1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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]Z^{-1}(\hat{F})=X^{-1}(Y^{-1}(\hat{F}))[/tex]
    [tex]=X^{-1}(({\tilde{\omega}\in\tilde{\Omega}:Y(\tilde{\omega})\in\hat{F}))[/tex]
    [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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Compositions of measurable mappings
  1. Measurable maps (Replies: 5)

  2. Composition of Maps (Replies: 3)

Loading...