$f证明$ $(1)$由${f_n}$依测度收敛于$f(x)$知，对任何自然数$k$，存在自然数${n_k}left( { > {n_{k - 1}}} ight)$，使得当$n ge {n_k}$时，有[mleft( {Eleft( {left| {{f_n} - f} ight| ge frac{1}{{{2^k}}}} ight)} ight) < frac{1}{{{2^k}}}]
记${E_k} = Eleft( {left| {{f_{{n_k}}} - f} ight| ge frac{1}{{{2^k}}}} ight)$，则$mleft( {{E_k}} ight) < frac{1}{{{2^k}}}$，令
[{F_k} = igcaplimits_{i = k}^infty {left( {Eackslash {E_i}} ight)} ]由于$Eackslash {E_i} = Eleft( {left| {{f_{{n_i}}} - f} ight| < frac{1}{{{2^i}}}} ight)$，所以我们有[{F_k} = Eleft( {left| {{f_{{n_i}}} - f} ight| < frac{1}{{{2^i}}},i = k,k + 1, cdots } ight)]
即函数列${f_{{n_i}}}left( x ight)$在${F_k}$上一致收敛于$f(x)$，于是${f_{{n_i}}}left( x ight)$在$F = igcuplimits_{k = 1}^infty {{F_k}} $上处处收敛于$f(x)$
$(2)$下面我们只需证明$mleft( {Eackslash F} ight) = 0$即可，由于[Eackslash F = igcaplimits_{k = 1}^infty {left( {Eackslash {F_k}} ight)} = igcaplimits_{k = 1}^infty {igcuplimits_{i = k}^infty {{E_i}} } = mathop {overline {lim } }limits_{i o infty } {E_i} subset igcuplimits_{i = 1}^infty {{E_i}} ]而[mleft( {igcuplimits_{i = 1}^infty {{E_i}} } ight) le sumlimits_{i = 1}^infty {mleft( {{E_i}} ight)} le sumlimits_{i = 1}^infty {frac{1}{{{2^i}}}} = 1]所以我们有$mleft( {Eackslash F} ight) = 0$
$f注1：$由上限集与下限集的定义知，[igcaplimits_{n = 1}^infty {{A_n}} subset mathop {underline {lim } }limits_{n o infty } {A_n} subset mathop {overline {lim } }limits_{n o infty } {A_n} subset igcuplimits_{n = 1}^infty {{A_n}} ]