Standard-Borel Real Codings and Determining Classes

For $0\le t\le 1\mathrm{/}2$ put $m_k=\lfloor 2^{k}t\rfloor$ for $k\ge 0$ and $b_k=m_k-2m_{k-1}$ for $k\ge 1$. Since $m_{k-1}\le 2^{k-1}t<m_{k-1}+1$, we have $b_k\in \{0,1\}$. Each $m_k$ is Borel: $\{m_k=l\}=[l{2}^{-k},(l+1)2^{-k})\cap [0,1\mathrm{/}2]$. Thus each digit is Borel.

Here $m_0=0$. Telescoping gives ${\sum }_{k=1}^n{b}_k{2}^{-k}=2^{-n}{m}_n$ and $0\le t-2^{-n}{m}_n<2^{-n}$, so $t={\sum }_{k\ge 1}{b}_k{2}^{-k}$. The digits cannot be eventually all ones: such a tail would make $t$ dyadic, whereas for dyadic $t$ the integers $2^{k}t$ are exact for all sufficiently large $k$, giving $b_k=0$. Thus there are infinitely many zeros, including when $t=0$ or $t=1\mathrm{/}2$.

Interleave cube digits by $a_{J(i,k-1)}=b_k(x_i\mathrm{/}2)$. The bijection in [F3] assigns exactly one digit to each positive position. For any binary sequence $a$, define $T(a)={\sum }_{j\ge 1}2a_j{3}^{-j}$. Its values lie in $[0,1]$. Agreement through position $n$ gives $\mathrm{\mid }T(a)-T(a^{\ast })\mathrm{\mid }\le 3^{-n}$. If the first differing position is $n$, its contribution has magnitude $23^{-n}$ and the remaining tail has magnitude at most $3^{-n}$, so $\mathrm{\mid }T(a)-T(a^{\ast })\mathrm{\mid }\ge 3^{-n}>0$. Thus $T$ is continuous and injective.

Conversely let $(b_k)$ be a binary sequence with infinitely many zeros and sum $t\le 1\mathrm{/}2$. Its tail after $n$, multiplied by $2^n$, lies in $[0,1)$: the all-one tail sums to one, and at least one digit is zero. Hence $\lfloor 2^{n}t\rfloor ={\sum }_{k=1}^n{b}_k{2}^{n-k}$, recovering exactly the digits of step 1.1. In binary sequence space $D=\{0,1{\}}^{{\mathbb{N}}_{\ge 1}}$, the allowable row set is $R={\bigcap }_{N\ge 1}{\bigcup }_{k\ge N}\{b:b_k=0\}\cap \{b:{\sum }_{k\ge 1}{b}_k{2}^{-k}\le 1\mathrm{/}2\}$. It is Borel: cylinders are clopen and the sum is continuous because the tail is at most $2^{-n}$.

For a binary word $w$ of length $n$, let $s_w={\sum }_{j=1}^{n}2w_j{3}^{-j}$ and $I_w=[s_w,s_w+3^{-n}]$. Distinct words of the same length give disjoint intervals separated by a positive gap. The closed set $K={\bigcap }_{n\ge 1}{\bigcup }_{\mathrm{\mid }w\mathrm{\mid }=n}{I}_w$ equals $T[D]$: each point of the intersection has a unique word at each length; nesting forces consistent prefixes; the resulting sequence has sum equal to the point since interval lengths tend to zero. Conversely each sum lies in every prefix interval. The inverse digits are continuous on $K$ because the finitely many cylinders at each length are separated. Hence $T:D\to K$ is a homeomorphism, without an appeal to product compactness.

The deinterleaving row maps $D\to D$, $a\mapsto (a_{J(i,k-1)}{)}_{k\ge 1}$, are continuous: a finite row-cylinder condition is a finite cylinder condition on $a$. Therefore $H=\{a:\text{every deinterleaved row belongs to }R\}$ is Borel in $D$. The homeomorphism gives $C=T[H]=(T^{-1}{)}^{-1}[H]$ Borel in $K$. Since $K$ is closed in $[0,1]$, a trace Borel set in $K$ is Borel in $[0,1]$: the trace sets form a sigma-algebra, and relative opens are traces of ambient opens.

The map $c(x)=T(a(x))$ is measurable: every interleaved digit is Borel by step 1.1 and every finite sum has finite range with Borel level sets (finite unions of intersections of digit level sets), so [F4] applies. Its inverse on $C$ is $x_i=2{\sum }_{k\ge 1}(T^{-1}z{)}_{J(i,k-1)}{2}^{-k}$. These coordinates are measurable by [F4]. They lie in $[0,1]$ and recover both compositions by the row characterization; thus $c$ is a bijection onto precisely $C$.

For completeness, rational intervals restricted to $[0,1]$ form a countable basis by density. Finite coordinate boxes from these intervals are countable explicitly. Enumerate rational endpoints by [F6]. A coordinate condition $x_i\in (p_j,p_k)\cap [0,1]$ has code $J(i,J(j,k))$. A list of condition codes $(z_0,\dots ,z_{l-1})$ has code $J(l,e_l)$, where $e_0=0$ and $e_{r+1}=J(e_r,z_r)$. Inverting the injective J recovers the length and every entry, so this encodes lists injectively. Each box is represented by such a finite list; assigning the least code of its representations injects the family of boxes into the naturals. The empty list represents the whole cube. Every open subset of the cube is a union of a subfamily of this countable basis. Thus the Borel sigma-algebra equals the coordinate-generated sigma-algebra, and coordinate measurability in step 5.1 proves measurability of the whole inverse. This establishes all assertions.

If $E=\mathrm{\varnothing }$, the empty bijection onto $\mathrm{\varnothing }$ is bimeasurable. Otherwise fix the single Polish presentation $h:E\to P$ of [F1]. Its separability and metrizability give a homeomorphic embedding $e:P\to Q=[0,1{]}^{\mathbb{N}}$ by [F2].

The interval $[0,1]$ is complete: a Cauchy sequence converges in $\mathbb{R}$ by [F7], and its limit stays between zero and one by the limit inequalities. Its metric is bounded by one. The metric $D(x,y)={\sum }_{n\ge 0}{2}^{-(n+1)}\mathrm{\mid }{x}_n-y_n\mathrm{\mid }$ makes $Q$ a metric space by [F3]. The image $Y=e[P]$ is completely metrizable, transporting a complete compatible metric from $P$. AC supplies DC as described in [F6], so [F4] yields that $Y$ is $G_{\delta }$ and consequently Borel in $Q$. The currently repaired supplier proves the ambient equality: points in every small open neighbourhood union lie within $1\mathrm{/}n$ of $Y$, hence in its closure, before the complete-metric limit argument.

Let $c:Q\to C$ be [F5]. Since $c^{-1}$ is measurable, $c[Y]=(c^{-1}{)}^{-1}[Y]$ is Borel in $C$, and therefore in $[0,1]$, because $C$ itself is Borel. Restricting $c$ and its inverse to $Y$ and $c[Y]$ preserves measurability. The homeomorphism $e$ is bimeasurable on trace Borel sets. Thus $c\circ e\circ h$ and $h^{-1}\circ e^{-1}\circ c^{-1}$ are mutually inverse measurable maps between $E$ and $c[Y]$.

Fix $f$ from [F1]. Enumerate the pullbacks $H_q=f^{-1}[B\cap (-\mathrm{\infty },q]]$ using [F2]. Let ${\mathcal{A}}_n$ be the Boolean algebra on the first $n$ pullbacks, with ${\mathcal{A}}_0=\{\mathrm{\varnothing },E\}$. Its atoms are the at most $2^n$ intersections obtained by taking each generator or its complement; every member is a union of atoms. Thus each ${\mathcal{A}}_n$ is finite, and $\mathcal{A}={\bigcup }_n{\mathcal{A}}_n$ is an algebra: any two elements lie in a common ${\mathcal{A}}_n$.

AC implies [F5], so [F4] makes $\mathcal{A}$ countable. This is the exact countable-choice use for enumerating the finite algebras. The trace of the generators of [F3] generates $\mathcal{B}(B)$, so bimeasurability of $f$ gives $\sigma (\mathcal{A})=\mathcal{S}$. If $x\mathrm{\ne }y$, injectivity gives different codes; [F8] provides a rational between them, and its pullback contains exactly the lower-coded point.

For finite $\mu ,\nu$ agreeing on $\mathcal{A}$, their total masses agree because $E\in \mathcal{A}$. The equality class $\mathcal{D}=\{A\in \mathcal{S}:\mu (A)=\nu (A)\}$ contains $E$, is closed under complements by subtracting from the common finite total, and under countable disjoint unions by countable additivity. It is a lambda-system containing the pi-system $\mathcal{A}$. By [F7], $\mathcal{S}\subseteq \mathcal{D}$. This also covers zero total mass and $E=\mathrm{\varnothing }$; finiteness prevents subtraction of infinite totals.

If $P$ is empty there is only the empty subset and the assertion holds. On nonempty $P$, fix a compatible complete metric and a countable basis using [F1], [F4] and [F5]. An open $U$ is $G_{\delta }$ (repeat $U$), so [F2] completely metrizes it; its closed complement is complete in the restricted original metric since a limit of a sequence in a closed set stays there. Both subspaces have countable trace bases and hence are Polish by [F4].

Bound each component metric by replacing $d$ with $\min (d,1)$; this preserves its topology and Cauchy sequences, hence completeness. On the disjoint union of $U$ and $P\setminus U$, retain those metrics within components and set cross-component distance equal to two. The triangle inequality holds within a component and across components (any cross-component path includes an edge of length two). A Cauchy sequence is eventually in one component and converges there. A union of the two countable bases is countable. This Polish topology is finer than $\tau$, makes $U$ clopen, and has the same Borel sets: each new open is the union of two old trace-open sets, hence old Borel. Empty components simply contribute no points.

Let $\mathcal{R}$ be the old Borel subsets that can be made clopen by such a refinement. Step 2.1 puts every open set in $\mathcal{R}$; closure under complements uses the same topology. Given $B_n\in \mathcal{R}$, [F5] selects a witnessing Polish topology ${\tau }_{n+1}$, complete bounded metric and countable basis for each $n$. Include ${\tau }_0=\tau$. In ${\prod }_{n\ge 0}(P,{\tau }_n)$ let $\mathrm{\Delta }=\{(x,x,\dots ):x\in P\}$.

The diagonal $\mathrm{\Delta }$ is closed. If two coordinates differ, disjoint neighbourhoods in the original metric topology pull back to open neighbourhoods in both refined coordinates; their product cylinder misses $\mathrm{\Delta }$. The product is completely metrized by [F3], so its closed subspace $\mathrm{\Delta }$ is complete. It has a countable basis of finite cylinders restricted to $\mathrm{\Delta }$; For each finite length the coordinate-index and basis-index lists form a countable set by [F8]; [F7] makes the union over lengths countable, with its countable-choice hypothesis supplied by [F5]. By [F4] it is Polish. Pull its topology back to $P$ along $x\mapsto (x,x,\dots )$. This refines every ${\tau }_n$. Each basic open is a finite intersection of old Borel sets, and every open is a union of a subfamily of the countable basis. Thus every new open is old Borel; the two Borel sigma-algebras coincide.

Each $B_n$ is clopen in the common refinement, so ${\bigcup }_n{B}_n$ is open. Apply the splitting construction of step 2.1 to that Polish topology, making the union clopen while preserving its Borel sets, hence the original Borel sets. Consequently $\mathcal{R}$ is a sigma-algebra containing $\tau$, and contains every old Borel set. For the specified $B$, restrict the resulting complete metric and countable basis to the closed set $B$. This is a Polish topology on $B$, finer than its original subspace topology; its Borel sets are precisely the old traces, since relative opens generate traces of Borel sets in either topology. The identity is the Polish presentation required by [F6].