# A trapping set with finite measure

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 $$r$$.

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 $$C\sum_{k>k_0} |T_k|. By the pigeonhole principle there exists a point $$r\in [a,b]$$ not covered by $$n^{-1} T_k$$ with $$k>k_0$$ and positive integer $$n$$.