## Tuesday, March 26, 2013

### Absolutely negligible sets

In my last post, I was thinking about arguments for why sets of cardinality less than that of the continuum should be taken to have measure zero. Here's the sort of argument I was thinking about:

1. The members of a family F of subsets of Rn should be counted as absolutely having measure zero when: (a) no invariant extension of Lebesgue measure assigns a non-zero measure to any of them and (b) every invariant extension of Lebesgue measure has a further invariant extension that assigns zero measure to every member of F.
The sets of cardinality less than that of the continuum satisfy (a) and (b). (The proof of (a) is here. The appendix in my post extends to prove (b) given (a).)

But (1) is actually false. Kharazishvili has introduced the notion of "absolutely negligible sets", which in the Lebesgue case are the sets whose singletons satisfy (a) and (b) (or a stronger condition involving quasi-invariant extensions), and has proved, inter alia, that Rn is a countable union of negligible subsets (see this paper and references therein; the same point applies to the circle R/N and any other solvable uncountable group.) Consequently, we should not accept (1).

There is a philosophical lesson here: Even where there is an apparently very reasonable way to make a nonmeasurable measurable, we can't simultaneously make measurable all the sets that there is a reaonable way of making measurable. And all this stuff on extensions of Lebesgue measure leads to a question that I have not seen addressed in the philosophy literature: Which invariant extension of Borel measure should we take to model cases of things like dart throws and to use as the basis for our physics?

Nonetheless, I think (1) is evidence that the members of F are reasonably to be taken as having measure zero, so something of my previous argument survives.

#### 1 comment:

Alexander R Pruss said...

The specific result for R^n was proved first I think by Ciesielski.