Topological complexity of ultrafilters in $2^\kappa$ for uncountable $\kappa$

Question

It is a well known fact that if $\mathcal{F}$ is a non-principal ultrafilter on $\omega$, then the set $\{ \alpha \in 2^\omega : \alpha \in \mathcal{F}\}$ (conflating binary strings with subsets of $\omega$) is not a Borel subset of $2^\omega$ with its standard product topology.

The proof of this that I am familiar with goes through showing that $\mathcal{F} \subseteq 2^\omega$ is not a measurable subset of $2^\omega$ by noting that if it were it would have density $\frac{1}{2}$ everywhere, contradicting the Lebesgue density theorem.

I am curious about the analogous statement with regards to ultrafilters on $\kappa$, considered as subsets of $2^\kappa$ with its compact product topology. I have difficulty imagining that a non-principal ultrafilter on $2^\kappa$ could be Borel (where by Borel I mean specifically an element of the $\sigma$-algebra generated by open sets, not just the $\sigma$-algebra generated by clopen sets), but I can't find a proof of this and the Lebesgue density theorem argument seems difficult to generalize to $2^\kappa$, even though there is a natural regular Borel measure on $2^\kappa$.

Answer 1

A Borel ultrafilter would have the property of Baire. Therefore either $\mathcal F$ or its complement $2^\kappa\setminus\mathcal F$ is comeager relative to some basic open set. Since $\mathcal F$ is invariant under finite changes, this would mean $\mathcal F$ or its complement is comeager. Since $\mathcal F$ is the image of $2^\kappa\setminus\mathcal F$ complement under a homeomorphism of $2^\kappa,$ this means both sets are comeager and must therefore intersect, which is impossible.

Fun exercise: show that a Borel map $2^\kappa\to[0,1]$ that vanishes on finite subsets of $\kappa$ cannot be a finitely additive probability measure.