2021年实变函数测试题与答案.pdf
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1、实变函数试题一,填空题1.设1,2nAn,1,2nL,则limnnA.2.,a b:,因为存在两个集合之间的一一映射为3.设E是2R中函数1cos,00,0 xyxx的图形上的点所组成的集合,则E,E.4.若集合nER满足EE,则E为集.5.若,是直线上开集G的一个构成区间,则,满足:,.6.设E使闭区间,a b中的全体无理数集,则mE.7.若()nmEfx()0f x,则说()nfx在E上.8.设nER,0nxR,若,则称0 x是E的聚点.9.设()nfx是E上几乎处处有限的可测函数列,()f x是E上几乎处处有限的可测函数,若0,有,则称()nfx在E上依测度收敛于()f x.10.设()
2、()nfxf x,xE,则()nfx的子列()jnfx,使得.精品w o r d 学习资料 可编辑资料-精心整理-欢迎下载-第 1 页,共 29 页二,判断题.正确的证明,错误的举反例.1.若,A B可测,AB且AB,则mAmB.2.设E为点集,PE,则P是E的外点.3.点集11,2,EnLL的闭集.4.任意多个闭集的并集是闭集.5.若nER,满足*m E,则E为无限集合.三,计算证明题1.证明:ABCABACUI2.设M是3R空间中以有理点(即坐标都是有理数)为中心,有理数为半径的球的全体,证明M为可数集.3.设nER,iEB且iB为可测集,1,2iL.根据题意,若有*0,imBEi,证明E
3、是可测集.4.设P是Cantor集,32ln 1,(),0,1xxPf xxxP.求10(L)()f x dx.5.设函数()f x在Cantor集0P中点x上取值为3x,而在0P的余集中长为13n的构成区间上取值为16n,1,2nL,求10()f x dx.精品w o r d 学习资料 可编辑资料-精心整理-欢迎下载-第 2 页,共 29 页文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU
4、2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S
5、3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU
6、2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S
7、3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU
8、2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S
9、3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y8文档编码:CU2X6W7G8P4 HI5U4H7X8S3 ZA5S6B4P9Y86.求极限:1
10、3230lim(R)sin1nnxnxdxn x.实变函数试题解答一填空题1.0,2.2.1(,)cos,0(0,)1x yyxyyxU;.3.闭集.4.ba.5.几乎处处收敛于()f x或a.e.收敛于()f x.6.对000,(,)Ux有0Ex.7.()()nfxf xa.e.于E.二判断题1.F.例如,(0,1)A,0,1B,则AB且AB,但1mAmB.2.F.例如,0(0,1),但 0 不是(0,1)的外点.3.F.由于0EE.4.F.例如,在1R中,11,1nFnn,3,4nL 是一系列的闭集,但是3(0,1)nnFU不是闭集.精品w o r d 学习资料 可编辑资料-精心整理-欢迎
11、下载-第 3 页,共 29 页文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8
12、X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J
13、4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5
14、W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3
15、B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M
16、1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1
17、X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X45.T.因为若E为有界集合,则存在有限区间I,I,使得EI,则*,m Em II于*m E.三,计算证明题.1.证明如下:2.M中任何一个元素可以由球心(,)x y z,半径为r唯一确定,x,y,z跑遍所有的正有理数,r跑遍所有的有理
18、数.因为有理数集于正有理数集为可数集都是可数集,故M为可数集.3.令1iiBBU,则iEBB且B为可测集,于是对于i,都有iBEBE,故*0imBEmBE,令i,得到*0mBE,故BE可测.从而EBBE可测.4.已知0mP,令0,1GP,则13202210130(L)()(L)ln 1(L)(L)()(L)(L)(R)()133PGGPGf x dxxdxx dxf x dxx dxx dxf x dxx.精品w o r d 学习资料 可编辑资料-精心整理-欢迎下载-第 4 页,共 29 页文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J
19、4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5
20、W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3
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22、1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1
23、X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1
24、H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE
25、6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X4文档编码:CE6M1J4T1X4 HP8N1X5W8W10 ZC3L1H3B8X45.将积分区间0,1分为两两不相交的集合:0P,1G,2GL,其中0P为Cantor集,nG是0P的余集中一切长为13n的构成区间(共有12n个)之并.由L积分的可数可加性,并且注意到题中的00mP,可得6.因为323sin1nxnxn x在0,1上连续,13230(R)sin1nxnxdxn x存在且与13230(L)sin1nxnxdxn x的值相等.易知由于1
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