞j∪j=1Ej=∪j=1∪k=1Ek;jnotethat∪k=1Ekisafiniteunionofsetsinthealgebraandishenceinthealgebra.5.)LetK⊂P(X)bethesetofallcountablesu

2、bsetsandshowthatN=∪F∈KNFisaσ-algebrawhereNFistheσ-algebrageneratedbyF.SupposethatEi∈NFi.Obviously,c∞∞Ei⊂NFi⊂NsinceNFiisanalgebra.Furthermore,∪j=1Ej⊂N∪∞Fj(notethat∪j=1Fjisj=1alsocountable).7).Thisfollowsintwosteps:(a)Ifµisameasureanda>0,thenitisobviousthata

3、µisameasure;(b)Ifµ1,µ2aremeasures,itisalsoobviousthatµ1+µ2isameasure.(Iwouldexpectyoutowritedowntheequationsforthis.)8).RecallthatliminfE=∪∞∩∞Eandliminfa=liminfa.Definekk=1n=knkk→∞n≥kkF=∩∞EsothatF⊂F⊂···.Bycontinuityfrombelow,kn=kn12∞µ(liminfEk)=µ(∪k=1Fk)=li

4、mµ(Fk)≤liminfµ(En)=liminfµ(Ek).k→∞k→∞n≥kTheinequalityusedmonotonicitysinceFk⊂Enifn≥k.(I’llleaveµ(limsupEk)foryou.)9).Usethesplittingpropertytwice:µ(E∪F)=µ([E∪F]∩E)+µ([E∪F]∩Ec)andµ(F)=µ(F∩E)+µ(F∩Ec).Rewritethefirstasµ(E∪F)=µ(E)+µ(F∩Ec)andthensubstituteinµ(F∩

5、Ec)=µ(F)−µ(F∩E)togetµ(E∪F)=µ(E)+µ(F)−µ(F∩E).10).SetµE(A)=µ(A∩E);we’llshowthatµEisameasure.ObviouslyµE(∅)=µ(∅)=0,sowejusthavetocheckσ-additivity.IfAiintheσ-algebraaredisjoint,thensoareE∩Ai.SinceE∩[∪iAi]=∪i[E∩Ai],theσ-additivityofµgivesXXµE(∪iAi)=µ(∪i[E∩Ai])

6、=µ(E∩Ai)=µE(Ai).ii14.)We’llarguebycontradiction:supposeµissemi-finite,µ(E)=∞,andsup{µ(F)

7、F⊂Eandµ(F)<∞}=C<∞.ChooseF⊂EwithC−2−i<µ(F)<∞.SetG=∪iFsoG⊂G⊂·,F⊂G⊂E,iiik=1k12PiiandC−2−i<µ(F)≤µ(G)(bymonotonicity).Bysub-additivity,µ(G)≤iµ(F)

8、vesµ(G)≤C.SetG=∪∞G;notethatG⊂E.Continuityii=1ifrombelowimpliesthatµ(G)=limi→∞µ(Gi)=C.Sinceµ(E)=∞,additivitygivesµ(EC)=∞.Sinceµissemi-finite,thereexistsF⊂(EG)withµ(F)>0.Byadditivity,µ(F∪G)=µ(F)+µ(G)>C.ThiscontradictsthatCwasthesupremum-weconcludethatthesup

9、remumhadtobe∞.ProfessorMinicozzi,Fall2002.1MATH605,HW2SOLUTIONSFolland’sRealAnalysis;Chapter1:18.)(a)Bydefinition,given>0,thereexistA∈A(j=1,2,...)withE⊂∪∞AandPj1jµ(A)≤µ∗(E)+.Subadditivityofµ∗andProposition1.