How to show this conclusion which concerns two distribution functions?

Question

If $F$ and $G$ are two distribution functions, how to prove that $$\int_{0}^{1} \lvert F^{-1}(t)-G^{-1}(t)\rvert dt=\int_{-\infty}^{+\infty}\lvert F(x)-G(x)\rvert dx$$ with $$F^{-1}(p)=\inf\{x\in\mathbb{R},F(x)\geq p\}$$ I have no idea to deal with the integral of the absolute value of the difference of two distribution functions. The left side is in fact equal to $\mathbb{E}[\lvert F^{-1}(U)-G^{-1}(U)\rvert]$ where $U$ is in the uniform distribution on $[0,1]$.\ This is an important exercice in my course and I really need someone who can help me solve it.

Answer 1

What you need to do is to express $|F(x) - G(x)|$ as $\int 1_{(F(x),G(x))} (t)dt$ then you just make use of Fubini theorem

Answer 2

I can answer this question under his help.

Let $\Phi(x,t)=\mathbb{1}_{\{F(x)\geq G(x),t\in [G(x),F(x)]\}}+\mathbb{1}_{\{F(x)<G(x),t\in [F(x),G(x)]\}}$, then we have $$\lvert F(x)-G(x)\rvert=\int_{0}^{1}\Phi(x,t)dt$$ so using the Fubini's theorem, we have \begin{equation} \begin{split} \int_{-\infty}^{+\infty}\lvert F(x)-G(x)\rvert dx&=\int_{-\infty}^{+\infty}\int_{0}^1 \Phi(x,t)dtdx\\ &=\int_0^1(\int_{-\infty}^{+\infty}\Phi(x,t) dx)dt\\ &=\int_0^1\lvert F^{-1}(t)-G^{-1}(t)\rvert dt \end{split} \end{equation} because $$\int_{-\infty}^{+\infty}\Phi(x,t) dx=\lvert F^{-1}(t)-G^{-1}(t)\rvert,\forall t\in[0,1]$$

Is it correct? I want to ask if it is correct.