Here's the problem:
Let $(X,\mathcal{F},\mu)$ be a probability space and $T:X \rightarrow X$ a measure preserving transformation. Show that if $f(Tx) \leq f(x)$ $\mu$ a.e., then its holds that $f(Tx) = f(x)$ $\mu$ a.e..
Intuitively I think I understand why this should be the case, but I'm struggling with finishing the details. This is how I've tried.
Let $A := \{x \in X: f(Tx) < f(x)\}$, and assume that $\mu(A) > 0$. Then by Poincare's reccurence theorem we have that for almost all $x \in X$ that there exists a $k \in \mathbb{N}$ such that $T^kx \in A$.
What I want to do is show that since $f(x) > f(Tx)$, and $f(Tx) \geq f(T^nx)$ for all $n \geq 2$, that $f(T^nx)$ will eventually be a point that is larger that $f(Tx)$. But since Poincare's recurrence theorem only guarantees we'll land in $A$ infinitely many times this might be a dead end.
Any ideas or tips guys?