# Macroscopic sets - a notion of largeness for Lebesgue null sets

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 $$r$$?

• 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 $\log_3(2)$. Commented Nov 28, 2023 at 5:20
• @WillieWong Could you ellaborate a bit ? I see the other inequality from my argument below, but I do not see that $\alpha \leq r$. Commented Nov 28, 2023 at 19:08
• Ok, see, I edited the answer so that it is complete. Commented Nov 28, 2023 at 20:14
• @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. Commented Nov 28, 2023 at 22:34

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.