A map, $f:(A,\mathcal{A})\rightarrow(F,\mathcal{F})$, is $\mathcal{A}/\mathcal{F}$ measurable if and only if for some generator $\mathcal{F}_0$ of $\mathcal{F}$ we have $\forall S\in\mathcal{F}_0[f^{-1}(S)\in\mathcal{A}]$

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Statement

A map from a $\sigma$-algebra $(A,\mathcal{A})$ to another $\sigma$-algebra $(F,\mathcal{F})$, $f:A\rightarrow F$, is $\mathcal{A}/\mathcal{F}$ measurable if and only if for some generator, $\mathcal{G}$, of $\mathcal{F}$ we have:

• $\forall S\in\mathcal{G}[f^{-1}(S)\in\mathcal{A}]$

Which we may alternatively write (for brevity, see: abuses of the implies-subset relation) as:

• $f^{-1}(\mathcal{G})\subseteq\mathcal{A}$

Proof

$\implies$: $f:A\rightarrow B$ is $\mathcal{A}/\mathcal{F}$-measurable $\implies$ for some generator $\mathcal{G}$ of $\mathcal{F}$ we have $\forall S\in\mathcal{G}[f^{-1}(S)\in\mathcal{A}]$

• Let $S\in\mathcal{G}$ be given
  • Note that $\mathcal{G}\subseteq\sigma(\mathcal{G})$, so by the implies-subset relation we see $S\in\mathcal{G}\implies S\in\sigma(\mathcal{G})$
  • By the definition of $\mathcal{A}/\mathcal{F}$-measurable:
    • $\forall S\in F[f^{-1}(S)\in\mathcal{A}]$
  • Thus $S\in\mathcal{G}\implies S\in\sigma(\mathcal{G})=\mathcal{F}$
    • But as we've just seen, if $S\in\mathcal{F}$ then $f^{-1}(S)\in\mathcal{A}$
• So $f^{-1}(S)\in\mathcal{A}$

This completes the proof

$\impliedby$:

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TODO: See ref page 6, also lemma 7.2 in

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