Irreducible convex paving for decomposition of multi-dimensional martingale transport plans
Hadrien De March Nizar Touzi





























 

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        CMAP, École Polytechnique, FranceThe authors gratefully acknowledge the financial support of the ERC 321111 Rofirm, and the Chairs Financial Risks (Risk Foundation, sponsored by Société Générale) and Finance and Sustainable Development (IEF sponsored by EDF and CA).TheoremTheorem[section]
        Lemma[Theorem]LemmaCorollary[Theorem]CorollaryProposition[Theorem]PropositionRemark[Theorem]RemarkExample[Theorem]ExampleDefinition[Theorem]DefinitionAssumption[Theorem]Assumption
        
        : Martingale transport plans on the line are known from Beiglböck & Juillet <cit.> to have an irreducible decomposition on a (at most) countable union of intervals. We provide an extension of this decomposition for martingale transport plans in ^d, d≥ 1. Our decomposition is a partition of ^d consisting of a possibly uncountable family of relatively open convex components, with the required measurability so that the disintegration is well-defined. We justify the relevance of our decomposition by proving the existence of a martingale transport plan filling these components. We also deduce from this decomposition a characterization of the structure of polar sets with respect to all martingale transport plans.
        Irreducible Convex Paving for Decomposition of Multi-dimensional Martingale Transport Plans
        
            Hadrien De Marchhadrien.de-march@polytechnique.org. Nizar Touzinizar.touzi@polytechnique.edu.
            Accepted ???. Received ???; in original form December 30, 2023
        =================================================================================================
        
         Key words.  Martingale optimal transport, irreducible decomposition, polar sets.
        
        
        
        § INTRODUCTION
        
        
        The problem of martingale optimal transport was introduced as the dual of the problem of robust (model-free) superhedging of exotic derivatives in financial mathematics, see Beiglböck, Henry-Labordère & Penkner <cit.> in discrete time, and Galichon, Henry-Labordère & Touzi <cit.> in continuous-time. The robust superhedging problem was introduced  by Hobson <cit.>, and was addressing specific examples of exotic derivatives by means of corresponding solutions of the Skorohod embedding problem, see <cit.>, and the survey <cit.>. 
        
        Given two probability measures μ,ν on ^d, with finite first order moment, martingale optimal transport differs from standard optimal transport in that the set of all coupling probability measures (μ,ν) on the product space is reduced to the subset (μ,ν) restricted by the martingale condition. We recall from Strassen <cit.> that (μ,ν)≠∅ if and only if μ≼ν in the convex order, i.e. μ(f)≤ν(f) for all convex functions f. Notice that the inequality μ(f)≤ν(f) is a direct consequence of the Jensen inequality, the reverse implication follows from the Hahn-Banach theorem.
        
        This paper focuses on the critical observation by Beiglböck & Juillet <cit.> that, in the one-dimensional setting d=1, any such martingale interpolating probability measure  has a canonical decomposition =∑_k≥ 0_k, where _k∈(μ_k,ν_k) and μ_k is the restriction of μ to the so-called irreducible components I_k, and ν_k := ∫_x∈ I_k(dx,·), supported in J_k, k≥ 0, is independent of the choice of _k. Here, (I_k)_k≥ 1 are open intervals, I_0:=∖(∪_k≥ 1 I_k), and J_k is an augmentation of I_k by the inclusion of either one of the endpoints of I_k, depending on whether they are charged by the distribution _k. Remarkably, the irreducible components (I_k,J_k)_k≥ 0 are independent of the choice of ∈(μ,ν). To understand this decomposition, notice that convex functions in one dimension are generated by the family f_x_0(x):=|x-x_0|,x_0∈, x_0∈. Then, in terms of the potential functions U^μ(x_0):=μ(f_x_0), and U^ν(x_0):=ν(f_x_0), x_0∈, we have μ≼ν if and only if U^μ≤ U^ν and μ,ν have same mean. Then, at any contact points x_0, of the potential functions, U^μ(x_0)=U^ν(x_0), we have equality in the underlying Jensen equality, which means that the singularity x_0 of the underlying function f_x_0 is not seen by the measure. In other words, the point x_0 acts as a barrier for the mass transfer in the sense that martingale transport maps do not cross the barrier x_0. Such contact points are precisely the endpoints of the intervals I_k, k≥ 1.
        
        The decomposition in irreducible components plays a crucial role for the quasi-sure formulation introduced by Beiglböck, Nutz, and Touzi <cit.>, and represents an important difference between martingale transport and standard transport. Indeed, while the martingale transport problem is affected by the quasi-sure formulation, the standard optimal transport problem is not changed. We also refer to Ekren & Soner <cit.> for further functional analytic aspects of this duality.
        
        Our objective in this paper is to extend the last decomposition to an arbitrary d-dimensional setting, d≥ 1. The main difficulty is that convex functions do not have anymore such a simple generating family. Therefore, all of our analysis is based on the set of convex functions. A first extension of the last decomposition to the multi-dimensional case was achieved by Ghoussoub, Kim & Lim <cit.>. Motivated by the martingale monotonicity principle of Beiglböck & Juillet <cit.> (see also Zaev <cit.> for higher dimension and general linear constraints), their strategy is to find a monotone set Γ⊂^d×^d, where the robust superhedging holds with equality, as a support of the optimal martingale transport in (μ,ν). Denoting Γ_x:={y:(x,y)∈Γ}, this naturally induces the relation x    x' if x∈ (Γ_x'), which is then completed to an equivalence relation ∼. The corresponding equivalence classes define their notion of irreducible components.
        
        Our subsequent results differ from <cit.> from two perspectives. First, unlike <cit.>, our decomposition is universal in the sense that it is not relative to any particular martingale measure in (μ,ν) (see example <ref>). Second, our construction of the irreducible convex paving allows to prove the required measurability property, thus justifying completely the existence of a disintegration of martingale plans. 
        
        Finally, during the final stage of writing the present paper, we learned about
        the parallel work by Jan Obłój and Pietro Siorpaes <cit.>.
        Although the results are close, our approach is different from theirs.
        We are grateful to them for pointing to us the notions of "convex
        face" and "Wijsmann topology" and the relative references, which allowed us to streamline our presentation.
        In an earlier version of this work we used instead a topology that we called the compacted Hausdorff distance, defined as the topology generated by the countable restrictions of the space to the closed balls centered in the origin with integer radii; the two are in our case the same topologies, as the Wijsman topology is locally equivalent to the Hausdorff topology in a locally compact set. We also owe Jan and Pietro special thanks for their useful remarks and comments on a first draft of this paper privately exchanged with them.
        
        The paper is organized as follows. Section <ref> contains the main results of the paper, namely our decomposition in irreducible convex paving, and shows the identity with the Beiglböck & Juillet <cit.> notion in the one-dimensional setting. Section <ref> collects the main technical ingredients needed for the statement of our main results, and gives the structure of polar sets. In particular, we introduce the new notions of relative face and tangent convex functions, together with the required topology on the set of such functions.  The remaining sections contain the proofs of these results. In particular, the measurability of our irreducible convex paving is proved in Section <ref>.
        
         Notation   We denote by :=∪{-∞,∞} the completed real line, and similarly denote _+:=_+∪{∞}. We fix an integer d≥ 1. For x∈^d and r≥ 0, we denote B_r(x) the closed ball for the Euclidean distance, centered in x with radius r. We denote for simplicity B_r := B_r(0). If x∈, and A⊂, where (, d) is a metric space, (x,A):=inf_a∈ A d(x,a). In all this paper, ^d is endowed with the Euclidean distance.
        
        If V is a topological affine space and A⊂ V is a subset of V, A is the interior of A, A is the closure of A, A is the smallest affine subspace of V containing A, A is the convex hull of A, (A):=( A), and A is the relative interior of A, which is the interior of A in the topology of A induced by the topology of V. We also denote by ∂ A:= A∖ A the relative boundary of A, and by _A the Lebesgue measure of A.
        
        The set  of all closed subsets of ^d is a Polish space when endowed with the Wijsman topology[The Wijsman topology on the collection of all closed subsets of a metric space (, d) is the weak topology generated by {(x,·):x∈}.] (see Beer <cit.>). As ^d is separable, it follows from a theorem of Hess <cit.> that a function F:^d⟶ is Borel measurable with respect to the Wijsman topology if and only if its associated multifunction is Borel measurable, i.e.
        *̱
        F^-(V):={x∈^d:F(x)∩ V≠∅}   
            V⊂^d.
        *
        The subset ⊂ of all the convex closed subsets of ^d is closed in  for the Wijsman topology, and therefore inherits its Polish structure. Clearly,  is isomorphic to := { K : K∈} (with reciprocal isomorphism cl). We shall identify these two isomorphic sets in the rest of this text, when there is no possible confusion.
        
        We denote Ω:=^d×^d and define the two canonical maps
        *̱
         X :(x,y)∈Ω⟼ x∈^d
               
         Y :(x,y)∈Ω⟼ y∈^d.
         *
        For φ,ψ:^d⟶, and h:^d⟶^d, we denote 
        *̱φ⊕ψ
        :=
        φ(X)+ψ(Y),
              
        h^⊗ := h(X)·(Y-X),
        *
        with the convention ∞-∞ = ∞.
        
        For a Polish space , we denote by () the collection of Borel subsets of , and () the set of all probability measures on (,()). For ∈(), we denote by _ the collection of all -null sets,  the smallest closed support of , and := the smallest convex closed support of . For a measurable function f:⟶, we denote f:={|f|<∞}, and we use again the convention ∞-∞ = ∞ to define its integral, and denote
         *̱[f]:=^[f] 
         = ∫_ f d 
         = ∫_ f(x) (dx)
               ∈().
         *
        Let  be another Polish space, and ∈(). The corresponding conditional kernel[The usual definition of a kernel requires that the map x↦_x[B] is Borel measurable for all Borel set B∈(^d). In this paper, we only require this map to be analytically measurable.]_x is defined μ-a.e. by:
        
            (dx,dy) = μ(dx)⊗_x(dy), where μ:=∘ X^-1.
        
        We denote by ^0(,) the set of Borel measurable maps from  to . We denote for simplicity ^0():=^0(,) and ^0_+():=^0(,_+). Let  be a σ-algebra of , we denote by ^(,) the set of -measurable maps from  to . For a measure m on , we denote ^1(,m):={f∈^0():m[|f|]<∞}. We also denote simply ^1(m):=^1(,m) and ^1_+(m):=^1_+(_+,m).
        
        We denote by  the collection of all finite convex functions f:^d⟶. We denote by ∂ f(x) the corresponding subgradient at any point x∈^d. We also introduce the collection of all measurable selections in the subgradient, which is nonempty by Lemma <ref>,
        
            ∂ f:={p∈Ł^0(^d,^d): p(x)∈∂ f(x) for all x∈^d}.
        
        We finally denote f_∞ := lim inf_n→∞f_n, for any sequence (f_n)_n≥ 1 of real numbers, or of real-valued functions.
        
        
        
        § MAIN RESULTS
        
        
        Throughout this paper, we consider two probability measures μ and ν on ^d with finite first order moment, and μ≼ν in the convex order, i.e. ν(f)≥μ(f) for all f∈. Using the convention ∞-∞=∞, we may define (ν-μ)(f)∈[0,∞] for all f∈. 
        
        We denote by (μ,ν) the collection of all probability measures on ^d×^d with marginals ∘ X^-1=μ and ∘ Y^-1=ν. Notice that (μ,ν)≠∅ by Strassen <cit.>.
        
        An (μ,ν)-polar set is an element of ∩_∈(μ,ν)_. A property is said to hold (μ,ν)-quasi surely (abbreviated as q.s.) if it holds on the complement of an (μ,ν)-polar set.
        
        
        
         §.§ The irreducible convex paving
        
        
        The next first result shows the existence of a maximum support martingale transport plan, i.e. a martingale interpolating measure  whose disintegration _x has a maximum convex hull of supports among all measures in (μ,ν).
        
        
        There exists ∈(μ,ν) such that
        ∈(μ,ν),    _X
         ⊂ _X,    μ-
         
         Furthermore _X is μ-a.s. unique, and we may choose this kernel so that
         
         (i)x⟼ _x is analytically measurable[Analytically measurable means measurable with respect to the smallest σ-algebra containing the analytic sets. All Borel sets are analytic and all analytic sets are universally measurable, i.e. measurable with respect to all Borel measures (see Proposition 7.41 and Corollary 7.42.1 in <cit.>).]^d⟶,
         (ii)x∈ I(x):=  _x, for all x∈^d, and {I(x),x∈^d} is a partition of ^d.
        
        
        This Theorem will be proved in Subsection <ref>. The (μ-a.s. unique) set valued map I(X) paves ^d by its image by (ii) of Theorem <ref>. By (<ref>), this paving is stable by all ∈(μ,ν):
        
        Y∈ I(X),   (μ,ν)-
        Finally, the measurability of the map I in the Polish space  allows to see it as a random variable, which allows to condition probabilistic events to X∈ I, even when these components are all μ-negligible when considered apart from the others. Under the conditions of Theorem <ref>, we call such I(X) the irreducible convex paving associated to (μ,ν).
        
        Now we provide an important counterexample proving that for some (μ,ν) in dimension larger than 1, particular couplings in (μ,ν) may define different pavings.
        
        
        In ^2, we introduce x_0:=(0,0), x_1:=(1,0), y_0:=x_0, y_-1:=(0,-1), y_1:=(0,1), and y_2:=(2,0). Then we set μ := 1/2(δ_x_0+δ_x_1) and ν := 1/8(4δ_y_0+δ_y_-1+δ_y_1+2δ_y_2). We can show easily that (μ,ν) is the nonempty convex hull of _1 and _2 where
        
            _1:=1/8(4δ_x_0,y_0+2δ_x_1,y_2+δ_x_1,y_1+δ_x_1,y_-1)
        
        and
        
            _2:=1/8(2δ_x_0,y_0+δ_x_0,y_1+δ_x_0,y_-1+2δ_x_1,y_0+2δ_x_1,y_2)
        
        
        (i)  The Ghoussoub-Kim-Lim <cit.> (GKL, hereafter) irreducible convex paving. Let c_1 = 1_{X=Y}, c_2 = 1-c_1 = 1_{X≠ Y}, and notice that _i is the unique optimal martingale transport plan for c_i, i=1,2. Then, it follows that the corresponding _i-irreducible convex paving according to the definition of <cit.> are given by 
         
            [  C__1(x_0)={x_0},
              C__1(x_1)= {y_1,y_-1,y_2},; C__2(x_0)= {y_1,y_-1},
              C__2(x_1)= {y_0,y_2}. ]
        
        Figure <ref> shows the extreme probabilities _1 and _2, and their associated irreducible convex pavings map C__1 and C__2.
        
        (ii)  Our irreducible convex paving. The irreducible components are given by 
         *̱
         I(x_0)= (y_1,y_-1)
               
         I(x_1)= (y_1,y_-1,y_2).
         *
        To see this, we use the characterization of Proposition <ref>. Indeed, as (μ,ν) = (_1,_2), for any ∈(μ,ν), ≪:=_1+_2/2, and supp _x
         ⊂ conv( supp _x) for x=x_0,x_1. Then I(x) =  conv( supp _x) for x=x_0,x_1 (i.e. μ-a.s.) by Proposition <ref>.
        In the one dimensional case, a convex paving which is invariant with respect to some ∈(μ, ν) is automatically invariant with respect to all ∈(μ, ν). Given a particular coupling ∈(μ, ν), the finest convex paving which is -invariant roughly corresponds to the GKL convex paving constructed in <cit.>. Then Example <ref> shows that this does not hold any more in dimension greater than two.
        
        Furthermore, in dimension one the "restriction" ν_I:=∫_I(dx,·) does not depend on the choice of the coupling ∈(μ,ν). Once again Example <ref> shows that it does not hold in higher dimension. Conditions guaranteeing that this property still holds in higher dimension will be investigated in <cit.>.
        
        
         §.§ Behavior on the boundary of the components
        
        
        For a probability measure  on a topological space, and a Borel subset A, |_A:=[·∩ A] denotes its restriction to A.
        
        
        We may choose ∈(μ,ν) in Theorem <ref> so that for all ∈(μ,ν) and y∈^d,
         *̱   μ[ _X[{y}]>0]
         ≤μ[ _X[{y}]>0],
            _X|_∂ I(X)⊂ _X|_∂ I(X), μ-*
        
         (i) The set-valued maps J(X):= I(X)∪{y∈^d:ν[y]>0,_X[{y}]>0}, and J̅(X):=I(X)∪ _X |_∂ I(X) are unique μ-a.s, and Y∈J̅(X), (μ,ν)-q.s.
        
         (ii) We may chose the kernel _X so that the map J̅ is convex valued, I⊂J⊂J̅⊂ I, and both J and J̅ are constant on I(x), for all x∈^d.
        
        
        The proof is reported in Subsection <ref>.
        
        
        
         §.§ Structure of polar sets
        
        
        Here we state the structure of polar sets that is a direct consequence, and will be made more precise by Theorem <ref>.
        
        
        A Borel set N∈(Ω) is (μ,ν)-polar if and only if
        
            N ⊂{X∈ N_μ}∪{Y∈ N_ν}∪{Y∉ J(X)},
        
         for some (N_μ,N_ν)∈_μ_ν and a set valued map J such that J⊂ J⊂J̅, the map J is constant on I(x) for all x∈^d, I(X)⊂(J(X)∖ N_ν'), μ-a.s. for all N_ν'∈_ν, and Y∈ J(X), (μ,ν)-q.s.
        
        
         §.§ The one-dimensional setting
        
        
        In the one-dimensional case, the decomposition in irreducible components and the structure of (μ,ν)-polar sets were introduced in Beiglböck & Juillet <cit.> and Beiglböck, Nutz & Touzi <cit.>, respectively. 
        
        Let us see how the results of this paper reduce to the known concepts in the one dimensional case. First, in the one-dimensional setting, I(x) consists of open intervals (at most countable number) or single points. Following <cit.> Proposition 2.3, we denote the full dimension  components (I_k)_k≥ 1. 
        
        We also have J=J̅ (see Proposition <ref> below) therefore, Theorem <ref> is equivalent to Theorem 3.2 in <cit.>. Similar to (I_k)_k≥ 1, we introduce the corresponding sequence (J_k)_k≥ 1, as defined in <cit.>. Similar to <cit.>, we denote by μ_k the restriction of μ to I_k, and ν_k:=∫_x∈ I_k[dx,·] is independent of the choice of ∈(μ,ν). We define the Beiglböck & Juillet (BJ)-irreducible components
        *̱(I^BJ,J^BJ):x↦
        (I_k,J_k)
            x∈ I_k, k≥ 1, 
        ({x},{x})
            x∉∪_k I_k.
        *
        
        
        Let d=1. Then I=I^BJ, and J̅=J = J^BJ, μ-a.s.
        By Proposition <ref> (i)-(ii), we may find ∈(μ,ν) such that
        _X
         =
          I(X), and _X|_∂ I(X) = J̅∖ I(X), μ-a.s. Notice that as J̅∖ I(^d) only consists of a countable set of points, we have J = J̅. By Theorem 3.2 in <cit.>, we have Y∈ J^BJ(X), (μ,ν)-q.s. Therefore, Y∈ J^BJ(X), -a.s. and we have J̅(X)⊂ J^BJ(X), μ-a.s.
        
        On the other hand, let k≥ 1. By the fact that u_ν-u_μ>0 on I_k, together with the fact that J_k∖ I_k is constituted with atoms of ν, for any N_ν∈_ν, J_k⊂(J_k∖ N_ν). As μ = ν outside of the components,
        
        J^BJ(X)⊂(J^BJ(X)∖ N_ν),    μ-
        Then by Theorem 3.2 in <cit.>, as {Y∉J̅(X)} is polar, we may find N_ν∈_ν such that J^BJ(X)∖ N_ν⊂J̅(X), μ-a.s. The convex hull of this inclusion, together with (<ref>) gives the remaining inclusion J^BJ(X)⊂J̅(X), μ-a.s.
        
        The equality I(X)=I^BJ(X), μ-a.s. follows from the relative interior taken on the previous equality.
        
        
        § PRELIMINARIES
        
        
        The proof of these results needs some preparation involving convex analysis tools.
        
        
        
         §.§ Relative face of a set
        
        
        For a subset A⊂^d and a∈^d, we introduce the face of A relative to a (also denoted a-relative face of A):
         
            _a A  
             :=
             {y∈ A: (a-(y-a),y+(y-a)) ⊂ A, for some >0}.
         Figure <ref> illustrates examples of relative faces of a square S, relative to some points. 
        For later use, we list some properties whose proofs are reported in Section <ref>.
        [_aA is equal to the only relative interior of face of A containing a, where we extend the notion of face to non-convex sets. A face F of A is a nonempty subset of A such that for all [a,b]⊂ A, with (a,b)∩ F≠∅, we have [a,b]⊂ F. It is discussed in Hiriart-Urruty-Lemaréchal <cit.> as an extension of Proposition 2.3.7 that when A is convex, the relative interior of the faces of A form a partition of A, see also Theorem 18.2 in Rockafellar <cit.>.
        ] (i) For A,A'⊂^d, we have _a(A∩ A') = _a(A)∩_a(A'), and _a A ⊂_a A' whenever A⊂ A'. Moreover, _a A ≠∅ iff a∈_a A iff a∈ A.
        
         (ii) For a convex A, _a A  =  A ≠∅ iff a∈ A. Moreover, _a A is convex relatively open, A∖_a A is convex, and if x_0∈ A∖_a A and y_0∈ A, then [x_0,y_0)⊂ A∖_a A. Furthermore, if a∈ A, then (_a A) =(A) if and only if a∈ A. In this case, we have _a A =   A =  A = _aA.
        
        
         §.§ Tangent Convex functions
        
        
        Recall the notation (<ref>), and denote for all θ:Ω→: 
         *̱_xθ   :=   _x  θ(x,·).
         *
        For θ_1,θ_2:Ω⟶, we say that θ_1=θ_2, , if 
         *̱_Xθ_1 = _Xθ_2,
               θ_1(X,·)=θ_2(X,·)
           _Xθ_1, μ-*
        
        The crucial ingredient for our main result is the following.
        
        
        A measurable function θ:Ω→_+ is a tangent convex function if 
         *̱   θ(x,·)
            θ(x,x)=0,
           
         x∈^d.
         *
        We denote by Θ the set of tangent convex functions, and we define
         *̱Θ_μ   :=   {θ∈Ł^0(Ω,_+):
               θ = θ', ,
                θ≥θ',  θ'∈Θ}.
         *
        
        
        In order to introduce our main example of such functions, let
        *̱_pf(x,y)
        :=
        f(y)-f(x)-p^⊗(x,y)≥ 0 ,
              
        f∈,  p∈∂ f.
        *
        Then, ():={_p f:f∈,p∈∂ f}⊂Θ⊂Θ_μ.
        
        
        The second inclusion is strict. Indeed, let d=1, and consider the convex function f:=∞1_(-∞,0). Then θ' := f(Y-X)∈Θ. Now let θ = θ'+√(|Y-X|). Notice that since _Xθ' = _Xθ = {X}, we have θ' = θ,  for any measure μ, and θ≥θ'. Therefore θ∈Θ_μ. However, for all x∈^d, θ(x,·) is not convex, and therefore θ∉Θ.
        
        In higher dimension we may even have X∈ θ(X,·), and θ(X,·) is not convex. Indeed, for d=2, let f:(y_1,y_2)⟼∞(1_{|y_1|> 1} + 1_{|y_2|> 1}), so that θ := f(Y-X)∈Θ. Let x_0:=(1,0) and θ := θ' + 1_{Y = X+x_0}. Then, θ = θ',  for any measure μ, and θ≥θ'. Therefore θ∈Θ_μ. However, θ∉Θ as θ(x,·) is not convex for all x∈^d.
         (i) Let θ∈Θ_μ, then _Xθ=_Xθ(X,·)⊂θ(X,·), μ-a.s.
         (ii) Let θ_1,θ_2∈Θ_μ, then _X(θ_1+θ_2) = _Xθ_1∩_Xθ_2, μ-a.s.
         (iii)Θ_μ is a convex cone.
         (i) It follows immediately from the fact that θ(X,·) is convex and finite on _Xθ, μ-a.s. by definition of Θ_μ. Then _Xθ⊂_Xθ(X,·). On the other hand, as θ(X,·)⊂ θ(X,·), the monotony of _x gives the other inclusion: _Xθ(X,·)⊂_Xθ.
        
         (ii) As θ_1,θ_2≥ 0, (θ_1+θ_2) =θ_1∩θ_2. Then, for x∈^d, (θ_1(x,·)+θ_2(x,·)) ⊂ θ_1(x,·)∩ θ_2(x,·). By Proposition <ref> (i),
         *̱_x(θ_1+θ_2) 
         ⊂_xθ_1∩_xθ_2,
               
         x∈^d.
         *
        As for the reverse inclusion, notice that (i) implies that _Xθ_1∩_Xθ_2⊂θ_1(X,·)∩θ_2(X,·) = (θ_1(X,·)+θ_2(X,·))⊂ (θ_1(X,·)+θ_2(X,·)), μ-a.s. Observe that _xθ_1∩_xθ_2 is convex, relatively open, and contains x. Then, 
         *̱_Xθ_1∩_Xθ_2
            =   _X(_Xθ_1∩_Xθ_2)
           ⊂   _X( (θ_1(X,·)+θ_2(X,·)))
           =   _X(θ_1+θ_2)
           μ-*
         (iii) Given (ii), this follows from direct verification.
        
        A sequence (θ_n)_n≥ 1⊂Ł^0(Ω) converges  to some θ∈Ł^0(Ω) if 
         *̱_X(θ_∞) =  _Xθ      θ_n(X,·) ⟶θ(X,·),  _Xθ,
          μ-*
        
        
        Notice that the -limit is  unique. In particular, if θ_n converges to θ, , it converges as well to θ_∞.
        
        
        Let (θ_n)_n≥ 1⊂Θ_μ, and θ:Ω⟶_+, such that θ_nn→∞⟶θ, ,
         (i)_Xθ⊂lim inf_n→∞_Xθ_n, μ-a.s.
         (ii) If θ_n' = θ_n, , and θ_n'≥θ_n, then θ_n'n→∞⟶θ, ;
         (iii)θ_∞∈Θ_μ.
         (i) Let x∈^d, such that θ_n(x,·) converges on _xθ to θ(x,·). Let y∈_xθ, let y'∈_xθ such that y' = x-ϵ(y-x), for some ϵ>0. As θ_n(x,y)n→∞⟶θ(x,y), and θ_n(x,y')n→∞⟶θ(x,y'), then for n large enough, both are finite, and y∈_xθ_n. y∈lim inf_n→∞_xθ_n, and _xθ⊂lim inf_n→∞_xθ_n. The inclusion is true for μ-a.e. x∈^d, which gives the result.
         
          (ii) By (i), we have _Xθ⊂lim inf_n→∞_Xθ_n = lim inf_n→∞_Xθ_n', μ-a.s. As θ_n≤θ_n', _Xθ'_∞⊂_Xθ_∞⊂lim inf_n→∞_Xθ_n, μ-a.s. We denote N_μ∈_μ, the set on which θ_n(X,·) does not converge to θ(X,·) on _Xθ(X,·). For x∉ N_μ, for y∈_xθ, θ_n(x,y)=θ_n'(x,y), for n large enough, and θ_n'(x,y)n→∞⟶θ(x,y)<∞. Then _Xθ = _Xθ'_∞, and θ_n'(X,·) converges to θ(X,·), on _Xθ, μ-a.s. We proved that θ_n'n→∞⟶θ, .
         
         (iii) has its proof reported in Subsection <ref> due to its length and technicality.
        
        
        
        The next result shows the relevance of this notion of convergence for our setting.
        
        
        Let (θ_n)_n≥ 1⊂Θ_μ. Then, we may find a sequence θ_n∈(θ_k,k≥ n), and θ_∞∈Θ_μ such that θ_n ⟶θ_∞,  as n→∞.
        
        
        The proof is reported in Subsection <ref>.
        
         (i) A subset ⊂Θ_μ is -Fatou closed if θ_∞∈ for all (θ_n)_n≥ 1⊂ converging  (in particular, Θ_μ is -Fatou closed by Proposition <ref> (iii)).
        
         (ii) The -Fatou closure of a subset A⊂Θ_μ is the smallest -Fatou closed set containing A:
         
            A
             :=
             ⋂{⊂Θ_μ: 
                          A⊂,
                           and   }.
        
        
        We next introduce for a≥ 0 the set _a:={f∈:(ν-μ)(f)≤ a}, and
        
            (μ,ν):=a≥ 0⋃ _a,
             where _a:=(_a),   (_a)
             :=
             {_p f: f∈_a,p∈∂ f}.
        (μ,ν) is a convex cone.
        
        We first prove that (μ,ν) is a cone. We consider λ,a>0, as we have λ_a=_λ a, and as convex combinations and inferior limit commute with the multiplication by λ, we have λ_a=_λ a. Then (μ,ν) =  cone(_1), and therefore it is a cone.
        
        We next prove that _a is convex for all a≥ 0, which induces the required convexity of (μ,ν) by the non-decrease of the family {_a,a≥ 0}. Fix 0≤λ≤ 1, a≥ 0, θ_0∈_a, and denote (θ_0):={θ∈_a:λθ_0+(1-λ)θ∈_a}. In order to complete the proof, we now verify that (θ_0)⊃(_a) and is -Fatou closed, so that (θ_0)=_a. 
        
        To see that (θ_0) is Fatou-closed, let (θ_n)_n≥ 1⊂(θ_0), converging . By definition of (θ_0), we have λθ_0+(1-λ)θ_n∈_a for all n. Then, λθ_0+(1-λ)θ_n⟶lim inf_n→∞λθ_0+(1-λ)θ_n, , and therefore λθ_0+(1-λ)θ_∞∈_a, which shows that θ_∞∈(θ_0). 
        
        We finally verify that (θ_0)⊃(_a). First, for θ_0∈(_a), this inclusion follows directly from the convexity of (_a), implying that (θ_0)=_a in this case. For general θ_0∈_a, the last equality implies that (_a)⊂(θ_0), thus completing the proof. 
        
        
        
        Notice that even though (_a)⊂Θ, the functions in (μ,ν) may not be in Θ as they may not be convex in y on (_xθ)^c for some x∈^d (see Example <ref>). The following result shows that some convexity is still preserved.
        
        
        For all θ∈(μ,ν), we may find N_μ∈_μ such that for x_1,x_2∉ N_μ, y_1,y_2∈^d, and λ∈[0,1] with := λ y_1 + (1-λ)y_2∈_x_1θ∩_x_2θ, we have:
         *̱λθ(x_1,y_1)+(1-λ)θ(x_1,y_2)-θ(x_1,)
            =   λθ(x_2,y_1)+(1-λ)θ(x_2,y_2)
              -θ(x_2,)
         ≥ 0.
         *
        
        
        The proof of this claim is reported in Subsection <ref>. We observe that the statement also holds true for a finite number of points y_1,...,y_k.[
        This is not a direct consequence of Proposition <ref>, as the barycentre  has to be in _x_1θ∩_x_2θ.
        ]
        
         §.§ Extended integral
        
        
        We now introduce the extended (ν-μ)-integral:
         *̱ν⊖μ[θ]
         :=
         inf{a≥ 0 :θ∈_a}      θ∈(μ,ν).
         *
        
         (i)[θ]≤ν⊖μ[θ]<∞ for all θ∈(μ,ν) and ∈(μ,ν).
         (ii)ν⊖μ[_p f] = (ν-μ)[f] for f∈∩Ł^1(ν) and p∈∂ f.
         (iii)ν⊖μ is homogeneous and convex.
         (i) For a>ν⊖μ[θ], set S^a:={F∈Θ_μ:[F]≤ a  ∈(μ,ν)}. Notice that S^a is -Fatou closed by Fatou's lemma, and contains (_a), as for f∈∩^1(ν) and p∈∂ f, [T_pf]= (ν-μ)[f] for all ∈(μ,ν). Then S^a contains _a as well, which contains θ. Hence, θ∈ S^a and [θ]≤ a for all ∈(μ,ν). The required result follows from the arbitrariness of a>ν⊖μ[θ].
        
         (ii) Let ∈(μ,ν). For p∈∂ f, notice that T_p f∈(_a)⊂_a for some a=(ν-μ)[f], and therefore (ν-μ)[f] ≥ν⊖μ[T_p f]. Then, the result follows from the inequality (ν-μ)[f] = [T_p f] ≤ν⊖μ[T_p f].
        
         (iii) Similarly to the proof of Proposition <ref>, we have λ_a=_λ a, for all  λ,a>0. Then with the definition of ν⊖μ we have easily the homogeneity.
        
        To see that the convexity holds, let 0<λ<1, and θ,θ'∈(μ,ν) with a>ν⊖μ[θ], a'>ν⊖μ[θ'], for some a,a'>0. By homogeneity and convexity of _1, λθ +(1-λ)θ'∈_λ a + (1-λ)a', so that ν⊖μ[λθ +(1-λ)θ']≤λ a + (1-λ)a'. The required convexity property now follows from arbitrariness of a>ν⊖μ[θ] and a'>ν⊖μ[θ'].
        
        
        
        The following compacteness result plays a crucial role.
        
        
        Let (θ_n)_n≥ 1⊂(μ,ν) be such that
        sup_n≥ 1 ν⊖μ(θ_n)<∞.
        Then we can find a sequence θ_n∈(θ_k,k≥ n) such that
        *̱θ_∞∈(μ,ν),   θ_n ⟶θ_∞,
         ,
             ν⊖μ(θ_∞)
        ≤lim inf_n→∞ν⊖μ(θ_n).
        *
        
        By possibly passing to a subsequence, we may assume that lim_n→∞(ν⊖μ)(θ_n) exists. The boundedness of ν⊖μ(θ_n) ensures that this limit is finite. We next introduce the sequence θ_n of Proposition <ref>. Then θ_n ⟶θ_∞, μ⊗ pw, and therefore θ_∞∈(μ,ν), because of the convergence θ_n ⟶θ_∞, . As (ν⊖μ)(θ_n)≤sup_k≥ n(ν⊖μ)(θ_k) by Proposition <ref> (iii), we have ∞>lim_n→∞(ν⊖μ)(θ_n)=lim_n→∞sup_k≥ n(ν⊖μ)(θ_k)≥lim sup_n→∞(ν⊖μ)(θ_n). Set l:= lim sup_n→∞ ν⊖μ(θ_n). For ϵ >0, we consider n_0∈ such that sup_k≥ n_0ν⊖μ(θ_k)≤ l+ϵ. Then for k≥ n_0, θ_k∈_l+2ϵ(μ,ν), and therefore θ_∞ = lim inf_k≥ n_0θ_k∈_l+2ϵ(μ,ν), implying ν⊖μ(θ)≤ l+2ϵ⟶ l, as ϵ→ 0. Finally, lim inf_n→∞(ν⊖μ)(θ_n)≥ν⊖μ(θ_∞).
        
        
         §.§ The dual irreducible convex paving
        
        
        Our final ingredient is the following measurement of subsets K⊂^d: 
         
            G(K) := (K) + g_K(K)
               
             g_K(dx) 
             :=
              e^-1/2|x|^2/(2π)^1/2 K λ_K(dx),
        
        Notice that 0 ≤ G≤ d+1 and, for any convex subsets C_1⊂ C_2 of ^d, we have 
        
        G(C_1) = G(C_2)
               C_1 = C_2
              C_1 =  C_2.
        
        For θ∈_+^0(Ω),A∈(^d), we introduce the following map from ^d to the set  of all relatively open convex subsets of ^d:
         
         K_θ,A(x)
         :=
         _x(θ(x,·)∖ A)
         =_X(θ+∞ 1_^d A),
            for all   
         x∈^d.
         
        We recall that a function is universally measurable if it is measurable with respect to every complete probability measure that measures all Borel subsets.
        
        
        For θ∈_+^0(Ω) and A∈(^d), we have:
        
         (i)θ(X,·):^d⟼, _Xθ:^d⟼, and K_θ,A:^d⟼ are universally measurable;
        
         (ii)G:⟶ is Borel measurable;
        
         (iii) if A∈_ν, and θ∈(μ,ν), then up to a modification on a μ-null set, K_θ,A(^d)⊂ is a partition of ^d with x∈ K_θ,A(x) for all x∈^d.
        
        
        The proof is reported in Subsections <ref> for (iii), <ref> for (ii), and <ref> for (i). The following property is a key-ingredient for our dual decomposition in irreducible convex paving.
        
        
        
        For all (θ,N_ν)∈(μ,ν)×_ν, we have the inclusion Y∈ K_θ,N_ν(X), (μ,ν)-q.s.
        
        For an arbitrary ∈(μ,ν), we have by Proposition <ref> that [θ]<∞. Then, [θ∖(^d× N_ν)] = 1 i.e. [Y∈ D_X] = 1 where D_x := (θ(x,·)∖ N_ν). By the martingale property of , we deduce that
         *̱
         X    =   ^[Y1_Y∈ D_X|X]
          = 
         (1-Λ)E_K + Λ E_D, 
           μ-*
        Where Λ := _X[Y∈ D_X∖K_θ,N_ν(X)], E_D := ^P_X[Y|Y∈ D_X∖K_θ,N_ν(X)], E_K := ^_X[Y|Y∈K_θ,N_ν(X)], and _X is the conditional kernel to X of . We have E_K∈_XD_X⊂ D_X and E_D∈ D_X∖_XD_X because of the convexity of D_X∖_XD_X given by Proposition <ref> (ii) (D_X is convex). The lemma also gives that if Λ≠ 0, then ^[Y|X]=Λ E_D + (1-Λ) E_K∈ D_X∖K_θ,N_ν(X). This implies that
         *̱{Λ≠ 0} ⊂ {^[Y|X]∈ D_X∖K_θ,N_ν(X)}   ⊂   {^[Y|X]∉K_θ,N_ν(X)}
           ⊂   {^[Y|X]≠ X}.
         *
        Then [Λ≠ 0] = 0, and therefore [Y∈ D_X∖K_θ,N_ν(X)] = 0. Since [Y∈ D_X] = 1, this shows that [Y∈K_θ,N_ν(X)] = 1.
        
        
        In view of Proposition <ref> and Lemma <ref> (iii), we introduce the following optimization problem which will generate our irreducible convex paving decomposition: 
         (θ,N_ν)∈(μ,ν)_νinfμ[G(K_θ,N_ν)].
         
        The following result gives another possible definition for the irreducible paving.
        
         (i) We may find a μ-a.s. unique universally measurable minimizer K:=K_θ,N_ν:^d→ of (<ref>), for some (θ,N_ν)∈(μ,ν)_ν; 
        
         (ii) for all θ∈(μ,ν) and N_ν∈_ν, we have K(X)⊂ K_θ,N_ν(X), μ-a.s;
        
         (iii) we have the equality K(X) = I(X), μ-a.s.
        
        
        In item (i), the measurability of I is induced by Lemma <ref> (i). Existence and uniqueness, together with (ii), are proved in Subsection <ref>. finally, the proof of (iii) is reported in Subsection <ref>, and is a consequence of Theorem <ref> below. Proposition <ref> provides a characterization of the irreducible convex paving by means of an optimality criterion on ((μ,ν),_ν).
        
        
        We illustrate how to get the components from optimization Problem (<ref>) in the case of Example <ref>. A (μ,ν) function minimizing this problem (with N_ν:=∅∈_ν) is θ:=lim inf_n →∞_p_nf_n, where f_n := n f, p_n:=n p for some p∈∂ f, and
        
            f(x):=(x,(y_1,y_-1))+(x,(y_1,y_2))+(x,(y_2,y_-1)).
        
        One can easily check that μ[f] = ν[f] for any n≥ 1: f,f_n∈_0. These functions separate I(x_0), I(x_1) and (I(x_0)∪ I(x_1))^c.
        
        Notice that in this example, we may as well take θ := 0, and N_ν := {y_-1,y_0,y_1,y_2}^c, which minimizes the optimization problem as well.
        
         §.§ Structure of polar sets
        
        
        Let θ∈(μ,ν), we denote the set valued map J_θ(X):= θ(X,·)∩J̅(X), where J̅ is introduced in Proposition <ref>.
        
        
        Let θ∈(μ,ν), up to a modification on a μ-null set, we have
         
        Y∈ J_θ(X), (μ,ν)-  J⊂ 
         J_θ⊂J̅,
          J_θ  I(x),
             
         x∈^d.
         
        These claims are a consequence of Proposition <ref> together with Lemma <ref>.
        
        Our second main result shows the importance of these set-valued maps:
        
        
        A Borel set N∈(Ω) is (μ,ν)-polar if and only if
        
            N ⊂{X∈ N_μ}∪{Y∈ N_ν}∪{Y∉ J_θ(X)},
        
        for some (N_μ,N_ν)∈_μ_ν and θ∈(μ,ν).
        
        
        
        The proof is reported in Section <ref>. This Theorem is an extension of the one-dimensional characterization of polar sets given by Theorem 3.2 in <cit.>, indeed in dimension one J = J_θ = J̅ by Proposition <ref>, together with the inclusion in Remark <ref>.
        
        We conclude this section by reporting a duality result which will be used for the proof of Theorem <ref>. We emphasize that the primal objective of the accompanying paper De March <cit.> is to push further this duality result so as to be suitable for the robust superhedging problem in financial mathematics. 
        
        Let c:^d×^d⟶_+, and consider the martingale optimal transport problem:
         _μ,ν(c)
            :=   sup_∈(μ,ν)[c].
         
        
        Notice from Proposition <ref> (i) that _μ,ν(θ) ≤ν⊖μ(θ) for all θ∈. We denote by _μ,ν^mod(c) the collection of all (φ,ψ,h,θ) in Ł^1_+(μ)Ł^1_+(ν)Ł^0(^d,^d)(μ,ν) such that
        *̱_μ,ν(θ) = ν⊖μ(θ),
              φ⊕ψ+h^⊗+θ ≥  c, 
          {Y∈ K_θ,{ψ = ∞}(X)}.
        *
        The last inequality is an instance of the so-called robust superhedging property. The dual problem is defined by:
        *̱_μ,ν^mod(c)
           :=   (φ,ψ,h,θ)∈_μ,ν^mod(c)inf μ[φ]+ν[ψ]+ν⊖μ(θ).
        *
        Notice that for any measurable function c:Ω⟶_+, any ∈(μ,ν), and any (φ,ψ,h,θ)∈_μ,ν^mod(c), we have [c]≤μ[φ]+ν[ψ]+[θ]≤μ[φ]+ν[ψ]+_μ,ν(θ), as a consequence of the above robust superhedging inequality, together with the fact that Y∈ K_θ,{ψ = ∞}(X), (μ,ν)-q.s. by Proposition <ref> This provides the weak duality:
         _μ,ν(c)   ≤   ^mod_μ,ν(c).
         
        The following result states that the strong duality holds for upper semianalytic functions. We recall that a function f:^d→ is upper semianalytic if {f≥ a} is an analytic set for any a∈. In particular, a Borel function is upper semianalytic.
        
        
        Let c:Ω→_+ be upper semianalytic. Then we have
        
         (i)_μ,ν(c) = _μ,ν^mod(c);
        
         (ii) If in addition _μ,ν(c)<∞, then existence holds for the dual problem _μ,ν^mod(c).
        
        By allowing h to be infinite in some directions, orthogonal to K_θ,{ψ = ∞}(X), together with the convention ∞-∞ = ∞, we may reformulate the robust superhedging inequality in the dual set as φ⊕ψ+h^⊗+θ≥ c pointwise.
        
        
         §.§ One-dimensional tangent convex functions
        
        
        For an interval J⊂, we denote (K) the set of convex functions on K. 
        
        
        Let d=1, then
         *̱(μ,ν) 
         = 
         {∑_k1_{X∈ I_k}_p_k f_k:
                f_k∈(J_k), p_k∈∂ f_k, ∑_k(ν_k-μ_k)[f_k]<  ∞},
         *
         (μ,ν)-q.s. Furthermore, for all such θ∈(μ,ν) and its corresponding (f_k)_k, we have
         ν⊖μ(θ) 
         = 
         ∑_k(ν_k-μ_k)[f_k].
        
        As all functions we consider are null on the diagonal, equality on ∪_kI_k J_k implies (μ,ν)-q.s. equality by Theorem 3.2 in <cit.>. Let  be the set on the right hand side. 
        
         Step 1: We first show ⊂, for a≥ 0, we denote _a:={θ∈:∑_k(ν_k-μ_k)[f_k]≤ a}. Notice that _a contains (_a) modulo (μ,ν)-q.s. equality. We intend to prove that _a is -Fatou closed, so as to conclude that _a⊂_a, and therefore (μ,ν)⊂ by the arbitrariness of a≥ 0. 
        
        Let θ_n=∑_k1_{X∈ I_k}_p_k^n f^n_k∈_a converging . By Proposition <ref>, θ_n⟶θ:=θ_∞, . For k≥ 1, let x_k∈ I_k be such that θ_n(x_k,·)⟶θ(x_k,·) on _x_kθ, and set f_k:=θ(x_k,·). By Proposition 5.5 in <cit.>, f_k is convex on I_k, finite on J_k, and we may find p_k∈∂ f_k such that for x∈ I_k, θ(x,·)=_p_kf_k(x,·). Hence, θ=∑_k1_{X∈ I_k}_p_k f_k, and ∑_k(ν_k-μ_k)[f_k]≤ a by Fatou's Lemma, implying that θ∈_a, as required.
        
         Step 2: To prove the reverse inclusion ⊃, let θ=∑_k1_{X∈ I_k}_p_k f_k∈. Let f_k^ϵ be a convex function defined by f_k^ϵ:=f_k on J_k^ϵ = J_k∩{x∈  J_k:(x, J^c_k)≥ϵ}, and f_k^ϵ affine on ∖ J_k^ϵ. Set ϵ_n:=n^-1, f̅_n = ∑_k=1^nf_k^ϵ_n, and define the corresponding subgradient in ∂f̅_n:
         *̱p̅_n
         :=
         p_k+∇(f̅_n-f^_n_k)
             J_k^_n, k≥ 1,
               p̅_n := ∇f̅_n
             ∖(∪_k J_k^_n).
         *
        We have (ν-μ)[f̅_n]=∑_k=1^n(ν_k-μ_k)[f_k^ϵ_n]≤∑_k(ν_k-μ_k)[f_k]<∞. By definition, we see that _p̅_nf̅_n converges to θ pointwise on ∪_k(I_k)^2 and to θ_*(x,y):=lim inf_→ yθ(x,) on ∪_kI_k× I_k where, using the convention ∞-∞ =∞, θ':=θ-θ_*≥ 0, and θ'=0 on ∪_k(I_k)^2. For k≥ 1, set Δ^l_k:=θ'(x_k,l_k), and Δ^r_k:=θ'(x_k,l_k) where I_k = (l_k,r_k), and we fix some x_k∈ I_k. For positive ϵ<r_k-l_k/2, and M≥ 0, consider the piecewise affine function g_k^ϵ,M with break points l_k+ϵ and r_k-ϵ, and:
         
            g_k^ϵ,M(l_k) = M∧Δ_k^l,
               
             g_k^ϵ,M(r_k) = M∧Δ_k^r,     
             g_k^ϵ,M(l_k+ϵ) = 0,
               
             g_k^ϵ,M(r_k-ϵ) = 0.
        
        Notice that g_k^ϵ,M is convex, and converges pointwise to g_k^M:=M∧θ'(l_k+r_k/2,·) on J_k, as ϵ→ 0, with
         *̱
         (ν_k-μ_k)(g_k^M) 
            =   ν_k[{l_k}](M∧Δ_k^l)
         +ν_k[r_k](M∧Δ_k^r)
         
           ≤   
        (ν_k-μ_k)[f_k]-(ν_k-μ_k)[(f_k)_*]
          ≤ 
         (ν_k-μ_k)[f_k],
         *
        where (f_k)_* is the lower semi-continuous envelop of f_k. Then by the dominated convergence theorem, we may find positive ϵ_k^n,M<r_k-l_k/2n such that 
        *̱
        (ν_k-μ_k)(g_k^ϵ_k^n,M,M) ≤ (ν_k-μ_k)[f_k] +2^-k/n.
        *
        Now let g̅_n = ∑_k=1^ng_k^ϵ_k^n,n,n, and p̅_n'∈∂g̅_n. Notice that _p̅_n'g_n⟶θ' pointwise on ∪_kI_k× J_k, furthermore, (ν-μ)(g̅_n)≤∑_k(ν_k-μ_k)[f_k]+1/n≤∑_k(ν_k-μ_k)[f_k]+1<∞.
        
        Then we have θ_n:=_p̅_nf̅_n+_p̅_n'g̅_n converges to θ pointwise on ∪_kI_k× J_k, and therefore (μ,ν)-q.s. by Theorem 3.2 in <cit.>. Since (ν-μ)(f̅_n+g̅_n) is bounded, we see that (θ_n)_n≥ 1⊂(_a) for some a≥ 0. Notice that θ_n may fail to converge . However, we may use Proposition <ref> to get a sequence θ_n∈(θ_k,k≥ n), and θ_∞∈Θ_μ such that θ_n ⟶θ_∞,  as n→∞, and satisfies the same (μ,ν)-q.s. convergence properties than θ_n. Then θ_∞∈(μ,ν), and θ_∞ = θ, (μ,ν)-q.s. 
        
        
        
        
        
        
        
        
        
        
        
        
        
        § THE IRREDUCIBLE CONVEX PAVING
        
        
         §.§ Existence and uniqueness
         Proof of Theorem <ref>
        (i) The measurability follows from Lemma <ref>. We first prove the existence of a minimizer for the problem (<ref>). Let m denote the infimum in (<ref>), and consider a minimizing sequence (θ_n,N_ν^n)_ n∈⊂(μ,ν)_ν with μ[G(K_θ_n,N^n_ν)]≤ m+1/n. By possibly normalizing the functions θ_n, we may assume that ν⊖μ(θ_n)≤ 1. Set
         *̱   θ := ∑_n≥ 1 2^-nθ_n
            and   N_ν :=∪_n≥ 1 N_ν^n∈_ν.   *
        Notice that θ is well-defined as the pointwise limit of a sequence of the nonnegative functions θ_N:=∑_n≤ N2^-nθ_n. Since ν⊖μ[θ_N]≤∑_n≥ 12^-n<∞ by convexity of ν⊖μ, θ_N⟶θ, pointwise, and θ∈(μ,ν) by Lemma <ref>, since any convex extraction of (θ_n)_n≥ 1 still converges to θ. Since θ^-1_n({∞})⊂θ^-1({∞}), it follows from the definition of N_ν that
         m+1/n
         ≥μ[G(K_θ_n,N^n_ν)]
          ≥ μ[G(K_θ,N_ν)], hence μ[G(K_θ,N_ν)]= m as θ∈(μ,ν), N_ν∈_ν.
        
        (ii) For an arbitrary (θ,N_ν)∈(μ,ν)×_ν, we define θ̅:=θ+θ∈(μ,ν) and N̅_ν:=N_ν∪ N_ν, so that K_θ̅,N̅_ν⊂ K_θ,N_ν. By the non-negativity of θ and θ, we have m≤μ[G(K_θ̅,N̅_ν)]≤μ[G(K_θ,N_ν)] = m. Then G(K_θ̅,N̅_ν)=G(K_θ,N_ν), μ-a.s. By (<ref>), we see that, μ-a.s. K_θ̅,N̅_ν=K_θ,N_ν and K_θ̅,N̅_ν=K_θ,N_ν=I. This shows that I⊂ K_θ,N_ν, μ-a.s.
        
        
         §.§ Partition of the space in convex components
        
        
        This section is dedicated to the proof of Lemma <ref> (iii), which is an immediate consequence of Proposition <ref> (ii).
        
        
        Let θ∈(μ,ν), and A∈(^d). We may find N_μ∈_μ such that:
        
         (i) for all x_1,x_2∉ N_μ with K_θ,A(x_1)∩ K_θ,A(x_2)≠∅, we have K_θ,A(x_1)=K_θ,A(x_2);
         (ii) if A∈_ν, then x∈  K_θ,A(x) for x∉ N_μ, and up to a modification of K_θ,A on N_μ, K_θ,A(^d) is a partition of ^d such that x∈ K_θ,A(x) for all x∈^d.
         (i) Let N_μ be the μ-null set given by Proposition <ref> for θ. For x_1,x_2∉ N_μ, we suppose that we may find ∈ K_θ,A(x_1)∩ K_θ,A(x_2). Consider y∈ K_θ,A(x_1). As K_θ,A(x_1) is open in its affine span, y':=  + ϵ/1-ϵ(-y)∈ K_θ,A(x_1) for 0<ϵ<1 small enough. Then = ϵ y+(1-ϵ)y', and by Proposition <ref>, we get
        
            ϵθ(x_1,y)+(1-ϵ)θ(x_1,y')-θ(x_1,)=ϵθ(x_2,y)+(1-ϵ)θ(x_2,y')-θ(x_2,)
        
        By convexity of _x_iθ, K_θ,A(x_i)⊂_x_iθ⊂θ(x_i,·). Then θ(x_1,y'), θ(x_1,), θ(x_2,y'), and θ(x_2,) are finite and
         *̱θ(x_1,y)<∞      θ(x_2,y)<∞.
         *
        Therefore K_θ,A(x_1)∩θ(x_1,·)⊂θ(x_2,·). We have obviously K_θ,A(x_2)∩θ(x_2,·)⊂θ(x_2,·) as well. Subtracting A, we get
        
            ( K_θ,A(x_1)∩θ(x_1,·)∖ A)∪( K_θ,A(x_2)∩θ(x_2,·)∖ A)⊂θ(x_2,·)∖ A.
        
        Taking the convex hull and using the fact that the relative face of a set is included in itself, we see that (K_θ,A(x_1)∪ K_θ,A(x_2))⊂(θ(x_2,·)∖ A). Notice that, as K_θ,A(x_2) is defined as the x_2-relative face of some set, either x_2∈ K_θ,A(x) or K_θ,A(x)=∅ by the properties of _x_2. The second case is excluded as we assumed that K_θ,A(x_1)∩ K_θ,A(x_2)≠∅. Therefore, as K_θ,A(x_1) and K_θ,A(x_2) are convex sets intersecting in relative interior points and x_2∈ K_θ,A(x_2), it follows from Lemma <ref> that x_2∈ (K_θ,A(x_1)∪ K_θ,A(x_2)). Then by Proposition <ref> (ii),
         *̱_x_2(K_θ,A(x_1)∪ K_θ,A(x_2))
            =    (K_θ,A(x_1)∪ K_θ,A(x_2)) 
         
           =   (K_θ,A(x_1)∪ K_θ,A(x_2)).
         *
        Then, we have (K_θ,A(x_1)∪ K_θ,A(x_2))⊂_x_2(θ(x_2,·)∖ A) = K_θ,A(x_2), as _x_2 is increasing. Therefore K_θ,A(x_1)⊂ K_θ,A(x_2) and by symmetry between x_1 and x_2, K_θ,A(x_1)= K_θ,A(x_2).
        
         (ii) We suppose that A∈_ν. First, notice that, as K_θ,A(X) is defined as the X-relative face of some set, either x∈ K_θ,A(x) or K_θ,A(x)=∅ for x∈^d by the properties of _x. Consider ∈(μ,ν). By Proposition <ref>, [Y∈ K_θ,A(X)]=1. As (_X)⊂ K_θ,A(X), μ-a.s., K_θ,A(X) is non-empty, which implies that x∈ K_θ,A(x). Hence, {X∈ K_θ,A(X)} holds outside the set N_μ^0:={(_X)⊄ I(X)}∈_μ. Then we just need to have this property to replace N_μ by N_μ∪ N_μ^0∈_μ.
        
        Finally, to get a partition of ^d, we just need to redefine K_θ,A on N_μ. If x∈x'∉ N_μ⋃K_θ,A(x') then by definition of N_μ, the set K_θ,A(x') is independent of the choice of x'∉ N_μ such that x∈ K_θ,A(x'): indeed, if x_1',x_2'∉ N_μ satisfy x∈ K_θ,A(x_1')∩ K_θ,A(x_2'), then in particular K_θ,A(x_1')∩ K_θ,A(x_2') is non-empty, and therefore K_θ,A(x_1')= K_θ,A(x_2') by (i). We set K_θ,A(x):=K_θ,A(x'). Otherwise, if x∉x'∉ N_μ⋃K_θ,A(x'), we set K_θ,A(x):={x} which is trivially convex and relatively open. With this definition, K_θ,A(^d) is a partition of ^d.
        
        
        § PROOF OF THE DUALITY
        
        
        For simplicity, we denote (ξ):=μ[φ]+ν[ψ]+ν⊖μ(θ), for ξ:=(φ,ψ,h,θ)∈^mod_μ,ν(c).
        
        
        
         §.§ Existence of a dual optimizer
        
        Let c,c_n:Ω⟶_+, and ξ_n∈^mod_μ,ν(c_n), n∈, be such that
        
            c_n ⟶ c,  (ξ_n)⟶_μ,ν(c)<∞  n→∞.
        
        Then there exists ξ∈^mod_μ,ν(c) such that
        (ξ_n)⟶(ξ) as n→∞.
        
        Denote
        ξ_n := (φ_n,ψ_n, h_n,θ_n), and observe that the convergence of (ξ_n) implies that the sequence (μ(φ_n),ν(ψ_n),ν⊖μ(θ_n))_n is bounded, by the non-negativity of φ_n,ψ_n and ν⊖μ(θ_n). We also recall the robust superhedging inequality
         φ_n⊕ψ_n + h_n^⊗ + θ_n ≥ c_n,
            on {Y∈ K_θ_n,{ψ_n = ∞}(X)},   n≥ 1.
          Step 1. By Komlòs Lemma together with Lemma <ref>, we may find a sequence (φ_n,ψ_n,θ_n)∈{(φ_k,ψ_k,θ_k),k≥ n} such that 
         
            [ φ_n⟶φ:=φ_∞,
              μ-ψ_n⟶ψ:=ψ_∞, ν-;     θ_n⟶θ:=
             θ_∞∈(μ,ν), μ⊗ pw. ]
        
        Set φ:=∞ and ψ:=∞ on the corresponding non-convergence sets, and observe that μ[φ]+ν[ψ]<∞, by the Fatou Lemma, and therefore N_μ:={φ=∞}∈_μ and N_ν:={ψ=∞}∈_ν. We denote by (h_n,c_n) the same convex extractions from {(h_k,c_k),k≥ n}, so that the sequence ξ_n:=(φ_n,ψ_n,h_n,θ_n) inherits from (<ref>) the robust superhedging property, as for θ_1,θ_2∈(μ,ν), ψ_1,ψ_2∈_+^1(^d), and 0<λ<1, we have K_λθ_1+(1-λ)θ_2,{λψ_1+(1-λ)ψ_2=∞}⊂ K_θ_1,{ψ_1=∞}∩ K_θ_2,{ψ_2=∞}: 
        φ_n⊕ψ_n+θ_n+ h_n^⊗≥c_n≥ 0,   pointwise on  K_θ_n,{ψ_n = ∞}(X).
         Step 2. Next, notice that l_n:=(h_n^⊗)^-:=max(-h_n^⊗,0)∈Θ for all n∈. By the convergence Proposition <ref>, we may find convex combinations l_n:=∑_k≥ nλ_k^nl_k⟶ l:= l_∞, μ⊗ pw. Updating the definition of φ by setting φ:=∞ on the zero μ-measure set on which the last convergence does not hold on (∂^x l)^c, it follows from (<ref>), and the fact that K_θ̅,{ψ=∞}⊂lim inf_n→∞ K_θ_n,{ψ_n = ∞}, that 
         
            l
             =
             l_∞≤lim inf_n ∑_k≥ nλ_k^n
                                     (φ_k⊕ψ_k+θ_k
                                     )
             ≤φ⊕ψ+θ̅,
              {Y∈ K_θ̅,{ψ=∞}(X)}.
        
         where θ̅:=lim inf_n∑_k≥ nλ_k^nθ_k∈(μ,ν). As {φ = ∞}∈_μ, by possibly enlarging N_μ, we assume without loss of generality that {φ = ∞}⊂ N_μ, we see that l⊃ (N_μ^c× N_ν^c) ∩θ̅∩{Y∈ K_θ̅,{ψ=∞}(X)}, and therefore
          
            K_θ̅,{ψ=∞}(X)⊂_X l'⊂  l'(X,·),  μ
         Step 3.
        Let h_n:=∑_k≥ nλ_k^n h_k. Then b_n:=h^⊗_n+l_n =∑_k≥ nλ_k^n (h_k^⊗)^+ defines a non-negative sequence in Θ. By Proposition <ref>, we may find a sequence b_n=:h_n^⊗+l_n∈(b_k,k≥ n) such that b_n⟶ b:=b_∞, μ⊗ pw, where b takes values in [0,∞]. b_n(X,·)⟶ b(X,·) pointwise on _X b, μ-a.s. Combining with (<ref>), this shows that
         *̱h_n^⊗(X,·)
         ⟶(b-l)(X,·)
               _X b∩ K_θ̅,{ψ=∞}(X),  μ-*
        (b-l)(X,·)=lim inf_nh_n^⊗(X,·), pointwise on K_θ̅,{ψ=∞}(X) (where l is a limit of l_n), μ-a.s. Clearly, on the last convergence set, (b-l)(X,·)>-∞ on K_θ̅,{ψ=∞}(X), and we now argue that (b-l)(X,·)<∞ on K_θ̅,{ψ=∞}(X), therefore K_θ̅,{ψ=∞}(X)⊂_X b, so that we deduce from the structure of h_n^⊗ that the last convergence holds also on K_θ̅,{ψ=∞}(X):
         h_n^⊗(X,·)
         ⟶(b-l)(X,·)=:h^⊗(X,·)
               
         K_θ̅,{ψ=∞}(X),  μ-
        Indeed, let x be an arbitrary point of the last convergence set, and consider an arbitrary y∈ K_θ̅,{ψ=∞}(x). By the definition of K_θ̅,{ψ=∞}, we have x∈ K_θ̅,{ψ=∞}(x), and we may therefore find y'∈ K_θ̅,{ψ=∞}(x) with x=py+(1-p)y' for some p∈(0,1). Then, p h_n^⊗(x,y)+(1-p)h_n^⊗(x,y')=0. Sending n→∞, by concavity of the lim inf, this provides p(b-l)(x,y)+(1-p)(b-l)(x,y')≤0, so that (b-l)(x,y')>-∞ implies that (b-l)(x,y)<∞.
        
         Step 4. Notice that by dual reflexivity of finite dimensional vector spaces, (<ref>) defines a unique h(X) in the vector space K_θ̅,{ψ=∞}(X)-X, such that (b-l)(X,·)=h^⊗(X,·) on K_θ̅,{ψ=∞}(X). At this point, we have proceeded to a finite number of convex combinations which induce a final convex combination with coefficients (λ̅_n^k)_k≥ n≥ 1. Denote ξ̅_n:=∑_k≥ nλ̅_n^kξ_k, and set θ:=θ̅_∞. Then, applying this convex combination to the robust superhedging inequality (<ref>), we obtain by sending n→∞ that (φ⊕ψ+h^⊗+θ)(X,·)≥ c(X,·) on K_θ̅,{ψ=∞}(X), μ-a.s. and φ⊕ψ+h^⊗+θ=∞ on the complement μ null-set. As θ is the liminf of a convex extraction of (θ_n), we have θ≥θ_∞=θ̅, and therefore K_θ,{ψ=∞}⊂ K_θ̅,{ψ=∞}. This shows that the limit point ξ:=(φ,ψ,h,θ) satisfies the pointwise robust superhedging inequality
         φ⊕ψ+θ+ h^⊗≥ 
         c,
            on {Y∈ K_θ,{ψ=∞}(X)}.
          Step 5. By Fatou's Lemma and Lemma <ref>, we have
         
            μ[φ]+ν[ψ]+ν⊖μ[θ]
             ≤lim inf_n μ[φ_n]+ν[ψ_n]+ν⊖μ[θ_n]
             =
             _μ,ν(c).
        
        By (<ref>), we have μ[φ]+ν[ψ]+[θ]≥[c] for all ∈(μ,ν). Then, μ[φ]+ν[ψ]+_μ,ν[θ]≥_μ,ν[c]. By Proposition <ref> (i), we have _μ,ν[θ]≤ν⊖μ[θ], and therefore
         
            _μ,ν[c]
             ≤μ[φ]+ν[ψ]+_μ,ν[θ] 
             ≤μ[φ]+ν[ψ]+ν⊖μ[θ]
             ≤_μ,ν(c),
        
        by (<ref>).
        Then we have (ξ)=μ[φ]+ν[ψ]+ν⊖μ[θ] = _μ,ν(c) and _μ,ν[θ]=ν⊖μ[θ], so that ξ∈^mod_μ,ν(c).
        
        
         §.§ Duality result
        
        
        We first prove the duality in the lattice _b of bounded upper semicontinuous fonctions Ω⟶_+. This is a classical result using the Hahn-Banach Theorem, the proof is reported for completeness.
        
        
        Let f∈_b, then _μ,ν(f) = ^mod_μ,ν(f)
        We have _μ,ν(f) ≤^mod_μ,ν(f) by weak duality (<ref>), let us now show the converse inequality _μ,ν(f)≥^mod_μ,ν(f). By standard approximation technique, it suffices to prove the result for bounded continuous f. We denote by _l(^d) the set of continuous mappings ^d→ with linear growth at infinity, and by _b(^d,^d) the set of continuous bounded mappings ^d⟶^d. Define 
         
            (f) 
             :=  
             { (φ̅,ψ̅,h̅)
                     ∈_l(^d)_l(^d)_b(^d,^d): φ̅⊕ψ̅+h̅^⊗≥ f 
             },
        
        and the associated _μ,ν(f) :=inf_(φ̅,ψ̅,h̅)∈(f)μ(φ̅)+ν(ψ̅). By Theorem 2.1 in Zaev <cit.>, and Lemma <ref> below, we have
         *̱_μ,ν(f) 
            =   _μ,ν(f)
          = inf_(φ̅,ψ̅,h̅)∈(f)μ(φ̅)+ν(ψ̅)
          ≥ ^mod_μ,ν(f),
         *
        which provides the required result.
        
         Proof of Theorem <ref> The existence of a dual optimizer follows from a direct application of the compactness Lemma <ref> to a minimizing sequence of robust superhedging strategies.
        
        As for the extension of duality result of Lemma <ref> to non-negative upper semi-analytic functions, we shall use the capacitability theorem of Choquet, similar to <cit.> and <cit.>. Let [0,∞]^Ω denote the set of all nonnegative functions Ω→ [0,∞], and _+ the sublattice of upper semianalytic functions. Note that _b is stable by infimum.
        
        Recall that a _b-capacity is a monotone map 𝐂:[0,∞]^Ω⟶ [0,∞], sequentially continuous upwards on [0,∞]^Ω, and sequentially continuous downwards on _b. The Choquet capacitability theorem states that a _b-capacity 𝐂 extends to _+ by:
         *̱ C(f)
          = sup{ C(g): g∈_b  g≤ f}      
         f∈_+.
         *
        In order to prove the required result, it suffices to verify that _μ,ν and ^mod_μ,ν are _b-capacities. 
        As (μ,ν) is weakly compact, it follows from similar argument as in Prosposition 1.21, and Proposition 1.26 in Kellerer <cit.> that _μ,ν is a _b-capacity.
        We next verify that ^mod_μ,ν is a _b-capacity. Indeed, the upwards continuity is inherited from _μ,ν together with the compactness lemma <ref>, and the downwards continuity follows from the downwards continuity of _μ,ν together with the duality result on _b of Lemma <ref>.
        
        
        Let c:Ω→_+, and (φ̅,ψ̅,h̅)∈(c). Then, we may find ξ∈^mod_μ,ν(c) such that (ξ) = μ[φ̅]+ν[ψ̅].
        
        Let us consider (φ̅,ψ̅,h̅)∈(c). Then φ̅⊕ψ̅+h̅^⊗≥ c ≥ 0, and therefore
        
            ψ̅(y)≥ f(y) := x∈^dsup-φ̅(x)-h̅(x)·(y-x).
        
        Clearly, f is convex, and f(x)≥ -φ̅(x) by taking value x=y in the supremum. Hence ψ̅-f≥ 0 and φ̅+f≥ 0, implying in particular that f is finite on ^d. 
        As φ̅ and ψ̅ have linear growth at infinity, f is in ^1(ν)∩^1(μ). We have f∈_a for a=ν[f]-μ[f]≥ 0. Then we consider p∈∂ f and denote θ:=_pf. θ∈(_a)⊂(μ,ν). Then denoting φ:=φ̅+f, ψ:=ψ̅-f, and h:=h̅+p, we have ξ:=(φ,ψ, h,θ)∈_μ,ν^mod(c) and
        
            μ[φ̅]+ν[ψ̅] = μ[φ]+ν[ψ] + (ν-μ)[f]=  μ[φ]+ν[ψ] + ν⊖μ[θ] = (ξ).
        
        
        § POLAR SETS AND MAXIMUM SUPPORT MARTINGALE PLAN
        
        
         §.§ Boundary of the dual paving
        
        
        Consider the optimization problems:
         
            (θ,N_ν)∈(μ,ν)_νinfμ[G(R_θ,N_ν)],
               
             R_θ,N_ν:=  (θ(X,·)∩∂K(X)∩ N_ν^c),
        
        and for all y∈^d we consider
         (θ,N_ν)∈(μ,ν)_νinf μ[y∈∂K(X)
                       ∩θ(X,·)∩ N_ν^c
                  ].
         
        These problems are well defined by the following measurability result, whose proof is reported in Subsection <ref>.
        
        
        Let F:^d⟶, γ-measurable. Then we may find N_γ∈_γ such that 1_Y∈ F(X)1_X∉ N_γ is Borel measurable, and if X∈ F(X) convex, γ-a.s., then 1_Y∈∂ F(X)1_X∉ N_γ is Borel measurable as well.
        
        
        By the same argument than that of the proof of existence and uniqueness in Proposition <ref>, we see that the problem (<ref>), (resp. (<ref>) for y∈^d) has an optimizer (θ^*,N^*_ν)∈(μ,ν)_ν, (resp. (θ^*_y,N^*_ν,y)∈(μ,ν)_ν). Furthermore, D:=R_θ^*,N_ν^*, (resp D_y(x):={y} if y∈∂K(x)∩θ^*_y(x,·)∩ N_ν,y^*, and ∅ otherwise, for x∈^d) does not depend on the choice of (θ^*,N^*_ν), (resp. θ_y^*) up to a μ-negligible modification. 
         
        We define K̅:=D∪K, and K_θ(X) := θ(X,·)∩K̅(X) for θ∈(μ,ν). Notice that if y∈^d is not an atom of ν, we may chose N_ν,y containing y, which means that Problem (<ref>) is non-trivial only if y is an atom of ν. We denote atom(ν), the (at most countable) atoms of ν, and define the mapping
        K:= (∪_y∈ atom(ν)D_y)∪K, 
        
        
        Let θ∈(μ,ν). Up to a modification on a μ-null set, we have
         (i)K̅ is convex valued, moreover Y∈K̅(X), and Y∈ K_θ(X), (μ,ν)-q.s.
         (ii)K⊂K⊂ K_θ⊂K̅⊂K,
         (iii)K, K_θ, and K̅ are constant on K(x), for all x∈^d.
         (i)
        For x∈^d, K̅(x)=D(x)∪K(x). Let y_1,y_2∈K̅(x), λ∈ (0,1), and set y:=λ y_1+(1-λ)y_2. If y_1,y_2∈K(x), or y_1,y_2∈ D(x), we get y∈K̅(x) by convexity of K(x), or D(x). Now, up to switching the indices, we may assume that y_1∈K(x), and y_2∈ D(x)∖K(x). As D(x)∖K(x)⊂∂K(x), y∈K(x), as λ>0. Then y∈K̅(x). Hence, K̅ is convex valued.
        
        Since θ^*(X,·)∖ N_ν^*∩K∖K⊂ R_θ^*,N_ν^*, we have θ^*(X,·)∖ N_ν^*∩K⊂ R_θ^*,N_ν^*∪K = K̅. Then, as Y∈θ^*(X,·)∖ N_ν^*, and Y∈K(X), Y∈K̅(X), (μ,ν)-q.s.
        
        Let θ∈(μ,ν), then Y∈θ(X,·), (μ,ν)-q.s. Finally we get Y∈θ(X,·)∩K̅(X)=K_θ(X), (μ,ν)-q.s.
        
         (ii)
        As R_θ,N_ν(X)⊂∂K(X) = K(X), K̅⊂K. By definition, K_θ⊂K̅, and K⊂K. For y∈ atom(ν), and θ_0∈(μ,ν), by minimality,
        
        D_y(X)⊂θ_0(X,·)∩∂K(X),   μ-
        
        Applying (<ref>) for θ_0 = θ, we get D_y⊂θ(X,·), and for θ_0 = θ^*, D_y(X)⊂K̅(X), μ-a.s. Taking the countable union: K⊂ K_θ, μ-a.s. (This is the only inclusion that is not pointwise). Then we change K to K on this set to get this inclusion pointwise.
        
         (iii) For θ_0∈(μ,ν), let N_μ∈_μ from Proposition <ref>. Let x∈ N_μ^c, y∈∂K(x), and y':= x+y/2∈K(x). Then for any other x'∈K(x)∩ N_μ^c, 1/2θ_0(x,y)-θ_0(x,y') = 1/2θ_0(x',x)+1/2θ_0(x',y)-θ_0(x',y'), in particular, y∈θ(x,·) if and only if y∈θ(x',·). Applying this result to θ, θ^*, and θ^*_y for all y∈ atom(ν), we get N_μ such that for any x∈^d, K̅, K_θ, and K are constant on K(x)∩ N_μ^c. To get it pointwise, we redefine these mappings to this constant value on K(x)∩ N_μ, or to K(x), if K(x)∩ N_μ^c = ∅. The previous properties are preserved.
        
        
         §.§ Structure of polar sets
        
        A Borel set N∈(Ω) is (μ,ν)-polar if and only if for some (N_μ,N_ν)∈_μ_ν and θ∈(μ,ν), we have
         
            N ⊂{X∈ N_μ}∪{Y∈ N_ν}∪{Y∉ K_θ(X)}
             .
        
        One implication is trivial as Y∈ K_θ(X), (μ,ν)-q.s. for all θ∈(μ,ν), by Proposition <ref>. We only focus on the non-trivial implication. For an (μ,ν)-polar set N, we have _μ,ν(∞1_N) = 0, and it follows from the dual formulation of Theorem <ref> that 0 = (ξ) for some ξ=(φ,ψ,h,θ)∈_μ,ν^mod(∞1_N). Then, 
         *̱φ <∞, μ-, 
            ψ <∞, ν-   θ∈(μ,ν),
         *
         
        As h is finite valued, and φ,ψ are non-negative functions, the superhedging inequality φ⊕ψ+θ+h^⊗≥∞1_N on {Y∈ K_θ,{ψ = ∞}(X)} implies that
        1_{φ=∞}⊕1_{ψ=∞}+1_{(θ)^c}≥1_N      {Y∈ K_θ,{ψ = ∞}(X)}
        
        By Proposition <ref> (ii), we have K(X)⊂ K_θ,{ψ=∞}(X), μ-a.s. Then K̅(X)⊂K(X)⊂ K_θ,{ψ=∞}(X), which implies that
        
            K_θ(X)
            :=
            θ(X,·)∩K̅(X) ⊂θ(X,·)∩ K_θ,{ψ=∞}(X), μ-
        
        We denote N_μ := {φ=∞}∪{K_θ(X)⊄θ(X,·)∩ K_θ,{ψ=∞}(X)}∈_μ, and N_ν:={ψ = ∞}∈_ν. Then by (<ref>), 1_N =0 on ({φ =∞}^c{ψ = ∞}^c)∩{Y∈θ(X,·)∩ K_θ,{ψ=∞}(X)}, and therefore by (<ref>),
        N⊂{X∈ N_μ}∪{Y∈ N_ν}∪{Y∉ K_θ(X)}.
        
        
         §.§ The maximal support probability
        
        
        In order to prove the existence of a maximum support martingale transport plan, we introduce the maximization problem.
         
         M:=∈(μ,ν)supμ[G(_X)].
         
        where we rely on the following measurability result whose proof is reported in Subsection <ref>.
        
        
        For ∈(Ω), the map _X is analytically measurable, and the map (_X|_∂K(X)) is μ-measurable.
        
        
        Now we prove a first Lemma about the existence of a maximal support probability.
        
        
        There exists ∈(μ,ν) such that for all ∈(μ,ν) we have the inclusion _X
         ⊂ _X, μ-a.s.
        
        We proceed in two steps:
        
         Step 1: We first prove existence for the problem <ref>. Let (^n)_n≥ 1⊂(μ,ν) be a maximizing sequence. Then the measure := ∑_n≥ 12^-n ^n∈(μ,ν), and satisfies ^n_X⊂_X for all n≥ 1. Consequently μ[G(_X^n_X)]≤μ[G(_X)], and therefore M=μ[G(_X)].
        
         Step 2: We next prove that _X⊂_X, μ-a.s. for all ∈(μ,ν). Indeed, the measure :=+/2∈(μ,ν) satisfies M≥μ[G (_X)]≥μ[G (_X)] = M, implying that G(_X)=G(_X), μ-a.s. The required result now follows from the inclusion _X⊂_X.
        
        
         Proof of Proposition <ref> (iii)
        Let ∈(μ,ν) from Lemma <ref>, if we denote S(X):=_X, then we have (_X)⊂ S(X), μ-a.s. Then {Y∉ S(X)} is (μ,ν)-polar. By Lemma <ref>, {Y∉ S(X)}∪{X∉ N_μ'} is Borel for some N_μ'∈_μ. By Theorem <ref>, we see that {Y∉ S(X)}⊂{Y∉ S(X)}∪{X∉ N_μ'}⊂{X∈ N_μ}∪{Y∈ N_ν}∪{Y∉ K_θ(X)}, and therefore
         *̱{Y∈ S(X)}   ⊃   {X∉ N_μ}∩{Y∈ K_θ(X)∖ N_ν},
         *
        for some N_μ∈_μ, N_ν∈_ν, and θ∈(μ,ν). The last inclusion implies that K_θ(X)∖ N_ν⊂ S(X), μ-a.s. However, by Proposition <ref> (ii), K(X)⊂(θ(X,·)∖ N_ν), μ-a.s. Then, since S(X) is closed and convex, we see that K(X)⊂ S(X). 
        
        To obtain the reverse inclusion, we recall from Proposition <ref> (i) that {Y∈K(X)}, (μ,ν)-q.s. In particular [Y∈K(X)]=1, implying that S(X) ⊂K(X), μ-a.s. as K(X) is closed convex. Finally, recall that by definition I:= S and therefore K(X) =  I(X), μ-a.s.
        
        We may choose ∈(μ,ν) in Theorem <ref> so that for all ∈(μ,ν) and y∈^d,
         
            μ[ _X[{y}]>0]
             ≤μ[ _X[{y}]>0],
                _X|_∂ I(X)⊂ _X|_∂ I(X), μ-
        
         In this case the set-valued maps J(X):= I(X)∪{y∈^d:ν[y]>0_X[{y}]>0}, and J̅(X):=I(X)∪ _X |_∂ I(X) are unique μ-a.s. Furthermore J(X)=K(X), J̅(X)=K̅(X), and J_θ(X) = K_θ(X), μ-a.s. for all θ∈(μ,ν).
         Step 1:
        By the same argument as in the proof of Lemma <ref>, we may find '∈(μ,ν) such that 
         
         M':=
          ∈(μ,ν)supμ[G((_X|_∂K(X)))]=μ[G((_X'|_∂K(X)))].
         
        We also have similarly that (_X|_∂K(X))⊂(_X'|_∂K(X)), μ-a.s. for all ∈(μ,ν). Then we prove similarly that S'(X):=(_X'|_∂K(X))=D(X), μ-a.s., where recall that D is the optimizer for (<ref>). Indeed, by the previous step, we have (_X|_∂K(X))⊂ S'(X), μ-a.s. Then {Y∉ S'(X)∪K(X)} is (μ,ν)-polar. By Theorem <ref>, we see that {Y∉ S'(X)∪K(X)}⊂{X∈ N_μ}∪{Y∈ N_ν}∪{Y∉ K_θ(X)∪K(X)}, or equivalently,
         {Y∈ S'(X)∪K(X)}   ⊃   {X∉ N_μ}∩{Y∈ K_θ(X)∖ N_ν},
         
        for some N_μ∈_μ, N_ν∈_ν, and θ∈(μ,ν). Similar to the previous analysis, we have K_θ(X)∖ N_ν∖K(X)⊂ S'(X), μ-a.s. Then, since S'(X) is closed and convex, we see that D(X)⊂ S'(X). 
        
        To obtain the reverse inclusion, we recall from Proposition <ref> that {Y∈K̅(X)}, (μ,ν)-q.s. In particular '[Y∈K(X)∪ D(X)]=1, implying that S'(X) ⊂ D(X), μ-a.s. By Proposition <ref> (iii), we have J̅(X) = (I∪ S')(X)=(K∪ D)(X) = K̅(X), μ-a.s.
        
        Finally, +'/2 is optimal for both problems (<ref>), and (<ref>). By definition, the equality J_θ(X) = K_θ(X), μ-a.s. for θ∈(μ,ν) immediately follows.
        
         Step 2: Let y∈ atom(ν), if y is an atom of γ_1∈(^d) and γ_2∈(^d), then y in an atom of λγ_1+(1-λ)γ_2 for all 0<λ<1. By the same argument as in Step 1, we may find ^y∈(μ,ν) such that
         
         M_y:=
          ∈(μ,ν)supμ[_X[{y}∩K(X)]>0] = μ[_X^y[{y}∩K(X)]>0].
         
        
        We denote S_y(X):=_X^y|_K(X)∩{y}. Recall that D_y is the notation for the optimizer of problem (<ref>). We consider the set N:={Y∉(K(X)∖{y})∪ S_y(X)}. N is polar as Y∈K(X), q.s., and by definition of S_y. Then N⊂{X∈ N_μ}∪{Y∈ N_ν}∪{Y∉ K_θ(X)}, or equivalently,
         {Y∉(K(X)∖{y})∪ S_y(X)}   ⊃   {X∉ N_μ}∩{Y∈ K_θ(X)∖ N_ν},
         
        for some N_μ∈_μ, N_ν∈_ν, and θ∈(μ,ν). Then D_y(X)⊂ K_θ(X)∖ N_ν⊂K(X)∖{y}∪ S_y(X), μ-a.s. Finally D_y(X)⊂ S_y(X), μ-a.s.
        
        On the other hand, S_y⊂ D_y, μ-a.s., as if _X^y[{y}]>0, we have θ(X,y)<∞, μ-a.s. at the corresponding points. Hence, D_y(X) = S_y(X), μ-a.s. Now if we sum up the countable optimizers for y∈ atom(ν), with the previous optimizers, then the probability  we get is an optimizer for (<ref>), (<ref>), and (<ref>), for all y∈^d (the optimum is 0 if it is not an atom of ν). Furthermore, the μ-a.e. equality of the maps S_y and D_y for these countable y∈ atom(ν) is preserved by this countable union, then together with Proposition <ref> (iii), we get J = K, μ-a.s.
        
        
        
        As a preparation to prove the main Theorem <ref>, we need the following lemma, which will be proved in Subsection <ref>.
        
        
        Let F:^d⟶ be a γ-measurable function for some γ∈(^d), such that x∈ F(x) for all x∈^d, and {F(x):x∈^d} is a partition of ^d. Then up to a modification on a γ-null set, F can be chosen in addition to be analytically measurable.
         Proof of Theorem <ref>
        Existence holds by Lemma <ref> above, (i) is a consequence of Lemma <ref>, and (ii) directly stems from Lemma <ref> (iii) together with Proposition <ref> (iii). Now we need to deal with the measurability issue. Lemma <ref> allows to modify _X to get (ii) while preserving its analytic measurability, we denote I its modification. However, we need to modify _X to get the result. As _X is analytically measurable by Lemma <ref>, the set of modification N_μ:={_X≠ I(X)}∈_μ is analytically measurable. Then we may redefine _X on N_μ, so as to preserve a kernel for . By the same arguments than the proof of Lemma <ref> (ii), the measure-valued map κ_X:= g_I(X) is a kernel thanks to the analytic measurability of I, recall the definition of g_K given by (<ref>). Furthermore, κ_X = I(X) pointwise by definition. Then a suitable kernel modification from which the result follows is given by
        
            _X':= 1_{X∈ N_μ}κ_X + 1_{X∉ N_μ}_X.
        
         Proof of Proposition <ref>
        The existence and the uniqueness are given by Lemma <ref> and the other properties follow from the identity between the J maps and the K maps, also given by the Lemma, together with Proposition <ref>.
        
         Proof of Theorem <ref>
        We simply apply Lemma <ref> to replace K_θ by J_θ in Proposition <ref>.
        
        
        § MEASURABILITY OF THE IRREDUCIBLE COMPONENTS
        
        
         §.§ Measurability of G
         Proof of Lemma <ref> (ii)
        As ^d is locally compact, the Wijsman topology is locally equivalent to the Hausdorff topology[The Haussdorff distance on the collection of all compact subsets of a compact metric space (, d) is defined by d_H(K_1,K_2)=sup_x∈|(x,K_1)-(x,K_2)|, for K_1,K_2⊂, compact subsets.], i.e. as n→∞, K_n⟶ K for the Wijsman topology if and only if K_n∩ B_M⟶ K∩ B_M for the Hausdorff topology, for all M≥ 0.
        
        We first prove that K⟼ K is a lower semi-continuous map →. Let (K_n)_n≥ 1⊂ with dimension d_n≤ d'≤ d converging to K. We consider A_n:= K_n. As A_n is a sequence of affine spaces, it is homeomorphic to a d+1-uplet. Observe that the convergence of K_n allow us to chose this d+1-uplet to be bounded. Then up to taking a subsequence, we may suppose that A_n converges to an affine subspace A of dimension less than d'. By continuity of the inclusion under the Wijsman topology, K⊂ A and K≤ A ≤ d'.
        
        We next prove that the mapping K↦ g_K(K) is continuous on { K = d'} for 0≤ d'≤ d, which implies the required measurability. Let (K_n)_n≥ 1⊂ be a sequence with constant dimension d', converging to a d'-dimensional subset, K in . Define A_n:= K_n and A:= K, A_n converges to A as for any accumulation set A' of A_n, K⊂ A' and A' =  A, implying that A'=A. Now we consider the map ϕ_n:A_n→ A, x↦ proj_A(x). For all M>0, it follows from the compactness of the closed ball B_M that ϕ_n converges uniformly to identity as n→∞ on B_M. Then, ϕ_n(K_n)∩ B_M⟶ K∩ B_M as n→∞, and therefore _A[ϕ_n(K_n∩ B_M)∖ K]+_A[K∖ϕ_n(K_n)∩ B_M]⟶ 0. As the Gaussian density is bounded, we also have
        
            g_A[ϕ_n(K_n∩ B_M)]⟶  g_A[K∩ B_M].
        
        We next compare g_A[ϕ_n(K_n∩ B_M)] to g_K_n(K_n∩ B_M). As (ϕ_n) is a sequence of linear functions that converges uniformly to identity, we may assume that ϕ_n is a ^1-diffeomorphism. Furthermore, its constant Jacobian J_n converges to 1 as n→∞. Then, 
        *̱∫_K_n∩ B_Me^-|ϕ_n(x)|^2/2/(2π)^d'/2_K_n(dx) 
           =   ∫_ϕ_n(K_n∩ B_M)e^-|y|^2/2J_n^-1/(2π)^d'/2_A(dy)
        
           =   
        J_n^-1 g_A[ϕ_n(K_n∩ B_M)].
        *
        As the Gaussian distribution function is 1-Lipschitz, we have
         
            |∫_K_n∩ B_Me^-|ϕ_n(x)|^2/2/(2π)^d'/2_K_n(dx)
                    - g_K_n(K_n∩ B_M)
             |
              ≤ _K_n[K_n∩ B_M]|ϕ_n-Id_A|_∞,
        
        where |·|_∞ is taken on K_n∩ B_M. Now for arbitrary ϵ>0, by choosing M sufficiently large so that g_V[V∖ B_M]≤ϵ for any d'-dimensional subspace V, we have
        *̱| g_K_n[K_n]- g_K[K]|   ≤   | g_K_n[K_n∩ B_M]- g_A[K∩ B_M]|+2ϵ
           ≤   | g_K_n[K_n∩ B_M]-∫_K_n∩ B_MCexp(-|ϕ_n(x)|^2/2)_K_n(dx)|
              
        +|J_n^-1 g_A[ϕ_n(K_n∩ B_M)]- g_A[K∩ B_M]|+2ϵ
           ≤   
        4ϵ,
        *
        for n sufficiently large, by the previously proved convergence. Hence G_d':=G|_^-1{d'} is continuous, implying that G :K⟼∑_d'=0^d1_^-1{d'}(K)G_d'(K) is Borel-measurable.
        
        
         §.§ Further measurability of set-valued maps
        
        
        This subsection is dedicated to the proof of Lemmas <ref> (i), <ref>, and <ref>. In preparation for the proofs, we start by giving some lemmas on measurability of set-valued maps. Let  be a σ-algebra of ^d. In practice we will always consider either the σ-algebra of Borel sets, the σ-algebra of analytically measurable sets, or the σ-algebra of universally measurable sets.
        
        
        Let (F_n)_n≥ 1⊂Ł^(^d,). Then ∪_n≥ 1F_n and ∩_n≥ 1F_n are -measurable.
        
        The measurability of the union is a consequence of Propositions 2.3 and 2.6 in Himmelberg <cit.>. The measurability of the intersection follows from the fact that ^d is σ-compact, together with Corollary 4.2 in <cit.>.
        
        Let F∈Ł^(^d,). Then, F, F, and _X  F are -measurable.
        
        The measurability of F is a direct application of Theorem 9.1 in <cit.>.
        
        We next verify that F is measurable. Since the values of F are closed, we deduce from Theorem 4.1 in Wagner <cit.>, that we may find a measurable x⟼ y(x), such that y(x)∈ F(x) if F(x)≠∅, for all x∈^d. Then we may write F(x) =    ∪_q∈(y(x)+q(F(x)-y(x))) for all x∈^d. The measurability follows from Lemmas <ref>, together with the first step of the present proof.
        
        We finally justify that _X  F is measurable. We may assume that F takes convex values. By convexity, we may reduce the definition of _x to a sequential form:
        *̱_x F(x) 
           =   ∪_n≥ 1{y∈^d , y+1/n(y-x)∈ F(x)  and  x-1/n(y-x)∈ F(x)}
           =   ∪_n≥ 1[{y∈^d , y+1/n(y-x)∈ F(x)}∩{y∈^d , x-1/n(y-x)∈ F(x)}]
           =   ∪_n≥ 1[(1/n+1x+n/n+1F(x)) ∩(-(n+1)x - nF(x))],
        *
        so that the required measurability follows from Lemma <ref>.
        
        
        
        We denote  the set of finite sequences of positive integers, and Σ the set of infinite sequences of positive integers. Let s∈, and σ∈Σ. We shall denote s<σ whenever s is a prefix of σ. 
        
        
        Let (F_s)_s∈ be a family of universally measurable functions ^d⟶ with convex image. Then the mapping (∪_σ∈Σ∩_s<σF_s) is universally measurable.
        
        Let  the collection of universally measurable maps from ^d to  with convex image. For an arbitrary γ∈(^d), and F:^d⟶, we introduce the map
         *̱γ G^*[F]:=inf_F⊂ F'∈γ G[F'],
               γ G[F']:= γ[G(F'(X))]  F'∈.
         *
        Clearly, γ G and γ G^* are non-decreasing, and it follows from the dominated convergence theorem that γ G, and thus γ G^*, are upward continuous.
        
         Step 1: In this step we follow closely the line of argument in the proof of Proposition 7.42 of Bertsekas and Shreve <cit.>. Set F:=(∪_σ∈Σ∩_s<σF_s), and let (F̅_n)_n a minimizing sequence for γ G^*[F]. Notice that F⊂F̅ := ∩_n≥ 1F̅_n∈, by Lemma <ref>. Then F̅ is a minimizer of γ G^*[F].
        
        For s,s'∈ S, we denote s≤ s' if they have the same length |s|=|s'|, and s_i≤ s_i' for 1≤ i≤ |s|. For s∈ S, let
         *̱
         R(s):= ∪_s'≤ s∪_σ>s'∩_s”<σF_s”      
         K(s):= ∪_s'≤ s∩_j=1^|s'|F_s_1',...,s_j'.
         *
        Notice that K(s) is universally measurable, by Lemmas <ref> and <ref>, and
         
            R(s)⊂ K(s),
              ∪_s_1≥ 1R(s_1) = F,
               ∪_s_k≥ 1R(s_1,...,s_k) = R(s_1,...,s_k-1).
        
        By the upwards continuity of γ G^*, we may find for all ϵ>0 a sequence σ^ϵ∈Σ s.t.
         *̱γ G^*[F]≤γ G^*[R(σ^ϵ_1)]+2^-1ϵ,
               γ G^*[R(σ_k-1)]
         ≤γ G^*[R(σ_k)]+2^-kϵ,
         *
        for all k≥ 1, with the notation σ^_k:=(σ_1^ϵ,…,σ_k^). Recall that the minimizer F and K(s) are in  for all s∈. We then define the sequence K_k^ϵ:=F∩ K(σ^ϵ_k)∈, k≥ 1, and we observe that
         
            (K_k^ϵ)_k≥ 1   F^ϵ:=∩_k≥ 1K_k^ϵ⊂ F,
               γ G[K_k^ϵ]≥γ G^*[F]-ϵ = γ G[F]-ϵ,
        
        by the fact that R(σ^ϵ_k)⊂ K_k^ϵ. We shall prove in Step 2 that, for an arbitrary α>0, we may find =(α)≤α such that (<ref>) implies that 
         γ G[F^ϵ]
         ≥inf_k≥ 1γ G[K_k^ϵ] - α≥γ G[F]-ϵ-α.
         
        Now let α=α_n:=n^-1, _n:=ϵ(α_n), and notice that F:= ∪_n≥ 1F^ϵ_n∈, with F^ϵ_n⊂F⊂ F ⊂F, for all n≥ 1. Then, it follows from (<ref>) that γ G[F]= γ G[F], and therefore F= F =F, γ -a.s. In particular, F is γ-measurable, and we conclude that F∈ by the arbirariness of γ∈(^d).
        
        
         Step 2: It remains to prove that, for an arbitrary α>0, we may find =(α)≤α such that (<ref>) implies (<ref>). This is the point where we have to deviate from the argument of <cit.> because γ G is not downwards continuous, as the dimension can jump down. 
        
        Set A_n:={G(F(X))-F(X)≤ 1/n}, and notice that ∩_n≥ 1A_n=∅. Let n_0≥ 1 such that γ[A_n_0]≤1/2α/d+1, and set ϵ := 1/21/n_0α/d+1>0. Then, it follows from (<ref>) that
         γ[inf_n G(K^ϵ_n)-F≤ 0]
            ≤   γ[inf_n G(K^ϵ_n)-G(F)≤ n_0^-1]
         
              
         + γ[G(F)-F≤ -n_0^-1]
         
           ≤   
         n_0(γ[G(F)]
                 -γ[inf_n G(K^ϵ_n)]
            )
         +γ[A_n_0]
         
           =   
         n_0(γ[G(F)]
                 -inf_n γ[G(K^ϵ_n)]
            )
         +γ[A_n_0]
         
           ≤   
         n_0ϵ+1/2α/d+1 = α/d+1,
         
        where we used the Markov inequality and the monotone convergence theorem. Then:
         *̱γ[inf_n G(K^ϵ_n)-G(F^ϵ)]
            ≤   γ[_{inf_n G(K^ϵ_n)-F≤ 0 }(inf_n G(K^ϵ_n)-G(F^ϵ))
         
               + _{inf_n G(K^ϵ_n)-F>0 }(inf_n G(K^ϵ_n)-G(F^ϵ))
               ]
         
           ≤   γ[(d+1)_{inf_n G(K^ϵ_n)-F≤ 0 }
               + _{inf_n G(K^ϵ_n)-F>0 }(inf_n G(K^ϵ_n)-G(F^ϵ))
               ].
         *
        We finally note that inf_n G(K^ϵ_n)-G(F^ϵ)=0 on {inf_n G(K^ϵ_n)-F>0 }. Then (<ref>) follows by substituting the estimate in (<ref>).
         
         Proof of Lemma <ref> (i)
        We consider the mappings θ:Ω→_+ such that θ = ∑_k=1^n λ_k 1_C^1_k C^2_k where n∈, the λ_k are non-negative numbers, and the C^1_k,C^2_k are closed convex subsets of ^d. We denote the collection of all these mappings . Notice that  for the pointwise limit topology contains all ^0_+(Ω). Then for any θ∈^0_+(Ω), we may find a family (θ_s)_s∈⊂, such that θ = inf_σ∈Σsup_s<σθ_s. For θ∈_+^0(Ω), and n≥ 0, we denote F_θ :x⟼  θ(x,·), and F_θ,n :x⟼  θ(x,·)^-1([0,n]). Notice that F_θ = ∪_n≥ 1F_θ,n. Notice as well that F_θ,n is Borel measurable for θ∈, and n≥ 0, as it takes values in a finite set, from a finite number of measurable sets. Let θ∈_+^0(Ω), we consider the associated family (θ_s)_s∈⊂, such that θ = inf_σ∈Σsup_s<σθ_s. Notice that F_θ,n = (∪_σ∈Σ∩_s<σF_θ_s,n) is universally measurable by Lemma <ref>, thus implying the universal measurability of F_θ = θ(X,·) by Lemma <ref>.
        
        In order to justify the measurability of _Xθ, we now define
         *̱
         F_θ^0 := F_θ      
         F_θ^k 
         := 
         (θ(X,·)∩ _XF_θ^k-1),
          k ≥ 1.
         *
        Note that F_θ^k = ∪_n≥ 1(∪_σ∈Σ∩_s<σF_θ_s,n∩ _xF_θ^k-1). Then, as F_θ^0 is universally measurable, we deduce that (F_θ^k)_k≥ 1 are universally measurable, by Lemmas <ref> and <ref>.
        
        As _Xθ is convex and relatively open, the required measurability follows from the claim:
         *̱
         F_θ^d   =   _Xθ.
         *
        To prove this identity, we start by observing that F_θ^k(x) ⊃_xθ. Since the dimension cannot decrease more than d times, we have _x F^d_θ(x) =  F^d_θ(x) and
         *̱
         F^d+1_θ(x)
            =   (θ(x,·)∩ _xF_θ^d(x))
         
           =   (θ(x,·)∩_x F_θ^d-1(x))
          = 
         F_θ^d(x).
         *
        i.e. (F^d+1_θ)_k is constant for k≥ d. Consequently,
        *̱_x(θ(x,·)∩ _xF_θ^d(x))    =    F_θ^d(x)
           ≥   (θ(x,·)∩ _xF_θ^d(x)).
        *
        As (θ(x,·)∩ _xF_θ^d(x))≥_x(θ(x,·)∩ _xF_θ^d(x)), we have equality of the dimension of (θ(x,·)∩ _xF_θ^d(x)) with its _x. Then it follows from Proposition <ref> (ii) that x∈ (θ(x,·)∩ _xF_θ^d(x)), and therefore:
         *̱
         F_θ^d(x)
            =   (θ(x,·)∩ _xF_θ^d(x))
         
           =    (θ(x,·)∩ _xF_θ^d(x))
         
           =   _x(θ(x,·)∩ _xF_θ^d(x))
          ⊂ _xθ.
         *
        Hence F_θ^d(x)= _xθ.
        
        Finally, K_θ,A = _X(θ+∞ 1_^d A) is universally measurable by the universal measurability of _X.
        
         Proof of Lemma <ref>
        We may find (F_n)_n≥ 1, Borel-measurable with finite image, converging γ-a.s. to F. We denote N_γ∈_γ, the set on which this convergence does not hold. For ϵ>0, we denote F_k^ϵ(X) := {y∈^d:dist(y,F_k(X))≤ϵ}, so that
        *̱
        F(x) = ∩_i≥1lim inf_n→∞F_n^1/i(x),      x∉ N_γ.
        *
        Then, as 1_Y∈ F(X)1_X∉ N_γ = inf_i≥1lim inf_n→∞1_Y∈ F_n^1/i(X)1_X∉ N_γ, the Borel-measurability of this function follows from the Borel-measurability of each 1_Y∈ F_n^1/i(X).
        
        Now we suppose that X∈ F(X) convex, γ-a.s. Up to redefining N_γ, we may suppose that this property holds on N_γ^c, then ∂ F(x) = ∩_n≥ 1F(x)∖(x+n/n+1(F(x)-x)), for x∉ N_γ. We denote a := 1_Y∈ F(X)1_X∉ N_γ. The result follows from the identity 1_Y∈∂ F(X)1_X∉ N_γ = a - sup_n≥ 1a(X,X+n/n+1(Y-X)).
        
         Proof of Lemma <ref>
        Let _:={K=(x_1,…,x_n): n∈,(x_i)_i≤ n⊂^d}. Then
         
            _x
             =
              ∪_N≥ 1∩{K∈_:_x∩ B_N⊂ K}
             =
             ∪_N≥ 1∩_K∈_F_K^N(x),
        
        where F_K^N(x) := K if _x[B_N∩ K] =_x[B_N], and F_K^N(x) :=^d otherwise.
        As for any K∈_ and N≥ 1, the map _X[B_N∩ K] -_X[B_N] is analytically measurable, then F_K^N is analytically measurable. The required measurability result follows from lemma <ref>.
        
        Now, in order to get the measurability of (_X|_∂ I(X)), we have in the same way
         *̱(_X|_∂ I(X))
            =   ∪_n≥ 1∩_K∈_F_K'^N(x),
         *
        where F_K'^N(x) := K if _x[∂ I(x)∩ B_N∩ K] =_x[∂ I(x)∩ B_N], and F_K'^N(x) :=^d otherwise. As _X[∂ I(X)∩ B_N∩ K] = _X[1_Y∈∂ I(X)1_X∉ N_μ1_Y∉ B_N∩ K], μ-a.s., where N_μ∈_μ is taken from Lemma <ref>, _X[∂ I(X)∩ B_N∩ K] is μ-measurable, as equal μ-a.s. to a Borel function. Then similarly, _X[∂ I(X)∩ B_N∩ K] -_X[∂ I(X)∩ B_N] is μ-measurable, and therefore (_X|_∂ I(X)) is μ-measurable.
        
         Proof of Lemma <ref>
        By γ-measurability of F, we may find a Borel function F_B:^d⟶ such that F = F_B, γ-a.s. Let a Borel N_γ∈_γ such that F = F_B on N_γ^c. By the fact that  is Polish, we may find a sequence (F_n)_n≥ 1 of Borel functions with finite image converging pointwise towards F_B when n⟶∞. We will give an explicit expression for F_n that will be useful later in the proof. Let (K_n)_n≥ 1⊂ a dense sequence,
        
        F_n(x)    :=   _K∈(K_i)_i≤ ndist(F_B(x),K),
        
        
        Where dist is the distance on  that makes it Polish, and we chose the K with the smallest index in case of equality.
        
        We fix n≥ 1, let K∈ F_n(N_γ^c), the image of F_n outside of N_γ, and A_K:= F_n^-1({K}). We will modify the image of F_n so that it is the same for all x'∈ F_B(x)= F(x), for all x∈ N_γ^c∩ A_K. Then we consider the set A'_K:= ∪_x∈ N_γ^c∩ A_K F_B(x), we now prove that this set in analytic. By Theorem 4.2 (b) in <cit.>, Gr F_B:={Y∈ F_B(X)} is a Borel set. Let λ >0, we define the affine deformation f_λ:Ω⟶Ω by f_λ(X,Y) := (X,X+λ(Y-X)). By the fact that for k≥ 1, f_1-1/k(Gr F_B) is Borel together with the fact that x∈ f_B(x) for x∉ N_γ, we have
        
            {Y∈ F_B(X)}∩{X∉ N_γ} = ∪_k≥ 1f_1-1/k(Gr F_B)∩{X∉ N_γ}.
        
        Therefore, {Y∈ F_B(X)}∩{X∉ N_γ} is Borel, and so is {Y∈ F_B(X)}∩{X∈ N_γ^c∩ A_K}. Finally,
        
            A'_K = Y({Y∈ F_B(X)}∩{X∈ N_γ^c∩ A_K}),
        
        therefore, A'_K is the projection of a Borel set, which is one of the definitions of an analytic set (see Proposition 7.41 in <cit.>). Now we define a suitable modification of F_n by F_n'(x) := K for all x∈ A'_K, we do this redefinition for all K∈ F_B(N_γ^c). Notice that thanks to the definition (<ref>) and the fact that F_B(x) = F_B(x') if x,x'∉ N_γ and x'∈ F_B(x) = F(x), we have the inclusion A'_K⊂ A_K∪ N_γ. Then the redefinitions of F_n only hold outside of N_γ, furthermore for different K_1, K_2∈ F_n(N_γ^c), A'_K_1∩ A'_K_2 = ∅ as the value of F_n(x) only depends on the value of F_B(x) by (<ref>). Notice that
        
        N_γ' := (∪_K∈ F_n(N_γ^c)A'_K)^c =  (∪_x∉ N_γF_B(x))^c⊂ N_γ
        is analytically measurable, as the complement of an analytic set, and does not depend on n. For x∈ N_γ', we define F_n'(x):= {x}. Notice that F_n' is analytically measurable as the modification of a Borel Function on analytically measurable sets.
        
        Now we prove that F_n' converges pointwise when n⟶∞. For x∈ N_γ', F_n'(x) is constant equal to {x}, if x∉ N_γ', by (<ref>)x∈∪_x∉ N_γF_B(x), and therefore F_n'(x) = F_B(x')=F(x') for some x∈ N_γ^c, for all n≥ 1. Then as F_n'(x') converges to F(x'), F_n'(x) converges to F(x). Let F' be the pointwise limit of F_n'. the maps F_n' are analytically measurable, and therefore, so does F'. For all n≥ 1, F_n' = F_n, γ-a.e. and therefore F' = F_B = F, γ-a.e. Finally, F'(N_γ^c) = F(N_γ^c), and ∪ F(N_γ^c) = (N_γ')^c. By property of F, F'(N_γ^c) is a partition of (N_γ')^c such that x∈ F'(x) for all x∉ N_γ'. On N_γ', this property is trivial as F'(x) = {x} for all x∈ N_γ'.
        
        
        § PROPERTIES OF TANGENT CONVEX FUNCTIONS
        
        
         §.§ x-invariance of the y-convexity
        
        
        We first report a convex analysis lemma.
        
        
        
        Let f:^d→ be convex finite on some convex open subset U⊂^d. We denote f_*:^d→ the lower-semicontinuous envelop of f on U, then
        *̱
        f_*(y)=lim_ϵ↘ 0f(ϵ x+(1-ϵ)y),      (x,y)∈ U U.
        *
        f_* is the lower semi-continuous envelop of f on U, i.e. the lower semi-continuous envelop of f':=f+∞1_U^c. Notice that f' is convex ^d⟶∪{∞}. Then by Proposition 1.2.5 in Chapter IV of <cit.>, we get the result as f=f' on U.
        
         Proof of Proposition <ref>
        The result is obvious in (_1), as the affine part depending on x vanishes. We may use N_ν = ∅. Now we denote  the set of mappings in Θ_μ such that the result from the proposition holds. Then we have (_1)⊂.
        
        We prove that  is -Fatou closed. Let (θ_n)_n be a sequence in  converging  to θ∈Θ_μ. Let n≥ 1, we denote N_μ, the set in _μ from the proposition applied to θ_n, and let N_μ^0∈_μ corresponding to the  convergence of θ_n to θ. We denote N_μ:=∪_n∈N_μ^n∈_μ. Let x_1,x_2∉ N_μ, and ∈_x_1θ∩_x_2θ. Let y_1,y_2∈_x_1θ, such that we have the convex combination = λ y_1 + (1-λ) y_2, and 0≤λ≤ 1. Then for i=1,2, θ_n(x_1,y_i)⟶θ(x_1,y_i), and θ_n(x_1,)⟶θ(x_1,), as n→∞. Using the fact that θ_n∈, for all n, we have
        
            Δ_n
             :=
             λθ_n(x_i,y_1)+(1-λ)θ_n(x_i,y_2)-θ_n(x_i,)
             ≥
             0,
               
             i=1,2.
        
        Taking the limit n→∞ gives that θ_∞(x_2,y_i)<∞, and y_i∈θ_∞(x_2,·).  is interior to _x_1θ, then for any y∈_x_1θ, y':=  + ϵ/1-ϵ(-y)∈_x_1θ for 0<ϵ<1 small enough. Then = ϵ y+(1-ϵ)y'. As we may chose any y∈_x_1θ, we have _x_1θ⊂θ_∞(x_2,·). Then, we have
         
            _x_2(_x_1θ∪_x_2θ)
             ⊂_x_2 (θ_∞(x_2,·)
                                 ) 
             = 
             _x_2θ.
        
        By Lemma <ref>, as _x_1θ∩_x_2θ≠∅, (_x_1θ∪_x_2θ)= (_x_1θ∪_x_2θ). In particular, (_x_1θ∪_x_2θ) is relatively open and contains x_2, and therefore _x_2(_x_1θ∪_x_2θ) = (_x_1θ∪_x_2θ). Finally, by (<ref>), _x_1θ⊂_x_2θ. As there is a symmetry between x_1, and x_2, we have _x_1θ = _x_2θ. Then we may go to the limit in equation (<ref>):
        
            Δ_∞
             :=
             λθ(x_i,y_1)+(1-λ)θ(x_i,y_2)-θ(x_i,)
             ≥
             0,
               
             i=1,2
        
        Now, let y_1,y_2∈^d, such that we have the convex combination = λ y_1 + (1-λ) y_2, and 0≤λ≤ 1. we have three cases to study.
        
         Case 1:y_i∉_x_1θ for some i=1,2.
        Then, as the average  of the y_i is in _x_1θ, by Proposition <ref> (ii), me may find i'=1,2 such that y_i'∉ θ(x_1,·), thus implying that θ(x_1,y_i)=∞. Then λθ(x_1,y_1)+(1-λ) θ(x_1,y_2)-θ(x_1,) = ∞≥ 0. As _x_1θ = _x_2θ, we may apply the same reasoning to x_2, we get λθ(x_1,y_1)+(1-λ) θ(x_2,y_2)-θ(x_2,) = ∞≥ 0. We get the result.
        
         Case 2:y_1,y_2∈_x_1θ.
        This case is (<ref>).
        
         Case 3:y_1,y_2∈_x_1θ.
        The problem arises here if some y_i is in the boundary ∂_x_1θ. Let x∉ N_μ, we denote the lower semi-continuous envelop of θ(x,·) in _xθ, by θ_*(x,y) := lim_ϵ↘ 0θ(x,ϵ x+(1-ϵ)y'), for y∈_xθ, where the latest equality follows from Lemma <ref> together with that fact that θ(x,·) is convex on _xθ. Let y∈_x_1θ, for 1≥ϵ>0, y^ϵ :=ϵ x_1 + (1-ϵ )y∈_x_1θ. By (<ref>), (1-ϵ )θ_n(x_1,y)-θ_n(x_1,y^ϵ) = (1-ϵ )θ_n(x_2,y)-θ_n(x_2,y^ϵ). Taking the lim inf, we have (1-ϵ )θ(x_1,y)-θ(x_1,y^ϵ) = (1-ϵ )θ(x_2,y)-θ(x_2,y^ϵ). Now taking ϵ↘ 0, we have θ(x_1,y)-θ_*(x_1,y) = θ(x_2,y)-θ_*(x_2,y). Then the jump of θ(x,·) in y is independent of x=x_1 or x_2. Now for 1≥ϵ>0, by (<ref>)
            λθ(x_1,y_1^ϵ)+(1-λ)θ(x_1,y_2^ϵ)-θ(x_1,^ϵ)
             =
            λθ(x_2,y_1^ϵ)+(1-λ)θ(x_2,y_2^ϵ)-θ(x_2,^ϵ)
             ≥ 0.
        
         By going to the limit ϵ↘ 0, we get
        
            λθ_*(x_1,y_1)+(1-λ)θ_*(x_1,y_2)-θ_*(x_1,)
             =
            λθ_*(x_2,y_1)+(1-λ)θ_*(x_2,y_2)-θ_*(x_2,)
             ≥ 0.
        
        As the (nonnegative) jumps do not depend on x= x_1 or x_2, we finally get
        
            λθ(x_1,y_1)+(1-λ)θ(x_1,y_2)-θ(x_1,)
             =
            λθ(x_2,y_1)+(1-λ)θ(x_2,y_2)-θ(x_2,)
             ≥ 0.
        
         Finally,  is -Fatou closed, and convex. _1⊂. As the result is clearly invariant when the function is multiplied by a scalar, the Result is proved on (μ,ν).
        
        
         §.§ Compactness
           
          
         Proof of Proposition <ref>
        We first prove the result for θ=(θ_n)_n≥ 1⊂Θ. Denote (θ) := {θ'∈Θ^: θ_n'∈(θ_k,k≥ n), n∈}. Consider the minimization problem:
         
         m
            :=   inf_θ'∈(θ)μ[G(_Xθ'_∞)],
         
        where the measurability of G(_Xθ'_∞) follows from Lemma <ref>.
        
         Step 1: We first prove the existence of a minimizer. Let (θ'^k)_k∈∈(θ)^ be a minimizing sequence, and define the sequence θ∈(θ) by:
         *̱   θ_n 
         := 
         (1-2^-n)^-1∑_k=1^n 2^-kθ'^k_n,
          n≥ 1.
         *
        Then, (θ_∞)⊂⋂_k≥ 1(θ'^k_∞) by the non-negativity of θ', and we have the inclusion {θ_nn→∞⟶∞}⊂{θ_n'^kn→∞⟶∞  k≥ 1}. Consequently,
         *̱   _xθ_∞ ⊂ (⋂_k≥ 1θ'^k_∞(x,·))
          ⊂ ⋂_k≥ 1_xθ'^k_∞     
         x∈^d.
         *
        Since (θ'^k)_k is a minimizing sequence, and θ∈(θ), this implies that μ[G(_Xθ_∞)]= m.
        
         Step 2: We next prove that we may find a sequence (y_i)_i≥ 1⊂^0(^d,^d) such that 
         
         y_i(X)∈(_Xθ_∞)
                
         (y_i(X))_i≥ 1  _Xθ_∞,
           μ-
        Indeed, it follows from Lemmas <ref>, and <ref> that the map x↦(_xθ_∞) is universally measurable, and therefore Borel-measurable up to a modification on a μ-null set. Since its values are closed and nonempty, we deduce from the implication (ii)  (ix) in Theorem 4.2 of the survey on measurable selection <cit.> the existence of a sequence (y_i)_i≥ 1 satisfying (<ref>).
        
         Step 3: Let m(dx,dy):=μ(dx)⊗∑_i≥ 02^-iδ_{y_i(x)}(dy). By the Komlòs lemma (in the form of Lemma A1.1 in <cit.>, similar to the one used in the proof of Proposition 5.2 in <cit.>), we may find θ∈(θ) such that θ_n⟶θ_∞∈Ł^0(Ω), m-a.s. Clearly, _xθ_∞⊂_xθ_∞, and therefore μ[G(_Xθ_∞)]≤μ[G(_xθ_∞)], for all x∈^d. This shows that 
         
         G(_Xθ_∞)
          = 
         G(_Xθ_∞),  μ-
        so that θ is also a solution of the minimization problem (<ref>). Moreover, it follows from (<ref>) that
         *̱ _Xθ_∞
         =
          _Xθ_∞,
                _Xθ_∞
         =
          _Xθ_∞,
           μ-a.s.
         *
         Step 4: Notice that the values taken by θ_∞ are only fixed on an m-full measure set. By the convexity of elements of Θ in the y-variable, _Xθ_n has a nonempty interior in (_Xθ_∞). Then as μ-a.s., θ_n(X,·) is convex, the following definition extends θ_∞ to Ω:
         *̱θ_∞(x,y)
         :=
         sup{ a· y + b: (a,b)∈^d×, a· y_n(x)+b≤θ_∞(x,y_n(x))  n≥ 0
             }.
         *
        This extension coincides with θ_∞, in (x,y_n(x)) for μ-a.e. x∈^d, and all n≥ 1 such that y_n(x)∉∂_Xθ_k for some k≥ 1 such that _xθ_n has a nonempty interior in (_xθ_∞). As for k large enough, ∂_Xθ_k is Lebesgue negligible in (_xθ_∞), the remaining y_n(x) are still dense in (_xθ_∞). Then, for μ-a.e. x∈^d, θ_n(x,·) converges to θ_∞(x,·) on a dense subset of (_xθ_∞). We shall prove in Step 6 below that 
          θ_∞(X,·)
           (_Xθ_∞),
            μ-
        Then, by Theorem <ref>, θ_n(X,·)⟶θ_∞(X,·) pointwise on (_Xθ_∞)∖∂θ_∞(X,·), μ-a.s. Since _Xθ_∞ = _Xθ_∞, and θ converges to θ_∞ on _Xθ_∞, μ-a.s., θ converges to θ_∞∈Θ, .
        
         Step 5: Finally for general (θ_n)_n≥ 1⊂Θ_μ, we consider θ_n', equal to θ_n, , such that θ_n'≤θ_n, for n≥ 1, from the definition of Θ_μ. Then we may find λ_n^k, coefficients such that θ_n':=∑_k≥ nλ_n^kθ_k'∈(θ') converges  to θ_∞∈Θ. We denote θ_n:=∑_k≥ nλ_n^kθ_k∈(θ), θ_n = θ_n', , and θ_n ≥θ_n'. By Proposition <ref> (iii), θ converges to θ_∞, . The Proposition is proved.
        
         Step 6: In order to prove (<ref>), suppose to the contrary that there is a set A such that μ[A]>0 and θ_∞(x,·) has an empty interior in (_xθ_∞) for all x∈ A. Then, by the density of the sequence (y_n(x))_n≥ 1 stated in (<ref>), we may find for all x∈ A an index i(x)≥ 0 such that 
         y(x):=y_i(x)(x)∈ _xθ_∞,
               θ_∞(x,y(x))=∞.
         
        Moreover, since i(x) takes values in , we may reduce to the case where i(x) is a constant integer, by possibly shrinking the set A, thus guaranteeing that y is measurable. Define the measurable function on Ω:
         θ^0_n(x,y)
         :=
         (y,L^n_x)
               
         L^n_x:={y∈^d:θ_n(x,y)<θ_n(x,y(x))}.
         
        Since L^n_x is convex, and contains x for n sufficiently large by (<ref>), we see that 
         θ^0_n     θ^0_n(x,y)≤ |x-y|,     (x,y)∈Ω.
         
        In particular, this shows that θ^0_n∈Θ. By Komlòs Lemma, we may find
         *̱   θ^0_n:=∑_k≥ nλ^n_k θ^0_k∈(θ^0)      θ^0_n⟶θ^0_∞,  m-*
        for some non-negative coefficients (λ^n_k,k≥ n)_n≥ 1 with ∑_k≥ nλ^n_k=1. By convenient extension of this limit, we may assume that θ^0_∞∈Θ. We claim that 
         
            θ^0_∞>0
               
             H_x:={h(x)·(y-y(x))>0},
               
             h(x)∈^d.
        
        We defer the proof of this claim to Step 7 below and we continue in view of the required contradiction. By definition of θ^0_n together with (<ref>), we compute that
         *̱θ^1_n(x,y)
          := ∑_k≥ nλ^n_kθ_k(x,y)
            ≥   ∑_k≥ nλ^n_kθ_k(x,y(x)) _{θ^0_n>0}
           ≥   ∑_k≥ nλ^n_kθ_k(x,y(x)) θ^0_k(x,y)/|x-y|
           ≥   θ^0_n(x,y)/|x-y|inf_k≥ nθ_k(x,y(x)).
         *
        By (<ref>) and (<ref>), this shows that the sequence θ^1∈(θ) satisfies
         *̱θ^1_n(x,·) ⟶∞,
               
         H_x,
             
        x∈ A.
         *
        We finally consider the sequence θ^1:=1/2(θ+θ^1)∈(θ). Clearly, θ^1_∞(X,·)⊂θ_∞(X,·), and it follows from the last property of θ^1 that θ^1_∞(x,·)⊂ H_x^c∩θ_∞(x,·) for all x∈ A. Notice that y(x) lies on the boundary of the half space H_x and, by (<ref>), y(x)∈_xθ_∞. Then G(_xθ^1_∞)<
        G(_xθ_∞) for all x∈ A and, since μ[A]>0, we deduce that μ[G(_Xθ^1_∞)]<
         μ[G(_Xθ_∞)], contradicting the optimality of θ, by (<ref>), for the minimization problem (<ref>).
        
         Step 7: It remains to justify (<ref>). Since θ_n(x,·) is convex, it follows from the Hahn-Banach separation theorem that:
         *̱θ_n(x,·) ≥θ_n(x,y(x))
           
         H^n_x := {y∈^d: h^n(x)·(y-y(x))>0},
         *
        for some h^n(x)∈^d, so that it follows from (<ref>) that L^n_x⊂ (H^n_x)^c, and
         *̱θ^0_n(x,y)
         ≥(y,(H^n_x)^c)
         =
         [(y-y(x))· h^n(x)]^+.
         *
        Denote g_x:=g__xθ_∞ the Gaussian kernel restricted to the affine span of _xθ_∞, and B_r(x_0) the corresponding ball with radius r, centered at some point x_0. By (<ref>), we may find r^x so that B^x_r:=B_r(y(x))⊂ _xθ_∞ for all r≤ r^x, and
         *̱∫_B_r^xθ^0_n(x,y)g_x(y)dy
            ≥   ∫_B_r^x[(y-y(x))· h^n(x)]^+g_x(y)dy
         
           ≥   min_B_r^xg_x∫_B_r(0) (y· e_1)^+ dy
         =:b^r_x >0,
         *
        where e_1 is an arbitrary unit vector of the affine span of _xθ_∞. Then we have the inequality ∫_B_r^xθ^0_n(x,y)g_x(y)dy
         ≥ b^r_x, and since θ^0_n has linear growth in y by (<ref>), it follows from the dominated convergence theorem that ∫_B^r_xθ^0_∞(x,y)g(dy)≥ b^r_x>0, and therefore θ^0_∞(x,y^r_x)>0 for some y^r_x∈ B^r_x. From the arbitrariness of r∈(0,r_x), We deduce (<ref>) as a consequence of the convexity of θ^0(x,·). 
         
         Proof of Proposition <ref> (iii)
         We need to prove the existence of some
         θ'∈Θ      θ_∞= θ',  ,    θ_∞≥θ'.
         
        For simplicity, we denote θ:=θ_∞. Let 
         *̱   F^1 :=   θ(X,·),
            
         F^k 
         := 
          (θ(X,·)∩ _XF^k-1),
          k ≥ 2,   
               
         F := ∪_n≥ 1(F^n∖_X F^n)∪_Xθ.
            *
        Fix some sequence _n↘ 0, and denote θ_*:=lim inf_n→∞θ(X,_n X+(1-_n)Y), and
         
            θ':= [∞1_Y∉ F(X)+1_Y∈_Xθθ_*]1_X∉ N_μ,
        
        where N_μ∈_μ is chosen such that 1_Y∈ F^k(X)1_X∉ N_μ are Borel measurable for all k from Lemma <ref>, and θ(x,·) (resp. θ_n(x,·)) is convex finite on _xθ (resp. _xθ_n), for x∉ N_μ. Consequently, θ' is measurable. In the following steps, we verify that θ' satisfies (<ref>).
        
        
        
         Step 1: We prove that θ'∈Θ. Indeed, θ'∈^0_+(Ω), and θ'(X,X) = 0. Now we prove that θ'(x,·) is convex for all x∈^d. For x∈ N_μ, θ'(x,·)=0. For x∉ N_μ, θ(x,·) is convex finite on _xθ, then by the fact that _xθ is a convex relatively open set containing x, it follows from Lemma <ref> that θ_*(x,·)=lim_n→∞θ(x,_n x+(1-_n)·) is the lower semi-continuous envelop of θ(x,·) on _xθ. We now prove the convexity of θ'(x,·) on all ^d. We denote F(x) := F(x)∖_xθ so that ^d = F(x)^c∪F(x) ∪_xθ. Now, let y_1,y_2∈^d, and λ∈(0,1). If y_1∈ F(x)^c, the convexity inequality is verified as θ'(x,y_1)=∞. Moreover, θ'(x,·) is constant on F(x), and convex on _xθ. We shall prove in Steps 4 and 5 below that
         
         F(x)
           _xF(x) = _xθ.
         
        In view of Proposition <ref> (ii), this implies that the sets F(x) and _xθ are convex. Then we only need to consider the case when y_1∈F(x), and y_2∈_xθ. By Proposition <ref> (ii) again, we have [y_1,y_2)⊂F(x), and therefore λ y_1 + (1-λ)y_2∈F(x), and θ'(x,λ y_1 + (1-λ)y_2) = 0, which guarantees the convexity inequality.
        
         Step 2: We next prove that θ=θ', . By the second claim in (<ref>), it follows that θ_*(X,·) is convex finite on _Xθ, μ-a.s. Then as a consequence of Proposition <ref> (ii), we have _Xθ' = _X(∞1_Y∉ F(X))∩_X(θ_*1_Y∈_Xθ), μ-a.s. The first term in this intersection is _XF(X) = _Xθ. The second contains _Xθ, as it is the _X of a function which is finite on _Xθ, which is convex relatively open, containing X. Finally, we proved that _Xθ=_Xθ', μ-a.s. Then θ'(X,·) is equal to θ_*(X,·) on _Xθ, and therefore, equal to θ(X,·), μ-a.s. We proved that θ=θ', .
        
         Step 3: We finally prove that θ'≤θ pointwise. We shall prove in Step 6 below that 
         θ(X,·)
            ⊂    F.
         
        Then, ∞1_Y∉ F(X)1_X∉ N_μ≤θ, and it remains to prove that
         *̱θ(x,y)
         ≥θ_*(x,y)
               
         y∈_xθ,  x∉ N_μ.
         *
        To see this, let x∉ N_μ. By definition of N_μ, θ_n(x,·)⟶θ(x,·) on _xθ. Notice that θ(x,·) is convex on _xθ, and therefore as a consequence of Lemma <ref>,
        *̱θ_*(x,y)
        =
        lim_ϵ↘ 0θ(x,ϵ x + (1-ϵ)y),      y∈_xθ.
        *
        Then y^ϵ := (1-ϵ)y + ϵ x∈_xθ_n, for ∈(0,1], and n sufficiently large by (i) of this Proposition, and therefore (1-ϵ)θ_n(x,y)-θ_n(x,y_ϵ)≥ (1-ϵ)θ_n'(x,y)-θ_n'(x,y_ϵ)≥ 0, for θ_n'∈Θ such that θ_n'=θ_n, , and θ_n≥θ_n'. Taking the lim inf as n→∞, we get (1-ϵ)θ(x,y)-θ(x,y_ϵ)≥ 0, and finally θ(x,y)
         ≥lim_ϵ↘ 0θ(x,ϵ x + (1-ϵ)y) 
         = 
         θ'(x,y), by sending ϵ↘ 0.
        
         Step 4: (First claim in (<ref>)) Let x_0∈^d, let us prove that F(x_0) is convex. Indeed, let x,y∈ F(x_0), and 0<λ<1. Since _xθ is convex, and F^n(x_0)∖_X F^n(x_0) is convex by Proposition <ref> (ii), we only examine the following non-obvious cases:
        
        ∙ Suppose x∈ F^n(x_0)∖_x_0 F^n(x_0), and y∈ F^p(x_0)∖_x_0 F^p(x_0), with n<p. Then as F^p(x_0)∖_x_0 F^p(x_0)⊂_x_0 F^n(x_0), we have λ x + (1-λ)y∈ F^n(x_0)∖_x_0 F^n(x_0) by Proposition <ref> (ii).
        
        ∙ Suppose x∈ F^n(x_0)∖_x_0 F^n(x_0), and y∈_x_0θ, then as _x_0θ⊂_x_0 F^n(x_0), this case is handled similar to previous case.
        
         Step 5: (Second claim in (<ref>)). We have _Xθ⊂ F(X), and therefore _Xθ⊂_XF(X). Now we prove by induction on k≥ 1 that _XF(X)⊂∪_n≥ k(F^n∖_X F^n)∪_Xθ. The inclusion is trivially true for k=1. Let k≥ 1, we suppose that the inclusions holds for k, hence _XF(X)⊂∪_n≥ k(F^n∖_X F^n)∪_Xθ. As ∪_n≥ k(F^n∖_X F^n)∪_Xθ⊂ F^k. Applying _X gives
        *̱_XF(X)   ⊂   _X[∪_n≥ k(F^n∖_X F^n)∪_Xθ]
           =   _X[F^k∩∪_n≥ k(F^n∖_X F^n)∪_Xθ]
           =   _X F^k∩_X[∪_n≥ k(F^n∖_X F^n)∪_Xθ]
           ⊂   _X F^k∩∪_n≥ k(F^n∖_X F^n)∪_Xθ
           ⊂   ∪_n≥ k+1(F^n∖_X F^n)∪_Xθ.
        *
        Then the result is proved for all k. In particular we apply it for k=d+1. Recall from the proof of Lemma <ref> that for n≥ d+1, F^n is stationary at the value _Xθ. Then ∪_n≥ d+1(F^n∖_X F^n)=∅, and _XF(X)⊂_X_Xθ = _Xθ. The result is proved.
        
         Step 6: We finally prove (<ref>). Indeed, θ(X,·)⊂ F^1 by definition. Then
        *̱θ(X,·)   ⊂   F^1∖ F^1∪(∪_2≤ k≤ d+1(θ(X,·)∩ _XF^k-1)∖ F^k)∪ F^d+1
           ⊂   F^1∖ F^1∪(∪_k≥ 2  (θ(X,·)∩ _XF^k-1)∖ F^k)∪ _Xθ
              
        =∪_k≥ 1F^k∖ F^k∪_Xθ
        =F.
        *
         
        
        § SOME CONVEX ANALYSIS RESULTS
        
        
        As a preparation, we first report a result on the union of intersecting relative interiors of convex subsets which was used in the proof of Proposition <ref>. We shall use the following characterization of the relative interior of a convex subset K of ^d:
          K 
            =   {x∈^d : x-ϵ(x'-x)∈ Kϵ>0,x'∈ K}
           =   {x∈^d : x∈ (x',x_0],x_0∈ K,x'∈ K }.
         
         
        We start by proving the required properties of the notion of relative face.
         Proof of Proposition <ref> (i) The proofs of the first properties raise no difficulties and are left as an exercise for the reader. We only prove that _a A  =  A ≠∅ iff a∈ A. We assume that _a A  =  A≠∅. The non-emptiness implies that a∈ A, and therefore a∈_aA =  A. Now we suppose that a∈ A. Then for x∈ A, [x,a-ϵ (x-a)]⊂ A⊂ A, for some ϵ>0, and therefore A⊂_a A. On the other hand, by (<ref>), A = {x∈^d : x∈ (x',x_0],x_0∈ A,x'∈ A}. Taking x_0 := a∈ A, we have the remaining inclusion _a A⊂ A.
        
         (ii) We now assume that A is convex.
        
           Step 1: We first prove that _aA is convex. Let x,y∈_aA and λ∈[0,1]. We consider ϵ >0 such that (a-ϵ(x-a),x+ϵ(x-a))⊂ A and (a-ϵ(y-a),y+ϵ(y-a))⊂ A. Then if we write z=λ x + (1-λ)y, (a-ϵ(z-a),z+ϵ(z-a))⊂ A by convexity of A, because a,x,y∈ A.
        
           Step 2: In order to prove that _aA is relatively open, we consider x,y∈_aA, and we verify that (x-ϵ(y-x),y+ϵ(y-x))⊂_aA for some ϵ>0. Consider the two alternatives:
        
             Case 1: If a,x,y are on a line. If a=x=y, then the required result is obvious. Otherwise,
        
            (a-ϵ(x-a),x+ϵ(x-a))∪(a-ϵ(y-a),y+ϵ(y-a))⊂_a A
        
        This union is open in the line and x and y are interior to it. We can find ϵ'>0 such that (x-ϵ'(y-x),y+ϵ'(y-x))⊂_aA.
        
             Case 2: If a,x,y are not on a line. Let ϵ>0 be such that (a-2ϵ(x-a),x+2ϵ(x-a))⊂ A and (a-2ϵ(y-a),y+2ϵ(y-a))⊂ A. Then x+ϵ(x-a)∈_a A and a-ϵ(y-a)∈_a A. Then, if we take λ := ϵ/1+2ϵ,
        
            λ (a-ϵ(y-a))+(1-λ)(x+ϵ(x-a))
            =
             (1-λ)(1+ϵ)x-λϵ y
            =
             x + λϵ (x-y)
        
        Then x + λϵ (x-y)∈_a A and symmetrically, y + λϵ (y-x)∈_a A by convexity of _aA. And still by convexity, (x-ϵ'(y-x),y+ϵ'(y-x))⊂_aA for  ϵ':=ϵ^2/1+2ϵ>0.
        
           Step 3: Now we prove that A∖_a A is convex, and that if x_0∈ A∖_a A and y_0∈ A, then [x_0,y_0)⊂ A∖_a A. We will prove these two results by an induction on the dimension of the space d. First if d = 0 the results are trivial. Now we suppose that the result is proved for any d'<d, let us prove it for dimension d.
        
             Case 1:a∈ A. This case is trivial as _aA =  A and A⊂ A = _aA because of the convexity of A. Finally A∖_aA = ∅ which makes it trivial.
        
             Case 2:a∉ A. Then a∈∂ A and there exists a hyperplan support H to A in a because of the convexity of A. We will write the equation of E, the corresponding half-space containing A, E:c· x ≤ b with c∈^d and b∈. As x∈_aA implies that [a-ϵ(x-a),x+ϵ(x-a)]⊂ A for some ϵ >0, we have (a-ϵ(x-a))· c ≤ b and (x+ϵ(x-a))· c ≤ b. These equations are equivalent using that a∈ H and thus a· c = b to -ϵ(x-a)· c≤ 0 and (1+ϵ)(x-a)· c ≤ 0. We finally have (x-a)· c = 0 and x∈ H. We proved that _aA⊂ H.
        
        Now using (i) together with the fact that _aA⊂ H and a∈ H affine, we have
        
            _a(A∩ H) = _a A∩_a H = _a A∩ H = _aA.
        
        Then we can now have the induction hypothesis on A∩ H because H = d-1 and A∩ H⊂ H is convex. Then we have A∩ H ∖_aA which is convex and if x_0∈ A∩ H∖_a (A∩ H), y_0∈ A∩ H and if λ∈ (0,1] then λ x_0 + (1-λ)y_0∈ A∖_a (A∩ H).
        
        First A∖_aA = (A∖ H) ∪((A∩ H) ∖_aA), let us show that this set is convex. The two sets in the union are convex (A∖ H = A∩ (E∖ H)), so we need to show that a non trivial convex combination of elements coming from both sets is still in the union. We consider x∈ A∖ H, y∈ A∩ H ∖_aA and λ >0, let us show that z:= λ x + (1-λ)y∈ (A∖ H) ∪ (A∩ H ∖_aA). As x,y∈ A (_a A⊂ A because A is closed), z∈ A by convexity of A. We now prove z∉ H,
        
            z· c
            =
            λ x· c + (1-λ)y· c
            =
            λ x· c + (1-λ)b
            <
            λ b  + (1-λ)b = b.
        
        Then z is in the strict half space: z∈ E∖ H. Finally z∈ A∖ H and A∖_aA is convex.
        
        Let us now prove the second part: we consider x_0∈ A∖_a A, y_0∈_a A and λ∈ (0,1] and write z_0 := λ x_0 + (1-λ)y_0.
        
             Case 2.1:x_0,y_0∈ H. We apply the induction hypothesis.
        
             Case 2.2:x_0,y_0∈ A∖ H. Impossible because _aA⊂ H and _aA⊂ H = H. y_0∈ H.
        
             Case 2.3:x_0 ∈ A∖ H and y_0∈ H. Then by the same computation than in Step 1,
        
            z_0∈ A∖ H⊂ A∖_aA.
           Step 4: Now we prove that if a∈ A, then (_a A) =(A) if and only if a∈ A, and that in this case, we have _a A =   A =  A = _aA. We first assume that a∈ A. As by the convexity of A, A =   A, _a A =   A, and therefore _a A =  A. Finally, taking the dimension, we have (_a A) =(A). In this case we proved as well that _a A =   A =  A = _aA, the last equality coming from the fact that A=_aA as a∈ A.
        
        Now we assume that a∉ A. Then a∈∂ A, and _a A⊂∂ A. Taking the dimension (in the local sense this time), and by the fact that ∂ A=∂ A< A, we have (_a A) <(A) (as _a A is convex, the two notions of dimension coincide).
        
        Let K_1,K_2⊂^d be convex with K_1∩ K_2≠∅. Then ( K_1∪ K_2) =  (K_1∪ K_2).
         We fix y∈ K_1∩ K_2.
        
        Let x∈( K_1∪ K_2), we may write x=λ x_1+(1-λ)x_2, with x_1∈ K_1, x_2∈ K_2, and 0≤λ≤ 1. If λ is 0 or 1, we have trivially that x∈ (K_1∩ K_2). Let us now treat the case 0<λ<1. Then for x'∈(K_1∪ K_2), we may write x'=λ' x_1'+(1-λ')x_2', with x_1'∈ K_1, x_2'∈ K_2, and 0≤λ'≤ 1. We will use y as a center as it is in both the sets. For all the variables, we add a bar on it when we subtract y, for example := x-y. The geometric problem is the same when translated with y,
        
            -ϵ('-)
             = λ(_1-ϵ(λ'/λ_1'- _1)) + (1-λ)(_2-ϵ(1-λ'/1-λ_2'-_2)).
        
        However, as _1 and _1' are in K_1-y, as 0 is an interior point, ϵ(λ'/λ_1'- _1)∈ K_1-y for ϵ small enough. Then as _1 is interior to K_1-y as well, _1-ϵ(λ'/λ_1'- _1)∈ K_1-y as well. By the same reasoning, _2-ϵ(1-λ'/1-λ_2'-_2)∈ K_2-y. Finally, by (<ref>), for ϵ small enough, x-ϵ(x'-x)∈(K_1∪ K_2). By (<ref>), x∈ (K_1∪ K_2).
        
        Now let x∈ (K_1∪ K_2). We use again y as an origin with the notation := x-y. As  is interior, we may find ϵ >0 such that (1+ϵ)∈(K_1∪ K_2). We may write (1+ϵ)=λ_1+(1-λ)_2, with _1∈ K_1-y, _2∈ K_2-y, and 0≤λ≤ 1. Then = λ1/1+ϵ_1+(1-λ)1/1+ϵ_2. By (<ref>), 1/1+ϵ_1∈ K_1, and 1/1+ϵ_2∈ K_2. ∈( (K_1-y)∪ (K_2-y)), and therefore x∈( K_1∪ K_2).
        
        
        
        Now we use the measurable selection theory to establish the non-emptiness of ∂ f.
        
        
        For all f∈, we have ∂ f≠∅.
         By the fact that f is continuous, we may write ∂ f(x) = ∩_n≥ 1F_n(x) for all x∈^d, with F_n(x):={p∈^d:f(y_n)-f(x)≥ p· (y_n-x)} where (y_n)_n≥ 1⊂^d is some fixed dense sequence. All F_n are measurable by the continuity of (x,p)⟼ f(y_n)-f(x)-p· (y_n-x) together with Theorem 6.4 in <cit.>. Therefore the mapping x⟼∂ f(x) is measurable by Lemma <ref>. Moreover, the fact that this mapping is closed nonempty-valued is a well-known property of the subgradient of finite convex functions in finite dimension. Then the result holds by Theorem 4.1 in <cit.>. 
        
        
        
        We conclude this section with the following result which has been used in our proof of Proposition <ref>. We believe that this is a standard convex analysis result, but we could not find precise references. For this reason, we report the proof for completeness.
        
        
        Let f_n,f:^d→ be convex functions with f ≠∅. Then f_n ⟶ f pointwise on ^d∖∂ f if and only if f_n ⟶ f pointwise on some dense subset A⊂^d∖∂ f.
        
        We prove the non-trivial implication "if". We first prove the convergence on f. f_n converges to f on a dense set. The reasoning will consist in proving that the f_n are Lipschitz, it will give a uniform convergence and then a pointwise convergence. First we consider K⊂  f compact convex with nonempty interior. We can find N∈ and x_1,...x_N∈ A∩(  f ∖ K) such that K⊂ (x_1,...,x_N). We use the pointwise convergence on A to get that for n large enough, f_n(x)≤ M for x∈(x_1,...,x_N), M>0 (take M=max_1≤ k≤ Nf(x_k)+1). Then we will prove that f_n is bounded from below on K. We consider a∈ A∩ K and δ_0:=sup_x∈ K|x-a|. For n large enough, f_n(a)≥ m for any a∈ A (take for example m=f(a)-1). We write δ_1 := min_(x,y)∈ K∂(x_1,...,x_N)|x-y|. Finally we write δ_2:= sup_x,y∈(x_1,...,x_N)|x-y|. Now, for x∈ K, we consider the half line x+_+(a-x), it will cut ∂(x_1,...,x_N) in one only point y∈∂(x_1,...,x_N). Then a∈[x,y], and therefore a= |a-y|/|x-y|x+|a-x|/|x-y|y. By the convex inequality, f_n(a)≤|a-y|/|x-y| f_n(x)+|a-x|/|x-y|f_n(y). Then f_n(x)
        ≥ -|a-x|/|a-y| M + |x-y|/|a-y| m
        ≥ -δ_0/δ_1 M + δ_2/δ_1 m. Finally, if we write m_0:=-δ_0/δ_1 M + δ_2/δ_1 m,
        *̱
        M≥ f_n≥ m_0,   K.
        *
        This will prove that f_n is M-m0/δ_1-Lipschitz. We consider x∈ K and a unit direction u∈^d-1 and f_n'∈∂ f_n(x). For a unique λ > 0, y:=x+λ u∈∂(x_1,...,x_N). As u is a unit vector, λ = |y-x|≥δ_1. By the convex inequality, f_n(y)≥ f_n(x)+f_n'(x)· (y-x).
        Then M-m_0≥δ_0 |f_n'· u| and finally |f_n'· u|≤M-m0/δ_1 as this bound does not depend on u, |f_n'|≤M-m0/δ_1 for any such subgradient. For n large enough, the f_n are uniformly Lipschitz on K, and so in f. The convergence is uniform on K, it is then pointwise on K. As this is true for any such K, the convergence is pointwise on f.
        
        Now let us consider x∈ ( f)^c. The set (x,  f)∖ f has a nonempty interior because (x, f)>0 and f≠∅. As A is dense, we can consider a∈ A∩(x,  f)∖ f. By definition of (x,  f), we can find y∈  f such that a=λ y + (1-λ) x. We have λ <1 because a∉ f. If λ = 0, f_n(x)=f_n(a)n→∞⟶∞. Otherwise, by the convexity inequality, f_n(a)≤λ f_n(y) + (1-λ) f_n(x). Then, as f_n(a)n→∞⟶∞, and f_n(y)n→∞⟶f(y)<∞, we have f_n(x)n→∞⟶∞.
        
        
        § ACKNOWLEDGEMENTS
        
        
        We are grateful to the two anonymous referees, whose fruitful remarks and comments contributed to enhance deeply this paper.
        
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