实变函数与泛函分析基础第三章习题答案

? ? 1.?E?mE 0,?mE?c,?E? ?E1,?mE1= c. ? a = inf xE x,b = sup xE x,?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) = 0 f(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 n S i=1 Si) = n P i=1 m(T Si).?T = n S i=1 Ei,?T Si= ( n S j=1 Ej) Si= Ei,T ( n S i=1 Si) = n S i=1 Ei,?m( n S i=1 Ei) = m(T ( n S i=1 Si) = n P i=1 m(T Si) = n P i=1 mEi. 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? ? 1 3, ? 2 9, , ?n? 2n1 3n , .?P?0,1? P n=1 2n1 3n = 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, ,? S i=1 Ii E,? P i=1 | Ii| mE + .?G = S i=1 Ii,?G?G E,?mE mG P i=1 mIi= P i=1 | Ii| mE + ,?mG mE ,?m(G E) . ?mE = ?E?E = S n=1 En(mEn ), ?En?Gn,?Gn En?m(GnEn) 2n. ?G = S i=1 Gn,G ?G E,?G E = S n=1 Gn S n=1 En S n=1(G n En). ? m(G E) n=1 m(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, T n=1 Bn Bi,? T n=1 Bn E Bi E.?E Ai,Bi E Bi Ai,? ?i, m n=1 Bn E ! m(Bi E) m(Bi Ai) = m(Bi Ai). ?i ,?m(Bi Ai) 0,?m ? T n=1 Bn E ? = 0.? T n=1 Bn E?Bn? ? T n=1 Bn?E = T n=1 Bn ? T n=1 Bn 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) 1 n. ?F = S n=1 Fn,? F?F E.?n,