Let $A$ be a set of all irrational numbers $\rho \in (0, 1)$ represented as a continued fraction $\rho=[a_{1}, a_{2},...,a_{n},...],$ such that $a_{n}\leq \text{const}\cdot n^{\epsilon}$ for some $\epsilon \in (0, 1/2).$ ($\epsilon$ and const are fixed.)

Question: Is the Lebesgue measure of the set $A$ positive? How can I find this measure?


1 Answer 1


Your set has measure zero by theorems of Khinchin. First a theorem of Khinchin shows that for almost all real numbers $x$ (i.e. outside a set of measure zero) one has $$ \lim_{n\to \infty} \frac{\log q_n}{n}= C $$ for a positive constant $C$. Here $p_n/q_n$ are the convergents of $x$. So almost surely, the denominators $q_n$ are exponential in $n$. In fact the constant $C$ here is known to be $\pi^2/(12 \log 2)$ -- the Levy or Khinchin-Levy constant.

Next another Theorem of Khinchin shows that for almost all $x$ one has infinitely many approximations $p/q$ with $$ \Big| x-\frac{p}{q}\Big| \le \frac{1}{q^2\log q}. $$ Here what is used is that $\sum_q 1/(q\log q)$ diverges.

Putting both results together (and since continued fractions give best rational approximations) we see that for almost all $x$ we have infinitely many $n$ such that $$ \Big| x-\frac{p_n}{q_n} \Big | \le \frac{C}{n q_n^2}, $$ for some positive constant $C$. Now recall that for any $x$ we have $|x-p_n/q_n|$ is about size $1/(q_nq_{n+1})$ and $q_{n+1}$ is about size $a_{n+1}q_n$. Therefore it follows from the above inequality that for infinitely many $n$ one has $a_n \ge cn$ for some constant $c>0$.

Edit: I just got the chance to look at Khinchin's book Continued Fractions, and he discusses this question explicitly in Theorem 30 of that book (page 63 in the Dover edition). Let $\phi(n)$ be an arbitrary positive function ($n\in {\Bbb N}$). Then the inequality $a_n \ge \phi(n)$ is satisfied for infinitely many $n$ for almost all real numbers $x$, provided $\sum 1/\phi(n)$ diverges. On the other hand if $\sum 1/\phi(n)$ converges, then this inequality is satisfied for only finitely many $n$ (and almost all $x$).

  • $\begingroup$ Thank you very much for your answer. How is it follow $a_{n}\geq cn$ from last inequality? Can you explain it please. $\endgroup$
    – sokho
    Mar 26, 2014 at 3:50
  • $\begingroup$ From properties of continued fractions one knows that $|x-p_n/q_n|$ is roughly $1/(q_nq_{n+1})$ and also $q_{n+1}$ is roughly $a_{n+1}q_n$. So the LHS of the last inequality is about $C/(q_n^2 a_{n+1})$ and now you get that $a_{n+1} \ge c n$. Does that make sense? $\endgroup$
    – Lucia
    Mar 26, 2014 at 4:08

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