Closed witness codings and completion measurability of Borel projections

Give $\mathbb{N}$ the discrete topology and write $W={\mathbb{N}}^{\mathbb{N}}$ with its product topology. Define $d_W(f,g)=0$ if $f=g$, and $d_W(f,g)=2^{-k}$ if $k$ is the first coordinate where they differ. If $k$ is the first coordinate where $f$ and $h$ differ, at least one of $f,g$ or $g,h$ differs at a coordinate no later than $k$; hence $d_W(f,h)\le \max \{d_W(f,g),d_W(g,h)\}$, so $d_W$ is an ultrametric. For $m\ge 1$, the prefix cylinder $C_m(f)=\{g:g(i)=f(i)\text{ for }i<m\}$ is the metric ball $B(f,2^{-(m-1)})$. Every product-basic neighborhood of $f$ contains a sufficiently long prefix cylinder, and every metric ball is a union of prefix cylinders: around a point $g\mathrm{\ne }f$ in the ball, fixing through the first coordinate where $g$ differs from $f$ stays inside the ball; around $f$, choose $m$ with $2^{-m}$ smaller than its radius. Thus $d_W$ induces the product topology. A Cauchy sequence eventually stabilizes at every coordinate, and the stabilized function is its limit, so $d_W$ is complete. By [F2,F4], the set ${\bigcup }_{m\in \mathbb{N}}{\mathbb{N}}^m$ of finite words is at most countable: apply finite-power countability and then the countable-union theorem, with Countable Choice supplied by AC. Choose a surjective enumeration $e:\mathbb{N}\to {\mathbb{N}}^{<\mathbb{N}}$ and give each word $s$ its least preimage $c(s)=\min \{n:e(n)=s\}$; distinct words have distinct indices. Every eventually-zero function has a unique finite prefix ending at its least zero-tail cutoff; these indices make this family countable, and extending any prefix by zeros proves it dense. The finite-prefix cylinders are a countable basis. A fixed bijection $\beta :\mathbb{N}\times \mathbb{N}\to \mathbb{N}$ gives a homeomorphism $W^{\mathbb{N}}\cong W$ by $(w_n{)}_n\mapsto w$, $w(\beta (n,k))=w_n(k)$, since it and its inverse send finite-coordinate basic open sets to finite-coordinate basic open sets.

If $P,Q$ are Polish and nonempty, choose complete compatible metrics and countable dense sets. Truncate each metric at $1$; balls of radius less than $1$ are unchanged, and a Cauchy sequence for the truncated metric is Cauchy in the original metric at every tolerance less than $1$, so truncation preserves topology and completeness. Sum the two truncated metrics on $P\times Q$. A Cauchy sequence for the sum is Cauchy in each coordinate, and the two limits give its product-metric limit. The sum metric induces the product topology: a sum-metric ball is contained in the product of coordinate balls of the same radius, while the product of coordinate balls of radius $\epsilon \mathrm{/}2$ lies in the sum-metric ball of radius $\epsilon$. The product of countable dense sets is countable and dense, so $P\times Q$ is Polish. Enumerate each countable dense set and the positive rationals (obtained from an enumeration of $\mathbb{Q}$ by replacing nonpositive values by $1$). In each factor, balls centered at dense points with positive rational radii form a countable base: given $x\in U$ open, choose $\delta >0$ with $B(x,\delta )\subseteq U$, a dense center $p$ with $d(x,p)<\delta \mathrm{/}4$, and a rational $r$ with $d(x,p)<r<\delta -d(x,p)$; then $x\in B(p,r)\subseteq U$. Product rectangles form a countable base. Enumerating that base, each product-open set is the countable union of its rectangles contained in it, hence belongs to the product sigma-algebra. Conversely, for open $V\subseteq Q$, the class of $A\subseteq P$ with $A\times V$ product-Borel is a sigma-algebra containing every open $A$, because open rectangles are product-open; thus it contains every Borel $A$. For each such Borel $A$, the class of $B\subseteq Q$ with $A\times B$ product-Borel is a sigma-algebra containing every open $V$ by the preceding sentence. Thus all Borel rectangles are product-Borel, and the product Borel sigma-algebra equals the product sigma-algebra. The product of two Polish presentations therefore gives a measurable isomorphism with the product Polish presentation, proving the finite-product standard-Borel claim in [F1]. Empty factors are immediate.

Let $\mu$ be a finite Borel measure on Polish $X$. For any $E\subseteq X$ put $a(E)=\inf \{\mu (B):B\text{ Borel},E\subseteq B\}$; the family is nonempty since it contains $X$, and its values are bounded between $0$ and $\mu (X)$. By the infimum property and AC, choose Borel $B_n\supseteq E$ with $\mu (B_n)<a(E)+1\mathrm{/}(n+1)$. Then $H={\bigcap }_n{B}_n$ is Borel, contains $E$, and satisfies $\mu (H)=a(E)$: monotonicity gives $\mu (H)\le \mu (B_n)<a(E)+1\mathrm{/}(n+1)$ for every $n$, hence $\mu (H)\le a(E)$ as $1\mathrm{/}(n+1)\to 0$, while the definition of $a(E)$ gives the reverse inequality. If $D\supseteq E$ is Borel, then $H\cap D$ is another Borel superset of $E$, so $\mu (H\cap D)\ge a(E)=\mu (H)$; monotonicity gives equality, and finite additivity yields $\mu (H\setminus D)=0$. Thus every set has a Borel envelope $H$ with this null-difference property.

Let $Z$ be a nonempty Polish space and choose a bounded complete compatible metric $d$ and a countable dense set $D$. Since $D$ is nonempty at most countable, fix a surjection $q:\mathbb{N}\to D$. Let $r_0:\mathbb{N}\to \mathbb{Q}$ be a surjection and define $r(j)=r_0(j)$ if $r_0(j)>0$ and $r(j)=1$ otherwise; then $r$ surjects onto ${\mathbb{Q}}_{>0}$. Pair the indices $(i,j)$ by a fixed bijection $\mathbb{N}\times \mathbb{N}\cong \mathbb{N}$. For a nonempty open $U\subseteq Z$ and depth $n$, declare $(i,j)$ admissible when $\bar{B}(q_i,r_j)\subseteq U$ and $\mathrm{diam}(B(q_i,r_j))\le 1\mathrm{/}(n+1)$. Each open ball is open: for $x\in B(q_i,r_j)$, the positive radius $r_j-d(q_i,x)$ gives a ball around $x$ contained in it by the triangle inequality. Each closed ball is closed: if $d(q_i,x)>r_j$, the positive radius $d(q_i,x)-r_j$ gives a ball around $x$ disjoint from it by the reverse triangle inequality; hence the closure of the open ball lies in the closed ball. These balls cover $U$: given $z\in U$, choose $\delta >0$ with $B(z,\delta )\subseteq U$, then choose $q_i$ with $d(z,q_i)<\min (\delta \mathrm{/}4,1\mathrm{/}(8n+8))$ and a rational radius $r_j$ with $d(z,q_i)<r_j<\min (\delta -d(z,q_i),1\mathrm{/}(2n+2))$. The strict upper bound puts the closed ball inside $B(z,\delta )$, and any two points in $B(q_i,r_j)$ are at distance less than $2r_j<1\mathrm{/}(n+1)$ by the triangle inequality. For each child index $k$, use its paired code to set $U_{s^{⌢}k}=B(q_i,r_j)$ when the decoded pair is admissible, and set it empty otherwise. The children cover each nonempty parent; empty parents have only empty children. Thus closures lie inside parents and child diameters tend to zero with depth.

Every closed subspace $C$ of a Polish space $Z$ is Polish. Restrict a compatible complete metric to $C$: a Cauchy sequence in $C$ converges in $Z$ by completeness, and closedness puts its limit in $C$. A countable base of $Z$ is given by the dense-centre rational balls constructed in step 1.2, so its intersections with $C$ form a countable base of $C$. If $C$ is nonempty, AC supplies one point in each nonempty basic intersection. These countably many points are dense, since every nonempty open subset of $C$ contains a nonempty basic intersection. Thus $C$ is separable and completely metrizable; the empty subspace is Polish with its empty metric and dense set.

Every nonempty Polish space $Z$ is a continuous image of $W$. Use the bounded complete metric and dense-centre rational-ball candidates of step 1.4, but index only admissible children. For each nonempty open parent $U_s$ at depth $n$, its set of admissible paired indices is infinite: choose a dense centre in $U_s$ and a positive margin whose closed ball lies inside $U_s$; infinitely many distinct positive rational radii below that margin and $1\mathrm{/}(2n+2)$ are admissible. Enumerate the admissible indices increasingly, using the least element and then the least index greater than its predecessor. Define $U_{s^{⌢}k}$ by the $k$th admissible pair, starting with $U_{\mathrm{\varnothing }}=Z$. Every child is nonempty, its closure is contained in its parent, the children cover the parent by step 1.4, and diameters at depth $n\ge 1$ are at most $1\mathrm{/}n$. For any branch $f\in W$ and $n\in \mathbb{N}$, let $c_n$ be the centre of $U_{f\mathrm{\mid }(n+1)}$. These centres form a Cauchy sequence because all later centres lie in each earlier ball; let $p(f)$ be its complete-metric limit. The limit lies in every $U_{f\mathrm{\mid }n}$, since the closure of the next ball is contained in that ball. A common prefix of length $n$ therefore places two image points in one ball of diameter at most $1\mathrm{/}n$, proving continuity of $p$ in the prefix-cylinder topology. For each fixed $z\in Z$, recursively take the least child containing $z$, possible because the children cover each parent. Its branch centres converge to $z$, so $p$ is onto. The limit and least-child constructions require no choice indexed by the branches; only the initial metric/dense enumeration and the inherited countable choices are used.

For Polish $P$, call $A\subseteq P$ closed-coded if $A={\mathrm{proj}}_P(C)$ for some closed $C\subseteq P\times W$. Closed $A$ is closed-coded by $A\times W$. If $A_n={\mathrm{proj}}_P(C_n)$, code ${\bigcup }_n{A}_n$ by the closed set of $(x,w)$ for which $(x,w_{\ge 1})\in C_{w(0)}$, where $w_{\ge 1}(k)=w(k+1)$. It is closed because each first-coordinate slice is clopen and the corresponding tail condition is closed.

Let $(A_s{)}_{s\in {\mathbb{N}}^{<\mathbb{N}}}$ be a decreasing Borel scheme, and let $E_s$ be the union of branch intersections over branches extending $s$. Then $E_s\subseteq A_s$ and $E_s={\bigcup }_k{E}_{s^{⌢}k}$. AC chooses a Borel envelope $H_s$ from step 1.3 for each finite word $s$. Define $B_s=A_s\cap {\bigcap }_{t⪯s}{H}_t$. Then $B_s$ is Borel, $E_s\subseteq B_s\subseteq H_s$, and $B_s\setminus D$ is null for every Borel $D\supseteq E_s$. In particular $C_s=B_s\setminus {\bigcup }_k{B}_{s^{⌢}k}$ is Borel null, since the child union contains $E_s$. For each natural-number code, let $C_m$ be the exceptional set for its unique decoded word if it is a valid code, and otherwise let $C_m=\mathrm{\varnothing }$; then $C={\bigcup }_m{C}_m$ is Borel and null by countable subadditivity. If $x\in B_{\mathrm{\varnothing }}\setminus C$, then at every node containing $x$ some child also contains $x$; recursion taking the least such child yields a branch $f$ with $x\in A_{f\mathrm{\mid }n}$ for all $n$. Conversely every branch point lies in $B_{\mathrm{\varnothing }}$. Thus $B_{\mathrm{\varnothing }}\setminus C\subseteq E_{\mathrm{\varnothing }}\subseteq B_{\mathrm{\varnothing }}$, so $E_{\mathrm{\varnothing }}=(B_{\mathrm{\varnothing }}\setminus C)\cup (E_{\mathrm{\varnothing }}\cap C)$ belongs to the completion and differs from a Borel set only inside the Borel null set $C$.

If $A_n={\mathrm{proj}}_P(C_n)$, use $W^{\mathbb{N}}\cong W$ to define the closed set of $(x,(w_n))$ satisfying $(x,w_n)\in C_n$ for every $n$. Its projection is ${\bigcap }_n{A}_n$: one inclusion is immediate, and for the other AC chooses one witness $w_n$ for each $n$. Thus closed-coded sets are closed under countable unions and intersections. If $U\subseteq P$ is open and $P\setminus U\mathrm{\ne }\mathrm{\varnothing }$, set $F=P\setminus U$ and define $d(x,F)=\inf \{d(x,y):y\in F\}$. The triangle inequality gives $\mathrm{\mid }d(x,F)-d(x^{\mathrm{\prime }},F)\mathrm{\mid }\le d(x,x^{\mathrm{\prime }})$, so each set $\{x:d(x,F)\ge 1\mathrm{/}(n+1)\}$ is closed. Each $x\in U$ has positive distance from $F$, so these sets cover $U$; if $x\in F$, its distance is $0$. For $U=P$ the assertion is immediate. The class of sets whose members and complements are closed-coded is therefore a sigma-algebra containing all open sets. Every Borel subset of $P$ is closed-coded.

If $Z=\mathrm{\varnothing }$ or $F=\mathrm{\varnothing }$, its projection is empty and the branch scheme with all terms empty has empty branch union. Otherwise start with $U_{\mathrm{\varnothing }}=Z$, use the tree from step 1.4 and put $A_s=\overline{{\mathrm{proj}}_X(F\cap (X\times U_s))}$, a closed decreasing scheme on $X$. Its branch union is exactly ${\mathrm{proj}}_X(F)$. If $(x,z)\in F$, recursively choose the least child at each depth containing $z$, which exists because the children cover their parent; this gives a branch with $z\in U_{f\mathrm{\mid }n}$, hence $x\in A_{f\mathrm{\mid }n}$ for every $n$. Conversely, if $x$ lies in a branch intersection, then every open ball around $x$ meets ${\mathrm{proj}}_X(F\cap (X\times U_{f\mathrm{\mid }n}))$: otherwise its closed complement would be a closed superset of that projection omitting $x$, contrary to the definition of $A_{f\mathrm{\mid }n}$. For each $n\in \mathbb{N}$ choose $x_n$ within $1\mathrm{/}(n+1)$ of $x$ and $z_n\in U_{f\mathrm{\mid }(n+1)}$ with $(x_n,z_n)\in F$. For $m\ge n\ge 0$, both $z_m,z_n\in U_{f\mathrm{\mid }(n+1)}$, whose diameter is at most $1\mathrm{/}(n+1)$ by step 1.4, so $(z_n{)}_{n\in \mathbb{N}}$ is Cauchy; completeness gives $z_n\to z$. Since $x_n\to x$ and $F$ is closed in the product topology, $(x,z)\in F$. Step 2.4 therefore gives completion-measurability and a Borel version for the projection of every closed $F$ under a finite Borel measure.

For sigma-finite $\mu$, choose a Borel cover $(E_n)$ by finite-measure sets and make it disjoint by $D_n=E_n\setminus {\bigcup }_{k<n}{E}_k$. For each $n$, ${\mu }_n(A)=\mu (A\cap D_n)$ is a finite Borel measure by countable additivity. For each Borel relation $R$, its piece $R\cap (D_n\times Y)$ is Borel; step 3.1 gives it a closed witness, and step 3.2 gives a finite-measure Borel version for its projection under ${\mu }_n$. For a branch union $S$ of a decreasing scheme $(A_s)$, the piece $S\cap D_n$ is the branch union of the Borel scheme $(A_s\cap D_n)$, so step 2.4 gives the same finite-measure conclusion. In either case intersect each Borel version and its Borel null exceptional set with $D_n$, obtaining $G_n,C_n\subseteq D_n$ with $G_n\setminus C_n\subseteq S_n\subseteq G_n$ and $\mu (C_n)=0$. Countable subadditivity shows that $G={\bigcup }_n{G}_n$ is Borel, $C={\bigcup }_n{C}_n$ is Borel null, and $G\setminus C\subseteq S\subseteq G$; therefore $S$ is measurable in the completion and agrees with $G$ off $C$. By [F4] the completion is a complete measure space. Finally, for Borel $R\subseteq X\times Y$, step 3.1 gives closed $F\subseteq X\times Y\times W$ with ${\mathrm{proj}}_{X\times Y}(F)=R$, so ${\mathrm{proj}}_X(R)={\mathrm{proj}}_X(F)$. This proves all the claims. The source’s Theorem A.C.6 is a conull selector statement with its proof referred out; no selector is used here.