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295565656

$\bf证明$  $(1)$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}}}

m(E(|fn?f|12k))<12k
<script id="MathJax-Element-1" type="math/tex; mode=display">m\left( {E\left( {\left| {{f_n} - f} \right| \ge \frac{1}{{{2^k}}}} \right)} \right) < \frac{1}{{{2^k}}}</script>

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)}

Fk=?i=k (E?Ei)
<script id="MathJax-Element-2" type="math/tex; mode=display">{F_k} = \bigcap\limits_{i = k}^\infty {\left( {E\backslash {E_i}} \right)} </script>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)
Fk=E(fni?f<12i,i=k,k+1,?)
<script id="MathJax-Element-3" type="math/tex; mode=display">{F_k} = E\left( {\left| {{f_{{n_i}}} - f} \right| < \frac{1}{{{2^i}}},i = k,k + 1, \cdots } \right)</script>

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}}

E?F=?k=1 (E?Fk) =?k=1 ?i=k Ei =limˉˉˉˉˉiEi??i=1 Ei
<script id="MathJax-Element-4" type="math/tex; mode=display">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}} </script>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
<script id="MathJax-Element-5" type="math/tex; mode=display">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</script>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}}

<script id="MathJax-Element-6" type="math/tex; mode=display">\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}} </script>