Measurable selection of geodesics

Question

Let $(X, d)$ be Polish and geodesic (i.e. for all $x,y \in X$ there exists a so called constant speed geodesic curve $\gamma :[0,1] \rightarrow X$, s.t. $$\gamma (0) = x, \gamma (1) = y$$ and $$d(\gamma(t), \gamma(s)) = |t-s|d(\gamma (0), \gamma (1)) = |t-s|d(x, y)$$ for all $t,s \in [0,1]$). The set of constant speed geodesics is denoted by $Geod(X)$ and is equipped with the sup-norm to form a complete metric space.
I am looking for a measurable selection theorem which ensures the existence of a Borel measurable map $GeodSel: X^2 \rightarrow Geod(X)$ such that $GeodSel(x,y)$ is a constant speed geodesic from $x$ to $y$.
My question arises from the proof of Theorem 2.10, page 35 in Ambrio's, Gigli's 'A user's guide to optimal transport' (https://pdfs.semanticscholar.org/c453/e5ea0b6061d87cfa98f5b260a26f8497d3a1.pdf). In the mentioned book they refer to the following lemma and state that it would, together with classical measurable selection theorems, imply the existence of such a map.

The multivalued map from $G:X^2 \rightarrow Geod(X)$ which associates to each pair $(x,y)$ the set $G(x,y)$ of constant speed geodesics connecting $x$ to $y$ has closed graph.

I've searched for some theorems on my own and came across the Kuratowski and Ryll-Nardzewski measurable selection theorem (https://en.wikipedia.org/wiki/Kuratowski_and_Ryll-Nardzewski_measurable_selection_theorem). But then I struggled to show the condition of weak measurability.

This was also discussed on MathOverflow some time ago https://mathoverflow.net/questions/145351/x-polish-geodesic-implies-p-2x-w-2-geodesic). But it seems to me that my question is still unanswered if $X$ isn't compact.

I appreciate any help and hints!

Answer 1

I think that the solution of this problem is Aumann's selection theorem, which can be found in "Measure Theory Vol 2" by V.I. Bogachev, Pag. 40

Theorem 6.9.13 Let $(\Omega,A,\mu)$ be a complete probability space, let $X$ be a Souslin space, and let $\Psi$ be a multivalued mapping from $\Omega$ to the set of nonempty subsets of $X$ such that its graph $\Gamma_\Psi$ belongs to $A\otimes B(X)$. Then, there exists an $A$-$B(X)$-measurable mapping $f: \Omega \to X$ such that $f(\omega) \in \Psi(\omega)$ for all $\omega \in \Omega$.

In our case $\Omega =X \times X$ is a complete probability space (up to a slight modification to the measure), $\mathrm{Geod}(X)$ is a Souslin space and the graph of $G$ is closed, thus in $A\otimes B(X)$.

Answer 2

Just to complete the discussion, let me mention that the (seemingly) implied claim in that book on optimal transport is actually false.

Wrong claim. If $X,Y$ are Polish spaces and $F:X\to\mathcal{P}(Y)$ is a multifunction with closed graph in $X\times Y$ (and $F(x)\neq\emptyset$ for every $x\in X$), then $F$ admits a Borel selection.

This does not hold: we can find a continuous, surjective map $g:Y\to X$ between Polish spaces such that there exists no Borel section $f$ (meaning $f:X\to Y$ with $g\circ f=\operatorname{id}_X$), according to Bogachev's Measure Theory, vol. 2, Theorem 6.9.8, where a section is called a "selection" (I didn't check the proof but the book looks reliable). We can define a multifunction $F$ by $F(x):=g^{-1}(x)$, for $x\in X$. Since $g$ is continuous, its graph is closed in $Y\times X$, and hence also the graph of $F$ is closed in $X\times Y$. However, Borel sections for $g$ are exactly Borel selections for $F$.

On the other hand, in the setting of the optimal transport book, we just need a selection $f$ for which every probability measure $\mu$ on $X$ has a well-defined pushforward $f_*\mu$. Let $\mathcal{A}$ be the $\sigma$-algebra generated by analytic subsets of $X$; recall that it includes the Borel $\sigma$-algebra, and more importantly it is always included in the completion of the Borel $\sigma$-algebra with respect to $\mu$. Let $\mathcal{B}$ be the Borel $\sigma$-algebra of $Y$. It is enough to show the following.

True claim. There exists an $(\mathcal{A},\mathcal{B})$-measurable selection $f$.

This is easy to achieve: write $Y=\bigcup_{i\in\mathbb N}Y_i$, with $Y_i$ closed sets of diameter $\le 1$, then for each $i\in\mathbb N$ write $Y_i=\bigcup_{j\in\mathbb N}Y_{ij}$, with $Y_{ij}$ closed of diameter $\le 1/2$, and so on. Pick $y_i\in Y_i$, $y_{ij}\in Y_{ij}$, and so on.

The set $X_i':=\{x\in X:F(x)\cap Y_i\neq\emptyset\}$ is analytic, since it is the projection of the closed set $\operatorname{graph}(F)\cap (X\times Y_i)\subseteq X\times Y$ on $X$. We then let $$ X_i:=X_i'\setminus\bigcup_{k<i}X_k' $$ (this set is not guaranteed to be analytic, but certainly $X_i\in\mathcal{A}$) and define $$ f_0(x):=y_i \quad\text{on }X_i $$ (note that the sets $X_i$ are a partition of $X$). We similarly define $X_{ij}'$ and we build the finer partition $$ X_{ij}:=X_i\cap\Big(X_{ij}'\setminus\bigcup_{k<j}X_{ik}'\Big), $$ and define $$ f_1(x):=y_{ij} \quad\text{on }X_{ij}, $$ and so on. We thus obtain a sequence of $(\mathcal{A},\mathcal{B})$-measurable functions $f_n$. Since $d_Y(f_m(x),f_n(x))\le 2^{-\operatorname{min}(m,n)}$, they converge pointwise to a measurable function $f$. We claim that $f$ is a selection. Indeed, given $x\in X$, take $i$ such that $x\in X_i$; since $F(x)$ intersects $Y_i$, which contains $f_0(x)=y_i$, we have $d_Y(f_0(x),F(x))\le\operatorname{diam}(Y_i)\le 1$. Similarly, $d_Y(f_n(x),F(x))\le 2^{-n}$. In the limit we have $d_Y(f(x),F(x))=0$, which gives $f(x)\in F(x)$ since $F(x)$ is closed.