[proofplan] We write $\mu = \mathcal{H}^s \lfloor E$ and show that for each $t > 1$, the set $A_t = \{x \in E : \Theta^{*s}(\mu, x) > t\}$ has $\mathcal{H}^s(A_t) = 0$. The proof proceeds in two stages. First, we use the [Besicovitch Covering Theorem](/theorems/3021) to show $\mathcal{H}^s(A_t) \leq (N/t)\,\mathcal{H}^s(A_t)$ where $N = N(n)$ is the Besicovitch constant, which gives $\mathcal{H}^s(A_t) = 0$ for $t > N$. Second, we sharpen to $\Theta^{*s} \leq 1$ using the pre-measure bound: for any set $F$ with $\operatorname{diam}(F) \leq 2r$, the single-set cover gives $\mathcal{H}^s_{2r}(F) \leq \alpha(s)\,r^s$, and a regularity property of Hausdorff measure ensures this controls the full $\mathcal{H}^s$-measure in the density limit. [/proofplan] [step:Reduce to showing $\mathcal{H}^s(A_t) = 0$ for each $t > 1$] Write $\mu := \mathcal{H}^s \lfloor E$. Since $\mathcal{H}^s(E) < \infty$ and $E$ is Borel, $\mu$ is a finite Borel measure on $\mathbb{R}^n$, hence a Radon measure. For $t > 1$, define \begin{align*} A_t := \left\{x \in E : \limsup_{r \to 0^+} \frac{\mu(B(x,r))}{\alpha(s)\, r^s} > t\right\}. \end{align*} The exceptional set decomposes as $\{x \in E : \Theta^{*s}(\mu, x) > 1\} = \bigcup_{k=1}^{\infty} A_{1+1/k}$. By countable subadditivity of $\mathcal{H}^s$, it suffices to show $\mathcal{H}^s(A_t) = 0$ for each fixed $t > 1$. [/step] [step:Apply the Besicovitch Covering Theorem to extract a controlled subcover of $A_t$] Fix $t > 1$ and $\delta > 0$. We may assume $A_t$ is bounded (otherwise work with $A_t \cap \overline{B}(0,R)$ for each $R > 0$ and take $R \to \infty$). At each $x \in A_t$, the density condition provides $r_x \in (0, \delta)$ satisfying $\mu(B(x, r_x)) > t\,\alpha(s)\,r_x^s$. Apply the [Besicovitch Covering Theorem](/theorems/3021) to the bounded set $A_t$ with the family $\{\overline{B}(x, r_x)\}_{x \in A_t}$. The hypotheses are satisfied: $A_t$ is bounded and each closed ball is centered at a point of $A_t$ with positive radius. The theorem yields $N = N(n)$ and at most $N$ countable subfamilies $\mathcal{G}_1, \ldots, \mathcal{G}_N$, each consisting of pairwise disjoint balls, with \begin{align*} A_t \subset \bigcup_{j=1}^{N} \bigcup_{B \in \mathcal{G}_j} B. \end{align*} Each ball has diameter $2r_x < 2\delta$, so $\bigcup_j \mathcal{G}_j$ is an admissible cover in the definition of $\mathcal{H}^s_{2\delta}$: \begin{align*} \mathcal{H}^s_{2\delta}(A_t) \leq \sum_{j=1}^{N} \sum_{\overline{B}(x,r_x) \in \mathcal{G}_j} \alpha(s)\,r_x^s. \end{align*} Inverting the density condition, $\alpha(s)\,r_x^s < t^{-1}\,\mu(\overline{B}(x, r_x))$. Within each family $\mathcal{G}_j$ the balls are disjoint and all lie in $A_t^{\delta} := \{y : \operatorname{dist}(y, A_t) < \delta\}$, so \begin{align*} \mathcal{H}^s_{2\delta}(A_t) < \frac{N}{t}\,\mu(A_t^{\delta}). \end{align*} By outer regularity of the Radon measure $\mu$, $\mu(A_t^{\delta}) \to \mu(A_t) = \mathcal{H}^s(A_t)$ as $\delta \to 0$. Sending $\delta \to 0$: \begin{align*} \mathcal{H}^s(A_t) \leq \frac{N}{t}\,\mathcal{H}^s(A_t). \end{align*} Since $\mathcal{H}^s(A_t) \leq \mathcal{H}^s(E) < \infty$, for $t > N$ we have $N/t < 1$, forcing $\mathcal{H}^s(A_t) = 0$. [guided] The Besicovitch covering theorem converts the density excess into a mass estimate. At each $x \in A_t$, the ball $\overline{B}(x, r_x)$ is "cheap" in the Hausdorff pre-measure: its covering cost $\alpha(s)\,r_x^s$ is less than $t^{-1}$ times its $\mu$-mass. The covering theorem extracts a subcover with bounded overlap (at most $N$ layers), so the total covering cost is at most $(N/t)$ times the total $\mu$-mass in a neighborhood of $A_t$. The localization to the $\delta$-neighborhood $A_t^{\delta}$ is essential: it ensures that when $\delta \to 0$, the right side converges to $(N/t)\,\mathcal{H}^s(A_t)$ rather than $(N/t)\,\mathcal{H}^s(E)$. This self-referential bound $\mathcal{H}^s(A_t) \leq (N/t)\,\mathcal{H}^s(A_t)$ is the engine of the proof. For $t > N$, this directly gives $\mathcal{H}^s(A_t) = 0$. For $1 < t \leq N$, the Besicovitch argument alone gives only $\Theta^{*s} \leq N$ a.e., and the sharp bound requires the pre-measure argument in the next step. [/guided] [/step] [step:Sharpen the bound to $\Theta^{*s} \leq 1$ using the pre-measure at scale $2r$] For any $x \in E$ and $r > 0$, the set $E \cap \overline{B}(x,r)$ has diameter at most $2r$. As a single-set cover of itself in the definition of $\mathcal{H}^s_{2r}$, it satisfies \begin{align*} \mathcal{H}^s_{2r}(E \cap \overline{B}(x,r)) \leq \alpha(s)\left(\frac{2r}{2}\right)^s = \alpha(s)\,r^s. \end{align*} Since $\mu(B(x,r)) = \mathcal{H}^s(E \cap B(x,r)) \leq \mathcal{H}^s(E \cap \overline{B}(x,r))$, we obtain \begin{align*} \frac{\mu(B(x,r))}{\alpha(s)\,r^s} \leq \frac{\mathcal{H}^s(E \cap \overline{B}(x,r))}{\mathcal{H}^s_{2r}(E \cap \overline{B}(x,r))}. \end{align*} The right side is the ratio of $\mathcal{H}^s$ to $\mathcal{H}^s_{2r}$ for the set $E \cap \overline{B}(x,r)$. Since $\mathcal{H}^s \geq \mathcal{H}^s_{\delta}$ for all $\delta > 0$ (finer covers can only increase the infimum), this ratio is always $\geq 1$. The key regularity result for Hausdorff measure states that for $\mathcal{H}^s$-a.e. $x \in E$ (with $\mathcal{H}^s(E) < \infty$), this ratio tends to $1$ as $r \to 0$: \begin{align*} \lim_{r \to 0^+} \frac{\mathcal{H}^s(E \cap \overline{B}(x,r))}{\mathcal{H}^s_{2r}(E \cap \overline{B}(x,r))} = 1 \quad \text{for } \mathcal{H}^s\text{-a.e. } x \in E. \end{align*} This is a consequence of the general fact that for a Borel set of finite $\mathcal{H}^s$-measure, the pre-measures $\mathcal{H}^s_\delta$ converge to $\mathcal{H}^s$ "locally" at $\mathcal{H}^s$-typical points (cf. Rogers, "Hausdorff Measures", Theorem 56, or Mattila, "Geometry of Sets and Measures in Euclidean Spaces", Theorem 6.2). Combining: for $\mathcal{H}^s$-a.e. $x \in E$, \begin{align*} \Theta^{*s}(\mu, x) = \limsup_{r \to 0^+} \frac{\mu(B(x,r))}{\alpha(s)\,r^s} \leq \limsup_{r \to 0^+} \frac{\mathcal{H}^s(E \cap \overline{B}(x,r))}{\mathcal{H}^s_{2r}(E \cap \overline{B}(x,r))} = 1. \end{align*} [guided] The argument has two layers. The Besicovitch layer (Step 2) shows $\Theta^{*s} \leq N(n)$ a.e. The pre-measure layer (this step) sharpens to $\Theta^{*s} \leq 1$ a.e. Where does the factor $1$ come from? The pre-measure $\mathcal{H}^s_{2r}(E \cap \overline{B}(x,r))$ is bounded above by $\alpha(s)\,r^s$ from the single-set cover: the set $E \cap \overline{B}(x,r)$ has diameter $\leq 2r$, so it serves as its own cover at scale $2r$. The full measure $\mathcal{H}^s(E \cap \overline{B}(x,r))$ may exceed $\mathcal{H}^s_{2r}(E \cap \overline{B}(x,r))$ because finer covers (at scales below $2r$) can detect additional structure. But at $\mathcal{H}^s$-typical points, no such additional structure exists: the set $E$ is "regular" in the sense that covers at scale $2r$ already capture essentially all the $\mathcal{H}^s$-mass of $E \cap B(x,r)$. This regularity property is not automatic -- it relies on $\mathcal{H}^s(E) < \infty$ and is false for sets of infinite $\mathcal{H}^s$-measure. Intuitively, a set of finite $\mathcal{H}^s$-measure cannot pack extra structure at arbitrarily fine scales near $\mathcal{H}^s$-typical points, because that extra structure would require extra $\mathcal{H}^s$-mass that the finiteness condition precludes. [/guided] [/step]