CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)

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.

L² convergence alone does not control exponential tilts

Statement refuted

Under the Axiom of Choice and its countable-family consequence, on a probability space, if real random variables $U_{n}$ converge to $0$ in $L^{2}$ , then their exponential normalizers $E [e^{- U_{n}}]$ are uniformly bounded.

Statement references

Stable $P (φ)_{2}$ construction contract

Facts & Assumptions

Given: The probability space $([0, 1], B, ℓ)$ with Lebesgue probability measure $ℓ$ , the construction-contract meaning of an exponential normalizer, and integers $n \geq 1$ .

[A1]

The Axiom of Choice and its countable-family consequence are inherited from the contract context; no selection is used in the explicit interval construction. The Axiom of Choice The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F1]

The proposed continuum interaction contract requires a finite positive exponential normalization as a separate proof obligation; it does not infer one from an $L^{2}$ limit. Stable $P (φ)_{2}$ construction contract

[F2]

For every fixed nonnegative integer power and positive scale, the exponential dominates that power at infinity. The exponential dominates every fixed nonnegative integer power at $+ \infty$

Counterexample

Given: [A1], [F1]–[F2], and $n \geq 1$ . Let $A_{n} = [0, e^{- n / 2}]$ and set $U_{n} (x) = - n$ for $x \in A_{n}$ , while $U_{n} (x) = 0$ for $x \notin A_{n}$ .

1.1A1F2givenalgebra

The set $A_{n}$ has Lebesgue measure $e^{- n / 2}$ , so $∥ U_{n} ∥_{2}^{2} = \int_{A_{n}} n^{2} d ℓ = n^{2} e^{- n / 2}$ . By [F2] with power $2$ and scale $1 / 2$ , this tends to $0$ ; hence $U_{n} \to 0$ in $L^{2}$ .

2.1F1F2step 1.1givenalgebra∎

On $A_{n}$ , $e^{- U_{n}} = e^{n}$ , and on its complement $e^{- U_{n}} = 1$ . Therefore $E [e^{- U_{n}}] = e^{n} e^{- n / 2} + 1 - e^{- n / 2} = e^{n / 2} + 1 - e^{- n / 2}$ , which tends to $+ \infty$ by [F2]. Thus these $L^{2}$ -convergent variables have unbounded exponential normalizers, contradicting the refuted implication.

L² convergence alone does not control exponential tilts

Statement refuted

Statement references

Facts & Assumptions

Counterexample

Depends on

Dependency tree · 2 levels

Sources