Monge transport problem. Let X, Y be Radon spaces. Given measures $\mu \in \mathscr {P}(X)$ , $\nu \in \mathscr {P}(Y)$ and a fixed Borel cost function $c: X \times Y \to [0, \infty ]$ , minimize

$$ \begin{align*} T \mapsto \int_{X} c(x, T(x)) ~d\mu \end{align*} $$

among all maps T such that $T_{*}\mu = \nu $ .

Monge–Kantorovich transport problem. Let X, Y be Radon spaces. Given measures $\mu \in \mathscr {P}(X)$ , $\nu \in \mathscr {P}(Y)$ and a fixed Borel cost function $c: X \times Y \to [0, \infty ]$ , minimize

(1) $$ \begin{align} \gamma \mapsto \int_{X \times Y} c(x, y) ~d\gamma(x, y) \end{align} $$

among all measures $\gamma \in \mathscr {P}(X \times Y)$ with marginals $\mu $ and $\nu $ , i.e. $\gamma $ satisfying $(\mathrm {proj}_{X})_{*}\gamma =\mu $ and $(\mathrm {proj}_{Y})_{*}\gamma =\nu $ , where $\mathrm {proj}_{X}$ and $\mathrm {proj}_{Y}$ are the canonical projections $(x, y) \mapsto x$ and $(x, y) \mapsto y$ , respectively.

Proof. We shall first consider the disintegration of measures on compact metric spaces. Then, we tackle the general case as it is stated.

Step 1: To get started, consider X to be a compact metric space with its Borel $\sigma $ -algebra $\mathcal {B}(X)$ and let $\mu $ be a Radon measure on X. If $(Y, \mathcal {E})$ is a measurable space, we define a measurable map $q: \mathcal {B}(X) \to \mathcal {E}$ so that we set $\nu =q_{*}\mu $ . Let $C(X)$ be the set of all continuous real functions $\omega :X \to \mathbb {R}$ . Then, for each $\omega \in C(X)$ , we associate a measure $\unicode{x3bb} $ given by

$$ \begin{align*} \unicode{x3bb}(A)= \int\limits_{q^{-1}(A)} \omega (x) ~d\mu (x) \end{align*} $$

for every $A \in \mathcal {E}$ . The measure $\unicode{x3bb} $ is absolutely continuous w.r.t. $\nu $ . Indeed, for every $A \in \mathcal {E}$ , we have that $\nu (A) = \mu (q^{-1}(A)) = 0$ implies $\unicode{x3bb} (A)=0$ . Therefore, since $\nu $ and $\unicode{x3bb} $ are positive measures and $\unicode{x3bb} \ll \nu $ , by the Radon–Nikodym theorem, there exists $h: Y \to [0, \infty ]$ , the density of $\unicode{x3bb} $ w.r.t. $\nu $ , such that $\unicode{x3bb} (A) = \int _{A} h ~d\nu $ for $A \in \mathcal {E}$ . Thus,

(2) $$ \begin{align} \int\limits_{A} h(y) ~d\nu(y) = \int\limits_{q^{-1}(A)} \omega(x) ~d\mu(x). \end{align} $$

Recall that $C(X)$ is a separable space with the supremum norm. Let $\mathcal {H} = \{ \omega _1 \equiv 1, \omega _2, \omega _3, \ldots \}$ be a dense subset of $C(X)$ . Suppose, without loss of generality, that $\mathcal {H}$ is a vector space over $\mathbb {Q}$ . Then, for each $n \in \mathbb {N}$ , we consider the Radon–Nikodym density $h_{n}$ associated with $\omega _n$ given by equation (2) so that for each $A \in \mathcal {E}$ ,

(3) $$ \begin{align} \int\limits_{A} h_{n}(y) ~d\nu(y) = \int\limits_{q^{-1}(A)} \omega_n(x) ~d\mu(x). \end{align} $$

Note that $h_{n} \geq 0$ almost always (a.a.), if $\omega _n \geq 0$ , since $\nu =q_{*}\mu $ .

Step 2: We will use the associated densities to construct a linear functional which will be extended using the Hahn–Banach theorem. To do so, we denote by $\mathbb {A}$ the set of all $y \in Y$ , such that, if $\omega _i= \alpha \omega _j + \beta \omega _k$ , then we have for their associated densities that the relation $h_{i}(y) = \alpha h_{j}(y) + \beta h_{k}(y)$ holds true, where $\alpha , \beta \in \mathbb {Q}$ and the associated density to $\omega _{1}$ is set to $h_{1}(y) = 1$ . The set $\mathbb {A}$ is measurable and $\nu (\mathbb {A})=1$ . Indeed,

$$ \begin{align*} \int_{q^{-1}(Y)} \omega_i(x) ~d\mu &= \int_{q^{-1}(Y)} (\alpha \omega_j + \beta \omega_k)(x)\,d\mu \\&= \alpha \bigg(\! \int_{q^{-1}(Y)}\! \omega_j(x) \,d\mu \!\bigg) + \beta \bigg(\! \int_{q^{-1}(Y)} \!\omega_k(x) ~d\mu \!\bigg),\! \text{ which by equation }(3),\\&= \alpha \bigg( \int_Y h_{j}(y) ~d\nu \bigg) + \beta \bigg( \int_Y h_{k}(y) ~d\nu \bigg) \\&= \int_Y \alpha h_{j}(y) + \beta h_{k}(y) ~d\nu, \end{align*} $$

so $ \int h_{i}(y)\,d\nu = \int \alpha h_{j}(y) + \beta h_{k}(y)\,d\nu $ . Consequently, $h_{i}(y) = \alpha h_{j}(y) + \beta h_{k}(y) \nu $ -a.a. Therefore, whenever $\omega _{n}$ is a linear combination of elements of $\mathcal {H}$ , then the associated densities defined by equation (3) can also be written point-wise as linear combinations of Radon–Nikodym densities.

Step 3: For each $y \in \mathbb {A}$ , we define the functional $\tilde {\varphi _y}: \mathcal {H} \to \mathbb {R}$ , given by $\tilde {\varphi _y}(\omega _n):=h_{n}(y)$ . Note that $\tilde {\varphi _y}: \mathcal {H} \to \mathbb {R}$ is $\mathbb {Q}$ -linear with $\| \tilde {\varphi _y} \| \leq 1$ and, since $\tilde {\varphi _y} (1)=1$ , we have actually that $\| \tilde {\varphi _y} \| = 1$ . Therefore, by the Hahn–Banach theorem, $\tilde {\varphi _y}$ can be extended to a continuous positive linear functional $\varphi _y: C(X) \to \mathbb {R}$ , with $\| \varphi _y \| = 1$ . Furthermore, the Riesz–Markov–Kakutani representation theorem assures that there exists a unique Radon measure $\mu _y$ on X such that $\varphi _y(\omega )= \int \omega ~d \mu _y$ for every $\omega \in C(X)$ and $\varphi _y (1)=1$ . Note that $\int \omega ~d \mu _y \leq 1$ , since $\| \varphi _y \| = 1$ . Thus, we conclude that $\mu _y$ is a probability measure. Note also that $\mu _y$ is supported on $q^{-1} \{ y \in \mathbb {A}\}$ . For $y \notin \mathbb {A}$ , consider $\mu _y = 0$ .

Step 4: Observe that $y \mapsto \int _{X} \omega _n\,d\mu _y$ is $\mathcal {E}$ -measurable for every $\omega _n \in \mathcal {H}$ and

$$ \begin{align*} \int_{Y} \int_{q^{-1}(y)} \omega_n(x) \,d\mu_y ~d\nu = \int_{Y} \varphi_y(\omega_n) \,d\nu = \int_{Y} h_{n}(y)\,d\nu = \int_{X} \omega_n(x)\,d \mu. \end{align*} $$

By the definition of $\mathcal {H}$ , we have that for each $\omega \in C(X)$ , there exists a sequence $(\omega _{i})_{i}$ , with $\omega _i \to \omega $ uniformly. So, by uniform convergence, we have that $y \mapsto \int _{X} \omega ~d\mu _y$ is $\mathcal {E}$ -measurable for every $\omega \in C(X)$ . Furthermore,

$$ \begin{align*} \int_{Y} \int_{q^{-1}(y)} \omega(x)\,d\mu_y \,d\nu = \int_{X} \omega(x)\,d \mu. \end{align*} $$

The same holds true for any bounded and $\mathcal {B}(X)$ -measurable function $\omega $ . Indeed, denote by $\mathscr {C}$ the class of functions such that $y \mapsto \int _{X} \omega \,d\mu _y$ is $\mathcal {E}$ -measurable and $\int _{Y} \int _{q^{-1}(y)} \omega (x)\,d\mu _y \,d\nu = \int _{X} \omega (x)\,d \mu $ . Note that $C(X) \subset \mathscr {C}$ , from what was shown before. If $A \subset X$ is an open set, then . So, let $\mathcal {D}$ be the set of all characteristic functions in $\mathscr {C}$ . If for $n \in \mathbb {N}$ , we have that . If , we have that . Thus, the class of measurable sets whose characteristic functions are in $\mathcal {D}$ is $\mathcal {B}(X)$ . Therefore, the result follows by monotone convergence. This completes the proof of Theorem A for compact spaces.

Step 5: Let X be a locally compact and separable metric space. Then, the set of continuous real-valued functions with compact support on X, denoted by $C_c(X)$ , is a vector space. Such a vector space can be seen as the union of the spaces $C_c(K_{i})$ of continuous functions with support on compact sets $K_{i}$ . Since $\mu $ is a Radon measure on X, the map $\varphi : C_c(X) \to \mathbb {R}$ , such that, $\omega \mapsto \int _{X} \omega (x)\,d\mu $ , is a continuous positive linear map. Note also that $\mu $ is supported on a set $\tilde {\mathcal {K}}$ , which is a countable union of compact subsets $K_{i} \subset X$ . So, we can imbed X into a compact metric space $\mathcal {K}$ and identify $\mu $ with a measure on $\mathcal {K}$ with support on $\tilde {\mathcal {K}}$ , and construct the measures $\mu _y$ as above. Consider $\mu _{y}=0$ for y such that $\pi ^{-1}(y) \notin \tilde {\mathcal {K}}$ . Thus, there exist probability measures $\mu _{y}$ on X such that each $\mu _{y}$ is supported on $\pi ^{-1}(y)$ and, for every $\omega \in C_c(X)$ ,

(4) $$ \begin{align} \int_{Y} \int_{\pi^{-1}(y)} \omega(x)\,d\mu_y \,d\nu = \int_{X} \omega(x)\,d \mu. \end{align} $$

Since Y is a locally compact and separable metric space and $\pi : X \to Y$ is a Borel map, then $y \mapsto \mu _y$ is a Borel map for $\nu $ -a.e. $y \in Y$ . Furthermore, note that equation (4) is equivalent to say that $\mu (A)= \int _{Y} \mu _{y} (A)\,d\nu (y)$ for every $A \in \mathcal {B}(X)$ and $\mu _{y}$ is concentrated on $\pi ^{-1}(y)$ for $\nu $ -a.e. $y \in Y$ . This concludes the proof.

Proof. Let $\eta _{x}$ and $\eta _{x}'$ be measures satisfying conditions (1) and (2). Let $(A_{n})_{n}$ be a sequence of open sets such that the finite intersection is also an open set which generates $\mathcal {B}(X \times Y)$ , the Borel $\sigma $ -algebra of $X \times Y$ . Consider $B \in \mathcal {B}(X \times Y)$ and $A=A_n \cap ~{\mathrm {proj}}_{X}^{-1}(B)$ for any $n \in \mathbb {N}$ . By condition (1), we have that

$$ \begin{align*} \gamma(A)=\int_{X} \eta_{x}(A)\,d\nu = \int_{B} \eta_{x}(A_n)\,d \nu \end{align*} $$

and

$$ \begin{align*} \gamma(A)=\int_{X} \eta_{x}' (A)\,d\nu = \int_{B} \eta_{x}' (A_n)\,d \nu. \end{align*} $$

Therefore,

$$ \begin{align*} \int_{B} \eta_{x}(A_n)\,d \nu = \int_{B} \eta_{x}' (A_n)\,d \nu. \end{align*} $$

Given that B is arbitrary, then $\eta _{x}(A_n) = \eta _{x}'(A_n)$ for $\nu $ -a.e. $x \in X$ and for any $n \in \mathbb {N}$ . So, there exists a set $N \subset X$ , with $\nu (N)=0$ such that $\eta _{x}(A_n) = \eta _{x}'(A_n)$ for any $n \in \mathbb {N}$ , $x \in (X-N)$ . Hence, for $\nu $ -a.e. $x \in X$ , $\eta _{x} = \eta _{x}'$ .

Let us denote $C=\{ x \in X : \eta _x(X \times Y)>0 \}$ . Let $\mathcal {G} \subset C$ be such that $\mu (\mathcal {G})=0$ . So, ${\mathrm {proj}}_{X}^{-1}(\mathcal {G})$ is such that $\gamma ({\mathrm {proj}}_{X}^{-1}(\mathcal {G}))=0$ . Therefore, condition (2) implies

$$ \begin{align*} 0=\int_{X} \eta_{x}({\mathrm {proj}}_{X}^{-1}(\mathcal{G}))\,d\nu = \int_{\mathcal{G}} \eta_{x} (X \times Y)\,d\nu. \end{align*} $$

Since $\eta _{x} (X \times Y)>0$ in $C \supset \mathcal {G}$ , we have that $\nu (\mathcal {G})=0$ . That is, $\nu |_{C} \ll \mu $ . Moreover, writing $\nu |_{C}=f \mu $ implies that $\gamma = f \mu \otimes \eta _{x}$ . However, by Corollary 3.1, $\gamma = \mu \otimes \gamma _{x}$ and then $({\nu |_{C}}/{\mu }) \eta _{x} = \gamma _{x}$ .

Proof. Denote by $\text {Lip}_{1}(Y)$ the set of Lipschitz functions whose Lipschitz constants are less than or equal to $1$ . By the Arzelà–Ascoli theorem, the space $\text {Lip}_{1}(Y)$ is compact with respect to the uniform convergence [Reference Ambrosio, Gigli and SavaréAGS05, Proposition 3.3.1]. Let $D \subset \text {Lip}_{1}(Y)$ be a countable dense subset and take $\varphi \in \text {Lip}_{1}(Y)$ . If the measures $\nu _{1}$ , $\nu _{2} \in \mathscr {P}(Y)$ have bounded support, we can use the duality formula to obtain

$$ \begin{align*} W_{1}(\nu_{1}, \nu_{2}) = \sup\limits_{\varphi \in Lip_{1}(Y)} \bigg\{\! \int_{Y} \varphi ~d(\nu_{1} -\nu_{2}) \bigg\} = \sup\limits_{\varphi \in D} \int_{Y} \varphi \,d(\nu_{1}-\nu_{2}); \end{align*} $$

see [Reference Ambrosio, Gigli and SavaréAGS05, §7.1] for details. For all $\varphi \in \text {Lip}_{1}(Y)$ ,

$$ \begin{align*} \psi_{\varphi}: X &\to \mathbb{R} \\ x &\mapsto \int_{Y} \varphi \,d(\nu-f(x)) \end{align*} $$

is Borel. For f to be a Borel map, it suffices that

$$ \begin{align*} f^{-1}(B(\nu, r))= \bigcap\limits_{\varphi \in D} \psi_{\varphi}^{-1}((-r, r)):=A, \end{align*} $$

where $B(\nu , r)$ is an open ball of radius r centred at $\nu $ in $(\mathscr {P}(Y), W_{1})$ . In fact, if $x \in A$ , then $|\psi _{\varphi }(x)|<r$ for all $\varphi \in D$ and $W_{1}(\nu , f{(x)})<r$ (by the definition of $\psi _{\varphi }$ ), so that $f{(x)} \in B(\nu , r)$ . In the same way, if $f{(x)} \in B(\nu , r)$ , then $W_{1}(\nu , f{(x)})<r$ and, by the duality formula,

$$ \begin{align*} |\psi_{\varphi}(x)|:=\Big|\int_{Y} \varphi\,d(\nu-f{(x)})\Big|<r \end{align*} $$

for every $\varphi \in D$ , so that $x \in A$ . Thus, one way is proven. Conversely, let $A \subset Y$ be an open subset. Note that

Let $I_{\varphi }$ be a function given by

(5) $$ \begin{align} I_{\varphi}: (\mathscr{P}(Y), W_{1}) & \to \mathbb{R} \nonumber \\ \unicode{x3bb} & \mapsto \int_{Y} \varphi(y) \,d\unicode{x3bb}, \end{align} $$

where $\varphi $ is a lower semicontinuous function over Y. Since $W_{1}$ metrizes the weak* topology of $\mathscr {P}(Y)$ [Reference Ambrosio, Gigli and SavaréAGS05, Ch. 7], the function $I_{\varphi }$ is lower semicontinuous. For every $x \in X$ , one obtains $\int _{Y} \varphi (y) ~df{(x)}=I_{\varphi }(f(x))$ . By assumption, f is Borel and then $f(\cdot )(A):X \to \mathbb {R}$ is a composition of a lower semicontinuous function and a Borel map, so it is a Borel map. Therefore, f is a disintegration map, which concludes the proof.

Proof. Define $\varphi (y)=\psi (\delta _{y})$ , where $\psi \in C((\mathscr {P}(Y), W_{1}))$ is arbitrarily chosen. Then,

$$ \begin{align*} \int_{Y} \psi d(f_{*}\mu) &= \int_{X} \psi(f(x)) \,d\mu \\ &= \int_{X} \psi (\delta_{T(x)})\,d\mu \\ &= \int_{X} \varphi(T(x))\,d\mu \\ &= \int_{X} \varphi(S(x))\,d\mu \\ &= \int_{Y} \psi\,d(g_{*}\mu). \end{align*} $$

Given the arbitrary choice of $\psi $ , it follows that $f_{*}\mu =g_{*}\mu $ . Conversely, consider $\varphi \in C(Y)$ and the application $I_{\varphi }$ defined by equation (5). Note that

$$ \begin{align*} \int_{\mathscr{P}(Y)} I_{\varphi}(\unicode{x3bb}) \,d(f_{*}\mu)=\int_{\mathscr{P}(Y)} I_{\varphi}(\unicode{x3bb})\,d(g_{*}\mu) \end{align*} $$

if and only if

$$ \begin{align*} \int_{X} I_{\varphi}(f{(x)})\,d\mu = \int_{X} I_{\varphi}(g{(x)}) \,d\mu, \end{align*} $$

which in turn occurs if and only if

$$ \begin{align*} \int_{X} \bigg( \int_{Y} \varphi (y)\,df{(x)} \bigg)\,d\mu = \int_{X} \bigg( \int_{Y} \varphi (y) \,dg{(x)} \bigg) \,d\mu. \end{align*} $$

Since

$$ \begin{align*} \int_{Y} \varphi (y)\,df{(x)}=\varphi (T(x)) \end{align*} $$

and

$$ \begin{align*} \int_{Y} \varphi (y)\,dg{(x)}=\varphi (S(x)), \end{align*} $$

the last equation can be written as

$$ \begin{align*} \int_{X} \varphi (T(x))\,d\mu = \int_{X} \varphi (S(x))\,d\mu. \end{align*} $$

Due to the arbitrary choice of $\varphi $ , it follows that $T_{*}\mu =S_{*}\mu $ .

Proof. By the approximation theorem [Reference AmbrosioAmb00, Theorem 9.3] and by the definition of equivalent by disintegration, there exist a sequence of Borel functions $S_{n}: X \to Y$ such that $\eta = \lim \limits _{n \to \infty } \mu \otimes \delta _{S_{n}(x)}$ and $(S_{n})_{*}\mu =\nu $ for every $n \in \mathbb {N}$ . So $\mu \otimes \delta _{S_{n}(x)} \in [\gamma ]$ for every $n \in \mathbb {N}$ . Consider $\varphi (y)=|y|^2$ and observe that

$$ \begin{align*} \int_{X} |S_{n}(x)|^2\,d\mu = \int_{X} \varphi (S_{n}(x))\,d\mu = \int_{Y} \varphi(y)\,d\nu = \int_{Y} |y|^2\,d\nu < \infty. \end{align*} $$

Let $\psi \in C((\mathscr {P}(Y), W_{1}))$ be a function given by $\psi (\delta _{y})=|y|^2$ and let $\Delta \subset \mathscr {P}(Y)$ be the set of Dirac measures. Observe that $\psi $ is Lipschitz over $\Delta $ w.r.t. $W_1$ . So, take $\psi $ any Lipschitz extension over $\mathscr {P}(Y)$ . Since $(\delta _{S_{n}})_{*}\mu =(\delta _{T})_{*}\mu $ , we have that

$$ \begin{align*} \int_{X}|S_{n}(x)|^2\,d\mu = \int_{X} \psi(\delta_{S_{n}(x)})\,d\mu = \int_{X} \psi (\delta_{T(x)})\,d\mu = \int_{X} |T(x)|^{2} \,d\mu \end{align*} $$

for every $n \in \mathbb {N}$ . Therefore, moving on to a subsequence, we can assume that $S_{n}$ is weakly convergent to S. By [Reference AmbrosioAmb00, Lemma 9.1], $\mu \otimes \delta _{S_{n}(x)}$ converges weakly to $\mu \otimes \delta _{S(x)}$ . This concludes the proof.

Proof. Let $\mathcal {D} \subset X$ be a countable and dense subset. Consider the space $\mathcal {M}$ defined by $\mathcal {M} := \{ \nu \in \mathscr {P}(X) : \nu = \sum _j a_j \delta _{x_j}, ~\text {with}~ a_j \in \mathbb {Q} ~\text {and}~ x_j \in \mathcal {D} \}.$ We want to show that $\mathcal {M}$ is dense in $(\mathscr {P}(X), W_2)$ .

Given $\mu \in (\mathscr {P}(X), W_{2})$ , then for any $\varepsilon> 0$ and $x_0 \in \mathcal {D}$ , there exists a compact set $K \subset X$ such that

$$ \begin{align*} \int_{X \backslash K}\,d(x_0, x)^{2} \,d\mu \leq \varepsilon^{2}. \end{align*} $$

Since K is compact, we may cover it with a family $\{ B(x_k, {\varepsilon }/{2}):1 \leq k \leq N, x_k \in \mathcal {D} \}$ and define

$$ \begin{align*} B_{k}':= B(x_k, \varepsilon) \backslash \bigcup_{j<k} B(x_j, \varepsilon) \end{align*} $$

so that $\{ B_{k}' \}$ are disjoint and still cover K. Define $\varphi $ on X by

$$ \begin{align*} \begin{cases} \varphi(B_{k}' \cap K) = \{ x_k \}, \\ \varphi(X-K)=\{ x_0 \}. \end{cases} \end{align*} $$

So $d(x, x_k) \leq \varepsilon $ for every $x \in K$ . Therefore,

$$ \begin{align*} \int_{X}\,d(x, \varphi(x))^{2} \,d\mu \leq \varepsilon^2 \int_{K}\,d\mu + \int_{X \backslash K} d(x, x_0)^2 \,d\mu \leq 2 \varepsilon^2 \end{align*} $$

and then $W_2 (\mu , \varphi _{*}\mu ) \leq 2 \varepsilon ^2$ . Note that $\varphi _{*}\mu $ can be written as $\sum _{0 \leq j \leq N} \alpha _j \delta _{x_j}$ , that is, $\mu $ might be approximated by a finite combination of Dirac masses. Moreover, the coefficients $\alpha _j$ might be replaced by rational coefficients (up to a small error in Wasserstein distance): since Wasserstein distances are controlled by weighted total variation [Reference VillaniVil09, Theorem 6.15],

$$ \begin{align*} W_{2} \bigg( \sum_{0 \leq j \leq N} \alpha_j \delta_{x_j}, \sum_{0 \leq j \leq N} \beta_j \delta_{x_j} \bigg) \leq 2^{{1}/{2}} \bigg[ \max_{k, l}\,d(x_k, x_l) \bigg] \sum_{0 \leq j \leq N} |\alpha_j - \beta_j|^{{1}/{2}}, \end{align*} $$

which can become small, taking suitable coefficients $\beta _j \in \mathbb {Q}$ . Thus, it follows that $\mathcal {M}$ is dense in $\mathscr {P}(X)$ . Consequently, $(\mathscr {P}(X), W_2)$ is separable.

Proof. Consider the $2$ -Wasserstein distance $W_{2}$ on $\mathscr {P}(X)$ , with d a complete bounded metric for X. Here, $W_{2}$ metrizes the weak convergence of $\mathscr {P}(X)$ [Reference VillaniVil09, Theorem 6.9]. Furthermore, $(\mathscr {P}(X), W_{2})$ is a separable space. Moreover, from Theorem A, the map $y \mapsto \mu _{y} \in \mathscr {P}(X)$ is measurable and $\nu $ is a Borel measure on Y, since $\nu \ll \operatorname {\mathrm {vol}}_Y$ . Then, by Lusin’s theorem for Y and $\varepsilon> 0$ , there exists $\mathcal {K} \subset Y$ with $\nu (Y\backslash \mathcal {K}) < \varepsilon $ such that the disintegration map restricted to $\mathcal {K}$ is weakly continuous.

Proof. On the one hand,

$$ \begin{align*} \mu(B) =& \int_{Y} \mu_y (B) \,d\nu \\ =& \int_{Y} \mu_y (B \cap \pi^{-1}(y)) \,d\nu. \end{align*} $$

On the other hand,

Then,

$$ \begin{align*} \int_{Y} \mu_y (B \cap \pi^{-1}(y)) \,d\nu = \int_{Y} \delta_{\pi^{-1}(y)} (B \cap \pi^{-1}(y)) \,d\nu. \end{align*} $$

Moreover, since $\mu _y$ is a probability and $B \cap \pi ^{-1}(y)$ is either a singleton or an empty set, we have that

$$ \begin{align*} \delta_{\pi^{-1}(y)} (B \cap \pi^{-1}(y)) \geq \mu_y (B \cap \pi^{-1}(y)). \end{align*} $$

Then, $\mu _y (B \cap \pi ^{-1}(y)) = \delta _{\pi ^{-1}(y)} (B \cap \pi ^{-1}(y))$ and it follows that $\mu _y = \delta _{\pi ^{-1}(y)}$ .

Suppose $y_n \longrightarrow y$ . Since $\pi $ is continuous, $\pi ^{-1}$ is continuous and then

$$ \begin{align*} \int g\,d\mu_{y_n} \longrightarrow \int g \,d\mu_y \end{align*} $$

for every bounded uniformly continuous function g. Therefore, f is weakly continuous.

Proof. Given a metric measure foliation of X, we have for every $y,{\kern-1pt} y' {\kern-1pt}\in{\kern-1pt} X^*, W_2(\mu _y, \mu _{y'}) {\kern-1pt}= d^* (y, y')$ , where $\mu _{y} = f(y)$ and $\mu _{y'} = f(y')$ , and the disintegration map is denoted by f. Consider a sequence $(y_n)_n$ in $X^*$ such that $y_n \longrightarrow y$ . Note that $W_2(\mu _{y_n}, \mu _{y}) \longrightarrow 0$ , since $d^*(y_n, y) \longrightarrow 0$ . Therefore, we have the weak continuity of f.

Proof. (i) This is a direct consequence of the weak continuity of the disintegration map, which will be denoted by f throughout the demonstration. Consider $y, y' \in Y$ . Let $\psi $ be a continuous curve in Y connecting y and $y'$ , that is, $\psi = \{ y_{t} : t \in [0, 1], \psi (0)=y, \psi (1)=y' \}$ . Taking $y_t \longrightarrow \bar {y} \in Y$ , we have $f(y_t) \stackrel {w}{\longrightarrow } f(\bar {y})$ , that is, $W_2(\mu _{y_{t}}, \mu _{\bar {y}}) \longrightarrow 0$ . Then, $\zeta = \{ \mu _{y_{t}} : t \in [0, 1] \}$ , where $\mu _{y_{t}}$ is the conditional measure associated with $y_t$ via f, for every $t \in [0, 1]$ , is a weakly continuous curve in $(\mathscr {P}(X), W_2)$ connecting $\mu _{y}$ and $\mu _{y'}$ . This proves the first part of our theorem.

The proof of part (ii) will be done in a few steps. The idea is to use a sort of ‘time-dependent’ version of optimal transport. See [Reference VillaniVil09, Ch. 7], for example. In short, we will consider the curve $\zeta $ given by the disintegration map as an interpolation between probability measures, called displacement interpolation. To this end, we consider a transport problem, and we associate $\zeta $ with a random curve $\xi $ in X.

Step 1: A minimizing curve in the space of measures. Consider the path $\zeta = \{ \mu _{y_{t}} : t \in [0, 1] \}$ as constructed in part (i). Taking $\psi $ as a minimizing curve in Y, $\zeta $ is a minimizing curve in $\mathscr {P}(X)$ , since f is minimizing invariant. By abuse of notation, we use the weak continuous path $\zeta : [0,1] \to (\mathscr {P}(X), W_{2})$ while referring to this path in $(\mathscr {P}(X), W_{2})$ given by the disintegration map f evaluated at the minimal curve $\psi $ joining $[y, y']$ in Y. More precisely, we consider the composition $f \circ \psi $ to describe $\zeta $ .

Step 2: A random curve in X. We want to associate $\zeta $ with a random curve $\xi{\kern-1pt} :{\kern-1pt} [0,{\kern-1pt}1] \to X$ . To set some notation, let $e_t$ be the evaluation map, given by $e_{t}(\xi ) = \xi (t) := \xi _{t}$ , meaning the evaluation of $\xi $ at t. We will also use some usual concepts related to geodesics in Riemannian manifolds throughout the demonstration. We suggest [Reference JostJost11] for a comprehensive reading.

Consider the curve $\zeta : [0,1] \to \mathscr {P}_2(X)$ joining $\mu _{0}$ and $\mu _{1}$ (from Step 1), denoting $\mu _0 = \mu _y$ and $\mu _1 = \mu _{y'}$ . We also denote $\mu _{t} = \mu _{y_t}$ . Suppose that there is a transport problem associated with this curve, whose respective spatial distributions are modelled by these probability measures. Assume that the cost function for the transport between the initial point $x_0 \in X$ (at time $0$ ) and the final point $x_1 \in X$ (at time $1$ ), denoted by $c^{0, 1} (x_0, x_1)$ , is associated with a family of functionals parametrized by the initial and the final times. Denote by $\mathcal {A}^{0, 1}$ the functional on the set of curves $[0, 1] \to X$ , such that,

$$ \begin{align*} c^{0, 1} (x_0, x_1)= \inf \{ \mathcal{A}^{0, 1} (\xi) : \xi_{0} = x_0,~\xi_{1} = x_1, ~\xi \in \mathcal{C}([0, 1]; X) \}. \end{align*} $$

In other words, $c^{0, 1}(x_0, x_1)$ is the minimal cost needed to go from point $x_0$ at initial time $0$ , to point $x_1$ at final time $1$ . Moreover, let $C^{0,1}(\mu _{0}, \mu _{1})$ be the optimal transport cost between $\mu _{0}$ and $\mu _{1}$ for the cost $c^{0, 1} (x_0, x_1)$ . For $t_1, t_2 \in [0, 1]$ , define

$$ \begin{align*} \mathcal{A}^{t_1, t_2}(\xi) = \frac{L(\xi)^{2}}{t_2 - t_1}, \end{align*} $$

where $L(\xi )$ is the length of $\xi $ , so that

$$ \begin{align*} c^{t_1, t_2} (x_0, x_1)=\frac{d(x_0, x_1)^2}{t_2 - t_1} \end{align*} $$

and

(10) $$ \begin{align} C^{t_1, t_2} (\mu_{0}, \mu_{1}) = \frac{W_2 (\mu_{0}, \mu_{1})^2}{t_2 - t_1}. \end{align} $$

We want to show that there exists a random minimizer $\xi : [0,1] \to X$ , such that, law $(\xi _t) = \mu _t$ for every $t \in [0,1]$ . In other words, we want to show that $\zeta $ is a curve in the space of measures which interpolates all possible measures along the minimizing path joining $\mu _{0}$ and $\mu _{1}$ . Such a curve is called displacement interpolation.

Step 3: $\zeta $ is a displacement interpolation. In this step, $\xi $ will be constructed by dyadic approximation, according to couplings of measures in $\zeta $ associated with times $t = {1}/{2^k}$ . To achieve this, we will use the iterative construction in line with [Reference VillaniVil09, Theorem 7.21].

Let $\Gamma $ be the set of minimizing curves in X. It will be necessary throughout the text to consider subsets of $\Gamma $ in which the geodesics are defined for certain time intervals and endpoints (or endpoints regions). So, for $s,t \in [0,1]$ , $x_s, x_t \in X$ , let $\Gamma _{x_s \to x_t}^{s, t}$ be the set of minimizing curves in X starting at $x_s$ at time s and ending at $x_t$ at time t. Similarly, for any two compact sets $K_s, K_t \subset X$ , let $\Gamma _{K_s \to K_t}^{s, t}$ be the set of minimizing curves starting in $K_s$ at time s and ending in $K_t$ at time t.

Considering the measures along $\zeta $ , for $t_1, t_2, t_3 \in [0, 1]$ , let $\gamma _{t_1 \to t_2}$ be an optimal transference plan between $\mu _{t_1}$ and $\mu _{t_2}$ for $c^{t_1, t_2}(x_{t_1}, x_{t_2})$ , and let $\gamma _{t_2 \to t_3}$ be an optimal transference plan between $\mu _{t_2}$ and $\mu _{t_3}$ for $c^{t_2, t_3}(x_{t_2}, x_{t_3})$ . By Lemma 2.5, it is possible to take random variables $(\xi _{t_1}, \xi _{t_2}, \xi _{t_3})$ such that law $(\xi _{t_1}, \xi _{t_2}) = \gamma _{t_1 \to t_2}$ , law $(\xi _{t_2}, \xi _{t_3}) = \gamma _{t_2 \to t_3}$ and law $(\xi _{t_i})=\mu _{t_i}$ for $i=1, 2 ,3$ . Since $\zeta $ is minimizing in $\mathscr {P}_2(X)$ (see Step 1) and $\mathscr {P}_2(X)$ is a geodesic space, it follows from equation (10) that

$$ \begin{align*} C^{t_1, t_2}(\mu_{t_1}, \mu_{t_2}) + C^{t_2, t_3}(\mu_{t_2}, \mu_{t_3}) = C^{t_1, t_3}(\mu_{t_1}, \mu_{t_3}). \end{align*} $$

This, in particular, implies:

1. (a) $(\xi _{t_1}, \xi _{t_3})$ is an optimal coupling of $(\mu _{t_1}, \mu _{t_3})$ for $c^{t_1, t_3}(\xi _{t_1}, \xi _{t_3})$ ;

2. (b) $c^{t_1, t_3} (\xi _{t_1}, \xi _{t_3}) =c^{t_1, t_2} (\xi _{t_1}, \xi _{t_2}) + c^{t_2, t_3} (\xi _{t_2}, \xi _{t_3})$ almost surely.

Let $(\xi _{0}, \xi _{1})$ be an optimal coupling of $(\mu _{0}, \mu _{1})$ . Consider optimal transference plans $\gamma _{0 \to 1/2}$ , $\gamma _{1/2 \to 1}$ , as above, and construct random variables $(\xi _{0}^{(1)}, \xi _{1/2}^{(1)}, \xi _{1}^{(1)})$ such that $(\xi _{0}^{(1)}, \xi _{1/2}^{(1)})$ is an optimal coupling of $(\mu _{0}, \mu _{1/2})$ for $c^{0, 1/2}(\xi _{0}^{(1)}, \xi _{1/2}^{(1)})$ ; $(\xi _{1/2}^{(1)}, \xi _{1}^{(1)})$ is an optimal coupling of $(\mu _{1/2}, \mu _{1})$ for $c^{1/2, 1}(\xi _{1/2}^{(1)}, \xi _{1}^{(1)})$ , and law $(\xi _{i}^{(1)}) = \mu _{i}$ for $i=0, \tfrac 12 ,1$ . Moreover, item (a) implies that $(\xi _{0}^{(1)}, \xi _{1}^{(1)})$ is an optimal coupling of $(\mu _{0}, \mu _{1})$ , and item (b) implies

$$ \begin{align*} c^{0, 1}(\xi_{0}^{(1)}, \xi_{1}^{(1)}) = c^{0, {1}/{2}} \big(\xi_{0}^{(1)}, \xi_{{1}/{2}}^{(1)} \big) + c^{{1}/{2}, 1} \big(\xi_{{1}/{2}}^{(1)}, \xi_{1}^{(1)}\big) \end{align*} $$

almost surely. Iterating this process, at the step k, we have random variables $(\xi _{0}^{(k)}, \xi _{{1}/{2^{k}}}^{(k)}, \xi _{{2}/{2^{k}}}^{(k)}, \ldots , \xi _{1}^{(k)})$ , so that for any two $i, j \leq 2^{k}$ , $(\xi _{{i}/{2^{k}}}^{(k)}, \xi _{{j}/{2^{k}}}^{(k)})$ is an optimal coupling of $(\mu _{{i}/{2^{k}}}, \mu _{{j}/{2^{k}}})$ . Furthermore, for $i_1, i_2, i_3 \leq 2^{k}$ ,

$$ \begin{align*} c^{{i_1}/{2^k}, {i_3}/{2^k}} \big(\xi_{{i_1}/{2^k}}^{(k)}, \xi_{{i_3}/{2^k}}^{(k)}\big) = c^{{i_1}/{2^k}, {i_2}/{2^k}}\big(\xi_{{i_1}/{2^k}}^{(k)}, \xi_{{i_2}/{2^k}}^{(k)}\big) + c^{{i_2}/{2^k}, {i_3}/{2^k}}\big(\xi_{{i_2}/{2^k}}^{(k)}, \xi_{{i_3}/{2^k}}^{(k)}\big) \end{align*} $$

almost always.

We want to extend the random variables $\xi ^{(k)}$ , defined for times ${i}/{2^{k}}$ , $i \leq 2^k$ , to continuous curves $(\xi ^{(k)})_{0 \leq t \leq 1}$ . For this, note that for all times $s, t \in [0, 1]$ , $s < t$ , there exists a Borel map $S_{s \to t}: X \times X \to \mathcal {C}([s, t];X)$ such that for all $x, z \in X$ , $S(x, z)$ belongs to $\Gamma _{x \to z}^{s, t}$ [Reference VillaniVil09, Proposition 7.16]. Indeed, let $E_{s,t}$ be the function given by $E_{s,t}(\xi ) := (\xi _{s}, \xi _{t})$ for $\xi : [s, t] \to X$ minimizing curve. Since X is a geodesic space, any two points of X can be joined by at least one minimizing curve, so $E_{s, t}$ is onto $X \times X$ . Moreover, $E_{s, t}$ is a continuous map between complete separable metric spaces, and $E_{s,t}^{-1}(x, z)$ is compact for every $x, z$ . Therefore, $E_{s,t}$ admits a measurable right-inverse $S_{s \to t}$ [Reference DellacherieDel75], that is, $E_{s,t} \circ S_{s \to t} = \mathrm {Id}$ . Thus, $S_{s \to t}$ is a measurable recipe to join two points $x,z$ by a minimizing curve, which was to be proved.

For $t \in ({i}/{2^k}, ({i+1})/{2^k})$ , define $\xi _{t}^{(k)}$ by $e_t (S_{{i}/{2^k} \to ({i+1})/{2^k}} ( \xi _{{i}/{2^{k}}}, \xi _{({i+1})/{2^{k}}} ) )$ . Then, the law of $(\xi _{t}^{(k)})_{0 \leq t \leq 1}$ is a probability on $C(X)$ . Let us denote it by $\Theta ^{(k)}$ . Note that $(e_t)_{*}\Theta ^{(k)}=\mu _{t}$ for every $t={i}/{2^k}$ , $i \leq 2^k$ , and $\Theta ^{(k)}$ is concentrated on $\Gamma $ .

In short, from $\zeta $ , we iteratively constructed probability measures $\Theta ^{(k)}$ on $\Gamma $ up to a step k. We want to pass to the limit as $k \longrightarrow \infty $ . Given $\varepsilon>0$ , since $\mu _{0}$ and $\mu _{1}$ are Radon measures, there exist compact sets $K_0$ and $K_1$ such that $\mu _{0}(X \backslash K_0) \leq \varepsilon $ , $\mu _{1} (X \backslash K_1) \leq \varepsilon $ . Also, the set $\Gamma _{K_0 \to K_1}^{0, 1}$ is compact [Reference VillaniVil09, Definition 7.13 and Example 7.15] and

$$ \begin{align*} \Theta^{(k)}(\Gamma ~\backslash~ \Gamma_{K_0 \to K_1}^{0, 1}) &= \mathbb{P}((\xi_{0}, \xi_{1}) \notin K_0 \times K_1) \\ & \leq \mathbb{P}(\xi_{0} \notin K_0) + \mathbb{P}(\xi_{1} \notin K_1) \\ & = \mu_{0}(X \backslash K_0) + \mu_{1} (X \backslash K_1) \\ & \leq 2 \varepsilon. \end{align*} $$

Then, we can take a subsequence of $(\Theta ^{(k)})_{k}$ that converges weakly to $\Theta $ . Since $\Gamma $ is closed in the topology of uniform convergence [Reference VillaniVil09, Theorem 7.16(v)], $\Theta $ is supported in $\Gamma $ . Moreover, given a constant a, for every $t={1}/{2^{a}} \in [0, 1]$ , if $k> a$ , we have $(e_t)_{*}\Theta ^{(k)} = \mu _{t}$ and, passing to the limit $k \longrightarrow \infty $ , $(e_t)_{*}\Theta = \mu _{t}$ . Finally, since $\mu _{t}$ depends continuously on t, to show that $(e_t)_{*}\Theta = \mu _{t}$ for every $t \in [0, 1]$ , it suffices to show that $(e_t)_{*}\Theta $ is continuous as a function of t. In other words, we need to show that, given u a bounded continuous function on X, $U(t) = \mathbb {E} u (\xi _{t})$ is a continuous function of t if $\xi $ is random geodesic with law $\Theta $ . In fact, since $t \mapsto \xi _{t}$ is continuous and the composition of continuous functions is also continuous, $t \mapsto u(\xi _{t})$ is continuous. Moreover, let $\{ u_n \}$ be Lebesgue integrable functions such that $u_n \longrightarrow u$ . Since u is bounded, $|u_n| \leq g$ for some integrable function g; by Lebesgue’s dominated convergence theorem, $\mathbb {E}u_n \longrightarrow \mathbb {E}u$ . From these results, the continuity of $U(t)$ follows, as we wanted.

Accordingly, we constructed $\xi $ such that for each $t \in [0, 1]$ , $\mu _{t}$ is the law of $\xi _{t}$ , where $(\xi _{t})_{0 \leq t \leq 1}$ is a dynamical optimal coupling of $(\mu _{0}, \mu _{1})$ . In other terms, we say that $\zeta $ is displacement interpolation.

Step 4: An important observation about displacement interpolation. By [Reference VillaniVil09, Theorem 8.5], if $\{ \mu _t \}$ is a displacement interpolation between two compactly supported probability measures on X, and $t_0 \in (0, 1)$ is given, then, for every $t \in [0, 1]$ , the transport map $T_{t_0 \to t}$ between the points $\xi (t_0)$ and $\xi (t)$ is well defined $\mu _{t_0}$ -almost everywhere and it is Lipschitz continuous. In other words, $T_{t_0 \to t}$ is a solution of the Monge problem between $\mu _{t_0}$ and $\mu _t$ . For the completeness of the text, we will comment briefly on the proof.

Note that $(e_0, e_1, e_0, e_1)_{*}(\Theta \otimes \Theta ) = \gamma _{0 \to 1} \otimes \gamma _{0 \to 1}$ . So, if one property holds true $\gamma _{0 \to 1} \otimes \gamma _{0 \to 1}$ -a.a. for quadruples, this property, for the endpoints of pairs of curves, holds true $\Theta \otimes \Theta $ -a.a. Since $\gamma _{0 \to 1}$ is optimal, it has a property named c-cyclical monotonicity [Reference VillaniVil09, Theorem 5.10], so that $c(x, y) + c(\tilde {x}, \tilde {y}) \leq c(x, \tilde {y}) + c(\tilde {x}, y)$ , $\Theta \otimes \Theta (dx,\,dy,\,d\tilde {x},\,d\tilde {y})$ -a.a. Thus, $c(\xi (0), \xi (1)) + c(\tilde {\xi }(0), \tilde {\xi }(1)) \leq c(\xi (0), \tilde {\xi }(1)) + c(\tilde {\xi }(0), \xi (1))$ , $\Theta \otimes \Theta (d\xi , d\tilde {\xi })$ -a.a.

Moreover, by Mather’s shortening lemma [Reference VillaniVil09, Theorem 8.1 and Corollary 8.2],

(11) $$ \begin{align} \sup_{0 \leq t \leq 1} d(\xi_t, \tilde{\xi_t}) \leq C_{K} d(\xi_{t_0}, \tilde{\xi_{t_0}}), \end{align} $$

where $C_K$ is a constant. Suppose that $\Theta $ is supported on a compact set S. Equation (11) defines a closed set for all pairs of curves $\xi , \tilde {\xi } \in S \otimes S$ . Let $e_{t_0}(S)$ be the union of all $\xi (t_0)$ , when $\xi $ varies over S, and $T_{t_0 \to t}$ by $T_{t_0 \to t}(\xi (t_0)) = \xi (t)$ . Note that if $\xi $ , $\tilde {\xi }$ in S are such that $\xi (t_0) = \tilde {\xi }(t_0)$ , then equation (11) implies $\xi = \tilde {\xi }$ . Moreover, $T_{t_0 \to t}$ is Lipschitz-continuous. So, $(\xi (t_0), T_{t_0 \to t}(\xi (t_0)))$ is a Monge coupling of $(\mu _{t_0}, \mu _{t})$ .

Step 5: Absolute continuity of measures $\mu _t$ on $\zeta $ with respect to the volume measure on X, when $\mu _0$ and $\mu _1$ are compactly supported. Without loss of generality, let us suppose that $\mu _{1}$ is absolutely continuous with respect to the volume measure on X, vol. If $\mu _0$ and $\mu _1$ are compactly supported, then $\zeta $ has a compact support. Indeed, let $A_0, A_1 \subset X$ be the compact supports of $\mu _0, \mu _1$ and $\gamma _{0 \to 1}$ be the transference plan with marginals $\mu _0$ and $\mu _1$ . Consider the canonical projections $(\mathrm {proj}_1)$ , $(\mathrm {proj}_2)$ on the first and second components, respectively. Since $(\mathrm {proj}_1)_{*}\gamma _{0 \to 1} =\mu _0$ and $(\mathrm {proj}_2)_{*}\gamma _{0 \to 1} =\mu _1$ ,

$$ \begin{align*} (\mathrm {proj}_1)_{*}\gamma_{0 \to 1}(X \times X) = \mu_{0}(X) = \mu_{0}(A_0) = \gamma_{0 \to 1}(A_0 \times X), \end{align*} $$

$$ \begin{align*} (\mathrm {proj}_2)_{*}\gamma_{0 \to 1}(X \times X) = \mu_{1}(X) = \mu_{1}(A_1) = \gamma_{0 \to 1}(X \times A_1). \end{align*} $$

Therefore, $\gamma _{0 \to 1}$ is concentrated in a compact set $A_0 \times A_1$ . Moreover, since $\gamma _{0 \to 1} = (e_0, e_1)_{*}\Theta $ and the evaluation map is continuous, $\Theta $ is concentrated in a compact set. The compactness of the $\zeta $ support follows from $(e_t)_{*} \Theta = \mu _{t}$ .

We can use Step 4, and there is a Lipschitz map T solving the Monge problem between $\mu _{t}$ and $\mu _{1}$ , $t \in (0, 1)$ . Let N be a set such that the volume measure is zero and consider $T(N)$ . If $T(N)$ is not Borel measurable, consider a negligible Borel set that contains $T(N)$ (which by abuse of notation, we will continue denoting $T(N)$ ). Note that $N \subset T^{-1}(T(N))$ , so

$$ \begin{align*} \mu_{t}(N) \leq \mu_{t}(T^{-1}(T(N))) = (T_{*} \mu_{t}) (T(N)) = \mu_{1} (T(N)) \end{align*} $$

and then $\mu _{t} (N)=0$ , since $\text {vol}(T(N)) \leq \| T \|_{\text {Lip}} \text {vol}(N) = 0$ (this inequality occurs since T is Lipschitz and $\| T \|_{\text {Lip}}$ stands for the Lipschitz constant) and $\mu _1 \ll \text {vol}$ by hypothesis. So, $\mu _{t}(N)=0$ for every Borel set N such that vol $(N)=0$ , that is, $\mu _{t} \ll \text {vol}$ .

Step 6: Absolute continuity of measures $\mu _t$ of $\zeta $ with respect to the volume measure on X: the general case. Without loss of generality, suppose $\mu _{1} \ll \text {vol}$ . Let us assume that there is some case in which neither $\mu _0$ nor $\mu _1$ is compactly supported. We will prove our statement (ii) by contradiction. Suppose that $\mu _{\tau }$ , for $\tau \in (0,1)$ , is not absolutely continuous with respect to the volume measure on X. Then, there exists a set $Z_{\tau } \subset X$ , such that vol $(Z_{\tau })=0$ and $\mu _{\tau }(Z_{\tau })> 0$ . Consider $\mathcal {Z}:=\{ \xi \in \Gamma : \xi _{\tau } \in Z_{\tau } \}$ so that $\Theta (\mathcal {Z}) = \mathbb {P}(\xi _{s\tau } \in Z_{\tau }) = \mu _{\tau }(Z_{\tau })> 0$ . Since $\Theta $ is a regular measure, there exists $\mathcal {K} \subset \mathcal {Z}$ compact such that $\Theta (\mathcal {K})> 0$ . So, if we set

and consider $\gamma _{0 \to 1}':= (e_0, e_1)_{*}\Theta '$ and $\mu _{t}'=(e_t)_{*}\Theta '$ , we have

$$ \begin{align*} \mu_t' \leq \frac{(e_t)_{*}\Theta}{\Theta (\mathcal{K})} =\frac{\mu_{t}}{\Theta (\mathcal{K})}. \end{align*} $$

In this way, $(\mu _{t}')$ is a displacement interpolation and, considering the previous equation for $t=1$ , $\mu _{1}' \ll \mu _1 \ll \text {vol}$ . Note that now, $\mu _{\tau }'$ is concentrated on $e_{\tau }(\mathcal {K}) \subset e_{\tau }(\mathcal {Z}) \subset Z_{\tau }$ and then $\mu _{\tau }'$ is singular. However, $\mu _0'$ is supported in $e_0(\mathcal {K})$ and $\mu _{1}'$ is supported in $e_1(\mathcal {K})$ , which are compact. This is the case of Step 5. Then, $\mu _{\tau }' \ll \text {vol}$ , which is a contradiction.

(iii) In this item, we denote $Y:=\Omega $ , where $\Omega $ is the subset of the quotient space $X^*$ on which the metric measure foliation is defined (see Definition 5.6), and $\pi := p$ , where p is the quotient map. The weak continuity of f in this case was proved in Proposition 5.8. Let $\psi $ be a minimizing curve on Y connecting y and $y'$ . Consider the path $\zeta = \{ \mu _{y_{t}} : t \in [0, 1] \}$ as constructed in item (i). Since $\psi $ was taken as a minimizing curve and f is an isometry, $\zeta $ is minimizing.

Observe that every optimal transference plan between $\mu _y$ and $\mu _{y'}$ is supported on $\{ (x, x') \in \pi ^{-1}(y) \times \pi ^{-1}(y') ~:~ d^*(y, y') = d(x, x') \}$ . In fact, let $\gamma _{y \to y'}$ be an optimal transference plan for $\mu _y$ , $\mu _{y'}$ . Since supp $(\mu _y) \subset \pi ^{-1}(y)$ , supp $(\mu _{y'}) \subset \pi ^{-1}(y')$ and $\pi $ is 1-Lipschitz, we have $\gamma _{y \to y'}$ supported on $\{ (x, x') \in \pi ^{-1}(y) \times \pi ^{-1}(y') ~:~ d^*(y, y') \leq d(x, x') \}$ . Consider the set $\Upsilon := \{ (x, x') \in \pi ^{-1}(y) \times \pi ^{-1}(y') ~:~ d^*(y, y') < d(x, x') \}$ . If $\gamma _{y \to y'} (\Upsilon )>0$ , then

$$ \begin{align*} \gamma_{y \to y'} (\Upsilon)\,d^{*} (y, y')^2 \leq & ~\gamma_{y \to y'} (\Upsilon) \,d(x, x')^2 \\ < & \int_{\Upsilon} d(x, x')^2 \,d\gamma_{y \to y'}. \end{align*} $$

So,

$$ \begin{align*} d^{*} (y, y')^2 < & \int_{\Upsilon}\,d(x, x')^2 \,d\gamma_{y \to y'} + \gamma_{y \to y'} (X \times X \backslash \Upsilon) \,d^{*} (y, y')^2 \\ = & \int_{\Upsilon}\,d(x, x')^2 \,d\gamma_{y \to y'} + \int_{X \times X \backslash \Upsilon} d^*(y, y')^2 \,d\gamma_{y \to y'} \\ \leq & \int_{\Upsilon}\,d(x, x')^2 \,d\gamma_{y \to y'} + \int_{X \times X \backslash \Upsilon}\,d(x, x')^2 \,d\gamma_{y \to y'} \\ =& \int_{X \times X}\,d(x, x')^2\,d\gamma_{y \to y'}. \end{align*} $$

Then, $d^{*} (y, y')^2 < W_2 (\mu _{y}, \mu _{y'})^2$ , which is a contradiction.

Therefore, we have the transport between $\mu _y$ and $\mu _{y'}$ orthogonal to the leaves. Furthermore, since $\zeta $ is minimizing, the existence of the optimal transport plan is guaranteed.

Suppose without loss of generality that $\mu _y$ is absolutely continuous with respect to the volume measure of the leaf y. Then, the support of $\mu _y$ contains more than one point. Considering the transport problem described above, for each x in the support of $\mu _y$ such that $(x, x')$ is in the support of $\gamma _{y \to y'}$ for some $x'$ , each one of the intermediate leaves $y_t$ must contain a corresponding point $x_{y_t}$ . Thus, each leaf will have the distribution $\mu _{y(t)}$ absolutely continuous with respect to the volume measure of the respective leaf.