TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-21

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.

The limsup of the measures is at most the measure of the set limsup under a finite-union bound

Statement

Let $(E_{k})$ be measurable and suppose $μ (⋃_{k} E_{k}) < + \infty$ . Then

${lim sup}_{k \to \infty} μ (E_{k}) \leq μ ({lim sup}_{k \to \infty} E_{k}),$

where the numerical limsup is taken in $[0, + \infty]$ .

Facts & Assumptions

Given: A measure $μ$ , measurable sets $(E_{k})$ , and $μ (⋃_{k} E_{k}) < + \infty$ .

[L1]

For decreasing measurable sets, if one has finite measure, the measure of their intersection is the infimum of their measures (Continuity from above when one set has finite measure).

[L2]

Measures are monotone under inclusion (Measures are monotone).

[L3]

The set limsup is $⋂_{N} ⋃_{k \geq N} E_{k}$ (Limit superior and limit inferior of a sequence of sets).

[L4]

For a nonnegative extended sequence $(a_{k})$ , ${lim sup}_{k} a_{k} = \inf_{N} \sup_{k \geq N} a_{k}$ (Limit superior and limit inferior of a nonnegative extended-real sequence), and all these bounds exist (Every subset of $\overline{R}$ has a least upper bound and a greatest lower bound in $\overline{R}$ , agreeing with the real supremum and infimum on nonempty sets bounded in $R$ ).

Proof

technique · direct

1.1

Put $G_{N} : = ⋃_{k \geq N} E_{k}$ . Then $G_{N + 1} \subseteq G_{N}$ , $⋂_{N} G_{N} = {lim sup}_{k} E_{k}$ , and $G_{0} = ⋃_{k} E_{k}$ has finite measure.

givenL3

1.2

For every $N$ and $k \geq N$ , $E_{k} \subseteq G_{N}$ , so $μ (E_{k}) \leq μ (G_{N})$ and hence $\sup_{k \geq N} μ (E_{k}) \leq μ (G_{N})$ .

givenL2L4

2.1

Continuity from above and step 1.2 give ${lim sup}_{k} μ (E_{k}) = \inf_{N} \sup_{k \geq N} μ (E_{k}) \leq \inf_{N} μ (G_{N}) = μ ({lim sup}_{k} E_{k})$ .

step 1.1step 1.2L1L4∎

Depends on

Used by

Measures converge for a convergent sequence of sets contained in one finite-measure set Corollary

Dependency tree · two levels

16 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

G. Folland, Real Analysis, 2nd ed., §1.3, Exercise 8 (standard reference, not scraped)