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$\bf命题2：$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

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