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Assuming countable choice, every source-algebra set is measurable for the induced outer measure

Statement

Assume the Axiom of Countable Choice. Every member of the source algebra is Carathéodory measurable for the induced outer measure.

Facts & Assumptions

Given: Countable choice, a premeasure $μ_{0}$ on an algebra $A_{0}$ , its induced outer set function $μ^{*}$ , a set $E \in A_{0}$ , and a test set $A \subseteq X$ .

[F1]

A premeasure on an algebra $A_{0}$ vanishes at the empty set and is countably additive whenever a disjoint sequence in $A_{0}$ has its union in $A_{0}$ . (Premeasures on algebras of sets)

[F2]

The set function induced by $μ_{0}$ assigns $E \subseteq X$ the infimum of $\sum_{k} μ_{0} (A_{k})$ over all countable algebra covers $E \subseteq ⋃_{k} A_{k}$ . (The outer set function induced by a premeasure)

[L1]

For every outer measure $μ^{*}$ and all $A, E \subseteq X$ , $μ^{*} (A) \leq μ^{*} (A \cap E) + μ^{*} (A ∖ E)$ . (In the Carathéodory identity, the subadditive inequality is automatic)

[L2]

Assuming countable choice, the outer set function induced by a premeasure is an outer measure. (Assuming countable choice, the outer set function induced by a premeasure is an outer measure)

Proof

technique · direct

1.1F1F2choosecases

If $μ^{*} (A) = + \infty$ , the reverse Carathéodory inequality is automatic. If $μ^{*} (A) < + \infty$ , then for any $ε > 0$ choose an algebra cover $(C_{k})$ of $A$ with $\sum_{k} μ_{0} (C_{k}) < μ^{*} (A) + ε$ ; since $E \in A_{0}$ , the two sequences $(C_{k} \cap E)$ and $(C_{k} ∖ E)$ lie in the algebra, cover $A \cap E$ and $A ∖ E$ , and [F1] gives $μ_{0} (C_{k}) = μ_{0} (C_{k} \cap E) + μ_{0} (C_{k} ∖ E)$ .

2.1step 1.1L1L2algebra∎

By the defining infima, step 1.1 gives $μ^{*} (A \cap E) + μ^{*} (A ∖ E) < μ^{*} (A) + ε$ ; since this holds for every positive $ε$ , the left side is at most $μ^{*} (A)$ . Countable choice makes $μ^{*}$ an outer measure by [L2], so [L1] applies to it and gives the reverse inequality. Thus the Carathéodory identity holds for every $A$ .

Assuming countable choice, every source-algebra set is measurable for the induced outer measure

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources