# More Measure Theory

• April 9th 2010, 10:34 AM
southprkfan1
More Measure Theory
I need help with this question, I think I'm doing something wrong in my answer

Let h be a diffeomorphism from R to R such that 1 <= h'(x) <= 2 for all x.

Show that for all measurable sets E, m(h(E)) <= 2*m(E)

First note, since 1 <= h'(x) <= 2, then by the Mean Value Thm, (h(b) - h(a)) <= 2*(b-a) for all a, b in R. Or (b - a) <=2*( $h^{-1}(b) - h^{-1}(a)$) Also, note h is increasing.

m(h(E)) = inf{ $\sum I_k$; $h(E) \subset \cup I_k$} = inf{ $\sum I_k$; $E \subset \cup h^{-1}(I_k)$} <= inf{ $\sum 2*h^{-1}(I_k)$; $E \subset \cup h^{-1}(I_k)$} = 2*m(E).

I feel like I'm skipping a lot of steps in terms of h being onto and can map to any open interval from any open interval.....thoughts?
• April 9th 2010, 11:05 AM
Focus
Quote:

Originally Posted by southprkfan1
I need help with this question, I think I'm doing something wrong in my answer

Let h be a diffeomorphism from R to R such that 1 <= h'(x) <= 2 for all x.

Show that for all measurable sets E, m(h(E)) <= 2*m(E)

First note, since 1 <= h'(x) <= 2, then by the Mean Value Thm, (h(b) - h(a)) <= 2*(b-a) for all a, b in R. Or (b - a) <=2*( $h^{-1}(b) - h^{-1}(a)$) Also, note h is increasing.

m(h(E)) = inf{ $\sum I_k$; $h(E) \subset \cup I_k$} = inf{ $\sum I_k$; $E \subset \cup h^{-1}(I_k)$} <= inf{ $\sum 2*h^{-1}(I_k)$; $E \subset \cup h^{-1}(I_k)$} = 2*m(E).

I feel like I'm skipping a lot of steps in terms of h being onto and can map to any open interval from any open interval.....thoughts?

I assume you mean that m is a Lebesgue measure on the Borel sets.

First off, h(E) is measurable as h^-1 is continuous. And you need it again in the final equality and you also need the bijection. Think about the definition of the Lebesgue measure, you need the infemum, over all open intervals.

As for your argument, you are using the fact that it is a bijection to have that $b-a\leq 2(h^{-1}(b)-h^{-1}(a))$ as for every point x you need a unique y such that $f^{-1}(y)=x$ (otherwise the statement may not be true).

I also suggest that you write $\sum |I_k|$ because the I_k are sets and the sum notation does not make any sense (in the old days some people used it for union).