Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-31
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Two Radon-Nikodym derivatives can differ on a null set

Statement refuted

Assume the Axiom of Countable Choice. The Radon-Nikodym derivative is a uniquely determined function.

Facts & Assumptions

Given: Countable choice, the zero measure on (R,B(R)), and the Cantor set C.

[L1]

The Cantor set is Lebesgue measurable and Lebesgue null. (The Cantor set is an uncountable subset of R of Lebesgue measure zero)

[L2]

The integral of a nonnegative function over a null set vanishes. (A nonnegative integral over a null set vanishes)

[L3]

A Radon-Nikodym derivative is only an almost-everywhere equivalence class of representing functions. (The Radon-Nikodym derivative as an almost-everywhere equivalence class)

Counterexample

technique · direct
1.1

Let h0:=0 and h1:=χC. For every measurable set E, [L1] and [L2] give Eh1dλ=λ(EC)=0=Eh0dλ. Thus both h0 and h1 represent the zero measure relative to λ.

L1L2construct
2.1

The functions h0 and h1 are not equal pointwise because h1=1 on C, but [L3] says only almost-everywhere equality is required. Hence pointwise uniqueness fails.

step 1.1L3

Depends on

Used by

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