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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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If μ is sigma-finite and A is countably generated, then Lp(μ) is separable for 1p<

Statement

Assume the Axiom of Countable Choice.

Let (X,A,μ) be a sigma-finite measure space with countably generated sigma-algebra, and let 1p<. Then Lp(μ) is separable.

Facts & Assumptions

Given: The Axiom of Countable Choice, a sigma-finite countably generated measure space, and an exponent 1p<.

[L1]

Simple functions with finite-measure support are dense in Lp(μ) (Simple functions with finite-measure support are dense in Lp(μ) for 1p<).

[L3]

The quotient Lp(μ) is the space in question, and separability means the existence of a countable dense subset (The space Lp(μ) as the quotient by null functions, Separability: the existence of an at most countable dense subset).

Proof

technique · direct
1.1

Let A0 be the countable algebra from [L2], and let S [L2, L3, given, algebra] be the set of all finite linear combinations j=1mqj1Aj with qjQ(i) and AjA0 satisfying μ(Aj)< for every j. Because both the coefficient set and the finite-measure members of A0 are countable, S is countable.

L2L3givenalgebra
1.2

To prove density, start with fLp(μ) and ε>0. By [L1, L2, given, choose, algebra] [L1], choose a finite-support simple function s=j=1mcj1Ej with fsp<ε/2. For each j, [L2] gives AjA0 with μ(Aj)< and μ(EjAj) arbitrarily small, and each coefficient cj can be approximated by qjQ(i). The resulting t:=jqj1Aj lies in S and satisfies stp<ε/2.

L1L2givenchoosealgebra
2.1

Then [L3, step 1.1, step 1.2] ftpfsp+stp<ε. So S is countable and dense in Lp(μ); by [L3], the space is separable.

L3step 1.1step 1.2

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