Dependence and second Borel-Cantelli lemma. I'll put the problem and then I'll explain my problem.
Problem: Let ${A_n}$ be events such as
$\operatorname{Cov}(I_{A_i},I_{A_j})=E[I_{A_i}I_{A_j}]-E[I_{A_i}]E[I_{A_j}]\leq 0,\ \forall i\neq j\tag{1}$
If $\sum P(A_i)=\infty$ then $P[\lim \sup A_n]=1$.
Answer: By (1) we have that $\operatorname{Cov}(I_{A^c_i},I_{A^c_j})\leq 0$ too.
So $P(\lim \inf A_n^c)=P(\lim \bigcap A_n^c)=\lim P(\bigcap A_n^c)\overset{Q!} {\leq} \lim \prod P(A_n^c)= \lim \prod (1-P(A_n))\leq \lim e^{-\sum P(A_n)}=0$
The detail is that in the inequality marked with a "Q!" I used that $P(\bigcap A_n)\leq \prod P(A_n)$. It is intuitive but i couldn't prove through the problem statement. But it is interesting result. We could use B-C lemma even for correlated events, since they are negatively correlated. What do you guys think about it?
 A: I think Fristedt & Gray's book on probability proves a version of Borel--Cantelli that assumes only nonpositive correlation rather than independence.  In particular, that means pairwise independence is enough.
Later edit: Here's what I find in A Modern Approach to Probability Theory by Bert Fristedt and Lawrence Gray, page 79:

Lemma 5. [Borel-Cantelli] Let $(A_1,A_2,\ldots)$ be a sequence of events in a probability space $(\Omega,\mathcal{F},P)$.  Assume that for each $i\ne j$, the events $A_i$ and $A_j$ are either negatively correlated or uncorrelated.  Let $A=\lim\sup_{n\to\infty} A_n$.  If $\sum_{n=1}^\infty P(A_n)=\infty,$ then $P(A)=1$.

A: The result you mention as being intuitive does not hold. To see this, one can recall the classical example of three uncorrelated dependent events, which is $A_1=[U=1]$, $A_2=[V=1]$ and $A_3=[UV=1]$, where $U$ and $V$ are i.i.d. $\pm1$ symmetric Bernoulli random variables. 
Then the events $A_i$ are pairwise independent, each $A_i$ has probability $\frac12$, and $A_1\cap A_2\cap A_3=[U=V=1]$ hence $\mathrm P(A_1\cap A_2\cap A_3)=\frac14\gt\frac18=\mathrm P(A_1)\mathrm P(A_2)\mathrm P(A_3)$.
A: The answer is given in Theorem 5 below. This post is essentially a reiteration of [1]'s first section ("Generalization of the Borel-Cantelli lemma").
Let $(\Omega, \mathcal{F}, P)$ be a probability space. Let $A_1, A_2, \dots \in \mathcal{F}$ be an infinite sequence of positive events (i.e. $P(A_n) > 0$ for every $n \in \{1, 2, \dots\}$). Note that this implies, in particular, that for every $k \in \{1, 2, \dots\}$, $E\big(\mathbb{1}_{A_k}\big) > 0$. For every $n \in \{1, 2, \dots\}$ define $X_n:\Omega\rightarrow\mathbb{R}$ to be the random variable $X_n := \sum_{k = 1}^n \mathbb{1}_{A_k}$.
Lemma 1
For every $n \in \{1, 2, \dots\}$,
$$
\frac{\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)}{\big(\sum_{k=1}^nP(A_k)\big)^2} = \frac{E\big(X_n^2\big)}{E^2(X_n)}.
$$
Proof
Let $n \in \{1, 2, \dots\}$. Then
$$
\begin{align}
\sum_{k = 1}^nP(A_i) &= \sum_{k = 1}^nE\big(\mathbb{1}_{A_i}\big) \\
&= E\bigg(\sum_{i = 1}^n\mathbb{1}_{A_i}\bigg) \\
&= E(X_n) \\
\sum_{i = 1}^n\sum_{j = 1}^nP(A_i\cap A_j) &= \sum_{i = 1}^n\sum_{j = 1}^nE\big(\mathbb{1}_{A_i\cap A_j}\big) \\
&= \sum_{i = 1}^n\sum_{j = 1}^nE\big(\mathbb{1}_{A_i}\mathbb{1}_{A_j}\big) \\
&= E\Bigg(\sum_{i = 1}^n\sum_{j = 1}^n\mathbb{1}_{A_i}\mathbb{1}_{A_j}\Bigg) \\
&= E\Bigg(\Bigg(\sum_{k = 1}^n\mathbb{1}_{A_k}\Bigg)^2\Bigg) \\
&= E\big(X_n^2\big)
\end{align}
$$
Q.E.D.
Lemma 2
For every $n \in \{1, 2, \dots\}$,
$$
\frac{E\big(X_n^2\big)}{E^2(X_n)} = 1 + \frac{V(X_n)}{E^2(X_n)}.
$$
Proof
Let $n \in \{1, 2, \dots\}$. The conclusion is a consequence of the fundamental variance identity $V(X_n) = E\big(X_n^2\big) - E^2(X_n)$.
Q.E.D.
Lemma 3
If $\sum_{n = 1}^\infty P(A_n) = \infty$ and
$$
\underset{n\rightarrow\infty}{\underline{\lim}}\frac{V(X_n)}{E^2(X_n)} = 0,
$$
then $P\bigg(\underset{n\rightarrow\infty}{\overline{\lim}}A_n\bigg) = 1$.
Proof
Choose an increasing sequence $n_1 < n_2 < \dots$ in $\{1, 2, \dots\}$ such that
$$
\sum_{k=1}^\infty \frac{V\big(X_{n_k}\big)}{E^2\big(X_{n_k}\big)} < \infty.
$$
Then for every $\varepsilon \in (0,1)$,
$$
\begin{align}
\sum_{k = 1}^\infty P\Big((1-\varepsilon)E\big(X_{n_k}\big) \geq X_{n_k}\Big) &= \sum_{k = 1}^\infty P\big(E\big(X_{n_k}\big) - X_{n_k} \geq \varepsilon E\big(X_{n_k}\big)\big) \\
&\leq \sum_{k = 1}^\infty P\Big(\big|X_{n_k} - E\big(X_{n_k}\big)\big| \geq \varepsilon E\big(X_{n_k}\big)\Big) \\
&\overset{\text{Chebyshev}}{\leq} \sum_{k = 1}^\infty \frac{V\big(X_{n_k}\big)}{\varepsilon^2\ E^2\big(X_{n_k}\big)} \\
&< \infty.
\end{align}
$$
Hence by the Borel-Cantelli lemma
$$
P\bigg(X_{n_k} > (1-\varepsilon)E\big(X_{n_k}\big)\ \text{eventually}\bigg) = 1. \tag{*}\label{star}
$$
Since
$$
\begin{align}
\sup_{n \in \{1, 2, \dots\}}E\big(X_n\big) &= \sup_{n \in \{1, 2, \dots\}}E\bigg(\sum_{k = 1}^n\mathbb{1}_{A_k}\bigg)\\
&= \sup_{n \in \{1, 2, \dots\}} \sum_{k = 1}^nE\big(\mathbb{1}_{A_k}\big) \\
&= \sup_{n \in \{1, 2, \dots\}} \sum_{k = 1}^nP(A_k) \\
&= \sum_{k = 1}^\infty P(A_k) \\
&= \infty,
\end{align}
$$
and since $X_1 \leq X_2 \leq \cdots$, we have $\sup_{k \in \{1, 2, \dots\}}E\big(X_{n_k}\big) = \infty$. Therefore \eqref{star} implies that
$$
P\bigg(\sup_{k\in\{1, 2, \dots\}} X_{n_k} = \infty\bigg) = 1.
$$
It remains to observe that
$$
\bigg\{\sup_{k\in\{1, 2, \dots\}} X_{n_k} = \infty\bigg\} = \underset{k\rightarrow\infty}{\overline{\lim}}A_{n_k} \subseteq \underset{n\rightarrow\infty}{\overline{\lim}}A_n.
$$
Q.E.D.
Corollary 4


*

*For every $n \in \{1, 2, \dots\}$,
$$
 \frac{\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)}{\big(\sum_{k=1}^nP(A_k)\big)^2} \geq 1.
 $$

*If $\sum_{n = 1}^\infty P(A_n) = \infty$ and
$$
 \underset{n\rightarrow\infty}{\underline{\lim}} \frac{\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)}{\big(\sum_{k=1}^nP(A_k)\big)^2} = 1,
 $$
then $P\Big(\underset{n\rightarrow\infty}{\overline{\lim}}A_n\Big) = 1$.
Proof


*

*By Lemma 1 and Lemma 2.

*By Lemma 1, Lemma 2, and Lemma 3.
Q.E.D.
Theorem 5
If $\sum_{n = 1}^\infty P(A_n) = \infty$ and if the $A_n$'s are negatively correlated in the sense that for every $i, j \in \{1, 2, \dots\}$ such that $i\neq j$, $P(A_i\cap A_j) \leq P(A_i)P(A_j)$, then $P\Big(\underset{n\rightarrow\infty}{\overline{\lim}}A_n\Big) = 1$.
Proof
By Corollary 4 it suffices to show that
$$
\underset{n\rightarrow\infty}{\underline{\lim}} \frac{\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)}{\big(\sum_{k=1}^nP(A_k)\big)^2} \leq 1.
$$
This is a consequence of the following observations.
$$
\begin{align}
\frac{\sum_{i=1}^n\sum_{j=1}^nP(A_i\cap A_j)}{\big(\sum_{k=1}^nP(A_k)\big)^2} &\leq \frac{\sum_{\substack{i,j\in\{1,\dots,n\}\\i\neq j}}P(A_i)P(A_j) + \sum_{k=1}^nP(A_k)}{\big(\sum_{k=1}^nP(A_k)\big)^2} \\
&= \frac{\big(\sum_{k = 1}^nP(A_k)\big)^2}{\big(\sum_{k=1}^nP(A_k)\big)^2} + \frac{\sum_{k = 1}^nP(A_k)\big(1 - P(A_k)\big)}{\big(\sum_{k=1}^nP(A_k)\big)^2} \\
&\leq 1 + \frac{\sum_{k = 1}^nP(A_k)}{\big(\sum_{k=1}^nP(A_k)\big)^2} \\
&= 1 + \frac{1}{\sum_{k=1}^nP(A_k)}.
\end{align}
$$
Q.E.D.

Works cited
[1] P. Erdős and A. Rényi. On Cantor's Series with Convergent $\sum\frac{1}{q_n}$. Mathematical Institute, Eötvös Loránd University, Budapest. Received 6 April, 1959.
