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$\bf证明$  $(1)$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

\[{F_k} = \bigcap\limits_{i = k}^\infty  {\left( {E\backslash {E_i}} \right)} \]由于$E\backslash {E_i} = E\left( {\left| {{f_{{n_i}}} - f} \right| < \frac{1}{{{2^i}}}} \right)$，所以我们有\[{F_k} = E\left( {\left| {{f_{{n_i}}} - f} \right| < \frac{1}{{{2^i}}},i = k,k + 1, \cdots } \right)\]

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\[E\backslash F = \bigcap\limits_{k = 1}^\infty  {\left( {E\backslash {F_k}} \right)}  = \bigcap\limits_{k = 1}^\infty  {\bigcup\limits_{i = k}^\infty  {{E_i}} }  = \mathop {\overline {\lim } }\limits_{i \to \infty } {E_i} \subset \bigcup\limits_{i = 1}^\infty  {{E_i}} \]而\[m\left( {\bigcup\limits_{i = 1}^\infty  {{E_i}} } \right) \le \sum\limits_{i = 1}^\infty  {m\left( {{E_i}} \right)}  \le \sum\limits_{i = 1}^\infty  {\frac{1}{{{2^i}}}}  = 1\]所以我们有$m\left( {E\backslash F} \right) = 0$

$\bf注1：$由上限集与下限集的定义知，\[\bigcap\limits_{n = 1}^\infty  {{A_n}}  \subset \mathop {\underline {\lim } }\limits_{n \to \infty } {A_n} \subset \mathop {\overline {\lim } }\limits_{n \to \infty } {A_n} \subset \bigcup\limits_{n = 1}^\infty  {{A_n}} \]