# Sequence of events; infinitely often and sufficiently large $n$ example

Let ($A_n : n \in \mathbb{N}$) be a sequence of events in some probability space $( \Omega, \mathcal{F}, \mathbb{P} )$. Set

$A = \{ \omega \in \Omega : \omega \in A_n \text{ infinitely often} \}$ , $B = \{ \omega \in \Omega : \omega \in A_n \text{ for all sufficiently large } n \}$

Show that $B = \cup_{n=1}^{\infty} \cap_{k=n}^{\infty} A_k$

I tried getting my head round what this question means, or is even asking, but this is too wacky...

• Just out of curiosity, since $A$ doesn't feature in the problem, are you also supposed to show that you get $A$ if you reverse the union and intersection? – Arthur Feb 10 '17 at 9:25
• The next wants you to show $\mathbb{P}(A) \leq \sum_{k=n}^{\infty} \mathbb{P}(A_k)$, and if $\sum_{n=1}^{\infty}$ converges then that $\mathbb{P}(A) = 0$. Hmmm... – Christopher Turnbull Feb 10 '17 at 14:51

Whatever the sets $\{U_n\}$ are, the following is true (by definition)
$$\bigcup_{n=1}^{\infty} U_n =\{\omega\in\Omega: \text{ there is an } n \text{ for which } \omega \in U_n\}.$$
$$U_n=\bigcap_{k=n}^{\infty}A_k=\{\omega\in\Omega:\omega \in U_k \text{ for all } k\geq n \text{ } \ \}.$$
then $$\bigcup_{n=1}^{\infty} U_n=\bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k=$$ $$=\{\omega\in\Omega: \text{ there is an } n \text{ for which } \omega \in A_k\text{ for all } k\geq n\}=$$ $$=\{\omega\in\Omega: \text{ for all sufficiently large } n, \ \omega \in A_n \}=B.$$