# Almost all non-negative real numbers have only finitely many multiples lying in a measurable set with finite measure

Let $$A$$ be Lebesgue measurable subset of $$[0,\infty)$$ such that Lebesgue measure of $$A$$ is positive i.e. $$0<\lambda(A)<\infty$$. Let $$S$$ be the set defined as follows: $$S:=\{t\in [0,\infty):nt\in A\text{ for infinitely many }n\in\Bbb N\}$$

What can we conclude about the measure of $$S$$?

I can guess that $$\lambda (S)=0$$ for when $$A$$ is an open set but can't prove it. More particular case, when $$A$$ is open with finitely many components then I can conclude that $$\lambda(S)=0$$

• cf. the problem from 2018 Moscow State University olympiad: if $f:(0,\infty)\to \mathbb{R}$ is continuous and $\int_{0}^\infty f(x) dx<\infty$, then there exists $x>0$ such that $\sum_{n=1}^\infty f(nx)<\infty$, with exactly the same solution as in Mateusz Kwaśnicki's answer. – Fedor Petrov Dec 1 '19 at 11:25

Let $$f(t) = 1$$ if $$t \in A$$ and $$f(t) = 0$$ otherwise. Suppose that $$a > 0$$. Then \begin{aligned} \int_a^{2 a} \operatorname{card} \{n : n t \in A\} dt & = \int_a^\infty \biggl(\sum_{n = 1}^\infty f(n t) \biggr) dt \\ & = \sum_{n = 1}^\infty \int_a^{2 a} f(n t) dt \\ & = \sum_{n = 1}^\infty \frac{1}{n} \int_{n a}^{2 n a} f(s) ds \\ & = \int_a^\infty \biggl( \sum_{n = \lceil s/(2 a) \rceil}^{\lfloor s/a \rfloor} \frac{1}{n} \biggr) f(s) ds \\ & \leqslant \int_a^\infty \frac{\lfloor s/a \rfloor - \lceil s/(2 a) \rceil + 1}{\lceil s/(2 a) \rceil} \, f(s) ds \\ & \leqslant \int_a^\infty \frac{s/a - s/(2 a) + 1}{s/(2 a)} \, f(s) ds \\ & \leqslant \int_a^\infty 3 f(s) ds \leqslant 3 \lambda(A) < \infty . \end{aligned} Therefore, $$\operatorname{card}\{n : n t \in A\}$$ is finite for almost all $$t \in (a, 2 a)$$. Since $$a$$ is arbitrary, we conclude that $$\operatorname{card}\{n : n t \in A\}$$ is finite for almost all $$t > 0$$, that is, $$\lambda(S) = 0$$.