Let $(\Omega,\mathcal{F})$ be a measurable space and $\mathcal{F}_1 \supseteq \mathcal{F}_2 \supseteq…$ be a sequence of decreasing subsets of $\mathcal{F}$. For $\mathcal{A} \subseteq \mathcal{F}$ let $\sigma(\mathcal{A})$ denote the sigma algebra generated by $\mathcal{A}$.
Does $\sigma(\bigcap_{n=1}^\infty \mathcal{F}_n)=\bigcap_{n=1}^\infty \sigma(\mathcal{F}_n)$ hold?
I can not find a counter example but I also didn't find any proof for this. I would be grateful for any hints.
Best Answer
Let $A \in \cap_n\mathcal{F}_n$. Then $A \in \mathcal{F}_n,\,\forall n$ which implies that $A \in \sigma(\mathcal{F}_n),\forall n$. Therefore $$\cap_n\mathcal{F}_n\subseteq \cap_n \sigma(\mathcal{F}_n)\implies \sigma(\cap_n\mathcal{F}_n)\subseteq \cap_n \sigma(\mathcal{F}_n)$$ Now consider $\mathcal{F}=\mathcal{B}(\mathbb{R})$ and an enumeration of the rationals $(q_n)_{n \in \mathbb{N}}$. Consider $$\mathcal{F}_n=\{(-\infty,q_k],\,k\geq n\}$$ We have $\cap_n\mathcal{F}_n=\emptyset$ so $\sigma(\cap_n\mathcal{F}_n)=\{\emptyset,\mathbb{R}\}$. However, $\sigma(\mathcal{F}_n)=\mathcal{B}(\mathbb{R}),\,\forall n$. So the converse is not valid.