ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31

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The Lebesgue decomposition of one half Lebesgue plus one half Cantor measure

Example

Assume the Axiom of Countable Choice. Let $ν : = \frac{1}{2} λ ⁣ ↾_{[0, 1]} + \frac{1}{2} μ_{c},$ where $μ_{c}$ is the Cantor measure. Then the Lebesgue decomposition of $ν$ relative to $λ$ is $ν_{a} = \frac{1}{2} λ ⁣ ↾_{[0, 1]}, ν_{s} = \frac{1}{2} μ_{c} .$

Facts & Assumptions

Given: The measure $ν = \frac{1}{2} λ ⁣ ↾_{[0, 1]} + \frac{1}{2} μ_{c}$ .

[L1]

The Cantor measure is singular with respect to Lebesgue measure. (The Cantor measure is a singular atomless probability measure concentrated on the Cantor set)

[A1]

The restriction $λ ⁣ ↾_{[0, 1]}$ is absolutely continuous with respect to $λ$ because intersection with $[0, 1]$ preserves Lebesgue-null sets.

[L3]

Under a common finite exhaustion, the Lebesgue decomposition relative to a fixed positive measure is unique (The Lebesgue decomposition of a sigma-finite signed measure is unique).

Verification

technique · direct

1.1

The measure $\frac{1}{2} λ ⁣ ↾_{[0, 1]}$ is absolutely continuous with respect to $λ$ by [A1], while $\frac{1}{2} μ_{c}$ is singular with respect to $λ$ by [L1]. Their sum is $ν$ by definition.

A1L1given

2.1

The sets $X_{n} = [- n, n]$ form a common finite exhaustion for $λ$ and the finite measure $ν$ . Since $ν = \frac{1}{2} λ ⁣ ↾_{[0, 1]} + \frac{1}{2} μ_{c}$ is already a decomposition into an absolutely continuous part and a singular part, [L3] forces it to be the Lebesgue decomposition of $ν$ relative to $λ$ .

step 1.1L3∎

Depends on

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