Problem about $\sigma$-algebra Space $\Omega$, $\mathcal C$ is a algebra,$\mathcal F=\sigma(\mathcal C)$ is a $\sigma$-algebra.
define:$\mathcal F_\omega=\{B\in\mathcal F|\omega\in B\}$,$\mathcal C_\omega=\{B\in\mathcal C|\omega\in B\}$
Then:
$$\bigcap_{B\in\mathcal F_\omega}B=\bigcap_{B\in\mathcal C_\omega}B$$
Obviously we have $$\bigcap_{B\in\mathcal F_\omega}B\subset\bigcap_{B\in\mathcal C_\omega}B$$
now I prove the opposite direction.
$\forall \omega_0\in $RHS,assume $\omega_0\notin$ LHS,
$\exists B_0\in \mathcal F,\omega\in B_0,\omega_0\notin B_0$,here I want to find a relationship between $\mathcal C_\omega $ and $B_0$(Wish $B_0$ can be expressed by  $\mathcal C_\omega $)
 A: The idea is that the generating class $\mathcal{C}$ should be sufficient to say when you can distinguish points in $\Omega$ by sets in $\mathcal{F}$. There is no simple way to express elements of $\mathcal{F}$ in terms of elements of $\Omega$, so we must proceed in a different way.
Let's stick to the idea of separating points by elements of $\mathcal{C}$. Let $\omega\in \Omega$ be fixed, and let $x\in \bigcap_{B\in \mathcal{C}_\omega}B$. Define the class
$$\mathcal{A}=\left\{B\subseteq\Omega:\omega\not\in B\text{ or }x\in B\right\}.$$
Let's show that $\mathcal{A}$ is a $\sigma$-algebra which contains $\mathcal{C}$:
First, let's show that $\mathcal{C}\subseteq\mathcal{A}$. Let $C\in\mathcal{C}$. If $\omega\not\in C$, then $C\in\mathcal{A}$. If $\omega\in C$, then $C\in \mathcal{C}_\omega$, thus $x\in\bigcap_{B\in\mathcal{C}_\omega}B\subseteq C$, so $C\in\mathcal{A}$ again. Either way, we obtained $\mathcal{C}\subseteq\mathcal{A}$.
Now, let's show that $\mathcal{A}$ is closed by complements: Let $B\in\mathcal{A}$. We have two possibilities:


*

*$\omega\in B$. In this case, $\omega\not\in B^c$, thus $B^c\in\mathcal{A}$.

*$\omega\not\in B$. Then $\omega\in B^c$, so $B^c\in\mathcal{C}_\omega$ (remember that $\mathcal{C}$ is an algebra, thus closed by complements!), thus $x\in\bigcap_{C\in\mathcal{C}_\omega}C\subseteq B^c$, therefore $B^c\in\mathcal{A}$.
Hence $\mathcal{A}$ is closed by complements.
I will leave the proof that $\mathcal{A}$ is closed by countable unions to you (if $(B_n)_{n=1}^\infty\subseteq \mathcal{A}$, we have two cases: either some $B_n$ contains $\omega$ or none of the $B_n$ contains $\omega$. In each case, conclude that $\bigcup_{n=1}^\infty B_n\in\mathcal{A}$).
Since $\mathcal{F}$ is the smallest $\sigma$-algebra containing $\mathcal{C}$, then $\mathcal{F}\subseteq\mathcal{A}$. Now we are able to show that $x\in\bigcap_{B\in\mathcal{F}_\omega}B$. Let $B\in\mathcal{F}_\omega$. Then $\omega\in B$. But also $B\in\mathcal{F}\subseteq\mathcal{A}$, so, by definition of $\mathcal{A}$, it is necessary that $x\in B$.
Since $B\in\mathcal{F}_\omega$ was arbitrary, we conclude that $x\in\bigcap_{B\in\mathcal{F}_\omega}B$, as we wanted.
Thus, we have showed that $\bigcap_{B\in\mathcal{C}_\omega}B\subseteq\bigcap_{B\in\mathcal{F}_\omega}$, which is what was left.
