SOME DUALITY RESULTS FOR EQUIVALENCE COUPLINGS AND TOTAL VARIATION

. Let (Ω , F ) be a standard Borel space and P ( F ) the collection of all probability measures on F . Let E ⊂ Ω × Ω be a measurable equivalence relation, that is, E ∈ F⊗F and the relation on Ω deﬁned as x ∼ y ⇔ ( x,y ) ∈ E is reﬂexive, symmetric and transitive. It is shown that there are two σ -ﬁelds G 0 and G 1 on Ω such that, for all µ, ν ∈ P ( F ),


Introduction
Throughout, (Ω, F ) is a measurable space, P(F ) the collection of all probability measures on F , and E ⊂ Ω × Ω a measurable equivalence relation.This means that E ∈ F ⊗ F and the relation on Ω defined as x ∼ y ⇔ (x, y) ∈ E is reflexive, symmetric and transitive.
The following notion of duality has been recently introduced by Jaffe [5].Given a sub-σ-field G ⊂ F , the pair (E, G) is said to satisfy strong duality if min P ∈Γ(µ,ν) for all µ, ν ∈ P(F ).
Here, as usual, Γ(µ, ν) is the set of all probability measures on F ⊗F with marginals µ and ν and the notation "min" asserts that the infimum is actually achieved.Moreover, is the total variation between µ and ν on G.
In addition to be intriguing from the foundational point of view, strong duality is useful in some probabilistic frameworks.Examples concern stochastic calculus, point processes, and random sequence simulation; see Section 2 of [5].
Say that E is strongly dualizable if (E, G) satisfies strong duality for some sub-σfield G ⊂ F .Various conditions for E to be strongly dualizable are given in [5] but no measurable equivalence relation which fails to be strongly dualizable is known to date.This suggests the conjecture that, under mild conditions on (Ω, F ) (say (Ω, F ) is a standard Borel space), every measurable equivalence relation is strongly dualizable.
This note focus on strong duality and includes three results.Let Such a G 0 is a sub-σ-field of F which plays a special role as regards strong duality.
In fact, E is strongly dualizable if and only if (E, G 0 ) satisfies strong duality; see [5,Prop. 3.15].Our first result is that, for all µ, ν ∈ P(F ), there is ν 0 ∈ P(F ) satisfying ν 0 = ν on G 0 and min Roughly speaking, the above condition means that strong duality is always true up to changing one between µ and ν out of G 0 .This is quite reasonable, in a sense, for µ − ν G0 only involves the restrictions of µ and ν on G 0 .
Next, suppose (Ω, F ) is a standard Borel space and denote by F the collection of those subsets of Ω which are universally measurable with respect to F ; see Section 2. Define also This time, G 1 is not a sub-σ-field of F .However, by our second result, one obtains inf In addition, the inf is achieved if P is allowed to be finitely additive.Precisely, min where M (µ, ν) is the collection of finitely additive probabilities on F ⊗ F with marginals µ and ν.
Finally, for each B ⊂ Ω, define Then, for all µ, ν ∈ P(F ), there is a set B ∈ F such that µ(B) = ν(B) = 1 and inf If compared with (1), the latter result has the advantage that G B is a sub-σ-field of F but the disadvantage that G B is not universal, for it depends on the pair (µ, ν).Note also that µ(B) = ν(B) = 1 and E ∩ (B × B) is a measurable equivalence relation on B. Therefore, for fixed (µ, ν), one can replace Ω with B and E with E ∩ (B × B).After doing this, everything works as regards the total variation side of strong duality.

Preliminaries
In this section, we introduce some further notation and recall a few known facts.
Let (S, E) be a measurable space.Then, P(E) denotes the set of probability measures on E and M (E) the set of bounded E-measurable functions f : S → R. We write µ(f ) = f dµ whenever µ ∈ P(E) and f ∈ M (E).
Moreover, we let where E µ is the completion of E with respect to µ.The elements of E are usually called universally measurable with respect to E. With a slight abuse of notation, for each µ ∈ P(E), the unique extension of µ to E is still denoted by µ.
If T is any topological space, B(T ) denotes the Borel σ-field.We say that T is Polish if its topology is induced by a distance d such that (T, d) is a complete separable metric space.Moreover, the measurable space (S, E) is a standard Borel space if E = B(S) for some Polish topology on S.
In a sense, perfectness is a non-topological analogous of the notion of tightness.In fact, if S is separable metric and E = B(S), then µ is perfect if and only if it is tight.
In particular, each element of P(E) is perfect whenever (S, E) is a standard Borel space.We refer to [9] for more on perfect probability measures.
Moreover, let c : S × S → R be a bounded measurable cost function.(Boundedness of c is generally superfluous and has been assumed for the sake of simplicity only).A primal minimizer, or an optimal coupling, is a probability measure P ∈ Γ(µ, ν) For a primal minimizer to exist, it suffices that S is separable metric, E = B(S), µ and ν are perfect, and the cost c is lower semi-continuous.To state the duality result, we denote by L the set of pairs (f, g) satisfying Then, in view of [10], one obtains inf P ∈Γ(µ,ν) provided at least one between µ and ν is perfect.
Finally, let D ⊂ E be a sub-σ-field and µ, ν ∈ P(E).For any measure γ on E, we write γ|D to denote the restriction of γ on D. The total variation between µ and ν on D is For our purposes, two remarks are in order.First, • D can be written as Letting A = f > g , one obtains A ∈ D (since f and g are D-measurable) and The second remark consists in the following lemma, which slightly improves some well known facts; see e. Lemma 1.Let D ⊂ E be a sub-σ-field and µ, ν ∈ P(E).Then, there are a probability space (Φ, A, P) and two measurable maps X, Y : (Φ, A) → (S, E) such that and to take P as the only extension of Q to A such that P(X = Y ) = 0.
Suppose now that µ|D = ν|D.Define λ, f and g by (3) and such a γ is a probability measure on E. Let (Φ, C, Q) be any probability space which supports three independent random variables U, X, Z with U uniformly distributed on (0, 1) and Moreover, for each A ∈ E, In particular, since Finally, denoting by Q * and Q * the inner and outer measures corresponding to Q, one obtains Therefore, to conclude the proof, it suffices to take (Φ, A, P) as the completion of (Φ, C, Q).
Regarding Lemma 1, we note that Y can be taken such that P(Y ∈ A) = ν(A) for all A ∈ E (and not only for all A ∈ D) provided µ|D = ν|D and µ|D ≪ ν|D.In this case, in fact, λ can be replaced by ν in the previous proof.Hence, g = 1 and

Results
It is quite intuitive that, when investigating strong duality, the partition of Ω in the equivalence classes of E plays a role.Let Π denote such a partition, i.e.
The σ-fields G 0 and G 1 , introduced in Section 1, can be written as where F denotes the universally measurable σ-field with respect to F .Another useful fact is provided the set A ⊂ Ω is a union of elements of Π.
Our starting point is the following.
Since (Ω, F ) is standard Borel, µ and ν are perfect.Therefore, inf On noting that sup f + sup g = sup Next, fix ǫ > 0 and take (f, g) ∈ L such that 0 ≤ f ≤ 1 and f (y) and note that h(x) + g(y) ≤ c(x, y) for all (x, y).Letting y = x, one obtains for all x ∈ Ω.
Since h(x) = h(y) whenever (x, y) ∈ E, for each a ∈ R the set h > a is a union of elements of Π.Moreover, since (Ω, F ) is standard Borel, the projection theorem yields h > a = x ∈ Ω : (x, y) ∈ E and f (y) > a for some y ∈ Ω ∈ F ; see e.g.Theorem A1.4, page 562, of [6].Hence, h > a ∈ G 1 .To sum up, Therefore, Finally, fix P ∈ Γ(µ, ν) and A ∈ G 1 .Since A is a union of elements of Π, inequality (4) yields
If regarded as a tool to get strong duality, Theorem 2 has two gaps: • Theorem 2 involves the inf and not the min over Γ(µ, ν).
The rest of this note focus on these two points.
3.1.G 1 is not a sub-σ-field of F .The fact that G 1 is not a sub-σ-field of F may be seen as unsuitable.Hence, we now prove the following result.
Theorem 3. If (Ω, F ) is a standard Borel space, then, for all µ, ν ∈ P(F ), there is a set B ∈ F such that Proof.Let µ, ν ∈ P(F ).By (2) and Theorem 2, there is D ∈ G 1 such that inf P ∈Γ(µ,ν) Since D is universally measurable with respect to F , there is A ∈ F such that µ + ν 2 A∆D = 0, or equivalently µ(A∆D) = ν(A∆D) = 0. Let Since D is a union of elements of Π, then 1 D (x) = 1 D (y) for all (x, y) ∈ E. Hence, where the first equality is because µ(A∆D) = ν(A∆D) = 0. Since (Ω, F ) is standard Borel and P (T ) = 0 for all P ∈ Γ(µ, ν), there is see [4] and [12, p. 2345].Therefore A ∈ G B , which in turn implies inf To prove the reverse inequality, fix any C ∈ G B and P ∈ Γ(µ, ν).Then, Hence, As already noted, the advantage of Theorem 3 with respect to Theorem 2 is that G B is a sub-σ-field of F while G 1 is not.The disadvantage is that G B is not universal, for it depends on the pair (µ, ν).However, for fixed (µ, ν), since µ(B) = ν(B) = 1 and E ∩ (B × B) is a measurable equivalence relation on B, it is reasonable to replace Ω with B and E with E ∩ (B × B).In other terms, for fixed (µ, ν), it makes sense to involve G B in the notion of strong duality.

3.2.
Existence of primal minimizers.Quite surprisingly, in mass transportation theory, existence of primal minimizers seems to have received only a little attention to date; see e.g.[1] and [5].To our knowledge, when the cost c is not lower semi-continuous, the only available results are in [7] and require c to be suitably approximable by regular costs.However, such results do not apply to our case where c = 1 − 1 E .
Let (Ω, F ) be a standard Borel space and c = 1 − 1 E .Then, c lower semicontinuous if and only if E is closed, and in this case E is strongly dualizable.
Similarly, E is strongly dualizable if the elements of the partition Π are the atoms of a countably generated sub-σ-field of F ; see [5] and [8].As noted above, however, we are not aware of any (reasonable) condition for a primal minimizer to exist.In the sequel, we discuss two strategies for circumventing this problem.
The first strategy is possibly expected and lies in using finitely additive probabilities.Let M (µ, ν) = finitely additive probabilities on F ⊗ F with marginals µ and ν .Moreover, for all µ, ν ∈ P(F ) there is B ∈ F such that µ(B) = ν(B) = 1 and min Proof.Just apply Theorems 2 and 3 and note that, by Theorem 2 of [13], min A remark on Theorem 4 is in order.Let P be a finitely additive primal minimizer, in the sense that P ∈ M (µ, ν) and 1 − P Moreover, let R be the field generated by the measurable rectangles A × B with A, B ∈ F .Since µ and ν are perfect (due to (Ω, F ) is standard Borel), the restriction P |R is σ-additive; see e.g.[11].Hence, it is tempting to define P ′ as the only σ-additive extension of P |R to σ(R) = F ⊗ F .Then, P ′ ∈ Γ(µ, ν) but it is not necessarily true that P ′ (E) = P (E).Hence, P ′ needs not be a primal minimizer.
The second strategy for dealing with primal minimizers is summarized by the next result.
Proof.By Lemma 1, applied with S = Ω, E = F and D = G 0 , there are a probability space (Φ, A, P) and two measurable maps X, Y : (Φ, A) → (Ω, F ) such that Up to replacing (Φ, A, P) with its completion, it can be assumed that (Φ, A, P) is complete.Let P * and P * be the inner and outer measures corresponding to P.
To conclude the proof, note that (X, Y ) ∈ H ∈ A for each H ∈ F ⊗ F and define ν 0 (A) = P(Y ∈ A) and P (H) = P (X, Y ) ∈ H for all A ∈ F and H ∈ F ⊗F .
• In Theorem 5, unlike the previous results, (Ω, F ) is not required to be a standard Borel space.

Proof.
Suppose first µ|D = ν|D.Let Φ = S × S, C = E ⊗ E and X(a, b) = a and Y (a, b) = b for all (a, b) ∈ S × S. Define also