SlowGuess/NeurIPS_2023_OCR huggingface dataset:
In the main document, see Section 2, we say that "a path measure  $\mathbb{P}$  is associated with  $\mathrm{d}\mathbf{X}_t = b(t,\mathbf{X}_t)\mathrm{d}t + \sigma (t,\mathbf{X}_t)\mathrm{d}\mathbf{B}_t$  with  $(\mathbf{B}_t)_{t\geq 0}$  a  $d$ -dimensional Brownian motion" if  $\mathbb{P}$  solves the martingale problem associated with  $\mathcal{A}$  given by (4). Unless specified, we always assume that such a path measure exists and is unique. Below, we recall the following theorem, see (Stroock & Varadhan, 1997, Theorem 10.2.2), which gives sufficient conditions for the existence and uniqueness of solutions to the martingale problem.
Theorem 11. Assume that for any  $x\in \mathbb{R}^d$  we have
$$
\inf  \{\langle \theta , \sigma \sigma^ {\top} (s, x) \theta \rangle : \theta \in \mathbb {R} ^ {d}, \| \theta \| = 1, s \in [ 0, T ] \} > 0,
$$
$$
\lim  _ {y \to x} \sup  \{\| \sigma (s, x) - \sigma (s, y) \|: s \in [ 0, T ] \} = 0.
$$
In addition, assume that there exists  $C > 0$  such that for any  $x\in \mathbb{R}^d$
$$
\sup  \{\| \sigma \sigma^ {\top} (t, x) \|: s \in [ 0, T ] \} + \sup  \{\langle x, b (t, x) \rangle : s \in [ 0, T ] \} \leq C (1 + \| x \| ^ {2}).
$$
Then, there exists a unique solution to the martingale problem with initialization  $x_0 \in \mathbb{R}^d$ .
D Theoretical results on Tree Schrödinger Bridges
We respectively provide in Appendix D.1, Appendix D.2 and Appendix D.3 the proofs of the results of the main manuscript presented in Section 3, Section 4 and Section 5. Finally, we make a detailed comparison between our setting and the framework of Haasler et al. (2021) in Appendix D.4. In the rest of this section, we consider an undirected tree  $\mathsf{T} = (\mathsf{V},\mathsf{E})$ , where  $|\mathsf{V}| = \ell +1$ , and some subset  $\mathsf{S}\subset \mathsf{V}$  which we denote by  $\mathsf{S} = \{i_0,\dots ,i_{K - 1}\}$ . We define  $\mathsf{S}^c = \mathsf{V}\backslash \mathsf{S}$ .
D.1 Proofs of Section 3
Proposition 1 is straightforward to obtain by combining the definition of the Brownian motion with the definition of  $\pi^0$  given in (2). The following lemma details the recursion relation between the (mIPF) iterates, which is key to prove Proposition 2.
Lemma 12. Let  $(\pi^n)_{n\in \mathbb{N}}$  be the sequence given by (mIPF). Let  $n\in \mathbb{N}$ ,  $k_{n} = (n - 1)\bmod (K)$ ,  $k_{n} + 1 = n\bmod (K)$ . Denote by  $\mathsf{T}_{k_n}$ , respectively  $\mathsf{T}_{k_n + 1}$  with edges  $\mathsf{E}_{k_n + 1}$ , the directed version of  $\mathsf{T}$  rooted in  $i_{k_n}$ , respectively in  $i_{k_n + 1}$ . We have:
(i)  $\pi^n\in \mathcal{P}_{\mathsf{T}_{k_n}}$
(ii)  $\pi^{n + 1} = \mu_{i_{k_n + 1}}\bigotimes_{(v,v')\in \mathsf{E}_{k_n + 1}}\pi_{v'|v'}^n$  . In particular, for any  $(v,v^{\prime})\in \mathsf{E}_{k_n + 1}$ $\pi_{v^{\prime}|v}^{n + 1} = \pi_{v^{\prime}|v}^{n}$
Proof. We show the result (i) by recursion on  $n \in \mathbb{N}$ , and will deduce (ii) from the proof. Using (2), we first have  $\pi^0 \in \mathcal{P}_{\mathsf{T}_r}$ , where  $r$  is chosen as  $i_{K-1}$ , see Section 3. Thus, we obtain the result (i) at step  $n = 0$ . Assume now that  $\pi^n \in \mathcal{P}_{\mathsf{T}_{k_n}}$  for some  $n \in \mathbb{N}$ .
Consider the paths  $\mathsf{P}_n = \mathrm{path}_{\mathsf{T}_{k_n}}(i_{k_n},i_{k_n + 1})$  and  $\mathsf{P}_{n + 1} = \mathrm{path}_{\mathsf{T}_{k_n + 1}}(i_{k_n + 1},i_{k_n})$ . Note that these two paths have the same length, denoted by  $J$ , and contain the same vertices, denoted by  $\mathsf{V}_n$ . Let  $\pi \in \mathcal{P}^{(\ell +1)}$  such that  $\mathrm{KL}(\pi |\pi^n) < +\infty$ . We have the following decomposition
$$
\mathrm {K L} (\pi | \pi^ {n}) = \mathrm {K L} \left(\pi_ {\mathbb {V} _ {n}} | \pi_ {\mathbb {V} _ {n}} ^ {n}\right) + \int_ {(\mathbb {R} ^ {d}) ^ {J + 1}} \mathrm {K L} \left(\pi_ {| \mathbb {V} _ {n}} | \pi_ {| \mathbb {V} _ {n}} ^ {n}\right) \mathrm {d} \pi_ {\mathbb {V} _ {n}} \left(x _ {\mathbb {V} _ {n}}\right).
$$
Hence, the  $(n + 1)$ -th iterate of (mIPF) is given by  $\pi^{n + 1} = \pi_{\mathsf{V}_n}^{n + 1}\otimes \pi_{|\mathsf{V}_n}^n$ , with
$$
\pi_ {\mathrm {V} _ {n}} ^ {n + 1} = \operatorname {a r g m i n} \left\{\mathrm {K L} \left(\pi \mid \pi_ {\mathrm {V} _ {n}} ^ {n}\right): \pi \in \mathcal {P} ^ {(J + 1)}, \pi_ {i _ {k _ {n} + 1}} = \mu_ {i _ {k _ {n} + 1}} \right\}.
$$
Since  $\pi^n\in \mathcal{P}_{\mathsf{T}_{k_n}}$  , we have (i)  $\pi_{|\mathsf{V}_n}^n = \bigotimes_{(v,v')\in \mathsf{E}_{k_n}\setminus \mathsf{P}_n}\pi_{v'|_v}^n$  and (ii)  $\pi_{\mathsf{V}_n}^n\in \mathcal{P}_{\mathsf{P}_{n + 1}}$  by Lemma 9, where  $\mathsf{P}_{n + 1}$  is viewed as a directed tree rooted in  $i_{k_n + 1}$  . Defining  $\mathsf{V}_{n + 1} = \mathsf{V}_n\backslash \{i_{k_n + 1}\}$  , we thus have  $\pi_{\mathsf{V}_n}^n = \pi_{i_{k_n + 1}}^n\otimes \pi_{\mathsf{V}_{n + 1}|i_{k_n + 1}}^n$  where  $\pi_{\mathsf{V}_{n + 1}|i_{k_n + 1}}^n = \bigotimes_{(v,v')\in \mathsf{P}_{n + 1}}\pi_{v'|_v}^n.$
Let  $\pi \in \mathcal{P}^{(J + 1)}$  such that  $\pi_{i_{k_n + 1}} = \mu_{i_{k_n + 1}}$  and  $\mathrm{KL}(\pi |\pi_{\mathrm{V}_n}^n) < +\infty$ . Similarly to the previous computation, we have the following decomposition
$$
\begin{array}{l} \mathrm {K L} (\pi | \pi_ {\mathsf {V} _ {n}} ^ {n}) = \mathrm {K L} (\pi_ {i _ {k _ {n} + 1}} | \pi_ {i _ {k _ {n} + 1}} ^ {n}) + \int_ {\mathbb {R} ^ {d}} \mathrm {K L} (\pi_ {| i _ {k _ {n} + 1}} | \pi_ {\mathsf {V} _ {n + 1} | i _ {k + 1}} ^ {n}) \mathrm {d} \pi_ {i _ {k _ {n} + 1}} (x _ {i _ {k _ {n} + 1}}) \\ = \mathrm {K L} (\mu_ {i _ {k _ {n} + 1}} | \pi_ {i _ {k _ {n} + 1}} ^ {n}) + \int_ {\mathbb {R} ^ {d}} \mathrm {K L} (\pi_ {| i _ {k _ {n} + 1}} | \pi_ {\bigvee_ {n + 1} | i _ {k _ {n} + 1}} ^ {n}) \mathrm {d} \mu_ {i _ {k _ {n} + 1}} (x _ {i _ {k _ {n} + 1}}). \\ \end{array}
$$
$$
\pi_ {\mathbf {V} _ {n}} ^ {n + 1} = \mu_ {i _ {k _ {n} + 1}} \otimes \pi_ {\mathbf {V} _ {n + 1} | i _ {k + 1}} ^ {n} = \mu_ {i _ {k + 1}} \bigotimes_ {(v, v ^ {\prime}) \in \mathsf {P} _ {n + 1}} \pi_ {v ^ {\prime} | v} ^ {n}.
$$
Noting that  $\mathsf{E}_{k_n} \backslash \mathsf{P}_n = \mathsf{E}_{k_n + 1} \backslash \mathsf{P}_{n + 1}$  and recalling that  $\pi^{n + 1} = \pi_{\mathsf{V}_n}^{n + 1} \otimes \pi_{|\mathsf{V}_n}^n$ , it finally comes
$$
\pi^ {n + 1} = \mu_ {i _ {k _ {n} + 1}} \bigotimes_ {(v, v ^ {\prime}) \in \mathsf {P} _ {n + 1}} \pi_ {v ^ {\prime} | v} ^ {n} \bigotimes_ {(v, v ^ {\prime}) \in \mathsf {E} _ {k _ {n} + 1} \backslash \mathsf {P} _ {n + 1}} \pi_ {v ^ {\prime} | v} ^ {n} = \mu_ {i _ {k _ {n} + 1}} \bigotimes_ {(v, v ^ {\prime}) \in \mathsf {E} _ {k _ {n} + 1}} \pi_ {v ^ {\prime} | v} ^ {n}. \tag {5}
$$
Therefore,  $\pi^{n + 1}\in \mathcal{P}_{\mathrm{T}_{k_n + 1}}$  , which achieves the recursion for (i), and we obtain (ii) by (5).
Hence, Lemma 12 shows that the (mIPF) iterates admit a Markovian factorization on  $\mathsf{T}$ , and can be defined recursively using the edges of  $\mathsf{T}$ . We now provide the proof of Proposition 2.
Proof of Proposition 2. We will prove this result by recursion on  $n \in \mathbb{N}$ . Observe that the initialisation is directly given by Proposition 1. Assume now that the result of Proposition 2 stands for some  $n \in \mathbb{N}$ . Let  $k_{n} = (n - 1) \bmod(K)$ ,  $k_{n} + 1 = n \bmod(K)$ . Denote by  $\mathsf{T}_{k_n}$  with edges  $\mathsf{E}_{k_n}$ , respectively  $\mathsf{T}_{k_n + 1}$  with edges  $\mathsf{E}_{k_n + 1}$ , the directed version of  $\mathsf{T}$  rooted in  $i_{k_n}$ , respectively in  $i_{k_n + 1}$ . For any vertex  $v$  of  $\mathsf{T}_{k_n + 1}$ , we define  $p(v)$  as the (unique) parent of  $v$  and  $c(v)$  as the unique child of  $v$  when it exists. Consider the  $(n + 1)$ -th dynamic iterate defined by (a) and (b), i.e.,  $(\mathbb{P}_{(v,v')}^{n + 1})(v,v') \in \mathsf{E}_{k_n + 1}$ . To prove that this iterate has the properties stated in Proposition 2, we proceed by recursion on the edges of  $\mathsf{T}_{k_n + 1}$ , following the bread-first order in  $\mathsf{T}_{k_n + 1}$ . In this order, the edge  $(i_{k_n + 1}, c(i_{k_n + 1}))$  is the first to be considered. Remark that  $c(i_{k_n + 1})$  is well defined since  $i_{k_n + 1}$  is a leaf of  $\mathsf{T}$ .
Here, we denote  $T_{c(i_{k_n + 1}),i_{k_n + 1}}$  by  $T$ . By construction, we have  $\mathbb{P}_{(i_{k_n + 1},c(i_{k_n + 1}))}^{n + 1} = \mu_{i_{k_n + 1}}\otimes (\mathbb{P}_{(c(i_{k_n + 1}),i_{k_n + 1})}^n)_{|0}^R$ . By recursion assumption,  $\mathbb{P}_{(c(i_{k_n + 1}),i_{k_n + 1})}^n\in \mathcal{P}(\mathrm{C}([0,T],\mathbb{R}^d))$  since
$(c(i_{k_n + 1}),i_{k_n + 1})\in \mathsf{E}_{k_n}$ . Then,  $\mathbb{P}_{(i_{k_n + 1},c(i_{k_n + 1}))}^{n + 1}$  is a well defined path measure on  $[0,T]$ . By definition of the (mIPF) sequence, we have  $\mu_{i_{k_n + 1}} = \pi_{i_{k_n + 1}}^{n + 1}$ . By recursion assumption, we also have that  $\mathrm{Ext}(\mathbb{P}_{(c(i_{k_n + 1}),i_{k_n + 1})}^n) = \pi_{c(i_{k_n + 1}),i_{k_n + 1}}^n$ . Hence, it comes that  $(\mathbb{P}_{(c(i_{k_n + 1}),i_{k_n + 1})}^n)_{T|0}^R = \pi_{c(i_{k_n + 1})|i_{k_n + 1}}^n = \pi_{c(i_{k_n + 1})|i_{k_n + 1}}^{n + 1}$ , where the last equality comes from Lemma 12. Finally, we obtain that  $\mathrm{Ext}(\mathbb{P}_{(i_{k_n + 1},c(i_{k_n + 1}))}^{n + 1}) = \pi_{i_{k_n + 1},c(i_{k_n + 1})}^{n + 1}$ , which proves the initialisation.
Assume now that  $\mathbb{P}^{n + 1}$  is well defined and has the right properties, up to some edge in  $\mathsf{T}_{k_n + 1}$ . Consider the following edge, denoted by  $(v,v^{\prime})\in E_{k_n + 1}$ , in the breadth-first order. By edge recursion, we have that  $\mathrm{Ext}(\mathbb{P}_{(p(v),v)}^{n + 1}) = \pi_{p(v),v}^{n + 1}$ , and thus  $\mathbb{P}_{(p(v),v),T_{p(v)},v}^{n + 1} = \pi_v^{n + 1}$ . Define the path  $\mathsf{P}_n = \mathrm{path}_{\mathsf{T}_{k_n}}(i_{k_n},i_{k_n + 1})$ . Then, we face two cases.
(i) Either  $(v,v^{\prime})\in \mathsf{E}_{k_n}\backslash \mathsf{P}_n$  . Then, we have by (a) that
$$
\mathbb {P} _ {(v, v ^ {\prime})} ^ {n + 1} = \mathbb {P} _ {(p (v), v), T _ {p (v), v}} ^ {n + 1} \otimes \mathbb {P} _ {(v, v ^ {\prime}) | 0} ^ {n} = \pi_ {v} ^ {n + 1} \otimes \mathbb {P} _ {(v, v ^ {\prime}) | 0} ^ {n}
$$
In particular,  $\mathbb{P}_{(v,v^{\prime})}^{n + 1}$  is a well defined path measure on  $[0,T_{v,v^{\prime}}]$ . Since  $(v,v^{\prime})\in \mathsf{E}_{k_n}$ ,  $\operatorname {Ext}(\mathbb{P}_{(v,v^{\prime})}^{n}) = \pi_{v,v^{\prime}}^{n}$  by recursion assumption. In particular,  $\mathbb{P}_{(v,v^{\prime}),T_{v,v^{\prime}}|0}^{n} = \pi_{v^{\prime}|v}^{n} = \pi_{v^{\prime}|v}^{n + 1}$  where the last equality comes from Lemma 12. We thus have  $\operatorname {Ext}(\mathbb{P}_{(v,v^{\prime})}^{n + 1}) = \pi_{v,v^{\prime}}^{n + 1}$ .
(ii) Or  $(v', v) \in \mathsf{P}_n$ . Then, we have by (b) that
$$
\mathbb {P} _ {(v, v ^ {\prime})} ^ {n + 1} = \mathbb {P} _ {(p (v), v), T _ {v, v ^ {\prime}}} ^ {n + 1} \otimes (\mathbb {P} _ {(v ^ {\prime}, v)} ^ {n}) _ {| 0} ^ {R} = \pi_ {v} ^ {n + 1} \otimes (\mathbb {P} _ {(v ^ {\prime}, v)} ^ {n}) _ {| 0} ^ {R}
$$
In particular,  $\mathbb{P}_{(v,v^{\prime})}^{n + 1}$  is a well defined path measure on  $[0,T_{v,v^{\prime}}]$ . Here,  $(v^{\prime},v)\in \mathsf{E}_{k_n}$  and thus,  $\mathrm{Ext}(\mathbb{P}_{(v^{\prime},v)}^{n}) = \pi_{v^{\prime},v}^{n}$  by recursion assumption. In particular,  $(\mathbb{P}_{(v^{\prime},v)}^{n})_{T_{v^{\prime},v}|0}^{R} = \pi_{v^{\prime}|v}^{n} = \pi_{v^{\prime}|v}^{n + 1}$  where the last equality comes from Lemma 12. We thus have  $\mathrm{Ext}(\mathbb{P}_{(v,v^{\prime})}^{n + 1}) = \pi_{v,v^{\prime}}^{n + 1}$ .
This achieves the recursion.
D.2 Proofs of Section 4
Remark on assumption A1. Although A1 is not needed to establish the result of Proposition 3, Corollary 4 and Proposition 5, it is however crucial in the proof of convergence of (mIPF) stated in Proposition 6. Nevertheless, we choose to keep A1 as an assumption in the statement of every theoretical result presented in Section 4 for sake of clarity.
Additional definitions. We define the set  $\mathcal{P}_{\mathsf{S}} = \cap_{i\in \mathsf{S}}\mathcal{P}_i$ , where  $\mathcal{P}_i = \{\pi \in \mathcal{P}^{(\ell +1)}:\pi_i = \mu_i\}$ , i.e.,  $\mathcal{P}_{\mathsf{S}}$  is the set of all probability measures  $\pi \in \mathcal{P}^{(\ell +1)}$  which verify
$$
\int_ {(\mathbb {R} ^ {d}) ^ {\ell + 1}} f _ {i} \left(x _ {i}\right) \mathrm {d} \pi \left(x _ {0: \ell}\right) = \int_ {\mathbb {R} ^ {d}} f _ {i} \left(x _ {i}\right) \mathrm {d} \mu_ {i} \left(x _ {i}\right),
$$
for any family of bounded measurable functions  $\{f_i\}_{i\in \mathbb{S}}\in \mathrm{C}(\mathbb{R}^d,\mathbb{R})^K$ . Since  $\mathbb{R}^d$  is separable, there exists a dense family of functions  $\{f_i^k\}_{k\in \mathbb{N}^*,i\in \mathbb{S}}$ , with  $f_{i}^{k}\in \mathrm{L}^{\infty}(\mu_{i})$  for any  $k\in \mathbb{N}^*$  and any  $i\in \mathbb{S}$  such that  $\pi \in \mathcal{P}_S$  if and only if
$$
\int_ {(\mathbb {R} ^ {d}) ^ {\ell + 1}} f _ {i} ^ {k} \left(x _ {i}\right) \mathrm {d} \pi \left(x _ {0: \ell}\right) = \int_ {\mathbb {R} ^ {d}} f _ {i} ^ {k} \left(x _ {i}\right) \mathrm {d} \mu_ {i} \left(x _ {i}\right)
$$
or equivalently, upon centering  $f_{i}^{k}$
$$
\int_ {(\mathbb {R} ^ {d}) ^ {\ell + 1}} f _ {i} ^ {k} \left(x _ {i}\right) \mathrm {d} \pi \left(x _ {0: \ell}\right) = 0.
$$
In the rest of the section, we consider such family  $\{f_i^k\}_{k\in \mathbb{N}^*,i\in \mathbb{S}}$
For any  $n \in \mathbb{N}^*$ , we also define  $\mathcal{P}_{\mathsf{S}}^{n} = \cap_{i \in \mathsf{S}} \mathcal{P}_{i}^{n}$ , where  $\mathcal{P}_{i}^{n} = \{\pi \in \mathcal{P}^{(\ell + 1)} : \int_{(\mathbb{R}^{d})^{\ell + 1}} f_{i}^{k}(x_{i}) \mathrm{d}\pi(x_{0: \ell}) = 0, \forall k \in \{1, \dots, n\}\}$ . In particular, we have
$$
\mathcal {P} _ {\mathbb {S}} = \cap_ {n \in \mathbb {N} ^ {*}} \mathcal {P} _ {\mathbb {S}} ^ {n}. \tag {6}
$$
Finally, (static-mSB) can be rewritten as
$$
\pi^ {\star} = \operatorname {a r g m i n} \left\{\mathrm {K L} \left(\pi \mid \pi^ {0}\right): \pi \in \mathcal {P} _ {\mathrm {S}} \right\}. \tag {7}
$$
Proof of Proposition 3 and Corollary 4. In this part of the section, we present an extension of the theoretical results from Nutz (2021) to the multi-marginal setting. We first present two technical results, Lemma 13 and Lemma 14, which are respectively adapted from (Nutz, 2021, Lemma 2.10.) and (Nutz, 2021, Lemma 2.11.).
Lemma 13. Let  $\{\tilde{\mu}_j\}_{j\in \mathbb{S}^c}$  be a family of probability measures defined on  $(\mathbb{R}^d,\mathcal{B}(\mathbb{R}^d))$ . We define  $\tilde{\pi}^0 = \bigotimes_{i\in \mathbb{S}}\mu_i\bigotimes_{j\in \mathbb{S}^c}\tilde{\mu}_j$ . Let  $\mathsf{A}\in \bigotimes_{m = 0}^{\ell}\mathcal{B}(\mathbb{R}^{d})$  such that  $\tilde{\pi}^0 (\mathsf{A}) = 1$ . Then, for  $\tilde{\pi}^0$ -almost any  $x^{\star}\in \mathsf{A}$ , there exists a family of sets  $\{\mathsf{X}_m^0\}_{m = 0}^\ell \subset (\mathbb{R}^d)^{\ell +1}$  such that
(a)  $\mu_{i}(\mathsf{X}_{i}^{0}) = 1$  for any  $i\in \mathbb{S}$ , and  $\tilde{\mu}_j(\mathsf{X}_j^0) = 1$  for any  $j\in \mathbb{S}^c$
(b)  $\mathsf{A}^0 = \mathsf{A}\cap (\prod_{m = 0}^\ell \mathsf{X}_m^0)$  satisfies  $x^{\star}\in \mathsf{A}^{0}$  and
$$
\left(x _ {0} ^ {\star}, \dots , x _ {m - 1} ^ {\star}, x _ {m}, x _ {m + 1} ^ {\star}, \dots , x _ {\ell} ^ {\star}\right) \in \mathsf {A} ^ {0}, \forall x \in \mathsf {A} ^ {0}, \forall m \in \{0, \dots , \ell \}.
$$
Proof. Consider such set  $\mathsf{A}$ . We define for any  $m\in \{0,\dots ,\ell \}$  the set
$$
\mathsf {X} _ {m} = \left\{u \in \mathbb {R} ^ {d}: \tilde {\pi} _ {- m} ^ {0} (\mathsf {A} _ {m} ^ {u}) = 1 \right\},
$$
where  $\mathsf{A}_m^u = \{y\in (\mathbb{R}^d)^\ell :(y_0,\ldots ,y_{m - 1},u,y_m,\ldots ,y_{\ell -1})\in \mathsf{A}\}$
Take  $i \in \mathbb{S}$ . Assume that  $\mu_i(\mathsf{X}_i) < 1$ . We recall that  $\tilde{\pi}^0 = \tilde{\pi}_{-i}^0 \otimes \mu_i$ . Using Fubini's theorem and that  $\int_{\mathsf{A}_i^{x_i}} \mathrm{d}\tilde{\pi}_{-i}^0(x_{-i}) < 1$  for any  $x_i \notin \mathsf{X}_i$ , we have
$$
\begin{array}{l} 1 = \tilde {\pi} ^ {0} (\mathsf {A}) = \int_ {\mathsf {A}} \mathrm {d} \tilde {\pi} _ {- i} ^ {0} (x _ {- i}) \otimes \mathrm {d} \mu_ {i} (x _ {i}) \\ = \int_ {\mathbb {R} ^ {d}} \left\{\int_ {\mathsf {A} _ {i} ^ {x _ {i}}} \mathrm {d} \tilde {\pi} _ {- i} ^ {0} (x _ {- i}) \right\} \mathrm {d} \mu_ {i} (x _ {i}) \\ = \int_ {\mathsf {X} _ {i}} \left\{\int_ {\mathsf {A} _ {i} ^ {x _ {i}}} \mathrm {d} \tilde {\pi} _ {- i} ^ {0} (x _ {- i}) \right\} \mathrm {d} \mu_ {i} (x _ {i}) + \int_ {\mathsf {X} _ {i} ^ {\mathrm {c}}} \left\{\int_ {\mathsf {A} _ {i} ^ {x _ {i}}} \mathrm {d} \tilde {\pi} _ {- i} ^ {0} (x _ {- i}) \right\} \mathrm {d} \mu_ {i} (x _ {i}) \\ <   \mu_ {i} \left(X _ {i}\right) + \mu_ {i} \left(X _ {i} ^ {\mathrm {c}}\right) = 1, \\ \end{array}
$$
which is absurd. Therefore, we obtain  $\mu_i(\mathsf{X}_i) = 1$ , and similarly, we have  $\tilde{\mu}_j(\mathsf{X}_j) = 1$  for any  $j \in \mathbb{S}^c$ . For any  $y \in (\mathbb{R}^d)^\ell$ , any  $m \in \{0, \dots, \ell\}$ , we define the set
$$
\bar {\mathsf {A}} _ {m} ^ {y} = \left\{u \in \mathbb {R} ^ {d}: (y _ {0}, \ldots , y _ {m - 1}, u, y _ {m}, \ldots , y _ {\ell - 1}) \in \mathsf {A} \right\}.
$$
Let  $i\in \mathbb{S}$  .We have by Fubini's theorem
$$
\begin{array}{l} 1 = \tilde {\pi} ^ {0} (\mathsf {A}) = \int_ {\mathsf {A}} \mathrm {d} \mu_ {i} (x _ {i}) \otimes \mathrm {d} \tilde {\pi} _ {- i} ^ {0} (x _ {- i}) \\ = \int_ {(\mathbb {R} ^ {d}) _ {\ell}} \left\{\int_ {\tilde {\mathrm {A}} _ {i} ^ {x - i}} \mathrm {d} \mu_ {i} (x _ {i}) \right\} \mathrm {d} \tilde {\pi} _ {- i} ^ {0} (x _ {- i}) \\ = \int_{\substack{\prod_{\substack{m = 0\\ m\neq i}}^{\ell}\times_{i}\left\{\int_{\bar{\mathsf{A}}_{i}^{x - i}}\mathrm{d}\mu_{i}(x_{i})\right\} \mathrm{d}\tilde{\pi}_{-i}^{0}(x_{-i})  , \\ \end{array}
$$
where the last equality comes from the fact that  $\mu_i(\mathsf{X}_i) = 1$  for any  $i\in \mathbb{S}$ ,  $\tilde{\mu}_j(\mathsf{X}_j) = 1$  for any  $j\in \mathbb{S}^c$  and that  $\tilde{\pi}^{0} = \bigotimes_{i\in \mathbb{S}}\mu_{i}\bigotimes_{j\in \mathbb{S}^{\mathrm{c}}}\tilde{\mu}_{j}$ . Consequently, there exists a measurable set  $\mathsf{A}_{-i}\subset \prod_{\substack{m = 0\\ m\neq i}}^{\ell}\mathsf{X}_i$  such that the following properties hold: (a)  $\mu_i(\bar{\mathsf{A}}_i^y) = 1$  for any  $y\in \mathsf{A}_{-i}$ , (b)  $\tilde{\pi}_{-i}^{0}(\mathsf{A}_{-i}) = 1$ . Similarly, this result holds for any  $j\in \mathbb{S}^c$ , i.e., there exists a measurable set  $\mathsf{A}_{-j}\subset \prod_{\substack{m = 0\\ m\neq j}}^{\ell}\mathsf{X}_i$  such that the following properties hold: (a)  $\tilde{\mu}_j(\bar{\mathsf{A}}_j^y) = 1$  for any  $y\in \mathsf{A}_{-j}$ , (b)  $\tilde{\pi}_{-j}^{0}(\mathsf{A}_{-j}) = 1$ . We consider such sets  $\{\mathsf{A}_{-m}\}_{m = 0}^{\ell}$  for the rest of the proof and finally define the set
$$
\tilde {\mathsf {A}} = \cap_ {m = 0} ^ {\ell} \tilde {\mathsf {A}} _ {m},
$$
where  $\tilde{\mathsf{A}}_m = \mathsf{A}_{-m}\times \{u\in \bar{\mathsf{A}}_m^y:y\in \mathsf{A}_{-m}\}$ . By definition, we have  $\tilde{\mathsf{A}}\subset \mathsf{A}\cap \prod_{m = 0}^{\ell}\mathsf{X}_m$ , using the fact that  $\tilde{\mathsf{A}}_m\subset \mathsf{A}$  for any  $m\in \{0,\ldots ,\ell \}$ . In addition, for any  $i\in S$ , we get by Fubini's theorem
$$
\tilde {\pi} ^ {0} (\tilde {\mathsf {A}} _ {i}) = \int_ {\tilde {\mathsf {A}} _ {i}} \mathrm {d} \mu_ {i} (x _ {i}) \otimes \mathrm {d} \tilde {\pi} _ {- i} ^ {0} (x _ {- i}) = \int_ {\mathsf {A} _ {- i}} \{\int_ {\tilde {\mathsf {A}} _ {i} ^ {x _ {- i}}} \mathrm {d} \mu_ {i} (x _ {i}) \} \mathrm {d} \tilde {\pi} _ {- i} ^ {0} (x _ {- i}) = \tilde {\pi} _ {- i} ^ {0} (\mathsf {A} _ {- i}) = 1,
$$
and similarly, we get  $\tilde{\pi}^0 (\tilde{\mathsf{A}}_j) = 1$  for any  $j\in \mathbb{S}^c$ . We can deduce that  $\tilde{\pi}^0 (\tilde{\mathsf{A}}) = 1$  since  $\tilde{\pi}^0 (\tilde{\mathsf{A}}^c)\leq \sum_{m = 0}^{\ell}\tilde{\pi}^{0}(\tilde{\mathsf{A}}_{m}^{c}) = 0$ .
Let  $x^{\star} \in \tilde{\mathsf{A}}$ . In particular,  $x^{\star} \in \mathsf{A}$ . We define the set  $\mathsf{A}^{0} = \mathsf{A} \cap (\prod_{m=0}^{\ell} \mathsf{X}_{m}^{0})$ , where  $\mathsf{X}_{m}^{0} = \mathsf{X}_{m} \cap \bar{\mathsf{A}}_{m}^{x^{\star}_{-m}}$  for any  $m \in \{0, \dots, \ell\}$ . We now establish the result of Lemma 13.
We first prove (a). Let  $i \in \mathbb{S}$ . Since  $x^{\star} \in \tilde{\mathsf{A}}$ , we have  $x^{\star} \in \tilde{\mathsf{A}}_i$  and therefore  $x_{-i}^{\star} \in \mathsf{A}_{-i}$ . By definition of  $\mathsf{A}_{-i}$ , we obtain that  $\mu_i(\bar{\mathsf{A}}_i^{x_{-i}^{\star}}) = 1$  and thus,
$$
\mu_ {i} (\{\mathrm {X} _ {i} ^ {0} \} ^ {\mathrm {c}}) \leq \mu_ {i} (\mathrm {X} _ {i} ^ {\mathrm {c}}) + \mu_ {i} (\{\bar {\mathrm {A}} _ {i} ^ {x _ {- i} ^ {\star}} \} ^ {\mathrm {c}}) = 0,
$$
which gives  $\mu_{i}(\mathsf{X}_{i}^{0}) = 1$ , and similarly, we have  $\tilde{\mu}_j(\mathsf{X}_j^0) = 1$  for any  $j\in S^c$ .
We now prove (b). Let  $m \in \{0, \ldots, \ell\}$ . Since  $x^{\star} \in \tilde{\mathsf{A}} \subset \mathsf{A}$ , we get  $x_{m}^{\star} \in \bar{\mathsf{A}}_{m}^{x_{-m}^{\star}}$ . Using that  $\tilde{\mathsf{A}} \subset \mathsf{A} \cap_{m=0}^{\ell} \mathsf{X}_{m}$ , we get  $x^{\star} \in \mathsf{A}^{0}$ . Let  $x \in \mathsf{A}^{0}$ . We denote  $x^{m} = (x_{0}^{\star}, \ldots, x_{m-1}^{\star}, x_{m}, x_{m+1}^{\star}, \ldots, x_{\ell}^{\star})$ . We need to show that  $x^{m} \in \mathsf{A}$  and  $x^{m} \in \prod_{j=1}^{\ell} \mathsf{X}_{j}^{0} = \prod_{j=1}^{\ell} (\mathsf{X}_{j} \cap \bar{\mathsf{A}}_{j}^{x_{-m}^{\star}})$ . First, since  $x_{j}^{m} = x_{j}$  or  $x_{j}^{\star}$  for any  $j \in \{0, \ldots, \ell\}$ , and  $x \in \mathsf{A}^{0}$  and  $x^{\star} \in \mathsf{A}^{0}$ , we get that for any  $j \in \{0, \ldots, \ell\}$ ,  $x_{j}^{m} \in \mathsf{X}_{j}$ . Similarly, for any  $j \in \{0, \ldots, \ell-1\}$ ,  $x_{j}^{m} \in \bar{\mathsf{A}}_{j}^{x_{m}^{\star}}$ . Therefore, we get that  $x^{m} \in \prod_{j=1}^{\ell} (\mathsf{X}_{j} \cap \bar{\mathsf{A}}_{j}^{x_{-m}^{\star}})$ . Since  $x_{m} \in \mathsf{A}_{m}^{x_{-m}^{\star}}$  (because  $x \in \prod_{j=1}^{\ell} (\mathsf{X}_{j} \cap \bar{\mathsf{A}}_{j}^{x_{-m}^{\star}})$ ), we get that  $x \in \mathsf{A}$ , which concludes the proof.
Lemma 14. Let  $\mathsf{A}^0\subset (\mathbb{R}^d)^{\ell +1}$ . For any  $m\in \{0,\ldots ,\ell \}$ , we denote  $\mathsf{X}_m^0 = \mathrm{proj}_m(\mathsf{A}^0)$ . We make the following assumptions.
(a) Assume there exists  $x^{\star} \in \mathsf{A}^{0}$  such that for any  $x \in \mathsf{A}^{0}$ , for any  $m \in \{0, \dots, \ell\}$ , we have  $(x_0^{\star}, \ldots, x_{m-1}^{\star}, x_m, x_{m+1}^{\star}, \ldots, x_\ell^{\star}) \in \mathsf{A}^{0}$ .
(b) Assume there exists a family of functions  $\{\varphi_{i_k}^n\}_{n\in \mathbb{N}^*,k\in \{0,\ldots ,K - 1\}}$  with  $\varphi_{i_k}^n:\mathsf{X}_{i_k}^0\to [-\infty , + \infty)$  such that for any  $n\in \mathbb{N}^*$  and any  $k\in \{0,\dots ,K - 2\}$ , we have  $\varphi_{i_k}^n (x_{i_k}^\star) = 0$
(c) Denote  $F^n(x) = \sum_{k=0}^{K-1} \varphi_{i_k}^n(x_{i_k})$  for any  $x \in \mathsf{A}^0$ . Assume that for any  $x \in \mathsf{A}^0$ ,  $F(x) = \lim_{n \to \infty} F^n(x)$  exists and is such that  $F(x) \in [-\infty, +\infty)$  with  $F(x^\star) \in \mathbb{R}$ .
Then, for any  $i \in \mathsf{S}$ , for any  $x_i \in \mathsf{X}_i^0$ ,  $\varphi_i(x_i) = \lim_{n \to \infty} \varphi_i^n(x_i)$  exists and is such that  $\varphi_i(x_i) \in [-\infty, +\infty)$ .
Proof. Consider  $\mathsf{A}^0\subset (\mathbb{R}^d)^{\ell +1}$  such that assumptions (a), (b) and (c) hold. Remark that we have  $F^n (x^\star) = \varphi_{i_{K - 1}}^n (x_{i_{K - 1}}^\star)$ .
Let  $x \in \mathsf{A}^0$ . We denote  $x^{m} = (x_{1}^{\star},\ldots ,x_{m - 1}^{\star},x_{m},x_{m + 1}^{\star},\ldots ,x_{l}^{\star})$  for any  $m \in \{0,\dots ,\ell \}$ . In particular, we have  $x^{m} \in \mathsf{A}^{0}$  by assumption (a). Let us define
$$
\varphi_ {i _ {k}} \left(x _ {i _ {k}}\right) = F \left(x ^ {i _ {k}}\right) - F \left(x ^ {\star}\right), \quad \forall k \in \{0, \dots , K - 2 \},
$$
$$
\varphi_ {i _ {K - 1}} \left(x _ {i _ {K - 1}}\right) = F \left(x ^ {i _ {K - 1}}\right).
$$