Pre-Image Sigma-Algebra on Domain 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 $X'$.

Then $f^{-1} \left({\Sigma'}\right)$, the pre-image $\sigma$-algebra on the domain of $f$, 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 $X \in f^{-1} \left({\Sigma'}\right)$.

Axiom $(2)$
Let $S = f^{-1} \left({E'}\right) \in f^{-1} \left({\Sigma'}\right)$.

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


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

As $X' \setminus E' \in \Sigma'$, this means $X \setminus S \in f^{-1} \left({\Sigma'}\right)$.

Axiom $(3)$
Let $\left({S_i}\right)_{i \in \N}$ be a sequence in $f^{-1} \left({\Sigma'}\right)$.

Let, for all $i \in \N$, $E'_i \in \Sigma'$ such that $S_i = f^{-1} \left({E'}\right)$.

By Mapping Preimage of Union: General Result, have:


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

Now $\displaystyle \bigcup_{i \mathop \in \N} E'_i \in \Sigma'$ as $\Sigma'$ is a $\sigma$-algebra.

Hence $\displaystyle \bigcup_{i \mathop \in \N} S_i \in f^{-1} \left({\Sigma'}\right)$.