I've always taken for granted that if $(X, \tau)$ and $(Y, \tau^{'})$ are topological spaces with Borel sigma algebras $\sigma$ and $\sigma^{'}$, repectively, then if $f:X\rightarrow Y$ is continuous, it is Borel measurable. The proof seems straightforward:

Let $Z=\{A \in Y \mid f^{-1}(A) \in \sigma \}. $

Then, by continuity it holds that $\tau^{'} \subseteq Z$, and once can check that $Z$ is a sigma-algebra. Because $Z$ is itself a sigma-algebra, it holds that $\sigma^{'} \subseteq Z$. Therefore, for any Borel measurable set $B \in \sigma^{'}$, $f^{-1}(B) \in \sigma$, so that $f$ is Borel measurable.

However, there seems to be some posts concerned with whether the topology is Hausdorff or not (see https://mathoverflow.net/questions/181752/is-every-continuous-function-measurable and Is every continuous function measurable? )

For the life of me, given the proof above, I cannot see why this would be a concern. Can anybody give me any points of reference here? Thanks!

  • $\begingroup$ If you define a Borel sigma algebra as a sigma algebra which is generated by the open sets then every continuous function is measurable. In these posts you can read that they are defining Borel sigma algebra in a different way in non Hausdorff spaces. $\endgroup$ – Mark May 10 '19 at 18:22
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    $\begingroup$ Borel means generated by open sets. Continuous means preimage of open set is open. $\endgroup$ – why May 10 '19 at 18:23

Define $\mathcal{S} = \{A \subseteq Y: f^{-1}[A]\in \textrm{Bor}(X)\}$ (note the notational improvements) and note that $\mathcal{S}$ is a $\sigma$-algebra on $Y$ as $f^{-1}[\cdot]$ preserves set operations and $\textrm{Bor}(X)$ is a $\sigma$-algebra.

Indeed $\tau' \subseteq \mathcal{S}$ by continuity of $f$ and so by minimality of the Borel $\sigma$-algebra, we have $\textrm{Bor}(Y) \subseteq \mathcal{S}$ which just says that $f^{-1}[B]$ is Borel in $X$ for any Borel subset $B$ of $Y$, ergo $f$ is $(\textrm{Bor}(X),\textrm{Bor}(Y)$)-measurable.

So far, so good, and this proof is valid if we define the Borel sets of $X$ as the smallest $\sigma$-algebra containing the open sets. The linked answers you provided however define Borel sets in a different way, namely the smallest $\sigma$-algebra containing both all open sets and all compact sets that are intersections of open sets (which are all compact sets in a $T_1$ space).

Of course this invalidates the proof as given. I'm no expert on this, but the other questions imply that this is relevant in topologies induced on a poset (Scott topologies etc.). In most cases you will not encounter this; in analysis (where most of measure theory is done) all spaces will be Hausdorff and the modified definition becomes irrelevant.

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