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$\bf证明$  由于$\left\{ {{f_n}\left( x \right)} \right\}$几乎处处收敛于$f(x)$，则存在零测集$E_0$，使得$\lim \limits_{n \to \infty } {f_n}\left( x \right) = f\left( x \right)$在$E_1=E\backslash {E_0}$上成立，

于是对任给的$\varepsilon  > 0$，我们有\[{E_1} = \bigcup\limits_{m = 1}^\infty  {\bigcap\limits_{n = m}^\infty  {{E_1}\left( {\left| {{f_n} - f} \right| < \varepsilon } \right)} } \]

即${E_1} = \mathop {\underline {\lim } }\limits_{n \to \infty } {E_1}\left( {\left| {{f_n} - f} \right| < \varepsilon } \right)$，从而由测度的性质知\[m\left( {{E_1}} \right) \le \mathop {\underline {\lim } }\limits_{n \to \infty } m\left( {{E_1}\left( {\left| {{f_n} - f} \right| < \varepsilon } \right)} \right)\]

由$m\left( E \right) < \infty $，我们得到\[\mathop {\overline {\lim } }\limits_{n \to \infty } m\left( {{E_1}\left( {\left| {{f_n} - f} \right| \ge \varepsilon } \right)} \right) = m\left( {{E_1}} \right) - \mathop {\underline {\lim } }\limits_{n \to \infty } m\left( {{E_1}\left( {\left| {{f_n} - f} \right| < \varepsilon } \right)} \right) \le 0\]所以对任给的$\varepsilon  > 0$，我们有$\lim \limits_{n \to \infty } m\left( {{E_1}\left( {\left| {{f_n} - f} \right| \ge \varepsilon } \right)} \right) = 0$

$\bf注1：$设$\left\{ {{E_n}} \right\}$是一列可测集，记$\mathop {\underline {\lim } }\limits_{n \to \infty } {E_n} = \bigcup\limits_{n = 1}^\infty  {\bigcap\limits_{k = n}^\infty  {{E_k}} } $，则\[m\left( {\mathop {\underline {\lim } }\limits_{n \to \infty } {E_n}} \right) \le \mathop {\underline {\lim } }\limits_{n \to \infty } m\left( {{E_n}} \right)\]