Show that if $A$ is a nonempty, rectifiable open set in $R^n$ , then $v(A) > 0$.

Question

Show that if $A$ is a nonempty, rectifiable open set in $R^n$ , then $v(A) > 0$.

I know that being $A$ rectifiable, we have that $\int_{A}1$ exists and $v(A)=\int_{A}1\geq 0$, but how can I conclude that $v(A)>0$?, what happens if I reason assuming that $\int_{A}1=0$, what contradiction can I get? Thank you very much.

Answer 1

This problem is Exercise 3 on p.120 in "Analysis on Manifolds" by James R. Munkres.

Since $A$ is rectifiable, $v(A)=\int_A 1$ exists.
Since $A$ is a nonempty set in $\mathbb{R}^n$, there is a point $a\in\mathbb{R}^n$ such that $a\in A$.
Since $A$ is an open set in $\mathbb{R}^n$ and $a\in A$, there exists a closed cube $C_\varepsilon:=\{x\in\mathbb{R}^n\mid |x-a|\leq\varepsilon\}$ such that $C_\varepsilon\subset A$.
$C_\varepsilon$ is rectifiable and $v(C_\varepsilon)=(2\varepsilon)^n$.
By Theorem 14.2(b) on p.112, $v(C_\varepsilon)\leq v(A)$.
So, $0<(2\varepsilon)^n\leq v(A)$.