$\bf命题2：$设$f\left( x \right)$在$\left( {0,1} \right)$上单调,且无界广义积分$\int_0^1
{f\left( x \right)dx} $收敛,则\[\mathop {\lim }\limits_{n \to \infty }
\frac{{f\left( {\frac{1}{n}} \right) + f\left( {\frac{2}{n}} \right) + \cdots +
f\left( {\frac{{n - 1}}{n}} \right)}}{n} = \int_0^1 {f\left( x \right)dx}
\]
证明：我们不妨只讨论$f\left( x \right)$单调增加的情况,则有不等式
\[\int_0^{1 - \frac{1}{n}} {f\left( x \right)dx} \le \frac{{f\left(
{\frac{1}{n}} \right) + f\left( {\frac{2}{n}} \right) + \cdots + f\left(
{\frac{{n - 1}}{n}} \right)}}{n} \le \int_{\frac{1}{n}}^1 {f\left( x \right)dx}
\]
令$n \to \infty $,则由夹逼原理即证
原文地址：http://www.cnblogs.com/ly758241/p/3706297.html
