$\alpha\in[0,1]$, and $$|\alpha-\frac pq|\lt\frac1{4q^2}$$ has infinitely solutions $p, q\in\Bbb Z$, $\gcd(p,q)=1$.
Let $E$ be the set of all such $\alpha\in[0,1]$, that is
$$E=\{\alpha\in[0,1]\colon |\alpha-\frac pq|\lt\frac1{4q^2} \text{ has infinitely solutions}\; p, q\in\Bbb Z, \gcd(p,q)=1\}$$
What is the measure of $E$?
Of course, $E\ne [0,1]$; some irrationals don't belong to $E$. I don't know how to move on.
Generally, we can consider $$E_r=\{\alpha\in[0,1]\colon |\alpha-\frac pq|\lt\frac1{rq^2} \text{ has infinitely solutions}\; p, q\in\Bbb Z, \gcd(p,q)=1\}$$ for $r\gt\sqrt5$
Thanks a lot!
from Khinchin
The integers $a_k$ are called the elements, with the possible exception of $a_0$ they are positive. Here we go, on page 7 we find his numbering, $$ q_{k+1} = a_{k+1} q_k + q_{k-1}. $$ On page 9 we have Theorem 9, $$ \left| \alpha - \frac{p_k}{q_k} \right| < \frac{1}{q_k q_{k+1}} < \frac{1}{ a_{k+1} q_k^2} $$ If the elements are unbounded, every time $a_{k+1} > r$ your inequality is satisfied.
Theorem 29, page 60: the set of all numbers in $(0,1)$ with bounded elements is of measure zero.
THEREFORE the set of all numbers in $(0,1)$ with an unbounded sequence of elements is of measure one.
Theorem 23, page 36: for every real number $\alpha$ with an unbounded sequence of elements and arbitrary $c > 0,$ there are infinitely many solutions to $$\left| \alpha - \frac{p}{q} \right| < \frac{c}{q^2} $$