Mapping Measurable iff Measurable on Generator

Theorem
Let $\left({X, \Sigma}\right)$ and $\left({X', \Sigma'}\right)$ be measurable spaces.

Suppose that $\Sigma'$ is generated by $\mathcal{G}'$.

Then a mapping $f: X \to X'$ is $\Sigma \, / \, \Sigma'$-measurable iff:


 * $\forall G' \in \mathcal{G}': f^{-1} \left({G'}\right) \in \Sigma$

That is, iff the preimage of every generator under $f$ is a measurable set.

Necessary Condition
Let $f$ be $\Sigma \, / \, \Sigma'$-measurable.

By definition of generated $\sigma$-algebra $\mathcal{G}' \subseteq \Sigma'$.

Hence, in particular, $f$ satisfies:


 * $\forall G' \in \mathcal{G}': f^{-1} \left({G'}\right) \in \Sigma$

Sufficient Condition
Suppose that:


 * $\forall G' \in \mathcal{G}': f^{-1} \left({G'}\right) \in \Sigma$

Consider the pre-image $\sigma$-algebra $\Sigma''$ on $X'$.

The supposition precisely states that $\mathcal{G}' \subseteq \Sigma''$.

By Generated Sigma-Algebras Respect Subset, $\sigma \left({\mathcal{G}'}\right) \subseteq \sigma \left({\Sigma''}\right)$.

Now we know that $\sigma \left({\mathcal{G}'}\right) = \Sigma'$, and $\sigma \left({\Sigma}\right) = \Sigma$.

By definition of $\Sigma''$, this precisely means:


 * $\forall E' \in \Sigma': f^{-1} \left({E'}\right) \in \Sigma$

That is, $f$ is $\Sigma \, / \, \Sigma'$-measurable.