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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 $0$.

Community
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sharpe
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    One way is to use the Lebesgue density theorem and the fact that no point on the graph can be a point at which the graph has Lebesgue density $1.$ (Because at each point of the graph there will be a conical sector with vertex at that point whose interior does not contain any points in the graph.) – Dave L. Renfro Aug 22 '16 at 14:36
  • This definition of Lipschitz domain seems simpler and more intuitive than the one given on Wikipedia. – Akira Sep 08 '23 at 22:15

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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)

Thompson
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  • Thank you for your reply. There exists a way to show $\partial D$ has Lebesgue measure zero? – sharpe Aug 22 '16 at 14:03
  • Yes. The bit I'm answering specifically is the question you ask: "but can we show.... Is countable?" – Thompson Aug 22 '16 at 14:13
  • OK. I admit the set I write down is not countable. But I am also concerned with how to prove Lebesgue measure of $\partial D$ is zero. What is $B_{r_{x_i}}$? – sharpe Aug 22 '16 at 14:25
  • Probably, I understand what you want to say. – sharpe Aug 22 '16 at 14:58
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    More to the point, $R^n$ is hereditarily Lindelof: If $S\subset R^n$ (e,g. $S=\partial D$) and $C$ is an open family in $R^n$ with $\cup C\supset S$ (e.g. $C={Q(x,r_x):x\in \partial D}$ ) then there exists a countable $C'\subset C$ with $\cup C'\supset S$. – DanielWainfleet Aug 24 '16 at 18:27