Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Conditional expectation is unique almost surely

Statement

If Y,Z are conditional-expectation versions of the same real integrable X given G, then Y=Z almost surely.

Facts & Assumptions

Given: A probability space, a sub-sigma-algebra G, an integrable real X, and two versions Y,Z with all its G-event integrals.

[F1]

Both versions have the same event integrals. (Conditional expectation given a sigma algebra)

[F2]

The difference is integrable and its integral is the difference of integrals. (The Lebesgue integral is linear on L1(μ))

[F3]

Differences and their positive and negative parts are measurable. (Closure properties of measurable functions used by the integral)

[F4]

Zero integral of a nonnegative function implies it vanishes almost everywhere. (A nonnegative measurable function has integral 0 exactly when it vanishes almost everywhere)

Proof

technique · direct
1.1

The difference D=YZ is G-measurable and integrable, and ADdP=0 for every AG. In particular the sets A+={D>0} and A={D<0} belong to G.

F1F2F3
2.1

On A+, D1A+=D+0 has zero integral; on A, D1A=D0 also has zero integral. By [F4], both parts vanish almost surely. Off the union of their two null exceptional sets, D=D+D=0, proving Y=Z almost surely.

step 1.1F4

Source notes

Durrett §4.1, uniqueness paragraph, printed p.206; van der Vaart Theorem 1.3, printed p.2.

Depends on

Used by

Dependency tree · two levels

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Sources