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\hypersetup{pdftitle={A Boolean L\'evy group with a free nonsingular action}}
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\newcommand{\N}{\mathbb N}
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\title{A Boolean L\'evy group with a free nonsingular action}
\author{Konstantin Slutsky}
\address{Department of Mathematics, Iowa State University,
  Ames, Iowa 50011, USA}
\email{kslutsky@iastate.edu}
\subjclass[2020]{Primary 22F10; Secondary 22A05, 37A40, 03E15}
\keywords{Polish groups, Boolean groups, L\'evy groups, concentration of measure,
  nonsingular actions.}
\date{October 27, 2026}
\begin{document}
\begin{abstract}
  We construct a Boolean L\'evy group with a free ergodic spatial nonsingular action on a standard
  probability space, answering a question of Glasner, Tsirelson and Weiss. The action admits no
  equivalent invariant $\sigma$-finite measure.  The group is a quotient of a weighted summable
  Boolean group by a discrete subgroup. By the nonsingular fixed-point theorem of Avraham-Re'em and
  Peterzil, the resulting group is not locally Wasserstein.
\end{abstract}
\maketitle
\enlargethispage{3pt}
\setcounter{tocdepth}{1}
\tableofcontents

\section{Introduction}

A topological group is \emph{Polish} if its topology is separable and admits a complete compatible
metric. A group is \emph{Boolean} if every element has order at most two.  It is automatically
abelian: $xy=(xy)^{-1}=y^{-1}x^{-1}=yx$.  Thus we write its operation additively. A \emph{standard
  Borel space} is a measurable space whose \(\sigma\)-algebra is the \(\sigma\)-algebra of Borel
sets for some Polish topology. Our reference
for descriptive set theory is Kechris \cite{Kechris1995}.

Let $G$ be a Polish group and $(X,\mu)$ a standard Borel probability space. A \emph{spatial
  nonsingular action} is a jointly Borel map $G\times X\to X$, denoted $(g,x)\mapsto gx$, satisfying
the action identities at every point and such that $g_*\mu\sim\mu$ for every $g$.  Here
$g_*\mu(A)=\mu(g^{-1}A)$, and equivalence of measures, denoted by $\sim$, means that they have the
same null sets. Such a measure is also called \emph{quasi-invariant}. The action is
\emph{measure-preserving} if $g_*\mu=\mu$ for every $g$. The term spatial distinguishes an
actual pointwise action from an action on the measure algebra, whose elements are measurable sets
modulo null sets.  An action is \emph{free} if $gx=x$ implies that $g$ is the identity, and
\emph{ergodic} if every Borel set invariant modulo null sets has measure zero or one. A measure is
\emph{nonatomic} if it has no atoms; for a probability measure on a standard Borel space this is
equivalent to every singleton having measure zero. A set is \emph{conull} if its complement is null. For a Borel map $T$
and a measure $\alpha$, we use $T_*\alpha$ for the pushforward measure
$T_*\alpha(A)=\alpha(T^{-1}(A))$.

\paragraph{Concentration and L\'evy groups.}
The concentration framework and its connection with fixed-point properties originate in
Gromov--Milman \cite{GromovMilman1983}.  For each $t\in\N$, let $(K_t,d_t)$ be a compact metric
space and let $h_t$ be a Borel probability measure on $K_t$. For a nonempty set $A\subseteq K_t$, write
$d_t(x,A)=\inf_{a\in A}d_t(x,a)$ and $B_{d_t}(A,\varepsilon)=\{x:d_t(x,A)<\varepsilon\}$.  The
sequence $(K_t,d_t,h_t)$ is a \emph{L\'evy family} if for every sequence of Borel sets
$A_t\subseteq K_t$ with $\liminf_t h_t(A_t)>0$ and every $\varepsilon>0$,
\[
 h_t\bigl(B_{d_t}(A_t,\varepsilon)\bigr)\longrightarrow1.
\]
A Polish group $G$ is a \emph{L\'evy group} if there is an increasing sequence \((K_{t})_{t}\) of compact subgroups with
dense union, \(\overline{\bigcup_{t} K_{t}} = G\), and a compatible right-invariant metric $d$ on $G$ such that $(K_t,d|_{K_t},h_t)$ is a
L\'evy family, where $h_t$ is the normalized Haar measure of $K_t$. For a finite group, normalized Haar measure is the uniform probability measure. The construction below uses finite compact
subgroups and a two-sided invariant metric.

A nontrivial L\'evy group is never locally compact.  Indeed, L\'evy groups are extremely
amenable: every continuous action on a nonempty compact Hausdorff space has a fixed point
\cite{GromovMilman1983}; see also \cite[Remark~1.4]{GTW}. On the other hand, Veech's theorem gives
a free continuous action on a nonempty compact Hausdorff space for every locally compact group
\cite{Veech1977}.

\paragraph{The fixed-point question.}
Glasner, Tsirelson and Weiss proved that every measure-preserving spatial action of a L\'evy group
is concentrated on its global fixed points \cite[Theorem~1.1]{GTW}. They asked whether such a group
can nevertheless have a nontrivial nonsingular Borel action \cite[Question~1.2]{GTW}. Pestov
included this question in his list of open problems about large topological
groups~\cite[Question~21]{Pestov}.

Extensions of~\cite[Theorem~1.1]{GTW} using spatial Poisson suspensions were obtained by
Avraham-Re'em and Roy \cite[Theorem~5]{AvrahamRoy2025} and, independently in the measure-preserving
case, Hoareau and Le Ma\^itre \cite[Theorem~E]{HoareauLeMaitre2026}. In particular, a continuous
action of a L\'evy group on a Polish space preserving an atomless $\sigma$-finite locally finite
Borel measure is supported on fixed points. Here locally finite means that every point has a
neighborhood of finite measure. Avraham-Re'em and Roy also treat nonsingular actions under an
additional regularity hypothesis: a positive pointwise Radon--Nikodym cocycle that is continuous in
the space variable; see \cite[Definition~6.2 and Theorem~5]{AvrahamRoy2025}.  These results require
topological regularity beyond the Borel and quasi-invariance assumptions of the original question.

We answer~\cite[Question~1.2]{GTW} affirmatively with the following theorem.

\getkeytheorem{boolean-main}

Avraham-Re'em and Peterzil introduced the class of \emph{locally Wasserstein groups} and proved a
nonsingular fixed-point theorem for this class~\cite[Theorem~1]{AP}. Together with their result,
Theorem~\ref{thm:gtw-main} gives an example of a
L\'evy group that is not locally Wasserstein.

\paragraph{Acknowledgement of AI use.}
The example presented in this paper was constructed by AI using OpenAI GPT-6-Astra and
OpenAI GPT-6.1-sol. These models also assisted with the writing, and
Claude Fable-5.1 and Claude Opus-5.5 were used for proofreading. I have substantially
revised and edited the draft to improve readability.

\section{The quotient group}\label{sec:group}
Throughout, $\N=\{0,1,2,\ldots\}$, and all sequences are indexed from $0$.
In this section we define the group $Q$ and show that it is a Boolean L\'evy group.  All additions
of subsets are symmetric differences, or equivalently coordinatewise additions of their indicators
in the field $\mathbb F_2=\{0,1\}$.

\subsection{Blocks and weights}
We first specify the features of the construction that the proof uses.  The numerical choices will
be made in the next subsection. There are two kinds of indices: $i$ denotes a coordinate, whereas
$n$ labels a block and also the coordinate that is the head of the relation introduced below.
In particular, the weight $w_n$ of the head is generally \emph{not} the common weight of the coordinates in $B_n$.

\paragraph{Block structure.}
Choose positive integers $m_n$, and set
\[
 M_{-1}=0,\qquad M_n=\sum_{k=0}^n m_k,\qquad
 B_n=\{M_{n-1}+1,\ldots,M_n\}.
\]
Thus $m_n=|B_n|$ is the length of $B_n$, and $M_n$ is its right endpoint. The blocks are
pairwise disjoint finite intervals that partition $\N\setminus\{0\}$. Because $m_k\geq1$,
$M_{n-1}\geq n$, so every element of $B_n$ is strictly larger than $n$:
\begin{equation}\label{eq:boolean-block-properties}
  \N=\{0\}\sqcup\bigsqcup_{n\geq0}B_n,\qquad
  |B_n|<\infty,\qquad \min B_n>n.
\end{equation}
We call $n$ the parent of each coordinate $i\in B_n$. Coordinate $0$ has no parent. Every other coordinate has exactly one parent, and repeatedly
passing to the parent leads to $0$ because indices decrease by~\eqref{eq:boolean-block-properties}.

Figure~\ref{fig:boolean-blocks} shows the blocks, their lengths, and their endpoints.
\begin{figure}[H]
\centering
\begin{tikzpicture}[x=.90cm,y=1cm,font=\small]
 \draw[gtwarr,black!60] (0,0)--(14.2,0) node[right] {$i$};
 \fill (.25,0) circle (2pt);
 \node[below=5pt] at (.25,0) {$0$};
 \draw[gtwblock,fill=blue!7] (1.1,.35) rectangle (4.1,1.2);
 \draw[gtwblock,fill=teal!8] (4.8,.35) rectangle (7.8,1.2);
 \draw[gtwblock,fill=orange!10] (10,.35) rectangle (13.5,1.2);
 \node at (2.6,.775) {$B_0$};
 \node at (6.3,.775) {$B_1$};
 \node at (8.9,.775) {$\cdots$};
 \node at (11.75,.775) {$B_n$};
 \draw[gtwbrace] (1.1,1.4)--(4.1,1.4) node[midway,above=5pt] {$m_0$};
 \draw[gtwbrace] (4.8,1.4)--(7.8,1.4) node[midway,above=5pt] {$m_1$};
 \draw[gtwbrace] (10,1.4)--(13.5,1.4) node[midway,above=5pt] {$m_n$};
 \foreach \x in {1.1,4.1,4.8,7.8,10,13.5}
   \draw[black!40] (\x,.35)--(\x,-.12);
 \node[below] at (1.1,-.12) {$1$};
 \node[below] at (4.1,-.12) {$M_0$};
 \node[below] at (4.8,-.65) {$M_0+1$};
 \node[below] at (7.8,-.12) {$M_1$};
 \node[below] at (10,-.65) {$M_{n-1}+1$};
 \node[below] at (13.5,-.12) {$M_n$};
\end{tikzpicture}
\caption{The blocks $B_n=\{M_{n-1}+1,\ldots,M_n\}$ on the coordinate axis,
with lengths $m_n=M_n-M_{n-1}$ and right endpoints $M_n$ (not to scale).}
\label{fig:boolean-blocks}
\end{figure}

\paragraph{Weights and block sums.}
Set $w_0=1$. On each block $B_n$ choose a constant weight $a_n$, so $w_i=a_n$ for $i\in B_n$. Define
the total weight $W_n$ and the squared-weight sum $V_n$:
\[
 W_n\coloneqq \sum_{\mathclap{i\in B_n}}w_i=m_na_n,\qquad
 V_n\coloneqq \sum_{\mathclap{i\in B_n}}w_i^2=m_na_n^2=W_na_n.
\]

We impose the following requirements on the weights \(a_{n}\).
\begin{enumerate}[label=\textup{(W\arabic*)},ref=\textup{(W\arabic*)},leftmargin=*]
\item\label{eq:boolean-small-weights} \textbf{Bounded positive weights:}
$0<a_n\leq1$ for all $n\in\N$.
\item\label{eq:boolean-drift-scale} \textbf{Total block weight and drift:}
$W_n\geq2$ for all $n\in\N$, and $W_n\sqrt{a_n}\to1$.
\item\label{eq:boolean-square-budget} \textbf{Squared-weight budget:}
$\sum_{n=0}^{\infty}V_n<\infty$.
\end{enumerate}

The requirements imply $V_n\to0$ and
\[
 V_nW_n=(W_n\sqrt{a_n})^2\longrightarrow1.
\]
Hence $W_n\to\infty$, and $a_n=V_n/W_n\to0$. Since the blocks
are finite and partition $\N\setminus\{0\}$, it follows that
\begin{equation}\label{eq:boolean-weight-limits}
 w_i\to0,\qquad W_n\to\infty,\qquad
 \sum_iw_i=\infty,\qquad \sum_iw_i^2=1+\sum_nV_n<\infty.
\end{equation}

\subsection{Explicit parameters}
The specific values of $a_n$ and $m_n$ do not matter as long as they satisfy the requirements
above. One possible choice is
\[
 m_n=8^{n+1},\qquad a_n=4^{-(n+1)}.
\]
The endpoints and blocks are then
\[
 M_n=\sum_{k=0}^n8^{k+1}=\frac{8(8^{n+1}-1)}7,
 \qquad B_n=\{M_{n-1}+1,\ldots,M_n\}.
\]
As above, $w_0=1$ and $w_i=a_n$ for $i\in B_n$.
The first three blocks illustrate the distinction between a head's
weight and its children's weight:
\begingroup
\setlength{\topsep}{12pt plus 3pt minus 2pt}
\renewcommand{\arraystretch}{1.18}
\begin{center}
\begin{tabular}{@{}ccccc@{}}
\toprule
Head $n$ & Block $B_n$ & $m_n$ & Head weight $w_n$ & Child weight $a_n$\\
\midrule
$0$ & $\{1,\ldots,8\}$ & $8$ & $1$ & $1/4$\\
$1$ & $\{9,\ldots,72\}$ & $64$ & $1/4$ & $1/16$\\
$2$ & $\{73,\ldots,584\}$ & $512$ & $1/4$ & $1/64$\\
\bottomrule
\end{tabular}
\end{center}
\endgroup
In particular, $1,2\in B_0$, but their own child blocks $B_1,B_2$
lie farther along the coordinate axis and are disjoint from $B_0$ and from each other. We have
\[
 W_n=2^{n+1}\geq2,\qquad V_n=2^{-(n+1)},\qquad W_n\sqrt{a_n}=1.
\]
Moreover, $\sum_nV_n=1$, so all three requirements hold. In particular,
$\sum_iw_i^2=2$. We use this explicit choice for the rest of the
construction, indicating where the general properties are used.

\subsection{The weighted group}
Define
\[
 G_w=\Bigl\{A\subseteq\N:
                  L(A)\coloneqq \sum_{i\in A}w_i<\infty\Bigr\},
 \qquad d_w(A,A')=L(A+A').
\]
Finite subsets are dense in this metric. For a Cauchy sequence $(A_k)$, each coordinate eventually stabilizes,
defining a set $A$. Fatou's lemma gives $L(A_k+A)\leq\liminf_m L(A_k+A_m)$, so $L(A_k+A)\to0$ and
$A\in G_w$. Thus $G_w$ is a Polish group.

Put $e_i=\{i\}$ and define
\[
 r_n=e_n+\sum_{\mathclap{j\in B_n}}e_j,
 \qquad N=\operatorname{span}_{\mathbb F_2}\{r_n:n\geq0\},
 \qquad Q=G_w/N.
\]
The vector $r_n$ has value $1$ at the head $n$ and at every child
in $B_n$, and zero elsewhere. In the quotient it becomes the relation
$e_n+N=\sum_{j\in B_n}(e_j+N)$.
Figure~\ref{fig:boolean-tree} illustrates how a child coordinate is itself the head of a later block.

\begin{figure}[H]
\centering
\begin{tikzpicture}[x=1cm,y=1cm,font=\small]
 \draw[gtwblock,fill=blue!7] (1,.1) rectangle (6.1,1.1);
 \draw[gtwblock,fill=teal!8] (8,.1) rectangle (13.1,1.1);
 \node[circle,draw,fill=white,inner sep=4pt] (n) at (3.55,2.65) {$n$};
 \node[above=4pt] at (n.north) {parent of $B_n$};
 \node[circle,draw,fill=white,inner sep=4pt] (j) at (3.55,.6) {$j$};
 \fill (1.55,.6) circle (2pt); \fill (5.55,.6) circle (2pt);
 \node at (2.35,.6) {$\cdots$}; \node at (4.75,.6) {$\cdots$};
 \fill (8.55,.6) circle (2pt); \fill (12.55,.6) circle (2pt);
 \node at (10.55,.6) {$\cdots$};
 \draw[gtwarr] (n.south)--(j.north);
 \draw[gtwarr,teal!70!black] (j.north east)
    .. controls (6.6,2.6) and (10.55,2.65) .. (10.55,1.1);
 \node[fill=white,inner sep=3pt,text=teal!70!black] at (9.5,2.65)
    {$j$ is the parent of $B_j$};
 \node at (3.55,-.3) {$B_n$: children of $n$};
 \node at (10.55,-.3) {$B_j$: children of $j$};
 \node at (7,-.9) {$n<j<\min B_j$,\qquad $B_n\cap B_j=\varnothing$};
\end{tikzpicture}
\caption{Parenthood refers to coordinates, not to inclusion of blocks.
The coordinate $j\in B_n$ is a child of $n$ and the parent of its own
later block $B_j$. Vertical placement is schematic; coordinate order
is expressed by the inequalities below the diagram.}
\label{fig:boolean-tree}
\end{figure}

We call the generators $r_n$ rows.

\begin{proposition}\label{prop:gtw-kernel}
The subgroup $N$ is countable and closed in $G_w$.
\end{proposition}
\begin{proof}
Every element of $N$ is a sum of finitely many rows, so $N$ is
countable. For a nonempty finite set $F\subseteq\N$, put
\[
 z_F=\sum_{k\in F}r_k,\qquad n=\max F.
\]
The row $r_n$ contains every coordinate of $B_n$. No other selected
row can cancel any of these coordinates: its child block is disjoint
from $B_n$, and its head is at most $n$, whereas every coordinate of
$B_n$ is larger than $n$. Thus $B_n\subseteq z_F$, and
\[
 L(z_F)\geq\sum_{\mathclap{j\in B_n}}w_j=W_n\geq2.
\]
In particular, no nonempty finite sum of rows vanishes, so the rows
are linearly independent. Distinct elements of $N$ have distance at
least $2$ in $d_w$. The group \(N\) is therefore discrete and closed.
\end{proof}

The quotient $Q=G_w/N$ is a Polish group in the quotient topology. Write
$q:G_w\to Q$ for the quotient map. We use the compatible invariant
quotient metric
\[
 \bar d\bigl(q(A),q(A')\bigr)=\inf_{z\in N}L(A+A'+z).
\]

\begin{proposition}\label{prop:gtw-group}
The quotient $Q=G_w/N$ is a Boolean L\'evy group.
\end{proposition}
\begin{proof}
Only the L\'evy property remains to be proved.
For $t\geq0$ set
\[
 J_t=\Bigl(\bigcup_{n\leq t}B_n\Bigr)\setminus\{0,\ldots,t\},
 \qquad K_t=\operatorname{span}_{\mathbb F_2}
                         \{q(e_j):j\in J_t\}.
\]
Here $J_t=\{t+1,\ldots,M_t\}$, and
$J_{t+1}=(J_t\setminus\{t+1\})\cup B_{t+1}$.
Figure~\ref{fig:boolean-boundary} shows this replacement step.
\begin{figure}[H]
\centering
\begin{tikzpicture}[x=1cm,y=1cm,font=\small]
 \filldraw[fill=black!8] (0,0) rectangle (2.4,.7);
 \filldraw[fill=orange!18] (2.4,0) rectangle (3.7,.7);
 \filldraw[fill=blue!8] (3.7,0) rectangle (8,.7);
 \draw[dashed,fill=teal!6] (8.2,0) rectangle (12.4,.7);
 \node at (1.2,.35) {$0,\ldots,t$};
 \node at (3.05,.35) {$t+1$};
 \node at (5.85,.35) {$t+2,\ldots,M_t$};
 \node at (10.3,.35) {$B_{t+1}$: parent $t+1$};
 \draw[gtwbrace] (2.4,1)--(8,1) node[midway,above=5pt] {$J_t$};
 \filldraw[fill=black!8] (0,-2) rectangle (3.7,-1.3);
 \filldraw[fill=blue!8] (3.7,-2) rectangle (8,-1.3);
 \filldraw[fill=teal!14] (8.2,-2) rectangle (12.4,-1.3);
 \node at (1.85,-1.65) {$0,\ldots,t+1$};
 \node at (5.85,-1.65) {retained coordinates};
 \node at (10.3,-1.65) {$M_t+1,\ldots,M_{t+1}$};
 \draw[gtwarr] (3.05,-.1)--(9.1,-1.2)
    node[pos=.5,above,fill=white,inner sep=2pt] {replace head by $B_{t+1}$};
 \draw[gtwbrace] (12.4,-2.3)--(3.7,-2.3)
    node[midway,below=5pt] {$J_{t+1}=(J_t\setminus\{t+1\})\cup B_{t+1}$};
\end{tikzpicture}
\caption{One step in the boundary construction. Gray coordinates
have been processed. The head $t+1$ leaves the boundary, its children
enter, and the blue coordinates remain. Although the sets $J_t$ are
not nested, their generated quotient subgroups $K_t$ are nested.}
\label{fig:boolean-boundary}
\end{figure}

The relation $q(e_{t+1})=\sum_{j\in B_{t+1}}q(e_j)$ therefore
shows that $K_t\subseteq K_{t+1}$.
The set $J_t$ is the remaining boundary after replacing heads $0,\ldots,t$.
Every $q(e_i)$ with $i\leq t$ can be expressed using $J_t$: recursively
replace indices at most $t$ by their children.
Thus $\bigcup_tK_t$ is dense in $Q$.

\paragraph{The finite probability space and the quotient map.}
Fix $t$ and give
\[
 \Omega_t=\{0,1\}^{J_t}
\]
the uniform probability measure $\mathbb P_t$, under which the
coordinate bits are independent and each takes the values $0$ and $1$
with probability $1/2$. Define the surjective group homomorphism
\[
 \phi_t:\Omega_t\longrightarrow K_t,\qquad
 \phi_t(v)=\sum_{\mathclap{j\in J_t}}v_jq(e_j).
\]
Each fiber of $\phi_t$ is a coset of its kernel, so all fibers have
the same cardinality. Therefore $(\phi_t)_*\mathbb P_t=h_t$, where
$h_t$ is the normalized Haar measure, or equivalently the uniform
probability measure, on the finite group $K_t$.
For the remainder of this calculation, $\mathbb P$ and $\mathbb E$
refer to $\mathbb P_t$ and its expectation.

Let $A\subseteq K_t$ satisfy $h_t(A)\geq b>0$, and define the
nonnegative random variable
\[
 f:\Omega_t\longrightarrow\R,\qquad
 f(v)=\dist_{\bar d}(\phi_t(v),A).
\]
Let $j\in J_t$ and let $v,v'\in\Omega_t$ differ only in the $j$th
bit. Then $\phi_t(v)$ and $\phi_t(v')$ differ by $q(e_j)$.
The triangle inequality for $\bar d$ implies that distance to $A$
is a $1$-Lipschitz function, so
\begin{equation}\label{eq:boolean-coordinate-bound}
 |f(v)-f(v')|
 \leq\bar d\bigl(\phi_t(v),\phi_t(v')\bigr)
 \leq L(e_j)=w_j.
\end{equation}
This bound holds for every choice of the other bits.

\paragraph{An elementary variance bound.}
We claim that
\begin{equation}\label{eq:boolean-variance-bound}
 \Var(f)\leq\frac14\sum_{\mathclap{j\in J_t}}w_j^2
          =\frac14\sigma_t^2,
 \qquad \sigma_t^2\coloneqq\sum_{\mathclap{j\in J_t}}w_j^2.
\end{equation}
This is the bounded-differences form of the Efron--Stein inequality.
We prove this bound by induction on the number of bits, assuming only the coordinate bounds.
With no bits, the function is constant and its variance is zero. For the induction step, fix
$j\in J_t$ and split the cube into the two equally likely halves
$v_j=0$ and $v_j=1$. Expanding the definition of variance gives
\[
 \Var(f)
 =\frac12\Var(f\mid v_j=0)+\frac12\Var(f\mid v_j=1)
  +\frac14\bigl(\mathbb E[f\mid v_j=0]
                    -\mathbb E[f\mid v_j=1]\bigr)^2.
\]
Here the conditional means and variances are simply those on the
indicated halves with their uniform measures. On each half the
coordinate bounds~\eqref{eq:boolean-coordinate-bound} still hold
for all the remaining bits, so induction bounds each conditional
variance by $\frac14\sum_{i\in J_t\setminus\{j\}}w_i^2$.
Pairing points in the two halves that agree in every other bit, we use
\eqref{eq:boolean-coordinate-bound} and the triangle inequality
to bound the difference between the two conditional means by $w_j$.
Substitution proves~\eqref{eq:boolean-variance-bound}.

\paragraph{From variance to concentration.}
Because $(\phi_t)_*\mathbb P=h_t$, we have
$\mathbb P(f=0)=h_t(A)\geq b$. Since $f$ vanishes outside $\{f>0\}$,
the Cauchy--Schwarz inequality gives
\[
 (\mathbb Ef)^2
 =\bigl(\mathbb E[f\mathbf1_{\{f>0\}}]\bigr)^2
 \leq\mathbb P(f>0)\,\mathbb E[f^2]
 \leq(1-b)\mathbb E[f^2].
\]
Consequently, $\Var(f)=\mathbb E[f^2]-(\mathbb Ef)^2\geq b\mathbb E[f^2]$.
For $r>0$, Markov's inequality applied to $f^2$, together with~\eqref{eq:boolean-variance-bound}, gives
\begin{equation}\label{eq:boolean-concentration}
 \begin{aligned}
 h_t\bigl(K_t\setminus B_{\bar d}(A,r)\bigr)
 &=\mathbb P(f\geq r)\\
 &\leq\frac{\mathbb E[f^2]}{r^2}
 \leq\frac{\Var(f)}{br^2}
 \leq\frac{\sigma_t^2}{4br^2}.
 \end{aligned}
\end{equation}
Finally, the inclusion $J_t\subseteq\{j:j>t\}$ and the squared-weight budget
\ref{eq:boolean-square-budget} imply
\[
 \sigma_t^2\leq\sum_{j>t}w_j^2\longrightarrow0.
\]
For any sequence $A_t\subseteq K_t$ with
$\liminf_t h_t(A_t)>0$, choose $b>0$ such that $h_t(A_t)\geq b$
for all sufficiently large $t$. The bound in~\eqref{eq:boolean-concentration} then tends to
zero for each fixed $r>0$, which is the L\'evy-family condition.
Since the finite subgroups $K_t$ are nested and have dense union,
$Q$ is a L\'evy group.
\end{proof}

\section{The product measure}\label{sec:product}
Let $Y=\{0,1\}^{\N}$, with its product topology and Borel sets.
The group $G_w$ acts on $Y$ by $(A,y)\mapsto y+A$.
Given a finite binary block, its bias is the imbalance between the
number of zeros and the number of ones. We use the weighted, signed score
\begin{equation}\label{eq:boolean-score}
 S_n(y)=\sum_{\mathclap{j\in B_n}}w_j(2y_j-1).
\end{equation}
Thus $S_n$ counts weights corresponding to $1$ positively and those
corresponding to $0$ negatively. Since all weights in $B_n$ equal $a_n$,
it equals $a_n$ times the difference between the number of ones and the number of zeros.

We choose a probability measure on $Y$ under which these scores converge almost surely
to $-\sqrt2$. The negative limit records a persistent
weighted excess of zeros. For the later likelihood argument, what matters is that the limit is
strictly negative. Define the Borel set
\begin{equation}\label{eq:boolean-conull-domain}
 D=\{y\in Y:S_n(y)\longrightarrow-\sqrt2\}.
\end{equation}
\begin{lemma}\label{lem:gtw-product}
There is a nonatomic product probability measure $\mu$ on $Y$
under which every translation by $G_w$ is nonsingular. Moreover, the
set $D$ satisfies $\mu(D)=1$ and $D+A=D$ for every $A\in G_w$.
\end{lemma}
\begin{proof}
For $u>0$ define
\[
 b(u)=\sqrt{1-e^{-2u}},\qquad p(u)=\frac{1-b(u)}2.
\]
The choice of $p(u)$ makes the affinity of a Bernoulli law and its flipped
law especially simple:
\[
 2\sqrt{p(u)(1-p(u))}=\sqrt{1-b(u)^2}=e^{-u}.
\]
The product \(\prod_{i \in A}e^{-w_{i}}\) over a summable set $A$ is $e^{-L(A)}>0$.
This positivity is what matters for nonsingularity; other marginal choices
giving positive affinity products over all summable sets are possible.
For the rest of the construction, the chosen marginals also provide the
drift and variance estimates below.

Put $p_i=p(w_i)$ and let
\[
 \mu_i=(1-p_i)\delta_0+p_i\delta_1,\qquad
 \mu=\bigotimes_i\mu_i.
\]
Thus $\mu\{y_i=1\}=p_i$, and $p_i\to1/2$ since $w_i\to0$.  The action $(A,y)\mapsto y+A$ of $G_w$ on
$Y$ is jointly continuous and hence Borel.

Fix $A\in G_w$ and let $g_{A,i}(t)=t+\mathbf1_A(i)$ modulo two.
The marginals of $A_*\mu$ are
\[
 \nu_i=(g_{A,i})_*\mu_i
 =\begin{cases}
   \mu_i,&i\notin A,\\
   p_i\delta_0+(1-p_i)\delta_1,&i\in A.
  \end{cases}
\]
All these laws assign positive mass to both points. Their Radon--Nikodym
derivatives are
\[
 \frac{d\nu_i}{d\mu_i}(t)
 =\begin{cases}
  1,&i\notin A,\\
  \bigl(p_i/(1-p_i)\bigr)^{1-2t},&i\in A,
 \end{cases}
 \qquad t\in\{0,1\}.
\]
Kakutani's criterion \cite{Kakutani1948} states that, for equivalent
probability measures $\mu_i\sim\nu_i$ on each coordinate space,
\[
 \bigotimes_i\mu_i\sim\bigotimes_i\nu_i
 \quad\Longleftrightarrow\quad
 \prod_i\int\sqrt{\frac{d\nu_i}{d\mu_i}}\,d\mu_i>0.
\]
The integrals are the Hellinger affinities of the marginal laws.
Here they equal $1$ for $i\notin A$ and
$2\sqrt{p_i(1-p_i)}=e^{-w_i}$ for $i\in A$. Kakutani's criterion gives
$A_*\mu\sim\mu$, since
\[
 \prod_i\int\sqrt{\frac{d\nu_i}{d\mu_i}}\,d\mu_i
 =\prod_{i\in A}e^{-w_i}=e^{-L(A)}>0.
\]

\paragraph{A conull $G_w$-invariant domain.}
Under $\mu$, the variables $y_j$ are independent Bernoulli variables.
For $j\in B_n$ they all have parameter $p(a_n)$, so
\[
 \mathbb E(2y_j-1)=2p(a_n)-1=-b(a_n),\qquad
 \Var(2y_j-1)=4p(a_n)(1-p(a_n))=e^{-2a_n}.
\]
Summing the weighted expectations gives
$\mathbb E S_n=-m_na_nb(a_n)=-W_nb(a_n)$.
Independence gives
$\Var(S_n)=m_na_n^2e^{-2a_n}=V_ne^{-2a_n}$. Thus the scores
in~\eqref{eq:boolean-score} satisfy
\begin{equation}\label{eq:boolean-score-moments}
 \mathbb E S_n=-W_n\sqrt{1-e^{-2a_n}}\longrightarrow-\sqrt2,
 \qquad \operatorname{Var}(S_n)=V_ne^{-2a_n}\leq V_n=2^{-(n+1)}.
\end{equation}
The mean limit follows from $W_n\sqrt{a_n}\to1$ and
$\sqrt{1-e^{-2a_n}}/\sqrt{a_n}\to\sqrt2$. The variance bound is
summable by~\ref{eq:boolean-square-budget}.
By Tonelli's theorem,
\[
 \mathbb E\Bigl[\sum_n(S_n-\mathbb ES_n)^2\Bigr]
 =\sum_n\operatorname{Var}(S_n)<\infty.
\]
The sum is therefore finite almost surely, so $S_n-\mathbb ES_n\to0$ almost surely. Together with
the mean limit in~\eqref{eq:boolean-score-moments}, this shows that $D$ has measure one.

For every $A\in G_w$ and every $y\in Y$,
\begin{equation}\label{eq:boolean-score-stability}
 |S_n(y+A)-S_n(y)|\leq2L(A\cap B_n)\longrightarrow0.
\end{equation}
The last limit follows from the disjointness of the blocks and $L(A)<\infty$.
Thus $y\in D$ if and only if $y+A\in D$, proving $D+A=D$ for every
$A\in G_w$.

The probabilities $p(w_i)$ tend to $1/2$, so the product measure is nonatomic.
\end{proof}

\begin{proposition}\label{prop:gtw-product-no-sigmafinite}
No $G_w$-invariant $\sigma$-finite Borel measure on $Y$ is equivalent
to the measure $\mu$ of Lemma~\ref{lem:gtw-product}.
\end{proposition}
\begin{proof}
It suffices to exclude invariance under finite-support flips, since
every such flip belongs to $G_w$. For a Bernoulli product with parameters $0<p_i\leq1/2$,
Moore's criterion
\cite[Theorem~2]{Moore1967} says that there is no equivalent
$\sigma$-finite Borel measure invariant under all finite-coordinate
flips if and only if
\begin{equation}\label{eq:boolean-moore-criterion}
 \sum_i p_i\min\Bigl\{\frac{1-p_i}{p_i}-1,\,1\Bigr\}^2=\infty.
\end{equation}
This is the two-point case of Moore's criterion with the larger marginal
mass being $1-p_i\geq1/2$.

Here $p_i=p(w_i)\to1/2$. Thus $(1-p_i)/p_i-1\to0$, so the
truncation by $1$ is inactive for all sufficiently large $i$.
Using $1-2p(w_i)=b(w_i)$, the summand then becomes
\[
 p_i\Bigl(\frac{1-p_i}{p_i}-1\Bigr)^2
 =\frac{(1-2p_i)^2}{p_i}
 =\frac{1-e^{-2w_i}}{p(w_i)}
 \sim4w_i.
\]
By~\eqref{eq:boolean-weight-limits}, $\sum_iw_i=\infty$, so the series in~\eqref{eq:boolean-moore-criterion} diverges. Moore's criterion therefore excludes an equivalent $\sigma$-finite Borel
measure invariant under $G_w$.
\end{proof}

The Radon--Nikodym derivatives of the translates of $\mu$ have an explicit form. For $u>0$ put
\[
 c(u)=\log\frac{1-p(u)}{p(u)}=2\operatorname{artanh}b(u).
\]
Figure~\ref{fig:boolean-log-odds} shows $c$ on the range $0\leq u\leq1$
containing all the coordinate weights, with the continuous extension $c(0)=0$.
The curve is increasing and has square-root behavior near zero:
$c(u)\sim2\sqrt{2u}$.
\begin{figure}[htbp]
\centering
\begin{tikzpicture}[x=7cm,y=1.15cm,font=\small]
 \draw[gtwarr] (0,0)--(1.16,0) node[right] {$u$};
 \draw[gtwarr] (0,0)--(0,3.85) node[above] {$c(u)$};
 \foreach \u/\lab in {0.25/{1/4},0.5/{1/2},0.75/{3/4},1/{1}} {
   \draw (\u,.045)--(\u,-.045) node[below=3pt] {$\lab$};
 }
 \node[below left=3pt] at (0,0) {$0$};
 \foreach \v in {1,2,3} {
   \draw (.008,\v)--(-.008,\v) node[left=3pt] {$\v$};
 }
 % Parametrize by u=t^2 to resolve the square-root behavior at the origin.
 % The equivalent formula c(u)=2 log(e^u+sqrt(e^{2u}-1)) avoids artanh.
 \draw[blue!70!black,very thick]
   plot[domain=0:1,samples=121,variable=\t]
     ({\t*\t},{2*ln(exp(\t*\t)+sqrt(exp(2*\t*\t)-1))});
 \draw[black!55,dashed,thick]
   plot[domain=0:1,samples=61,variable=\t]
     ({\t*\t},{2*sqrt(2)*\t});
 \node[right,text=blue!70!black] at (1,3.3149) {$c(u)$};
 \node[right,text=black!65] at (1,2.8284) {$2\sqrt{2u}$};
\end{tikzpicture}
\caption{The increasing log-odds function $c(u)$ (solid) and its
approximation $2\sqrt{2u}$ near zero (dashed). In particular,
$c(w_i)\leq c(1)$ for all coordinate weights.}
\label{fig:boolean-log-odds}
\end{figure}

\section{The quotient action}\label{sec:selector}
We now realize the orbit quotient by $N$ as a standard Borel space.
An \emph{orbit selector} on $D$ is a map $R:D\to D$ with
$R(y)\in y+N$ and $R(y+n)=R(y)$ for all $n\in N$.
Its fixed-point set contains exactly one representative of each orbit;
such a set is called a \emph{transversal}.
Every $z\in N$ has finite support; write $z_i\in\{0,1\}$ for its
indicator at $i$. Flipping coordinate $i$ multiplies its marginal
density by
\[
 \Bigl(\frac{p_i}{1-p_i}\Bigr)^{1-2y_i}
 =e^{-c(w_i)(1-2y_i)}.
\]
Thus the finite product of these factors defines an everywhere positive
version of the likelihood ratio $\rho_z=d(z_*\mu)/d\mu$:
\[
 \rho_z(y)=\exp(-E_z(y)),\qquad
 E_z(y)=\sum_i z_i c(w_i)(1-2y_i).
\]
These formulas satisfy, for every $u,z\in N$ and every $y\in Y$,
\begin{equation}\label{eq:boolean-likelihood-cocycle}
 \rho_z(y+u)=\frac{\rho_{z+u}(y)}{\rho_u(y)}.
\end{equation}

\begin{lemma}\label{lem:gtw-finite-likelihood}
For every $y\in D$, the set $\{z\in N:\rho_z(y)\geq1\}$ is finite.
\end{lemma}
\begin{proof}
The function $c$ is increasing, since both $b$ and
$\operatorname{artanh}$ are increasing. Put $C=c(1)$; then
$c(w_i)\leq C$ because $w_i\leq1$. Define
\[
 T_n(y)=\sum_{\mathclap{j\in B_n}}c(w_j)(1-2y_j)
       =-\frac{c(a_n)}{a_n}S_n(y).
\]
If $F\subseteq\N$ is a finite set of row indices, put
$z_F=\sum_{n\in F}r_n$. The disjoint child blocks contribute
$\sum_{n\in F}T_n(y)$. Adding each selected head changes the energy
by at most $C$. Consequently,
\begin{equation}\label{eq:boolean-energy-bound}
 E_{z_F}(y)\geq\sum_{n\in F}\bigl(T_n(y)-C\bigr).
\end{equation}
For $y\in D$, we have $-S_n(y)\to\sqrt2$. Since
$c(u)\sim2\sqrt{2u}$ as $u\to0$ and $a_n\to0$,
\[
 T_n(y)=-\frac{c(a_n)}{a_n}S_n(y)
       \sim\frac4{\sqrt{a_n}}\longrightarrow\infty.
\]
Thus the numbers $T_n(y)-C$ tend to $+\infty$.
Consequently, there are only finitely many $z\in N$ with
$E_z(y)\leq0$, or equivalently $\rho_z(y)\geq1$.
\end{proof}

\begin{proposition}\label{prop:gtw-action}
On $D$, the action of $N$ has a Borel orbit selector $R$. The induced
action of $Q=G_w/N$ on the Borel set $Z=\{y\in D:R(y)=y\}$ is free,
jointly Borel, and nonsingular for $\nu=R_*(\mu|D)$.
\end{proposition}
\begin{proof}
  By Lemma~\ref{lem:gtw-finite-likelihood}, for each $y\in D$ there are only finitely many $z\in N$
  with $\rho_z(y)\geq1$. This set is nonempty because $\rho_0(y)=1$. The global maximum of the
  $\rho_z(y)$ is therefore attained in this finite set, and there are only finitely many maximizers.
  We can pick one such maximizer in a Borel way.  For instance, order binary sequences
  lexicographically, comparing them at their first differing coordinate with $0<1$. The Borel
  map \(R\) is given by taking \(R(y)\) to be the least
  point in the finite set
\[
 \{y+z:z\in N,\ \rho_z(y)=\max_{v\in N}\rho_v(y)\}.
\]
The cocycle identity~\eqref{eq:boolean-likelihood-cocycle} ensures
that the set of maximizing points is unchanged along each $N$-orbit. Hence
\[
 R(y+u)=R(y)\quad(u\in N),\qquad R(y)\in y+N.
\]
In particular, $R$ is a Borel orbit selector on $D$, and
\[
 Z=\{y\in D:R(y)=y\}
\]
is a Borel subset of the standard Borel space $D$, hence a standard
Borel transversal; see \cite[Section~12]{Kechris1995}.

\paragraph{Descent to a free spatial nonsingular $Q$-action.}
For $g\in G_w$ and $z\in Z$ put $U_g(z)=R(z+g)$.
This is jointly Borel. Since $R(z+h)$ differs from $z+h$ by an element
of $N$, its invariance gives
\[
 U_g(U_h(z))=R(R(z+h)+g)=R(z+h+g)=U_{g+h}(z).
\]
Moreover, $U_{g+n}=U_g$ for $n\in N$, since $R(z+g+n)=R(z+g)$. Thus $N$ acts trivially, and the action
descends to an action $(q(g),z)\mapsto U_g(z)$ of $Q$, with all action identities holding
pointwise. To verify joint Borelness on $Q\times Z$, write $\pi=q\times\mathrm{id}:G_w\times Z\to
Q\times Z$ for the continuous surjection between standard Borel spaces, and let $U:Q\times Z\to Z$ be
the action map. For a Borel set $E\subseteq Z$ put $S=U^{-1}(E)$. Then
$\pi^{-1}(S)=\{(g,z):R(z+g)\in E\}$ is Borel, so $S=\pi(\pi^{-1}(S))$ and
$(Q\times Z)\setminus S=\pi(\pi^{-1}((Q\times Z)\setminus S))$ are both analytic. By Souslin's
theorem $S$ is Borel \cite[Theorem~14.11]{Kechris1995}.

This action of $Q$ is free. If $R(z+g)=z$, then $z+g=z+n$ for some
$n\in N$, and cancellation of binary sequences gives $g=n$.
Finally, let $\nu=R_*(\mu|D)$. For a Borel set $A\subseteq Z$,
\[
 R^{-1}(U_g^{-1}A)=R^{-1}(A)+g
\]
inside the invariant set $D$. The quasi-invariance of $\mu$ under $G_w$
therefore makes $\nu$ quasi-invariant under every $U_g$, and hence under
every element of $Q$. This completes the construction.
\end{proof}

\section{Ergodicity and invariant measures}\label{sec:obstruction}
It remains to check that the action of \(Q\) on \((Z,\nu)\) is ergodic and admits no equivalent
\(\sigma\)-finite invariant measures.

\begin{proposition}\label{prop:gtw-ergodic}
The action on $(Z,\nu)$ is ergodic and $\nu$ is nonatomic.
\end{proposition}
\begin{proof}
Let $\Gamma\leq G_w$ be the countable group of finite-support flips.
If a Borel set $A\subseteq Z$ is invariant modulo $\nu$-null sets, put
$B=R^{-1}(A)\subseteq D$. Pointwise equivariance of $R$ gives
$\mu((B+f)\mathbin\triangle B)=0$ for every $f\in\Gamma$.
Consequently
\[
 B^{\mathrm{inv}}=\bigcap_{f\in\Gamma}(B+f)
\]
is pointwise $\Gamma$-invariant and agrees with $B$ modulo a null set.
Membership in $B^{\mathrm{inv}}$ is
unchanged by arbitrary alterations of any finite initial segment, so
$B^{\mathrm{inv}}$ belongs, for every $k$, to $\sigma(y_i:i>k)$, the sigma-algebra
generated by the coordinates after $k$. Independence of the Bernoulli
coordinates makes $B^{\mathrm{inv}}$ independent of every finite-coordinate cylinder,
hence of every Borel set by a monotone class argument. In particular,
$\mu(B^{\mathrm{inv}})=\mu(B^{\mathrm{inv}})^2$, proving $\nu(A)=\mu(B)\in\{0,1\}$.

To check that \(\nu\) is nonatomic, note that Lemma~\ref{lem:gtw-product} implies that $\mu$ assigns zero
mass to each point.  Every fiber of $R$ is one countable $N$-orbit; therefore
$\nu(\{z\})=\mu(R^{-1}(\{z\}))=0$ for all $z\in Z$.
\end{proof}

\begin{proposition}\label{prop:gtw-no-sigmafinite}
No $Q$-invariant $\sigma$-finite Borel measure on $Z$ is equivalent to $\nu$.
\end{proposition}
\begin{proof}
Suppose $\lambda\sim\nu$ were such an invariant measure. We first
lift it to $D$ by counting points in each $N$-orbit. For Borel
$E\subseteq D$, define
\[
 F_E(z)=\sum_{n\in N}\mathbf1_E(z+n),\qquad
 \widetilde\mu(E)=\int_Z F_E(z)\,d\lambda(z).
\]
The functions $F_E$ are Borel, and countable additivity follows from
Tonelli's theorem. Choose Borel sets $Z_k$ covering $Z$ with
$\lambda(Z_k)<\infty$. Since $Z$ meets each orbit once,
$\widetilde\mu(Z_k+n)=\lambda(Z_k)$, and the sets $Z_k+n$, for
$k\in\N$ and $n\in N$, cover $D$. Hence $\widetilde\mu$ is
$\sigma$-finite.

For Borel $E\subseteq D$, the set
$R(E)=\bigcup_{n\in N}((E+n)\cap Z)$ is Borel, and
$R^{-1}(R(E))=\bigcup_{n\in N}(E+n)$. Since $F_E$ is positive
exactly on $R(E)$, we have
\[
 \widetilde\mu(E)=0
 \ \Longleftrightarrow\ \lambda(R(E))=0
 \ \Longleftrightarrow\ \nu(R(E))=0
 \ \Longleftrightarrow\ \mu(E)=0.
\]
The last equivalence uses the quasi-invariance of $\mu$ under the
countable group $N$. Thus $\widetilde\mu\sim\mu|D$.
For $g\in G_w$, the points $U_gz$ and $z+g$ differ by an element of
$N$, so reindexing the sum over $N$ gives $F_{E+g}(z)=F_E(U_gz)$.
Invariance of $\lambda$ therefore yields
\[
 \widetilde\mu(E+g)=\int_Z F_E(U_gz)\,d\lambda(z)
                  =\widetilde\mu(E).
\]
Extending $\widetilde\mu$ by zero off the invariant conull set $D$ gives a
$G_w$-invariant $\sigma$-finite Borel measure on $Y$ equivalent to
$\mu$, contradicting Proposition~\ref{prop:gtw-product-no-sigmafinite}.
\end{proof}

The preceding propositions now give the theorem announced in the introduction.

\begin{theorem}[store=boolean-main,label=thm:gtw-main]
There is a Boolean L\'evy group $Q$ with a
free ergodic spatial nonsingular action on a standard Borel
probability space $(Z,\nu)$. There is no $Q$-invariant $\sigma$-finite
Borel measure equivalent to $\nu$.
\end{theorem}
\begin{proof}
Proposition~\ref{prop:gtw-group} shows that $Q$ is a Boolean L\'evy
group. Proposition~\ref{prop:gtw-action} supplies its free spatial
nonsingular action on the standard Borel probability space $(Z,\nu)$.
This action is ergodic by Proposition~\ref{prop:gtw-ergodic}, and
Proposition~\ref{prop:gtw-no-sigmafinite} excludes an equivalent
$Q$-invariant $\sigma$-finite Borel measure.
\end{proof}

\begin{remark}
A related construction also yields a Boolean L\'evy group with a free ergodic spatial
action preserving a nonatomic infinite $\sigma$-finite Borel measure on a standard
Borel space. The construction uses the same pattern of block relations, with different
block metrics and measures. Details will be presented in a subsequent version of this paper.
\end{remark}

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\end{thebibliography}
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