Lemma 6. Let d≥2 be an integer. If α and β are absolutely continuous Borel probability measures on ^d , there is a Borel map Q:^d→^d such that Q_#α=β .

[Proof of Proposition 5] Write K_0= g , and let K_1,…,K_4 be its images under the four maps in . These are nonempty compact sets in the five disjoint components, and μ(K)=1 for K=⋃_ a=0 ^4K_a . In particular, η=_ a≠b dist(K_a,K_b)>0. The optimal coupling π_0 constructed in Section is supported on products of three distinct component sets. Fix δ>0 . Partition each K_a into finitely many Borel cells of diameter at most δ , for example by a finite grid of half-open cubes. Discard the cells of zero μ -mass and denote the remaining cells by A_1,…,A_N . They partition a set of full μ -measure and each belongs to a single K_a . Put q_ ijk =π_0(A_i×A_j×A_k). The marginal constraints on π_0 read ∑_ j,k q_ ijk =μ(A_i), ∑_ i,k q_ ijk =μ(A_j), ∑_ i,j q_ ijk =μ(A_k). For each fixed i , split A_i into Borel sets B_ ijk of masses q_ ijk , ignoring zero-mass pieces. Such a split can be made by successive first-coordinate slabs: the first-coordinate marginal of μ|_ A_i is absolutely continuous, so its cumulative distribution is continuous and takes every value between zero and μ(A_i) . Endpoints of the slabs have zero mass; any unused null set is included in a fixed piece. For each q_ ijk >0 , Lemma 6 supplies maps on B_ ijk sending the probability q_ ijk ^ -1 μ|_ B_ ijk , respectively, to μ|_ A_j μ(A_j) μ|_ A_k μ(A_k) . Reset any exceptional values outside the indicated target cell to a fixed point in that cell. These changes are Borel and occur on null sets. Glue the maps over the finitely many pieces to obtain T_2,T_3 , and give them a fixed value on the remaining Borel null set in ^3 . Different source pieces may use the same normalized target law. Their weights q_ ijk add to the full mass of the target cell, so no target partition is needed. Indeed, for any Borel set D , μ(T_2^ -1 (D)) =∑_ i,j,k:q_ ijk >0 q_ ijk (μ(D∩A_j))/(μ(A_j)) =∑_jμ(D∩A_j)=μ(D), and the same calculation proves (T_3)_#μ=μ . The graph coupling π_δ=(,T_2,T_3)_#μ gives each product A_i×A_j×A_k exactly the mass q_ ijk . Whenever q_ ijk >0 , the three cells lie in three distinct component sets. Thus all pairwise distances in their product are at least η , including after the redistribution within the cells. If x and y are two triples in the same such product, then |x_ℓ-y_ℓ|≤δ for each coordinate. Consequently |1 |x_ℓ-x_r| -1 |y_ℓ-y_r| | ≤(2δ)/(η^2), |c(x)-c(y)|≤(6δ)/(η^2). Both costs are finite, bounded by 3/η on all active products. Since π_δ and π_0 have identical masses on these products, summing their oscillation bounds gives |∫cπ_δ-∫cπ_0| ≤(6δ)/(η^2). Letting δ0 proves the required upper bound on the Monge infimum. The reverse bound follows because every admissible graph plan is a coupling. Strict nonattainment is Theorem 1.1.