CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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A free ultrafilter induces a finitely additive zero-one probability that is not countably additive

Statement refuted

A finitely additive set function of total mass $1$ need not be countably additive, even when it takes only the values $0$ and $1$ . Assuming the Axiom of Choice, a free ultrafilter on $N$ gives such a set function.

Facts & Assumptions

Given: The Axiom of Choice and the tails $B_{n} = {k \in N : n \leq k}$ .

[L1]

A filter base is nonempty, excludes $\emptyset$ , and is downward directed; its upward closure is the filter it generates (Filter base and the filter it generates, The upward closure of a filter base is the smallest filter containing it).

[L2]

Under the Axiom of Choice, every filter is contained in an ultrafilter (The Axiom of Choice, The ultrafilter lemma, from the Axiom of Choice: every filter extends to an ultrafilter), and an ultrafilter contains exactly one of each set and its complement (Ultrafilter, Characterisation of ultrafilters: every set or its complement).

[L3]

The natural numbers begin at $0$ (The natural numbers $N$ (von Neumann)), their order is total (Order on the natural numbers, $\leq$ is a linear order on $N$ ), and $n + 1$ is the immediate successor of $n$ (Discreteness: $σ (n)$ is the immediate successor).

[L4]

A finitely additive nonnegative set function vanishes at the empty set and is additive on disjoint pairs (Finitely additive nonnegative set functions).

Counterexample

technique · direct

1.1

The family $B = {B_{n} : n \in N}$ is a filter base: $B_{0} = N$ , no tail is empty, and $B_{\max (m, n)} \subseteq B_{m} \cap B_{n}$ . Let $F = ⟨ B ⟩$ .

givenL1L3

2.1

By [L2], extend $F$ to an ultrafilter $U$ . It is free: if ${m} \in U$ , then $B_{m + 1} \in F \subseteq U$ , so closure under intersections would put $\emptyset = {m} \cap B_{m + 1}$ in the filter.

step 1.1L2L3

3.1

Define $q (A) = 1$ when $A \in U$ and $q (A) = 0$ otherwise. Then $q (\emptyset) = 0$ and $q (N) = 1$ .

step 2.1L2

4.1

If $A, B$ are disjoint, then $A \cup B \in U$ exactly when one of $A, B$ lies in $U$ : the reverse implication is upward closure; for the forward implication, if $A \notin U$ , then $A^{c} \in U$ and $(A \cup B) \cap A^{c} = B$ lies in $U$ . Disjointness prevents both from lying in $U$ . Thus $q (A \cup B) = q (A) + q (B)$ , so $q$ is finitely additive.

step 2.1step 3.1L2L4

4.2

Every singleton has value $0$ because $U$ is free, but their disjoint union is $N$ and has value $1$ . Hence countable additivity fails.

step 2.1step 3.1L3

5.1

Steps 4.1 and 4.2 give a finitely additive zero-one probability that is not countably additive, refuting the proposed implication.

step 4.1step 4.2∎

Depends on

Used by

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Dependency tree · two levels

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Sources

D. Galvin, Ultrafilters, with Applications to Analysis, Social Choice and Combinatorics, §2 (standard reference, not scraped)