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?

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
$\lim_{m \to \infty} P(E_m) = \sum_n P(\Delta_n) = 1$.
