$T:X\to X$ is a continuous function on a compact manifold $X$ and let $\text{Leb}$ be a Lebesgue measure normalized so that $\text{Leb}(X)= 1$. We denote by $\mathcal{M}(X)$ the space of all Borel probabilty measure on $X$ with weak$^{\ast}$ topology.

For any point $x \in X$ we denote by $V(x)$ the set of all Borel probabilities on $X$ that are limits in the weak$^{\ast}$ topology of convergent subsequences of the sequence \begin{equation}\label{empirical} \sigma_{n, x}:=\frac{1}{n} \sum_{j=0}^{n-1} \delta_{T^{j}(x)} \end{equation} where $\delta_{y}$ is the Dirac delta probability measure supported at $y \in X.$

Let $\mu \in \mathcal{M}(X)$. For any $\varepsilon>0$ the set $B_{\epsilon}(\mu)=\{x \in X: V(x)\cap N_{\epsilon}(\mu)\neq \emptyset\}$ has positive Lebesgue measure, where $N_{\epsilon}(\mu)$ is the $\epsilon$-neighborhood of $\mu$ under the metric dist∗.

For some countable dense subset $\varphi_n$ in $C^0(X, \mathbb{R})$, $$dist*(\nu, \mu)=\sum_{n=1}^{\infty}\frac{|\int \varphi_n d\mu - \int \varphi_n d\nu |}{2^n \sup_x |\varphi_n(x)|}.$$

$\textbf{Question:}$ Is the support of $\mu$ the subset $B_{\epsilon}(\mu)?$