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Carathéodory's theorem: measurable sets form a sigma-algebra carrying a complete measure

Statement

For an outer measure μ on X, the Carathéodory measurable subsets form a sigma-algebra, and the restriction of the outer measure to it is a complete measure.

Facts & Assumptions

Given: An outer measure μ on X and its family Mμ of Carathéodory measurable sets.

[L1]

The Carathéodory measurable subsets of X form an algebra of subsets. (Carathéodory measurable sets form an algebra)

[L2]

A countable disjoint union of Carathéodory measurable sets is Carathéodory measurable, and μ(AkEk)=kμ(AEk) for every AX. (Countable disjoint unions of Carathéodory measurable sets are measurable and split every test set)

[L3]

Every set of outer measure zero, and every subset of it, is Carathéodory measurable and has outer measure zero. (Every outer-null set is Carathéodory measurable)

[L4]

Let A be an algebra of subsets of X. If the union of every pairwise disjoint sequence in A belongs to A, then A is a sigma-algebra on X. (An algebra closed under countable disjoint unions is a sigma-algebra)

Proof

technique · direct
1.1

The family Mμ is an algebra by [L1] and is closed under countable disjoint unions by [L2], so [L4] makes it a sigma-algebra.

L1L2L4
1.2

For a pairwise disjoint sequence (Ek) in Mμ, use A=kEk in [L2]; then AEk=Ek, so μ(kEk)=kμ(Ek), including infinite values, and the restriction is a measure.

L2algebra
2.1

If NMμ has restricted measure zero and SN, then [L3] makes S measurable with outer measure zero; hence the restricted measure is complete.

L3algebra

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