\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 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. 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