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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∗(A∩⋃kEk)=∑kμ∗(A∩Ek) for every A⊆X. (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
1.1L1L2L4

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

1.2L2algebra

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

2.1L3algebra∎

If N∈Mμ∗ has restricted measure zero and S⊆N, then [L3] makes S measurable with outer measure zero; hence the restricted measure is complete.

Depends on

Used by

Dependency tree · two levels

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Sources