PropositionStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Dominated families are uniformly integrable

Statement

Let $(X, A, μ)$ be a measure space. If $F \subseteq L^{1} (μ)$ and there is a nonnegative function $g \in L^{1} (μ)$ such that $∣ f ∣ \leq g$ almost everywhere for every $f \in F$ , then $F$ is uniformly integrable.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , a family $F \subseteq L^{1} (μ)$ , and a nonnegative integrable function $g$ with $∣ f ∣ \leq g$ almost everywhere for every $f \in F$ .

[L1]

If $g \in L^{1} (μ)$ and $ε > 0$ , then there is $δ > 0$ such that $μ (E) < δ \Rightarrow \int_{E} g d μ < ε$ . (Absolute continuity of the integral)

[L2]

If $h : X \to [0, + \infty]$ is measurable and $t > 0$ , then $μ ({h \geq t}) \leq t^{- 1} \int h d μ$ . (Chebyshev-Markov inequality for the integral)

[L3]

If two integrable functions are equal almost everywhere, then their integrals over every measurable set agree. (Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree)

[L4]

For a measurable set $E$ and a nonnegative measurable function $u$ , $\int_{E} u d μ = \int u χ_{E} d μ$ . (Integral over a measurable subset)

[L5]

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

Proof

technique · direct

1.1

Let $ε > 0$ . Use [L1] for $g$ to choose $δ > 0$ such that $μ (E) < δ$ implies $\int_{E} g d μ < ε$ . Since $g \in L^{1} (μ)$ , choose $M > δ^{- 1} \int g d μ$ . Then [L2] applied to $h : = g$ gives $μ ({g \geq M}) \leq M^{- 1} \int g d μ < δ,$ and so [L1] yields $\int_{{g \geq M}} g d μ < ε .$

L1L2choosealgebra

2.1

Fix $f \in F$ , and choose a measurable null set $N_{f}$ such that $∣ f ∣ \leq g$ on $X ∖ N_{f}$ . Put $u_{f} : = ∣ f ∣ χ_{X ∖ N_{f}} .$ Then $u_{f}$ is integrable, $u_{f} = ∣ f ∣$ almost everywhere, and $u_{f} χ_{{∣ f ∣ > M}} \leq g χ_{{g \geq M}}$ pointwise. Therefore [L3], [L4], and [L5] give $\int_{{∣ f ∣ > M}} ∣ f ∣ d μ = \int_{{∣ f ∣ > M}} u_{f} d μ \leq \int_{{g \geq M}} g d μ < ε .$ Since the same $M$ works for every $f \in F$ , this is exactly uniform integrability.

step 1.1L3L4L5construct

3.1

The family $F$ is uniformly integrable.

step 2.1L1∎

Depends on

Used by

Dominated convergence is a Vitali corollary Corollary

Dependency tree · two levels

15 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

Terence Tao, 245A Notes 4: Modes of convergence, Exercise 21.2 (standard reference, not scraped)