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$\bf命题2：$设$f\left( x \right) \in C\left( { - \infty , + \infty } \right)$，令\[{f_n}\left( x \right) = \sum\limits_{k = 0}^{n - 1} {\frac{1}{n}} f\left( {x + \frac{k}{n}} \right)\]

证明：对任意$x \in \left[ {a,b} \right] \subset \left( { - \infty , + \infty } \right)$，有${f_n}\left( x \right)$一致收敛于$\int_0^1 {f\left( {x + t} \right)dt}$

证明：由$f\left( x \right) \in C\left( { - \infty , + \infty } \right)$知，$f\left( x \right) \in C\left[ {a,b} \right]$，则

由$\bf{Cantor定理}$知，$f\left( x \right)$在$\left[ {a,b} \right]$上一致连续，即对任意$\varepsilon > 0$，存在$\delta > 0$，使得对任意的$x,y \in \left[ {a,b} \right]$满足$\left| {x - y} \right|
< \delta $时，有\[\left| {f\left( x \right) - f\left( y \right)} \right| < \varepsilon \]
取$N = \frac{1}{\delta }$，则当$n > N$时，对任意$x \in \left[ {a,b} \right]$，$t \in \left[ {\frac{k}{n},\frac{{k + 1}}{n}} \right]$，有
\[\left| {\left( {x + \frac{k}{n}} \right) - \left( {x + t} \right)} \right| < \delta \]

从而有\[\left| {f\left( {x + \frac{k}{n}} \right) - f\left( {x + t} \right)} \right| < \varepsilon \]

所以对任意$\varepsilon > 0$，存在$N = \frac{1}{\delta } > 0$，使得当$n > N$时，对任意$x \in \left[ {a,b} \right]$，有
\[\left| {{f_n}\left( x \right) - \int_0^1 {f\left( {x + t} \right)dt} } \right| = \left| {\sum\limits_{k = 0}^{n - 1} {\int_{\frac{k}{n}}^{\frac{{k + 1}}{n}} {\left[ {f\left( {x + \frac{k}{n}} \right) - f\left( {x + t} \right)} \right]dt} } } \right| < \varepsilon \]
从而由函数列一致收敛的定义即证

$\bf注1：$$\int_0^1 {f\left( {x + t} \right)dt} {\rm{ = }}\sum\limits_{k = 0}^{n - 1} {\int_{\frac{k}{n}}^{\frac{{k + 1}}{n}} {f\left( {x + t} \right)dt} } $