Let $\Sigma$ be a $\sigma$-algebra in $X$, let $Y$ be a topological space, and let $f$ map $X$ into $Y$. Theorem 1.12 of Rudin's Real and Complex Analysis says
If $f$ is measurable and $E$ is a Borel set in $Y$, then $f^{-1}(E) \in \Sigma$.
In the preceding part we showed that the collection $\Omega$ of sets $E$ in $Y$ whose pre-images are in $\Sigma$ is a $\sigma$-algebra in $Y$ (this is a set-theoretic result regardless of conditions on $f$). In the proof of this part, Rudin says that if $f$ is measurable, then $\Omega$ contains all the open sets in $Y$, and so must contain the Borel sets by the minimality of the Borel $\sigma$-algebra.
I understand every sentence except “if $f$ is measurable, then $\Omega$ contains all the open sets in $Y$.'' I thought measurability means that the pre-images of open sets are measurable. That doesn't mean that the images of measurable sets are open, right?
Best Answer
The definiton of $\Omega$ is
$$\Omega = \{A \subseteq Y: f^{-1}[A] \in \Sigma\}$$
If $f$ is measurable, and $O$ is open, $f^{-1}[O]$ is in $\Sigma$ by the definition of measurability. So by this definition, this means $O \in \Sigma$. This is what Rudin was saying, and it's true.
Now, you have to convince yourself that because $\Sigma$ is a $\sigma$-algebra, so is $\Omega$. Now $\Omega$ is some $\sigma$-algebra on $Y$ that contains all open sets as we saw. And $\text{Bor}(Y)$ is by definition the smallest such $\sigma$-algebra so
$$\text{Bor}(Y) \subseteq \Omega$$
which is exactly the conclusion of the theorem by the definition of $\Omega$ again.