SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Proof of Theorem 1. Since function  $(x,y)\mapsto c(x,y)$  is continuous, there exists a measurable selection  $T^{*}:\mathcal{X}\to \mathcal{Y}$  from the set-valued map  $x\mapsto \arg \min_{y\in \operatorname {Supp}(\mathbb{Q})}c(x,y)$ , see [2, Theorem 18.19]. This map for all  $x\in \mathcal{X}$  satisfies  $c\bigl (x,T(x)\bigr) = \min_{y\in \operatorname {Supp}(\mathbb{Q})}c(x,y) = c^{*}(x)$ . As a result,  $\int_{\mathcal{X}}c\bigl (x,T(x)\bigr)d\mathbb{P}(x) = \int_{\mathcal{X}}c^{*}(x)d\mathbb{P}(x)$  and (7) is tight.
Lemma 1. (Distinctness) Let  $\mu, \nu \in \mathcal{M}(\mathcal{Y})$ . Then  $\mu \leq \nu$  holds if and only if for every  $f \in \mathcal{C}(\mathcal{Y})$  satisfying  $f \leq 0$  it holds that  $\int_{\mathcal{Y}} f(y) d\mu(y) \geq \int_{\mathcal{Y}} f(y) d\nu(y)$ .
Proof of Lemma 1. If  $\mu \leq \nu$ , then the inequality  $\int_{\mathcal{Y}} f(y) d(\mu - \nu)(y) \geq 0$  for every (measurable)  $f \leq 0$  follows from the definition of the Lebesgue integral. Below we prove the statement in the other direction.
Assume the opposite, i.e.,  $\int_{\mathcal{V}}f(y)d(\mu -\nu)(y)\geq 0$  for every continuous  $f\leq 0$  but still  $\nu \neq \mu$ . The latter means there exists a measurable  $A\subset \mathcal{V}$  satisfying  $\mu (A) > \nu (A)$ . Let  $\epsilon = \frac{1}{2}\bigl (\mu (A) - \nu (A)\bigr) > 0$ . Consider the negative indicator function  $f_{A}(y)$  which equals  $-1$  if  $y\in A$  and 0 when  $y\notin A$ . Consider a variation measure  $|\nu -\mu |\in \mathcal{M}_{+}(\mathcal{V})$ . Thanks to [23, Proposition 7.9], the continuous functions  $C(\mathcal{V})$  are dense in the space  $\mathcal{L}^1 (|\mu -\nu |)$ . Therefore, there exists a function  $f_{A,\epsilon}\in C(\mathcal{V})$  satisfying  $\int_{\mathcal{X}}|f_A(y) - f_{A,\epsilon}(y)|d|\mu -\nu |(y) < \epsilon$ . We define  $f_{A,\epsilon}^{-}(y)\stackrel {def}{=}\min \{0,f_{A,\epsilon}(y)\} \leq 0$ . This is a non-positive continuous function, and for  $y\in \mathcal{V}$  it holds that  $|f_{A}(y) - f_{A,\epsilon}(y)|\geq |f_{A}(y) - f_{A,\epsilon}^{-}(y)|$  because  $f_{A}$  takes only non-positive values  $\{0, - 1\}$ . We derive
$$
\begin{array}{l} \int_ {y} f _ {A, \epsilon} ^ {-} (y) d (\mu - \nu) (y) = \overbrace {\int_ {\mathcal {Y}} f _ {A} (y) d (\mu - \nu) (y)} ^ {= \nu (A) - \mu (A) = - 2 \epsilon} + \int_ {\mathcal {Y}} \left(f _ {A, \epsilon} ^ {-} (y) - f _ {A} (y)\right) d (\mu - \nu) (y) \leq \\ - 2 \epsilon + \int_ {\mathcal {Y}} | f _ {A, \epsilon} ^ {-} (y) - f _ {A} (y) | d | \mu - \nu | (y) \leq - 2 \epsilon + \int_ {\mathcal {Y}} | f _ {A, \epsilon} (y) - f _ {A} (y) | d | \mu - \nu | (y) \leq \\ - 2 \epsilon + \epsilon = - \epsilon <   0, \\ \end{array}
$$
which is a contradiction to the fact that  $\int_{\mathcal{Y}}f(y)d(\mu -\nu)(y)\geq 0$  for every continuous  $f\leq 0$ . Thus,  $\mu \leq \nu$ .
Proof of Proposition 2. To begin with, we prove that  $\Pi^w (\mathbb{P},\mathbb{Q})$  is a weak-\* compact set. Pick any sequence  $\pi_n\in \Pi^w (\mathbb{P},\mathbb{Q})$ . It is bounded as all  $\pi_{n}$  are probability measures  $(\| \pi_n\| _1 = 1)$ . Hence by the Banach-Alaoglu theorem [62, Box 1.2], there exists a subsequence  $\pi_{n_k}$  weakly-\* converging to some  $\pi \in \mathcal{M}(\mathcal{X}\times \mathcal{Y})$ . It remains to check that  $\pi \in \Pi^w (\mathbb{P},\mathbb{Q})$ . Let  $(\mu_{n_k},\nu_{n_k})$  denote the marginals
of  $\pi_{n_k}$  and  $(\mu, \nu)$  be the marginals of  $\pi$ . Pick any  $f \in \mathcal{C}(\mathcal{Y})$  with  $f \leq 0$ . Since  $\pi_{n_k} \in \Pi^w(\mathbb{P}, \mathbb{Q})$ , it holds that  $0 \leq \nu_{n_k} \leq w\mathbb{Q}$ ,  $w \int_{\mathcal{Y}} f(y)d\mathbb{Q}(y) \leq \int_{\mathcal{Y}} f(y)d\nu_{n_k}(y) \leq 0$  (Lemma 1). We have
$$
\int_ {\mathcal {Y}} f (y) d \nu (y) = \int_ {\mathcal {X} \times \mathcal {Y}} f (y) d \pi (x, y) = \lim _ {k \to \infty} \int_ {\mathcal {X} \times \mathcal {Y}} f (y) d \pi_ {n _ {k}} (x, y) = \lim _ {k \to \infty} \int_ {\mathcal {Y}} f (y) d \nu_ {n _ {k}} (y).
$$
The latter limit is  $\geq w\int_{\mathcal{Y}}f(y)d\mathbb{Q}(y)$  and  $\leq 0$ . As this holds for every continuous  $f\leq 0$ , we conclude that  $0\leq \nu \leq w\mathbb{Q}$ . By the analogous analysis one may prove that  $\mu = \mathbb{P}$  and  $\pi \geq 0$ . Thus,  $\pi \in \Pi^w (\mathbb{P},\mathbb{Q})$  and  $\Pi^w (\mathbb{P},\mathbb{Q})$  is a weak-\* compact.
The functional  $\pi \mapsto \int_{\mathcal{X} \times \mathcal{Y}} c(x, y) d\pi(x, y)$  is continuous in the space  $\mathcal{M}(\mathcal{X} \times \mathcal{Y})$  equipped with weak- $*$  topology because  $c: \mathcal{X} \times \mathcal{Y} \to \mathbb{R}$  is continuous. Since  $\Pi^w(\mathbb{P}, \mathbb{Q})$  is a compact set, there exists  $\pi^* \in \Pi^w(\mathbb{P}, \mathbb{Q})$  attaining the minimum on  $\Pi^w(\mathbb{P}, \mathbb{Q})$ . This follows from the Weierstrass extreme value theorem and ends the proof.
Bibliographical remark. The results showing the existence of minimizers  $\pi^{*}$  in partial OT (4) already exist, see [10, Lemma 2.2] or [22, §2]. They also provide the existence of minimizers in our IT problem (12). Yet, the authors study the particular case when  $\mathbb{P},\mathbb{Q}$  have densities on  $\mathcal{X},\mathcal{Y}\subset \mathbb{R}^D$ . For completeness, we include a separate proof of existence as we do not require the absolute continuity assumption. The proof is performed via the usual technique based on weak-\* compactness in dual spaces and is analogous to [62, Theorem 1.4] which proves the existence of minimizers for OT problem (2). Our proof is slightly more technical due to the inequality constraint.
Proof of Proposition 3. Let  $\pi^{*}\in \Pi^{w}(\mathbb{P},\mathbb{Q})$  be an IT plan. Consider the OT problem between  $\mathbb{P}$  and  $\pi_y^*$ :
$$
\min  _ {\pi \in \Pi \left(\mathbb {P}, \pi_ {y} ^ {*}\right)} \int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi (x, y). \tag {18}
$$
It turns out that  $\pi^*$  is a minimizer here. Assume the contrary, i.e., that there exists a more optimal  $\pi' \in \Pi(\mathbb{P}, \pi_y^*)$  satisfying
$$
\int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi^ {\prime} (x, y) <   \int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi^ {*} (x, y) \quad \left(= \operatorname {C o s t} _ {w} (\mathbb {P}, \mathbb {Q})\right). \tag {19}
$$
This plan by the definition of  $\Pi (\mathbb{P},\pi_y^*)$  satisfies  $\pi_x' = \mathbb{P}$  and  $\pi_y' = \pi_y^*\leq w\mathbb{Q}$ , i.e.,  $\pi^{\prime}\in \Pi^{w}(\mathbb{P},\mathbb{Q})$ . However, (19) contradicts the fact that  $\pi^*$  is an IT plan in (18) as  $\pi^\prime$  provides smaller cost. Thus, min in (18) equals  $\mathrm{Cost}_w(\mathbb{P},\mathbb{Q})$  in (12).
Thanks to [62, Theorem 1.33], problem (18) has the same minimal value as the inf in the Monge's problem
$$
\inf  _ {T \sharp \mathbb {P} = \pi_ {y} ^ {*}} \int_ {\mathcal {X}} c (x, T (x)) d \mathbb {P} (x), \tag {20}
$$
i.e., for every  $\epsilon > 0$  there exists  $T_{\epsilon}: \mathcal{X} \to \mathcal{Y}$  satisfying  $T_{\epsilon} \sharp \mathbb{P} = \pi_y^*$  and  $\int_{\mathcal{X}} c(x, T_{\epsilon}(x)) d\mathbb{P}(x) < \mathrm{Cost}_w(\mathbb{P}, \mathbb{Q}) + \epsilon$ . It remains to substitute this  $T_{\epsilon}$  to Monge's IT problem (11) to get an  $\epsilon$ -close transport cost to Kantorovich's IT cost (12). As this works for every  $\epsilon > 0$ , we conclude that min in (12) is the same as inf in (11).
Proof of Theorem 2. The fact that  $w \mapsto \operatorname{Cost}_w(\mathbb{P}, \mathbb{Q})$  is non-increasing follows from the inclusion  $\Pi^{w_1}(\mathbb{P}, \mathbb{Q}) \subset \Pi^{w_2}(\mathbb{P}, \mathbb{Q})$  for  $w_1 \leq w_2$ . This inclusion means that for larger values of  $w$ , the minimization in (12) is performed over a larger set of admissible plans. As for convexity, take any IT plans  $\pi^{w_1} \in \Pi^{w_1}(\mathbb{P}, \mathbb{Q}), \pi^{w_2} \in \Pi^{w_2}(\mathbb{P}, \mathbb{Q})$  for  $w_1, w_2$ , respectively. For any  $\alpha \in [0,1]$  consider the mixture  $\pi' = \alpha \pi^{w_1} + (1 - \alpha) \pi^{w_2}$ . Note that  $\pi_y' = \alpha \pi_y^{w_1} + (1 - \alpha) \pi_y^{w_2} \leq \alpha w_1 \mathbb{Q} + (1 - \alpha) w_2 \mathbb{Q} = (\alpha w_1 + (1 - \alpha) w_2) \mathbb{Q}$ . Therefore,  $\pi' \in \Pi^{\alpha w_1 + (1 - \alpha) w_2}(\mathbb{P}, \mathbb{Q})$ . We derive
$$
\begin{array}{l} \operatorname {C o s t} _ {\alpha w _ {1} + (1 - \alpha) w _ {2}} (\mathbb {P}, \mathbb {Q}) \leq \int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi^ {\prime} (x, y) = \\ \alpha \int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi^ {w _ {1}} (x, y) + (1 - \alpha) \int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi^ {w _ {2}} (x, y) = \\ \alpha \operatorname {C o s t} _ {w _ {1}} (\mathbb {P}, \mathbb {Q}) + (1 - \alpha) \operatorname {C o s t} _ {w _ {2}} (\mathbb {P}, \mathbb {Q}), \tag {21} \\ \end{array}
$$
which shows the convexity of  $w\mapsto \operatorname{Cost}_w(\mathbb{P},\mathbb{Q})$
Now we prove that  $\lim_{w\to \infty}\mathrm{Cost}_w(\mathbb{P},\mathbb{Q}) = \mathrm{Cost}_\infty (\mathbb{P},\mathbb{Q})$ . For every  $w\geq 1$  and  $\pi \in \Pi^w (\mathbb{P},\mathbb{Q})$ , it holds that  $\pi_y\leq w\mathbb{Q}$ . This means that  $\mathrm{Supp}(\pi_y)\subset \mathrm{Supp}(\mathbb{Q})$ . As a result, we see that  $\Pi^w (\mathbb{P},\mathbb{Q})\subset \Pi^\infty (\mathbb{P},\mathbb{Q})$ , i.e.,  $\mathrm{Cost}_w(\mathbb{P},\mathbb{Q})\geq \mathrm{Cost}_\infty (\mathbb{P},\mathbb{Q})$ . We already know that  $w\mapsto \mathrm{Cost}_w(\mathbb{P},\mathbb{Q})$  is nonincreasing, so it suffices to show that for every  $\epsilon >0$  there exists  $w = w(\epsilon)\in [1, + \infty)$  such that  $\mathrm{Cost}_w(\mathbb{P},\mathbb{Q})\leq \mathrm{Cost}_\infty (\mathbb{P},\mathbb{Q}) + \epsilon$ . This will provide that  $\lim_{w\to \infty}\mathrm{Cost}_w(\mathbb{P},\mathbb{Q}) = \mathrm{Cost}_\infty (\mathbb{P},\mathbb{Q})$ .
Pick any  $\epsilon > 0$ . Consider the set  $\mathcal{S} \stackrel{\text{def}}{=} \{(x, y) \in \mathcal{X} \times \mathcal{Y}\}$  such that  $y \in \mathrm{NN}(x)$ . It is a compact set. To see this, we pick any sequence  $(x_n, y_n) \in \mathcal{S}$ . It is contained in compact  $\mathcal{X} \times \mathcal{Y}$ . Therefore, it has a sub-sequence  $(x_{n_k}, y_{n_k})$  converging to some  $(x, y) \in \mathcal{X} \times \mathcal{Y}$ . It remains to check that  $(x, y) \in \mathcal{S}$ . Since  $\operatorname{Supp}(\mathbb{Q})$  is compact and  $y_{n_k} \in \mathrm{NN}(x_{n_k}) \subset \operatorname{Supp}(\mathbb{Q})$ , we have  $y \in \operatorname{Supp}(\mathbb{Q})$  as well. At the same time, by the continuity of  $c^*$  (Proposition 1) and  $c$ , we have
$$
c (x, y) - c ^ {*} (x) = \lim  _ {k \rightarrow \infty} \left\{c \left(x _ {n _ {k}}, y _ {n _ {k}}\right) - c ^ {*} \left(x _ {n _ {k}}\right)\right\} = \lim  _ {k \rightarrow \infty} 0 = 0,
$$
which means that  $y \in \mathrm{NN}(x)$  and  $(x, y) \in S$ , i.e.,  $S$  is compact.
Since  $(x,y)\mapsto c(x,y) - c^{*}(x)$  is a continuous function, for each  $(x,y)\in S$  there exists an open neighborhood  $U_{x}\times V_{y}\subset \mathcal{X}\times \mathcal{Y}$  of  $(x,y)$  such that for all  $(x^{\prime},y^{\prime})\in U_{x}\times V_{y}$  it holds that  $c(x^{\prime},y^{\prime}) - c^{*}(x^{\prime}) < \epsilon$  or, equivalently,  $c(x^{\prime},y^{\prime}) < c^{*}(x^{\prime}) + \epsilon$ . Since  $\bigcup_{(x,y)\in S}U_x\times V_y$  is an open coverage of the compact set  $S$ , there exists a finite sub-coverage  $\bigcup_{n = 1}^{N}U_{x_n}\times V_{y_n}$  of  $S$ . In particular,  $\mathcal{X} = \bigcup_{n = 1}^{N}U_{x_n}$ . For convenience, we simplify the notation and put  $U_{n}\stackrel {def}{=}U_{x_{n}}$  and  $V_{n}\stackrel {def}{=}V_{y_{n}}$ . Now we put  $U_1^\prime \stackrel {def}{=}U_1$  and iteratively define  $U_{n}^{\prime}\stackrel {def}{=}U_{n}\setminus U_{n - 1}^{\prime}$  for  $n\geq 2$ . By the construction, it holds that the entire space  $\mathcal{X}$  is a disjoint union of  $U_{n}^{\prime}$ , i.e.,  $\mathcal{X} = \bigsqcup_{n = 1}^{N}U_{n}^{\prime}$ . Some of  $U_{n}^{\prime}$  may be empty, so we just remove them from the sequence and for convenience assume that each  $U_{n}^{\prime}$  is not empty. Now consider the measure  $\pi \in \mathcal{P}(\mathcal{X}\times \mathcal{Y})$  which is given by
$$
\pi \stackrel {\text {d e f}} {=} \sum_ {n = 1} ^ {N} \left[ \mathbb {P} \mid_ {U _ {n} ^ {\prime}} \times \frac {\mathbb {Q} \mid_ {V _ {n}}}{\mathbb {Q} (V _ {n})} \right]. \tag {22}
$$
Here for  $\mu, \nu \in \mathcal{M}_{+}(\mathcal{X}), \mathcal{M}_{+}(\mathcal{Y})$ , we use  $\times$  to denote their product measure  $\mu \times \nu \in \mathcal{M}(\mathcal{X} \times \mathcal{Y})$ . In turn, for a measurable  $A \subset \mathcal{X}$ , we use  $\mu|_{A}$  to denote the restriction of  $\mu$  to  $A$ , i.e., measure  $\mu' \in \mathcal{M}(\mathcal{X})$  satisfying  $\mu'(B) = \mu(A \cap B)$  for every measurable  $B \subset \mathcal{X}$ . Note that  $\sum_{n=1}^{N} \mathbb{P}|_{U_n'} = \mathbb{P}$  and  $\sum_{n=1}^{N} \mathbb{P}(U_n') = \sum_{n=1}^{N} \mathbb{P}|_{U_n'}(U_n') = 1$  by the construction of  $U_n'$ . At the same time, for each  $n$  it holds that  $\frac{\mathbb{Q}|_{V_n}}{\mathbb{Q}(V_n)}$  is a probability measure because of the normalization  $\mathbb{Q}(V_n)$ . Note that this normalization is necessarily positive because  $V_n$  is a neighborhood of a point in  $\operatorname{Supp}(\mathbb{Q})$ . Therefore, since sets  $U_n'$  are disjoint and cover  $\mathcal{X}$ , we have  $\pi_x = \sum_{n=1}^{N} \mathbb{P}|_{U_n'} = \mathbb{P}$ . Now let us show that there exists  $w$  such that  $\pi_y \leq w\mathbb{Q}$ . It suffices to take  $w = \sum_{n=1}^{N} \frac{\mathbb{P}(U_n')}{\mathbb{Q}(V_n)}$ . Indeed, in this case for every measurable  $A \subset \mathcal{Y}$  we have
$$
\pi_ {y} (A) = \sum_ {n = 1} ^ {N} \mathbb {P} \left(U _ {n} ^ {\prime}\right) \frac {\mathbb {Q} \left(A \cap V _ {n}\right)}{\mathbb {Q} \left(V _ {n}\right)} \leq \sum_ {n = 1} ^ {N} \mathbb {P} \left(U _ {n} ^ {\prime}\right) \frac {\mathbb {Q} (A)}{\mathbb {Q} \left(V _ {n}\right)} \leq \mathbb {Q} (A) \sum_ {n = 1} ^ {N} \frac {\mathbb {P} \left(U _ {n} ^ {\prime}\right)}{\mathbb {Q} \left(V _ {n}\right)} \leq w \mathbb {Q} (A),
$$
which yields  $\pi_y \leq w\mathbb{Q}$  and means that  $\pi \in \Pi^w (\mathbb{P},\mathbb{Q})$  for our chosen  $w$ . Now let us compute the cost of  $\pi$ :
$$
\int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi (x, y) = \sum_ {n = 1} ^ {N} \int_ {U _ {n} ^ {\prime}} \left\{\frac {1}{\mathbb {Q} \left(V _ {n}\right)} \int_ {V _ {n}} c (x, y) d \mathbb {Q} | _ {V _ {n}} (y) \right\} d \mathbb {P} | _ {U _ {n} ^ {\prime}} (x) \leq
$$
$$
\begin{array}{l} \sum_ {n = 1} ^ {N} \int_ {U _ {n} ^ {\prime}} \underbrace {\left\{\frac {1}{\mathbb {Q} (V _ {n})} \int_ {V _ {n}} \big (c ^ {*} (x) + \epsilon \big) d \mathbb {Q} | _ {V _ {n}} (y) \right\}} _ {= c ^ {*} (x) + \epsilon} d \mathbb {P} | _ {U _ {n} ^ {\prime}} (x) = \sum_ {n = 1} ^ {N} \int_ {U _ {n} ^ {\prime}} \left(c ^ {*} (x) + \epsilon\right) d \mathbb {P} | _ {U _ {n} ^ {\prime}} (x) = \\ \int_ {\mathcal {X}} \left(c ^ {*} (x) + \epsilon\right) d \mathbb {P} (x) = \operatorname {C o s t} _ {\infty} (\mathbb {P}, \mathbb {Q}) + \epsilon . \tag {23} \\ \end{array}
$$
To finish the proof it remains to note that this plan is not necessarily a minimizer for (12), i.e.,  $\int_{\mathcal{X}\times \mathcal{Y}}c(x,y)d\pi (x,y)$  is an upper bound on  $\mathrm{Cost}_w(\mathbb{P},\mathbb{Q})$  . Therefore, we have  $\mathrm{Cost}_w(\mathbb{P},\mathbb{Q})\leq$ $\mathrm{Cost}_{\infty}(\mathbb{P},\mathbb{Q}) + \epsilon$  for our chosen  $w = w(\epsilon)$
Bibliographical remark. There exist seemingly related but actually different results in the fundamental OT literature, see [22, Lemma 2.1] or [10, §3]. There the authors study partial OT problem (4) and study how the partial OT plan and OT cost evolve when the marginals  $w_0\mathbb{P}$  and  $w_{1}\mathbb{Q}$  are fixed and the required mass amount to transport changes from 0 to  $\min \{w_0,w_1\}$ . In our study, the first marginal  $w_0\mathbb{P}$  and the amount of mass to transport  $m$  are fixed  $(w_{0} = m = 1)$ , and we study how the OT cost changes when  $w_{1}\rightarrow \infty$  in the IT problem.
Proof of Theorem 3. Note that  $\mathcal{P}(\mathcal{X}\times \mathcal{Y})$  is (weak-*) compact. This can be derived from the Banach-Alaoglu theorem analogously to the compactness of  $\Pi^w (\mathbb{P},\mathbb{Q})$  in the proof of Theorem (2). Therefore, any sequence in  $\mathcal{P}(\mathcal{X}\times \mathcal{Y})$  has a converging sub-sequence. In our case, for brevity, we assume that  $\pi^{w_n}\in \Pi^{w_n}(\mathbb{P},\mathbb{Q})$  itself weakly-\* converges to some  $\pi^*\subset \mathcal{P}(\mathcal{X}\times \mathcal{Y})$ . Since  $\pi_x^{w_n} = \mathbb{P}$  for all  $n$ , we also have  $\pi_x^* = \mathbb{P}$ . As  $\lim_{n\to \infty}w_n = \infty$ , we conclude from Theorem 2 that
$$
\operatorname {C o s t} _ {\infty} (\mathbb {P}, \mathbb {Q}) = \lim  _ {n \rightarrow \infty} \operatorname {C o s t} _ {w _ {n}} (\mathbb {P}, \mathbb {Q}) = \lim  _ {n \rightarrow \infty} \int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi^ {w _ {n}} (x, y) = \int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi^ {*} (x, y), \tag {24}
$$
where the last equality holds since  $\pi^{w_n}$  (weakly-\*) converges to  $\pi^*$ . From (24), we see that the cost of  $\pi^*$  is perfect and it remains to check that  $\operatorname{Supp}(\pi_y^*) \subset \operatorname{Supp}(\mathbb{Q})$ . Assume the opposite and pick any  $y^* \in \operatorname{Supp}(\pi_y^*)$  such that  $y^* \notin \operatorname{Supp}(\mathbb{Q})$ . By the definition of the support, there exists  $\epsilon > 0$  and a neighborhood  $U = \{y \in \mathcal{V}\}$  such that  $\| y - y^* \|_{\mathcal{V}} < \epsilon$  of  $y^*$  satisfying  $\pi_y^*(U) > 0$  and  $U \cap \operatorname{Supp}(\mathbb{Q}) = \emptyset$ . Let  $h(y) \stackrel{\text{def}}{=} \max\{0, \epsilon - \| y - y^* \|_{\mathcal{V}}\}$ . From  $\pi_y^{w_n} \leq w_n \mathbb{Q}$  (for all  $n$ ), it follows that  $\operatorname{Supp}(\pi_y^{w_n}) \subset \operatorname{Supp}(\mathbb{Q})$ . Therefore,  $\pi_y^{w_n}(U) = 0$  for all  $n$ . Since  $\pi^{w_n}$  converges to  $\pi^*$ , we have
$$
\lim  _ {n \rightarrow \infty} \int_ {\mathcal {Y}} h (y) d \pi^ {w _ {n}} (y) = \int_ {\mathcal {Y}} h (y) d \pi^ {*} (y). \tag {25}
$$
The left part is zero because  $h(y)$  vanishes outside  $U$  and  $\int_{U} h(y) d\pi^{w_n}(y) = 0$  as  $\pi_y^{w_n}(U) = 0$ . The right part equals  $\int_{U} h(y) d\pi^*(y)$  and is positive as  $\pi^*(U) > 0$  and  $h(y) > 0$  for  $y \in U$ . This is a contradiction. Therefore,  $\operatorname{Supp}(\pi_y^*) \subset \operatorname{Supp}(\mathbb{Q})$ . Now we see that  $\pi^* \in \Pi^\infty(\mathbb{P}, \mathbb{Q})$  is a perfect plan as its cost matches the perfect cost.
Proof of Corollary 1. Assume the inverse. Then  $\exists \varepsilon$  such that  $\forall w(\varepsilon) \exists w \geq w(\varepsilon)$  and  $\exists$  IT plan  $\pi^w \in \Pi_w(\mathbb{P},\mathbb{Q})$  solving (12) such that  $\forall$  ET plan  $\pi^*$ , it holds that  $\mathbb{W}_1(\pi^w,\pi^*) \geq \varepsilon$ . This means that there exists an increasing sequence  $w_1, w_2, \ldots, w_n, \dots \to \infty$  and the corresponding sequence of IT plans  $\pi^{w_1}, \pi^{w_2}, \ldots, \pi^{w_n}, \ldots$  such that for every ET plan  $\pi^*$  it holds that  $\mathbb{W}_1(\pi^{w_n},\pi^*) \geq \varepsilon$  for all  $n$ . At the same time, from Theorem 3, this sequence of plans must have a sub-sequence  $\{\pi^{w^{n_k}}\}$  which is (weakly-*') converging to some ET plan  $\pi^* \colon \pi^{w^{n_k}} \to \pi^*$ . Recall that the convergence in  $\mathbb{W}_1$  coincides with the weak-* convergence (for compact  $\mathcal{X},\mathcal{Y}$ ), see [62, Theorem 5.9]. Hence, the sub-sequence should also converge to  $\pi^*$  in  $\mathbb{W}_1$  but it is not since  $\mathbb{W}_1(\pi^{w_n},\pi^*) \geq \varepsilon$ . This is a contradiction.
Proof of Theorem 4. Let  $\Pi(\mathbb{P}) \subset \mathcal{P}(\mathcal{X} \times \mathcal{Y})$  denote the subset of probability measures  $\pi \in \mathcal{P}(\mathcal{X} \times \mathcal{Y})$  satisfying  $\pi_x = \mathbb{P}$ . Consider a functional  $I: \Pi(\mathbb{P}) \to \{0, +\infty\}$  defined by  $I(\pi) \stackrel{\text{def}}{=} \sup_{f \leq 0} \int_{\mathcal{Y}} f(y) d(w\mathbb{Q} - \pi_y)(y)$ , where the sup is taken over non-positive  $f \in \mathcal{C}(\mathcal{Y})$ . From Lemma 1, we have  $I(\pi) = 0$  when  $\pi \in \Pi^w(\mathbb{P}, \mathbb{Q})$  and  $I(\pi) = +\infty$  otherwise. Indeed, if there exists a non-positive function satisfying  $\int_{\mathcal{Y}} f(y) d(w\mathbb{Q} - \pi_y)(y) > 0$ , then function  $C_f$  (for  $C > 0$ ) also satisfies this condition and provides  $C$ -times bigger value which tends to  $\infty$  with  $C \to \infty$ . We use  $I(\pi)$  incorporate the right constraint  $\pi_y \leq w\mathbb{Q}$  in  $\pi \in \Pi^w(\mathbb{P}, \mathbb{Q})$  to the objective and obtain the equivalent to (12) problem:
$$
\begin{array}{l} \min  _ {\pi \in \Pi^ {w} (\mathbb {P}, \mathbb {Q})} \int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi (x, y) = \min  _ {\pi \in \Pi (\mathbb {P})} \left\{\int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi (x, y) + I (\pi) \right\} = \\ \min  _ {\pi \in \Pi (\mathbb {P})} \left\{\int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi (x, y) + \sup  _ {f \leq 0} \int_ {\mathcal {Y}} f (y) d \left(w \mathbb {Q} - \pi_ {y}\right) (y) \right\} = \\ \min  _ {\pi \in \Pi (\mathbb {P})} \sup  _ {f \leq 0} \left\{\int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi (x, y) + \int_ {\mathcal {Y}} f (y) d \left(w \mathbb {Q} - \pi_ {y}\right) (y) \right\} = \tag {26} \\ \end{array}
$$
$$
\sup  _ {f \leq 0} \min  _ {\pi \in \Pi (\mathbb {P})} \left\{\int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi (x, y) + \int_ {\mathcal {Y}} f (y) d \left(w \mathbb {Q} - \pi_ {y}\right) (y) \right\} = \tag {27}
$$
$$
\sup  _ {f \leq 0} \left\{\min  _ {\pi \in \Pi (\mathbb {P})} \left\{\int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi (x, y) - \int_ {\mathcal {Y}} f (y) d \pi_ {y} (y) \right\} + w \int_ {\mathcal {Y}} f (y) d \mathbb {Q} (y) \right\} =
$$
$$
\sup  _ {f \leq 0} \left\{\min  _ {\pi \in \Pi (\mathbb {P})} \left\{\int_ {\mathcal {X} \times \mathcal {Y}} c (x, y) d \pi (x, y) - \int_ {\mathcal {X} \times \mathcal {Y}} f (y) d \pi (x, y) \right\} + w \int_ {\mathcal {Y}} f (y) d \mathbb {Q} (y) \right\} = \tag {28}
$$
$$
\begin{array}{l} \sup  _ {f \leq 0} \left\{\min  _ {\pi \in \Pi (\mathbb {P})} \left\{\int_ {\mathcal {X}} \int_ {\mathcal {Y}} c (x, y) d \pi (y | x) \underbrace {d \mathbb {P} (x)} _ {= d \pi_ {x} (x)} - \int_ {\mathcal {X}} \int_ {\mathcal {Y}} f (y) d \pi (y | x) \underbrace {d \mathbb {P} (x)} _ {= d \pi_ {x} (x)} \right\} + w \int_ {\mathcal {Y}} f (y) d \mathbb {Q} (y) \right\} = (29) \\ \sup  _ {f \leq 0} \left\{\min  _ {\pi \in \Pi (\mathbb {P})} \left\{\int_ {x} \int_ {y} (c (x, y) - f (y)) d \pi (y | x) d \mathbb {P} (x) \right\} + w \int_ {y} f (y) d \mathbb {Q} (y) \right\} (30) \\ \end{array}
$$
In transition from (26) to (27) we use the minimax theorem to swap sup and min [64, Corollary 2]. This is possible because the expression in (26) is a bilinear functional of  $(\pi ,f)$ . Thus, it is convex in  $\pi$  and concave in  $f$ . At the same time,  $\Pi (\mathbb{P})$  is a convex and (weak-\*) compact set. The latter can be derived analogously to the compactness of  $\Pi^w (\mathbb{P},\mathbb{Q})$  in the proof of Theorem 2. In transition from (28) to (29), we use the measure disintegration theorem to represent  $d\pi (x,y)$  as the marginal  $d\pi_{x}(x) = d\mathbb{P}(x)$  and a family of conditional measures  $d\pi (y|x)$  on  $\mathcal{V}$ . We note that
$$
\min  _ {\pi \in \Pi (\mathbb {P})} \int_ {\mathcal {X}} \int_ {\mathcal {Y}} \left(c (x, y) - f (y)\right) d \pi (y | x) d \mathbb {P} (x) \} \geq \int_ {\mathcal {X}} \underbrace {\min  _ {y \in \mathcal {Y}} \left(c (x , y) - f (y)\right)} _ {= f ^ {c} (x)} d \mathbb {P} (x). \tag {31}
$$