實變函數(shù)與泛函分析基礎(程其襄張奠宙著)高等教育出版社課后答案.doc

_1.A (B C) = (A B) (A C).-可編輯修改-x (A (B C).x A,x A B, x A C,x (A B) (A C).x B C,x A Bx 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 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 (B C) = (A B) (A C);(5)(A B) (C D) = (A C) (B D);(6)A (A B) = A B.(1)A (A B) = A s(A B) = A (sA sB) = (A sA) (A sB) = A B;(A B) B = (A B) sB = (A sB) (B sB) = A B;(2)(A B) (A C) = (A B) s(A C) = (A B) (sA sC) = (A B sA) (A B 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) = A s(B sC) = A (sB C) = (A sB) (A C) =(A B) (A C);(5)(A B) (C D) = (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);(A B) (A C) = (A sB) (A sC) = A sB sC = A s(B C) = A (B C). 4.s(Ai) =sAi.i=1i=1x s(i=1Ai),x S,x i=1Ai,i,x Ai,x sAi,1x i=1sAi.x i=1sAi,i,x sAi,x S,x Ai,x S,x i=1Ai,x s(i=1Ai).s(i=1Ai) =i=1sAi.5.(1) (A) B =(A B);(2)(A) B =(A B).(1)A B = (A) sB =(A sB) =(A B);(2)A B = (A) sB =(A sB) =(A B).n16.AnnnB1 = A1, Bn = An (=1A), n > 1.Bn=1A =1B, 1 n .i = j,i < j.Bi Ai(1 i n).j1Bi Bj Ai (Aj n=1nAn) = Ai Aj sA1 sA2 · · · sAi · · · sAj1 = .nBi Ai(1 = i = n)i=1Bi i=1Ai.nnx i=1Ai,x A1,x B1 i=1Bi.x A1,inx Ain,in1in1nnnx i=1Aix Ain.x Ain i=1Ai = Bin i=1Bi.i=1Ai =i=1Bi.7.A2n1 = 0, n1 , A2n = (0, n), n = 1, 2, · · · ,n An = (0, );AnNx (0, ),N,x < N,xn>NAn,0 < x < n,x n An,xn An (0, ),n An = (0, ).n An = ;x n An = ,N,n > N,x An.2n 1 > Nx A2n1,0 <x<1n 0 < x 0,n An = .8.lim An =nn=1 m=nAm.x n An,N,n > N,x An,x m=n+1Am n=1 m=nAm,n An n=1 m=nAm.x n=1 m=nAm,n,x m=nAm,m n,x An,x n An.lim An =Am.nn=1 m=n2limx A2n,limlimlimlimlimn.limlimlimlim9.(1, 1)(, +)(, ) : (1, 1) (, +).x (1, 1), (x) = tan 2 x. (1, 1)10.(0, 0, 1)(x, y, z) S(0, 0, 1),xOyM(x, y, z) =x y,1 z 1 z M.SM11.AAzrzGG = z|zz,Gzrz,12.Annn = 1, 2, · · · ,A =n=0An.Ann +1nn +106,An = a,13.A§44,A = a.()§4AA: (x, y, r).(x, y)rx, yr0A = a.14.f(, )E,(1)(2)x Ex (, ),xlim0+ f(x + x) = f(x + 0)f(x + 0) > f(x 0).xlim f(x + x) = f(x 0)(3)x E,15.x1, x2 E,(0, 1)x1 < x2,110, 13(3) E xS : x2 + y2 + (z 12)2 = ( 12 )20f(x1 0) < f(x1 + 0) f(x2 0) < f(x2 + 0),(f(x 0), f(x + 0),16.(0, 1)0,1 (0,1)AR = r1, r2, · · · , (1) = r2,AAn.A = x1, x2, · · · , AAn 2nA =A.An = x1, x2, · · · , xn, AnAn, AAn=1A17.0, 1c.0,1B =A,0,12 2,2 3, ··· ,2n, ···r1, r2, · · · , A(2)=2n22n + 12,n +1) = rn,n = 1, 2, · · ·n = 1, 2, · · ·(x) = x,x B.18.xiA0,1Ac0,1Ac,c.Ac.A = ax1x2x3··· ,EAi R i.ax1x2x3··· A.(ax1x2x3···) = (1(x1), 2(x2), 3(x3), · · · ).A(ax1x2x3···) = (ax1x2x3···),i, i(xi) = i(xi).ixi =xi, ax1x2x3··· = ax1x2x3···.i(xi) = ai.ax1x2x3··· A, (ax1x2···) = (1(x1), 2(x2), · · · ) =A E c.i19.Anc,n0,An0c.n=1E = c,n=1An = E.An < c, n = 1, 2, · · · .PiERx = (x1, x2, · · · , xn, · · · ) E,Pi(x) = xi.Ai = Pi(Ai), i = 1, 2, · · · ,4 (0) = r1, (rn) = rn+2, n = 1, 2, · · · (x) = x, x (0, 1)R),xi Ai, Ai = c, i = 1, 2, · · · .xi Ai,(a1, a2, · · · ),(a1, a2, a3, · · · ) E, ai R, i = 1, 2, · · · , A < A< c, i = 1, 2, · · · .An.n=1 n=1An,i,i RAi ,i, = (i, 2, · · · , n, · · · ) E.i = Pi() Pi(Ai) = Ai , RAi n=1An = E, Ei0,Ai0 = c.20.01T ,Tc.T = 1, 2, · · · | i = 0 or 1, i = 1, 2, · · · .Tx (0, 1(0,1 TE f(0, 1)(0,1 2i 0 1,TE (T )f(x) = 1, 2, · · · ,A = c.f5 Ai,i i : 1, 2, · · · 2, 3, · · · ,A E = c,x = 0.12 · · · ,T (0, 1 = c.EoEE¯1.P0P0P0 EP1U(P, )(E (P0P0P1)U(P, ) (oU(P, ) E.P0)P0P0P1 E U(P0) E U(P, )E.U(P, ),P1 = P ,P0P0U(P0) U(P, ),P1P0P1P0E,P0P1E,P0U(P0)P0 Eo,U(P0) E.P0 U(P, ) E,U(P0) U(P, ) E,P0 Eo.2.E10, 1E1R1E1 , E1o, E¯1.E1 = 0, 1,E1o = ,E¯1 = 0, 1.3.E2 = (x, y)|x2 + y2 < 1.E2R2E2 , E2o, E¯2.E2 = (x, y)|x2 + y2 1,E1o = (x, y)|x2 + y2 < 1,E¯1 = (x, y)|x2 + y2 1.4.E3R2y =0,xE3x = 0,x =0E3 E3.E3 = E3 (0, y)| 1 y 1,E3o = .5.R22E1 , E1o, E¯1E1 = (x, 0)|0 x 1, , E1o = , E¯1 = E1 .6.FF¯ = F.FFF F,F¯ = F F = F.F¯ = F,F F F = F¯ = F,7.GF G = F GFGFG F = G F8.f(x)E = x|f(x) a1a, E = x|f(x) > aP0 E),P0 E,P0 E . sin 1 , o(, )x0 E,(, ),|x x0| < f(x0) > a.f(x) > a,f(x)x U(x0, )x E, > 0,U(x0, ) E,Ex xn E,xn x0(n ).f(xn) a,f(x)f(x0) = n f(xn) a,x0 E,9.Ey0 F ,F11n11d(x0, F ) = inf d(x0, y) 1d(x0, F ) <1n). =1n > 0,x U(x0, ), d(x0, x) < .d(x, y0) d(x0, x) + d(x0, y0) < + = +1n =1n.d(x, F ) = inf d(x, y) d(x, y0) <1x Gn.U(x0, ) Gn,Gnx n=1Gn,n, x Gn, d(x, F ) <1n ,d(x, F ) = 0.Fx F (x F ,yn F,d(x, yn) 0,x F F ,),n=1Gn F.Gn F, n = 1, 2, · · · ,n=1Gn F ,n=1Gn = F ,FGGGn,G =Gn,n=1GnGG = (G) = (n=1Gn) =n=1Gn,10.0,10,10,177(0.7,0.8).7······(0.07, 0.08) (0.17, 0.18)···(0.97, 0.98).0,1n7(0.a1a2 · · · an17, 0.a1a2 · · · an18),ai(i = 1, 2, · · · , n 1)n 1097a1, a2, · · · , an1Ann=12limGn = x|d(x, F ) <,Gny F, d(x0, y) d(x0, y0) = <(n.n,x0 Gn, d(x0, F ) <n,n,yFn,yFn.0,17n=1An (, 0) (1, ) .An, (, 0), (1, )0,1711. f(x)E1 = x|f(x) ca, bc,E = x|f(x) cf(x)a, b8EE1EE1x0 a, b. f(x)f(xn) f(x0) 0,f(x0) < f(x0) + 0 = c),x0E0 > 0, xn x0, f(xn) f(x0) + 0c = f(x0) + ,f(x) a, b12.§25:E = , E = Rn,E(E = ).P0 = (x1, x2, · · · , xn) E, P1 = (y1, · · · , yn) E.(1 t)x2, · · · , tyn + (1 t)xn), 0 t 1.t0 = supt|Pt E.Pt = (ty1 + (1 t)x1, ty2 +Pt0 E.t0, tn t0 Ptn E, Ptn Pt0, Pt0 E.t0 = 1.Pt E.tn, 1 > tn >Pt0 E. E = .t0 = 0,tn, 0 < tn < t0, tn t0, Ptn Pt0, Ptn E,P13.c.P1,P1 2,3 31 2,9 97 8,9 9······= (0.1, 0.2),= (0.01, 0.02),= (0.21, 0.22),(P ),n2n1(n)(n)= (0.a1a2 · · · an11, 0.a1a2 · · · an12),a1, a2, · · · , an11, P02.0, 1 P1,x P ,xx =a13a2 an+ 32 + ··· + 3n + ··· ,anA P .02.aiA,A P .1,0, 1 PA 0, 1,A0, 1 P3x0 E(xn E = x|f(x) c,Pt0 E. t 0, 1 t0 < t 1,Pt0 E,Ik , k = 1, 2, · · · , 2n1IkAB: : x =n=1an3nn=112n·an2,an = 0P¯ c ,2,P¯ = c.AB1-1Ac,A P ,P¯ c,41.EmE < +.EIE I.mE mI < +.2.RpE = xi | i = 1, 2, · · · .p 2i > 0,Ii),i=1Ii,Ii E,xii=1 Ii,|Ii| = .|Ii| =mE = 0.3.EmE > 0,mEc,EE1,a, ba = inf x, b = sup x,xEx > 0E a, b.Ex = a, x E, a x b, f(x) = mEx| 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)f(x)a, bx > 0, x 0f(a) = mEa = m(E a) = 0f(b) = m(E a, b) = mE.4.c,c < mE,mE1 = c.S1, S2, · · · , Snx0 a, bf(x0) = c.mEx0 = m(a, x0 E) = c., Ei Si, i = 1, 2. · · · , n,m(E1 E2 · · · En) = mE1 + mE2 + · · · + mEn.T , S1, S2, · · · , SnnSi) =i=1 i=1T =ni=1Ei,3T Si = (j=1n1,Ej) Si =Ei, T (i=1nSi) =ni=1Ei,nm(i=1Ei) = m(T (i=1nSi) =ni=1m(T Si) =ni=1mEi.5.mE = 0,ET ,T = (E T ) (T E),mT m(E T ) + m(T E).E T E, m(E T ) mE = 0.T E T, m(T E) mT,m(E T ) + m(T E) mT.1xi2i (m E1 = c.xEf x x) f(x),E1 = E a, x0 E.m (T m (T Si).§2nmT = m(T E) + m(T E),E6.(Cantor)P0, 1n13n, ······ .1P0, 1n=12n13n,n).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 RpmB < +.Am(AB) = mA+mB m(AB).Am(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) < +,m(B A) = mB m(A B),m(A B) = mA + mB m(A B).8. Em(G E) < , m(E F ) < . > 0,GF ,F E G,mE < > 0,Ii, i = 1, 2, · · · ,i=1Ii E,i=1| Ii |< mE + .G =i=1Ii,GG E,mE mG i=1mIi =i=1| Ii |<mE + ,mG mE < ,m(G E) < .mE = EE =n=1En(mEn < ),EnGn,Gn Enm(Gn En) <G =i=1Gn, GG E,G E =n=1Gn n=1En n=1(Gn En).m(G E) n=1m(Gn En) < .EE> 0G, G E,m(G E) < .G E = G E = E (G) = E G,F = G,Fm(E F ) = m(G E) < .E9.E Rq,An, Bn,An E Bnm(BnAn) 0(n ),223,9, · · · · · ·2= 1(2n .i,n=1Bn Bi,n=1Bn E Bi E.E Ai, Bi E Bi Ai,i,mn=1Bn E m(Bi E) m(Bi Ai) = m(Bi Ai).i ,m(Bi Ai) 0,mn=1Bn E= 0.n=1Bn EBnn=1BnE =n=1Bn n=1Bn E10.A, B Rp,m(A B) + m(A B) mA + mB.GG1mA = + mB = +,G2,mA < +mG1 = mA, mG2 = mB.mB < +m(A B) m(G1 G2), m(A B) m(G1 G2).m(A B) + m(A B) m(G1 G2) + m(G1 G2) = mG1 + mG2 = mA + mB.11.E Rp. > 0,F E,m(E F ) < ,En,Fn E,m(E Fn) <1F =n=1Fn,FF E.n,m(E F ) = 0,12.E Fm(E F ) m(E Fn) <E = F (E F )µ1n.M,µ M,µ M.µ M,µ = M.cc,3G1 A, G2 B,n.1.f(x)Er,Ef > rEf = rf(x)r,Ef > r,rnEf > a =Ef > rn,Ef > rnEf > f(x)En=1Ef >r,Ef = rx z, f(x) =2 = zf(x)3; x z, f(x) =f2,E = (, ), z (, )r,Ef = r = E2.f(x), fn(x)(n = 1, 2, · · · )fn(x)-7f(x)k=1a, bn E | fn f |<1kkA| fn(x) f(x) |<1Efnx A,1x n E | fn f |< k.k,N,n > Nkx k=1n E | fn f |<1k.x k=1 n E | fn f |<1k, > 0,k0,1k0< ,x n E | fn f |<1k0N,n > Nx E | fn f |<1k0,| fn(x) f(x) |<1k0< ,lim fn(x) = f(x),nx A.3.fnEA =k=1n E | fn f |<1k.§16, lim fn(x)nlim fn(x)nEEn fn = +