A Note on Duality Theorems in Mass Transportation

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 Ramachandran and Ruschendorf (Probab Theory Relat Fields 101:311–319, 1995) if c satisfies the quoted conditions and the cardinalities of $\mathcal {X}$ and $\mathcal {Y}$ do not exceed the continuum.

1 Introduction

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

$$\begin{aligned} \mathcal {H}=\mathcal {F}\otimes \mathcal {G} \end{aligned}$$

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

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

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

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

In this notation, we let

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

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

$$\begin{aligned} f_1+g_1\le c\le f_2+g_2\quad \quad \text {for some }f_1+g_1\in L\text { and }f_2+g_2\in L. \end{aligned}$$

(1)

For such a c, we define

$$\begin{aligned} \alpha (c)= & {} \inf \,\bigl \{P(c):\,P\in \Gamma (\mu ,\nu )\bigr \}, \\ \alpha ^*(c)= & {} \sup \,\bigl \{P(c):\,P\in \Gamma (\mu ,\nu )\bigr \}, \\ \beta (c)= & {} \sup \,\bigl \{\mu (f)+\nu (g):\,f+ g\in L,\,f+ g\le c\bigr \}, \\ \beta ^*(c)= & {} \inf \,\bigl \{\mu (f)+\nu (g):\,f+ g\in L,\,f+ g\ge c\bigr \}. \end{aligned}$$

It is not hard to see that

$$\begin{aligned} \beta (c)\le \alpha (c)\le \alpha ^*(c)\le \beta ^*(c). \end{aligned}$$

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

$$\begin{aligned} \alpha (c)=\beta (c)\quad \text {and}\quad \alpha ^*(c)=\beta ^*(c). \end{aligned}$$

(2)

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) cannot 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 $\mu (\limsup _nA_n)=0$ or $\nu (\limsup _nB_n)=0$. Without some extra conditions, 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$(\Omega )\le \,$card$(\mathbb {R})$. Thus, perfectness implies separability, but not conversely, if card$(\Omega )\le \,$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

$$\begin{aligned} M=\bigl \{\mathcal {H}\text {-measurable functions }c:\mathcal {X}\times \mathcal {Y}\rightarrow \mathbb {R}\text { satisfying condition }(1)\bigr \}, \end{aligned}$$

and note that

$$\begin{aligned} \alpha ^*(c)=-\alpha (-c)\quad \text {and}\quad \beta ^*(c)=-\beta (-c)\quad \text {for all }c\in M. \end{aligned}$$

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 $0\le c_n\le c\le k$ for some integer k. Under this assumption, for each n, there is $f_n+ g_n\in L$ such that

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

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

$$\begin{aligned} f_{m_n}\rightarrow f\text { weakly in }L_1(\mu )\quad \text {and}\quad g_{m_n}\rightarrow g\text { weakly in }L_1(\nu ). \end{aligned}$$

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 $\{(f_{m_j},g_{m_j}):j\ge n\}$ 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, $\phi _n+\psi _n\ge c_{m_n}$. Hence, after modifying f and g on null sets, one obtains $f+ g\ge c$. On noting that $(\beta ^*(c_n))$ is a monotone sequence, it follows that

$$\begin{aligned} \mu (f)+\nu (g)\ge \beta ^*(c)\ge & {} \lim _n\beta ^*(c_n)=\lim _n\beta ^*(c_{m_n}) \\= & {} \lim _n\bigl \{\mu (f_{m_n})+\nu (g_{m_n})\bigr \}=\mu (f)+\nu (g). \end{aligned}$$

This concludes the proof if $0\le c_n\le c\le k$. To deal with the general case, fix $p+q\in L$ such that $p+q\le c_1$ and define $b_n=c_n-(p+q)$ and $b=c-(p+q)$. Then, $0\le b_n\le b$. 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

$$\begin{aligned} f_k+ g_k\ge b\wedge 2k\quad \text {and}\quad \mu (f_k)+\nu (g_k)<\beta ^*\bigl (b\wedge 2k)+1/k. \end{aligned}$$

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

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

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

$$\begin{aligned} b\le & {} b\,1_{\{b\le 2k\}}+(f+ g)\,1_{\{f+ g>2k\}} \\\le & {} f_k+ g_k+(f+ g)\,\bigl (1_{\{f>k\}}+1_{\{g>k\}}\bigr ) \\\le & {} f_k+ g_k+3f\,1_{\{f>k\}}+3g\,1_{\{g>k\}}. \end{aligned}$$

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

$$\begin{aligned} \beta ^*(b)\le & {} \mu (f_k)+\nu (g_k)+3\mu \bigl [f\,1_{\{f>k\}}\bigr ]+3\nu \bigl [g\,1_{\{g>k\}}\bigr ] \\< & {} \beta ^*(b\wedge 2k)+(1/k)+3\mu \bigl [f\,1_{\{f>k\}}\bigr ]+3\nu \bigl [g\,1_{\{g>k\}}\bigr ]. \end{aligned}$$

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,

$$\begin{aligned} \beta ^*(b)<\beta ^*(b\wedge 2k)+\epsilon =\lim _n\beta ^*(b_n\wedge 2k)+\epsilon \le \lim _n\beta ^*(b_n)+\epsilon . \end{aligned}$$

This concludes the proof. $\square$

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 $f_1+ g_1\le c$; see [14, Proposition 3]. Define $h=c-(f_1+g_1)$ and fix $t>1$ and $P\in \Gamma (\mu ,\nu )$. Then,

$$\begin{aligned} P(h)= & {} P\bigl [h\,1_{\{h\le t^{-1}\}}\bigr ]+P\bigl [h\,1_{\{t^{-1}<h\le 2t\}}\bigr ]+P\bigl [h\,1_{\{h>2t\}}\bigr ] \\\le & {} t^{-1}+2t\,P(h>t^{-1})+P\bigl [h\,1_{\{h>2t\}}\bigr ]. \end{aligned}$$

Take $f_2+ g_2\in L$ such that $f_2+ g_2\ge c$ and define

$$\begin{aligned} f=f_2-f_1\quad \text {and}\quad g=g_2-g_1. \end{aligned}$$

Since $h\le f+ g$,

$$\begin{aligned} P\bigl [h\,1_{\{h>2t\}}\bigr ]\le & {} P\bigl [(f+g)\,1_{\{f+ g>2t\}}\bigr ]\\\le & {} P\bigl [(f+g)\,1_{\{f>t\}}\bigr ]+P\bigl [(f+g)\,1_{\{g>t\}}\bigr ] \\= & {} \mu \bigl [f\,1_{\{f>t\}}\bigr ]+\nu \bigl [g\,1_{\{g>t\}}\bigr ]+P\bigl [f\,1_{\{g>t\}}+g\,1_{\{f>t\}}\bigr ]. \end{aligned}$$

Arguing as in the proof of Lemma 1,

$$\begin{aligned} P\bigl [f\,1_{\{g>t\}}+g\,1_{\{f>t\}}\bigr ]\le 2\,P\bigl [f\,1_{\{f>t\}}+g\,1_{\{g>t\}}]=2\mu \bigl [f\,1_{\{f>t\}}\bigr ]+2\nu \bigl [g\,1_{\{g>t\}}\bigr ]. \end{aligned}$$

Hence,

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

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 $\beta (H)\le Q(H)$ for all $H\in \mathcal {H}$ and

$$\begin{aligned} Q(h)=Q(c)-Q(f_1+g_1)=\beta (c)-\mu (f_1)-\nu (g_1)=0. \end{aligned}$$

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

$$\begin{aligned} \alpha (h>t^{-1})=\beta (h>t^{-1})\le Q(h>t^{-1})\le t\,Q(h)=0. \end{aligned}$$

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

$$\begin{aligned} \alpha (c)\le & {} P_t(c)=P_t(f_1+ g_1)+P_t(h)=\mu (f_1)+\nu (g_1)+P_t(h) \\\le & {} \beta (c)+3\,\bigl \{1/t+\mu \bigl [f\,1_{\{f>t\}}\bigr ]+\nu \bigl [g\,1_{\{g>t\}}\bigr ]\bigr \}\quad \text {for all }t>1. \end{aligned}$$

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. $\square$

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

$$\begin{aligned} \sup _{y\in \mathcal {Y}}\,|c(x,y)-c(z,y)|\le \epsilon \quad \text {whenever }x,\,z\in A_i\text { and }i>0. \end{aligned}$$

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

$$\begin{aligned} \mathcal {F}_0=\sigma \bigl (A_0\cap A,\,A_i:A\in \mathcal {F},\,i>0\bigr ),\quad \mu _0=\mu |\mathcal {F}_0, \\c_0(x,y)=1_{A_0}(x)c(x,y)+\sum _{i>0}1_{A_i}(x)c(x_i,y). \end{aligned}$$

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 $f_1+ g_1\le c\le f_2+ g_2$. Since $|c-c_0|\le \epsilon$, then $f_1+ g_1-\epsilon \le c_0\le f_2+ g_2+\epsilon$. Further, $\sup _{A_i}f_1<+\infty$ and $\inf _{A_i}f_2>-\infty$ for each $i>0$. Define

$$\begin{aligned} \phi _1=-\epsilon +1_{A_0}f_1+\sum _{i>0}1_{A_i}\,\Bigl (\sup _{A_i}f_1\Bigr )\quad \text {and}\quad \phi _2=\epsilon +1_{A_0}f_2+\sum _{i>0}1_{A_i}\,\Bigl (\inf _{A_i}f_2\Bigr ). \end{aligned}$$

Then, $\phi _1,\,\phi _2\in L_1(\mu _0)$ and

$$\begin{aligned} \phi _1+ g_1\le c_0\le \phi _2+ g_2. \end{aligned}$$

Because of such inequality and since $c_0$ is $\mathcal {F}_0\otimes \mathcal {G}$-measurable, one can define

$$\begin{aligned} \alpha _0=\inf _{T\in \Gamma (\mu _0,\nu )}T(c_0)\quad \text {and}\quad \beta _0=\sup _{(f,g)}\bigl [\mu _0(f)+\nu (g)], \end{aligned}$$

where $\sup$ is over the pairs (f, g) such that $f\in L_1(\mu _0)$, $g\in L_1(\nu )$ and $f+ g\le c_0$.

Since $\mu _0$ is an atomic probability measure, then $\mu _0$ is perfect, which in turn implies $\alpha _0=\beta _0$. Since $|c-c_0|\le \epsilon$, then $\beta _0\le \beta (c)+\epsilon$. Hence, there is $T\in \Gamma (\mu _0,\nu )$ such that

$$\begin{aligned} T(c_0)<\alpha _0+\epsilon =\beta _0+\epsilon \le \beta (c)+2\epsilon . \end{aligned}$$

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

$$\begin{aligned} \alpha (c)\le P(c)\le \epsilon +P(c_0)=\epsilon +T(c_0)<3\epsilon +\beta (c). \end{aligned}$$

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

$$\begin{aligned} \mu _i(A)=\mu (A\mid A_i)\quad \text { and }\quad \nu _i(B)=T\bigl (\mathcal {X}\times B\mid A_i\times \mathcal {Y}\bigr ), \end{aligned}$$

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

$$\begin{aligned} P=\sum _i\mu (A_i)\,(\mu _i\times \nu _i), \end{aligned}$$

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

$$\begin{aligned} P(A\times B)=\sum _i\mu (A_i)\mu _i(A)\nu _i(B)=\sum _i\mu (A\mid A_i)T(A_i\times B)=T(A\times B). \end{aligned}$$

Therefore, $P=T$ on $\mathcal {F}_0\otimes \mathcal {G}$. $\square$

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

$$\begin{aligned} \sup _{x\in \mathcal {X}}\,|c(x,y)-c(x,z)|\le \epsilon \quad \text {whenever }y,\,z\in B_i\text { and }i>0. \end{aligned}$$

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.,

$$\begin{aligned} \sup _{y\in \mathcal {Y}}\,|c(x,y)-c(z,y)|\le u\,d(x,z)\quad \text {for all }x,\,z\in \mathcal {X}, \end{aligned}$$

(3)

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 $n\ge 1$ and $A_i\in \mathcal {F}$, $B_i\in \mathcal {G}$ such that $A_i\cap A_j=\emptyset$ for $i\ne j$. 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 $H=\cup _nR_n$ and $K=\cap _nR_n$ where $R_n\in \mathcal {R}$ for each n. Then,

$$\begin{aligned} \alpha ^*(H)=\beta ^*(H)\quad \text {and}\quad \alpha (K)=\beta (K). \end{aligned}$$

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

$$\begin{aligned} \mu (\limsup _nA_n)=0\quad \text {or}\quad \nu (\limsup _nB_n)=0. \end{aligned}$$

(Here, $\limsup _nA_n=\cap _n\cup _{j>n}A_j$ and $\limsup _nB_n=\cap _n\cup _{j>n}B_j$.)

Proof

Let $H_n=\cup _{i=1}^nR_i$. Since $\alpha ^*(H_n)=\beta ^*(H_n)$, Lemma 1 implies

$$\begin{aligned} \alpha ^*(H)= & {} \sup _{P\in \Gamma (\mu ,\nu )}P(H)=\sup _{P\in \Gamma (\mu ,\nu )}\sup _nP(H_n)=\sup _n\sup _{P\in \Gamma (\mu ,\nu )}P(H_n) \\= & {} \sup _n\alpha ^*(H_n)=\sup _n\beta ^*(H_n)=\beta ^*(H). \end{aligned}$$

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

$$\begin{aligned} \alpha (K)= & {} 1-\alpha ^*(K^c)=1-\alpha ^*\bigl (\cup _nR_n^c\bigr ) \\= & {} 1-\beta ^*\bigl (\cup _nR_n^c\bigr )=1-\beta ^*(K^c)=\beta (K). \end{aligned}$$

Next, suppose $H=\cup _n (A_n\times B_n)$ and $\mu (\limsup _nA_n)=0$. Let $V_n=\cup _{i=1}^n(A_i\times B_i)$. Given $\epsilon >0$, take $n\ge 1$ 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

$$\begin{aligned} \alpha (H)\le & {} P(H)\le P(V_n)+P\bigl (\cup _{i>n}(A_i\times B_i)\bigr )\le P(V_n)+\mu \bigl (\cup _{i>n}A_i\bigr ) \\< & {} \alpha (V_n)+2\epsilon =\beta (V_n)+2\epsilon \le \beta (H)+2\epsilon . \end{aligned}$$

The proof is exactly the same if $\nu (\limsup _nB_n)=0$. $\square$

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 $\mu (\limsup _nA_n)=0$ or $\nu (\limsup _nB_n)=0$. 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

$$\begin{aligned} \text {for each }n\ge & {} 1\text { there is a measurable function }\phi _n:\mathcal {X}\rightarrow \mathcal {Y}\text { such that} \nonumber \\ \nu= & {} \mu \circ \phi _n^{-1}\quad \text {and}\quad \mu \bigl \{x:(x,\phi _n(x))\in H\bigr \}< 1/n. \end{aligned}$$

(4)

Define in fact $P_n(U)=\mu \bigl \{x:(x,\phi _n(x))\in U\bigr \}$ for each $n\ge 1$ and $U\in \mathcal {H}$. Then, $P_n\in \Gamma (\mu ,\nu )$ and $\alpha (H)\le P_n(H)< 1/n$. 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

$$\begin{aligned} \alpha (H\cap \Delta ^c)=\beta (H\cap \Delta ^c)\quad \text {and}\quad \alpha ^*(H\cap \Delta ^c)=\beta ^*(H\cap \Delta ^c). \end{aligned}$$

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, $\alpha (H\cap \Delta )\le P_1(H\cap \Delta )\le P_1(\Delta )=0$, 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

$$\begin{aligned} \alpha ^*(H\cap \Delta )\le \beta ^*(H\cap \Delta )\le \mu \bigl (\cup _n(A_n\cap B_n)\bigr )=P_2(H\cap \Delta )\le \alpha ^*(H\cap \Delta ). \end{aligned}$$

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

$$\begin{aligned} H_n\in \mathcal {H}_0\text { and }H_n\subset H_{n+1}\text { for each }n\quad \Longrightarrow \quad \alpha \bigl (\cup _nH_n\bigr )=\beta (\cup _nH_n\bigr ). \end{aligned}$$

(5)

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. $\square$

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 $c\ge 0$ 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 $c\ge 0$ 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

$$\begin{aligned} P\bigl (\cup _i R_i\bigr )=\sup _n P\bigl (\cup _{i=1}^n R_i\bigr )=\sup _n Q\bigl (\cup _{i=1}^n R_i\bigr )\le Q\bigl (\cup _i R_i\bigr ) \end{aligned}$$

provided $R_i\in \mathcal {R}$ for all i. Hence, $P(c>\epsilon )\le Q(c>\epsilon )=0$ 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.,

$$\begin{aligned} \alpha (c)=\inf _{P\in \Gamma (\mu ,\nu )}\overline{P}(c)\quad \text {where }\overline{P}\text { is the completion of }P. \end{aligned}$$

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

$$\begin{aligned} c_n(x,y)=\inf _{z\in D}\bigl \{n\,d(x,z)+c(z,y)\bigr \}, \end{aligned}$$

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

Since c is bounded below, $c_n$ is real-valued, and a direct calculation shows that

$$\begin{aligned} \sup _{y\in \mathcal {Y}}\,|c_n(x,y)-c_n(z,y)|\le n\,d(x,z)\,\,\,\text { for all }x,\,z\in \mathcal {X}. \end{aligned}$$

(6)

Since D is countable, $y\mapsto c_n(x,y)$ is Borel measurable. Hence, $c_n\,1_A$ is $\mathcal {H}$-measurable. In addition, $c_n\le c_{n+1}$ 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

$$\begin{aligned} c(x,y)=\sup _nc_n(x,y)\quad \text {for all }(x,y)\in \overline{A}\times \mathcal {Y}. \end{aligned}$$

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 $n\ge 1$ 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,

$$\begin{aligned} \alpha ^*(c)= & {} \alpha ^*(c\,1_A)\le \beta ^*(c\,1_A)=\lim _n\beta ^*(c_n\,1_A) \\= & {} \lim _n\alpha ^*(c_n\,1_A)\le \alpha ^*(c\,1_A)=\alpha ^*(c). \end{aligned}$$

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 $\psi +g\le c$. Define

$$\begin{aligned} f(x)=\inf _{y\in \mathcal {Y}}\bigl \{c(x,y)-g(y)\bigr \},\quad x\in \mathcal {X}, \end{aligned}$$

and note that $f\in L_1(\mu )$, $f+g\le c$ and f is upper-semicontinuous. Define also

$$\begin{aligned} c_n(x,y)=\inf _{z\in \mathcal {X}}\bigl \{n\,d(x,z)+c(z,y)-f(z)\bigr \}. \end{aligned}$$

Again, condition (6) holds, $c_n$ meets condition (1) (since $g\le c_n\le c-f$) 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,

$$\begin{aligned} \beta ^*(c)-\mu (f)= & {} \beta ^*(c-f)=\lim _n\beta ^*(c_n) \\= & {} \lim _n\alpha ^*(c_n)\le \alpha ^*(c-f)=\alpha ^*(c)-\mu (f). \end{aligned}$$

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

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

$$\begin{aligned} c_n(x,y)=\inf _{z\in \mathcal {Y}}\bigl \{n\,\rho (y,z)+c(x,z)-g(z)\bigr \}, \end{aligned}$$

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. $\square$

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,

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

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

• (iii) $\alpha (c)=\beta (c)$ 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

  $$\begin{aligned} \text {card}(\mathcal {X})\le \text {card}(\mathbb {R})\quad \text {and}\quad \text {card}(\mathcal {Y})\le \text {card}(\mathbb {R}). \end{aligned}$$

  Under such cardinality assumption, in fact, perfectness implies separability but not conversely; see Sect. 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, Ramachandran and Ruschendorf [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 Sect. 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

  $$\begin{aligned} \int d(x,x_0)\,\mu (dx)+\int d(x,x_0)\,\nu (dx)<\infty \quad \text {for some }x_0\in \mathcal {X}. \end{aligned}$$

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

  $$\begin{aligned} \beta (d)=\sup _f\,|\mu (f)-\nu (f)|, \end{aligned}$$

  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.