# 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.

• The wikipedia page on this distance (Lévy–Prokhorov distance) says this is equivalent to weak convergence of measures (congergence in the weak-* topology) if the space is separable – Nick Alger May 6 '16 at 16:19
• @NickAlger Is this really the same distance? – Rhjg May 6 '16 at 16:22
• @NickAlger This is the en.wikipedia.org/wiki/L%C3%A9vy_metric – Rhjg May 6 '16 at 16:23
• I'm not an expert so there could be a subtlety I'm missing, but it looks the same and the linked article says it is the same – Nick Alger May 6 '16 at 16:35

"$$\Leftarrow$$": 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 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.

• Hi, in "$\implies$" direction, for $x<x_1$, $F_\mu (x_1)<\frac{\epsilon}{2}$ but why is $F_{\mu_{n}} (x_1)- F_\mu (x_1)<\frac{\epsilon}{2}$ because from (3), $F_{\mu_{n}} (x_1)- F_\mu (x_1)<\epsilon$? Thanks – manifolded Mar 13 '19 at 11:08
• @manifolded I changed "$\leq \epsilon$" to "$\leq \epsilon/2$" in (3); this should fix your problem – saz Mar 13 '19 at 11:18