首页 > 代码库 > 2626

2626

2024-07-07 04:47:23   225人阅读

$\bf证明$  由于$m\left( {E\left( {{f_n} \nrightarrow f} \right)} \right) = 0$，则我们不妨设$\left\{ {{f_n}\left( x \right)} \right\}$处处收敛于$f(x)$，此时E = \bigcup\limits_{m = 1}^\infty  {\bigcap\limits_{n = m}^\infty  {E\left( {\left| {{f_n} - f} \right| < \frac{1}{k}} \right)} } ,k \in {N_ + }

<script id="MathJax-Element-1" type="math/tex; mode=display">E = \bigcup\limits_{m = 1}^\infty {\bigcap\limits_{n = m}^\infty {E\left( {\left| {{f_n} - f} \right| < \frac{1}{k}} \right)} } ,k \in {N_ + }</script>记{B_{m,k}} = \bigcap\limits_{n = m}^\infty  {{E_{n,k}}}  = E\left( {\left| {{f_n} - f} \right| < \frac{1}{k},n \ge m} \right)
<script id="MathJax-Element-2" type="math/tex; mode=display">{B_{m,k}} = \bigcap\limits_{n = m}^\infty {{E_{n,k}}} = E\left( {\left| {{f_n} - f} \right| < \frac{1}{k},n \ge m} \right)</script>其中${E_{n,k}} = E\left( {\left| {{f_n} - f} \right| < \frac{1}{k}} \right)$，则对于固定的$k$，${B_{m,k}}$是单调递增的集合列，并且$E = \bigcup\limits_{m = 1}^\infty  {{B_{m,k}}} $，所以我们有$m\left( E \right) = \lim \limits_{m \to \infty } m\left( {{B_{m,k}}} \right)$，而$m\left( E \right) < \infty $，则对任给$\delta  > 0$，存在${n_k}\left( { > {n_{k - 1}}} \right)$，使得m\left( E \right) - m\left( {{B_{{n_k},k}}} \right) < \frac{\delta }{{{2^k}}}
<script id="MathJax-Element-3" type="math/tex; mode=display">m\left( E \right) - m\left( {{B_{{n_k},k}}} \right) < \frac{\delta }{{{2^k}}}</script>

令F = \bigcap\limits_{k = 1}^\infty  {{B_{{n_k},k}}}  = E\left( {\left| {{f_n} - f} \right| < \frac{1}{k},n \ge {n_k}} \right)

<script id="MathJax-Element-4" type="math/tex; mode=display">F = \bigcap\limits_{k = 1}^\infty {{B_{{n_k},k}}} = E\left( {\left| {{f_n} - f} \right| < \frac{1}{k},n \ge {n_k}} \right)</script>

则我们有m\left( {E\backslash F} \right) = m\left( {\bigcup\limits_{k = 1}^\infty  {\left( {E\backslash {B_{{n_k},k}}} \right)} } \right) \le \sum\limits_{k = 1}^\infty  {m\left( {E\backslash {B_{{n_k},k}}} \right)}  < \delta

<script id="MathJax-Element-5" type="math/tex; mode=display">m\left( {E\backslash F} \right) = m\left( {\bigcup\limits_{k = 1}^\infty {\left( {E\backslash {B_{{n_k},k}}} \right)} } \right) \le \sum\limits_{k = 1}^\infty {m\left( {E\backslash {B_{{n_k},k}}} \right)} < \delta </script>

以及对任给的$\varepsilon  > 0$，存在${k_0} > \frac{1}{\varepsilon }$，使得当$n \ge {n_{{k_0}}}$时，对任意的$x \in F$，有\left| {{f_n}\left( x \right) - f\left( x \right)} \right| < \frac{1}{{{k_0}}} < \varepsilon

<script id="MathJax-Element-6" type="math/tex; mode=display">\left| {{f_n}\left( x \right) - f\left( x \right)} \right| < \frac{1}{{{k_0}}} < \varepsilon </script>

所以$\left\{ {{f_n}\left( x \right)} \right\}$在$F$上一致收敛于$f(x)$


联系
我们