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Let $X$ be a Polish space and $T\colon X\to X$ be a continuous map. We say that a point $x\in X$ is quasi-regular if for every bounded continous function $\varphi\colon X\to\mathbb{R}$ the sequence $A_n\varphi(x)$ of Birkhoff averages converges as $n\to\infty$, where $$A_n\varphi(x)=\frac1n\sum_{j=0}^{n-1} \varphi(T^j(x)).$$

Setting $$L_x(\varphi)=\lim_{n\to\infty}A_n\varphi(x)$$ we obtain a bounded linear functional on the space $\mathcal{C}_b(X)$ of bounded continous functions $\varphi\colon X\to\mathbb{R}$. Clearly, $L_x(\varphi)=L_x(\varphi\circ T)$, but does $L_x$ have to be given by $$L_x(\varphi)=\int_X\varphi\,\text{d}\mu$$ for some $T$-invariant Borel probability measure $\mu$ on $X$?

Comments

Clearly, the answer is yes if we assume that $X$ is compact (by the Riesz theorem). Furthermore, if $\hat{X}$ is any compactification of $X$, the $L_x$ must be given by some Borel probability measure $\hat{\mu}$ on $\hat{X}$, but since it may be impossible to extend $T$ to $\hat{X}$ we cannot claim that $\hat{\mu}$ is $T$-invariant.

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$A_n \varphi (x) = \int_X \varphi \mathrm{d} \mu_n$ for a Borel probability measure $\mu_n$, in fact a measure with finite support $\{ T^j(x) \mid j=0,1,\ldots,n-1\}$. By the assumption about the point $x$, the sequence of $\mu_n$ is weakly Cauchy, where "weakly" refers to the duality between the space $M_\sigma(X)$ of finite Borel measures on $X$ and the space $C_b(X)$. The space $M_\sigma(X)$ is weakly sequentially complete (Bogachev, ``Measure Theory'', Theorem 8.7.1). Therefore there is a measure $\mu\in M_\sigma(X)$ such that $\int_X\varphi\mathrm{d}\mu=\lim_n\int_X\varphi\mathrm{d}\mu_n$ for every $\varphi\in C_b(X)$. Then $L_x (\varphi) = \int_X \varphi \mathrm{d} \mu$. The measure $\mu$ is $T$-invariant because $L_x(\varphi)=L_x(\varphi\circ T)$ for every $\varphi\in C_b(X)$.

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    $\begingroup$ To use completeness, it is necessary to specify a metric, and prove that a sequence is Cauchy in that metric. $\endgroup$ Aug 2, 2022 at 16:23
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    $\begingroup$ @YuvalPeres Here, weak sequential completeness means that every fundamental sequence converges, where a sequence $\mu_n$ is said to be fundamental if for every bounded continuous function $f$, the sequence $\int f\,\mathrm{d}\mu_n$ is fundamental ($=$ Cauchy). [Bogachev, Sections 8.1 and 8.7] $\endgroup$
    – Algernon
    Aug 2, 2022 at 21:37
  • $\begingroup$ I have added more words to the answer, to spell out the steps of the argument. $\endgroup$
    – user95282
    Aug 3, 2022 at 10:28

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