# A Kakeya-like problem: must a union of annuli fill the plane?

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.

-
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.) –  Terry Tao Oct 1 '12 at 16:23
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). –  Terry Tao Oct 1 '12 at 16:26
Thanks so much, Terry! I thought the point of my "translation" construction was that, even after you add an additional degree of freedom to the lines problem (by saying that there must be line segments in every direction arbitrary close to any given line), the set S can still have arbitrarily-small Lebesgue measure. But I suppose the catch is that the degree of freedom we've added doesn't "count" for measurability purposes, because of the "arbitrarily close" proviso? –  Scott Aaronson Oct 1 '12 at 17:12
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. –  Sean Eberhard Oct 1 '12 at 20:11
Though admittedly, as far as I can see, this doesn't rule out the possibility of a nonmeasurable such $S$. –  Sean Eberhard Oct 1 '12 at 20:28