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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 '12 at 18:45
I don't see it. Could you elaborate? – Alexander Pruss Sep 26 '12 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 '12 at 21:07
@Alexander: In the title, it should be $(a, b]$ not $(b, a]$. – Michael Albanese Sep 26 '12 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 '12 at 1:41
up vote 6 down vote accepted

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 '12 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 '12 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 $0<s<1$ and try to get $\delta_s$ times the $s$-dimensional Hausdorff measure when we have a set of dimension $s$ (up to smaller-size infinitesimals). And, for that matter, when we do Hausdorff measures there is no reason to use only constant gauge functions. Maybe we would want to do this with come other nonarchimedean extension of the reals, rather than NSA. – Gerald Edgar Sep 27 '12 at 17:57
@Gerald Edgar: Indeed, we can do much much more with the same technique, which boils down to replacing a single ultraproduct with an iterated ultraproduct to get more control. (In this case, a two-step iteration was sufficient.) To my mind, the most salient barriers are finite obstructions like the Banach-Tarski obstruction to congruency invariance in three dimensions. I don't see a finite obstruction to what you propose regarding Hausdorff measure. – David Milovich Sep 27 '12 at 20:27

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

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Perhaps more simply, if $a<b$ then $$\mu(\lbrace a\rbrace) + (b-a) = \mu(\lbrace a\rbrace\cup(a,b]) = \mu([a,b]) = \mu([a,b)\cup\lbrace b\rbrace) = (b-a) + \mu(\lbrace b\rbrace).$$ – Sean Eberhard Sep 26 '12 at 22:18
Sean, I agree, and I have edited. – Joel David Hamkins Sep 26 '12 at 22:46
I stand corrected about my comment above. I kind of forgot that I also required $\mu((a,b])=b-a$. (My initial thinking about the problem only required $\mu([a,b))=b-a$. I still don't know the answer to that one, either.) – Alexander Pruss Sep 27 '12 at 12:58

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