Suppose we are given two sets, $S$ and $A$ in the plane, such that $S$ is finite, with a special point, $s_0$, while neither $A$ nor its complement is a null-set, i.e., the outer Lebesgue measure of $A$ and $A^c=\mathbb R^2\setminus A$ are both non-zero. Can we find two congruent copies of $S$, $f_1(S)$ and $f_2(S)$, such that $f_1^{-1}(f_1(S)\cap A)\Delta f_2^{-1}(f_2(S)\cap A)=\{s_0\}$, i.e., $s_0$ is the only element of $S$ that goes in to/out of $A$ when we go from $S_1$ to $S_2$?

My motivation is to extend a Falconer-Croft proof that works for measurable sets, see the details here. It is easy to see that this can be done when the elements of $S$ are the three vertices of an equilateral triangle; in fact, the same trick works whenever $S\setminus\{s_0\}$ falls on a line, which doesn't contain $s_0$.

The problem seems quite similar to the finite Steinhaus problem/Jackson sets, as Laczkovich has pointed out.

  • $\begingroup$ You reverted my tag edit, but this question has little to do with real analysis. $\endgroup$ – YCor Aug 15 '19 at 20:44
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    $\begingroup$ @YCor Yes, because I think that many such problems are tagged real-anal, like mathoverflow.net/questions/219860, and I also had such problems on my real-anal class at the uni. While this is debatable, I'm sure that this question has nothing to do with discrete-geometry, which you've tagged it. $\endgroup$ – domotorp Aug 15 '19 at 20:55
  • $\begingroup$ Agreed on discrete geometry. I don't think the linked post fits real-analysis either. On the other hand both clearly lack measure-theory. $\endgroup$ – YCor Aug 15 '19 at 21:02
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    $\begingroup$ @YCor Yes, that's my main problem too, that if I ask a measure-theory type question about non-measurable sets, then it's hard to tag it. Btw, I've also tagged my previous, similar question real-anal: mathoverflow.net/questions/337847 $\endgroup$ – domotorp Aug 15 '19 at 21:05
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    $\begingroup$ "Translate" in title, "congruent" in body. I think you mean "congruent". $\endgroup$ – bof Aug 15 '19 at 22:09

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