$\int_{F^{-1}(O)}F'(x)dx = m(O)$, $F$ is absolutely continuous

Question

I'm solving the following question.

$F$ is absolutely continuous on $[a,b]$ and increasing. Let $A = F(a)$ and $B = F(b)$. Show that $\int_{F^{-1}(O)}F'(x)dx = m(O)$, where $O$ is any open set in $[A,B]$.

The following is my trial.

Let $\chi$ be a characteristic function.

$\chi_{F^{-1}(O)}(x) = 1 \iff x \in F^{-1}(O) \iff F(x) \in O \iff \chi_O (F(x)) = 1$

So, I argued that $\chi_{F^{-1}(O)}(x) = \chi_O (F(x))$.

Then, $\int_{F^{-1}(O)}F'(x)dx = \int_{[a,b]} \chi_{F^{-1}(O)}(x)F'(x)dx = \int_{[a,b]} \chi_O (F(x))F'(x)dx$

This is where I'm stuck, because I don't think I can use the change of variable right away (the characteristic function is not necessarily Riemann integrable as far as I know).

Is there some way to finish the proof?

Thank you.

Answer 1

Since $O$ is a disjoint union of open intervals it is enough to show this when $O=(a,b)$. When $a$ and $b$ are finite this follows from the fact that $F^{-1}(a,b)$ is an interval; if its end points are $\alpha$ and $\beta$ then left side becomes $\int_{\alpha} ^{\beta} F'(x)dx=F(\beta)-F(\alpha)=b-a$ which is $m(O)$. I will leave the case of infinte intervals to you.