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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

1 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 $1$.

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