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Thread: restriction of sigma-algebra

  1. #1
    Senior Member Dinkydoe's Avatar
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    restriction of sigma-algebra

    I hope some of you can help me out a little.

    I read the following in a article, like it's an obvious fact. (http://research.microsoft.com/en-us/...papers/poi.pdf page 18, lemma 11)

    Let $\displaystyle \mathcal{F}$ be a $\displaystyle \sigma$-algebra on the probability space $\displaystyle \Omega=\mathbb{R}$, and let
    $\displaystyle S=[-r,r]$ be some bounded interval.

    Let $\displaystyle \mathcal{F}_S$ be the $\displaystyle \sigma$-algebra, generated by the restriction of $\displaystyle \mathcal{F}$ to $\displaystyle S$. Let $\displaystyle A\in \mathcal{F} = \mathcal{F}_{\mathbb{R}}$. Why does there exist for every $\displaystyle \epsilon>0$ some $\displaystyle r=r(\epsilon)$ and an event $\displaystyle A_{\epsilon}\in \mathcal{F}_{[-r,r]}$ such that $\displaystyle \mathbb{P}(A\Delta A_{\epsilon}) < \epsilon$. ??

    In otherwords, $\displaystyle A$ can sufficiently be approximated by $\displaystyle A_{\epsilon}$. Is this obvious?
    Is it because $\displaystyle \lim_{r\to\infty}\mathcal{F}_{[-r,r]}=\mathcal{F}$...(and continuity of the probability measure $\displaystyle \mathbb{P}$?)

    Also is claimed that $\displaystyle \mathcal{F}_{(-2r,0]}\subset \mathcal{F}_{(-\infty,0]}$ Is this true?

    Something general like $\displaystyle \mathcal{F}_A\subset \mathcal{F}_B$ if $\displaystyle A\subset B$, does not hold right?

    Really looking forward to your suggestions.
    Last edited by Dinkydoe; Oct 15th 2011 at 07:39 AM.
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