If we consider a vicsek fractal of width 1 and height 1 (in other words constructed from the unit square) its easy to show that the hausdorff dimension of this is $\frac{\ln(5)}{\ln(3)}$ through the scale by 3 and observe 5 copies occur argument.
Now my question is how do we actually measure the Hausdorff Measure of this set in the dimension $\frac{\ln(5)}{\ln(3)}$?
Now I'm aware there is a definition of Hausdorff Measure of a set $S$ given here which i'll reproduce (taking the gamma scaling convention to agree with the Lebesgue measure) as
$$ H^d(s) = 2^{-d} \frac{\pi^{\frac{d}{2}}}{\Gamma \left( \frac{d}{2} + 1 \right)} \lim_{\delta \rightarrow 0} \left[ \inf \left\lbrace \sum_{i=1}^{\infty} \left( \text{diam} U_i\right)^d : S \subseteq \cup_{i=1}^{\infty} U_i : \text{diam} U_i < \delta \right\rbrace \right]$$
And so we just want to compute $H^{\frac{\ln(5)}{\ln(3)}}$.
Now the problem with using this definition directly is I don't think I have the tools to just take an infinimum over uncountably many countable sized covers of a set and I don't see how to algorithmically go after that either. Also given that the Vicsek Fractal is one of the most basic fractals describable I assume this has already been studied.
So how does one actually go about computing the hausdorff measure of the unit vicsek fractal?
Conjecture: The unit Vicsek fractal has $\frac{\ln(5)}{\ln(3)}$ - Hausdorff Measure 1.
Besides the fact the Vicsek Fractal is almost identically constructed to the Cantor Sets we can also motivate this a lot more concretely. The Unit Vicsek Fractal can be described as the limit of the following sequence of definitions
- 2 line segments of length 1 intersecting at a cross + the rest of it (see below):
- 10 line segments of length 1/3 intersecting the usual way + the rest of it
- 50 line segments of length 1/9 + the rest of it
(lazy to draw it but you can imagine...)
- 250 line segments of length 1/27 + the rest of it
and then in general 2*5^n segments of length 1/3^n and the rest of it.
So basically our Vicsek Fractal lives in the limit of all of these. Now if you try to naively compute the 1-Hausdorff Measure of it you get the sequence
$$ 2 \times 1 \rightarrow 10 \times \frac{1}{3} \rightarrow 50 \times \frac{1}{9} \rightarrow ... \frac{2*5^n}{3^n} $$
And it's easy to see that this tends to infinity as we would expect.
Now if you were to raise the length of segments to the power $\frac{\ln(5)}{\ln(3)} = \log_3(5)$ then you end up computing its Hausdorff Measure as
$$ 2 \rightarrow 10 \times \frac{1}{5} = 2 \rightarrow 50 \times \frac{1}{25} = 2 \rightarrow ... \ 2 \ ... $$
It's just always 2. Now we need to recall something, this description that we have counts the middle point twice (both line segments intersect at the middle and we measure both). It also counts the 4 outer crosses twice at the second level. It also counts the 16 next biggest crosses twice. In fact it counts every intersection in the Vicsek Fractal twice so basically this scheme ends up overlapping a set dense in the Vicsek Fractal twice. We might through complete hand waving then say that "this is covering the Vicsek Fractal twice so lets cut the measure in half!"
And thus we arrive at the number $1$.
So that gives us a conjectured value of 1 for two different reasons. But I guess i'm still looking for a proof or a refutation.