Measurability of Brjuno numbers A positive irrational number $\alpha\in{\mathbb R}\setminus {\mathbb Q}$ is said to be a Brjuno number if $$\sum_{i=1}^\infty\frac{\log q_{i+1}}{q_i} < \infty$$ where $q_i>0$ is the denominator of the $i$th convergent $\frac{p_i}{q_i}$ of the continued fraction expansion of $\alpha$.
Is the set $B$ of Brjuno numbers measurable, and if yes, what is the measure of $B\cap[0,1]$?
 A: It is of full measure because it is implied by a Diophantine condition. Unfortunately, I don't remember a reference for these facts at the moment. The Brjuno condition appears in the work of Yoccoz concerning the linearization of circle diffeomorphism. The Diophantine condition appears in a previous work of Siegel, so searching for "Siegel linearization theorem" should lead you to the diophantine codition and to a proof of the fact that it is of full measure. Try there for example.
A: I replace $\alpha$ by $x$ below.
Restricting ourselves to positive numbers is useless, since the property considered does not depend on $a_0(x) = \lfloor x \rfloor$. Therefore, $B$ is the union of all images of $B \cap [0,1[$ by integer translations.
Now, call $G$ the Gauss-Kuzmin map from $[0,1[ \setminus \mathbb{Q}$ to itself: $G(x) = 1/x-\lfloor 1/x \rfloor$.
This map preserves the probability measure $\mu$ with density $x \mapsto (\ln 2)^{-1}(1+x)^{-1}$ on $[0,1[ \setminus \mathbb{Q}$ and is ergodic.
The partial quotients $(a_n(x))_{n \ge 1}$ of $\alpha$ are the integer parts of the iterates $(G^n(x))_{n \ge 1}$. Hence $(a_n)_{n \ge 1}$ and $(q_n)_{n \ge 1}$ are Borel functions, so $B$ is a Borel set.
One checks that $\ln(a_1+1)$ is integrable with regard to $\mu$. Birkhoff theorem applies, so when $n \to +\infty$
$$\frac{\ln q_n}{n} \le \frac1n\sum_{k=1}^n \ln(a_k+1) \to C:= \int\ln(a_1+1)\mathrm{d}\mu \quad \mu \text{-a.s..}$$
On the other hand, $(q_n)_{n \ge 1}$  is bounded below by Fibonacci sequence. Therefore, by comparison, the series $\sum_n \frac{\ln q_{n+1}}{q_n} $ converges $\mu$-almost surely, and also almost surely for the Lebesgue measure.
