How to prove $Pleft(cup_{i=1}^{infty}A_iright)=1$ implies that $P({A_i i.o.})=1$

Question

Suppose that ${A_i}$ is a sequence of independent events with $Pleft(bigcup_{i=1}^infty A_iright) = 1$ and $P(A_i)<1$ for all $iin mathbb{N}$. Show that
$$
P(A_i text{ occurs infinitely often})=1 
$$

My attempt:
We only need to show $Pleft(cap_{i=1}^{infty} A_i^c right)=0 Longrightarrow P(A_i i.o.)=1$. Note that
$$
begin{aligned}
Pleft(cap_{i=1}^{infty} A_i^c right)&= prod_{i=1}^{infty}P(A_i^c)&&text{(independence)}\
&= prod_{i=1}^{infty}(1-P(A_i))
end{aligned}
$$
For any $k$, we have
begin{aligned}
Pleft(cap_{i=1}^{k} A_i^c right)&= prod_{i=1}^{k}P(A_i^c)\
&= prod_{i=1}^{k}(1-P(A_i))\
&leq prod_{i=1}^k e^{-P(A_i)}quad(1-xleq e^{-x}) \
&=e^{-sum_{i=1}^kP(A_i)}
end{aligned}
Let $k to infty$, then $0=Pleft(cap_{i=1}^{infty} A_i^c right)leq e^{-sum_{i=1}^{infty}P(A_i)}$. If we can show $e^{-sum_{i=1}^{infty}P(A_i)}=0$, which implies that $sum_{i=1}^{infty}P(A_i)=infty$, then the result follows by the second Borel-Cantelli Lemma.
My question is how to show $e^{-sum_{i=1}^{infty}P(A_i)}=0$. If we cannot, is there any other way to prove this result? I would appreciate if you could explain in details.

Peter · Answer

One way of solving this is to show that $mathbb{P}left(bigcup_{i=n}^infty A_iright) = 1$ for all $ninmathbb{N}$ (Let's refer to this statement as $(star)$). If this has been shown, we can conclude $$mathbb{P}left( A_i text{ infinitely often}right) = mathbb{P} left(bigcap_{n=1}^infty bigcup_{i=n}^infty A_iright) = lim_{nto infty} mathbb{P}left(bigcup_{i=n}^infty A_iright) = 1,$$ where we have used continuity from above in the second step. To show $(star)$, we use the following statement.

If $A$ and $B$ are two independent events and $mathbb{P}(Acup B) = 1$, then $mathbb{P}(A) = 1$ or $mathbb{P}(B) = 1$.

This follows easily from $0 = 1 - mathbb{P}(Acup B) = mathbb{P}left(A^c cap B^cright) = mathbb{P}(A^c) mathbb{P}(B^c)$.

Now in our case, if $ninmathbb{N}$, then by assumption, $mathbb{P}left( bigcup_{i=1}^{n-1}A_i cup bigcup_{i=n}^infty A_iright) = mathbb{P}left(bigcup_{i=1}^infty A_i right) = 1$, so we only need to show that $mathbb{P}left(bigcup_{i=1}^{n-1}A_iright) < 1$. This, however, follows from the assumption that $mathbb{P}(A_i) < 1$ and thus $mathbb{P}(A_i^c) > 0$ for all $iin mathbb{N}$, since begin{align*} mathbb{P}left(bigcup_{i=1}^{n-1} A_iright) &= 1 - mathbb{P}left(bigcap_{i=1}^{n-1} A_i^cright) = 1 - underbrace{prod_{i=1}^{n-1} mathbb{P}(A_i^c)}_{>0} < 1. end{align*}

How to prove $Pleft(cup_{i=1}^{infty}A_iright)=1$ implies that $P({A_i i.o.})=1$

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