How would you visualize Levy metric? From Wikipedia:

Let $F, G : \mathbb{R} \to [0, 1]$ be two cumulative distribution functions. Define the Lévy distance between them to be
  :$$L(F, G) := \inf \{ \varepsilon > 0 | F(x - \varepsilon) - \varepsilon \leq G(x) \leq F(x + \varepsilon) + \varepsilon \mathrm{\,for\,all\,} x \in \mathbb{R} \}.$$
Intuitively, if between the graphs of $F$ and $G$ one inscribes squares with sides parallel to the coordinate axes (at points of discontinuity of a graph vertical segments are added), then the side-length of the largest such square is equal to $L$($F$, $G$).

I was wondering how to picture "between the graphs of $F$ and $G$ one inscribes squares with sides parallel to the coordinate axes (at points of discontinuity of a graph vertical segments are added)"?
Why "the side-length of the largest such square is equal to $L$($F$, $G$)"?
How is the Levy metric a special case of the Lévy–Prokhorov metric?
Thanks and regards!
 A: For the picture, (first assuming that both $F$ and $G$ are continuous) for every $x$, I would start out from the point $P_G(x):=(x,G(x))$ on the graph of $G$ and draw both lines from that with tangents $\pm45^\circ$. Where these lines meet the graph of $F$ will determine the square on the picture around $P_G(x)$ which meets the graph of $F$ on both left and right side.
With this property we roll these squares around $P_G(x)$ for each $x$ and now have to take the supremum of their side lengths, because that will be the infimum of the $\varepsilon$'s that satisfy the criterium for all $x$.

For the other question, a distribution function $F:\Bbb R\to [0,1]$ determines a Borel measure $\mu_F$ on $\Bbb R$, defined as
$$\mu_F((a,b]):=F(b)-F(a) \ . $$
A: *

*In the domain of $x$ where $G(x)\geqslant F(x)$, it is obviuous that $G(x)\geqslant F(x-\varepsilon)-\varepsilon$ for any $\varepsilon>0$.
See following figure for the inscribes square with side length $r$ bwtween $F$ and $G$. The square touches the point $(x_0,G(x_0))$ and
$
G(x_0)=F(x_0+r)+r.
$
For any $\tilde{r}>r$, we have
$$G(x_0)< F(x_0+\tilde{r})+\tilde{r}.\tag{*}$$


*

*In the domain of $x$ where $F(x)\geqslant G(x)$, it is obviuous that $G(x)\geqslant F(x+\varepsilon)+\varepsilon$ for any $\varepsilon>0$.
See following figure for the inscribes square bwtween $F$ and $G$. The square touches the point $(x_0,G(x_0))$ and
$
G(x_0)=F(x_0-r)-r.
$
For any $\tilde{r}>r$, we have
$$G(x_0)> F(x_0+\tilde{r})+\tilde{r}.\tag{**}$$


*

*Equations (*) and (**) explain why we take $\inf_{\varepsilon>0}$ in the definition of Levy metric.


*There are infinitely many such inscribes squares, and the side-length of the largest such square is equal to the Levy's metric.
