# Comptes Rendus
Mathématique

Thibaut Le Gouic

A note on flatness of non separable tangent cone at a barycenter

Volume 358, issue 4 (2020), p. 489-495

Published online: 28 July 2020

https://doi.org/10.5802/crmath.66

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Geometry, Probability Theory / Géométrie, Probabilité

# A note on flatness of non separable tangent cone at a barycenter

*Une note sur la platitude du cône tangeant à un barycentre*

Thibaut Le Gouic<sup>a</sup>

<sup>a</sup> Massachusetts Institute of Technology, Department of Mathematics and Centrale
Marseille, I2M, UMR 7373, CNRS, Aix-Marseille univ., Marseille, 13453, France.

E-mail: thibaut.le_gouic@math.cnrs.fr.

**Abstract.** Given a probability measure $\mathbf{P}$ on an Alexandrov space $S$ with curvature bounded below, we prove that the support of the pushforward of $\mathbf{P}$ on the tangent cone $T_{b^*}S$ at its (exponential) barycenter $b^*$ is a subset of a Hilbert space, without separability of the tangent cone.

**Résumé.** Étant donné une mesure de probabilité $\mathbf{P}$ sur un espace d'Alexandrov $S$ avec courbure minorée, nous prouvons que le support de la mesure poussée de $\mathbf{P}$ sur le cône tangent $T_{b^*}S$ à son barycentre (exponentiel) $b^*$ est un sous-ensemble d'un espace de Hilbert, sans condition de séparabilité du cône tangent.

**Funding.** Supported by ONR grant N00014-17-1-2147.

Manuscript received 16th September 2019, revised 31st December 2019 and 20th April 2020, accepted 29th April 2020.

## 1. Introduction

Barycenter of a probability measure $\mathbf{P}$ (a.k.a. Fréchet means) provides an extension of expectation on Euclidean space to arbitrary metric spaces. We present here a useful tool for the study of barycenters on Alexandrov spaces with curvature bounded below: the support of $\log_{b^*} \mathbf{P}$ in the tangent cone at the barycenter is included in a Hilbert space. This rigidity result has been stated in [9] as Theorem 45, however the proof is not written. Moreover, there is an extra assumption of support of $\log_{b^*} \mathbf{P}$ being separable, which does not even seem to be a consequence of the support of $\mathbf{P}$ being separable. As pointed out by [7], it is not clear if even $S$ being proper ensures that the tangent cone is separable. This paper presents a proof of this rigidity result, without this extra separable assumption on the tangent cone. For measurability purposes (see Lemma 6), we suppose however that $S$ is separable. The proof is essentially the one of Theorem 45 of [9], with needed approximations dealt with a bit differently.
## 2. Setting and main result

We use a classical notion of curvature bounded below for geodesic spaces, referred to as Alexandrov curvature. We recall several notions whose formal definitions can be found for instance in [3] or in the work in progress [2].

For a metric space $(S, d)$, we denote by $\mathcal{P}_1(S)$ the set of probability measures on $S$ with finite moment of order 1 (i.e. such that there exists $x \in S$ such that $\int d(x, y)d\mathbf{P}(y) < \infty$). The support of a measure $\mathbf{P}$ will be denoted by $\text{supp} \mathbf{P}$. We use both notation $\int f d\mathbf{P}$ and $\mathbf{P}f$ for the integral of $f$ w.r.t. $\mathbf{P}$.

A geodesic space is a metric space $(S, d)$ such that every two points $x, y \in S$ at distance is connected by a curve of length $d(x, y)$. Such shortest curves are called geodesics. For $\kappa \in \mathbf{R}$, the model space $(\mathbf{M}_\kappa, d_\kappa)$ denotes the 2-dimensional simply connected complete surface of constant Gauss curvature $\kappa$. A geodesic space $(S, d)$ is an Alexandrov space with curvature bounded below by $\kappa \in \mathbf{R}$ if for every triangle (3-uple) $(x_0, x_1, y) \in S$, and a constant speed geodesic $(x_t)_{t \in [0;1]}$ there exists an isometric triangle $(\tilde{x}_0, \tilde{x}_1, \tilde{y}) \in \mathbf{M}_\kappa$, such that the geodesic $(\tilde{x}_t)_{t \in [0;1]}$ satisfies for all $t \in [0; 1]$,

$$d(y, x_t) \ge d_\kappa(\tilde{y}, \tilde{x}_t).$$

For such spaces, angles between two unit-speed geodesics $\gamma_1, \gamma_2$ starting at the same point $p \in S$ can be defined as follows:

$$\cos \angle_p(\gamma_1, \gamma_2) = \lim_{t \to 0} \frac{d^2(\gamma_1(t), p) + d^2(\gamma_2(t), p) - d^2(\gamma_1(t), \gamma_2(t))}{2d(p, \gamma_1(t))d(p, \gamma_2(t))},$$

where angle $\angle_p(\gamma_1, \gamma_2) \in [0; \pi]$. Denote by $\Gamma_p$ the set of all unit-speed geodesics emanating from $p$. Using angles, we can define the tangent cone $T_p S$ at $p \in S$ as follows. First define $T'_p S$ as the (quotient) set $\Gamma_p \times \mathbf{R}^+$, equipped with the (pseudo-)metric defined by

$$\| (\gamma_1, t) - (\gamma_2, s) \|_p^2 := s^2 + t^2 - 2s.t \cos \angle_p(\gamma_1, \gamma_2).$$

Then, the tangent cone $T_p S$ is defined as the completion of $T'_p S$ equipped with the metric $\| \cdot \|_p$. We will use the notation for $u, v \in T_p S$,

$$\langle u, v \rangle_p := \frac{1}{2} (\|u\|_p^2 + \|v\|_p^2 - \|u - v\|_p^2),$$

We will often identify a point $\gamma(t) \in S$ with $(\gamma, t) \in T_p S$. Although such $\gamma$ might not be unique, we will assume a choice of a map $\log_p : S \to T_p S$, called logarithmic map, such that for all $x \in S$, there exists a unit-speed geodesic $\gamma$ emanating from $p$ such that, for some $t > 0$, $\gamma(t) = x$ and

$$\log_p(x) = (\gamma, t).$$

This map can be chosen to be $\mathfrak{S}_B$-measurable, where $\mathfrak{S}_B$ denotes the $\sigma$-algebra generated by open balls on the tangent cone $T_p S$ (see Lemma 6) and this weak measurability is enough for our results to hold and will be assumed for the rest of the paper. Then the pushforward of $\mathbf{P}$ by $\log_p$ will be denoted by $\log_p \# \mathbf{P}$.

The tangent cone is not necessarily a geodesic space (see [4]), however, it is included in a geodesic space, namely the ultratangent space (see for instance Theorem 14.4.2 and 14.4.1 in [2]) that is an Alexandrov space with curvature bounded below by 0.

The tangent cone $T_p S$ contains the subspace $\text{Lin}_p$ of all points with an opposite, formally defined as follows. A point $u$ belongs to $\text{Lin}_p \subset T_p S$ if and only if there exists $v \in T_p S$ such that $\|u\|_p = \|v\|_p$ and

$$\langle u, v \rangle_p = - \|u\|_p^2.$$

Our main result is based on the following Theorem.

**Theorem (Theorem 14.5.4 in [2]).** The set $\text{Lin}_p$ equipped with the induced metric of $T_p S$ is a Hilbert space.
A point $b^* \in S$ is a barycenter of the probability measure $\mathbf{P} \in \mathcal{P}_1(S)$ if for all $b \in S$

$$0 \le \int d^2(x, b) - d^2(x, b^*) \, d\mathbf{P}(x).$$

Such barycenter might not be unique, neither exists. However, when they exist, they satisfy

$$\int \langle x, y \rangle_{b^*} d\mathbf{P} \otimes \mathbf{P}(x, y) = 0. \qquad (1)$$

A point $b^* \in S$ satisfying (1) is called an exponential barycenter of $\mathbf{P}$.

We can now state our main result.

**Theorem 1.** Let $(S, d)$ be an Alexandrov space with curvature bounded below by some $\kappa \in \mathbf{R}$ and $\mathbf{P} \in \mathcal{P}_1(S)$. If $b^* \in S$ is an exponential barycenter of $\mathbf{P}$, then $\supp \log_{b^*} \#\mathbf{P} \subset \text{Lin}_{b^*} S$. In particular, $\supp \log_{b^*} \#\mathbf{P}$ is included in a Hilbert space.

This result allows to prove the following Corollary, that has been implicitly used in [1].

**Corollary 2 (Linearity).** Let $b \in T_{b^*} S$. Then, the map $\langle \cdot, b \rangle_{b^*} : \text{Lin}_{b^*} \to \mathbf{R}$ is continuous and linear. In particular, if $b^*$ is an exponential barycenter of $\mathbf{P}$, then

$$\int \langle x, b \rangle_{b^*} d\mathbf{P}(x) = 0.$$

### 3. Proofs

Recall that we always identify a point in $S$ and its image in the tangent cone $T_pS$ by the $\log_p$ map.

**Proof of Corollary 2.** Linearity is obvious from the definition of $\langle \cdot, b \rangle_{b^*}$. We check that $x \mapsto \langle x, b \rangle_{b^*}$ is a convex and concave function in $\text{Lin}_{b^*} S$. Let $t \in (0, 1)$, $x_0, x_1$ in $\text{Lin}_{b^*} S$, and set $x_t = (1-t)x_0 + tx_1$. Since the tangent cone is included in an Alexandrov space with curvature bounded below by 0 on one hand, and $\text{Lin}_{b^*}$ is a Hilbert space on the other hand,

$$\begin{align*}
\langle x_t, b \rangle_{b^*} &= \frac{1}{2} (\|x_t\|_{b^*}^2 + \|b\|_{b^*}^2 - \|x_t - b\|^2) \\
&\le \frac{1}{2} \left( (1-t)(\|x_0\|_{b^*}^2 - \|x_0 - b\|_{b^*}^2) + t(\|x_1\|_{b^*}^2 - \|x_1 - b\|^2) + \|b\|_{b^*}^2 \right) \\
&= (1-t)\langle x_0, b \rangle_{b^*} + t\langle x_1, b \rangle_{b^*}.
\end{align*}$$

The same lines applied to $-x_0$ and $-x_1$ gives the converse inequality

$$\langle -x_t, b \rangle_{b^*} \le (1-t)\langle -x_0, b \rangle_{b^*} + t\langle -x_1, b \rangle_{b^*}.$$

The second statement follows from the fact that $b^*$ is a Pettis integral of the pushforward of $\mathbf{P}$ onto $\text{Lin}_{b^*} \subset T_{b^*} S$, as a direct consequence of Theorem 1. $\square$

**Proof of Theorem 1.** Let $L \subset \{x \in S | \int \langle x, \cdot \rangle_{b^*} d\mathbf{P} = 0\}$ be a measurable set such that $\mathbf{P}(L) = 1$ given by Lemma 3. Let $x \in L$. For $U = \{x\}$, use Lemma 5 with $\mathbf{Q} = \mathbf{P}$ and $B_\delta$ a ball of radius $\delta$ around $x$ in $T_{b^*} S$, to get a sequence $(y_\delta^n)_n \subset T_{b^*} S$ such that,

$$\begin{align*}
\limsup_n \cos \angle (\uparrow_{b^*}^x, \uparrow_{b^*}^{y_\delta^n}) &= \limsup_n \frac{\langle x, y_\delta^n \rangle_{b^*}}{d(b^*, x)d(b^*, y_\delta^n)} \\
&= \frac{1}{d(b^*, x)} \limsup_n \frac{1}{\|x, y_\delta^n\|_{b^*}} \lim_n \frac{1}{d(b^*, y_\delta^n)} \\
&\le \frac{1}{d(b^*, x)} \frac{\int_{B_\delta^c} \langle x, y \rangle_{b^*} d\mathbf{P}(x)}{\mathbf{P}(B_\delta)} \frac{\mathbf{P}(B_\delta)}{\left(\int_{B_\delta} \int_{B_\delta} \langle x, y \rangle_{b^*} d\mathbf{P} \otimes \mathbf{P}(x, y)\right)^{1/2}}.
\end{align*}$$
Then, since $\int \langle x, y \rangle_{b^*} d\mathbf{P}(y) = 0$, letting $\delta \to 0$, one gets

$$ \frac{1}{\mathbf{P}(B_\delta)} \int_{B_\delta^c} \langle x, y \rangle_{b^*} d\mathbf{P}(y) = - \frac{1}{\mathbf{P}(B_\delta)} \int_{B_\delta} \langle x, y \rangle_{b^*} d\mathbf{P}(y) \to -d^2(b^*, x). $$

and

$$ \frac{\left( \int_{B_\delta} \int_{B_\delta} \langle x, y \rangle_{b^*} d\mathbf{P} \otimes \mathbf{P}(x, y) \right)^{1/2}}{\mathbf{P}(B_\delta)} \to d(b^*, x) $$

Thus,

$$ \lim_{\delta \to 0^+} \limsup_n \cos \angle(\uparrow_{b^*}^x, \uparrow_{b^*}^{y_\delta^n}) = -1 $$

One can thus choose $(\bar{y}^n)_n$ a sequence in $(y_\delta^n)_{n,\delta}$ such that $\cos \angle(\uparrow_{b^*}^x, \uparrow_{b^*}^{\bar{y}_\delta^n}) \to -1$. Since $T_{b^*} S$ is a subspace of an Alexandrov space of curvature bounded below by 0, we also have

$$ \angle(\uparrow_{b^*}^{\bar{y}_\delta^n}, \uparrow_{b^*}^{\bar{y}_\delta^k}) \le 2\pi - \angle(\uparrow_{b^*}^{\bar{y}_\delta^n}, \uparrow_{b^*}^x) - \angle(\uparrow_{b^*}^x, \uparrow_{b^*}^{\bar{y}_\delta^k}) \to 0, $$

as $n, k \to \infty$. Thus $(\bar{y}^n)_n$ corresponds to a Cauchy sequence in the space of direction, and thus admits a limit in $T_{b^*} S$, since its “norm” also admits a limit $d(b^*, x)$. Its limit $\bar{y}$ satisfies $\cos \angle(\uparrow_{b^*}^x, \uparrow_{b^*}^{\bar{y}}) = -1$, and therefore, it is the opposite $\bar{y} = -x$.

Finally, by definition of the support, for $x \in \text{supp}(\log_{b^*} \mathbf{P})$, every ball centered at $x$ have a positive probability, and thus there exists a sequence $(x_n)_{n \ge 1} \subset L$ such that $x_n \to x$. We conclude with the completeness of $\text{Lin}_{b^*}$. $\square$

**Lemma 3 (Proposition 1.7 of [8] for non separable metric space).** Suppose $(S, d)$ is an Alexandrov space with curvature bounded below. Then, for any probability measure $\mathbf{Q} \in \mathcal{P}_1(S)$, and any $b^* \in S$,

$$ \int \langle x, y \rangle_{b^*} d\mathbf{Q} \otimes \mathbf{Q}(x, y) \ge 0. $$

Moreover, if $b^*$ is an exponential barycenter of $\mathbf{Q}$, then for $\mathbf{Q}$-almost all $x \in S$,

$$ \int \langle x, y \rangle_{b^*} d\mathbf{Q}(y) = 0. $$

**Proof.** For brevity, we will adopt the notation $\mathbf{Q}g$ for $\int g d\mathbf{Q}$.

The result for $\mathbf{Q}$ finitely supported is the Lang-Schroeder inequality (Proposition 3.2 in [5]). Thus, we just need to approximate $\mathbf{Q} \otimes \mathbf{Q}(\cdot, \cdot)_{b^*}$ by some $\mathbf{Q}_n \otimes \mathbf{Q}_n(\cdot, \cdot)_{b^*}$ for some finitely supported $\mathbf{Q}_n$.

To approximate $\mathbf{Q} \otimes \mathbf{Q}(\cdot, \cdot)_{b^*}$, draw two independent sequences of i.i.d. random variables $(X_i^1)_i$ and $(X_i^2)_i$ of common law $\mathbf{Q}$, and denote $\mathbf{Q}_n^1$ and $\mathbf{Q}_n^2$ the corresponding empirical measures. In particular, $\mathbf{Q}_n^1 \otimes \mathbf{Q}_n^2$ and $\mathbf{Q}_n^2 \otimes \mathbf{Q}_n^1$ are both empirical measures of $\mathbf{Q} \otimes \mathbf{Q}$. Since $S$ is not separable, we can not apply the fundamental theorem of statistics that ensures almost sure weak convergence of $\mathbf{Q}_n^1 \otimes \mathbf{Q}_n^1$ to $\mathbf{Q} \otimes \mathbf{Q}$. However, for a measurable function $f: S \times S \to \mathbf{R}$, such that $\mathbf{Q} \otimes \mathbf{Q}f < \infty$, the law of large number ensures that almost surely

$$ \mathbf{Q}_n^1 \otimes \mathbf{Q}_n^2 f \to \mathbf{Q} \otimes \mathbf{Q}f $$

and

$$ \mathbf{Q}_n^2 \otimes \mathbf{Q}_n^1 f \to \mathbf{Q} \otimes \mathbf{Q}f. $$

Since the sequence $(X_1^1, X_1^2, X_2^1, X_2^2, X_3^1, X_3^2, \dots)$ is also an i.i.d. sequence of random variables of common law $\mathbf{Q}$, the subsequence of the associated empirical measures $(\mathbf{Q}_n^3)_n$ defined by

$$ \mathbf{Q}_n^3 := \frac{1}{2}(\mathbf{Q}_n^1 + \mathbf{Q}_n^2) $$

also satisfies the almost sure convergence

$$ \mathbf{Q}_n^3 \otimes \mathbf{Q}_n^3 f \to \mathbf{Q} \otimes \mathbf{Q}f. $$
Now, since

$$Q_n^3 \otimes Q_n^3 = \frac{1}{4}(Q_n^1 \otimes Q_n^2 + Q_n^1 \otimes Q_n^1 + Q_n^2 \otimes Q_n^1 + Q_n^2 \otimes Q_n^2),$$

we proved that almost surely

$$Q_n^1 \otimes Q_n^1 f + Q_n^2 \otimes Q_n^2 f \to 2Q \otimes Qf.$$

And since $(Q_n^1)_n$ and $(Q_n^2)_n$ are independent and with same law, it implies that both $Q_n^1 \otimes Q_n^1 f$ and $Q_n^2 \otimes Q_n^2 f$ converge to $Q \otimes Qf$ almost surely. In particular, since $Q_n^1$ is supported on $n$ points, there exists a sequence of finitely supported measures (that we rename $(Q_n)_n$) such that $Q_n \otimes Q_n f \to Q \otimes Qf$. We thus proved the first result applying $f = \langle \cdot, \cdot \rangle_{b^*}$.

Now, for any $x \in S$, applying this first result to the measure $Q_\epsilon := \frac{1}{1+\epsilon}Q + \frac{\epsilon}{1+\epsilon}\delta_x$, we get

$$0 \le (1 + \epsilon)Q_\epsilon \otimes Q_\epsilon \langle \cdot, \cdot \rangle_{b^*} \\ = Q \otimes Q \langle \cdot, \cdot \rangle_{b^*} + 2\epsilon Q \langle x, \cdot \rangle_{b^*} + \epsilon^2 \|x\|_{b^*}^2.$$

Letting $\epsilon \to 0^+$, we get

$$Q \langle x, \cdot \rangle_{b^*} \ge 0.$$

Then equality follows from the hypothesis $Q \otimes Q \langle \cdot, \cdot \rangle_{b^*} = 0$ meaning that $b^*$ is an exponential barycenter. $\square$

**Lemma 4 (Subadditivity, Lemma A.6 of [5]).** Let $(S, d)$ be an Alexandrov space with curvature bounded below. Take $b^* \in S$. Let $x_1, \dots, x_n \in T_{b^*} S$ and $U \subset T_{b^*} S$ finite. Then, for all $\epsilon > 0$, there exists $y \in T_{b^*} S$ such that for all $u \in U$,

$$\langle y, u \rangle_{b^*} \le \sum_{i=1}^{n} \langle x_i, u \rangle_{b^*} + \epsilon,$$

and

$$\|y\|^2 \le \sum_{i,j=1}^{n} \langle x_i, x_j \rangle_{b^*} + \epsilon.$$

**Lemma 5 (Approximation).** Let $U \subset T_{b^*} S$ finite. Take $B \subset S$ measurable and a probability measure $P \in \mathcal{P}_1(S)$ such that $P \otimes P(\langle \cdot, \cdot \rangle_{b^*} = 0)$ and $P(B) > 0$. Then, there exists a sequence $(y^n)_n$ such that for all $u \in U$

$$\frac{1}{P(B)} \int_{B^c} \langle u, x \rangle_{b^*} dP(x) \ge \limsup_n \langle u, y^n \rangle_{b^*} \quad (2)$$

and

$$\frac{1}{\mathbf{P}(B)^2} \int_{B} \int_{B} \langle x, y \rangle_{b^*} d\mathbf{P} \otimes \mathbf{P}(x, y) = \lim_{n} d^2(b^*, y^n). \quad (3)$$

**Proof.** Using the same arguments as in Lemma 3, we see that the empirical measures $(P_n)_n$ satisfy

$$P_n \otimes P_n f \to P \otimes P f,$$

almost surely, for any $f: S \times S \to \mathbf{R} \in L^1(S \times S, P \otimes P)$. In particular, taking $f(x, y) = \langle x, y \rangle_{b^*} \mathbf{1}_{B \times B}(x, y)$, the following convergence holds in $L^2(P^{\otimes \infty})$,

$$\int_B \int_B \langle \cdot, \cdot \rangle_{b^*} d\mathbf{P}_n \otimes \mathbf{P}_n \to \int_B \int_B \langle \cdot, \cdot \rangle_{b^*} d\mathbf{P} \otimes \mathbf{P}, \quad (4)$$

and similarly for $B^c$. Also, the law of large number ensures that almost surely, for all $u \in U$,

$$\int_B \langle \cdot, u \rangle_{b^*} d\mathbf{P}_n \to \int_B \langle \cdot, u \rangle_{b^*} d\mathbf{P}, \quad (5)$$

and again, the same for $B^c$. Thus, there exists a subsequence (of a deterministic realization of) $P_n$ (that we rename $P_n$) such that (4) and (5) both hold for all $u \in U$.
Then, applying Lemma 4 to finite sum

$$ \frac{1}{\mathbf{P}(B)} \int_{B^c} \langle \cdot, u \rangle_{b^*} d\mathbf{P}_n, $$

shows that there exists a sequence $(y^n)_n \in T'_{b^*} S$ such that (2) holds and for a sequence $(\epsilon_n)_n$ s.t. $\epsilon_n \to 0$,

$$ \|y^n\|_{b^*}^2 \le \frac{1}{\mathbf{P}(B)^2} \int_{B^c} \int_{B^c} \langle \cdot, \cdot \rangle_{b^*} d\mathbf{P}_n \otimes \mathbf{P}_n + \epsilon_n. $$

Then, applying the same Lemma 4 again shows that there exists a sequence $(z^n)_n \subset T'_{b^*} S$, such that

$$ \begin{aligned} & 0 - \frac{1}{\mathbf{P}(B)^2} \iint \langle x, y \rangle_{b^*} d\mathbf{P}_n \otimes \mathbf{P}_n(x, y) \\ &= \frac{1}{\mathbf{P}(B)^2} \left( \int_B \int_B + \int_{B^c} \int_{B^c} + 2 \int_B \int_{B^c} \right) \langle x, y \rangle_{b^*} d\mathbf{P}_n \otimes \mathbf{P}_n(x, y) \\ &\ge \|z^n\|_{b^*}^2 + \|y^n\|_{b^*}^2 + 2 \langle y^n, z^n \rangle_{b^*} - \epsilon_n. \end{aligned} \quad (6) $$

Letting $n \to \infty$, one obtains

$$ \begin{aligned} 0 &\ge \lim_n \|z^n\|_{b^*}^2 + 2 \langle y^n, z^n \rangle_{b^*} + \|y^n\|_{b^*}^2 \\ &\ge \lim_n \|z^n\|_{b^*}^2 - 2 \|y^n\|_{b^*} \|z^n\|_{b^*} + \|y^n\|_{b^*}^2 \\ &= \lim_n (\|z^n\|_{b^*} - \|y^n\|_{b^*})^2 \ge 0. \end{aligned} $$

and which shows $\lim_n \|y^n\| = \lim_n \|z^n\|$ and also that (6) becomes an equality at the limit and therefore

$$ \lim_n \|z^n\|_{b^*}^2 = \frac{1}{\mathbf{P}(B)^2} \int_B \int_B \langle x, y \rangle_{b^*} d\mathbf{P} \otimes \mathbf{P}(x, y) \quad \square $$

This Lemma appears in a remark of [6].

**Lemma 6 (Measurability of the log map).** Let $(S, d)$ be a separable Alexandrov space. Let $p \in S$. Then $\log_p: S \to T_pS$ can be chosen to be $\mathfrak{S}_B$-measurable.

**Proof.** Denote $G_p$ the space of all constant speed geodesics emanating from $p$ equipped with the sup distance $\| \cdot \|_{\infty}$. Then $(G_p, \| \cdot \|_{\infty})$ is separable and complete too. Using Kuratowski and Ryll-Nardzewski measurable selection theorem, one can choose a Borel map $g: S \to G_p$ such that $g$ maps $x$ to a geodesic from $p$ to $x$. Then, using the (proof of) Lemma 4.2 of [7], the map $l: G_p \to T_pS$ is measurable $T_pS$ is equipped with the $\sigma$-algebra $\mathfrak{S}$ generated by open balls. Therefore, $\log_p := l \circ g$ is $\mathfrak{S}_B$-measurable. $\square$

## Acknowledgements

This note originated from technical issues emanating from the problem considered in [1] and [6]. I would like to thank Quentin Paris and Philippe Rigollet for fruitful discussions that lead to this article. I am also indebted to the anonymous referee for their careful review and comments.

## References

[1] A. Ahidar-Coutrix, T. Le Gouic, Q. Paris, “On the rate of convergence of empirical barycentres in metric spaces: curvature, convexity and extendible geodesics”, https://arxiv.org/abs/1806.02740v1, 2018.

[2] S. Alexander, V. Kapovitch, A. Petrunin, “Alexandrov geometry”, http://arxiv.org/abs/1903.08539, 2019.

[3] D. Burago, Y. Burago, S. Ivanov, *A course in metric geometry*, Graduate Studies in Mathematics, vol. 33, American Mathematical Society, 2001.

[4] S. Halbeisen, “On tangent cones of Alexandrov spaces with curvature bounded below”, *Manuscr. Math.* **103** (2000), no. 2, p. 169-182.
[5] U. Lang, V. Schroeder, "Kirszbraun's theorem and metric spaces of bounded curvature", *Geom. Funct. Anal.* **7** (1997), no. 3, p. 535-560.

[6] T. Le Gouic, Q. Paris, P. Rigollet, A. J. Stromme, "Fast convergence of empirical barycenters in Alexandrov spaces and the Wasserstein space", https://arxiv.org/abs/1908.00828, 2019.

[7] S.-I. Ohta, "Barycenters in Alexandrov spaces of curvature bounded below", *Adv. Geom.* **12** (2012), no. 4, p. 571-587.

[8] K.-T. Sturm, "Metric spaces of lower bounded curvature", *Expo. Math.* **17** (1999), no. 1, p. 35-47.

[9] T. Yokota, "A rigidity theorem in Alexandrov spaces with lower curvature bound", *Math. Ann.* **353** (2012), no. 2, p. 305-331.