If $A$ is a subset of $\mathbb{R}^d$, we denote the diameter of $A$ by $\Delta(A)=\sup \{|x-y|: x, y \in A\} \in[0, \infty]$. For every $\alpha>0$ and every $\varepsilon>0$, we set, for every subset $A$ of $\mathbb{R}^d$, $$ \mu_{\alpha, \varepsilon}(A)=\inf _{\left(U_k\right)_{k \in \mathbb{N}} \in \mathcal{R}_{\varepsilon}(A)} \sum_{k \in \mathbb{N}} \Delta\left(U_k\right)^\alpha, $$ where $\mathcal{R}_{\varepsilon}(A)$ is the set of all countable coverings of $A$ by open sets of diameter smaller than $\varepsilon$. (1) Observe that $\mu_{\alpha, \varepsilon}(A) \geq \mu_{\alpha, \varepsilon^{\prime}}(A)$ if $\varepsilon<\varepsilon^{\prime}$, and thus one can define a mapping $A \mapsto \mu_\alpha(A)$ by $$ \mu_\alpha(A):=\lim _{\varepsilon \downarrow 0} \mu_{\alpha, \varepsilon}(A) \in[0, \infty] $$ (2) Prove that $\mu_\alpha$ is an outer measure on $\mathbb{R}^d$. (3) Verify that, for every subset $A$ of $\mathbb{R}^d$, there exists a (unique) real number $\operatorname{dim}(A) \in[0, d]$, called the Hausdorff dimension of $A$, such that, for every $\alpha>0$, $$ \mu_\alpha(A)= \begin{cases}0 & \text { if } \alpha>\operatorname{dim}(A) \\ \infty & \text { if } \alpha<\operatorname{dim}(A)\end{cases} $$ (4) Let $\alpha>0$ and let $A$ be a Borel subset of $\mathbb{R}^d$. Assume that there exists a measure $\mu$ on $\left(\mathbb{R}^d, \mathcal{B}\left(\mathbb{R}^d\right)\right.$ ) such that $\mu(A)>0$ and $\mu(B) \leq r^\alpha$ for every open ball $B$ of radius $r>0$ centered at a point of $A$. Prove that $\operatorname{dim}(A) \geq \alpha$.
Can someone help me with part 4 of this.