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? 



Measure of a set of irrational numbers
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](http://en.wikipedia.org/wiki/Levy%27s_constant) constant.


Next another [Theorem of Khinchin](http://en.wikipedia.org/wiki/Khinchin%27s_theorem_on_Diophantine_approximations#Khinchin.27s_theorem_and_extensions) 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 $x$).


