Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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.

In the Carathéodory identity, the subadditive inequality is automatic

Statement

For every outer measure μ∗ and all A,E⊆X, μ∗(A)≤μ∗(A∩E)+μ∗(A∖E). Consequently, to prove that E is Carathéodory measurable (Carathéodory measurable sets), it suffices to prove the reverse inequality for every A⊆X.

Facts & Assumptions

Given: An outer measure μ∗ on X and subsets A,E⊆X.

[F1]

An outer measure on a set X is a function μ∗:P(X)→[0,+∞] that vanishes at the empty set, is monotone, and is countably subadditive. (Outer measures)

Proof

technique · direct
1.1givenalgebra

The test set decomposes as A=(A∩E)∪(A∖E).

2.1step 1.1F1algebra∎

Apply countable subadditivity in [F1] to the two sets in step 1.1 followed by empty sets. This gives μ∗(A)≤μ∗(A∩E)+μ∗(A∖E), including 0=0+0 when A=∅ and without subtracting when a value is +∞.

Depends on

Used by

Dependency tree · two levels

4 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