TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)

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Every finite Borel measure on R has a unique absolutely continuous, discrete, and singular-continuous decomposition

Statement

Assume the Axiom of Choice. Let $μ$ be a finite Borel measure on $R$ . Then there exist unique finite Borel measures $μ_{a c}, μ_{d}, μ_{s c}$ such that $μ = μ_{a c} + μ_{d} + μ_{s c},$ where $μ_{a c} ≪ λ$ , $μ_{d}$ is discrete, and $μ_{s c}$ is atomless and singular with respect to Lebesgue measure $λ$ .

Facts & Assumptions

Given: AC and a finite Borel measure $μ$ on $R$ .

[L1]

Under countable choice a finite Borel measure splits as $μ_{d} (E) = μ (E \cap A)$ and $η (E) = μ (E ∖ A)$ , where $A = {x : μ ({x}) > 0}$ is countable and $η$ is atomless. (Every finite Borel measure on $R$ splits as an atomic part plus an atomless part, An atom of a measure on $R$ )

[L2]

Lebesgue measure has $λ ([- n, n]) = 2 n$ and $λ ({x}) = 0$ . (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included)

[L3]

Under AC, common-finite-exhaustion Lebesgue decomposition exists, with positive components dominated by the original measure in the positive case. Given two decompositions, their components are unique without any choice hypothesis. (Every sigma-finite signed measure admits a Lebesgue decomposition relative to a sigma-finite positive measure, The Lebesgue decomposition of a sigma-finite signed measure is unique)

[L4]

AC supplies [L3] and implies the countable choice required by [L1] and [L2]. (The Axiom of Choice)

Proof

technique · direct

1.1L1L4given

Apply [L1] to write $μ = μ_{d} + η$ . The measure $μ_{d}$ is discrete, meaning concentrated on a countable set, and $η$ is atomless. These measures are finite positive Borel measures.

2.1L2L3L4step 1.1construct

On $X_{n} = [- n, n]$ both $λ$ and $η$ are finite by [L2] and finiteness of $η$ . Under AC, [L3] decomposes $η = μ_{a c} + μ_{s c}$ with positive finite components, $μ_{a c} ≪ λ$ and $μ_{s c} ⊥ λ$ . Since $η ({x}) = 0$ , positivity gives $μ_{a c} ({x}) = μ_{s c} ({x}) = 0$ for every $x$ . Thus $μ_{s c}$ is atomless, proving existence of the claimed three parts.

3.1L2L3step 1.1step 2.1algebra∎

For uniqueness, consider any such three-part decomposition. The absolutely continuous part has zero mass on singletons by [L2], and the singular-continuous part does by atomlessness. Thus the discrete part has singleton mass $μ ({x})$ at every $x$ . A measure concentrated on a countable set is determined by these masses through countable additivity, so this discrete part equals $μ_{d}$ from step 1.1. The other two parts therefore sum to the same finite remainder $η$ . The uniqueness clause of [L3] identifies them with $μ_{a c}$ and $μ_{s c}$ .

Depends on

Used by

Dependency tree · two levels

43 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

Gerald B. Folland, Real Analysis, 2nd ed., §1.5 (standard reference, not scraped)
John K. Hunter, Measure Theory, Example 2.37 and §6.8 (standard reference, not scraped)