Adaptation of Resnick Exercise 4.9: $P(\bigcup_{n=1}^\infty A_n)=1$ iff $P(A_n i.o.)=1$ $P(\bigcup_{n=1}^\infty A_n)=1$ iff $P(A_n i.o.)=1$
I have always struggled with the "infinitely often" and "all but finitely often" concepts for sequences. My first time through Resnick, I don't think I was equipped with the tools necessary for this exercise using only what I had learned in the text previously, and so I wanted to come back to a few problems.
I still am not sure how to proceed at all here. 
I think I understand the question to basically state "The probability that a union of a sequence of events is 1 if and only if the probability of that sequence occurring infinitely often is also 1.
As per the comments below, an additional assumption is that each of the events has $P(A_i)<1$ and each event is independent.
So, to start out, I can define:
$\{A_n i.o.\}=\limsup_{n\to\infty}A_n=\bigcap_{n=1}^\infty\bigcup_{k=n}A_k$
but from here I am not sure where to go (or if I've even begun correctly).
Feel free to be as explicit as possible as these ideas ($\limsup$, etc.) have always seemed to elude me. Thank you for the help.
 A: Hint Show that (1) $\iff$ (2) $\iff$ (3) $\iff$ (4) $\iff$ (5) $\iff$ (6), where:
(1) $\mathbb  P\left(\bigcup\limits_{n\geqslant1}A_n\right)=1$.
(2) $\mathbb  P\left(\bigcap\limits_{n\geqslant1}(\Omega\setminus A_n)\right)=0$. (3) $\prod\limits_{n\geqslant1}(1-\mathbb P(A_n))=0$. (4) $\sum\limits_{n\geqslant1}\mathbb P(A_n)=+\infty$. (5) $\mathbb P\left(\limsup A_n\right)=1$. (6) $\mathbb P\left(A_n\ \text{i.o.}\right)=1$.
If a step is unclear, please say so.
A: While did's solution is correct and complete, I thought that it needs a little justification that why from the step $\prod_{n \ge 1} (1 - P(A_n))$ we can conclude that $P(lim sup A_n) = 1$. 
Here is my solution:
We need to show: $\{A_n i.o.\}=\limsup_{n\to\infty}A_n=\bigcap_{n=1}^\infty\bigcup_{k=n}A_k$
Consider $B_i = \bigcup_{k=i}A_k $. We need to show that $\forall i : p(B_i) = 1$ to do that we have:
Based on the assumption $P(\bigcup_{n=1}^\infty A_n)=1 \rightarrow P(\bigcup_{n=1}^{i} A_n \cup \bigcup_{n= i + 1}^\infty A_n)=1 \rightarrow \text{we expand the prob} \rightarrow P(\bigcup_{n=1}^{i} A_n) + P(\bigcup_{n= i + 1}^\infty A_n) - P(\bigcup_{n=1}^{i} A_n)P(\bigcup_{n= i + 1}^\infty A_n) \text{the third part is because events are independent} \rightarrow 1 - P(\bigcup_{n=1}^{i} A_n) = P(\bigcup_{n= i + 1}^\infty A_n)(1 - P(\bigcup_{n=1}^{i} A_n)) \rightarrow \text{not that since $P(A_i)<1$ then $P({A_i}^{c})\ne 1$ so $(1 - P(\bigcup_{n=1}^{i} A_n)) = P({(\bigcup_{n=1}^{i} A_n)}^c) \ne 0$} \rightarrow P(\bigcup_{n= i + 1}^\infty A_n) = p(B_i) = 1 \text{ $\forall i$}$
Now note that because of continuity from above of $p(B_i) = P(\bigcup_{n= i + 1}^\infty A_n)$, we know that $P(B_i) \downarrow p(lim sup A_n)$ and note that $P(B_i) = 1 \text{ $\forall i$}$ $\rightarrow P(\limsup_{n\to\infty}A_n) = 1 \rightarrow P(A_n i.o.) = 1$
Hope it helps.
