实变函数试题集锦.pdf

该【实变函数试题集锦 】是由【小屁孩】上传分享，文档一共【9】页，该文档可以免费在线阅读，需要了解更多关于【实变函数试题集锦 】的内容，可以使用淘豆网的站内搜索功能，选择自己适合的文档，以下文字是截取该文章内的部分文字，如需要获得完整电子版，请下载此文档到您的设备，方便您编辑和打印。实变函数测试题集锦一、填空题?1?A,2???limA????????????????nnn设??,n?1,2?,则n??.??????????????a,b???,??,因为存在两个集合之间的一一映射为?????????????????????????????????????????????.?1?cos,x?0y??x?0,????????x?0设E是R2中函数?的图形上的点所组成的集合,则E?????????????????????????E?????????????????????????,.???????????????????????若集合E?Rn满足E??E,则E为集.????????若,是直线上开集G的一个构成区间,则,满足:????????????????????????????????????????????????????????????????????????????????,.??mE?????????????????设E使闭区间a,b中的全体无理数集,?f(x)?f(x)??0???f(x)?????????????????若n,则说n在E上????????????????????????????????????????.nnx?R????????????????????????????????????????x设E?R,0,若,则称0是E的聚点.?f(x)?f(x)设n是E上几乎处处有限的可测函数列,是E上几乎处处有?0??????????????????????????????????f(x)?限的可测函数,若??,有,则称nf(x)在E上依测度收敛于.??f(x)f(x)?f(x)?f(x)?x?En设n,,则?n的子列j,使得????????????????????????????????????????.??1???,1?11.?n?=.n?1??n?1n?1???,??nn??1=.13.(0,1)到(a,a?b).[0,1]?代数记为B,*(AB)?*A?0,则对任意的点集B,必有?.,m*E?.(x)在可测集E上几乎处处有限,若对任意给定的??0,存在E中的一个闭集F,使m(E\F)??,且f(x)在F上连续,则f(x)(是或不是选其一填写).,如果则称E是开集,如果E??E则称E是.?G??GG{G}:i?1,则称为.?F??F{F}:i?1,.?a,b?R(b?a),f在E上可测,则E(f?a)?E(f?b)=..{f}q26.(Fatou引理)设n是可测集E?R上一列非负可测函数,、,,B若可测,A?B且A?B,则mA?,P?E,则P是E的外点.?1?E??1,2,?,??n点集???Rn,满足m*E???,,,若A???(E?F)?,则存在?型集F,使得F?E,且.(L)f(x)dx(R)bf(x)dxf(x)?a,b?f(x)?a,b????,则在R可积且?a,b?a三、计算证明题A??B?C???A?B???A?C?:(即坐标都是有理数)为中心,有理数为半径的球的全体,?BBi?1,2??R,i且i为可测集,.根据题意,若有m*?B?E??0,???i???i,证明E是可测集.?3???ln1?x,?????????x?Pf(x)???x2,?????????????????????x??0,1??P设P是Cantor集,?.(L)?1f(x)(x)CantorPxx3P设函数在集0中点上取值为,而在0的余集中长为11?n?1,2??3n的构成区间上取值为6n,,求?1f(x)(R)?sin3nxdx23求极限:n??01?,.{f}ff?hf?,,(xt)f(x)dx0f(x)?a??,b???????,则t?(L)???01?mx.??limA=??Anm12、证明n??n?1m?n。???x?limAx??A???An?Nn?Nx?Amm证明:设n??,则,使一切,n,所以m?n?1n?1m?n,?????limA???Ax???Ax??Anmmnm则可知n??n?1m?n。设n?1m?n,则有,使m?n,所以??x?limAlimA??Annmn??。因此,n??=n?1m?n。?22?0E??x,y?x?y?1'E2E2EE13、设。求2在R内的2,2,2。?E????x,y?x2?y2?1?E???x,y?x2?y2?1?解:2,2,E???x,y?x2?y2?1?2。14、若E?Rn,对???0,存在开集G,使得E?G且满足m*(G?E)??,证明E是可测集。1m*?G?E??G?En证明:对任何正整数n,由条件存在开集n,使得。?1G??Gm*?G?E??m*?G?E??令n,则G是可测集,又因nn,n?1对一切正整数n成立,因而m?(G?E)=0,即M?G?E是一零测度集,故可测。由E?G?(G?E)知E可测。证毕。1mE?15、试构造一个闭的疏朗的集合E?[0,1],2。157(,)解:在[0,1]中去掉一个长度为6的开区间1212,接下来在剩下的两个闭区间11?分别对称挖掉长度为63的两个开区间,以此类推,一般进行到第n次时,11?一共去掉2n?1个各自长度为63n?1的开区间,剩下的2n个闭区间,如此重复下去,这样就可以得到一个闭的疏朗集,去掉的部分的测度为11212n?11?????????66363n?12。111mE?1??mE?所以最后所得集合的测度为22,即2。f(x)?f(x)f(x)?f()x16、设在E上n,且nn?1几乎处处成立,n?1,2,3,?,则{f(x)}f(x)。f(x)?f(x){f}?{f}f(x)f(x)证明因为,则存在nn,。niiEf(x)f(x)E?E[f?f]mE?0,mE?0设是n不收敛到的点集。,则。因0innn?10n???m(?E)??mE?0E??Ennnf(x)f(x)此。在上,n收敛到f(x),且是单调n?0n?0n?1inf(x)f(x)的。因此n收敛到(单调序列的子列收敛,则序列本身收敛到同一极限)。??Enf(x)f(x)f(x)f(x)即除去一个零集n?1外,n收敛于,。??17、设E?R1,。证明存在定义于R1上的一列limg(x)?f(x){g(x)},使得n??于。证明:因为f(x)在E上可测,由鲁津定理,对任何正整数n,存在E的可测子1m?E?E??Enn1g(x)集n,使得,同时存在定义在R上的连续函数n,使得当x?Eg(x)f(x)??0E[f?g??]?E?En时有n=。所以对任意的,成立nn,??1mE??f?g?????mE?E?由此可得nnn。limmE[f?g??]?0?g?x??ng(x)?f(x)因此n??,即n,由黎斯定理存在n的?g?x??n子列k,使得limg(x)?f(x)nk??。mE??,?f?18、,证明:f(x)lim?ndx?0n??E1?f(x)nf(x)?0的充要条件是n。?f?fE?n????E?f???n?0?n?f(x)??1?f?1?f证明:若n0,由于n,则n。f(x)0?n?11?f(x)?n?1,2,3??mE??E又n,,,常函数1在上可积分,由f(x)lim?ndx??0dx?0n??E1?f(x)E勒贝格控制收敛定理得n。f(x)f(x)?ndx?0n?0E1?f(x)1?f(x)???0反之,若n(n??),而且n,对,xy?e?E?f???x??1令n?n?,由于函数1?x,当时是严格增加函数,?f(x)f(x)me??ndx??ndx?0n1??e1?f(x)E1?f(x)因此nnn。limE?f????0nf(x)?0所以n,即n。?x2?(R)?1dx?1(1?x2)n19、试求n?1。x2f(x)?,x?[?1,1]n(1?x2)nf(x)解令,则n为非负连续函数,从而非负可积。根据L积分逐项积分定理,于是,?x2?x2?(R)?1dx??(L)?dx?1(1?x2)n[?1,1](1?x2)nn?1n?1?x2?(L)??dx[?1,1](1?x2)nn?1?(L)?1dx[?1,1]?2。。mE??f(x)g(x)n?1,2,3,?20、设,,,分别依f(x)g(x)f(x)?g(x)?f(x)?g(x)测度收敛于和,证明nn。f?x??g?x??f?x??g?x??f?x??f?x??g?x??g?x?证明:因为nnnn于是???0,成立??E[|(f?g)?(f?g)|??]?E[|f?f|?]?E[|g?g|?]nnn2n2,所以??mE[|(f?g)?(f?g)|??]?mE[|f?f|?]?mE[|g?g|?]nnn2n2??limmE[|(f?g)?(f?g)|??]?limmE[|f?f|?]?limmE[|g?g|?]?0nnnnn??n??2n??2g?f?g?f即nn1??1?x???x2?x3???,0?x?1,21、试从1?x求证111ln2?1?????234。证明:在x?[0,1]时,xn?xn?1?0,n?1,2,3,?,由L逐项积分定理,??(L)???x2n?x2n?1?dx??(L)??x2n?x2n?1?dx[0,1][0,1]n?0n?0?1??(R)??x2n?x2n?1?dx0n?0??11???????2n?12n?2?n?0111?1?????234另一方面11(L)?dx?(R)?1dx?ln2[0,1]1?x01?x因此可得:111ln2?1?????234。1nx2lim(R)???01?nx.??A?(?CB)??(A?B){A}{B},n、n是两个集列,证明:n?1n?(x)?f(x){f}[a,b],m??,?m([a,b]\?E)?0E?[a,b]n?1,2,?n证明:存在一列可测集n,,使得n?1,{f}(R)?e???01??11txlim?dx05???.(Lusin)?R?,f(x),证明:存在定义在R?上的limg(x)?f(x).{g}n一列连续函数n,使得n??于E.