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An uncountable class can have a nonmeasurable empirical supremum

Statement

Assume Countable Choice. Let $Ω = [0, 1]$ , let $A$ be the countable-cocountable sigma-algebra, and define

$P (A) : = {\begin{cases} 0, & A is countable, \\ 1, & Ω ∖ A is countable . \end{cases}$

For $U : = [0, 1 / 2]$ , take the uncountable class of functions $F : = {f_{t} : t \in U}$ with $f_{t} (ω) : = 1_{{t}} (ω)$ . Every $f_{t}$ is measurable and $P f_{t} = 0$ , but for one observation $Z (ω) = ω$ ,

$\sup_{t \in U} ∣ f_{t} (Z (ω)) - P f_{t} ∣ = 1_{U} (ω),$

which is not measurable because $U$ is neither countable nor cocountable.

Facts & Assumptions

Given: The Axiom of Countable Choice and the stated $Ω, A, P, U, F$ .

[A1]

Countable Choice implies that a countable union of at most countable sets is at most countable (The Axiom of Countable Choice ( ${A C}_{ω}$ ), Countable unions of at most countable sets, assuming ${A C}_{ω}$ ). The sigma-algebra closure of the stated family is established locally in step 1.1.

[F1]

Every nondegenerate real interval is uncountable (Every nondegenerate interval of $R$ is uncountable).

[F2]

Every singleton is countable and every subset of a countable set is countable (Finite, countably infinite, countable, uncountable, Every subset of an at most countable set is at most countable).

[F3]

A probability measure is a countably additive measure of total mass one (Measures on sigma-algebras, Probability measures and probability spaces).

[F4]

For one observation, the empirical average is $P_{1} f = f (Z)$ and $P f$ is the integral of $f$ (Empirical, centered, and Rademacher processes).

[F5]

A family containing the empty set and closed under complements and countable unions is a sigma-algebra (Sigma-algebras).

Proof

Given: The Axiom of Countable Choice and the objects of the Counterexample.

1.1A1F1F2F5given

By [F1], $Ω$ is uncountable. The empty set is at most countable, and complementation exchanges the countable and cocountable alternatives, so $A$ contains the empty set and is complement-closed. For a sequence $(A_{n})$ in $A$ , if every $A_{n}$ is countable then [A1] makes their union countable; otherwise some $A_{j}$ is cocountable, and $Ω ∖ ⋃_{n} A_{n} \subseteq Ω ∖ A_{j}$ is countable by [F2]. Thus the union lies in $A$ in either case, proving all sigma-algebra axioms in [F5]. The assignment $P$ is well-defined: a set and its complement cannot both be countable, since their union would make $Ω$ countable.

2.1A1F2F3step 1.1givenchoose

Let $(A_{n})$ be pairwise disjoint members of $A$ . If every $A_{n}$ is countable, [A1] makes their union countable; then $P (⋃_{n} A_{n}) = 0 = \sum_{n} P (A_{n})$ . Otherwise choose $j$ with $A_{j}$ cocountable. For $n \neq j$ , disjointness gives $A_{n} \subseteq Ω ∖ A_{j}$ , a countable set, so [F2] makes $A_{n}$ countable. The union is cocountable and its measure is $1 = P (A_{j}) + \sum_{n \neq j} P (A_{n})$ . Also $P (Ω) = 1$ and $P$ is nonnegative, so [F3] shows that $P$ is a probability measure.

3.1F1F2F3F4givenstep 2.1

Regard $(Ω, A, P)$ as the observation space and let $Z$ be its identity random element, which has law $P$ . For each $t \in U$ , the singleton ${t}$ is in $A$ by [F2], so $f_{t} = 1_{{t}}$ is measurable and $P f_{t} = P ({t}) = 0$ . If $s \neq t$ , then $f_{s} (s) = 1$ and $f_{t} (s) = 0$ , so $t \mapsto f_{t}$ is injective; since $U$ is uncountable by [F1], $F$ is uncountable.

4.1F1step 3.1given∎

For any $ω \in Ω$ , if $ω \in U$ then the index $t = ω$ gives $f_{t} (ω) = 1$ , while if $ω \notin U$ every $t \in U$ differs from $ω$ and all the function values are zero. Hence the displayed supremum equals $1_{U} (ω)$ . Both $U$ and $Ω ∖ U$ contain a nondegenerate interval and are uncountable by [F1]; therefore $U \notin A$ , so the supremum is nonmeasurable.

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Sources

David Pollard, Probability Tools, Tricks, and Miracles, 2026 manuscript, Section 13.3 (standard reference, not scraped)
Jon A. Wellner, Empirical Processes: Theory and Applications, Chapter 1, Sections 5-6 (standard reference, not scraped)