Does $P(A^c \text{ i.o.}) = 0$ imply that $P(A \text{ i.o.})=1$? I am studying on the notion of "infinitely often" in probability theory.
Does $P(A^c \text{ i.o.}) = 0$ imply that $P(A \text{ i.o.})=1$ ("i.o." stands for "infinitely often")?
 A: Yes. The complement of the event "$A_n^c$ infinitely often" is "$A_n$ eventually." To be precise,
$$ \{ A_n^c \text{ i.o.}) = \bigcap_{m \geq 1} \bigcup_{n \geq m} A_n^c $$
means that for every $m$, you can always find an integer $n \geq m$ such that $A_n^c$ occurs. The complement of this event is
$$ \{ A_n \text{ ev.}) = \bigcup_{ m \geq 1} \bigcap_{n \geq m} A_n ,$$
which means that there is some $m$ past which all the events $A_n$, with $n \geq m$, occur. Using de Morgan's laws, it's easy to check that these events are complements of one another. It's also intuitive (and easy to check) that $\{ A_n \text{ ev.}) \subseteq \{ A_n \text{ i.o.})$, because if the events $A_n$ eventually occur, then infinitely many of these events occur.
What this all amounts to is that $P(A_n^c \text{ i.o.}) = 0$ is equivalent to $P(A_n \text{ ev.}) = 1$, and this latter statement implies that $P(A_n \text{ i.o.}) = 1$.
A: It does imply that, but the converse is false. I.e. it is possible that both $A$ and $A^\text{C}$ occur infinitely often. For example, the probability of getting both infinitely many "heads" and infinitely many "tails" in a sequence of coin tosses is $1.$
