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)



A question about Borel sets on the unit interval
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](https://en.wikipedia.org/wiki/Universally_measurable_set). In that case, the answer to your question is No: there are universally measurable sets which are not Borel. 


For instance, in most any descriptive set theory text (such as Kechris's *Classical Descriptive Set Theory*), you should be able to find the result that every [analytic](https://en.wikipedia.org/wiki/Analytic_set) set is universally measurable, as well as a construction of an analytic set that is not Borel.


