This is a duplicate of this question that has not got an answer. I am going to try to improve my question that is probably missworded since I do not believe it to be difficult, even though I can't seem to solve it.
The definition of random variable I am using is:
Given a probability triple $(\Omega, F, P) $, a random variable is a function $X$ from $\Omega$ to the real numbers $R$, such that:
$$\{ w \in \Omega : X(\omega) \le x \} \in F, x \in R$$
Take $Z_1, Z_2 ... $ random variables such that the $\lim_{n \to \infty}Z_n(w)$ exists for all $w$ and $Z(w)=\lim_{n \to \infty}Z_n(w)$.
And suppose $w \in \{Z \le z \}$, for $z \in R$, i.e., $w \in Z^{-1}((-\infty, z ])$.
Then, and this is what I am trying to prove, for every $m \in N$, there exists an $n \in N$ s.t. for all $k \ge n$ we have $Z_k(w) \le z + \frac{1}{m}$, $k \in N$. Why is this?
I know that by the definition of convergence of a sequence we have that for every $m > 0$ there exists an $k \in N$ s.t. $|Z_n(w) - Z(w)| < m$ whenever $n \ge k $. If I could drop the absolute value I would be very close to finding what I want but there is no guarantee that $|Z_n(w) - Z(w)|$ will be positive, and how would I fit $z$ in the inequality?
Since $\lim\limits_{n\to\infty}Z_n(\omega){}={}Z(\omega)$ then, by definition, for $\dfrac{1}{m}{}>{}0$ there exists $n(m)$ such that, for all $n{}>{}n(m)$, we have $$ \left|Z_n(\omega){}-{}Z(\omega)\right|{}\leq{}\dfrac{1}{m}{}\iff{}-\dfrac{1}{m}\leq Z_n(\omega){}-{}Z(\omega) {}\leq{}\dfrac{1}{m}\,. $$
In particular, since $Z(\omega){}\leq{}z$, use the last inequality on the right to see that
$$ Z_n(\omega){}\leq{}Z(\omega){}+{}\dfrac{1}{m}{}\leq{}z{}+{}\dfrac{1}{m}\,. $$