Let ($X, \mathscr{F}, \mu)$ be a measure space and, $T:X \to X$ $\mathscr{F}/\mathscr{F}$ measurable map such that: $\mu(C)=\mu(T^{-1}(C))$ for all $C\in \mathscr{F}$, and let $u\in\mathscr{L}^1(\mu).$
Prove that if $u \le u\circ T, \mu$ a.e. then $u=u\circ T \mu$ a.e..
My attempt: We know the following: $\mu({u\gt u\circ T})=0$, by assumption. We want show $\int |u \circ T-u|d\mu=0$ almost everywhere. Note that by assumption: $\int |u \circ T-u|d\mu=\int u \circ T-ud\mu= \int u \circ T d\mu-\int u d\mu=*$,
Since $T$ preserves the measure we get: $\int u\circ Td\mu=\int u d\mu$, hence:
$* = 0$. Thus $|u \circ T-u|=0, \mu$ a.e.. Conclusion: $u\circ T=u, \mu$ a.e.
Is this argument correct?
Your argument is not correct. 'This coresponds with ...' does not make sense and you did not even make use of the fact that $T$ is measure preserving. Use the fact that $\int [u\circ T-u]=0$ because $\int u\circ T=\int u$ and $u\circ T-u$ is non-negative to conclude that $u\circ T=u$ a.e.