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)

My answer

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*($\displaystyle h^{-1}(b) - h^{-1}(a)$) Also, note h is increasing.

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