Show intersection of sigma-algebras is again a sigma-algebra

**ecnanif** · November 20th 2010, 06:06 AM

Hi

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

Solution(?)

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

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

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

Thanks for your help!

**DrSteve** · November 20th 2010, 07:21 AM

Looks like you have the idea, but your notation is a little off. For example you mean $\emptyset \in \mathcal G_{\alpha}$ .

In your last line you can't use A - it's already being used for the index set. Perhaps use X instead. Your last line should look something like this:

$X\in \mathcal H \Leftrightarrow (\forall \alpha \in A) X\in G_{\alpha} \Leftrightarrow (\forall \alpha \in A) \Omega - X \in G_{\alpha} \Leftrightarrow \Omega - X \in \mathcal H$ .

Of course you still have to check closure under countable unions and intersections, but each part of the proof is quite straightforward.

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