Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedverified 2026-09-23 (gpt-6-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.

An indicator function is measurable exactly when its set is measurable

Statement

Let (X,A) be a measurable space and let E⊆X. The indicator function

1E(x):={1,x∈E,0,x∉E

is measurable as a map X→R if and only if E∈A.

Facts & Assumptions

Given: A measurable space (X,A), a subset E⊆X, and the indicator function 1E.

[L1]

A function is measurable exactly when the preimage of every measurable set in the codomain is measurable in the domain. (A measurable function between measurable spaces)

Proof

technique · direct
1.1givenL1

If E∈A, then the preimage of any Borel set B⊆R under 1E is one of ∅, X, E, or X∖E, because 1E takes only the values 0 and 1. Each of those sets lies in A, so 1E is measurable by [L1].

1.2givenL1

If 1E is measurable, then E=1E−1((1/2,∞)). The interval (1/2,∞) is Borel, so [L1] gives E∈A.

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove the equivalence.

Depends on

Used by

Dependency tree · two levels

2 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