Is $Z_n→Z$ in distribution and if $z_n→0$ then $z_nZ_n→0$ in distribution I'm trying to prove the statement made by Did in the comments:

Is $Z_n→Z$ in distribution and if $z_n→0$ then $z_nZ_n→0$ in distribution.

So we need to prove that $$\forall t>0: F_{z_nZ_n}(t)\to1$$ $$\forall t<0: F_{z_nZ_n}(t)\to0$$
Assuming $z_n >0$, for $t>0$:
$$F_{z_nZ_n}(t)= P \left(Z_n \leq \frac{t}z_n \right)$$
How do we make use of $Z_n \xrightarrow{\text{in dist.}} Z$ when we have $z_n$ on the RHS? And what about $F_Z$ being continuous? 
 A: Hint: First note that $|Z_n|$ converges in distribution to $|Z|$ (absolute value is continuous, and convergence in distribution is preserved when applying a continuous function).
Let $\epsilon>0$. Let $\eta>0$ be a continuity point of cdf $F_{|Z|}$. For large enough $n$ we have $|z_n|<\epsilon/\eta$. Then:
$$P(|z_nZ_n|>\epsilon) \leq P(|Z_n|>\eta) \to P(|Z|>\eta).$$
Taking $\eta$ large enough, $P(|Z|>\eta)$ can be made arbitrarily small. So, $P(|z_nZ_n|>\epsilon)\to 0$.
A: For any $\varepsilon>0$, choose $T$ large enough that $P(Z\le T)\ge 1-\varepsilon/2$.  Since $z_n\rightarrow 0$, you can choose $N$ large enough that $t/z_n > T$ for all $n\ge N$.  Moreover, since $Z_n\xrightarrow{\text{in dist.}} Z$, you can choose $M$ large enough that $P(Z_n \le T) \ge P(Z \le T)-\varepsilon/2 \ge 1-\varepsilon$ for all $n\ge M$.
Then $F_{z_n Z_n}(t) = P(Z_n \le t/z_n) \ge P(Z_n \le T)\ge 1-\varepsilon$ for all $n\ge \max\{M,N\}$.  Since $\varepsilon$ is arbitrary, you can conclude that $P(Z_n \le t/z_n)\rightarrow 1$ as $n\rightarrow \infty$.  (The proof for negative $t$ is analogous.)
