The following is a conclusion in a Chinese textbook "Lecture of Measure Theory" and the proof of it is left as an exercise.
$(\Omega,\mathcal{F})$ is a measurable space, and $\mathcal{C}$ is an algebra consisting of subsets of $\Omega$ such that $\sigma(\mathcal{C})=\mathcal{F}$. Given $\omega\in\Omega$, define
$$\mathcal{F}_{\omega}=\{B\in\mathcal{F}\mid \omega\in B\}\quad\text{and}\quad\mathcal{C}_{\omega}=\{B\in\mathcal{C}\mid \omega\in B\}.$$
It is obvious that $A(\omega):=\bigcap_{B\in\mathcal{F}_{\omega}}B\subset\bigcap_{B\in\mathcal{C}_{\omega}}B$, but how to prove the opposite inclusion relation, i.e.
$$\bigcap_{B\in\mathcal{C}_{\omega}}B\subset\bigcap_{B\in\mathcal{F}_{\omega}}B.$$
Best Answer
With the help of my friend, here is a solution of this question:
Define $$\mathcal{G}=\left\{A\in\mathcal{F}\mid \bigcap_{B\in\mathcal{C}_{\omega}}B\subset A\quad\text{or}\quad\bigcap_{B\in\mathcal{C}_{\omega}}B\subset A^c\right\}.$$ We can verify that $\mathcal{C}\subset\mathcal{G}$ and $\mathcal{G}$ is a $\pi$-system as well as a $\lambda$-system, then $$\mathcal{F}=\sigma(\mathcal{C})\subset\sigma(\mathcal{G})=\lambda(\mathcal{G})=\mathcal{G}.$$ For any $A\in\mathcal{F}$ and $\omega\in A$, it follows from $\mathcal{F}=\mathcal{G}$ that $A\in\mathcal{G}$, therefore $\bigcap_{B\in\mathcal{C}_{\omega}}B\subset A$ and thus $$\bigcap_{B\in\mathcal{C}_{\omega}}B\subset\bigcap_{B\in\mathcal{F}_{\omega}}B.$$