Let $\mathcal{G}$ and $\mathcal{H}$ be two $\sigma$-algebras on a set $\Omega$.
Does it hold that $$\sigma(\mathcal{H},\mathcal{G}) = \sigma\{G\cap H: G\in\mathcal{G}, H\in \mathcal{H}\}?$$
How to prove it or disapprove it?
$\endgroup$1 Answer
$\begingroup$Let $\mathcal{G}$ and $\mathcal{H}$ be two $\sigma$-algebras on a set $\Omega$. Let $S=\{G\cap H: G\in\mathcal{G}, H\in \mathcal{H}\}$.
Since $\Omega \in \mathcal{G}$, we have that, for all $H\in \mathcal{H}$, $H = \Omega \cap H \in S$. So $\mathcal{H} \subseteq S$. In a similar way, since $\Omega \in \mathcal{H}$, we have that $\mathcal{G} \subseteq S$.
So we have $\mathcal{G} \cup \mathcal{H} \subseteq S$. Then we have
$$ \sigma(\mathcal{H},\mathcal{G}) = \sigma(\mathcal{H} \cup \mathcal{G}) \subseteq \sigma(S)\tag{1} $$
On the other hand, we have that $S \subseteq \sigma(\mathcal{H},\mathcal{G})$, so we have:
$$ \sigma(S) \subseteq \sigma(\mathcal{H},\mathcal{G}) \tag{2} $$
From $(1)$ and $(2)$, we have
$$\sigma(\mathcal{H},\mathcal{G}) = \sigma(S) $$
that is$$\sigma(\mathcal{H},\mathcal{G}) = \sigma\{G\cap H: G\in\mathcal{G}, H\in \mathcal{H}\}$$
$\endgroup$
