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$f证明$  $(1)$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}} ]