# If $x\notin\mathbb{Q}$, then $\left|x-\frac{p}{q}\right|<\frac{1}{q^{2+\epsilon}}$ for infinitely many $\frac{p}{q}$?

This appears on Problem 1 of Chapter 1 in Stein & Shakarchi's Real Analysis:

Given an irrational $x$, one can show (using the pigeon-hole principle, for example) that there are infinitely many fractions $p/q$, with relatively prime integers $p$ and $q$ such that $$\left|x-\frac{p}{q}\right|\leq\frac{1}{q^2}.$$ However, prove that the set of all $x\in\mathbb R$ such that there exist infinitely many fractions $p/q$, with relatively prime integers $p$ and $q$ such that $$\left|x-\frac{p}{q}\right|\leq\frac{1}{q^{3}} \text{ (or} \leq 1/q^{\epsilon+2})$$ is a set of measure zero.

By the hint, I am trying to prove this using the Borel-Cantelli lemma. As my family of countable sets, I'm considering $E_{q} = \{x : \exists p. |x - p/q| < 1/q^{2+\epsilon} \}$.

I'm having trouble showing that the sum of the measures of $E_{q}$ converges. I can show that $E_q$ is the union of intervals of length $2/q^{2 + \epsilon}$, but I can't put a bound on the total number of intervals because the problem statement says $x \in \mathbb{R}$ so $p$ could be arbitrarily large.

I found a similar problem statement in this question, but there the bound comes from the fact that $x \in [0,1]$.

Am I considering the wrong family of sets for this problem, mis-interpreting the problem statement, or encountering some other issue?

• Related (but different): math.stackexchange.com/questions/556320/… Jan 17, 2014 at 0:47
• If you can read french there was a solution to this problem in the exercise notes of one of my couses mathaa.epfl.ch/prob/enseignement/mesure/series/cor3a.pdf (Exercise 4). I have to go to bed though now so I don't have time to outline/translate it right now : /. (But this subforum is very competent so someone else is bound to have explained it before I wake anyways) Jan 17, 2014 at 1:00

What you want for a $x$ is the property that is hit by infinite rational numbers with the desired property.

The $E_i$ is good to count when this property happens once so to avoid overcounting.

Take an enumeration of rationals $a_i$, and write $a_i = \frac{p_i}{q_i}$

This gives the intuition to define $E_i = \{x| |x- \frac{p_i}{q_i}|< \frac{1}{q_i ^{2+ \epsilon}} \}$

But now $$\sum _{i=1}^{\infty} \mu (E_i) = \sum _{i=1}^{\infty}2 \frac{1}{q_i ^{2+ \epsilon}} <\infty$$

So borel canteli lemma can be applied.

Another way to see borel canteli lemma is by writing

$$f(x)= \sum _{i=1}^{\infty}\chi _{(a_i -\frac{1}{q_i ^{2+ \epsilon}} , a_i +\frac{1}{q_i ^{2+ \epsilon}}) } (x)$$

which $f(x_0)$ counts the number which the desired property for $x_0$ is achivied. We want to show that the set $x_0$ hit by infinite number of intervals has measure zero. Equivalently we show $$\int f(x) \rm{d}x <\infty$$ Which is true by monotone convergence.

• Hi, wouldn't this argument work for the power 2 as well? In fact any $1+\epsilon$ power? Where does it break down?
– user172377
Feb 22, 2017 at 3:43
• I think you forgot to range over the possible p's which are many for a given qi.
– user172377
Feb 22, 2017 at 4:50

As @user172377 has pointed out, there is a problem in @clark's answer. The same $$q_i$$ in fact should appear an infinite number of times in the sum, because there are an infinite number of $$p$$'s that are relative primes to $$q_i$$. We need to also interate over these $$p$$'s, for each $$q_i$$.

(Furthermore, @clark's argument would have worked for $$\epsilon=0$$. But we know it's not true.)

Instead, we can focus on the intervals $$[i,i+1)$$, where $$i$$ is an integer, one at a time.

Define $$E_{i,q} \equiv E_q \cap [i,i+1)$$. You can show that $$\cup_q E_{i,q}$$ has measure 0 for each $$i$$. Therefore $$\cup_i (\cup_q E_q)$$ also has measure zero by Theorem 3.2 in Stein and Shakarchi.