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

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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]

The finite-measure theorem proves the epsilon-delta criterion only under a finiteness hypothesis. (For finite signed or complex measures, absolute continuity is equivalent to the epsilon-delta small-set condition)

Counterexample

technique · direct

1.1

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

L1given

2.1

Let $ε : = 1$ and let $δ > 0$ . Put $E : = (0, δ / 2)$ . Then $λ (E) = δ / 2 < δ$ , but $ν (E) = \int_{0}^{δ / 2} \frac{d x}{x} = + \infty > 1.$ So the epsilon-delta conclusion fails for this absolutely continuous sigma-finite measure.

step 1.1L2algebra∎

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

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