TITLE: Generator of sigma field that is generated by an intersection of a decreasing sequence of generators.
QUESTION [5 upvotes]: Let $\Omega$ be a non-empty set and let $(E_n)_{n\in\mathbb{N}}$ be 
sequence of intersection stable systems over $\Omega$ (i.e. $E_n$ is a set consisting of subsets of $\Omega$ satisfying $A, B\in E_n$ $\Longrightarrow$ $A \cap B\in E_n$)  such that $E_{n+1} \subset E_n$ for all $n\in\mathbb{N}$.
Is it true that 
$$
\sigma\left(\bigcap_{n\in\mathbb{N}} E_n\right)= \bigcap_{n\in\mathbb{N}} \sigma(E_n),
$$
where $\sigma(E)$ denotes the smallest $\sigma$-field over $\Omega$ that contains $E$?

REPLY [4 votes]: The statement is wrong: We consider a Brownian Motion $B= (B_t)_{t\in [0,1]}$ defined on the space $\Omega = C([0,1])$ of continuous functions from [0,1] to $\mathbb{R}$. Consider the canonical filtration $(\mathcal{F}_t)_{t\in [0,1]}$. 
It is a known fact that the filtration 
is not right continuous, i.e one has $\mathcal{F}_0 \not= \bigcap_{n\in\mathbb{N}}\mathcal{F}_{1/n}$.  
This can now be used to give an counter-example to the above statement. 
For $t\in [0,1]$ we denote by $E_t$ the system of all sets
$\{B_{t_1}\in A_1, \dots, B_{t_n}\in A_n\}$
where $n\in\mathbb{N}$, $0\le t_1\le \dots \le t_n\le t$ and $A_1,\dots, A_n$ are arbitrary Borel-sets of $\mathbb{R}$.
Then, $\bigcap_{n\in\mathbb{N}} E_{1/n} = E_0$ and 
$\sigma(E_0) = \mathcal{F}_0 \not= \bigcap_{n\in\mathbb{N}}\mathcal{F}_{1/n} = \bigcap_{n\in\mathbb{N}} \sigma(E_{1/n})$