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?



Does positive density imply existence of the density for some part of a decomposition?
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 $0$.


