I don't think this answer is fundamentally different from Joel's, but perhaps the differing exposition may help.

Every irrational real number in $[0,1]$ has a unique binary expansion, and so every irrational real number in ${}^\ast[0,1]$ has a unique $\ast$-binary expansion. The set of irrational real numbers whose $N$-th binary digit is 0 (where $N$ is an infinite nonstandard natural number) is not Lebesgue measurable. This follows from the Lebesgue Differentiation Theorem.

I'm slowly beginning to grasp that adjectives like "countable" and "measurable", although applied to individual sets, actually describe the ambient set theory and *not* the set itself.

Edit: The weakest form of the Lebesgue Differentiation Theorem is this: Let $A\subseteq[0,1]$ be measurable. For almost all $x\in A$,
$$ \lim_{r\to0^+} \frac{\lambda(A \cap (x-r,x+r))}{2r} = 1.$$
Such $x$ are often called "points of density". More involved statements allow one to consider more general measure spaces than $[0,1]$, to integrate functions (instead of taking the measure of a set), and most interestingly, to consider more general types of neighborhoods than balls of radius $r$. For "nice" neighborhoods this generalizes, and there is still some issue as to whether particular neighborhoods are too ugly to be nice, or not.

The take-away is this: a measurable set has its positive measure in clumps. The set described above is *too* uniform to be measurable.

in its present form- as David Roberts' comment points out, it is not clear which part of Connes' anecdote the original poster wants to understand – Yemon Choi Mar 2 '11 at 5:56