limit of probability Let $X_1,X_2,...Y_1,Y_2,...,Z_1,Z_2,...$ be random variables and $a,b \in \mathbb{R}$.
Assume that $X_n \leq Y_n \leq Z_n$ for any $n\geq 1$. 
Assume that $P($ lim$_{n\rightarrow\infty}X_n = a)=1$ and  $P($ lim$_{n\rightarrow\infty}Z_n = a)=1$.
Prove that $P($ lim$_{n\rightarrow\infty}Y_n = a)=1$
I know if without probability, this claim just follows the squeeze lemma. However, I'm not very clear how to prove this with probability.
Thanks in advance. 
 A: Hint: first convince yourself that $\{\omega\colon\lim_{n}X_{n}(\omega)=a\}$ is measurable whenever $X_{n}$ is measurable for each $n$ since 
$$
\bigcup_{r\in(0,\infty)\cap\mathbb{Q}}\,\bigcap_{N \in \{1,2,\ldots\}}\,\bigcup_{n\in \{N,N+1,\ldots\}}\left\{ \omega\colon\left|X_{n}(\omega)-a\right|\geq r\right\} 
$$
is measurable. The statements
$$
P(\lim_{n}X_{n}=a)=P(\lim_{n}Z_{n}=a)=1
$$
are equivalent to the existence of a measurable set $\Xi$ such that $$\lim_{n}X_{n}(\omega)=\lim_{n}Z_{n}(\omega)=a \text{ for each } \omega \in \Xi.$$ Can you finish the argument by applying the squeeze theorem?

Hint (w/o measure theory): without measure theoretic probability, we can still establish the result for the case in which our sample space $\Omega$ is finite.
Remark: You can think of a sample $\omega \in \Omega$ as a realization of the universe. For example, if you were rolling a six-sided die, you may take your realizations to be $\Omega \equiv \{1,2,\ldots,6\}$. A random variable in this context is simply a function $X$ from $\Omega$ to real numbers $\mathbb{R}$. For example, if you bet a dollar that the die is even, your outcome in this game is described by $X(2j+1)=-1$ (lose a dollar) and $X(2j)=1$ (win a dollar).
Remark: without loss of generality, we may assume that each sample $\omega$ has positive probability (i.e., $P(\{\omega\})>0$). If a sample has zero probability, we can simply remove it from our sample space.
In this setting, the statements
$$
P(\lim_{n}X_{n}=a)=P(\lim_{n}Z_{n}=a)=1
$$
are equivalent to saying that $$\lim_{n}X_{n}(\omega)=\lim_{n}Z_{n}(\omega)=a \text{ for each } \omega \in \Omega.$$ Can you finish the argument by applying the squeeze theorem?
A: Define events $A=\{ X_n \to a\},B=\{ Y_n \to a \},C=\{ Z_n \to a \}$. Try to show that $A \cap C \subseteq B$ (this essentially follows from the squeeze theorem as you mentioned) and that $P(A \cap C)=1$, so that $P(B)$ must also be $1$.
There are some measure-theoretic details swept under the rug here, having to do with the fact that $A,B,C$ are in fact well-defined events. But these details also go through.
