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

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

    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!
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  2. #2
    Senior Member
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    Staten Island, NY
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    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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