Pinsker $\sigma$-algebra is $T$ invariant Let $(X,B,\nu )$ be a separable probability space and $T:X\to X$ be a measure-preserving transformation.
We define $D(T)$ as the collection of all $A\in B$ for which there exists a partition $Q$ of $X$ such that $A\in Q$, $H(Q)$ is finite and $h(T,Q)=0$ (entropy of $T$ with respect to $Q$).
The Pinsker $\sigma$-algebra $P(T)$ is defined as the $\sigma$-algebra generated by $D(T)$.
I want to show that $T^{-1}(P(T))=P(T)$.
I was able to show that $T^{-1}(P(T))\subset P(T)$, but so far all my attempts to show the converse inclusion fails.
I am using as bibliography: Topics in Ergodic Theory by W. Parry
Any help will be appreciated.
 A: There are often multiple ways of deriving the general entropy statements like the one you are asking based on what one is allowed to use as fact/definition, as such I note below the identities in Parry's book one can use to substantiate the argument Parry gives on p.64 that the Pinsker $\sigma$-algebra is invariant.

Let us define $\mathfrak{Z}(\nu,T)$ to be the collection of all countable $\nu$-ae partitions $Q$ of $X$ with $H_\nu(Q)<\infty$ and $h_\nu(T;Q)=0$, so that $D(T)=\bigcup_{Q\in\mathfrak{Z}(\nu,T)}Q$. First note that if $Q\in\mathfrak{Z}(\nu,T)$, then so is $T^{-1}(Q)$. Indeed, $H_\nu(T^{-1}(Q))=H_{T_\ast(\nu)}(Q)=H_\nu(Q)<\infty$ (by the first identity on p.39) and by definition (again on p.39) $h_\nu(T;T^{-1}(Q))$ $= H_\nu(T^{-1}(Q)\,|\, \bigvee_{n\geq 1} T^{-n}(T^{-1}(Q)))$ $= H_\nu(T^{-1}(Q)\,|\, T^{-1}(\bigvee_{n\geq 1}T^{-n}(Q)))$ $= H_{T_\ast(\nu)}(Q\,|\, \bigvee_{n\geq 1}T^{-n}(Q))$ $= H_{\nu}(Q\,|\, \bigvee_{n\geq 1}T^{-n}(Q))$ $= h_\nu(T;Q)=0$. Note that since $T^{-1}$ is a morphism of $\sigma$-algebras, this shows that $T^{-1}(P(T))\subseteq P(T)$.
Next, if $Q_1,Q_2\in \mathfrak{Z}(\nu,T)$, then (e.g. by the basic identity on p.35 and monotonicity) $H_\nu(Q_1\vee Q_2)= H_\nu(Q_1)+H_\nu(Q_2\,|\, Q_1)\leq H_\nu(Q_1)+H_\nu(Q_2)<\infty$, and (by Thm.7 on p.39 and again subadditivity)
\begin{align*}
0\leq h_\nu(T;Q_1\vee Q_2) 
&= \lim_{n\to \infty} \dfrac{1}{n} H_\nu\left(\bigvee_{k=0}^{n-1} T^{-n}(Q_1\vee Q_2)\right)\\
&= \lim_{n\to \infty} \dfrac{1}{n} H_\nu\left(\bigvee_{k=0}^{n-1} T^{-n}(Q_1) \vee \bigvee_{k=0}^{n-1} T^{-n}(Q_2)\right)\\
&\leq \lim_{n\to \infty} \dfrac{1}{n} H_\nu\left(\bigvee_{k=0}^{n-1} T^{-n}(Q_1) \right) +\lim_{n\to \infty} \dfrac{1}{n} H_\nu\left(\vee \bigvee_{k=0}^{n-1} T^{-n}(Q_2)\right)\\
&= h_\nu(T;Q_1) +h_\nu( Q_2) \leq 0, 
\end{align*}
whence we have that $Q_1\vee Q_2\in \mathfrak{Z}(\nu,T)$. In particular, for any $n\in\mathbb{Z}_{\geq1}$ and for any $Q\in\mathfrak{Z}(\nu,T)$, we have that $\bigvee_{k=0}^{n-1} T^{-k}(Q)\in\mathfrak{Z}(\nu,T)$. Consequently the $\sigma$-algebra $\bigvee_{k\geq0} T^{-k}(Q)$ generated by $\bigcup_{n\geq1}\bigvee_{k=0}^{n-1} T^{-k}(Q)$ is a sub-$\sigma$-algebra of $P(T)$. (Note that the partition corresponding to $\bigvee_{k\geq0} T^{-k}(Q)$ may fail to be countable, as such it is  not an element of $\mathfrak{Z}(\nu,T)$, but it's certainly measurable, that is, the associated factor space is a separable probability space when endowed with the pushforward measure.)
To show $T^{-1}(P(T))\supseteq P(T)$, let $Q\in\mathfrak{Z}(\nu,T)$. Then $0=h_\nu(T;Q)$ $= H_{\nu}(Q\,|\, \bigvee_{n\geq 1}T^{-n}(Q))$. Then since for $R,S$ two partitions, $H_\nu(R\,|\, S)=0$ iff $S$ refines $R$ iff the $\sigma$-algebra generated by $R$ is a sub-$\sigma$-algebra of the $\sigma$-algebra generated by $S$ (by the first emphasized sentence on p.34 (alternatively one can directly use the definition of conditional entropy (again on p.34) and observe that this is a consequence of properties of conditional expectation) (observe that the heuristic is: "$H_\nu(R\,|\, S)=0$ means that if one has the knowledge of $S$, then on average w/r/t $\nu$ knowing which cell of $R$ is hit is not surprising.")), we have that the $\sigma$-algebra generated by $Q$ is a sub-$\sigma$-algebra of $\bigvee_{n\geq 1}T^{-n}(Q)=T^{-1}(\bigvee_{n\geq 0} T^{-n}(Q))$, which in turn is a sub-$\sigma$-algebra of $T^{-1}(P(T))$ by the previous paragraph. Taking unions and generating the $\sigma$-algebra we are done.
