# Definition of Wasserstein distance through cumulative distribution

Let $$X$$ and $$Y$$ be random variables on the same probability space. The $$\infty$$-Wasserstein distance between $$X$$ and $$Y$$ is defined as

$$d_{\infty}(X, Y) = \inf \|X_1 - Y_1\|_{L_{\infty}},$$

where the infimum is over all random variables $$X_1$$ and $$Y_1$$ with same distribution as $$X$$ and $$Y$$, respectively.

Let $$F_X$$ and $$F_Y$$ be the cumulative distributions of $$X$$ and $$Y$$. I need help proving that

$$d_{\infty}(X, Y) = \inf\{\epsilon > 0 : F_X(t - \epsilon) \leq F_Y(t) \leq F_X(t + \epsilon) \mbox{ for all } t \in \mathbb{R}\}$$

Can someone help me?

Reference: Aubrun and Szarek, Alice and Bob Meet Banach: The Interface of Asymptotic Geometric Analysis and Quantum Information Theory, pg 161.

$$\newcommand\ep\epsilon\newcommand\R{\mathbb R}$$Let $$d:=d_\infty(X,Y)$$, $$\|\cdot\|:=\|\cdot\|_{L_\infty}$$, $$E:=\{\ep>0\colon F_X(t-\ep)\le F_Y(t)\le F_X(t+\ep) \text{ for all } t \in \R\},$$ $$D:=\inf E.$$ We need to show that $$d=D$$.
Take any real $$c>d$$. Then for some random variables (r.v.'s) $$X_1$$ and $$Y_1$$ with the same distributions as $$X$$ and $$Y$$, respectively, we have $$\|X_1-Y_1\| and hence $$X_1-c almost surely. So, for all real $$t$$ $$F_Y(t)=P(Y_1\le t)\le P(X_1-c\le t)=F_X(t+c)$$ and similarly $$F_X(t-c)\le F_Y(t)$$, so that $$c\in E$$ and hence $$D\le c$$, for any real $$c>d$$. So, $$D\le d.\tag{1}$$
On the other hand, let $$X_1:=F_X^{-1}(U)\quad\text{and}\quad Y_1:=F_Y^{-1}(U),$$ where $$U$$ is any r.v. uniformly distributed on the interval $$(0,1)$$ and $$F^{-1}(u):=\inf\{x\in\R\colon F(x)\ge u\}=\min\{x\in\R\colon F(x)\ge u\}$$ for any cumulative distribution function $$F$$ and any $$u\in(0,1)$$. Then $$X_1$$ and $$Y_1$$ have the same distributions as $$X$$ and $$Y$$, respectively. Take any $$\ep\in E$$. Then it is easy to see that $$|F_X^{-1}(u)-F_Y^{-1}(u)|\le\ep$$ for all $$u\in(0,1)$$, whence $$\|X_1-Y_1\|\le\ep$$. So, $$d\le\|X_1-Y_1\|\le\ep$$, for any $$\ep\in E$$. So, $$d\le\inf E=D.$$ In view of (1), we get $$d=D$$, as desired.