实变函数与泛函分析习题解答

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1、100081§1.11.(1)A1∪A2=(A∩C)∪(A∩D)=A∩(C∪D)=A∩(A∪B)=A.(2)A∩C=∅,A∪B=C∪DA⊂DC⊂B.D⊂AB⊂C.Æ2.
Bc=(D∩A)∪(Dc∪A)c=(D∩A)∪(D∩Ac)=D∩(A∪Ac)=D.c
B=D(D∩A)c∩(Dc∪A)=(Dc∪Ac)∩(Dc∪A)=Dc=B.3.E=ABÆ∞∞4.x∈An∩Bn,∃m,n∈Nx∈Amx∈Bn.

2、m≤n,n=1n=1∞{An}x∈An,x∈An∩Bn⊆(An∩Bn).n=1∞∞∞x∈(An∩Bn),n∈Nx∈An∩Bn,x∈Anx∈Bn,n=1n=1n=1∞∞x∈An∩Bn,Æn=1n=15. {Bn}Æm,n∈N,m1cn−1n−1B=A∩A=A∩Ac,nninii=1i=1x∈Bn,x/∈Am,x/∈Bm.x∈Bm,x∈Am,x/∈Bn.{Bn}nnÆ n,An=Bn.i=1i=1∞

3、∞∞6.A=Amk.m=1n=1k=n127.An={x

4、fn(x)>a}A=xlimfn(x)>a.n→∞∞fn(x)>a
nlim→∞fn(x)>a.An⊂A,An⊂A.x∈A,∃n0∈N,n=1∞fn0(x)>a,x∈An0,A⊂An.n=11,8.(1)∀y∈{x

5、f(x)>a},f(y)>a.n∈Nf(y)≥a+n∞11y∈{x

6、f(x)≥a+}⊆x∈{f(x)≥a+}.nnn=1∞1{x

7、f(x)>a}⊆x∈{f(x)≥a+}.nn=1∞1∀y∈n=1{xA#:f(x)≥a+n

8、},n∈N,1y∈{x

9、f(x)≥a+},n1f(y)≥a+n>a.y∈{x

10、f(x)>a}.∞1x

11、f(x)≥a+⊆{x

12、f(x)>a}.nn=1∞1x

13、f(x)≥a+={x

14、f(x)>a}.nn=11,12∀y∈{x

15、f(x)≥a},f(y)≥a.∀n∈Nf(y)>a−ny∈{x

16、f(x)>a−n}∞1Æy∈x

17、f(x)>a−n.n=1∞1{x

18、f(x)≥a}⊆x

19、f(x)>a−.nn=1∞x

20、f(x)>a−1,11y∈n∀n∈N,y∈x

21、f(x)>a−n,f(y)>a−n

22、n=1Æf(y)≥a,y∈{x

23、f(x)≥a}.∞1x

24、f(x)>a−⊆{x

25、f(x)≥a}.nn=1∞1x

26、f(x)>a−={x

27、f(x)≥a}.nn=19.limAn=limAn=[−1,1].n→∞n→∞310.E⊂limEn,limEn⊂E.limEn=E.n→∞n→∞n→∞12.(0,1]⊂limAn,limAn⊂(0,1]Æx∈(0,1],n∈N,n→∞n→∞k≥n,ak

28、x∈limAn⇐⇒∀n∈N,∃k≥n,x∈Akn→∞⇐⇒∀n∈N,∃k≥n,0n0x∈An.n>n0n→∞χAn(x)=1.limχAn(x)=1=χlimAn(x);n→∞n→∞,x/∈limAn,∀n∈N,n∈Nn>n,x/∈AnχAn(x)=0.n→∞limχAn(x)n→∞2)∀x,x∈limAn,∀n∈N,n0∈N,

29、x∈An0.χAn(x)=1.n→∞0limχAn(x)=1=χlimAn(x);n→∞n→∞x/∈nlim→∞An,n0∈N,n>n0x/∈An,χAn(x)=0.limχAn(x)=0=χlimAn(x).n→∞n→∞§1.21.(1)f−1(YE)=f−1(Y∩Ec)=f−1(Y)∩f−1(Ec)=f−1(Y)∩(f−1(E))c=f−1(Y)f−1(E).(2)X=[0,2π],A=[0,π/2],f(x)=sinx.f(X)f(A)=[−1,0),f(XA)=[−1,1).2.(1)A=(A

30、B)∪(A∩B),B=(BA)∪(A∩B)(BA)∩(A