Given $x \in \mathbb{R}$, define its set of good exponents by $$G_{x}=\left\{\lambda \in [1, \infty) : 0<\left|x-\frac{p}{q}\right| \leq \frac{1}{q^{\lambda}} \ \text{admits infinitely many integral solutions} \ (p,q), q \neq 0 \right\}$$
The irrationality measure $\mu=\mu(x)$ of $x$ can be defined by $\mu = \sup E_x$, and this is either a real number $\geq 1$ or $+\infty$. Let $\mathcal{B} \subset \mathbb{R}$ denote the set of $x \in \mathbb{R}$ with finite irrationality measure (i.e., the non-Liouville numbers). It is well-known that this set has full measure in $\mathbb{R}$.
Now, for any $x \in \mathcal{B}$, the set $G_x$ can only possibly look like a bounded interval of the form $G_x=[1, \mu)$ or $G_x=[1, \mu]$
Conjecture: the first case is necessarily impossible, hence for any $x \in \mathcal{B}$ we always have $G_x$ a bounded compact interval $G_x=[1,\mu]$.
My question asks for an affirmative proof of the above conjecture, or a counterexample.
One observation is that the conjecture holds for almost all $x$, in the sense of Lebesgue measure. The reason for this is that it's well-known that $\mu(x)=2$ for almost all irrational $x$, and Dirichlet's approximation theorem implies that $\lambda=2$ is a good exponent for all irrational $x$.
Best Answer
It is well-known that if $\lambda > 2$, then for all sufficiently large $q$, the inequality $\left| x-\frac{p}{q} \right| \le \frac1{q^\lambda}$ implies that $\frac{p}{q} = \frac{p_n}{q_n}$ is a convergent of the continued fraction expansion of $x$, in which case $\frac{1}{2 q_n q_{n+1}} \le \left| x-\frac{p_n}{q_n} \right| \le \frac{1}{q_n q_{n+1}}$. This means that $q_{n+1} \ge \frac{q_n^{\lambda-1}}{2}$. From the recursive formula $q_{n+1} = a_{n+1} q_n + q_{n-1} \le (a_{n+1}+1)q_n$ for the denominators of the continued fraction convergents (where $a_n$ are the coefficients of the continued fraction) this means the inequality implies that $a_{n+1}+1 \ge \frac{q_n^{\lambda-2}}{2}$. On the other hand, if $a_{n+1} \ge q_n^{\lambda-2}$, then $\left| x-\frac{p_n}{q_n} \right| \le \frac{1}{q_n^\lambda}$.
Now with these inequalities one can easily construct a continued fraction expansion such that $\mu = \sup G_x = 3$, but $3 \notin G_x$. Roughly, the idea is to recursively define something like $a_{n+1} = \frac{q_n^{1-1/n}}{3}$ (for $n \ge n_0$, and rounded down to the next integer), which then implies that for every $\epsilon>0$ one has $a_{n+1} \ge q_n^{1-\epsilon}$ for $n$ sufficiently large, but always $a_{n+1} \le \frac{q_n}{3} < \frac{q_n}2$.