# Support of a measure

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 $$$$\label{empirical} \sigma_{n, x}:=\frac{1}{n} \sum_{j=0}^{n-1} \delta_{T^{j}(x)}$$$$ where $$\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 $$\text{supp}(\mu)\cap B_{\epsilon}(\mu)\neq \emptyset$$?

• Do you mean by Lebesgue measure the measure coming from the top-degree form? Nov 8 '21 at 3:48
• @LSpice: What do you mean top-degree form?
Nov 8 '21 at 10:57
• '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. Nov 8 '21 at 11:13
• @LSpice: Ah, you're right
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.
• 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?