$f证明$ 由于$mleft( {Eleft( {{f_n} rightarrow f} ight)} ight) = 0$，则我们不妨设$left{ {{f_n}left( x ight)} ight}$处处收敛于$f(x)$，此时[E = igcuplimits_{m = 1}^infty {igcaplimits_{n = m}^infty {Eleft( {left| {{f_n} - f} ight| < frac{1}{k}} ight)} } ,k in {N_ + }]记[{B_{m,k}} = igcaplimits_{n = m}^infty {{E_{n,k}}} = Eleft( {left| {{f_n} - f} ight| < frac{1}{k},n ge m} ight)]其中${E_{n,k}} = Eleft( {left| {{f_n} - f} ight| < frac{1}{k}} ight)$，则对于固定的$k$，${B_{m,k}}$是单调递增的集合列，并且$E = igcuplimits_{m = 1}^infty {{B_{m,k}}} $，所以我们有$mleft( E ight) = lim limits_{m o infty } mleft( {{B_{m,k}}} ight)$，而$mleft( E ight) < infty $，则对任给$delta > 0$，存在${n_k}left( { > {n_{k - 1}}} ight)$，使得[mleft( E ight) - mleft( {{B_{{n_k},k}}} ight) < frac{delta }{{{2^k}}}]
令[F = igcaplimits_{k = 1}^infty {{B_{{n_k},k}}} = Eleft( {left| {{f_n} - f} ight| < frac{1}{k},n ge {n_k}} ight)]
则我们有[mleft( {Eackslash F} ight) = mleft( {igcuplimits_{k = 1}^infty {left( {Eackslash {B_{{n_k},k}}} ight)} } ight) le sumlimits_{k = 1}^infty {mleft( {Eackslash {B_{{n_k},k}}} ight)} < delta ]
以及对任给的$varepsilon > 0$，存在${k_0} > frac{1}{varepsilon }$，使得当$n ge {n_{{k_0}}}$时，对任意的$x in F$，有[left| {{f_n}left( x ight) - fleft( x ight)} ight| < frac{1}{{{k_0}}} < varepsilon ]
所以$left{ {{f_n}left( x ight)} ight}$在$F$上一致收敛于$f(x)$