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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)

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

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 set is universally measurable, as well as a construction of an analytic set that is not Borel.

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