## Probability Exercise: If it must happen once it happens infinitely often.

Exercise: Let $\{A_n\}$ be a sequence of independent events with $\mathbb{P}(A_n) < 1$ for all $n$. Show that $\mathbb{P}(\bigcup_n A_n) = 1 \implies \mathbb{P}(A_n \text{ i.o. }) = 1$.

Solution:

$\mathbb{P}(\bigcup_n A_n) = 1 \implies \mathbb{P}(\bigcap_n A_n^{c}) = 0$ which by independence means that $\prod_n\mathbb{P}(A_n^c) = 0$. Now since $\mathbb{P}(A_n^c) > 0 \; \forall n$, we know that for any $M, \prod_{n \geq M}$ $\mathbb{P}(A_n^c) = 0$, which means that $\mathbb{P}(\bigcap_{n \geq M} A_n^c) = 0.$ Now suppose $\textit{w} \in \{A_n\}$ f.o. Then for sufficiently large $M, \textit{w} \in \bigcap_{n \geq M} A_n^{c}$. Thus the set $\{A_n\}$ f.o., can be written as $W = \bigcup_{M}$ $\bigcap_{n \geq M}$ $A_n^{c}$, which is a countable union of sets of measure $0$ by the above reasoning. Thus $\mathbb{P}(W) = 0 \implies \mathbb{P}(A_n \text{ i.o. }) = 1$.