$X_n\rightarrow_{a.s.}Y $ implies $\frac{Z}{X_n}\rightarrow_{a.s.}\frac{Z}{Y} $

Question

Consider a sequence of real-valued random variables $\{X_n\}_{n\in \mathbb{N}}$. Consider two other real-valued random variables $Y,Z$.

Suppose that $$ X_n\rightarrow_{a.s.}Y \text{ }\text{ as $n\rightarrow \infty$} $$

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Is it true that $$ \frac{Z}{X_n}\rightarrow_{a.s.}\frac{Z}{Y} \text{ }\text{ as $n\rightarrow \infty$} $$

Could you help me to prove it? Is this an application of continuous mapping theorem?

Answer 1

If there is a set $A$ with $P(A)=1$ such that all $X_n$ and $Y$ are $\neq 0$ on $A$, then it is true and can be proved directly. Otherwise it cannot be true because the fractions would not be even defined a.s.

Answer 2

Let $\omega \in \{ \lim X_n(\omega) = Y(\omega)\}$. Then

$$\lim_n \frac{Z(\omega)}{X_n(\omega)}= \frac{\lim_n Z(\omega)}{\lim_n X_n(\omega)}=\frac{Z(\omega)}{Y(\omega)} $$

Thus, $$\{ \lim X_n(\omega) = Y(\omega)\} \subseteq \{ \lim_n \frac{Z(\omega)}{X_n(\omega)}= \frac{\lim_n Z(\omega)}{\lim_n X_n(\omega)}=\frac{Z(\omega)}{Y(\omega)} \}$$

$$\to 1 = P(\{ \lim X_n(\omega) = Y(\omega)\}) \le P(\{ \lim_n \frac{Z(\omega)}{X_n(\omega)}= \frac{\lim_n Z(\omega)}{\lim_n X_n(\omega)}=\frac{Z(\omega)}{Y(\omega)} \})$$

QED