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$\bf命题：$设连续函数$f,g：$$\left[ {0,1} \right] \to \left[ {0,1} \right]$，且$f(x)$单调递增，则$$\int_0^1 {f\left( {g\left( x \right)} \right)dx} \le \int_0^1 {f\left( x \right)dx} + \int_0^1 {g\left( x \right)dx} $$
证明：由积分中值定理知，存在$\xi \in \left[ {0,1} \right]$，使得
\[\int_0^1 {\left[ {f\left( {g\left( x \right)} \right) - g\left( x \right)} \right]dx} = f\left( {g\left( \xi \right)} \right) - g\left( \xi \right) = f\left( u \right) - u\]
其中$u = g\left( \xi \right) \in \left[ {0,1} \right]$，而由$f\left( x \right)$的取值范围与单调性知
\[\int_0^1 {f\left( x \right)dx} \ge \int_u^1 {f\left( x \right)dx} \ge f\left( u \right)\left( {1 - u} \right) \ge f\left( u \right) - u\]
从而命题成立