[proofplan] We show $\mathcal{H}^s(A) > 0$, which forces $\dim_{\mathcal{H}}(A) \geq s$. The positive upper density on $A$ is used to extract a subset $B \subset A$ with $\mu(B) > 0$ carrying a uniform upper ball-mass bound $\mu(B(x,r)) \leq M r^s$ for all $x \in B$ and small $r$. The [Mass Distribution Principle](/theorems/3063) applied to $\mu \lfloor B$ then gives $\mathcal{H}^s(B) > 0$. [/proofplan] [step:Extract a subset with $\mu(B) > 0$ and a uniform upper ball-mass bound] For each $m \in \mathbb{N}$, define $A_m := \{x \in A : \Theta^{*s}(\mu, x) \geq 1/m\}$. Since $\Theta^{*s}(\mu, x) > 0$ for all $x \in A$, we have $A = \bigcup_{m=1}^\infty A_m$. Since $\mu(A) > 0$, there exists $m_0$ with $\mu(A_{m_0}) > 0$. For each $k \in \mathbb{N}$, define \begin{align*} B_k := \left\{x \in A_{m_0} \cap \overline{B}(0, k) : \mu(B(x,r)) \leq k \cdot \alpha(s) \cdot r^s \;\text{ for all }\; r \in \left(0, \frac{1}{k}\right)\right\}. \end{align*} Since $\mu$ is Radon, for each $x \in A_{m_0}$ with $\Theta^{*s}(\mu, x) < \infty$, the ratio $\mu(B(x,r))/(\alpha(s) r^s)$ is bounded on $(0, 1/k)$ for large enough $k$, so $x \in B_k$ eventually. Since $\mu(A_{m_0}) > 0$, there exists $k_0$ with $\mu(B_{k_0}) > 0$. Write $B := B_{k_0}$, $M := k_0 \alpha(s)$, and $r_0 := 1/k_0$. [/step] [step:Apply the Mass Distribution Principle to conclude $\mathcal{H}^s(B) > 0$] Define $\nu := \mu \lfloor B$. Then $\nu(B) = \mu(B) > 0$, and for all $x \in B$ and $r \in (0, r_0)$: \begin{align*} \nu(B(x,r)) \leq \mu(B(x,r)) \leq M \cdot r^s. \end{align*} The hypotheses of the [Mass Distribution Principle](/theorems/3063) are satisfied: $\nu$ is a Borel measure with $\nu(B) > 0$, and $\nu(B(x,r)) \leq M \cdot r^s$ for all $x \in B$ and $r \in (0, r_0)$. The principle gives \begin{align*} \mathcal{H}^s(B) \geq \frac{\nu(B)}{M} = \frac{\mu(B)}{k_0 \cdot \alpha(s)} > 0. \end{align*} [/step] [step:Conclude $\dim_{\mathcal{H}}(A) \geq s$] Since $B \subset A$, monotonicity gives $\mathcal{H}^s(A) \geq \mathcal{H}^s(B) > 0$. By the definition of Hausdorff dimension, $\dim_{\mathcal{H}}(A) = \inf\{t \geq 0 : \mathcal{H}^t(A) = 0\} \geq s$, since $\mathcal{H}^s(A) > 0$. [/step]