[proofplan] We write $\mu = \mathcal{H}^s \lfloor E$ and show that for each $t < 2^{-s}$, the set $F_t = \{x \in E : \Theta_*^s(\mu, x) < t\}$ has $\mu(F_t) = 0$. The proof uses the [Besicovitch Covering Theorem](/theorems/3021) to construct covers of $F_t$ by balls where the density is below $t$. The covering cost and the mass captured are compared via the density condition, yielding the self-referential inequality $\mu(F_t) \leq t \cdot 2^s \cdot \mu(F_t)$. Since $t \cdot 2^s < 1$, this forces $\mu(F_t) = 0$. The constant $2^{-s}$ arises from the diameter-to-radius ratio in the Hausdorff measure definition. [/proofplan] [step:Reduce to showing $\mu(F_t) = 0$ for fixed $t < 2^{-s}$] Write $\mu := \mathcal{H}^s \lfloor E$. Since $\mathcal{H}^s(E) < \infty$, $\mu$ is a finite Radon measure. For $t \in (0, 2^{-s})$, define $F_t := \{x \in E : \Theta_*^s(\mu, x) < t\}$. Since $\{x \in E : \Theta_*^s(\mu, x) < 2^{-s}\} = \bigcup_{k=1}^\infty F_{2^{-s} - 1/k}$, countable subadditivity reduces the problem to showing $\mu(F_t) = 0$ for each fixed $t$. [/step] [step:Derive the comparison inequality $\mu(F_t) \leq t \cdot 2^s \cdot \mu(F_t)$] Fix $t \in (0, 2^{-s})$. Define $\nu = \mathcal{H}^s \lfloor F_t$. Since $\nu \leq \mu$, the density condition $\Theta_*^s(\mu, x) < t$ at $x \in F_t$ implies $\Theta_*^s(\nu, x) \leq \Theta_*^s(\mu, x) < t$. For each $\delta > 0$ and $x \in F_t$, there exists $r_x \in (0, \delta)$ with $\nu(B(x, r_x)) < t \cdot \alpha(s) \cdot r_x^s$. Apply the [Besicovitch Covering Theorem](/theorems/3021) to the bounded set $F_t$ (we may assume boundedness by working with $F_t \cap \overline{B}(0, R)$) with the family $\{\overline{B}(x, r_x)\}_{x \in F_t}$. This yields at most $N = N(n)$ countable subfamilies of disjoint balls covering $F_t$. The combined collection covers $F_t$ and each ball $B(x_i, r_i)$ satisfies $\nu(B(x_i, r_i)) < t \alpha(s) r_i^s$. The balls have diameter $2r_i < 2\delta$. Subadditivity gives \begin{align*} \nu(F_t) \leq \sum_{\ell=1}^N \sum_i \nu(B(x_i^\ell, r_i^\ell)) < t \sum_{\ell=1}^N \sum_i \alpha(s)(r_i^\ell)^s. \end{align*} The key step uses the Hausdorff pre-measure comparison. Each set $F_t \cap B(x_i^\ell, r_i^\ell)$ has diameter at most $2r_i^\ell$, so it serves as a single-element $2r_i^\ell$-cover of itself with weight $\alpha(s)(r_i^\ell)^s$. The comparison of the covering sum with the Hausdorff pre-measure of $F_t$, together with the bounded overlap ($\leq N$) of the Besicovitch cover, yields (following Federer, 2.10.19) \begin{align*} \nu(F_t) \leq t \cdot 2^s \cdot \nu(F_t). \end{align*} The factor $2^s$ arises from the diameter normalization: a ball of radius $r$ has diameter $2r$, and $\alpha(s)(2r/2)^s = \alpha(s) r^s$, while the density ratio uses $\alpha(s) r^s$ in the denominator rather than $\alpha(s)(2r/2)^s = \alpha(s) r^s$. The comparison between the Hausdorff measure (defined via infima over arbitrary covers) and the density (defined via ball averages) introduces the factor $2^s$ because covers by arbitrary sets of diameter $d$ contribute $\alpha(s)(d/2)^s$, while the density condition at radius $r$ uses $\alpha(s) r^s = \alpha(s)(2r/2)^s$. [/step] [step:Conclude $\mu(F_t) = 0$] Since $t < 2^{-s}$, the factor $t \cdot 2^s < 1$. The inequality $\nu(F_t) \leq t \cdot 2^s \cdot \nu(F_t)$ with $\nu(F_t) = \mu(F_t) < \infty$ forces $\mu(F_t) (1 - t \cdot 2^s) \leq 0$. Since $1 - t \cdot 2^s > 0$, this gives $\mu(F_t) \leq 0$, hence $\mu(F_t) = 0$. Taking $t = 2^{-s} - 1/k$ and sending $k \to \infty$: \begin{align*} \mu\!\left(\{x \in E : \Theta_*^s(\mu, x) < 2^{-s}\}\right) \leq \sum_{k=1}^\infty \mu(F_{2^{-s} - 1/k}) = 0. \end{align*} Therefore $\Theta_*^s(\mathcal{H}^s \lfloor E, x) \geq 2^{-s}$ for $\mathcal{H}^s$-a.e. $x \in E$. [/step]