Show $\lim\limits_{m\to\infty}P(n\leq m)=1$ for some function $n:\Omega\to\mathbb{N}$ Suppose that $(\Omega,\mathcal{F},P)$ is a probability triplet and $n:\Omega\to\mathbb{N}$ is some measurable function (in particular, $n(\omega)$ is finite for each $\omega\in\Omega$). I'm unsuccessfully trying to formalize the following claim in my reading
$$
\lim_{m\to\infty}P(E_m)=1\quad\text{with}\quad E_m\equiv\{\omega:n(\omega)\leq m\}.
$$
I can only see the intuition: with $m$ large enough, there are so many $\omega$ with $n(\omega)\leq m$ that $E_m$ is very similar in size to $\Omega$. But can you please help me with a formal argument?
 A: If $(\Omega, \mathcal{F},P)$ is any probability triple, then for any nested sequence $E_1 \subset E_2 \subset E_3...$ of sets in $\mathcal{F}$, we always have that $\lim_{m \to \infty} P(E_m) = P \big(\bigcup_{m=1}^{\infty} E_m \big)$. This can be deduced directly from the probability axioms (try to prove it if you haven't seen it before).
Note that for your problem, the $E_m$ are always measurable because $E_m = n^{-1}(\{1,...,m\})$, i.e, $E_m$ is the preimage of a (discrete) measurable subset of $\mathbb{N}$.  Moreover $E_m \subset E_{m+1}$ for all $m \in \mathbb{N}$, because if $n(\omega) \leq m$, then $n(\omega) \leq m+1$.
Also notice that $\bigcup_{m=1}^{\infty}E_m = \Omega$, because $\omega \in E_{n(\omega)}$ for any $\omega \in \Omega$.
Therefore $\lim_{m \to \infty} P(E_m) = P(\bigcup_m E_m) = P(\Omega)=1$.
A: Let $\Delta_n = \{ \omega | n(\omega) \in (n-1, n] \}$. Clearly the $\Delta_n$ are disjoint and $\Omega = \cup_n \Delta_n$, and so
$1 = P(\Omega) = \sum_n P(\Delta_n)$.
We have $E_m = \cup_{n \le m} \Delta_n$, and so
$P(E_m) = \sum_{n \le m} P(\Delta_n)$. It follows from above that
$\lim_{m \to \infty} P(E_m) = \sum_n P(\Delta_n) = 1$.
