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 $F\cap G=\underbrace{A\cap A'}_{\in\mathcal A}\cap \underbrace{B\cap B'}_{\in\mathcal B}$.

  • $\begingroup$ @0xbadf00d Yes. I have edited. $\endgroup$ – Davide Giraudo Jan 8 at 12:46

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