CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-31

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Lebesgue plus counting measure has no Lebesgue decomposition relative to Lebesgue measure

Statement refuted

The measure $λ ⁣ ↾_{[0, 1]} + c$ on $[0, 1]$ admits a Lebesgue decomposition relative to Lebesgue measure, where $c$ is counting measure.

Facts & Assumptions

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

[L1]

Counting measure gives value $1$ to every singleton. (Counting measure on an arbitrary set, Counting measure is a measure)

[L2]

A Lebesgue decomposition would have the form $ν = ν_{a} + ν_{s}$ with $ν_{a} ≪ λ$ and $ν_{s} ⊥ λ$ . (Every sigma-finite signed measure admits a Lebesgue decomposition relative to a sigma-finite positive measure)

Counterexample

technique · direct

1.1

Suppose $ν = ν_{a} + ν_{s}$ were such a decomposition. For each $x \in [0, 1]$ , absolute continuity gives $ν_{a} ({x}) = 0$ , so $ν_{s} ({x}) = ν ({x}) = 1.$

L1L2assume-contraalgebra

2.1

If $ν_{s} ⊥ λ$ , then $ν_{s}$ is concentrated on some Lebesgue-null set $N$ . But every $x \in [0, 1] ∖ N$ satisfies ${x} \subseteq [0, 1] ∖ N$ , so concentration would force $ν_{s} ({x}) = 0$ , contradicting step 1.1. Therefore no such decomposition exists.

step 1.1contradiction: concentration off a null setdischarge-contradiction∎

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