ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-30

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Cantor measure minus Lebesgue measure on [0,1] is already in Jordan form

Example

Assume the Axiom of Countable Choice. Let $μ_{c}$ be the Cantor measure and let $λ_{0} (E) : = λ (E \cap [0, 1])$ . Then $ν : = μ_{c} - λ_{0}$ is a signed measure whose Jordan decomposition is already $ν^{+} = μ_{c}, ν^{-} = λ_{0} .$

Facts & Assumptions

Given: The Cantor measure $μ_{c}$ and the restricted Lebesgue measure $λ_{0}$ on $[0, 1]$ .

[L1]

The Cantor measure is a singular probability measure concentrated on the Cantor set $C$ , and $λ (C) = 0$ . (The Cantor measure is a singular atomless probability measure concentrated on the Cantor set)

[L2]

Jordan decomposition is the unique decomposition of a signed measure into mutually singular positive parts. (Jordan decomposition of a signed measure into unique mutually singular positive parts)

Verification

technique · direct

1.1

By [L1], $μ_{c}$ vanishes on measurable subsets of $R ∖ C$ , [L1, L2] while $λ_{0}$ vanishes on measurable subsets of $C$ because $λ (C) = 0$ . Thus $μ_{c} ⊥ λ_{0}$ . Both are positive measures, so $ν = μ_{c} - λ_{0}$ is a signed measure already written as a difference of mutually singular positive measures.

2.1

The uniqueness clause in [L2] now forces [L2, step 1.1] ∎ $ν^{+} = μ_{c}, ν^{-} = λ_{0} .$ Hence $μ_{c} - λ_{0}$ is already in Jordan form.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

11 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

John K. Hunter, Measure Theory, Example 6.20 (standard reference, not scraped)
Richard F. Bass, Real Analysis for Graduate Students, Example 12.7 (standard reference, not scraped)