Mapping Measurable iff Measurable on Generator

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.

$\blacksquare$

Sources

• 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $7.2$