周民强《实变函数论》课后答案

1、? ? Ex 1. fj(x)Rn x : fj(x) 1 k ,(j,k = 1,2,) ! ? x : lim jfj(x) 0 ? . #$ ? x : lim jfj(x) 0 ? = k=1 ? x : lim jfj(x) 1 k ? = k=1 ? x : lim n sup jn fj(x) 1 k ? = k=1 n=1 j=n ? x : fj(x) 1 k ? Ex 2: fn(x)a,b E a,b% . x lim nEn, 6 n,k n,78fk(x) 1/2,9:E lim nEn = .; ? lim n En= a,b E. Ex 4: f : X Y,A

2、X,B Y , ABCDEFHG (i)f1(Y B) = f1(Y ) f1(B); (ii)f(X A) = f(X) f(A). #$ (i):IJKL (ii)MNOMPJK QRST f(x) = x2,X = 1,1,Y = 0,1,A = 0,1. Ex 5: UVW (x,y) : x2+ y2 1XY W (x,y) : x2+ y2 1Z L #$_ba (x,y) : x2+ y2 1 k k=1 x : f(x) 1 k g x : f(x) 1 k ? L Ex 9: E R3 u % E u vw 0 ? f(x + ) f(x ) = 0 l x0 Ec,6 07

3、8f(x0+)f(x0) = 0, ( f (x0,x0+)9:y (x0,x0+), 5 = miny(x0),(x0+)y), ? f(y +)f(y ) = 0,9:(x0,x0+) Ec, x0 Ec b Ec d 01E e Ex 14: F Rn 0, 6 S xF S(x,r) e bS(x,r)5x d r d L Ex 18: f C(R1),Fk R1 u ? | f k=1 Fk ! = k=1 f(Fk). b $,+ ? MN ? f ? T k=1 Fk ? T k=1 f(Fk)?JKL M l y0 T k=1 f(Fk),6lk,xk Fk78f(xk) =

4、y0. . xk ? O 6k0, 2 k k0 3 xk= xk0, y0= f(xk0) f ? T k=1 Fk ? . . xk 3 大学课程网w w w .u c o u r s e .n e t （提示：文档修改密码：u c o u r s e ） 第 3 页，共 19 页6/10/2009 ? O + ( ? e ), xk ? M x0,x0 T k=1 Fk, f + f - f(x0) = y0, y0 f ? T k=1 Fk ? .45f ? T k=1 Fk ? T k=1 f(Fk).18 f k=1 Fk ! = k=1 f(Fk). Ex 21: E Rn. E

5、 6= , % E 6= Rn, | E 6= . $ ( ? / Rn /e/ / 078O(x0,0) Ec, ? =sup O(x0,)Ec . = +, 6 Ec= Rn,E = g E 6= / 9/: 0,78O(x,x) Ec, S(x0,) xS(x0,) O(x,x) 5S(x0,) ?e + Heine-Borel ?$?% Pc ? xi78 S(x0,) m i=1 O(xi,xi) Ec ? 0= minxi,i = 1,2,m 6 O(x0,+ 0) Ec g P? d ? x .? (x0,y0) 0 A dB 9 : ? ay2 0= 0,by20+cy0= 0

6、,dy0= 0, (x0,y0) 0 f M (x0,y1),y16= y0, C ? P(x,y1) = ax2y2 1+ bxy21+ cxy1+ dy1 d ? x .?/M x /A d B 9: ? ay2 1 = 0,by2 1+ cy1 = 0,dy1= 0,?a = b = c = d = 0,? 34?D E R2,F?G ?H /?I?J?K4?D O ?b L Ex 25: f : R1 R1, ( G1= (x,y) : y f(x). | f C(R1)L%?M?LG1 X G2 V L b $ f C(R1), b G1 G2 ?4?N b L(x0,y0) G1,

7、 6 + G1 P?lPc G O ?d i nm?o ?d i?p?lJ M Y ;?q_ E ! F M 9:R1 M?r ?lJ ?s O?t?uc. M x R1,(x,x + 1) M C?K18/?s / /O Y c.? ? R1 M?r ?lJ ?s d c. $#7 b+ 4?v J wBorel 4 d c,9:R1 M?r ?s ? O?t?uc. Ex 27: F Rn u ?h ? ? Email: , ?Z 2007.4.9 ) 8 大学课程网w w w .u c o u r s e .n e t （提示：文档修改密码：u c o u r s e ） 第 8 页，共

8、 19 页6/10/2009 ? Lebesgue? ? Ex 1: E R1, q (0,1),(a,b),In: E (a,b) n=1 In, X n=1 m(In) (b a)q. m(E) = 0. ! #%$#5565?55A55B5C5259m(E) P n=1 m(E In) q P n=1 |In|, 34 2 m(E) qm(E) 34 mE = 0, D mE = 0. Ex 2: A1,A2 Rn,A1 A2,A1 EFGH m(A1) = m(A2) , A2 E FGHI ! #%$#J A1+=K9MLCaratheodory Z6_ mT m(T A2) + m

9、(T Ac 2) ( mT = m(T A2) + m(T Ac 2) ab A2+=K I c . ed5f 65g5h5+5i5j5RCaratheodory A1=qrsA26=C I 1 大学课程网w w w .u c o u r s e .n e t （提示：文档修改密码：u c o u r s e ） 第 9 页，共 19 页6/10/2009 Ex 3: A,B Rn EFGH 9 m(A B) + m(A B) = m(A) + m(B). t uwv m(A),m(B) N 2yxyz U , Xy| y yyyy I 3y45y m(A) ,m(B) . QRA,B6=CC

10、aratheodory LF +K9F E, m(E F) /2, O G2= Fc, G2+/KG2 Ec, m(G1 G2) = m(G1 F) m(G1 E) + m(E F) 0,LG+/K9ME G,F+K9M F E, b m(E F) . m5nk= 1/k,55L5/505155AGk5505155AFk r59Fk E Gk,m(GkFk) 1/k, G = T k=1 Gk,F = S k=1 Fk, m(G F) 1/k,-.k9MDm(G F) = 0,mn s F E G, ( E= I Ex II(6): A,B Rn,A B EFGH 9M m(A B) , m(

11、A B) = m(A) + m(B), A,B FGI ! # T B 6 =H, q H B,H=9 $#J A B=9 34 H A B, O H (A B) TU B 6 = I QRH6=C9 Y Caratheodory %1%4%6,%9%1 4 / I, ? I%5%2 f(x) = ? 1,x = 0,A%B C%3 S(x) = supf(x) : I = I(x)9%0%1%4%5%2%6 Ex 3: f(x) $ a,b) %D $%E%F % ,!%G%H E%F % f0 +(x) a,b) % 6 I%J * %=%K%L%5%2 f0 +(x) = lim h0+

12、 f(x + h) f(x) h = lim n nf(x + 1/n) f(x) Ex 6: fk(x) E Rn %M%N%O% m(E) 0 lim j m(x E : sup kj|f k(x)| ) = 0. YZ *,%_%a 6 ?%b%c 0,dfe E(sup kj|f k(x)| ) k=j E(|fk(x)| /2) d Lebesgueg%h%i + m(E) , lim k fk(x) = 0,a.e. x E 3 lim j m( k=j E(|fk(x)| /2) = 0. %j%k%a 6ldfe E(fk(x)9%m%n%e0) = i=1 j=1 k=j E

13、(|fk(x)| 1/i) o lim j m(x E : sup kj|f k(x)| ) = 0. 8 i N, lim j m( k=j E(|fk(x)| 1/i) = 0. 1 大学课程网w w w .u c o u r s e .n e t （提示：文档修改密码：u c o u r s e ） 第 12 页，共 19 页6/10/2009 p m(E) ,df4%q %r%s%a%t%u m( j=1 k=j E(|fk(x)| 1/i) = lim j m( k=j E(|fk(x)| 1/i) = 0. v df4%q %w 1%x a%u m(E(fk(x)9%m%n%e0)

14、 = 0. Ex 7: f(x),f1(x),f2(x),fk(x), a,b %y%z% %| % , lim k fk(x) = 0,a.e. x a,b, !%G%H%D $ En a,b (n = 1,2,),% m a,b n=1 En ! = 0 fk(x)$%En % f(x). I%J * %= = 1/n,% Egoroff ;%h%6 Ex 8: fk(x) $ E % f(x),gk(x) $ E % g(x), !%G H fk(x) + gk(x) $ E % f(x) + g(x). I%J * %= x E : |fk(x)+gk(x)(f(x)+g(x)| x

15、E : |fk(x)f(x)| /2x E : |gk(x)g(x)| /2 Ex 9: f(x),f1(x),f2(x),fk(x), E yz | m(E) .%$fk(x) W %O fki(x)% D $ y%z% f(x) %O fkij(x), ! G%H fk(x) $ E % f(x). *,%+ e Riesz;%h :% * f(x),f1(x),f2(x),fk(x), E m(E) 0,% lim k m(E(|fk(x) f(x)| 0) 6= 0. 8%0 0,k N,nk % % m(E(|fnk f(x)|) 0) 0.() C3 fnk91 u mnef(x)

16、 78 o fnki mnef(x), 8dLebesgueg%hfnki%4%q%m%n%ef(x), lim i m(E(|fnki(x) f(x)| 0) = 0. C% () % 6 2 大学课程网w w w .u c o u r s e .n e t （提示：文档修改密码：u c o u r s e ） 第 13 页，共 19 页6/10/2009 Ex 13: fk(x),gk(x) $ E % 0, !%G%H fk(x) gk(x) $ E % % 0. I%J * 0, D $ E % F,m(E F) ,%f(x) $ F % ,!%G%H f(x) E % 6 *,% %

17、0Lusin;%h % ;%h%6 I%J * %= = 1/n,8%Fn E,m(E Fn) 0, R lim j m k=j x : |fk(x) f(x)| = 0. !%G%H fk(x) $ E % f(x). I%J *,% % Egoroff ;%h YZ 6 (% *,% 2 % Email: ,%2007.5.16 22:13) 3 大学课程网w w w .u c o u r s e .n e t （提示：文档修改密码：u c o u r s e ） 第 14 页，共 19 页6/10/2009 ? Lebesgue? ? Ex 1: f(x) E Rn ! # R Ef(x

18、)dx = 0, $%:66= Ek E,m(EEk) ?A lim k Z Ek f(x)dx = $%&f(x) = E B9 )(+*-C E1= E1, Ek= Ek k1 S j=1 Ej(k = 2,3,),.1D Ei Ej= (i 6= j), k S j=1 Ej= Ek, (k = 1,2,). E E = S j=1 Ej,.Dm(E E) m(E k S j=1 Ej) = m(E Ek) fk+1(x)opMqr /0st S k=1 Ek(pMq r )uvVwra fk(x)oJKxyNOz - HILevi| 9 Ex 7:i = Rn f(x), de 0,

19、? Z Rn h(x) g(x)dx 0, Z 0,t f(x)dx = Z 0,tE f(x)dx+ Z 0,tE f(x)dx Z 0,tE f(x)dx+ Z t,1E f(x)dx = Z E f(x)dx. Ex 11.1:$%& 1 () Z (0,) x1 ex 1dx = X n=1 n( 1). * () = R+ 0 ett1dt. (*- 1 ex 1 = ex 1 ex = ex 1 1 ex = X n=1 enx. HIJKLMNOVRS 1 () Z (0,) x1 ex 1dx = 1 () X n=1 Z (0,) x1enxdx = 1 () X n=1 Z

20、 (0,) nx1exdx = X n=1 n. Ex 13: f(x) L(R1),p 0,$%& lim n npf(nx) = 0,a.e.x R1. (*-, F(x) = P n=1 |npf(nx)|,HIJKLMNOVRSD 9 Z R1 F(x)dx = Z R1 X n=1 |npf(nx)|dx = X n=1 Z R1 |npf(nx)|dx = X n=1 np Z R1 |f(nx)|dx = X n=1 n1p Z R1 |f(x)|dx + 34F(x) R1aD 34V lim n npf(nx) = 0,a.e.x R1. Ex 16: f L(0,1), l

21、im n Z 0,1 nln ? 1 + |f(x)|2 n2 ? dx = 0. (*- NOf(x) = ln(1 + x2) x,f0(x) = 2x 1+x2 1 = (x1)2 1+x2 0,x 0,/0 ln(1 + x2) x(x 0) nln ? 1 + |f(x)|2 n2 ? |f(x)| 2 大学课程网w w w .u c o u r s e .n e t （提示：文档修改密码：u c o u r s e ） 第 16 页，共 19 页6/10/2009 Y Lebesgue lim n Z 0,1 nln ? 1 + |f(x)|2 n2 ? dx = Z 0,1 li

22、m n nln ? 1 + |f(x)|2 n2 ? dx. 8 N8Og(x) = ln(1+x2)x2,g0(x) = 2x 1+x2 2x = 2x( 1 1+x21) 0(x 0), /80ln(1+x2) x2 nln ? 1 + |f(x)|2 n2 ? 0 (x E)$%& lim k Z E f(x)1/kdx = m(E). 3 大学课程网w w w .u c o u r s e .n e t （提示：文档修改密码：u c o u r s e ） 第 17 页，共 19 页6/10/2009 (*, E0= Ef = +,E1= E1 f +,E2= Ef 1,/80 E = E0E1E2, Y Q f L(E) /0f EaD - _ m(E0) = 0,/0 Z E f(x)1/kdx = Z E1 f(x)1/kdx + Z E2 f(x)1/kdx, E1 a Y Q f(x)1/k f(x), Y Lebesgue lim k Z E1 f(x)1/kdx = Z E1 lim kf(x) 1/kdx = Z E1 1dx = m(E1). E2 a Y Q 0 f(x) 1 /0f(x)1/k|Q1, Y Levi| lim k Z E2 f(x)1/kdx = Z E2 lim kf(x) 1/kdx = Z E2 1dx = m(E2)