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The claim is that for every Lebesgue measure $0$ set $N$, there exists $x \in \mathbb R$ s.t. $N + x \cap N = \emptyset$.

I came up with this question while trying to figure out another proof of the uniqueness of Lebesgue measure (as a translation invariant Borel measure on $\mathbb R$ that is $1$ on $(0,1)$). For that, I think it suffices to prove the claim above. Effectively this should allow you to show that any (locally finite) translation invariant measure has a density with respect to Lebesgue.

I'm not sure whether this is true, but no counterexamples readily come to mind (it's possible a tiling of $\mathbb R$ by the Cantor set is a counterexample, but I can't immediately show it). Note that in $\mathbb R^2$ there is a natural counterexample, take the $x$ and $y$ axes. This readily generalizes to all higher dimensions too. Note also it clearly suffices to consider Borel sets. So does the claim hold?

Physical Mathematics
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    A wonderful question! :) – paul garrett Jan 31 '21 at 23:18
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    @HallaSurvivor Yes. Thanks for finding it (I looked somewhat before posting but clearly not hard enough). Apparently I can't unilaterally close my own question as a duplicate for some reason, and it seems perhaps worthwhile to keep up as a "signpost" to the other question. I did vote to close at least. – Physical Mathematics Jan 31 '21 at 23:26
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    @PhysicalMathematics: if it wasn’t for your question I wouldn’t have found those answers. So thank you for asking this, even if it is a duplicate. It’s a very interesting question. By the way, you were right, a tiling of $\mathbb R$ by Cantor sets indeed is a counterexample. – Giuseppe Negro Jan 31 '21 at 23:32

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