Application of the Borel-Cantelli Lemma Let $p > 2$ and $c > 0$. Then the set
$$
 \left \{ x \in [0,1] : \left | x-\frac aq\right | \leq \frac c {q^p} \text{ for infinitly many $a,q \in \mathbb N^*$ } \right \}
$$ has measure zero. Recall the Borel-Cantelli Lemma:
If $\Omega_1,\Omega_2,\cdots$ are measurable subsets of $\mathbb R^n$ and $\sum_{n=1}^\infty m(\Omega_n) < \infty$ then the set
$$
 \left \{ x \in \mathbb R^n : x \in \Omega_n \text{ for infinitly many } n \right \}
$$ has measure zero.
This is by the way Tao Exercise 19.2.7.
 A: For any pair $(a,q)$ of positive integers, let
$$A(a,q) := \left[\frac{a}{q}-\frac{c}{q^p},\, \frac{a}{q} + \frac{c}{q^p}\right]$$
the interval consisting of those $x$ that satisfy the inequality
$$\left\lvert x - \frac{a}{q}\right\rvert \leqslant \frac{c}{q^p}\tag{1}$$
for that particular pair. Then
$$\Omega_q = \bigcup_{a = 1}^q A(a,q) \cap [0,\,1]$$
is the set of $x \in [0,\,1]$ that satisfy $(1)$ for $q$ and some positive $a$, since for $a > q$, we have $\bigl(A(a,q)\cap[0,\,1]\bigr) \subset \bigl(A(q,q)\cap[0,\,1]\bigr)$.
Then $$\left \{ x \in [0,1] : \left | x-\frac aq\right | \leq \frac c {q^p} \text{ for infinitely many $a,q \in \mathbb N^*$ } \right \}$$
is the set of $x \in [0,\,1]$ belonging to infinitely many $\Omega_q$,
$$\bigcap_{n=1}^\infty \left(\bigcup_{q=n}^\infty \Omega_q\right).$$
Now, $m\left(A(a,q)\right) \leqslant \frac{2c}{q^p}$, and hence $m(\Omega_q) \leqslant \frac{2cq}{q^p} = \frac{2c}{q^{p-1}}$, and
$$\sum_{q=1}^\infty m(\Omega_q) \leqslant 2c \sum_{q=1}^\infty \frac{1}{q^{p-1}} < \infty,$$
since $p > 2$.
