= Solution
If $\theta$ is measure-preserving, equality holds on every measurable set and hence on the specified $\pi$-system. Conversely, let $\mathcal D=\{A\in\mathcal E:\mu_\theta(A)=\mu(A)\}$. Both finite <measures> have equal total mass. Therefore $E\in\mathcal D$, complements of members remain in $\mathcal D$ by subtracting from the finite total mass, and disjoint countable <unions> remain in $\mathcal D$ by <countable additivity>. Thus $\mathcal D$ is a <Dynkin system>.
It contains the generating <pi-system> $\mathcal A$, so <Dynkin lemma> gives $\mathcal E=\sigma(\mathcal A)\subseteq\mathcal D$. Consequently
$$
\boxed{\mu(\theta^{-1}(A))=\mu(A)\quad\text{for every }A\in\mathcal E.}
$$
This is exactly the measure-preserving property.
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