实变函数期末考试重点

实变函数考试重点题目第一章：求极限Eg：求1(,)nAnn的上下极限下极限1111liminf(,)(,)(0,)nnmnmmAnmnm上极限1111limsup(,)(,)(0,)nnmnmmAnmnmP24页第5题5、设F是]1,0[上全体实函数所构成的集合,cF2.证明：(1)设)(xE为E的示性函数,]}1,0[|{EEA,FExBE]}1,0[|)({,显然BA~,于是FBAc2；(2)设]}1,0[|))(,{(xxfxGf,}|{FfGCf,}]1,0[|{RPPD,显然DCF~,于是cDCF2,总之,cF2.P30页定理1定理2P35页第212题2.设一元实函数)()(RCxfRa,})(|{axfxG是开集,})(|{axfxF是闭集.证明：(1)Gx0,取0)(0axf,因)()(0xCxf,那么对于0,0,..ts||0xx时,|)()(|0xfxf,即axfxf)()(0,从而GxN),(0,所以G是开集.(2)Fx0,互异点列Fxk}{..ts0xxk,显然axfk)(,因)()(0xCxf,有axfxfkk)(lim)(0,即Fx0,于是FF,所以所以F是闭集.12、设实函数)()(nCxfROG,O)(1Gf.证明：“”OG,)(10Gfx,因OGxf)(0,0..tsGxfNxf)),(()(00,那么对于0,0,..ts),(0xNx,均有GxfNxf)),(()(0,从而)(1Gfx,于是)(),(10GfxN,所以O)(1Gf.“”nxR0,0,由于O)),((0xfNG,那么O)(10Gfx,这样0..ts)(),(10GfxN,从而)(),(10GfxNx,均有)),(()(0xfNxf,即)()(nCxfR.P42页定理4P44页定理2定理3定理2：非空nER,0d,}),(|{dExxUOUE.证明：显然UE.Ux,取0),(Exd,),(xUy,有dExExxyEy),(),(),(),(可见Uy,这样UxUx),(,OUE.P45页第5.6题5、设非空nER,则),(EP在nR上一致连续.证明：0,取,nQPR,,只要),(QP,由于),(),(),(EQQPEP,),(),(),(EPPQEQ,有),(|),(),(|QPEQEP,所以,),(EP在nR上一致连续.6、非空C21,FF)()(nCPfR..ts1)(0Pf,且0)(Pf,1FP;1)(Pf,2FP.证明：显然)(),(),(),()(211nCFPFPFPPfR,1)(0Pf,且0)(Pf,1FP;1)(Pf,2FP.P54页定理（3）（4）P57页第57题5、设实函数)(xf在],[ba上连续,}),(|),{(bxaxfyyxE,证明0*Em.证明：因为],[)(baCxf,于是)(xf在],[ba上一致连续,那么0,0,..ts当||ts,时,|)()(|sftf.取nab,将],[ba进行n等分,其分点为bxxxan10,记],[1iiixxI,])(,)([iiixfxfJ,显然,)(}),(|),{(11niiiniiJIIxxfyyxE,niiiniiiJmImJImEm11*)]()([)(0)(2)2(1abnabni,于是,由的任意性,知0*Em.7、0*Em,证明必Ex,..ts0,都有0)),((*xNEm.证明：反证.假设Ex,0x,使得0)),((*xxNEm,当然存在以有理数为端点的区间xI..ts),(xxxNIx,由于}{xI至多有可数个,记作}{kJ,有)(1kkJEE那么0)(01**kkJEmEm,这与条件0*Em不符,说明必Ex,..ts0,都有0)),((*xNEm.P65页定理5定理6P68页第45911题4、设M}{mE,证明mmmmmEEminflim)inflim(.又)(1mmEm,证明mmmmmEEmsuplim)suplim(.证明：因mmkkEE,有mmmkkmmmkkmmmEEEmEminflimlim)()inflim(1.又因mmkkEE,)(1mmEm,有mmmkkmmmkkmmmEEEmEmsuplimlim)()suplim(1.5、设M}{mE,1)(mmEm,证明0suplimmmmE.证明：因mmkkEE,11)()(mmmmEmEm,有0)(lim)(lim)()suplim(01mkkmmkkmmmkkmmEmEmEmEm,所以0suplimmmmE.P103页第2题2、证明当)(xf既是1E上又是2E上的非负可测函数时,)(xf也是21EE上的非负可测函数.证明：由条件知Ra,nExaxfxEM],)(;[1,nExaxfxEM],)(;[2,于是],)(;[21EExaxfxEnExaxfxEExaxfxEM],)(;[],)(;[11所以)(xf也是21EE上的非负可测函数.P104页第611题6、设实函数)()(nCxfR,证明:ME,均有)()(ExfM.证明：ME,Ra,显然O),(aG,下面证明M)(1Gf.},)(|{)(10nxaxfxGfxR,因OGxf)(0,0..tsGxfNxf)),(()(00,这样对于0,0,..ts),(0xNx,均有GxfNxf)),(()(0,从而)(1Gfx,于是)(),(10GfxN,那么MO)(1Gf.由于M)(},)(|{)(11GfEExaxfxGf,所以)()(ExfM.11、设)(xf是E上的可测函数,)(yg是R上的连续函数,证明)]([xfg是E上的可测函数.证明：Ra,因)()(RCyg,若O),(aG,有O})(|{)(1aygyGg由于})]([|{axfgxxaxfg)]([)()(1Ggxf)]([11Ggfx,于是M)]([})]([|{11Ggfaxfgx,所以)()]([ExfgM.P117页第2题2、设Kxfk|)(|..eaE,)()(xfxfmkEx,证明Kxf|)(|..eaE.证明：Nm,当mxfxfk1|)()(|,Kxfk|)(|时,mKxfxfxfxfkk1|)(||)()(||)(|,于是]1|)(|;[mKxfxmmEm]|)(|;[]1|)()(|;[Kxfxmmxfxfxmkk0]1|)()(|;[mxfxfxmk,k,有0mmE,因}{mE,有0lim]|)(|;[mmEKxfxm所以Kxf|)(|..eaE.课件第四章第四节倒数第2~5题3、定理：设)()(xfxfmk,)()(xgxfmkEx,则)(~)(xgxfE.证明：Nkm,,若mxfxfk21|)()(|,mxgxfk21|)()(|,有mxgxfxfxfxgxfkk1|)()(||)()(||)()(|,于是]1|)()(|;[mxgxfxE]21|)()(|;[]21|)()(|;[mxgxfxEmxfxfxEkk,从而]1|)()(|;[mxgxfxmE]21|)()(|;[]21|)()(|;[mxgxfxmEmxfxfxmEkk000,又因1]1|)()(|;[)]()(;[mmxgxfxExgxfxE,有0)]()(;[xgxfxmE,所以)(~)(xgxfE.1、设)()(xfxfmk,)()(xgxgmk,Ex,证明)()()()(xgxfxgxfmkk.证明：已知,0,当2|)()(|xfxfk,2|)()(|xgxgk,时,|)()(||)()(||)]()([)]()([|xgxgxfxfxgxfxgxfkkkk,由于)()(xfxfmk,)()(xgxgmk,Ex,有]|)]()([)]()([|;[0xgxfxgxfxmkk0]2|)()(|;[]2|)()(|;[xgxgxmxfxfxmkk,所以)()()()(xgxfxgxfmkk.2、设)()(xfxfmk,)()(ExgM且几乎处处有限,证明)()()()(xgxfxgxfmk.证明：已知,)()(xfxfmk,)(xg在E上几乎处处有限,那么0,0,0K..ts2]|)()(|;[Kxfxfxmk,2]|)(|;[Kxgxm]|)()()()(|;[xgxfxgxfxmk]]|)(||)()(|;[xgxfxfxmk]|)(|;[]|)()(|;[KxgxmKxfxfxmk]|)(|;[]|)()(|;[KxgxmKxfxfxmk,所以)()()()(xgxfxgxfmk.3、设0)(mkxf,证明0)(2mkxf.证明：已知,0)(mkxf,那么0,0,..ts]|)()(|;[xfxfxmk,有]|)(|;[]|0)(|;[2xfxmxfxmkk,所以0)(2mkxf.