$\displaystyle \cap_{\alpha\in A}\mathcal G_{\alpha}$ is a $\sigma$-algebra Let $(\mathcal G_{\alpha})_{\alpha \in A}$ be an arbitrary family of $\sigma$-algebra defined on an abstract space $\Omega$. Show that $\displaystyle \cap_{\alpha\in A}\mathcal G_{\alpha}$ is also a $\sigma$-algebra. 
 A: A $\sigma$- algebra has three properties:
1.) Nonempty
2.) Closed under complementation
3.) Closed under countable unions
If $\displaystyle \cap_{\alpha\in A}\mathcal G_{\alpha}$ satisfies these properties, then it is a $\sigma$- algebra. Check each
1.) $\displaystyle \emptyset \in \mathcal G_{\alpha}\forall \alpha\in A \Rightarrow \emptyset \in \cap_{\alpha\in A}\mathcal G_{\alpha}$
2.) If $\displaystyle K\in \cap_{\alpha\in A}\mathcal G_{\alpha}$, then $\displaystyle K\in\mathcal G_{\alpha}\forall \alpha\in A \Rightarrow K^{c}\in\mathcal G_{\alpha}\forall \alpha \in A$ (Since each
$\displaystyle \mathcal G_{\alpha}$ is a $\sigma$- algebra.) So $K^{c}\in \cap_{\alpha\in A}\mathcal G_{\alpha}$
3.) Let $\displaystyle\left(A_n\right)_{n\geq 1}$ be a sequence in $\displaystyle \cap_{\alpha\in A}\mathcal G_{\alpha}$. Since each $\displaystyle A_n\in \mathcal G_{\alpha}$, $\cup_{n=1}^{\infty}A_n$
is in $\displaystyle\mathcal G_{\alpha}$ since $\displaystyle \mathcal G_{\alpha}$ is a $\sigma$- algebra for each $\displaystyle \alpha \in A$. Hence $\cup_{n=1}^{\infty}A_n\in\cap_{\alpha\in A}\mathcal G_{\alpha}$.
All three conditions are met, hence we have shown that $\displaystyle \cap_{\alpha\in A}\mathcal G_{\alpha}$ is a
$\sigma$-algebra. $\blacksquare$
