How to prove $P\left(\cup_{i=1}^{\infty}A_i\right)=1$ implies that $P(\{A_i\ i.o.\})=1$ Suppose that $\{A_i\}$ is a sequence of independent events with $P\left(\bigcup_{i=1}^\infty A_i\right) = 1$ and $P(A_i)<1$ for all $i\in \mathbb{N}$. Show that
$$
P(A_i \text{ occurs infinitely often})=1 
$$

My attempt:
We only need to show $P\left(\cap_{i=1}^{\infty} A_i^c \right)=0 \Longrightarrow P(A_i\ i.o.)=1$. Note that
$$
\begin{aligned}
P\left(\cap_{i=1}^{\infty} A_i^c \right)&= \prod_{i=1}^{\infty}P(A_i^c)&&\text{(independence)}\\
&= \prod_{i=1}^{\infty}(1-P(A_i))
\end{aligned}
$$
For any $k$, we have
\begin{aligned}
P\left(\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-x\leq e^{-x}) \\
&=e^{-\sum_{i=1}^kP(A_i)}
\end{aligned}
Let $k \to \infty$, then $0=P\left(\cap_{i=1}^{\infty} A_i^c \right)\leq e^{-\sum_{i=1}^{\infty}P(A_i)}$. 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.
 A: One way of solving this is to show that $\mathbb{P}\left(\bigcup_{i=n}^\infty A_i\right) = 1$ for all $n\in\mathbb{N}$ (Let's refer to this statement as $(\star)$). If this has been shown, we can conclude $$\mathbb{P}\left( A_i \text{ infinitely often}\right) = \mathbb{P} \left(\bigcap_{n=1}^\infty \bigcup_{i=n}^\infty A_i\right) = \lim_{n\to \infty} \mathbb{P}\left(\bigcup_{i=n}^\infty A_i\right) = 1,$$ 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 $\mathbb{P}(A\cup B) = 1$, then $\mathbb{P}(A) = 1$ or $\mathbb{P}(B) = 1$.

This follows easily from $0 = 1 - \mathbb{P}(A\cup B) = \mathbb{P}\left(A^c \cap B^c\right) = \mathbb{P}(A^c) \mathbb{P}(B^c)$.
Now in our case, if $n\in\mathbb{N}$, then by assumption, $\mathbb{P}\left( \bigcup_{i=1}^{n-1}A_i \cup \bigcup_{i=n}^\infty A_i\right) = \mathbb{P}\left(\bigcup_{i=1}^\infty A_i \right) = 1$, so we only need to show that $\mathbb{P}\left(\bigcup_{i=1}^{n-1}A_i\right) < 1$. This, however, follows from the assumption that $\mathbb{P}(A_i) < 1$ and thus $\mathbb{P}(A_i^c) > 0$ for all $i\in \mathbb{N}$, since
\begin{align*}
\mathbb{P}\left(\bigcup_{i=1}^{n-1} A_i\right)
&= 1 - \mathbb{P}\left(\bigcap_{i=1}^{n-1} A_i^c\right) = 1 - \underbrace{\prod_{i=1}^{n-1} \mathbb{P}(A_i^c)}_{>0} < 1.
\end{align*}
A: Your approach is fine up to the point where you need to show that $\sum_{i=1}^\infty P(A_i) = +\infty$ in order to apply the second Borel-Cantelli lemma.
Since $P(\bigcup_{i=1}^\infty A_i) = 1$ and the events $A_i$ are pairwise independent , we have
$$\tag{1}\prod_{i=1}^\infty[1- P(A_i)]= P\left(\bigcap_{i=1}^\infty A_i^c\right) = 0,$$
and you have shown that
$$\tag{2} P\left(\bigcap_{i=1}^kA_i^c\right) = \prod_{i=1}^k[1-P(A_i)]\leqslant \exp\left(-\sum_{i=1}^k P(A_i)\right)$$
We will assume that $\sum_{i=1}^\infty P(A_i) = S < +\infty$ and show that this contradicts (1).
Since $P(A_i) < 1$, we have $$\mathcal{P_k} := \prod_{i=1}^k [1-P(A_i)] = \prod_{i=1}^{k-1} P(A_i) \cdot [1-P(A_k)]< \prod_{i=1}^k [1-P(A_i)] = \mathcal{P_{k-1}},$$
and the sequence of partial products $(\mathcal{P}_k)$ is decreasing. Now we will show it is bounded below by a positive number.  Since $\sum P(A_i)$ is convergent, given some fixed $\epsilon \in (0,1)$ there exists $N \in \mathbb{N}$ such that for all $k > N$
$$\sum_{i=N+1}^k P(A_i) < \epsilon$$
Thus,
$$\frac{\mathcal{P}_k}{\mathcal{P}_N}=\prod_{i=N+1}^kP(A_i) \geqslant 1 - \sum_{i=N+1}^kP(A_i)> 1 - \epsilon > 0$$
Hence, $\mathcal{P}_k \geqslant \min((1-\epsilon)\mathcal{P}_N,\mathcal{P}_1,\mathcal{P_2},\ldots, \mathcal{P}_{N})$ for all $k$ and the sequence of partial products converges to a positive limit, that is
$$ \lim_{k\to \infty}\mathcal{P_k}=\prod_{i=1}^\infty [1-P(A_i)]  >0$$
This contradicts (1), and , therefore we have $\sum_{i=1}^\infty P(A_i) = + \infty$.
A: Here is a route chart for proving the conclusion you want.
\begin{gather*}
 \mathsf{P}\Big(\bigcup_{i=1}^\infty A_i\Big)=1\\
 \hphantom{\Big(\bigcup_{i=1}^\infty A_i\Big)^c=\bigcap_{i=1}^\infty A^c_i} \Downarrow  \Big(\bigcup_{i=1}^\infty A_i\Big)^c=\bigcap_{i=1}^\infty A^c_i \\ 
    \mathsf{P}\Big(\bigcap_{i=1}^\infty A^c_i\Big)=0\\
    \hphantom{\{A_i\}\; \text{is independent}}  \Downarrow  \{A_i\}\; \text{is independent}\\ 
    \prod_{i=1}^\infty[1-\mathsf{P}(A_i)]=0\\
    \hphantom{\mathsf{P}(A_i)<1 \text{ and } (\ast)}  \Downarrow \mathsf{P}(A_i)<1 \text{ and } (\ast)  \\
    \sum_{i=1}^{\infty}\mathsf{P}(A_i)=\infty \\
    \hphantom{\text{2nd B-C Lemma}} \Downarrow \text{2nd B-C Lemma}\\
    \mathsf{P}(A_i  \text{ i.o.})=1,    
\end{gather*}
where (*) is a ready-made fact in infinite product, it could be proved as following:
Let $a_i=\mathsf{P}(A_i)$, then $0\le a_i<1$ and
\begin{equation*}
  \lim_{n\to\infty}\prod_{i=1}^n[1-\mathsf{P}(A_i)]=0 \iff
  \lim_{n\to\infty}\sum_{i=1}^n [-\log(1-a_i)]=\infty,
\end{equation*}
where $\sum\limits_{i=1}^\infty [-\log(1-a_i)] $ is a infinite series with non-negative terms. If $\varlimsup\limits_{n\to\infty}a_n>0$, then
\begin{equation*}
 \lim_{n\to\infty}\sum_{i=1}^n \mathsf{P}(A_i)=\infty \quad\text{and} \quad
 \lim_{n\to\infty}\sum_{i=1}^n [-\log(1-a_i)]=\infty.
\end{equation*}
If $\lim\limits_{n\to\infty}a_n=0$, then
\begin{equation*}
 \lim_{n\to\infty}\frac{-\log(1-a_i)}{a_i}=1,
\end{equation*}
and
\begin{equation*}
  \sum_{i=1}^\infty [-\log(1-a_i)]=\infty \iff  \sum_{i=1}^\infty a_i=\infty.
\end{equation*}
