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Fixed error in comment about rho
Joel David Hamkins
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I think this is a very interesting question.

In response to your comment, let me argue that if 1 holds and the measure is additive, then the singleton values are all the same. This is the sense in which the strong form of 2 follows from the weak form of 2.

To see this, following Sean's comment, observe that $\mu (\{a\})+\mu((a,b])=\mu([a,b])=\mu([a,b))+\mu(\{b\})$, and so $\mu(\{a\})=\mu(\{b\})$. So all singletons must have the same measure, and so the strong form of 2 follows from the weak form of 2.

In particular, the proposed function $\rho$ in your comment to the question does not exhibit the desired properties, in light of the decomposition $[0,\frac{1}{2}]=\{0\}\cup(0,\frac12]=[0,\frac12)\cup\{\frac12\}$.

Joel David Hamkins
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