(Tao Vol.2, P.200, Exercise 8.2.7) Let $p > 2$ and $c > 0$. Using the Borel-Cantelli lemma, show that the set $$\left\{x \in [0,1] : \left|x - \frac{a}q\right| \le \frac{c}{q^p} \quad\text{for infinitely many positive integers $a, q$}\right\}$$ has measure zero. (Hint: one only has to consider those integers $a$ in the range $0 \le a \le q$. Show that the sum $\sum_{q=1}^\infty \frac{c(q+1)}{q^p}$ is finite.)
Borel-Cantelli lemma: Let $\Omega_1, \Omega_2, ...$ be measurable subsets of $\mathbb{R}^n$ such that $\sum_{n=1}^\infty m(\Omega_n)$ is finite. Then the set $$\{x \in \mathbb{R}^n : x \in \Omega_n \quad\text{for infinitely many $n$}\}$$ is a set of measure zero. In other words, almost every point belongs to only finitely many $\Omega_n$.
We only need to consider $0\le a \le q$ since when $a>q$, $\frac{c}{q^p} + \frac{a}{q} >1$ for any $c>0$ and $p >2$, and the sum suggested in the hint is finite by $p$-test. Let $\Omega_{q} =\bigcup_{a=0}^q \{x \in [0,1]: |x - \frac{a}{q}| \le \frac{c}{q^p}\}$. Then by subadditivity
$$m(\Omega_q) \le \sum_{a=0}^q m\left(\left\{x \in [0,1]: \left|x - \frac{a}{q}\right| \le \frac{c}{q^p}\right\}\right).$$
I wish that the right side is less than equal to $\frac{c(q+1)}{q^p}$, which means that $m(\{x \in [0,1]: |x - \frac{a}{q}| \le \frac{c}{q^p}\}) \le \frac{c}{q^p}$. But, I don't think that this is true.
Is there any way to reach $\sum_{q=1}^\infty \frac{c(q+1)}{q^p}$ as suggested hint?
Thanks in advance.
Hint: $|x-x_0| \leq r$ implies $x \in (x_0-r, x_0+r)$ so $m(\{x:|x-x_0| \leq r\}) \leq 2r$.