Pre-Image Sigma-Algebra on Domain is Generated by Mapping

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Theorem

Let $X, X'$ be sets, and let $f: X \to X'$ be a mapping.

Let $\Sigma'$ be a $\sigma$-algebra on $X'$.

Let $f: X \to X'$ be a mapping.


Then:

$\sigma \left({f}\right) = f^{-1} \left({\Sigma'}\right)$

where

$\sigma \left({f}\right)$ denotes the $\sigma$-algebra generated by $f$
$f^{-1} \left({\Sigma'}\right)$ denotes the pre-image $\sigma$-algebra under $f$


Proof

By Characterization of Sigma-Algebra Generated by Collection of Mappings:

$\sigma \left({f}\right) = \sigma \left({f^{-1} \left({\Sigma'}\right)}\right)$

where the latter $\sigma$ denotes a $\sigma$-algebra generated by a collection of subsets.

By Pre-Image Sigma-Algebra on Domain is Sigma-Algebra, $f^{-1} \left({\Sigma'}\right)$ is a $\sigma$-algebra.

Hence $\sigma \left({f^{-1} \left({\Sigma'}\right)}\right) = f^{-1} \left({\Sigma'}\right)$.

$\blacksquare$


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