I'm given two sequences of events $\{A_n\}$ and $\{B_n\}$ with $ \mathbb{P}(\liminf_n A_n) = \mathbb{P}(\limsup_n B_n) = 1$. Can we deduce that $\mathbb{P}(\limsup_n A_n \cap B_n) = 1$?
My attempt was to write $$ \mathbb{P}(\liminf_n A_n) = 1 \Rightarrow \mathbb{P}\left( \left\{ w : \sum_n 1_{A_n}(w) \text{ finite} \right\} \right) = 0 $$ Additionally, we know that $$ \mathbb{P}(\limsup_n (A_n \cap B_n)) = \mathbb{P}\left(\left\{w : \sum_{n=1}^{\infty} 1_{A_n}(w) \cdot 1_{B_n}(w) =\infty \right\}\right) \\= \mathbb{P}\left( \left\{ w : \sum_n 1_{A_n}(w) = \infty, \sum_n 1_{A_n}(w) \cdot 1_{B_n}(w) = \infty \right\} \cup \left\{ w : \sum_n 1_{A_n}(w) < \infty, \sum_n 1_{A_n}(w) \cdot 1_{B_n}(w) = \infty \right\} \right) \\ = \mathbb{P}\left(\left\{ w: \sum_{k=1}^{\infty} 1_{B_{n_k}}(w) = \infty \right\}\right) + 0 = \mathbb{P}(\limsup_n B_n) = 1 $$ where in the above we made use of the fact that since $w \in \liminf_n A_n$, the terms in the sum $\sum_n 1_{A_n} 1_{B_n}(w)$ have $1_{A_n} = 0$ for only a finite set of indices. Is that correct, or am I missing something?
Additionally, does this hold when I change the condition of $\mathbb{P}(\liminf_n A_n) = 1$ to $\mathbb{P}(\limsup_n A_n) = 1$? I've been trying to find a counterexample to no avail.
Note that $\{\liminf A_n\} \cap \{\limsup B_n\} \subset \limsup (A_n \cap B_n)$. If two events have probability 1 then their intersection also has probability 1. Hence $P(\{\liminf A_n\} \cap \{\limsup B_n\})=1$ which gives $P\{\limsup (A_n \cap B_n)\}=1$.