Assume that $d\mu=f\,dx$ for a nonnegative increasing $f$ such that $\mu$ is a probability measure on $[0,1]$, does it follows that \begin{align*} \int_{0}^{1}F'\,d\mu\leq F(1)-F(0) \end{align*} for $F\in C^{1}[0,1]$ with $F'\geq 0$?
There is a theorem by Lebesgue saying that \begin{align*} \int_{0}^{1}F'\,d\mu_{L}\leq F(1)-F(0), \end{align*} where $\mu_{L}$ is the Lebesgue measure and $F\in C^{1}[0,1]$ with $F'\geq 0$. I tried to mimic the lengthy proof, it seems that my guess is correct. But is there any reference to that?
Suppose $d\mu = f\,dx$ and $f$ is a smoothed version of the function $$ \hat{f} = \begin{cases} 0 & \text{ if } x\leq 1/2 \\ 2 &\text{ if } x\geq 1/2 \end{cases} $$ Then $$ \int_{[0,1]} F' d\mu \sim 2\int_{[1/2,1]} F' \,dx = 2\left(F(1)-F(1/2)\right)$$