Pre-Image Sigma-Algebra of Generated Sigma-Algebra

Theorem
Let $f: X \to Y$ be a mapping.

Let $\mathcal G \subseteq \mathcal P \left({Y}\right)$ be a collection of subsets of $Y$.

Then the following equality of $\sigma$-algebras on $X$ holds:


 * $f^{-1} \left({\sigma \left({\mathcal G}\right)}\right) = \sigma \left({f^{-1} \left({\mathcal G}\right)}\right)$

where


 * $\sigma$ denotes a generated $\sigma$-algebra
 * $f^{-1} \left({\sigma \left({\mathcal G}\right)}\right)$ denotes the pre-image $\sigma$-algebra
 * $f^{-1} \left({\mathcal G}\right)$ is the preimage of $\mathcal G$ under $f$

Proof
Since $\mathcal G \subseteq \sigma \left({\mathcal G}\right)$, it follows immediately that:


 * $f^{-1} \left({\mathcal G}\right) \subseteq f^{-1} \left({\sigma \left({\mathcal G}\right)}\right)$

By Pre-Image Sigma-Algebra on Domain is Sigma-Algebra, the latter is a $\sigma$-algebra, and so by Generated Sigma-Algebra Preserves Subset, it follows that:


 * $\sigma \left({f^{-1} \left({\mathcal G}\right)}\right) \subseteq f^{-1} \left({\sigma \left({\mathcal G}\right)}\right)$

Conversely, by definition of generated $\sigma$-algebra, we have:


 * $f^{-1} \left({\mathcal G}\right) \subseteq \sigma \left({f^{-1} \left({\mathcal G}\right)}\right)$

Hence from Mapping Measurable iff Measurable on Generator, it follows that $f$ is $\sigma \left({f^{-1} \left({\mathcal G}\right)}\right) \, / \, \sigma \left({\mathcal G}\right)$-measurable.

But this by definition of measurable mapping means that:


 * $f^{-1} \left({\sigma \left({\mathcal G}\right)}\right) \subseteq \sigma \left({f^{-1} \left({\mathcal G}\right)}\right)$

Hence the result, by Equality of Sets.