答：不一定，如???1??n?1???11?,?1?????1,1? nn?18. 可测集E上的可测函数与简单函数有什么关系？
答：简单函数必是可测函数但可测函数不一定是简单函数，可测函数一定可表示成简单函数列的极限形式. 19. ?a,b?上的有界变差函数与单调函数有什么关系？
答：单调函数必为有界变差函数但有界变差函数不一定为单调函数，有界变差函数可表示成单调函数之差. 20. 简述无穷多个闭集的并集是否必为闭集？ 答：不一定 如???1??n?1???11?,?1?????1,1? nn?21. 可测集E上的可测函数与连续函数有什么关系？
答：E上连续函数必为可测函数但E上的可测函数不一定时连续函数，E上可测函数在E上是“基本上”连续的函数 22. ?a,b?上的绝对连续函数与有界变差函数有什么关系？
答：绝对连续函数必为有界变差函数但有界变差函数不一定为绝对连续函数
六、计算题
2??x1. 设f?x???3??xx?Ex??0,1?\\E，其中E为?0,1?中有理数集，求
?0,1??f?x?dx.
x3dx，
?0,1?解：因为mE?0，所以f?x??x3,a.e于?0,1?，于是
?0,1?3?f?x?dx??而x在?0,1?上连续，从而黎曼可积，故由黎曼积分与勒贝格积分的关系，
x411xdx??R??xdx?|0? ?044?0,1?313因此
?0,1??f?x?dx?1. 4x??r1,r2,?rn???12. 设?rn?为?0,1?中全体有理数，fn?x???，求lim?fn?x?dx.
n??0x?0,1\\r,r,?r????12n???0,1?解：显然fn?x?在?0,1?上可测，另外由fn?x?定义知，fn?x??0,a.e于?0,1??n?1? 所以
?0,1??f?x?dx??0dx?0
n?0,1?因此limn???0,1??f?x?dx?0.
n3. 设f?x???x?P?sinx，P为康托集，求?f?x?dx. 2xx?0,1\\P????0,1?2解：因为mP?0，所以f?x??x,a.e于?0,1?
于是
?0,1?2?f?x?dx??0,1??x2dx
而x在?0,1?上连续，所以
x311xdx??R??xdx?|0? ?033?0,1?212因此
?0,1??1f?x?dx?.
3nxsin?nx?,x??0,1?，求lim?fn?x?dx. 4. 设fn?x??n??1?n2x2?0,1?解：因为fn?x?在?0,1?上连续，所以可测?n?1,2,?? 又fn?x??而limnxsin?nx?nxnx1???,x??0,1?,n?1,2,? 22221?nx1?nx2nx2nx?0，所以limfn?x??0.
n??n??1?n2x2因此由有界控制收敛定理
limn???0,1??f?x?dx??limf?x?dx??0dx?0
n?0,1?n??n?0,1??x3?5. 设f?x????cosx?x?E??f?x?dx. ???，E为?0,?中有理数集，求
x??0,?\\E2?????2?0,??????2?解：因为mE?0，所以f?x??cosx,a.e于?0,1? 于是
???0,2?????f?x?dx???cosxdx
???0,2???而cosx在?0,???上连续，所以黎曼可积，由牛顿莱布尼公式 ?2????20?0,1??cosxdx??R?????0,2???cosxdx?sinx|02?1
因此
??f?x?dx?1
nxcos?nx?,x??0,1?，求lim?fn?x?dx.
n??1?n2x2?0,1?6. 设fn?x??解：因为fn?x?在?0,1?上连续，所以可测?n?1,2,??
又fn?x??而limnxcos?nx?nxnx1???,x??0,1?,n?1,2,?
1?n2x21?n2x22nx2nx?0，所以limfn?x??0.
n??n??1?n2x2因此由有界控制收敛定理
limn???0,1??f?x?dx??limf?x?dx??0dx?0
n?0,1?n??n?0,1?3??sinx7. 设f?x?????xx?Px??0,1?\\P，P为康托集，求
?0,1??f?x?dx.
解：因为mP?0，所以f?x??x,a.e于?0,1? 于是
?0,1??f?x?dx??xdx
?0,1?而x在?0,1?上连续，所以
x211xdx??R??xdx?|0? ?022?0,1?12因此
?0,1??f?x?dx?1. 2ln?x?n??xecosxdx. 8. 求lim?n??n?0,n?解：令fn?x????0,n??x?ln?x?n??xecosx n显然fn?x?在?0,???上可测，且
ln?x?n??xecosxdx??fn?x?dx ?n?0,n??0,???ln?x?n??xln?x?n?ecosx?,?x??0,???,n?1,2,? 因为fn?x??nnln?x?n?不难验证gn?x??，当n足够大时，是单调递减非负函数，且
nlimgn?x??0，所以
n??limln?x?n?dx?lim?gn?x?dx??limgn?x???0dx?0
n???n??n??n?0,????0,????0,????0,???由勒贝格控制收敛定理
limn???0,????fn?x?dx?0
故limln?x?n??xecosxdx?0.
n???n?0,n?9. 设D?x????1??1x为?0，上的有理点，求?D?x?dx.
0x为0，上的无理点1???1???0，证明 记E1是?0,1?中有理数集，E2是?0,1?中无理数集，则
?0,1??E1?E2,E1?E2??，mE1?0,mE2?1，且D?x??1?E所以
?0,1?1?0?E2，
?D?x?dx?1mE?0mE12?0.
10 求limn??0???ln?x?n??xecosxdx. nln?x?n??xecosx?0 证明 易知limn??n对任意x?0,n?1，
ln?x?n??xln?x?n? ecosx?nny?ln?x?y?ln?x?y?x?y设f(y)?，y?0，则f?(y)?， 2yy当y?3时，
y?1?ln?x?y?，f?(y)?0. x?y则f(n)?ln?x?n?是单调减函数且非负（n?3）； n又limln?x?n?1?lim?0，由Levi单调收敛定理得
n??n??nx?n??????ln?x?n?ln?x?n?ln?x?n?dx??limdx??0dx?0，即?L(E)，
0n??0nnnlim?n??0再由Lebsgue控制收敛定理得
??????ln?x?n??xln?x?n??xecosxdx??limecosxdx??0dx?0
0n??0nnlim?n??02??x11. 设f?x???3??xx?Px??0,1??P，其中P为康托集，求
1??0，?f?x?dx.
解：因为P为康托集，故mP?0，m?0,1?\\P?1 所以f?x??x3??0,1??P?x2?P 所以
???0,1??f?x?dx?xmP?xm??0,1??P??x
23312. 求fn?x??解：易知：limnx,E??0,1?，求lim?fn?x?dx.
n??1?n2x2Enx?0?x??0,1??
n??1?n2x2nx1,gx?令fn?x??， ??2221?nxx21nx1?n2x2?nx31?nx?n?x????0 则g?x??fn?x??2?nxnxx1?n2x222?x?x22221?nx1?nx所以0?fn?x??g?x?x??0,1?,n?1 又因为g?x?在?0,1?上Lebesgue可积， 所以由控制收敛定理，得 lim
七、证明题
1．证明集合等式：(A\\B)?B?A?B 证明
??nxdx??0dx?0
n???1?n2x2EE(A\\B)?B?(A?Bc)?B?(A?Bc)?(A?B)?B?A?(B?Bc)?B?A?B
2．设E是[0,1]中的无理数集，则E是可测集，且mE?1
c*证明 设F是[0,1]中的有理数集，则F是可数集，从而mF?0，因此F是可测集，从而F可测，又
E?[0,1]\\F?[0,1]?Fc，故E是可测集.由于E?F??，所以
1?m[0,1]?m(E?F)?mE?mF?0?mF，故mF?1
3．设f(x),g(x)是E上的可测函数，则E[x|f(x)?g(x)]是可测集 证明 设{rn}为全体有理数所成之集，则
E[x|f(x)?g(x)]??E[x|f(x)?rn?g(x)]???E[x|f(x)?rn]?E[x|g(x)?rn]?
n?1n?1??因为f(x),g(x)是E上的可测函数，所以E[x|f(x)?rn]，E[x|g(x)?rn]是可测集，n?1,2,?，于是由可测