If we know that the sequence of events $A_n$ converges, what can we say about the infinite union and intersection? Suppose that $\lim_{n\to\infty} A_n = A$. It is intuitive to me for the infinite intersection, that it is equal to $\bigcap_{n\geq1}A_n = A$. However, I am failing to see if there does indeed exist anything for the infinite union. Is there a nice result for that?
 A: By $\lim_{n\to\infty}A_n=A$, I will assume you mean that the indicator functions ${\bf 1}_{A_n}$ converge pointwise to ${\bf 1}_A$. Then the statement
$$
\bigcap_{n\geq 1}A_n=A
$$
is not necessarily true: for example, you can make the intersection empty by setting $A_1$ to be the empty set and leaving all other terms in the limit unchanged (limits are unaffected by finitely many changes to the sequence).
A: From the definitions, we easily get
$$
\bigcap A_n \subset A \subset \bigcup A_n. 
$$
To see that the inclusions can be strict, note that modifying finitely many terms of the sequence $(A_n)_n$ does not change the limit $A$. 
Thus, by setting $A_1=\emptyset$, we can achieve $\bigcap A_n =\emptyset$ and by setting $A_2=\Omega$, we can achieve $\bigcup A_n =\Omega$, where $\Omega$ is our base set. 
A: $A \subseteq \bigcap_{n=1}^{\infty} A_n?$ No.

Suppose we have a measure space $(S, \Sigma, \mu)$ and measurable sets $A_1, A_2, ...$
'$\lim A_n = A$' is an abbreviation for
$$\limsup A_n = \liminf A_n = A$$
Definitions:
$$\limsup A_n = \bigcap_{m \ge 1} \bigcup_{n \ge m} A_n$$
$$\liminf A_n = \bigcup_{m \ge 1} \bigcap_{n \ge m} A_n$$
If $\omega \in \limsup A_n$, then $\forall m \ge 1, \exists n \ge m$ s.t. 
$\omega \in A_n$ or:
$$\forall n \ge 1, \omega \in \bigcup_{m=n}^{\infty} A_n$$
If $\omega \in \liminf A_n$, then $\exists m \ge 1, \forall n \ge m$ s.t. $\omega \in A_n$ or:
$$\exists m \ge 1 s.t. \omega \in \bigcap_{n=m}^{\infty} A_n$$
In the latter case, we don't necessarily have $m=1$. Hence, it is not necessarily the case that
$$\omega \in \bigcap_{n=1}^{\infty} A_n \tag{*}$$

$A \supseteq \bigcap_{n=1}^{\infty} A_n?$ Yes.

Now obviously, if we have $(*)$, then we know that indeed that
$$\exists m \ge 1 s.t. \omega \in \bigcap_{n=m}^{\infty} A_n$$
namely $m = 1$ and hence
$$\omega \in \liminf A_n$$
Since $\liminf A_n \subseteq \limsup A_n$,
we have
$$\omega \in \limsup A_n$$
Hence
$$\omega \in \lim A_n = A$$
