$P_n \Rightarrow P$ if and only if $P_n\{x\} \to P\{x\}$ for all $x$ if $S$ is a countable discrete space.

Question

This is problem 2.3 from Billingsley's Convergence of Probability Measures.

If $S$ is countable and discrete, then $P_n \Rightarrow P$ if and only if $P_n\{x\} \to P\{x\}$ for each singleton. Show that in this case $\sup_{A \in \mathscr{S}} |P_n A - PA| \to 0$.

I proved the only if direction, but I am stuck on the if direction. How can we show that $P_n$ converges weakly to $P$ from the convergence on each singleton? And finally how do we show $\sup_{A \in \mathscr{S}} |P_n A - PA| \to 0$ from this?

Answer 1

It suffices that $\int f \textrm{d} P_n\to \int f\textrm{d} P$ for every continuous $f$ on $S$ with compact support.

In this case, of course, compact support means finite support, so assume that $f$ is supported on $s_1,...s_N$ and note that $\int f\textrm{d}P_n=\sum_{j=1}^N f(s_j) P_n(\{s_j\})\to \sum_{j=1}^N f(s_j) P(\{s_j\})=\int f\textrm{d} P.$

As for the second part, you can use that weak convergence implies tightness of the family $\{P_n\}_{n\in\mathbb{N}}\cup \{P\}$ to conclude that for every $\varepsilon$ there exists a compact set $K_{\varepsilon}\subseteq S$ such that $P_n(S\setminus K_{\varepsilon})\leq \varepsilon$ for every $n$ and also $P(S\setminus K_{\varepsilon})\leq \varepsilon$. This implies that $|P(A)-P_n(A)|\leq 2\varepsilon$ for every $A\subseteq S\setminus K_{\varepsilon}$, but $K_{\varepsilon}$ is finite, so for general $A$, we have $$ |P_n(A)-P(A)|\leq |P_n(A\setminus K_{\varepsilon})-P(A\setminus K_{\varepsilon})|+|P_n(A\cap K_{\varepsilon})-P(A\cap K_{\varepsilon})| $$ $$ \leq 2\varepsilon +\sum_{s\in K_{\varepsilon}} |P_n(\{s\})-P(\{s\})|, $$ and the latter sum tends to $0$, implying that for $n$ sufficiently large, $|P_n(A)-P(A)|\leq 3\varepsilon$. As $\varepsilon$ was arbitrary, the desired follows.

Answer 2

For the first part you can use the fact that every open $G$ set is the countable union of finite sets. Actually, every set. If $G = \{x_1,x_2,\dots\}$ we set $G_k= \cup^k_{i=1}\{x_i\}$ and $G_k \uparrow G$ then $\lim_kP(G_k) \to P(G)$. So for every $\epsilon>0$ exists $G_k$ such that $P(G-G_k)<\epsilon$ and $$P(G) < \epsilon + P(G_k) = \epsilon + \liminf_n P_n(G_k) \leq \epsilon + \liminf P_n(G).$$ Since $\epsilon$ was arbitrary we can conclude with Portmateu theorem.