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 


*

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