## Hyperreal finitely-additive measure on [0,1) assigning $b-a$ to $[a,b)$ or $(a,b]$ and infinitesimals to singletons

Is there a hyperreal-valued finitely additive measure on all the subsets of [0,1), or at least the Borel ones, that

1. assigns $b-a$ to $[a,b)$ and to $(a,b]$ for all $a\lt b,$ and

2. assigns an infinitesimal--ideally, the same one--to each singleton?

It's (1) that's a problem. The Bernstein-Wattenberg construction yields a finitely-additive measure that gives (1) up to infinitesimals. But it would be nice to have (1) exactly.

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We get the "same one" property of 2 from 1. – Gerald Edgar Sep 26 at 18:45
I don't see it. Could you elaborate? – Alexander Pruss Sep 26 at 21:02
Actually, now that I think about it, we don't get the "same one" property from 1, unless it does so trivially because there is no measure satisfying 1. For suppose that $\mu$ satisfies 1 and 2. Let $\nu(A) = \mu(A) - \mathrm{st} \mu(A)$ be the infinitesimal part of $\mu$. Let $\rho(A) = \mu(A) + \nu(A \cap [0,1/2)) + 2\nu(A \cap [1/2,1))$. Then $\rho(A)$ satisfies 1 but not 2. – Alexander Pruss Sep 26 at 21:07
@Alexander: In the title, it should be $(a, b]$ not $(b, a]$. – Michael Albanese Sep 26 at 22:36
I can throw some buzzwords around, but I'm out of my depth here. If you take a nonstandard extension that is an enlargement (or polysaturated) then there is a hyperfinite set $b\subseteq *[0,1]$ with $[0,1] \subseteq b$. This feels relevant, but I'm not sure how exactly. – Kevin O'Bryant Sep 27 at 1:41

Yes, by compactness.

Let $R$ denote your favorite hyperreal ordered field and let $\delta\in R$ be a positive infinitesimal. Let $\mathcal{E}$ denote the set of all (standard) finite Boolean subalgebras of $\mathcal{P}([0,1))$. For every $A\in\mathcal{E}$, let $\lambda_A(I)$ be the (exact) length of $I$ for all half-open intervals $I\in A$; for all open or closed intervals $I\in A$, respectively subtract or add $\delta$ to the length of $I$ to define $\lambda_A(I)$; let $\lambda_A(S)=\delta$ for all singletons $S\in A$. Extend $\lambda_A$ to a probability measure $\mu_A$ on $A$.

(Specifically, for each minimal finite union of intervals $F\in A$, let the connected components of $F$ be $[a_0,b_0),\ldots,[a_k,b_k)$ with $p$ elements of $\{a_i,b_i:i\leq k\}$ added and $q$ removed. Partition $F$ into its atomic subsets $H_0,\ldots,H_n$. Choose a positive $\mu_A(H_i)\in R$ for each $i$, such that $\sum_{i\leq n}\mu_A(H_i)=(p-q)\delta+\sum_{i\leq k}(b_i-a_i)$. Now extend $\mu_A$ from the atoms to all of $A$.)

Let $U$ be a fine ultrafilter on $\mathcal{E}$. ("Fine" means that $\{B\in\mathcal{E}:A\subseteq B\}\in U$ for all $A\in\mathcal{E}$.) The ultraproduct measure $\mu_U$ is $R^U$-valued and has the two properties you seek.

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 +1. Very nice! – Joel David Hamkins Sep 27 at 16:09 I just minorly corrected the answer: the parenthetical paragraph now correctly handles cases such as $[0,1/3)\cup[2/3,1)\in A$ but $[0,1/3)\not\in A$. – David Milovich Sep 27 at 16:58 But probably we want more than the OP asks for. In addition to assigning lengths to intervals and $\delta$ to points, maybe we want to pick a certain infinitesimal $\delta_s$ for each $0 I think this is a very interesting question. In response to your comment, let me argue that if 1 holds and the measure is additive, then the singleton values are all the same. This is the sense in which the strong form of 2 follows from the weak form of 2. To see this, following Sean's comment, observe that$\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\}$. -  Perhaps more simply, if$a