Integral conditions of measure preserving transformation Let $T$ be a transformation of probability space $(X,\mathcal{B},\mu)$. We know that $T$ is measure preserving iff
$$\forall f\in L^1(\mu):\int fd\mu=\int f\circ T\,d\mu.$$
Can $L^1$ in the above condition be replaced by $L^\infty$?
I searched for a similar question, which is weaker than this one and left no answer. "We know" is given directly from the book, so I also don't know the proof.

Suppose $\forall f\in L^{\infty}(\mu):\int fd\mu=\int f\circ T\,d\mu$. Then for any measurable set $E$ the equation holds with $f=\chi_E$ so $\mu (E)=\mu (T^{-1}(E))$. So $T$ is measure preserving.
Conversely, if $T$ is measure preserving then the equation golds for simple functions. Now use the fact that $f\in L^{\infty}(\mu)$ implies $f$ is a limit in $L^{\infty}(\mu)$ of simple, functions.
For $L^{1}(\mu)$ the proof is same except that you approximate $f$ by simple functions in $L^{1}(\mu)$.
