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Total variation controls expectations of bounded functions

Statement

Let (E,E) be a measurable space, let P and Q be probability measures on it, and let f:E→R be measurable with ∣f∣≤M pointwise for some finite M≥0. With the convention ∥P−Q∥TV:=sup⁡A∈E∣P(A)−Q(A)∣, one has ∣∫f dP−∫f dQ∣≤2M∥P−Q∥TV. For every measurable g:E→R with 0≤g≤M pointwise, one also has ∣∫g dP−∫g dQ∣≤M∥P−Q∥TV. Both integral differences are zero when M=0.

Facts & Assumptions

Given: A measurable space (E,E), probability measures P,Q, and measurable real functions f,g satisfying ∣f∣≤M<∞ and 0≤g≤M.

[F1]

A probability measure has total mass one. Probability measures and probability spaces

[F2]

Total variation here is the event supremum sup⁡A∈E∣P(A)−Q(A)∣ and has no half-factor. Total variation distance for probability laws

[F3]

On nonnegative simple functions, the nonnegative Lebesgue integral agrees with the simple integral. The nonnegative integral agrees with the simple integral on simple functions

[F4]

The simple integral is the sum of the coefficients times the measures of the level sets; in particular, an indicator integrates to the measure of its set. The integral of a nonnegative simple function

[F5]

The nonnegative integral is additive, monotone, and nonnegatively homogeneous. Additivity of the nonnegative Lebesgue integral Monotonicity and nonnegative homogeneity of the nonnegative integral

[F6]

A real measurable function is integrable when its absolute value has finite nonnegative integral; the integral is linear on integrable functions. Integrable real and complex functions, and their integrals The Lebesgue integral is linear on L1(μ)

[F7]

For every η>0 there is an integer n≥1 with 1/n<η. For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε

Proof

Given: The hypotheses in the Statement.

1.1givenconstruct

If M=0, then f=0 and g=0 pointwise, so both claims hold. To prove the nonnegative claim for an arbitrary finite bound L>0, let w:E→R be measurable with 0≤w≤L, fix an integer n≥1, put Δ=L/n, and define Ak={x:w(x)≥kΔ} for 1≤k≤n and sn=Δ∑k=1n1Ak. The sets are measurable, sn is nonnegative simple, and 0≤w−sn≤Δ pointwise.

2.1F1F2F3F4F5step 1.1algebra

By [F3]–[F5], ∫sn dP=Δ∑k=1nP(Ak) and likewise for Q, so ∣∫sn dP−∫sn dQ∣≤nΔ∥P−Q∥TV=L∥P−Q∥TV. Also 0≤∫(w−sn) dP≤Δ and 0≤∫(w−sn) dQ≤Δ, because w−sn≤Δ and both measures have mass one. The triangle inequality therefore gives ∣∫w dP−∫w dQ∣≤L∥P−Q∥TV+2L/n.

3.1F7step 2.1algebra

Given any η>0, choose n by [F7] with 1/n<η/(2L). The estimate in step 2.1 then gives ∣∫w dP−∫w dQ∣<L∥P−Q∥TV+η. If the bound by L∥P−Q∥TV failed, choosing η to be half the positive excess would contradict this inequality. Taking w=g and L=M proves the stated nonnegative case when M>0.

4.1F1F3F4F5F6step 3.1algebra∎

For the signed case with M>0, let h=f+M. Then h is measurable and 0≤h≤2M, so step 3.1 with w=h and L=2M gives ∣∫h dP−∫h dQ∣≤2M∥P−Q∥TV. Also ∣f∣≤M, so [F1], [F3]–[F5] give ∫∣f∣ dP≤M and ∫∣f∣ dQ≤M; hence f is integrable under both measures by [F6]. The constant M has integral M under each probability measure, so linearity [F6] gives ∫f dP−∫f dQ=∫h dP−∫h dQ. This proves the signed case.

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