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 $\alpha, r \in [0, 1]$,
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What is the infimal/supremal Hausdorff dimension of a set of macroscopic order $\alpha$?
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What is the infimal/supremal macroscopic order of a set of Hausdorff dimension $r$?
Best Answer
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 $f(E)=[0,1]$.
EDIT
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.