Generalizing the second Borel-Cantelli Lemma Let $(A_n)_{n\geq 1}$ and $(B_n)_{n\geq 1}$ be two sequences of events. Assume that:


*

*The two sequences $(A_n)_{n\geq 1}$ and $(B_n)_{n\geq 1}$ are independent.

*$\sum_{n\geq 1} \mathbb{P} (B_n) = \infty$ and the $B_n$ are mutually independent.

*$\mathbb{P}(A_n)\longrightarrow 1$ as $n\to +\infty$.


My question is: do we have $$\mathbb{P}(A_n\cap B_n \,\, \text{i.o.})=1$$ i.e. is $A_n\cap B_n$ realized infinitely often almost surely?
I have the intution that this could be true but I cannot prove it or find a counter-example.
 A: Thanks a lot for your answer Michael! Let me try to give a formal proof even if you already made everythings:
By Borel-Cantelli, almost surely $B_n$ is realized infinitely often. So, as you said, for almost every $\omega$, there exists a subsequence $N_k(\omega)$ such that $\omega\in B_{N_k(\omega)}$ for every every $k\in\mathbb{N}^*$. Then, as you suggest, we define the event $C_k=\{\omega\in A_{N_k(\omega)}\}$. According to your Claim 1, In order to prove that $A_n\cap B_n$ occurs infinitely often, it is enough to prove that $\mathbb{P}(C_k)\underset{k\to\infty}\longrightarrow 1$. To do this, we write:
\begin{align*}
\mathbb{P}(C_k)&=\sum_{l\geq 1}\mathbb{P}(\{\omega\in A_{N_k(\omega)}\}\cap \{N_k(\omega)=l\})\\
          &=\sum_{l\geq 1}\mathbb{P}(A_l\cap \{N_k(\omega)=l\}).
\end{align*}
The sequences $A_n$ and $B_n$ being independent, we deduce that the sequences $A_n$ and $N_k$ are also independent. And then, we get:
$$\mathbb{P}(C_k)=\sum_{l\geq 1}\mathbb{P}(A_l)\cdot\mathbb{P} (N_k=l).$$
Let $\varepsilon>0$. By the third hypothesis, there exist $l_0\in\mathbb{N}^*$ such that for every $l\geq l_0$, we have: $\mathbb{P}(A_l)\geq 1-\varepsilon$. Take $k\geq l_0$, we have:
$$\mathbb{P}(C_k)=\sum_{l= 1}^{l_0-1}\mathbb{P}(A_l)\cdot\mathbb{P} (N_k=l)+\sum_{l\geq l_0}^{\infty}\mathbb{P}(A_l)\cdot\mathbb{P} (N_k=l).$$
As $(N_k)$ is a subsequence, we have $N_k\geq k$. So if $l\leq l_0-1$ and $k\geq l_0$, we have: $\mathbb{P} (N_k=l)=0$. Hence we get for $k\geq l_0$:
$$\mathbb{P}(C_k)\geq (1-\varepsilon)\cdot \sum_{l\geq l_0}^{\infty}\mathbb{P} (N_k=l)=1-\varepsilon.$$
Do you think the proof is correct? 
A: Yes. The independence between $\{A_n\}_{n=1}^{\infty}$ and $\{B_n\}_{n=1}^{\infty}$ is crucial here. Here is a useful claim.
Claim 1:
If $\{C_k\}_{k=1}^{\infty}$ is a sequence of events such that $P[C_k]\rightarrow 1$, then $C_k$ occurs infinitely often with prob 1.
Proof: We have as $M\rightarrow\infty$:
$$ \cap_{k=1}^M \left(\cup_{r \geq k} C_r\right)\searrow \cap_{k=1}^{\infty} \left(\cup_{r \geq k} C_r\right) = \{C_k \: i.o.\} $$
So:
$$P\left[\cap_{k=1}^M\left(\cup_{r\geq k} C_r\right)\right] \searrow P[C_k \: i.o] \quad (*)$$
However, for all positive integers $M$:
$$ \cap_{k=1}^M \left(\cup_{r \geq k} C_r\right)  \supseteq C_M $$
So:
$$ \lim_{M\rightarrow\infty}P[ \cap_{k=1}^M \left(\cup_{r \geq k} C_r\right) ] \geq \lim_{M\rightarrow\infty}P[C_M]=1$$
Hence, $P[ \cap_{k=1}^M \left(\cup_{r \geq k} C_r\right) ]\searrow 1$, and so (*) implies $P[C_k \: i.o.]=1$. $\Box$

Here is an informal way of using Claim 1 to prove your result is true:  We know $B_n$ occurs infinitely often with prob 1. So (with prob 1) we can let $B_{N[k]}$ be an infinite subsequence such that $B_{N[k]}$ occurs for each $k\in \{1, 2, 3, \ldots\}$.  Now define $C_k = A_{N[k]}$, so
$P[C_k] = P[A_{N[k]}]$. Since the sequences $\{A_n\}_{n=1}^{\infty}$ and $\{B_n\}_{n=1}^{\infty}$ are independent, the sequences $\{A_n\}_{n=1}^{\infty}$ and $\{N[k]\}_{k=1}^{\infty}$ are independent, so  $N[k]$ looks like an independently chosen subsequence that does not depend on the $\{A_n\}$ realizations.  Since $P[A_n]\rightarrow 1$, it can be  shown that $P[C_k]\rightarrow 1$.  Thus, claim 1 implies $C_k$ occurs infinitely often (with prob 1), meaning $A_{N[k]}\cap B_{N[k]}$ occurs infinitely often.
