$\limsup$ and probability I cant do this:
assume that $P(\limsup A_n)=1$ and $P(\liminf B_n)=1$. Prove that $ P(\limsup (A_n \cap B_n))=1$.
The most I get is that if $P(\liminf A_n)=1  $ this is right.
 A: Since it's a homework question, I will only give two intermediate steps. 


*

*Notice that $\mathbb P(\limsup_nA_n\cap \liminf_n B_n)=1$.

*Show that $\limsup_nA_n\cap \liminf_n B_n\subset\limsup_n(A_n\cap B_n)$.
A: (Not the best nor the shortest proof). By the definitions of $\lim \sup$ and $\lim \inf$ you have that $$\lim \sup (A_n\cap B_n):=\{x \in X: x \in A_n\cap B_n \text{ for infinitely many $n$}\}$$ From the given conditions you have that there exist two set of measure zero $N_A$ and $N_B$ such that $$x\in\lim\sup A_n=\{x \in X: x \in A_n\text{ for infinitely many $n$}\}$$ for all $x\in X\backslash N_A$ and $$x\in\lim\inf B_n=\{x \in X: x \in B_n \text{ finally for every $n$}\}$$ for all $x\in X\backslash N_B$. Then every $x\in X\backslash(N_A\cup N_B)$ (which is still a set of measure $1$) belongs to infinitely many $A_n$ and finally to every $B_n$. So it belongs to infinitely many sets of the form $A_n\cap B_n$ and thus it belongs to the set $$\lim \sup (A_n\cap B_n)$$ which proves that the measure of the latter set is still $1$.  
