Let $(X,\Sigma, \mu)$ be a probability space, and $T:X\rightarrow X$ be a measure-preserving transformation. We say $\mu$ is ergodic with respect to $T$ if
for every $E\in \Sigma $ with $T^{-1}(E) = E$, either $\mu(E)=1$ or $0$.
I have a very fundamental problem about this definition:
the result "$\mu(E)=1$ or $0$" does not have any information about $T$ or does not say anything about $T$. So is the condition "$T^{-1}(E)=E, \forall E\in \Sigma$" necessary?
$T^{-1}(E)=E$ should depend on the choice of $T$; I pick a particular $T$ such that $T^{-1}(E)=E$. However, $T$ does not show up in "$\mu(E)=1$ or $0$"
For any $E\in \Sigma$, let $P(E)$ be the implication "$\left(T^{-1}E=E\right)\Rightarrow \left(\mu(E)\in \{0,1\}\right)$". Ergodicity of $\mu$ with respect to $T$ means that fro all $E\in\Sigma$, proposition $P(E)$ is true. Since the first part of the implication involves $T$, the definition of ergodicity also implies $T$.
Note also that "$T^{-1}(E)=E, \forall E\in \Sigma$" does not hold in general. For example, if $\Sigma$ contains the singletons, this would force $T$to be the identity.