Does positive density imply existence of the density for some part of a decomposition?

Question

Suppose a $\mathcal{H}^{1}$ measurable set $A\subset \mathbb{R}^{n}$ has positive Hausdorff density $\Theta^{1}(\mathcal{H}^{1},A,x)=c>0$ in a point $x\in A$.

If we have a decomposition $A=B\cup C$ with disjoint $B$ and $C$, can it happen that $\Theta^{1}(\mathcal{H}^{1},B,x)$ and $\Theta^{1}(\mathcal{H}^{1},C,x)$ do not exists? Can it even happen that the lower densities of these sets are both zero? Are there conditions on $A$ to prevent this?

More specifically: Let $C_{s,\epsilon}(x)$ denote the double cone in direction $s\in\mathbb{S}^{n-1}$ with opening angle $\epsilon$ at the point $x$.

Is it possible that $\Theta^{1}(\mathcal{H}^{1},A\cap C_{s,\epsilon}(x),x)$ does not exists for all $s\in \mathbb{S}^{n-1}$ and all $\epsilon\in (0,\epsilon_{s})$? ($\epsilon_{s}$ is some maximal angle depending on the direction $s$) Is it possible that the lower Hausdorff density vanishes for all these double cones?

Answer 1

Today at our problem coffee someone (I don't know if he wants his name mentioned) showed me a counterexample to the first part of my question, in case anyone else is interested.

We take $X:= [0,1]\subset \mathbb{R}^{2}$ - for $\mathbb{R}^{n}$ the same argument should work, only with slightly more complicated notation - and $x=0$. Let $\lambda_{k}\in (0,1)$ be a positive sequence converging monotonically to zero. We define the monotonically decreasing sequence $a_{n}=\prod_{i=1}^{n}\lambda_{i}$ and for $n\in\mathbb{N}$ we set $A_{0}:=[1,a_{1}], A_{n}:=[a_{2n+1},a_{2n}], B_{n}:=(a_{2n},a_{2n-1})$ and finally $A:=\bigcup_{n\in\mathbb{N}\cup\lbrace 0\rbrace} A_{n}\cup\lbrace 0\rbrace$, $B:=\bigcup_{n\in\mathbb{N}}B_{n}$. Since $a_{n}\leq \lambda_{1}^{n}\rightarrow 0$ we have $A\cup B=X$. If we choose $r_{n}=a_{2n}$ and $R_{n}=a_{2n+1}$ we get

$\mathcal{H}^{1}(A\cap \overline B_{R_{n}}(x))\leq a_{2n+2}=\prod_{i=1}^{2n+2}\lambda_{i}\leq \lambda_{2n+2}R_{n}$ and $\mathcal{H}^{1}(B\cap \overline B_{r_{n}}(x))\leq a_{2n+1}=\prod_{i=1}^{2n+1}\lambda_{i}\leq \lambda_{2n+1}r_{n}$, which shows that both lower densities are $0$.