Today i was reading Tao's book on measure theory and there is an exercise where the pushforward of a measure is defined. So i started to think if the following fact is true:
Given a $C^1(\mathbb{R}^d)$ diffeomorphism $\varphi$, such that the Jacobian determinant never vanishes, is it true that $\varphi_{*}m(E) = \int_{E} \frac{1}{|\text{Jac}(\varphi)|}dm$?
I didn't think too much of that, so I confined my attention to the case where $d=1$. I thought this:
let $g: \mathbb{R} \to \mathbb{R}$ be a $C^1$ diffeomorphism with $g'(x) \ne 0$, so we can assume $g'>0$, thus we get $g_{*}m([a, b]) = [g^{-1}(a), g^{-1}(b)]$ for every $a
Note that $g_{*}m[a, b] = g^{-1}(b) - g^{-1}(a) = \int_{[a, b]} \frac{1}{g'} dm$ thanks to the fundamental theorem of calculus. (Similar it can be done for half-open and open intervals). Note that if $E$ is a finite union of almost disjoint closed, open, half-open intervals, the claim is still true. Now take $A$ open, we have that $A$ can be expressed as countable union of increasing elementary sets, thus the claim is easily obtained also in this case. If $E$ is measurable with finite measure, express $E$ as a decreasing sequence of open sets with finite measure, and then derive the general case in an obvious manner.
Doing so we can prove that if $f: \mathbb{R} \to [0, \infty]$ is borel measurable, we have:
$\int_{\mathbb{R}} f \circ g dm = \int_{\mathbb{R}} \frac{f}{g'}dm$
Is my reasoning correct?
Is it possible to prove that $g$ is a lebesgue morphism?