[proofplan] We first verify that $\mathcal{H}^s$ is an outer measure by checking the three defining properties (empty set, monotonicity, countable subadditivity) directly from the definition as a supremum of $\mathcal{H}^s_\delta$. We then prove that every Borel set is $\mathcal{H}^s$-measurable by verifying the Caratheodory criterion for open sets and invoking [Caratheodory's Theorem](/theorems/3010). Finally, we establish Borel regularity by constructing, for each $A \subset \mathbb{R}^n$, a Borel set $B \supset A$ with $\mathcal{H}^s(B) = \mathcal{H}^s(A)$, using a double-limit construction over rational $\delta$ values and near-optimal covers. [/proofplan]

[step:Verify that $\mathcal{H}^s(\varnothing) = 0$] For every $\delta > 0$, the empty collection is a valid $\delta$-cover of $\varnothing$ (the union of zero sets contains $\varnothing$). The corresponding covering sum is $0$, so $\mathcal{H}^s_\delta(\varnothing) \leq 0$. Since $\mathcal{H}^s_\delta$ takes values in $[0, +\infty]$, we have $\mathcal{H}^s_\delta(\varnothing) = 0$ for all $\delta > 0$. Therefore \begin{align*} \mathcal{H}^s(\varnothing) = \sup_{\delta > 0} \mathcal{H}^s_\delta(\varnothing) = 0. \end{align*} [/step]

[step:Verify monotonicity: $A \subset B$ implies $\mathcal{H}^s(A) \leq \mathcal{H}^s(B)$] Fix $\delta > 0$. Every countable $\delta$-cover of $B$ is also a countable $\delta$-cover of $A$ (since $A \subset B$). Therefore the infimum over $\delta$-covers of $A$ is taken over a larger collection of covers than for $B$, giving $\mathcal{H}^s_\delta(A) \leq \mathcal{H}^s_\delta(B)$. Taking the supremum over $\delta > 0$ yields $\mathcal{H}^s(A) \leq \mathcal{H}^s(B)$. [/step]

[step:Verify countable subadditivity of $\mathcal{H}^s$] Let $(A_j)_{j=1}^\infty$ be a sequence of subsets of $\mathbb{R}^n$. Fix $\delta > 0$ and $\varepsilon > 0$. For each $j \geq 1$, choose a countable $\delta$-cover $\{C_{j,k}\}_{k=1}^\infty$ of $A_j$ such that \begin{align*} \sum_{k=1}^\infty \alpha(s) \left(\frac{\operatorname{diam}(C_{j,k})}{2}\right)^s \leq \mathcal{H}^s_\delta(A_j) + \frac{\varepsilon}{2^j}. \end{align*} (If $\mathcal{H}^s_\delta(A_j) = +\infty$ for some $j$, the right-hand side of the subadditivity inequality is $+\infty$ and there is nothing to prove.) The doubly-indexed collection $\{C_{j,k}\}_{j,k \geq 1}$ is a countable $\delta$-cover of $\bigcup_{j=1}^\infty A_j$, since every element of the union belongs to some $A_j$ and is therefore covered by $\{C_{j,k}\}_{k \geq 1}$. Moreover, each $C_{j,k}$ has $\operatorname{diam}(C_{j,k}) \leq \delta$. Therefore \begin{align*} \mathcal{H}^s_\delta\!\left(\bigcup_{j=1}^\infty A_j\right) &\leq \sum_{j=1}^\infty \sum_{k=1}^\infty \alpha(s) \left(\frac{\operatorname{diam}(C_{j,k})}{2}\right)^s \leq \sum_{j=1}^\infty \left(\mathcal{H}^s_\delta(A_j) + \frac{\varepsilon}{2^j}\right) = \sum_{j=1}^\infty \mathcal{H}^s_\delta(A_j) + \varepsilon. \end{align*} Since $\varepsilon > 0$ is arbitrary, $\mathcal{H}^s_\delta\!\left(\bigcup_j A_j\right) \leq \sum_j \mathcal{H}^s_\delta(A_j)$. Taking the supremum over $\delta > 0$ on the left-hand side and using that $\mathcal{H}^s_\delta(A_j) \leq \mathcal{H}^s(A_j)$ for each $j$ and $\delta$, we obtain \begin{align*} \mathcal{H}^s\!\left(\bigcup_{j=1}^\infty A_j\right) \leq \sum_{j=1}^\infty \mathcal{H}^s(A_j). \end{align*} [/step]

[step:Prove the metric additive condition: $\operatorname{dist}(A, B) > 0$ implies $\mathcal{H}^s(A \cup B) = \mathcal{H}^s(A) + \mathcal{H}^s(B)$] Suppose $A, B \subset \mathbb{R}^n$ with $d := \operatorname{dist}(A, B) > 0$. Countable subadditivity gives $\mathcal{H}^s(A \cup B) \leq \mathcal{H}^s(A) + \mathcal{H}^s(B)$. We prove the reverse inequality. Fix $\delta \in (0, d)$. Let $\{C_j\}_{j=1}^\infty$ be any $\delta$-cover of $A \cup B$. Since $\operatorname{diam}(C_j) \leq \delta < d = \operatorname{dist}(A, B)$, each $C_j$ can intersect at most one of $A$ and $B$. Define \begin{align*} J_A = \{j : C_j \cap A \neq \varnothing\}, \qquad J_B = \{j : C_j \cap B \neq \varnothing\}. \end{align*} Then $J_A \cap J_B = \varnothing$, $\{C_j\}_{j \in J_A}$ is a $\delta$-cover of $A$, and $\{C_j\}_{j \in J_B}$ is a $\delta$-cover of $B$. Therefore \begin{align*} \sum_{j=1}^\infty \alpha(s)\left(\frac{\operatorname{diam}(C_j)}{2}\right)^s \geq \sum_{j \in J_A} \alpha(s)\left(\frac{\operatorname{diam}(C_j)}{2}\right)^s + \sum_{j \in J_B} \alpha(s)\left(\frac{\operatorname{diam}(C_j)}{2}\right)^s \geq \mathcal{H}^s_\delta(A) + \mathcal{H}^s_\delta(B). \end{align*} Taking the infimum over all $\delta$-covers of $A \cup B$ gives $\mathcal{H}^s_\delta(A \cup B) \geq \mathcal{H}^s_\delta(A) + \mathcal{H}^s_\delta(B)$. Taking $\delta \to 0^+$ (through values less than $d$) yields $\mathcal{H}^s(A \cup B) \geq \mathcal{H}^s(A) + \mathcal{H}^s(B)$. [/step]

[step:Deduce that every Borel set is $\mathcal{H}^s$-measurable] The previous steps establish that $\mathcal{H}^s$ is an outer measure on $\mathbb{R}^n$ satisfying the metric additivity condition: $\mathcal{H}^s(A \cup B) = \mathcal{H}^s(A) + \mathcal{H}^s(B)$ whenever $\operatorname{dist}(A, B) > 0$. We now verify the Caratheodory condition for every closed set $F \subset \mathbb{R}^n$. Let $F \subset \mathbb{R}^n$ be closed and let $E \subset \mathbb{R}^n$ be an arbitrary test set. We must show \begin{align*} \mathcal{H}^s(E) \geq \mathcal{H}^s(E \cap F) + \mathcal{H}^s(E \setminus F). \end{align*} (The reverse inequality follows from countable subadditivity.) If $\mathcal{H}^s(E) = +\infty$, the inequality is immediate. Assume $\mathcal{H}^s(E) < +\infty$. For each $k \geq 1$, define $E_k = \{x \in E \setminus F : \operatorname{dist}(x, F) \geq 1/k\}$. Since $F$ is closed and $x \notin F$, we have $\operatorname{dist}(x, F) > 0$ for every $x \in E \setminus F$, so $E \setminus F = \bigcup_{k=1}^\infty E_k$. Moreover, $\operatorname{dist}(E \cap F, E_k) \geq 1/k > 0$ for each $k$, so by metric additivity: \begin{align*} \mathcal{H}^s(E) \geq \mathcal{H}^s((E \cap F) \cup E_k) = \mathcal{H}^s(E \cap F) + \mathcal{H}^s(E_k). \end{align*} Since $E_k \nearrow E \setminus F$, we need to show $\mathcal{H}^s(E_k) \to \mathcal{H}^s(E \setminus F)$ as $k \to \infty$. Define the annular layers $D_k = E_{k+1} \setminus E_k = \{x \in E \setminus F : 1/(k+1) \leq \operatorname{dist}(x, F) < 1/k\}$ for $k \geq 1$. Then $E \setminus F = E_k \cup \bigcup_{j=k}^\infty D_j$, and by countable subadditivity: \begin{align*} \mathcal{H}^s(E \setminus F) \leq \mathcal{H}^s(E_k) + \sum_{j=k}^\infty \mathcal{H}^s(D_j). \end{align*} Since $\mathcal{H}^s(E \setminus F) \leq \mathcal{H}^s(E) < +\infty$ and $E \setminus F = E_1 \cup \bigcup_{j=1}^\infty D_j$, the series $\sum_{j=1}^\infty \mathcal{H}^s(D_j)$ converges (its partial sums are bounded by $\mathcal{H}^s(E)$). Therefore the tail $\sum_{j=k}^\infty \mathcal{H}^s(D_j) \to 0$ as $k \to \infty$, giving $\mathcal{H}^s(E \setminus F) \leq \lim_{k \to \infty} \mathcal{H}^s(E_k)$. The reverse inequality $\mathcal{H}^s(E_k) \leq \mathcal{H}^s(E \setminus F)$ holds by monotonicity, so $\mathcal{H}^s(E_k) \to \mathcal{H}^s(E \setminus F)$. Taking $k \to \infty$ in the inequality $\mathcal{H}^s(E) \geq \mathcal{H}^s(E \cap F) + \mathcal{H}^s(E_k)$ yields \begin{align*} \mathcal{H}^s(E) \geq \mathcal{H}^s(E \cap F) + \mathcal{H}^s(E \setminus F). \end{align*} Thus every closed set $F$ is $\mathcal{H}^s$-measurable. Since the $\sigma$-algebra of $\mathcal{H}^s$-measurable sets contains all closed sets and is closed under complements and countable unions (by [Caratheodory's Theorem](/theorems/3010)), it contains the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R}^n)$. [/step]

[step:Construct a Borel set $B \supset A$ with $\mathcal{H}^s(B) = \mathcal{H}^s(A)$ (Borel regularity)] Let $A \subset \mathbb{R}^n$. If $\mathcal{H}^s(A) = +\infty$, take $B = \mathbb{R}^n$, which is Borel and satisfies $\mathcal{H}^s(B) \geq \mathcal{H}^s(A) = +\infty$. Assume $\mathcal{H}^s(A) < +\infty$. For each $k \geq 1$, set $\delta_k = 1/k$. For each pair $(k, m)$ with $k, m \geq 1$, choose a countable $\delta_k$-cover $\{C_{k,m,j}\}_{j=1}^\infty$ of $A$ such that \begin{align*} \sum_{j=1}^\infty \alpha(s)\left(\frac{\operatorname{diam}(C_{k,m,j})}{2}\right)^s \leq \mathcal{H}^s_{\delta_k}(A) + \frac{1}{m}. \end{align*} Define the open set \begin{align*} U_{k,m} = \bigcup_{j=1}^\infty \left\{x \in \mathbb{R}^n : \operatorname{dist}(x, C_{k,m,j}) < \frac{1}{m}\right\}. \end{align*} Each set in the union is open (as an open $1/m$-neighborhood of $C_{k,m,j}$), so $U_{k,m}$ is open, and $A \subset \bigcup_j C_{k,m,j} \subset U_{k,m}$. Define the $G_\delta$ set \begin{align*} B = \bigcap_{k=1}^\infty \bigcap_{m=1}^\infty U_{k,m}. \end{align*} Then $B$ is a countable intersection of open sets, hence a Borel set ($G_\delta$), and $A \subset B$. We claim $\mathcal{H}^s(B) = \mathcal{H}^s(A)$. By monotonicity, $\mathcal{H}^s(B) \geq \mathcal{H}^s(A)$. For the reverse, fix $k \geq 1$. For every $m \geq 1$, the set $B \subset U_{k,m}$ is covered by the open $1/m$-neighborhoods of the sets $C_{k,m,j}$. Each neighborhood satisfies \begin{align*} \operatorname{diam}\!\left(\{x : \operatorname{dist}(x, C_{k,m,j}) < 1/m\}\right) \leq \operatorname{diam}(C_{k,m,j}) + 2/m. \end{align*} For any $\delta > 0$, choosing $m$ large enough so that $\delta_k + 2/m \leq \delta + \delta_k$ and $1/m < \delta/2$, these neighborhoods form a cover of $B$ by sets of diameter at most $\delta_k + 2/m$. As $m \to \infty$, the diameters of the covering sets approach those of the original $C_{k,m,j}$, and the covering sums approach $\mathcal{H}^s_{\delta_k}(A)$. Specifically, for each fixed $k$: \begin{align*} \mathcal{H}^s_{\delta_k + 2/m}(B) \leq \sum_{j=1}^\infty \alpha(s)\left(\frac{\operatorname{diam}(C_{k,m,j}) + 2/m}{2}\right)^s. \end{align*} As $m \to \infty$, the right-hand side approaches $\mathcal{H}^s_{\delta_k}(A)$, because $\operatorname{diam}(C_{k,m,j}) \leq \delta_k$ and the extra $2/m$ vanishes. Therefore $\mathcal{H}^s_{\delta_k}(B) \leq \mathcal{H}^s_{\delta_k}(A)$ (since $\delta_k + 2/m \to \delta_k$). Taking $k \to \infty$ (equivalently $\delta_k \to 0$) gives $\mathcal{H}^s(B) \leq \mathcal{H}^s(A)$. Combining, $\mathcal{H}^s(B) = \mathcal{H}^s(A)$. [/step]