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Compositions of measurable mappings

  1. Mar 18, 2010 #1
    1. The problem statement, all variables and given/known data
    I have to prove that compositions of measurable mappings are measurable.

    i.e. If X is [tex]F/\widetilde{F}[/tex] measurable and Y is [tex]\widetilde{F}/\widehat{F}[/tex] measurable, then Z:=YoX:[tex]\Omega\rightarrow\widehat{\Omega}[/tex] is [tex]F/\widehat{F}[/tex] measurable.

    2. Relevant equations
    X is [tex]F/\widetilde{F}[/tex] measurable if [tex]X^{-1}(\widetilde{F})=(\omega \in \Omega: X(\omega)\in\widetilde{F})\in F}[/tex]
    (that last F is meant to be a curly F, sigma algebra, and the brackets before the little omega and before the last "element of" are meant to be braces.)

    3. The attempt at a solution
    I know you're not supposed to help if I haven't attempted it, but I've never been great at proofs and honestly don't know where to start. Can anyone give me a nudge in the right direction?
  2. jcsd
  3. Mar 18, 2010 #2
    Okay, so I came up with something that seems wrong, but can someone tell me if it holds?

    [tex]=(\omega\in\Omega:X(\omega)\in\tilde{F})\in F[/tex]
    [tex]=(\omega\in\Omega:Z(\omega)\in\hat{F})\inF[/tex] (since Z:YoX)

    Again, some of those brackets are meant to be braces...
    Last edited: Mar 18, 2010
  4. Mar 18, 2010 #3
    I'm going to relabel the domains as X, Y, Z and the functions f:X -> Y, g:Y -> Z.

    You need to show

    (f o g):X -> Z

    has the property that for every measurable set E in Z, the preimage (f o g)^{-1}(E) is a measurable set in X.

    You know this property holds for both f and g (as they are measurable). What remains is show it holds for the composition.
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