A Function is Measureable iff Preimages of Generating Sets are in the Source Sigma-Algebra

Last updated Nov 1, 2022

Let $(X, \mathcal{M})$, $(Y, \mathcal{N})$ be Measure Spaces and let $f: X \to Y$. Suppose $\mathcal{N} = \sigma(\mathcal{E})$ for some $\mathcal{E} \subset \mathcal{P}(Y)$ (i.e. $\mathcal{N}$ is generated by $\mathcal{E}$). Then $f$ is $(\mathcal{M}, \mathcal{N})$-measureable If and Only If $f^{-1}(\mathcal{E}) \subset \mathcal{M}$.

$(\Rightarrow)$: If $f$ is $(\mathcal{M}, \mathcal{N})$-measureable, then, by definition, $$f^{-1}(\mathcal{E}) \subset f^{-1}(\mathcal{N}) \subset \mathcal{M}$$ $(\Leftarrow)$: Suppose $f^{-1}(\mathcal{E}) \subset \mathcal{M}$. Then $\sigma^{}(f) \supset \mathcal{E}$. Therefore, by definition of Sigma Algebra Generated by Set, we have that $\sigma^{}(f) \supset \mathcal{N} = \sigma(\mathcal{E})$. Then $f^{-1}(\mathcal{N}) \subset f^{-1}(\sigma^{*}(f)) \subset \mathcal{M}$ and $f$ is $(\mathcal{M}, \mathcal{N})$-measureable. $\blacksquare$

1. $f^{-1}(\mathcal{E}) := {f^{-1}(E) : E \in \mathcal{E}}$.

• Axiom Schema of Specification
• Sigma Algebra produced by a Function