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4367

$\bf(Lusin定理)$设$f\left( x \right)$是可测集$E$上几乎处处有限的可测函数,

则对任给$\delta  > 0$,存在闭集$F \subset E$,使得$m\left( {E\backslash F} \right) < \delta $,且$f\left( x \right)$在$F$上连续

$\bf证明$  由于$m\left( {E\left( {\left| f \right| =  + \infty } \right)} \right) = 0$,我们不妨设$f\left( x \right)$是处处有限的

   $\bf(1)$首先,我们考虑$f\left( x \right)$是简单函数的情况,于是

f(x)=i=1nciχEi(x),xE=?i=1nEi
<script id="MathJax-Element-1" type="math/tex; mode=display">f\left( x \right) = \sum\limits_{i = 1}^n {{c_i}{\chi _{{E_i}}}\left( x \right)} ,x \in E = \bigcup\limits_{i = 1}^n {{E_i}} </script>由于每个${E_i}$是可测的,则对任给$\delta  > 0$,存在闭集${F_i} \subset {E_i}$,使得
m(Ei?Fi)<δ/n
<script id="MathJax-Element-2" type="math/tex; mode=display">m\left( {{E_i}\backslash {F_i}} \right) < \delta /n</script>

又由于$f\left( x \right)$在每个${F_i}$上是常值函数,从而在${F_i}$上连续;而${F_1}, \cdots ,{F_n}$互不相交,令

F=?i=1nFi
<script id="MathJax-Element-3" type="math/tex; mode=display">F = \bigcup\limits_{i = 1}^n {{F_i}} </script>则闭集$F \subset E$,使得$m\left( {E\backslash F} \right) = \sum\limits_{i = 1}^n {m\left( {{E_i}\backslash {F_i}} \right)}  < \delta $,且$f\left( x \right)$在$F$上连续

   $\bf(2)$其次,我们考虑$f\left( x \right)$是一般可测函数的情况,由于可作变换

g(x)=f(x)1+|f(x)|
<script id="MathJax-Element-4" type="math/tex; mode=display">g\left( x \right) = \frac{{f\left( x \right)}}{{1 + \left| {f\left( x \right)} \right|}}</script>因此我们不妨设$f\left( x \right)$是有界可测函数,于是存在可测的简单函数列$\left\{ {{\varphi _k}\left( x \right)} \right\}$在$E$上一致收敛于$f\left( x \right)$,从而由$\bf(1)$知,对任给$\delta  > 0$,存在闭集${F_k} \subset {E}$,使得$m\left( {E\backslash {F_k}} \right) < \frac{\delta }{{{2^k}}}$,且${{\varphi _k}\left( x \right)}$在${F_k} $上连续,令
F=?k=1 Fk
<script id="MathJax-Element-5" type="math/tex; mode=display">{F} = \bigcap\limits_{k = 1}^\infty {{F_k}} </script>则${F }$为闭集,且

m(E?F)=m(E??k=1Fk)=m(?k=1(E?Fk))k=1m(E?Fk) <δ 
<script id="MathJax-Element-6" type="math/tex; mode=display">\begin{array}{l} m\left( {E\backslash {F }} \right) &= m\left( {E\backslash \bigcap\limits_{k = 1}^\infty {{F_k}} } \right)\\ &= m\left( {\bigcup\limits_{k = 1}^\infty {\left( {E\backslash {F_k}} \right)} } \right) \le \sum\limits_{k = 1}^\infty {m\left( {E\backslash {F_k}} \right)} < \delta \end{array}</script>由于${{\varphi _k}\left( x \right)}$在${F }$上连续,且一致收敛于$f\left( x \right)$,所以$f\left( x \right)$在${F }$连续