Uniform null-code measurability makes the Raisonnier filter rapid

Fix an arbitrary strictly increasing sequence $⟨n_i:i<\omega ⟩$ of natural numbers and let $r$ code this sequence. Put $y=x\oplus r$ and $M=L[y]$. Since $L[x]\subseteq M\subseteq V$, the inequalities ${\omega }_1^{L[x]}\le {\omega }_1^M\le {\omega }_1$ and the Given equality show that ${\omega }_1^M={\omega }_1$. For $f\in M\cap 2^{\omega }$ put $\bar{f}(i)=f\upharpoonright n_i$, using the canonical natural-number code for the finite word. Both $f$ and the sequence $⟨n_i⟩$ belong to $M$, so $\bar{f}\in M$ and [F2] puts the code of $N_{\bar{f}}$ in $M$.

By the uniform measurability hypothesis, $A(y)$ is measurable. Hence [F1] makes the union of all null Borel sets coded in $M=L[y]$ null, and that union contains every $N_{\bar{f}}$ from step 1.1. By the definition of the completed coin measure, the union is contained in a Borel null set $Z$. Apply [F5] to $2^{\omega }\setminus Z$ and choose a compact $K$ there with $\nu (K)>0$; then $U=2^{\omega }\setminus K$ is open, contains the union, and has $\nu (U)<1$. For every $f\in M\cap 2^{\omega }$ there is $i_f$ such that $f\upharpoonright n_i=\bar{f}(i)\in {\phi }_U(i)$ for all $i\ge i_f$, by [F2]'s capture clause.

Let ${\mathrm{\Psi }}_i$ be the elements of ${\phi }_U(i)$ that decode binary strings of length $n_i$. Define $a\subseteq \omega$ by declaring $k\in a$ iff, for $i=\min \{j:k\le n_j\}$, there are distinct $s,t\in {\mathrm{\Psi }}_i$ whose first differing prefix length $h(s,t)$ is $k$. Thus the positive lengths in the block $(n_{i-1},n_i]$, with $n_{-1}:=0$, are assigned to level $i$, exactly as required by the prefix-length convention of [F3].

The values assigned to level $i$ lie in $(n_{i-1},n_i]$ and are the first-difference prefix lengths realized by pairs from ${\mathrm{\Psi }}_i$. If a finite set of equal-length binary strings has $k\ge 1$ members, its prefix tree has at most $k-1$ branching levels; if $k=0$, it realizes no first differences. Thus level $i$ contributes at most $\max \{\mathrm{\mid }{\mathrm{\Psi }}_i\mathrm{\mid }-1,0\}\le 2^{i+1}-1$ values. In particular, even with the harmless overcount of a possible endpoint $n_i$, $\mathrm{\mid }a\cap n_i\mathrm{\mid }\le {\sum }_{j\le i}(2^{j+1}-1)\le 2^{i+2}-2.$

For $k<\omega$ and $s\in 2^{n_k}$ put $F_{k,s}=\{f\in M\cap 2^{\omega }:f\upharpoonright n_k=s\text{ and }f\upharpoonright n_i\in {\mathrm{\Psi }}_i\text{ for every }i\ge k\}\mathrm{.}$ These countably many sets cover $M\cap 2^{\omega }$ by step 1.2. If distinct $f,g\in F_{k,s}$ and $i$ is least with $f\upharpoonright n_i\mathrm{\ne }g\upharpoonright n_i$, then $i>k$, both length-$n_i$ prefixes lie in ${\mathrm{\Psi }}_i$, and their first differing prefix length equals $h(f,g)$ and lies in $(n_{i-1},n_i]$. Thus $h(f,g)\in a$ by step 2.1, so $H(F_{k,s})\subseteq a$. The cover therefore witnesses $a\in F(y)$. Since $L[x]\cap 2^{\omega }\subseteq L[y]\cap 2^{\omega }$, the same cover also witnesses $a\in F(x)$.

Given the arbitrary strictly increasing sequence $⟨n_i⟩$, step 3.2 produced $a\in F(x)$ with $\mathrm{\mid }a\cap n_i\mathrm{\mid }\le 2^{i+2}-2<2^{i+2}$ for every $i$. The same conclusion for a merely nondecreasing sequence follows by replacing it with a pointwise larger strictly increasing one. Since $i\mapsto 2^{i+2}$ is one fixed bound, the uniform-bounding form of [F3] makes $F(x)$ rapid.

The steps above establish the rapidity of $F(x)$ under the stated hypotheses; this is the Statement.