Let's start with the definition of a Liouville number:
A Liouville number is a real number $\xi$ such that for any
$m\in\mathbb N_>0$ there exists a pair of coprime integers
$(p,q)\in\mathbb Z^2$ with $q>0$ such that: $$
0<\left\vert\xi-\frac{p}{q}\right\vert< q^{-m} $$
Then let's give the definition of irrationality measure:
The irrationality measure of $\xi\in\mathbb R$ is: $$
\mu(\xi):=\sup_{t\in \mathbb R}\left\{\text{$\exists$ infinitely many coprime couples $(p,q)\in\mathbb Z^2$ s. t. }
0<\left\vert\xi-\frac{p}{q}\right\vert< \vert q\vert^{-t}\right\} $$
It is obvious the fact that if $\mu(\xi)=+\infty$ then it must be a Liouville number.
But it seems to me that it is not true that any Liouville number that irrationality measure equal to $+\infty$. Altough I don't have any example.
There are some authors (for instance Waldschmidt, Bugeaud) that as definition of Liouville numbers take exactly those numbers $\xi\in\mathbb R$ such that $\mu(\xi)=+\infty$. This makes me suspect that my previous claim is perhaps wrong. So what is the truth behind this story? Are Liouville number exactly those numbers with infinite irrational measure?
Best Answer
First, I think your first definition is slightly off since for any irrational number $\xi$ we have
$$0<\left|\xi -\frac{\lfloor \xi\rfloor}{1}\right|<1=1^{-m}$$
for all $m$. I think your definition has to be changed to $q>1$. Continuing:
You are trying to show the equivalence (or lack thereof) of two supposed definitions for Liouville numbers. You are satisfied that an irrationality measure of $+\infty$ means the number is a Liouville number, but not that being a Liouville number (according to your first definition) implies an infinite irrationality measure.
Assume by way of contradiction that we have a Liouville number $\xi$ with a finite irrationality measure $\mu$. By definition, there are only finitely many coprime-pairs $(p,q)$ such that
$$\left|\xi-\frac{p}{q}\right|<\frac{1}{q^{\chi}}$$
where $\chi=\lceil \mu\rceil +1>\mu$. Let $(P,Q)$ be given such that $Q$ is the largest amongst such $Q$. Such a $Q$ must exist since
$$\left|\xi-\frac{\lfloor\xi\rfloor}{1}\right|<1=\frac{1}{1^{\chi}}$$
Now, let $(p_n,q_n)$ be a sequence of coprime-integers such that
$$\left|\xi-\frac{p_n}{q_n}\right|<\frac{1}{q_n^n}$$
Next, we shall show that $q_n$ goes to infinity. If it didn't and was bounded by some $M\in\mathbb{N}$ then
$$\left|\xi M!-\frac{M!p_n}{q_n}\right|\geq \min\{M!\xi-\lfloor M!\xi\rfloor,\lceil M!\xi\rceil-M!\xi\}$$
since $\frac{M!p_n}{q_n}$ is always integral. But this is a finite number, which is a contradiction since
$$0\leq\lim_{n\to\infty}\frac{1}{q_n^n}\leq \lim_{n\to\infty}\frac{1}{2^n}=0$$
(note that this is the step where we use $q>1$). We conclude that $q_n\to\infty$. This implies that there exists $r$ such that $q_r>Q$. Consider what happens at $N=\max\{r,\chi\}$: we have
$$\left|\xi-\frac{p_N}{q_N}\right|<\frac{1}{q_N^N}\leq \frac{1}{q_N^\chi}$$
However, this is a contradiction since by our construction $q_N>Q$. We conclude both definitions are equivalent (after fixing the slight issue with the first one).