How to show that the space of probability measures on $\mathbb{R}$ is separable under Lévy metric The Lévy metric between distribution functions $F$ and $G$ is given by:
$$\rho(F,G) = \inf\left\{\epsilon : F(x-\epsilon)-\epsilon\leq G(x)\leq F(x+\epsilon)+\epsilon\right\}.$$ 
Another way to write this is:  
$$\rho(F,G) = \inf\left\{\epsilon : F(x)\leq G(x+\epsilon)+\epsilon ; G(x)\leq F(x+\epsilon)+\epsilon\right\}.$$
I also know that $\rho(F_n,F)\rightarrow 0 \iff F_n\rightarrow F$ weakly.

The question is to show that the space of probability measures, say $P$, on $\mathbb{R}$ is separable with the Lévy metric.

In other words, have to show $\exists$  a countable dense subset of $P$. I was thinking about discretizing the distribution function of any given probability measure, along $x$, and estimate the value in an interval by $\sup\{q: q\leq F(x), q\in Q\}$ i.e. approximating $F(x)$ by a rational number. 
But I am not sure as to how use Lévy metric to show that in any $\epsilon$ neighborhood of $F$, $\exists F_Q$. Thanks and appreciate a hint!
 A: Since you already know that $\varrho$-convergence is equivalent to weak convergence, it suffices to find a countable family $Q$ of probability measures such that for any probability measure $\mathbb{P} \in P$ there exists a sequence $(\mathbb{P}_{\ell})_{\ell \in \mathbb{N}} \subseteq Q$ with $\mathbb{P}_{\ell} \to \mathbb{P}$ weakly. We will show that
$$Q := \left\{ \sum_{k=1}^K q_k \delta_{y_k}; y_k \in \mathbb{Q}, q_k \in \mathbb{Q}, K \in \mathbb{N}\right\}$$
does the job. For the proof I will use the fact that $\mathbb{P}_{\ell}$ converges weakly to $\mathbb{P}$ if and only if $\int f \, d\mathbb{P}_{\ell} \to \int f \, d\mathbb{P}$ for any uniformly continuous bounded function $f$.

Construction of the approximating sequence: Fix $\mathbb{P} \in P$. For $k \in \mathbb{Z}$ and $\ell \geq 1$ set
$$I_k^{\ell} := [k2^{-\ell}, (k+1) 2^{-\ell}) \quad \text{and} \quad a_k^{\ell} := \mathbb{P}(I_k^{\ell}).$$
Since $\mathbb{P}$ is a probability measure (hence a finite measure) we can choose $N_{\ell} \in \mathbb{N}$ sufficiently large such that
$$\mathbb{P} \left( K_{\ell}^c \right) \leq \frac{1}{\ell} \quad \text{for} \, \, K_{\ell} := \bigcup_{k=-N_{\ell}}^{N_{\ell}} I_{k}^{\ell}. \tag{1}$$
For $k \in \{-N_{\ell},\ldots,N_{\ell}\}$ we choose $q_k^{\ell} \in \mathbb{Q}$, such that $q_k^{\ell} \leq a_k^{\ell}$ and $$|q_k^{\ell}-a_k^{\ell}| \leq \frac{1}{2 \ell N_{\ell}} \tag{2}$$ and we set $$\mathbb{P}_{\ell} :=(1-r^{\ell}) \delta_0+ \sum_{k=-N_{\ell}}^{N_{\ell}} q_k^{\ell} \delta_{k 2^{-\ell}}$$ where $r^{\ell} := \sum_{k=-N_{\ell}}^{N_{\ell}} q_{k}^{\ell}$. Since $q_k^{\ell} \leq a_k^{\ell}$, we have $r^{\ell} \leq 1$, and it follows easily that each $\mathbb{P}_{\ell}$ is a probability measure.
Proof of the weak convergence: Fix $\epsilon>0$ and a uniformly continuous bounded function $f$. Since $f$ is uniformly continuous, we can choose $L \geq 1$ sufficiently large such that
$$|x-y| \leq 2^{-L} \implies |f(x)-f(y)| \leq \epsilon. \tag{3}$$
Using that
\begin{align*}\int f \, d\mathbb{P}_{\ell}- \int f \, d\mathbb{P} &= -\int_{K_{\ell}^c} f \, d\mathbb{P} + (1-r^{\ell}) f(0) - \sum_{k=-N_{\ell}}^{N_{\ell}} \int_{I_k^{\ell}} (f(x)-f(k2^{-\ell})) \, d\mathbb{P}(x) \\ &\quad + \sum_{k=-N_{\ell}}^{N_{\ell}} (q_k^{\ell}-a_k^{\ell}) f(k 2^{-\ell}) \end{align*}
it follows from the boundedness of $f$ and (3) that
$$\left| \int f \, d\mathbb{P}_{\ell}- \int f \, d\mathbb{P}\right| \leq \|f\|_{\infty} \left[ \mathbb{P}(K_{\ell}^c) + (1-r^{\ell}) \right] + \epsilon \underbrace{\mathbb{P}(K_{\ell})}_{\leq 1} + \|f\|_{\infty} \sum_{k=-N_{\ell}}^{N_{\ell}} |q_k^{\ell}-a_k^{\ell}|$$ for all $\ell \geq L$. Because of $(2)$ we have
$$\sum_{k=-N_{\ell}}^{N_{\ell}} |q_k^{\ell}-a_k^{\ell}| \leq \frac{1}{\ell},$$
and on the other hand it is not difficult to see that $r_{\ell} \to 1$ as $\ell \to \infty$, i.e. $|r^{\ell}-1| \leq \epsilon$ for $\ell$ sufficiently large. As, by construction, $\mathbb{P}(K^c_{\ell}) \leq 1/\ell$ we conclude that
$$\left| \int f \, d\mathbb{P}_{\ell}- \int f \, d\mathbb{P}\right| \leq \|f\|_{\infty} \left( \frac{2}{\ell}+\epsilon \right) + \epsilon$$
for all $\ell$ sufficiently large which clearly proves $\int f \, d\mathbb{P}_{\ell} \to \int f \, d\mathbb{P}$ as $\ell \to \infty$.
