Hi

Let $\displaystyle (\mathcal{G}_{\alpha})_{\alpha \in A} $ be an arbitrary familiy of $\displaystyle \sigma$-algebras defined on an abstract space $\displaystyle \Omega $. Show that $\displaystyle \mathcal{H} = \cap_{\alpha \in A} \mathcal{G}_{\alpha} $ as also a $\displaystyle \sigma$-algebra.

Solution(?)

Since $\displaystyle \left\{\emptyset\right\},\left\{\Omega\right\} \in \mathcal{G}_{\alpha},\alpha \in A $, we have that $\displaystyle \cap_{\alpha \in A} \mathcal{G}_{\alpha} = \mathcal{H} $ as always "at least" $\displaystyle \left\{\emptyset,\Omega\right\}$, which is the trivial $\displaystyle \sigma$-field.

In fact, if $\displaystyle \mathcal{G}_{\alpha_{i}} \cap \mathcal{G}_{\alpha_{j}} = \emptyset $, $\displaystyle \alpha_{i},\alpha_{j} \in A $, then $\displaystyle \mathcal{H} = \left\{\emptyset,\Omega\right\} $.

Otherwise, if say $\displaystyle A \subset \Omega $ is in $\displaystyle \mathcal{G}_{\alpha} $, $\displaystyle \alpha \in A$, then $\displaystyle A^{c} \in \mathcal{G}_{\alpha} $ $\displaystyle \Rightarrow A,A{c} \in \mathcal{H} $, and $\displaystyle \mathcal{H} $ is a $\displaystyle \sigma$-algebra.

Thanks for your help!