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As part of an exercise in a graduate measure theory course, I need to prove that if a function $f: \mathbb{R}\rightarrow \mathbb{R}$ is continuous and $B$ is a Borel set in $\mathbb{R}$, then $f^{-1}(B)$ is also a Borel set. If $B$ can somehow be expressed using countable unions and intersections of open and closed sets, this proof would be (relatively) easy.

Here is my intuition for why this is true. Let $\mathcal{F}$ be the set of countable unions of countable intersections of closed and open sets. Let $\mathcal{G}$ be the set of countable intersections of countable unions of closed and open sets. I claim that $\mathcal{B} = \mathcal{F}\cup\mathcal{G}$.

Note $ \mathcal{F}\cup\mathcal{G}\subset \mathcal{B}$ because $\mathcal{B}$ is a $\sigma$-algebra. If $ \mathcal{F}\cup\mathcal{G}$ is a $\sigma$-algebra, then $\mathcal{B} \subset \mathcal{F}\cup\mathcal{G}$ and $\mathcal{B} = \mathcal{F}\cup\mathcal{G}$. Thus, its sufficient to show that $ \mathcal{F}\cup\mathcal{G}$ is a $\sigma$-algebra.

Take an arbitrary set $S \in \mathcal{F}\cup\mathcal{G}.$ If $S \in \mathcal{F},$ De Morgan's law implies that $S^c \in \mathcal{G}.$ Similarly, if $S \in \mathcal{G}$, De Morgan's law implies that $S^c \in \mathcal{F}$. Thus, $S^c \in \mathcal{F} \cup\mathcal{G}.$

My intuition tells me that the countable union of sets in $\mathcal{F}\cup\mathcal{G}$ is also in $\mathcal{F}\cup\mathcal{G}$, but I can't think of a rigorous argument. Is this true, or is this where my intuition failed me?

Peachy
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    This is basically never the way you want to approach $\sigma$ algebras. Call $h$ measurable if $h^{-1}(U)\in \mathcal F$ for every open set. Show that this condition implies that $h^{-1}(B)\in\mathcal F$ for every open set (hint: ${E:h^{-1}(E)\in \mathcal F}$ is a $\sigma$-algebra.). It is evident now, I hope, that continuity implies measurability. – Andrew Oct 08 '24 at 04:14
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    See also here: https://math.stackexchange.com/a/2854237/151552 – PhoemueX Oct 08 '24 at 04:58
  • @PhoemueX That thread is great! But it is also a bit silly. It's much easier to just put $f(x) = x/a$, and note that $f^{-1}(A)=aA$. – Andrew Oct 08 '24 at 05:48
  • @Andrew: Indeed. But it illustrates the "good set principle" proof technique, which was what I was looking for in this case. – PhoemueX Oct 08 '24 at 06:52
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    Your $\mathcal{F}\cup\mathcal{G}$ doesn't even reach all of the sets at either of the 3rd Borel levels ${\mathcal F}{\sigma \delta \sigma}$ and ${\mathcal G}{\delta \sigma \delta}.$ See pp. 115-117 in A Course on Borel Sets by S. M. Srivastava (1998). Also, Robin Pemantle's notes Transfinite Induction includes useful exposition on the hierarchy of Borel sets. – Dave L. Renfro Oct 08 '24 at 11:28

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