A note on duality theorems in mass transportation

TL;DR

This paper explores duality theorems in the Monge-Kantorovich mass transportation problem within an abstract measure theoretic framework, extending existing results to broader conditions and less restrictive assumptions.

Contribution

It proves duality theorems without requiring perfect measures and establishes conditions under which duality holds for various classes of cost functions.

Findings

Duality holds for upper-semicontinuous costs on metric spaces.

Duality is valid for bounded continuous costs.

Results extend previous theorems to non-perfect measures and broader spaces.

Abstract

The duality theory of the Monge-Kantorovich transport problem is investigated in an abstract measure theoretic framework. Let $(\mathcal{X},\mathcal{F},\mu)$ and $(\mathcal{Y},\mathcal{G},\nu)$ be any probability spaces and $c:\mathcal{X}\times\mathcal{Y}\rightarrow\mathbb{R}$ a measurable cost function such that $f_1+g_1\le c\le f_2+g_2$ for some $f_1,\,f_2\in L_1(\mu)$ and $g_1,\,g_2\in L_1(\nu)$. Define $\alpha(c)=\inf_P\int c\,dP$ and $\alpha^*(c)=\sup_P\int c\,dP$, where $\inf$ and $\sup$ are over the probabilities $P$ on $\mathcal{F}\otimes\mathcal{G}$ with marginals $\mu$ and $\nu$. Some duality theorems for $\alpha(c)$ and $\alpha^*(c)$, not requiring $\mu$ or $\nu$ to be perfect, are proved. As an example, suppose $\mathcal{X}$ and $\mathcal{Y}$ are metric spaces and $\mu$ is separable. Then, duality holds for $\alpha(c)$ (for $\alpha^*(c)$) provided $c$ is upper-semicontinuous (lower-semicontinuous). Moreover, duality holds for both $\alpha(c)$ and $\alpha^*(c)$ if the maps $x\mapsto c(x,y)$ and $y\mapsto c(x,y)$ are continuous, or if $c$ is bounded and $x\mapsto c(x,y)$ is continuous. This improves the existing results in \cite{RR1995} if $c$ satisfies the quoted conditions and the cardinalities of $\mathcal{X}$ and $\mathcal{Y}$ do not exceed the continuum.

Equations

$$\mathcal{H}=\mathcal{F}\otimes \mathcal{G}$$

$$P(A\times \mathcal{Y})=\mu (A)\text{and}P(\mathcal{X}\times B)=\nu (B)\text{for all }A\in \mathcal{F}\text{ and }B\in \mathcal{G}.$$

$$(f+g)(x,y)=f(x)+g(y)\text{for all }(x,y)\in \mathcal{X}\times \mathcal{Y}.$$

$$L=\{f+g:f\in L_1(\mu ),g\in L_1(\nu )\}.$$

$$f_1+g_1\le c\le f_2+g_2\text{for some }{f}_1+g_1\in L\text{ and }{f}_2+g_2\in L.$$

$$\displaystyle\alpha(c)=\inf\,\bigl{\{}P(c):\,P\in\Gamma(\mu,\nu)\bigr{\}},$$

$$\displaystyle\alpha^{*}(c)=\sup\,\bigl{\{}P(c):\,P\in\Gamma(\mu,\nu)\bigr{\}},$$

$$\displaystyle\beta(c)=\sup\,\bigl{\{}\mu(f)+\nu(g):\,f+g\in L,\,f+g\leq c\bigr{\}},$$

$$\displaystyle\beta^{*}(c)=\inf\,\bigl{\{}\mu(f)+\nu(g):\,f+g\in L,\,f+g\geq c\bigr{\}}.$$

$$\beta (c)\le \alpha (c)\le {\alpha }^{\ast }(c)\le {\beta }^{\ast }(c).$$

$$\alpha (c)=\beta (c)\text{and}{\alpha }^{\ast }(c)={\beta }^{\ast }(c).$$

$$\displaystyle M=\bigl{\{}\mathcal{H}\text{-measurable functions }c:\mathcal{X}\times\mathcal{Y}\rightarrow\mathbb{R}\text{ satisfying condition \eqref{new7}}\bigr{\}}$$

$${\alpha }^{\ast }(c)=-\alpha (-c)\text{and}{\beta }^{\ast }(c)=-\beta (-c)\text{for all }c\in M.$$

$$f_n+g_n\ge c_n,0\le f_n,g_n\le k,\mu (f_n)+\nu (g_n)<{\beta }^{\ast }(c_n)+1/n;$$

$$f_{m_n}\to f\text{ weakly in }{L}_1(\mu )\text{and}{g}_{m_n}\to g\text{ weakly in }{L}_1(\nu ).$$

$$\mu (f)+\nu (g)\ge {\beta }^{\ast }(c)\ge \underset{n}{\lim }{\beta }^{\ast }(c_n)=\underset{n}{\lim }{\beta }^{\ast }(c_{m_n})$$

$$\displaystyle=\lim_{n}\bigl{\{}\mu(f_{m_{n}})+\nu(g_{m_{n}})\bigr{\}}=\mu(f)+\nu(g).$$

$$\displaystyle f_{k}+g_{k}\geq b\wedge 2k\quad\text{and}\quad\mu(f_{k})+\nu(g_{k})<\beta^{*}\bigl{(}b\wedge 2k)+1/k.$$

$$f{1}_{\{g>k\}}=f{1}_{\{f\le k,g>k\}}+f{1}_{\{f>k,g>k\}}\le g{1}_{\{g>k\}}+f{1}_{\{f>k\}}.$$

$$b\le b{1}_{\{b\le 2k\}}+(f+g)1_{\{f+g>2k\}}$$

$$\displaystyle\leq f_{k}+g_{k}+(f+g)\,\bigl{(}1_{\{f>k\}}+1_{\{g>k\}}\bigr{)}$$

$$\le f_k+g_k+3f{1}_{\{f>k\}}+3g{1}_{\{g>k\}}.$$

$$\displaystyle\beta^{*}(b)\leq\mu(f_{k})+\nu(g_{k})+3\mu\bigl{[}f\,1_{\{f>k\}}\bigr{]}+3\nu\bigl{[}g\,1_{\{g>k\}}\bigr{]}$$

$$\displaystyle<\beta^{*}(b\wedge 2k)+(1/k)+3\mu\bigl{[}f\,1_{\{f>k\}}\bigr{]}+3\nu\bigl{[}g\,1_{\{g>k\}}\bigr{]}.$$

$${\beta }^{\ast }(b)<{\beta }^{\ast }(b\wedge 2k)+ϵ=\underset{n}{\lim }{\beta }^{\ast }(b_n\wedge 2k)+ϵ\le \underset{n}{\lim }{\beta }^{\ast }(b_n)+ϵ.$$

$$\displaystyle P(h)=P\bigl{[}h\,1_{\{h\leq t^{-1}\}}\bigr{]}+P\bigl{[}h\,1_{\{t^{-1}<h\leq 2t\}}\bigr{]}+P\bigl{[}h\,1_{\{h>2t\}}\bigr{]}$$

$$\displaystyle\leq t^{-1}+2t\,P(h>t^{-1})+P\bigl{[}h\,1_{\{h>2t\}}\bigr{]}.$$

$$f=f_2-f_1\text{and}g=g_2-g_1.$$

$$\displaystyle P\bigl{[}h\,1_{\{h>2t\}}\bigr{]}\leq P\bigl{[}(f+g)\,1_{\{f+g>2t\}}\bigr{]}\leq P\bigl{[}(f+g)\,1_{\{f>t\}}\bigr{]}+P\bigl{[}(f+g)\,1_{\{g>t\}}\bigr{]}$$

$$\displaystyle=\mu\bigl{[}f\,1_{\{f>t\}}\bigr{]}+\nu\bigl{[}g\,1_{\{g>t\}}\bigr{]}+P\bigl{[}f\,1_{\{g>t\}}+g\,1_{\{f>t\}}\bigr{]}.$$

$$\displaystyle P\bigl{[}f\,1_{\{g>t\}}+g\,1_{\{f>t\}}\bigr{]}\leq 2\,P\bigl{[}f\,1_{\{f>t\}}+g\,1_{\{g>t\}}]=2\mu\bigl{[}f\,1_{\{f>t\}}\bigr{]}+2\nu\bigl{[}g\,1_{\{g>t\}}\bigr{]}.$$

$$\displaystyle P(h)\leq t^{-1}+2t\,P(h>t^{-1})+3\,\bigl{\{}\mu\bigl{[}f\,1_{\{f>t\}}\bigr{]}+\nu\bigl{[}g\,1_{\{g>t\}}\bigr{]}\bigr{\}}.$$

$$Q(h)=Q(c)-Q(f_1+g_1)=\beta (c)-\mu (f_1)-\nu (g_1)=0.$$

$$\alpha (h>t^{-1})=\beta (h>t^{-1})\le Q(h>t^{-1})\le tQ(h)=0.$$

$$\alpha (c)\le P_t(c)=P_t(f_1+g_1)+P_t(h)=\mu (f_1)+\nu (g_1)+P_t(h)$$

Full text

1. Introduction

Throughout, $(\mathcal{X},\mathcal{F},\mu)$ and $(\mathcal{Y},\mathcal{G},\nu)$ are probability spaces and

[TABLE]

is the product $\sigma$-field on $\mathcal{X}\times\mathcal{Y}$. Further, $\Gamma(\mu,\nu)$ is the collection of probability measures $P$ on $\mathcal{H}$ with marginals $\mu$ and $\nu$, namely,

[TABLE]

For any probability space $(\Omega,\mathcal{A},Q)$, we write $L_{1}(Q)$ to denote the class of $\mathcal{A}$-measurable and $Q$-integrable functions $\phi:\Omega\rightarrow\mathbb{R}$ (without identifying maps which agree $Q$-a.s.). We also write $Q(\phi)=\int\phi\,dQ$ for $\phi\in L_{1}(Q)$.

With a slight abuse of notation, for any maps $f:\mathcal{X}\rightarrow\mathbb{R}$ and $g:\mathcal{Y}\rightarrow\mathbb{R}$, we still denote by $f$ and $g$ the functions on $\mathcal{X}\times\mathcal{Y}$ given by $(x,y)\mapsto f(x)$ and $(x,y)\mapsto g(y)$. Thus, $f+g$ is the map on $\mathcal{X}\times\mathcal{Y}$ defined as

[TABLE]

In this notation, we let

[TABLE]

Let $c:\mathcal{X}\times\mathcal{Y}\rightarrow\mathbb{R}$ be an $\mathcal{H}$-measurable function satisfying

[TABLE]

For such a $c$, we define

[TABLE]

It is not hard to see that

[TABLE]

A duality theorem (for both $\alpha(c)$ and $\alpha^{*}(c)$) is the assertion that

[TABLE]

Indeed, duality theorems arise in a plenty of frameworks. The main one is possibly mass transportation, where $c(x,y)$ is regarded as the cost for moving a unit of good from $x\in\mathcal{X}$ into $y\in\mathcal{Y}$. However, duality results play a role even in risk theory, optimization problems and dependence modeling. See e.g. [1], [3], [4], [5], [6], [9], [11], [12], [13], [16], [17] and references therein.

Starting from Kantorovich himself [8], there is a long line of research on duality theorems; see again [3], [4], [17] and references therein. To our knowledge, under the present assumptions on $c$, the best result is due to Ramachandran and Ruschendorf [14]. According to the latter, one obtains both $\alpha(c)=\beta(c)$ and $\alpha^{*}(c)=\beta^{*}(c)$ provided $c$ is $\mathcal{H}$-measurable, it satisfies condition (1), and at least one between $\mu$ and $\nu$ is perfect.

Now, some form of condition (1) can not be dispensed while removing measurability leads to involve inner and outer measures; see [9, Section 2]. Instead, whether the perfectness assumption can be dropped is still an open problem. Thus, if $c$ is measurable and meets (1) but $\mu$ and $\nu$ are both non-perfect, it is currently unknown whether condition (2) is true or false. See points (2)-(3), page 355, of [15].

This paper provides duality theorems not requiring perfectness.

Suppose $\mathcal{X}$ and $\mathcal{Y}$ are metric spaces and $\mathcal{F}$ and $\mathcal{G}$ the Borel $\sigma$-fields. Then, condition (2) is shown to be true if at least one of $\mu$ and $\nu$ is separable, $c$ meets (1) and all the $c$-sections are continuous. Or else, condition (2) holds if $\mu$ and $\nu$ are both separable, $c$ is bounded and measurable, and at least one of the $c$-sections is continuous. These results improve [14] when $c$ satisfies the quoted assumptions and the cardinalities of $\mathcal{X}$ and $\mathcal{Y}$ do not exceed the continuum. Under the latter condition, in fact, a perfect probability measure is separable but not conversely. Note also that, if $\mathcal{X}$ and $\mathcal{Y}$ are separable metric spaces (so that separability of $\mu$ and $\nu$ is automatic) the scope of our results is to replace assumptions on $\mu$ or $\nu$ (required by [14]) with assumptions on $c$.

Various conditions for $\alpha(c)=\beta(c)$ or $\alpha^{*}(c)=\beta^{*}(c)$, but not necessarily for both, are given as well. For instance, if $c$ meets (1) and at least one of $\mu$ and $\nu$ is separable, then $\alpha^{*}(c)=\beta^{*}(c)$ or $\alpha(c)=\beta(c)$ provided $c$ is lower or upper semicontinuous. As another example, $\alpha^{*}(1_{H})=\beta^{*}(1_{H})$ if $H=\cup_{n}(A_{n}\times B_{n})$ with $A_{n}\in\mathcal{F}$ and $B_{n}\in\mathcal{G}$. Further, $\alpha(1_{H})=\beta(1_{H})$ if $\mu(\limsup_{n}A_{n})=0$ or $\nu(\limsup_{n}B_{n})=0$. Without some extra condition, however, we do not know whether $\alpha(1_{H})=\beta(1_{H})$.

2. Preliminaries

For any topological space $S$, the Borel $\sigma$-field on $S$ is denoted by $\mathcal{B}(S)$.

Let $(\Omega,\mathcal{A},Q)$ be a probability space. Then, $Q$ is perfect if, for any $\mathcal{A}$-measurable $\phi:\Omega\rightarrow\mathbb{R}$, there is $B\in\mathcal{B}(\mathbb{R})$ such that $B\subset\phi(\Omega)$ and $Q(\phi\in B)=1$.

An important special case is $\Omega$ a metric space and $\mathcal{A}=\mathcal{B}(\Omega)$. In that case, $Q$ is separable if $Q(A)=1$ for some separable $A\in\mathcal{A}$ and $Q$ is tight if $Q(A)=1$ for some $\sigma$-compact $A\in\mathcal{A}$. Clearly, tightness implies separability but not conversely. Furthermore, tightness is equivalent to perfectness provided $\Omega$ satisfies the following condition:

• The power set of $\Omega$ does not support any 0-1-valued probability measure $T$ such that $T\{\omega\}=0$ for each $\omega\in\Omega$;

see [10, Theorem 3.2].

Two remarks are in order. First, the above condition on $\Omega$ is automatically true if card$(\Omega)\leq\,$card$(\mathbb{R})$. Thus, perfectness implies separability, but not conversely, if card$(\Omega)\leq\,$card$(\mathbb{R})$ (in particular, if $\Omega$ is a separable metric space). Second, it is consistent with the usual axioms of set theory (ZFC) that, for any metric space $\Omega$, any probability measure on $\mathcal{B}(\Omega)$ is separable.

Note also that a simple example of non perfect probability measure is any non tight probability measure on the Borel sets of a separable metric space. For instance, take $Q$ the outer Lebesgue measure on $\mathcal{B}(\Omega)$, where $\Omega$ is a subset of $[0,1]$ with outer Lebesgue measure 1 and inner Lebesgue measure 0. Then, $Q$ is not perfect.

Let us come back to duality theorems. Define

[TABLE]

and note that

[TABLE]

Thus, to get condition (2), it suffices to show $\alpha(c)=\beta(c)$ under some conditions which hold true for both $c$ and $-c$.

Two preliminary lemmas are needed. The first is inspired to [7, Lemma 1].

Lemma 1**.**

Let $c\in M$. Then, $\beta^{*}(c)=\lim_{n}\beta^{*}(c_{n})$ whenever $(c_{n})\subset M$ is an increasing sequence such that $c_{n}\uparrow c$ pointwise.

Proof.

We first suppose $0\leq c_{n}\leq c\leq k$ for some integer $k$. Under this assumption, for each $n$, there is $f_{n}+g_{n}\in L$ such that

[TABLE]

see e.g. [9, Lemma 1.8].

Since the sequences $(f_{n})$ and $(g_{n})$ are uniformly bounded, there are $f\in L_{1}(\mu)$, $g\in L_{1}(\nu)$ and a subsequence $(m_{n})$ such that

[TABLE]

In turn, this implies the existence of a sequence $(\phi_{n},\psi_{n})$ such that $\phi_{n}\rightarrow f$ in $L_{1}(\mu)$, $\psi_{n}\rightarrow g$ in $L_{1}(\nu)$ and $(\phi_{n},\psi_{n})$ is a convex combination of $\{(f_{m_{j}},g_{m_{j}}):j\geq n\}$ for each $n$. By taking a further subsequence, it can be also assumed $\mu(\phi_{n}\rightarrow f)=\nu(\psi_{n}\rightarrow g)=1$. Since $(c_{n})$ is increasing, $\phi_{n}+\psi_{n}\geq c_{m_{n}}$. Hence, after modifying $f$ and $g$ on null sets, one obtains $f+g\geq c$. On noting that $(\beta^{*}(c_{n}))$ is a monotone sequence, it follows that

[TABLE]

This concludes the proof if $0\leq c_{n}\leq c\leq k$. To deal with the general case, fix $p+q\in L$ such that $p+q\leq c_{1}$ and define $b_{n}=c_{n}-(p+q)$ and $b=c-(p+q)$. Then, $0\leq b_{n}\leq b$. Further, since $\beta^{*}(h+p+q)=\beta^{*}(h)+\mu(p)+\nu(q)$ for each $h\in M$, it suffices to show that $\beta^{*}(b)=\lim_{n}\beta^{*}(b_{n})$.

Given $k$, take $f_{k}+g_{k}\in L$ such that

[TABLE]

Take also $f+g\in L$ such that $f+g\geq b$ and note that

[TABLE]

Similarly, $g\,1_{\{f>k\}}\leq g\,1_{\{g>k\}}+f\,1_{\{f>k\}}$. Hence,

[TABLE]

Since $f_{k}+g_{k}+3f\,1_{\{f>k\}}+3g\,1_{\{g>k\}}$ belongs to $L$, it follows that

[TABLE]

Fix $\epsilon>0$ and take $k$ such that (1/k)+3\mu\bigl{[}f\,1_{\{f>k\}}\bigr{]}+3\nu\bigl{[}g\,1_{\{g>k\}}\bigr{]}<\epsilon. By what already proved, $\beta^{*}(b\wedge 2k)=\lim_{n}\beta^{*}(b_{n}\wedge 2k)$. Therefore,

[TABLE]

This concludes the proof. ∎

In the second lemma, and in the rest of the paper, we write $\alpha(H)=\alpha(1_{H})$ whenever $H\in\mathcal{H}$. The same notation is adopted for $\beta$, $\alpha^{*}$ and $\beta^{*}$.

Lemma 2**.**

Let $c\in M$. Then, condition (2) holds provided $\alpha(H)=\beta(H)$ for each $H\in\mathcal{H}$.

Proof.

It suffices to show $\alpha(c)=\beta(c)$. To this end, we first note that $\beta(c)$ is attained, i.e., $\beta(c)=\mu(f_{1})+\nu(g_{1})$ for some $f_{1}+g_{1}\in L$ such that $f_{1}+g_{1}\leq c$; see [14, Proposition 3]. Define $h=c-(f_{1}+g_{1})$ and fix $t>1$ and $P\in\Gamma(\mu,\nu)$. Then,

[TABLE]

Take $f_{2}+g_{2}\in L$ such that $f_{2}+g_{2}\geq c$ and define

[TABLE]

Since $h\leq f+g$,

[TABLE]

Arguing as in the proof of Lemma 1,

[TABLE]

Hence,

[TABLE]

Next, by Theorem 2.1.1 and Remark 2.1.2(b) of [13], there is a finitely additive probability $Q$ on $\mathcal{H}$, with marginals $\mu$ and $\nu$, such that $Q(c)=\beta(c)$. Since $Q$ has marginals $\mu$ and $\nu$, then $\beta(H)\leq Q(H)$ for all $H\in\mathcal{H}$ and

[TABLE]

Finally, since $h\geq 0$ and $\alpha(H)=\beta(H)$ for all $H\in\mathcal{H}$, one obtains

[TABLE]

Hence, there is $P_{t}\in\Gamma(\mu,\nu)$ such that $P_{t}(h>t^{-1})<t^{-2}$. It follows that

[TABLE]

Since 1/t+\mu\bigl{[}f\,1_{\{f>t\}}\bigr{]}+\nu\bigl{[}g\,1_{\{g>t\}}\bigr{]}\rightarrow 0 as $t\rightarrow\infty$, this concludes the proof.

∎

3. Duality theorems without perfectness

It is convenient to distinguish two cases.

3.1. The abstract case

Theorem 3**.**

Let $c\in M$. Then, condition (2) holds provided

• (*)

For each $\epsilon>0$, there is a countable partition $\{A_{0},A_{1},\ldots\}\subset\mathcal{F}$ of $\mathcal{X}$ such that $\mu(A_{0})=0$ and

[TABLE]

Proof.

Again, it suffices to show $\alpha(c)=\beta(c)$. Given $\epsilon>0$, fix a point $x_{i}\in A_{i}$ for each $i>0$, and define

[TABLE]

Let $\Gamma(\mu_{0},\nu)$ be the set of probability measures on $\mathcal{F}_{0}\otimes\mathcal{G}$ with marginals $\mu_{0}$ and $\nu$.

Take $f_{1}+g_{1}\in L$ and $f_{2}+g_{2}\in L$ such that $f_{1}+g_{1}\leq c\leq f_{2}+g_{2}$. Since $\lvert c-c_{0}\rvert\leq\epsilon$, then $f_{1}+g_{1}-\epsilon\leq c_{0}\leq f_{2}+g_{2}+\epsilon$. Further, $\sup_{A_{i}}f_{1}<+\infty$ and $\inf_{A_{i}}f_{2}>-\infty$ for each $i>0$. Define

[TABLE]

Then, $\phi_{1},\,\phi_{2}\in L_{1}(\mu_{0})$ and

[TABLE]

Because of such inequality and since $c_{0}$ is $\mathcal{F}_{0}\otimes\mathcal{G}$-measurable, one can define

[TABLE]

where $\sup$ is over the pairs $(f,g)$ such that $f\in L_{1}(\mu_{0})$, $g\in L_{1}(\nu)$ and $f+g\leq c_{0}$.

Since $\mu_{0}$ is an atomic probability measure, then $\mu_{0}$ is perfect, which in turn implies $\alpha_{0}=\beta_{0}$. Since $\lvert c-c_{0}\rvert\leq\epsilon$, then $\beta_{0}\leq\beta(c)+\epsilon$. Hence, there is $T\in\Gamma(\mu_{0},\nu)$ such that

[TABLE]

If $T$ can be extended to a probability measure $P\in\Gamma(\mu,\nu)$, then

[TABLE]

Thus, to conclude the proof, it suffices to show that $T$ can be actually extended to a probability measure $P\in\Gamma(\mu,\nu)$.

For each $i$ with $\mu(A_{i})>0$, define

[TABLE]

where $A\in\mathcal{F}$ and $B\in\mathcal{G}$. Define also

[TABLE]

where $\mu_{i}\times\nu_{i}$ is the product measure of $\mu_{i}$ and $\nu_{i}$ (so that $\mu_{i}\times\nu_{i}$ is a probability measure on $\mathcal{H}$). It is straightforward to see that $P\in\Gamma(\mu,\nu)$. Fix $A\in\mathcal{F}_{0}$ and $B\in\mathcal{G}$. For $i>0$, either $A\cap A_{i}=\emptyset$ or $A\cap A_{i}=A_{i}$, so that

[TABLE]

Therefore, $P=T$ on $\mathcal{F}_{0}\otimes\mathcal{G}$.

∎

In Theorem 3, clearly, the roles of $\mu$ and $\nu$ can be interchanged. Accordingly, condition (*) can be replaced by

• (**)

For each $\epsilon>0$, there is a countable partition $\{B_{0},B_{1},\ldots\}\subset\mathcal{G}$ of $\mathcal{Y}$ such that $\nu(B_{0})=0$ and

[TABLE]

As an example, condition (*) holds (with $A_{0}=\emptyset$) if $\mathcal{X}$ is a separable metric space and the function $x\mapsto c(x,y)$ is Lipschitz uniformly with respect to $y$, i.e.,

[TABLE]

where $u>0$ is a constant and $d$ the distance on $\mathcal{X}$. Fix in fact $\epsilon>0$. Because of separability, $\mathcal{X}$ can be partitioned into sets $A_{1},A_{2},\ldots$ whose diameter is less than $\epsilon/u$. Hence, condition (*) follows trivially from (3). Similarly, condition (**) holds if $\mathcal{Y}$ is a separable metric space and $y\mapsto c(x,y)$ is Lipschitz uniformly with respect to $x$. Further, as shown in the proof of Theorem 7, separability of $\mathcal{X}$ (of $\mathcal{Y}$) can be weakened into separability of $\mu$ (of $\nu$).

Another example is the following. Let $\mathcal{R}$ be the field of subsets of $\mathcal{X}\times\mathcal{Y}$ generated by the measurable rectangles. Each $R\in\mathcal{R}$ can be written as $R=\cup_{i=1}^{n}(A_{i}\times B_{i})$ for some $n\geq 1$ and $A_{i}\in\mathcal{F}$, $B_{i}\in\mathcal{G}$ such that $A_{i}\cap A_{j}=\emptyset$ for $i\neq j$. Thus, when $c=1_{R}$ with $R\in\mathcal{R}$, condition (*) is trivially true and Theorem 3 yields $\alpha(R)=\beta(R)$ and $\alpha^{*}(R)=\beta^{*}(R)$. We next prove duality of certain sets related to $\mathcal{R}$.

Theorem 4**.**

Let $H=\cup_{n}R_{n}$ and $K=\cap_{n}R_{n}$ where $R_{n}\in\mathcal{R}$ for each $n$. Then,

[TABLE]

In addition, $\alpha(H)=\beta(H)$ provided $H$ can be written as $H=\cup_{n}(A_{n}\times B_{n})$ with $A_{n}\in\mathcal{F}$, $B_{n}\in\mathcal{G}$, and

[TABLE]

(Here, $\limsup_{n}A_{n}=\cap_{n}\cup_{j>n}A_{j}$ and $\limsup_{n}B_{n}=\cap_{n}\cup_{j>n}B_{j}$).

Proof.

Let $H_{n}=\cup_{i=1}^{n}R_{i}$. Since $\alpha^{*}(H_{n})=\beta^{*}(H_{n})$, Lemma 1 implies

[TABLE]

Thus, $\alpha^{*}$ and $\beta^{*}$ agree on countable unions of elements of $\mathcal{R}$. Since $R_{n}^{c}\in\mathcal{R}$, this implies

[TABLE]

Next, suppose $H=\cup_{n}(A_{n}\times B_{n})$ and $\mu(\limsup_{n}A_{n})=0$. Let $V_{n}=\cup_{i=1}^{n}(A_{i}\times B_{i})$. Given $\epsilon>0$, take $n\geq 1$ such that \mu\bigl{(}\cup_{i>n}A_{i}\bigr{)}<\epsilon, and then take $P\in\Gamma(\mu,\nu)$ satisfying $P(V_{n})<\alpha(V_{n})+\epsilon$. Since $\alpha(V_{n})=\beta(V_{n})$, one obtains

[TABLE]

The proof is exactly the same if $\nu(\limsup_{n}B_{n})=0$. ∎

Because of Theorem 4, a (classical) question raised by Arveson [2] admits a positive answer for countable unions of measurable rectangles.

Arveson’s problem: If $H\in\mathcal{H}$ satisfies $P(H)=0$ for all $P\in\Gamma(\mu,\nu)$, are there $A\in\mathcal{F}$ and $B\in\mathcal{G}$ such that $\mu(A)=\nu(B)=0$ and $H\subset(A\times\mathcal{Y})\cup(\mathcal{X}\times B)$ ?

Indeed, it is not hard to see that $\beta^{*}(H)=\mu(A)+\nu(B)$ for some $A\in\mathcal{F}$ and $B\in\mathcal{G}$ with $H\subset(A\times\mathcal{Y})\cup(\mathcal{X}\times B)$; see e.g. [7, Lemma 1]. If $H$ is a countable union of measurable rectangles, Theorem 4 implies $\beta^{*}(H)=\alpha^{*}(H)=0$ so that $\mu(A)=\nu(B)=0$.

In addition, exploiting Theorem 4, duality for $H=\cup_{n}(A_{n}\times B_{n})$ can be obtained under various conditions. One such condition is $\mu(\limsup_{n}A_{n})=0$ or $\nu(\limsup_{n}B_{n})=0$. A similar condition is that $H^{c}$ is also a countable union of measurable rectangles. In this case, in fact, $\alpha^{*}(H^{c})=\beta^{*}(H^{c})$ or equivalently $\alpha(H)=\beta(H)$. A last condition is

[TABLE]

Define in fact P_{n}(U)=\mu\bigl{\{}x:(x,\phi_{n}(x))\in U\bigr{\}} for each $n\geq 1$ and $U\in\mathcal{H}$. Then, $P_{n}\in\Gamma(\mu,\nu)$ and $\alpha(H)\leq P_{n}(H)<1/n$. Thus $\alpha(H)=0$, which in turn implies $\alpha(H)=\beta(H)$. Here is a simple example.

Example 5*.*

Suppose $(\mathcal{X},\mathcal{F})=(\mathcal{Y},\mathcal{G})$ and $\mu=\nu$, with $\mathcal{X}$ a separable metric space and $\mathcal{F}=\mathcal{B}(\mathcal{X})$. (Up to some technicalities, separability of $\mathcal{X}$ could be weakened into separability of $\mu$). Let $\Delta=\{(x,x):x\in\mathcal{X}\}$ be the diagonal and $H$ a countable union of measurable rectangles. Then, duality holds for $H\cap\Delta^{c}$, and it holds for $H\cap\Delta$ provided $\mu$ vanishes on singletons. In fact, $H\cap\Delta^{c}$ is a countable union of measurable rectangles and \mu\bigl{\{}x:(x,x)\in H\cap\Delta^{c}\bigr{\}}=0. Letting $\phi_{n}(x)=x$, Theorem 4 and condition (4) yield

[TABLE]

To deal with $H\cap\Delta$, suppose $\mu$ null on singletons and define $P_{1}=\mu\times\mu$ and P_{2}(U)=\mu\bigl{\{}x:(x,x)\in U\bigr{\}} for each $U\in\mathcal{H}$. Then, $P_{1},\,P_{2}\in\Gamma(\mu,\mu)$. Since $\mu$ is null on singletons, $\alpha(H\cap\Delta)\leq P_{1}(H\cap\Delta)\leq P_{1}(\Delta)=0$, which in turn implies $\alpha(H\cap\Delta)=\beta(H\cap\Delta)$. Finally, writing $H$ as $H=\cup_{n}(A_{n}\times B_{n})$, one obtains

[TABLE]

We close this Subsection with two remarks. The first (stated as a lemma) suggests a possible strategy for proving a general duality theorem.

Lemma 6**.**

Let \mathcal{H}_{0}=\bigl{\{}H\in\mathcal{H}:\alpha(H)=\beta(H) and \alpha^{*}(H)=\beta^{*}(H)\bigr{\}}. Then, condition (2) holds for each $c\in M$ if and only if

[TABLE]

Proof.

By Lemma 2, it suffices to show that $\mathcal{H}_{0}=\mathcal{H}$. In turn, since $\mathcal{R}\subset\mathcal{H}_{0}$, it suffices to see that $\mathcal{H}_{0}$ is a monotone class. Also, since $\mathcal{H}_{0}$ is closed under complements, it is enough to prove that $H\in\mathcal{H}_{0}$ provided $H$ is the union of an increasing sequence of elements of $\mathcal{H}_{0}$. Let $H=\cup_{n}H_{n}$ where $H_{n}\in\mathcal{H}_{0}$ and $H_{n}\subset H_{n+1}$ for each $n$. For such $H$, arguing as in the proof of Theorem 4, one obtains $\alpha^{*}(H)=\beta^{*}(H)$. Thus, under (5), $\mathcal{H}_{0}$ is actually a monotone class. ∎

The second remark briefly compares the arguments underlying Theorem 3 and the usual duality theorems. The latter are summarized into the result by Ramachandran and Ruschendorf [14].

For definiteness, we aim to prove $\alpha(c)=\beta(c)$. By (1) and since $\beta(c)$ is attained, it can be assumed $c\geq 0$ and $\beta(c)=0$. As noted in the proof of Lemma 2, there is a finitely additive probability $Q$ on $\mathcal{H}$, with marginals $\mu$ and $\nu$, satisfying $Q(c)=\beta(c)$. Since $c\geq 0$ and $\beta(c)=0$, it must be $Q(c>\epsilon)=0$ for each $\epsilon>0$. A basic intuition in [14] is that, if one of $\mu$ and $\nu$ is perfect, then $Q$ is $\sigma$-additive on $\mathcal{R}$; see [13, Theorem 2.1.3] and recall that $\mathcal{R}$ is the field generated by the measurable rectangles. Hence, there is $P\in\Gamma(\mu,\nu)$ such that $P=Q$ on $\mathcal{R}$. With such a $P$, one obtains

[TABLE]

provided $R_{i}\in\mathcal{R}$ for all $i$. Hence, $P(c>\epsilon)\leq Q(c>\epsilon)=0$ if the set $\{c>\epsilon\}$ is a countable union of measurable rectangles. Up to some technicalities, suitable versions of this argument work even if $\{c>\epsilon\}$ fails to be a countable union of measurable rectangles. This provides a rough sketch of the proof of $\alpha(c)=\beta(c)$ under the assumption that one of $\mu$ and $\nu$ is perfect. We now turn to Theorem 3. Here, instead of proving that $Q$ is $\sigma$-additive on $\mathcal{R}$, one requires that $c$ can be suitably approximated by $\mathcal{R}$-simple functions. For instance, conditions ()-(**) are trivially true if $c$ is the uniform limit of a sequence of $\mathcal{R}$-simple functions, and in this case no assumptions on $\mu$ or $\nu$ are needed. Apparently, conditions ()-(**) are too restrictive to be useful in real problems. Instead, they allow to get duality in various situations, including Theorem 4, Example 5, and the results in the next subsection.

3.2. The metric case

In this subsection, $\mathcal{X}$ and $\mathcal{Y}$ are metric spaces, $\mathcal{F}=\mathcal{B}(\mathcal{X})$ and $\mathcal{G}=\mathcal{B}(\mathcal{Y})$. The sections of $c$ are the functions $x\mapsto c(x,y)$ and $y\mapsto c(x,y)$, with $y$ fixed in the first map and $x$ fixed in the second.

A remark is in order. All claims made so far are still valid, even if $c$ is not $\mathcal{H}$-measurable, provided $c\,1_{A\times B}$ is $\mathcal{H}$-measurable for some $A\in\mathcal{F}$ and $B\in\mathcal{G}$ with $\mu(A)=\nu(B)=1$. In fact, $\alpha(c)=\alpha(c\,1_{A\times B})$ whenever $\alpha(c)$ is defined in the obvious way, i.e.

[TABLE]

Similarly, $\alpha^{*}(c)=\alpha^{*}(c\,1_{A\times B})$, $\beta(c)=\beta(c\,1_{A\times B})$ and $\beta^{*}(c)=\beta^{*}(c\,1_{A\times B})$.

In the next result, $c\,1_{A\times B}$ is actually $\mathcal{H}$-measurable for some $A\in\mathcal{F}$ and $B\in\mathcal{G}$ such that $\mu(A)=\nu(B)=1$ (with possibly $A=\mathcal{X}$ or $B=\mathcal{Y}$).

Theorem 7**.**

Suppose $c$ satisfies condition (1), the map $x\mapsto c(x,y)$ is continuous for each $y\in\mathcal{Y}$ and the map $y\mapsto c(x,y)$ is $\mathcal{G}$-measurable for each $x\in\mathcal{X}$. Then,

• (i)

$\alpha^{*}(c)=\beta^{*}(c)$* if $c$ is bounded below and $\mu$ is separable;*

• (ii)

$\alpha(c)=\beta(c)$* if $c$ is bounded above and $\mu$ is separable;*

• (iii)

$\alpha(c)=\beta(c)$* and $\alpha^{*}(c)=\beta^{*}(c)$ if $c$ is bounded and $\mu$ is separable;*

• (iv)

$\alpha(c)=\beta(c)$* and $\alpha^{*}(c)=\beta^{*}(c)$ if all the sections of $c$ are continuous and at least one of $\mu$ and $\nu$ is separable.*

Proof.

Since (ii) and (iii) are consequences of (i), it suffices to prove (i) and (iv).

Let $\mu$ and $c$ be as in (i). Since $\mu$ is separable, there is a separable set $A\in\mathcal{F}$ with $\mu(A)=1$. Since $x\mapsto c(x,y)$ is continuous, $y\mapsto c(x,y)$ is Borel measurable and $A$ is separable, the restriction of $c$ on $A\times\mathcal{Y}$ is measurable with respect to $\mathcal{B}(A)\otimes\mathcal{B}(\mathcal{Y})$. Therefore, $c\,1_{A}$ is $\mathcal{H}$-measurable.

Take a countable set $D\subset A$ such that $\overline{D}=\overline{A}$ and define

[TABLE]

where $n\geq 1$, $(x,y)\in\mathcal{X}\times\mathcal{Y}$ and $d$ is the distance on $\mathcal{X}$.

Since $c$ is bounded below, $c_{n}$ is real-valued, and a direct calculation shows that

[TABLE]

Since $D$ is countable, $y\mapsto c_{n}(x,y)$ is Borel measurable. Hence, $c_{n}\,1_{A}$ is $\mathcal{H}$-measurable. In addition, $c_{n}\leq c_{n+1}$ and $c_{n}$ meets condition (1) (since $c$ meets (1) and is bounded below). Finally, since $x\mapsto c(x,y)$ is continuous, one obtains

[TABLE]

Next, $\overline{D}=\overline{A}$ implies $A\subset\bigcup_{x\in D}B(x,\delta)$ for each $\delta>0$, where $B(x,\delta)$ is the $\mathcal{X}$-ball of radius $\delta$ around $x$. Given $n\geq 1$ and $\epsilon>0$, it follows that $A$ can be partitioned into sets $A_{1},A_{2},\ldots\in\mathcal{F}$ whose diameter is less than $\epsilon/n$. Hence, $c_{n}$ meets condition (*) (with $A_{0}=A^{c}$) because of (6). By Lemma 1 and Theorem 3,

[TABLE]

This concludes the proof of (i).

Let us turn to (iv). Suppose that all the $c$-sections are continuous. Since the sections of $-c$ are continuous as well, it suffices to prove $\alpha^{*}(c)=\beta^{*}(c)$. We first assume $\mu$ separable.

By (1), there are $\psi\in L_{1}(\mu)$ and $g\in L_{1}(\nu)$ such that $\psi+g\leq c$. Define

[TABLE]

and note that $f\in L_{1}(\mu)$, $f+g\leq c$ and $f$ is upper-semicontinuous. Define also

[TABLE]

Again, condition (6) holds, $c_{n}$ meets condition (1) (since $g\leq c_{n}\leq c-f$) and $y\mapsto c_{n}(x,y)$ is Borel measurable (it is in fact upper-semicontinuous). On noting that $x\mapsto c(x,y)-f(x)$ is lower-semicontinuous, it is not hard to see that $c_{n}\uparrow c-f$ pointwise as $n\rightarrow\infty$. Because of (6) and $\mu$ separable, $c_{n}$ meets condition (*). Thus,

[TABLE]

Hence, $\alpha^{*}(c)=\beta^{*}(c)$ if $\mu$ is separable.

Finally, if $\nu$ is separable, it suffices to let

[TABLE]

where now $g$ is upper-semicontinuous and $\rho$ is the distance on $\mathcal{Y}$. Arguing as above and using separability of $\nu$, it follows that $c_{n}$ meets condition (**) and $c_{n}\uparrow c-g$ pointwise as $n\rightarrow\infty$. Hence, $\alpha^{*}(c)=\beta^{*}(c)$ and this concludes the proof. ∎

Once again, the roles of $\mu$ and $\nu$ can be interchanged in Theorem 7.

Theorem 8**.**

Suppose $c$ satisfies condition (1), the map $x\mapsto c(x,y)$ is $\mathcal{F}$-measurable for each $y\in\mathcal{Y}$ and the map $y\mapsto c(x,y)$ is continuous for each $x\in\mathcal{X}$. Then,

• (j)

$\alpha^{*}(c)=\beta^{*}(c)$* if $c$ is bounded below and $\nu$ is separable;*

• (jj)

$\alpha(c)=\beta(c)$* if $c$ is bounded above and $\nu$ is separable;*

• (jjj)

$\alpha(c)=\beta(c)$* and $\alpha^{*}(c)=\beta^{*}(c)$ if $c$ is bounded and $\nu$ is separable.*

Note that, if $\mu$ and $\nu$ are both separable, then $\alpha(c)=\beta(c)$ and $\alpha^{*}(c)=\beta^{*}(c)$ provided $c$ is bounded, $\mathcal{H}$-measurable, and at least one of the $c$-sections is continuous. Further, the argument underlying Theorems 7-8 yields other similar results. As an example, we state (without a proof) the following.

Theorem 9**.**

Suppose $c$ satisfies condition (1) and at least one of $\mu$ and $\nu$ is separable. Then, $\alpha^{*}(c)=\beta^{*}(c)$ if $c$ is lower-semicontinuous (with respect to the product topology on $\mathcal{X}\times\mathcal{Y}$) and $\alpha(c)=\beta(c)$ if $c$ is upper-semicontinuous.

Finally, we list some consequences of Theorems 7-9. Indeed, they unify and slightly improve some known results.

• •

Theorems 7-8 improve [14], the result by Ramachandran and Ruschendorf, provided $c$ satisfies some conditions and

[TABLE]

Under such cardinality assumption, in fact, perfectness implies separability but not conversely; see Section 2.

• •

As an example, suppose $c\in M$ and $\mathcal{X}$ and $\mathcal{Y}$ are separable metric spaces (so that $\mu$ and $\nu$ are both separable and the cardinality assumption is satisfied). Then, [14] implies $\alpha(c)=\beta(c)$ and $\alpha^{*}(c)=\beta^{*}(c)$ provided at least one between $\mu$ and $\nu$ is perfect. Instead, Theorems 7-8 lead to the same conclusions whenever all the $c$-sections are continuous, or whenever $c$ is bounded and at least one of the $c$-sections is continuous.

• •

By Theorems 7-8, it is consistent with the usual axioms of set theory (ZFC) that condition (2) holds for every $c\in M$ with continuous sections, or for every bounded $c\in M$ with at least one continuous section. In fact, as noted in Section 2, it is consistent with ZFC that any Borel probability on any metric space is separable.

• •

Let $\mathcal{X}=\mathcal{Y}$ and $c=d$, where $d$ is the distance on $\mathcal{X}$. Suppose $d$ measurable with respect to $\mathcal{B}(\mathcal{X})\otimes\mathcal{B}(\mathcal{X})$ and

[TABLE]

Then, $\alpha(d)$ reduces to Wasserstein distance between $\mu$ and $\nu$ while $\beta(d)$ can be written as

[TABLE]

where $\sup$ is over the 1-Lipschitz functions $f:\mathcal{X}\rightarrow\mathbb{R}$. In this case, it is well known that $\alpha(d)=\beta(d)$ if $\mathcal{X}$ is separable; see e.g. [9, page 400]. This known fact is generalized by Theorems 7-9 under two respects: separability of $\mathcal{X}$ can be weakened into separability of at least one of $\mu$ and $\nu$, and $d$ can be replaced by any upper-semicontinuous function or by any function with continuous sections.

• •

By Theorem 9, Arveson’s question has a positive answer if $H$ is open and one of $\mu$ and $\nu$ is separable.