On the Existence of Monge Maps for the Gromov–Wasserstein Problem

Abstract

The Gromov–Wasserstein problem is a non-convex optimization problem over the polytope of transportation plans between two probability measures supported on two spaces, each equipped with a cost function evaluating similarities between points. Akin to the standard optimal transportation problem, it is natural to ask for conditions guaranteeing some structure on the optimizers, for instance, if these are induced by a (Monge) map. We study this question in Euclidean spaces when the cost functions are either given by (i) inner products or (ii) squared distances, two standard choices in the literature. We establish the existence of an optimal map in case (i) and of an optimal 2-map (the union of the graphs of two maps) in case (ii), both under an absolute continuity condition on the source measure. Additionally, in case (ii) and in dimension one, we numerically design situations where optimizers of the Gromov–Wasserstein problem are 2-maps but are not maps. This suggests that our result cannot be improved in general for this cost. Still in dimension one, we additionally establish the optimality of monotone maps under some conditions on the measures, thereby giving insight into why such maps often appear to be optimal in numerical experiments.

Appendices

Appendix A. Proofs of Lemmas 3.3 and 3.4

In this section, we prove Lemmas 3.3 and 3.4. Let us first recall the statement of Lemma 3.3:

Lemma 3.3

Let $\mu ,\nu \in \mathcal {P}(E)$ and let $\psi _1,\psi _2:E\rightarrow F$ be homeomorphisms. Let ${\tilde{c}}:F\times F\rightarrow \mathbb {R}$ and consider the cost $c(x,y)= c( \psi _1(x),\psi _2(y))$. Then a map is optimal for the cost c between $\mu$ and $\nu$ if and only if it is of the form $\psi _2^{-1}\circ T\circ \psi _1$ with $T:F\times F$ optimal for the cost ${\tilde{c}}$ between $\psi _{1\#}\mu$ and $\psi _{2\#}\nu$.

Proof

Remark that the continuity of $\psi _1$, $\psi _2$ and their inverse ensures their measurability. We have the following equalities:

$$\begin{aligned} \mathop {\mathrm {arg\,min}}\limits _{\pi \in \Pi (\mu ,\nu )} \int {\tilde{c}}(\psi _1 (x), \psi _2(y))\mathop {}\!\mathrm d\pi (x,y)&=\mathop {\mathrm {arg\,min}}\limits _{\pi \in \Pi (\mu ,\nu )} \int \tilde{c}( u,v)\mathop {}\!\mathrm d(\psi _1,\psi _2)_\#\pi (u,v) \\&=(\psi _1 ^{-1} ,\psi _2^{-1} )_\#\mathop {\mathrm {arg\,min}}\limits _{{\tilde{\pi }} \in \Pi (\psi _{1\#} \mu , \psi _{2\#}\nu )} \int \tilde{c}(u,v)\mathop {}\!\mathrm d{\tilde{\pi }}(u,v) \end{aligned}$$

since the mapping $(\psi _1 ^{-1},\psi _2^{-1} )_\#$ is a one-to-one correspondence from $\Pi (\psi _{1\#} \mu , \psi _{2\#}\nu )$ to $\Pi (\mu ,\nu )$ by bijectivity of $\psi _1$ and $\psi _2$. This bijectivity ensures that any optimal deterministic transport plan ${\tilde{\pi }}^\star$ between $\psi _{1\#}\mu$ and $\psi _{2\#}\nu$ induces an optimal deterministic transport plan $\pi ^\star$ between $\mu$ and $\nu$, and vice versa. Writing ${\tilde{\pi }}^\star =(\mathop {\textrm{id}}\limits ,T)_\#(\psi _{1\#}\mu )$, this plan $\pi ^\star$ is given by

$$\begin{aligned} \pi ^\star&= (\psi _1 ^{-1} ,\psi _2^{-1} )_\#{\tilde{\pi }} ^\star \\&=(\psi _1 ^{-1} ,\psi _2^{-1} )_\#(\mathop {\textrm{id}}\limits ,T)_\#\psi _{1\#} \mu \\&=(\mathop {\textrm{id}}\limits ,\psi _2^{-1}\circ T\circ \psi _1)_\#\mu . \square \end{aligned}$$

We now recall the statement of Lemma 3.4:

Lemma 3.4

Let $h\ge 1$ and $\mu ,\nu \in \mathcal {P}(\mathbb {R}^h)$ with compact supports and such that $\mu \ll {\text {Leb}}_h$. Consider the cost $c(x, y)= -\langle \psi _1(x),\psi _2(y)\rangle$ where $\psi _1,\psi _2:\mathbb {R}^h\rightarrow \mathbb {R}^h$ are diffeomorphisms. Then there exists a unique optimal transport plan between $\mu$ and $\nu$ for the cost c, and it is induced by a map $t:\mathbb {R}^h\rightarrow \mathbb {R}^h$ of the form $t=\psi _2^{-1}\circ \nabla f\circ \psi _1$, with $f:\mathbb {R}^h\rightarrow \mathbb {R}$ convex.

Proof

As $\psi _{1\#}\mu$ has a density with respect to the Lebesgue measure since $\psi _1$ is a diffeomorphism and $\psi _{1\#}\mu$ and $\psi _{2\#}\nu$ have compact support, Brenier’s theorem states that there exists a unique optimal transport plan between $\psi _{1\#}\mu$ and $\psi _{2\#}\nu$ and that it is induced by a map $\nabla f$, where $\phi$ is a convex function. Using Lemma 3.3 then yields the result. $\square$

Remark A.1

(Hypothesis of Lemma 3.4) In the proof of Lemma 3.4, we only needed (i) $\psi _1$, $\psi _2$ and their inverse to be measurable, (ii) $\psi _{1\#}\mu$ to have a density with respect to the Lebesgue measure, and (iii) $\psi _{1\#}\mu$ and $\psi _{2\#}\nu$ to have compact support. Imposing $\psi _1$ to be a diffeomorphism and $\psi _2$ to be a homeomorphism ensures both (i) and (ii) and is natural to expect.

Appendix B. Measurable selection of maps in the manifold setting

1.1 B.1 Measurability of set-valued maps

Let X and U be two topological spaces, and let $\mathcal {B}$ denote the Borel $\sigma$-algebra on X. A set-valued map S is a map from X to P(U) (the set of subsets of U) and will be denoted by $S: X \rightrightarrows U$. The idea is to introduce a notation that is consistent with the case where $S(x) = \{u\}$ for all x in X, where we want to retrieve the standard case of maps $X \rightarrow U$. Definitions are taken from Rockafellar and Wets [49], where measurability is studied when $U = \mathbb {R}^n$. Most results and proofs adapt to a more general setting, in particular when U is a complete Riemannian manifold M, which we shall assume in the following. For the sake of completeness, we provide all the proofs and highlight those that require specific care when replacing $\mathbb {R}^n$ by a manifold. Of importance for our proofs, we define the pre-image of a set $B \subset U$ by S,

$$\begin{aligned} S^{-1}(B) = \{ x \in X \mid S(x) \cap B \ne \varnothing \}, \end{aligned}$$

and the domain of S, defined as $S^{-1}(U)=\{ x \in X \mid S(x) \ne \varnothing \}$. We will often use the following relation: If a set A can be written as $A = \bigcup _k A_k$, then $S^{-1}(A) = \bigcup _k S^{-1}(A_k)$. Indeed,

$$\begin{aligned}{} & {} x \in S^{-1}(A) \iff S(x) \cap A \ne \varnothing \iff \exists k,\, S(x) \cap A_k \ne \varnothing \iff \exists k,\,\\{} & {} x \in S^{-1}(A_k) \iff x \in \textstyle \bigcup _k S^{-1}(A_k). \end{aligned}$$

A set-valued map $S: X \rightrightarrows U$ is said to be measurable if, for any open set $O \subset U$,

$$\begin{aligned} S^{-1}(O) \in \mathcal {B}. \end{aligned}$$

Note that if S is measurable (as a set-valued map), then its domain must be measurable as well (as an element of $\mathcal {A}$). We say that $S: X \rightrightarrows U$ is closed-valued if S(x) is a closed subset of U for all $x \in X$.

Proposition B.1

[49, Theorem 14.3.c] A closed-valued map $S: X \rightrightarrows U$ is measurable if and only if the property “$S^{-1}(B) \in \mathcal {B}$” holds either:

• (a) for all $B \subset U$ open (the definition);

• (b) for all $B \subset U$ compact;

• (c) for all $B \subset U$ closed.

Proof

• (a) $\Longrightarrow$ (b): For a compact $B \subset U$, let $B_k = \{ x \in U \mid d(x,B) < k^{-1} \}$ for $k \ge 0$ (which is open). Note that

  $$\begin{aligned} x \in S^{-1}(B) \iff S(x) \cap B \ne \varnothing \iff S(x) \cap B_k \ne \varnothing \text { for all { k},} \end{aligned}$$

  because S(x) is a closed set. Hence, $S^{-1}(B) = \bigcap _k S^{-1}(B_k)$. All the $S^{-1}(B_k)$ are measurable, and so is $S^{-1}(B)$ as a countable intersection of measurable sets.

• (b) $\Longrightarrow$ (a): Fix O an open set of U. As we assumed U to be a complete separable Riemannian manifold, O can be written as a countable union of compact balls: $O = \bigcup _n \overline{B(x_n, r_n)}$.

• (b) $\Longrightarrow$ (c): Immediate.

• (c) $\Longrightarrow$ (b): A closed set B can be obtained as a countable union of compact sets by letting $B = \bigcup _n B \cap \overline{B(x_0, n)}$ for some $x_0$. Hence, $S^{-1}(B) = \bigcup _n S^{-1}(B \cap \overline{B(x_0,n)})$ is in $\mathcal {B}$.

$\square$

Now, we state a result on operations that preserve the measurability of closed–set-valued maps. The proof requires adaptation from the one of [49] because the latter uses explicitly the fact that one can compute Minkowski sums of sets (which may not make sense on a manifold).

Proposition B.2

([49, Proposition 14.11], adapted to the manifold case) Let $S_1$ and $S_2: X \rightrightarrows U$ be two measurable closed–set-valued maps. Then

• $P: x \mapsto S_1(x) \times S_2(x)$ is measurable as a closed-valued map in $U \times U$ (equipped with the product topology).

• $Q: x \mapsto S_1(x) \cap S_2(x)$ is measurable.

Proof

The first point can be proved in the same spirit as the proof proposed by Rockafellar and Wets. Namely, let $O'$ be an open set in $U \times U$. By definition of the product topology, $O'$ can be obtained as $\bigcup _n O_1^{(n)} \times O_2^{(n)}$ where $O_1^{(n)}$ and $O_2^{(n)}$ are open sets in U. Then $P^{-1}(O') = \bigcup _n P^{-1}(O_1^{(n)} \times O_2^{(n)})$. Now, observe that

$$\begin{aligned}{} & {} P^{-1}(A \times B) = \{ x\mid S_1(x) \times S_2(x) \in A \times B\} = \{x\mid S_1(x) \in A \text { and } S_2(x) \in B \}\\{} & {} = S_1^{-1}(A) \cap S_2^{-1}(B), \end{aligned}$$

so that finally, $P^{-1}(O') = \bigcup _n S_1^{-1}(O_1^{(n)}) \cap S_2^{-1}(O_2^{(n)})$ is measurable as a countable union of (finite) intersection of measurable sets (given that $S_1$ and $S_2$ are measurable). Note that this does not require $S_1$ and $S_2$ to be closed-valued.

Now, let us focus on the second point, which requires more attention. Thanks to Proposition B.1, it is sufficient to show that $Q^{-1}(C) \in \mathcal {B}$ for any compact set $C \subset U$. In [49], this is done by writing

$$\begin{aligned} Q^{-1}(C) = \{ x\mid S_1(x) \cap S_2(x) \cap C \ne \varnothing \} = \{x\mid R_1(x) \cap R_2(x) \ne \varnothing \} \\= \{x\mid 0 \in (R_1(x) - R_2(x)) \} = (R_1 - R_2)^{-1}(0), \end{aligned}$$

where $R_j(x) = S_j(x) \cap C$ (which is also closed-valued), and using the fact that the (Minkowski) difference of measurable closed-valued maps is measurable as well [49, Proposition 14.11.c]. To adapt this idea (we cannot consider Minkowski differences in our setting), we introduce the diagonal $\Delta = \{(u,u)\mid u \in U\} \subset U \times U$. Now, observe that

$$\begin{aligned} R_1(x) \cap R_2(x) \ne \varnothing \iff (R_1(x) \times R_2(x)) \cap \Delta \ne \varnothing , \end{aligned}$$

that is $x \in R^{-1}(\Delta )$, where $R(x) = R_1(x) \times R_2(x)$. Since the maps $R_1$ and $R_2$ are measurable closed-valued maps (inherited from $S_1,S_2$), so is R according to the previous point. And since $\Delta$ is closed, $R^{-1}(\Delta ) = Q^{-1}(C)$ is measurable. $\square$

1.2 B.2 Proof of Proposition 2.10

Let us first recall the result.

Proposition 2.10

(Measurable selection of maps, manifold case) Let M be a complete Riemannian manifold and $(B, \Sigma , m)$ a measure space. Consider a measurable function $B\ni u \mapsto (\mu _u, \nu _u) \in P(M)^2$. Assume that for m-almost every $u\in B$, $\mu _u\ll {\text {vol}}_M$ and $\mu _u$ and $\nu _u$ have a finite transport cost. Let $T_u$ denote the (unique by Theorem 1.8) optimal transport map induced by the squared distance cost $d_M^2$ on M between $\mu _u$ and $\nu _u$. Then there exists a measurable function $(u,x)\mapsto T(u,x)$ such that for m-almost every u, $T(u,x)=T_{u}(x)$, $\mu _{u}$-a.e.

The proof is essentially an adaptation of the one of Theorem 1.1 from Fontbona et al. [18], with additional care required since we do not have access to a linear structure on the manifold M. It relies on the measurability of set-valued maps (see [49, Chapters 5 and 14] and “Appendix B.1” for a summary). In the following, we consider a partition $(A_{nk})_n$ of M made of cells with diameter $\le 2^{-k}$ and such that $(A_{nj})_n$ is a refinement of $(A_{nk})_n$ in sense that each $A_{nk}$ is itself partitioned by some $A_{n'j}$. The crucial point regarding measurability is the following proposition.

Proposition B.3

The set $B_{nk} = \{ (u,x)\mid T_u(x) \in A_{nk} \}$ is measurable.

Its proof relies on a core lemma:

Lemma B.4

Let $F \subset M$ be a closed set. Then the set $B_F = \{ (u,x)\mid T_u(x) \in F \}$ is measurable.

The key will be to identify this set as the domain of a measurable set-valued map, see Section “Appendix B.1.”

Proof of Lemma B.4

Observe that

$$\begin{aligned} B_F = \{(u,x)\mid (\{x\} \times F) \cap \mathop {\textrm{gph}}\limits (T_u) \ne \varnothing \}, \end{aligned}$$

where $\mathop {\textrm{gph}}\limits (T_u) :=\{(x,T_u(x))\mid x \in M\}$ denotes the topological closure of the graph of the optimal transport map $T_u$ that pushes $\mu _u$ onto $\nu _u$. Let $S_1: (u,x) \mapsto \{x\} \times F$ and $S_2: (u,x) \mapsto \mathop {\textrm{gph}}\limits (T_u)$, so that $B_F = \textrm{dom}(S)$, where $S(x) = S_1(x) \cap S_2(x)$. According to Proposition B.2, given that $S_1$ and $S_2$ are closed-valued, if they are measurable, so is S, and so is $B_F$ as the domain of a measurable map. Establishing the measurability of these two maps can be easily done by adapting the work of [18]. We give details here for the sake of completeness.

• Measurability of $S_1$: Let $O \subset M \times M$ be open. Observe that

  $$\begin{aligned} S_1^{-1}(O) = \{(u,x) \mid \{x\} \times F \cap O \ne \varnothing \} = B \times \{x \mid \{x\} \times F \cap O \ne \varnothing \}. \end{aligned}$$

  Fix $z \in F$. There exists $\epsilon > 0$ such that $B(x,\epsilon ) \times \{z\} \subset O$ (since O is open), and thus $B(x,\epsilon ) \times F \cap O \ne \varnothing$, proving that there is a neighborhood of x included in $\{x \mid \{x\} \times F \cap O \ne \varnothing \}$ which is thus open, making in turn $S_1^{-1}(O)$ open (hence measurable), hence the measurability of $S_1$.

• Measurability of $S_2$: Given that $u \mapsto (\mu _u,\nu _u)$ is measurable by assumption, and that measurability is preserved by composition, we want to show that

  • (i) the map $S: (\mu ,\nu ) \mapsto \Pi ^\star (\mu ,\nu )$ (the set of optimal transport plans between $\mu$ and $\nu$ for the squared distance cost on M) is measurable, and

  • (ii) the map $\mathfrak {U}: \pi \in P(M^2) \mapsto \mathop {\textrm{supp}}\limits \pi$ satisfies that $\mathfrak {U}^{-1}(O)$ is open for any open set $O \subset P(M^2)$.

  From these two points, we obtain the measurability of $(U \circ S)^{-1}(O)$, thus the measurability of $S_2$. To get (i), observe first that S is closed-valued, so that it is sufficient to prove that $S^{-1}(C)$ is measurable for any closed set $C \subset P(M^2)$ according to Proposition B.1. Let $C \subset P(M^2)$ be closed. Then, $S^{-1}(C) = \{(\mu ,\nu )\mid \Pi ^\star (\mu ,\nu ) \cap C \ne \varnothing \}$. Consider a sequence $(\mu _n, \nu _n)_n$ in $S^{-1}(C)$ that converges to $(\mu ,\nu )$ for the weak topology. Let $\pi _n \in \Pi ^\star (\mu _n,\nu _n) \cap C$. According to [60, Theorem 5.20], $(\pi _n)_n$ admits a weak limit $\pi$ in $\Pi ^\star (\mu ,\nu )$; but since C is closed, $\pi \in C$, so $(\mu ,\nu ) \in S^{-1}(C)$, which is closed (hence measurable), proving the measurability of S. (ii) simply follows from the fact that

  $$\begin{aligned} \mathfrak {U}^{-1}(O) = \{ \pi \mid \mathop {\textrm{supp}}\limits \pi \cap O \ne \varnothing \} = \{ \pi \mid \pi (O) > 0 \}, \end{aligned}$$

  which is open. Indeed, the Portmanteau theorem shows that if $\pi _n \rightarrow \pi$ (weakly) and $\pi _n(O) = 0$, then $0 = \liminf \pi _n(O) \ge \pi (O) \ge 0$, so $\pi (O) = 0$. The complementary set of $U^{-1}(O)$ is closed, that is $\mathfrak {U}^{-1}(O)$ is open.

$\square$

Proof of Proposition B.3

Using that $A_{nk}$ can be inner-approximated by a sequence of closed set $F_j \subset A_{nk}$, we obtain a sequence of sets $(B_{F_j})_j$ such that $\bigcup _j B_{F_j} = A_{nk}$. By Lemma B.4, the $(B_{F_j})_j$ are measurable, so is $A_{nk}$ as the (countable) union of measurable sets. $\square$

We can now prove our main proposition on measurable selections of maps.

Proof of Proposition 2.10

Recall that we assume that $M = \bigsqcup _n A_{nk}$. For each (n, k), select (in a measurable way) an element $a_{nk}\in A_{nk}$. Then, define the map

$$\begin{aligned} T^{(k)}: (u,x) \mapsto a_{nk},\ \text { where } T_u(x) \in A_{nk}\,. \end{aligned}$$

This map is measurable: for any open $O \subset M$,

$$\begin{aligned} T^{(k),-1}(O) = \{(u,x) \mid T_u(x) \in A_{nk} \} = B_{nk}, \end{aligned}$$

which is measurable by Proposition B.3. Now, for two maps $f,g: B \times M \rightarrow M$, let $D_1$ denotes the natural $L^1$ distance on M, that is

$$\begin{aligned} D_1(f,g) = \int _B \int _M d\big (f(u,x), g(u,x)\big ) \mathop {}\!\mathrm d\mu _u(x) \mathop {}\!\mathrm dm(u). \end{aligned}$$

This yields a complete metric space [12], and $(T^{(k)})_k$ is a Cauchy sequence for this distance. Indeed, for $k \le j$ two integers, recall that we assumed that $(A_{nj})_n$ is a refinement of $(A_{nk})_n$, yielding

$$\begin{aligned} D_1(T^{(k)}, T^{(j)})&= \int _{B} \int _M d\big (T^{(k)}(u,x), T^{(j)}(u,x)\big ) \mathop {}\!\mathrm d\mu _u (x) \mathop {}\!\mathrm dm(u) \\&= \int _{B} \int _M \sum _n \sum _{n' :\,A_{n'j} \subset A_{nk}} \mathbb {1}_{B_{n'j}}(u,x) d(a_{nk}, a_{n'j}) \mathop {}\!\mathrm d\mu _u(x) \mathop {}\!\mathrm dm(u) \\&= \int _B \int _M \sum _n \sum _{n' :\,A_{n'j} \subset A_{nk}} d(a_{nk},a_{n'j}) \mathop {}\!\mathrm d\nu _u(A_{n'j}) \mathop {}\!\mathrm dm(u) \\&\le 2^{-k} \end{aligned}$$

where we successively used (i) that for all u, $\int _{M} \mathbb {1}_{B_{n'j}}(u,x) \mathop {}\!\mathrm d\mu _u(x) = \nu _u(A_{n'j})$ by construction, (ii) that the diameter of the partition $A_{nk}$ is less than or equal to $2^{-k}$ and (iii) that $\nu _u$ and m are probability measures. Note that claim (i) is given by

$$\begin{aligned} (u,x) \in B_{n'j} \iff T_u(x) \in A_{n'j} \iff x \in \mu _u(T_u^{-1}(A_{n'j})) = T_{u \#} \mu _u(A_{n'j})=\nu _u(A_{n'j}). \end{aligned}$$

Now, let T denote the limit of $(T^{(k)})_k$ (which is measurable). It remains to show that $T(u,x) = T_u(x)$, m-a.e. This can be obtained by proving that

$$\begin{aligned} \int _M g(x) f(T(x,u)) \mathop {}\!\mathrm d\mu _u(x) = \int _M g(x) f(T_u(x)) \mathop {}\!\mathrm d\mu _u(x), \end{aligned}$$

for any pair $f,g: M \rightarrow \mathbb {R}$ of bounded Lipschitz-continuous functions [57, Lemma 2.24]. As in [18], let $\Vert f\Vert :=\sup _{x \ne y} \frac{|f(x) - f(y)|}{d(x,y)} + \sup _x |f(x)|$. The difference between these two terms can be bounded using the partition $(A_{nk})_n$. We have for m-a.e. u:

$$\begin{aligned}{} & {} \left| \int _M g(x) f(T_u(x)) \mathop {}\!\mathrm d\mu _u(x) - \int _M g(x) f(T(u,x)) \mathop {}\!\mathrm d\mu _u(x) \right| \\{} & {} \quad \le \left| \int _M g(x) f(T_u(x)) \mathop {}\!\mathrm d\mu _u(x) - \int _M g(x) f(T^{(k)}(u,x)) \mathop {}\!\mathrm d\mu _u(x) \right| \\{} & {} \qquad + \Vert g\Vert \Vert f\Vert \int _M d\big ( T^{(k)}(u,x), T(u,x) \big ) \mathop {}\!\mathrm d\mu _u(x). \end{aligned}$$

Since $T^{(k)} \rightarrow T$ in $D_1$, it implies that up to a subsequence, for m-a.e. u, the second term tends to zero:

$$\begin{aligned} \int _M d\big (T^{(k)}(u,x), T(u,x)\big ) \mathop {}\!\mathrm d\mu _u(x)\xrightarrow []{k\rightarrow \infty }0. \end{aligned}$$

To treat the first term and show that it tends to 0 as $k \rightarrow \infty$ for a subset of B with full m-measure, we write for m-a.e. u:

$$\begin{aligned}&\left| \int _M g(x) f(T_u(x)) \mathop {}\!\mathrm d\mu _u(x) - \int _M g(x) f(T^{(k)}(u,x)) \mathop {}\!\mathrm d\mu _u(x) \right| \\&\quad \le \int _M |g(x)| \big | f(T_u(x)) - f(T^{(k)}(u,x))\big | \mathop {}\!\mathrm d\mu _u(x) \\&\quad \le \Vert g\Vert \Vert f\Vert \int _M d\big (T_u(x),T^{(k)}(u,x)\big ) \mathop {}\!\mathrm d\mu _u(x) \\&\quad \le \Vert g\Vert \Vert f\Vert \ 2^{-k} \sum _n \nu _u(A_{nk})\xrightarrow []{k\rightarrow \infty }0, \end{aligned}$$

which concludes the proof. $\square$

Appendix C. Measure disintegration

The following definition and theorem are taken from Ambrosio et al. [1].

Definition C.1

(Measure disintegration) Let $\mathcal {X}$ and $\mathcal {Z}$ be two Radon spaces, $\mu \in \mathcal {P}(\mathcal {X})$ and $\varphi : \mathcal {X}\rightarrow \mathcal {Z}$ a Borel-measurable function. A family of probability measures $\{\mu _{u}\}_{u \in \mathcal {Z}} \subset \mathcal {P}(\mathcal {X})$ is a disintegration of $\mu$ by $\varphi$ if:

• (i) the function $u \mapsto \mu _{u}$ is Borel-measurable;

• (ii) $\mu _{u}$ lives on the fiber $\varphi ^{-1}(u)$, that is for $\varphi _\#\mu$-a.e. $u \in \mathcal {Z}$,

  $$\begin{aligned} \mu _{u}\big (\mathcal {X}\setminus \varphi ^{-1}(u)\big )=0, \end{aligned}$$

  and therefore, $\mu _{u}(B)=\mu _{u}(B \cap \varphi ^{-1}(u))$ for any Borel $B\subset \mathcal {X}$;

• (iii) for every measurable function $f: \mathcal {X}\rightarrow [0, \infty ]$,

  $$\begin{aligned} \int _{\mathcal {X}} f(x) \mathop {}\!\mathrm d\mu (x)=\int _{\mathcal {Z}} \Big (\int _{\varphi ^{-1}(u)} f(x) \mathop {}\!\mathrm d\mu _{u}(x)\Big ) \mathop {}\!\mathrm d(\varphi _\#\mu )(u). \end{aligned}$$

  In particular, for any Borel $B \subset \mathcal {X}$, taking f to be the indicator function of B,

  $$\begin{aligned} \mu (B)=\int _{\mathcal {Z}} \mu _{u}(B) \mathop {}\!\mathrm d(\varphi _\#\mu )(u). \end{aligned}$$

Theorem C.2

(Disintegration theorem) Let $\mathcal {X}$ and $\mathcal {Z}$ be two Radon spaces, $\mu \in \mathcal {P}(\mathcal {X})$ and $\varphi : \mathcal {X}\rightarrow \mathcal {Z}$ a Borel-measurable function. There exists a $\varphi _\#\mu$-a.e. uniquely determined family of probability measures $\{\mu _{u}\}_{u \in \mathcal {Z}} \subset \mathcal {P}(\mathcal {X})$ that provides a disintegration of $\mu$ by $\varphi$.