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 $X'$.

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

As $f$ is a mapping, it is immediate that $f^{-1} \sqbrk {X'} = X$.

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

Hence $X' \in \Sigma'$.

$\Box$

Let $S' \in \Sigma'$.

$f^{-1} \sqbrk {X' \setminus S'} = f^{-1} \sqbrk {X'} \setminus f^{-1} \sqbrk {S'}$

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

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

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

$\Box$

Let $\sequence {S'_i}_{i \mathop \in \N}$ be a sequence in $\Sigma'$.

$\ds f^{-1} \sqbrk {\bigcup_{i \mathop \in \N} S'_i} = \bigcup_{i \mathop \in \N} f^{-1} \sqbrk {S'_i}$

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

Now $\ds \bigcup_{i \mathop \in \N} f^{-1} \sqbrk {S'_i} \in \Sigma$ as $\Sigma$ is a $\sigma$-algebra.

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

$\blacksquare$