Let $X\subset\mathbb{R}^n$ be a compact subset, $\mathcal{B}(X)$ be the Borel $\sigma$-algebra on $X$. Denote $\mathcal{M}(X)$ the collection of all Borel probability measures on $X$. It is clear that $\mathcal{M}(X)$ is compact under the $\text{weak}^*-\text{topology}$.
$\textbf{Lower Hausdorff dimension}$ of $\mu\in\mathcal{M}(X)$ is defined by $$\underline{\dim}_{\text{H}}(\mu)=\sup\{\alpha:\underline{\dim}_{\text{loc}}(\mu,x)\geq\alpha\text{ for }\mu-\text{a.e. }x\in X\},$$ where the $\textbf{lower local dimension}$ of $\mu$ at $x$ is defined by $$\underline\dim_{\text{loc}}(\mu,x)=\liminf_{r\to 0}\frac{\log\mu(B_r(x))}{\log r},$$ and $B_r(x)$ denote the closed ball with radius $r$ centered at $x$.
So my $\textbf{question}$ is: can we prove that $\underline{\dim}_{\text{H}}(\cdot):\mathcal{M}(X)\to\mathbb{R}^+$ is a Borel measurable function on $\mathcal{M}(X)$?
Given $x\in X$ and $r>0$, we have $\limsup_{n\to\infty}\mu_n(B_r(x))\leq\mu(B_r(x))$ whenever $\mu_n$ converges weakly to $\mu$ (this is because $B_r(x)$ is closed). So we conclude that the map $\mu\mapsto\mu(B_r(x))$ is upper semi-continuous and hence Borel measurable when $r>0$ and $x\in X$ are fixed. Therefore, $\underline{\dim}_{\text{loc}}(\cdot,x):\mathcal{M}(X)\to\mathbb{R}$ is Borel measurable when $x\in X$ is fixed (this is because we can replace the limit process $r\to 0$ by $r_n\downarrow 0$ for some appropriate $r_n$). That's all what I can do and I cannot use this result to show that $\underline{\dim}_{\text{H}}(\cdot)$ is Borel.
Can anyone help me to give me some ideas? Thank you very much!