# Lebesgue measure of a set of irrational numbers

Let $I_{\lambda},$ $\lambda>0$ be a subset of all irrational numbers $\rho=[a_{1},a_{2},...,a_{n},...]\in(0,1)$ such that $a_{n}\leq \text{const}\cdot n^{\lambda}.$

Here, $[a_{1},a_{2},...,a_{n},...]$ is the continued fraction with partial quotients $a_1,a_2,\dots$.

My question is: under what values of $\lambda$ the set $I_{\lambda}$ has a positive Lebesgue measure?

• Your notation seems to imply you want $I_\lambda$ to be countable, which would make your question trivial. Could you please clarify if this is really what you want? – Yemon Choi Jul 16 '14 at 22:41
• @Yemon Choi: It looks like the question is about the set of numbers whose simple continued fraction coefficients don't have early large values. – Douglas Zare Jul 16 '14 at 22:56
• @DouglasZare Ah, that makes much more sense. Thanks. – Yemon Choi Jul 16 '14 at 22:59
• I don't think @EmilJeřábek is right here. The fact that the geometric mean of the $a_i$'s is something doesn't tell you how big the largest one is. – Anthony Quas Jul 17 '14 at 1:28
• See my answer to mathoverflow.net/questions/161441/… which gives a reference to Khinchin's book, and a Theorem of Khinchin that discusses general such problems. – Lucia Jul 17 '14 at 5:04

The condition $\lambda>1$ is sufficient and, at least almost, necessary:

To clarify, the space of irrational numbers $(0,1)-\mathbb Q$ is homeomorphic to $\omega^\omega$ under the map that sends $\frac{1}{a_1+\frac{1}{a_2+\cdots}}$ to a function $f$ satisfying $f(n)=a_{n+1}-1$. (As is well known.)

This way Lebesgue measure on $(0,1)$ induces a "Gaussian" measure on $\omega^\omega$.

Under this measure $\mu$, I claim that the following set has measure one: $$\{f:f(n)\text{ is eventually bounded by n^t for t>1, but not for t=1}\}$$

Proof: With $\delta_n=\sum_{k\le n^t}\epsilon_{nk}$ and $\Delta_n=\delta_n\log 2$ and $\log=\log_e$, we have

$$\log\prod_{n=1}^\infty \sum_{k\le n^t}\frac{\log\left(1+\frac{1}{k(k+2)}\right)}{\log 2}+\epsilon_{nk}=$$ $$\log\prod_{n=1}^\infty \frac{ \log 2+(\log (n^t+1)-\log(n^t+2))}{\log 2}+\delta_n =$$ $$\sum_{n=1}^\infty \log\left[ \log 2+(\log (n^t+1)-\log(n^t+2)) +\Delta_n \right]-\log\log 2$$

Here $\epsilon_{nk}=\pm\frac{A}{k(k+1)}e^{-\lambda\sqrt{n-1}}$ for certain constants $A,\lambda$.

By comparison of the negative of this series with $\sum \frac{1}{n^t}$, where we get a constant limit, the series converges iff $t>1$:

$$\lim_{n\to\infty}\frac{\log\log 2-\log(\log 2+(\log (n^t+1)-\log (n^t+2)+\Delta_n))}{n^{-t}}=$$ $$\lim_{n\to\infty} \frac{(\log 2+(\log (n^t+1)-\log (n^t+2)+\Delta_n))^{-1}[\frac{tn^{t-1}}{n^t+1}-\frac{tn^{t-1}}{n^t+2}+\Delta'(n)]} {-tn^{-t-1}}=$$

$$\lim_{n\to\infty}\frac{-n^{t+1}[\frac{tn^{t-1}}{n^t+1}-\frac{tn^{t-1}}{n^t+2}+\Delta'(n)]}{t(\log 2+(\log (n^t+1)-\log (n^t+2)+\Delta_n))}=$$

$$\lim_{n\to\infty}\frac{-n^{t+1}[\frac{n^{t-1}}{n^t+1}-\frac{n^{t-1}}{n^t+2}+\Delta'(n)/t]}{\log 2+(\log (n^t+1)-\log (n^t+2)+\Delta_n)}=$$

$$\lim_{n\to\infty}\frac{-n^{2t}[\frac{1}{n^t+1}-\frac{1}{n^t+2}+\frac{\Delta'(n)}{tn^{t-1}}]}{\log 2+(\log (n^t+1)-\log (n^t+2)+\Delta_n)}=$$

$$\lim_{n\to\infty}\frac{-n^{2t}[\frac{1}{(n^t+1)(n^t+2)}+\frac{\Delta'(n)}{tn^{t-1}}]}{\log 2+(\log (n^t+1)-\log (n^t+2)+\Delta_n)}=$$ $$\frac{1}{\log 2}$$

provided $n^{t+1}\Delta'(n)\to 0$.

And indeed

$$n^{t+1}\Delta'(n)=(\pm A)(\log 2)n^{t+1}\frac{d}{dn}[e^{-\lambda\sqrt{n-1}}\sum_{k\le n^t}\frac{1}{k(k+1)}]$$

Ignoring the constant and using $f(n)=\sum_{k\le n^t}g(k)\approx \int_1^{n^t}g(k)dk$, so $f'(n)\approx g(n^t)-g(1)$, we have

$$n^{t+1} \left[ e^{-\lambda\sqrt{n-1}}[-\lambda\frac{1}{2\sqrt{n-1}}]f(n)+e^{-\lambda\sqrt{n-1}}f'(n) \right]=$$

$$e^{-\lambda\sqrt{n-1}} n^{t+1} \left[ \left(-\lambda\frac{1}{2\sqrt{n-1}}\right)f(n)+f'(n) \right]$$

and this clearly goes to 0 as $n\to\infty$.

So $\log\prod=-\infty$ and hence $\prod=0$. This completes the proof.

• Thank you very much. Could you suggest me books regarding these facts. – sokho Jul 16 '14 at 23:07
• Actually my coauthor Bonnie S. Huggins and I just proved it from scratch. But there should be some books about continued fractions and the Gaussian measure on the irrational numbers. – Bjørn Kjos-Hanssen Jul 16 '14 at 23:12

A precise solution to this problem is known. In Khintchine's book on continued fractions it says:

$\mathbf{Theorem~30}$ Suppose that $\varphi(n)$ is an arbitrary positive function with natural argument $n$. The inequality $$a_n = a_n(\alpha) \geq \varphi(n)$$ is, for almost all $\alpha$, satisfied by an infinite number of values $n$ if the series $\sum_n 1/\varphi(n)$ diverges. On the other hand, this inequality is, for almost all $\alpha$, satisfied by only a finite number of values of $n$ if the series $\sum_n 1 /\varphi(n)$ converges.

(quoting from: A. Ya. Khintchine. Continued Fractions. University of Chicago Press, 1964)