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 $0<s<1$ 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 $d>1$.

  • $\begingroup$ an option would be to construct a set $A$ with $0<H^s(A)<\infty$ and then scale to make the measure equal 1. $\endgroup$
    – Skeeve
    Oct 24, 2019 at 15:42
  • 4
    $\begingroup$ @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...) $\endgroup$
    – Dirk
    Oct 24, 2019 at 15:51

1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – Dirk
    Sep 18, 2023 at 21:12
  • $\begingroup$ @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. $\endgroup$ Sep 19, 2023 at 13:33
  • $\begingroup$ Actually, my question was just out of curiosity, so you don't need to put in a lot of effort… $\endgroup$
    – Dirk
    Sep 19, 2023 at 14:41

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