Conditional expectation is $T$-invariant in a measure preserving system. Let $(X,\mathcal{A}, \mu, T)$ be a measure preserving system and $\mathcal{B}$ be a sub-sigma algebra of $\mathcal{A}$. Show that
$$E(f|\mathcal{B})= E(f\circ T|T^{-1}\mathcal{B}).$$
My attempt:
We have $$\int_{T^{-1}B}f\circ T\:d\mu= \int_{T^{-1}B}E(f\circ T|T^{-1}\mathcal{B})\:d\mu\quad \text{for all }B\in \mathcal{B},$$
and
$$\int_{B}f\:d\mu= \int_{B}E(f|\mathcal{B})\:d\mu \quad \text{for all }B\in \mathcal{B}.$$
As $\mu$ is $T$-invariant, we have
$$\int_Bf\:d\mu=\int_{T^{-1}B}f\circ T\:d\mu.$$
Therefore
$$\int_{T^{-1}B}E(f\circ T|T^{-1}\mathcal{B})\:d\mu=\int_{B}E(f|\mathcal{B})\:d\mu \quad \text{for all }B\in \mathcal{B}.$$
Please give me some hint on how to proceed.
 A: This is not true. If $f=I_B$ with $B \in \mathcal B$ the LHS is $I_B$ and RHS is $I_{T^{-1}B}$. The stated result is true if $f$ is $T-$ invariant: i.e. $f\circ T=f$ a.e..
A: Kavi Rama Murthy's answer demonstrates that the formula is not correct in general. On the other hand, categorical principles point to the correct version of it. In a slightly more general setting, the correct formula is:
Lemma: Let $T:(X,\mathcal{B}(X))\to (Y,\mathcal{B}(Y))$ be a measurable map between measurable spaces. Then for any probability measure $\mu$ on $X$, for any sub-$\sigma$-algebra $\mathcal{A}$ of $\mathcal{B}(Y)$, and for any measurable $f:Y\to \mathbb{R}$, we have:
$$\overleftarrow{T}\left(\mathbb{E}_{\overrightarrow{T}(\mu)}(f\,|\, \mathcal{A})\right) = \mathbb{E}_{\mu}\left(\overleftarrow{T}(f)\,\left|\, \overleftarrow{T}(\mathcal{A})\right.\right).$$
Here $\mathbb{E}_\nu(g\,|\, \mathcal{B})$ is the conditional measure of the measurable function $g$ conditioned on the $\sigma$-algebra $\mathcal{B}$ w/r/t the probability measure $\nu$; the third $\overleftarrow{T}$ is pullback of $\sigma$-algebras, the first $\overleftarrow{T}$ is pullback that transforms $\mathcal{A}$-measurable functions to $\overleftarrow{T}(\mathcal{A})$-measurable functions, the second $\overleftarrow{T}$ is pullback that transforms $\mathcal{B}(Y)$-measurable functions to $\mathcal{B}(X)$-measurable functions, and the $\overrightarrow{T}$ is pushforward acting on measures. In short, all decorated $T$'s are maps functorially induced by $T$ ($\overleftarrow{T}$ is typically denoted by $T^{-1}$ or $T^\ast$ and $\overrightarrow{T}$ is typically denoted by $T_\ast$). (I'll leave it as an exercise that all this is indeed syntactic). As a diagram it is arguably easier to convey what is going on:


The proof is essentially a version of the algebraic manipulations the OP displays. The important point is that by the characterizing property of conditional expectations what needs to be shown is that
$$\forall A\in\mathcal{A}: \int_{\overleftarrow{T}(A)} \overleftarrow{T}\left(\mathbb{E}_{\overrightarrow{T}(\mu)}(f\,|\, \mathcal{A})\right) \, d\mu = \int_{\overleftarrow{T}(A)} \overleftarrow{T}(f)\, d\mu.$$
It is straightforward to adapt the formula to the special case of $T:X\to X$ measure preserving. Concisely (using the OP's notation), it says:
$$E(f\,|\, \mathcal{A})\circ T= E(f\circ T\,|\, T^{-1}(\mathcal{A})).$$
A: The statement in the OP is false in general since $\mathbb{E}[f|\mathcal{B}]$ and $\mathbb{E}[f\circ T|T^{-1}(\mathcal{B})]$ are $\mathcal{B}$ and $T^{-1}(\mathcal{B})$ measurable respectively, and $\mathcal{B}$ and $T^{-1}(\mathcal{B})=\{T^{-1}(B):B\in\mathcal{B}\}$ are different $\sigma$-algebras in general.
There are however a few interesting identities that relate the quantities in the OP under some minor assumptions.

*

*For simplicity assume that $\mu$ is a probability measure on $(X,\mathcal{B})$ and that $T:(X,\mathcal{B})\rightarrow(X,\mathcal{B})$ (i.e., $T$ is $\mathcal{B}/\mathcal{B}$-measurable).

*For any $f\in L_1(\mu)$, define $\mu_f(A):=\int_A f\,d\mu$.

*Define the push forward measure  $\mu\circ T^{-1}(A):=\mu(T^{-1}(A))$.

If $\mu\circ T^{-1}\ll \mu$ (this holds for example when $T$ is $\mu$-invariant), then $\mu_f\circ T^{-1}\ll \mu\circ T^{-1}$ and $\|\mu_f\circ T^{-1}\|_{TV}=|\mu_f|(T^{-1}(X))=\|f\|_1<\infty$. Indeed,
if $\mu(T^{-1}(A))=0$, then
$$\mu_f(T^{-1}(A))=\int\mathbb{1}_{T^{-1}(A)}f\,d\mu=0$$
The last statement follows by definition of the push-forward and the definition of total variation. It follows that the  Radon-Nikodym derivative $P_Tf:=\frac{d\mu_f\circ T^{-1}}{d\mu\circ T^{-`1}}$ is well defined. Hence
\begin{align}
\int\mathbb{1}_B\,P_Tf\,d(\mu\circ T^{-1})&=\int\mathbb{1}_B\,d(\mu_f\circ T^{-1})\\
&=\int\big(\mathbb{1}_B\circ T\big)\,d\mu_f=\int \mathbb{1}_{T^{-1}(B)}f\,d\mu\\
&=\int\big(\mathbb{1}_{B}\circ T\big)\,\mathbb{E}_\mu[f|T^{-1}(\mathcal{B})]\,d\mu
\end{align}
As $\mathbb{E}_\mu[f|T^{-1}(\mathcal{B})]$ is $T^{-1}(\mathcal{B})$-measurable, there is a $\mathcal{B}$-measurable function $h$ such that $\mathbb{E}_\mu[f|T^{-1}(\mathcal{B})]=h\circ T$. Hence
\begin{align}
\int\mathbb{1}_B\,P_Tf\,d(\mu\circ T^{-1})=\int(\mathbb{1}_B\circ T)\,h\circ T\,d\mu=\int\mathbb{1}_B h\,d(\mu\circ T^{-1})\tag{1}\label{one}
\end{align}
As this holds for all $B\in\mathcal{B}$, it follows that $P_Tf=h$ $\mu\circ T^{-`1}$-a.s. and so,
$$P_Tf\circ T=\mathbb{E}_\mu[f|T^{-1}(\mathcal{B})]$$
The operator $P_T:f\mapsto\frac{d\mu_f\circ T^{-1}}{d\mu\circ T^{-1}}$ îs known în the literature as the Ruelle-Perron-Frobenius transfer operator. When $\mu\circ T^{-1}=\mu$,  identity \erqref{one} implies that $P_T$ is the dual operator of the transform $f\mapsto f\circ T$ on $L_1(\mu)$.
