Measure Invariant on Generator is Invariant

Theorem
Let $\struct {X, \Sigma, \mu}$ be a $\sigma$-finite measure space.

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

Suppose that $\Sigma$ is generated by $\GG \subseteq \powerset X$.

Also, let $\GG$ satisfy the following:


 * $(1): \quad \forall G, H \in \GG: G \cap H \in \GG$
 * $(2): \quad$ There exists an exhausting sequence $\sequence {G_n}_{n \mathop \in \N} \uparrow X$ in $\GG$ such that:
 * $\quad \forall n \in \N: \map \mu {G_n} < +\infty$

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


 * $(3): \quad \map \mu {\map {\theta^{-1} } G} = \map \mu G$

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

Proof
Consider the pushforward measure $\theta_* \mu$ on $\struct {X, \Sigma}$.

By definition, this makes equation $(3)$ come down to:


 * $\theta_* \mu \restriction_\GG = \mu \restriction_\GG$

where $\restriction$ denotes restriction.

The suppositions $(1)$, $(2)$ and $(3)$ together constitute precisely the prerequisites to Uniqueness of Measures.

Hence $\theta_* \mu = \mu$, that is, $\mu$ is $\theta$-invariant.