Let $\{E_n\}$ be arbitrary events in $\mathscr{F}$. If for each $m$, $\sum_{n>m}\mathscr{P}\{E_n\mid E_m^c\cap\cdots\cap E_{n-1}^c\}=\infty$, then $\mathscr{P}\{E_n\mbox{ i.o.}\}=1.$
$\bullet$ Proof.
Consider the probability of the complement of $\{E_n\mbox{ i.o.}\}$,
$$\begin{array}{rl}\mathscr{P}\{E_n\mbox{ i.o.}\}^c&=\mathscr{P}\left\{\bigcup_{m=1}^\infty\bigcap_{n=m}^\infty E_n^c\right\}\\ &=\underset{m\rightarrow\infty}{\lim}\mathscr{P}\left\{\bigcap_{n=m}^\infty E_n^c\right\}\\ &=\underset{m\rightarrow\infty}{\lim}\underset{n_0\rightarrow\infty}{\lim}\mathscr{P}\left\{\bigcap_{n=m}^{n_0} E_n^c\right\}. \end{array}$$Then $$\begin{array}{rl}\mathscr{P}\left\{\bigcap_{n=m}^{n_0} E_n^c\right\}&= \mathscr{P}\left\{E_{n_0}^c\mid\bigcap_{n=m}^{n_0-1} E_n^c\right\}\mathscr{P}\left\{\bigcap_{n=m}^{n_0-1} E_n^c\right\}\\ &= \mathscr{P}\left\{E_{n_0}^c\mid\bigcap_{n=m}^{n_0-1} E_n^c\right\}\mathscr{P}\left\{E_{n_0-1}^c\mid\bigcap_{n=m}^{n_0-2} E_n^c\right\}\mathscr{P}\left\{\bigcap_{n=m}^{n_0-2} E_n^c\right\}\\ &\vdots\\ &=\prod_{n=m+1}^{n_0}\mathscr{P}\left\{E_{n}^c\mid\bigcap_{j=m}^{n-1} E_j^c\right\}\\ &=\prod_{n=m+1}^{n_0}\left(1-\mathscr{P}\left\{E_{n}\mid\bigcap_{j=m}^{n-1} E_j^c\right\}\right) \\ & \leq\prod_{n=m+1}^{n_0}\exp{\left\{-\mathscr{P}\left\{E_{n}\mid\bigcap_{j=m}^{n-1} E_j^c\right\}\right\}}\\ &=\exp{\left\{\sum_{n=m+1}^{n_0}-\mathscr{P}\left\{E_{n}\mid\bigcap_{j=m}^{n-1} E_j^c\right\}\right\}}\\ &\rightarrow0\mbox{ as }n_0\rightarrow\infty\end{array}$$because $\sum_{n>m}\mathscr{P}\{E_n\mid E_m^c\cap\cdots\cap E_{n-1}^c\}=\infty$. Thus, we have $$\mathscr{P}\{E_n\mbox{ i.o.}\}^c=\underset{m\rightarrow\infty}{\lim}\underset{n_0\rightarrow\infty}{\lim}\mathscr{P}\left\{\bigcap_{n=m}^{n_0} E_n^c\right\}\leq\underset{m\rightarrow\infty}{\lim}0=0,$$and hence, $$\mathscr{P}\{E_n\mbox{ i.o.}\}=1.$$
$\Box$
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