# About unions of $\sigma$-algebra being sigma algebras

Let $\Omega$ be a set and $\mathcal{A}$ and $\mathcal{B}$ be two sigma-algebras on $\Omega$. Put $$\mathcal{F}=\{A\cap B:A\in\mathcal{A}\;\text{and}\;B\in\mathcal{B}\}.$$

I have two question which seem intuitively true, but I am unable to prove them, since I am not a mathematician, but an engineer with an interest in probability theory:

1. Is it true that the sigma-algebra generated by $\mathcal{F}$ equals the sigma-algebra generated by $\mathcal{A\cup B}$, i.e. do we have $$\sigma(\mathcal{F})=\sigma(\mathcal{A}\cup\mathcal{B})?$$
2. Does $\mathcal{F}$ satisfy the property $$F,G\in\mathcal{F}\implies F\cap G\in\mathcal{F}?$$

1. $$\mathcal F$$ certainly contains $$\mathcal A$$ and $$\mathcal B$$, hence $$\sigma(\mathcal F)\supset \sigma(\mathcal A\cup\mathcal B)$$. If $$F\in \mathcal F$$, then $$F=A\cap B$$ for some $$A\in\mathcal A$$ and $$B\in\mathcal B$$. Sets of this form belong to the $$\sigma$$-algebra generated by $$\mathcal A\cup\mathcal B$$, as finite intersection of elements of $$\mathcal A\cup\mathcal B$$.
2. Yes, since $$\mathcal A$$ and $$\mathcal B$$ are stable under finite intersections: write $$F=A\cap B$$, $$G=A'\cap B'$$, with $$A,A'\in\mathcal A$$ and $$B,B'\in\mathcal B$$. Then $$F\cap G=\underbrace{A\cap A'}_{\in\mathcal A}\cap \underbrace{B\cap B'}_{\in\mathcal B}$$.