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The computation of the Hausdorff measure is extremely difficult due to the infimum appearing in its definition. This has made the calculation of the Hausdorff measure for nearly all fractals difficult or impossible. However, there are some good bounds, given by Baoguo Jia, for some exactly self-similar fractals such as the Sierpiński Triangle and Koch Curve.

One exception to this trend are the Cantor sets. Their Hausdorff measure has been explicitly computed.

My question is this. Are there any other non-trivial fractals that have known exact Hausdorff measures? Could any of these other fractals be used to help aid in the calculation of other Hausdorff measures?

Edit: I'm still looking for answers, but now I know that techniques that can link the Hausdorff measure of specific fractals to more general ones will have very sophisticated arguments.

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    $\begingroup$ The logarithmic capacity is related to the notion of Hausdorff measure and Hausdorff dimension. If $p(z)=a_{0}+...+z^{n}$ is a monic polynomial with complex coefficients, then the logarithmic capacity for the Julia set that corresponds to the polynomial $p$ is always $1$. $\endgroup$
    – Joseph Van Name
    Commented Apr 29, 2015 at 15:26

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There are higher-dimensional generalizations of the cantor sets, such as the Sierpinski triangle, and Sierpinski carpet, and there are some results, see e.g McMullens paper.

There is also a general result regarding iterated function systems (IFS), where the fractal is invariant under say $k$ similarity maps, and that the images if these do not overlap. Then, the dimension $s$ is given by the solution of $\sum_{i=1}^k c_k^s = 1$, and $c_i$ is the contraction factor of the $i$'th map. (This is mentioned on wikipedias page with fractals listed after Hausdorff dimension).

The trouble is of course when we have overlap of the similarity maps, so some points are counted "twice" in some sense, which completely messes up everything.

I don't have a good explanation for why we cannot consider general non-overlapping affine maps, but it has to do with that magnification in certain directions affect the dimension more, in some sense (make a line segment twice as thick, and its still a line segment, but magnification in the other direction makes it longer).

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    $\begingroup$ I ask about Hausdorff Measure not Hausdorff Dimension... $\endgroup$
    – Zachary W. Robertson
    Commented Apr 28, 2015 at 21:47
  • $\begingroup$ Ah, I read too quickly! My apologies. I guess that one first need to know the dimension, in order to compute the measure... hm... I guess this is a good start: ams.org/journals/tran/1999-351-09/S0002-9947-99-01982-0/… Looks like for dimension greater than 1 is quite difficult... and linear IFS's are the most simple ones, so this is probably all that is known.. $\endgroup$
    – Per Alexandersson
    Commented Apr 28, 2015 at 21:57
  • $\begingroup$ Also, looks like the area of Julia sets can be computed via some converging method: math.stackexchange.com/questions/632911/… $\endgroup$
    – Per Alexandersson
    Commented Apr 28, 2015 at 22:03

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