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Context: these lecture notes, exercise 47. Not a homework assignment.

$(X,\mathcal{F},\mu)$ is a probability space, $T:X\rightarrow X$ is a probability preserving map. Define $$\mathcal{G}:= \sigma (\{A\in \mathcal{F}:T^{-1}(A)=A\}).$$ Prove that $\mathcal{G}$-measurable RV's are $T$-invariant (i.e. $f\circ T = f,$ $\mu$ almost everywhere).

My attempt: $$T^{-1}(A) = A \implies A = T(A)$$ $$\implies f(T(A)) = f(A).$$

So when restricted to the generators of $\mathcal{G}$, $f\circ T=f$. How can I extend this to the whole $\sigma$-algebra?

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    Please check my editing. – 2017-01-06

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Actually, we have $$ \mathcal G=\{A\in\mathcal F:T^{-1}A=A\}. $$ If you prefer: the collection of all measurable $T$-invariant sets is a $\sigma$-algebra (simply verify the axioms). Because of this, you have in fact already considered all sets in $\mathcal G$.

PS: One can show that $f$ is $\mathcal G$-measurable if and only if it is $T$ invariant almost everywhere.

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    Oh ok, so that collection of sets is already a $\sigma$-algebra, hence I've already checked it for the whole $\sigma$-algebra – 2017-01-06