[proofplan]
We compare $\mu$ with the probability measure obtained by assigning to each measurable set $A\in\mathcal E$ the value $\mu(T^{-1}(A))$. The observable invariance assumption says that every [test function](/page/Test%20Function) in the measure-determining class has the same integral against these two probability measures, once the defining integral identity for this image measure is used. The measure-determining property then forces the two measures to be equal, and this equality is exactly probability preservation for the measurable self-map $T$.
[/proofplan]
[step:Form the pushforward probability measure induced by $T$]
Define the measure $\nu$ on $(E,\mathcal E)$ by
\begin{align*}
\nu(A):=\mu(T^{-1}(A))
\end{align*}
for every $A\in\mathcal E$. Since $T:E\to E$ is $\mathcal E/\mathcal E$-measurable, $T^{-1}(A)\in\mathcal E$ for every $A\in\mathcal E$, so $\nu$ is well-defined. Countable additivity follows from countable additivity of $\mu$ and preservation of countable disjoint unions under preimage. Moreover,
\begin{align*}
\nu(E)=\mu(T^{-1}(E))=\mu(E)=1,
\end{align*}
so $\nu$ is a probability measure on $(E,\mathcal E)$. By definition, $\nu=T_\#\mu$.
[/step]
[step:Use the pushforward identity to transfer observable invariance to $\nu$]
Fix $f\in\mathcal C$. Then $f:E\to\mathbb R$ is bounded and $\mathcal E/\mathcal B(\mathbb R)$-measurable, and $f\circ T:E\to\mathbb R$ is $\mathcal E/\mathcal B(\mathbb R)$-measurable because $T$ and $f$ are measurable. By the defining integral identity for the pushforward measure $T_\#\mu=\nu$,
\begin{align*}
\int_E f\,d\nu(x)=\int_E f\circ T\,d\mu(x).
\end{align*}
The hypothesis gives
\begin{align*}
\int_E f\circ T\,d\mu(x)=\int_E f\,d\mu(x).
\end{align*}
Therefore
\begin{align*}
\int_E f\,d\nu(x)=\int_E f\,d\mu(x).
\end{align*}
This holds for every $f\in\mathcal C$.
[guided]
Fix an arbitrary test observable $f\in\mathcal C$. The point of introducing $\nu=T_\#\mu$ is that integrals against $\nu$ are exactly integrals of pullbacks along $T$ against $\mu$. Since $f:E\to\mathbb R$ is bounded and $\mathcal E/\mathcal B(\mathbb R)$-measurable, and since $T:E\to E$ is $\mathcal E/\mathcal E$-measurable, the composition
\begin{align*}
f\circ T:E\to\mathbb R
\end{align*}
is $\mathcal E/\mathcal B(\mathbb R)$-measurable and bounded. Thus all integrals below are finite [real numbers](/page/Real%20Numbers).
By the defining integral identity for the pushforward measure $\nu=T_\#\mu$,
\begin{align*}
\int_E f\,d\nu(x)=\int_E f\circ T\,d\mu(x).
\end{align*}
The theorem hypothesis says that this pulled-back observable has the same $\mu$-integral as the original observable:
\begin{align*}
\int_E f\circ T\,d\mu(x)=\int_E f\,d\mu(x).
\end{align*}
Combining the two equalities gives
\begin{align*}
\int_E f\,d\nu(x)=\int_E f\,d\mu(x).
\end{align*}
Because $f\in\mathcal C$ was arbitrary, the same equality holds for every observable in the class $\mathcal C$. This is exactly the input needed for the measure-determining property.
[/guided]
[/step]
[step:Apply the measure-determining property to identify the measures]
Both $\nu$ and $\mu$ are probability measures on $(E,\mathcal E)$. The preceding step proves that
\begin{align*}
\int_E f\,d\nu(x)=\int_E f\,d\mu(x)
\end{align*}
for every $f\in\mathcal C$. By the measure-determining property of $\mathcal C$, applied with $\rho=\nu$ and $\eta=\mu$, we obtain
\begin{align*}
\nu=\mu.
\end{align*}
Since $\nu=T_\#\mu$, this says
\begin{align*}
T_\#\mu=\mu.
\end{align*}
[/step]
[step:Translate equality of pushforwards into probability preservation]
For every $A\in\mathcal E$, the definition of $\nu=T_\#\mu$ gives
\begin{align*}
(T_\#\mu)(A)=\mu(T^{-1}(A)).
\end{align*}
Since $T_\#\mu=\mu$, it follows that
\begin{align*}
\mu(T^{-1}(A))=\mu(A)
\end{align*}
for every $A\in\mathcal E$. Hence $T$ is probability-preserving.
[/step]
