# Is every continuous function measurable?

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, for a continuous function $f : X \to Y$ between topological spaces, is $f$ necessarily measurable?

This question is equivalent to the following. If we write $\uparrow y$ for the intersection of all open subsets containing $y$, which happens to be compact saturated, is it true that $f^{-1}(\uparrow y)$ is measurable for all $y \in Y$?

Thank you very much for your help. Paul

Edit: this question has now been migrated to MathOverflow, see here.

• Inverse image under a continuous map preserves openness (definitionally) and inverse image under a function preserves arbitrary intersections, so the only remaining criterion is properness (inverse image of a compact set being compact). Given that continuous functions aren't usually proper, I expect this not to be the case. – jdc May 24 '14 at 20:57
• Dear jdc, since indeed the inverse image of a compact by a compact needs not to be compact, I do not expect this to happen. Without being compact, the inverse image of a compact saturated subset could still be measurable. – polmath May 25 '14 at 10:18
• Not to be nitpicky, but migration is when the question is moved. What you did is to crosspost the question on MathOverflow. – Asaf Karagila Sep 26 '14 at 12:58
• @AsafKaragila: thanks. Indeed I asked for "official" migration" but the post was too old for that; talking about "crossposting" is certainly more correct. – polmath Sep 26 '14 at 13:19

Let $X$ be a topological space with a compact, saturated, non-open set $S$ and $Z$ a space that is not itself the union of countably many compact subsets. Consider the projection $X \times Z \to X$. I suspect the inverse image $S \times Z$ will not behave as hoped. I think it would be as you wanted under the product Borel algebra but not under that of the Borel algebra associated to the product topology.

Or I'm missing something.

I would like to complete my question with the following elements.

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? (sufficient conditions on $X$, $Y$ or $f$ in order for $f$ to be measurable?)