Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a Lebesgue integrable function and $\varphi: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous differentiable bijection. Question: How to prove that: $$ \int_{\mathbb{R}} f \circ \varphi |\varphi'|\mathrm{d}m = \int_{\mathbb{R}} f\mathrm{d}m \quad (Lebesgue)$$
I want to prove this for the simple case first where $f = \chi_A$ but get stucked: I turn the equation to be proved into $ \int_{\mathbb{R}} \chi_{\varphi^{-1}(A)}|\varphi'|\mathrm{d}x = \int_{\mathbb{R}} f\mathrm{d}m$ and don't know how to proceed.
Since $\phi$ is a bijection it is either strictly increasing or decreasing. Suppose $\phi$ is strictly increasing for definiteness.
If $a<b$, then $\int 1_{[a,b]} (\phi(x)) \phi'(x) dx = \int 1_{[\phi^{-1}(a), \phi^{-1}(b)]}(x) \phi'(x) dx = \phi(x) \mid_{\phi^{-1}(a)}^{\phi^{-1}(b)} = b-a$.
The remainder is a standard line of argument in measure theory.
Hence the result is true for intervals. Since an open set can be written as a countable collection of disjoint intervals, the dominated convergence theorem shows that the result is true for open sets.
Outer regularity along with the dominated convergence theorem shows that the result is true for bounded measurable sets.
Linearity shows that it is true for simple functions and the dominated convergence theorem shows that it is true for integrable functions.