# Trying to show an implication for a measure preserving $T$ on a probability space.

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?

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Consider $g(x)=f(x)-f(Tx)$, then $g\geqslant0$ $\mu$-almost everywhere and $\int\limits g\,\mathrm d\mu=0$ since $T$ preserves $\mu$. Now, it is a general fact that such a function $g$ must be zero $\mu$-almost everywhere.
To show this, consider $A_t=\{g\geqslant t\}$ for some $t\gt0$ and note that $g\geqslant t\mathbf 1_{A_t}$ $\mu$-almost everywhere (this is where the hypothesis that $g\geqslant0$ $\mu$-almost everywhere is used) hence $0=\int\limits g\,\mathrm d\mu\geqslant t\mu(A_t)\geqslant0$, that is, $\mu(A_t)=0$. The union over every $n\geqslant1$ of the measurable sets $A_{1/n}$ has $\mu$-measure zero and is the set $\{g\gt0\}$, hence $\{g=0\}$ has full measure.
From where do you get that $\int g\,d\mu = 0$? Isn't it true that if $g \geq 0$ and $\int g \,d\mu =0$ this already implies that $g = 0$ $\mu$ a.e. ? – BallzofFury Nov 4 '12 at 10:43
You assumed that $T$ is a measure preserving transformation. And $T_*\mu=\mu$ means exactly that $\int f(Tx)\mathrm d\mu(x)=\int f(x)\mathrm d\mu(x)$ for every integrable function $f$. – Did Nov 4 '12 at 10:51
Ah, thats the missing piece I was looking for. Just for the streamlining of the proof, isn't the $A_t$ part unnecessary? Can't we just state (in shorthand) $\mu\{g \geq 0\}=0$ by assumption, and $\int g\,d\mu = 0$ by $T$ measure preserving. Thus $\mu\{g = 0\}=1$ and $\mu\{f(x) = f(Tx)\}=1$? – BallzofFury Nov 4 '12 at 10:56