How to show $\int_{h(g(A))} f=\int_{g(A)} (f\circ h)|\det h'|$ holds when $\int_{h(B)} f=\int_{B} (f\circ h)|\det h'|$ Calculus on Manifolds Spivak.

Question

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. If the theorem is true for $g:A\to\mathbb{R}^n$ and for $h:B\to\mathbb{R}^n$, where $g(A)\subset B$, then it is true for $h\circ g:A\to\mathbb{R}^n$.

Proof of (3).
$\int_{h\circ g(A)} f = \int_{h(g(A))} f=\int_{g(A)} (f\circ h)|\det h'|=\int_A [(f\circ h)\circ g][|\det h'|\circ g]\cdot |\det g'|=\int_A f\circ (h\circ g) |\det (h\circ g)'|.$

Since the theorem is true for $h:B\to\mathbb{R}^n$, $\int_{h(B)} f=\int_{B} (f\circ h)|\det h'|$ holds for any $f:h(B)\to\mathbb{R}$ which is integrable.
But how to show $\int_{h(g(A))} f=\int_{g(A)} (f\circ h)|\det h'|$ holds?
Can we say $\int_{h(g(A))} f=\int_{g(A)} (f\circ h)|\det h'|$ automatically holds because $g(A)\subset B$?