实变函数与泛函分析基础第三版(程其襄)+课后答案.pdf

?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 (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)=(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 Ai,?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.?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 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.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=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,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,?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,?P0 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?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(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?=(0.1,0.2),?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,?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)=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=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 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 ),?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?,?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 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|=,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=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),?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)fn+1(x)?n=1,2,?fn(x)?f(x).?fn(x)f(x),?fni fn,?fni(x)?E?a.e.?f(x).?E0?fni(x)?f(x)?En=Efn 0,E|f gn|(Sn=1En)E|f fn|.?mE|f gn|m(n=1En)+mE|f fn|=mE|f fn|.3?fn(x)f(x),?0 limmE|f gn|limmE|f fn|=0?gn(x)f(x).12.?mE 0,?mE|fn f|0?0 0,?fnk,?mE|fnk f|0 0 0.(1)?fnk?f(x)?fnkj?E?a.e.?f,?mE +,?E?fnkj f(x),?(1)?13.?mE ,?fn(x)?gn(x),n=1,2,?f(x)?g(x),?(1)fn(x)gn(x)f(x)g(x);(2)fn(x)+gn(x)f(x)+g(x);(3)minfn(x),gn(x)minf(x),g(x);maxfn(x),gn(x)maxf(x),g(x).?(1)?f(x)a.e.?mE|f|=0.?Tn=0E|f|n=E|f|=,?E|f|n E|f|n+1?E|f|1 E,mE|f|1 mE 0,0,?k,mE|f|k 5?mE|g|k N?mE|gn g|0 5,mE|fn f|0 5?E|gn|k+1 E|g|k E|gn g|1 E|g|k E|gn g|0.mE|gn|k+1 mE|g|k+mE|gn g|0 5+5=25.?Eh|gnfn gnf|2i E|gn|k+1E?|fn f|2(k+1)?E|gn|k+1E|fnf|0.?mEh|gnfn gnf|2i mE|gn|k+1+mE|fn f|0 25+5=35.?Eh|fgn fg|2i E|f|k+1E?|gn g|2(k+1)?E|f|kE|gng|0.4?mEh|fgn fg|2i mE|f|k+mE|gn g|0 5+5=25.?E|gnfn gf|Eh|gnfn gnf|2i Eh|fgn fg|2i,?mE|gnfn gf|mEh|gnfn gnf|2i+mEh|fgn fg|2i 0,0,?N,?n N?mE|gnfn gf|,?gnfn gf.(2)E|(fn+gn)(f+g)|Eh|fn f|2i Eh|gn g|2i?mE|(fn+gn)(f+g)|mEh|fn f|2i+mEh|gn g|2i,limnmE|(fn+gn)(f+g)|limnmEh|fn f|2i+limnmEh|gn g|2i.?fn+gn f+g.(3)?fn f,?|fn|f|.?E|fn f|E|fn|f|.?limnmE|fn|f|limnmE|fn f|=0,?|fn|f|.?fn f,?a 6=0,afn af.?E|afn af|=E?|fn f|a|?,?limnmE|afn af|=limnmE?|fn f|a|?=0.minfn(x),gn(x)=fn(x)+gn(x)|fn(x)gn(x)|2.?(2),fn(x)+gn(x)f+g,fn(x)gn(x)f g.?|fn(x)gn(x)|f(x)g(x)|,?(2),fn(x)+gn(x)|fn(x)gn(x)|f(x)+g(x)|f(x)g(x)|.?fn(x)+gn(x)|fn(x)gn(x)|2f(x)+g(x)|f(x)g(x)|2.5?minfn(x),gn(x)minf(x),g(x).?maxfn(x),gn(x)=fn(x)+gn(x)+|fn(x)gn(x)|2,?maxfn(x),gn(x)maxf(x),g(x).6?1.?Lebesgue?Darboux?Darboux?f(x)?E?E?D:E1,E2,En,?max1inmEi 0?S(D,f)ZEf(x)dx,S(D,f)ZEf(x)dx.?0,1?f(x)=1,x?0,1?,0,x?0,1?.?n,?0,1?Dn=Eni,?Eni=hi1n,in?,i=1,2,n 1,Enn=hn1n,1i.?max1inmEni=1n 0(n ).?S(D,f)=nXi=1supxEnimEni=nXi=11 1n=1.?f(x)?0,1?Z0,1f(x)dx=Z0,1f(x)dx=0.?Darboux?2.?Cantor?P0?f(x)=0,?P0?13n?n(n=1,2,),?f(x)?f(x)?En?P0?13n?mEn=2n13n,?Z0,1f(x)dx=Xn=1ZEnf(x)dx=Xn=1nmEn=Xn=1n 2n13n=3.?f(x)?3.3.?f(x)?E?en=E|f|n,?limnn men=0.?f(x)?E?E?a.e.?mE|f|=0.?en en+1,me1 mE 0,?0,?e E?me?Ze|f(x)|dx 0,?N,?n N?men,?n menZen|f(x)|dx .?limnn men=0.4.?mE ,f(x)?E?En=En 1 f n,?f(x)?E?P|n|mEn.?f(x)?E?|f(x)|?E?n 1?En?n 1|f(x)|=f(x)ZE|f(x)|dx=Xn=1ZEn|f|dx+Xn=0ZEn|f|dx Xn=1(n 1)mEn+Xn=0|n|mEn=Xn=1|n|mEn+Xn=0|n|mEnXn=1mEn=X|n|mEnXn=1mEn,?En?E?Xn=1mEn=m(Xn=1En)mE ,?P|n|mEn.?P|n|mEn,?ZE|f(x)|dx=Xn=1ZEn|f|dx+Xn=0ZEn|f|dx Xn=1nmEn+Xn=0|n 1|mEn=Xn=1|n|mEn+Xn=0|n|mEn+Xn=0mEnX|n|mEn+mE .?|f(x)|?f+(x)?f(x)?f(x)=f+(x)f(x)?5.?f(x)?a,b?R?(?),?f(x)?a,b?L?|f(x)|?a,b?R?(?),?Za,bf(x)dx=(R)Zbaf(x)dx.?f(x)?a,b?R?(?),?0 0,?fn?mE|fn|ZE|fn|fn(x)dx ZEfndx.?mE|fn|1ZEfn(x)dx,limmE|fn|=limn1ZEfn(x)dx=0.?fn(x)0.7.?mE ,fn?a.e.?limnZE|fn(x)|1+|fn(x)|dx=0?fn(x)0.?fn 0?E|fn|1+|fn|#E|fn|,limmE|fn|1+|fn|#limmE|fn|=0.?|fn|1+|fn|0,?0|fn|1+|fn|1(n=1,2,),mE 1?E|fn|1+|fn|1+#=E|fn|,?limnmE|fn|=limnmEh|fn|1+|fn|1+i=0,?fn 0.8.?f(x)=sin1xx,0 x 1,?f(x)?0,1?L?3?1?Z10|f(x)|dx Z101x|sin1x|dx Z101x|sin1x|dx=Z|sinyy|dy=,?1?f(x)L?0,?(x)?Ef?mEf ZEff(x)dx=ZEf(x)(x)dx=0,?mEf =0.?mEf =0,?mE|f|=0.?Ef,0=n=1E|f|1n#.?mEf,0 Pn=1mEh|f|1ni=0,?f(x)=0a.e.?E.11.?limnZ(0,)dt?1+tn?nt1n=1.?t (0,1)?1?1+tn?nt1n1t1n1t(n 2);?t 1,)?n 2?1?1+tn?nt1n=1?1+t+n12nt2+?t1n2nt2(n 1)4t2.4?F(t)=1t,t (0,1),4t2,t 1,),?Z(0,)F(x)dx=Z10dtt+Z14dtt2=6,?F(x)?(0,)?limnZ(0,)dt?1+tn?nt1n=Z(0,)limn1?1+tn?nt1ndt=Z(0,)dtet=1.12.?11+x=(1 x)+(x2 x3)+,0 x 1,?ln2=1 12+1314+.?0,1?xn xn+1 0,?5?3?Z1011+xdx=Xn=0Z10(x2n x2n+1)dx=Xn=0 12n+112n+2!=1 12+1314+,?R1011+xdx=ln2,?ln2=1 12+1314+.13.?f(x,t)?|t t0|?x?a,b?K,?tf(x,t)?K,a x b,|t t0|,?ddtZbaf(x,t)dt=Zbaft(x,t)dx.?hn,?limnhn=0?hn,0.?ddtZbaf(x,t)dt=limn1hnZbaf(x,t+hn)f(x,t)dx,?f(x,t+hn)f(x,t)hn?=?ft(x,t+hn)hnhn?=|ft(x,t+hn)|K,?0 1,a x b,t0 t+hn 1).5?xp1 xln1x=(Xn=0 xn)xpln1x=Xn=0 xn+pln1x,?x (0,1)?xn+pln1x 0,?Z10 xp1 xln1xdx=Xn=0Z10 xn+pln xdx=Xn=01(n+p+1)2=Xn=11(n+p)2.15.?fn?E?limnfn(x)=f(x)a.e.?E,?ZE|fn(x)|dx 0,?a ,b+?(x),?Zb+a|f(x)(x)|dx 0(?),?x,x a,b+,?|x x|,?|(x)(x)|3(b a).?0 t?x a,b,?|(x+t)(t)|3(b a),?Zba|f(x+t)f(x)|dx Zba|f(x)(x)|dx+Zba|f(x+t)(x+t)|dx+Zba|(x+t)(x)|dx3+3+3(b a)(b a)=.?limt0Zba|f(x+t)f(x)|dx=0.617.?f(x),fn(x)(n=1,2,)?E?limnfn(x)=f(x)a.e.?E,?limnZE|fn(x)|dx=ZE|f(x)|dx,?e E,?limnZe|fn(x)|dx=Ze|f(x)|dx.?Ze|f(x)|dx=Zelimn|fn(x)|dx limnZe|fn(x)|dx.?Ze|f(x)|dx limnZe|fn(x)|dx.?Ze|f(x)|dx Ze|f(x)|dx,?limiZEe|fni(x)|dx=limiZE|fni(x)|dx limiZe|fni(x)|dx 0?xn(0,),?limnxn=,?|f(xn)|0.?f(x)?(0,)?0,?x,x(0,),?|x x|?|f(x)f(x)|02.?x (xn,xn+)?|f(x)f(xn)|02.?|f(x)|f(xn)|0202.?xn,?xni,?xni 2,i=1,2,.?Eni=(xni,xni+),?Eni?Z(0,)|f(x)|dx ZSi=1Eni|f(x)|dx=Xi=1ZEni|f(x)|dx 02Xi=1mEni=Xi=10=,7?|f(x)|?(0,)?limxf(x)=0.19.?f(x)?Rp?g(y)?Rq?f(x)g(y)?Rp Rq?f(x)?Rp?Rp?Rp Rq?g(y)?Rp Rq?f(x)g(y)?Rp Rq?f(x),g(y)?6?4,f(x)g(y)?Rp Rq?ZZRpRqf(x)g(x)dxdy=ZRpdxZRqf(x)g(y)dy=ZRpf(x)dx ZRqg(y)dy 0?Z+0F(y)dy=ZEf(x)g(x)dx.?f(x),g(x)?E?y 0,Ey?F(y)=REyf(x)dx?(?+)?F(y)0.?Z+0F(y)dy=Z+0ZEyf(x)dxdy=Z0 ZEEy(x)f(x)dx!dy=ZEf(x)Zg(x)0dy!dx=ZEf(x)g(x)dx.8?1.?(a,b)?f(x),g(x)?(a,b)?E?(a,b)?E?f(x)=g(x).?M?f(x)?N?g(x)?x0 M,?f(x0+0)=f(x0)=f(x0 0).?E?(a,b)?xn E,?xn x0?limnxn=x0,?f(x0 0)=limnf(xn)=limng(xn)=g(x0 0).?f(x0+0)=g(x0+0),?x0?g(x)?x0 E,?M N.?N M,?M=N,?f(x)?g(x)?2.?fn?a,b?fn(x)f(x)(n ),?f(x)?bWa(fn)K,(n=1,2,),?f(x)?a,b?T:a=x0 x1 xm=b,mXi=1|fn(xi)fn(xi1)|b_a(fn)0)?,?0,1?x0=0,xk=h(n 1)k)+2i1,k=1,2,n 1,xn=1.?nXk=1|f(xk)f(xk1)|=nXk=1?xksin1xk xk1sin1xk1?n1Xk=1(?1(n 1)k)+2?+?1(n k)+2?)n1Xk=1?1(n 1)k)+2+1(n k)+2?n1Xk=1?1(n k)+1(n k+1)?n1Xk=12(n k+1)=2nXk=21k.1?nPk=21k=,?1_0(f)=supXk=1|f(xk)f(xk1)|=,?f(x)?0,1?0.?f(x)=x1sin1x x1cos1x,?|f(x)|x1+x1.?0?1 1,1 1.?|f(x)|?0,1?R?L?f(x)?0,1?(L)Zx0f(t)dt=lim0(L)Zxf(t)dt=lim0(R)Zxf(t)dt=lim0?tsin1t?x=xsin1x0=f(x)f(0).?f(x)=f(0)+Zx0f(t)dt.?f(x)?0,1?4.?f(x)?a,b?f(x)0a.e.?a,b,?f(x)?x1,x2 a,b,x2 x1.?f(x)?a,b?f(x1)=f(a)+Zx1af(t)dt;f(x2)=f(a)+Zx2af(t)dt.?f(x2)f(x1)=Zx2x1f(t)dt.?f(x)0a.e.?a,b,?Rx2x1f(t)dt 0,?f(x2)f(x1),?f(x)?5.?f(x)?a,b?M 0,?0?b_a+(f)M,?f(x)?a,b?x (a,b),?|f(x)f(b)|b_x(f)M,?|f(x)|M+|f(b)|,?a,b?T,T:a=x0 x1 x2 xn=b,2?T?v=nXi=1|f(xi)f(xi1)|=|f(x1)f(a)|+nXi=2|f(xi)f(xi1)|f(x1)|+|f(a)|+b_x1(f)2M+|f(b)|+|f(a)|,?b_a(f)2M+|f(b)|+|f(a)|,?f(x)?a,b?6.?fn?a,b?f(x)=Pn=1fn(x)?a,b?f(x)?a,b?n,?fn(x)?a,b?fn(x)=fn(a)+Zxafn(t)dt.?fn(x)?fn(x)0,a.e.?a,b.?f(x)?a,b?f(x)?a,b?L?f(x)=Xn=1fn(x)=Xn=1fn(a)+Xn=1Zxafn(t)dt.?Xn=1Zxafn(t)dt=ZxaXn=1fn(t)dt.?Pn=1fn(x)?Xn=1Zbafn(t)dt Xn=1fn(b)fn(a)x,?|f(x)f(x)|=?Zxxg(t)dt?K|x x|,?f(x)?a,b?Lipschitz?(1)?(2).?f(x)?a,b?Lipschitz?f(x)?a,b?f(x)=f(a)+Zxaf(t)dt.?x,y a,b,?Zxyf(t)dt?=|f(x)f(y)|K|x y|,?K?Lipschitz?|f(x)|K,a.e.?a,b.?g(x)=f(x),?|f(x)|K,K,?|f(x)|K,?g(x)?a,b?f(x)=f(0)+Zx0g(t)dt.8.?”?”?S?”?”?f(x),(x)?a,b?a,b?T:a=x0 x1 xn=b,?Mi?mi?f(x)?xi1,xi?i=1,2,n,?S(T,f,)=nXi=1Mi(xi)(xi1),s(T,f,)=nXi=1mi(xi)(xi1).?Zbaf(x)d(x)=infTS(T,f,),Zbaf(x)d(x)=supTs(T,f,),Rbaf(x)d(x)=Rbaf(x)d(x),?f(x)?(x)S?”?”?”?”?f(x)=0,x?1,12?12,1?,1,x?12,12?,4(x)=0,x?1,12?12,1?,1,x?12,12?.?1,1?T:1=x0 xi1 12 xi xj112 xj 12,j12,0,i 12,j12?i12,1,i 12,j12,?f(x)?1,1?(x)?S?”?”?f(x)?S?T:1=x0 x1 xn=1,12,12?(xi1,xi)?(x0)=(x1)=(xn)=0.?S(T,f,)=0,s(T,f,)=0;12,12?12 xi1,xi,12 xj1,xj,?S(T,f,)=nXk=1Mk(xk)(xk1)=1 1=0,s(T,f,)=nXk=1mk(xk)(xk1)=0,?R11f(x)d(x)=R11f(x)d(x),?f(x)?S?8.?(x)?(,)?(?)?L S?E R1,mE=inf(Xi=1m(ai,bi),i=1(ai,bi)E).?(bi)(bi 0),(ai)(ai+0),?|Ii|=(bi)(ai)(bi 0)(ai+0)=m(ai,bi),?i=1Ii=i=1(ai,bi)E,Xi=1m(ai,bi)Xi=1|Ii|,?inf(Xi=1m(ai,bi)inf(Xi=1|Ii|)=mE,5?Si=1(ai,bi)E,?Xi=1m(ai,bi)m i=1(ai,bi)!mE,?inf(Xi=1m(ai,bi),i=1(ai,bi)E)mE.?(x)?(x).?x,?(x0)=(x0),(x+0)=(x+0),?m(ai,bi)=m(ai,bi).?E R1,?mE=inf(Xi=1m(ai,bi),i=1(ai,bi)E)=inf(Xi=1(bi 0),(ai+0),i=1(ai,bi)E)=inf(Xi=1(bi 0),(ai+0),i=1(ai,bi)E)=inf(Xi=1m(ai,bi),i=1(ai,bi)E)=mE.?L S?L S?L S?L S?L S?(x)?L S?6