A question about Borel sets on the unit interval

It is known that each non-decreasing continuous function $\phi$ induces a $\sigma$-additive measure $d\phi$ such that $\int_0^1 f(x) d\phi(x)$ exists for every bounded real-valued Baire function $f$. This follows because every Borel set is measurable with respect to $d\phi(x)$, and for every Baire subset $A \subseteq [0,1]$ the characteristic function $\chi_A$ is a Baire function.

Conversely, suppose now that $A \subseteq [0,1]$ is such that $A$ is $d\phi$-measurable for every $\phi$ as above. Does it follow that $A$ is a Borel set?

(Question formulated by Prof. Jan-Erik Björk at Stockholm University)

• In your first paragraph, didn't you mean to say "every Baire subset $A$"? (Or equivalently "Borel")? – Nate Eldredge Feb 5 '16 at 14:33
• Yes, thanks. And thanks for the reference in your answer. – godelian Feb 5 '16 at 15:04

Unless I'm missing something, your measures $d\phi$ are precisely the atomless finite Borel measures (equivalently, Baire measures) on $[0,1]$. Then your condition on $A$ is that it is universally measurable. In that case, the answer to your question is No: there are universally measurable sets which are not Borel.