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This question resisted attempts on MSE.

Let $E \subset \mathbb R^n$ be a Lebesgue measurable set. We say that $x \in \mathbb R^n$ is a turbulent point of E if both the following conditions hold:

$$\liminf_{r \to 0_+} \frac{|E \cap B_r (x)|}{|B_r (x)|} = 0$$

$$\limsup_{r \to 0_+}\frac{|E \cap B_r (x)|}{|B_r (x)|} = 1.$$

In other words, the density of $E$ oscillates maximally between $0$ and $1$.

How large can this set be? More specifically,

Question: What is the supremal Hausdorff dimension of the set of turbulent points of a subset of $\mathbb R^n$? And is the supremum achieved?

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  • $\begingroup$ Presumably $| \cdot |$ denotes Lebesgue measure, and $B_r(x)$ is a ball of radius $r$ centred at $x$, but you should say so. $\endgroup$ Commented Oct 9 at 15:56
  • $\begingroup$ Naive question: what can be said about the Hausdorff dimension of the set where lower and upper densities differ? $\endgroup$ Commented Oct 9 at 18:58
  • $\begingroup$ @PavelGubkin Also a good question… I feel it might be significantly bigger dimension wise, even equal to $n$ possibly. I will be chewing on this today haha. $\endgroup$
    – Nate River
    Commented Oct 9 at 20:42

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The Hausdorff dimension of the turbulent points can be $n.$

For $n=1$ (the reals), this is a consequence of the proof of Theorem 1 in On Lebesgue’s density theorem by Casper Goffman (1950):

THEOREM 1. If $Z$ is any set of measure $0,$ there is a measurable set $S$ whose metric density does not exist at any point of $Z.$

Remark 1: An examination of Goffman's proof shows that he actually proved the stronger result that each point of $Z$ is a turbulent point of $S.$ Moreover, the construction by Goffman shows that $S$ can be an $F_{\sigma}$ set. Note that we're not claiming the set of turbulent points of $S$ is equal to $Z.$ Indeed, the former is a Borel set (surely in Borel class $3,$ if not actually in Borel class $2),$ and $Z$ can be a non-Borel set.

Remark 2: Goffman's result (turbulent point version) holds more generally for ${\mathbb R}^n,$ but I don't recall to what extent modifications of his proof might be needed. Thus, in ${\mathbb R}^n$ the Hausdorff dimension can be $n.$

Remark 3: Goffman uses non-centered (i.e. not necessarily symmetric) density, and thus his result is weaker than the corresponding result for centered/symmetric Lebesgue density, since we have $0 \; \leq \; \underline{d} \; \leq \; \underline{d^{s}} \; \leq \; \overline{d^{s}} \; \leq \; \overline{d} \; \leq \; 1$ where $d$ & $d^{s}$ refer to non-centered & centered/symmetric Lebesgue density notions. However, Christian Remling showed in this mathoverflow answer that Goffman's result can be strengthened to centered Lebesgue density in ${\mathbb R}^n.$ Other discussions related to this are given in this Mathematics SE answer and this 21 November 2006 sci.math post.

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    $\begingroup$ That is an extremely surprising result. Thank you for the answer and resources! $\endgroup$
    – Nate River
    Commented Oct 9 at 22:04

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