[proofplan]
We show that for $t > s$, the $\delta$-approximation $\mathcal{H}^t_\delta(E)$ is bounded by a constant times $\delta^{t-s}$ times $\mathcal{H}^s_\delta(E)$. Since $\mathcal{H}^s(E) < +\infty$ implies $\mathcal{H}^s_\delta(E) \leq \mathcal{H}^s(E) < +\infty$ for all $\delta$, and $t - s > 0$ makes $\delta^{t-s} \to 0$ as $\delta \to 0$, the product vanishes in the limit, giving $\mathcal{H}^t(E) = 0$.
[/proofplan]
[step:Bound each covering summand at scale $t$ in terms of the summand at scale $s$]
Fix $t > s$, $\delta > 0$, and a countable $\delta$-cover $\{C_j\}_{j=1}^\infty$ of $E$ (so $E \subset \bigcup_j C_j$ and $\operatorname{diam}(C_j) \leq \delta$ for all $j$). For each $j$, since $\operatorname{diam}(C_j) \leq \delta$, we have $\operatorname{diam}(C_j)/2 \leq \delta/2$, so
\begin{align*}
\left(\frac{\operatorname{diam}(C_j)}{2}\right)^t = \left(\frac{\operatorname{diam}(C_j)}{2}\right)^{t-s} \cdot \left(\frac{\operatorname{diam}(C_j)}{2}\right)^s \leq \left(\frac{\delta}{2}\right)^{t-s} \left(\frac{\operatorname{diam}(C_j)}{2}\right)^s.
\end{align*}
Multiplying both sides by $\alpha(t)$ and summing over $j$:
\begin{align*}
\sum_{j=1}^\infty \alpha(t)\left(\frac{\operatorname{diam}(C_j)}{2}\right)^t \leq \frac{\alpha(t)}{\alpha(s)} \left(\frac{\delta}{2}\right)^{t-s} \sum_{j=1}^\infty \alpha(s)\left(\frac{\operatorname{diam}(C_j)}{2}\right)^s.
\end{align*}
[/step]
[step:Take the infimum over covers to compare $\mathcal{H}^t_\delta(E)$ and $\mathcal{H}^s_\delta(E)$]
The inequality derived in the previous step holds for every $\delta$-cover $\{C_j\}$ of $E$. Define the constant $c(s,t) := \alpha(t)/\alpha(s)$, where $\alpha(s) = \pi^{s/2}/\Gamma(s/2 + 1)$. Taking the infimum over all $\delta$-covers on both sides:
\begin{align*}
\mathcal{H}^t_\delta(E) \leq c(s,t) \left(\frac{\delta}{2}\right)^{t-s} \mathcal{H}^s_\delta(E).
\end{align*}
Since $\delta \mapsto \mathcal{H}^s_\delta(E)$ is non-decreasing as $\delta$ decreases, and $\mathcal{H}^s(E) = \sup_{\delta > 0} \mathcal{H}^s_\delta(E) < +\infty$ by hypothesis, we have $\mathcal{H}^s_\delta(E) \leq \mathcal{H}^s(E) < +\infty$ for all $\delta > 0$. Therefore
\begin{align*}
\mathcal{H}^t_\delta(E) \leq c(s,t) \left(\frac{\delta}{2}\right)^{t-s} \mathcal{H}^s(E).
\end{align*}
[/step]
[step:Send $\delta \to 0$ to conclude $\mathcal{H}^t(E) = 0$]
Since $t > s$, the exponent $t - s > 0$, and $(\delta/2)^{t-s} \to 0$ as $\delta \to 0^+$. The bound from the previous step gives
\begin{align*}
\mathcal{H}^t(E) = \lim_{\delta \to 0^+} \mathcal{H}^t_\delta(E) \leq \lim_{\delta \to 0^+} c(s,t) \left(\frac{\delta}{2}\right)^{t-s} \mathcal{H}^s(E) = 0.
\end{align*}
Since $\mathcal{H}^t(E) \geq 0$ by definition, we conclude $\mathcal{H}^t(E) = 0$.
[guided]
The mechanism of the jump is the interplay between the covering constraint $\operatorname{diam}(C_j) \leq \delta$ and the exponent gap $t - s > 0$. When we compute $\mathcal{H}^t_\delta(E)$, each covering set $C_j$ contributes $(\operatorname{diam}(C_j)/2)^t$ to the sum. Since $\operatorname{diam}(C_j) \leq \delta$, we can factor out the "extra" power:
\begin{align*}
\left(\frac{\operatorname{diam}(C_j)}{2}\right)^t = \underbrace{\left(\frac{\operatorname{diam}(C_j)}{2}\right)^{t-s}}_{\leq (\delta/2)^{t-s}} \cdot \left(\frac{\operatorname{diam}(C_j)}{2}\right)^s.
\end{align*}
The first factor is bounded by $(\delta/2)^{t-s}$, which is a universal constant for the entire cover. The second factor is the same as the summand for $\mathcal{H}^s_\delta$. Summing over $j$, we get a bound of the form $\mathcal{H}^t_\delta(E) \lesssim \delta^{t-s} \cdot \mathcal{H}^s_\delta(E)$.
The crucial point is that $\mathcal{H}^s_\delta(E) \leq \mathcal{H}^s(E) < +\infty$ stays bounded as $\delta \to 0$, while $\delta^{t-s} \to 0$. The product vanishes, giving $\mathcal{H}^t(E) = 0$. This is why the transition from positive (or infinite) to zero is instantaneous: for any $t$ even slightly larger than $s$, the extra factor $\delta^{t-s}$ kills the entire sum.
Note also the contrapositive: if $\mathcal{H}^t(E) > 0$ for some $t$, then $\mathcal{H}^s(E) = +\infty$ for all $s < t$. This follows because $\mathcal{H}^s(E) < +\infty$ would force $\mathcal{H}^t(E) = 0$, contradicting $\mathcal{H}^t(E) > 0$. Together, these two statements establish that $s \mapsto \mathcal{H}^s(E)$ jumps from $+\infty$ to $0$ at a single critical value, which is the Hausdorff dimension.
[/guided]
[/step]
