I have a question about the boundary of Lipschitz domain and its measure.
Although there is a similar question in enter link description here, I don't really understand the answer in this question.
In the following, $D$ denotes a connected open subset of $\mathbb{R}^d$ with Lipschitz boundary (not necessary bounded). We say $\partial D$ is Lipschitz if for each $x \in \partial D$, there exist $r_x>0$ and a Lipschitz function $F^x:\mathbb{R}^{d-1} \to \mathbb{R}$ such that (upon rotating and relabeling the coordinate axes if necessary) we have
\begin{align*}
D \cap Q(x,r_x)=\{y \in \mathbb{R}^d:F^x(y_{1},\ldots,y_{d-1})<y_{d}\} \cap Q(x,r_x),
\end{align*}
where $Q(x,r)=\{y \in \mathbb{R}^d : |y_{i}-x_{i}|<r, i=1,\ldots,d\}$.
In other words, near $x$, $\partial D$ is a graph of a Lipschitz function.
My question
I admit that Lebesgue measure of the graph of a Lipschitz function $F: \mathbb{R}^{d-1} \to \mathbb{R}$ has zero. But can we show that $\{(r_{x},F_{x})\}_{x \in \partial D}$ is countable set? If not, I am not convinced that Lebesgue measure of $\partial D$ is $0$. How do we show that Lebesgue measure of $\partial D$ is $0$?
Thank you in advance.
ADD: Proof of Lebesgue measure of $\partial D$ is zero.
Define $A=\bigcup_{x \in \partial D}Q(x,r_x)$. Since $A$ is open subset of $\mathbb{R}^{d}$, there are countably many point $x_i \in \partial D$ such that $A=\bigcup_{i=1}^{\infty}Q(x_{i},r_{x_i})$. Clearly, $\partial D \subset A$ and Lebesgue measure of $\partial D \cap Q(x_i,r_i)$ is $0$. Therefore, Lebesgue measure of $\partial D$ is $0$.
Best Answer
Euclidean space is what's called "Lindelof", which means that every open cover has a countable subcover.
So there are countably many points $x_i \in D$, $i=1,2,3...$ such that
$\bigcup_{i=1}^{\infty} B_{r_{x_i}} \supset \partial D$
(The set you write down isn't countable)