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 $X$?


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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.



Are all quasi-regular points on Polish spaces generic points?
$A\_n \varphi (x) = \int\_X \varphi \mathrm{d} \mu\_n$ 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 $\varphi\in C\_b(X)$.


