# 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.

• $$\lim_{n\to\infty}\sum_{i=1}^nP(A_i)=\infty\implies \lim_{n\to \infty}e^{-\sum_{i=1}^nP(A_i)}=0.$$ This because of the fact $x\to \infty\implies e^{-x}\to 0+$. But note that there is no real number $r$ such that $e^{-r}=0$. Sep 13 '20 at 9:59
• math.stackexchange.com/questions/359404/probability-of-limsup Sep 13 '20 at 10:28

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*}