For a.e. $x \in [0, 1]$, there are finitely many $p/q$ such that $\left| x - p/q \right| < 1 / \left( q \log q \right)^2$

Question

I am stuck on a qualifying exam problem and was hoping to get some help.

Show for a.e. $x \in [0, 1]$ that there are finitely many $p/q \in \mathbf{Q}$ in reduced form such that $q \geq 2$ and $\left| x - p/q \right| < 1 / \left( q \log q \right)^2$.

There is a hint, which is to consider the intervals of length $2/\left(q \log q \right)^2$ centered at the $p/q$.

To simplify notation, let $\mathbf{Q}_x$ be the set of such $p/q$ corresponding to each $x$.

I've been working for awhile...I have been examining things like $$ \int_0^1 \sum_{p/q \in \mathbf{Q}_x} \frac{dx}{ \left( q \log q \right)^2} > \int_0^1 \sum_{p/q \in \mathbf{Q}_x} \left| x - p/q \right| dx = \sum_{p/q \in \mathbf{Q}} \int_0^1 \left| x - p/q \right| \chi_{\mathbf{Q}_x} dx.$$

I proved for $p/q \in (0, 1)$, $q \geq 3$, that $\left[ p/q, p/q + 1/ \left( q \log q \right) \right] \subset [0, 1]$.

I feel like I could really use some guidance. I have no idea how to capture what goes wrong if $\mathbf{Q}_x$ is infinite on a set of positive measure.

Thanks for your help.

Answer 1

You can show the contrary, i.e. the set of $x \in [0, 1]$ that there are infinitely many $p/q \in \mathbf{Q}$ in reduced form such that $q \geqslant 2$ and $\left| x - p/q \right| < 1 / \left( q \log q \right)^2$ is of measure $0$.

Edit: Let $a_i=p_i/q_i \:(p_i<q_i,\:2\leqslant q_i\in\Bbb{N})$ be the enumeration of the rational numbers in $[0, 1]$ and $A_i=(a_i-1/(q_i\log q_i)^2, a_i+1/(q_i\log q_i)^2)$. Then clearly, there are at most $q_i-1$ intervals that have the same length of $A_i$, which is $2/(q_i\log q_i)^2$. So $$ \sum_{i=1}^{\infty}q_i\mu(A_i)\leqslant 2\sum_{i=1}^{\infty}\frac{q_i}{(q_i\log q_i)^2}=2\sum_{i=1}^{\infty}\frac{1}{q_i\log^2 q_i}<\infty\tag1 $$ Thus $$ \sum_{i=1}^{\infty}\mu(A_i)<\sum_{i=1}^{\infty}q_i\mu(A_i)<\infty $$ The set $C$ of $x\in [0, 1]$ with infinitely many $p/q \in \mathbf{Q}$ such that $q \geqslant 2$ and $\left| x - p/q \right| < 1 / \left( q \log q \right)^2$ is the set that $x$ is in the infinite many $A_i$, i.e. infinite often of $A_i$. By Borel-Cantelli lemma and $(1)$ $$ \mu(C)=\mu(A_i\:i.o)=\mu(\bigcap_{i=1}^{\infty}\bigcup_{k=i}^{\infty}A_k)=0 $$

Answer 2

Just to elaborate on the method proposed by user1952009:

Let $Q(x)$ be the number of rationals $p/q \in \mathbf{Q} - \left\{0, 1\right\}$ in reduced form with $q \geq 2$ satisfying $$ \left| x - \frac{p}{q} \right| < \frac{1}{\left( q \log q \right)^2} $$ and let $\mathbf{Q}(x)$ be the set of such rational numbers. Since $$\int_0^1 Q(x) dx < \infty \Longleftrightarrow Q(x) < \infty \text{ a.e.} $$ it suffices to prove the former. We have by Tonelli's theorem $$\int_0^1 Q(x) dx = \sum_{p/q} \int_0^1 \chi_{\mathbf{Q}(x)}\left( p/q \right) dx \leq \sum_{q=2} \frac{q-1}{q} \frac{2}{q \log^2(q)} < \infty.$$ So we are done.