Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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Total variation duality, kernel contraction, and coupling bound

Statement

Let μ,ν be probability measures on (X,X) and let P be a probability kernel on that space. Then ∥μ−ν∥TV=sup⁡f:X→[0,1] X-measurable∣∫f dμ−∫f dν∣, ∥μP−νP∥TV≤∥μ−ν∥TV. If (U,V) is a coupling of μ and ν on a probability space and E is a measurable event on which U=V, then ∥μ−ν∥TV≤P(Ec).

Facts & Assumptions

Given: Two probability measures μ,ν on the same measurable space, a probability kernel P, and, for the coupling conclusion, random elements (U,V) with laws μ,ν and a measurable event E such that U=V on E.

[F1]

Total variation distance for probability laws: total variation is sup⁡A∈X∣μ(A)−ν(A)∣.

[F2]

Measure kernel and probability kernel: P(x,⋅) is a probability measure and x↦P(x,A) is measurable for each A.

[F3]

Tonelli's theorem for nonnegative measurable functions on a sigma-finite product: nonnegative product-measurable functions may be integrated in either order.

Proof

Given: [F1]–[F3] and the hypotheses in the Statement.

1.1F1F3givenalgebra

For measurable f:X→[0,1], put At={x:f(x)>t} for t∈[0,1]. The identity f(x)=∫011At(x) dt and [F3] give ∫f dμ−∫f dν=∫01[μ(At)−ν(At)] dt. Its absolute value is at most ∫01∣μ(At)−ν(At)∣ dt≤∥μ−ν∥TV by [F1]. Conversely, each indicator 1A is an admissible f and attains the setwise difference for its set A. Taking suprema proves the dual formula.

2.1F1F2step 1.1given

For each A∈X, the function x↦P(x,A) is measurable and takes values in [0,1] by [F2]. Apply step 1.1 to that function to obtain ∣μP(A)−νP(A)∣≤∥μ−ν∥TV. Taking the supremum over A proves kernel contraction.

3.1F1givenalgebra∎

For any A∈X, ∣μ(A)−ν(A)∣=∣E[1A(U)−1A(V)]∣≤E∣1A(U)−1A(V)∣≤P(Ec), because the two indicators agree on E. Taking the supremum over A proves the stated bound.

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Sources