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 $\alpha, r \in [0, 1]$,

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

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