Find a Borel set $F$ such that $\lim_{k \to 0} \frac{m_N(F \cap B_k(x))}{m_N(B_k(x))}$ does not exist at some $x$.

Question

Let $F$ a Borel set of $\mathbb{R}^N$ and we define, if exists,

$$ D_F(x) := \lim_{k \to 0} \frac{m_N(F \cap B_k(x))}{m_N(B_k(x))}, $$

where $B_k(x)$ is the open ball of radius $k$ and center $x$.

Problem. Find a Borel set $F$ such that $D_F(x)$ does not exists at some point $x$.

I hope that someone can help me to find a solution. Here is the original image.

Answer 1

Choose $r_n, \alpha_n$ so that

1. $\alpha_n \in (0, 1)$ for all $n$ and $\alpha_n$ does not converge as $n\to\infty$.
2. $0 < r_{n+1} < \alpha_n r_n$ and $r_{n+1}/(\alpha_n r_n) \to 0$ as $n\to\infty$.

Now define $F$ by $F = \bigcup_{n=1}^{\infty} F_n$, where $F_n = \{ x \in \mathbb{R}^N : \alpha_n r_n \leq \|x\| \leq r_n \}$. In other words, $F$ is a disjoint union of concentric annuli with extremely different scales $r_n$ and relative thickness $\alpha_n$. Then

$$ 1-\alpha_n^N = \frac{m_N(F_n)}{m_N(B_{r_n}(0))} \leq \frac{m_N(F \cap B_{r_n}(0))}{m_N(B_{r_n}(0))}, $$

and likewise,

$$ \frac{m_N(F \cap B_{r_n}(0))}{m_N(B_{r_n}(0))} \leq \frac{m_N(F_n) + m_N(B_{r_{n+1}}(0))}{m_N(B_{r_n}(0))} = 1-\alpha_n^N + \left( \frac{r_{n+1}}{\alpha_n r_n} \right)^N. $$

So if $D_F(0)$ exists, then it must follow that $1-\alpha_n^N$ converge, which is impossible by the assumption.