# Does the set of real numbers with bounded partial quotients have positive measure?

We say a real number $$x$$ has bounded partial quotients if its continued fraction expansion $$[a_0; a_1, a_2 \cdots]$$ is bounded by some constant $$M=M(x)$$. The set $$A$$ consisting of those numbers whose partial quotients are bounded forms a dense, uncountable subset of $$\mathbb{R}$$ which includes the algebraic numbers of degree $$\leq 2$$. It appears that it is an open problem whether or not $$A$$ contains any algebraic numbers of degree $$>2$$.

Question : Is it known whether or not the set $$A$$ has measure zero?

The reason this question is interesting has to do with Diophantine approximation. We say a real number $$x$$ is badly-approximable if there exists a positive constant $$C=C(x)$$ such that $$\left|x-\frac{p}{q}\right|>\frac{C}{q^2}$$

for all rational $$p/q \neq x$$. Here, the name fits, since for any irrational $$x$$, $$|x-p/q|<\frac{1}{q^2}$$ for infinitely many pairs $$(p,q)$$, and badly-approximable numbers are precisely those for which we cannot do better than this, i.e., merely scaling the numerator by a certain constant $$C$$ ruins everything.

It turns out that the property defining the set $$A$$ discussed earlier (i.e., bounded partial quotients) is in fact completely equivalent to this property of being badly-approximable. Thus asking about the size of $$A$$ is essentially asking about the size of the set of worst-approximable real numbers.

• It's known that for some specific $C$, the set is very small; there are only countably many numbers that satisfy the conditions with $C=\frac13$, for instance. You might be interested in en.wikipedia.org/wiki/Lagrange_number, en.wikipedia.org/wiki/Markov_constant and en.wikipedia.org/wiki/Markov_spectrum (assuming you aren't already familiar with these aspects of the theory, of course). This is a really interesting question, though; I'm curious as to the answer! May 17, 2020 at 4:15
• @StevenStadnicki I think I have the answer. See below! May 17, 2020 at 8:10

I think I have a proof that the claim is true. It seems to follow from a theorem of Khinchin in $$1924$$. This theorem asserts the following. Consider an arbitrary function $$\psi:\mathbb{N} \to \mathbb{R}_{\geq 0}$$ such that $$\{q \psi(q)\}_{q \in \mathbb{N}}$$ is decreasing, and let $$\mathcal{K}$$ denote the set of real numbers $$\alpha$$ for which $$\left|\alpha-\frac{p}{q}\right|<\frac{\psi(q)}{q}$$

has infinitely many solutions $$p/q \neq \alpha$$. Then if $$\sum_{q \geq 1} \psi(q)=+\infty$$, $$\mathcal{K}$$ has full measure. Note that it follows that $$\mathbb{R} \setminus \mathcal{K}$$ has measure zero, and this complement set consists precisely of those $$\alpha$$ such that $$\left|\alpha-\frac{p}{q}\right|<\frac{\psi(q)}{q}$$ has at most finitely many solutions $$p/q \neq \alpha$$.

I have taken the formulation of Khinchin's theorem from this paper by Dimitris Koukoulopoulos and James Maynard (not quite verbatim). In their paper, everything was done in the interval $$[0,1]$$, but replacing $$[0,1]$$ with $$\mathbb{R}$$ is immaterial—I've reworded things accordingly.

Now, going back to the original problem, our set $$A$$ can be equivalently written as $$A=\bigcup_{N \geq 1} A_N$$ where $$A_N=\left\{\alpha \in \mathbb{R} : \left|\alpha-\frac{p}{q}\right| \geq \frac{1}{Nq^2} \ \text{for all rational} \ p/q \neq \alpha \right\}$$

It thus suffices to prove each $$A_N$$ has measure zero, by the subadditivity of measure. We note that $$A_N \subset A'_N$$, where we define $$A'_N$$ similar to $$A_N$$, but with "for all rational $$p/q$$" replaced by "for all but possibly finitely many rational $$p/q$$". It suffices to prove $$A'_N$$ has measure zero. Here, taking $$\psi(q)=\frac{1}{qN}$$, we have by Khinchin's theorem that $$A'_N$$ is exactly $$\mathbb{R} \setminus \mathcal{K}$$, where $$\mathcal{K}$$ is as defined earlier. Moreover $$\sum_{q \geq 1} \frac{1}{qN}=+\infty$$. It follows that $$A'_N$$ has measure zero. This proves the result.

Note we also have a simple corollary. For fixed $$\lambda>2$$, say a real number $$\alpha$$ is $$\lambda$$-badly-approximable if $$\left|\alpha-\frac{p}{q}\right| > \frac{C}{q^\lambda}$$ for all rational $$p/q \neq \alpha$$ and some positive constant $$C=C(\lambda, \alpha)$$. Let $$A_{\lambda}$$ denote the set of $$\lambda$$-badly-approximable numbers. Then $$A_{\lambda}$$ has full measure. In other words, almost all real numbers are $$\lambda$$-badly-approximable. This is in stark contrast to the case $$\lambda=2$$, which we've shown has zero measure. The reason this holds is that the full statement of Khinchin's theorem states not only that if $$\sum_{q \geq 1} \psi(q)=+\infty$$, $$\mathcal{K}$$ has full measure, but it also states that if $$\sum_{q \geq 1} \psi(q)<+\infty$$, $$\mathcal{K}$$ has zero measure. Thus we can mimick the argument earlier, this time using $$\psi(q)=\frac{1}{Nq^{\lambda-1}}$$, and using that $$\mathbb{R} \setminus \mathcal{K} = A_{\lambda}$$ and $$\sum_{q} \frac{1}{Nq^{\lambda-1}} < + \infty$$.

I also suspect that Khinchin's theorem can be used to prove the fact that almost all real numbers have irrationality measure equal to $$2$$ (unless of course Khinchin used this fact in the original proof!)

• AlohaSine, you might consider to "accept" this answer of yourself, to signal to the other readers that you've "closed/solved-the-case" . Oct 14, 2021 at 9:31
• @GottfriedHelms Done Oct 14, 2021 at 9:50

Khinchin, in his lovely monograph, "Continued Fractions," University of Chicago Press, 1964, proves in Theorem 29 that the set A has measure 0.

• maybe this is more suitable as a comment. May 23, 2021 at 2:33