Let {fn} be a sequence of real-valued measurable functions d

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Let {fn} be a sequence of real-valued measurable functions de\x0cned on [0,1].Show that there exists a sequence of positive real numbers {an} such that anfn->0 a.e.

给定n>0,fn^(-1)[-N,N]对所有N=1,2,.的并 = [0,1]
存在 Nn>n 使得 An = fn^(-1)[-Nn ,Nn] 的测度 > 1-1/(2^n)
取 an = 1/Nn^2,于是对任何 x 属于 fn^(-1)[-Nn ,Nn],anfn(x)= 1 - 1/(2^n)-1/(2^(n+1))-1/(2^(n+2))-...= 1- 2/2^n
并且对一切 x属于En,m>=n,amfm(x) 0
令E=En 对n=1,2,...求并.
则任给x属于E,anfn(x) ---->0
同时 E的测度 > En的测度 > 1- 2/2^n -----> 1.
即 E的测度=1.所以结论成立.
再问：这里应该是anfn(x) En的测度 > 1- 2/2^n -----> 1." 这只说明E的测度大于1啊
再答：对 anfn(x)= 1 - 1/(2^n)-1/(2^(n+1))-1/(2^(n+2))-...怎么得出的？
再答： En = An 交A(n+1) 交 A(n+2) 交....， n=1,2,... 则 En的测度 >= 1- （An在[0,1]中补集的测度 + A(n+1)在[0,1]中补集的测度 + A(n+2)在[0,1]中补集的测度+。。。） =1 - 1/(2^n)-1/(2^(n+1))-1/(2^(n+2))-...= 1- 2/2^n
再问：这里：存在 Nn>N (是大N,不是小n吧？)使得 An = fn^(-1)[-Nn , Nn] 的测度 > 1-1/(2^n) 是大N,不是小n吧？还有，为什么要证明 En的测度为1？
再答：我被你问糊涂啦。 Nn>N 是什么意思？我上面好像没有证明En的测度=1吧。

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