Measure Invariant on Generator is Invariant

Theorem
Let $\left({X, \mathcal A, \mu}\right)$ be a measure space.

Let $\theta: X \to X$ be an $\mathcal A / \mathcal A$-measurable mapping.

Suppose that $\mathcal A$ is generated by $\mathcal G \subseteq \mathcal P \left({X}\right)$.

Suppose that, for all $G \in \mathcal G$, $\mu$ satisfies:


 * $\mu \left({\theta^{-1} \left({G}\right) }\right) = \mu \left({G}\right)$

Then $\mu$ is a $\theta$-invariant measure.