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The induced outer measure agrees with the premeasure on the source algebra

Statement

For every $A \in A_{0}$ , the outer measure induced by a premeasure satisfies $μ^{*} (A) = μ_{0} (A)$ .

Facts & Assumptions

Given: A premeasure $μ_{0}$ on an algebra $A_{0}$ and its induced outer measure $μ^{*}$ .

[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)

[L1]

Every countable algebra cover of an algebra set disjointifies inside that set into algebra members subordinate to the original cover. (A countable algebra cover disjointifies inside the covered algebra set)

[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)

Proof

technique · direct

1.1

The sequence $(A, \emptyset, \emptyset, \dots)$ covers $A$ at cost $μ_{0} (A)$ , so $μ^{*} (A) \leq μ_{0} (A)$ .

F2construct

1.2

For any algebra cover $(A_{k})$ of $A$ , [L1] gives disjoint $B_{k} \subseteq A_{k}$ with union $A$ ; [F1] gives $μ_{0} (A) = \sum_{k} μ_{0} (B_{k})$ , and finite additivity applied to $A_{k} = B_{k} ⊔ (A_{k} ∖ B_{k})$ gives $μ_{0} (B_{k}) \leq μ_{0} (A_{k})$ , hence $μ_{0} (A) \leq \sum_{k} μ_{0} (A_{k})$ .

F1L1algebra

2.1

Since step 1.2 bounds every covering cost below by $μ_{0} (A)$ , [F2] gives $μ_{0} (A) \leq μ^{*} (A)$ ; combining this with step 1.1 proves equality, including infinite values without subtraction.

step 1.1step 1.2F2algebra∎

Remarks

No choice principle is used here. Steps 1.1, 1.2 and 2.1 read the defining infimum of The outer set function induced by a premeasure, disjointify one given cover, and apply countable additivity of the premeasure; none of them selects a cover for each index. The name outer measure is the one that Assuming countable choice, the outer set function induced by a premeasure is an outer measure earns for the induced set function under countable choice, and the identity proved here holds for the set function whether or not that hypothesis is in force.

Depends on

Used by

Assuming countable choice, a premeasure extends through its induced outer measure Theorem

Dependency tree · two levels

7 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

G. Folland, Real Analysis, 2nd ed., Proposition 1.13(a) (standard reference, not scraped)