Tuesday, December 11, 2012

Uniform measure and nonmeasurable sets, without the Axiom of Choice

Given the Axiom of Choice, there is no translation invariant probability measure on the interval [0,1) (the relevant translation is translation modulo 1). But this fact really does need something in the way of the Axiom of Choice. Moreover, the fact only obtains for countably additive measures. Interestingly, however, if we add the assumption that our measure assigns non-zero (presumably infinitesimal) weight to each point of [0,1), then the non-existence of a translation invariant finitely additive measure follows without the Axiom of Choice. I got the proof of this from Paul Pedersen who thinks he got it from the classic Bernstein and Wattenberg piece (I don't have their paper at hand). I am generalizing trivially.

Theorem: Let P be any finitely additive measure taking values in a partially ordered group G and defined on a collection of subsets of [0,1) such that every countable subset has a measure in G. Suppose P({x})>0 for some x in G. Then P is not translation invariant (modulo 1).

Proof: To obtain a contradiction, suppose P is translation invariant. Then P({x})>0 for every x in [0,1). Let r be any irrational number in (0,1), and let R be the set of numbers of the form nr modulo 1, as n ranges over the positive integers. Let R' be the set of numbers of the form nr modulo 1, as n ranges over the integers greater than 1. Then R' is a translation of R by r, modulo 1. Observe that r is not a member of R' since there is no natural number n greater than 1 such that r=nr modulo 1, since if there were, we would have (n−1)r=0 modulo 1, and hence r would be a rational number with denominator n−1. Thus by finite additivity P(R)=P(R')+P({r})>P(R'). Hence, R is a counterexample to translation invariance, contradicting our assumption.

Note 1: On the assumption that the half-open intervals are all measurable and the measurable sets form an algebra (the standard case), translation invariance modulo 1 follows from ordinary translation invariance within the interval, namely the condition that P(A)=P(A+x) whenever both A and A+x={y+x:y in A} are subsets of [0,1).

Note 2: The proof above shows that if P({x})>0 for every x in [0,1), then the set of all positive integral multiples of any fixed irrational number (modulo 1) is nonmeasurable. It is interesting to note that this nonmeasurable set is actually measurable using standard Lebesgue measure. Thus, by enforcing regularity using infinitesimals, one is making some previously measurable sets nonmeasurable if one insists on translation invariance.

Note 3: Bernstein and Wattenberg construct a hyperreal valued measure that is almost translation invariant: the difference between the measure of a set and of a translation of the set is infinitesimal.


Alexander R Pruss said...

Slight generalization.

Let P be a finitely additive measure on the countable subsets of some set Omega taking values in some partially ordered group G. Let H be a group that acts on Omega such that there are g in H and w in Omega with { g^n w : n in N } infinite.

Then (a) P assigns null weight to some non-empty sets (i.e., P is not regular) or (b) P is not H-invariant.

If H acts transitively on Omega, then (a) implies P assigns null weight to all singletons.

Matt said...

Could you please explain the first sentence? Surely Lebesgue measure is translation-invariant in the requisite sense, no?

Matt said...
This comment has been removed by the author.
Alexander R Pruss said...

I meant: on all subsets of [0,1).