Ultrafilter limits of asymptotic density are not universally measurable

Jörg Brendle  Paul B. Larson

December 15, 2009

Given a nonprincipal ultrafilter U on ω and a sequence x̄ = ⟨x_{n} : n ∈ ω⟩ consisting of members of a compact Hausdorff space X, the U-limit of x̄ (written lim_{n→U} x_{n}) is the unique y ∈ X such that for every open set O ⊆ X containing y, {n | x_{n} ∈ O} ∈ U. Letting X be the unit interval [0, 1], this operation defines a finitely additive measure μ_{U} on P(ω) in terms of asymptotic density, letting μ_{U}(A) = lim_{n→U} |A ∩ n|/n, for each A ⊆ ω. A medial limit is a finitely additive measure on P(ω), giving singletons measure 0 and ω itself measure 1, such that for each open set O ⊆ [0, 1], the collection of A ⊆ ω given measure in O is universally measurable, i.e., is measured by every complete finite Borel measure on P(ω) (see [1] for more on medial limits and universally measurable sets). If there could consistently be a nonprincipal ultrafilter U such that measure given by the U-limit of asymptotic density were universally measurable, this would give a relatively simple example of a medial limit. We show here, however, that this cannot be the case.

Theorem 0.1. If U is a nonprincipal ultrafilter on ω, then the function

μ_{U}: P(ω) → [0, 1]

defined by letting μ_{U}(A) = lim_{n→U} |A ∩ n|/n is not universally measurable.

Proof. Let I_{0} = {0}, and for each positive n ∈ ω let

I_{n} = {5^{n−1}, 5^{n−1} + 1, …, 5^{n} − 1}.

Either the union of the I_{n}’s for n even is in U, or the corresponding union for n odd is. In the first case, let J_{0} = I_{0} ∪ I_{1} ∪ I_{2}, and for each positive n, let J_{n} = I_{2n+1} ∪ I_{2n+2}. In the second case, let J_{n} = I_{2n} ∪ I_{2n+1} for all n ∈ ω. In either case, let S be the set of A ⊆ ω such that A ∩ J_{n} ∈ {∅, J_{n}} for all n ∈ ω. Then S is a perfect subset of P(ω), and the mapping H: S → P(ω) sending A ∈ S to {n | A ∩ J_{n} = J_{n}} is a homeomorphism. Let F_{0} be the set of A ⊆ ω such that μ_{U}(A) ∈ [0, 1/4), and let F_{1} be the set of A ⊆ ω such that μ_{U}(A) ∈ (3/4, 1] (so F_{1} is the set of complements of elements of F_{0}). It will be enough to show that F_{1} ∩ S is not a universally measurable subset of S.

We claim for each A ∈ S, the set of n such that |A∩n|/n ∈ [0, 1/4)∪(3/4, 1] is in U. This follows from the fact that all (but possibly one) of the J_{n}’s are unions of two consecutive I_{m}’s, and that the union of the larger members of these pairs is in U. Each such consecutive pair (for n > 0) has the form I_{m} = {5^{m−1}, 5^{m−1} +1, …, 5^{m} −1} and I_{m+1} = {5^{m}, 5^{m} +1, …, 5^{m+1} −1}, and if A ∈ S, then A either contains or is disjoint from I_{m} ∪ I_{m+1}. If it contains both, then for each k ∈ I_{m+1},

|A ∩ k|/k ≥ (5^{m} − 5^{m−1})/5^{m} = 1 − 1/5 = 4/5 > 3/4,

and if it is disjoint from both then

|A ∩ k|/k ≤ 5^{m−1}/5^{m} = 1/5 < 1/4.

This establishes the claim. It follows that S ⊆ F_{0} ∪ F_{1}. Since μ_{U} is a finitely additive measure, the intersection of two sets of μ_{U}-measure greater than 3/4 cannot be less than 1/4, so F_{1} ∩ S is closed under finite intersections. It follows that H maps F_{1} ∩ S homoemorphically to a nonprincipal ultrafilter, and thus that F_{1} ∩ S is not universally measurable.

A version of the proof just given, in the special case {5^{n} : n ∈ ω} ∈ U, led to the proof in [1] that consistently there are no medial limits.

References

[1] P.B. Larson, The Filter Dichotomy and medial limits, in preparation

Graduate School of Engineering Kobe University Rokko-dai 1-1, Nada Kobe 657-8501 JAPAN brendle@kurt.scitec.kobe-u.ac.jp

Department of Mathematics Miami University Oxford, Ohio 45056 USA larsonpb@muohio.edu