Does a sequence of pairwise disjoint event converge to the empy set, $\emptyset$? Let $(A_n)$ be any sequence of pairwise disjoint events, is
$$
\lim_{n \to \infty} A_n = \emptyset.
$$
always true?
Conceptually I would say yes, since if we partition $\Omega$ an infinite amount of times, I guess that the events would get smaller and smaller.

One can say that a sequence of sets $(A_n)_{n\geqslant 1}$ converges to $A$ if
$\limsup_{n\to\infty}A_n=\liminf_{n\to\infty}A_n=A$, where
$$
\limsup_{n\to\infty}A_n=\bigcap_{N\geqslant 1}\bigcup_{n\geqslant N}A_n
$$
and $$
\liminf_{n\to\infty}A_n=\bigcup_{N\geqslant 1}\bigcap_{n\geqslant N}A_n.
$$
One can show that $\omega$ belongs to $\limsup_{n\to\infty}A_n$ if and only if the set $\{n\in\mathbb N, \omega\in A_n\}$ is infinite and that $\omega$ belongs to $\liminf_{n\to\infty}A_n$ if and only if there exists an integer $N$ such that $\omega\in A_n$ for all $n\geqslant N$.
The inclusion $\liminf_{n\to\infty}A_n\subset \limsup_{n\to\infty}A_n$ always hold. In the particular case where $(A_n)_{n\geqslant 1}$ is pairwise disjoint, $\limsup_{n\to\infty}A_n$ is empty, because no $\omega$ belongs to more than one $A_n$.
