Does there exist a measurable subset $T$ of $[0, \infty)$ with finite measure and some $\epsilon > 0$ such that for every $r$ with $0 < r < \epsilon$, $nr$ is in $T$ for infinitely many positive integers $n$?


Note: The integers $n$ such that $nr$ lie in $T$ can depend on $r$.



A trapping set with finite measure
No. Denote $T\_k=T\cap [k,k+1)$. Then $\sum |T\_k|<\infty$ (where $|X|$ stands for the measure of $X\subset \mathbb{R}$). Choose a segment $[a,b]\subset (0,\epsilon)$. Note that if $r\in [a,b]$ and $nr\in T\_k$, then $na\leqslant nr< k+1$ and $nb\geqslant nr\geqslant k$, thus $n\in [k/b,(k+1)/a]$. The union of $n^{-1}T\_k$ over all positive integers $n\in [k/b,(k+1)/a]$ has measure at most $$|T\_k|\cdot \sum\_{n\in [k/b,(k+1)/a]} n^{-1}\leqslant C|T\_k|,$$
where $C$ depends only on $a$ and $b$. Now choose $k\_0$ so that $C\sum\_{k>k\_0} |T\_k|<b-a$. By the pigeonhole principle there exists a point $r\in [a,b]$ not covered by $n^{-1} T\_k$ with $k>k\_0$ and positive integer $n$. 


