TITLE: boundary density of the Von Koch flake
QUESTION [5 upvotes]: Given a measurable set $K\subset \mathbb{R}^d$ we consider the occupation ratio $$f_r(x)=vol(K\cap B(x,r))/r^d$$ and especially the asymptotics when $r\to 0$. When $K$ has a fractal boundary and $x$ is on the boundary, it is not clear whether most $x$'s have a large occupation ratio.
Has anyone already seen this quantity? Is it studied somewhere? 
If you want a precise question, let $K$ be a non-negligible set with self-similar boundary with dimension $s$ and positive lower Minkowski content. Do we have $$\liminf_r\int_{\partial K}f_r(x)d\nu(x)>0$$ where $\nu$ is the $s$-dimensional Hausdorff measure? For the Von Koch flake i think the answer is yes but I don't see a general scheme.
EDIT: I insist on the fact that the set $K$ itself is not fractal, but its boundary is.

REPLY [3 votes]: I don't have a reference for such quantities, but I think the answer to your precise question is negative.
Consider the Sierpiński triangle $S$ as the boundary of $K$. Then if you define $K$ as the union $S \cup \text{blue triangle}$ (see the picture below), then 
$$\liminf_r\int_{\partial K}f_r(x)d\nu(x) = 0.$$
This is (for example) because the $\partial (\text{int}K)$ is the boundary of the blue triangle which has lower dimension than the dimension of $S$.
 (source)
If you would like to require that $\partial (\text{int}K) = S$ still the answer would be negative. To see this, you can consider $K = S \cup \text{red triangles}$ where we have an infinite number of "removed" (red) triangles as shown in the second picture. By having sufficiently large gaps between the scales at which you attach the removed triangles, you still have
$$\liminf_r\int_{\partial K}f_r(x)d\nu(x) = 0.$$
The next question could be what happens if you consider
$$\int_{\partial K}\liminf_rf_r(x)d\nu(x),$$
but even for this I think the second example, if correctly tuned, should give zero.
One could go further with inventing question and ask for example what happens if you require the boundary to be a self-similar type Jordan curve. Then the answer should be positive.
As a final remark: I guess you could do the above examples in $\mathbb R$ with a Cantor set, but I preferred a planar case for easier illustration.