CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)

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$x^{- 1} d λ$ on $(0, 1)$ shows finiteness is needed in the epsilon-delta criterion

Statement refuted

The epsilon-delta small-set condition characterises absolute continuity for every sigma-finite measure.

Facts & Assumptions

Given: The measure $ν (E) = \int_{E} x^{- 1} χ_{(0, 1)} (x) d λ (x)$ on $R$ .

[L1]

A nonnegative measurable density defines a measure (The measure with density $f$ relative to $μ$ ), and its integral over every Lebesgue-null set vanishes (A nonnegative integral over a null set vanishes), so the resulting measure is absolutely continuous with respect to $λ$ .

[L2]

Under its stated Axiom of Choice, the finite-measure theorem proves the epsilon-delta criterion only under a finiteness hypothesis. This contextual observation is not used to construct the counterexample (For finite signed or complex measures, absolute continuity is equivalent to the epsilon-delta small-set condition).

Counterexample

technique · direct

1.1L1given

By [L1], the measure $ν$ is absolutely continuous with respect to $λ$ .

2.1step 1.1givenalgebra∎

Let $ε : = 1$ and let $δ > 0$ . Put $a : = \min (δ / 2, 1 / 2)$ and $E : = (0, a) \subseteq (0, 1)$ . Then $λ (E) = a < δ$ , but $ν (E) = \int_{0}^{a} \frac{d x}{x} = + \infty > 1.$ The measure $ν$ is sigma-finite: $(0, 1) = ⋃_{m \geq 2} (1 / m, 1)$ and every displayed interval has finite $ν$ -mass, while $ν$ vanishes off $(0, 1)$ . Thus the epsilon-delta conclusion fails for an absolutely continuous sigma-finite measure.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 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

Richard F. Bass, Real Analysis for Graduate Students, Proposition 13.2 discussion (standard reference, not scraped)

x−1dλx−1dλ on (0,1)(0,1) shows finiteness is needed in the epsilon-delta criterion