Borel quasi-isometry between proper geodesic spaces

Question

Let $(S,d_{1})$ and $(S',d_{2})$ be two proper geodesic metric spaces. If there exists a quasi-isometric embedding $f: S\rightarrow S'$, does there exist a $\textbf{Borel}$ quasi-isometric embedding $g: S\rightarrow S'$ as well (with respect to the corresponding Borel $\sigma$-algebras)?.

I asked an analogous question, Continuous quasi-isometry between Riemannian manifolds, regarding continuity. In said post I was told that in general a continuous quasi-isometry cannot be expected solely from the existence of a quasi-isometry (although, the uniform contractibility on the counterdomain ensures a positive answer, as was answered as well).

Just to recall, a $\textbf{quasi-isometry}$ $f: S\rightarrow S'$ is a function for which there exist positive constants $L,C$ such that $$L^{-1}d_{1}(x,y)-C\leq d_{2}\big(f(x),f(y)\big)\leq Ld_{1}(x,y)+C\hspace{0.5cm} \forall x,y\in S.$$ Moreover, if the preimage of a Borel set on $S'$ under $f$ is a Borel set on $S$ , $f$ is a $\textbf{Borel quasi-isometry}$.

Answer 1

Here is an outline (I do not need the geodesic property, only properness of your metric space). If you are interested in metric geometry, you should be able to fill in the details.

Definition. A separated net in a metric space $(X,d)$ is a subset $N\subset X$ such that for some $r>0, \epsilon>0$ the following hold:

(a) $\bigcup_{x\in N} B(x,r)=X$ (i.e. $N$ is an net).

(b) $B(x,\epsilon)\cap B(y,\epsilon)=\emptyset$ for all distinct points $x, y\in N$ (i.e. $N$ is separated).

I leave it to you as an exercise to prove that:

(1) A separated net exists in every metric space.

(2) If $(X,d)$ is proper, then every ball $B(x,R)\subset X$ contains only finitely many elements of $N$.

(3) The inclusion map $\iota: (N,d|_N)\to (X,d)$ is a quasi-isometry.

The key is to prove that for every separated net $N\subset X$ in a proper metric space, there exists a Borel map $q: X\to N$ which is a quasi-inverse to the quasi-isometry $\iota$. Once you have this, then for every quasi-isometric embedding $$ f: (X,d)\to (X',d') $$ you take $g:= f\circ q$, a new quasi-isometric embedding which is Borel and is at a finite distance from $f$.

Now, I will define $q$. Given a separated net $N\subset X$, consider the corresponding Voronoi tiling of $X$ corresponding to $N$. The (closed) tile corresponding to a point $x\in N$ will be denoted $V_x$: $$ V_x=\{y\in X: d(x,y)\le d(z,y) \quad \forall z\in N\}. $$ In view of properness of $(X,d)$, each point in $X$ belongs only to finitely many tiles. Moreover, the diameter of each tile is $\le 2r$.

Each tile is a Borel subset and so are the intersections of the tiles $$ V_{x_1,...,x_k}=V_{x_1}\cap .... \cap V_{x_k}, $$ (the "closed strata" of the tiling) as well as the following subsets, "open strata" of the tiling: $$ V^{'}_{x_1,...,x_k}:= V_{x_1,...,x_k}- \bigcup_{x\in N - \{x_1,...,x_k\}} V_x. $$ For instance, for $x_1\in N$, $$ V^{'}_{x_1}= \{y\in X: d(x_1,y)< d(z,y) \quad \forall z\in N-\{x_1\}\}. $$

Then $X$ is a disjoint union of the open strata as above. I will order arbitrarily all the finite subsets $\{x_1,...,x_k\}$ of $N$ for which $V^{'}_{x_1,...,x_k}$ is nonempty and let $x_1$ denote the smallest element of this finite subset.

Define the map $$ q: X\to N, q(x)=x_1, $$ where $x\in V^{'}_{x_1,...,x_k}$.

(4) I leave it to you to check that $q$ is a Borel quasi-isometry as required.

Answer 2

This is a consequence of:

Fact: let $X,Y$ be metric spaces and $f$ a coarse map $X\to Y$ (where coarse means that $\forall x_1,x_2, d_Y(f(x_1),f(x_2))\le u(d_X(x_1,x_2))$ for some fixed increasing function $u$. Suppose that for some $R$, $X$ is countable union of open $R$-balls. Then $f$ is at bounded distance to a Borel function $g$.

Proof: write $X=\bigcup B_n$ where each $B_n$ is an $R$-ball. Define $B'_n=B_n\smallsetminus \bigcup_{k<n}B_k$. Extracting if necessary, we can suppose that no $B'_n$ is empty (unless $X$ is bounded, but this case is trivial.) Choose $x_n\in B'_n$. Define $g(x)=f(x_n)$ for $x\in B'_n$. Then $f,g$, are at bounded distance. I claim that $g$ is Borel. Namely, $g^{-1}(Y')$ is Borel for every subset $Y'$ of $Y$. Indeed, $g^{-1}(Y')$ equals $\bigcup_{n\in I}B'_n$ for some subset $I$ of integers. So this is indeed Borel. $\Box$

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The assumption on $X$ is clearly satisfied if $X$ is proper. And the condition on $f$ is clearly satisfied by a QI-embedding, and the latter is preserved when passing to a map at bounded distance ($f,g$ at bounded distance means $\sup_x d_Y(f(x),g(x))<\infty$).