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$\bf证明$  由于$f_n$几乎处处收敛于$f$,且$\displaystyle|{f_n}|\mathop { \le} \limits_{a.e.} F$,则令$n \to \infty $,有$\displaystyle|{f}|\mathop { \le} \limits_{a.e.} F$,从而由$F$可积得到$f$是可积的

  $(1)$首先考虑$m\left( E \right) < \infty $的情况

由于$F$可积,则由积分的全连续性知,对任给的$\varepsilon  > 0$,存在$\delta  > 0$,使得对任意可测集$e \subset E$满足$m\left( e \right) < \delta $时,有\[\int_e {F\left( x \right)dx}  < \frac{\varepsilon }{4}\]

又由于$f_n$几乎处处收敛于$f$,则由$\bf{Egoroff定理}$知,对上述的$\delta  > 0$,存在可测集${E_\delta } \subset E$,使得\[m\left( {E\backslash {E_\delta }} \right) < \delta \]且$f_n$在${E_\delta }$上一致收敛于$f$,即对任给的$\varepsilon  > 0$,存在$N$,使得当$n \ge N$时,对任意$x \in {E_\delta }$,有\[\left| {{f_n}\left( x \right) - f\left( x \right)} \right| < \frac{\varepsilon }{{2m\left( E \right)}}\]

从而可知\begin{align*}\left| {\int_E {{f_n}\left( x \right)dx} - \int_E {f\left( x \right)dx} } \right| &\le \int_E {\left| {{f_n}\left( x \right) - f\left( x \right)} \right|dx} \\
&= \int_{E\backslash {E_\delta }} {\left| {{f_n}\left( x \right) - f\left( x \right)} \right|dx} + \int_{{E_\delta }} {\left| {{f_n}\left( x \right) - f\left( x \right)} \right|dx} \\
&< \int_{E\backslash {E_\delta }} {2F\left( x \right)dx} + \frac{\varepsilon }{{2m\left( E \right)}} \cdot m\left( E \right) < \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon 
\end{align*}

  $(2)$其次考虑一般的$E$的情况

设$\left\{ {{E_n}} \right\}$是$E$的测度有限的单调覆盖,则由$F$可积的定义知\[\mathop {\lim }\limits_{n \to \infty } \int_{{E_n}} {{{\left[ F \right]}_n}\left( x \right)dx}  = \int_E {F\left( x \right)dx} \]即对任给的$\varepsilon  > 0$,存在$N$,使得当$n \ge N$时,有\[0 \le \int_E {F\left( x \right)dx}  - \int_{{E_n}} {{{\left[ F \right]}_n}\left( x \right)dx}  < \frac{\varepsilon }{4}\]于是当$n \ge N$时,我们有

\begin{align*}
\left| {\int_{E\backslash {E_n}} {\left[ {{f_n}\left( x \right) - f\left( x \right)} \right]dx} } \right| &\le \int_{E\backslash {E_n}} {2F\left( x \right)dx} \\
&= 2\left( {\int_E {F\left( x \right)dx} - \int_{{E_n}} {F\left( x \right)dx} } \right)\\
&\le 2\left( {\int_E {F\left( x \right)dx} - \int_{{E_n}} {{{\left[ F \right]}_n}\left( x \right)dx} } \right) < \frac{\varepsilon }{2}
\end{align*}又对于测度有限的${E_n}$,由$(1)$可知,当$n \ge N$时,有\[\left| {\int_{{E_n}} {\left[ {{f_n}\left( x \right) - f\left( x \right)} \right]dx} } \right| < \frac{\varepsilon }{2}\]所以对任给的$\varepsilon  > 0$,存在$N$,使得当$n \ge N$时,有

\begin{align*}
\left| {\int_E {\left[ {{f_n}\left( x \right) - f\left( x \right)} \right]dx} } \right|& \le \left| {\int_{E\backslash {E_n}} {\left[ {{f_n}\left( x \right) - f\left( x \right)} \right]dx} } \right| + \left| {\int_{{E_n}} {\left[ {{f_n}\left( x \right) - f\left( x \right)} \right]dx} } \right|\\
&< \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon
\end{align*}

$\bf注1:$$\bf(引理)$设$f$是$E$上的可积函数,$g$是$E$上的可测函数,若$\left| g \right| \le f$,则$g$也是可积的

方法一

$\bf注2:$$\bf(积分的全连续性)$设$f$是$E$上的可积函数,则对任给的$\varepsilon  > 0$,存在$\delta  > 0$,使得对任意可测集$e \subset E$满足$m\left( e \right) < \delta $时,有\[\int_e {\left| {f\left( x \right)} \right|dx}  < \frac{\varepsilon }{4}\]

方法一

$\bf注3:$

 

 

 

posted on 2014-06-03 11:06  一阴一阳之谓道  阅读(176)  评论(0编辑  收藏  举报

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