Rapid filters are not Lebesgue measurable

Assume for contradiction that $F$ is measurable for the coin measure $\nu$. Since $F$ extends the Fréchet filter, it is closed under finite changes: if $a\in F$ and $a^{\mathrm{\prime }}$ differs from $a$ only below $k$, then $a\cap (\omega \setminus k)\in F$ and is contained in $a^{\mathrm{\prime }}$, so $a^{\mathrm{\prime }}\in F$. Membership in $F$ therefore depends only on the coordinates at and above any fixed level $n$.

To contradict nullity, fix an arbitrary closed $B\subseteq 2^{\omega }$ with $\nu (B)>0$. The density theorem supplies a finite nonempty family $T_0$ of binary strings such that every $s\in T_0$ satisfies $\nu (B\cap [s])>(1-2^{-2})\nu ([s])\mathrm{.}$ For example, enumerate canonically the positive-length strings satisfying the display and take the first one whose cylinder meets $B$. Put $n_0=\max \{\mathrm{lh}(s):s\in T_0\}$.

Zero-one law: for each $n$, the measurable tail event $F$ is independent of the sigma-algebra generated by the first $n$ coordinates. Hence the family of measurable $A$ satisfying $\nu (F\cap A)=\nu (F)\nu (A)$ contains every finite-coordinate cylinder. It is a lambda-system, while the cylinders are a generating pi-system, so [F5] makes the family the whole Borel sigma-algebra and then its completion. Taking $A=F$ gives $\nu (F)=\nu (F{)}^2$, hence $\nu (F)\in \{0,1\}$.

Recursively, after finite nonempty $T_i$ and $n_i$ are fixed, let $S_i$ be the canonically enumerated set of strings $t$ with $\mathrm{lh}(t)>n_i$ and $\nu (B\cap [t])>(1-2^{-(i+3)})\nu ([t])\mathrm{.}$ The density theorem gives $\nu (B\setminus {\bigcup }_{t\in S_i}[t])=0$. Let $T_{i+1}$ be the first finite initial segment of this enumeration for which $\nu (B\setminus {\bigcup }_{t\in T_{i+1}}[t])<2^{-(n_i+i+2)},$ and put $n_{i+1}=\max \{\mathrm{lh}(t):t\in T_{i+1}\}$. These choices are canonical, and every length in $T_{i+1}$ exceeds $n_i$, so $n_0<n_1<\cdots$.

The value is not one. The complement map $T(a)=\omega \setminus a$ is measure preserving, and $T[F]\cap F=\mathrm{\varnothing }$: otherwise a filter would contain both $a$ and its complement and hence their empty intersection. If $\nu (F)=1$, then $\nu (T[F])=1$, contradicting additivity. Thus the assumed measurable filter is null.

Apply rapidity to the increasing function $i\mapsto n_{i+1}$. There is $a\in F$ such that $\mathrm{\mid }a\cap n_{i+1}\mathrm{\mid }\le i$ for every $i$.

Choose any $s_0\in T_0$ whose cylinder meets $B$; the canonical first such member suffices. Recursively suppose $s_i\in T_i$ has been chosen and has value $1$ on every coordinate in $a\cap \mathrm{lh}(s_i)$. Put $H_i=\{z\in 2^{\omega }:z(m)=1\text{ for every }m\in a\cap [\mathrm{lh}(s_i),n_{i+1})\}\mathrm{.}$ The coordinates defining $H_i$ lie above those defining $[s_i]$, so the two events are independent, and step 3.2 gives $\nu (H_i)\ge 2^{-i}$. The density bound for $s_i$ and $\mathrm{lh}(s_i)\le n_i$ therefore give $\begin{array}{rl}{\displaystyle \nu (H_i\cap [s_i]\cap B)}& {\displaystyle \ge \nu (H_i)\nu ([s_i])-\nu ([s_i]\setminus B)}\\ {\displaystyle }& {\displaystyle >(2^{-i}-2^{-(i+2)})2^{-\mathrm{lh}(s_i)}}\\ {\displaystyle }& {\displaystyle \ge 3\cdot 2^{-(n_i+i+2)}\mathrm{.}}\end{array}$ For $i=0$ use the $T_0$ bound in step 1.2; for $i>0$ the defining bound on $T_i$ in step 2.2 is exactly the displayed $2^{-(i+2)}$ conditional error. The uncovered part of $B$ at level $T_{i+1}$ has measure below $2^{-(n_i+i+2)}$, so choose $z_i\in H_i\cap [s_i]\cap B\cap {\bigcup }_{t\in T_{i+1}}[t]$ and then the canonical $s_{i+1}\in T_{i+1}$ with $z_i\in [s_{i+1}]$. Since both strings are initial segments of $z_i$ and $\mathrm{lh}(s_{i+1})>n_i\ge \mathrm{lh}(s_i)$, we have $s_i⊊s_{i+1}$; the definition of $H_i$ preserves the induction invariant.

Let $b={\bigcup }_i{s}_i$. The points $z_i\in B$ converge to $b$, because both $z_i$ and $b$ extend $s_{i+1}$ and the string lengths tend to infinity; closedness gives $b\in B$. The induction invariant and unbounded lengths give $a\subseteq b$, so $b\in F$ by upward closure. Thus every closed positive-set $B$ meets $F$.

Hence $F$ has positive outer measure: if it had outer measure zero, an open $O\supseteq F$ with $\nu (O)<1$ would have a closed positive complement missing $F$, contrary to step 5.1. This contradicts $\nu (F)=0$ from step 3.1. The assumption of measurability is false, and [F2] transfers the conclusion to Lebesgue measure under the standard coding.