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$\bf(Lusin定理)$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

$$f\left( x \right) = \sum\limits_{i = 1}^n {{c_i}{\chi _{{E_i}}}\left( x \right)} ,x \in E = \bigcup\limits_{i = 1}^n {{E_i}}$$ 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 $$m\left( {{E_i}\backslash {F_i}} \right) < \delta /n$$

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

$$F = \bigcup\limits_{i = 1}^n {{F_i}}$$ 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

$$g\left( x \right) = \frac{{f\left( x \right)}}{{1 + \left| {f\left( x \right)} \right|}}$$ 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 $${F} = \bigcap\limits_{k = 1}^\infty {{F_k}}$$ 则${F }$为闭集，且

$$\begin{array}{l} m\left( {E\backslash {F }} \right) &= m\left( {E\backslash \bigcap\limits_{k = 1}^\infty {{F_k}} } \right)\\ &= m\left( {\bigcup\limits_{k = 1}^\infty {\left( {E\backslash {F_k}} \right)} } \right) \le \sum\limits_{k = 1}^\infty {m\left( {E\backslash {F_k}} \right)} < \delta \end{array}$$ 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

 

 

 

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