Measure of a subset of $\mathbb{R_{>0}}$ that intersects every $\mathbb{N}_{>0}x$ Imagine you're looking at a light indicator that flashes periodically, but with an unknown period. The goal is to keep your eyes open for a finite amount of time to see it flash at least once.
Formally, let $A$ be a subset of $\Bbb{R}_{>0}$, such that:
$$ \forall x \in \Bbb{R}_{>0}, \exists n \in \mathbb{N}_{>0},nx \in A. $$
Another way to put it is:
$$ \Bbb{R}_{>0}= \bigcup_{n \in \mathbb{N}_{>0}} \frac{1}{n} A. $$
Assume that $A$ is Lebesgue-measurable. Can it be of finite measure?
 A: The set $A$ necessarily has infinite measure. Let's prove it by contradiction, assuming that it has finite measure.
Since Lebesgue measure is outer regular, $A$ can be expanded to an open set with finite measure.  Then,
$$A=\bigcup_{k=0}^{+\infty}{B(x_k,r_k)}$$
is a disjoint union of open balls with non-zero radii (its connected components). Moreover,
$$\mu(A)=\sum_{k=0}^{+\infty}{2r_k}$$
so the sequence $(r_k)$ converges towards zero and is bounded, say, by $R \in \mathbb{R}_{>0}$.
For each $k$ in $\mathbb{N}$, let $I_k=\bigcup_{n=1}^{+\infty}{\frac{1}{n}B(x_k,r_k)}$. Then, $\mathbb{R}_{>0}=\bigcup_{k=0}^{+\infty}{I_k}$. By the way, $\frac{1}{n}B(x_k,r_k) =B(\frac{x_k}{n},\frac{r_k}{n})$.
Let $a$ and $b$ be real numbers, with $b > a > 2R$. For each $k$ in $\mathbb{N}$, define
$$S_k=\left\{n \in \mathbb{N}_{>0} \bigg| \frac{1}{n}B(x_k,r_k) \text{ meets } \left] a,b \right[ \right\}.$$
The supremum of $\frac{1}{n}B(x_k,r_k)$ is $\frac{x_k+r_k}{n}$, which converges to zero as $n$ goes to infinity, so it is less than $a$ for big enough $n$, because $a>0$. As a consequence, the set $S_k$ is finite. Moreover, $S_k$ is an integer interval.
If $S_k$ is non-empty, call its minimum and maximum $m_k$ and $M_k$, respectively. Then, \begin{align*}
\frac{x_k-r_k}{m_k} & < b, \\
\frac{x_k+r_k}{M_k} & > a.
\end{align*}
Dividing these two inequalities yields:
$$\frac{M_k}{m_k} \cdot \frac{x_k-r_k}{x_k+r_k} \leqslant \frac{b}{a}.$$
Then,
\begin{align*}
\mu(I_k\cap\left] a,b \right[) & \leqslant \sum_{n\in S_k}{\mu \left( \frac{1}{n}B(x_k,r_k) \right)} \\
& = \sum_{n\in S_k}{\frac{2r_k}{n}} \\
& = 2r_k\sum_{n=m_k}^{M_k}{\frac{1}{n}}.
\end{align*}
Let $h_p = \sum_{i=1}^{p} \frac{1}{i}$ be the $p$-th partial sum of the harmonic series. By simple comparisons between sums and integrals, for all $p\in \mathbb{N}_{>0}$,
$$ \ln(p+1) \leqslant h_p \leqslant \ln(p) +1,$$
and the left inequality also holds for $p=0$.
As a consequence,
\begin{align*}
\sum_{n=m_k}^{M_k}{\frac{1}{n}} & = h_{M_k} - h_{m_k-1} \\
& \leqslant \ln(M_k) + 1  - \ln(m_k) \\
& = \ln \left( \frac{M_k}{m_k} \right) + 1 \\
& \leqslant \ln \left( \frac{b}{a} \cdot \frac{x_k+r_k}{x_k-r_k} \right) + 1 \\
& = \ln \left( \frac{b}{a} \right) + \ln \left( \frac{x_k+r_k}{x_k-r_k} \right) + 1.
\end{align*}
Indeed, $x_k+r_k \geq \frac{x_k+r_k}{M_k} > a$ so $x_k -r_k > a - 2r_k \geq a-2R > 0$, so it is possible to divide by $x_k-r_k$. The function from $[a-R,+\infty] \times [0,R]$ defined by $(x,r) \mapsto \frac{x+r}{x-r}$ and $(+\infty,r) \mapsto 1$ is well-defined since $a>2R$, and is continuous on a compact domain. As a consequence, it is bounded, say, by $Q \in \mathbb{R}_{>0}$. Then,
\begin{align*}
\mu(I_k\cap\left] a,b \right[) & \leqslant 2 r_k \left( \ln \left( \frac{b}{a} \right) + \ln \left( Q \right) + 1 \right) \\
& \leqslant 2 r_k \left( \ln \left( b \right) + P \right),
\end{align*}
where $P=\ln \left( Q \right) + 1 - \ln(a)$ depends on $a$ but not on $b$. If $S_k$ is empty, then $I_k$ does not meet $\left] a,b \right[$ so $\mu(I_k\cap\left] a,b \right[)=0$. As a consequence, the above inequality still holds trivially in this case. Finally, as all the $I_k$ together cover $\mathbb{R}_{>0}$,
\begin{align*}
b-a & = \mu(\left] a,b \right[) \\
& \leqslant \sum_{k=0}^{+\infty}{\mu(I_k\cap\left] a,b \right[)} \\
& \leqslant \sum_{k=0}^{+\infty}{2 r_k \left( \ln \left( b \right) + P \right)} \\
& = \mu(A) \left( \ln \left( b \right) + P \right).
\end{align*}
With a fixed $a$, this inequality cannot hold as $b$ goes to infinity, so this is a contradiction.
