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A free ultrafilter on $N$ , viewed as a subset of ${0, 1}^{N}$ and hence of $[0, 1]$ , is not Lebesgue measurable

Statement

Assume the Axiom of Countable Choice. Let $U$ be a free ultrafilter on $N$ , let $D$ be the dyadic rationals in $[0, 1]$ , and put $X : = [0, 1] ∖ D$ . Every $x \in X$ has a unique binary expansion $x = 0. b_{0} b_{1} b_{2} \dots$ ; write $S_{x} : = {n \in N : b_{n} = 1}$ . Define

$E_{U} : = {x \in X : S_{x} \in U} .$

Then $E_{U}$ is not Lebesgue measurable. This is what the title means by viewing $U$ as a subset of ${0, 1}^{N}$ and hence of $[0, 1]$ : the dyadic ambiguity is removed on the null set $D$ .

Facts & Assumptions

Given: The Axiom of Countable Choice, a free ultrafilter $U$ on $N$ , and the associated sets $D$ , $X$ , and $E_{U} \subseteq X$ .

[L1]

A measurable subset of $[0, 1]$ that is invariant under changing finitely many binary digits has measure $0$ or $1$ (A Lebesgue measurable subset of $[0, 1]$ that is invariant under changing finitely many binary digits has measure $0$ or $1$ ).

[L2]

In an ultrafilter on a set $X$ , for every $A \subseteq X$ exactly one of $A$ and $X ∖ A$ lies in the ultrafilter (Characterisation of ultrafilters: every set or its complement).

[L3]

A free ultrafilter is a non-principal ultrafilter (Ultrafilter).

[L4]

Every at most countable subset of $R$ has measure zero (Every at most countable subset of $R$ has measure zero).

[L5]

Assuming countable choice, every interval with any endpoint convention is Lebesgue measurable with its usual length (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[L6]

Assuming countable choice, Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation (Lebesgue outer measure, Lebesgue measurability and Lebesgue measure are unchanged by translation).

[L7]

Assuming countable choice, reflection in the origin preserves Lebesgue measurability and Lebesgue measure (For a nonzero real $c$ , dilation by $c$ multiplies Lebesgue outer measure by $∣ c ∣^{n}$ , and reflection in the origin preserves it).

Proof

technique · contradiction

1.1

No singleton belongs to $U$ : if ${n} \in U$ , then every set containing $n$ lies in $U$ by upward closure, and [L2] excludes every set omitting $n$ , so $U$ is principal at $n$ , contradicting [L3]. Consequently no finite set belongs to $U$ , by induction on the size of the finite set using [L2] and the implication $A \cup B \in U \Rightarrow A \in U$ or $B \in U$ . Therefore every cofinite subset of $N$ belongs to $U$ .

L2L3algebra

2.1

If $A, B \subseteq N$ differ by finitely many points, then $A \in U$ if and only if $B \in U$ . Indeed, with $F : = N ∖ (A △ B)$ , step 1.1 gives $F \in U$ ; if $A \in U$ then $A \cap F = B \cap F \in U$ , hence $B \in U$ by upward closure, and the converse is symmetric. A binary expansion represents a dyadic point exactly when it is eventually $0$ or eventually $1$ , and finite digit changes preserve that property. Thus two points represented by expansions differing at finitely many indices are either both in $D$ , hence both outside $E_{U}$ , or both in $X$ , where their unique expansions give sets with finite symmetric difference. Therefore $E_{U}$ is invariant under changing finitely many binary digits.

step 1.1L2algebra

3.1

Suppose, for contradiction, that $E_{U}$ is Lebesgue measurable. Then step 2.1 and [L1] give $λ (E_{U}) \in {0, 1}$ .

step 2.1L1assume-contra

4.1

The set $D$ is countable, hence null by [L4]. For $x \in X$ , the unique binary expansion of $r (x) : = 1 - x$ is obtained by complementing every digit, so $S_{r (x)} = N ∖ S_{x}$ . Therefore $r$ maps $X$ to itself and [L2] gives $x \in E_{U}$ if and only if $r (x) \notin E_{U}$ . For measurable $M \subseteq X$ , one has $r [M] = (- M) + 1$ , so reflection invariance [L7] followed by translation invariance [L6] gives $λ (r [M]) = λ (M)$ . Applying this to $M = E_{U}$ and using $λ (D) = 0$ and $λ ([0, 1]) = 1$ from [L5] yields $λ (E_{U}) = λ (r [E_{U}]) = λ (X ∖ E_{U}) = 1 - λ (E_{U}),$ so $λ (E_{U}) = 1 / 2$ .

step 3.1L2L4L5L6L7

5.1

The value $1 / 2$ from step 4.1 contradicts the dichotomy of step 3.1. Therefore $E_{U}$ is not Lebesgue measurable.

step 3.1step 4.1discharge-contradiction∎

Depends on

Used by

What the Vitali set, Bernstein sets and free ultrafilters cost in choice Remark

Dependency tree · two levels

57 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Jacek Cichoń, Aleksander Kharazishvili, and Bogdan Węglorz, Subsets of the Real Line, Theorem 8.13 (standard reference, not scraped)
S. Sierpiński, Sur un problème concernant les ensembles mesurables superficiellement, Fund. Math. 1 (1920) (standard reference, not scraped)

A free ultrafilter on NN, viewed as a subset of {0,1}N{0,1}N and hence of [0,1][0,1], is not Lebesgue measurable