TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27

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Absolute continuity of the integral

Statement

Let $f \in L^{1} (μ)$ and let $ε > 0$ . Then there is $δ > 0$ such that for every measurable $E$ , $μ (E) < δ ⟹ \int_{E} ∣ f ∣ d μ < ε .$

Facts & Assumptions

Given: An integrable function $f$ and a real number $ε > 0$ .

[L1]

The truncations $∣ f ∣ \land n$ increase pointwise to $∣ f ∣$ , so their integrals converge to $\int ∣ f ∣ d μ$ by monotone convergence (Monotone convergence for the integral).

[L2]

The integral over a measurable set is defined by $\int_{E} h d μ = \int h χ_{E} d μ$ (Integral over a measurable subset).

[L3]

The nonnegative integral is monotone and homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[L4]

Integrability means $\int ∣ f ∣ d μ < + \infty$ (Integrable real and complex functions, and their integrals).

Proof

technique · direct

1.1

Choose $n$ so large that [L1, L4, choose] $\int (∣ f ∣ - ∣ f ∣ \land n) d μ < ε / 2,$ which is possible by [L1] and [L4]. Put $δ : = ε / (2 n + 1)$ .

2.1

If $μ (E) < δ$ , then [step 1.1, L2, L3, algebra] ∎ $\int_{E} ∣ f ∣ d μ \leq \int_{E} (∣ f ∣ \land n) d μ + \int (∣ f ∣ - ∣ f ∣ \land n) d μ \leq n μ (E) + ε / 2 < ε,$ using [L2] and [L3] for the first term.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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 Notes, Proposition 4.16 (standard reference, not scraped)