I want to prove exercise 4.3.5 of Brin & Stuck
Let $T$ be a measure preserving transformation of measure space $(X,\mathcal{B},\mu)$ and let $f\in L^1(X,\mathcal{B},\mu)$ satisfy $f(T(x))\leq f(x)$ for a.e. $x$.
Prove that $f(T(x))= f(x)$ for a.e. $x$.
My (wrong) attempt (mentioned by John B.) :
$$ f(T(x))\leq f(x),f\in L^1(X,\mathcal{B},\mu) \implies \int_Xf(T(x))d\mu\leq \int_Xf(x)d\mu\leq \int_X|f(x)|d\mu<\infty $$ The measure preserving of $T$ on $\mu$ means $\forall B\in \mathcal{B}$, $ \mu(T^{-1}(B))=\mu(B)$ for a.e $x\in B$.
How do I include the measure preserving to come to some kind of conclusion?
Attempt 2:
for $f=\chi_A$ and $A$ measurable so $T^{-1}(A)$ is measurable:
$$ f(T(x))\leq f(x) \implies \chi_{T^{-1}(A)}(x)=\chi_A(T(x))\leq \chi_A(x) $$ As the indicator fuction of measurable sets are integrable:
$$ \int_X\chi_{T^{-1}(A)}(x)d\mu=\int_X\chi_A(T(x))d\mu\leq \int_X\chi_A(x)d\mu $$ $$ \mu(T^{-1}(A))\leq\mu(A) $$ and by assumption of measure preserving $$ \mu(A)=\mu(T^{-1}(A))\leq\mu(A) $$ Thus, all are equal.
Then the next step for $f=\sum_{i=0}^N \alpha_i\chi_{A_i}$ with $\alpha_i\in \mathbb{R_{\geq0}}$ and then $f=f^+-f^-$
$\int |f(Tx)|d\mu =\int |f| d\mu < \infty$. Now $g(x)=f(x)-f(T(x))$ is non-negative and its integral is 0. Hence $g=0$ a.e. which is what you want.