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4 Fixed error in comment about rho

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 commentparticular, the point would be that proposed function $\rho$ is in your comment to the question does not additiveexhibit the desired properties, since $[0,1]=\{0\}\cup(0,1]=[0,1)\cup\{1\}$, but in light of the decomposition $\rho$ does not add properly on these unions.[0,\frac{1}{2}]=\{0\}\cup(0,\frac12]=[0,\frac12)\cup\{\frac12\}$. 3 added 22 characters in body I think this is a very interesting question. In response to your comment, let me argue that if 1 holds and the measure is additive, then the singleton values are all the same. This is the sense in which that part the strong form of 2 follows from 1the weak form of 2. To see this, following Sean's comment, observe that$\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$does not add properly on these unions. 2 Simplified the argument via Sean's comment; added 45 characters in body I think this is a very interesting question, which deserves more votes. (My vote is currently the only +1.) But meanwhile, in In response to your comment, let me argue that if 1 holds and all the singletons get infinitesimal valuemeasure is additive, then these the singleton values are all the same. This is the sense in which that part of 2 follows from 1. To see this, suppose$\mu$has property 1. Let$\epsilon=\mu(\{1\})$, a nonzero infinitesimal number. Observe following Sean's comment, observe that$\mu([0,1])=\mu([0,1)\cup\{1\})=1+\epsilon$. For any$x\in \mu (0,1)$, let$\tau=\mu(\{x\})$, \{a\})+\mu((a,b])=\mu([a,b])=\mu([a,b))+\mu(\{b\})$, and observe that $[0,1]=[0,x)\cup\{x\}\cup(x,1]$, which has measure so $x+\tau+(1-x)=1+\tau$. \mu(\{a\})=\mu(\{b\})$. So$\tau=\epsilon$. Similarly,$[0,1]=\{0\}\cup(0,1]$, which shows that$\mu(\{0\})=\epsilon$also. Thus, all the singletons must have the same measure, and so the strong form of 2 holdsfollows from the weak form of 2. In your comment, the point would be that$\rho$is not additive, since$[0,1]=\{0\}\cup(0,1]=[0,1)\cup\{1\}$, but$\rho\$ does not add properly on these unions.

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