$f证明$ 由于$left{ {{f_n}left( x ight)} ight}$几乎处处收敛于$f(x)$，则存在零测集$E_0$，使得$lim limits_{n o infty } {f_n}left( x ight) = fleft( x ight)$在$E_1=Eackslash {E_0}$上成立，
于是对任给的$varepsilon > 0$，我们有[{E_1} = igcuplimits_{m = 1}^infty {igcaplimits_{n = m}^infty {{E_1}left( {left| {{f_n} - f} ight| < varepsilon } ight)} } ]
即${E_1} = mathop {underline {lim } }limits_{n o infty } {E_1}left( {left| {{f_n} - f} ight| < varepsilon } ight)$，从而由测度的性质知[mleft( {{E_1}} ight) le mathop {underline {lim } }limits_{n o infty } mleft( {{E_1}left( {left| {{f_n} - f} ight| < varepsilon } ight)} ight)]
由$mleft( E ight) < infty $，我们得到[mathop {overline {lim } }limits_{n o infty } mleft( {{E_1}left( {left| {{f_n} - f} ight| ge varepsilon } ight)} ight) = mleft( {{E_1}} ight) - mathop {underline {lim } }limits_{n o infty } mleft( {{E_1}left( {left| {{f_n} - f} ight| < varepsilon } ight)} ight) le 0]所以对任给的$varepsilon > 0$，我们有$lim limits_{n o infty } mleft( {{E_1}left( {left| {{f_n} - f} ight| ge varepsilon } ight)} ight) = 0$
$f注1：$设$left{ {{E_n}} ight}$是一列可测集，记$mathop {underline {lim } }limits_{n o infty } {E_n} = igcuplimits_{n = 1}^infty {igcaplimits_{k = n}^infty {{E_k}} } $，则[mleft( {mathop {underline {lim } }limits_{n o infty } {E_n}} ight) le mathop {underline {lim } }limits_{n o infty } mleft( {{E_n}} ight)]