Let $E$ be a measurable subset of $\mathbb R$. We say $E$ is $\alpha$-macroscopic, for $0 \leq \alpha \leq 1$, if there exists an $\alpha$-Holder continuous function $f: \mathbb R \to \mathbb R$ such that $f(E)$ is of nonzero Lebesgue measure, where by convention, a $0$-Holder continuous function will simply be a continuous function.

We define the macroscopic order of $E$ to be the supremum of all $\alpha \in [0, 1]$ such that $E$ is $\alpha$-macroscopic.

Note that there exist macroscopic sets of zero Lebesgue measure.

Example 1 (Zero set of Brownian motion): As the primary and motivating example, the local time of Brownian motion maps the set of zeroes of Brownian motion to an interval, and is $\alpha$-Holder continuous for all $\alpha < \frac{1}{2}$.

Example 2 (Middle thirds Cantor set): The Cantor function maps the middle thirds Cantor set to an interval and is continuous, thus the middle thirds Cantor set is a $0$-macroscopic set.

Question: What is the relation between the the order of $E$ as a macroscopic set and its Hausdorff dimension, if any? Specifically, the following two questions are of interest - for $\alpha, r \in [0, 1]$,

  1. What is the infimal/supremal Hausdorff dimension of a set of macroscopic order $\alpha$?

  2. What is the infimal/supremal macroscopic order of a set of Hausdorff dimension $r$?

  • 1
    $\begingroup$ The definition seems to immediately imply that $\alpha \leq r$. // The cantor function is $C^\alpha$ with $\alpha = \log_3(2)$, this shows that the Cantor function has macroscopic order $\log_3(2)$. $\endgroup$ Commented Nov 28, 2023 at 5:20
  • $\begingroup$ @WillieWong Could you ellaborate a bit ? I see the other inequality from my argument below, but I do not see that $\alpha \leq r $. $\endgroup$ Commented Nov 28, 2023 at 19:08
  • $\begingroup$ Ok, see, I edited the answer so that it is complete. $\endgroup$ Commented Nov 28, 2023 at 20:14
  • $\begingroup$ @an_ordinary_mathematician: your argument is what I had in mind. The bound turns out to be sharp if you allow domains to be general metric spaces. $\endgroup$ Commented Nov 28, 2023 at 22:34

1 Answer 1


By Frostman's lemma, if $E$ is a compact set of positive $\alpha$-Hausdorff content, then there exists a probability Borel measure $\mu$ supported in $E$ such that $\mu(I) \leq c |I|^\alpha $ for every interval $I$ and $c$ is a constant. Then the distribution function of the measure $\mu$ must be $\alpha$-Holder continuous and $f(E)=[0,1]$.


In the other direction, if such a function $f$ exists and if $I_i$ is a cover of $E$ by open intervals, then $f(I_i)$ is a cover of $f(E)$ by intervals, hence $ \sum_{i}|f(I_i)| \geq |f(E)|>0$, but $|f(I_i)| \leq C(f) |I_i|^\alpha $ by Holder continuity, therefore $E$ must have positive $\alpha$- Hausdorff content.

  • $\begingroup$ Oh, so they are the same. Tbh, that is a bit disappointing… but thank you for the slick answer! $\endgroup$
    – Nate River
    Commented Nov 28, 2023 at 23:02

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