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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) 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 ?

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    $\begingroup$ Given the question, the title should replace "continuous" with "measurable", don't you think? $\endgroup$ – Benoît Kloeckner Jun 22 '20 at 9:01
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    $\begingroup$ Yes indeed. Fixed :) $\endgroup$ – dohmatob Jun 22 '20 at 9:01
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    $\begingroup$ What you call "Polish space" is usually called "complete metric space". In a Polish space the distance is not fixed: it's just assumed it exists. The distinction is very useful, I think. $\endgroup$ – YCor Jun 22 '20 at 9:25
  • $\begingroup$ Good point. Fixed. $\endgroup$ – dohmatob Jun 22 '20 at 9:31
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    $\begingroup$ @coudy Sorry, I meant "separable complete metric space", as opposed to "Polish space". Of course separability is essential (maybe not in OP's question who mentions possibly non-separable CAT(0) spaces), it was just not the point of my comment and I forgot it. $\endgroup$ – YCor Jun 22 '20 at 11:15
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We will use the Kuratowski–Ryll-Nardzweski 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 and thus universally measurable. So if we are given a Borel measure $\mu$ on $X \times X$, we get a $\mu$-measurable midpoint function.

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    $\begingroup$ Ah, this one is looking like the "book proof". Thanks! Anecdotally, a few hours ago I stumbled on the "selection theorems" page on wikipedia, and eventually stumbled upon KRN selection theorem, I felt this could eventually give a solution after some very hard work (and learning after learning some brand new stuff too). $\endgroup$ – dohmatob Jun 22 '20 at 12:42
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    $\begingroup$ Nice argument! Although I must say I was curious how you would do without the local compactness... I think there should be a different argument in the case where the space is not locally compact and the midpoint is unique, but I cannot quite think it through. $\endgroup$ – Pierre PC Jun 22 '20 at 14:39
  • $\begingroup$ @GeraldEdgar Thanks for the nice answer. I think everyone will agree that this is the "book proof". $\endgroup$ – dohmatob Jun 22 '20 at 14:59
  • $\begingroup$ @GeraldEdgar: You mean Borel measure $\mu$ on $X^2$, right ? $\endgroup$ – dohmatob Jun 22 '20 at 15:41
  • $\begingroup$ @dohmatob ... yes, fixed. $\endgroup$ – Gerald Edgar Jun 22 '20 at 15:56
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If $(X,d)$ is a complete metric space with the algebraic midpoint property (i.e. for all $x$ and $y$ in $X$, there exists $z\in X$ such that $d(x,z)=d(y,z)=d(x,y)/2$) then $X$ is a path metric space. Indeed, for all $x,y\in X$ one can iteratively construct a map $\gamma$ from $[0,1]\cap\mathbb D$ to $X$ such that $d(\gamma(s),\gamma(t))=|t-s|$, and extend it using completeness of $X$. Here $\mathbb D$ is the set of dyadic numbers. This result is Theorem 1.8 in [G].

Suppose that $(X,d)$ is complete, locally compact, and has the algebraic midpoint property. Then $X$ has the measurable midpoint property.

Of course, this does not include general closed convex subsets of Banach spaces, but it covers for instance any complete manifold.

Let us $(U_k)_{k\geq0}$ construct a countable basis of $X$; in particular, it will show that $X$ is second countable. I want the diameter of $U_k$ to tend to zero as $k$ goes to infinity, and every fixed $x$ to be contained in an infinite number of $U_k$. By [G, Theorem 1.10 (Hopf-Rinow)], the closed metric balls of $X$ are compact; then one can take a finite open cover of $B(x_0,1)$ by open balls of radius $1/1$, then a finite open cover of $B(x_0,2)$ by open balls of radius $1/2$, etc. Let also $(z_k)_{k\geq0}$ be a sequence such that $z_k\in U_k$.

Note that for any closed set $F$, the set of pairs $(x,y)$ such that $F$ contains at least one midpoint of $\lbrace x,y\rbrace$ is closed, using the compacity of closed bounded sets. Let $k_0(x,y)$ be the first $k$ such that the closure $\overline U_k$ contains at least one midpoint of $\lbrace x,y\rbrace$, and iteratively $k_{n+1}(x,y)$ is the first $k>k_n(x,y)$ such that the closed intersection $$ \overline U_k\cap\bigcap_{0\leq m\leq n}\overline U_{k_m(x,y)} $$ contains at least one midpoint of $\lbrace x,y\rbrace$.

Note that $k_n(x,y)$ is measurable, since the set of $(x,y)$ such that $k_n(x,y)\leq K$ is a finite union of closed sets. Then obviously $f_n:(x,y)\mapsto z_{k_n(x,y)}$ is measurable as well. Since the diameter of $U_k$ tends to zero, $(f_n(x,y))_{n\geq0}$ is a Cauchy sequence for all $(x,y)$, and $f(x,y):=\lim_{n\to\infty}f_n(x,y)$ is a well-defined midpoint of $\lbrace x,y\rbrace$. As a limit of measurable functions, it is measurable as well.

[G] M. Gromov, Metric structures for Riemannian and non-Riemannian spaces. 3rd printing (2007).

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  • $\begingroup$ Thanks. You need separability of $(X,d)$, in order to ensure there existence of a countable basis $(U_k)_k$. It might be nice to make this more explicit in your write-up (btw separability is part of my assumptions). $\endgroup$ – dohmatob Jun 22 '20 at 12:12
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    $\begingroup$ @dohmatob The separability is a consequence of completeness, local compacity, and the algebraic midpoint property, as mentioned. Indeed, it makes $(X,d)$ into a complete, locally compact path space, hence the closed balls are compact. So $(X,d)$ is a $\sigma$-compact metric space, and it is second countable using the argument I gave in the answer. $\endgroup$ – Pierre PC Jun 22 '20 at 12:26
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Maybe the unit circle embedded in the euclidean plane is an example of a space that has several measurable midpoints structures but no continuous such structure?

Let us choose as the middle point of two points which are not on a diameter the point in the middle of the shortest arc connecting the two points.

When two points are on a diameter, we may choose one of the middle points on the two arcs as our middle point, but this cannot be made continuously.

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  • $\begingroup$ Nice example, thanks. Interestingly, your example seems to be continuous almost everywhere on $X^2$. $\endgroup$ – dohmatob Jun 22 '20 at 9:35
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    $\begingroup$ In any complete Riemannian manifold $X$, if a measurable midpoint exists, it will be continuous almost everywhere on $X^2$. Indeed, it will be continuous at all $(x,y)$ such that $y$ is not in the cut locus of $x$ (there is no choice for any such pair). The cut locus of $x$ has measure zero in $X$ (and in fact it has Hausdorff dimension at most $\operatorname{dim}M-1$), so the set of singularities $(x,y)$ has measure zero in $X^2$ using Fubini's theorem. $\endgroup$ – Pierre PC Jun 22 '20 at 10:14
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Let us construct an example of a measurable midpoint space that is not a continuous midpoint space. The idea is to create a "jump" of midpoints somewhere. One way to do that is to consider a slit rectangle, e.g. $$ \tilde E = [0,1]\times[0,1]\setminus \{\frac12\}\times(0,1) $$ endowed with the length metric induced by the canonical euclidean scalar product and $E$ is the completion of $\tilde E$ (obtained by adding two copies of the slit interval, one on the right of the slit and one on its left). Any two points in the same half (e.g. $[0,\frac12]\times [0,1]$ have an obvious midpoint, the Euclidean one; two points in different halves are connected by one or two shortest paths, hence have a midpoint. The midpoints are explicit enough to be easily seen to depend measurably on the endpoints. However, the midpoint between $x=(\frac12,1)$ and $y_t=(t,0)$ jumps at $t=\frac12$, from the center of the left copy of the slit interval to the center of its right copy, which are at distance $1$ one from the other. Hence $E$ does not have the continuous midpoint property.

(I must admit have trouble imagining a polish space with the midpoint property, but without the measurable midpoint property.)

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  • $\begingroup$ Nice example, thanks. Concerning your last point, maybe you could make it a theorem (possibly with an additional contion on $X$) ? ;) $\endgroup$ – dohmatob Jun 22 '20 at 9:33
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We are ultimately looking at the complexity of the multivalued function $\mathrm{MidPoint}_\mathbf{X} : \mathbf{X} \times \mathbf{X} \rightrightarrows \mathbf{X}$ assigning some midpoint to the points here. This includes what choice functions there are, but need not be limited to it. The framework to study the complexity of such operations is Weihrauch reducibility.

Just by definition of the midpoint, it follows that the map from a pair of points to the closed set of midpoints (equipped with the lower Vietoris topology) is continuous. This tells us that $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{C}_\mathbf{X}$, where $\mathrm{C}_\mathbf{X}$ is closed choice on $\mathbf{X}$, which maps non-empty closed sets to some element. With $\mathrm{UC}_\mathbf{X}$ I denote the restriction of $\mathrm{C}_\mathbf{X}$ to singletons.

Everything we need about closed choice for this is found here.

Since we are assuming $\mathbf{X}$ to be Polish1, we immediately get the following:

  1. If $\mathbf{X}$ is sigma-compact and midpoints are unique, then $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{C}_\mathbb{N}$. This implies that the midpoint map is piecewise continuous, ie that there is a countable cover of $\mathbf{X} \times \mathbf{X}$ by closed sets, such that on each piece the map is continuous.
  2. If $\mathbf{X}$ is sigma-compact, then $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{C}_\mathbb{R}$. This guarantees that there is a Baire class 1 selection function for midpoints, but we get more [Baire class 1 is equivalent to "preimages of opens are $\Sigma^0_2$, so this is much simpler than Borel measurable]. For example, there always is a midpoint which is low (in the computability-theoretic sense) relative to the space.
  3. If midpoints are unique, then $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{UC}_{\mathbb{N}^\mathbb{N}}$. Since the domain of $\mathrm{MidPoint}_\mathbf{X}$ is a Polish space, this already implies that $\mathrm{MidPoint}_\mathbf{X}$ is Borel measurable.
  4. Without any restrictions, we just get that $\mathrm{MidPoint}_\mathbf{X} \leq_{\mathrm{W}} \mathrm{C}_{\mathbb{N}^\mathbb{N}}$. This doesnt rule out that we could avoid Borel measurable selection functions for the midpoint, but any construction would need to be very weird. The best starting point I can think of is using diagonally non-arithmetic functions.

1 We don't need that the metric defining our midpoints is complete, we just need some equivalent complete metric around.

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  • $\begingroup$ Thanks. Upvoted. There are many things which seem to be missing. You seem to be using the fact that Polishness of the space implies many things, but it's not clear by what mechanism these conclusions are drawn. Because we're in a metric space, the assumption in case 1 is equivalent to saying that $X$ is locally compact (at this point, I'm fine with such an assumption, since the other posted solutions require it too). Finally, is every 'Baire class 1" function also Borel-measurable ? $\endgroup$ – dohmatob Jun 26 '20 at 5:32
  • $\begingroup$ Also, my 2 cents is that a version of the arguments you use here can be in principle used to prove the Kuratowski-Ryll-Nardzewski measurable selection theorem used in Gerald Edgar's answer. $\endgroup$ – dohmatob Jun 26 '20 at 8:59
  • $\begingroup$ Indeed, this is done in Theorem 4.1 of cca-net.de/vasco/publications/borel.pdf. $\endgroup$ – dohmatob Jun 26 '20 at 9:11
  • $\begingroup$ I've clarified that Baire class 1 is much simpler than Borel measurable. The Polish-requirement is what makes sure that Baire space occurs as subscript where it does. $\endgroup$ – Arno Jun 30 '20 at 12:35
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    $\begingroup$ Yes, every Baire class 1 function is Borel measurable. The set of Borel measurable functions is the union of the Baire class $\alpha$ functions where $\alpha$ ranges over the countable ordinals. Hence, the "much simpler" statement. $\endgroup$ – Arno Jun 30 '20 at 12:54

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