Erdős Problem #1197

Let \(E\subset (0,\infty)\) be a set of positive measure. Is it true that, for almost all \(x>0\), for all sufficiently large (depending on \(x\)) integers \(n\) there exists an integer \(r\geq 1\) such that \(nx\in r\cdot E\)? A problem of Haight, who constructed a set \(E\subset (0,\infty)\) of infinite measure such that, for all \(x\in E\), \(x\not\in r\cdot E\) if \(r\geq 2\), and also for all \(x>0\), if \(n\) is large enough then \(nx\not\in E\) (see [1195](https://www.erdosproblems.com/1195)). This is trivially true if \(E\) contains an interval \((a,b)\) with \(a<b\), since for any \(x>0\), for all large \(n\), the interval \((\frac{nx}{b},\frac{nx}{a})\) has length \(>1\) so contains at least one integer \(r\geq 1\). Buczolich and Mauldin [BuMa99] proved that there exists an open set \(E\subset (0,\infty)\) and two intervals \(I,J\subset [1/2,1)\) such that, for all \(x\in I\), \(x\in \frac{1}{n}\cdot E\) for infinitely many \(n\geq 1\), and for almost all \(x\in J\), \(x\not\in \frac{1}{n}\cdot E\) for all sufficiently large (depending on \(x\)) \(n\). This was solved in the negative by ebarschkis in the comments, who constructed a counterexample using a variant of the Buczolich-Mauldin construction.