I am reading "Analysis on Manifolds" by James R. Munkres.
Theorem 20.1. Let $A$ be an $n$ by $n$ matrix. Let $h:\mathbb{R}^n\to\mathbb{R}^n$ be the linear transformation $h(x)=A\cdot x$. Let $S$ be a rectifiable set in $\mathbb{R}^n$, and let $T=h(S)$. Then $$v(T)=|\det A|\cdot v(S).$$
Proof.
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Consider now the case where $A$ is singular; then $\det A=0$. We show that $v(T)=0$. Since $S$ is bounded, so is $T$. 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 you can check. Then $\overline{T}$ is closed and bounded and has measure zero in $\mathbb{R}^n$.
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I wondered why $\overline{T}$ has measure zero in $\mathbb{R}^n$.
I proved this as follows:
Proposition 1.
Let $V$ be a linear subspace of $\mathbb{R}^n$.
Then, $V$ is closed in $\mathbb{R}^n$.
Proof.
If $V=\mathbb{R}^n$, then $V$ is closed in $\mathbb{R}^n$.
If $V\neq\mathbb{R}^n$, then we can write $V=\{x\in\mathbb{R}^n: B\cdot x=0\}$ for some $p$ by $n$ matrix whose row vectors are linearly independent. Since $x\mapsto B\cdot x$ is continuous, $V_1:=\{x\in\mathbb{R}^n: B\cdot x>0\}$ and $V_2:=\{x\in\mathbb{R}^n: B\cdot x<0\}$ are open in $\mathbb{R}^n$.
Therefore $V=\mathbb{R}^n-\left(V_1\cup V_2\right)$ is closed in $\mathbb{R}^n$.
Proposition 2.
$\overline{T}$ has measure zero in $\mathbb{R}^n$.
Proof:
$h(\mathbb{R}^n)$ is a subspace of $\mathbb{R}^n$ whose dimension is less than $n$.
So, by Sub-dimensional linear subspaces of $\mathbb{R}^{n}$ have measure zero., $h(\mathbb{R}^n)$ has measure zero in $\mathbb{R}^n$.
Since $h(\mathbb{R}^n)$ is a linear subspace of $\mathbb{R}^n$, $h(\mathbb{R}^n)$ is closed in $\mathbb{R}^n$ by the preceding proposition.
Since $T=h(S)\subset h(\mathbb{R}^n)$, $\overline{T}=\overline{h(S)}\subset h(\mathbb{R}^n)$.
Since $\overline{T}$ is a subset of $h(\mathbb{R}^n)$ and $h(\mathbb{R}^n)$ has measure zero, $\overline{T}$ also has measure zero.
I proved any linear subspace of $\mathbb{R}^n$ is closed in $\mathbb{R}^n$.
Is this proposition necessary to prove that $\overline{T}$ has measure zero?
I found a fatal error in my proof of Proposition 1.
But I believe the result is true.
