Conull Borel uniformizations and Borel versions of measured suprema

If $X=\mathrm{\varnothing }$, take $X_0=\mathrm{\varnothing }$; all selector assertions are vacuous and any bounded scalar function has the zero Borel version agreeing on $X_0$. Otherwise choose Polish presentations of $X$ and $Y$ by [F1] and transport $R$ to those presentations. Let $W={\mathbb{N}}^{\mathbb{N}}$ and $Z:=Y\times W$. Since $X$ is nonempty and every $R_x$ is nonempty, $Y$ and $Z$ are nonempty. By [F7,F14], $Z$ is Polish, so choose a compatible complete metric $d$ on $Z$. By [F2], $R$ is the projection of a closed witness in $(X\times Y)\times W$. The coordinate reassociation $((x,y),w)\mapsto (x,(y,w))$ is a homeomorphism because both product topologies have bases of open rectangles [F14]; transporting the witness gives a closed $F\subseteq X\times Z$ with ${\mathrm{proj}}_{X\times Y}(F)=R$. Every fibre $F_x:=\{z:(x,z)\in F\}$ is nonempty because every $R_x$ is nonempty.

Fix a bounded real Borel function $\phi$ and a relation $R$ as in the statement. The bounded set $\{\phi (x,y):(x,y)\in R\}$ is nonempty for every $x$, so its supremum $m(x)$ exists by [F8]. For every rational $q$, the set $\{x:m(x)>q\}$ equals ${\mathrm{proj}}_X(R\cap {\phi }^{-1}((q,\mathrm{\infty })))$: a supremum exceeds $q$ exactly when some value exceeds $q$. The relation inside this projection is Borel by [F9], so [F2] makes every strict rational superlevel set completion-measurable. By the completion description [F3], AC [F10] chooses Borel representatives $B_q$ and Borel null sets $N_q$ for all rational $q$.

Fix a countable dense sequence $(a_i{)}_{i\in \mathbb{N}}$ in $Z$, an enumeration $(r_j{)}_{j\in \mathbb{N}}$ of the positive rationals, and a bijection $\beta :{\mathbb{N}}^2\to \mathbb{N}$ using [F5]. Set $U_{\mathrm{\varnothing }}=Z$. For every finite word $s$ of length $n$ and every pair $(i,j)$, define $U_{s^{⌢}\beta (i,j)}=B(a_i,r_j)$ if the closed ball $\bar{B}(a_i,r_j)$ is contained in $U_s$ and $r_j<2^{-(n+2)}$, and define it to be empty otherwise. Recursion [F6] defines this family. The children cover each $U_s$: for $z\in U_s$, choose $\epsilon >0$ with $B(z,\epsilon )\subseteq U_s$, then choose $a_i$ close enough to $z$ and a positive rational $r_j$ with $d(a_i,z)<r_j<\min \{\epsilon -d(a_i,z),2^{-(n+2)}\}$ by [F5]. The triangle inequality puts $\bar{B}(a_i,r_j)$ inside $U_s$ and $z$ inside the child ball. Each child lies in its parent and has diameter at most $2r_j<2^{-(n+1)}$.

Remove the Borel null union $N_{\phi }:={\bigcup }_{q\in \mathbb{Q}}{N}_q$ and put $X_{\phi }=X\setminus N_{\phi }$. On $X_{\phi }$, $x\in B_q$ iff $m(x)>q$. Define $\tilde{m}(x):=\sup \{q\in \mathbb{Q}:x\in B_q\}$ on $X_{\phi }$ and $\tilde{m}(x):=0$ off $X_{\phi }$. The rational density [F5,F8] gives $\tilde{m}=m$ on $X_{\phi }$. For each real $a$, $\{\tilde{m}>a\}$ is the union of $X_{\phi }\cap B_q$ over rationals $q>a$, together with $X\setminus X_{\phi }$ when $0>a$; hence it is Borel. Also $\{\tilde{m}<b\}={\bigcup }_{q\in \mathbb{Q},q<b}(X\setminus \{\tilde{m}>q\})$ is Borel. Rational open intervals form a basis by [F5], so $\tilde{m}$ is Borel by [F4,F9].

For each finite word $s$, let $P_s:={\mathrm{proj}}_X(F\cap (X\times U_s))$. The set inside the projection is Borel, so [F2] makes $P_s$ completion-measurable and gives a Borel representative outside a Borel null set. Since $F$ projects onto $R$ and every section $R_x$ is nonempty, $P_{\mathrm{\varnothing }}=X$. The child-cover property in step 2.1 gives $P_s={\bigcup }_k{P}_{s^{⌢}k}$.

Define completion-measurable prefix cells by $C_{\mathrm{\varnothing }}=X$ and $C_{s^{⌢}k}:=C_s\cap P_{s^{⌢}k}\setminus {\bigcup }_{j<k}{P}_{s^{⌢}j}$; these select the least child containing $x$ and partition each parent cell, using the least-element property in [F2]. By [F3], the completion domain is a sigma-algebra, so every $C_s$ is completion-measurable. AC [F10] chooses a Borel set $B_s$ and Borel null set $N_s$ with $C_s\mathrm{△}{B}_s\subseteq N_s$ for each finite word $s$. The finite-word indices [F13] index these exceptional sets by naturals, using the empty set for unused codes; the countable-union fact [F12] makes $N:={\bigcup }_s{N}_s$ Borel and null. Put $X_0:=X\setminus N$. On $X_0$, membership in every prefix cell agrees with membership in its Borel representative, and at each length those representatives partition $X_0$.

For $x\in X_0$ and $n\ge 1$, let $s_n(x)$ be its unique selected word of length $n$ and let $c_n(x)$ be the centre of $U_{s_n(x)}$. Set $c_0=c_1$. Each $c_n$ is Borel because it is constant on the countable Borel partition $\{B_s\cap X_0:\mathrm{\mid }s\mathrm{\mid }=n\}$: preimages of Borel sets are unions of the corresponding Borel cells by [F4,F9,F13]. The selected balls are nested and their diameters tend to zero, so $(c_n(x){)}_{n\in \mathbb{N}}$ is Cauchy; let $z(x)\in Z$ be its limit by completeness [F7]. Since $x\in P_{s_n(x)}$, each $F_x\cap U_{s_n(x)}$ is nonempty. AC [F10] chooses a point $w_n$ in each such set for $n\ge 1$ and this fixed $x$; set $w_0=w_1$. then $d(w_n,c_n(x))\le 2r_{s_n(x)}\to 0$, so $w_n\to z(x)$. The fibre $F_x$ is closed: if $z\mathrm{\notin }{F}_x$, the open complement of $F$ contains a basic product rectangle $U\times V$ around $(x,z)$, and $V$ is disjoint from $F_x$. Thus $z(x)\in F_x$.

The limit map $z:X_0\to Z$ is Borel. For a fixed nonempty closed $C\subseteq Z$ and $m\ge 1$, put $V_m:={\bigcup }_{y\in C}B(y,1\mathrm{/}m)$ and $E_{n,m}:=\{x\in X_0:c_n(x)\in V_m\}$ for $n\ge 1$. The set $V_m$ is open; each $E_{n,m}$ is Borel because $c_n$ is constant on a countable Borel partition. Since $c_n(x)\to z(x)$ and $C$ is closed, $z(x)\in C$ exactly when, for every $m$, $c_n(x)\in V_m$ eventually: if $z(x)\mathrm{\notin }C$, choose $\epsilon >0$ with $B(z(x),\epsilon )\cap C=\mathrm{\varnothing }$, then choose $m$ with $1\mathrm{/}m<\epsilon \mathrm{/}2$ and take $n$ large enough that $d(c_n(x),z(x))<\epsilon \mathrm{/}2$ and $c_n(x)\in V_m$. Membership in $V_m$ gives $y\in C$ with $d(c_n(x),y)<1\mathrm{/}m$, so the triangle inequality puts $y$ inside $B(z(x),\epsilon )$, a contradiction. Thus $z^{-1}(C)={\bigcap }_{m\ge 1}{\bigcup }_{N\ge 1}{\bigcap }_{n\ge N}{E}_{n,m}$ is Borel. The empty closed set has empty preimage, and closed-set preimages being Borel implies Borel measurability by [F4,F9]. Project $z(x)$ to $Y$ and undo the chosen Polish presentation. The projection is continuous, hence Borel by [F9], so this gives a Borel selector $s:X_0\to Y$ and $(x,s(x))\in R$ by the defining property of $F$.

For countably many relations and bounded Borel functions on the same base, repeat steps 4.1–6.1 for each relation and step 2.2 for each scalar function, then remove the union of their Borel null exceptions. The indices are countable: finite prefixes have the injective indices in [F13], rational levels are countable by [F5], and pairs of natural indices are coded by [F5]. AC [F10] supplies the countable family of representatives; the union is Borel and null by [F12]. Restrict each selector and each Borel version to this common Borel conull set.