# How do sets with unit fractional Hausdorff measure of dimension $>1$ look like?

Triggered by the recent question How can we not know the measure of the Sierpiński triangle? I would like to ask:

Let $$s>1$$ and $$s$$ not be an integer. How to construct a set $$A$$ with $$\mathfrak{H}^s(A) = 1$$, i.e. an $$s$$-dimensional set with $$s$$-dimensional Hausdorff measure $$1$$? Is there a set which in some sense "as simple as possible"?

For $$0 this is not so difficult, since covering with disjoint intervals makes the respective terms as large as possible, but I do not see how a similar argument can be made in higher dimensions.

Phrased differently: What should "the unit cube in $$d$$-dimensions" be for non-integer $$d>1$$.

• an option would be to construct a set $A$ with $0<H^s(A)<\infty$ and then scale to make the measure equal 1. Oct 24, 2019 at 15:42
• @skeeve Sure, that's why I aaked about measure 1 in the first place. The thing is: You need to know the measure! (From the linked question it seems like we do not know any of these...)
– Dirk
Oct 24, 2019 at 15:51

Let $$s>1$$ be aby number. The unit interval $$[0,1]$$ with the metric $$d(x,y)=|x-y|^{1/s}$$ has poisitive and finite $$s$$-dimensional Hausdorff measure. While, it is an abstract metric space, it can be embedded in a bi-Lipschitz way into a Euclidean space by Assouad's theorem and the embedded set will have positive and finite $$s$$-dimensional Hausdorff measure.

We say that a metric space $$(X,d)$$ is doubling if there is $$M$$ such that every ball $$B$$ in $$X$$ can be covered by at most $$M$$ balls with half the radius of $$B$$.

Theorem (Assouad). Let $$(X,d)$$ be a doubling metric space. Then for any $$\alpha\in (0,1)$$, there is $$N$$ and a bi-Lipschitz embedding $$\Phi:(X,d^\alpha)\to\mathbb{R}^N$$, i.e. a mapping such that $$C_1d(x,y)^\alpha\leq |\Phi(x)-\Phi(y)|\leq C_2d(x,y)^\alpha$$ for some $$C_1,C_2>0$$ and all $$x,y\in X$$.

You simply apply the result to $$X=[0,1]$$ and $$\alpha=1/s$$. There is a quite extensive literature on the Assouad theorem. You can find the proof in the book by Heinonen.

Heinonen, Juha, Lectures on analysis on metric spaces, Universitext. New York, NY: Springer (2001). ZBL0985.46008.

• Thanks, that's interesting, and I did not know this theorem before. However, it is not really what I was looking for, as neither the embedding is concrete (not even $N$ seems to be clear…) nor do I see what the value of the measure of $\Phi(X)$ is.
– Dirk
Sep 18, 2023 at 21:12
• @Dirk I am not sure what you are looking for. There are several different constructions and I will mention some more when I have time. Is it just out of curiosity or you need it for some specific problem? Knowing that I could try to tailor he example to your needs. Sep 19, 2023 at 13:33
• Actually, my question was just out of curiosity, so you don't need to put in a lot of effort…
– Dirk
Sep 19, 2023 at 14:41