Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)
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Absolute continuity of the integral

Statement

Let f∈L1(μ) and let ε>0. Then there is δ>0 such that for every measurable E, μ(E)<δ⟹∫E∣f∣ dμ<ε.

Facts & Assumptions

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

[L1]

The truncations ∣f∣∧n increase pointwise to ∣f∣, so their integrals converge to ∫∣f∣ dμ by monotone convergence (Monotone convergence for the integral).

[L2]

The integral over a measurable set is defined by ∫Eh dμ=∫hχE dμ (Integral over a measurable subset).

[L3]

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

[L4]

Integrability means ∫∣f∣ dμ<+∞ (Integrable real and complex functions, and their integrals).

[L5]

The nonnegative integral is additive (Additivity of the nonnegative Lebesgue integral).

Proof

technique · direct
1.1L1L4L5choose

Put hn=∣f∣−∣f∣∧n≥0. The pointwise identity ∣f∣=(∣f∣∧n)+hn and [L5] give ∫hn=∫∣f∣−∫(∣f∣∧n); the subtraction is valid because both integrals are finite by [L4]. By [L1] choose n so large that ∫hn dμ<ε/2, and put δ:=ε/(2n+1).

2.1step 1.1L2L3L5algebra∎

If μ(E)<δ, then [L2], [L3], and [L5] give ∫E∣f∣ dμ=∫E(∣f∣∧n) dμ+∫Ehn dμ≤nμ(E)+∫hn dμ<nδ+ε/2<ε. The last inequality follows from n/(2n+1)<1/2.

Depends on

Used by

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