Are the Borel sets precisely those whose measure is fixed by finite additivity?

Question

Let $\mu:\mathscr P(\mathbb R)\to[0,\infty]$ satisfy:

1. (Finite additivity) $A\cap B=\varnothing\implies$$\mu(A\cup B)=\mu(A)+\mu(B)$
2. (Interval property) $a<b\in\mathbb R\implies\mu\big((a,b)\big)=b-a$.

These conditions fix the measures of every interval (open or closed, infinite or finite), and of many other sets (e.g., finite unions of intervals). Are the sets whose measures are fixed by these properties precisely the Borel sets, or perhaps the Lebesgue sets?

To be more precise, define $\mu^-$ and $\mu^+$ as follows:

$$\mu^-(A)=\sup \left\{\left. \sum_{k=1}^n \mu(I_k)\; \right|\; I_k\text{ are disjoint intervals and }\bigcup_kI_k\subseteq A \right\}$$

$$\mu^+(A)=\inf \left\{\left. \sum_{k=1}^n \mu(I_k)\; \right|\; I_k\text{ are intervals and }A\subseteq \bigcup_kI_k \right\}.$$

Then $\mu^-(A)\leqslant\mu(A)\leqslant\mu^+(A).$ If $\mu^-(A)=\mu^+(A)$, then $\mu(A)$ is fixed. Let $\mathscr A=\{A\subseteq \mathbb R\;|\; \mu^-(A)=\mu^+(A)\}$. Is $\mathscr A$ the Borel or Lebesgue $\sigma$-algebra over $\mathbb R$?

Answer 1

Unless I'm missing something, you've hit upon Jordan measure. Jordan measure applies to a much narrower collection of sets even than the Borel sets, as the example of a fat Cantor set demonstrates; on the other hand, since regular Cantor sets are Jordan-measurable, it's a bit wider than the algebra generated by intervals. In particular, there are non-Borel sets which are Jordan-measurable (take any non-Borel subset of the Cantor set)!

I think the class of Jordan-measurable sets is rather complicated, at the end of the day. As evidence of this, note that even pinning down the $\sigma$-algebra generated by the Jordan-measurable sets is nontrivial.