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.
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.