I am reading "Analysis on Manifolds" by James R. Munkres.
I cannot understand the last part of Step 5 in the proof of change of variables theorem (Lemma 19.1)
https://archive.org/details/MunkresJ.R.AnalysisOnManifoldsTotal/page/n177/mode/2up
The author wrote as follows:
We apply the induction hypothesis; we have, for fixed $t$, the equation $$\int_{y\in V_t} f(y,t)=\int_{x\in U_t} f(k(x,t),t)\left|\det\frac{\partial k}{\partial x}\right|.$$ For $x\in U_t$, the integrand on the right equals $$f(h(x,t))|\det Dh|=F(x,t).$$ The lemma follows.
The author defined $F$ as follows:
let us define a function $F:\mathbb{R}^n\to\mathbb{R}$ by letting it equal $(f\circ h)|\det Dh|$ on $\operatorname{Int}S$ and vanish elsewhere.
$U_t$ is strictly larger than $S_t$, where $S_t$ is the set such that $S_t\times t= S\cap (\mathbb{R}^{n-1}\times t)$. (Please see Figure 19.3 above.)
Why does $$f(h(x,t))|\det Dh|=F(x,t)$$ hold for $x\in U_t$?
If $x \in S_t$ then $f(h(x,t)) |\!\det Dh| = F(x,t)$ by definition. If $x \in U_t - S_t$, then $(x,t) \in U - S$ and $h(x,t) \in V - Q$, so $F(x,t) = 0$ and $f(h(x,t)) = 0$. So again, we have $f(h(x,t)) |\!\det Dh| = F(x,t)$ .