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Sigma Algebras

  1. Sep 8, 2009 #1
    Let f be a function mapping [tex]\Omega[/tex] to another space E with a sigma-algebra[/tex] E. Let A = {A C [tex]\Omega[/tex]: there exists B [tex]\epsilon[/tex] E with A = [tex]f^{-1}(B)[/tex]}. Show that A is a sigma-algebra on [tex]\Omega[/tex].

    Okay, so I should start by showing that [tex]\Omega[/tex] is in A. I wasn't sure if this was as easy as saying that since A is made up of all subsets of [tex]\Omega[/tex], then clearly, [tex]\Omega[/tex] must be in A since it is a subset of itself.


    Next, I would have to show it is closed under complement. Here is what I tried doing.


    [tex]A = f^{-1}(B)[/tex]
    [tex]A^c = (f^{-1}(B))^c = f^{-1}(B^c).[/tex] Since E is a sigma-algebra, [tex]B^c[/tex] is in E, thus by the definition of A, [tex]f^{-1}(B^c)[/tex] is in A so it is closed under complement.


    The last thing would be to show it is closed under countable union. I'm sort of unsure how to set this up, but here is what I tried doing.


    [tex]A_i \epsilon[/tex]A. Then, [tex]A_i = f^{-1}(B_i)[/tex] where [tex]B_i \epsilon[/tex] E. So, [tex]\bigcup_{i=1}^{\infty}A_i = \Bigcup_{i=1}^{\infty}f^{-1}(B_i)=f^{-1}(\bigcup_{i=1}^{\infty}B_i).[/tex] And the union of the [tex]B_i[/tex]'s is in E since it is a sigma-algebra. Therefore, can I conclude that A is closed under countable union and thus, a sigma-algebra?
    1. The problem statement, all variables and given/known data



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    3. The attempt at a solution
     
  2. jcsd
  3. Sep 8, 2009 #2

    LeonhardEuler

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    But A is not made up of all subsets of [tex]\Omega[/tex], just those that are the inverse image of some B[tex]\epsilon[/tex] E. However, we know that E is a sigma algebra, so can you think of a set whose inverse image is [tex]\Omega[/tex]?

    Everything else you did looks ok.
     
  4. Sep 8, 2009 #3
    Ohh okay! So, wouldn't [tex]f^{-1}(E)= \Omega[/tex]? And E is in E, so I think this works.
     
  5. Sep 8, 2009 #4

    LeonhardEuler

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    That works!
     
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