Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck pass
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.

Integration against a Dirac measure evaluates at its point

Statement

Let A be a sigma-algebra on a set X, let x0∈X, and let δx0 be the Dirac probability measure on (X,A). For every nonnegative A-measurable f:X→[0,+∞], ∫Xf dδx0=f(x0).

Facts & Assumptions

Given: A measurable space, a point x0 in it, its Dirac measure, and a nonnegative measurable function f.

[F1]

δx0(A)=1 if x0∈A and 0 otherwise, and this set function is a probability measure on (X,A). (The Dirac set function at a point, A Dirac set function is a probability measure)

[F2]

A nonnegative simple function has a finite representation by disjoint measurable sets, and its simple integral is the sum of its values times the set measures. (Nonnegative simple measurable functions, The integral of a nonnegative simple function, The nonnegative integral agrees with the simple integral on simple functions)

[F3]

Every nonnegative measurable function is the pointwise increasing limit of nonnegative simple functions. (Every nonnegative measurable function is the increasing limit of simple measurable functions)

[F4]

Increasing limits pass through the nonnegative Lebesgue integral. (Monotone convergence for the integral)

Proof

technique · direct
1.1F1F2

If s=∑j=1mcj1Aj is a nonnegative simple function with pairwise disjoint Aj∈A, then [F1] and [F2] give ∫s dδx0=∑jcjδx0(Aj)=∑jcj1Aj(x0)=s(x0).

2.1F3F4step 1.1choose∎

Choose the increasing simple approximation sn↑f from [F3]. By [F4] and step 1.1, ∫f dδx0=lim⁡n∫sn dδx0=lim⁡nsn(x0)=f(x0), including when the value is +∞.

Depends on

Used by

Dependency tree · two levels

23 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