# How to prove that $\mu(A_q) \leq 2/q^{1+\epsilon}$ with $A_q = [0,1] \bigcap \bigcup_{p=0}^q [p/q-1/q^{2+\epsilon},p/q+1/q^{2+\epsilon}]$?

Let $(E,A,\mu)$ be a measured space. Let $(A_n)_{n \geq 1}$ be a sequence of sets of $A$.

Let $\epsilon > 0$ be a real number. For $q$ an integer $\geq 1$, we consider in $\mathbb{R}$ :

$A_q = [0,1] \bigcap \bigcup_{p=0}^q [p/q-1/q^{2+\epsilon},p/q+1/q^{2+\epsilon}]$.

We want to prove that $\mu(A_q) \leq 2/q^{1+\epsilon}$ and $\sum_{q=1}^{\infty} \mu(A_q) < \infty$.

I really don't know how to do it, especially the first point... Someone could help me ? Thank you in advance...

For almost all real numbers, the irrationality measure is $2$.
We have by subadditivity of the measure $\mu$, $$\mu(A_q) \leq \mu([0,1/q^{2+\epsilon}])+\sum_{p=1}^{q-1} \mu( [p/q-1/q^{2+\epsilon},p/q+1/q^{2+\epsilon}]) +\mu([1-1/q^{2+\epsilon},1])$$ $$= \frac {2q}{q^{2+\epsilon}} = \frac 2{q^{1+\epsilon}}.$$
Then it goes on to prove that $\sum_q \frac 2{q^{1+\epsilon}}$ is finite. Then by Borel-Cantelli lemma, the result follows.