Alphabeta Math
CounterexampleConstruction: AI-generatedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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.

Conditioning does not preserve strict inequalities

Statement refuted

Even when X(ω)<Y(ω) at every point, arbitrary versions of their conditional expectations need not satisfy that strict inequality at every point. This is a pointwise-version counterexample; strict almost-sure inequalities are preserved.

Facts & Assumptions

Given: The claim that pointwise strict input order must hold for every pair of conditional versions at every point; a two-atom witness will be constructed.

[F1]

The class is specified by event integrals, not by fixed values at null points. (Conditional expectation as an ae class)

[F2]

Strict almost-sure order is preserved by conditional expectation. (Basic algebra and order properties of conditional expectation)

Counterexample

technique · direct
1.1

On Ω={a,b} take the full sigma-algebra G=F and P(A)=1{b∈A}. Countable additivity holds because at most one member of a disjoint sequence contains b, and total mass is one. Let X=0 and Y=1 everywhere. Then X<Y at both points. Set S(a)=2,S(b)=0 and T=1 everywhere; these are measurable and integrable.

given
2.1

For every A, ∫AS=0=∫AX and ∫AT=P(A)=∫AY, so [F1] makes S,T conditional versions. At a, however, S(a)=2 is larger than T(a)=1. At the mass-one point b, S(b)=0<T(b)=1. Hence the failed strict pointwise inequality is fully consistent with the strict almost-sure order theorem [F2].

step 1.1F1F2∎

Source notes

Durrett §4.1 version convention, printed p.206. The local basic-properties theorem proves strict almost-sure preservation. The stable requested ID retains the Step-3-approved pointwise interpretation.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

12 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