Compositions of measurable mappings

[muso07]

Homework Statement

I have to prove that compositions of measurable mappings are measurable.

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

Homework Equations

X is F/\widetilde{F} measurable if X^{-1}(\widetilde{F})=(\omega \in \Omega: X(\omega)\in\widetilde{F})\in F}
(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?

 

[muso07]

Okay, so I came up with something that seems wrong, but can someone tell me if it holds?

Z^{-1}(\hat{F})=X^{-1}(Y^{-1}(\hat{F}))
=X^{-1}(({\tilde{\omega}\in\tilde{\Omega}:Y(\tilde{\omega})\in\hat{F}))
=(\omega\in\Omega:X(\omega)\in\tilde{F})\in F
=(\omega\in\Omega:Z(\omega)\in\hat{F})\inF (since Z:YoX)

Again, some of those brackets are meant to be braces...

 

[some_dude]

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.