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Assuming countable choice, every measure space has a unique complete extension to its completion

Statement

Assume the Axiom of Countable Choice. Let $(X, A, μ)$ be a measure space, and let $(X, \overline{A}, \overline{μ})$ be its completion construction. Then $\overline{μ}$ is a complete measure on $\overline{A}$ extending $μ$ . It is the unique complete measure on $\overline{A}$ that extends $μ$ .

Facts & Assumptions

Given: A measure space $(X, A, μ)$ and the Axiom of Countable Choice.

[L1]

The completion domain is a sigma-algebra containing $A$ (Assuming countable choice, the completion domain is a sigma-algebra).

[L2]

The value $\overline{μ} (E) = μ (A)$ is independent of the measurable core in a completed representation (The completed measure is independent of the representing measurable set).

[L3]

Countable unions of measurable null sets are null, and completeness makes all their subsets measurable and null (Null sets are closed under countable unions and, in a complete space, under arbitrary subsets).

[L4]

A measure vanishes at the empty set and is countably additive on disjoint measurable sequences (Measures on sigma-algebras).

[L5]

Countable choice selects witnesses from each nonempty natural-number-indexed family (The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L6]

Every member of the completion domain has a representation $E = A \cup N$ with $A$ measurable and $N$ contained in a measurable null set, and the proposed completed value is $μ (A)$ (The completion domain and proposed completed set function of a measure space).

Proof

technique · direct

1.1givenL1L2L4

By [L1] and [L2], $\overline{μ}$ is a function on the sigma-algebra $\overline{A}$ , and $\overline{μ} (A) = μ (A)$ for $A \in A$ by the representation $A = A \cup \emptyset$ ; in particular $\overline{μ} (\emptyset) = 0$ .

1.2givenL3L5L6choose

Let $(E_{k})$ be disjoint in $\overline{A}$ . By [L6] each family of completed representations is nonempty, so [L5] chooses $E_{k} = A_{k} \cup N_{k}$ with measurable $A_{k} \subseteq E_{k}$ and $N_{k}$ contained in a measurable null set $Z_{k}$ . Then the $A_{k}$ are disjoint, $⋃_{k} E_{k} = (⋃_{k} A_{k}) \cup N$ with $N \subseteq ⋃_{k} Z_{k}$ , and $⋃_{k} Z_{k}$ is null by [L3].

1.3givenL2L3L6

If $E \in \overline{A}$ and $\overline{μ} (E) = 0$ , choose a representation $E = A \cup N$ from [L6] with $N \subseteq Z$ null. Then $A \cup Z$ is measurable and null, and every $S \subseteq E$ is represented by the empty measurable core plus the sub-null set $S \subseteq A \cup Z$ .

1.4givenL2L3L4L6

Let $λ$ be any complete measure on $\overline{A}$ extending $μ$ . For a representation $E = A \cup N$ from [L6], with $N \subseteq Z$ and $μ (Z) = 0$ , one has $λ (Z) = 0$ and completeness gives $λ (N) = 0$ . Replacing $N$ by $N ∖ A$ makes the union disjoint without changing $E$ , so finite additivity gives $λ (E) = λ (A) = μ (A) = \overline{μ} (E)$ .

2.1step 1.1step 1.2L2L4

Countable additivity of $μ$ applied to the measurable cores in step 1.2 gives $\overline{μ} (⋃_{k} E_{k}) = μ (⋃_{k} A_{k}) = \sum_{k} μ (A_{k}) = \sum_{k} \overline{μ} (E_{k})$ ; with step 1.1, this proves that $\overline{μ}$ is a measure.

2.2step 1.3L2L3

Step 1.3 shows that every subset of every $\overline{μ}$ -null set belongs to $\overline{A}$ and has value $0$ , so the completed measure space is complete.

3.1step 2.1step 2.2step 1.4∎

Steps 2.1, 2.2 and 1.4 prove respectively that $\overline{μ}$ is an extending measure, is complete, and is the unique complete extension on $\overline{A}$ .

Assuming countable choice, every measure space has a unique complete extension to its completion

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources