# A strengthened implication of the ergodicity

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?

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

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