1
$\begingroup$

Let $( X , Σ , μ )$ be a probability space, and $T : X → X$ be a measure-preserving transformation. We say that $T$ is ergodic with respect to $μ$ if for every $E ∈ Σ$ with $T^{-1}(E)=E$ either $μ ( E ) = 0$ or $μ ( E ) = 1$.

One of the implications (actually an equivalent definition) says

Every measurable function $f:X\to \mathbb R$ such that $f\circ T=f$ a.e. is constant almost everywhere.

However, when $T:X\to X$ is ergodic, the distinction between the role $T(x)$ and $x$ is not so obvious (my feeling). I wonder if we also have

Every measurable function $f:X\to \mathbb R$ such that $f\circ T \ge f$ a.e. is constant almost everywhere. Because I think somehow $f\circ T \le f$ a.e. will also follow from the ergodicity.

Is it correct? How to prove it if so?

$\endgroup$
1
  • $\begingroup$ Peter Walters An Introduction to Ergodic theTheory Page 29 $\endgroup$ – Neil hawking Sep 8 '18 at 22:15
1
$\begingroup$

Your intuition is right, but for a simple reason. Note that, since $\mu$ is invariant under $T$, we have \begin{equation} \int f\circ T\, \mathrm{d}\mu = \int f\, \mathrm{d}(T\mu) = \int f\, \mathrm{d}\mu \;, \end{equation} hence, \begin{equation} \int (f\circ T - f)\, \mathrm{d}\mu = 0 \;. \tag{*} \end{equation} Now, if the integrand in (*) is almost surely non-negative, it cannot be non-zero on a set of positive measure.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.