Let $\beta>2$ be a real number, is it always possible to construct a real number $x$ such that $\mu(x)=\beta$? where $\mu$ denotes the irrationality measure.
The problem as I see it is to find $(a_n)_n$ an unbounded sequence of positive integers such that $\limsup \ln_{q_n}(q_{n+1})=\beta-1$ where $q_n$ is the numerator of the $n$-th convergent of $[a_0;a_1,a_2,\dots]$. But I was not able to continue.
Of course it is possible. Irrationality measure more than 2 means having continued fraction with terms that grow mighty fast, so we'll have more than enough freedom for every next term.
Since $q_{n+1}=a_{n+1}q_n+q_{n-1}$, we might simply choose $$a_{n+1}=\left\lceil\left(q_n^{\beta-1}-q_{n-1}\over q_n\right)^{1/(\beta-1)}\right\rceil$$
As $n$ and $q_n\to\infty$, so does $a_{n+1}$, which quickly makes the rounding error negligible.