Why does the set of real numbers with irrationality measure $\gt 2$ have zero measure?

Question

Recall the irrationality measure of a real number $r$ is $$ \mu(r)= \inf \left\{ \lambda\colon \left\lvert r-\frac{x}{y}\right\rvert\lt \frac{1}{y^{\lambda}} \text{ has only finitely many solutions} \right\}$$

Does anyone have a reference or proof?

Answer 1

A proof of the fact that the set of reals with irrationality measure $\gt 2$ has Lebesgue measure $0$ can be found in several places. I think the result is due to Khinchin, and is in his book on continued fractions.

Here is an online proof by Beukers. It is towards the end of the paper, but independent of the rest, which anyway is worth reading.

It follows that the set of reals in the interval $[0,1]$ with irrationality measure $2$ has Lebesgue measure $1$.

Answer 2

This can be established using the Borel-Cantelli Lemma. First note that is suffices to assume that $r \in (0,1)$ as otherwise we could just consider rational approximations shifted by an integer. Now define sets $E_n$ as follows: $$ E_n = (0,1) \cap\bigcup_{k=0}^n \left(\frac{k}{n}-\frac{1}{n^{2+\epsilon}}, \frac{k}{n}+\frac{1}{n^{2+\epsilon}} \right), \; n \geq 1 $$ for some $\epsilon>0$. Then the statement that there exist infinitely many solutions to $$ |r-\frac{x}{y}|<\frac{1}{y^{2+\epsilon}} $$ is equivalent to the statement that $r \in E_n$ for infinitely many of the $E_n$. But $m(E_n) = \frac{2}{n^{1+\epsilon}}, \; n \geq 1$ and so for all $\epsilon>0$ the sum $ \sum_{n=1}^{\infty} m(E_n) $ converges. The Borel-Cantelli Lemma now applies and we conclude the set $$ E = \{ r \in (0,1) | r \in E_n \text{ for infinitely many } n\} $$ has measure zero.