I wonder the following property is true or not: $$ \sigma(\bigcap_n \mathcal C_n)=\bigcap_n \sigma (\mathcal C_n) $$ where $\mathcal C_n$ are decreased collections of sets in a measurable space. And "σ " denotes the generated σ-algebra.
The motivation of this question is that I want to prove or disprove the following question in probability theory:
Suppose $\mathcal F_t$ is a right-continuous filtration, and $\mathcal G$ is a σ-algebra. Then is the filtration $\mathcal F_t\vee \mathcal G$ is still right-continuous? Actually we need to prove $\bigcap_{h>0}\sigma(\mathcal F_{t+h}\cup \mathcal G)=\sigma(\bigcap_{h>0}\mathcal \{\mathcal F_{t+h}\cup \mathcal G\})$. If we let $\mathcal C_n=\mathcal F_{t+1/n}\cup \mathcal G$ then we will get the result.
Below are my attempts:
The relation "$\subset$" is trivial, and it is also trivial when $\mathcal C_n$ are σ-algebra. The hard part is the relation "$\supset$".I try to find a counterexample that $\bigcap_n \mathcal C_n=\emptyset$ with some proper measurable set $A\in \sigma (\mathcal C_n)\forall n$. I've tried some simple examples like $\mathcal C_n=\{n,n+1,...\}\subset\mathbb N$, but the relation turn out to be right.
About the filtration question, I proved when $\mathcal G$ is the σ-algebra generated by a countable partition of the whole space: Since every set $A$ in $\mathcal F_t\vee \mathcal G$ can be writen as $A=\bigcup_n(G_n\cap F_n^t)$, where $\{G_n\}\subset \mathcal G$ is the partition and $F_n^t\in\mathcal F_t\forall n$. Then for all $h>0, A=\bigcup_n(G_n\cap F_n^{t+h})$. Notice that $G_n$ are disjoint, thus $A=\bigcap_h\bigcup_n(G_n\cap F_n^{t+h})=\bigcup_n(G_n\cap \bigcap_hF_n^{t+h})\in\mathcal F_{t+}\vee\mathcal G=\mathcal F_t\vee\mathcal G$.
Let $\langle q_n\rangle$ be an enumeration of the rational numbers. For each $n$, let $\mathcal{C}_n$ be the family of all intervals with rational endpoints different from $q_1,\ldots,q_{n-1}$. This gives you a counterexample.