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Compositions of measurable mappings

  • Thread starter muso07
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  • #1
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Homework Statement


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.

Homework 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.)

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?
 

Answers and Replies

  • #2
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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:
  • #3
93
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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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