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

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?

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Would you mind terribly accepting answers to some of the other questions you have asked? – Gerry Myerson Aug 20 '12 at 23:25
Concerning accepting answers - please see meta.math.stackexchange.com/questions/3399/… – Gerry Myerson Aug 20 '12 at 23:27
I think it will not be $0$, more like (in the interval $[0,1]$ measure $1$. – André Nicolas Aug 20 '12 at 23:40
Gerry: I accepted some answers – hello Aug 20 '12 at 23:49
Andre, yes youre right. Sorry, I meant the set of real numbers that have irrationality measure larger than 2. – hello Aug 21 '12 at 0:05

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