# Are continuous functions measurable with respect to abstract Borel sigma algebras?

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!

• 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. – Mark May 10 '19 at 18:22
• Borel means generated by open sets. Continuous means preimage of open set is open. – 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.