Mapping Measurable iff Measurable on Generator

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Theorem

Let $\struct {X, \Sigma}$ and $\struct {X', \Sigma'}$ be measurable spaces.

Suppose that $\Sigma'$ is generated by $\GG'$.


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

$\forall G' \in \GG': \map {f^{-1} } {G'} \in \Sigma$

That is, if and only if the preimage of every generator under $f$ is a measurable set.


Proof

Necessary Condition

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

By definition of generated $\sigma$-algebra $\GG' \subseteq \Sigma'$.

Hence, in particular, $f$ satisfies:

$\forall G' \in \GG': \map {f^{-1} } {G'} \in \Sigma$

$\Box$


Sufficient Condition

Suppose that:

$\forall G' \in \GG': \map {f^{-1} } {G'} \in \Sigma$

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

The supposition precisely states that $\GG' \subseteq \Sigma''$.

By definition of generated $\sigma$-algebra, $\map \sigma {\GG'} \subseteq \Sigma''$.


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

$\forall E' \in \Sigma': \map {f^{-1} } {E'} \in \Sigma$


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

$\blacksquare$


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