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Let $S$ be a subset of $\mathbb{R}^2$ with the following property. For all $x \in \mathbb{R}^2$ and $\varepsilon \gt 0$, there exists a nontrivial interval $[a,b] \subseteq [1-\varepsilon,1]$, such that $S$ contains all circles centered at $x$ whose radius is in $[a,b]$. Then my question is:

Can the complement of $S$ have positive Lebesgue measure?

Update: Thanks so much to Terry Tao for sketching an answer to the above question! Alas, on further reflection, I realized the condition "there exists a nontrivial interval $[a,b] \subseteq [1-\varepsilon,1]$ such that $S$ contains all circles centered at $x$ whose radius is in $[a,b]$" was a bit stronger than I intended. What if we only know that, for all $x \in \mathbb{R}^2$ and $\varepsilon \gt 0$, the set $S$ contains a set of circles, centered at $x$, that has positive Lebesgue measure within the annulus $ \{ y : | y-x | \in [1-\varepsilon,1] \}$? (But the set of radii could, for example, be a fat Cantor set, which contains no nontrivial intervals?) Or, even weaker, what if we only know (as in my comment below) that for all $x \in \mathbb{R}^2$ and $\varepsilon \gt 0$, the set $S$ contains a circle centered at $x$ whose radius is between $1-\varepsilon$ and $1$? Is that already enough to force $S$ to have full measure?


This question could be seen as another variant of the Kakeya/Besicovitch problem in the plane. We know, from Besicovitch, that a subset of the plane can have measure zero, yet still contain a translate of every unit line segment. From that, I think it follows that a Lebesgue-measurable subset $S$ of the plane can have arbitrarily-small positive measure, yet still, for every unit line segment $L$ and every $\varepsilon \gt 0$, contain all the horizontal translates of $L$ by distances in $[a,b]$, for some nontrivial interval $[a,b] \subseteq [0,\varepsilon]$. (For example, let $S$ contain all the translates of the Besicovitch set by distances that are most $\delta 2^{-b}$ from the $b^{th}$ rational number in some ordering, for arbitrarily-small $\delta \gt 0$.) If this works, then certainly the complement of $S$ can have positive (in fact, arbitrarily large) Lebesgue measure.

On the other hand, Marstrand (sorry, behind a paywall) proved that, if a subset $S \subseteq \mathbb{R}^2$ contains a circle centered about $x$ for every $x \in \mathbb{R}^2$, then $S$ must have positive Lebesgue measure. So the situation might be interestingly different for circles than for lines.

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    $\begingroup$ I believe Marstrand's arguemnt can be adapted to show that S has positive Lebesgue measure in every ball, and hence (by the Lebesgue differentiation theorem) must have full measure. (Note that one can assume wlog that S is the union of open annuli, hence is open, hence is measurable, so there are no issues with measurability here.) $\endgroup$ – Terry Tao Oct 1 '12 at 16:23
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    $\begingroup$ Also, while curvature is of course an important factor in this subject, there is a more fundamental difference between the linear problem and the circular problem here, which is that the dimensions are different: even though lines and circles are both one-dimensional, the space of all possible directions for lines in the plane is only one-dimensional, but the space of all possible centres for circles in the plane is two-dimensional, which already makes it MUCH more likely to expect a Marstrand-type result in this setting due to the surplus dimensions (1+2 > 2, whereas 1+1=2). $\endgroup$ – Terry Tao Oct 1 '12 at 16:26
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    $\begingroup$ Reiterating one of Terry's points, wlog $S$ is open. For instance, if the complement $S^c$ of the set $S$ has positive measure, then by the inner regularity of Lebesgue measure there exists compact $K\subset S^c$ also of positive measure. Thus (also using the compactness of the circle) your latest formulation of the problem is, possibly surprisingly, no more general. $\endgroup$ – Sean Eberhard Oct 1 '12 at 20:11
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    $\begingroup$ Though admittedly, as far as I can see, this doesn't rule out the possibility of a nonmeasurable such $S$. $\endgroup$ – Sean Eberhard Oct 1 '12 at 20:28
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    $\begingroup$ So, is this question answered? If so, could someone kindly assemble a clean answer from the comments into the "Answer" category below? (If you do not wish to get credit for the contribution, you may mark it "community wiki"). $\endgroup$ – Boaz Tsaban Apr 3 '16 at 22:13

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