I am reading "Calculus on Manifolds" by Michael Spivak.
I tried to solve Problem 3-39 but I could not solve it.
My attempt:
By Theorem 3-14, $g(B)$ has measure $0$.
Since $g$ in Theorem 3-13 is injective, I guess $B$ also has measure $0$.
If $B$ has measure $0$, then for almost all $x\in A$, $\det g'(x)\neq 0$.
I guess the following proposition holds:
3-13' Theorem. Let $A\subset\mathbb{R}^n$ be an open set and $g:A\to\mathbb{R}^n$ a 1-1, continuously differentiable function such that $\det g'(x)\neq 0$ for all $x\in A-B$, where $B$ is a subset of $A$ which has measure $0$. If $f: g(A)\to\mathbb{R}$ is integrable, then $$\int_{g(A)}f=\int_A(f\circ g)|\det g'|.$$
Problem 3-39. Use Theorem 3-14 to prove Theorem 3-13 without the assumption $\det g'(x)\neq 0$.
3-13 Theorem. Let $A\subset\mathbb{R}^n$ be an open set and $g:A\to\mathbb{R}^n$ a 1-1, continuously differentiable function such that $\det g'(x)\neq 0$ for all $x\in A$. If $f: g(A)\to\mathbb{R}$ is integrable, then $$\int_{g(A)}f=\int_A(f\circ g)|\det g'|.$$
3-14. Theorem (Sard's Theorem). Let $g:A\to\mathbb{R}^n$ be continuously differentiable, where $A\subset\mathbb{R}^n$ is open, and let $B=\{x\in A:\det g'(x)=0\}$. Then $g(B)$ has measure $0$.