I have a question about a property of Lipschitz domain.
Let $D \subset \mathbb{R}^d$ be a bounded domain (connected open subset ). $D$ is called a bounded Lipschitz domain if there exist positive constants $\delta$, $M$ such that for each $x_0 \in \partial \Omega$ there exist a neighborhood $U_{x_0}$ of $x_0$, local coordinates $y=(y',y_d) \in \mathbb{R}^{d-1} \times \mathbb{R}$, with $y=0$ at $x_0$, and a Lipschitz continous function $f_{x_0}:\mathbb{R}^{d-1} \to \mathbb{R}$, such that \begin{equation*} D \cap U_{x_0}=\{(y',y_N) \in D \cap U_{x_0} \mid y' \in B(0,\delta),\, y_N>f(y') \},\quad \text{Lip}(f) \le M, \end{equation*} where we define $\text{Lip}(f)=\inf \{L \ge 0 \mid |f(x)-f(y)| \le L|x-y|,\, x,y \in B(0,\delta) \}$.
In the above definition, $B(0,\delta)$ denotes the open ball in $\mathbb{R}^d$ centered at the origin with radius $1$.
My question
We denote by $m$ the Lebesgue measure on $D$.
Let $D \subset \mathbb{R}^d$ be a bounded Lipschitz domain. Then, can we show the following?
There exists a positive constant $C \ge 1$ such that $$C^{-1}r^d \le m(\bar{D} \cap B(x,r)) \le Cr^d$$ for any $x \in \bar{D}$, $r \in (0,\text{diam}(D)]$.
It is clear that $m(\bar{D} \cap B(x,r)) \le Cr^d$ for any $x \in \bar{D}$, $r \in (0,\text{diam}(D)]$.
If you know a proof or references, please let me know.