How to prove that the sequence of random variables converges to a random variable? If $Z_1,Z_2,\cdots,Z_n$ are random variables such that $Z(\omega)=\lim_{n \to \infty}Z_n(\omega)$ exists $\forall \omega \in \Omega$, then $Z$ is also a random variable.
I was reading a book on probability which provided a very terse proof which I cannot follow. Here is what the author writes. 
$$\{Z\leq x\}=\bigcap_{m \geq1} \bigcup_{n \geq 1} \bigcap_{k\geq n}\left \{Z_k \leq x+ \frac{1}{m}\right\}$$
So the author claims that as $Z_k$ is measurable, so is $Z$ as the set $\{Z\leq x\}$ can be written as countable union/intersection of measurable sets.I have spent considerable time trying to figure out why the two sets a equal.
Can someone give me a hint as to how should I go about proving it? It turns out that this "clue" or "solution" is incomprehensible to me.
Thanks.
 A: What this is saying is that the limit $Z(\omega)$ is at most $x$ is the same as, for all $\epsilon>0$, the sequence $Z_n(\omega)$ being eventually less than $x+\epsilon$. They used $\frac1m$ instead of $\epsilon$ to allow for a countable union.
Here's how you can prove it rigorously. Suppose $Z(\omega)\le x$. Then for all $\epsilon>0$ (in particular, for all $1/m$, $m\in\mathbb{N}$), there is an number $n$ so $k\ge n$ implies $|Z_k(\omega)-Z(\omega)|<\frac1m$, implying $Z_k(\omega)\le Z(\omega)+\frac1m\le x+\frac1m$. This shows $\omega\in\bigcap_m\bigcup_n\bigcap_{k\ge n}\{Z_k\ge x+\frac1m\}$. 
Now, suppose $\omega$ is in the RHS. Assume (by way of contradiction) that $x<Z(\omega)$. Choose $m$ large enough so $x<x+\frac1m<Z(\omega)$. Since $Z_n(\omega)\to Z(\omega)$, for large enough $n$, $Z_n(\omega)$ is eventually greater than $x+\frac1m$. But this is impossible, since $\omega$ being in RHS implies $Z_n(\omega)$ is eventually $less$ than $x+\frac1m$ for large enough $n$. Thus, we must have $Z(\omega)\le x$, proving the other inclusion.
