I want to prove a stronger version of the second Borel-Cantelli lemma, where only pairwise independence is given. Namely, I want to prove the following:
Lemma: Let $(\Omega, \mathcal{F}, P)$ be a probability space and $\{A_n: n \in \mathbb{N}\} \subseteq \mathcal{F}$ with $A_i$ and $A_j$ independent for all $i\neq j$. If $\sum_{n=1}^\infty P(A_i) = \infty$ then $P(\limsup_{n\geq 1}A_n)=1$.
My proof: First let $$S_n = \sum_{i=1}^n \mathbf{1}_{A_i} \;\text{ and } \; s_n = \mathbb{E}[S_n]=\sum_{i=1}^n P(A_i)$$ It's easy to prove that $$\mathbb{E}[S_n^2] = s_n + s_n^2 - \sum_{i=1}^nP(A_i)^2 \leq s_n + s_n^2$$
Next, by Paley-Zygmund inequality we have for all $\varepsilon > 0$ $$P(S_n \geq \varepsilon s_n) \geq (1-\varepsilon)^2\frac{s_n^2}{s_n+s_n^2}$$ Finally, we let $S=\sum_{i=1}^n \mathbf{1}_{A_i}$ and notice that it suffices to prove that $P(S=\infty)=1$. To do so, we let $\varepsilon > 0$ and write $$\{S=\infty\} = \bigcap_{n=1}^\infty \{S\geq \varepsilon s_n\}$$ Also, $\{S\geq \varepsilon s_n\} \supseteq \{S_n\geq \varepsilon s_n\}$ so putting everything together we have $$P(S=\infty) = \lim_{n\to\infty}P(S\geq \varepsilon s_n) \geq \lim_{n\to\infty} P(S_n \geq \varepsilon s_n) \geq \lim_{n\to\infty} (1-\varepsilon)^2 \frac{s_n^2}{s_n+s_n^2} = (1-\varepsilon)^2$$ To conclude, we just let $\varepsilon \to 0^+$.
Is this proof correct? If not, then where is the mistake? If yes, can it be improved?
This is correct, except that in the last display, the limit $$\lim_{n\to\infty} P(S_n \geq \varepsilon s_n)$$ need not exist, but replacing it by limsup or liminf works.
This stronger version was proved several times, e.g. see the Erdos-Renyi Theorem [1] and the Kochen-Stone lemma [2].
[1] P. Erdos,A. R enyi,On Cantor’s series with convergent $1/qn, Ann. Univ. Sci. Budap. Rolando Eotvos, Sect. Math. 2 (1959) 93–109.
[2] https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-8/issue-2/A-note-on-the-Borel-Cantelli-lemma/10.1215/ijm/1256059668.full