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$f(Lusin定理)$设$fleft( x ight)$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$mleft( {Eackslash F} ight) < delta $，且$fleft( x ight)$在$F$上连续

$f证明$  由于$mleft( {Eleft( {left| f ight| =  + infty } ight)} ight) = 0$，我们不妨设$fleft( x ight)$是处处有限的

   $f(1)$首先，我们考虑$fleft( x ight)$是简单函数的情况，此时[fleft( x ight) = sumlimits_{i = 1}^n {{c_i}{chi _{{E_i}}}left( x ight)} ,x in E = igcuplimits_{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[mleft( {{E_i}ackslash {F_i}} ight) < delta /n]

又由于$fleft( x ight)$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 = igcuplimits_{i = 1}^n {{F_i}} ]则闭集$F subset E$，使得$mleft( {Eackslash F} ight) = sumlimits_{i = 1}^n {mleft( {{E_i}ackslash {F_i}} ight)}  < delta $，且$fleft( x ight)$在$F$上连续

   $f(2)$其次，我们考虑$fleft( x ight)$是一般可测函数的情况，由于可作变换[gleft( x ight) = frac{{fleft( x ight)}}{{1 + left| {fleft( x ight)} ight|}}]因此我们不妨设$fleft( x ight)$是有界可测函数，于是存在可测的简单函数列$left{ {{varphi _k}left( x ight)} ight}$在$E$上一致收敛于$fleft( x ight)$，从而由$f(1)$知，对任给$delta  > 0$，存在闭集${F_k} subset {E}$，使得$mleft( {Eackslash {F_k}} ight) < frac{delta }{{{2^k}}}$，且${{varphi _k}left( x ight)}$在${F_k} $上连续，令[{F} = igcaplimits_{k = 1}^infty  {{F_k}} ]则闭集$F subset E$，使得

[egin{array}{l}
mleft( {Eackslash {F }} ight) &= mleft( {Eackslash igcaplimits_{k = 1}^infty {{F_k}} } ight)\
&= mleft( {igcuplimits_{k = 1}^infty {left( {Eackslash {F_k}} ight)} } ight) le sumlimits_{k = 1}^infty {mleft( {Eackslash {F_k}} ight)}  < delta 
end{array}]且${{varphi _k}left( x ight)}$在${F }$上连续，而$left{ {{varphi _k}left( x ight)} ight}$一致收敛于$fleft( x ight)$，所以$fleft( x ight)$在${F }$连续