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 $\lambda(S)=0$

  • 4
    $\begingroup$ 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. $\endgroup$ – 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, $\lambda(S) = 0$.


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