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Let $O$ be a fixed point of the plane $\mathbb R^2$. A set $S$ of points in the plane has the property that for any real number $r > 0$, at most one point of $S$ lies on the circle centered at $O$ with radius $r$.

I am hoping that this might be enough to show that $S$ is Lebesgue measurable with measure zero. Is this true?

(This question came up in a discussion with a student of mine, so I don't know if it's easily answerable.)

Evan Chen
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  • I believe you can construct a nonmeasurable set with no two points at the same distance ffrom the origin. I think the usual construction of a Bernstein can be adapted to have that property. Haven't thought about it carefully though. – bof Nov 12 '19 at 02:37
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    Exclude the origin, then $S$ is the graph, in polar coordinates, of a (partial) function that sends distance to the origin to angles in $(-\pi, \pi]$. If the graph is measurable it will have measure zero. It can't contain a set of positive measure, but it can be non-measurable with positive outer measure – conditionalMethod Nov 12 '19 at 02:39
  • Thanks! Figured it was too optimistic. If you post this comment as an answer I'll accept it. – Evan Chen Nov 12 '19 at 03:30

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