The most even partition of $\mathbb R$ into measure dense sets

Question

Notation: $\mu$ denotes the Lebesgue measure. Let $\mathcal D$ be the set of Lebesgue measurable subsets of $\mathbb R$ such that itself and its complement have nonzero Lebesgue measure in every interval.

We define the thinness $T_D (x)$ of a set $D \subset \mathbb R$ at $x \in \mathbb R$ by

$$T_D (x) := \sup \{r > 0 \,| \, \limsup_{\delta \to 0_+} \frac{\mu(D \cap B_d (x))}{\delta^r} < \infty.\},$$

and the overall thinness $T_D$ by

$$T_D = \sup_{x \in \mathbb R} T_D (x). $$

Question: What is the infimum, over $D \in \mathcal D$ of the greater of the thinness of $D, D^c$? That is, what is the value of the quantity

$$\inf_{D \in \mathcal D} \max(T_D, T_{D^c})?$$

Answer 1

The infimum is 1. Here is a construction of a set $E$ and its complement $O$, both of which have thinness 1.

If $\omega\in [0,1]$ is distributed according to Lebesgue measure, define a sequence $(X_n(\omega))$ of independent $0,1$-valued random variables by $X_n(\omega)=1$ if the $2^n$th to $(2^n+n-1)$st binary digits of $\omega$ are all 0; and $X_n(\omega)=0$ otherwise. Since binary digits are independent, and distinct $X_n$'s are functions of disjoint collections of digits, the $(X_n)$ are mutually independent.

Let $\alpha=\prod_{n=1}^\infty (1-2^{-n})$. Using the independence, we see

$$\mathbb P(X_{n+1}=X_{n+2}=\ldots=0|X_n=1)=\mathbb P(X_{n+1}=X_{n+2}=\ldots=0)=\prod_{i=n+1}^\infty (1-2^{-i})\ge\alpha.$$

Note that by the first Borel-Cantelli lemma, almost surely only finitely many $X_n$'s are non-zero.

Let $E=\{\omega\colon \text{ the largest $n$ with $X_n=1$ is even}\}$ and $O=\{\omega\colon\text{ the largest $n$ with $X_n=1$ is odd}\}$. $E$ and $O$ are disjoint sets with union of measure 1 by the above observation.

We claim that $E$ and $O$ are both maximally thick. Let $J$ be a dyadic sub-interval of $[0,1]$, that is an interval of the form $[i/2^k,(i+1)/2^k]$. We will show that $\mu(J\cap E)$ and $\mu(J\cap O)$ are at least $\alpha 2^{-k}/(4k)$.

To see this, let $n=\lceil \log_2 k\rceil$ so that $2^n>k$. Assume for now that $n$ is even. Then $\mu(J\cap \{X_n=1\})=2^{-n}\mu(J)$, so that $\mu(J\cap E)\ge 2^{-n}\alpha\mu(J)\ge \alpha 2^{-(k+\log_2 k+1)}$.

Similarly, $\mu(J\cap \{X_{n+1}=1\})=2^{-(n+1)}\mu(J)$ so that $\mu(J\cap O)\ge \alpha 2^{-(k+\log_2 k+2)}$. Both are at least $\alpha 2^{-k}/(4k)$. If $n$ is odd, the roles of $E$ and $O$ are reversed, but the conclusion remains the same.

Finally if $I$ is any interval, it contains a dyadic interval $J$ of length at least $|I|/4$. It follows that

$$\mu(I\cap E),\mu(I\cap O)\ge \alpha\mu(I)2^{-\log_2(-\log_2\mu(I))}/16$$

$$= \alpha\mu(I)/(-16\log_2\mu(I)).$$

This gives the required thickness.