Munkres-Analysis on Manifolds: Theorem 20.1 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!

As Munkres explains, since $\det A=0$, $h(S)(=T)$ is contained in a vector space of dimension smaller than $n$. But any subset of such a vector space has measure $0$.
