ExampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedprecheck pendingverified 2026-10-23 (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.

Unseen input invalidates a training-law guarantee

Example

Let $d_{s} = 2$ , $D = (0, 1)^{2}$ , and $I = [0, 1]$ . Let $X$ consist of the constant product-measurable fields $c e_{1} : D \times I \to R^{2}$ , where $c \in R$ and $e_{1} = (1, 0)$ , with norm $∥ c e_{1} ∥_{X} = ∣ c ∣$ and its Borel $σ$ -algebra; thus $X$ is isometric to $R$ and is a separable real Banach space. Let $Z = {0, 1}$ with its power-set $σ$ -algebra and define $S (z) = 0$ , $\tilde{S} (0) = 0$ , $\tilde{S} (1) = M e_{1}$ for arbitrary $M > 0$ , and $J (c e_{1}) = c$ , which is $1$ -Lipschitz. An $X$ -valued simple map means a map with finite range and measurable fibers. A map is strongly measurable with respect to a probability law if, outside a measurable set of measure zero, it is a pointwise norm limit of a sequence of simple maps. Here $S$ and $\tilde{S}$ are themselves simple maps, so their constant approximating sequences establish strong measurability for either law below. For $z \in Z$ , define the Dirac law $δ_{z}$ by $δ_{z} (E) = 1$ if $z \in E$ and $0$ otherwise. Take training law $μ = δ_{0}$ and target law $ν = δ_{1}$ . Then the training-law mean-square state error is zero, while the target-law observable bias is $M$ . Absolute continuity $ν ≪ μ$ means that $μ (E) = 0$ implies $ν (E) = 0$ for every measurable $E \subseteq Z$ . The target law is not absolutely continuous with respect to the training law, so the bounded-density hypothesis of the shift certificate fails; as $M$ is arbitrary, training-law error alone supplies no finite uniform target-bias bound.

Example references

Masashi Sugiyama, Matthias Krauledat, and Klaus-Robert Müller, “Covariate Shift Adaptation by Importance Weighted Cross Validation,” §2.1.

Dive into Deep Learning, §3.7.3.1, “Covariate Shift Correction”.

Facts & Assumptions

Given: The field domain $D \times I$ , the normed space $X$ , the finite input set and its power-set $σ$ -algebra, the maps $S, \tilde{S}$ , the observable $J$ , the laws $μ, ν$ , and $M > 0$ stated in the Example.

[A1]

$μ = δ_{0}$ and $ν = δ_{1}$ are probability measures on $Z$ .

[F1]

Lipschitz observable bias under bounded law shift: the stated target-bias certificate requires a target density $g \leq C$ with respect to the training law and a population mean-square training error bound.

Verification

Given: The data in [A1] and the Example.

1.1givenA1F1algebra

The norm identifies $X$ isometrically with $R$ , the maps have finite range and all their fibers are measurable because every subset of $Z$ is measurable, so they are simple maps and their constant approximating sequences prove strong measurability as defined in the Example, and $J$ is $1$ -Lipschitz; thus the state-space and measurability hypotheses of [F1] hold. Under $μ = δ_{0}$ , both $S (0)$ and $\tilde{S} (0)$ equal zero, so $E_{μ} ∥ S - \tilde{S} ∥_{X}^{2} = 0$ ; the separate state second moments are also finite because the maps have finite range.

2.1givenA1step 1.1algebra

Under $ν = δ_{1}$ , $J (S (1)) = 0$ and $J (\tilde{S} (1)) = M$ , hence $∣ E_{ν} J (S) - E_{ν} J (\tilde{S}) ∣ = M$ . Since $M$ is arbitrary, the target bias can be arbitrarily large while the training-law error remains zero.

3.1givenA1F1step 2.1∎

The set ${1}$ has $μ$ -measure zero and $ν$ -measure one. If a target density $g$ existed, then $ν ({1}) = \int_{{1}} g d μ = 0$ , a contradiction; hence $ν ≪̸ μ$ and the density hypothesis in [F1] fails.

Depends on

Lipschitz observable bias under bounded law shift