Let $(X, \mathcal{B}, \mu, T)$ be a measure-preserving system, with $T$ invertible, where the $\sigma$-algebra $\mathcal{B}$ is a Borel algebra arising from a topology which makes $T$ continuous, and such that $X$ is compact Hausdorff. Then there's a well-defined action of the semigroup $\beta \mathbb{Z}$ on $X$ :


$ \forall p \in \beta \mathbb{Z}, \forall x \in X,\quad T^p x := p\lim\limits\_{n \in \mathbb{Z}} T^n x$


If $p$ is a principal ultrafilter, it is clear that $T^p$ is measure-preserving. Is it still true 
for *any* ultrafilter on $\mathbb{Z}$ ?


The measure preserving property of a $T^p$ would be a consequence of the corresponding dominated convergence theorem along the ultrafilter $p$ ; as far as I can see, this is not a direct consequence of the (usual) dominated convergence theorem, and I didn't succeed in proving or disproving it. Any ideas ?



Does the "measure-preserving property" commute with ultralimits ?
Looks false to me. Let $X$ be the product of countably many copies of $\{0,1\}$ with the standard Bernoulli measure (each point has measure $1/2$). Index these copies with ${\bf Z}$, and let $T$ be the shift operator. I claim that evaluating along any free ultrafilter won't be measure-preserving. To see this, let $A\_0 \subset X$ be the set of sequences $x = (\ldots, x\_{-1}, x\_0, x\_1, \ldots)$ with $x\_0 = 0$ and let $A\_1$ be its complement, the set of sequences $x$ with $x\_0 = 1$. Both of these sets have measure $1/2$. However, it is easy to check that $T^p(A\_0) = T^p(A\_1) = T^p(X)$, using the fact that the limit of a sequence along a free ultrafilter doesn't change if you modify the sequence in finitely many coordinates. So there's no way $T^p$ can be measure preserving. (Also, it isn't a homeomorphism.)


