Show that the Levy distance is a metric The Levy metric is defined on the space of cumulative distribution functions as $$\rho(F,G):=\inf\{\varepsilon>0\mid F(x-\varepsilon)-\varepsilon\leq G(x)\leq F(x+\varepsilon)+\varepsilon,\quad \forall x\in \mathbb R\},$$where $F$ and $G$ are cumulative distribution functions. I would like to show that it is a metric space.
I think I have shown that $\rho(F,G)=\rho(G,F)$ but I also need to show

*

*$\rho(F,G)=0\iff F=G$

*$\rho(F,H)\leq \rho(F,G)+\rho (G,H)$
For the first one I first managed to show it when I supposed that $F$ and $G$ are continuous but this hypothesis is obviously not required.
For the second one I don't know...
Thank you.
 A: $\textbf{1}$ . $\rho(F,G)\ge0$ by definition .
$\textbf{2}$ . To show Symmetric ,
$\rho(F,G)=\epsilon_1$ and $\rho(G,F)=\epsilon_2$
So, $F(x-\epsilon_1)-\epsilon_1 \le G(x) \le F(x+\epsilon_1)+\epsilon_1 ,$ $\forall x\in \mathbb{R}$
Set x-$\epsilon_1$=y ,  So , $F(y)\le G(y+\epsilon_1)+\epsilon_1 ,\forall y \in \mathbb{R}$
and for x+$\epsilon_1$=z ,  $G(z-\epsilon_1)-\epsilon_1 \le F(z) , \forall z \in \mathbb{R}$
Joining this , we get  $G(x-\epsilon_1)-\epsilon_1 \le F(x) \le G(x+\epsilon_1)+\epsilon_1 ,$ $\forall x\in \mathbb{R}$
So , $\epsilon_1 \in \{\epsilon>0 :G(x-\epsilon)-\epsilon \le F(x) \le G(x+\epsilon)+\epsilon ,$ $\forall x\in \mathbb{R} \}$
But , $\epsilon_2$ is in the infrimum of this set . So, $\epsilon_1 \ge \epsilon_2$
Conversely , we also get $\epsilon_2 \ge \epsilon_1$ in the similar way with reversing F and G .
So, $\epsilon_1=\epsilon_2$ and hence $\rho(F,G)=\rho(G,F)$
Hence , $\rho$ is a symmetric .
$\textbf{3}$ . $\rho(F,G)=0$ .To show ,F $\equiv$G ,
As $\rho(F,G)=0$, $\exists \{\epsilon_n\} , \lim\epsilon_n\to0$ such that ,
$\epsilon_n \in  \{\epsilon>0 :F(x-\epsilon)-\epsilon \le G(x) \le F(x+\epsilon)+\epsilon ,$ $\forall x\in \mathbb{R} \},\forall n \in \mathbb{N}$
Hence , $\lim [F(x - \epsilon_n ) - \epsilon_n] \le G(x) \le \lim [F(x+\epsilon_n) + \epsilon_n]$ .
Limits can be broken as $\lim \epsilon_n = 0$ and  $\lim F(x - \epsilon_n)$ is bounded by 1 .
So , $\lim F(x - \epsilon_n ) - \lim\epsilon_n \le G(x) \le \lim F(x+\epsilon_n) + \lim\epsilon_n$
$\Rightarrow \lim F(x - \epsilon_n ) \le G(x) \le  F(x) $ [ As F is right continuous]
Similarly , $\rho(G,F)=0$ ,so we get
$\Rightarrow \lim G(x - \epsilon_n ) \le F(x) \le  G(x) $ [ As G is right continuous]
Combining , $F(x) \le  G(x)$ and $G(x) \le  F(x)$ , we get F(x)=G(x) $\forall x \in \mathbb{R}$
$\textbf{4}$ .   To Show , $\rho(F,H)\leq \rho(F,G)+\rho (G,H)$
Set $\rho(F,G)=\epsilon_1 , \rho (G,H)=\epsilon_2 , \rho(F,H)=\epsilon_3 $
$F(x-\epsilon_1-\epsilon_2)-\epsilon_1 \le G(x-\epsilon_2)$
As, $\rho(F,G)= \epsilon_1$ and here we have taken x=x-$\epsilon_2$ as it is true for all real numbers .
So , $F(x-\epsilon_1-\epsilon_2)-\epsilon_1-\epsilon_2 \le G(x-\epsilon_2)-\epsilon_2$
Similarly ,we also get  $ G(x+\epsilon_2)+\epsilon_2 \le F(x+\epsilon_1+\epsilon_2)+\epsilon_1 +\epsilon_2$
And as $  \rho (G,H)=\epsilon_2 $ ,
So , $G(x-\epsilon_2)- \epsilon_2 \le H(x) \le G(x + \epsilon_2) + \epsilon_2$
Hence , $F(x-\epsilon_1-\epsilon_2)-\epsilon_1-\epsilon_2 \le G(x-\epsilon_2)-\epsilon_2 \le H(x) \le G(x + \epsilon_2) + \epsilon_2 \le F(x+\epsilon_1+\epsilon_2)+\epsilon_1 +\epsilon_2$
So , $(\epsilon_1+\epsilon_2) \in \{\epsilon>0 :F(x-\epsilon)-\epsilon \le H(x) \le F(x+\epsilon)+\epsilon ,$ $\forall x\in \mathbb{R} \} $
But $\epsilon_3$ is infrimum of the set . So , $\epsilon_1+\epsilon_2 \ge \epsilon_3$
Hence , $\rho(F,H)\leq \rho(F,G)+\rho (G,H)$
So, as all 4 the properties of metric space are satisfied , hence this Levy-Metric Space is indeed a metric space .   [Q.E.D.]
