I have a question about Ahlfors regular space.
Let $U$ be a bounded open subset in $\mathbb{R}^d$. We denote by $m$ the Lebesgue measure on $U$. Then, can we show the following?
There exists a positive constant $C>0$ such that $$C^{-1}r^d \le m(U \cap B(x,r)) \le Cr^d$$ for any $x \in U$ and $0<r<\text{diam}(U)$. Here $B(x,r)$ denotes the open ball centered at $x$ with radius $r>0$.
It is easy to prove $ m(U \cap B(x,r)) \le Cr^d$. Can we show $ m(U \cap B(x,r)) \ge C^{-1}r^d$?
The answer is negative.
The idea is to find $U$ with a "local density" decreasing to zero.
Consider in $\mathbb R$ $$U = \bigcup_{n \in \mathbb N} (\frac{1}{n}, \frac{1}{n}+\frac{1}{n^3}).$$
Then for $n \ge 4$:$$m(U \cap B\left(\frac{1}{n}+\frac{1}{2n^3},\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right)\right)) = \frac{1}{n^3}$$
And $$\lim\limits_{n \to \infty}\dfrac{ \frac{1}{n^3}}{\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+1}\right)}=0$$
in contradiction with $$\dfrac{m(U \cap B(x,r))}{r} \ge C^{-1}>0$$ if such a $C$ would exist.
