Proof Assume that $\mu $ is continuous. Take $(x,y)\in X \times Y.$ It follows that

$$\begin{align*}\mu(\{x\})=\gamma\big (\{x\} \times Y\big)\geq \gamma\big(\{x\} \times \{y\}\big),\end{align*}$$

from which the desired result follows. The proof is similar if $\nu $ is continuous.

Proof of Theorem 1.2

We assume that $\mu $ is a continuous measure. It follows from Lemma 2.7 that $\gamma $ is also continuous. It follows from Theorem 2.6 that the Borel measurable spaces $(X, \mathcal {B}(X), \mu )$ and $(X \times Y, \mathcal {B}(X \times Y), \gamma )$ are isomorphic. Thus, there exists an isomorphism $T=(T_1, T_2)$ from $(X, \mathcal {B}(X), \mu )$ onto $(X \times Y, \mathcal {B}(X \times Y), \gamma )$ . It can be easily deduced that $T_1 : X\to X$ and $T_2: X \to Y$ are surjective maps and

$$\begin{align*}{(T_1)}_\# \mu=\mu \quad \& \quad (T_2)_\# \mu=\nu.\end{align*}$$

Consider the convex set

$$\begin{align*}\mathcal{M}(T_1, \mu)=\big\{ \lambda \in \mathcal{P}(X); \, (T_1)_\# \lambda=\mu\big \},\end{align*}$$

and note that $\mu \in \mathcal {M}(T_1, \mu ).$ Since $ \mu \in \mathcal {M}(T_1, \mu )$ , it follows from Theorem 2.4 that there exists a probability measure $\xi $ on $\sum _{\mathrm{ext} \, M(T_1, \mu )}$ such that for each $B \in \mathcal {B}(X)$ ,

(1) $$ \begin{align} \mu(B)=\int_{\mathrm{ext} \, M(T_1, \mu)} \varrho(B)\, d\xi(\varrho), \qquad \big (\varrho \to \varrho(B) \text{ is measurable}\big).\end{align} $$

Since $\Gamma =\mathrm{spt}(\gamma ),$ it follows that $T^{-1}(\Gamma )$ is a measurable subset of X with ${\mu \big (T^{-1}(\Gamma ) \big )=1.}$ Let $A_\gamma \in \mathcal {B}(X)$ be the set such that $A_\gamma \subseteq T^{-1}(\Gamma )$ and for all $ x \in A_\gamma $ , the cardinality of the set $\Gamma _x$ does not exceed m. It follows from the assumption that $\mu (A_\gamma )=1.$ Since $\mu (X\,{\backslash}\, A_\gamma )=0,$ it follows from (1) that

$$\begin{align*}\int_{\mathrm{ext} \, M(T_1, \mu)} \varrho(X_1\,{\backslash}\, A_\gamma)\, d\xi (\varrho)=\mu(X\,{\backslash}\, A_\gamma)=0,\end{align*}$$

and therefore there exists a $\xi $ -full measure subset $K_\gamma $ of $\mathrm{ext} \, M(T_1, \mu )$ such that ${\varrho (X\,{\backslash}\, A_\gamma )=0}$ for all $\varrho \in K_\gamma .$ Let $\mathcal {S}(T_1)$ be the set of all sections of $T_1$ and define

$$\begin{align*}\mathcal{K}:=\big \{F\in \mathcal{S}(T_1);\,\, \exists \varrho \in K_\gamma \text{ with } \mu=F_\# \varrho \big \}.\end{align*}$$

Let $\mathcal {G}(\mathcal {K})$ be the set of all measurable sections of $T_1$ generated by $\mathcal {K}$ as in Definition 2.2. By Proposition 2.1, there exists a sequence $\{F_i\}_{i=1}^k \subset \mathcal {G}(\mathcal {K})$ with $ k \in \mathbb {N} \cup \{\infty \}$ satisfying assertions (1)–(3) in that proposition. Let $B_\gamma :=\cap _{i=1}^{k}F_i^{-1}(A_\gamma ),$ and for each $k \in \mathbb {N}\cup \{\infty \}$ , define

Let $\varrho _i:={F_i}_\# \mu $ for each $i \in \mathbb {N}_k.$ We shall now proceed with the proof in several steps.

Step I: In this step, we show that $\mu \big (B_\gamma \big )=1$ and

(2) $$ \begin{align} \big (x, T_2 \circ F_i( x) \big ) \in \Gamma, \qquad \,\,\forall x \in B_\gamma, \, \, \, \forall i \in \mathbb{N}_k. \end{align} $$

Note first that $\varrho _i(X\,{\backslash}\, A_\gamma )=0$ for each $i \in \mathbb {N}_k$ . In fact, for a fixed $i\in \mathbb {N}_k,$ since ${F_i \in \mathcal {G}(\mathcal {K})}$ there exists a sequence $\{F_{\sigma _j}\}_{j=1}^{\infty } \subset \mathcal {K}$ such that $X=\cup _{j=1}^{\infty }A_j$ , where

$$\begin{align*}A_j=\{x \in X; \, \, F_i(x)=F_{\sigma_j}\}. \end{align*}$$

Let $\sigma _j \in K_{\gamma }$ be such that the map $F_{\sigma _j}$ is a push-forward from $\sigma _j$ to $\mu .$ It follows that

$$ \begin{align*} \varrho_i(X\,{\backslash}\, A_\gamma)=\mu\big (F_i^{-1}(X\,{\backslash}\, A_{\gamma})\big )&=\mu\big ((\cup_{j=1}^{\infty}A_j) \cap F_i^{-1}(X\,{\backslash}\, A_{\gamma})\big )\\ &\leq \sum_{j=1}^\infty \mu\big (A_j \cap F_i^{-1}(X\,{\backslash}\, A_{\gamma})\big )\\ &=\sum_{j=1}^\infty \mu\big (A_j \cap F_{\sigma_j}^{-1}(X\,{\backslash}\, A_{\gamma})\big )\\ &\leq \sum_{j=1}^\infty \mu\big ( F_{\sigma_j}^{-1}(X\,{\backslash}\, A_{\gamma})\big )=\sum_{j=1}^\infty \sigma_j(X\,{\backslash}\, A_{\gamma})=0. \end{align*} $$

This proves that $\varrho _i(X\,{\backslash}\, A_\gamma )=0.$ Since $\varrho _i$ is a probability measure, we have that $\varrho _i(A_\gamma )=1$ for every $i \in \mathbb {N}_k.$ Therefore, $\mu \big (F_i^{-1}(A_\gamma )\big )=\varrho _i(A_\gamma )=1$ . This implies that $\mu (B_\gamma )=\mu \big (\cap _{i=1}^{k}F_i^{-1}(A_\gamma )\big )=1.$ We shall now prove that

$$\begin{align*}\big (x, T_2 \circ F_i (x)\big ) \in \Gamma, \qquad \,\,\forall x \in B_\gamma, \, \, \, \forall i \in \mathbb{N}_k. \end{align*}$$

Since for all $x \in A_\gamma $ , we have $T(x)=(T_1 x, T_2x) \in \Gamma $ , it follows that for each $i \in \mathbb {N}_k,$

$$\begin{align*}\big (T_1\circ F_i (x), T_2 \circ F_i (x)\big ) \in \Gamma, \qquad \forall x \in F_i^{-1}(A_\gamma), \end{align*}$$

from which together with $T_1 \circ F_i=Id_X$ one obtains

(3) $$ \begin{align} \big (x, T_2 \circ F_i (x) \big ) \in \Gamma, \qquad \forall x \in F_i^{-1}(A_\gamma). \end{align} $$

Thus,

$$\begin{align*}\big (x, T_2 \circ F_i (x)\big ) \in \Gamma, \qquad \,\,\forall x \in B_\gamma, \, \, \, \forall i \in \mathbb{N}_k. \end{align*}$$

This completes the proof of Step I.

Step II: In this step, we assume that assumption (1) of the theorem holds. In this case, we show that $k \leq m.$

To do this, let us assume that $k>m. $ It follows from Step I that

(4) $$ \begin{align} \big (x, T_2 \circ F_i (x)\big ) \in \Gamma, \qquad \,\,\forall x \in B_\gamma, \,\, \, \forall i\in \{1,\dots,m+1\}. \end{align} $$

Note that by assertion $(1)$ in Proposition 2.1, we have $\mu (B_{m+1})>0.$ Since $\mu (B_\gamma )=1$ and $\mu (B_{m+1})>0,$ it follows that $B_\gamma \cap B_{m+1}\not =\emptyset .$ Take $x \in B_\gamma \cap B_{m+1}.$ We have that the cardinality of the set $\Gamma _x$ is at most m. On the other hand, it follows from (4) that $T_2 \circ F_i(x) \in \Gamma _x$ for all $i \in \{1,2,\dots ,m+1\}.$ Thus, there exist $i, j \in \{1,2,\dots ,m+1\}$ with $i <j$ such that $T_2 \circ F_i(x)=T_2 \circ F_j(x).$ Since $T_1 \circ F_i=T_1 \circ F_j=Id_X$ and the map ${T=(T_1,T_2)}$ is injective, it follows that $F_{i}(x)=F_j(x).$ On the other hand, ${x \in B_{m+1} \subseteq B_j}$ from which we have $F_j(x)\not \in \{F_1(x),\dots ,F_{j-1}(x)\}.$ This leads to a contradiction and therefore $k\leq m$ in this case.

Step III: In this step, we assume that assumption (2) of the theorem holds. In this case, we prove that if $k=\infty $ , then $\mu (\cap _{i=1}^\infty B_i)=0.$

To prove this, let us assume that $ k=\infty $ and $\mu (\cap _{i=1}^\infty B_i)>0.$ By Step I, we have that $\mu (B_\gamma )=1$ and

(5) $$ \begin{align} \big (x, T_2 \circ F_i (x)\big ) \in \Gamma, \qquad \,\,\forall x \in B_\gamma, \, \, \, \forall i \in \mathbb{N.} \end{align} $$

Take $x \in \big ( \cap _{i=1}^\infty B_i\big ) \cap B_\gamma. $ It follows from (5) that $T_2 \circ F_i x \in \Gamma _x$ for each $i \in \mathbb {N}.$ On the other hand, by assumption, we have that $card(\Gamma _x)<\infty .$ Thus, there exist $i, j$ with $i <j$ such that $T_2 \circ F_i(x)=T_2 \circ F_j(x).$ As in Step II, since $T_1 \circ F_i=T_1 \circ F_j=Id_X$ and the map ${T=(T_1,T_2)}$ is injective, it follows that $F_{i}(x)=F_j(x).$ On the other hand, $x \in \cap _{i=1}^\infty B_i \subseteq B_j$ from which we have $F_j(x)\not \in \{F_1(x),\dots ,F_{j-1}(x)\}.$ This leads to a contradiction and Step III follows.

It now follows from Steps II and III that either $k\not =\infty $ or, if $k=\infty $ , then ${\mu (\cap _{i=1}^\infty B_i)=0}$ . On the other hand, Proposition 2.1 yields that if either $k\not =\infty $ or, $k=\infty $ and $\mu (\cap _{i=1}^\infty B_i)=0$ , then for every $F \in \mathcal {G}(\mathcal {K})$ , the measure $\varrho _F=F_\#\mu $ is absolutely continuous with respect to the measure $\sum _{i=1}^k \varrho _i $ , where $\varrho _i={F_i}_\#\mu $ for $i \in \mathbb {N}_k.$ This together with the representation

$$ \begin{align*} \mu(B)=\int_{\mathrm{ext} \, M(T_1, \mu)} \varrho(B)\, d\xi(\varrho)= \int_{K_{\gamma}} \varrho(B)\, d\xi(\varrho), \qquad \big (\forall B \in \mathcal{B}(X)\big), \end{align*} $$

imply that $\mu $ is absolutely continuous with respect to $\sum _{i=1}^k \varrho _i.$ It then follows that there exists a nonnegative measurable function such that

$$\begin{align*}\frac{d \mu}{d \big(\sum_{i=1}^k \varrho_i\big)}=\alpha. \end{align*}$$

Define $\alpha _i=\alpha \circ F_i$ for $i\in \mathbb {N}_k.$ We show that $ \sum _{i=1}^k\alpha _i(x)=1$ for $\mu $ -almost every $x \in X.$ In fact, for each $B \in \mathcal {B}(X),$ we have

$$ \begin{align*} \mu(B)&=\mu(T_1^{-1}(B))\\[5pt]&=\sum_{i=1}^k\int_{T_1^{-1}(B)} \alpha(x) \, d\varrho_i =\sum_{i=1}^k\int_{F_i^{-1}\circ T_1^{-1}(B)} \alpha(F_ix) \, d \mu =\sum_{i=1}^k\int_{B} \alpha_i(x) \, d \mu, \end{align*} $$

from which we obtain $\mu (B)=\sum _{i=1}^k\int _{B} \alpha _i(x) \, d \mu .$ Since this holds for all $B \in \mathcal {B}(X),$ we have

$$\begin{align*}\sum_{i=1}^k\alpha_i(x)=1, \qquad \quad \mu-a.e. \end{align*}$$

It now follows from Proposition 2.2 that each $F_i$ is $\mu $ -a.e. equal to a $(\mathcal {B}(X),\mathcal {B}(X))$ -measurable function for which we still denote it by $F_i.$ For each $i \in \mathbb {N}_k,$ let $G_i=T_2 \circ F_i.$ We now show that $\gamma =\sum _{i=1}^k (\text {Id}\times G_i)_\# (\alpha _i \mu )$ . For each bounded continuous function , it follows that

$$ \begin{align*} \int_{X \times Y} f(x,y) \, d \gamma=\int_X f(T_1x,T_2x) \, d\mu&= \sum_{i=1}^k\int_X \alpha(x)f(T_1x,T_2x) \, d\varrho_i\\[5pt] &= \sum_{i=1}^k\int_X \alpha\big (F_i (x)\big ) f\big (T_1\circ F_i (x),T_2 \circ F_i (x) \big ) \, d\mu\\[5pt] &= \sum_{i=1}^k\int_X \alpha_i(x) f\big (x,G_i (x) \big ) \, d\mu. \end{align*} $$

Therefore,

$$\begin{align*}\gamma =\sum_{i=1}^k (\text{Id}\times G_i)_\# (\alpha_i \mu).\\[-42pt]\end{align*}$$

Proof Let $Y=X_2\times \cdots \times X_n$ and $\nu $ be the projection of $\gamma $ on $Y.$ It follows that ${\gamma \in \Pi (\mu _1, \nu ).}$ Since $\mu _1$ is continuous the desired result follows from Theorem 1.2.

Proof By standard results in the theory of optimal transportation, there exist measurable functions and with

(7) $$ \begin{align}\psi(y)=\inf_{x \in \mathcal{T}}\{c(x,y)-{\varphi}(x)\} \qquad and \qquad {\varphi}(x)=\inf_{y \in \mathcal{T} }\{c(x,y)-\psi(y)\}, \end{align} $$

such that for any optimal plan $\gamma $ of (6),

$$\begin{align*}\mathrm{spt}(\gamma)\subseteq \big \{(x,y) \in \mathcal{T} \times \mathcal{T}\; :\, {\varphi}(x)+\psi(y)= c(x,y) \big \}.\end{align*}$$

Since $\mathcal {T}$ is bounded, it follows from Lemma C.1 in [Reference Gangbo and McCann3] that ${\varphi }$ is locally Lipschitz on $\mathcal {T}$ . Let $M=\mathrm{Dom} (D {\varphi }).$ It follows from Rademacher’s theorem together with the absolute continuity of $\mu $ that $\mu (M)=1.$ For $x_0 \in M$ , if there exist $y_0, y \in \mathcal {T}$ with $(x_0,y_0)$ and $ (x_0,y) \in \mathrm{spt}(\gamma ),$ then we must have $D_1c(x_0,y_0)=D_1c(x_0,y).$ Let $\vec {N}(x_0)$ be the outward normal vector at $x_o.$ If

$$\begin{align*}D_1c(x_0,y_0)=D_1c(x_0,y),\end{align*}$$

then $y-y_0=\alpha \vec {N}(x_0) $ for some This implies that $y=y_0+ \alpha \vec {N}(x_0).$ The latter argument shows that all the points in the set

$$\begin{align*}\Big\{y \in \mathcal{T};\, D_1c(x_0,y_0)=D_1c(x_0,y)\Big\},\end{align*}$$

live on a straight line through $y_0$ and parallel to the normal vector $\vec {N}(x_0).$ On the other hand, one can easily observe that any straight line can intersect the manifold $\mathcal {T}$ in at most eight points. This proves that $\text {card} \big (\Gamma _x \big )\leq 8 $ is for $\mu $ -a.e. $x\in \mathcal {T}$ where ${\Gamma _x=\big \{y \in \mathcal {T}; \, (x,y)\in \mathrm{spt} (\gamma )\big \}.}$ Therefore, by Theorem 1.2, there exist $k\in \{1,2,\dots ,8\}$ and a sequence of Borel measurable maps $\{G_i\}^k_{i=1}$ from $\mathcal {T}$ to $\mathcal {T}$ such that the measure $\gamma $ is concentrated on their graphs.