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Let E be a measureable subset of [0,1] with positive measure. Let $E-E= \{ x-y | x,y \in E\}$Show that this set contains a neighborhood of 0. (hint use inner approximations)

I think : this set is measurable as it is a union of translates of E, 0 is contained in the set as $x-x \in E-E $ ( $ \forall x \in E$),and I suspect that E-E is contained in E but am not sure about this. Thanks for any help or hints

Jmaff
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  • Try to do the case that $E = [a,b]$ for some $0<a<b<1$. What is $E - E$? Now $E$ might not contain an interval, but it should have something close. Edit: I made a stupid mistake, sorry. Thanks for the catch guys! – minimalrho Sep 18 '13 at 21:10
  • @minimalrho $0\notin E$ – Hagen von Eitzen Sep 18 '13 at 21:11
  • yeah we didn't stipulate 0 is in E. Yet since the reals numbers are ordered, negative numbers should be in E-E. So there being an interval about zero seems plausible I guess – Jmaff Sep 18 '13 at 21:12
  • Sorry folks , I didn't see that other post when I did my query. Thanks you – Jmaff Sep 19 '13 at 04:37

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If $\epsilon\notin E-E$ then $(E+\epsilon)\cap E=\emptyset$, i.e. $\lambda((E+\epsilon)\cup E)=2\lambda(E)$ ($\lambda$ is the Lebesgue measure). We can suppose that $E$ is compact, as we can replace it with a compact subset of positive measure. Let $\epsilon_n\to0$, $\epsilon_n\notin E-E$ (i.e. suppose it's not true). On one hand $\lambda((E+\epsilon_n)\cup E)=2\lambda(E)$. On the other hand, $\lambda((E+\epsilon_n)\cup E)\to\lambda(E)$ -this follows from the compactness of $E$, as in the definition of the external measure we can replace countable collections of intervals covering $E$ with finite collections. So we have a contradiction.

user8268
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