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:
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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.
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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.
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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?
Thank you for your help.
Best Answer
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.