Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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.

Support of a continuous pushforward is the image closure

Statement

Let X⊆Rk and Y⊆Rd be closed Euclidean subspaces. Let μ be a finite nonnegative Borel measure on X, define spt⁡μ={x:μ(U)>0 for every relatively open neighborhood U of x}, and let f:X→Y be continuous. Define the pushforward by f#μ(A)=μ(f−1(A)) for Borel A. Then

spt⁡(f#μ)=f(spt⁡μ)‾.

Facts & Assumptions

Given: μ and f as in the Statement.

Proof

1.1givenconstruct

If y=f(x) for x∈spt⁡μ and V is any open neighborhood of y, continuity makes f−1(V) an open neighborhood of x. Hence f#μ(V)=μ(f−1(V))>0, so y∈spt⁡(f#μ). Taking the closure relative to Y, since support is closed, f(spt⁡μ)‾⊆spt⁡(f#μ).

2.1givenstep 1.1choose∎

The complement of spt⁡μ has μ-measure zero: every point in it lies in a zero-measure open neighborhood, and the countable relative Euclidean basis gives a countable cover by zero-measure basis sets. If y∉f(spt⁡μ)‾, choose an open neighborhood V disjoint from that closure. Then f−1(V)⊆X∖spt⁡μ, so f#μ(V)=0 and y∉spt⁡(f#μ). Thus spt⁡(f#μ)⊆f(spt⁡μ)‾.

Used by

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources