实变函数论课后答案第三章2
第三章第二节习题
1. 举例说明两个不可测集合的并，交，差即可以是不可测集合也可以是可测集.
解：先回忆P56知：?x??0,1?.令
Rx???;0???1,??x为有理数?则Rx?Ry??,若x?y
?x,Rx是无穷点集，?0,1???Rx.
x??0,1?在每一个Rx中取出一点Zx?Rx，令
S??Zx;x?Rx?.
?则我们从书上的论证已知S是不可测集，S??0,1?.
??因此?0,1??S也不可测.（否则?0,1?可测且S??0,1????0,1??S??应可测，
??????得矛盾）
但S与?0,1??S这两个不可测集的并?0,1?却是可测的.
??S?S?S不可测，S??S?10??不可测. ????????注意：E可测，??a,E?a??x?a;x?E?可测.
??S,S?3这两个不可测集的交为??S??S?3??. ?????????S?S?S不可测，S?S??可测，S??S?2??S不可测. ????????如同P56将??1,1?中全体有理数排成序列?1,?2?,?n?
E可测，??a,E?a??x?a;x?E?可测.
证明：若?a?Rn,E?a可测，则取a?0?E可测
为证反面的结果，须证 （1）
?T,E?Rn,a?RnT?E?a??T?a???E?a?c（2）?E?a??Ec?a
证明：（1）?y?T?E?a??z?T?a,y?z?a
?y?E?a,y?T?a
?y??T?a???E?a?
反过来，?y??T?a???E?a?,?y?T?E?a??z?T,w?E 使y?z?a,y?w?a，?w?z?E?T?y?E?T?a. （2）?y?Ec?a?y?a?Ec.
?y?a?E,?y?a?E?y??a?E?c
c反过来，?y??a?E?,y?a?E?y?a?E,?y?a?Ec.
?y?Ec?a.
若E可测，则?T?Rn有
m?T??m?T?E??m?T?Ec?
由m的平移不变性
m?T??m?T?a??m??T?a??E??T?a??Ec??m??T?a??E?a??m??T?a??Ec?a??m??T?a?a???E?a???m?T?a?a???E?a?c??
?m?T??E?a???mT??E?a??c?故E?a可测.
2. 试在二维空间R2中作出一不可测集合来.
解：设A?Rp,B?Rq为一不可测集，则A?B?Rp?q必不可测
3. 举例说明定理6的结果对任m?T??的T可以不成立. 解：令An??n,??,T?R1????,??，则A1?A2??An?An?1?
???E??An??,m??An??0 n?1?n?1??m??T?E??m??E??m?????0
m??T?An??limm??An??? 而limn??n???Th6中m?T??是必需的.
4. 证明对任意可测集合A和B都有
m?A?B??m?A?B??m?A??m?B? （*）
证明：若m?A?B???，则A?B?A,B
?m?A?B??0,m?A???,m?B???
???m?A?B??m?A?B??m?A??m?B???成立.
若m?A?B???则（*）等价于
m?A?B??m?A??m?B??m?A?B?
注意到A?B?A??B?A?,A??B?A???
且A,B可测?B?A可测
m?A?B??m?A??m?B?A?
?A可测
m?B??m?A?B??m?Ac?B??m?A?B??m?B?A?
?m?A?B???,m?B?A??m?B??m?A?B? ?m?A?B??m?A??m?B??m?A?B?
5. 证明：对任意E?Rn，及任意??0都有Rn中的开集G，使G?E且
m?G?m?E??
证明：???0,?Ii使E??Ii
i?1????i?1?Ii为开区间?
?Ii?1?ii?m?E??
?令G??Ii，则G为开集，且
i?1???mG?m??Ii???m??Ii???Ii?m?E??
?i?1?i?1i?1????6. 证明对于任意R中的可测集合序列
n?Ek??k?1，都有
mliminfEk?liminfm?Ek?
k??k??????若存在k0，使m??Ek????，则还有
?k?k0?mlimsupEk?limsupm?Ek?
k??k????infEk???Ei，limsupEk???Ei 证明：由P9定理8，limk??k??m?1i?mm?1i?m????infEk,limsupEk都是可测的. 故从P61定理3，P64定理4知limk??k??令?Ei?Sm,则S1?S2???Sm，
i?m?mliminfEkk????????????m???Ei??m??Sm??m?1i?m??m?1?????limm?Sm??limm??Ei??liminfm?Ei? m??m???i?m?m??i?m?supinfm?Ei??liminfm?Ek?m?1i?mk??????m??Ei??m?Ej?,?j?m?i?m?（） ????m??Ei??infm?Ej??i?m?j?m令 Km??Ei，则K1?K1??Km，由条件
i?m??K0,KK0??Ei满足mKK0??
i?K0???由P65定理4
mlimsupEkk?????????????????m???Ei??m??Km??m?KK0???Km???m?1i?m??m?1??m?1???????limmKK0?Km?limm?Km??limm??Ei??limsupm?Ei?? m??m??m???i?m?m??i?mlimsupm?Ek??limsupm?Ek???m??i?mk??7. 设E是Rn中的可测集，mE??，证明如果E的可测子集的序列
?Ek??kk?1?k?1，使
k?m?E???，则m?limsupE??0
k???supEk???Ek 证明：limk??m?1k?m?令Sm??Ek，则Sm可测（?Ek可测）且S1?S2???Sm??m?
k?m????m?S1??m??Ek???m?Ek??0?m???
?k?1?k?m?从S1?S2???Sm??m?，m?S1???
???????由P65定理6有mlimsupEk?m???Ei??m??Sm??limm?Sm??0. k???m?1i?m??m?1?m????证毕.