Author 
Message 
LpSm2126 New poster
Post Number: 5
 Posted on Wednesday, 02 April, 2014  01:04 pm:  
If [inline]f: X \to \mathbb{R}[/inline] is an arbitrary function and [inline]\Sigma = \sigma([x\in X  f(x) > a ])[/inline], (should be set brackets  can't seem to get them to show), show that if [inline]g:\mathbb{R} \to \mathbb{R}[/inline] is bounded and [inline]\Sigma[/inline]measurable then there is a Borelmeasurable [inline]G[/inline] such that [inline]g=G \circ f[/inline]. The hint is to use the monotone class theorem, but I can't seem to get started. It's not clear to me what I'm supposed to show using the monotone class theorem, or what the vector space H is in the statement (here  http://en.wikipedia.org/wiki/Monotone_class_theorem). If someone could just point me in the right direction, that'd be great. Thanks. 
Dominic Yeo Veteran poster
Post Number: 419
 Posted on Sunday, 13 April, 2014  11:07 pm:  
Is there a class of functions g for which the result is easier to show (perhaps by definition)? These could be defined independently or by making relation to f and [inline]\Sigma[/inline]. Then you might be able to write some general g as a suitable monotone limit of the g's you can deal with, and use the monotone class theorem. 
