CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-31

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Counting measure on $[0, 1]$ shows the dominating measure needs sigma-finiteness

Statement refuted

Lebesgue measure on $[0, 1]$ has a Radon-Nikodym density with respect to counting measure on $[0, 1]$ .

Facts & Assumptions

Given: Counting measure $c$ on $[0, 1]$ and Lebesgue measure $λ$ on $[0, 1]$ .

[L1]

Counting measure is a measure, and for a singleton one has $c ({x}) = 1$ . (Counting measure on an arbitrary set, Counting measure is a measure)

[L2]

A density representation would mean $λ (E) = \int_{E} f d c$ for every measurable set $E$ . (The measure with density $f$ relative to $μ$ )

Counterexample

technique · direct

1.1

Suppose $λ (E) = \int_{E} f d c$ for every measurable set $E \subseteq [0, 1]$ . Applying this to a singleton ${x}$ and using [L1] gives $0 = λ ({x}) = \int_{{x}} f d c = f (x) (x \in [0, 1]) .$ Thus $f = 0$ pointwise.

L1L2assume-contraalgebra

2.1

Step 1.1 and [L2] then give $λ ([0, 1]) = \int_{[0, 1]} f d c = 0,$ contradicting the fact that the interval has Lebesgue mass $1$ . Therefore no such density exists.

step 1.1L2contradiction: total mass onedischarge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

13 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 6.23 and Example 6.28 (standard reference, not scraped)

Counting measure on [0,1][0,1] shows the dominating measure needs sigma-finiteness