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 $f^{-1} \sqbrk {\Sigma'}$, the pre-image $\sigma$-algebra on the domain of $f$, 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 $X \in f^{-1} \sqbrk {\Sigma'}$.

$\Box$

Let $S = f^{-1} \sqbrk {E'} \in f^{-1} \sqbrk {\Sigma'}$.

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

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

$\Box$

Let $\sequence {S_i}_{i \mathop \in \N}$ be a sequence in $f^{-1} \sqbrk {\Sigma'}$.

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

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

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

Hence $\ds \bigcup_{i \mathop \in \N} S_i \in f^{-1} \sqbrk {\Sigma'}$.

$\blacksquare$