Cfrgtkky

Let's $X$ and $Y$ have CDF functions admitting moment of order $1$.
Let's be $F$ cdf of $X$ and $G$ cdf of $Y$.

I want to show that $$int_{mathbb R} mid F(x)-G(x)mid dx = int_{0}^{1} mid F^{-1}(u)-G^{-1}(u)mid du,.$$

If $mu$ is 2D Lebesgue measure, then interpreting the integral as the unsigned area$^*$ between $F$ and $G$,
$$int |F(x) - G(x)| dx = mubig[(x,y)inmathbb Rtimes [0,1]:G(x)le y<F(x)big] + mubig[(x,y)inmathbb Rtimes [0,1]:F(x)le y<G(x)big ] $$

Then note that

$$G(x) le y < F(x) iff x le G^{-1}(y) , F^{-1}(y)<x iff F^{-1}(y)<x le G^{-1}(y)$$
and similarly $ F(x) le y < G(x) iff G^{-1}(y) < x le F^{-1}(y)$.
thus

$$int |F(x) - G(x)| dx = mubig[(x,y)inmathbb Rtimes [0,1]:F^{-1}(y)<x le G^{-1}(y)big] + mubig[(x,y)inmathbb Rtimes [0,1]:G^{-1}(y) < x le F^{-1}(y)big ] $$

returning to the 1D integral notation, this is saying that
$$ int_{mathbb R} |F(x) - G(x)| dx = int_0^1 |F^{-1}(y) - G^{-1}(y) | dy $$

Finally a graph - this indicates that the result should be true even for some functions without an inverse. ( desmos link )

$^*$ For a positive function $f$, $int_A f(x) dx = int_A int_0^{f(x)} dydx = mu( (x,y) in Atimes operatorname{im}f : 0le yle f(x)).$ Appropriate case analysis leads to the above expression.

This answer is inspired by Calvin Khor's solution. Here, we do not assume that $F$ and $G$ possess inverse functions. In this answer, we define $T^{-1}:(0,1)to mathbb{R}$ as
$$T^{-1}(u):=supbig{vinmathbb{R},|,T(v)leq ubig}$$
for any cumulative distribution function $T:mathbb{R}to[0,1]$. Since $T^{-1}$ is nondecreasing, it is a measurable function. We first note that, if $T$ admits the first moment, then $$int_{-infty}^0,T(x),text{d}x+int_0^{+infty},big(1-T(x)big),text{d}x=int_mathbb{R},|x|,text{d}T(x)<infty,,$$
so we have
$$int_{-infty}^0,T(x),text{d}x<inftytext{ and }int_0^{+infty},big(1-T(x)big),text{d}x<infty,.tag{*}$$

Now, because $F$ and $G$ admit the first moments, the integral
$$I:=int_mathbb{R},big|F(x)-G(x)big|,text{d}x$$
is finite due to (*). From Calvin Khor's answer, we have
$$I=mu(E^+)+mu(E^-),,$$ where $mu$ is the Lebesgue measure on $mathbb{R}^2$,
$$E^+:=big{(x,y)inmathbb{R}times (0,1),|,G(x)leq y<F(x)big},,$$
and
$$E^-:=big{(x,y)inmathbb{R}times (0,1),|,F(x)leq y<G(x)big},.$$
Observe that
$$E^+subseteq S^+:=big{(x,y)inmathbb{R}times (0,1),|,F^{-1}(y)leq x
leq G^{-1}(y)big}$$
and
$$E^-subseteq S^-:=big{(x,y)inmathbb{R}times (0,1),|,G^{-1}(y)leq x
leq F^{-1}(y)big},.$$
Note that $$S^+setminus E^+subseteq big{(x,y)inmathbb{R}times (0,1),|,xtext{ is the unique solution to }F(x)=ybig}$$ and $$S^-setminus E^-subseteq big{(x,y)inmathbb{R}times (0,1),|,xtext{ is the unique solution to }G(x)=ybig}$$ are of Lebesgue measure $0$. Therefore,
$$I=mu(S^+)+mu(S^-)=int_0^1,big|F^{-1}(u)-G^{-1}(u)big|,text{d}u,.$$

Since you use the notation $F^{-1}$, resp. $G^{-1}$, for the quantile function you tacitly assume that $F$ and $G$ are continuous and strictly increasing on some interval $Jsubset{mathbb R}$. I suggest you draw a figure showing two reasonable such functions. The left hand side of the claimed formula then represents the unsigned area enclosed between the graphs of $F$ and $G$. Turning the figure $90^circ$ you then can verify that the right hand side of the claimed formula is the same area.

This means that one has to prove that
$$A:=bigl{(x,u)in Jtimes[0,1]bigm|min{F(x),G(x)}leq uleqmax{F(x),G(x)}bigr}$$
and
$$A':=bigl{(x,u)in Jtimes[0,1]bigm|min{F^{-1}(u),G^{-1}(u)}leq xleqmax{F^{-1}(u),G^{-1}(u)}bigr}$$
are in fact the same sets. This is "pure logic"; one just has to go through the motions.