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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 ?

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    $\begingroup$ There should exist a non-normal pure state on l^{\infty}(\mathbb{Z}), and that should provide you with a counterexample. I hope this helps. $\endgroup$
    – Jon Bannon
    Commented May 10, 2012 at 13:35
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    $\begingroup$ A non-normal pure state on $l^\infty({\bf Z})$ is the same as a non-principal ultrafilter on ${\bf Z}$. $\endgroup$
    – Nik Weaver
    Commented May 10, 2012 at 15:25
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    $\begingroup$ @Nik: I suspected so. How does one get that every nonprincipal ultrafilter isn't normal (a reference will do)? $\endgroup$
    – Jon Bannon
    Commented May 10, 2012 at 15:44
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    $\begingroup$ Let $f_n \in l^\infty({\bf Z})$ be the bi-infinite sequence which is zero between $-n$ and $n$, and one outside that interval. This sequence converges weak* to the zero sequence, but each $f_n$ evaluates to 1 along any nonprincipal ultrafilter. So evaluating on any nonprincipal ultrafilter cannot be a normal linear functional on $l^\infty({\bf Z})$. $\endgroup$
    – Nik Weaver
    Commented May 10, 2012 at 15:51

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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.)

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    $\begingroup$ Thanks for the neat counterexample. It also disproves the "dominated convergence theorem" for convergence along a free ultrafilter. $\endgroup$
    – js21
    Commented May 10, 2012 at 19:31

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