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Thread: sigma algebra over R

  1. #1
    Member Maccaman's Avatar
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    sigma algebra over R

    Consider a $\displaystyle \sigma $-algebra $\displaystyle \mathcal{F} $ on $\displaystyle \mathbb{R} $ containing all intervals $\displaystyle (-\infty, x], x \in \mathbb{R} $. Show that it must contain the sets of the form $\displaystyle (a,b], (a,b) $ and $\displaystyle \lbrace a \rbrace $ $\displaystyle ( $ where $\displaystyle a < b), $ and the countable unions thereof.

    In class my lecturer said that I should keep de Morgan's laws in mind, as well as the fact that $\displaystyle \mathcal{F} $ is closed under complements and under countably infinite solutions.
    Last edited by mr fantastic; Aug 8th 2009 at 05:52 AM. Reason: Re-titled
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  2. #2
    Moo
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    Hello,

    You just have to play with the axioms of a sigma algebra.
    By combining the property of complements and the property of countable unions, we get, by de Morgan's law, that a sigma algebra is closed under countable intersections.

    The complement in $\displaystyle \mathbb{R}$ of $\displaystyle (-\infty,x]$ is $\displaystyle (x,\infty)$ and belongs to the sigma algebra.
    Consider $\displaystyle z>x$
    $\displaystyle (x,\infty)\cap (-\infty,z]=(x,z]$
    (1) So for z>x, (x,z] belongs to the sigma algebra.

    Okay, what about this... ?
    Let's consider, for y>x, $\displaystyle U_n=\left(-\infty,y-\tfrac 1n\right]$
    A sigma algebra is closed under countable union.
    So $\displaystyle U=\bigcup_{n\geq 1} U_n$ belongs to the sigma algebra.
    Thus $\displaystyle U=(-\infty,y)$ belongs to the sigma algebra.

    Intersect $\displaystyle (x,\infty)$ with U to get $\displaystyle (x,y)$ (2)

    Let's take a<b. So by (1) and (2), (a,b] and (a,b) belong to the sigma algebra.
    Intersect (a,b] with the complement of (a,b) and you will get $\displaystyle \{b\}$
    So for any x in $\displaystyle \mathbb{R}$, $\displaystyle \{x\}$ belongs to the sigma algebra.

    I hope all of this is clear... The letters are quite confusing. I suggest you do it on your own once you've catched the tricks...


    Edit : and sorry for the mistakes.
    Last edited by Moo; Aug 9th 2009 at 12:24 AM.
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    Quote Originally Posted by Moo View Post
    Hello,

    You just have to play with the axioms of a sigma algebra.
    By combining the property of complements and the property of countable unions, we get, by de Morgan's law, that a sigma algebra is closed under countable intersections.

    Let's take $\displaystyle y<x$
    $\displaystyle (-\infty,y]\cap (-\infty,x]=[y,x]$
    [snip]
    Ahem (clears throat).

    $\displaystyle (-\infty,y]\cap (-\infty,x]=\color{red}(-\infty,y]$
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    Moo
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    Quote Originally Posted by awkward View Post
    Ahem (clears throat).

    $\displaystyle (-\infty,y]\cap (-\infty,x]=\color{red}(-\infty,y]$
    Oh sorry !

    And this isn't even the only mistake...
    I'll correct them, thanks
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  5. #5
    Member Maccaman's Avatar
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    I apologize for posting this in the wrong forum. Thanks for moving Mr Fantastic. Thank-you awkward and Moo for your help. Greatly appreciated
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