Is every continuous function measurable?

This question has already been asked on Math StackExchange here, but was too old to be migrated, and I think will be more appropriate to MathOverflow.

In non-Hausdorff topology it is standard to define the Borel algebra of a topological space $X$ as the $\sigma$-algebra generated by the open subsets and the compact saturated subsets. Recall that a subset is saturated if it is an intersection of open subsets, and that compact saturated subsets play the role of compact subsets when the space $X$ is not $T_1$ (which is typically the case for a partially ordered set equipped with the Scott topology for instance).

In this situation, one may ask whether every continuous function $f : X \to Y$ between topological spaces is measurable, or equivalently whether every continuous function $f : X \to Y$ is such that $f^{-1}(\uparrow y)$ is measurable for all $y \in Y$, where I write $\uparrow y$ for the intersection of all open subsets containing $y$, which happens to be compact saturated.

I do not expect this to be true, but I am rather looking for sufficient conditions on $X$, $Y$ or $f$ in order for $f$ to be measurable. For instance:

• If $Y$ is $T_1$, then every subset $\uparrow y$ is closed (it coincides with the singleton $\{ y \}$ which itself coincides with its closure), so the continuous map $f : X \to Y$ is measurable.

• If $Y$ is first-countable, then every subset $\uparrow y$ can be written as a countable intersection of open subsets, so again $f$ is measurable.

• If $f$ is open and bijective, one can show that the inverse image of $\uparrow y$ is of the form $\uparrow x$, $x \in X$, so $f$ is measurable.

Do we know other such situations?

• Well the last example does not tell much as every open bijection is actually a homeomorphism... Oct 11, 2014 at 22:42
• True, thanks, I did not pay attention to this. Oct 12, 2014 at 7:47
• For finding further conditions, it might be helpful to see an example continuous function that is not measurable. Oct 13, 2014 at 12:14

If $X, Y$ are topological spaces such that for every continuous map $f: X\to Y$ and any $K\subseteq Y$ compact, $f^{-1}(K)$ is a compact of subset of $X$, then every continuous function is measurable.
• dominiczypen: if you read carefully the question above, I am not considering the $\sigma$-algebra generated by the topology (i.e. by the open sets), but rather the one generated by both the open sets and the compact saturated sets. Oct 9, 2014 at 10:44