If $E$ is a Lebesgue measurable set in $\mathbb{R}$ with positive measure, the set $E-E = \{ x-y : x,y \in E \}$ is also measurable?

Question

I know that the set contains an interval (the Steinhaus theorem), but I can't use this for answer the question, maybe someone have a hint to prove that's true or a counterexample for show that's false.

Possible duplicate of The set of differences for a set of positive Lebesgue measure — Lev Bahn, Nov 12 '18 at 19:41

Robert Israel · Accepted Answer · 2018-11-12T20:55:25.297

8

The statement is false. Compare the results of Ciesielski, Fejzić, and Freiling: you can have a subset $A$ of a compact set of measure $0$ (which therefore is Lebesgue measurable) such that $A + A$ is non-measurable. WLOG take $A \subset [0,1]$. Let $E = A \cup (10 - A) \cup [100,101]$ which is measurable with measure $1$. Then $$(E - E) \cap [-10, -8] = A - (10-A) = -10 + (A + A)$$ and this is nonmeasurable.

edited Nov 12 '18 at 20:55

answered Nov 12 '18 at 19:49

Robert Israel

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Interesting article, of course, it's a counterexample. Thanks. – Ludwik Nov 13 '18 at 15:48

If $E$ is a Lebesgue measurable set in $\mathbb{R}$ with positive measure, the set $E-E = \{ x-y : x,y \in E \}$ is also measurable?

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