Measure Invariant on Generator is Invariant

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Theorem

Let $\left({X, \Sigma, \mu}\right)$ be a $\sigma$-finite measure space.

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


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

Also, let $\mathcal G$ satisfy the following:

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


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

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


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


Proof

Consider the pushforward measure $\theta_* \mu$ on $\left({X, \Sigma}\right)$.

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

$\theta_* \mu \restriction_{\mathcal G} = \mu \restriction_{\mathcal G}$

where $\restriction$ denotes restriction.


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

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

$\blacksquare$


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