# How can we not know the $s$-measure of the Sierpiński triangle?

I'm preparing a presentation that would enable high-school level students to grasp that the (self-similarity) dimension of an object needs not be an integer. The first example we look at is the Sierpiński triangle and with some effort we learn that its dimension is $$s := \log(3)/\log(2) \approx 1.585.$$ After this I thought that it would be nice to mention what the actual Hausdorff $$s$$-measure of the triangle is, but all I found was measure estimates for a certain class of Sierpiński carpets and some estimates of the Sierpiński triangle.

I am literally shocked to learn that we apparently do not know the exact value of Hausdorff $$s$$-measure of the Sierpiński triangle! Especially since it's such a concrete and symmetric object. To comply to the idea of this site hosting questions instead of rants, I formulate my bafflement as follows:

Why is the $$s$$-measure of the Sierpiński triangle and other self-similar fractals so hard to calculate?

Are we missing a link to some complicated machinery or is the problem connected to some deep problem that one would expect to remain unsolved?

• That's an interesting question! But I want to add that I would not know what to make of the fact that some specific set would have $0.7664$-dimenaional measure $1$. I suspect that the same is true for others and this may explain that some people would not put too much effort in such calculations. – Dirk Oct 21 at 18:52
• I don't know this subject at all, but is there any reason to expect the answer to be a closed-form number? If the answer is not a closed-form number, what would one mean by "knowing the exact value"? – Timothy Chow Oct 21 at 20:46
• @TimothyChow One could ask for a relatively fast algorithm for computing the digits of the number. Based on the reference above, it looks like it's not even known to one significant digit, so "closed-form" isn't really the issue.. – verret Oct 21 at 20:55
• @verret : Granted. I was reacting to the sentence, "I am literally shocked to learn that we apparently do not know the exact value." – Timothy Chow Oct 21 at 21:53
• Do you know exactly how many disks of radius 1 it would take to cover a disc of radius 1000000? I think this is the flavour of the question you’re asking. (Yes I know with Hausdorff dimension you’re allowed to have sets of different diameters, but morally...) – Anthony Quas Oct 22 at 4:44

where the best values are given as $$0.77 \le \mathscr{H}^s(\Lambda) \le 0.81794 .$$ He also proves an upper estimate $$\mathscr{H}^s(\Lambda) \le 0.819161232881177$$ of which he says "everybody can check it easily".
He also explains why this is harder and more technical than the reasons found in the comments here; for instance, the obvious upper bound gives only $$\mathscr{H}^s(\Lambda) \le 1$$.