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

  1. #1
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    Show intersection of sigma-algebras is again a sigma-algebra

    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!
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  2. #2
    Senior Member
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    Looks like you have the idea, but your notation is a little off. For example you mean $\displaystyle \emptyset \in \mathcal G_{\alpha}$.

    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:

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

    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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