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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.

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  • $\begingroup$ 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 $\endgroup$
    – Nick Alger
    May 6 '16 at 16:19
  • $\begingroup$ @NickAlger Is this really the same distance? $\endgroup$
    – Rhjg
    May 6 '16 at 16:22
  • $\begingroup$ @NickAlger This is the en.wikipedia.org/wiki/L%C3%A9vy_metric $\endgroup$
    – Rhjg
    May 6 '16 at 16:23
  • $\begingroup$ 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 $\endgroup$
    – Nick Alger
    May 6 '16 at 16:35
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"$\Leftarrow$": 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.

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  • $\begingroup$ 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 $\endgroup$
    – manifolded
    Mar 13 '19 at 11:08
  • $\begingroup$ @manifolded I changed "$\leq \epsilon$" to "$\leq \epsilon/2$" in (3); this should fix your problem $\endgroup$
    – saz
    Mar 13 '19 at 11:18
  • $\begingroup$ Does $F_{\mu}(x-\epsilon)-\epsilon \leq F_{\mu_n}(x),$ automatically for follow from $F_{\mu_n}(x) \leq F_{\mu}(x+\epsilon)+\epsilon$ by substituting $x = x-\epsilon$? $\endgroup$ Feb 15 '20 at 17:17
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    $\begingroup$ @kasa Well, no, I don't think so. If you substitute it, then you get $F_{\mu_n}(x-\epsilon) \leq F_{\mu}(x)+\epsilon$ $\endgroup$
    – saz
    Feb 15 '20 at 18:28
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    $\begingroup$ @Kolmogorov There is a countable set $D \subseteq \mathbb{R}$ such that $F_{\mu}$ is continuous at every $x \in \mathbb{R} \setminus D$. $\endgroup$
    – saz
    Nov 27 '20 at 6:18

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