As I was reading an article on the HRT conjecture, namely 'Proof of the HRT conjecture for almost every (1,3) configuration', by Wencai Liu, I stumbled upon the assertion :
$E = \{ x \in \mathbb{R} : (\exists c > 1) \liminf n^c ||nx|| < \infty \}$ has zero Lebesgue measure.
Here, ||x|| denotes the distance to the nearest integer.
Till the present moment, I don't know how to prove it.
I tried to write $E \subset F$ where $F^c$ is such that if $x \in F^c$, $(\exists \epsilon >0) (\exists k \in \mathbb{N}) (\forall n \geq k) : ||nx|| \geq n^{\epsilon - c}$.
$E$ is invariant by translation by $\mathbb{Z}$, so it suffices to show that $\lambda(F \cap [0,1])=0$, but I can't get through the computations.
What I showed :
One can simplify ||nx|| when $\frac{i}{2x} \leq n < \frac{i+1}{2x}$ and $i \in \mathbb{N}$, depending on the parity of
$i$.$F \cap [0,1]$ can be expressed by a formula of the type $\cap_{\epsilon > 0} \cap_{k \in \mathbb{N}} \cup_{n \geq k} \cup ...$
For $\epsilon \in \{1,\frac{1}{2},\frac{1}{3},\dots\}$ and $C,N \in \mathbb{N}$, let $$E_{\epsilon,C,N} = \left\{x \in [0,1] : ||Nx|| \le \frac{C}{N^{1+\epsilon}}\right\} = [0,1] \cap \bigcup_{0 \le k \le N} \left(\frac{k}{N}-\frac{C}{N^{2+\epsilon}},\frac{k}{N}+\frac{C}{N^{2+\epsilon}}\right).$$ Clearly $$|E_{\epsilon,C,N}| \le \sum_{k=0}^N \frac{2C}{N^{2+\epsilon}} \le \frac{4C}{N^{1+\epsilon}},$$ so $$0 = \lim_{N_0 \to +\infty} \sum_{N \ge N_0} |E_{\epsilon,C,N}| \ge \lim_{N_0 \to +\infty} \left|\bigcup_{N \ge N_0} E_{\epsilon,C,N}\right|,$$ meaning $$|\{x \in [0,1] : \liminf_{n \to \infty} n^{1+\epsilon}||nx|| \le C\}| = 0$$ for each $\epsilon,C$, meaning $$|\{x \in [0,1] : \liminf_{n \to \infty} n^{1+\epsilon} ||nx|| < \infty\}| = 0$$ for each $\epsilon > 0$. This gives your result, since the sets $$\{x \in [0,1] : \liminf_{n \to \infty} n^{1+\epsilon} ||nx|| < \infty\}$$ are increasing as $\epsilon$ goes down to $0$, and due to the noted translation invariance of the question.