265985
2024-07-06 21:49:34 225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$\left\{ {{f_n}\left( x \right)} \right\}$处处收敛与$f(x)$，此时
E=?
m=1∞ ?n=m∞ E(|fn?f|<1k),k∈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Bm,k=?n=m∞ En,k =E(|fn?f|<1k,n≥m) <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)$，使得<script id="MathJax-Element-3" type="math/tex; mode=display">m\left( E \right) - m\left( {{B_{{n_k},k}}} \right) < \frac{\delta }{{{2^k}}}</script>令
F=?k=1∞ Bnk,k =E(|fn?f|<1k,n≥nk) <script id="MathJax-Element-4" type="math/tex; mode=display">F = \bigcap\limits_{k = 1}^\infty {{B_{{n_k},k}}} = E\left( {\left| {{f_n} - f} \right| < \frac{1}{k},n \ge {n_k}} \right)</script>则我们有
m(E?F)=m(?k=1∞ (E?Bnk,k))≤∑k=1∞ m(E?Bnk,k) <δ <script id="MathJax-Element-5" type="math/tex; mode=display">m\left( {E\backslash F} \right) = m\left( {\bigcup\limits_{k = 1}^\infty {\left( {E\backslash {B_{{n_k},k}}} \right)} } \right) \le \sum\limits_{k = 1}^\infty {m\left( {E\backslash {B_{{n_k},k}}} \right)} < \delta </script>以及对任给的$\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$，有
<script id="MathJax-Element-6" type="math/tex; mode=display">\left| {{f_n}\left( x \right) - f\left( x \right)} \right| < \frac{1}{{{k_0}}} < \varepsilon </script>所以$\left\{ {{f_n}\left( x \right)} \right\}$在$F$上一致收敛于$f(x)$
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