Let $(\Omega, \mathcal{A})$ a measurable space; $\mathcal{B} \subseteq \mathcal{A}$ a countable algebra, such that $\sigma(\mathcal{B}) = \mathcal{A}$. For $\omega \in \Omega$;
$$A(\omega) = \cap \{A:\, A\in \mathcal{A}, \, \omega \in A\} $$
Show that $A(\omega) = \cap \{B:\, B\in \mathcal{B}, \, \omega \in B\}$.
I was trying to do this question, the part "$\subseteq$" is easy, since $\{B:\, B\in \mathcal{B}, \, \omega \in B\} \subseteq \{A:\, A\in \mathcal{A}, \, \omega \in A\}$, but I don't know how to prove the "$\supseteq$". I tried a proof by contradiction considering $x\in \cap \{B:\, B\in \mathcal{B}, \, \omega \in B\}\setminus A(\omega)$ and considering the set "$A(x)$", I tried to construct a "good" $\sigma-$algebra since our $\sigma-$algebra is the "smallest" that contains $\mathcal{B}$, but none of my attempts worked well.
So, my question is, how to prove $A(\omega) \supseteq \cap \{B:\, B\in \mathcal{B}, \, \omega \in B\}$?
Best Answer
Define an equivalence relation $\sim$ on $\Omega$ such that $\omega\sim\omega'$ if and only if $\omega$ and $\omega'$ lie in exactly the same elements of $\mathcal{B}.$ Note that $A(\omega)$ is simply the equivalence class containing $\omega$. Let $\mathcal{F}$ consist of arbitrary unions of such equivalence classes. Verify that $\mathcal{F}$ is a $\sigma$-algebra containing $\mathcal{B}$ and that $$A(\omega) = \cap \{A:\, A\in \mathcal{F}, \, \omega \in A\}.$$