实变函数与泛函分析基础第三版(程其襄)+课后答案.pdf
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1、?1.?A (B C)=(A B)(A C).?x (A(B C).?x A,?x AB,x AC,?x (AB)(AC).?x B C,?x A B?x A C,?x (A B)(A C),?A (B C)(A B)(A C).?x (A B)(A C).?x A,?x A (B C).?x 6 A,?x A B?x A C,?x B?x C,?x B C,?x A (B C),?(A B)(A C)A (B C).?A (B C)=(A B)(A C).2.?(1)A B=A (A B)=(A B)B;(2)A (B C)=(A B)(A C);(3)(A B)C=A (B C);(4)A
2、 (B C)=(A B)(A C);(5)(A B)(C D)=(A C)(B D);(6)A (A B)=A B.?(1)A(AB)=As(AB)=A(sAsB)=(AsA)(AsB)=AB;(A B)B=(A B)sB=(A sB)(B sB)=A B;(2)(AB)(AC)=(AB)s(AC)=(AB)(sAsC)=(AB sA)(AB sC)=A (B sC)=A (B C);(3)(A B)C=(A sB)sC=A s(B C)=A (B C);(4)A(B C)=A(B sC)=As(B sC)=A(sB C)=(AsB)(AC)=(A B)(A C);(5)(A B)(C D)=
3、(A sB)(C sD)=(A C)s(B D)=(A C)(B D);(6)A (A B)=A s(A sB)=A (sA B)=A B.3.?(A B)C=(A C)(B C);A (B C)=(A B)(A C).?(A B)C=(A B)sC=(A sC)(B sC)=(A C)(B C);(AB)(AC)=(AsB)(AsC)=AsB sC=As(B C)=A(B C).4.?s(Si=1Ai)=Ti=1sAi.?x s(Si=1Ai),?x S,?x 6Si=1Ai,?i,x 6 Ai,?x sAi,?1x Ti=1sAi.?x Ti=1sAi,?i,x sAi,x S,x 6 A
4、i,?x S,?x 6Si=1Ai,?x s(Si=1Ai).?s(Si=1Ai)=Ti=1sAi.5.?(1)(SA)B=S(A B);(2)(TA)B=T(A B).?(1)SA B=(SA)sB=S(A sB)=S(A B);(2)TA B=(TA)sB=T(A sB)=T(A B).6.?An?B1=A1,Bn=An(n1S=1A),n 1.?Bn?nS=1A=nS=1B,1 n .?i 6=j,?i j.?Bi Ai(1 i n).Bi Bj Ai(Ajj1n=1An)=Ai Aj sA1 sA2 sAi sAj1=.?Bi Ai(1 6=i 6=n)?nSi=1BinSi=1Ai.
5、?x nSi=1Ai,?x A1,?x B1nSi=1Bi.?x 6 A1,?in?x Ain,x 6in1Si=1Ai?x Ain.?x Ainin1Si=1Ai=BinnSi=1Bi.?nSi=1Ai=nSi=1Bi.7.?A2n1=?0,1n?,A2n=(0,n),n=1,2,?An?limnAn=(0,);?x (0,),?N,?x N?0 x N,?x An.?2n 1 N?x A2n1,0 x 1n.?n?0 N,x An,?x Tm=n+1AmSn=1Tm=nAm,?limnAnSn=1Tm=nAm.?x Sn=1Tm=nAm,?n,?x Tm=nAm,?m n,?x An,?x
6、 limnAn.?limnAn=Sn=1Tm=nAm.29.?(1,1)?(,+)?:(1,1)(,+).?x (1,1),(x)=tan2x.?(1,1)?(,)?10.?S:x2+y2+(z 12)2=(12)2?(0,0,1)?xOy?M?(x,y,z)S(0,0,1),(x,y,z)=?x1 z,y1 z?M.?S?M?11.?A?A?G=z|z?,?z?rz,?z?rz?z?G?G?12.?An?n?n=1,2,?A=Sn=0An.An?n+1?n?n+1?0?4?6,An=a,?4?4,A=a.13.?A?(?)?A?A?:(x,y,r).?(x,y)?r?x,y?r?0?A=a.
7、14.?f?(,)?E,?(1)?x (,),limx0+f(x+x)=f(x+0)?limx0f(x+x)=f(x 0)?(2)x E?f(x+0)f(x 0).(3)?x1,x2 E,?x1 x2,?f(x10)f(x1+0)f(x20)f(x2+0),?x E,?(f(x 0),f(x+0),?(3)?E?x?11?15.?(0,1)?0,1?3?(0,1)?R=r1,r2,?(0)=r1,(1)=r2,(rn)=rn+2,n=1,2,(x)=x,x (0,1)R),?0,1?(0,1)?16.?A?A?A=x1,x2,A?eA.An=x1,x2,xn,An?eAn.?eAn?2n?eA
8、=Sn=1eAn,?eA?A?eA?17.?0,1?c.?0,1?A,0,1?r1,r2,?B=(22,23,2n,)A?(22n)=2n+1,n=1,2,(22n+1)=rn,n=1,2,(x)=x,x 6 B.?A?0,1?0,1?c,?A?c.18.?A?A=ax1x2x3,?xi?c?A?c.?xi Ai,Ai=c,i=1,2,.?Ai?R?i.?A?E?ax1x2x3 A.(ax1x2x3)=(1(x1),2(x2),3(x3),).?(ax1x2x3)=(ax1x2x3),?i,i(xi)=i(xi).?i?xi=xi,?ax1x2x3=ax1x2x3.?(a1,a2,a3,)E,
9、ai R,i=1,2,?i?xi Ai,?i(xi)=ai.?ax1x2x3 A,?(ax1x2)=(1(x1),2(x2),)=(a1,a2,),?A?E?c.19.?Sn=1An?c,?n0,?An0?c.?E=c,?Sn=1An=E.?An c,n=1,2,.?PiE?R?x=(x1,x2,xn,)E,?Pi(x)=xi.?Ai=Pi(Ai),i=1,2,4?Ai Ai c,i=1,2,.?i,?i RAi,?=(i,2,n,)E.?6Sn=1An.?Sn=1An,?i,?Ai,?i=Pi()Pi(Ai)=Ai,?RAi?6Sn=1An=E,?E?i0,?Ai0=c.20.?0?1?T,
10、?T?c.?T=1,2,|i=0or1,i=1,2,.?T?E?:1,2,2,3,?T?E?(T)?A E=c,?(0,1?2?x (0,1?x=0.12,?i0?1,?f(x)=1,2,?f?(0,1?T?f(0,1)?T (0,1=c.?A=c.5?Eo?E?E?1.?P0 E?P0?U(P,)(?P0?)?P0?P1?E(?P1?),?P0 Eo?P0?U(P,)(?P0?)?U(P,)E.?P0 E,?P0?U(P,),?P0?U(P0)U(P,),?P1 E U(P0)E U(P,)?P16=P,?P0?P1?P0?E.?P0?P0?P1?E,?P0?U(P0)?P0?P1?E,?P
11、0 E.?P0 Eo,?U(P0)E.?P0 U(P,)E,?U(P0)U(P,)E,?P0 Eo.2.?E1?0,1?E1?R1?E1,Eo1,E1.?E1=0,1,Eo1=,E1=0,1.3.?E2=(x,y)|x2+y2 1.?E2?R2?E2,Eo2,E2.?E2=(x,y)|x2+y2 1,Eo1=(x,y)|x2+y2 a?E=x|f(x)a?1?x0 E,?f(x0)a.?f(x)?0,?x(,),|xx0|a,?x U(x0,)?x E,?U(x0,)E,E?xn E,?xn x0(n ).?f(xn)a,?f(x)?f(x0)=limnf(xn)a,?x0 E,?E?9.?F
12、?Gn=?x|d(x,F)1n?,Gn?x0 Gn,d(x0,F)1n,?y0 F,?d(x0,y0)=1n.(?y F,d(x0,y)1n,?d(x0,F)=infyFd(x0,y)1n,?d(x0,F)0,?x U(x0,),d(x0,x).d(x,y0)d(x0,x)+d(x0,y0)+=+1n =1n.?d(x,F)=infyFd(x,y)d(x,y0)1n,?x Gn.?U(x0,)Gn,?Gn?x Tn=1Gn,?n,x Gn,d(x,F)0,xn x0,f(xn)f(x0)+0?f(xn)f(x0)0,?c=f(x0)+,?xn E=x|f(x)c,?x06 E(?f(x0)f(
13、x0)+0=c),?E?f(x)?a,b?12.?2?5:?E 6=,E 6=Rn,?E?(?E 6=).?P0=(x1,x2,xn)E,P1=(y1,yn)6 E.?Pt=(ty1+(1 t)x1,ty2+(1 t)x2,tyn+(1 t)xn),0 t 1.t0=supt|Pt E.?Pt0 E.?Pt0 E.?t06=1.?t 0,1?t0 tnt0,tn t0?Ptn E,?Ptn Pt0,?Pt0 E.?Pt06 E,?t06=0,?tn,0 tn t0,tn t0,Ptn Pt0,Ptn E,?Pt0 E.?E 6=.13.?P?1,?P?c.?P?(?P),?13,23?=(0.
14、1,0.2),?19,29?=(0.01,0.02),?79,89?=(0.21,0.22),?n?2n1?I(n)k,k=1,2,2n1?I(n)k=(0.a1a2an11,0.a1a2an12),?a1,a2,an1?0?2.?0,1 P?1,?P?1,?x P,?x?x=a13+a232+an3n+,?an?0?2.?A,?A P.?A 0,1,?0,1 P?ai?1,?0,1 P?A?A P.3?A?B?:x=Xn=1an3nXn=112nan2,?an=0?2,?A?B?1-1?A?c,?A P,?P c,?P c,?P=c.4?1.?E?mE +.?E?I?E I.?mE mI 0
15、,?Ii,?xi Ii,?|Ii|=2i(?Rp?pp2i?xi?Ii),?Si=1Ii E,?Pi=1|Ii|=.?mE=0.3.?E?mE 0,?mE?c,?E?E1,?mE1=c.?a=infxEx,b=supxEx,?E a,b.?Ex=a,x E,a x b,f(x)=mEx?a,b?x 0?|f(x+x)f(x)|=|mEx+x mEx|m(Ex+E)|m(x,x+x=x.?x 0?f(x+x)f(x),?f(x)?x 0,x 0?f(x x)f(x),?f(x)?a,b?f(a)=mEa=m(E a)=0f(b)=m(E a,b)=mE.?c,c mE,?x0 a,b?f(x0)
16、=c.?mEx0=m(a,x0 E)=c.?E1=E a,x0 E.?mE1=c.4.?S1,S2,Sn?,Ei Si,i=1,2.,n,?m(E1 E2 En)=mE1+mE2+mEn.?S1,S2,Sn?2?3?1,?T,?m(T nSi=1Si)=nPi=1m(T Si).?T=nSi=1Ei,?T Si=(nSj=1Ej)Si=Ei,T (nSi=1Si)=nSi=1Ei,?m(nSi=1Ei)=m(T (nSi=1Si)=nPi=1m(T Si)=nPi=1mEi.5.?mE=0,?E?T,T=(E T)(T E),?mT m(E T)+m(T E).?E T E,?m(E T)mE
17、=0.T E T,m(T E)mT,?m(E T)+m(T E)mT.1?mT=m(T E)+m(T E),?E?6.?(Cantor)?P?0,1?13,?29,?n?2n13n,.?P?0,1?Pn=12n13n=1(?).?m0,1=m(P (0,1 P)=mP+m(0,1 P).?mP=m0,1 m(0,1 P)=1 1=0,?0.7.A,B Rp?mB +.?A?m(AB)=mA+mBm(AB).?A?m(A B)=m(A B)A)+m(A B)A)=mA+m(B A).?mB=m(B A)+m(B A),?mB +,?m(B A)0,?G?F,?F E G,?m(G E),m(E
18、F).?mE 0,?Ii,i=1,2,?Si=1Ii E,?Pi=1|Ii|mE+.?G=Si=1Ii,?G?G E,?mE mG Pi=1mIi=Pi=1|Ii|mE+,?mG mE ,?m(G E).?mE=?E?E=Sn=1En(mEn),?En?Gn,?Gn En?m(GnEn)2n.?G=Si=1Gn,G?G E,?G E=Sn=1GnSn=1EnSn=1(Gn En).?m(G E)n=1m(Gn En)0?G,G E,?m(G E).?G E=G E=E (G)=E G,?F=G,?F?m(E F)=m(G E).9.E Rq,?An,Bn,?An E Bn?m(BnAn)0(n
19、 ),?E?2?i,Tn=1Bn Bi,?Tn=1Bn E Bi E.?E Ai,Bi E Bi Ai,?i,m n=1Bn E!m(Bi E)m(Bi Ai)=m(Bi Ai).?i ,?m(Bi Ai)0,?m?Tn=1Bn E?=0.?Tn=1Bn E?Bn?Tn=1Bn?E=Tn=1Bn?Tn=1Bn E?10.A,B Rp,?m(A B)+m(A B)mA+mB.?mA=+?mB=+,?mA +?mB 0,?F E,?m(E F),?E?n,?Fn E,?m(E Fn)1n.?F=Sn=1Fn,?F?F E.?n,?m(E F)m(E Fn)r?Ef=r?f(x)?r,Ef r?,?
20、rn?Ef a=Sn=1Ef rn,?Ef rn?Ef?f(x)?E?r,Ef=r?f(x)?E=(,),z?(,)?x z,f(x)=3;x 6 z,f(x)=2,?r,Ef=r=?Ef 2=z?f?2.?f(x),fn(x)(n=1,2,)?a,b?k?k=1limnE?|fn f|N?|fn(x)f(x)|1k,?x limnE?|fn f|1k?.?k?x k=1limnE?|fn f|1k?.?x Tk=1limnE?|fn f|0,?k0,?1k0,?x limnEh|fn f|N?x Eh|fn f|1k0i,?|fn(x)f(x)|1k0,?limnfn(x)=f(x),?x
21、A.?A=k=1limnE?|fn f|limnfn?fn?fn(x)?E?E F limnfn=+E limnfn=E limnfn limfn.1?E limnfn=+E limnfn=E limnfn limnfn?4.?E?0,1?f(x)=(x,x E,x,x 0,1 E.?f(x)?0,1?|f(x)|?f(x)?0 E,?Ef 0=E?0 6 E,?Ef 0=E?f(x)?x 0,1?|f(x)|=x?|f(x)|?0,1?5.?fn(x)(n=1,2,)?E?a.e.?|fn|a.e.?f.?0?c?E0 E,m(EE0),?E0?n?|fn(x)|c.?mE .?E|fn|=
22、,Efn6 f?n=0,1,2,.?E1=Efn6f(Sn=0E|fn|=),?mE1=0.?EE1?fn(x)?f(x).?E2=EE1,?x E2,supn|fn(x)|.?E2=k=1E2supn|fn|k,E2supn|fn|k E2supn|fn|k+1.?mE2=limkmE2supn|fn|k.?k0?mE2 mE2supn|fn|k0 .?E0=E2supn|fn|k0,c=k0.?E0?n,|fn(x)|c,?m(E E0)=m(E E2)+m(E2 E0).6.?f(x)?(,)?g(x)?a,b?f(g(x)?E1=(,),E2=a,b.?f(x)?E1?c,E1f c=
23、Sn=1(n,n),?(n,n)?(?n?,n?+).?E2f(g)c=Sn=1E2n g n E2g n,E2g c?7.?fn(x),(n=1,2,)?E?”?”?f(x),?fna.e.?f.?fn(x)?E?”?”?f(x),?0,?E E,?m(E E)?fn?E?f(x).?E0?E?fn?,E0 E E(?E?fn?),?mE0 m(E E0)0,?E E?f(x)?E?m(E E),?f(x)?E?a.e.?1/n,?En E,?f(x)?En?m(E En)1n.?E0=E Sn=1En,?n,?mE0=m(E Sn=1En)m(E En)a=E0f a (Sn=1Enf a
24、),?f?En?Enf a?m(E0f a)mE0=0,?E0f a?Ef a?f?f?En?Sn=1En?f(x)a.e.?9.?fn?E?f,?fn(x)g(x)a.e.?E,n=1,2,.?f(x)g(x)?E?fn(x)f(x),?fni fn,?fni(x)?E?a.e.?f(x).?E0?fni(x)?f(x)?En=Efn g.?mE0=0,mEn=0.m(Sn=0En)Pn=0mEn=0.?E Sn=0En?fni(x)g(x),fni(x)?f(x),?f(x)=limfni(x)g(x)?E Sn=0En?f(x)g(x)?E?10.?E?fn(x)f(x),?fn(x)f
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