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Math Help - sigma algebra over R

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

    Consider a  \sigma -algebra  \mathcal{F} on  \mathbb{R} containing all intervals  (-\infty, x], x \in \mathbb{R} . Show that it must contain the sets of the form  (a,b], (a,b) and \lbrace a \rbrace ( where  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  \mathcal{F} is closed under complements and under countably infinite solutions.
    Last edited by mr fantastic; August 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 \mathbb{R} of (-\infty,x] is (x,\infty) and belongs to the sigma algebra.
    Consider z>x
    (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, U_n=\left(-\infty,y-\tfrac 1n\right]
    A sigma algebra is closed under countable union.
    So U=\bigcup_{n\geq 1} U_n belongs to the sigma algebra.
    Thus U=(-\infty,y) belongs to the sigma algebra.

    Intersect (x,\infty) with U to get (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 \{b\}
    So for any x in \mathbb{R}, \{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; August 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 y<x
    (-\infty,y]\cap (-\infty,x]=[y,x]
    [snip]
    Ahem (clears throat).

    (-\infty,y]\cap (-\infty,x]=\color{red}(-\infty,y]
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  4. #4
    Moo
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    Quote Originally Posted by awkward View Post
    Ahem (clears throat).

    (-\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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