TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28

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On a finite measure space, uniform integrability is equivalent to L^1-boundedness plus uniform absolute continuity

Statement

Let $(X, A, μ)$ be a measure space with $μ (X) < + \infty$ , and let $F \subseteq L^{1} (μ)$ be a family of integrable real-valued functions. Then the following are equivalent:

$F$ is uniformly integrable;
$\sup_{f \in F} \int ∣ f ∣ d μ < + \infty$ and for every $ε > 0$ there is $δ > 0$ such that $μ (E) < δ \Rightarrow \sup_{f \in F} \int_{E} ∣ f ∣ d μ < ε$ .

Facts & Assumptions

Given: A finite measure space $(X, A, μ)$ and a family $F \subseteq L^{1} (μ)$ .

[L1]

Uniform integrability means that for every $ε > 0$ there is $M > 0$ such that $f \in F \Rightarrow \int_{{∣ f ∣ > M}} ∣ f ∣ d μ < ε$ . (A uniformly integrable family)

Proof

technique · direct

1.1L1algebra

Assume $F$ is uniformly integrable. Applying [L1] with $ε : = 1$ gives $M > 0$ such that $\int_{{∣ f ∣ > M}} ∣ f ∣ d μ < 1$ for every $f \in F$ . Then $\int ∣ f ∣ d μ \leq \int_{{∣ f ∣ \leq M}} ∣ f ∣ d μ + \int_{{∣ f ∣ > M}} ∣ f ∣ d μ \leq M μ (X) + 1,$ so $F$ is $L^{1}$ -bounded.

1.2L1algebra

Conversely, assume $F$ is $L^{1}$ -bounded by some constant $C$ , and assume the stated uniform absolute continuity. Let $ε > 0$ , and choose $δ > 0$ such that $μ (E) < δ \Rightarrow \sup_{f \in F} \int_{E} ∣ f ∣ d μ < ε$ . Choose $M > C / δ$ . For $f \in F$ , put $E_{f} : = {∣ f ∣ > M}$ . Then $M μ (E_{f}) \leq \int ∣ f ∣ d μ \leq C$ , so $μ (E_{f}) < δ$ and hence $\int_{{∣ f ∣ > M}} ∣ f ∣ d μ = \int_{E_{f}} ∣ f ∣ d μ < ε .$ Since this bound is uniform in $f$ , [L1] holds.

2.1step 1.1L1algebra

Still under step 1.1, let $ε > 0$ . Use [L1] with $ε / 2$ to choose $M > 0$ such that $\int_{{∣ f ∣ > M}} ∣ f ∣ d μ < ε / 2$ for every $f \in F$ , and put $δ : = ε / (2 M + 1)$ . If $μ (E) < δ$ , then for every $f \in F$ , $\int_{E} ∣ f ∣ d μ \leq \int_{E \cap {∣ f ∣ \leq M}} ∣ f ∣ d μ + \int_{{∣ f ∣ > M}} ∣ f ∣ d μ \leq M μ (E) + ε / 2 < ε .$ So $F$ has the stated uniform absolute continuity.

3.1step 1.1step 2.1step 1.2∎

Steps 1.1 and 2.1 prove that uniform integrability implies clause 2, and step 1.2 proves the converse implication.

Depends on

Used by

Dependency tree · two levels

6 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, Exercise 7.21 (standard reference, not scraped)
Terence Tao, 245A Notes 4: Modes of convergence, Exercises 23 and 24 (standard reference, not scraped)