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1、总复习题 1.证明:.(P.3)Proof.Letthenforanythereexistsasuchthat,thuscannotbefinite,conversely,ifisinfinitethenforanythereexistsasuchthat,thusand,showingthefirstformula.Againletthenthereexistsasuchthat,thusforany,,therebyisfinite,converselyifisfinite,thenlet,forany,,so
2、,and,showingthesecondformula.2.证明:.(P.11)Proof.isfinite.isinfinite.3.证明Cantor集是可测的,并且其测度为0.（P.66）Proof.BytheconstructionofCantor’sset,firststeptakeawayoneopenintervaloflength,secondsteptakeawaytwoopenintervalsandoflength,thirdsteptakeawayopenintervalsandoflengt
3、h,-thsteptakeawayopenintervalsoflength,weseeismeasurableand.64.(1)设是中的开集,是零测集.证明:(提示:Letthensince,thereexistsasuchthat,thusand).(2)写出Caratheodory条件.()(3)Letbeanoutermeasuregeneratedbythemeasureofringspannedbytheleftclosedandrightopenintervalsonstraightline.Show
4、foranyand,.(4)设是直线上的Lebusgue测度,证明对任意可测集,,有.(P.67)(1)Proof.Obviously,.Letthenthereexistsamaking.Sinceandforanythereexistsaand.Thusand.Since,and,showing.(3)Proof.Firstlyweknowthesetofthedisjointunionoffiniteleftclosedandrightopenintervals.Letthen.Obviously,.Letwh
5、erethenthereexistsasuchthat,thusforsome,,and,showing.Forany,,since,,thereby,showing.6(4)Proof.Obviously,forand,,.Sinceforany.SoisL-measurableand.5.设,证明.(P.82)Proof.Wemightassumeaswellinandinwhere.Inwehave.Since,anda.e.6.设=讨论:(1)是否几乎处处收敛?(2)是一致收敛?(3)是否依测度收敛?(P.8
6、2)Solution.(1)Letthenin,.(2)Ifisuniformlyconvergenttothennaturallyit’salsoconvergenttoandsoby(1).Forandany,letbearationalnumbergreaterthanthen,acontradiction.Soisnotuniformlyconvergentin.(3)Ifthenthereexistsasequencemaking.6(2)Sinceand.Butforany,,,contraryto.7.
7、设在上可积并且一致连续,证明.(P.103)Proof.SinceisL-measurableon,alsois.If,thenthereexistsasuchthatforanythereexistsa.Selectingwemightassume.Letthenthereexistsasuchthataslongso,.Then,and,acontradiction.8.设在上可积.对任意上的有界函数,有,证明(a.e)(提示:Define,then).(P.103)Proof.Letthenisboundeda
8、nd.Itfollowsthata.eanda.e.9.设在上可测,,问是否一定可积?(P.103).Answer.Maybenot.Forexample,for,isobviouslyintegrableonbutisnotmeasurableon.Alsolet,then,obviously,but6for.10.设,是可积函数,证明:若,
