Conditional expectation of pullback of sigma algebra

Let $T: X \to X$ be a measure preserving invertible transformation and let $F$ be as sub sigma algebra. My book said it is clear that $E(\chi_{T^{-1}A} | T^{-1}F)(x)=E(\chi_A | F)(Tx)$ but I am not sure why, especially considering how the right hand side doesn't seem to be $T^{-1}F$ measurable.

Edit: The book doesn't mention it, but I think it is implicitly assuming that $T^{-1}F \subset F$. This is from an ergodic theory textbook, and this result is used to prove that the conditional entropy satisfies $H(T^{-1}A | T^{-1}F)=H(A|F)$

Solution 1:

If $Y$ is $F-$ measurable the $Y\circ T$ is $T^{-1}(F)-$ measurable: $$(Y\circ T)^{-1}(E)=T^{-1}(Y^{-1}(E)) \in T^{-1}(F)$$ since $Y^{-1}(E) \in F$. Take $Y=E(\chi_A|F)$ in this.

Hence, there is no measurabilty problem and no extra assumption is necessary.