Uniform null G-delta sets capture block functions

For an open $O\subseteq 2^{\omega }$, let $P_O$ be the prefix-free set of shortest finite words $s$ with $[s]\subseteq O$, and put $\nu (O)={\sum }_{s\in P_O}{2}^{-\mathrm{\mid }s\mathrm{\mid }}$, the supremum of its finite partial sums in the fixed word order. Define the closed content $m(K)=1-\nu (2^{\omega }\setminus K)$ and call $E$ null when, for every $k$, it has an open cover of content below $2^{-k}$. Refining finitely many cylinders to one common length proves finite additivity on clopen sets, monotonicity, and countable subadditivity for open unions directly from binary-word counts. If $K\subseteq D$ with $K$ closed and $D$ clopen, then $m(K)\le \nu (D)$. Every definition uses a fixed enumeration or a real supremum and hence exists in ZF.

Using the canonical pairing from [F1], put $C_{n,m}=\{⟨⟨n,m⟩,j⟩:j\le n\}$ and $B_{n,m}=\{x:\mathrm{\forall }c\in C_{n,m}x(c)=1\}$. The coordinate groups are disjoint and have size $n+1$. Refining to a prefix above the finitely many coordinates shows $\nu (B_{n,m})=2^{-(n+1)},\nu \left({\bigcap }_{(n,m)\in J}{B}_{n,m}^c\right)={\prod }_{(n,m)\in J}(1-2^{-(n+1)})$ for every finite $J$; this is a finite pattern count, not a product-measure theorem.

Fix open $U$ with $\nu (U)<1$ and put $K_0=2^{\omega }\setminus U$, so $m(K_0)>0$. Enumerate the finite words as $(s_i)$. Let $Z$ be the union of those traces $K_0\cap [s_i]$ with closed content zero. Such a compact zero-content trace has, for each requested rational error, a finite clopen cover of smaller content: choose a finite clopen subset of its open complement whose content is sufficiently close to one and take the complement. Choose the least finite cover in the fixed code order. Assigning error $2^{-i-k-2}$ to the pair $(i,k)$ and taking the open union proves directly that $Z$ is null, without Countable Choice. Put $K=K_0\setminus Z$. The set $Z$ is $K_0$ intersected with the open union of the corresponding cylinders, so $K$ is closed. Monotonicity gives $m(K)\le m(K_0)$, while the arbitrarily small canonical open covers of $Z$ and the finite/open content inequalities give $m(K_0)\le m(K)+\epsilon$ for every positive rational $\epsilon$; hence $m(K)=m(K_0)>0$. Every nonempty trace $K\cap [s]$ has positive closed content, since otherwise the corresponding $K_0\cap [s]$ was removed.

For $f:\omega \to \omega$ put $N_f={\bigcap }_{k<\omega }{\bigcup }_{n\ge k}{B}_{n,f(n)}$. It is $G_{\delta }$ by [F2]. For every $k$, its displayed tail union is an open cover of content at most ${\sum }_{n\ge k}{2}^{-(n+1)}=2^{-k}$ by step 1.1, so $N_f$ is null by the local definition. The assignment is arithmetic in $f$ and the fixed blocks; therefore its code belongs to every transitive model containing $f$.

For $s\in T_K=\{s:[s]\cap K\mathrm{\ne }\mathrm{\varnothing }\}$ put $A_s(n)=\{m:K\cap [s]\cap B_{n,m}=\mathrm{\varnothing }\}$. For every finite $J\subseteq \{(n,m):m\in A_s(n)\}$, steps 1.1--2.2 give $0<m(K\cap [s])\le {\prod }_{(n,m)\in J}(1-2^{-(n+1)})\le \exp (-{\sum }_{(n,m)\in J}{2}^{-(n+1)})\mathrm{.}$ Taking canonical finite initial subsets shows that ${\sum }_n\mathrm{\mid }{A}_s(n)\mathrm{\mid }\mathrm{/}{2}^{n+1}$ converges. Hence every $A_s(n)$ is finite and $\mathrm{\mid }{A}_s(n)\mathrm{\mid }\mathrm{/}{2}^{n+1}\to 0$.

Assume $N_f\subseteq U$, so $K\cap N_f=\mathrm{\varnothing }$. If every $K\cap {\bigcup }_{n\ge m}{B}_{n,f(n)}$ met every nonempty basic open subset of $K$, these sets would be dense open. The least-branch construction in [F3] makes $K$ separable and complete in ZF, so the Baire theorem would make their intersection $K\cap N_f$ nonempty. Therefore some $s\in T_K$ and $m$ satisfy $K\cap [s]\cap {\bigcup }_{n\ge m}{B}_{n,f(n)}=\mathrm{\varnothing }$.

Let $i:2^{<\omega }\to \omega$ be the fixed bijection, and let $n(s)$ be the least threshold after which $\mathrm{\mid }{A}_s(n^{\mathrm{\prime }})\mathrm{\mid }\mathrm{/}{2}^{n^{\mathrm{\prime }}+1}\le 2^{-i(s)-1}$. Put ${\phi }_U(n)=\bigcup \{A_s(n):s\in T_K,n(s)\le n\}$. For any finite $E\subseteq {\phi }_U(n)$, assign to each $m\in E$ the least witnessing $s$ in the fixed word order. Then $\frac{\mathrm{\mid }E\mathrm{\mid }}{2^{n+1}}\le {\sum }_{s:n(s)\le n}\frac{\mathrm{\mid }{A}_s(n)\mathrm{\mid }}{2^{n+1}}\le {\sum }_s{2}^{-i(s)-1}\le 1.$ If ${\phi }_U(n)$ had more than $2^{n+1}$ elements, its first $2^{n+1}+1$ elements would contradict this bound. Thus it is finite and has the required size.

With $s,m$ as in step 4.1 and $\mathrm{\ell }=\max \{m,n(s)\}$, every $n\ge \mathrm{\ell }$ satisfies $K\cap [s]\cap B_{n,f(n)}=\mathrm{\varnothing }$, hence $f(n)\in A_s(n)\subseteq {\phi }_U(n)$. Together with steps 3.1 and 4.2 this proves capture, the size bound and model-membership of the codes.

The steps above provide the uniformly assigned null $G_{\delta }$ sets and the capture sets with all stated properties, which is the Statement.