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A real L^1 density defines a finite signed measure with its canonical Hahn and Jordan data

Statement

Let (X,A,μ) be a measure space and let f∈L1(μ) be real-valued. Define ν(E):=∫Ef dμ(E∈A). Then ν is a finite signed measure. Its canonical Hahn sets are P:={f>0},N:={f≤0}, its Jordan parts are ν+(E)=∫Ef+ dμ,ν−(E)=∫Ef− dμ, and its total variation is ∣ν∣(E)=∫E∣f∣ dμ.

These explicit constructions require no choice axiom. Hahn partitions are unique only up to signed-null sets; the Jordan measures are unique.

Facts & Assumptions

Given: A measure space (X,A,μ) and a real-valued function f∈L1(μ).

[L1]

A nonnegative measurable density defines a positive measure. (The indefinite integral of a nonnegative measurable function is a measure)

[L2]

For a real integrable function, the positive and negative parts satisfy f=f+−f− and ∣f∣=f++f−. (Integrable real and complex functions, and their integrals)

[L3]

Arithmetic and threshold operations preserve measurability. (Closure properties of measurable functions used by the integral)

[L4]

Given Hahn decompositions, their positive pieces differ by a signed-null set; given a Hahn partition, its Jordan measures are constructed by restriction and are unique. These are the separately stated choice-free conditional clauses, not the AC-qualified general existence assertions. (Hahn decomposition for signed measures, unique up to total-variation-null sets, Jordan decomposition of a signed measure into unique mutually singular positive parts)

[L5]

The nonnegative integral is additive. (Additivity of the nonnegative Lebesgue integral)

Proof

technique · direct
1.1givenL1L2L3L6algebra

Put α(E)=∫Ef+ dμ and β(E)=∫Ef− dμ. By [L1]–[L3] these are positive measures, and their total masses are finite because f is integrable. By [L2], ν(E)=α(E)−β(E) is finite. For disjoint measurable En, the series ∑nα(En) and ∑nβ(En) have finite sums, so they can be subtracted termwise to obtain countable additivity of ν; moreover ∑n∣ν(En)∣≤α(X)+β(X)<∞. Also ν(∅)=0. Thus [L6] makes ν a finite signed measure.

2.1L3L4L6step 1.1

By [L3], P={f>0} and N={f≤0} are measurable and partition X. The density f− vanishes on P and f+ vanishes on N, so on measurable subsets of P one has ν=α≥0, and on subsets of N one has ν=−β≤0. These signs prove directly that (P,N) is a Hahn partition. Only the conditional comparison in [L4] is needed to say that every other supplied Hahn partition differs by a signed-null set.

3.1L4step 1.1step 2.1algebra

Since α vanishes on N and β vanishes on P, the conditional construction in [L4] identifies them with the unique Jordan measures: ν(E∩P)=α(E) and −ν(E∩N)=β(E). No arbitrary Hahn partition is selected.

4.1L2L5L6step 2.1step 3.1algebra∎

For any countable measurable partition E=⨆jEj, one has ∑j∣ν(Ej)∣≤∑j(α(Ej)+β(Ej))=α(E)+β(E). Conversely, the explicit partition E∩P,E∩N (with empty later parts) has variation sum exactly α(E)+β(E) by step 2.1. Taking the supremum in [L6] proves ∣ν∣(E)=α(E)+β(E)=∫E∣f∣ dμ by [L2] and [L5]. Together with steps 1.1–3.1 this proves every assertion without using the general Hahn/Jordan existence clauses.

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