[proofplan]
We compare the entropy sums atom by atom. Measure preservation gives $\mu(T^{-1}(A_i))=\mu(A_i)$ for every atom $A_i$ of $\alpha$. The only minor point is that the pullback partition may omit empty preimages; those correspond to atoms of $\alpha$ with measure zero, so their entropy contribution is zero. Therefore the entropy sum is unchanged.
[/proofplan]
[step:Verify that the pullbacks form the relevant finite measurable partition]
Since $\alpha$ is a finite measurable partition of $X$, choose $m\in\mathbb N$ and measurable atoms $A_1,\dots,A_m\in\mathcal B$ such that $\alpha=\{A_1,\dots,A_m\}$. For each index $i\in\{1,\dots,m\}$, define the measurable set $B_i\in\mathcal B$ by
\begin{align*}
B_i := T^{-1}(A_i).
\end{align*}
The measurability follows because $T:X\to X$ is measurable and $A_i\in\mathcal B$. Because $\alpha$ is a partition of $X$, the sets $A_1,\dots,A_m$ are pairwise disjoint and satisfy
\begin{align*}
X=\bigcup_{i=1}^m A_i.
\end{align*}
Taking preimages under $T$ gives
\begin{align*}
X=T^{-1}(X)=T^{-1}\left(\bigcup_{i=1}^m A_i\right)=\bigcup_{i=1}^m T^{-1}(A_i)=\bigcup_{i=1}^m B_i.
\end{align*}
If $i\neq j$, then $A_i\cap A_j=\varnothing$. Therefore
\begin{align*}
B_i\cap B_j=T^{-1}(A_i)\cap T^{-1}(A_j)=T^{-1}(A_i\cap A_j)=T^{-1}(\varnothing)=\varnothing.
\end{align*}
Thus the nonempty members among $B_1,\dots,B_m$ form the finite measurable partition $T^{-1}\alpha$.
[/step]
[step:Use measure preservation to match the atom measures]
For every $i\in\{1,\dots,m\}$, measure preservation applied to the measurable set $A_i\in\mathcal B$ gives
\begin{align*}
\mu(B_i)=\mu(T^{-1}(A_i))=\mu(A_i).
\end{align*}
If $B_i=\varnothing$, then $\mu(B_i)=0$, and therefore $\mu(A_i)=0$.
[guided]
The entropy of a finite partition depends only on the measures of its atoms, so the central task is to compare the number $\mu(A_i)$ with the number attached to its pullback atom. For each $i\in\{1,\dots,m\}$, we have already defined $B_i:=T^{-1}(A_i)$. The hypothesis that $T$ is measure-preserving says precisely that for every measurable set $E\in\mathcal B$,
\begin{align*}
\mu(T^{-1}(E))=\mu(E).
\end{align*}
We may apply this with $E=A_i$, since each atom $A_i$ of the measurable partition $\alpha$ belongs to $\mathcal B$. Hence
\begin{align*}
\mu(B_i)=\mu(T^{-1}(A_i))=\mu(A_i).
\end{align*}
There is one bookkeeping issue: the definition of $T^{-1}\alpha$ removes empty preimages, because partitions are usually written using nonempty atoms. If $B_i=\varnothing$, then $\mu(B_i)=0$, so the equality above gives $\mu(A_i)=0$. Therefore an omitted empty pullback can only correspond to an atom $A_i$ whose entropy contribution is
\begin{align*}
-\mu(A_i)\log\mu(A_i)=0
\end{align*}
by the convention $0\log 0:=0$. Thus deleting empty pullbacks cannot change the entropy sum.
[/guided]
[/step]
[step:Compare the two entropy sums]
Let $I := \{i\in\{1,\dots,m\}: B_i\neq\varnothing\}$. Then $T^{-1}\alpha=\{B_i:i\in I\}$, and therefore
\begin{align*}
H_\mu(T^{-1}\alpha)=-\sum_{i\in I}\mu(B_i)\log\mu(B_i).
\end{align*}
Using $\mu(B_i)=\mu(A_i)$ for $i\in I$, we obtain
\begin{align*}
H_\mu(T^{-1}\alpha)=-\sum_{i\in I}\mu(A_i)\log\mu(A_i).
\end{align*}
For every $i\notin I$, we have $B_i=\varnothing$, hence $\mu(A_i)=0$, and the convention $0\log 0:=0$ gives
\begin{align*}
-\mu(A_i)\log\mu(A_i)=0.
\end{align*}
Consequently,
\begin{align*}
-\sum_{i\in I}\mu(A_i)\log\mu(A_i)=-\sum_{i=1}^m\mu(A_i)\log\mu(A_i)=H_\mu(\alpha).
\end{align*}
Combining the displayed equalities gives
\begin{align*}
H_\mu(T^{-1}\alpha)=H_\mu(\alpha).
\end{align*}
This proves the asserted entropy invariance under measure-preserving pullback.
[guided]
We now convert the atom-by-atom measure comparison into the entropy identity. Define $I := \{i\in\{1,\dots,m\}: B_i\neq\varnothing\}$. For every $i\in\{1,\dots,m\}$, the set $B_i=T^{-1}(A_i)$ is measurable because $T$ is measurable and $A_i\in\mathcal B$. The union identity $X=\bigcup_{i=1}^m B_i$ follows by taking preimages of $X=\bigcup_{i=1}^m A_i$, and pairwise disjointness follows from $B_i\cap B_j=T^{-1}(A_i\cap A_j)=\varnothing$ whenever $i\neq j$. Thus, after discarding exactly the empty sets, the atoms of $T^{-1}\alpha$ are the sets $B_i$ with $i\in I$. Therefore, by the definition of entropy for a finite measurable partition,
\begin{align*}
H_\mu(T^{-1}\alpha)=-\sum_{i\in I}\mu(B_i)\log\mu(B_i).
\end{align*}
For every $i\in I$, measure preservation gave $\mu(B_i)=\mu(A_i)$, so substituting equal atom measures into the entropy sum gives
\begin{align*}
H_\mu(T^{-1}\alpha)=-\sum_{i\in I}\mu(A_i)\log\mu(A_i).
\end{align*}
It remains only to account for the indices not in $I$. If $i\notin I$, then $B_i=\varnothing$, so the previous step gives $\mu(A_i)=\mu(B_i)=0$. With the entropy convention $0\log 0:=0$, each omitted index contributes
\begin{align*}
-\mu(A_i)\log\mu(A_i)=0.
\end{align*}
Hence adding those omitted zero terms does not change the sum:
\begin{align*}
-\sum_{i\in I}\mu(A_i)\log\mu(A_i)=-\sum_{i=1}^m\mu(A_i)\log\mu(A_i)=H_\mu(\alpha).
\end{align*}
Combining the two entropy computations yields
\begin{align*}
H_\mu(T^{-1}\alpha)=H_\mu(\alpha).
\end{align*}
This is exactly the asserted invariance of entropy under measure-preserving pullback.
[/guided]
[/step]
