Given a (separable complete) metric space $X=(X,d)$, let us say $X$ has the *measurable (resp. continuous) midpoint property* if there exists a measurable (resp. continuous) mapping $m:X \times X \to X$ such that $d(x,m(x,x')) = d(x',m(x,x')) = d(x,x') / 2$ for all $x,x' \in X$.


It seems to be known (e.g see section 6 of [this paper](https://www.researchgate.net/publication/333720656_Generalizations_of_Hyperconvex_Metric_Spaces_RNA2019)) that continuous midpoint spaces (i.e Polish spaces with the continuous midpoint property) include:


* Hilbert spaces.
* Closed convex subsets of Banach spaces.
* Hyperconvex spaces.
* CAT(0) spaces.


Hopefully, the collection of measurable midpoint spaces contains much more general examples (for the above list is quite restrictive).



> 
> **Question.** What are some examples of measurable midpoint spaces ?
> 
> 
> 



Examples of metric spaces with measurable midpoints
We will use the [Kuratowski–Ryll-Nardzweski selection theorem](https://en.wikipedia.org/wiki/Kuratowski_and_Ryll-Nardzewski_measurable_selection_theorem):  

Let $(\Omega, \mathscr{F})$ be a measurable space. Let $E$ be a Polish space. Let $\Gamma$ be a set-valued function from $\Omega$ to $E$; that is, for each $\omega \in \Omega$, let a set $\Gamma(\omega) \subseteq E$ be given. Assume that, for all $\omega \in \Omega$, the set $\Gamma(\omega)$ is nonempyty and closed in $E$. Assume that $\Gamma$ is $\mathscr F$-measurable in the sense:
$$
\text{for every open set }U\subseteq E,\qquad 
\{\omega\,:\,\Gamma(\omega) \cap U \ne \varnothing\} \in \mathscr F .
$$
Then there is a **measurable selection** $\gamma$ for $\Gamma$: that is, a
function $\gamma : \Omega \to E$ with


$\bullet $ $\gamma(\omega) \in \Gamma(\omega)$


$\bullet $ for every open set $U \subseteq E,\quad \gamma^{-1}(U) \in \mathscr F$.




---


Let $X$ be a locally compact complete separable metric space with the midpoint property. For $a,b \in X$, let $\Gamma(a,b)$ be the midpoint set,
$$
\Gamma(a,b) = \left\{m : d(a,m)=d(b,m) = \frac{d(a,b)}{2}\right\} .
$$
Then $\Gamma$ is a set-valued function from $X \times X$ to $X$. Note $\Gamma(a,b)$ is nonempty and closed. Let $\mathscr F$ be the sigma-algebra of Borel sets in $X \times X$. We will prove (see below) that $\Gamma$ is $\mathscr F$-measurable. An application of the Kuratowski–Ryll-Nardzweski selection theorem then establishes the existence of an $\mathscr F$-measurable $\gamma : X\times X \to X$ with $\gamma(a,b) \in \Gamma(a,b)$.


Proof that $\Gamma$ is $\mathscr F$-measurable:


Let $U \subseteq X$ be open. We have to show $T\_U \in \mathscr F$, where
$$
T\_U := \{(a,b) \in X \times X\,:\,\Gamma(a,b) \cap U \ne \varnothing\} .
$$
The set
$$
Q := \{(a,b,u) \in X \times X \times X \,:\, 
d(a,u) = \textstyle\frac{1}{2}d(a,b)\text{ and }
d(b,u) = \textstyle\frac{1}{2}d(a,b)\}
$$
is a closed set. Write $\pi$ for the continuous "projection" function $(x,y,u) \mapsto (x,y)$.
Then $T\_U$ is the projection
$$
T\_U = \pi(Q\cap(X \times X \times U)) = 
\bigcup\_{u \in U}\{(a,b) \in X \times X \,:\, 
d(a,u) = \textstyle\frac{1}{2}d(a,b)\text{ and }
d(b,u) = \textstyle\frac{1}{2}d(a,b)\} .
$$
Now in our case, any open set $U$ is a countable union of compact sets,
so the projection is sigma-compact, and therefore Borel.




---


**added**  

Without assuming locally compact, we do know that the [projection of
a Borel set is analytic](https://en.wikipedia.org/wiki/Projection_(measure_theory)) and thus universally measurable. So if we are given a Borel measure $\mu$ on $X \times X$, we get a $\mu$-measurable midpoint function.


