Relationship to weak toplogy (Lévy metric) By $P(\Omega)$, denote the space of all probability measures on $(\mathbb{R},\mathcal{B})$. Let $F_{\mu}$ denote the distribution function of $\mu\in P(\Omega)$. Let, 
$$
d_L(\mu,\nu):=\inf\left\{\varepsilon\geq 0: F_{\mu}(x-\varepsilon)-\varepsilon\leq F_{\nu}(x)\leq F_{\mu}(x+\varepsilon)+\varepsilon~\forall x\in\mathbb{R}\right\}
$$
This is the so-called Lévy metric.


Show that $\lim_{n\to\infty}F_{\mu_n}(x)=F_{\mu}(x)$ for all points $x\in\mathbb{R}$ in which $F$ is continuous $\Leftrightarrow d_L(\mu_n,\mu)\to 0$


Its a bit confusing to me, did not find a start.
 A: "$\Leftarrow$": Suppose that $d_L(\mu_n,\mu) \to 0$ as $n \to \infty$. Fix a sequence $(\varrho_n)_{n \in \mathbb{N}}$ such that $\varrho_n \to 0$ as $n \to \infty$ and $$d_L(\mu_n,\mu) = d_L(\mu,\mu_n) < \varrho_n.$$
By the very definition of the Lévy metric, there exists $\epsilon_n \in [0,\varrho_n]$ such that
$$F_{\mu}(x-\epsilon_n)-\epsilon_n \leq F_{\mu_n}(x) \leq F_{\mu}(x+\epsilon_n) + \epsilon_n, \qquad x \in \mathbb{R}. \tag{1} $$
Note that $\epsilon_n \to 0$ as $n \to \infty$. Now if $x$ is a continuity point of $F_{\mu}$, then
$$F_{\mu}(x-\epsilon_n)-\epsilon_n \to F_{\mu}(x) \quad \text{and} \quad F_{\mu}(x+\epsilon_n)+\epsilon_n \to F_{\mu}(x),$$
and therefore it follows from $(1)$ that $F_{\mu_n}(x) \to F_{\mu}(x)$ as $n \to \infty$.
"$\Rightarrow$": Suppose that $F_{\mu_n}(x) \to F_{\mu}(x)$ for all continuity points of $F_{\mu}$. Fix $\epsilon>0$. Since $F_{\mu}$ is monotone and bounded, it can have at most countably many jumps, and so it is continuous up to a countable set $D$ of discontinuity points. In particular, we can find continuity points $x_1 < \ldots < x_k$ such that $x_{i+1}-x_i < \epsilon$ for all $i$ and
$$F_{\mu}(x_1) < \frac{\epsilon}{2} \qquad \text{and} \qquad F_{\mu}(x_k)>1-\frac{\epsilon}{2}. \tag{2}$$
Choose $N \in \mathbb{N}$ sufficiently large such that
$$|F_{\mu_n}(x_i)-F_{\mu}(x_i)| \leq \frac{\epsilon}{2} \qquad \text{for all $n \geq N$, $i=1,\ldots,k$}. \tag{3}$$
For any $x \in [x_{i-1},x_i]$, we have by the monotonicity of $F_{\mu}$, $F_{\mu_n}$ and $(3)$
$$F_{\mu_n}(x) \leq F_{\mu_n}(x_i) < F_{\mu_n}(x_i)+ \frac{\epsilon}{2} + F_{\mu}(x_i)-F_{\mu}(x_i) \stackrel{(3)}{\leq} F_{\mu}(x_i)+\epsilon \leq F_{\mu}(x+\epsilon)+\epsilon$$
for all $n \geq N$. For $x<x_1$ it holds that
$$F_{\mu_n}(x) \leq F_{\mu_n}(x_1)-F_{\mu}(x_1)+F_{\mu}(x_1) \stackrel{(2),(3)}{<} \frac{\epsilon}{2} + \frac{\epsilon}{2} \leq F_{\mu}(x+\epsilon)+\epsilon$$
and for $x>x_k$
$$F_{\mu_n}(x) \leq 1 \stackrel{(2)}{\leq} F_{\mu}(x+\epsilon)+\epsilon.$$
This proves $F_{\mu_n}(x) \leq F_{\mu}(x+\epsilon)+\epsilon$ for all $x \in \mathbb{R}$ and $n \geq N$. A very similar reasoning shows
$$F_{\mu}(x-\epsilon)-\epsilon \leq F_{\mu_n}(x),$$
and since $\epsilon>0$ is arbitrary, this finishes the proof.
