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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)
<script id="MathJax-Element-1" type="math/tex; mode=display">{E_1} = \bigcup\limits_{m = 1}^\infty {\bigcap\limits_{n = m}^\infty {{E_1}\left( {\left| {{f_n} - f} \right| < \varepsilon } \right)} } </script>

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???nm(E1(|fn?f|<ε))
<script id="MathJax-Element-2" type="math/tex; mode=display">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)</script>

由$m\left( E \right) < \infty $,我们得到

limˉˉˉˉˉnm(E1(|fn?f|ε))=m(E1)?lim???nm(E1(|fn?f|<ε))0
<script id="MathJax-Element-3" type="math/tex; mode=display">\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</script>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???nEn)lim???nm(En)
<script id="MathJax-Element-4" type="math/tex; mode=display">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)</script>