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

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

2012-09-26T15:21:30Z

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

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

Answer by Joel David Hamkins for Hyperreal finitely-additive measure on [0,1) assigning $b-a$ to $[a,b)$ or $(a,b]$ and infinitesimals to singletons

2012-09-26T21:51:12Z

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

Answer by David Milovich for Hyperreal finitely-additive measure on [0,1) assigning $b-a$ to $[a,b)$ or $(a,b]$ and infinitesimals to singletons

2012-09-27T14:57:01Z

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.