Alphabeta Math
TheoremStatement: AI-adaptedProof: 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.

Continuous mapping theorem

Statement

Let S,T be metric spaces, g:S→T measurable, and μn⇒μ. If the discontinuity set Dg is μ-null, then g∗μn⇒g∗μ. Consequently Xn⇒X implies g(Xn)⇒g(X) whenever PX(Dg)=0.

Facts & Assumptions

[F1]

Portmanteau theorem: For Borel probabilities μn,μ on a metric space S, the following are equivalent: (i) μn⇒μ; (ii) integrals converge for all bounded uniformly continuous real tests; (iii) lim sup⁡nμn(F)≤μ(F) for every closed F; (iv) lim inf⁡nμn(G)≥μ(G) for every open G; (v) μn(A)→μ(A) for every Borel A with μ(∂A)=0.

[F2]

Laws commute with measurable maps: Let X:(Ω,F,P)→(S,Σ) be a random element, and let g:(S,Σ)→(T,T) be measurable. Then g∘X is a random element and for every B∈T, Pg∘X(B)=PX(g−1(B)).

Proof

Given: The objects, hypotheses and definitions in the statement. Its conclusions are to be established below.

1.1givenalgebra

For r≥1 let Ur be the union of all open subsets V of S such that the diameter of g(V) is less than 1/r. The continuity set is ⋂rUr: continuity at x supplies such a neighborhood by making all image points within 1/(3r) of g(x); conversely a neighborhood with image diameter below ε forces d(g(y),g(x))<ε there. Thus Dg is Borel.

1.2F1

If F is closed in T and x lies outside g−1(F)∪Dg, continuity at x and the open complement of F give a neighborhood disjoint from g−1(F). Hence g−1(F)‾⊆g−1(F)∪Dg. Applying F1 to this closed preimage closure gives lim sup⁡nμn(g−1(F))≤μ(g−1(F)‾)≤μ(g−1(F)).

2.1F1F2step 1.2∎

The inequality in step 1.2 is the closed-set bound for the pushforward probabilities, so F1 gives their weak convergence. F2 identifies these pushforwards with the laws of g(Xn) and g(X), proving the random-element formulation.

Depends on

Used by

Dependency tree · two levels

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Sources