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.
On
Hint: $T_\ast(\mu)=\mu$ iff the integral condition holds with $f$ an arbitrary characteristic function by definition. Then by taking linear combinations this is equivalent to the integral condition with $f$ a simple function. Finally by approximating an $L^\infty$ or $L^1$ function by an increasing sequence of simple functions this is equivalent to the integral condition with $f$ an $L^\infty$ or $L^1$ function.
The above works even when $\mu$ is not necessarily a finite measure. In the case of $\mu$ a finite measure one could alternatively use the fact that $L^\infty\subseteq L^1$, given the $L^1$ characterization.
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 $L^{1}(\mu)$.