Is the intersection of two $\sigma$-algebras the $\sigma$-algebra generated by the intersection of the generators? Let $\Omega$ be a set and $A, B \subseteq \mathcal{P}(\Omega)$ be two collections of subsets closed under intersection (sometimes called $\pi$-system).
I would like to know if
$$ \sigma(A\cap B) = \sigma(A) \cap \sigma(B) $$
It's easy to show that $\sigma(A\cap B) \subseteq \sigma (A) \cap \sigma(B)$: the RHS is a $\sigma$-algebra that contains $A\cap B$ and so it contains also $\sigma (A\cap B)$ by definition of $\sigma(\cdot)$.
It can also be shown that the statement holds if $A\subseteq B$ (so that $A\cap B = A$), but I can't prove the general case.
Any idea on how to proceed to prove it or find a counterexample?

Here's a counterexample: let $\Omega=\mathbb{R}$, $A=\{(-\infty,t]|t\in\mathbb{R}\}$ and $B=\{(a,b)|a,b\in\mathbb{R}\}\cup\{\emptyset\}$ - then $A$ and $B$ are closed under intersection and disjoint from each other but they generate the same $\sigma$-algebra (i.e. the standard Borel $\sigma$-algebra on $\mathbb{R}$).
