Convergence in law of random variables with values in a countable subset

Question

I'm studying convergence in law and I'm wondering about a statement where we assume that a sequene of random variables have values in a countable subset of $\mathbb{R}$: Assume we have a sequence $Z_j$ of random variables which converges in law to a r.v. $Z$ and all of them have values in a countable subset $S \subset \mathbb{R}$. Can we say that this is equivalent to $$ P(Z_n = s) \to P(Z = s) \ \ \forall s \in S? $$

Answer 1

One implication is false. Let $S=\{0,1/2,1/3,.. ./\},X_n=1/n$ and $X=0$. Then we have almost sure convergence but the conclusion fails when $s=0$. The other implication is true. Let us write $x^{+}$ for $\max \{x,0\}$ and $x^{-}$ for $-\min \{x,0\}$. Suppose $P(Z_n=s) \to P(Z=s)$ for every $s \in S$. Then $\sum (P(Z=s)-P(Z_n=s))^{+} \to 0$ by DCT since $ 0 \leq (P(Z=s)-P(Z_n=s))^{+} \leq P(Z=s)$ and $\sum P(Z=s) <\infty$. Also $P(Z=s)-P(Z_n=s)=(P(Z=s)-P(Z_n=s))^{+}-(P(Z=s)-P(Z_n=s))^{-}$ so (using the hypothesis) we get $\sum (P(Z=s)-P(Z_n=s))^{-} \to 0$. It follows now that $\sum |P(Z=s)-P(Z_n=s)| \to 0$ from which it is clear that $P(Z_n \leq x)=\sum_{s \leq x} P(Z_n \leq x) \to \sum_{s \leq x} P(Z \leq x)=P(Z\leq x)$ for every $x$. Hence $Z_n \to Z$ in law.