\label{thm} If $C$ is stably embedded in $\mathbf{M}$ and $\mu$ an $A$-valued Fubini measure on $\operatorname{Def}(C)$,
then $\mu$ extends uniquely to an $A$-valued Fubini measure on $\operatorname{Def}(C)^{\operatorname{f}}$.

\begin{proof} For every $p\in M^k$ the set $\cal{S}(p)$ has by Lemma~\ref{l5} an $(r+1)$-step fibering over $C$. As in the proof of Lemma~\ref{l5}, this yields a finite set $E$ of tuples
$$\big( f_1,\dots, f_{r+1}, f_{r+2}; m_1,\dots,m_{r+1}, n_1,\dots, n_{r+1},n_{r+2}\big)$$
where for $j=1,\dots, r+2$: $f_j$ is a definable map $ \Sigma_j\to M^{n_i}$ with (definable) domain
$\Sigma_j\subseteq M^l\times M^{m_1}\times\cdots\times M^{m_{j-1}}\times M^{m+n_1+\cdots + n_{j-1}}$, and for each $p\in M^k$ there is
a tuple in $E$ as displayed and $q\in M^{l}$
such that $\cal{S}(p)=\Sigma_1(q)$ and the tuple
$$ \big(f_1(q,-),\dots, f_{r+1}(q,-), f_{r+2}(q,-); m_1,\dots, m_{r+1}, n_1,\dots, n_{r+1}, n_{r+2}\big)$$
is an $(r+1)$-step fibering of the domain $\Sigma_1(q)\subseteq M^m$ of $f_1(q,-)$ over $C$.

Consider one such tuple $\tau\in E$ as displayed above. Let $Q$ be the definable
set of $q\in M^{l}$ such that
for some $p\in M^k$ we have $\cal{S}(p)=\Sigma_1(q)$ and
$$ \big(f_1(q,-),\dots, f_{r+1}(q,-), f_{r+2}(q,-); m_1,\dots, m_{r+1}, n_1,\dots, n_{r+1},n_{r+2}\big)$$
is an $(r+1)$-step fibering of the domain $\Sigma_1(q)\subseteq M^m$ of $f_1(q,-)$ over $C$.
Each $q\in Q$ yields a map
$f_1(q,-): \Sigma_1(q)\to M^{n_1}$ such that its preimages of points $y\in M^{n_1}$ are in $\cC_r$. The maps $f_1(q,-)$ with $q\in Q$ make up a definable family $\cal{F}_\tau$ of maps from subsets of $M^m$ into $M^{n_1}$ as described just before Lemma~\ref{l2+}. That lemma then tells us that $\{\mu\big(\Sigma_1(q)\big): q\in Q\}$ is finite, and that
$\{q\in Q:\ \mu\big(\Sigma_1(q)\big)=a\}$ is definable for every $a$. This holds for all $\tau\in E$, which yields the
desired result.
\end{proof}

\noindent
This finishes the inductive proof of Theorem~\ref{thm} in the case of $\omega$-saturated
$\cM$.

\subsection*{Reduction to the $\omega$-saturated case}  Here $C$ is stably embedded in $\cM$, but we do not assume $\cM$ is $\omega$-saturated.

Let $\cM^*$ be an elementary extension of $\cM$. We define an $A$-valued Fubini measure $\mu^*$ on  the full subcategory
$\Def(C^*)$ of $\Def(\cM^*)$: Let $Y\subseteq (C^*)^n$ be definable in $\cM^*$; take an $\cal{L}$-formula $\phi(x,y)$ with $x=(x_1,\dots, x_m)$, $y=(y_1,\dots, y_n)$, and a point $p_Y\in (C^*)^m$ such that $Y=\{q\in (C^*)^n: \cM^*\models \phi(p_Y,q)\}$. To simplify notation, set $\phi(p, C^n):=\{q\in C^n:\ \cM\models \phi(p,q)\}$ for $p\in C^m$, and likewise define
$\phi\big(p, (C^*)^n\big)\subseteq (C^*)^n$ for $p\in (C^*)^m$; in particular, $Y=\phi\big(p_Y, (C^*)^n\big)$.
Now $C^m$ is the disjoint union of the definable sets $X_a:=\{p\in C^m:\ \mu\big(\phi(p,C^m)\big)=a\}$ (which is empty for all but finitely many $a$).
Set $\mu^*(Y):=a$, where $a$ is unique such that $p_Y\in (X_a)^*$. It is routine to verify that this particular $a$ does not depend on the choice of $\phi$ and that it yields indeed a Fubini measure on $\Def(C^*)$.

Now take such an elementary extension $\cM^*$ to be $\omega$-saturated. Then $\mu^*$ extends uniquely to an $A$-valued Fubini measure $\mu^*$ on the full subcategory $\Def(C^*)^{\f}$ of $\Def(\cM^*)$.
For $X\in \Def(C)^{\f}$ we have $X^*\in \Def(C^*)^{\f}$; set $\mu(X):= \mu^*(X^*)$.
This extends the original $A$-valued Fubini measure $\mu$ on $\Def(C)$ to an $A$-valued Fubini measure $\mu$ on $\Def(C)^{\f}$.  As to uniqueness, suppose $X$ is fiberable over $C$, and take an $r$-step fibering $(f_1,\dots, f_r, f_{r+1}; m_1,\dots, m_{r}, n_1,\dots, n_{r+1})$
of $X$ over $C$.
For $r=0$ the definable map $f_1: X \to C^{n_1}$ is injective, and so by Fubini
$\mu(X)$ is uniquely determined by $\mu: \Def(C)\to A$. For $r\ge 1$ the definable map $f_1: X\to M^{n_1}$ is such that
$f_1(X)$ and every
$f_1^{-1}(y)$ with $y\in f_1(X)$ is $(r-1)$-step fiberable over $C$, and we can assume inductively that the values of
$\mu\big(f(X)\big)$ and $\mu\big(f_1^{-1}(y)\big)$ for
$y\in f_1(X)$ are uniquely determined by $\mu: \Def(C)\to A$, which again by Fubini uniquely deternines $\mu(X)$.
This concludes the proof of Theorem~\ref{thm}.

\section{Fiberability and Co-analyzability}\label{co}

\noindent
Fiberability is a somewhat ad hoc notion, but agrees under natural model-theoretic conditions with the more intrinsic notion of co-analyzability from \cite{HHM}, as we shall see. {\it Throughout this section $X\subseteq M^m$ is definable}.

\subsection*{Fiberability implies co-analyzability}
We recall here the recursion on $r$ that defines ``$X$ is co-analyzable in $r$ steps" (tacitly: relative to $C$ in $\cM$), where $\cM$ is assumed to be $\omega$-saturated: \begin{enumerate}
\item  $X$ is co-analyzable in $0$ steps iff $X$ is finite;
\item  $X$ is co-analyzable in $r+1$ steps iff for some definable set $R \subseteq M^m\times C$ we have: $\breve{R}(y)$ is co-analyzable in $r$ steps for every $y\in C$, and
$X:= \bigcup_{y\in C}\breve{R}(y)$.
\end{enumerate}
 See \cite[p. 35]{ADH} for how this extends to not necessarily $\omega$-saturated $\cM$ in such a way that
 for any elementary extension $\cM^*$ of $\cM$: $X$ is co-analyzable in $r$ steps relative to $C$ in $\cM$ iff $X^*$ is co-analyzable in $r$ steps relative to $C^*$ in $\cM^*$.  Call $X$ {\em co-analyzable\/} if $X$ is co-analyzable in $r$ steps for some $r$.

 By \cite{HHM} this notion can be characterized as follows:

 \begin{prop}\label{hhm}
Let $T$ be a complete $L$-theory, where $L$ is a countable one-sorted language. Let $y$ be a single variable and $D(y)$ an $L$-formula that defines in each model $\cal{N}=(N;\cdots)$ of $T$ a set $D(N)$ with $|D(N)|\ge 2$.
Then the following conditions on an $L$-formula $\varphi(x)$ with $x=(x_1,\dots,x_m)$ are equivalent:
\begin{enumerate}
\item[\textup{(i)}] for some  $\cal{N}\models T$, $\varphi(N^m)$ is co-analyzable relative to $D(N)$;
\item[\textup{(ii)}] for all $\cal{N}\models T$, $\varphi(N^m)$ is co-analyzable relative to $D(N)$;
\item[\textup{(iii)}] for all  $\cal{N}\models T$, if
$D(N)$ is countable, then so is $\varphi(N^m)$;
\item[\textup{(iv)}] for all $\cal{N}\preceq\cal{N}^*\models T$, if
$D(N)=D(N^*)$, then $\varphi(N^m)=\varphi((N^*)^m)$.
\end{enumerate}
\end{prop}

\noindent
The equivalences (i)$\Leftrightarrow$(ii)$\Leftrightarrow$(iv) hold without assuming that $L$ is countable:
For (i)$\Leftrightarrow$(ii)$\Rightarrow$(iv), reduce to countable $L$ by taking suitable reducts. For (iv)$\Rightarrow$(ii), use also that (iv) amounts to $T_2\models \sigma$ where $T_2$ is the theory of elementary pairs of models of $T$ and $\sigma$ is a certain sentence, depending on $\varphi$, in the language of $T_2$. Note that in addition we can restrict $\cal{N}$ and $\cal{N}^*$ in (iv) to be  $\omega$-saturated.

\begin{lemma}\label{fico} If $X$ is fiberable over $C$, then  $X$ is co-analyzable relative to $C$.
\end{lemma}
\begin{proof}
We arrange that $\cM$ is $\omega$-saturated.
Below we prove by induction on $r$:  if $X$ is $r$-step fiberable over $C$, then $X$ is co-analyzable relative to $C$.
For $r=0$, if $f: X\to C^n$ is definable and injective, then $X$ is co-analyzable in $n$ steps relative to $C$, by an easy induction on $n$. Thus the claimed implication holds for $r=0$. Assume this implication holds for a certain $r$, and let
$X$ be $(r+1)$-step fiberable over $C$.
To show $X$ is co-analyzable relative to $C$ in $\cM$ we
augment $\cal{L}$ by names for finitely many elements of $M$ to arrange that $X$ is $0$-definable, so $X=\varphi(M^m)$ with $\varphi(x)$ an $\cal{L}$-formula, $x=(x_1,\dots, x_m)$. We now apply (iv)$\Rightarrow$(ii) of Proposition~\ref{hhm} without assuming $L=\cal{L}$ is countable, with $T:=\Th(\cM)$ and $D(y)$ an $\cal{L}$-formula such that $D(M)=C$.
So let $\cal{N}\preceq \cal{N}^*\models T$ with $D(N)=D(N^*)$; it suffices to show that then $\varphi(N^m)=\varphi\big((N^*)^m\big)$.
We also assume $\cal{N}$ is $\omega$-saturated, as we may by a remark following Proposition~\ref{hhm}. Since $\cal{M}\equiv \cal{N}$, the set $\varphi(N^m)$ is $(r+1)$-step fiberable over $C^{\cal{N}}:= D(N)$ in $\cal{N}$. Take $f: \varphi(N^m)\to Y$ where $Y\subseteq N^n$ and $f$ are definable in $\cal{N}$, such that $Y$ is $r$-step fiberable over $C^{\cal{N}}$ and for all
$y\in Y$ the fiber $f^{-1}(y)$ is $r$-step fiberable over $C^{\cal{N}}$. Thus by our inductive assumption and Proposition~\ref{hhm}, $Y$ and the fibers
$f^{-1}(y)$ for $y\in Y$ do not change in passing from $\cal{N}$ to $\cal{N}^*$. Hence the same is true for $\varphi(N^m)$, that is,
$\varphi(N^m)=\varphi\big((N^*)^m\big)$.
\end{proof}

\label{ms}

\noindent
So far we assumed $\cM$ to be one-sorted. But many structures are more naturally viewed as many-sorted; for example, valued fields as $3$-sorted structures with residue field a

\begin{proof} Take definable $f: Y\to M^m$ with $Y\subseteq C^\nu$, $\nu\in \N$, such that $X=f(Y)$. Since $(C;\cM)$ has EI, we have a definable map $g:Y\to C^n $ such that for all $y,z\in Y$ we have $f(y)=f(z)\Leftrightarrow g(y)=g(z)$. This yields a definable bijection $$\qquad\qquad \qquad \qquad f(y)\mapsto g(y)\ :\  X\to g(Y),\qquad (y\in Y). \qquad \qquad \qquad \qquad \qedhere$$
\end{proof}

\subsection*{Analyzability}
The concept of {\em analyzability\/} is more popular in model theory than co-analyzability, but is usually defined for types
in the environment of a monster model, instead of definable sets. I have been told that for definable sets and $\omega$-saturated $\cM$ its recursive definition should be as follows:

{\em $X$ is $0$-step analyzable in $C$ iff $X$ is internal to $C$, and $X$ is $(r+1)$-step analyzable in $C$ iff there is a $\Def(\cM)$-morphism $f: X \to Y$ such that $Y$ is $r$-step analyzable over $C$ and all fibers $f^{-1}(y)$, $y\in Y$, are internal to $C$. Finally, $X$ is analyzable in $C$ iff $X$ is $r$-step analyzable in $C$ for some $r$}.

This extends to not necessarily $\omega$-saturated $\cM$ by requiring in the $(r+1)$-clause the existence of a definable
family of maps from subsets of $C^n$ (for a fixed $n$) into $M^m$ such that for each $y\in Y$ some member of the family has image $f^{-1}(y)$.
With this extended notion and any elementary extension $\cM^*$ of $\cM$,
$$X \text{ is $r$-step analyzable in }C\ \Longleftrightarrow\  X^* \text{ is $r$-step analyzable in }C^*.$$
Using this it follows from Proposition~\ref{hhm} that if $X$ is analyzable in $C$, then $X$ is co-analyzable relative to $C$. Ben Castle and Rahim Moosa have assured me that the converse should also hold under some natural model-theoretic conditions to be satisfied in particular by differentially closed fields (and probably by $\T$) with $C$ its constant field. I have not pursued this, but it would be good to have this confirmed. One motivation for this: it might help in showing that for $\T$ and for differentially closed fields the notions of  fiberability over $C$ and co-analyzability relative to $C$ do not collapse for any $r$ to $r$-step fiberability over $C$ and co-analyzability in $r$-steps relative to $C$, respectively. This is plausible in view of \cite{J} where Jin constructs in a ``big" differentially closed field for any $r\ge 1$ a $1$-type that is $r$-step analyzable but not $(r-1)$-step analyzable (in the constant field).

\section{Application to $\T$ and to differentially closed fields} \label{special}

\subsection*{The case of $\T$} We construe here $\T$ as an ordered differential field. Its subfield $\R$ is its constant field, and
a set $Y\subseteq \R^n$ is definable in $\T$ iff $X$ is semialgebraic in the sense of $\R$. In particular, $\R$ is stably embedded in
$\T$. Thus by Theorem~\ref{thm}, the $\N_{\trop}$-valued Fubini measure $Y\mapsto \dim Y$ on $\Def(\R)$ extends uniquely to an $\N_{\trop}$-valued Fubini measure on the full subcategory $\Def(\R)^{\f}$ of $\Def(\T)$. Likewise for the $\Z$-valued Fubini measure $Y\mapsto E(Y)$ on $\Def(\R)$.

All this holds also for any $\upo$-free, newtonian, Liouville closed $H$-field $K$ and its constant field instead of
$\T$ and its constant field $\R$; this includes all ordered differential fields $K \equiv\T$.  The extended Euler characteristic now yields an answer to a test question raised in \cite[p. 34, end of Section 5]{ADH}:

\begin{cor} With $K$ as above, let $X\subseteq K$ be definable in $K$. Then $X$ with the ordering induced by $K$
is not elementarily equivalent to the ordered set $\omega$.
\end{cor}
\begin{proof} Suppose $X\equiv \omega$. Then $X$ is discrete as a subspace of $K$ with its order topology. Hence $X$ is fiberable
over $C$ by \cite{ADH}. Now $X\equiv \omega$ also yields  a definable bijection between $X$ and $X\setminus \{\text{the least element of $X$}\}$, but the extended Euler characteristic then gives $E(X)=E(X)-1$, a contradiction.
\end{proof}

\label{int}

\noindent
Semialgebraic sets $Y\subseteq  {\bbold R} ^n$ have a dimension $\dim(Y)\in  {\bbold N} \cup\{-\infty\}$ and an Euler characteristic $E(Y)\in  {\bbold Z} $ with good properties; see \cite[Ch. 4]{D}. In \cite{ADH} we asked for a natural extension of this dimension and Euler characteristic to the definable sets $X\subseteq  {\bbold T} ^m$
that are ``fiberable by $ {\bbold R} $" as defined there; here $ {\bbold T} $ is the real closed differential field of transseries, which has its subfield $ {\bbold R} $ as constant field.

There are many similar situations, and so it makes sense to do the above as a special case of a general extension result for what we call {\em Fubini measures}, with ``Fubini'' signaling good behavior in families. For example, at the end of Section~\ref{special} we shall also apply this to differentially closed fields (models of $\operatorname{DCF}_0$).

\medskip\noindent
Let $\mathbf{M}=(M;\cdots)$ be a one-sorted $\mathcal{L}$-structure;
``definable" will mean ``definable in $\mathbf{M}$ with parameters" unless we indicate otherwise.  For definable $X\subseteq M^m$ and an elementary extension $\mathbf{M}^*=(M^*;\cdots)$ of $\mathbf{M}$ we let $X^*\subseteq (M^*)^m$ be defined in $\mathbf{M}^*$ by any $\mathcal{L}_M$-formula that defines $X$ in $\mathbf{M}$.
The (small) category $\operatorname{Def}(\mathbf{M})$ has as its objects the definable sets $X\subseteq M^m$; the number $m\in  {\bbold N} $ is part of  specifying $X$ as an object of $\operatorname{Def}(\mathbf{M})$.  For objects $X,Y$ of $\operatorname{Def}(\mathbf{M})$, the morphisms $X\to Y$ are the definable maps $X\to Y$,
and composition of morphisms is composition of maps, so an isomorphism $X\to Y$ is the same as a definable bijection $X\to Y$. We shall abuse language by
indicating the class of objects of a category $\mathcal{C}$ also by $\mathcal{C}$, so $X\in \mathcal{C}$ means that $X$ is an object of $\mathcal{C}$.

\medskip\noindent
Let $\cal{C}$ be a full subcategory of $\operatorname{Def}(\mathbf{M})$ such that for all $X,Y\in \cal{C}$: \begin{itemize}
\item[(a)] if $X, Y\subseteq M^m$, then $X\cup Y\subseteq M^m$ is an object of $\cal{C}$;
\item[(b)] every definable subset of $X$ is an object of $\cal{C}$;
\item[(c)]  $X\times Y\in \cal{C}$ (where $X\times Y\subseteq  M^{m+n}$ for $X\subseteq M^m, Y\subseteq M^n$).
\end{itemize}
Let $A$ be a commutative semiring, that is, a set with binary operations of addition and multiplication and distinguished elements
$0$ and $1$ such that $A$ is a commutative monoid with respect to both addition and multiplication, with $0$ and $1$ as neutral elements for addition and multiplication, respectively, such that multiplication is distributive with respect to addition, and such that also $0\cdot a=0$ for all $a\in A$.

Let  $\mu$ an $A$-valued Fubini measure on $\cal{C}$, that is, $\mu: \cal{C}\to A$, and for all $X,Y\in  \cal{C}$:\begin{enumerate}
\item[(d)] $\mu(X)=0$ whenever $X$ is empty; $\mu(X)=1$ whenever $X$ is a singleton;
\item[(e)] $\mu(X\cup Y)=\mu(X)+\mu(Y)$ whenever $X,Y\subseteq M^m$ are disjoint;
\item[(f)] every $\cal{C}$-morphism $f: X \to Y$ satisfies ``Fubini": \begin{enumerate}
\item[(F1)] the subset $\{\mu\big(f^{-1}(y)\big):\ y\in Y\}$ of $A$ is finite;
\item[(F2)] for all $a\in A$, the set
$Y_a := \{y\in Y:\ \mu\big(f^{-1}(y)\big)=a\}$ is definable;
\item[(F3)] if $\mu\big(f^{-1}(y)\big)$ takes the constant value $a$ on $Y$, then
$\mu(X)= a\mu(Y)$.
\end{enumerate}
\end{enumerate}
Thus for $X,Y\in  \cal{C}$:  $\mu(X\times Y)=\mu(X)\cdot \mu(Y)$; if $X$ and $Y$ are $\cal{C}$-isomorphic, then $\mu(X)=\mu(Y)$ (a ``motivic'' property);  for $X,Y\subseteq M^m$ we can represent $X\cup Y$ as the disjoint union of $X\setminus Y, X\cap Y$, and $Y\setminus X$ to obtain
$$\mu(X\cup Y)+\mu(X\cap Y)\ =\ \mu(X)+\mu(Y).$$
If $f: X\to Y$ is a $\cal{C}$-morphism, then $\mu(X)=\sum_a a\mu(Y_a)$, using  notation of (F2).

\medskip\noindent
{\bf Example.} Let $\mathbf{M}= {\bbold T} $ as in the beginning of the introduction. Then the subsets of $ {\bbold R} ^n$ definable in $\mathbf{M}$ are  semialgebraic in the sense of $ {\bbold R} $, so the full subcategory $\cal{C}$ of $\operatorname{Def}(\mathbf{M})$
whose objects are these semialgebraic sets satisfies conditions (a), (b), (c). The above Euler characteristic $Y\mapsto E(Y)$ is then a Fubini measure on $\cal{C}$ with values in the ring $ {\bbold Z} $ of integers.  Also $Y\mapsto \dim(Y)$ is a Fubini measure on $\cal{C}$, taking values in the tropical semiring $\N_{\operatorname{trop}}$, which has underlying set
$ {\bbold N} \cup\{-\infty\}$ and addition and multiplication given respectively by
$(p,q)\mapsto \max(p,q)$ and $(p,q)\mapsto p+q$; this semiring has $-\infty$ as its zero element for addition and $0$ as its multiplicative identity.

\bigskip\noindent
To state the main result of this paper, we consider a $0$-definable set $C\subseteq M$ with more than one element, and let $\operatorname{Def}(C)$ be the full subcategory of $\operatorname{Def}(\mathbf{M})$ whose objects are the definable subsets of the powers $C^n$.  Below we define the notion, for definable $X\subseteq M^m$, of {\em $X$ being fiberable over $C$}. These $X$ are the objects of a full subcategory
$\operatorname{Def}(C)^{\operatorname{f}}\supseteq  \operatorname{Def}(C)$ of $\operatorname{Def}(\mathbf{M})$. The categories $\operatorname{Def}(C)$ and $\operatorname{Def}(C)^{\operatorname{f}}$ satisfy conditions (a), (b), (c)  imposed earlier on $\cal{C}$.

There are many similar situations, and so it makes sense to do the above as a special case of a general extension result for what we call {\em Fubini measures}, with ``Fubini'' signaling good behavior in families. For example, at the end of Section~\ref{special} we shall also apply this to differentially closed fields (models of $\operatorname{DCF}_0$).

\medskip\noindent
Let $\mathbf{M}=(M;\cdots)$ be a one-sorted $\mathcal{L}$-structure;
``definable" will mean ``definable in $\mathbf{M}$ with parameters" unless we indicate otherwise. For definable $X\subseteq M^m$ and an elementary extension $\mathbf{M}^*=(M^*;\cdots)$ of $\mathbf{M}$ we let $X^*\subseteq (M^*)^m$ be defined in $\mathbf{M}^*$ by any $\mathcal{L}_M$-formula that defines $X$ in $\mathbf{M}$.
The (small) category $\operatorname{Def}(\mathbf{M})$ has as its objects the definable sets $X\subseteq M^m$; the number $m\in {\bbold N} $ is part of specifying $X$ as an object of $\operatorname{Def}(\mathbf{M})$. For objects $X,Y$ of $\operatorname{Def}(\mathbf{M})$, the morphisms $X\to Y$ are the definable maps $X\to Y$,
and composition of morphisms is composition of maps, so an isomorphism $X\to Y$ is the same as a definable bijection $X\to Y$. We shall abuse language by
indicating the class of objects of a category $\mathcal{C}$ also by $\mathcal{C}$, so $X\in \mathcal{C}$ means that $X$ is an object of $\mathcal{C}$.

\medskip\noindent
Let $\cal{C}$ be a full subcategory of $\operatorname{Def}(\mathbf{M})$ such that for all $X,Y\in \cal{C}$: \begin{itemize}
\item[(a)] if $X, Y\subseteq M^m$, then $X\cup Y\subseteq M^m$ is an object of $\cal{C}$;
\item[(b)] every definable subset of $X$ is an object of $\cal{C}$;
\item[(c)] $X\times Y\in \cal{C}$ (where $X\times Y\subseteq M^{m+n}$ for $X\subseteq M^m, Y\subseteq M^n$).
\end{itemize}
Let $A$ be a commutative semiring, that is, a set with binary operations of addition and multiplication and distinguished elements
$0$ and $1$ such that $A$ is a commutative monoid with respect to both addition and multiplication, with $0$ and $1$ as neutral elements for addition and multiplication, respectively, such that multiplication is distributive with respect to addition, and such that also $0\cdot a=0$ for all $a\in A$.

Let $\mu$ an $A$-valued Fubini measure on $\cal{C}$, that is, $\mu: \cal{C}\to A$, and for all $X,Y\in \cal{C}$:\begin{enumerate}
\item[(d)] $\mu(X)=0$ whenever $X$ is empty; $\mu(X)=1$ whenever $X$ is a singleton;
\item[(e)] $\mu(X\cup Y)=\mu(X)+\mu(Y)$ whenever $X,Y\subseteq M^m$ are disjoint;
\item[(f)] every $\cal{C}$-morphism $f: X \to Y$ satisfies ``Fubini": \begin{enumerate}
\item[(F1)] the subset $\{\mu\big(f^{-1}(y)\big):\ y\in Y\}$ of $A$ is finite;
\item[(F2)] for all $a\in A$, the set
$Y_a := \{y\in Y:\ \mu\big(f^{-1}(y)\big)=a\}$ is definable;
\item[(F3)] if $\mu\big(f^{-1}(y)\big)$ takes the constant value $a$ on $Y$, then
$\mu(X)= a\mu(Y)$.
\end{enumerate}
\end{enumerate}
Thus for $X,Y\in \cal{C}$: $\mu(X\times Y)=\mu(X)\cdot \mu(Y)$; if $X$ and $Y$ are $\cal{C}$-isomorphic, then $\mu(X)=\mu(Y)$ (a ``motivic'' property); for $X,Y\subseteq M^m$ we can represent $X\cup Y$ as the disjoint union of $X\setminus Y, X\cap Y$, and $Y\setminus X$ to obtain
$$\mu(X\cup Y)+\mu(X\cap Y)\ =\ \mu(X)+\mu(Y).$$
If $f: X\to Y$ is a $\cal{C}$-morphism, then $\mu(X)=\sum_a a\mu(Y_a)$, using notation of (F2).

\medskip\noindent
{\bf Example.} Let $\mathbf{M}= {\bbold T} $ as in the beginning of the introduction. Then the subsets of $ {\bbold R} ^n$ definable in $\mathbf{M}$ are semialgebraic in the sense of $ {\bbold R} $, so the full subcategory $\cal{C}$ of $\operatorname{Def}(\mathbf{M})$
whose objects are these semialgebraic sets satisfies conditions (a), (b), (c). The above Euler characteristic $Y\mapsto E(Y)$ is then a Fubini measure on $\cal{C}$ with values in the ring $ {\bbold Z} $ of integers. Also $Y\mapsto \dim(Y)$ is a Fubini measure on $\cal{C}$, taking values in the tropical semiring $\N_{\operatorname{trop}}$, which has underlying set
$ {\bbold N} \cup\{-\infty\}$ and addition and multiplication given respectively by
$(p,q)\mapsto \max(p,q)$ and $(p,q)\mapsto p+q$; this semiring has $-\infty$ as its zero element for addition and $0$ as its multiplicative identity.

\bigskip\noindent
To state the main result of this paper, we consider a $0$-definable set $C\subseteq M$ with more than one element, and let $\operatorname{Def}(C)$ be the full subcategory of $\operatorname{Def}(\mathbf{M})$ whose objects are the definable subsets of the powers $C^n$. Below we define the notion, for definable $X\subseteq M^m$, of {\em $X$ being fiberable over $C$}. These $X$ are the objects of a full subcategory
$\operatorname{Def}(C)^{\operatorname{f}}\supseteq \operatorname{Def}(C)$ of $\operatorname{Def}(\mathbf{M})$. The categories $\operatorname{Def}(C)$ and $\operatorname{Def}(C)^{\operatorname{f}}$ satisfy conditions (a), (b), (c) imposed earlier on $\cal{C}$.

\bigskip\noindent
To state the main result of this paper, we consider a $0$-definable set $C\subseteq M$ with more than one element, and let $\Def(C)$ be the full subcategory of $\Def(\cM)$ whose objects are the definable subsets of the powers $C^n$. Below we define the notion, for definable $X\subseteq M^m$, of {\em $X$ being fiberable over $C$}. These $X$ are the objects of a full subcategory
$\Def(C)^{\f}\supseteq \Def(C)$ of $\Def(\cM)$. The categories $\Def(C)$ and $\Def(C)^{\f}$ satisfy conditions (a), (b), (c) imposed earlier on $\cC$.

\noindent
Suppose $\cM$ is $\omega$-saturated, and let $X\subseteq M^m$ be definable. By recursion on $r\in \N$ we define ``$X$ is $r$-step fiberable over $C$": this means the existence of a definable $f: X \to M^n$ such that
if $r=0$, then $f$ is injective and $f(X)\subseteq C^n$, and if $r\ge 1$, then $f(X)$ and every fiber $f^{-1}(y)$, $y\in M^n$,
is $(r-1)$-step fiberable over $C$.

\begin{lemma}\label{smfub} If $\cM$ is strongly minimal $($which includes $M$ being infinite$)$, then $X\mapsto \operatorname{MR}(X)$ is an $\N_{\trop}$-valued Fubini measure on $\Def(\cM)$.
\end{lemma}

\subsection*{The case of $\T$} We construe here $\T$ as an ordered differential field. Its subfield $\R$ is its constant field, and
a set $Y\subseteq \R^n$ is definable in $\T$ iff $X$ is semialgebraic in the sense of $\R$. In particular, $\R$ is stably embedded in
$\T$. Thus by Theorem~\ref{thm}, the $\N_{\trop}$-valued Fubini measure $Y\mapsto \dim Y$ on $\Def(\R)$ extends uniquely to an $\N_{\trop}$-valued Fubini measure on the full subcategory $\Def(\R)^{\f}$ of $\Def(\T)$. Likewise for the $\Z$-valued Fubini measure $Y\mapsto E(Y)$ on $\Def(\R)$.

\begin{prop}\label{uniq} If $\operatorname{char}(\k)>0$, then there is no $\Z$-valued Fubini measure on $\Def(\k)$. If
$\operatorname{char}(\k)=0$, then there is a unique $\Z$-valued Fubini measure on $\Def(\k)$.
\end{prop}
\begin{proof} Suppose $\mu$ is a $\Z$-valued Fubini measure on $\Def(\k)$. Then $\mu(\k)=1$:
 if $\operatorname{char}(\k)\ne 2$, the fibers of
 $y\mapsto y^2: \k\to \k$
have size $2$ except at $0$, so
$$\mu(\k)\ =\ 1+2\big(\mu(\k)-1\big),$$
 and thus $\mu(\k)=1$; if $\operatorname{char}(\k)=2$, the map $y\mapsto y^3: \k\to \k$ gives likewise $\mu(\k)=1+3\big(\mu(\k)-1\big)$, again resulting in $\mu(\k)=1$.

\noindent
Let now $\k$ be a proper subfield of the algebraically closed field $K$, and $\cM:=(K, \k)$. Then
$\cM$ is $\omega$-stable with $\text{MR}(\k)=1$, $\text{MR}(K)=\omega$, and a set $Y\subseteq \k^n$ is definable in $\cM$ iff $Y$ is constructible in the sense of the algebraically closed field $\k$. Thus $\k$ is stably embedded in $\cM$.
Let $X\subseteq K^m$ be definable. Then
$$X\text{ is almost internal to }\k\
\Leftrightarrow\ X \text{ is co-analyzable relative to }\k\
 \Leftrightarrow\ \operatorname{MR}(X)<\omega.$$
These equivalences are in \cite{AD}, which also has an example of a definable $X\subseteq K$ that is almost internal to $\k$ but not internal to $\k$. Another proof of the first equivalence using \cite{P} was pointed out to me by Pillay. Lemmas~\ref{fico} and ~\ref{cofi} also give:
$$X \text{ is fiberable over }\k\ \Longleftrightarrow\ X \text{ is co-analyzable relative to }\k.$$
 By Lemma~\ref{smfub} the function $Y\mapsto \text{MR}(Y)$
on $\Def(\k)$ is a Fubini measure with values in $\N_{\trop}$. By Theorem~\ref{thm} it extends uniquely to
an $\N_{\trop}$-valued Fubini measures on $\Def(\k)^{\f}$; this extension is necessarily
$X\mapsto \text{MR}(X)$: the above relation to almost internality implies that it is a Fubini measure on $\Def(\k)^{\f}$.

\medskip\noindent
Now assume $\operatorname{char}(\k)=0$. Then we have the unique $\Z$-valued Fubini measure $Y\mapsto E(Y)$ on $\Def(\k)$ from the proof of Proposition~\ref{uniq}. By Theorem~\ref{thm} it extends uniquely to a $\Z$-valued Fubini measure on
$\Def(\k)^{\f}$.
Although "fiberable over $\k$" collapses here to ``almost internal to $\k$", we still need all of Section~\ref{prelim} and much of Section~\ref{proof} to obtain this extension result.

\medskip
{\em If $\cC$ is stably embedded in $\cM$ and $\mu$ is an $A$-valued Fubini measure on $\cC$,
then $\mu$ extends uniquely to an $A$-valued Fubini measure on $\cC^{\f}$}.

\label{prelim}

\noindent
It will be convenient to consider measures that do not necessarily satisfy (F3). We call  them {\em semi-Fubini measures\/} and they are defined in the first subsection below

\begin{lemma}  If $A, B$ are commutative semirings and $\mu: \cC\to A$ and $\nu: \cC\to B$ are Fubini measures, then $(\mu,\nu): \cC\to A\times B$ is a Fubini measure.
\end{lemma}

\section{Proof of the Theorem}\label{proof}

\noindent
{\it In this section $\mu$ is an $A$-valued Fubini measure on $\Def(C)$}. Till further notice we also assume $\cM$ is $\omega$-saturated, indicating at the end of the section how to reduce to that case.

\medskip\noindent
By recursion on $r$ we now introduce a sequence $(\cC_r)$ of full subcategories of $\Def(\cM)$, such that $ \cC_r\subseteq \cC_{r+1}$ for all $r$ and $\bigcup_r \cC_r=\Def(C)^{\f}$:

\medskip\noindent
 $\cC_0:= \Def(C)^{\iso}$, and the objects of $\cC_{r+1}$ are the definable $X\subseteq M^m$ for which there exists a definable
$f: X\to Y$ with $Y\in \cC_r$, such that $f^{-1}(y)\in  \cC_r$ for all $y\in Y$; such $f$ is called a $\cC_{r+1}$-witness. Thus the objects of $\cC_r$ are the definable $X\subseteq M^m$ that are $r$-step fiberable over $C$ as defined in the Introduction.
In particular, the objects of $\cC_0$ are the definable $X\subseteq M^m$ that admit a definable injective map $X\to C^n$. Also, if the definable set $X\subseteq M^m$ admits a definable map $X\to C^n$ all whose fibers are finite, then $X\in \cC_1$.

\medskip\noindent
 Induction on $r$ shows that $\cC_r$ satisfies conditions (a), (b), (c) for all $r$, and that $\cC_r^{\iso}=\cC_r$ for all $r$.  In the inductive step we note that if $f: X\to Y$ is a $\cC_{r+1}$-witness and $X'\subseteq X$ and $Y'\subseteq Y$ are definable with
$f(X')\subseteq Y'$, then $f|_{X'}: X'\to Y'$ is also a $\cC_{r+1}$-witness. Induction on $r$ also shows that there is at most one
$A$-valued Fubini measure on $\cC_r$ extending the given $A$-valued Fubini measure $\mu$ on $\Def(C)$. In the beginning of the previous section we have extended $\mu$ to an $A$-valued Fubini measure on $\cC_0$, also to be denoted by $\mu$.

\subsection*{More about $\cC_0$} {\it In this subsection we assume that $C$ is stably embedded in $\cM$}.
Let  $\cal{S}\subseteq M^k\times C^n$ be definable. Think of $\cal{S}$ as the definable family of sets
$\cal{S}(p)\subseteq C^n$ parametrized by the points $p\in M^k$.

\begin{lemma}\label{l1}  The set $\{\mu\big(\cal{S}(p)\big):\ p\in M^k\}$ is finite, and for every $a$ the set $\{p\in M^k:\ \mu\big(\cal{S}(p)\big)=a\}$ is definable.
\end{lemma}

\begin{theorem}\label{thm} If $C$ is stably embedded in $\cM$ and $\mu$ an $A$-valued Fubini measure on $\Def(C)$,
then $\mu$ extends uniquely to an $A$-valued Fubini measure on $\Def(C)^{\f}$.
\end{theorem}

\noindent
Suppose $\mathbf{M}$ is $\omega$-saturated, and let $X\subseteq M^m$ be definable. By recursion on $r\in  {\bbold N} $ we define ``$X$ is $r$-step fiberable over $C$":  this means the existence of a definable $f: X \to M^n$ such that
if $r=0$, then $f$ is injective and $f(X)\subseteq C^n$, and if $r\ge 1$, then $f(X)$ and every fiber $f^{-1}(y)$, $y\in M^n$,
is $(r-1)$-step fiberable over $C$.

Now drop the assumption that $\mathbf{M}$ is $\omega$-saturated, and let $X\subseteq M^m$ be definable.
Then the definition of ``$X$ is $r$-step fiberable
over $C$'' (with $\mathbf{M}$ as ambient structure) is more complicated, see end of Section~\ref{proof}, but will be such that it is equivalent to ``$X^*$ is $r$-step fiberable  over $C^*$" with $\mathbf{M}^*$ as ambient structure, where $\mathbf{M}^*=(M^*,\dots)$
is any $\omega$-saturated elementary extension of $\mathbf{M}$. Define $X$ to be {\em fiberable over $C$\/} if  it is $r$-step fiberable over $C$ for some $r$.

Using this equivalence the proof of Theorem~\ref{thm} reduces to the case where $\mathbf{M}$ is $\omega$-saturated, which is very convenient. The idea is to show by induction on $r$ that $\mu$ extends to an $A$-valued Fubini measure on the full subcategory of $\operatorname{Def}(\mathbf{M})$ whose objects are the definable $X\subseteq M^m$ that are $r$-step fiberable over $C$. In the step from $r$ to $r+1$ we consider an $(r+1)$-step fiberable $X\subseteq M^m$ over $C$. This is witnessed by a definable $f: X\to M^n$ such that $f(X)$ and all $f^{-1}(y)$ are $r$-step fiberable over $C$. A carefully chosen inductive assumption will tell us that $\mu\big(f(X)\big)$ is defined and that $\mu\big(f^{-1}(y)\big)$ is defined and takes only finitely many values for $y\in f(X)$. Then we define
$\mu_f(X)$ as a suitably weighted sum of these values and show that $\mu_f(X)$ does not depend on $f$, but only on $X$, and that the resulting $\mu(X)=\mu_f(X)$ defined for such $X$ preserves the Fubini property. We also have to show that the inductive assumption is preserved in this step from $r$ to $r+1$. All this is carried out in detail in Section~\ref{proof}.

Section~\ref{prelim} contains preliminary results. In section~\ref{co} we relate fiberability over $C$ to the more intrinsic notion of {\em co-analyzability relative to $C$\/}  from \cite{HHM}, and show their equivalence under various model theoretic conditions. These conditions are satisfied by $ {\bbold T} $ and by differentially closed fields, with $C$ their constant field. This is discussed in Section~\ref{special}, where it is used to answer questions left open in \cite{ADH}. In the short last section~\ref{ms} we state for the record a many-sorted version of Theorem~\ref{thm}.

\medskip\noindent
The  measures from  \cite{MS, AMSW} also take values in semirings and satisfy {\em Fubini}, but seem to be of a rather different nature. Nevertheless, some variant of Theorem~\ref{thm} might hold in that setting. One could also take a look at motivic measures as in ~\cite{HK}, and consider their ``Fubini" nature, or lack of it.

\medskip\noindent
I thank Ben Castle, Artem Chernikov, Rahim Moosa, Anand Pillay, and Charles Steinhorn for useful feedback after a talk I gave on an earlier version of this paper.

\subsection*{Notations and Terminology} We let $d,e, k, l, m,n,r$, sometimes subscripted, range over $ {\bbold N} =\{0,1,2,\dots\}$.  For sets $P, Q$, and $\mathcal{S}\subseteq P\times Q$ and $p\in P$ we set $$\mathcal{S}(p)\ :=\ \{q\in Q:\ (p,q)\in \mathcal{S}\}. $$  (Think of $\mathcal{S}$ as describing the family $\big(\mathcal{S}(p)\big)_{p\in P}$ of subsets of $Q$.) Throughout, $\mathbf{M}=(M;\dots)$ is a one-sorted $\mathcal{L}$-structure. When $C$ gets mentioned, we assume $C\subseteq M$ is $0$-definable and $|C|\ge 2$. Recall, for example from \cite{HHM}, that $C$ is said to be {\em stably embedded in $\mathbf{M}$\/} if for every $0$-definable $\mathcal{S}\subseteq M^{m}\times M^n$ there is a $0$-definable $\mathcal{T}\subseteq M^{\mu}\times M^n$, $\mu\in  {\bbold N} $, such that for all $p\in M^m$ there exists $c\in C^{\mu}$ for which $\mathcal{S}(p)\cap C^n = \mathcal{T}(c)\cap C^n$.
  For $\omega$-saturated $\mathbf{M}$, this is equivalent to all definable $X\subseteq C^n$ being $C$-definable.

Suppose $X\subseteq M^m$ and $Y\subseteq M^n$ are definable. Then a {\em definable family of maps from subsets of $X$ into $Y$\/} is a definable set
$\mathcal{F}\subseteq P\times X \times Y$, with definable $P\subseteq M^k$ such that for each ``parameter" $p\in P$ the set
$\mathcal{F}(p)\subseteq X\times Y$ is the graph of a map $f_p$ with domain a subset of $X$ and codomain $Y$.  We think of such $\mathcal{F}$ as the family $(f_p)_{p\in P}$. Abusing language we write $f\in \mathcal{F}$ to indicate that $f=f_p$ for some $p\in P$, and to state that the set of $f\in \mathcal{F}$ with a certain property is definable means that the set of $p\in P$ such that $f_p$ has that property is definable. Given $\mathcal{S}\subseteq P\times Q$, $p\in P$, and a map $f: \mathcal{S} \to Z$, we let $f(p,-)$ denote the map $q\mapsto f(p,q): \mathcal{S}(p)\to Z$.

Our definition of ``$r$-step fiberable over $C$'' differs from that of ``fiberable by $C$ in $r$ steps''
in \cite{ADH} for a Liouville closed $H$-field $K$ and its constant field $C$. But, in that case,  ``fiberable over $C$'' as defined here is  equivalent to ``fiberable by $C$ in $r$ steps for some $r$" as defined in \cite{ADH}.

Using the equivalence (for an $omega$-saturated elementary extension $\mathbf{M}^*$) between “$X$ is $r$-step fiberable over $C$” and “$X^*$ is $r$-step fiberable over $C^*$”, the proof of Theorem~\ref{thm} “reduces to the case where $\mathbf{M}$ is $\omega$-saturated”. One then proves the extension by “induction on $r$”, showing that $\mu$ extends to an $A$-valued Fubini measure on the full subcategory of $\operatorname{Def}(\mathbf{M})$ consisting of definable sets that are $r$-step fiberable over $C$.

For the step $r\to r+1$, take an $(r+1)$-step fiberable $X\subseteq M^m$, witnessed by a definable $f:X\to M^n$ with $f(X)$ and all fibers $f^{-1}(y)$ ($y\in M^n$) $r$-step fiberable over $C$. A “carefully chosen inductive assumption” gives that $\mu\big(f(X)\big)$ is defined and that $\mu\big(f^{-1}(y)\big)$ is defined and “takes only finitely many values for $y\in f(X)$”. Define $\mu_f(X)$ as a “suitably weighted sum of these values”, and show $\mu_f(X)$ “does not depend on $f$, but only on $X$”, so one may set $\mu(X)=\mu_f(X)$. Finally, verify that this $\mu$ “preserves the Fubini property” and that the inductive assumption is preserved from $r$ to $r+1$. (Details are said to be in Section~\ref{proof}.)

Using the equivalence (for an $\omega$-saturated elementary extension $\mathbf{M}^*$) between “$X$ is $r$-step fiberable over $C$” and “$X^*$ is $r$-step fiberable over $C^*$”, the proof of the main theorem reduces to the case where $\mathbf{M}$ is $\omega$-saturated. One then proves the extension by induction on $r$, showing that $\mu$ extends to an $A$-valued Fubini measure on the full subcategory of $\operatorname{Def}(\mathbf{M})$ consisting of definable sets that are $r$-step fiberable over $C$.

More precisely, one introduces by recursion on $r$ a sequence $(\cC_r)$ of full subcategories of $\Def(\cM)$, such that $ \cC_r\subseteq \cC_{r+1}$ for all $r$ and $\bigcup_r \cC_r=\Def(C)^{\f}$:

$\cC_0:= \Def(C)^{\iso}$, and the objects of $\cC_{r+1}$ are the definable $X\subseteq M^m$ for which there exists a definable
$f: X\to Y$ with $Y\in \cC_r$, such that $f^{-1}(y)\in  \cC_r$ for all $y\in Y$; such $f$ is called a $\cC_{r+1}$-witness. Thus the objects of $\cC_r$ are the definable $X\subseteq M^m$ that are $r$-step fiberable over $C$ as defined in the Introduction.
In particular, the objects of $\cC_0$ are the definable $X\subseteq M^m$ that admit a definable injective map $X\to C^n$. Also, if the definable set $X\subseteq M^m$ admits a definable map $X\to C^n$ all whose fibers are finite, then $X\in \cC_1$.

For the step $r\to r+1$, take an $(r+1)$-step fiberable $X\subseteq M^m$, witnessed by a definable $f:X\to Y$ with $Y\in \cC_r$ and all fibers $f^{-1}(y)$ ($y\in Y$) lying in $\cC_r$. The inductive setup yields that $\mu$ is already defined on $Y$ and on each fiber $f^{-1}(y)$. To ensure the finiteness/definability needed to form a Fubini-type sum, one uses the following lemma.

\begin{lemma}\label{l1}  The set $\{\mu\big(\cal{S}(p)\big):\ p\in M^k\}$ is finite, and for every $a$ the set $\{p\in M^k:\ \mu\big(\cal{S}(p)\big)=a\}$ is definable.
\end{lemma}

Using this, one obtains that $\mu\big(f^{-1}(y)\big)$ takes only finitely many values as $y$ ranges over $Y=f(X)$. One then defines $\mu_f(X)$ as the corresponding suitably weighted sum of these values and shows that $\mu_f(X)$ does not depend on the choice of witness $f$ but only on $X$, allowing one to set $\mu(X)=\mu_f(X)$. Finally, one checks that this definition preserves the Fubini property and that the inductive hypotheses propagate from $r$ to $r+1$.

The details are carried out later.

\begin{lemma}  If $A, B$ are commutative semirings and $\mu: \cC\to A$ and $\nu: \cC\to B$ are Fubini measures, then $(\mu,\nu): \cC\to A\times B$ is a Fubini measure.
\end{lemma}

\section{Proof of the Theorem}\label{proof}

\noindent
{\it In this section $\mu$ is an $A$-valued Fubini measure on $\Def(C)$}. Till further notice we also assume $\cM$ is $\omega$-saturated, indicating at the end of the section how to reduce to that case.

\medskip\noindent
By recursion on $r$ we now introduce a sequence $(\cC_r)$ of full subcategories of $\Def(\cM)$, such that $ \cC_r\subseteq \cC_{r+1}$ for all $r$ and $\bigcup_r \cC_r=\Def(C)^{\f}$:

\medskip\noindent
 $\cC_0:= \Def(C)^{\iso}$, and the objects of $\cC_{r+1}$ are the definable $X\subseteq M^m$ for which there exists a definable
$f: X\to Y$ with $Y\in \cC_r$, such that $f^{-1}(y)\in  \cC_r$ for all $y\in Y$; such $f$ is called a $\cC_{r+1}$-witness. Thus the objects of $\cC_r$ are the definable $X\subseteq M^m$ that are $r$-step fiberable over $C$ as defined in the Introduction.
In particular, the objects of $\cC_0$ are the definable $X\subseteq M^m$ that admit a definable injective map $X\to C^n$. Also, if the definable set $X\subseteq M^m$ admits a definable map $X\to C^n$ all whose fibers are finite, then $X\in \cC_1$.

\medskip\noindent
 Induction on $r$ shows that $\cC_r$ satisfies conditions (a), (b), (c) for all $r$, and that $\cC_r^{\iso}=\cC_r$ for all $r$.  In the inductive step we note that if $f: X\to Y$ is a $\cC_{r+1}$-witness and $X'\subseteq X$ and $Y'\subseteq Y$ are definable with
$f(X')\subseteq Y'$, then $f|_{X'}: X'\to Y'$ is also a $\cC_{r+1}$-witness. Induction on $r$ also shows that there is at most one
$A$-valued Fubini measure on $\cC_r$ extending the given $A$-valued Fubini measure $\mu$ on $\Def(C)$. In the beginning of the previous section we have extended $\mu$ to an $A$-valued Fubini measure on $\cC_0$, also to be denoted by $\mu$.

\subsection*{More about $\cC_0$} {\it In this subsection we assume that $C$ is stably embedded in $\cM$}.
Let  $\cal{S}\subseteq M^k\times C^n$ be definable. Think of $\cal{S}$ as the definable family of sets
$\cal{S}(p)\subseteq C^n$ parametrized by the points $p\in M^k$.

\begin{lemma}\label{l1}  The set $\{\mu\big(\cal{S}(p)\big):\ p\in M^k\}$ is finite, and for every $a$ the set $\{p\in M^k:\ \mu\big(\cal{S}(p)\big)=a\}$ is definable.
\end{lemma}

Question: Let \(\mathbf M\) be a one-sorted \(\mathcal L\)-structure, let \(C\subseteq M\) be a \(0\)-definable set with more than one element, and let \(A\) be a commutative semiring. Write \(\operatorname{Def}(C)\) for the full subcategory of definable sets whose objects are the definable subsets of powers \(C^n\). An \(A\)-valued Fubini measure on a full subcategory assigns to each object an element of \(A\), sends the empty set to \(0\) and singletons to \(1\), is additive on disjoint definable unions, and satisfies the stated Fubini fiber conditions for definable maps. Let \(\operatorname{Def}(C)^{\operatorname f}\) be the full subcategory of definable sets fiberable over \(C\), equivalently the union of the classes \(\mathcal C_r\) defined by \(\mathcal C_0=\operatorname{Def}(C)^{\operatorname{iso}}\) and \(X\in \mathcal C_{r+1}\) iff there is a definable map \(f:X\to Y\) with \(Y\in \mathcal C_r\) and every fiber \(f^{-1}(y)\in \mathcal C_r\). Assume that \(C\) is stably embedded in \(\mathbf M\), and let \(\mu\) be an \(A\)-valued Fubini measure on \(\operatorname{Def}(C)\). Which statement holds about extending \(\mu\)?

There exists exactly one \(A\)-valued Fubini measure \(\widetilde\mu\) on \(\operatorname{Def}(C)^{\operatorname f}\) such that \(\widetilde\mu(X)=\mu(X)\) for every object \(X\in \operatorname{Def}(C)\); equivalently, \(\mu\) extends uniquely to an \(A\)-valued Fubini measure on \(\operatorname{Def}(C)^{\operatorname f}\).

Choices:
B. There exists exactly one \(A\)-valued semi-Fubini measure \(\widetilde\mu\) on \(\operatorname{Def}(C)^{\operatorname f}\) such that \(\widetilde\mu(X)=\mu(X)\) for every object \(X\in \operatorname{Def}(C)\); in general one cannot require \(\widetilde\mu\) to satisfy the full Fubini axiom on all of \(\operatorname{Def}(C)^{\operatorname f}\).
C. There exists an \(A\)-valued Fubini measure \(\widetilde\mu\) on \(\operatorname{Def}(C)^{\operatorname f}\) such that \(\widetilde\mu(X)=\mu(X)\) for every object \(X\in \operatorname{Def}(C)\).
D. For every \(r\in \mathbb N\), there exists exactly one \(A\)-valued Fubini measure \(\widetilde\mu_r\) on \(\mathcal C_r\) extending \(\mu\), but in general these extensions need not glue to a single \(A\)-valued Fubini measure on the whole union \(\operatorname{Def}(C)^{\operatorname f}=\bigcup_r \mathcal C_r\).
E. There exists exactly one \(A\)-valued Fubini measure \(\widetilde\mu\) on the full subcategory of all definable sets \(X\subseteq M^m\) for which there is a definable map \(X\to C^n\) with all fibers finite, such that \(\widetilde\mu(X)=\mu(X)\) for every object \(X\in \operatorname{Def}(C)\); equivalently, \(\mu\) extends uniquely to \(\mathcal C_1\), and this already exhausts \(\operatorname{Def}(C)^{\operatorname f}\).

promotion from semi-Fubini to full Fubini in the inductive step

witness-independence and compatibility across the chain \(\mathcal C_r\)

all fiberable sets are already in \(\mathcal C_1\)


Question: Let $\mathbf{M}$ be a one-sorted structure, let $C\subseteq M$ be a $0$-definable set with more than one element, and let $A$ be a commutative semiring. Let $\operatorname{Def}(C)$ be the full subcategory whose objects are the definable subsets of powers $C^n$ (with definable maps as morphisms). Define recursively classes $\mathcal C_0,\mathcal C_1,\dots$ by letting $\mathcal C_0=\operatorname{Def}(C)^{\mathrm{iso}}$ (the definable sets definably bijective to an object of $\operatorname{Def}(C)$), and for $r\ge 0$, letting $\mathcal C_{r+1}$ consist of those definable sets $X$ for which there is a definable map $f:X\to Y$ with $Y\in \mathcal C_r$ and with every fiber $f^{-1}(y)\in \mathcal C_r$; set $\operatorname{Def}(C)^{\mathrm f}=\bigcup_{r\ge 0}\mathcal C_r$. Assume that $C$ is stably embedded in $\mathbf M$, meaning that every definable subset of $C^n$ is definable with parameters from $C$. Suppose $\mu:\operatorname{Def}(C)\to A$ is an $A$-valued Fubini measure, i.e. $\mu(\emptyset)=0$, $\mu(\{x\})=1$ for singletons, $\mu(X\cup Y)=\mu(X)+\mu(Y)$ for disjoint definable $X,Y\subseteq C^m$, and for every definable map $f:X\to Y$ between objects of $\operatorname{Def}(C)$: the set $\{\mu(f^{-1}(y)):y\in Y\}$ is finite, each level set $Y_a=\{y\in Y:\mu(f^{-1}(y))=a\}$ is definable, and if all fibers have the same measure $a$, then $\mu(X)=a\mu(Y)$. Under these assumptions, which statement about extending $\mu$ is valid?

There exists exactly one $A$-valued Fubini measure $\widetilde\mu$ on $\operatorname{Def}(C)^{\mathrm f}$ whose restriction to $\operatorname{Def}(C)$ is $\mu$.

Choices:
B. There exists exactly one $A$-valued Fubini measure $\widetilde\mu$ on $\bigcup_{r\le 1}\mathcal C_r$ whose restriction to $\operatorname{Def}(C)$ is $\mu$, and every such extension automatically extends uniquely to all of $\operatorname{Def}(C)^{\mathrm f}$ without any further assumption on $C$.
C. There exists at most one $A$-valued Fubini measure $\widetilde\mu$ on $\operatorname{Def}(C)^{\mathrm f}$ whose restriction to $\operatorname{Def}(C)$ is $\mu$.
D. For every definable set $X\in \operatorname{Def}(C)^{\mathrm f}$ and every definable witness $f:X\to Y$ with $Y\in \mathcal C_r$ and all fibers $f^{-1}(y)\in \mathcal C_r$, the values $\mu\big(f^{-1}(y)\big)$ are definable in $y$ and therefore determine an extension $\widetilde\mu(X)$; hence there exists an $A$-valued Fubini measure on $\operatorname{Def}(C)^{\mathrm f}$ extending $\mu$, but it need not be unique.
E. There exists exactly one $A$-valued semi-Fubini measure $\widetilde\mu$ on $\operatorname{Def}(C)^{\mathrm f}$ whose restriction to $\operatorname{Def}(C)$ is $\mu$; moreover this extension always satisfies the full Fubini axiom on $\operatorname{Def}(C)^{\mathrm f}$ even if $C$ is not stably embedded in $\mathbf M$.

full Fubini versus semi-Fubini and dependence on stable embeddedness


[\cite{HooshKoh2025}]
A nontrivial finite group $G$ satisfies the strong CFS property if and only if $G$ is either
\begin{itemize}
  \item a cyclic group of prime order, or
  \item one of the four groups $C_4$, $C_2^2$, $C_2^3$, or $C_3^2$.
\end{itemize}

\label{section:Introduction}
Let $G$ be a finite group.  A subset $A \subseteq G$ is called a \textit{Lagrange subset} if $\lvert A\rvert$ divides $\lvert G\rvert$.  If there exists a subset $B \subseteq G$ such that every element $x\in G$ can be written \textit{uniquely} in the form
\[
x = ab,
\quad a\in A,\; b\in B,
\]
then $A$ is said to be a (left) \textit{factor} of $G$.  In this situation one immediately has
\[
\lvert G\rvert = \lvert A\rvert \,\lvert B\rvert,
\]
and both $A$ and $B$ are Lagrange subsets of $G$.

The problem of factoring a finite group $G$ into two (or more) Lagrange subsets goes back to Haj\'os’s
combinatorial proof of Minkowski’s conjecture on lattice tilings by unit cubes \cite{Hajos1941}.
In particular, Haj\'os\ showed that any lattice tiling of $\mathbb{R}^n$ by unit $n$-cubes must contain two
cubes sharing a full $(n-1)$-face, and he recast that argument in group‑theoretic terms to characterize exactly
which cyclic groups admit factorizations by arithmetic progressions \cite{Hajos1941}.
In the broader abelian setting, subsequent work of R\'edei \cite{Redei1952},
Sands \cite{Sands1974}, Szab\'o \cite{Szabo2004}, and others has connected these set‑factorizations with tiling problems, zero‑sum sequences, and
related combinatorial phenomena.  A good survey of these developments can be found in
\cite{Szabo-Sands2009}.

Bernstein \cite{Bernstein1968} extended Haj\'os’s theorem to the non‑abelian setting, showing that many of the
combinatorial tiling arguments carry over to arbitrary finite groups.  A related thread concerns the
existence of \textit{minimal logarithmic signatures} (MLS): conjecturally, every finite group $G$ admits
a factorization
\[
G = A_1 A_2 \cdots A_k
\]
where each $\lvert A_i\rvert$ is prime or equals $4$.
It is known that every finite solvable group has an
MLS, and that if a normal subgroup $K\lhd G$ and the quotient $G/K$ both admit MLS then so does $G$
\cite{Rahimipour2018}.  Consequently, any counterexample to the MLS conjecture must be a non-abelian
simple group, reducing the problem to the finite simple groups.

Hooshmand \cite[Problem 19.35]{Kourovka2018}
(see also \cite{Hooshmand2014,Banakh2018multifactorizable, Hooshmand2021})
asked whether for every factorization $\lvert G\rvert = m_1\cdots m_k$, there exist subsets
$A_i\subseteq G$ with $\lvert A_i\rvert=m_i$ and $G=A_1\cdots A_k$.
The case $k=2$ has been studied in \cite{Bildanov2020, Hooshmand2021}, where it is shown that a finite
group is $2$‑factorizable if and only if every finite simple group is $2$‑factorizable,
and all simple groups of order up to $10~000$ are $2$-factorizable \cite{Bildanov2020}.

For $k=3$, Bergman \cite{Bergman2020} proved that the alternating group $A_4$ is not $3$‑factorizable.
Brunault \cite{Banakh2018Factorizable} used computer methods to show that $A_5$ has no
$(2,3,5,2)$\nobreakdash-factorization (in fact, $A_5$ does not even admit
a $(2,15,2)$\nobreakdash-factorization; this was shown in \cite{Kab2021} without computer calculations).
More generally, it is proved in \cite{Kab2021} that for every integer $k\geq3$,
there exist integers $m_1,m_2,\ldots,m_k$ all greater than $1$
and a finite group of order $m_1m_2\ldots m_k$ that does not have any
$(m_1,m_2,\ldots,m_k)$\nobreakdash-factorization.
Among groups of order at most $100$, there are only eight groups that do not admit a $k$-factorization
for at least one $k\geq2$; all other groups are $k$-factorizable for every admissible $k$ \cite{Kab2021}.
It was also shown that the simple groups of orders $168$ and $360$ are $k$‑factorizable
for every admissible $k$ \cite{Kab2024}.

Recently, Chin, Wang, and Wong \cite{Chin-Wang-Wong} introduced the notion of \textit{complete
factorization}. Subsets $A_1, \ldots, A_k$ of $G$ form a complete factorization if they are pairwise
disjoint and every $g \in G$ is uniquely represented as $g = a_1 \cdots a_k$ with $a_i \in A_i$.
They proved that if $G$ is a finite abelian group and $|G|=m_1\ldots m_k$
where $m_1,\ldots,m_k$ are integers greater than $1$ and $k>2$, then
there exist subsets $A_1,\ldots,A_k$ of $G$ (with $|A_i|=m_i$ for all $i=1,2,\ldots,k$)
which form a complete factorization of the group $G$.
This result was extended to nilpotent groups in \cite{Kab2023}.

A finite group $G$ is said to have the \textit{strong CFS} (converse of factorization by subsets)
property if every Lagrange subset of $G$ is a factor of $G$. The notions of a Lagrange subset and a
strong CFS group were introduced in \cite{HooshKoh2025}, where the following was proved:

The problem of factoring a finite group $G$ into two (or more) Lagrange subsets goes back to Haj\'os’s
combinatorial proof of Minkowski’s conjecture on lattice tilings by unit cubes \cite{Hajos1941}.
In particular, Haj\'os\ showed that any lattice tiling of $\mathbb{R}^n$ by unit $n$-cubes must contain two
cubes sharing a full $(n-1)$-face, and he recast that argument in group‑theoretic terms to characterize exactly
which cyclic groups admit factorizations by arithmetic progressions \cite{Hajos1941}.
In the broader abelian setting, subsequent work of R\'edei \cite{Redei1952},
Sands \cite{Sands1974}, Szab\'o \cite{Szabo2004}, and others has connected these set‑factorizations with tiling problems, zero‑sum sequences, and
related combinatorial phenomena. A good survey of these developments can be found in
\cite{Szabo-Sands2009}.

Bernstein \cite{Bernstein1968} extended Haj\'os’s theorem to the non‑abelian setting, showing that many of the
combinatorial tiling arguments carry over to arbitrary finite groups. A related thread concerns the
existence of \textit{minimal logarithmic signatures} (MLS): conjecturally, every finite group $G$ admits
a factorization
\[
G = A_1 A_2 \cdots A_k
\]
where each $\lvert A_i\rvert$ is prime or equals $4$.
It is known that every finite solvable group has an
MLS, and that if a normal subgroup $K\lhd G$ and the quotient $G/K$ both admit MLS then so does $G$
\cite{Rahimipour2018}. Consequently, any counterexample to the MLS conjecture must be a non-abelian
simple group, reducing the problem to the finite simple groups.

Hooshmand \cite[Problem 19.35]{Kourovka2018}
(see also \cite{Hooshmand2014,Banakh2018multifactorizable, Hooshmand2021})
asked whether for every factorization $\lvert G\rvert = m_1\cdots m_k$, there exist subsets
$A_i\subseteq G$ with $\lvert A_i\rvert=m_i$ and $G=A_1\cdots A_k$.
The case $k=2$ has been studied in \cite{Bildanov2020, Hooshmand2021}, where it is shown that a finite
group is $2$‑factorizable if and only if every finite simple group is $2$‑factorizable,
and all simple groups of order up to $10~000$ are $2$-factorizable \cite{Bildanov2020}.

For $k=3$, Bergman \cite{Bergman2020} proved that the alternating group $A_4$ is not $3$‑factorizable.
Brunault \cite{Banakh2018Factorizable} used computer methods to show that $A_5$ has no
$(2,3,5,2)$\nobreakdash-factorization (in fact, $A_5$ does not even admit
a $(2,15,2)$\nobreakdash-factorization; this was shown in \cite{Kab2021} without computer calculations).
More generally, it is proved in \cite{Kab2021} that for every integer $k\geq3$,
there exist integers $m_1,m_2,\ldots,m_k$ all greater than $1$
and a finite group of order $m_1m_2\ldots m_k$ that does not have any
$(m_1,m_2,\ldots,m_k)$\nobreakdash-factorization.
Among groups of order at most $100$, there are only eight groups that do not admit a $k$-factorization
for at least one $k\geq2$; all other groups are $k$-factorizable for every admissible $k$ \cite{Kab2021}.
It was also shown that the simple groups of orders $168$ and $360$ are $k$‑factorizable
for every admissible $k$ \cite{Kab2024}.

Recently, Chin, Wang, and Wong \cite{Chin-Wang-Wong} introduced the notion of \textit{complete
factorization}. Subsets $A_1, \ldots, A_k$ of $G$ form a complete factorization if they are pairwise
disjoint and every $g \in G$ is uniquely represented as $g = a_1 \cdots a_k$ with $a_i \in A_i$.
They proved that if $G$ is a finite abelian group and $|G|=m_1\ldots m_k$
where $m_1,\ldots,m_k$ are integers greater than $1$ and $k>2$, then
there exist subsets $A_1,\ldots,A_k$ of $G$ (with $|A_i|=m_i$ for all $i=1,2,\ldots,k$)
which form a complete factorization of the group $G$.
This result was extended to nilpotent groups in \cite{Kab2023}.

A finite group $G$ is said to have the \textit{strong CFS} (converse of factorization by subsets)
property if every Lagrange subset of $G$ is a factor of $G$. The notions of a Lagrange subset and a
strong CFS group were introduced in \cite{HooshKoh2025}, where the following was proved:

\begin{document}
\maketitle
\begin{abstract}
In a finite group, a subset is called a Lagrange subset if its size divides the group order, and a factor if it admits a complementary subset. We provide a new and comparatively direct proof of the classification of groups in which every Lagrange subset is a factor. We show that any nontrivial such group must be a cyclic group of prime order, the cyclic group of order 4, or an elementary abelian group of order 4, 8, or 9.
\end{abstract}

Hooshmand \cite[Problem 19.35]{Kourovka2018}
(see also \cite{Hooshmand2014,Banakh2018multifactorizable, Hooshmand2021})
asked whether for every factorization $\lvert G\rvert = m_1\cdots m_k$, there exist subsets
$A_i\subseteq G$ with $\lvert A_i\rvert=m_i$ and $G=A_1\cdots A_k$.
The case $k=2$ has been studied in \cite{Bildanov2020, Hooshmand2021}, where it is shown that a finite
group is $2$‑factorizable if and only if every finite simple group is $2$‑factorizable,
and all simple groups of order up to $10~000$ are $2$-factorizable \cite{Bildanov2020}.

For $k=3$, Bergman \cite{Bergman2020} proved that the alternating group $A_4$ is not $3$‑factorizable.
Brunault \cite{Banakh2018Factorizable} used computer methods to show that $A_5$ has no
$(2,3,5,2)$\nobreakdash-factorization (in fact, $A_5$ does not even admit
a $(2,15,2)$\nobreakdash-factorization; this was shown in \cite{Kab2021} without computer calculations).
More generally, it is proved in \cite{Kab2021} that for every integer $k\geq3$,
there exist integers $m_1,m_2,\ldots,m_k$ all greater than $1$
and a finite group of order $m_1m_2\ldots m_k$ that does not have any
$(m_1,m_2,\ldots,m_k)$\nobreakdash-factorization.
Among groups of order at most $100$, there are only eight groups that do not admit a $k$-factorization
for at least one $k\geq2$; all other groups are $k$-factorizable for every admissible $k$ \cite{Kab2021}.
It was also shown that the simple groups of orders $168$ and $360$ are $k$‑factorizable
for every admissible $k$ \cite{Kab2024}.

A finite group $G$ is said to have the \textit{strong CFS} (converse of factorization by subsets)
property if every Lagrange subset of $G$ is a factor of $G$. The notions of a Lagrange subset and a
strong CFS group were introduced in \cite{HooshKoh2025}, where the following was proved:

The authors of \cite{HooshKoh2025} used Cayley graph arguments to constrain the possible orders of $G$,
reducing the classification to an examination of about twenty concrete groups. In this paper, we offer a
streamlined proof of the "only if" direction, which breaks down into four independent steps:
\begin{enumerate}[label=(\arabic*)]
 \item Analysis of groups with "large" proper subgroups (Lemma \ref{lemma:main});
 \item Groups of even order containing elements of odd order (Lemma \ref{lemma:small subset});
 \item Investigation of groups of order $8$ (Lemma \ref{lemma:groups of order 8});
 \item A direct check for the three remaining groups: $D_4$, $C_8$, and $C_9$.
\end{enumerate}

\begin{lemma}
\label{lemma:small subset}
\begin{enumerate}[label=\textup{(\roman*)}]\!
 \item Let $G$ be a group of even order, and let $a\in G$, $a\neq e$.
 The subset $\{e,a\}$ is a factor of $G$ if and only if the order of $a$ is even.
 \item Let $G$ be an elementary abelian $2$-group.
 Every subset of size $4$ is a factor of $G$.
 \item\,Let $G$ be an elementary abelian $3$-group.
 Every subset of size $3$ is a factor of $G$.
\end{enumerate}
\end{lemma}
\begin{proof}[Proof of \textup{(i)}]
First suppose that the order of $a$ is even, say $|a| = 2m$ for some positive integer $m$.
Let $H = \langle a\rangle$.
Set
\[
B = \{e,a^2,\ldots,a^{2m-2}\}.
\]
Clearly $H = \{e,a\} B$, and it is straightforward to check that
$B \cap aB = \varnothing$.
Thus $\{e,a\}$ is a factor of $H$.
By Lemma~\ref{lemma:subset of subgroup}, $\{e,a\}$ is then a factor of $G$.

\begin{lemma}
\label{lemma:groups of order 8}
Let $G$ be a group of order $8$, and let $a,b\in G$ be two elements of order $4$ with $\langle a\rangle\neq\langle b\rangle$.
Then
\[
 A = \{a,a^2,a^3,b\}
\]
is not a left factor of $G$.
\end{lemma}
\begin{proof}
 Suppose to the contrary that there is an element $x\in G$ such that
 \[
 G=A\cup Ax \text{ and } A\cap Ax=\varnothing.
 \]
 Then $e\in Ax$ and hence $x=a^{-i}$ for some $i\in\{1,2,3\}$, or $x=b^{-1}$.
 If $x=a^{-i}$, then $a\in A\cap A x$, contradicting disjointness.
 If $x=b^{-1}=b^3$, then
 \[
 Ax=\{ab^3,a^2b^3,a^3b^3,e\}.
 \]
 Since $b^3\notin A$ and also $b^3\notin A x$,
 the element $b^3$ is not contained in $A\cup A x=G$, a clear contradiction.
\end{proof}

First, suppose that $|G|$ is even. By Lemma~\ref{lemma:small subset}\,(i),
every non-identity element of $G$ must have even order.
It follows that $G$ is a $2$-group.
Since $G$ cannot have a proper subgroup of order greater than $4$ (by Lemma~\ref{lemma:main}),
the order of $G$ must be $4$ or $8$.
If $|G|=4$, $G$ is either $C_4$ or $C_2^2$, both of which appear in the theorem's list.
If $|G|=8$, Lemma~\ref{lemma:groups of order 8} implies that $G$ cannot contain
two distinct cyclic subgroups of order $4$.
This eliminates the quaternion group $Q_8$ and the product $C_4 \times C_2$.
Since $C_2^3$ is a solution, the only remaining candidates to check are $D_4$ and $C_8$.

\begin{lemma}
\label{lemma:groups of order 8}
Let $G$ be a group of order $8$, and let $a,b\in G$ be two elements of order $4$ with $\langle a\rangle\neq\langle b\rangle$.
Then
\[
  A = \{a,a^2,a^3,b\}
\]
is not a left factor of $G$.
\end{lemma}

\begin{lemma}
\label{lemma:main}
Let $H$ be a proper subgroup of a finite group $G$ with $\lvert H\rvert \ge 5$.
Choose $h_0\in H$ with $h_0\neq e$ and choose $g\in G\setminus H$.
Let $h_1$ be an element of $H$ not in the set
$\{e,h_0,h_0^{-1}\}$,
and set
\[
H'= H\setminus\{e,h_0\}.
\]
Define
\begin{equation}\label{equ:subset-not-factor}
A= H'\cup\{g,h_1g\}.
\end{equation}
Then $\lvert A\rvert = \lvert H\rvert$ and $A$ is not a left factor of $G$.
\end{lemma}

\begin{lemma}
\label{lemma:small subset}
\begin{enumerate}[label=\textup{(\roman*)}]\!
    \item Let $G$ be a group of even order, and let $a\in G$, $a\neq e$.
    The subset $\{e,a\}$ is a factor of $G$ if and only if the order of $a$ is even.
    \item Let $G$ be an elementary abelian $2$-group.
    Every subset of size $4$ is a factor of $G$.
    \item\,Let $G$ be an elementary abelian $3$-group.
    Every subset of size $3$ is a factor of $G$.
\end{enumerate}
\end{lemma}

The authors of \cite{HooshKoh2025} used Cayley graph arguments to constrain the possible orders of $G$,
reducing the classification to an examination of about twenty concrete groups. In this paper, we offer a
streamlined proof of the "only if" direction, which breaks down into four independent steps:
\begin{enumerate}[label=(\arabic*)]
  \item Analysis of groups with "large" proper subgroups (Lemma \ref{lemma:main});
  \item Groups of even order containing elements of odd order (Lemma \ref{lemma:small subset});
  \item Investigation of groups of order $8$ (Lemma \ref{lemma:groups of order 8});
  \item A direct check for the three remaining groups: $D_4$, $C_8$, and $C_9$.
\end{enumerate}

\vspace{1ex}
\noindent\textbf{Notation.} Throughout, $G$ is a finite group with identity $e$. For any $A \subseteq G$ and $x \in G$, we denote $Ax = \{ax \mid a \in A\}$. We write $\langle g \rangle$ for the cyclic subgroup generated by $g$. A subset $A$ is a \textit{left factor} if $G = \bigsqcup_{b \in B} Ab$ for some $B \subseteq G$.

With these conventions in place, we turn in Section \ref{section:lemmas} to our lemmas,
and then in Section \ref{section:proof theorem} to the completion of the proof.

The post-theorem introduction explains that \cite{HooshKoh2025} used Cayley graph arguments to constrain the possible orders of $G$, reducing the classification to examining about twenty concrete groups. It then sketches a streamlined proof of the \emph{``only if''} direction, broken into four independent steps:
\begin{enumerate}[label=(\arabic*)]
  \item \emph{Analysis of groups with ``large'' proper subgroups} (Lemma \ref{lemma:main});
  \item \emph{Groups of even order containing elements of odd order} (Lemma \ref{lemma:small subset});
  \item \emph{Investigation of groups of order $8$} (Lemma \ref{lemma:groups of order 8});
  \item \emph{A direct check} for the three remaining groups: $D_4$, $C_8$, and $C_9$.
\end{enumerate}
It further indicates the paper proceeds by proving these lemmas in Section \ref{section:lemmas} and then completing the proof in Section \ref{section:proof theorem}.

The post-theorem introduction explains that \cite{HooshKoh2025} used Cayley graph arguments to constrain the possible orders of $G$, reducing the classification to examining about twenty concrete groups. It then sketches a streamlined proof of the \emph{``only if''} direction, broken into four independent steps:
\begin{enumerate}[label=(\arabic*)]
  \item \emph{Analysis of groups with ``large'' proper subgroups}. We first prove the following lemma.

\begin{lemma}
\label{lemma:main}
Let $H$ be a proper subgroup of a finite group $G$ with $\lvert H\rvert \ge 5$.
Choose $h_0\in H$ with $h_0\neq e$ and choose $g\in G\setminus H$.
Let $h_1$ be an element of $H$ not in the set
$\{e,h_0,h_0^{-1}\}$,
and set
\[
H'= H\setminus\{e,h_0\}.
\]
Define
\begin{equation}\label{equ:subset-not-factor}
A= H'\cup\{g,h_1g\}.
\end{equation}
Then $\lvert A\rvert = \lvert H\rvert$ and $A$ is not a left factor of $G$.
\end{lemma}

  \item \emph{Groups of even order containing elements of odd order}. We next prove the following lemma.

\begin{lemma}
\label{lemma:small subset}
\begin{enumerate}[label=\textup{(\roman*)}]\!
    \item Let $G$ be a group of even order, and let $a\in G$, $a\neq e$.
    The subset $\{e,a\}$ is a factor of $G$ if and only if the order of $a$ is even.
    \item Let $G$ be an elementary abelian $2$-group.
    Every subset of size $4$ is a factor of $G$.
    \item\,Let $G$ be an elementary abelian $3$-group.
    Every subset of size $3$ is a factor of $G$.
\end{enumerate}
\end{lemma}

  \item \emph{Investigation of groups of order $8$}. We then prove the following lemma.

\begin{lemma}
\label{lemma:groups of order 8}
Let $G$ be a group of order $8$, and let $a,b\in G$ be two elements of order $4$ with $\langle a\rangle\neq\langle b\rangle$.
Then
\[
  A = \{a,a^2,a^3,b\}
\]
is not a left factor of $G$.
\end{lemma}

  \item \emph{A direct check} for the three remaining groups: $D_4$, $C_8$, and $C_9$.
\end{enumerate}
It further indicates the paper proceeds by proving these lemmas next and then completing the proof later.

Biconditional or Equivalence
Classification or Bijection

Question: Let $G$ be a nontrivial finite group. A subset $A\subseteq G$ is called a Lagrange subset if $|A|$ divides $|G|$. Call $A$ a factor of $G$ if there exists a subset $B\subseteq G$ such that the multiplication map $A\times B\to G$, $(a,b)\mapsto ab$, is bijective. The group $G$ is said to have the strong CFS property if every Lagrange subset of $G$ is a factor of $G$. Which finite groups $G$ have the strong CFS property?

Exactly the cyclic groups of prime order, together with the four groups $C_4$, $C_2^2$, $C_2^3$, and $C_3^2$.

Choices:
B. Exactly the cyclic groups of prime order, together with the five groups $C_4$, $C_2^2$, $C_2^3$, $C_3^2$, and $C_9$.
C. At least the cyclic groups of prime order and the four groups $C_4$, $C_2^2$, $C_2^3$, and $C_3^2$ have the strong CFS property.
D. Exactly the finite abelian groups in which every nontrivial proper divisor of $|G|$ is $2$, $3$, or a prime, namely the cyclic groups of prime order together with $C_4$, $C_2^2$, $C_2^3$, $C_3^2$, and $C_9$.
E. Exactly the cyclic groups of prime order, together with the groups $C_4$, $C_2^2$, $C_2^3$, $C_3^2$, and all groups of order $8$ having a unique cyclic subgroup of order $4$.

dropped the 'only if' classification/exhaustiveness

confuses the specific classified groups with a broader abelian/order-divisibility condition

misuses the order-8 lemma by turning elimination of groups with two distinct order-4 cyclic subgroups into a positive classification criterion


Question: Let $G$ be a nontrivial finite group. Call a subset $A\subseteq G$ a Lagrange subset if $|A|$ divides $|G|$, and call $A$ a factor of $G$ if there exists a subset $B\subseteq G$ such that every element of $G$ can be written uniquely as $ab$ with $a\in A$ and $b\in B$. The group $G$ is said to have the strong CFS property if every Lagrange subset of $G$ is a factor of $G$. Which of the following statements is equivalent to $G$ having the strong CFS property? (Here $C_n$ denotes the cyclic group of order $n$, and $C_2^2$, $C_2^3$, $C_3^2$ denote elementary abelian groups of orders $4$, $8$, and $9$, respectively.)

$G$ is either a cyclic group of prime order or one of the four groups $C_4$, $C_2^2$, $C_2^3$, or $C_3^2$.

Choices:
B. $G$ is either a cyclic group of prime order or one of the five groups $C_4$, $C_2^2$, $C_2^3$, $C_3^2$, or $C_9$.
C. $G$ is either a cyclic group of prime order or an abelian group of order $4$, $8$, or $9$.
D. $G$ is either a cyclic group of prime order, a cyclic group of order $4$, or an elementary abelian group of order $2^n$ or $3^2$.
E. $G$ is either a cyclic group of prime order or one of the groups $C_4$, $C_2^2$, $D_4$, $C_2^3$, or $C_3^2$.

replace the exact order-$4,8,9$ abelian list by the broader abelian-order condition

small-subset lemmas for elementary abelian groups do not extend to all elementary abelian $2$-groups

direct check rules out $D_4$ among the final order-$8$ candidates