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The following exercise is taken from Fremlin book: measure theory.

For simplicity, we only consider probability measures. Vague convergence is defined in term of bounded and continuous functions which is equivalent to the convergence of distribution functions at any point of continuity.

I succeeded in proving that $\rho$ is a metric.

If $(P_k)_k,P$ are probability measures such that $\rho(P_k,P) \to_{k \to \infty}0,$ and fixing a continuity point $u \in \mathbb{R}^r,$ how to prove that $\lim_kP_k(]-\infty,u])=P(]-\infty,u])$ ?

Also, if $P_k$ converges vaguley to $P,$ how to prove $\rho(P_k,P) \to_{k \to \infty} 0$?

To be noted that "Cf. $274Yc.$" corresponds to $r=1,$ which was asked once on stack exchange.

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  • $\begingroup$ Have you heard of the Portmanteau theorem? As for your last question, it is not difficult to prove that if $P_n$ converges vaguely to $P$, and $P_n$, $P$ are probability measures, then $P_n$ converges weakly to $P$. The weak topology restricted to probability measures is metrizable by $\rho$. $\endgroup$ Jun 30 at 16:07
  • $\begingroup$ I know this theorem. Can you elaborate the last fact of yours "The weak topology restricted to probability measures is metrizable by $\rho$" (how to prove that $\rho(P_k,P) \to _{k \to \infty} 0$ using vague convergence, instead of weak convergence) $\endgroup$
    – john
    Jun 30 at 22:30
  • $\begingroup$ How about a reference: Dudley, R. Real Analysis and Probability, section. 11.3. $\endgroup$ Jun 30 at 22:36
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Let $F(u) = P( ( -\infty,u]),$ and $F_k(u) = P_k( ( -\infty,u])$, and suppose $u$ is a continuity point of $F$. Let $\epsilon>0$, and by the continuity of $F$ choose $\delta$ so that $\delta \le \epsilon$, and for all $0 < x < \delta$, $F(u+x) - F(u-x) < \epsilon$. Since $\rho_k := \rho(P_k,P) \to 0$, choose $N_0$ large enough so that for all $k \ge N_0$, $\rho_k < \delta$. By the infimum definition of $\rho_k$, there must exist a sequence $a_k >0$ so that $\rho_k < a_k < \delta$, and $$F(u-a_k) \le F_k(u ) + a_k \le F(u+a_k) + 2a_k$$.

Subtract $F(u)$ to get that

$$F(u-a_k) - F(u) \le F_k(u )- F(u) + a_k \le F(u+a_k)- F(u) + 2a_k.$$

It follows then from a little algebra that for all $k \ge N_0$,

$$|F_k(u )- F(u)| \le \max\{ |F(u-a_k) - F(u)| + a_k , |F(u+a_k) - F(u)| + a_k \} < 2 \epsilon$$ Hence $F_k(u) \to F(u)$. The idea is the same in $\mathbb{R}^d$, since the same monotonicity of $F$ holds: $F(u+x{\bf1}) - F(u-x{\bf1}) < \epsilon$ $\implies$ $F(u+x{\bf1}) - F(u ) < \epsilon,$ $F(u ) - F(u-x{\bf1}) < \epsilon$.

After I wrote this, I found another stackexchange article that goes into the details for general metric space: Lévy's metric on $\mathbb{R}^d$

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  • $\begingroup$ Is there an argumrent such that math.stackexchange.com/questions/1774319/…, to consider a discussion on the position of $x$? $\endgroup$
    – john
    Jun 23 at 0:44
  • $\begingroup$ I think that is only relevant in proving the other direction of the "if and only if" statement. The important point here is that if $u$ is a continuity point of $F$, then $F$ is monotone in a neighborhood of $u$. $\endgroup$ Jun 23 at 17:18
  • $\begingroup$ The question is updated to reflect the problem, how to prove the other implication. $\endgroup$
    – john
    Jun 23 at 17:32
  • $\begingroup$ perhaps there is another way other then the following : math.stackexchange.com/questions/3606510/…, to be noted that this exercise is question $285Y$ d) after the characteristic functions section, we already know the proof on the line, we can conclude the case of $\mathbb{R}^r,$ by considering the scalar product, in other word the pushforward measure by $f_y(x)=\sum_{k=1}^r x_ky_k,x,y \in \mathbb{R}^r$ (Lemma 285P and 285Y e) are very close to the problem) $\endgroup$
    – john
    Jun 23 at 17:54
  • $\begingroup$ To my knowledge there is no alternative proof, although the proof is not really more complicated in $\mathbb{R}^n$ than in $\mathbb{R}$, since the CDF's $F$ and $F_k$ must each have only countably many discontinuity points (each discontinuity may be associated with a box in $\mathbb{R}^n$, which must contain a rational point). $\endgroup$ Jun 23 at 17:55

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