$\bf证明$ 由于$m\left( {E\left( {{f_n} \nrightarrow f} \right)} \right) = 0$，则我们不妨设$\left\{ {{f_n}\left( x \right)} \right\}$处处收敛于$f(x)$，此时E = \bigcup\limits_{m = 1}^\infty {\bigcap\limits_{n = m}^\infty {E\left( {\left| {{f_n} - f} \right| < \frac{1}{k}} \right)} } ,k \in {N_ + }
令F = \bigcap\limits_{k = 1}^\infty {{B_{{n_k},k}}} = E\left( {\left| {{f_n} - f} \right| < \frac{1}{k},n \ge {n_k}} \right)
则我们有m\left( {E\backslash F} \right) = m\left( {\bigcup\limits_{k = 1}^\infty {\left( {E\backslash {B_{{n_k},k}}} \right)} } \right) \le \sum\limits_{k = 1}^\infty {m\left( {E\backslash {B_{{n_k},k}}} \right)} < \delta
以及对任给的$\varepsilon > 0$，存在${k_0} > \frac{1}{\varepsilon }$，使得当$n \ge {n_{{k_0}}}$时，对任意的$x \in F$，有\left| {{f_n}\left( x \right) - f\left( x \right)} \right| < \frac{1}{{{k_0}}} < \varepsilon
所以$\left\{ {{f_n}\left( x \right)} \right\}$在$F$上一致收敛于$f(x)$
原文地址：http://www.cnblogs.com/ly428571/p/3770644.html
