$f命题2：$设$fleft( x ight) in Cleft( { - infty , + infty } ight)$，令[{f_n}left( x ight) = sumlimits_{k = 0}^{n - 1} {frac{1}{n}} fleft( {x + frac{k}{n}} ight)]
证明：对任意$x in left[ {a,b} ight] subset left( { - infty , + infty } ight)$，有${f_n}left( x ight)$一致收敛于$int_0^1 {fleft( {x + t} ight)dt}$
证明：由$fleft( x ight) in Cleft( { - infty , + infty } ight)$知，$fleft( x ight) in Cleft[ {a,b} ight]$，则
由$f{Cantor定理}$知，$fleft( x
ight)$在$left[ {a,b}
ight]$上一致连续，即对任意$varepsilon > 0$，存在$delta > 0$，使得对任意的$x,y in left[ {a,b}
ight]$满足$left| {x - y}
ight|
< delta $时，有[left| {fleft( x
ight) - fleft( y
ight)}
ight| < varepsilon ]
取$N = frac{1}{delta }$，则当$n > N$时，对任意$x in left[ {a,b}
ight]$，$t in left[ {frac{k}{n},frac{{k + 1}}{n}}
ight]$，有
[left| {left( {x + frac{k}{n}}
ight) - left( {x + t}
ight)}
ight| < delta ]
从而有[left| {fleft( {x + frac{k}{n}} ight) - fleft( {x + t} ight)} ight| < varepsilon ]
所以对任意$varepsilon > 0$，存在$N = frac{1}{delta } > 0$，使得当$n > N$时，对任意$x in left[ {a,b}
ight]$，有
[left| {{f_n}left( x
ight) - int_0^1 {fleft( {x + t}
ight)dt} }
ight| = left| {sumlimits_{k = 0}^{n - 1} {int_{frac{k}{n}}^{frac{{k + 1}}{n}} {left[ {fleft( {x + frac{k}{n}}
ight) - fleft( {x + t}
ight)}
ight]dt} } }
ight| < varepsilon ]
从而由函数列一致收敛的定义即证
$f注1：$$int_0^1 {fleft( {x + t} ight)dt} { m{ = }}sumlimits_{k = 0}^{n - 1} {int_{frac{k}{n}}^{frac{{k + 1}}{n}} {fleft( {x + t} ight)dt} } $