Let $\mathscr{F}$ be a Borel field and $\{E_n\}_{n\geq1}\in\mathscr{F}$ are events. We have
(1) $\sum_{n=1}^\infty \mathscr{P}\{E_n\} < \infty \implies \mathscr{P}\{E_n\mbox{ i.o.}\}=0;$
(2) If $\sum_{n=1}^\infty \mathscr{P}\{E_n\} = \infty$ and $E_n$'s are independent. Then $\mathscr{P}\{E_n\mbox{ i.o.}\}=1.$
$\bullet$ Proof.
(1) Recall Boole's inequality,
$$\mathscr{P}\{\bigcup_n E_n\}\leq\sum_n \mathscr{P}\{E_n\}.$$ By Boole's inequality, we have
$$\mathscr{P}\{E_n\mbox{ i.o.}\}=\mathscr{P}\{\underset{m\geq1}{\bigcap}\underset{n\geq m}{\bigcup}E_n\}=\underset{m\rightarrow\infty}{\lim}\mathscr{P}\{\underset{n\geq m}{\bigcup}E_n\}\leq\underset{m\rightarrow\infty}{\lim}\sum_{n=m}^\infty \mathscr{P}\{E_n\}=0.$$
(2) Here we first show that for $0\leq x\leq1$, $e^{-x}\geq 1-x$ by Taylor's expansion:
$$e^{-x}\approx 1-x+x^2\implies e^{-x}\geq 1-x.$$ Consider the complement of $\{E_n\mbox{ i.o.}\}$, $$\begin{array}{ccl}\mathscr{P}\left\{\{E_n\mbox{ i.o.}\}^c\right\}
&=&\mathscr{P}\left\{\underset{m\geq1}{\bigcup}\underset{n\geq m}{\bigcap}E_n^c\right\}\;\mbox{(DeMorgan's Law)}\\
&=&\underset{m\rightarrow\infty}{\lim}\mathscr{P}\{\underset{n\geq m}{\bigcap}E_n^c\}\;\mbox{(monotone)}\\
&=&\underset{m\rightarrow\infty}{\lim}\prod_{n=m}^\infty\left(1-\mathscr{P}\{E_n\}\right)\;\mbox{(independent)}\\
&\leq&\underset{m\rightarrow\infty}{\lim}\prod_{n=m}^\infty e^{-\mathscr{P}\{E_n\}} \\
&=&\underset{m\rightarrow\infty}{\lim} e^{-\sum_{n=m}^\infty \mathscr{P}\{E_n\}}=0 \\
\end{array}$$ since $\sum_{n=1}^\infty \mathscr{P}\{E_n\} = \infty$. Thus, we have $\mathscr{P}\{E_n\mbox{ i.o.}\}=1.$
$\Box$
$\star$ See Extension of Borel-Cantelli Lemma
$\star$ See Application of Borel-Cantelli Lemma
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