If you meant
$$
P\left(\bigcap_{n=1}^{\infty}A_n\right)
$$
then that is the probability that given $\omega \in \Omega$,
$$\omega \in A_n \ \forall n \ge 1$$
or
$$\omega \in A_1, A_2, A_3, ...$$
That is, almost surely, all the events $A_1, A_2, A_3, ...$ occur.
Sometimes this arises in the case of independent events where we have
$$P\left(\bigcap_{n=1}^{\infty}A_n\right) = \prod_{n=1}^{\infty} P(A_n)$$
Sometimes we want to compute
$$
P\left(\bigcup_{n=1}^{\infty}A_n\right)
$$
We can do this by noting that:
$$
P\left(\bigcup_{n=1}^{\infty}A_n\right) = 1 -
P\left(\bigcap_{n=1}^{\infty}A_n^C\right)
$$
So, for example, if we have a bunch of random variables $X_1, X_2, ...$ in $(\Omega, \mathscr F, \mathbb P)$
If we want to compute the probability that at least 1 is greater than 5, we can compute the probability that all are less than 5:
$$
P\left(\bigcup_{n=1}^{\infty}(X_n \ge 5)\right) = 1 -
P\left(\bigcap_{n=1}^{\infty}(X_n < 5)\right)
$$
If the random variables are independent, we have
$$1 - P\left(\bigcap_{n=1}^{\infty}(X_n < 5)\right) = 1 - \prod_{n=1}^{\infty}P(X_n < 5) = 1 - \prod_{n=1}^{\infty}F_{X_n}(5)$$
Edit: You said
$$
P\left(\bigcap_{n=1}^{\infty}A_i\right) = P\left(\omega \in \Omega:\bigcap_{n=1}^{\infty}A_i(\omega)\right)
$$
Actually it's supposed to be
$$
P\left(\bigcap_{n=1}^{\infty}A_n\right) = P\left(\omega \in \Omega:\omega \in \bigcap_{n=1}^{\infty}A_n\right)
$$
'$A_i(\omega)$' is a notation we apply to random variables. For example if we have random variables $X_1, X_2, ...$,
define $X := \limsup X_n$ s.t. for a given $\omega \in \Omega$
$$X(\omega) := [\limsup X_n](\omega) := \limsup [X_n(\omega)]$$
Perhaps you meant to say '$\omega \in A_i$' as in
$$
P\left(\bigcap_{n=1}^{\infty}A_n\right) = P\left(\omega \in \Omega: \bigcap_{n=1}^{\infty}(\omega \in A_n)\right)
$$
?
I'm not sure, but maybe that could work. I don't think I've seen that notation a lot.