Measurability with respect to the $\sigma$-algebra of invariant sets.

Question

Let $(X, \mathcal F, \mu)$ be a measure space and $T:X\to X$ be a measure preserving transformation. Let $$\mathcal E=\{E\in \mathcal F:\ \mu(T^{-1}(E)\ \Delta\ E) = 0\}$$ where $\Delta$ denotes the symmetric difference. Note that $\mathcal E$ is also a $\sigma$-algebra.

Let $g:X\to \mathbb R$ be $\mathcal E$-measurable.

Is is necessarily true that $g$ is $T$-invairant, that is, $g(x)=g\circ T(x)$ for almost all $x\in X$.

Answer 1

Yes. If $g$ is a simple function on $(X,\mathcal E)$ this is easy: just note that $\mu(T^{-1} E \Delta E)=0$ iff $\int |I_E \circ T -I_E| \, d\mu=0$ iff $I_E \circ T =I_E$ almost everywhere. Now approximate $g^{+}$ and $g^{-}$ using simple functions to finish the proof.