Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-27
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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.

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An indicator function is measurable exactly when its set is measurable

Statement

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

1E(x):={1,xE,0,xE

is measurable as a map XR if and only if EA.

Facts & Assumptions

Given: A measurable space (X,A), a subset EX, 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
1.1

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

givenL1
1.2

If 1E is measurable, then

givenL1

E=1E1((1/2,)),

and (1/2,) is a Borel subset of R. Hence [L1] gives EA. [given, L1]

2.1

Steps 1.1 and 1.2 prove the equivalence.

step 1.1step 1.2

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