Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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Doubling is strongly mixing for Lebesgue measure

Statement

Assume countable choice. Doubling is strongly mixing for Borel Lebesgue probability and for its completion: λ(A∩D−nB)→λ(A)λ(B) for every pair of measurable circle sets.

Facts & Assumptions

[F1]

The strong-mixing theorem covers b=2 and both measure domains. Every integer-base circle map is strongly mixing.

[F2]

Doubling is the base-two fractional-part map. The circle, rotations and the doubling map.

Proof

Given: Assume countable choice. Doubling is strongly mixing for Borel Lebesgue probability and for its completion: λ(A∩D−nB)→λ(A)λ(B) for every pair of measurable circle sets.

1.1

The definition gives D=D_2. Taking b=2 in the integer-base theorem, with exactly its countable-choice assumption, yields the stated correlation limit for every Borel pair and every completed pair.

F1F2
2.1

For clarity, if A and B are dyadic intervals of depths r and s, the proof gives the exact value 2n−r2−(n+s)=2−r−s=λ(A)λ(B) for every n>=r. This includes the depth-zero whole circle; an empty test set gives zero. Thus the specialization retains the explicit dyadic correlation calculation as well as the general measurable-set conclusion.

step 1.1F1∎

Depends on

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Sources