Pre-Image Sigma-Algebra on Codomain is Sigma-Algebra

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

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

Denote with $\Sigma'$ the pre-image $\sigma$-algebra on the domain of $f$.

Then $\Sigma'$ is a $\sigma$-algebra on $X'$.

Proof
Verify the axioms for a $\sigma$-algebra in turn:

Axiom $(1)$
As $f$ is a mapping, it is immediate that $f^{-1} \left({X'}\right) = X$.

Also $X \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra.

Hence $X' \in \Sigma'$.

Axiom $(2)$
Let $S' \in \Sigma'$.

By Mapping Preimage of Set Difference and $f^{-1} \left({X'}\right) = X$, have:


 * $f^{-1} \left({X' \setminus S'}\right) = f^{-1} \left({X'}\right) \setminus f^{-1} \left({S'}\right)$

Now $f^{-1} \left({X'}\right) = X \in \Sigma$, and $f^{-1} \left({S'}\right) \in \Sigma$ was already assumed.

As $\Sigma$ is a $\sigma$-algebra, it follows that $f^{-1} \left({X' \setminus S'}\right) \in \Sigma$ as well.

Hence $X' \setminus S' \in \Sigma'$.

Axiom $(3)$
Let $\left({S'_i}\right)_{i \in \N}$ be a sequence in $\Sigma'$.

By Mapping Preimage of Union: General Result, have:


 * $\displaystyle f^{-1} \left({\bigcup_{i \in \N} S'_i}\right) = \bigcup_{i \in \N} f^{-1} \left({S'_i}\right)$

By assumption, $S'_i \in \Sigma'$ for all $i \in \N$; hence $f^{-1} \left({S'_i}\right) \in \Sigma$ for all $i \in \N$.

Now $\displaystyle \bigcup_{i \in \N} f^{-1} \left({S'_i}\right) \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra.

Hence $\displaystyle \bigcup_{i \in \N} S'_i \in \Sigma'$.