# To show a set is a set of positive Lebesgue measure in $\mathbb{R}$

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

• 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)? – David Handelman Oct 7 '20 at 21:27
• 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. – 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 $$x\notin\bigcap_{|t|<\varepsilon}(t+E)$$.