In a book I found the following exercise:
Let $F,G$ be two cumulative distribution functions. Then
$$\int_0^1 \vert F^{-1} (t) - G^{-1} (t)\vert \text d t = \int_\Bbb R \vert F(x) - G(x) \vert \text d x$$ where $H^{-1} (t) := \inf \{ x \in \Bbb R : H(x) > t\}$, for a CDF $H$.
I tried to write $\vert F(x) - G(x) \vert = \int_0^1 1_{(-\infty , \vert F(x) - G(x) \vert]} (t) \text d t$ and use Fubini afterwards, in order to rewrite the set $\{x : t\leq \vert F(x) - G(x) \vert\}$ for fixed $t$ and obtaining a set with mass $\vert F^{-1}(t) - G^{-1}(t) \vert$.
But I fail with the last part. Maybe this is even the wrong approach.
Clearly, $$\int_\Bbb R \vert F(x) - G(x) \vert \text d x = \int_\Bbb R \int_{\min \{F(x),G(x)\}}^{\max\{F(x),G(x)\}} \text d t \text d x \\ = \text{Lebesgue} (\{(x,t)\in \Bbb R\times [0,1] : \min \{F(x),G(x)\} < t <\max\{F(x),G(x)\} \})$$ But if w.l.o.g. $F(x) < t < G(x)$, then by definition holds: $$x\geq G^{-1}(t) \quad \text{and since $F^{-1}$ is non-decreasing:} \quad F^{-1}(t) \geq F^{-1}(F(x)) \geq x$$ Conversly, if $F^{-1}(t) > x > G^{-1}(t)$, then by definition $$F(x) \leq t \quad \text{and since $G^{-1}$ is non-decreasing:} \quad G(x) \geq G(G^{-1}(t)) \geq t$$ Thus, the quantity above equals by Fubini $$\text{Lebesgue}(\{(x,t)\in \Bbb R\times [0,1] : \min \{F^{-1}(t),G^{-1}(t)\} < x <\max\{F^{-1}(t),G^{-1}(t)\} \}) \\ = \int_0^1 \int_{\min \{F^{-1}(t),G^{-1}(t)\}}^{\max\{F^{-1}(t),G^{-1}(t)\}} \text d x \text d t = \int_0^1 \vert F^{-1} (t) - G^{-1} (t) \vert \text dt$$