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