$\bf证明$ $(1)$由${f_n}$依测度收敛于$f(x)$知，对任何自然数$k$，存在自然数${n_k}\left( { > {n_{k - 1}}} \right)$，使得当$n \ge {n_k}$时，有m\left( {E\left( {\left| {{f_n} - f} \right| \ge \frac{1}{{{2^k}}}} \right)} \right) < \frac{1}{{{2^k}}}
记${E_k} = E\left( {\left| {{f_{{n_k}}} - f} \right| \ge \frac{1}{{{2^k}}}} \right)$，则$m\left( {{E_k}} \right) < \frac{1}{{{2^k}}}$，令
{F_k} = \bigcap\limits_{i = k}^\infty {\left( {E\backslash {E_i}} \right)}
即函数列${f_{{n_i}}}\left( x \right)$在${F_k}$上一致收敛于$f(x)$，于是${f_{{n_i}}}\left( x \right)$在$F = \bigcup\limits_{k = 1}^\infty {{F_k}} $上处处收敛于$f(x)$
$(2)$下面我们只需证明$m\left( {E\backslash F} \right) = 0$即可，由于E\backslash F = \bigcap\limits_{k = 1}^\infty {\left( {E\backslash {F_k}} \right)} = \bigcap\limits_{k = 1}^\infty {\bigcup\limits_{i = k}^\infty {{E_i}} } = \mathop {\overline {\lim } }\limits_{i \to \infty } {E_i} \subset \bigcup\limits_{i = 1}^\infty {{E_i}}
$\bf注1：$由上限集与下限集的定义知，\bigcap\limits_{n = 1}^\infty {{A_n}} \subset \mathop {\underline {\lim } }\limits_{n \to \infty } {A_n} \subset \mathop {\overline {\lim } }\limits_{n \to \infty } {A_n} \subset \bigcup\limits_{n = 1}^\infty {{A_n}}
原文地址：http://www.cnblogs.com/ly758241/p/3764686.html
