[proofplan]
We prove both $\mathcal{L}^1(\mathcal{L}) = 0$ and $\dim_{\mathcal{H}}(\mathcal{L}) = 0$. The Lebesgue measure claim follows from the inclusion $\mathcal{L} \subset \bigcap_{m=1}^\infty \bigcup_{q \geq 2} \bigcup_{|p| \leq qR} B(p/q, q^{-m})$ on any bounded interval, where the $\mathcal{L}^1$-measure of the inner union tends to $0$ as the truncation parameter $Q \to \infty$. For the Hausdorff dimension, we fix an arbitrary $s > 0$, choose an approximation degree $m > 2/s$ so that the counting cost of rational approximations is dominated by the diameter savings, and construct for each $\delta > 0$ a cover of $\mathcal{L}$ by balls of diameter at most $\delta$ whose total $s$-cost tends to $0$. Since $\mathcal{H}^s(\mathcal{L}) = 0$ for every $s > 0$, the dimension is $0$.
[/proofplan]
[step:Express $\mathcal{L}$ as a limsup set of rational approximation balls]
Recall that $x \in \mathcal{L}$ if and only if for every $m \in \mathbb{N}$, there exist $p \in \mathbb{Z}$ and $q \in \mathbb{N}$ with $q \geq 2$ such that $|x - p/q| < q^{-m}$, i.e., $x \in B(p/q, q^{-m})$. Therefore
\begin{align*}
\mathcal{L} \subset \bigcap_{m=1}^\infty \bigcup_{q=2}^\infty \bigcup_{p \in \mathbb{Z}} B\!\left(\frac{p}{q},\, q^{-m}\right).
\end{align*}
For any fixed $R > 0$, the intersection $\mathcal{L} \cap [-R, R]$ is contained in
\begin{align*}
\mathcal{L} \cap [-R, R] \subset \bigcap_{m=1}^\infty \bigcup_{q=2}^\infty \bigcup_{\substack{p \in \mathbb{Z} \\ |p/q| \leq R+1}} B\!\left(\frac{p}{q},\, q^{-m}\right),
\end{align*}
where for a given $q$, the number of integers $p$ with $|p/q| \leq R + 1$ is at most $2q(R+1) + 1 \leq 3q(R+1)$ for $q \geq 2$.
[/step]
[step:Prove $\mathcal{L}^1(\mathcal{L}) = 0$]
Fix $R > 0$ and $m \geq 3$. Define the open set
\begin{align*}
V_m(Q) = \bigcup_{q=Q}^\infty \bigcup_{\substack{p \in \mathbb{Z} \\ |p/q| \leq R+1}} B\!\left(\frac{p}{q},\, q^{-m}\right).
\end{align*}
The $\mathcal{L}^1$-measure of $V_m(Q)$ satisfies
\begin{align*}
\mathcal{L}^1(V_m(Q)) \leq \sum_{q=Q}^\infty 3q(R+1) \cdot 2q^{-m} = 6(R+1) \sum_{q=Q}^\infty q^{1-m}.
\end{align*}
Since $m \geq 3$, the exponent $1 - m \leq -2$, so $\sum_{q=Q}^\infty q^{1-m} < \infty$ and this tail sum tends to $0$ as $Q \to \infty$. Since $\mathcal{L} \cap [-R, R] \subset V_m(Q)$ for every $Q \geq 2$ (because Liouville numbers are approximable at rate $m$ by rationals of arbitrarily large denominator), we get
\begin{align*}
\mathcal{L}^1(\mathcal{L} \cap [-R, R]) \leq \mathcal{L}^1(V_m(Q)) \to 0 \quad \text{as } Q \to \infty.
\end{align*}
Therefore $\mathcal{L}^1(\mathcal{L} \cap [-R, R]) = 0$. Taking $R \to \infty$ and using $\sigma$-subadditivity:
\begin{align*}
\mathcal{L}^1(\mathcal{L}) = \lim_{R \to \infty} \mathcal{L}^1(\mathcal{L} \cap [-R, R]) = 0.
\end{align*}
[/step]
[step:Fix $s > 0$ and choose $m > 2/s$ to control the $s$-dimensional cost]
Fix an arbitrary $s > 0$. Choose $m \in \mathbb{N}$ with $m > 2/s$, so that $ms > 2$, equivalently $1 - ms < -1$. This ensures the series $\sum_{q=2}^\infty q^{1-ms}$ converges.
[/step]
[step:Construct a $\delta$-fine cover of $\mathcal{L} \cap [-R,R]$ with vanishing $s$-cost]
Fix $R > 0$ and $\delta > 0$. Choose $Q \in \mathbb{N}$ large enough so that $2Q^{-m} < \delta$ (which is possible since $Q^{-m} \to 0$). Every Liouville number $x \in \mathcal{L} \cap [-R, R]$ admits, by definition, a rational $p/q$ with $q \geq 2$ and $|x - p/q| < q^{-m}$. Since Liouville numbers are approximable at rate $m$ by rationals of arbitrarily large denominator, we may require $q \geq Q$. This gives
\begin{align*}
\mathcal{L} \cap [-R, R] \subset \bigcup_{q=Q}^\infty \bigcup_{\substack{p \in \mathbb{Z} \\ |p/q| \leq R+1}} B\!\left(\frac{p}{q},\, q^{-m}\right).
\end{align*}
Each ball $B(p/q, q^{-m})$ has diameter $2q^{-m} \leq 2Q^{-m} < \delta$, so this is a $\delta$-admissible cover. The $s$-dimensional cost is
\begin{align*}
\sum_{q=Q}^\infty \sum_{\substack{p \in \mathbb{Z} \\ |p/q| \leq R+1}} (\operatorname{diam}(B(p/q, q^{-m})))^s = \sum_{q=Q}^\infty \sum_{\substack{p \in \mathbb{Z} \\ |p/q| \leq R+1}} (2q^{-m})^s = 2^s \sum_{q=Q}^\infty N(q) \cdot q^{-ms},
\end{align*}
where $N(q) \leq 3q(R+1)$ is the number of relevant numerators for denominator $q$. Therefore
\begin{align*}
\mathcal{H}^s_\delta(\mathcal{L} \cap [-R,R]) \leq 2^s \sum_{q=Q}^\infty 3q(R+1) \cdot q^{-ms} = 3 \cdot 2^s (R+1) \sum_{q=Q}^\infty q^{1-ms}.
\end{align*}
Since $ms > 2$ (from the choice $m > 2/s$), the exponent $1 - ms < -1$, and the tail sum $\sum_{q=Q}^\infty q^{1-ms}$ converges and tends to $0$ as $Q \to \infty$.
[/step]
[step:Conclude $\mathcal{H}^s(\mathcal{L}) = 0$ and $\dim_{\mathcal{H}}(\mathcal{L}) = 0$]
As $Q \to \infty$, $\delta = 2Q^{-m} \to 0$ and $\mathcal{H}^s_\delta(\mathcal{L} \cap [-R,R]) \to 0$. Since $\mathcal{H}^s(\mathcal{L} \cap [-R,R]) = \lim_{\delta \to 0} \mathcal{H}^s_\delta(\mathcal{L} \cap [-R,R])$, we conclude $\mathcal{H}^s(\mathcal{L} \cap [-R,R]) = 0$ for every $R > 0$.
By countable subadditivity of $\mathcal{H}^s$, writing $\mathcal{L} = \bigcup_{k=1}^\infty (\mathcal{L} \cap [-k, k])$:
\begin{align*}
\mathcal{H}^s(\mathcal{L}) \leq \sum_{k=1}^\infty \mathcal{H}^s(\mathcal{L} \cap [-k, k]) = 0.
\end{align*}
Since this holds for every $s > 0$, the definition of Hausdorff dimension gives
\begin{align*}
\dim_{\mathcal{H}}(\mathcal{L}) = \inf\{s > 0 : \mathcal{H}^s(\mathcal{L}) = 0\} = 0.
\end{align*}
Finally, the set $\mathcal{L}$ is uncountable because it is a dense $G_\delta$ subset of $\mathbb{R}$ (being the countable intersection $\bigcap_{m=1}^\infty \bigcup_{q=2}^\infty \bigcup_{p} B(p/q, q^{-m})$ of open dense sets), and by the Baire category theorem, a dense $G_\delta$ in a complete metric space is uncountable.
[/step]
