Munkres-Analysis on Manifolds: Theorem 20.1

Question

I am studying Analysis on Manifolds by Munkres. I have a problem with a proof in section 20:

It states that:

Let $A$ be an $n$ by $n$ matrix. Let $h:\mathbb{R}^n\to \mathbb{R}^n$ be the linear transformation $h(x)=A x$. Let $S$ be a rectifiable set (the boundary of $S=BdS$ has measure $0$) in $\mathbb{R}^n$. Then $v(h(S))=|\det A|v(S)$ ($v=$volume).

The author starts his proof by considering tha case of $A$ being a non-singular matrix (invertible).
I think I understand his steps in that case (I basically had to prove that $h(int S)=int$ $h(S)$ and $h(S)$ is rectifiable, if anybody knows a way this statements are proven autumatically please tell me).

He proceeds by considering the case where $A$ is singular, so $\det A=0$. He tries to show now that $v(T)=0$. He states that since $S$ is bounded so is $h(S)$ (I think thats true because $|h(x)-h(a)|\leq n|A||x-a|$ for each $x$ in $S$ and fixed a in $S$, if there is again a better explanation please tell me).

Then he says that $h(\mathbb{R}^n)=V$ with $\dim V=p<n$ and that $V$ has measure $0$ (for each $ε>0$ it can be covered by countably many open rectangles of total volume less than $ε$), a statemant that I have no clue how to prove. Then he says that the closure of $h(S)=cl(h(S))$ is closed and bounded and has neasure $0$ (of course $cl((h(S))$ is closed but why is it bounded with measure $0$?). Then makes an addition step (which I understand) and proves the theorem for that case too.

Cound someone help me clarify the points of the proof that I don't understand? Thank you in advance!

Answer 1

I am reading Theorem 20.1.

I think we need to use the result of Theorem 18.2 to understand the proof of Theorem 20.1.
Note that the sentence "These results also hold when $D$ is not compact, provided $\operatorname{Bd}D\subset A$ and $\operatorname{Bd}E\subset B$." is not included in the following pdf:
https://archive.org/details/MunkresJ.R.AnalysisOnManifoldsTotal/page/n165/mode/1up
But in my printed book, this sentence is added.
And we need to use Theorem 18.2 in the case in which $D:=S$ is not necessarily compact but $\operatorname{Bd} D\subset A:=\mathbb{R}^n$ and $E:=T$ is not necessarily compact but $\operatorname{Bd} E\subset B:=\mathbb{R}^n$.

Theorem 18.2 on p.154. Let $g:A\to B$ be a diffeomorphism of class $C^r$, where $A$ and $B$ are open sets in $\mathbb{R}^n$. Let $D$ be a compact subset of $A$, and let $E=g(D)$.
(a) We have $$g(\operatorname{Int}D)=\operatorname{Int}E\,\,\,\,\,\text{and}\,\,\,\,\,g(\operatorname{Bd}D)=\operatorname{Bd}E.$$
(b) If $D$ is rectifiable, so is $E$.
These results also hold when $D$ is not compact, provided $\operatorname{Bd}D\subset A$ and $\operatorname{Bd}E\subset B$.

My Proposition 1.
Let $V$ be a linear subspace of $\mathbb{R}^n$.
Then, $V$ is a closed set in $\mathbb{R}^n$.

Proof of my Proposition 1.
If $V=\mathbb{R}^n$, then $V$ is closed in $\mathbb{R}^n$.
If $V=\{0\}$, then $V$ is closed in $\mathbb{R}^n$.
We consider the case where $V\neq\{0\}$ and $V\neq\mathbb{R}^n$.
Let $\dim V=p$.
Then, $0<p<n$.
There is an orthonormal basis for $\mathbb{R}^n$ whose first $p$ elements form a basis for $V$. (Please see Lemma 21.1 on p.180.)
Let $y\in\mathbb{R}^n-V$.
Let $y=y_1v_1+\cdots+y_pv_p+y_{p+1}v_{p+1}+\cdots+y_nv_n$.
The distance between $y$ and $V$ is $\sqrt{y_{p+1}^2+\cdots+y_n^2}$.
Let $\varepsilon:=\frac{\sqrt{y_{p+1}^2+\cdots+y_n^2}}{2}$.
Then $B(y;\varepsilon)\cap V=\varnothing$.
So, $\mathbb{R}^n-V$ is open in $\mathbb{R}^n$.
So, $V$ is closed in $\mathbb{R}^n$.

Proof of Theorem 20.1. Consider first the case where $A$ is non-singular. Then $h$ is a diffeomorphism of $\mathbb{R}^n$ with itself; $h$ carries $\operatorname{Int}S$ onto $\operatorname{Int}T$ by Theorem 18.2(a) on p.154; and $T$ is rectifiable by Theorem 18.2(b) on p.154.
We have $$v(T)=v(\operatorname{Int}T)$$ by Theorem 14.2(e) on p.112.
We have $$v(\operatorname{Int}T)=\int_{\operatorname{Int}T}1$$ by the definition of $v$.
We have $$\int_{\operatorname{Int}T}1=\int_{\operatorname{Int}S}(1\circ h)|\det Dh|=\int_{\operatorname{Int}S}1|\det Dh|=\int_{\operatorname{Int}S}|\det Dh|$$ by the change of variables theorem. Hence $$v(T)=\int_{\operatorname{Int}S}|\det A|=|\det A|\cdot\int_{\operatorname{Int}S}1=|\det A|\cdot v(\operatorname{Int}S)=|\det A|\cdot v(S)$$ by Theorem 14.2(e) on p.112 and by linearlity (Theorem 13.3(a) on p.106).
Consider now the case where $A$ is singular; then $\det A=0$. We show that $v(T)=0$. Since $S$ is bounded, there exists a closed rectangle $Q$ such that $S\subset Q$. Since $h=(h_1,\dots,h_n)$ is continuous on $Q$, so each of $h_i$ has a maximum value and a minimum value on $Q$. So $T=h(S)$ is bounded. The transformation $h$ carries $\mathbb{R}^n$ onto a linear subspace $V$ of $\mathbb{R}^n$ of dimension $p$ less than $n$, which has measure zero in $\mathbb{R}^n$, as the answers to this question show.
$\overline{T}$ is closed because any closure of a set is closed.
Since $T$ is bounded, there exists a positive real number $M$ such that $||y||<M$ for any point $y\in T$.
Let $y'$ be an arbitrary point in $\overline{T}$. Then there exists a point $y\in T$ such that $||y-y'||<1$.
By the triangle inequality, $||y'||=||y'-y||+||y||<1+M$.
So, $\overline{T}$ is bounded.
By my Proposition 1, $h(\mathbb{R}^n)$ is a closed set in $\mathbb{R}^n$.
Since $T=h(S)\subset h(\mathbb{R}^n)$ and $h(\mathbb{R}^n)$ is closed in $\mathbb{R}^n$, $\overline{T}=\overline{h(S)}\subset h(\mathbb{R}^n)$.
So, $\overline{T}$ is a subset of $h(\mathbb{R}^n)$ and $h(\mathbb{R}^n)$ has measure zero in $\mathbb{R}^n$.
So, $\overline{T}$ also has measure zero in $\mathbb{R}^n$.
The function $1_T$ that equals $1$ on $T$ and $0$ outside $T$ is continuous on the open sets $\operatorname{Ext}T$ and $\operatorname{Int}T$. It fails to be continuous at each point of $\operatorname{Bd}T$. By Theorem 11.2 on p.93, the function $1_T$ is integrable over a rectangle $Q$ containing $T$ because $\operatorname{Bd}T\subset\overline{T}$ has measure zero. So, the integral $\int_T 1$ exists.
Since $T\subset h(\mathbb{R}^n)$ has measure zero, $\int_T 1=v(T)=0$ by Theorem 14.2(d) on p.112.