The theory of Radon measures relies a lot on the hypothesis that compact subsets of a topological space are Borel (i.e., in the $\sigma$-algebra generated by the open sets). This is an okay assumption in Hausdorff spaces (where the bulk of the introductory theory takes place) because all compact subsets are closed and hence Borel. However, this answer remarks that there ARE topological spaces containing non-Borel compact sets.
To flesh out the linked answer, let $X$ be any set containing more than one point. Then let $\tau \subseteq \mathscr{P}(X)$ be the trivial topology (e.g., $\tau = \{\varnothing, X\}$). Then the Borel $\sigma$-algebra is then $\mathcal{B}_X = \tau = \{\varnothing,X\}$. Any singleton $\{x\} \subseteq X$ is then clearly not Borel, yet it is compact since all singletons are so (in fact, all subsets in the trivial topology are trivially compact).
My question is this: What conditions must be placed on the topology to ensure compact subsets are or aren't Borel? We have the obvious extremes above. However, I am aware that "Hausdorff" is not a necessary condition. For an example from algebraic geometry, the Zariski topology on the prime spectrum of a ring is $T_0$ yet has the peculiar property that the basic open sets are compact. To be more specific, every subset of $\text{Spec}(\mathbb{Z})$ is Borel (every nonzero point is closed, and the whole space is countable, so zero is Borel too).
This gives us a few avenues of attack:
(1). Are all compact subsets of a $T_0$ topological space Borel? Or,
(2). Is there an example of a $T_0$ space having a non-Borel compact subset? (Note the trivial topology is $T_0$ iff it's also discrete.)
(3). Same questions, but for $T_1$ spaces instead.
Any insight is appreciated!
The first thing that occurred to me is to change the topology (and hence the meaning of compact) without changing the Borel $\sigma$-algebra. This leads to an example showing that $T_0$ is not enough.
Consider the real line with the non-trivial open subsets of the form $(-\infty,a)$ (= the left order topology). This generates the same Borel $\sigma$-algebra as the usual topology, because we can write the intervals $(-\infty,a]$ as countable intersections of open sets etc.
Let $S$ be a non Lebesgue measurable subset of $[0,1]$ (a Vitali set). Then $S\cup\{1\}$ is still non-measurable, and hence non-Borel. But it becomes compact in the left order topology.
The construction of Vitali sets relies (as far as I know) on the axiom of choice. I don't know what happens if a different set theory is used.