Extension of Borel-Cantelli Lemma II

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Let $\mathscr{F}$ be a Borel field and $\{E_n\}_{n\geq1}\in\mathscr{F}$ are events.  If $\{E_n\}$ are pairwise independent, then the conclusion
$$\sum_{n=1}^\infty \mathscr{P}\{E_n\} = \infty\implies\mathscr{P}\{E_n\mbox{ i.o.}\}=1$$remains true.

[See Borel-Cantelli Lemma]

$\bullet$ Proof.

Let $I_n=I\{\omega\in E_n\}$.  $$\{E_n\mbox{ i.o.}\}\iff\omega\in\limsup{E_n}\iff\sum_{n=1}^\infty I_n=\infty.$$Thus,  we need to show $$\mathscr{P}\left\{\sum_{n=1}^\infty I_n=\infty\right\}=1.$$Define the partial sum $$J_n=\sum_{k=1}^nI_k.$$Thus our goal is transformed to $$\mathscr{P}\left\{J_\infty=\infty\right\}=1.$$Consider the mean and variance of $J_n$, $$\begin{array}{ll}(1)\,\mathscr{E}(J_n) &=\sum_{k=1}^n\mathscr{E}(I_k)=\sum_{k=1}^n\mathscr{P}(E_k); \\ (2)\,\sigma^2(J_n) &=\sum_{k=1}^n\sigma^2(I_k)+\mbox{covariance term}\,(=0,\,\because\mbox{pairwise indep.}) \\
&=\sum_{k=1}^n\left(\mathscr{P}(E_k)-\mathscr{P}(E_k)^2\right) \\
&\leq\mathscr{E}(J_n).\end{array}$$We have $$\sigma^2(J_n)\leq\mathscr{E}(J_n)\implies\sigma(J_n)\leq\mathscr{E}(J_n)^{\frac{1}{2}}=o\left(\mathscr{E}(J_n)\right).$$For every $A>0$, by Chevbyshev's inequality $$\mathscr{P}\{|J_n-\mathscr{E}(J_n)|\leq A\sigma(J_n)\}\geq1-\frac{\sigma^2(J_n)}{A^2\sigma^2(J_n)}=1-\frac{1}{A^2}.$$The probability in left hand side of the inequality can be extended by $$\begin{array}{rl}\mathscr{P}\{|J_n-\mathscr{E}(J_n)|\leq A\sigma(J_n)\}
&=\mathscr{P}\{\mathscr{E}(J_n)-A\sigma(J_n)\leq J_n \leq \mathscr{E}(J_n)+A\sigma(J_n)\}\\
&\leq\mathscr{P}\{J_n\geq\mathscr{E}(J_n)-A\sigma(J_n)\}\\
&\leq\mathscr{P}\left\{J_n\geq\mathscr{E}(J_n)-A\mathscr{E}(J_n)^{\frac{1}{2}}\right\}\quad\left(\because\,\sigma(J_n)\leq\mathscr{E}(J_n)^{\frac{1}{2}}\right)\\
&=\mathscr{P}\left\{J_n\geq\mathscr{E}(J_n)\left[1-A\frac{\mathscr{E}(J_n)^{\frac{1}{2}}}{\mathscr{E}(J_n)}\right]\right\}. \end{array}$$Note here, we can omit the $A$ inside the probability since $$\mathscr{E}(J_n)^{\frac{1}{2}}=o\left(\mathscr{E}(J_n)\right)\implies\frac{\mathscr{E}(J_n)^{\frac{1}{2}}}{\mathscr{E}(J_n)}\rightarrow0\mbox{ as }n\rightarrow\infty.$$Then the following inequality holds, for $n\geq n_0(A)$,
$$\mathscr{P}\{J_n\geq r\mathscr{E}(J_n)\}\geq\mathscr{P}\{|J_n-\mathscr{E}(J_n)|\leq A\sigma(J_n)\}\geq1-\frac{1}{A^2}.$$where $0<r<1$.  Since $J_\infty\geq J_n$, $$\mathscr{P}\{J_\infty\geq r\mathscr{E}(J_n)\}\geq\mathscr{P}\{J_n\geq r\mathscr{E}(J_n)\}\geq1-\frac{1}{A^2}.$$Finally, as $n\rightarrow\infty$, $\mathscr{E}(J_\infty)=\sum_{n=1}^\infty\mathscr{E}(I_n)=\sum_{n=1}^\infty\mathscr{P}\{E_n\}=\infty$ by assumption, we have $$\mathscr{P}\{J_\infty=\infty\}\geq1-\frac{1}{A^2}$$for any $A>0$.  Thus, $J_\infty=\infty$ almost surely, i.e. $\mathscr{P}\{E_n\mbox{ i.o.}\}=1$.

$\Box$