[proofplan]
We prove the result by induction on the iterate index $k$. The base case is the identity map, whose preimage operation fixes every measurable set. In the induction step, measurability follows from the definition of measurability and the preimage identity for a composition, while measure preservation follows by applying first the measure-preservation hypothesis for $T$ and then the induction hypothesis for $T^k$.
[/proofplan]
[step:Verify the identity iterate preserves every measurable set]
For $k=0$, the map $T^0=\operatorname{id}_E:E\to E$ is measurable because, for every $A\in\mathcal E$, $(T^0)^{-1}(A)=\operatorname{id}_E^{-1}(A)=A\in\mathcal E$. The same identity of preimages gives $\mu((T^0)^{-1}(A))=\mu(A)$ for every $A\in\mathcal E$. Hence $T^0$ is measure-preserving.
[/step]
[step:Propagate measurability and measure preservation through one more iterate]
Assume that, for some $k\in\mathbb N\cup\{0\}$, the map $T^k:(E,\mathcal E)\to(E,\mathcal E)$ is measurable and satisfies
\begin{align*}
\mu((T^k)^{-1}(A))=\mu(A)
\end{align*}
for every $A\in\mathcal E$. We prove the same statement for $T^{k+1}=T^k\circ T$.
Let $A\in\mathcal E$. By the preimage identity for composition,
\begin{align*}
(T^{k+1})^{-1}(A)=T^{-1}((T^k)^{-1}(A)).
\end{align*}
Since $T^k$ is measurable, $(T^k)^{-1}(A)\in\mathcal E$. Since $T$ is measurable, it follows that $T^{-1}((T^k)^{-1}(A))\in\mathcal E$. Thus $(T^{k+1})^{-1}(A)\in\mathcal E$, so $T^{k+1}$ is measurable.
Using the same preimage identity and then applying measure preservation first for $T$ and then for $T^k$, we obtain
\begin{align*}
\mu((T^{k+1})^{-1}(A))=\mu(T^{-1}((T^k)^{-1}(A))).
\end{align*}
Because $(T^k)^{-1}(A)\in\mathcal E$ and $T$ is measure-preserving,
\begin{align*}
\mu(T^{-1}((T^k)^{-1}(A)))=\mu((T^k)^{-1}(A)).
\end{align*}
By the induction hypothesis applied to $A$,
\begin{align*}
\mu((T^k)^{-1}(A))=\mu(A).
\end{align*}
Therefore
\begin{align*}
\mu((T^{k+1})^{-1}(A))=\mu(A).
\end{align*}
Since $A\in\mathcal E$ was arbitrary, $T^{k+1}$ is measure-preserving.
[guided]
Assume that $T^k:(E,\mathcal E)\to(E,\mathcal E)$ is measurable and measure-preserving for some $k\in\mathbb N\cup\{0\}$. We prove that the next iterate $T^{k+1}=T^k\circ T$ is also measurable and measure-preserving.
To prove measurability, let $A\in\mathcal E$ and let $x\in E$ be arbitrary. By the definition of preimage, we have
\begin{align*}
x\in (T^{k+1})^{-1}(A)
\end{align*}
if and only if
\begin{align*}
T^{k+1}(x)\in A.
\end{align*}
Since $T^{k+1}=T^k\circ T$, this is equivalent to
\begin{align*}
T^k(T(x))\in A.
\end{align*}
That condition is equivalent to
\begin{align*}
T(x)\in (T^k)^{-1}(A),
\end{align*}
which is equivalent to
\begin{align*}
x\in T^{-1}((T^k)^{-1}(A)).
\end{align*}
Therefore
\begin{align*}
(T^{k+1})^{-1}(A)=T^{-1}((T^k)^{-1}(A)).
\end{align*}
Because $A\in\mathcal E$ and $T^k$ is measurable, the set $(T^k)^{-1}(A)$ belongs to $\mathcal E$. Because $T$ is measurable, its preimage under $T$ also belongs to $\mathcal E$. Hence $(T^{k+1})^{-1}(A)\in\mathcal E$, so $T^{k+1}$ is measurable.
To prove measure preservation, let $A\in\mathcal E$. The identity above gives
\begin{align*}
\mu((T^{k+1})^{-1}(A))=\mu(T^{-1}((T^k)^{-1}(A))).
\end{align*}
Again $(T^k)^{-1}(A)\in\mathcal E$, so the measure-preservation hypothesis for $T$ applies and yields
\begin{align*}
\mu(T^{-1}((T^k)^{-1}(A)))=\mu((T^k)^{-1}(A)).
\end{align*}
The induction hypothesis applied to $A$ gives
\begin{align*}
\mu((T^k)^{-1}(A))=\mu(A).
\end{align*}
Combining these equalities yields
\begin{align*}
\mu((T^{k+1})^{-1}(A))=\mu(A).
\end{align*}
Since $A$ was arbitrary, $T^{k+1}$ is measure-preserving.
[/guided]
[/step]
[step:Conclude the result for all iterates by induction]
The base case establishes the claim for $k=0$, and the induction step shows that validity for a fixed $k\in\mathbb N\cup\{0\}$ implies validity for $k+1$. By induction, for every $k\in\mathbb N\cup\{0\}$, the iterate $T^k:(E,\mathcal E)\to(E,\mathcal E)$ is measurable and satisfies $\mu((T^k)^{-1}(A))=\mu(A)$ for every $A\in\mathcal E$. This is exactly the statement that every iterate $T^k$ is measure-preserving.
[/step]
