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Let $T:X\to X$ be a continuous function on a compact manifold $X$ and let $\text{Leb}$ be the Lebesgue measure normalized so that $\text{Leb}(X)= 1$. We denote by $\mathcal{M}(X)$ the space of all Borel probabilty measures on $X$ with the 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∗, defined as follows.

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

$\textbf{Question:}$ Is the support of $\mu$ the subset $B_{\epsilon}(\mu)?$or $\text{supp}(\mu)\cap B_{\epsilon}(\mu)\neq \emptyset$?

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  • $\begingroup$ Do you mean by Lebesgue measure the measure coming from the top-degree form? $\endgroup$
    – LSpice
    Nov 8 '21 at 3:48
  • $\begingroup$ @LSpice: What do you mean top-degree form? $\endgroup$
    – Adam
    Nov 8 '21 at 10:57
  • $\begingroup$ 'The' element of $\bigwedge^n{\operatorname T}^*X$, where $n$ is the dimension. Basically, just guessing what Lebesgue measure means for an arbitrary compact manifold. $\endgroup$
    – LSpice
    Nov 8 '21 at 11:13
  • $\begingroup$ @LSpice: Ah, you're right $\endgroup$
    – Adam
    Nov 8 '21 at 12:25
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The answer, in general, is negative. That is, there exist continuous maps $T:X \to X$ of the type you describe and Borel probability measures $\mu$ such that, for small $\epsilon$, the set $B_\epsilon(\mu)$ is empty. For instance, this happens whenever $T$ is uniquely ergodic [1] (The canonical example of that is an irrational rotation on the circle.)

Indeed, if $T$ is uniquely ergodic, then there is a unique $T$-invariant Borel probability measure $\nu_T$ on $X$. Observe that for each $x$ the measures in $V(x)$ are all $T$-invariant, so $V(x)=\{\nu_T\}$. If $\mu \ne \nu_T$ then for small enough $\epsilon$, we have $\nu_T \notin N_\epsilon(\mu)$, so $B_\epsilon(\mu)$ is empty.

[1] https://en.wikipedia.org/wiki/Ergodicity#Unique_ergodicity

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  • $\begingroup$ Thank you very much for your answer, but I assumed that the set $B_{\epsilon}(\mu)$ is non-empty (I assumed that the set always has positive measure as I wanted to exclude examples like yours). In other words, Is it true that $\text{supp}(\mu) \cap B_{\epsilon}(\mu)\neq \emptyset$ when $B_{\epsilon}(\mu)$ is non-empty? $\endgroup$
    – Adam
    Nov 8 '21 at 10:56

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