3
$\begingroup$

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

$\endgroup$
2
  • 1
    $\begingroup$ 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
7
$\begingroup$

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 $x\notin\bigcap_{|t|<\varepsilon}(t+E)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.