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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 your comment, the point would be that $\rho$ is not additive, since $[0,1]=\{0\}\cup(0,1]=[0,1)\cup\{1\}$, but $\rho$ does not add properly on these unions.

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 $[0,\frac{1}{2}]=\{0\}\cup(0,\frac12]=[0,\frac12)\cup\{\frac12\}$.

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 that part of 2 follows from 1.

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 your comment, the point would be that $\rho$ is not additive, since $[0,1]=\{0\}\cup(0,1]=[0,1)\cup\{1\}$, but $\rho$ does not add properly on these unions.

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 your comment, the point would be that $\rho$ is not additive, since $[0,1]=\{0\}\cup(0,1]=[0,1)\cup\{1\}$, but $\rho$ does not add properly on these unions.

I think this is a very interesting question, which deserves more votes. (My vote is currently the only +1.)

But meanwhile, in response to your comment, let me argue that if 1 holds and all the singletons get infinitesimal value, then these values are all the same. This is the sense in which that part of 2 follows from 1.

To see this, suppose $\mu$ has property 1. Let $\epsilon=\mu(\{1\})$, a nonzero infinitesimal number. Observe that $\mu([0,1])=\mu([0,1)\cup\{1\})=1+\epsilon$. For any $x\in (0,1)$, let $\tau=\mu(\{x\})$, and observe that $[0,1]=[0,x)\cup\{x\}\cup(x,1]$, which has measure $x+\tau+(1-x)=1+\tau$. So $\tau=\epsilon$. Similarly, $[0,1]=\{0\}\cup(0,1]$, which shows that $\mu(\{0\})=\epsilon$ also. Thus, all the singletons must have the same measure, and so 2 holds.

In your comment, the point would be that $\rho$ is not additive, since $[0,1]=\{0\}\cup(0,1]=[0,1)\cup\{1\}$, but $\rho$ does not add properly on these unions.

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 that part of 2 follows from 1.

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 your comment, the point would be that $\rho$ is not additive, since $[0,1]=\{0\}\cup(0,1]=[0,1)\cup\{1\}$, but $\rho$ does not add properly on these unions.