3235656

$\bf证明$  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

E1=?m=1∞ ?n=m∞ E1(|fn?f|<ε)

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(E1)≤lim???n→∞m(E1(|fn?f|<ε))

由$m\left( E \right) < \infty $，我们得到

limˉˉˉˉˉn→∞m(E1(|fn?f|≥ε))=m(E1)?lim???n→∞m(E1(|fn?f|<ε))≤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(lim???n→∞En)≤lim???n→∞m(En)

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