LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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Finite unrolled domains and outputs are Borel

Statement

Let $n, m \geq 1$ , $X = R^{n}$ , and let $Ω$ be a Borel subset of $R^{m \times n} \times R^{m}$ with its relative Borel sigma-algebra. Equip $Ω \times X$ with the finite product sigma-algebra $B (Ω) \otimes B (X)$ . Fix an integer $N \geq 0$ , real parameters $τ_{0}, \dots, τ_{N - 1}$ , a Borel initialization $I : Ω \to X$ , Borel sets $D_{k} \subseteq Ω \times X$ , and Borel maps $B_{k} : D_{k} \to X$ for $0 \leq k < N$ , where each $D_{k}$ carries its relative product sigma-algebra. For $p = (A, y) \in Ω$ and $x \in X$ , define $c (p, x) = A^{T} (y - A x)$ . Set $E_{0} = Ω$ and $x_{0} = I$ ; recursively, for $0 \leq k < N$ , set

$E_{k + 1} : = {p \in E_{k} : (p, x_{k} (p) + τ_{k} c (p, x_{k} (p))) \in D_{k}}, x_{k + 1} (p) : = B_{k} (p, x_{k} (p) + τ_{k} c (p, x_{k} (p))) (p \in E_{k + 1}) .$

Then every $E_{k}$ is a Borel subset of $Ω$ and every $x_{k} : E_{k} \to X$ is Borel. In particular, $E_{N}$ is exactly the set of inputs where all $N$ updates are defined, and the finite output map $x_{N} : E_{N} \to X$ is Borel.

Statement references

Measurable spaces and relative measurable subsets.

Measurable maps are characterized by measurable preimages.

Finite product sigma-algebras are generated by measurable rectangles.

Continuous maps have Borel preimages of Borel sets.

Composition preserves measurability.

Facts & Assumptions

Given: The finite-dimensional Borel spaces, Borel update domains and maps, Borel initialization, fixed finite layer parameters, and the recursive definitions in the Statement.

[F1]

The product sigma-algebra and its finite iterates: the product sigma-algebra is generated by measurable rectangles.

[F2]

A measurable function between measurable spaces: a map is measurable when the preimage of each measurable codomain set is measurable.

[F3]

A continuous map has Borel preimages of Borel sets: a continuous map between topological spaces has Borel preimages of Borel sets.

[F4]

Composition with a Borel measurable outer map preserves measurability: the composition of measurable maps between measurable spaces is measurable.

Verification

technique · induction

1.1F1F3givenalgebra

The spaces $Ω$ and $X$ are second-countable subspaces of finite-dimensional Euclidean spaces, so rectangles from countable bases form a countable base for $Ω \times X$ . Every open set is therefore a countable union of open rectangles and belongs to $B (Ω) \otimes B (X)$ by [F1]. Conversely, for Borel $C \subseteq Ω$ and $F \subseteq X$ , $C \times F = π_{1}^{- 1} (C) \cap π_{2}^{- 1} (F)$ is Borel in the product topology because the coordinate projections are continuous and [F3] applies. Since such rectangles generate the product sigma-algebra by [F1], the product sigma-algebra is the Borel sigma-algebra of the product topology.

1.2givenbase

The base domain $E_{0} = Ω$ is Borel in itself, and the initial map $x_{0} = I$ is Borel by the Given clause.

1.3F1F2givenihassume-hypalgebra

Fix $k < N$ and assume as the induction hypothesis that $E_{k}$ is Borel and $x_{k} : E_{k} \to X$ is Borel. Define $ι_{k} : E_{k} \to Ω \times X$ by $ι_{k} (p) = (p, x_{k} (p))$ . For every measurable rectangle $C \times F$ , [F2] and the induction hypothesis give $ι_{k}^{- 1} (C \times F) = (E_{k} \cap C) \cap x_{k}^{- 1} (F)$ Borel in $E_{k}$ . The rectangles generate the target sigma-algebra by [F1], so preimages preserve the generated sigma-algebra and $ι_{k}$ is Borel.

2.1F3F4step 1.1step 1.3algebra

Define $Γ_{k} (A, y, x) = (A, y, x + τ_{k} A^{T} (y - A x))$ . Each coordinate of its last component is a polynomial in the matrix, data, and state coordinates, so $Γ_{k}$ is continuous. By [F3] it is Borel; hence $ψ_{k} : = Γ_{k} \circ ι_{k}$ is Borel by [F4].

3.1F2step 1.2step 2.1algebra

The next domain is $E_{k + 1} = ψ_{k}^{- 1} (D_{k})$ . Since $D_{k}$ is Borel and $ψ_{k}$ is Borel, [F2] makes $E_{k + 1}$ Borel in $E_{k}$ ; because $E_{k}$ is Borel in $Ω$ , $E_{k + 1}$ is Borel in $Ω$ .

4.1F2F4step 2.1step 3.1given

The restriction $ψ_{k} ∣_{E_{k + 1}} : E_{k + 1} \to D_{k}$ is Borel: the preimage of each relative-measurable set $D_{k} \cap H$ is $E_{k + 1} \cap ψ_{k}^{- 1} (H)$ , which is Borel. Since $B_{k} : D_{k} \to X$ is Borel, [F4] makes $x_{k + 1} = B_{k} \circ (ψ_{k} ∣_{E_{k + 1}})$ Borel.

5.1step 1.2step 1.3step 2.1step 3.1step 4.1discharge-induction: step 1.3∎

Step 1.2 supplies the base case, and steps 1.3, 2.1, 3.1, and 4.1 prove the implication from any Borel pair $(E_{k}, x_{k})$ to $(E_{k + 1}, x_{k + 1})$ . Finite induction gives the claim for every $k \leq N$ . By its recursive definition, $E_{N}$ contains exactly the inputs at which each successive update is defined, and step 4.1 gives Borel measurability of the final output.

Depends on

Used by

Unrolled regularization architecture Definition

Dependency tree · two levels

11 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

Terence Tao, An Introduction to Measure Theory, preliminary full-text book (standard reference, not scraped)