Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Working through some more measure theory, and I am totally stuck on this problem.

Given $\Omega =(x_0,\infty)$ and $g(y)=x_0+e^y,y \in \mathbb{R}$

I am trying to show that given any measure $\mu$ on $(\mathbb{R},B(\mathbb{R}))$ there is a measure $\mu_g$ on $(\Omega,B(\Omega))$ defined by:

$\mu_g(A)=\mu(\left \{ y\in\mathbb{R}: g(y)\in A \right \}), A \in B(\Omega)$

At first I thought this was some kind of extension question, but now I am not sure. One source of confusion is that I always associated the Borel sets with the real line, not just some subset of it, so I'm having a hard time visualizing the sigma-field as defined for the second measure space.

If this measure exists (and I can only assume it does or else why would the problem be asked!), can I use it in conjunction with $\Omega$ and $B(\Omega)$ to define $\mu((a,b]) = F(b)-F(a)$, assuming that F is continuous and increasing?

Also, does it change things if $\Omega$ is finite?

Thank you!

share|improve this question
    
In the line $\mu(A) = \mu\bigl(g^{-1}[A]\bigr)$ do you mean the very same $\mu$ on both sides? Or do we have to read this paragraph as: "Given any measure $\mu$ on $\bigl(\mathbb R, \mathcal B(\mathbb R)\bigr)$ there is a measure $\mu_g$ on $\bigl(\Omega, \mathcal B(\Omega)\bigr)$ defined by the above." –  martini May 22 '12 at 7:26
    
@martini...apologies, but I am trying to look at two different measures as you mention. I will correct for clarity –  Justin May 22 '12 at 14:22
    
Ok ... so now $\mu_g(A) := \mu(g^{-1}[A])$ defines (as you write) a measure. To check this, just prove that $\mu_g\colon \mathcal B(\Omega) \to [0,\infty]$ (1) is well-defined (i. e. $g^{-1}[A] \in \mathcal B(\mathbb R)$ for all $A \in \mathcal B(\Omega)$) and (2) has the properties a measure should have ($\sigma$-addivity, $\mu_g(\emptyset) = 0$). –  martini May 22 '12 at 14:26
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.