Let $\psi: \mathbb{N} \to \mathbb{R}^+$ be some function and denote by $W(\psi)$ the set of numbers $x \in [0,1]$ for which $$\left\lvert x - \frac{p}{q} \right\rvert < \frac{\psi(q)}{q} \quad \text{for infinitely many }(p,q) \in \mathbb{Z}\times\mathbb{N}.$$ In the context of metric Diophantine approximation, Jarník's Theorem can be stated as follows:
Let $\psi: \mathbb{N} \to \mathbb{R}^+$ be monotonically decreasing. Then $$\mathcal{H}^s(W(\psi)) = \begin{cases} 0 \quad \text{if }\sum_{q=1}^\infty q^{1-s}\psi(q)^s < \infty\\ \ \\ 1 \quad \text{if }\sum_{q=1}^\infty q^{1-s}\psi(q)^s = \infty, \end{cases}$$
where $\mathcal{H}^s$ denotes the Hausdorff measure with parameter $s \geq 0$. Note that this is not quite the same result (Theorem 10.3) given in Falconer's Fractal Geometry book that is also named Jarník's Theorem (which is commonly also called the Jarník-Besicovitch Theorem). But this is what I am working with.
I am trying to write a proof of the first case, when $\sum_{q=1}^\infty q^{1-s}\psi(q)^s$ converges, and so far it is as follows:
First observe that the set $W(\psi)$ is a limsup set. In particular, for some given $q$, consider the set $W_q := \bigcup_{p=0}^q B\left(\frac{p}{q}, \frac{\psi(q)}{q}\right) \cap [0,1]$. Then $W(\psi)$ may be expressed as $W(\psi) = \limsup_{q \to \infty}W_q$; as a limsup set of a sequence of balls.
For a fixed $q$, there are $q+1$ choices for the integer $p$ (since $p$ ranges from $0$ to $q$), and so we can bound this above by $2q$ (since $2q \geq q+1$ for $q \geq 1$).
Now let $\delta > 0$ be arbitrary. Then there is some integer $q_0 \geq 1$ for which the radius of a ball $\psi(q)/q < \delta$ for every $q \geq q_0$, and so taking the union over all $q \geq q_0$ gives us an appropriate $\delta$-cover for $W(\psi)$, namely $$\bigcup_{q=q_0}^{\infty} \bigcup_{p=0}^{\infty}B\left(\frac{p}{q}, \frac{\psi(q)}{q}\right).$$
Next we compute an upper bound for the quantity $\mathcal{H}_\delta^s(W(\psi))$. In particular, $$\mathcal{H}_\delta^s(W(\psi)) \leq \sum_{q \geq q_0} 2q \cdot \left(2\frac{\psi(q)}{q}\right)^s \leq \sum_{q =1}^\infty 2q \cdot \left(2\frac{\psi(q)}{q}\right)^s = 2^{s+1}\sum_{q=1}^\infty q^{1-s}\psi(q)^s < \infty,$$ where we have assumed the condition of the first case of the theorem to establish convergence.
As $\delta \to 0$, $\mathcal{H}_\delta^s(W(\psi)) \to \mathcal{H}^s(W(\psi))$.
[Some steps that I'm missing] then imply that $\mathcal{H}^s(W(\psi)) = 0$.
In the above, as indicated, there are a few steps (possibly just one) that are missing and that I can't seem to figure out. I get that the Hausdorff measure is finite since the pre-measure is also finite. What then guarantees that the Hausdorff measure is zero? I thought it could be anything positive (even infinite) if $s$ it at its "critical value" for Hausdorff dimension. What I would like is an explanation of where to go from here and why.