Let X be locally compact and Hausdorff, and let $f:X\rightarrow\mathbb R$ be a function. Suppose that for all finite regular (positive) Borel measures $\mu$, we know that $f$ is $\mu$-measurable. Does it follow that $f$ is Borel? If not, what's a good counter-example?


Definitions: The Borel sigma-algebra is generated by the open sets. So $f$ is Borel if $f^{-1}(U)$ is Borel for each open $U\subseteq\mathbb R$.


I think the definition of $\mu$-measurable is that for each open $U\subseteq\mathbb R$, we have that $f^{-1}(U)$ is in the completed sigma-algebra for $\mu$. That is, we can find Borel sets $A$ and $B$ with $A \subseteq f^{-1}(U) \subseteq B$ with $\mu(B\setminus A)=0$.


Remark: The "obvious" measures are the point mass measures, but then the completed sigma-algebra is $2^X$, so all functions are measurable! So the question is, in some sense, whether $X$ supports "enough" finite regular Borel measures.


Vague motivation: This old paper of Barry Johnson: Separate continuity and measurability.
Proc. Amer. Math. Soc. 20 1969 420--422 see <http://www.jstor.org/stable/2035668> But Barry's paper clearly gives me enough for what I want, so really this question is out of curiosity, not an attempt to understand the paper better!



Borel vs measure for all Borel measures
Let $f: \mathbb{R} \to \mathbb{R}$ be the characteristic function of a subset $A \subseteq \mathbb{R}$ which is analytic but not Borel. Then $f$ is universally measurable but not Borel.


