Let $B_k$ be the event $\cup_{i=k}^\infty A_i$ for $k=1,2,\ldots,$ . If $x$ is in the event $A_i$ 's i.o., then $x\in B_k$ for all $k$ . So $x\in \cap_{k=1}^\infty B_k$ .

Conversely, if $x\in B_k$ for all $k$ , then we can show that $x$ is in $A_i$ 's i.o. Indeed, $x\in B_1 = \cup_{i=1}^\infty A_i$ means that $x\in A_{j(1)}$ for some $j(1)$ . However $x\in B_{j(1)+1}$ implies that $x\in A_{j(2)}$ for some $j(2)$ that is strictly larger than $j(1)$ . Thus we can produce an infinite sequence of integer $j(1)<j(2)<j(3)<\ldots$ such that $x\in A_{j(i)}$ for all $i$ .

Let $E$ be the event $\{x:\, x\in A_i \mbox{ i.o.}\}$ . We have$$ E = \bigcap_{k=1}^\infty \bigcup_{i=k}^\infty A_i.$$

From $E\subseteq B_k$ for all $k$ , it follows that $P(E)\leq P(B_k)$ for all $k$ . By union bound, we know that $P(B_k)\leq \sum_{i=k}^\infty P(A_i)$ . So $P(B_k)\rightarrow 0$ , by the hypothesis that $\sum_{i=1}^\infty P(A_i)$ is finite. Therefore, $P(E)=0$ .