Let $E\subset \mathbb{R}$ be a set of positive Lebesgue measure. Can we find $l>0$ such that $$\bigcap_{-l\leq t \leq l}t+E$$ is a set of positive Lebesgue measure?

Notation: $t+E=\{t+e|e\in E\}$

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    $\begingroup$ There is possibly something wrong with the formulation of this question. First, you might as well assume that $l=1$. Second, if $E $ is the any proper subset of the open interval $(0,1)$, then the intersection is going to be empty. Perhaps you mean $E$ is a subset of full measure (that is, its complement has measure zero)? $\endgroup$ – David Handelman Oct 7 '20 at 21:27
  • $\begingroup$ You probably get it wrong. Given the set $E$ I am asking whether there is $l>0$(obviously depending on $E$) such that the above intersection is non-empty or not. For example if your $E=(0,1)$ then if we take $l=0.001$ then the above intersection will be non-empty. $\endgroup$ – Duplicate Oct 13 '20 at 7:28

No. Every set $E$ without interior points (e.g. the complements of the rationals) has the property that $$\bigcap_{|t|<\varepsilon}(t+E)=\emptyset$$ for every $\varepsilon>0$. Indeed, for every $x\in E$, there is $t$ with $|t|<\varepsilon$ and $x-t\notin E$, hence $x\notin\bigcap_{|t|<\varepsilon}(t+E)$.


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