Does squeeze theorem apply for almost sure convergence

Question

Suppose we are given that $X'_n \leq Z_n \leq X_n$ for random variables $X'_n,Z_n, X_n$. If we are told that $X_n,X'_n$ converges almost surely to some random variable $Y$, can we conclude that $Z_n \stackrel{a.s}{\to}Y$?

The closest variant of this question that I have seen is Squeeze theorem for convergence in distribution. I've considered $\{\omega: \lim_{n \to \infty}Z_n(\omega)=Y(\omega)\}$, and realized that this set is a superset of the intersection of $\{X_n(\omega)=Y\}$ and $\{X'_n(\omega)=Y\}$ (due to the inequality), but that doesn't allow us to conclude that $Pr(\{\omega: \lim_{n \to \infty}Z_n(\omega)=Y(\omega)\})=1$ unfortunately.

Answer 1

As you said, the event $\{Z_n \to Y\}$ is a superset of $\{X_n \to Y\} \cap \{X'_n \to Y\} \cap \{X'_n \le Z_n \le X_n\}$. This intersection has probability $1$, since the complement of this intersection has probability $$P(\{X_n \to Y\}^c \cup \{X'_n \to Y\}^c \cup \{X'_n \le Z_n \le X_n\}^c) \le (1-P(X_n \to Y)) + (1-P(X'_n\to Y)) + (1-P(X'_n \le Z_n \le X_n)) = 0 + 0 + 0.$$

Thus $P(Z_n \to Y)=1$ as well.