I am reading "Calculus on Manifolds" by Michael Spivak.
In the proof of Theorem 3-13 (Change of variable), the author wrote as follows:
We are now prepared to give the proof, which proceeds by induction on $n$. The remarks before the statement of the theorem, together with (1) and (2), prove the case $n=1$.
The remarks before the statement of the theorem are the following:
If $g:[a,b]\to\mathbb{R}$ is continuously differentiable and $f:\mathbb{R}\to\mathbb{R}$ is continuous, then, as is well known, $$\int_{g(a)}^{g(b)} f = \int_a^b (f\circ g)\cdot g'.$$
The proof is very simple: if $F'=f$, then $(F\circ g)'=(f\circ g)\cdot g'$; thus the left side is $F(g(b))-F(g(a))$, while the right side is $F\circ g(b)-F\circ g(a)=F(g(b))-F(g(a))$.
We leave it to the reader to show that if $g$ is 1-1, then the above formula can be written $$\int_{g((a,b))} f=\int_{(a,b)} f\circ g\cdot |g'|.$$
(Consider separately the cases where $g$ is increasing and where $g$ is decreasing.) The generalization of this formula to higher dimensions is by no means so trivial.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 $$\int_{g(A)}f=\int_A(f\circ g)|\det g'|.$$
(1) is the following:
- Suppose there is an admissible cover for $A$ such that for each $U\in\mathcal{O}$ and any integrable $f$ we have $$\int_{g(U)} f=\int_U (f\circ g)|\det g'|.$$ Then the theorem is true for all of $A$.
(2) is the following:
- It suffices to prove the theorem for the function $f=1$.
I wonder why we need (2) to prove the case $n=1$.
We want to prove the following:
Let $A\subset\mathbb{R}^1$ be an open set and $g:A\to\mathbb{R}^1$ a 1-1, continuously differentiable function such that $g'(x)\neq 0$ for all $x\in A$. If $f:g(A)\to\mathbb{R}$ is integrable, then $$\int_{g(A)}f=\int_A(f\circ g)|g'|.$$
There is an admissible cover $\mathcal{O}$ for $A$ such that each $U\in\mathcal{O}$ is $(a,b)$, where $a,b\in\mathbb{R}$ such that $a<b$.
For any $U\in\mathcal{O}$ and for any integrable $f$, we have $$\int_{g(U)} f=\int_U (f\circ g)|g'|$$ by the remarks before the statement of the theorem.
So, by (1), $\int_{g(A)}f=\int_A(f\circ g)|g'|$ holds.
Where do we need to use (2)?