Localizable locally determined measurable spaces with
negligibles]Localizable locally determined
measurable spaces with

On leave of absence from

Ph. Bouafia]Philippe Bouafia

Fédération de Mathématiques FR3487

3 rue Joliot Curie

91190 Gif-sur-Yvette

Th. De Pauw]Thierry De Pauw

School of Mathematical Sciences

Shanghai Key Laboratory of PMMP

East China Normal University

500 Dongchuang Road

Shanghai 200062

P.R. of China

and NYU-ECNU Institute of Mathematical Sciences at NYU Shanghai

3663 Zhongshan Road North

Shanghai 200062

Université Paris Diderot

Sorbonne Université

Institut de Mathématiques de Jussieu – Paris Rive Gauche, IMJ-PRG

F-75013, Paris

[2020]Primary 28A15; Secondary 28A75,28A05

The first author was partially supported by the Science and Technology Commission of Shanghai (No. 18dz2271000).

We study measurable spaces equipped with a
$\sigma$-ideal of negligible sets. We find conditions under which
they admit a localizable locally determined version – a kind of fiber
space that describes locally their directions – defined by a universal property in an appropriate category that we introduce. These methods allow to promote each measure space $(X, \calA,
\mu)$ to a strictly localizable version $(\hat{X}, \hat{\calA},
\hat{\mu})$, so that the dual of $\bL_1(X, \calA, \mu)$ is
$\bL_\infty(\hat{X}, \hat{\calA}, \hat{\mu})$. Corresponding to this
duality is a generalized Radon-Nikodým theorem. We also provide a
characterization of the strictly localizable version in special
cases that include integral geometric measures, when the negligibles are the purely unrectifiable sets in a given dimension.

§ FOREWORD

The Radon-Nikodým Theorem does not hold for every measure space

One way to phrase this precisely is to consider the canonical

\begin{equation*}
\Upsilon : \bL_\infty(X,\calA,\mu) \to \bL_1(X,\calA,\mu)^*.
\end{equation*}

The following hold.

(A) $\Upsilon$ is injective (this corresponds to the uniqueness
almost everywhere of Radon-Nikodým derivatives) if and only if
$(X,\calA,\mu)$ is semi-finite.

(B) $\Upsilon$ is surjective (this corresponds to the existence
of Radon-Nikodým derivatives) if and only if the Boolean algebra
$\calA / \calN_{\mu,\rmloc}$ is Dedekind complete (i.e. order
complete as a lattice).

While (A) is classical, see e.g. <cit.>, (B) is
recent and due to the second author, see <cit.>.

Let us recall the relevant definitions.

Given a measure space $(X,\calA,\mu)$, we abbreviate $\calA^f \defeq
\calA \cap \{ A : \mu(A) < \infty \}$ and $\calN_\mu \defeq \calA \cap
\{ N : \mu(N) = 0 \}$.

We say that $(X,\calA,\mu)$ is semi-finite if every $A \in
\calA$ of infinite measure contains some $F \in \calA^f \setminus
\calN_\mu$.

Equivalently, $\mu(A) = \sup \{ \mu(F) : A \supset F \in \calA^f \}$.

We further define the $\sigma$-ideal of locally $\mu$-null sets
as follows: $\calN_{\mu,\rmloc} \defeq \calA \cap \{ A : A \cap F \in
\calN_\mu \text{ for all } F \in \calA^f \}$.

It is easy to see <cit.> that $(X,\calA,\mu)$ is
semi-finite if and only if $\calN_{\mu,\rmloc}=\calN_\mu$.

Thus we obtain the following classical criterion, <cit.>.

(C) $\Upsilon$ is an isometric isomorphism if and only if
$(X,\calA,\mu)$ is semi-finite and the Boolean algebra $\calA /
\calN_\mu$ is Dedekind complete.

Though semi-finiteness is a natural property, Caratheodory's method
does not always provide it.

For instance, the measure spaces $(\R^2,\calA_{\calH^1},\calH^1)$ and
$(\R^2,\calB(\R^2),\calI^1_\infty)$ are not semi-finite – see
<cit.> and <cit.>.

Here, $\calH^1$ is the 1-dimensional Hausdorff measure in the
Euclidean plane <cit.>, $\calA_{\calH^1}$ is the
$\sigma$-algebra consisting of $\calH^1$-measurable sets in
Caratheodory's sense, $\calI^1_\infty$ is a 1-dimensional integral
geometric measure <cit.>, and $\calB(\R^2)$ is the
$\sigma$-algebra whose members are the Borel subsets of $\R^2$.

Both $\calH^1(\Gamma)$ and $\calI^1_\infty(\Gamma)$ coincide with the
usual Euclidean length of $\Gamma$ when this is a Lipschitz curve.

It is natural to want to associate, with an arbitrary
$(X,\calA,\mu)$, an improved version of itself – in a universal way
– ideally one for which the Radon-Nikodým Theorem holds.

This is one of our several achievements in this paper.

It is not difficult to modify slightly the measure $\mu$, keeping the
underlying measurable space $(X,\calA)$ untouched, in order to make it

Specifically, letting $\mu_{\mathrm{sf}}(A) = \sup \{ \mu(A \cap F) :
F \in \calA^f \}$, for $A \in \calA$, one checks that
$(X,\calA,\mu_{\mathrm{sf}})$ is semi-finite and that
$\calN_{\mu_{\mathrm{sf}}} = \calN_{\mu,\rmloc}$.

However, it appears to be a more delicate task to modify
$(X,\calA,\mu)$ in a canonical way in order for $\Upsilon$ to become

An idea for testing whether $\Upsilon$ is surjective is as

Given $\alpha \in \bL_1(X,\calA,\mu)^*$ we apply the Radon-Nikodým
Theorem “locally”, as it is valid on each finite measure subspace
$(F,\calA_F,\mu_F)$, $F \in \calA^f$, i.e. we represent by integration
the functional $\alpha \circ \iota_F \in \bL_1(F,\calA_F,\mu_F)^*$,
where $\iota_F \colon \bL_1(F, \calA_F, \mu_F) \to \bL_1(X, \calA,
\mu)$ is the obvious map.

This produces a family of Radon-Nikodým derivatives $\la f_F \ra_{F
\in \calA^f}$.

By the almost everywhere uniqueness of Radon-Nikodým derivatives in
finite measure spaces, this is a compatible family in the sense
that $F \cap F' \cap \{ f_F \neq f_{F'} \} \in \calN_\mu$, for every
$F,F' \in \calA^f$.

In order to obtain a globally defined Radon-Nikodým derivative, one
ought to be able to “glue” together the functions of this family.

A gluing of $\la f_F \ra_{F \in \calA^f}$ is, by definition, an
$\calA$-measurable function $f \colon X \to \R$ such that $F \cap \{ f
\neq f_F \} \in \calN_\mu$ for every $F \in \calA^f$.

The question whether such a gluing exists takes us away from the
realm of measure spaces, as it rather pertains to measurable
spaces with negligibles, abbreviated MSNs, i.e. triples
$(X,\calA,\calN)$ where $(X,\calA)$ is a measurable space and $\calN
\subset \calA$ is a $\sigma$-ideal.

The notion of compatible family $\la f_E \ra_{E \in \calE}$ of
$\calA_E$-measurable functions $E \to \R$ subordinated to an arbitrary
collection $\calE \subset \calA$ readily makes sense in this more
general setting, as does the notion of gluing of such a compatible

That each compatible family of partially defined measurable functions
admits a gluing is equivalent to the Boolean algebra $\calA / \calN$
being Dedekind complete.

In this case we say that $(X,\calA,\calN)$ is localizable.

Equivalently, $(X,\calA,\calN)$ is localizable if and only if every
collection $\calE \subset \calA$ admits an $\calN$-essential supremum
(see <ref> for a definition), which corresponds to taking
an actual supremum in the Boolean algebra $\calA / \calN$.

For a proof of these classical equivalences, see
e.g. <cit.>.

It will be convenient to call $\calN$-generating a collection
$\calE \subset \calA$ that admits $X$ as an $\calN$-essential

For instance, one easily checks that if $(X,\calA,\mu)$ is a
semi-finite measure space, then $\calA^f$ is $\calN_\mu$-generating.

Let $(X,\calA,\calN)$ be a localizable MSN, $\calE \subset \calA$
be $\calN$-generating, and $\la f_E \ra_{E \in \calE}$ be a compatible
family of partially defined measurable functions.

The problem of gluing this compatible family in our setting is
reminiscent of the fact that, for a topological space $X$, the functor
of continuous functions on open sets is a sheaf.

However, unlike in the case of continuous functions, in order to
define $f$ globally, we ought to make choices on the domains $E \cap
E'$, for $E,E' \in \calE$, because $f_E$ and $f_{E'}$ do not coincide
everywhere there, but merely almost everywhere.

In an attempt to avoid the issue, one can replace $\calE$ with an
almost disjointed refinement of itself, say $\calF$.

By this we mean that each member of $\calF$ is contained in a member
of $\calE$, that $\calF$ is $\calN$-generating, and that $F \cap F'
\in \calN$ whenever $F,F' \in \calF$ are distinct.

The existence of $\calF$ follows from Zorn's Lemma, see <ref>.

Still, $F \cap F'$ may not be empty whenever $F,F' \in \calF$ are
distinct and we are again in a position to make choices.

A step further along the road would be to produce from $\calF$ a
disjointed family $\calG$ whose union is conegligible.

From classical measure theory, we learn of two situations when this is

First, in the presence of a lower density of $(X,\calA,\calN)$ (see
<ref> for a definition), and second when $\rmcard \calF \leq
\mathfrak{c}$ (see the proof of <ref>).

In those cases, a gluing exists.

In fact, in the context of measure spaces, the existence of a lower
density yields a somewhat stronger structure than localizability.

In order to state this, we need one more definition.

We say that a measure space $(X,\calA,\mu)$ is locally
determined if it is semi-finite and if the following holds:

\begin{equation*}
\forall A \subset X : \big[ \forall F \in \calA^f : A \cap F \in
\calA \big] \Rightarrow A \in \calA.
\end{equation*}

A complete locally determined measure space $(X,\calA,\mu)$ admits a
lower density if and only if it is strictly localizable, which means,
by definition, that there exists a partition $\calG \subset \calA$ of
$X$ such that $\calA = \calP(X) \cap \{ A : A \cap G \in \calA \text{
for all } G \in \calG \}$ and $\mu(A) = \sum_{G \in \calG} \mu(A
\cap G)$, for $A \in \calA$.

See <cit.> for a proof.

The existence of a lower density for a strictly localizable measure
space follows from the case of finite measure spaces by gluing, and
the case of finite measure spaces is a consequence of a martingale
convergence theorem.

Even though the notion of a lower density makes sense for MSNs, their
existence does not hold for even the most natural generalization of
finite measure spaces, namely ccc MSNs (satisfying the countable chain
condition, <ref> and <ref>), see <cit.>.

Both notions of localizability (of an MSN) and local
determination (of a measure space) seem to express in different ways
the fact that “there are enough measurable sets”.

For instance, one easily checks that an MSN $(X,\calA,\{\emptyset\})$,
such that $\calA$ contains all singletons, is localizable if and only
if $\calA = \calP(X)$.

Thus, given an arbitrary MSN $(X,\calA,\calN)$, one may naively
attempt to “add measurable sets” in a smart way in order to obtain a
localizable MSN $(X,\hat{\calA},\hat{\calN})$, just as many as needed,
and that would be a “localizable version” of $(X,\calA,\calN)$.

Unfortunately, within ZFC this cannot always be done while “sticking
in the base space $X$”, as shown by the following, quoted from

Assume that:

(1) $C \subset [0,1]$ is some Cantor set of Hausdorff dimension 0;

(2) $X = C \times [0,1]$;

(3) $\calA$ is a $\sigma$-algebra such that $\calB(X) \subset
\calA \subset \calP(X)$;

(4) $\calN = \calN_{\calH^1}$ or $\calN = \calN_{\rmpu}$.

Then $(X,\calA,\calN)$ is consistently not localizable.

Here, $\calN_{\rmpu}$ consists of those subsets $S$ of $X$ that are purely 1-unrectifiable, i.e. $\calH^1(S \cap \Gamma) = 0$ for every
Lipschitz (or, for that matter, $C^1$) curve $\Gamma \subset \R^2$.

Thus, one may need to also add points to the base space $X$ and, in
particular cases such as the one above, we give a very specific way of
doing so, in the last section of this paper.

In the case of a general measure space $(X,\calA,\mu)$, we can get a
feeling of what needs to be done, when trying to define the gluing of
a compatible family $\la f_F \ra_{F \in \calA^f}$.

Indeed, each $x \in X$ may belong to several $F \in \calA^f$ and this
calls for considering an appropriate quotient of the fiber bundle
$\{(x,F) : x \in F \in \calA^f\}$.  One of the tasks that we
assign ourselves in this paper is to define a general notion of
“localization” of an MSN and to prove existence results in some

Since a definition of “localization” will involve a universal
property, it is critical to determine which category is appropriate
for our purposes.

As this offers unexpected surprises, we describe the several steps in
some detail.

The objects of our first category $\sfMSN$ are the saturated MSN
$(X,\calA,\calN)$, by what we mean that for every $N,N' \subset X$, if
$N \subset N'$ and $N' \in \calN$, then $N \in \calN$.

This is in analogy with the notion of a complete measure space.

In order to define the morphisms between two objects $(X,\calA,\calN)$
and $(Y,\calB,\calM)$, we say that a map $f \colon X \to Y$ is
$[(\calA,\calN),(\calB,\calM)]$-measurable if $f^{-1}(B) \in \calA$
for every $B \in \calB$ and $f^{-1}(M) \in \calN$ for every $M \in
\calM$.

For instance, if $X$ is a Polish space and $\mu$ is a diffuse
probability measure on $X$, there exists <cit.> a
Borel isomorphism $f \colon X \to [0,1]$ such that $f_\# \mu = \calL^1$,
where $\calL^1$ is the Lebesgue measure, thus $f$ is

We define an equivalence relation for such measurable maps $f,f' \colon X
\to Y$ by saying that $f \sim f'$ if and only if $\{ f \neq f' \} \in
\calN$.

The morphisms in the category $\sfMSN$ between the objects
$(X,\calA,\calN)$ and $(Y,\calB,\calM)$ are the equivalence classes of
$[(\calA,\calN),(\calB,\calM)]$-measurable maps.

At this stage, we need to suppose that $(X, \calA, \calN)$ is
saturated for the relation of equality almost everywhere to be
transitive <ref>. With this assumption, the composition of
measurable maps is also compatible with $\sim$, see <ref>.

We let $\sfLOC$ be the full subcategory of $\sfMSN$ whose objects
are the localizable MSNs.

We may be tempted to define the localization of a saturated MSN
$(X,\calA,\calN)$ as its coreflection (if it exists) along the
forgetful functor $\sfForget \colon \sfLOC \to \sfMSN$, and the question of
existence in general becomes that of the existence of a right adjoint
to $\sfForget$.

Specifically, we may want to say that a pair
$[(\hat{X},\hat{\calA},\hat{\calN}),\bp]$, where
$(\hat{X},\hat{\calA},\hat{\calN})$ is saturated localizable MSN and
$\bp$ is a morphism $\hat{X} \to X$, is a localization of
$(X,\calA,\calN)$ whenever the following universal property holds.

For every pair $[(Y,\calB,\calM),\bq]$, where $(Y,\calB,\calM)$ is a
saturated localizable MSN and $\bq$ is a morphism $Y \to X$, there
exists a unique morphism $\br \colon Y \to \hat{X}$ such that $\bq = \bp
\circ \br$.

\begin{equation}
\label{diagram}
\begin{tikzcd}
(Y, \calB, \calM) \arrow[rd, swap, "\bq"] \arrow[rr, dotted,
"\exists!\br"] & & (\hat{X}, \hat{\calA}, \hat{\calN})
\arrow[ld, "\bp"] \\ & (X, \calA, \calN) &
\end{tikzcd} \tag{$\bigtriangledown$}
\end{equation}

However, we now illustrate that the notion of morphism defined so
far is not yet the appropriate one that we are after.

We consider the MSN $(X,\calA,\{\emptyset\})$ where $X = \R$ and
$\calA$ is the $\sigma$-algebra of Lebesgue measurable subsets of

We recall that we want the localization of $(X,\calA,\{\emptyset\})$
to be $[(X,\calP(X),\{\emptyset\}),\bp]$ with $\bp$ induced by the
identity $\rmid_X$.

Assume if possible that this is the case.

In the diagram above we consider $(Y, \calB,\calM) =
(X,\calA,\calN_{\calL^1})$ and $\bq$ induced by the identity.

Note that this is, indeed, a localizable MSN since it is associated
with a $\sigma$-finite measure space (see <ref> and

Thus, there would exist a morphism $\br$ in $\sfMSN$ such that $\bp
\circ \br = \bq$.

Picking $r \in \br$, this implies that $X \cap \{ x: r(x) \neq x \}$
is Lebesgue negligible.

The measurability of $r$ would then imply that $r^{-1}(S) \in \calA$
for every $S \in \calP(X)$, contradicting the existence of non
Lebesgue measurable subsets of $\R$.

The problem with the example above is that the objects
$(X,\calA,\calN_{\calL^1})$ and $(X,\calA,\{\emptyset\})$ should not
be compared, in other words that $\bq$ should not be a legitimate

We say that a morphism $\mathbf{f} \colon (X,\calA,\calN) \to
(Y,\calB,\calM)$ of the category $\sfMSN$ is supremum preserving
if the following holds for (one, and therefore every) $f
\in\mathbf{f}$.

If $\calF \subset \calB$ admits an $\calM$-essential supremum $S \in
\calB$, then $f^{-1}(S)$ is an $\calN$-essential supremum of

It is easy to see that adding this condition to the definition of
morphism rules out the $\bq$ considered in the preceding paragraph.

We define the category $\sfMSNsp$ to be that whose objects are the
saturated MSNs and whose morphisms are those morphisms of $\sfMSN$
that are supremum preserving.

We define similarly $\sfLOCsp$.

We now define the localizable version (if it exists) of a
saturated MSN with the similar universal property illustrated in
(<ref>), except for we now require all morphisms to be in
$\sfMSNsp$, i.e. supremum preserving.

In other words, it is a coreflection of an object of $\sfMSNsp$ along
$\sfForget \colon \sfLOCsp \to \sfMSNsp$.

Unfortunately, this is not quite yet the right setting.

Indeed, we show in <ref> that if $X$ is uncountable
and $\calC(X)$ is the countable-cocountable $\sigma$-algebra of $X$,
then $[(X,\calP(X),\{\emptyset\}),\biota]$ (with $\biota$ induced by
$\rmid_X$) is not the localizable version of

This prompts us to introduce a new category.

We say that an object $(X,\calA,\calN)$ of $\sfMSN$ is locally determined if for every $\calN$-generating collection
$\calE \subset \calA$ the following holds:

\begin{equation*}
\forall A \subset X : \big[ \forall E \in \calE : A \cap E \in \calA
\big] \Rightarrow A \in \calA.
\end{equation*}

In case $(X,\calA,\calN)$ is the MSN associated with some complete
semi-finite measure space $(X,\calA,\mu)$, then it is locally
determined (in the sense of MSNs) if and only if $(X,\calA,\mu)$ is
locally determined (in the sense of measure spaces) – see
<ref>(F) – even though the latter sounds stronger because we
test with any generating family $\calE$.

We say that an object of $\sfMSN$ is lld if it is both
localizable and locally determined, and we let $\sfLLDsp$ be the
corresponding full subcategory of $\sfLOCsp$.

We now define the lld version of an object of $\sfMSNsp$ to be
its coreflection (if it exists) along $\sfForget \colon \sfLLDsp \to
\sfMSNsp$, i.e. it satisfies the corresponding universal property
illustrated in (<ref>) with $Y$ and $\hat{X}$ being lld, and
the morphisms being supremum preserving.

This definition is satisfactory in at least the simplest case,
<ref> : If $(X,\calA,\{\emptyset\})$ is so that $\calA$
contains all singletons, then it admits
$[(X,\calP(X),\{\emptyset\}),\biota]$ as its lld version.

Our general question has now become whether $\sfForget : \sfLLDsp
\to \sfMSNsp$ admits a right adjoint.

Freyd's Adjoint Functor Theorem <cit.> could prove
useful, however do not know whether it applies, mostly because we do
not know whether coequalizers exist in $\sfMSNsp$.

We gather in Table <ref> the information that we know about
limits and colimits in the three categories we introduced.

$\sfMSNsp$     $\sfLOCsp$     $\sfLLDsp$
equalizers      exist if $\{f=g\}$ is meas. <ref>(C)     $\boldsymbol{?}$     exist <ref>
products     (countable) exist <ref>     $\boldsymbol{?}$     $\boldsymbol{?}$
coequalizers     $\boldsymbol{?}$     $\boldsymbol{?}$ see <ref>     $\boldsymbol{?}$ see <ref>
coproducts     exist <ref>(D)     exist <ref>     exist <ref> and <ref>(D)

Limits and colimits in the three categories of MSNs.

In view of proving some partial existence result for lld
versions, we introduce the intermediary notion of a cccc
saturated MSN, short for coproduct (in $\sfMSNsp$) of ccc saturated

It is easy to see that cccc MSNs are lld, <ref> and <ref>.

The cccc version of an object of $\sfMSNsp$ is likewise defined
by its universal property in diagram (<ref>), using supremum
preserving morphisms.

Our main results are about locally ccc MSNs, i.e. those
saturated MSNs $(X,\calA,\calN)$ such that $\calE_{\rmccc} = \calA
\cap \{ Z : \text{ the subMSN } (Z,\calA_Z,\calN_Z) \text{ is ccc}\}$
is $\calN$-generating.

A complete semi-finite measure space $(X,\calA,\mu)$ is clearly
locally ccc, since $\calA^f$ is $\calN_\mu$-generating.

Similarly, one can define the more general locally localizable
objects in $\sfMSNsp$.

In <ref>, we give an example of an MSN which is not even locally

Let $(X,\calA,\calN)$ be a saturated locally ccc MSN. The following

(1) $(X,\calA,\calN)$ admits a cccc version,

(2) If furthermore $\calE_{\rmccc}$ contains an
$\calN$-generating subcollection $\calE$ such that $\rmcard \calE
\leq \mathfrak{c}$ and each $(Z,\calA_Z)$ is countably separated,
for $Z \in \calE$, then $(X,\calA,\calN)$ admits an lld version
which is also its cccc version.

By saying that a measurable space $(Z,\calA_Z)$ is countably
separated we mean that $\calA_Z$ contains a countable subcollection
that separates points in $Z$.

The cccc version $(\hat{X},\hat{\calA},\hat{\calN})$ is obtained as a
coproduct $\coprod_{Z \in \calE} (Z,\calA_Z,\calN_Z)$ where $\calE$ is
an $\calN$-generating almost disjointed refinement of
$\calE_{\rmccc}$, whose existence ensues from Zorn's Lemma.

In order to establish that this, in fact, is also the lld version
under the extra assumptions in (2), we need to build an appropriate
morphism $\br$ in diagram (<ref>), associated with an lld pair

It is obtained as a gluing of $\la q_Z \ra_{Z \in \calE}$ where $q_Z
\colon q^{-1}(Z) \to \hat{X}$ is the obvious map.

Since $\calE$ is almost disjointed, $\la q_Z \ra_{Z \in \calE}$ is
compatible and, since $(Y,\calB,\calM)$ in diagram (<ref>) is
localizable, the only obstruction to gluing is that $\hat{X}$ is not

Notwithstanding, $(\hat{X},\hat{\calA}) = \coprod_{Z \in \calE}
(Z,\calA_Z)$ is itself countably separated because $\rmcard \calE \leq
\mathfrak{c}$, <ref> so that the local determinacy of
$(Y,\calB,\calM)$ and the fact that $q^{-1}(\calE)$ is
$\calM$-generating (because $\calE$ is $\calN$-generating and $q$ is
supremum preserving) provides a gluing $r$, <ref>.

We now explain how this applies to associating, in a canonical
way, a strictly localizable measure space with any measure space

First, we recall that without changing the base space $X$ we can
render the measure space complete and semi-finite.

In that case, $\calA^f$ is $\calN_\mu$-generating and witnesses the
fact that the saturated MSN $(X,\calA,\calN_\mu)$ is locally ccc.

By the theorem above, it admits a cccc version

Let $(X,\calA,\mu)$ be a complete semi-finite measure space and
$[(\hat{X},\hat{\calA},\hat{\calN}),\bp]$ its corresponding cccc
version. Let $p \in \bp$. There exists a unique (and independent of
the choice of $p$) measure $\hat{\mu}$ defined on $\hat{\calA}$ such
that $p_\# \hat{\mu} = \mu$ and $\calN_{\hat{\mu}}=\hat{\calN}$.

Furthermore $(\hat{X},\hat{\calA},\hat{\mu})$ is a strictly
localizable measure space, and the Banach spaces
$\bL_1(X,\calA,\calN)$ and $\bL_1(\hat{X},\hat{\calA},\hat{\mu})$
are isometrically isomorphic.

Of course, the general process for constructing $\hat{X}$ is non
constructive, as it involves the axiom of choice to turn $\calA^f$
into an almost disjointed generating family.

This is why, in the last two sections of this paper, we explore a
particular case where we are able to describe explicitly $\hat{X}$ as
a quotient of a fiber bundle, all “hands on”.

We start with the measure space $(\Rm,\calB(\Rm),\calI^k_\infty)$
where $1 \leq k \leq m-1$ are integers, $\calB(\Rm)$ is the
$\sigma$-algebra of Borel subsets of $\Rm$, and $\calI^k_\infty$ is
the integral geometric measure described in <cit.> and

Note that it is not semi-finite, <cit.>.

Thus, we replace it with its complete semi-finite version

We let $\calE$ be the collection of $k$-dimensional submanifolds $M
\subset \Rm$ of class $C^1$ such that $\phi_M = \calH^k \hel M$ is
locally finite.

It follows from the Besicovitch Structure Theorem <cit.>
that $\calE$ is $\calN_{\tilde{\calI}^k_\infty}$-generating,

Now, for each $x \in \Rm$ we define $\calE_x = \calE \cap \{ M : x \in
M \}$ and we define on $\calE_x$ an equivalence relation as follows.

We declare that $M \sim_x M'$ if and only if

\begin{equation*}
\lim_{r \to 0^+} \frac{\calH^k(M \cap M' \cap \bB(x,r))}{\balpha(k)r^k}
= 1.
\end{equation*}

Letting $[M]_x$ denote the equivalence class of $M \in \calE_x$, we
prove <ref> that underlying set of the cccc, lld, and strictly
localizable version of the MSN
$(\Rm,\widetilde{\calB(\Rm)},\calN_{\tilde{\calI}^k_\infty})$ can be
taken to be

\begin{equation*}
\hat{X} = \{ (x,[M]_x ) : x \in \Rm \text{ and } M \in \calE_x \} \,.
\end{equation*}

This leads to an explicit description of the dual of
$\bL_1(\Rm,\widetilde{\calB(\Rm)},\tilde{\calI}^k_\infty)$ as
$\bL_\infty( \hat{X},\hat{\calA},\hat{\calN})$.

We are indebted to David Fremlin whose point of view on measure
theory – generously shared in his immense treatise <cit.> – influenced our
work in this paper. It is the second author's pleasure to record
useful conversations with Francis Borceux.

§ MEASURABLE SPACES WITH NEGLIGIBLES

Let $X$ be a set.  A $\sigma$-algebra on $X$ is a set $\calA
\subset \calP(X)$ such that

(1) $\emptyset \in \calA$;

(2) If $A \in \calA$ then $X \setminus A \in \calA$;

(3) If $\la A_n \ra_{n \in \N}$ is a sequence in $\calA$ then
$\bigcup_{n \in \N} A_n \in \calA$.

If $\calA$ is a $\sigma$-algebra on $X$ then $X \in \calA$ and
$\bigcap_{n \in \N} A_n \in \calA$ whenever $\la A_n \ra_{n \in \N}$ is
a sequence in $\calA$.  Clearly $\{\emptyset,X\}$ and $\calP(X)$ are
$\sigma$-algebras on $X$, respectively the coarsest and the finest.
If $\la \calA_i \ra_{i \in I}$ is a nonempty family of
$\sigma$-algebras on $X$ then $\bigcap_{i \in I} \calA_i$ is a
$\sigma$-algebra on $X$.  Thus each $\calE \subset \calP(X)$ is
contained in a coarsest $\sigma$-algebra on $X$ which we will denote
by $\sigma(\calE)$.  If $\calE,\calA \subset \calP(X)$ and $\calA =
\sigma(\calE)$ we say that the $\sigma$-algebra $\calA$ is
generated by $\calE$.  Clearly, if $\calE_1 \subset \calE_2
\subset \calP(X)$ then $\sigma(\calE_1) \subset \sigma(\calE_2)$.  A
measurable space is a couple $(X,\calA)$ where $X$ is a set
and $\calA$ is a $\sigma$-algebra on $X$.  In this case, if no
confusion is possible we call measurable the members of

[Measurable maps]
Let $(X,\calA)$ and $(Y,\calB)$ be measurable spaces and $f \colon X
\to Y$.  We say that $f$ is $(\calA,\calB)$-measurable (or
simply measurable if no confusion can occur) if $f^{-1}(B) \in
\calA$ whenever $B \in \calB$. Measurable spaces, together with
measurable maps, form a well-defined category, as one can check that
the composition of two measurable maps is measurable.

Let $(X, \calA)$ be a measurable space. A $\sigma$-ideal
$\calN$ of $\calA$ is a subset of $\calA$ that satisfies the
following requirements:

(1) $\emptyset \in \calN$;

(2) If $A \in \calA$, $N \in \calN$ and $A \subset N$ then $A \in \calN$;

(3) If $\la N_n\ra_{n \in \N}$ is a sequence in $\calN$, then
$\bigcup_{n \in\N} N_n \in \calN$.

[Measurable space with negligibles]
A measurable space with negligibles (abbreviated MSN) is
a triple $(X, \calA, \calN)$ where $(X, \calA)$ is a measurable
space and $\calN$ is a $\sigma$-ideal of $\calA$.  Given an MSN $(X,
\calA, \calN)$, elements belonging to $\calN$ are referred to as
$\calN$-negligible sets (simply negligible sets if no
confusion can occur). Complements of $\calN$-negligible sets
are called $\calN$-conegligible sets (or simply
conegligible sets).

We can associate to any measure space $(X, \calA, \mu)$ the MSN $(X,
\calA, \calN_\mu)$ where $\calN_\mu$ is the $\sigma$-ideal
$\calN_{\mu} = \calA \cap \{N : \mu(N) = 0\}$.  Conversely, any MSN
$(X, \calA, \calN)$ derives from a measure space: it suffices to
consider the measure $\mu \colon\calA \to [0,
\infty]$ that sends negligible sets to $0$ and the remaining sets
to $\infty$.

[Saturated MSNs]
An MSN $(X, \calA, \calN)$ is called saturated whenever the
following property holds: For all $N \in \calN$, any subset $N'
\subset N$ is $\calA$-measurable – therefore, in fact, $N' \in \calN$.
property is of purely technical nature, as an MSN $(X, \calA,
\calN)$ that does not have it can be turned into a saturated MSN
($X, \bar{\calA}, \bar{\calN})$, by setting:

\begin{equation*}
\begin{split}
\bar{\calN} & = \calP(X) \cap \{\bar{N} : \bar{N}
\subset N \text{ for some }  N \in \calN\} \\
\bar{\calA}&  = \calP(X) \cap \{\bar{A} : A \ominus \bar{A} \in \bar{\calN} \text{ for some } A \in \calA\} \\
& = \calP(X) \cap \{ A \ominus \bar{N} : A \in \calA \text{ and } \bar{N} \in \bar{\calN} \} .
\end{split}
\end{equation*}

Here, $\ominus$ denotes the symmetric difference of sets.  We call
$(X,\bar{\calA},\bar{\calN})$ the saturation of
$(X,\calA,\calN)$. In case the original MSN corresponds to a measure
space $(X,\calA,\mu)$, its saturation corresponds to the measure
space usually referred to as the completion of $(X,\calA,\mu)$. We
will denote the latter by $(X,\bar{\calA},\bar{\mu})$.

Let $(X, \calA, \calN)$ and $(Y, \calB, \calM)$ be two MSNs. We say
that a map $f \colon X \to Y$ is $[(\calA, \calN), (\calB,
\calM)]$-measurable (or simply measurable) if

(1) $f$ is $(\calA, \calB)$-measurable;

(2) $f^{-1}(M) \in \calN$ for every $M \in \calM$.

It is easy to check that measurability in the above sense is
preserved by composition.

[Morphisms of saturated MSNs]

Let $(X, \calA, \calN)$ and $(Y, \calB, \calM)$ be two saturated
MSNs. A morphism from $(X, \calA, \calN)$ to $(Y, \calB,
\calM)$ is an equivalence class of $[(\calA, \calN), (\calB,
\calM)]$-measurable maps under the relation $\sim$ of equality
almost everywhere: $f \sim f'$ whenever $\{f \neq f'\} \in \calN$.
In order to check that this relation is, indeed, transitive, it is
important to assume that $(X,\calA,\calN)$ is saturated for
otherwise we would not know that $\{f \neq f'' \} \in \calA$ when
$f,f'' : X \to Y$ are both
$[(\calA,\calN),(\calB,\calM)]$-measurable.  Also, in the special
case where $X$ is $\calN$-negligible and $Y = \emptyset$, we follow
the convention that there is unique morphism from $(X, \calA,
\calN)$ to $(\emptyset, \{\emptyset\}, \{\emptyset\})$.

Let $(X, \calA, \calN)$, $(Y, \calB, \calM)$ and $(Z, \calC, \calP)$
be MSNs and let $f, f' \colon X \to Y$ and $g, g' \colon Y
\to Z$ be maps. If

(A) $f, f'$ are $[(\calA, \calN), (\calB, \calM)]$-measurable;

(B) $g, g'$ are $[(\calB, \calM), (\calC, \calP)]$-measurable;

(C) $(X,\calA,\calN)$ is saturated;

(D) $f \sim f'$, $g \sim g'$,

then $g \circ f$, $g' \circ f'$ are $[(\calA, \calN), (\calC,
\calP)]$-measurable and $g \circ f \sim g' \circ f'$.

The first conclusion follows from hypotheses (A) and (B) and
Paragraph <ref>.  The second conclusion is a
consequence of $\{g \circ f \neq g' \circ f'\} \subset \{f \neq f'\}
\cup f^{-1}(\{g \neq g'\})$ and hypotheses (A), (C) and (D).

[Category $\sfMSN$]
Thanks to the preceding result, there is a notion of composition for
morphisms between saturated MSNs: if $\mathbf{f} \colon (X, \calA,
\calN) \to (Y, \calB, \calM)$ and $\bg \colon (Y, \calB, \calM) \to
(Z, \calC, \calP)$ are morphisms, we let $\bg \circ \mathbf{f}
\colon (X, \calA, \calN) \to (Z, \calC, \calP)$ be the equivalence
class of $g \circ f$ where $f \in \mathbf{f}$ and $g \in
\mathbf{g}$.

This allows to define the category $\sfMSN$ whose objects are
saturated MSNs and whose morphisms are described in the
paragraph <ref>. Additionally, we add the convention
that, for a negligible saturated MSN, i.e. an MSN of the form $(X,
\calP(X), \calP(X))$, there is a unique morphism from $(X, \calP(X),
\calP(X))$ to $(\emptyset, \{\emptyset\}, \{\emptyset\})$. This way,
negligible saturated MSNs are isomorphic to one another in the
category $\sfMSN$.

The categorical point of view is rarely considered in measure theory,
mainly due to the lack of a well-behaved notion of morphism between
measure spaces. The category
$\sfMSN$ also appears in the work <cit.> under the
name $\mathsf{StrictEMS}$. We start to investigate the existence of
limits and colimits in this category.

Let $(X, \calA, \calN)$ be an MSN and $Z \in \calP(X)$. We define
the subMSN $(Z, \calA_Z, \calN_Z)$, where $\calA_Z \defeq \{A
\cap Z: A \in \calA\}$ and $\calN_Z \defeq \{N \cap Z : N \in
\calN\}$. Note that in the special case where $Z$ is
$\calA$-measurable, we have $\calA_Z = \calA \cap \{A : A \subset
Z\}$ and likewise $\calN_Z = \calN \cap \{N : N \subset Z\}$.

The inclusion map $\iota_Z \colon Z \to X$ is $[(\calA_Z, \calN_Z),
(\calA, \calN)]$-measurable and induces a morphism
$\boldsymbol{\iota}_Z$ between $(Z, \calA_Z, \calN_Z)$ and $(X,
\calA, \calN)$.

Let $\mathbf{f}, \bg$ be a pair of
morphisms $(X, \calA, \calN) \to (Y, \calB, \calM)$ in the category $\sfMSN$, represented by
the maps $f \in \mathbf{f}$ and $g \in \bg$, and set $Z \defeq \{f =
g\}$. Then the equalizer of $\mathbf{f}, \bg$ is $[(Z, \calA_Z,
\calN_Z), \boldsymbol{\iota}_Z]$.

As $f \circ \iota_Z = g \circ \iota_Z$, we have clearly $\mathbf{f}
\circ \boldsymbol{\iota}_Z = \bg \circ \boldsymbol{\iota}_Z$. Let
$\bh \colon (T, \calC, \calP) \to (X, \calA, \calN)$ be any other
morphism that satisfies the relation $\mathbf{f} \circ \bh = \bg
\circ \bh$ and let $h \in \bh$. Then $h^{-1}(Z)$ is conegligible in
$T$. Up to modifying $h$, we can suppose that it has values in
$Z$. The restriction $h' \colon T \to Z$ of $h$ is $[(\calC, \calP),
(\calA_Z, \calN_Z)]$-measurable and we have $h = \iota_Z \circ
h'$, leading to a factorization $\bh = \boldsymbol{\iota}_Z \circ
\bh'$. This factorization is unique, as any morphism $\bh'$
satisfying $\bh = \boldsymbol{\iota}_Z \circ \bh'$ must derive from
a map $h' \colon T \to Z$ that coincides almost everywhere with $h$.

The category $\sfMSN$ has coproducts. Let $\la (X_i, \calA_i,
\calN_i)\ra_{i \in I}$ be a family of saturated MSNs. Its coproduct
is the MSN $(X, \calA, \calN)$ whose underlying set is $X =
\coprod_{i \in I} X_i$, and whose $\sigma$-algebra and
$\sigma$-ideal are defined by

\begin{gather*}
\calA = \calP(X) \cap \{A : A \cap X_i \in \calA_i \text{ for all }
i \in I\}, \\
\calN = \calP(X) \cap \{N : N \cap X_i \in \calN_i \text{ for all }
i \in I\}.
\end{gather*}

For $i \in I$, the canonical morphism $\boldsymbol{\iota}_i \colon
(X_i, \calA_i, \calN_i) \to (X, \calA, \calN)$ is the morphism
induced by the inclusion map $\iota_i \colon X_i \to X$.

Notice that, indeed, $(X,\calA,\calN)$ is a saturated MSN.  Let $(Y,
\calB, \calM)$ be a saturated MSN and $\la \mathbf{f}_i \ra_{i \in
I}$ be a collection of morphisms from $(X_i, \calA_i, \calN_i)$ to
$(Y, \calB, \calM)$, each $\mathbf{f}_i$ being represented by a
measurable map $f_i$. We set $f = \coprod_{i \in I} f_i$, the map
such that $f \circ \iota_i = f_i$ for any $i \in I$. It is clear
that $f$ is $[(\calA, \calN), (\calB, \calM)]$-measurable and
$\mathbf{f} \circ \boldsymbol{\iota}_i = \mathbf{f}_i$ holds for all
$i \in I$. We need to show that $\mathbf{f}$ is the unique morphism
$(X, \calA, \calN) \to (Y, \calB, \calM)$ with this property.

Suppose $\bg \colon (X, \calA, \calN) \to (Y, \calB, \calM)$ is
another morphism, represented by a measurable map $g \colon X \to
Y$, for which $\bg \circ \boldsymbol{\iota}_i = \mathbf{f}_i$ for
all $i \in I$. Then $f$ and $g$ coincide almost everywhere on each
$X_i$, which implies, due to the choice of $\calN$, that $f$ and $g$
are equal almost everywhere and $\mathbf{f} = \bg$.

The category $\sfMSN$ has countable products. Let $\la (X_i,
\calA_i, \calN_i) \ra_{i \in I}$ be a countable family of saturated
MSNs. Its product is the MSN $(X, \calA, \calN)$, whose underlying
set is the product $X = \prod_{i \in I} X_i$, whose $\sigma$-ideal is

\[
\calN = \calP(X) \cap \left\{ N : \exists \la N_i \ra_{i \in I} \in
\prod_{i \in I} \calN_i, \quad N \subset \bigcup_{i \in I} \pi_i^{-1}(N_i)
\right\},
\]

where $\pi_i \colon X \to X_i$ denotes the projection map, and whose
$\sigma$-algebra is the saturation of $\bigotimes_{i \in I}
\calA_i$:

\[
\calA = \left\{ A \ominus N : A \in \bigotimes_{i \in I} \calA_i
\text{ and } N \in \calN\right\}.
\]

For $i \in I$, the projection morphism $\boldsymbol{\pi}_i \colon
(X, \calA, \calN) \to (X_i, \calA_i, \calN_i)$ is the map induced by
the projection map $\pi_i$.

First note that the projection maps $\pi_i$ are $[(\calA, \calN),
(\calA_i, \calN_i)]$-measurable by construction of $\calA$ and
$\calN$. Let $(Y, \calB, \calM)$ be a saturated MSN and $\la
\mathbf{f}_i \ra_{i \in I}$ be a collection of morphisms from $(Y,
\calB, \calM)$ to $(X_i, \calA_i, \calN_i)$, each $\mathbf{f}_i$
being represented by a measurable map $f_i \colon Y \to X_i$. We
define $f = \prod_{i \in I} f_i \colon Y \to \prod_{i \in I} X_i$
that assigns $y \in Y$ to $\la f_i(y) \ra_{i \in I}$. Clearly, $f$
is $(\calB, \bigotimes_{i \in I} \calA_i)$-measurable. Moreover, for
any negligible set $N \in \calN$, we can find a sequence $\la N_i
\ra_{i \in I}$ such that $N \subset \bigcup_{i \in I}
\pi_i^{-1}(N_i)$. Thus $f^{-1}(N) \subset \bigcup_{i \in I} (\pi_i
\circ f)^{-1}(N_i) = \bigcup_{i \in I} f_i^{-1}(N_i)$. As $I$ is
countable and $f_i^{-1}(N_i) \in \calM$ for all $i \in I$, we find
that $f^{-1}(N) \in \calM$, which entails that the map $f$ is
$[(\calB, \calM), (\calA, \calN)]$-measurable.

Let $\bg \colon (Y, \calB, \calM) \to (X, \calA, \calN)$ be another
morphism satisfying the identities $\boldsymbol{\pi}_i \circ \bg =
\mathbf{f}_i$ for $i \in I$. Let $g \colon Y \to X$ be a
representative of $\bg$. The coordinate functions $\pi_i \circ g$
must coincide with $f_i$ almost everywhere. As there are only
countably many of them, we conclude that $f$ and $g$ are equal
almost everywhere, that is, $\mathbf{f} = \bg$.

In case $(X_i,\calA_i,\calN_i)$ are associated with measure spaces $(X_i,\calA_i,\mu_i)$, $i=1,2$, the $\sigma$-ideal $\calN$ considered in the above proposition may not coincide with $\calN_{\mu_1 \otimes \mu_2}$. This is the case, for instance, when $(X_i,\calA_i,\mu_i)=(\R,\calB(\R),\calL^1)$, $i=1,2$, since the diagonal $D = \R^2 \cap \{ (x,x) : x \in
\R \} \in \calB(\R) \otimes \calB(\R)$ is $\calL^2$-negligible but does not belong to $\calN$.

§ SUPREMUM PRESERVING MORPHISMS

One of the reasons we were led to introduce MSNs is that the
category of measure spaces and (equivalence classes of) measure
preserving measurable maps does not have good properties at
all. Roughly speaking, this can be attributed to the fact that it
has very few arrows. One way to increase their number is to define
as morphisms $(X, \calA, \mu) \to (Y, \calB, \nu)$ the $(\calA,
\calB)$-measurable maps $\varphi \colon X \to Y$ such that the
pushforward measure $\varphi_\# \mu$ is absolutely continuous with
respect to $\nu$. If we drop the measures and retain only which sets
have measure zero, we get the notion of $[(\calA, \calN_\mu),
(\calB, \calN_{\nu})]$-measurability
of <ref>. However, doing so, we may introduce some
“irregular” maps. For example, if $\calA$ is the $\sigma$-algebra
of Lebesgue measurable sets of the real line, $\calL^1$ the Lebesgue
measure and $\nu$ the counting measure on $(\R, \calA)$ then the
identity map induces a morphism $(\R, \calA, \calL^1) \to (\R,
\calA, \nu)$. But, $\calL^1$ does not really compare to $\nu$,
although it is absolutely continuous with respect to $\nu$. For
instance, $\calL^1$ has no Radon-Nikodým density with respect to
$\nu$, not even in the generalized sense of
Section <ref>. Forgetting the measures, the morphism of MSNs
$(\R,\calA, \calN_{\calL^1}) \to (\R, \calA, \{\emptyset\})$ is
still somehow inappropriate. To avoid this, we restrict our
attention to the supremum preserving morphisms introduced
below. This will allow us to define a new category $\sfMSNsp$ of
saturated MSNs with supremum preserving morphisms. Later, we will be
able to define localizable versions of MSNs and similar notions by
means of universal properties to be satisfied in $\sfMSNsp$.

[Boolean algebras]

Many of the properties that we will introduce underneath for MSNs
are related to their Boolean algebra, defined in the
following way: given an MSN $(X, \calA, \calN)$, we observe that the
$\sigma$-algebra $\calA$ is a Boolean algebra and $\calN$ is an
ideal of $\calA$ in the ring-theoretic sense; we then associate to
$(X, \calA, \calN)$ the quotient Boolean algebra $\calA/\calN$.

When we restrict our attention to saturated MSNs, this construction
becomes functorial. Call $\mathsf{Bool}(X, \calA, \calN) =
\calA/\calN$ the Boolean algebra of a saturated MSN. Given a
morphism $\mathbf{f} \colon (X, \calA, \calN) \to (Y, \calB, \calM)$
represented by a measurable map $f \colon X \to Y$, we define
$\mathsf{Bool}(f) \colon \mathsf{Bool}(Y, \calB,\calM) \to
\mathsf{Bool}(X, \calA, \calN)$ that maps the equivalence class of
$B \in \calB$ to the equivalence class of $f^{-1}(B)$. This map is
well-defined because of the $[(\calA,\calN),
(\calB,\calM)]$-measurability of $f$, it is a morphism of Boolean
algebras, and it does not depend on the representative of
$\mathbf{f}$, as one can easily check.

Let $(X, \calA, \calN)$ be an MSN and $\calE$ be a subcollection of
$\calA$. We say that $U \in \calA$ is an $\calN$-essential
upper bound of $\calE$ whenever $E \setminus U \in \calN$ for all
$E \in \calE$. Furthermore, a measurable set $S \in \calA$ is an
$\calN$-essential supremum of $\calE$ whenever

(1) $S$ is an $\calN$-essential upper bound of $\calE$;

(2) If $S'$ is an $\calN$-essential upper bound of
$\calE$, then $S \setminus S' \in \calN$.

In particular, if $S$, $S'$ are both $\calN$-essential suprema of
$\calE$, their symmetric difference $S \ominus S'$ is negligible. In
other words, an essential supremum, when it exists, is unique up to
negligible sets. In fact, it corresponds to a (unique) supremum in
$\mathsf{Bool}(X,\calA,\calN)$. A collection $\calE \subset \calA$
that admits $X$ as an $\calN$-essential supremum is called $\calN$-generating. We will use the following repeatedly. If
$\calE \subset \calA$ and $S \in \calA$ is an $\calN$-essential
supremum of $\calE$, then $\calE \cup \{ X \setminus S \}$ is

The next ubiquitous lemma expresses that $\cap$ is distributive over
the (partially defined) operation of taking essential suprema. It
implies the following fact, which we will use frequently: If $\calE$
is $\calN$-generating and $A \in \calA \setminus \calN$, then $E
\cap A \not \in \calN$ for some $E \in \calE$.

[Distributivity Lemma]

Let $(X, \calA, \calN)$ be an MSN, $\calE \subset \calA$ be a
collection that has an $\calN$-essential supremum $S$, and $C \in
\calA$. Then $C \cap S$ is an $\calN$-essential supremum of $\{C
\cap E : E \in \calE\}$.

Condition (1) in Definition <ref> is met because $C
\cap E \setminus C \cap S = C \cap (E \setminus S) \in \calN$ for
all $E \in \calE$. As for (2), we let $S'$ be an $\calN$-essential
upper bound for $\{C \cap E: E \in \calE\}$. We claim that $S''
\defeq S' \cup (X \setminus C)$ is an $\calN$-essential upper bound
for $\calE$. Indeed, for any $E \in \calE$, we have $E \setminus S''
= (C \cap E) \setminus S' \in \calN$. It follows that $(C \cap S)
\setminus S' = S \setminus S'' \in \calN$.

Note that if $(X,\calA,\calN)$ and $(Y,\calB,\calM)$ are MSNs, $f :
X \to Y$ is $[(\calA,\calN),(\calB,\calM)]$-measurable, $\calE
\subset \calB$, and $S \in \calB$ is an $\calM$-essential upper
bound of $\calE$, then $f^{-1}(S)$ is an $\calN$-essential upper
bound of $f^{-1}(\calE)$. However, if $S$ is an $\calM$-essential
supremum of $\calE$ then $f^{-1}(S)$ may not be an $\calN$-essential
supremum of $f^{-1}(\calE)$. Consider, for instance,
$(Y,\calB,\calM)=(\R,\calB(\R),\{\emptyset\})$, $f=\rmid_{\R}$, and
$\calE = \{ \{x\} : x \in \R \}$. Then $\R$ is an
$\{\emptyset\}$-essential supremum of $\calE$, $\emptyset$ is an
$\calN_{\calL^1}$-essential supremum of $\calE = f^{-1}(\calE)$, and
$\R \setminus \emptyset \not\in \calN_{\calL^1}$.

There are several objects that we can call supremum
preserving. For saturated MSNs $(X, \calA, \calN)$ and $(Y,
\calB, \calM)$, we define

* A morphism of Boolean algebras $\varphi \colon \frA \to \frB$
is called supremum preserving if, for any family $\frE \subset \frA$ that admits a
supremum, the family $\varphi(\frE)$ admits a supremum and $\varphi(\sup \frE) = \sup
\varphi(\frE)$.

* A morphism $\mathbf{f} \colon (X, \calA, \calN) \to (Y, \calB,
\calM)$ is called supremum preserving whenever
$\mathsf{Bool}(\mathbf{f})$ is.

* An $[(\calA, \calN), (\calB, \calM)]$-measurable map $f \colon
X \to Y$ is called supremum preserving if, for any collection $\calE
\subset \calB$ possessing an $\calM$-essential supremum $S$, then
$f^{-1}(S)$ is an $\calN$-essential supremum of $f^{-1}(\calE)
\defeq \{f^{-1}(E) : E \in \calE\}$.

For a morphism $\mathbf{f}$ represented by $f \in \mathbf{f}$, the
supremum preserving characters of $f$, $\mathbf{f}$ and
$\mathsf{Bool}(\mathbf{f})$ are all equivalent. Also, the
composition of two supremum preserving morphisms is supremum
preserving. We call $\sfMSNsp$ the subcategory of $\sfMSN$ that
consists of saturated MSNs and supremum preserving morphisms. In the
next proposition, we gather some basic facts about supremum
preserving morphisms and the category $\sfMSNsp$.

The following hold.

(A) Two saturated MSNs are isomorphic in $\sfMSN$ if and only
if they are isomorphic in $\sfMSNsp$.

(B) Let $(X, \calA, \calN)$ be a saturated MSN and $Z \in
\calA$. The morphism $\boldsymbol{\iota}_Z \colon (Z, \calA_Z,
\calN_Z) \to (X, \calA, \calN)$ induced by the inclusion map
$\iota_Z \colon Z \to X$ is supremum preserving.

(C) Let $\mathbf{f}, \bg \colon (X, \calA, \calN) \to (Y,
\calB, \calM)$ be a pair of morphisms in $\sfMSNsp$, represented
by $f \in \mathbf{f}$ and $g \in \mathbf{g}$. If $Z \defeq \{f =
g\}$ is $\calA$-measurable, then $((Z, \calA_Z, \calN_Z),
\boldsymbol{\iota}_Z)$ is the equalizer of $\mathbf{f}, \bg$ in $\sfMSNsp$.

(D) The category $\sfMSNsp$ has coproducts, which are
preserved by the forgetful functor $\sfMSNsp \to \sfMSN$.

(A) Let $\mathbf{f}$ be an isomorphism in $\sfMSN$. Then
$\mathsf{Bool}(\mathbf{f})$ is an isomorphism of Boolean
algebras. More specifically, it is an isomorphism of posets and for
this reason it preserves suprema.

(B) This is the content of Lemma <ref>.

(C) By Proposition <ref>, $((Z, \calA_Z, \calN_Z),
\boldsymbol{\iota}_Z)$ is the equalizer of $\mathbf{f}, \bg$ in
$\sfMSN$ and by (B) the morphism $\boldsymbol{\iota}_Z$ is a morphism of $\sfMSNsp$. Let $\bh \colon (T, \calC, \calP) \to (X, \calA, \calN)$
be a supremum preserving morphism that satisfies $\mathbf{f} \circ \bh = \bg
\circ \bh$. Recalling the proof of Proposition <ref>,
there is a representative $h \in \bh$ with values in $Z$, and its
restriction $h' \colon T \to Z$ induces the unique morphism $\bh'$
such that $\bh = \boldsymbol{\iota}_Z \circ \bh'$. The results
follows from the fact that $h'$ is easily checked to be supremum

(D) Let $\la (X_i, \calA_i, \calN_i)\ra_{i \in I}$ be a family of
saturated MSNs, $(X, \calA, \calN)$ be their coproduct in the
category $\sfMSN$, and $\la \mathbf{f}_i \ra_{i \in I}$ be a family
of supremum preserving morphisms from $(X_i, \calA_i, \calN_i)$ to a
saturated $(Y, \calB, \calM)$, each represented by $f_i \colon X_i
\to Y$. We need to show that $f \defeq \coprod_{i \in I} f_i \colon
X \to Y$ is supremum preserving. For this, let $\calE \subset \calB$
be a collection that has an $\calM$-essential supremum $S$. We
observe that $f^{-1}(S) = \coprod_{i \in I} f^{-1}_i(S)$ is an
$\calM$-essential upper bound of $f^{-1}(\calE)$. Let $U$ be a
second $\calN$-essential upper bound of $f^{-1}(\calE)$. Then $X_i
\cap U$ is an $\calN_i$-essential upper bound of $\{X_i \cap
f^{-1}(E) : E \in \calE\} = f_i^{-1}(\calE)$. It follows that
$f_i^{-1}(S) \setminus (X_i \cap U) \in \calN_i$. As this happens
for all $i \in I$, we get that $f^{-1}(S) \setminus U \in \calN$.

§ LOCALIZABLE, CCCC AND STRICTLY LOCALIZABLE MSNS

[Localizable MSN]
An MSN $(X, \calA, \calN)$ is localizable whenever each
collection $\calE \subset \calA$ admits an $\calN$-essential
supremum. Equivalently, $(X, \calA, \calN)$ is localizable whenever
its Boolean algebra $\calA / \calN$ is Dedekind complete, that
is, each subset of $\calA/ \calN$ has a supremum.

Originally, localizability was introduced by Segal <cit.> in
the context of measure spaces. Since then, many minor variations
over the definition in that context have been proposed (see <cit.> for
an overview). We will follow the definition
in <cit.>. A measure space $(X, \calA, \mu)$ is
called localizable whenever

(1) it is semi-finite, i.e. for all $A \in
\calA$ with $\mu(A) > 0$, there is a measurable set $A' \subset A$
such that $0 < \mu(A') < \infty$;

(2) the underlying MSN $(X, \calA, \calN_\mu)$ is

[Semi-finite measure space]

In the definition of localizable measure space, semi-finiteness
plays on important rôle. Let us rephrase it. Given $(X,\calA,\mu)$
a measure space, we abbreviate $\calA^f \defeq \calA \cap \{E :
\mu(E) < \infty\}$. We say that $N \in \calA$ is locally
$\mu$-negligible whenever $N \cap E \in \calN_\mu$ for every $E
\in \calA^f$. We let $\calN_{\mu,\rmloc}$ be the $\sigma$-ideal
consisting of locally $\mu$-negligible measurable sets. The
following are equivalent:

(1) $\calN_\mu = \calN_{\mu,\rmloc}$.

(2) $(X,\calA,\mu)$ is semi-finite.

(3) $\calA^f$ is $\calN_\mu$-generating.

The only non trivial part is $(3) \Rightarrow (1)$. If $N \in \calA$ then $N$ is an $\calN_\mu$-essential supremum of $\{ N \cap F : F \in \calA^f \}$, according to the Distributivity Lemma <ref>. If also $N \in \calN_{\mu,\rmloc}$, then it follows that $N \in \calN_\mu$. The notion of locally $\mu$-negligible sets will appear again in <ref>.

Next we introduce some classes of localizable MSNs that will appear
throughout the paper.

[Countable chain condition]

Let $(X, \calA, \calN)$ be an MSN. A family $\calE \subset \calA
\setminus \calN$ is called almost disjointed whenever $E \cap
E' \in \calN$ for any pair of distinct $E, E' \in \calE$. The MSN $(X,
\calA, \calN)$ is said to have the countable chain condition
(in short: is ccc) whenever an almost disjointed family in
$\calA \setminus \calN$ is at most countable.

The previous notions have counterparts in the realm of Boolean
algebras. Given a Boolean algebra $\frA$, a subset $\frE \subset
\frA$ is called disjointed whenever $x \wedge y = 0$ for any
pair of distinct elements $x, y \in \frE$. The Boolean algebra
$\frA$ has the countable chain condition (or: is ccc)
whenever each of its disjointed families is at most countable. Of
course, an MSN $(X, \calA, \calN)$ is ccc if and only if its Boolean
algebra $\calA/\calN$ is. In the following proposition, we show that
being ccc is stronger than localizability. It is related to the
fact, first established in <cit.>, that a Dedekind
$\sigma$-complete Boolean algebra (that is, a Boolean algebra where
countable collections have suprema) having the countable chain
condition is Dedekind complete.

If an MSN $(X, \calA, \calN)$ is ccc and $\calE \subset \calA$ is a
collection, then there is a countable subcollection $\calE' \subset
\calE$ such that $\bigcup \calE'$ is an $\calN$-essential supremum
of $\calE$. In particular, $(X, \calA, \calN)$ is localizable.

Suppose the existence of a collection $\calE \subset \calA$ for
which one cannot find a countable subcollection $\calE' \subset
\calE$ whose union is an $\calN$-essential supremum of $\calE$. This
assumption allows us to construct transfinitely a sequence $\la
E_\alpha \ra_{\alpha < \omega_1}$ with values in $\calE$ such that
for every $\alpha < \omega_1$, one has $F_\alpha \defeq E_\alpha
\setminus \bigcup_{\beta < \alpha} E_\beta \not\in \calN$. But the
disjointed family $\{F_\alpha : \alpha < \omega_1\}$ contradicts the
fact that $(X, \calA, \calN)$ is ccc.

Let $(X, \calA, \mu)$ be a finite measure space. Then $(X, \calA,
\calN_\mu)$ is ccc.

Let $\calE \subset \calA \setminus \calN$ be an almost disjointed
family. For each positive integer $n$, set $\calE_n = \calE \cap \{E
: \mu(E) > n^{-1}\}$. As $\mu(X) \geq \mu\left(\bigcup
\calE_n\right) \geq n^{-1}\rmcard \calE_n$, we have that $\calE_n$
is finite. Consequently, $\calE$ is at most countable.

A coproduct $\coprod_{i \in I} (X_i, \calA_i, \calN_i)$ of saturated
localizable MSNs is localizable.

Let $\calE$ be a collection of measurable sets of a coproduct
$\coprod_{i \in I} (X_i ,\calA_i, \calN_i)$. For each $i \in I$, the
collection $\calE_i \defeq \{X_i \cap E : E \in \calE\}$ has an
$\calN_i$-essential supremum $S_i \subset X_i$. We then routinely
check that $S \defeq \coprod_{i \in I} S_i \in \calA$ is an
$\calN$-essential supremum of $\calE$.

[Stronger notions of localizability]
An MSN is called strictly localizable if it is isomorphic to a
coproduct of the form $\coprod_{i \in I} (X_i, \calA_i,
\calN_{\mu_i})$, where $(X_i, \calA_i, \mu_i)$ are complete finite
measure spaces.  Examples of strictly localizable MSNs are provided
by MSNs associated to complete $\sigma$-finite measure spaces $(X,
\calA, \mu)$. Indeed, denoting $\la X_i \ra_{i \in I}$ a countable
partition of $X$ into measurable subsets of finite $\mu$ measure,
one can verify that $(X, \calA, \calN_\mu)$ is isomorphic to
$\coprod_{i \in I} (X_i, \calA_{X_i}, \calN_{\mu \shel X_i})$.

Likewise, we say that an MSN is cccc whenever it is isomorphic
to a coproduct of saturated ccc MSNs. We have the chain of

\[
\text{strictly localizable} \implies \text{cccc} \implies \text{localizable}.
\]

The first implication comes from Proposition <ref>, the
second one from Propositions <ref>
and <ref>. Examples of non localizable spaces are provided
by the next results.

Let $(X, \calA, \calN)$ be a localizable MSN, $\calE \subset \calA
\setminus \calN$ an almost disjointed family. Then $\rmcard (\calA /
\calN) \geq 2^{\rmcard \calE}$.

Consider the application $\calP(\calE) \to \calA / \calN$ which maps
each subcollection $\calE' \subset \calE$ to the equivalence class
of its $\calN$ essential supremum. We claim that this map is
injective. Indeed, suppose $\calE', \calE'' \subset \calE$ are
distinct. Call $S'$ (resp. $S''$) an $\calN$ essential supremum of
$\calE'$ (resp. $\calE''$). Without loss of generality, there is $F
\in \calE' \setminus \calE''$. By Lemma <ref>, $F \cap
S'$ (resp. $F \cap S''$) is an essential supremum of $\{F\cap E: E
\in \calE'\}$ (resp. $\{F \cap E : E \in \calE''\}$). We deduce that $F
\setminus S' \in \calN$ and, taking the
almost disjointed character of $\calE$ into account, that $F \cap S'' \in \calN$. This
implies that $S'$ and $S''$ do not induce the same equivalence class
in $\calA/\calN$.

We will use the following many times.

Let $(X,\calA,\calN)$ be an MSN and let $\calC \subset \calA$ be $\calN$-generating. There exists $\calE \subset \calA \setminus \calN$ with the following properties.

(A) $\calE$ is almost disjointed.

(B) For each $E \in \calE$, there exists $C \in \calC$ such that $E \subset C$.

(C) $\calE$ is $\calN$-generating.

There is no restriction to assume that $\calN \neq \calA$; in
particular, $\calC \neq \emptyset$. Consider the set $\bE$ consisting
of those $\calE \subset \calA \setminus \calN$ that satisfy conditions
(A) and (B) above, ordered by inclusion. Thus, $\bE$ is nonempty and
one readily checks that every chain in $\bE$ possesses a maximal
element. Therefore, $\bE$ admits a maximal element $\calE$, according
to Zorn's Lemma. We ought to show that $\calE$ is
$\calN$-generating. If this were not the case, there would exist an
$\calN$-essential upper bound $U \in \calA$ of $\calE$ such that $X
\setminus U \not \in \calN$. The latter, together with the fact that
$\calC$ is $\calN$-generating, implies the existence of $C \in \calC$
such that $C \cap (X \setminus U) \not \in \calN$. Then, $\calE \cup
\{ C \cap (X \setminus U) \}$ contradicts the maximality of $\calE$.

[$\mathsf{ZFC} + \mathsf{CH}$]
Let $X$ be a Polish space endowed with its Borel $\sigma$-algebra
$\calB(X)$ and $\mu \colon \calB(X) \to [0, \infty]$ be a
semi-finite Borel measure. Under the Continuum Hypothesis, one has
the following dichotomy: either $\mu$ is $\sigma$-finite, or the
MSN $(X, \calB(X), \calN_{\mu})$ is not localizable.

Let $\calE$ be associated with $\calC \defeq \calB(X) \cap \{ A :
\mu(A) < \infty \}$ in Lemma <ref>. Recall <ref>
that $\calC$ is $\calN_\mu$-generating.  If $\calE$ is countable,
then $\bigcup \calE$ is measurable and, accordingly, an
$\calN_\mu$-essential upper bound of $\calE$. Thus $X \setminus \bigcup
\calE \in \calN_\mu$, since $\calE$ is $\calN_\mu$-generating. We
have proven that $\mu$ is $\sigma$-finite.

On the other hand, if $\calE$ is uncountable, the Continuum
Hypothesis guarantees that it has cardinal greater or equal to
$\mathfrak{c}$. Assume if possible that $(X, \calB(X), \mu)$ is
localizable. As the map $\calB(X) \to \calB(X) / \calN_{\mu}$ is
onto, we deduce from Lemma <ref> that $\rmcard \calB(X) \geq
2^{\mathfrak{c}} > \mathfrak{c}$. However, Borel sets are Suslin,
and Suslin sets are continuous images of closed subsets of a
particular Polish space, the Baire space, see e.g
<cit.>. This gives the upper bound $\rmcard \calB(X)
\leq \mathfrak{c}$, contradicting the preceding inequality.

[$\calP$-version of an MSN]

Let $\calP$ be a property associated to MSNs. We suppose that the
property $\calP$ is hereditary: if $(X, \calA, \calN)$ has
$\calP$ then the MSNs $(Z, \calA_Z, \calN_Z)$ also has $\calP$ for
all $Z \in \calA$. “Being strictly localizable”, “being cccc” or
“being localizable” are examples of hereditary properties.

Let $(X, \calA, \calN)$ be a saturated MSN. We define a $\calP$-version of $(X, \calA, \calN)$ to be a couple $[(\hat{X},
\hat{\calA}, \hat{\calN}), \bp]$ consisting of a saturated MSN
$(\hat{X}, \hat{\calA}, \hat{\calN})$ with the property $\calP$ and
a supremum preserving morphism $\bp \colon (\hat{X}, \hat{\calA},
\hat{\calN}) \to (X, \calA, \calN)$ satisfying the following
property: For any saturated MSN $(Y, \calB, \calM)$ with the
property $\calP$ and any supremum preserving morphism $\bq \colon
(Y,\calB, \calM) \to (X, \calA, \calN)$, there is a unique supremum
preserving morphism $\br\colon (Y, \calB, \calM) \to (\hat{X},
\hat\calA, \hat{\calN})$ such that $\bq =\bp \circ \br$.

\[
\begin{tikzcd}
(Y, \calB, \calM) \arrow[rd, swap, "\bq"] \arrow[rr, dotted,
"\exists!\br"] & & (\hat{X}, \hat{\calA}, \hat{\calN})
\arrow[ld, "\bp"] \\ & (X, \calA, \calN) &
\end{tikzcd}
\]

By this definition, a $\calP$-version must satisfy a universal
property, and as such it is unique up to a unique isomorphism of the
category $\sfMSNsp$. More specifically, if $[(\hat{X}, \hat{\calA},
\hat{\calN}), \bp]$ and $[(\hat{X}', \hat{\calA}', \hat{\calN}'),
\bp']$ are two $\calP$-versions, then we easily check that there is
a unique isomorphism $\br \colon (\hat{X}, \hat{\calA}, \hat{\calN})
\to (\hat{X}', \hat{\calA}', \hat{\calN}')$ such that $\bp' \circ
\br = \bp$.

[Atomic MSNs]

One of our motivations in this article is to find a universal
construction that transforms an MSN into something with better
localizability properties. As such, it is wise to first have a look
at the not so easy case of MSNs $(X, \calA, \calN)$ such that all
singletons are $\calA$-measurable and $ \calN = \{\emptyset\}$. We
call such MSNs atomic.

In an atomic MSN $(X, \calA, \{\emptyset\})$, it is easy to see that
a subset $\calE \subset \calA$ has an $\{\emptyset\}$-essential
supremum if and only if $\bigcup \calE \in \calA$, in which case
$\bigcup \calE$ is the $\{\emptyset\}$-essential supremum. Therefore
the MSN $(X, \calA, \{\emptyset\})$ is localizable if and only if
$\calA = \calP(X)$. In other words, the non localizability of $(X,
\calA, \{\emptyset\})$ can only be due to the lack of measurable
sets; therefore it seems sensible to ask for $(X, \calP(X),
\{\emptyset\})$ to be the “localization” of $(X, \calA,
\{\emptyset\})$.

Unfortunately, Proposition <ref> gives a negative
result. It tells us that the localizable version of an MSN, as
defined in <ref>, is not the right notion of
“localization”. This issue will be addressed in
Section <ref> by introducing a notion of local determination
for MSNs.

Let $X$ be an uncountable set, $\calC(X)$ be its
countable-cocountable $\sigma$-algebra. Let $\boldsymbol{\iota}$ be
the morphism $(X, \calP(X), \{\emptyset\}) \to (X, \calC(X),
\{\emptyset\})$ induced by the identity map. Then $[(X, \calP(X),
\{\emptyset\}), \boldsymbol{\iota}]$ is not a localizable version of
$(X, \calC(X), \{\emptyset\})$.

That $\boldsymbol{\iota}$ is supremum preserving follows from the
discussion in Paragraph <ref>. Assume if possible that $((X,
\calP(X), \{\emptyset\}), \boldsymbol{\iota})$ is a localizable
version of $(X, \calA, \{\emptyset\})$. We will get a contradiction
if we manage to build a localizable saturated MSN $(Y, \calB,
\calM)$ and a function $q \colon Y \to X$ that is $[(\calB, \calM),
(\calC(X), \{\emptyset\})]$-measurable, supremum preserving, but
not $(\calB, \calP(X))$-measurable.

We choose $Y = X^2 \times \{0, 1\}$. For any subset $B \subset Y$,
we call $B[0]$ and $B[1]$ the subsets defined by

\[
B[i] \defeq X^2 \cap \{(x_1, x_2) : (x_1, x_2, i) \in B\} \quad
\text{for } i \in \{0, 1\}
\]

We let $\calB = \calP(Y) \cap \{B : B[0] \ominus B[1] \text{ is
countable}\}$. We claim that $\calB$ is a $\sigma$-algebra of
$Y$. The stability of $\calB$ under countable unions is a
consequence of the formula

\[
\left(\bigcup_{n \in \N} B_n\right)[0] \ominus \left(\bigcup_{n \in
\N} B_n\right)[1] \subset \bigcup_{n \in \N} B_n[0] \ominus B_n[1]
\]

that holds for any sequence $\la B_n \ra_{n \in \N}$ of subsets in $Y$,
we leave the other points to the reader. Finally, we define the
$\sigma$-ideal $\calM \defeq \calB \cap \{M : M[0] =
\emptyset\}$. Clearly $(Y, \calB, \calM)$ is a saturated MSN.

Let us show that $(Y, \calB, \calM)$ is localizable. Let $\calE
\subset \calB$ be any collection. We set $A \defeq \bigcup_{E \in
\calE} E[0]$ and $S \defeq A \times \{0, 1\}$. The set $S$ is
$\calB$-measurable, because $S[0] = S[1] = A$. For any $E \in
\calE$, we have $(E \setminus S)[0] = E[0] \setminus S[0] =
\emptyset$, meaning that $S$ is an $\calM$-essential upper bound of
$\calE$. Denoting by $U$ another essential upper bound of $\calE$,
then $(E \setminus U)[0] = E[0] \setminus U[0] = \emptyset$ for all
$E \in \calE$. It follows that $A \subset U[0]$ and $(S \setminus
U)[0] = S[0] \setminus U[0] = \emptyset$. Thus, $S$ is an
$\calM$-essential supremum, as we wanted.

Now, let $\sigma \colon X \to X$ be a bijection of $X$ without
fixed points.  For example, choose a partition $X = Z \cup Z'$ into
subsets $Z, Z'$ that have the same cardinality as $X$, choose a
bijection $f \colon Z \to Z'$ and set $\sigma$ so that $\sigma(x) =
f(x)$ for all $x \in Z$ and $\sigma(x) = f^{-1}(x)$ for all $x \in
Z'$. We define the map $q \colon Y \to X$ by

\[
\forall (x_1, x_2, i) \in Y, \quad q(x_1, x_2, i)
= \begin{cases} x_2 & \text{if } i = 1 \text{ and } x_2 =
\sigma(x_1) \\ x_1 & \text{otherwise}
\end{cases}
\]

First we show that $q$ is $[(\calB, \calM), (\calC(X),
\{\emptyset\})]$-measurable. It suffices to show that
$q^{-1}(\{x\}) \in \calB$ for all $x \in X$. But we have

\begin{align*}
q^{-1}(\{x\})[0] & = \{x\} \times X \\
q^{-1}(\{x\})[1] & =
\big(\{x\} \times \left(X \setminus \{\sigma(x)\}\right)\big) \cup
\{(\sigma^{-1}(x)\,, x)\}
\end{align*}

Consequently, $q^{-1}(\{x\})[0] \ominus q^{-1}(\{x\})[1]$ has only
two elements. By the definition of $\calB$, this ensures the
measurability of $q^{-1}(\{x\})$.

However, we claim that $q$ is not $(\calB, \calP(X))$-measurable. To
this end, we will show that $q^{-1}(Z) \not\in \calB$. We have

\begin{align*}
q^{-1}(Z)[0] & = Z \times X \\ q^{-1}(Z)[1] & = \{(x_1, x_2) : x_1
\in Z, x_2 \neq \sigma(x_1)\} \cup \{(\sigma^{-1}(x), x) : x \in
\end{align*}

It follows that $q^{-1}(Z)[0] \ominus q^{-1}(Z)[1] = \{(x, \sigma(x))
x \in Z\} \cup \{(\sigma^{-1}(x), x) : x \in Z\}$ is
uncountable. Thus, $q^{-1}(Z) \not\in \calB$.

It only remains to prove that $q$ is supremum preserving. Let $\calE
\subset \calC(X)$ be a collection that has an
$\{\emptyset\}$-essential supremum $S$. This implies that $S =
\bigcup \calE$. We wish to prove that $q^{-1}(S)$ is an
$\calM$-essential supremum of $q^{-1}(\calE) = \{q^{-1}\{x\} : x
\in S\}$. First suppose that $\calE$ consists
only of singletons. Of course, $q^{-1}(S)$ is an $\calM$-essential
upper bound of $q^{-1}(\calE)$. Let $U$ an arbitrary $\calM$-essential upper bound of
$q^{-1}(\calE)$.  For all $x \in S$, we have
$q^{-1}\{x\} \setminus U \in \calM$, meaning that $\{x\} \times X =
(q^{-1}\{x\})[0] \subset U[0]$. Thus $S \times X \subset U[0]$,
which implies $(q^{-1}(S) \setminus U)[0] = q^{-1}(S)[0] \setminus
U[0] = S\times X \setminus U[0] = \emptyset$. It means that
$q^{-1}(S) \setminus U \in \calM$. Thus, we have shown that
$q^{-1}(S)$ is an $\calM$-essential supremum of $\calE$.

Now we turn to the general case, where $\calE$ need not consist only
of singletons. Let $\calE' = \{ \{x\} : x \in E \in
\calE\}$. Clearly, $\calE$ and $\calE'$ have the same
$\{\emptyset\}$-essential supremum $S \defeq \bigcup \calE = \bigcup
\calE'$. By what precedes, $q^{-1}(S)$ is an $\calM$-essential
supremum of $q^{-1}(\calE')$ and it is an $\calM$-essential upper
bound of $q^{-1}(\calE)$. An $\calM$-essential upper bound $U$ of
$q^{-1}(\calE)$ is also an upper bound for $q^{-1}(\calE')$, as any
member of $q^{-1}(\calE')$ is a subset of a member of
$q^{-1}(\calE)$. Therefore, $q^{-1}(S) \setminus U \in \calM$,
showing that $q^{-1}(S)$ is an $\calM$-essential supremum of

[Example of an MSN with no localizable part]

Consider an MSN of the form $(X, \calP(X), \calK(X))$, where $X$ is
a set of cardinality $\aleph_1$ and $\calK(X)$ is the
$\sigma$-ideal of countable subsets. There is a bijection $\varphi
\colon X \to X \times X$ and we can use it to construct an
uncountable family of “horizontal lines” $H_x \defeq
\varphi^{-1}(X \times \{x\})$ indexed by $x \in X$ witnessing that
$(X, \calP(X), \calK(X))$ is not ccc. Actually, we can do better
and prove that it is not localizable. Suppose $\{H_x : x \in X\}$
has a $\calK(X)$-essential supremum $S$. For each $x \in X$ choose
a point $p_x \in S \cap H_x$. Then it is easy to see that $U \defeq
S \setminus \{p_x : x \in X\}$ is an essential upper bound for the
family of horizontal lines, however $S \setminus U = \{p_x : x \in
X\}$ is not negligible, contradicting that $S$ is an essential

Observe that the MSN $(X, \calP(X), \calK(X))$ is isomorphic to all
its non negligible subMSNs. In particular, it has no nontrivial ccc
or localizable part, an unpleasant situation that we will rule out
in the next paragraph by introducing the notions of “locally
localizable” and “locally ccc” MSN.

We will prove nonetheless that $(X, \calP(X), \calK(X))$ has a cccc
version, that is disappointingly the trivial MSN $(\emptyset,
\{\emptyset\}, \{\emptyset\})$ (with the only morphism from there
to $(X, \calP(X), \calK(X))$). To establish this fact, one needs to
prove that if $(Y, \calB, \calM)$ is a cccc MSN and $\mathbf{f}
\colon (Y, \calB, \calM) \to (X, \calP(X), \calK(X))$ is a supremum
preserving morphism, then $Y \in \calM$ (actually, the supremum
preserving character of $\mathbf{f}$ will not be used). We can
reduce to the case where $(Y, \calB, \calM)$ is ccc.

We reproduce an argument due to Ulam <cit.>, showing that
there is a family $\la A_{n,\alpha} \ra_{n \in \N, \alpha <
\omega_1}$ of subsets of $X$ such that

* for all $n \in \N$, the family $\la A_{n, \alpha} \ra_{\alpha
< \omega_1}$ is disjointed;

* for all $\alpha < \omega_1$, the union $\bigcup_{n \in \N}
A_{n, \alpha}$ is conegligible (that is, cocountable).

Any ordinal $\beta < \omega_1$ is countable, so we can select a
sequence $\la k_{\alpha, \beta} \ra_{\alpha < \beta}$ of distinct
integers. Let $\la x_\beta \ra_{\beta < \omega_1}$ be an enumeration
of all the elements in $X$. Set $A_{n, \alpha} \defeq \{x_\beta :
\beta > \alpha \text{ and } k_{\alpha, \beta} = n\}$ for every $n
\in \N$ and $\alpha < \omega_1$. For distinct $\alpha, \alpha' <
\omega_1$, there cannot be some $x_\beta \in A_{n, \alpha} \cap
A_{n, \alpha'}$, for otherwise we would have $k_{\alpha, \beta} =
k_{\alpha', \beta}$. In addition, one has $\bigcup_{n \in \N} A_{n,
\alpha} = \{x_\beta : \beta > \alpha\}$ whose complement in $X$ is
the countable set $\{x_\beta : \beta \leq \alpha\}$.

Now, fix a representative $f \in \mathbf{f}$. The family $\la
f^{-1}(A_{n, \alpha}) \ra_{\alpha < \omega_1}$ being disjointed,
the set $C_n \defeq \omega_1 \cap \{\alpha : f^{-1}(A_{n, \alpha})
\not\in \calM\}$ is countable for all $n \in \N$. Hence the
existence of some $\alpha < \omega_1 \setminus \bigcup_{n \in \N}
C_n$. Now we see that $f^{-1}\left( \bigcup_{n \in \N} A_{n, \alpha}
\right)$ is both negligible and conegligible in $Y$, which can
happen only if $Y \in \calM$.

Let $\calP$ be a hereditary property associated to MSNs. We say that
an MSN $(X, \calA, \calN)$ is locally $\calP$ whenever one of
the following equivalent statements holds:

(A) The collection $\calA \cap \{Z : (Z, \calA_Z, \calN_Z)
\text{ has the property } \calP\}$ is $\calN$-generating;

(B) for any $Y \in \calA \setminus \calN$ there is $Z \in
\calA \setminus \calN$ such that $Z \subset Y$ and $(Z, \calA_Z,
\calN_Z)$ has the property $\calP$.

(A) $\implies$ (B) For $Y \in \calA \setminus \calN$, an
application of Lemma <ref> gives that $Y$ is an
essential supremum of $\{Y \cap Z : Z \in \calA \text{ and } (Z,
\calA_Z, \calN_Z) \text{ has the property } \calP\}$. Therefore,
there must be some $Z \in \calA$ such that $Y \cap Z \not\in
\calN$ and $(Z, \calA_Z, \calN_Z)$ has the property $\calP$. The
subset $Y \cap Z \subset Y$ establishes (B).

(B) $\implies$ (A) Clearly, $X$ is an $\calN$-essential upper
bound of the collection $\calA \cap \{Z : (Z, \calA_Z, \calN_Z)
\text{ has the property } \calP\}$. Let $S$ be another upper
bound. If $X \setminus S$ were not negligible, (B) gives the
existence of some measurable $Z \in \calA \setminus \calN$ such
that $Z \subset X \setminus S$ and $(Z, \calA_Z, \calN_Z)$ has the
property $\calP$. But $Z \setminus S = Z \not\in \calN$, which
contradicts that $S$ is an essential upper bound.

For instance, in a semi-finite measure space $(X, \calA, \mu)$, any
non negligible set $A \in \calA \setminus \calN_\mu$ contains a
measurable subset $Z$ of nonzero finite measure. By (B), this
implies that the associated MSN $(X, \calA, \calN_\mu)$ is locally
strictly localizable.

We conclude this section with an important property of “local
isomorphism” that holds for $\calP$-versions.

Let $(X, \calA, \calN)$ be a saturated MSN and $[(\hat{X},
\hat{\calA}, \hat{\calN}), \bp]$ a $\calP$-version of it. Fix a
representative map $p \in \bp$. For any $F \in \calA$, we set
$\hat{F} \defeq p^{-1}(F)$ and we call $\bp_F \colon (\hat{F},
\hat{\calA}_{\hat{F}}, \hat{\calN}_{\hat{F}}) \to (F, \calA_F,
\calN_F)$ the morphism induced by the restriction $p_F \colon
\hat{F} \to F$ of $p$.

(A) $[(\hat{F}, \hat{\calA}_{\hat{F}},
\hat{\calN}_{\hat{F}}), \bp_F]$ is the $\calP$-version of $(F,
\calA_F, \calN_F)$;

(B) If $(F, \calA_F, \calN_F)$ has the property $\calP$, then
$\bp_F$ is an isomorphism.

(A) Since the property $\calP$ is hereditary, we can assert
$(\hat{F}, \hat{\calA}_{\hat{F}}, \hat{\calN}_{\hat{F}})$ has it. We
also readily check that $p_F$ is supremum preserving. Let $\bq
\colon (Y, \calB, \calM) \to (F, \calA_F, \calN_F)$ be a supremum
preserving morphism starting from a saturated MSN with the property
$\calP$. Then $\boldsymbol{\iota_F} \circ \bq$ is a supremum
preserving morphism ending in $(X, \calA, \calN)$. It has a lifting
$\br \colon (Y, \calB, \calM) \to (\hat{X}, \hat{\calA},
\hat{\calN})$. Let $r \in \br$ and $q \in \bq$ be
representatives. As $p(r(y)) = q(y)$ for $\calM$-almost all $y \in
Y$, we lose no generality in supposing that $r$ has values in
$\hat{F}$. Calling its restriction $r' \colon Y \to \hat{F}$, we see
that $p_F \circ r'$ and $q$ coincide $\calM$-almost everywhere. The
induced morphism $\br'$ provides a factorization of $\bq$ through
$(\hat{F}, \hat{\calA}_{\hat{F}}, \hat{\calN}_{\hat{F}})$.

To establish the uniqueness of this factorization, we proceed as
follows. For any morphism $\br''$ such that $\bp_F \circ \br'' =
\bq$ we notice that $\boldsymbol{\iota}_F \circ \bq =
\boldsymbol{\iota}_F \circ \bp_F \circ \br'' = \bp \circ
\boldsymbol{\iota}_{\hat{F}} \circ \br''$. Since this holds for
$\br'$ we obtain $\bp \circ \boldsymbol{\iota}_{\hat{F}} \circ \br'
= \bp \circ \boldsymbol{\iota}_{\hat{F}} \circ \br''$ and, by
uniqueness of the factorization relative to the universal property
of $(\hat{X},\hat{\calA},\hat{\calN})$,
$\boldsymbol{\iota}_{\hat{F}} \circ \br' =
\boldsymbol{\iota}_{\hat{F}} \circ \br''$. Thus, $r'$ and $r''$
coincide $\calM$-almost everywhere.

(B) If $(F, \calA_F, \calN_F)$ has property $\calP$, then obviously
$[(F, \calA_F, \calN_F), \boldsymbol{\mathrm{id}}]$ is a second
$\calP$-version. From the uniqueness of the $\calP$-version, we
obtain a isomorphism $\br \colon (F, \calA_F, \calN_F) \to (\hat{F},
\hat{\calA}_{\hat{F}}, \hat{\calN}_{\hat{F}})$ such that
$\boldsymbol{\mathrm{id}} = \bp_F \circ \br$, whence $\bp_F =
\br^{-1}$.

§ LOCALIZABLE LOCALLY DETERMINED MSNS

In order to motivate the main definition in this section, we start with the following result, of which we can think as a way of testing whether an
MSN has a property $\calP$. For instance, if each $F \in \calF$ corresponds to a ccc subMSN, then $(X,\calA,\calN)$ is cccc. The difficulty in applying this proposition stems with both hypotheses: conditions (1) and (2)
will be turned into a definition in <ref>, whereas condition (3), that $\calF$ is disjointed rather than merely almost disjointed, calls for techniques that transform almost disjointed generating families (whose existence, in applications, follows from Lemma <ref>) into partitions – see the proof of Theorem <ref> in case $\rmcard \calE \leq \mathfrak{c}$ and the notion of compatible family of densities introduced in Section <ref>.

Let $(X,\calA,\calN)$ be an MSN and $\calF \subset \calA$. Assume that

(1) For every $A \subset X$ the following holds:

\[
\big[ \forall F \in \calF : A \cap F \in \calA \big] \Rightarrow A \in \calA ;
\]

(2) For every $N \subset X$ the following holds:

\[
\big[ \forall F \in \calF : N \cap F \in \calN \big] \Rightarrow N \in \calN ;
\]

(3) $\calF$ is a partition of $X$.

Then the MSNs $(X,\calA,\calN)$ and $\coprod_{F \in \calF} (F,\calA_F,\calN_F)$ are isomorphic in $\sfMSNsp$.

We abbreviate $(Y,\calB,\calM)$ for $\coprod_{F \in \calF}
(F,\calA_F,\calN_F)$. Since $\calF$ is a partition of $X$, there is
a canonical bijection $\vphi : X \to Y$. Its inverse $\vphi^{-1}$ is
$[(\calB,\calM),(\calA,\calN)]$-measurable, by definition of
coproduct of MSNs. We now show that $\vphi$ is
$[(\calA,\calN),(\calB,\calM)]$-measurable. Given $B \in \calB$, we
note that $\vphi^{-1}(B) = \bigcup_{F \in \calF} B \cap F$, whence
$\vphi^{-1}(B) \cap F = B \cap F \in \calA$ for every $F \in \calF$,
by definition of $\calB$. We infer from hypothesis (1) that
$\vphi^{-1}(B) \in \calA$. Let $M \in \calM$. As above we infer from
the definition of $\calM$ that $\vphi^{-1}(M) \cap F \in \calN$ for
every $F \in \calF$, whence $\vphi^{-1}(M) \in \calN$, in view of
hypothesis (2). In other words, $(X,\calA,\calN)$ and
$(Y,\calB,\calM)$ are isomorphic in $\sfMSN$. The conclusion follows
from Proposition <ref>(A).

We borrow the following definition from <cit.>. A measure space $(X, \calA, \mu)$
is locally determined whenever it is semi-finite and, for every
subset $A \subset X$,

\[
\big[ \forall E \in \calA^f : A \cap E \in \calA \big] \Rightarrow A \in \calA,
\]

where, as usual, $\calA^f = \calA \cap \{ E : \mu(E) < \infty \}$.

The definition relies on the particular collection $\calA^f$ (which
is $\calN_\mu$-generating, recall <ref>). This makes sense
because we are dealing with a measure space. It is a rather
good surprise that we can define an analogous notion of locally
determined MSNs, by substituting for $\calA^f$ an arbitrary
generating collection. Namely, a saturated MSN $(X, \calA, \calN)$
is called locally determined whenever the following holds. For every $\calN$-generating collection $\calE \subset \calA$ and every $A \subset X$,

\[
\big[ \forall E \in \calE : A \cap E \in \calA\big] \Rightarrow A \in \calA .
\]

An MSN that
is both localizable and locally determined is called lld.

The following is useful as well. We say that a saturated MSN $(X,\calA,\calN)$ has locally determined negligible sets whenever the following holds. For every $\calN$-generating collection $\calE \subset \calA$ and every $N \subset X$,

\[
\big[ \forall E \in \calE : N \cap E \in \calN\big] \Rightarrow N \in \calN .
\]

We observe that if $(X,\calA,\calN)$ is locally determined, then it has
locally determined negligible sets. Indeed, let $\calE \subset \calA$ and
$N \subset X$ be as above, we first infer from the local determinacy of $(X,\calA,\calN)$ that $N \in \calA$ and, in turn from the Distributivity Lemma <ref>, that $N$ is an $\calN$-essential supremum of $\{ N \cap E : E \in \calE \}$. Therefore, $N \in \calN$.

Next we prove some elementary properties concerning locally
determined MSNs. In particular, the consistency between both notions
of local determination (for complete semi-finite measure spaces and MSNs) is established in
Proposition <ref>(F). Here, the semi-finiteness property of a measure space is critical as the following example shows. We consider $\calH^1$, the 1-dimensional Hausdorff measure in $\R^2$ and $\calA$ the $\sigma$-algebra consisting of $\calH^1$-measurable subsets of $\R^2$ in the sense of Carathéodory. The following hold:

(a) $\forall A \subset \R^2 : \big[ \forall F \in \calA^f : A \cap F \in \calA \big] \Rightarrow A \in \calA$;

(b) The measure space $(\R^2,\calA,\calH^1)$ is not semi-finite;

(c) The (saturated) MSN $(\R^2,\calA,\calN_{\calH^1})$ does not have locally determined negligible sets and, in particular, is not locally determined.

For (a), see for instance <cit.>. For (b), see <cit.>. Now (c) follows for example from <cit.>. It follows from <ref> that $\calA^f$ is not $\calN_{\calH^1}$-generating.

The following hold.

(A) Being locally determined is a hereditary property.

(B) Being locally determined is a property invariant under
isomorphisms in $\mathsf{MSN_{sp}}$.

(C) A saturated ccc MSN is locally determined.

(D) A coproduct of locally determined MSNs is locally determined.

(E) A cccc MSN is locally determined.

(F) A complete semi-finite measure space $(X, \calA, \mu)$ is
locally determined (as a measure space) if and only if the MSN
$(X, \calA, \calN_\mu)$ is locally determined.

(A) Let $(X, \calA, \calN)$ be a locally determined MSN and $Z \in
\calA$. Let $\calE$ be an $\calN_Z$-generating family in the subMSN
$(Z, \calA_Z, \calN_Z)$ and $A \subset Z$ be such that $E \cap A \in
\calA_Z$ for any $E \in \calE$. The family $\calE \cup \{X \setminus
Z\}$ is $\calN$-generating in $(X, \calA, \calN)$ and $E \cap A \in
\calA$ for all $E \in \calE \cup \{X \setminus Z\}$. It follows that
$A \in \calA$.

(B) Let $(X, \calA, \calN)$ and $(Y, \calB, \calM)$ be two
saturated MSNs, $f \colon X \to Y$ and $g \colon Y \to X$ be two
measurable supremum preserving maps that induce reciprocal
isomorphisms. Assume that $(X, \calA, \calN)$ is locally
determined. Let $\calE \subset \calB$ be an $\calM$-generating
collection and $B \subset Y$ be such that $E \cap B \in \calB$ for all
$E \in \calE$. Then $f^{-1}(E) \cap f^{-1}(B) = f^{-1}(E \cap B) \in
\calA$. As $f$ is supremum preserving, $f^{-1}(\calE)$ is
$\calN$-generating. And as $(X, \calA, \calN)$ is locally
determined, we infer that $f^{-1}(B) \in \calA$. Therefore
$g^{-1}(f^{-1}(B)) \in \calB$. But $B \ominus g^{-1}(f^{-1}(B)) \in
\calM$ and as $(Y, \calB, \calM)$ is saturated we conclude that $B
\in \calB$.

(C) Let $(X, \calA, \calN)$ be a saturated MSN, $\calE \subset
\calA$ an $\calN$-generating family and $A \in \calP(X)$ be such
that $E \cap A \in \calA$ for all $E \in \calE$. By
Proposition <ref>, there is a countable subset $\calE'
\subset \calE$ that is $\calN$-generating. Then

\[
A = \left(\bigcup_{E \in \calE'} E \cap A\right) \cup \left(A \setminus \bigcup
\calE' \right)
\]

Since $A \setminus
\bigcup \calE'$ is $\calN$-negligible and $(X, \calA, \calN)$ is saturated, we infer that $A \setminus
\bigcup \calE'$ is
$\calA$-measurable. Therefore, $A \in \calA$.

(D) Let $(X, \calA, \calN)$ be the coproduct of a family $\la(X_i,
\calA_i, \calN_i)\ra_{i \in I}$ of locally determined MSNs. It is readily saturated. Let
$\calE \subset \calA$ be an $\calN$-generating family and $A \subset
X$ such that $E \cap A \in \calA$ for all $E \in \calE$. For all $i
\in I$, the family $\calE_i \defeq \{E \cap X_i : E \in \calE\}$ is
$\calN_i$-generating in $(X_i, \calA_i, \calN_i)$ by
Lemma <ref>. This observation leads to the fact that $A
\cap X_i \in \calA_i$ for all $i \in I$, in other words, $A \in
\calA$.

(E) This obviously follows from (C) and (D).

(F) Suppose that the measure space $(X, \calA, \mu)$ is locally
determined. Let $\calE \subset \calA$ be an $\calN_\mu$-generating
family and $A \subset X$ be such that $E \cap A \in \calA$ for all
$E \in \calE$. Let $F \in \calA^f$. By Lemma <ref>, the
collection $\{F \cap E : E \in \calE\}$ is $\calN_F$-generating in
$(F, \calA_F, (\calN_\mu)_F)$ and of course $F \cap E \cap A \in
\calA_F$ for all $E \in \calE$. On top of that, $(F, \calA_F,
(\calN_\mu)_F)$ is a ccc MSN by Proposition <ref> and it is saturated. We
get from (B) above that $A \cap F$ is measurable. As this happens
for all $F \in \calA^f$, we conclude that $A \in \calA$.

Conversely, suppose the MSN $(X, \calA, \calN_\mu)$ is locally
determined. Owing to the semi-finiteness of $(X, \calA, \mu)$, the
collection $\calA^f$ is $\calN_\mu$-generating. Then $(X, \calA,
\mu)$ is easily seen to be locally determined: if $A \in \calP(X)$
satisfies $A \cap F \in \calA$ for all $F \in \calA^f$, then $A \in
\calA$.

[lld version of an atomic MSN]

As a first result, we mention that the lld version of an atomic MSN
$(X, \calA, \{\emptyset\})$ is $[(X, \calP(X), \{\emptyset\}),
\boldsymbol{\iota}]$, where $\boldsymbol{\iota}$ is the morphism
induced by the identity map (that $\boldsymbol{\iota}$ is
supremum preserving follows from <ref>). This amounts to
prove that, for any lld MSN $(Y, \calB, \calM)$, a $[(\calB, \calM),
(\calA, \{\emptyset\})]$-measurable supremum preserving map $q
\colon Y\to X$ is automatically $(\calB,
\calP(X))$-measurable. Indeed, let $S \in \calP(X)$. Then $q^{-1}(S)
\cap q^{-1}\{x\}$ is either $q^{-1}\{x\}$ or $\emptyset$, hence
$q^{-1}(S) \cap q^{-1}\{x\}$ is $\calB$-measurable for every $x \in X$.
Besides, $q$
is supremum preserving, thus the collection $\{q^{-1}\{x\} : x \in
X\}$ is $\calM$-generating. By local determination in $(Y, \calB,
\calM)$ we conclude that $q^{-1}(S) \in \calB$.

Call $\mathsf{LLD_{sp}}$ the full subcategory of $\sfMSNsp$ that
consists of lld MSNs, and consider the forgetful functor
$\mathsf{Forget} \colon \mathsf{LLD_{sp}} \to \sfMSNsp$. In
categorical terms, an lld version is the coreflection of an MSN $(X,
\calA, \calN)$ along the functor $\mathsf{Forget}$, see
<cit.>. In this paper, we do not answer the question whether there exists an lld version for
each saturated MSN. This is equivalent
to the existence of a right adjoint $\mathsf{R}$ of
$\mathsf{Forget}$. As a matter of fact, if such an adjoint exists,
there would be a natural transformation $\varepsilon \colon
\mathsf{Forget} \circ \mathsf{R} \implies \rmid_{\sfMSNsp}$ such
that the pair $[\mathsf{R}(X, \calA, \calN), \varepsilon_{(X, \calA,
\calN)}]$ gives the lld version of any saturated MSN $(X, \calA,
\calN)$.

In search for an abstract proof of the existence of $\mathsf{R}$,
one might think of using Freyd's Adjunction Theorem,
<cit.>. Following this path, one needs to
establish (setting aside the solution set condition) that:

(A) The category $\sfMSNsp$ is cocomplete;

(B) The forgetful functor $\mathsf{Forget}$ preserves small

Assertion (A) boils down to showing that $\sfMSNsp$ has two types of
small colimits: coproducts and coequalizers. The existence of the
former is shown in Proposition <ref>(D). We do not know
whether coequalizers exist in $\sfMSNsp$ and it is the main
difficulty here.

As for (B), which, regarding the existence of lld versions, is a
necessary condition even if (A) were to be false, we have already
proven in Propositions <ref> and <ref>(D) that
coproducts of lld MSNs are lld. Our next goal is
Proposition <ref> which states that coequalizers of lld MSNs
are lld. Before that, we need to introduce some notation and a

For a saturated MSN $(X, \calA, \calN)$ and an arbitrary $\calE \subset
\calA$, we define

\begin{align*}
\calA_{\calE} & \defeq \calP(X) \cap \{A : E \cap A \in \calA \text{
for all } E \in \calE\}, \\ \calN_{\calE} & \defeq \calP(X) \cap
\{N : E \cap N \in \calN \text{ for all } E \in \calE\}.
\end{align*}

It is clear that $(X, \calA_{\calE}, \calN_{\calE})$ is a saturated

Let $(X, \calA, \calN)$ be a localizable saturated MSN and $\calE
\subset \calA$ an $\calN$-generating family. Let
$\boldsymbol{\iota}$ be the morphism $(X, \calA_{\calE}, \calN_{\calE})
\to (X, \calA, \calN)$ induced by the identity map on $X$. Then
$\mathsf{Bool}(\boldsymbol{\iota})$ is an isomorphism. In
particular, $\boldsymbol{\iota}$ is supremum preserving and $(X,
\calA_{\calE}, \calN_{\calE})$ is localizable.

First we make the following observation, to be used later in the
proof: $\calA \cap \calN_{\calE} = \calN$. Indeed, if $N \in \calA$ is
such that $E \cap N \in\calN$ for all $E \in \calE$, then by the
distributivity lemma <ref>, we conclude that $N \in
\calN$. This proves the inclusion $\calA \cap \calN_{\calE} \subset
\calN$, the reciprocal being trivial.

The identity map $\iota \colon X \to X$ is $[(\calA_{\calE},
\calN_{\calE}), (\calA,\calN)]$-measurable because $\calA \subset
\calA_{\calE}$ and $\calN \subset \calN_{\calE}$. Let us show that
$\mathsf{Bool}(\boldsymbol{\iota}) \colon \calA /\calN \to
\calA_{\calE} / \calN_{\calE}$ is injective by inspecting its
kernel. Let $\boldsymbol{A} \in \calA / \calN$ be a class
represented by $A \in \calA$ such that
$\mathsf{Bool}(\boldsymbol{\iota})(\boldsymbol{A}) = 0$, in other
words $A = \iota^{-1}(A) \in \calN_{\calE}$. Then $A \in \calA \cap
\calN_{\calE} = \calN$. This means that $\boldsymbol{A} = 0$ in $\calA
/ \calN$. Therefore, $\mathsf{Bool}(\boldsymbol{\iota})$ is

Now let us show that $\mathsf{Bool}(\boldsymbol{\iota})$ is surjective.
this end, let $\boldsymbol{H} \in \calA_{\calE} / \calN_{\calE}$ be a
class represented by $H \in \calA_{\calE}$. We ought to prove
that $\boldsymbol{H}$ is in the range of
$\mathsf{Bool}(\boldsymbol{\iota})$. Set $\calF \defeq \{E \cap H :
E \in \calE\}$. Note that $\calF \subset \calA$. The localizability
of $(X, \calA, \calN)$ guarantees that $\calF$ has an
$\calN$-essential supremum $S \in \calA$. In particular, $E \cap H
\setminus S \in \calN$ for all $E \in \calE$, meaning that $H
\setminus S \in \calN_{\calE}$.

We also claim that $S \setminus H \in \calN_{\calE}$. Indeed, let $E_0
\in \calE$. Set $S' \defeq S \setminus (E_0 \cap S \setminus H)$. We
note that $S' \in \calA$. For all $E \in \calE$, we have $E \cap H
\setminus S' = E \cap H \setminus S \in \calN$, as $H \setminus S
\in \calN_{\calE}$. This means that $S'$ is an $\calN$-essential upper
bound of $\calF$. It follows that $S \setminus S' = E_0 \cap H
\setminus S \in \calN$. As $E_0\in \calE$ is arbitrary, we obtain $S
\setminus H \in \calN_{\calE}$, as required.

We proved that $H \ominus S \in \calN_{\calE}$. Calling
$\boldsymbol{S}$ the equivalence class of $S$ in $\calA / \calN$, we
have that $\mathsf{Bool}(\boldsymbol{\iota})(\boldsymbol{S}) =
\boldsymbol{H}$.

Consider the following diagram in $\sfMSNsp$, where $((Z, \calC,
\calP), \bh)$ is the coequalizer of $\mathbf{f}, \bg$.

\[
\begin{tikzcd}
(X, \calA, \calN) \arrow[r, shift left, "\mathbf{f}"] \arrow{r}[shift
right, below]{\bg} & (Y, \calB, \calM) \arrow{r}{\bh} &
(Z, \calC, \calP)
\end{tikzcd}
\]

(A) If $(Y, \calB, \calM)$ is
localizable, so is $(Z, \calC, \calP)$.

(B) If $(Y, \calB, \calM)$ is lld,
so is $(Z, \calC, \calP)$.

(A) Let us call $\boldsymbol{2}$ the special MSN $(\{0, 1\},
\calP(\{0,1\}), \{\emptyset\})$.  First we show the following
intermediate result: for any MSN $(X, \calA, \calN)$, there is a
one-to-one correspondence $\Upsilon_X$ between the Boolean algebra
$\mathsf{Bool}(X, \calA,\calN)$ and the set of morphisms $\rmHom((X,
\calA, \calN), \boldsymbol{2})$ (those are automatically supremum
preserving since the Boolean algebra of $\boldsymbol{2}$ is
finite). Given a class $\boldsymbol{A} \in \mathsf{Bool}(X, \calA,
\calN)$, represented by a set $A$, the characteristic function
$\ind_A \colon X \to \{0, 1\}$ induces a morphism $\boldsymbol{1_A}$
which only depends on the equivalence class
$\boldsymbol{A}$. Indeed, if $A'$ is another representative of
$\boldsymbol{A}$, then $\ind_A$ and $\ind_{A'}$ coincide
$\calN$-almost everywhere. We set $\Upsilon_X(\boldsymbol{A}) \defeq
\boldsymbol{1_A}$.

This map is surjective because each morphism $\boldsymbol{\varphi} \colon
(X, \calA, \calN) \to \boldsymbol{2}$ is represented by a map
$\varphi \in \boldsymbol{\varphi}$ which has the form $\varphi =
\ind_{\varphi^{-1}(\{1\})}$. It is injective because if $\ind_A$
coincides with $\ind_B$ almost everywhere, for measurable sets $A$
and $B$, then $A$ and $B$ yield the same equivalence class in
$\mathsf{Bool}(X, \calA, \calN)$.

Now we turn to the proof of conclusion (A). By naturality of
$\Upsilon$, the following diagram is commutative, where $\rmHom(\bh,
\mathbf{2})$, $\rmHom(\mathbf{f}, \mathbf{2})$ and $\rmHom(\bg,
\mathbf{2})$ denote the right composition with $\bh$, $\mathbf{f}$,
and $\bg$, respectively.

\[
\begin{tikzcd}
\rmHom((Z, \calC, \calP), \boldsymbol{2})  \arrow[rr, "{\rmHom(\bh, \boldsymbol{2})}"] & &
\rmHom((Y, \calB, \calM), \boldsymbol{2})  \arrow[rr, shift left, "{\rmHom(\mathbf{f},
\boldsymbol{2})}"] \arrow{rr}[shift right,
below]{\rmHom(\mathbf{g}, \boldsymbol{2})} & & \rmHom((X, \calA,
\calN), \boldsymbol{2})
\\ \mathsf{Bool}(Z, \calC, \calP)\arrow[u,
"{\Upsilon_Z}"] \arrow[rr,
"{\mathsf{Bool}(\bh)}"] & & \mathsf{Bool}(Y, \calB, \calM)\arrow[u,
\arrow[rr, shift left, "{\mathsf{Bool}(\mathbf{f})}"]
\arrow{rr}[shift right, below]{\mathsf{Bool}(\bg)}
\mathsf{Bool}(X,\calA,\calN)\arrow[u, "{\Upsilon_X}"]
\end{tikzcd}
\]

We show that $\rmHom(\bh,\mathbf{2})$ is injective. Indeed, if
$\boldsymbol{\vphi}$ and $\boldsymbol{\psi}$ are such that
then, upon letting $\bk = \boldsymbol{\vphi} \circ \bh =
\boldsymbol{\psi} \circ \bh$, we infer that $\bk \circ \mathbf{f} =
\bk \circ \bg$. By the universal property of $(Z,\calC,\calP)$,
there exists a unique $\boldsymbol{\ell} \in
\rmHom((Z,\calC,\calP),\mathbf{2})$ such that $\boldsymbol{\ell}
\circ \bh = \bk$. Since $\boldsymbol{\vphi}$ and $\boldsymbol{\psi}$
have the property of $\boldsymbol{\ell}$, we conclude that they
coincide. Similarly, the universal property of coequalizers tells us
that the range of $\rmHom(\bh,\mathbf{2})$ consists of those
morphisms $\bk$ such that $\rmHom(\mathbf{f},
\mathbf{2})(\bk) = \rmHom(\bg,
\mathbf{2})(\bk)$. On the second line of the
diagram, these two observations translate to the fact that
$\mathsf{Bool}(\bh)$ induces an isomorphism of Boolean algebras from
$\mathsf{Bool}(Z, \calC, \calP)$ onto the Boolean subalgebra

\[
\frA \defeq \mathsf{Bool}(Y, \calB, \calM) \cap \{ \xi :
\mathsf{Bool}(\mathbf{f})(\xi) = \mathsf{Bool}(\bg)(\xi)\}.
\]

It remains to prove that $\frA$ is Dedekind complete. Let $\frE
\subset \frA$ be a collection. It has a supremum $s$ in
$\mathsf{Bool}(Y, \calB, \calM)$, as $(Y, \calB, \calM)$ is
localizable. Since $\mathbf{f}$ and $\bg$ are supremum preserving,
we have $\mathsf{Bool}(\mathbf{f})(s) = \sup
\mathsf{Bool}(\mathbf{f})(\frE) = \sup
\mathsf{Bool}(\mathbf{g})(\frE) = \mathsf{Bool}(\bg)(s)$. Hence $s
\in \frA$ and $\frA$ is Dedekind complete.

(B) That $(Z, \calC, \calP)$ is localizable follows from (A). Let
$\calG \subset \calC$ be any $\calP$-generating family.  We wish to
prove that $\calC_{\calG} \subset \calC$. If we manage to do so,
then $(Z, \calC, \calP)$ is locally determined, as $\calG$ is

Let $h$ be a representative of $\bh$. By definition, it is $[(\calB,
\calM), (\calC, \calP)]$-measurable and supremum preserving. We
claim that it is, in fact, $[(\calB, \calM), (\calC_{\calG},
\calP_{\calG})]$-measurable. Indeed, let $C \in \calC_{\calG}$. For
all $G \in \calG$, we have $G \cap C \in \calC$ which implies that
$h^{-1}(G) \cap h^{-1}(C) = h^{-1}(G \cap C) \in \calB$. Moreover,
$h^{-1}(\calG)$ is $\calM$-generating, as $\calG$ is $\calP$-generating
and $h$ is
supremum preserving. Thus, since $(Y, \calB, \calM)$ is locally
determined, we have that $h^{-1}(C) \in \calB$. Next, if $P \in
\calP_{\calG}$, then $h^{-1}(P) \in \calB$ by what precedes and $h^{-1}(P) \cap
h^{-1}(G) = h^{-1}(P \cap G) \in \calM$ for all $G \in \calG$. By
the distributivity lemma <ref>, we obtain $h^{-1}(P) \in
\calM$.

Denote as $\bh' \colon (Y,\calB, \calM) \to (Z, \calC_{\calG},\calP_{\calG})$
the morphism induced by $h$, and denote as $\boldsymbol{\iota} \colon (Z,
\calC_{\calG},\calP_{\calG}) \to (Z, \calC, \calG)$ the morphism
induced by the identity map $\rmid_Z$. By Lemma <ref>, we have
$\mathsf{Bool}(\bh') = \mathsf{Bool}(\boldsymbol{\iota})^{-1} \circ
\mathsf{Bool}(\bh)$, which is the composition of two
supremum preserving morphisms of Boolean algebras. Thus, $\bh'$ is a
supremum preserving as well. Also, we recall $\bh \circ \mathbf{f} =
\bh \circ \bg$. As $\bh$ and $\bh'$ are induced by the same map, we
deduce that $\bh' \circ \mathbf{f} = \bh' \circ \bg$. By the
universal property of coequalizers, there is a morphism $\bk \colon
(Z, \calC, \calP) \to (Z, \calC_{\calG}, \calP_{\calG})$ such that $\bh'
= \bk \circ \bh$.

\[
\begin{tikzcd}
& & (Z, \calC_{\calG}, \calP_{\calG}) \arrow[d, shift left, "\boldsymbol{\iota}"]
\\ (X, \calA, \calN) \arrow[r, shift left, "\mathbf{f}"]
\arrow{r}[shift right, below]{\bg} & (Y, \calB, \calM)
\arrow{r}{\bh} \arrow[ru, "\bh'"] & (Z, \calC, \calP)
\arrow[u, "\bk"]
\end{tikzcd}
\]

Hence $\boldsymbol{\iota} \circ \bk \circ \bh = \boldsymbol{\iota}
\circ \bh' = \bh = \rmid_{(Z,\calC,\calP)} \circ \bh$. The
uniqueness in the universal property of equalizers implies that
$\bh$ is an epimorphism. Thus $\boldsymbol{\iota} \circ \bk =
\rmid_{(Z,\calC,\calP)}$. A representative $k \in \bk$ must satisfy
$z = \rmid_Z(k(z)) = k(z)$ for $\calP$-almost all $z \in Z$,
i.e. $P = Z \cap \{ z : z \neq k(z) \} \in \calP$. Let $C \in
\calC_{\calG}$. Since $k$ is $(\calC,\calC_{\calG})$-measurable, it
follows that $k^{-1}(C) \in \calC$. Since $k^{-1}(C) \ominus C
\subset P$, we deduce that $k^{-1}(C) \ominus C \in \calC$ and, in
turn, $C \in \calC$.

The last two results of this section show that the category $\mathsf{LLD_{sp}}$
has better categorical properties than $\sfMSNsp$: it has equalizers
in full generality. This result is reminiscent of

Let $(X, \calA, \calN)$ be an lld MSN and $Y \subset X$ any
subset. Then the subMSN $(Y, \calA_Y, \calN_Y)$ is lld and the
canonical morphism $\boldsymbol{\iota}_Y \colon (Y, \calA_Y,
\calN_Y) \to (X, \calA, \calN)$ is supremum preserving.

First we show that the map $\iota_Y \colon Y \to X$ is
supremum preserving. Let $\calE \subset \calA$ and assume $S \in \calA$
is an $\calN$-essential supremum of $\calE$. The set $S \cap Y = \iota_Y^{-1}(S)$ is an $\calN_Y$-essential
upper bound of $\iota_Y^{-1}(\calE)$.
Let $U \in \calA_Y$ be an arbitrary $\calN_Y$-essential upper bound of $\iota_Y^{-1}(\calE)$. We ought to show that
$S \cap Y \setminus U \in \calN_Y$. For all $E \in \calE$, one has
$E \cap S \cap Y \setminus U \subset E \cap Y \setminus U \in
\calN_Y$. As $(X, \calA, \calN)$ is saturated, $\calN_Y \subset \calN$ and, also, $E \cap S \cap Y \setminus U \in \calN$. Of course $(X \setminus S) \cap
S \cap Y \setminus U = \emptyset$ is also $\calN$-negligible. Since the
family $\calE \cup \{X \setminus S\}$ is $\calN$-generating and $(X,
\calA,\calN)$ is locally determined, we deduce that $S \cap Y
\setminus U$ is $\calA$-measurable and, in turn, that it is
$\calN$-negligible by the distributivity lemma <ref>. The proof that $\boldsymbol{\iota}_Y$ is
supremum preserving is complete.

Since the morphism $\mathsf{Bool}(\boldsymbol{\iota}_Y) \colon \calA
/ \calN \to \calA_Y / \calN_Y$ is onto, supremum preserving, and
$\calA / \calN$ is Dedekind complete, so is $\calA_Y / \calN_Y$,
meaning that $(Y, \calA_Y, \calN_Y)$ is localizable. It remains to show
that $(X,\calA,\calN)$ is locally determined.

We claim the following: If $\calE \subset \calA_Y$ is
$\calN_Y$-generating and $N \in \calP(Y)$ satisfies $E
\cap N \in \calN_Y$ for all $E \in \calE$, then $N \in \calN_Y$. By
definition of $\calA_Y$, any set $E \in \calE$ can be written as $E
= E' \cap Y$, for some $E' \in \calA$, so there is a subset $\calE'
\subset \calA$ such that $\calE = \iota_Y^{-1}(\calE')$. The
localizability of $(X, \calA, \calN)$ guarantees the existence of an
$\calN$-essential supremum $S$ of $\calE'$. For all $E' \in \calE'$
one has $E' \cap N = E' \cap Y \cap N \in \calN_Y \subset \calN$,
because $E' \cap Y \in \calE$. Also $S \cap Y = \iota_Z^{-1}(S)$ is
an $\calN_Y$-essential supremum of $\calE =
\iota_Z^{-1}(\calE')$, by the first paragraph. Recalling that $\calE$ is
$\calN_Y$-generating, we find that $Y \setminus S = Y \setminus (S \cap Y) \in \calN_Y$. Consequently, $(X \setminus S) \cap
N \subset Y \setminus S \in \calN_Y \subset \calN$. As $(X, \calA, \calN)$ is saturated, we
find that $(X \setminus S) \cap N \in \calN$. In conclusion, $E'
\cap N \in \calN$ for any $E'$ that belongs to the
$\calN$-generating family $\calE' \cup \{X \setminus S\}$. Since $(X,
\calA, \calN)$ is locally determined, we infer that $N \in \calA$
and then that $N \in \calN$ by the distributivity lemma <ref>. As $N
\subset Y$, we conclude that $N \in \calN_Y$.

Now let $\calE \subset \calA_Y$ be an $\calN_Y$-generating collection
and $A \in \calP(Y)$ be such that $E \cap A \in \calA_Y$ for all $E
\in \calE$. We want to prove that $A \in \calA_Y$. As $(Y, \calA_Y,
\calN_Y)$ is localizable, $\{E \cap A : E \in \calE\}$ has an
$\calN_Y$-essential supremum $S$. This implies that $E \cap A
\setminus S \in \calN_Y$ for all $E \in \calE$. By the claim above, $A \setminus S \in \calN_Y$.

Fix $E_0 \in \calE$. Note that $E_0 \cap (S \setminus A) = (E_0 \cap
S) \setminus (E_0 \cap A) \in \calA_Y$. Also,

\begin{align*}
E \cap A \setminus \left((S \setminus (E_0 \cap S \setminus A)
\right) & = E \cap A \cap ((Y \setminus S) \cup (E_0 \cap S
\setminus A))\\ & = E \cap A \setminus S \in \calN_Y,
\end{align*}

for all $E \in \calE$. In other words, $S \setminus (E_0
\cap S \setminus A)$ is an $\calN_Y$-essential upper bound of $\{E
\cap A : E \in \calE\}$. As $S$ is an $\calN_Y$-essential supremum
of this family, $S \setminus (S \setminus (E_0 \cap S \setminus A))
= E_0 \cap S \setminus A \in \calN_Y$. Applying again the claim above, we deduce that $S
\setminus A \in \calN_Y$ from the arbitrariness of $E_0$. Summing up, $A \ominus S \in \calN_Y$. As $S \in
\calA_Y$ we infer that $A
\in \calA_Y$. The proof that $(Y, \calA_Y, \calN_Y)$
is locally determined is now complete.

$\mathsf{LLD_{sp}}$ has equalizers preserved by the forgetful
functor $\mathsf{LLD_{sp}} \to \mathsf{MSN}$.

Consider a pair of supremum preserving morphisms $\mathbf{f}, \bg
\colon (X, \calA, \calN) \to(Y, \calB, \calM)$ in the category
$\mathsf{LLD_{sp}}$, represented by maps $f \in \mathbf{f}$ and $g
\in \mathbf{g}$. Let $\bh \colon (T, \calC, \calP) \to (X, \calA,
\calN)$ be another supremum preserving morphism in
$\mathsf{LLD_{sp}}$, such that $\mathbf{f} \circ \bh = \bg \circ
\bh$.

Set $Z = \{f = g\}$. We know since Proposition <ref>
that $((Z, \calA_Z, \calN_Z), \boldsymbol{\iota}_Z)$ is the
equalizer of the pair $\mathbf{f}, \bg$ in the category
$\mathsf{MSN}$, so there is a unique morphism $\bh' \colon (T,
\calC, \calP) \to (X, \calA, \calN)$ such that $\bh =
\boldsymbol{\iota}_Z \circ \bh'$. By the
proposition <ref>, $\boldsymbol{\iota}_Z$ is
supremum-preserving and $(Z, \calA_Z, \calN_Z)$ is lld.  It remains
to prove that $\bh'$ is supremum preserving. This follows from the
fact that $\mathsf{Bool}(\bh) = \mathsf{Bool}(\bh') \circ
\mathsf{Bool}(\boldsymbol{\iota}_Z)$, where $\mathsf{Bool}(\bh)$ is
supremum preserving and $\mathsf{Bool}(\boldsymbol{\iota}_Z)$ is
supremum preserving and surjective.

§ GLUING MEASURABLE FUNCTIONS

Let $(X, \calA, \calN)$ be an MSN and $(Y, \calB)$ a measurable
space. Let $\calE \subset \calA$ be a collection. A family
subordinated to $\calE$ is a family of functions $\la f_E \ra_{E
\in \calE}$ such that:

(1) $f_E \colon E \to Y$ is $(\calA_E, \calB)$-measurable
for every $E \in \calE$.

We further say that $\la f_E \ra_{E \in \calE}$ is compatible whenever

(2) for all pairs $E, E' \in \calE$ one has $E \cap E' \cap
\{f_E \neq f_{E'}\} \in \calN$.

A gluing of a compatible family $\la f_E \ra_{E \in \calE}$
subordinated to $\calE$ is a function $f \colon X \to Y$ such that

(3) $f$ is $(\calA, \calB)$-measurable;

(4) $E \cap \{f \neq f_{E}\} \in \calN$ for every $E \in
\calE$.

In this section, we will be mainly concerned about the existence of
gluings, as they will be of use in the construction of the cccc
version of a locally ccc MSN in Section <ref>. This turns
out to depend both on the domain and the target space. In case where
$(Y, \calB)$ is the real line equipped with its Borel
$\sigma$-algebra $(\R, \calB(\R))$, we can glue measurable functions
together if $(X, \calA, \calN)$ is localizable. In fact, this
important property is a characterization of localizability. The
interested reader may find a proof of this classical result
expressed in the language of MSNs in <cit.>. Only the measurable structure of $(\R,
\calB(\R))$ is involved, thus, the result holds in the more general case where
$(Y, \calB)$ is a standard Borel space, see <cit.>.

Many questions arise when we remove the condition that $(Y, \calB)$
is a standard Borel space. In this case, we need some additional
assumptions on $(X, \calA, \calN)$. We will focus on two cases: $(X,
\calA, \calN)$ is cccc or lld. But first, we prove that a gluing
inherits some of the properties of the functions $f_E$.

Let $(X, \calA, \calN)$ be a saturated MSN, $(Y, \calB, \calM)$
an MSN, and $\calE \subset \calA$ an $\calN$-generating collection. We let $\la f_E \ra_{E \in \calE}$ be
a compatible family of functions subordinated to $\calE$ and we
assume that:

(1) for every $E \in \calE$, the map $f_E$ is $[(\calA_E,
\calN_E), (\calB, \calM)]$-measurable;

(2) the family $\la f_E \ra_{E \in \calE}$ has a gluing $f$.

(A) the gluing $f$ is
$[(\calA, \calN), (\calB, \calM)]$-measurable;

(B) if $f_E$ is supremum preserving, for every $E \in \calE$,
then so is $f$.

We start with the following easy observation. For each $E \in \calE$ and $B \in \calB$ one has $f_E^{-1}(B) \ominus (E \cap f^{-1}(B)) \subset E \cap \{ f_E \neq f \} \in \calN$.

(A) As the gluing $f$ is $(\calA, \calB)$-measurable by definition,
we need only show that $f^{-1}(M) \in \calN$ for $M \in \calM$.  Since $f_E^{-1}(M)$ is $\calN$-negligible, the above observation applied with $B=M$ ensures that $E \cap f^{-1}(M) \in \calN$ for any $E
\in \calE$. We next use Lemma <ref> to assert that
$f^{-1}(M)$ is an $\calN$-essential supremum of $\{E \cap f^{-1}(M)
: E \in \calE\}$. This forces $f^{-1}(M)$ to be $\calN$-negligible.

(B) Let $\calF \subset \calB$ be a collection that admits an
$\calM$-essential supremum $S$. Since $f_E$ is supremum preserving
for every $E \in \calE$, $f_E^{-1}(S)$ is an $\calN$-essential supremum

of $\{ f_E^{-1}(F) : F \in \calF \}$ and it ensues from the observation above, applied with $B \in \{S\} \cup \calF$, that $E \cap
f^{-1}(S)$ is an $\calN$-essential supremum of $\{E \cap f^{-1}(F) : F
\in
\calF\}$. Therefore,

\begin{equation*}
\begin{split}
\quad\quad f^{-1}(S) & = \calN\text{-}\rmesssup_{E \in \calE} E \cap f^{-1}(S)
\intertext{(by Lemma \ref{supDistrib})}
& = \calN\text{-}\rmesssup_{E \in \calE} \left( \calN\text{-}\rmesssup_{F \in \calF} E \cap f^{-1}(F)\right)
\intertext{(from what precedes)}
& = \calN\text{-}\rmesssup_{F \in \calF} \left( \calN\text{-}\rmesssup_{E \in \calE} E \cap f^{-1}(F)\right) \\
& = \calN\text{-}\rmesssup_{F \in \calF} f^{-1}(F)
\intertext{(by Lemma \ref{supDistrib}).}
\end{split}
\end{equation*}

Let $(X, \calA, \calN)$ be a locally determined MSN and $(Y, \calB)$
be any nonempty measurable space. Let $\calE \subset \calA$ be an
$\calN$-generating collection. If a compatible family $\la f_E \ra_{E
\in \calE}$ has a gluing, then it is unique up to equality almost

Let $f, g \colon X \to Y$ be two gluings of $\la f_E \ra_{E \in
\calE}$. We warn the reader that the measurability of $\{f \neq g\}$
is not immediate, since the diagonal $\{(y, y) : y
\in Y\}$ may not be measurable in $(Y^2, \calB \otimes \calB)$. Notwithstanding, for
all $E \in \calE$, we have $E \cap \{f \neq g\} \subset (E \cap \{f
\neq f_E\}) \cup (E \cap \{g \neq f_E\})$. Since $f, g$ are gluings
and $(X, \calA, \calN)$ is saturated, it follows that $E \cap \{f
\neq g\} \in \calN$. This happens for any $E$ in the
$\calN$-generating set $\calE$. By local determination and the
distributivity lemma <ref>, $\{f \neq g\} \in \calN$.

Let $(X, \calA, \calN)$ be a cccc MSN and $(Y, \calB)$ be any
nonempty measurable space. Let $\calE \subset \calA$ be an
$\calN$-generating collection. Any compatible family $\la f_E \ra_{E
\in \calE}$ subordinated to $\calE$ admits a unique gluing $f$ up
to equality $\calN$-almost everywhere.

First observe that the uniqueness of the gluing up to almost
everywhere equality follows from Proposition <ref>, as
a cccc MSN is locally determined by Proposition <ref>(E).

Let us treat the special case where $(X, \calA, \calN)$ is a
saturated ccc MSN. According to Proposition <ref>, we can
find a sequence of sets $\la E_i \ra_{i \in \N}$ in $\calE$ such
that $\bigcup_{i \in \N} E_i$ provides an $\calN$-essential supremum
of $\calE$. We then define the $(\calA, \calB)$-measurable map $f
\colon X \to Y$ which, for all $i \in \N$, coincides with $f_i$ on
the set $E_i \setminus \bigcup_{j < i} E_j$, and maps the negligible
set $N \defeq X \setminus \bigcup_{i \in \N} E_i$ to some arbitrary
point. Let $E \in \calE$. For every $i \in \N$, we have $N_i \defeq
E \cap E_i \cap \{f_E \neq f_{E_i}\} \in \calN$, by hypothesis. Thus,
$E \cap \{f \neq f_{E}\} \subset N \cup \bigcup_{i \in \N} N_i$ is

Suppose now that $(X, \calA, \calN)$ is a coproduct $\coprod_{i \in
I}(X_i, \calA_i, \calN_i)$ of saturated ccc MSNs. For each $E \in
\calE$ and $i \in I$, call $f_{E,i}$ the restriction of $f_E$ to $E
\cap X_i$. By Lemma <ref>, $X_i$ is an $\calN_i$-essential supremum of the collection $\{E \cap X_i : E \in
\calE\}$. Also, $\la f_{E,i}\ra_{E\in \calE}$ is a compatible family
of measurable maps subordinated to $\la E \cap X_i \ra_{E \in
\calE}$. From what precedes, it admits a gluing $f_i \colon X_i
\to Y$. Define $f \defeq \coprod_{i \in I} f_i$. The verification
that $f$ is a gluing of $\la f_E \ra_{E\in \calE}$ is

It would be interesting to know whether cccc MSNs are the only lld
MSNs such that gluings can always be performed, with no restriction
on the target space $(Y, \calB)$. This property will be used in the
next section and justifies the special role played by cccc MSN.

[Countably separated measurable spaces]
A measurable space $(Y, \calB)$ is called countably separated
whenever there is a countable set $\calC \subset \calB$ such that
for any distinct $y_1, y_2 \in Y$ there exists $C \in \calC$ such
that $y_1 \in C \not\ni y_2$ or $y_1 \not\in C \ni y_2$. Here is a
well-known characterization of countably separated spaces.

Let $(Y, \calB)$ be a measurable space. The following statements are

(A) $(Y, \calB)$ is countably separated;

(B) there is an injective measurable map $(Y, \calB) \to (\R,
\calB(\R))$;

(C) there is an injective measurable map $(Y, \calB) \to (X,
\calB(X))$ to a Polish space $X$.

(A) $\implies$ (B) Let $\calC \subset \calB$ be a countable set
that separates the points of $Y$. Let $\la C_n \ra_{n \in \N}$ be a
enumeration of $\calC$ and $h \colon Y \to \R$ be the map $h =
\sum_{n=0}^\infty 3^{-n} \ind_{C_n}$. The sets $C_n$ are measurable,
therefore $h$ is measurable.

Let $y_1, y_2$ be distinct points in $Y$ and $n_0 \defeq \min \{n
\in \N : \ind_{C_n}(y_1) \neq \ind_{C_n}(y_2)\}$. Then

\[
|h(y_1) - h(y_2)| \geq 3^{-n_0} - \sum_{n=n_0+1}^\infty 3^{-n}
|\ind_{C_n}(y_1) - \ind_{C_n}(y_2)| \geq 3^{-n_0} -
\frac{3^{-n_0}}{2} > 0
\]

thus $h(y_1) \neq h(y_2)$, which shows that $f$ is injective.

(B) $\implies$ (C) is obvious.

(C) $\implies$ (A) Let $\calU$ be a countable basis for the
topology of $X$. If there is an injective measurable map $h \colon
(Y, \calB) \to (X, \calB(X))$, then $h^{-1}(\calU) \subset \calB$ is
a countable set that separates points.

Let $\la (Y_i, \calB_i) \ra_{i \in I}$ be a family of countably
separated measurable spaces. If $\rmcard I \leq \mathfrak{c}$, then
$\coprod_{i \in I} (Y_i, \calB_i)$ is countably separated.

For each $i \in I$, there is an injective $(\calB_i,\calB(\R))$-measurable map $h_i \colon
Y_i \to \R$ by Proposition <ref>.
Choose an arbitrary injective map $g \colon I \to \R$. Let $h \colon
\coprod_{i \in I} Y_i \to \R^2$ be the map defined by $h(y_i) \defeq
(h_i(y_i), g(i))$ for all $i \in I$ and $y_i \in Y_i$. Let $B
\subset \R^2$ be a Borel set.
Then, for any $i \in I$, we have $h^{-1}(B) \cap Y_i = h_i^{-1}(\R \cap \{x
: (x, g(i)) \in B\}) = h_i^{-1}(B^{g(i)})$. This last set is $\calB_i$-measurable
as $h_i$ is measurable and the horizontal section $B^{g(i)}$ is Borel. As $i$ is arbitrary, we conclude that
$h^{-1}(B)$ is measurable. This means that $h$ is measurable. By
Proposition <ref>, it follows that $\coprod_{i \in I} (Y_i,
\calB_i)$ is countably separated.

The restriction on the cardinal of $I$ is necessary, since a
countably measurable space must have cardinal less or equal than
$\mathfrak{c}$ by Proposition <ref>(B).

Let $(X, \calA, \calN)$ be an lld MSN and $(Y, \calB)$ be a nonempty
countably separated measurable space. Let $\calE \subset \calA$ be
$\calN$-generating. Any compatible family $\la f_E \ra_{E \in \calE}$
subordinated to $\calE$ admits a gluing, unique up to equality
almost everywhere.

Let $h$ be a measurable injective map $(Y, \calB) \to (\R,
\calB(\R))$, whose existence follows from Proposition <ref>.
Now, $\la h \circ f_E \ra_{E \in \calE}$ is still a compatible
family of measurable functions, this time with values in $(\R,
\calB(\R))$. As $(X, \calA, \calN)$ is localizable, it admits a
gluing $g \colon X \to \R$. For all $E \in \calE$, one has $E \cap
g^{-1}(\R \setminus h(Y)) \subset E \cap \{g \neq h \circ
f_E\}$. Therefore $E \cap g^{-1}(\R \setminus h(Y))$ is
negligible. This holds for any $E$ in the $\calN$-generating set
$\calE$. By local determination and Lemma <ref>, we deduce that $g^{-1}(\R \setminus
h(Y)) \in \calN$. Thus, we lose no generality in supposing, from now on, that $g$
takes values in $h(Y)$. Define $f \defeq h^{-1} \circ g$. We claim
that $f$ is a gluing. For $E \in \calE$, we observe that $E \cap \{f
\neq f_E\} \subset E \cap \{g \neq h \circ f_E\} \in \calN$, since
$h$ is injective. Therefore, condition (4) of <ref> is

Also, let $B \in \calB$, then $(E \cap f^{-1}(B)) \ominus
f_E^{-1}(B) = E \cap \{f \neq f_E\} \in \calN$. Since $f_E$ is
measurable, we have $f_E^{-1}(B) \in \calA$ and, in turn, $E \cap f^{-1}(B) \in \calA$. Since $E$ is arbitrary,
we deduce that $f^{-1}(B) \in \calA$, by local determination, showing
that $f$ is measurable. Of course, the uniqueness of the gluing is
given by Proposition <ref>.

In this paragraph, we exhibit an lld MSN $(X,\calA, \calN)$, a
measurable space $(Y, \calB)$ and, within this setting, a compatible
family of measurable maps that cannot be glued. With regards to
Proposition <ref>, it is natural to turn towards
Fremlin's example in <cit.> of a localizable,
locally determined but not strictly localizable[In the
context of measure spaces, we follow the terminology
of <cit.>: $(X, \calA, \mu)$ is strictly localizable
whenever there is a measurable partition $\la X_i \ra_{i \in I}$
such that a set $A \subset X$ is measurable whenever the sets $A
\cap X_i$ are, and in that case $\mu(A) = \sum_{i \in I} \mu(A \cap
X_i)$.] measure space $(X, \calA, \mu)$. Let us recall its
construction. Fix a set $Y$ with cardinal greater than
$\mathfrak{c}$ and we set $X \defeq \{0, 1\}^{\calP(Y)}$. For any $y
\in Y$, we define $x_y \in X$ by

\[
\forall Z \in \calP(Y), \qquad x_y(Z) = \begin{cases}
1 & \text{if } y \in Z \\
0 & \text{if } y \not\in Z
\end{cases}
\]

Let $\calK \subset \calP(\calP(Y))$ be the family of countable
subsets of $\calP(Y)$. For any $K \in \calK$ and $y \in Y$, we
define $F_{y, K} \defeq X \cap \{x : x(Z) = x_y(Z) \text{ for all }
Z \in K\}$. Then we define, for all $y \in Y$,

\[
\calA_y = \calP(X) \cap \{A : \text{there is } K \in \calK
\text{ such that } F_{y, K} \subset A \text{ or } F_{y, K}
\subset X \setminus A\}
\]

Let us prove that $\calA_y$ is a $\sigma$-algebra. Clearly
$\emptyset \in \calA_y$ and $\calA_y$ is closed under
complementations. Let $\la A_n \ra_{n \in \N}$ be a sequence in
$\calA_y$. Suppose there is some $n_0 \in \N$ and $K \in \calK$
such that $F_{y, K} \subset A_{n_0}$. Then $F_{y, K}
\subset \bigcup_{n \in \N} A_n$, which implies $\bigcup_{n \in \N}
A_n \in \calA_y$. Suppose on the contrary that for all $n \in
\N$, there is $K_n \in \calK$ such that $F_{y, K_n} \subset X
\setminus A_n$. Then $\bigcap_{n \in \N} F_{y, K_n} =
F_{y, \bigcup_{n \in \N} K_n} \subset X \setminus \bigcup_{n
\in \N} A_n$ which also gives that $\bigcup_{n \in \N} A_n \in
\calA_y$.

Finally set $\calA \defeq \bigcap_{y \in Y} \calA_y$ and
define the measure $\mu \colon \calA \to [0, \infty]$ by

\[
\forall A \in \calA, \qquad \mu(A) = \rmcard(Y \cap \{ y :
x_y \in A\})
\]

For the rest of the discussion, we admit that $(X, \calA, \mu)$ is
complete, localizable, locally determined and not strictly
localizable. The proof of the latter relies on a non trivial fact in
infinitary combinatorics; we refer to <cit.>
for more details. The associated MSN $(X, \calA, \calN_\mu)$ is
saturated, localizable, and it is locally determined, by
Proposition <ref>(E).

Define $E_y = X \cap \{x : x(\{y\}) = 1\}$ for all $y \in Y$. This
set is $\calA$ measurable, because $F_{y, \{\{y\}\}} = E_y$ (hence
$E_y \in \calA_y$), and for any $z \in Y \setminus \{y\}$, we have
$F_{z, \{\{y\}\}} = X \setminus E_y$ (hence $E_y \in
\calA_{z}$). Note that $Y \cap \{z : x_z \in E_y\} = \{y\}$.

We now choose $\calB$ to be the countable cocountable
$\sigma$-algebra of $Y$. For any $y \in Y$, we define the measurable
map $f_y \colon (E_y, \calA_{E_y}) \to (Y, \calB)$ that is constant
equal to $y$. We claim that $\la f_y \ra_{y \in Y}$ is a
compatible family of measurable maps subordinated to $\la E_y \ra_{y
\in Y}$. This ensues from the fact that $E_y \cap E_{z} \in
\calN_{\mu}$ for any distinct $y, z \in Y$. Assume by contradiction
that we can find a gluing $f \colon X \to Y$. We will use the
decomposition $\la f^{-1}(\{y\})\ra_{y \in Y}$ to show $(X, \calA,
\mu)$ is strictly localizable.

Let $A \subset X$ such that $A \cap f^{-1}(\{y\}) \in \calA$
for all $y$. We want to show that $A \in \calA$. For $y
\in Y$, we have

* Case $x_y \in A$: as $A \cap f^{-1}(\{y\})\in
\calA_y$ and $x_y \in A \cap f^{-1}(\{y\})$, there
is $K \in \calK$ such that $F_{y,K} \subset A \cap
f^{-1}(\{y\}) \subset A$ (because $x_y \in F_{y,K}$, this is the branch of the dichotomy, in the definition of $\calA_y$, that occurs). Therefore $A \in \calA_y$.

* Case $x_y \not\in A$: since $x_y \not\in A \cap
f^{-1}(\{y\}) \in \calA_y$, we can find $K \in \calK$
such that $F_{y, K} \subset X \setminus (A \cap
f^{-1}(\{y\}))$.  We deduce that $F_{y, K} \cap
f^{-1}(\{y\}) \subset X \setminus A$. But $x_y \in
f^{-1}(\{y\}) \in \calA_y$, therefore there is $K' \in
\calK$ such that $F_{y, K'} \subset
f^{-1}(\{y\})$. Whence $F_{y, K \cup K'} = F_{y, K}
\cap F_{y, K'} \subset F_{y, K} \cap f^{-1}(\{y\})
\subset X \setminus A$. It follows that $A \in \calA_y$.

In any case, we have shown that $A \in \calA_y$. As $y \in Y$ is
arbitrary, $A \in \calA$.

Now, one observes that the only $z \in Y$ such that $x_z \in
f^{-1}(\{y\})$ is $y$. Therefore $\mu(A \cap f^{-1}(\{y\}))$ equals
$1$ if $x_y \in A$ and $0$ otherwise. In consequence, we have
$\mu(A) = \sum_{y \in Y} \mu(A \cap f^{-1}(\{y\}))$ as desired.

§ EXISTENCE OF CCCC AND LLD VERSIONS

Let $(X, \calA, \calN)$ be a saturated MSN and $\calE \subset \calA
\setminus \calN$. We suppose that

(1) $(Z, \calA_Z, \calN_Z)$ is cccc for every $Z \in \calC$;

(2) $\calE$ is almost disjointed;

(3) $\calE$ is $\calN$-generating.

Then the pair consisting of the MSN

\[
(\hat{X}, \hat{\calA}, \hat{\calN}) = \coprod_{Z \in \calE} (Z,
\calA_Z, \calN_Z)
\]

and the morphism $\bp = \coprod_{Z \in \calE} \boldsymbol{\iota}_Z$
is the cccc version of $(X, \calA, \calN)$ (as usual
$\boldsymbol{\iota}_Z$ is the morphism induced by the inclusion map
$\iota_Z \colon Z \to X$).

The MSN $(\hat{X}, \hat{\calA}, \hat{\calN})$ is cccc as a coproduct
of cccc MSNs (this is a general fact, in any category, a coproduct
of coproducts is a coproduct,
see <cit.>), and $\bp$ is supremum
preserving, according to <ref>(B) and (D). Observe that each
$Z \in \calE$ is also a subset of $\hat{X}$ and we denote by
$\hat{\iota}_Z \colon Z \to \hat{X}$ the corresponding inclusion

Let $(Y, \calB, \calM)$ be a cccc MSN and $\bq \colon (Y, \calB,
\calM) \to (X, \calA, \calN)$ be a supremum preserving morphism,
represented by $q \in \bq$. For all $Z \in \calE$, call $q_Z \defeq
\hat{\iota}_Z \circ (q \hel q^{-1}(Z)) \colon q^{-1}(Z) \to
\hat{X}$. Because $\bq$ is supremum preserving, $Y = q^{-1}(X)$ is
an $\calM$-essential supremum of the family $\la q^{-1}(Z)\ra_{Z \in
\calE}$. The family $\calE$ being almost disjointed and $q$ being
measurable, $q^{-1}(Z) \cap q^{-1}(Z') = q^{-1}(Z \cap Z') \in
\calM$ for any distinct $Z, Z' \in \calE$. As a result, the family
$\la q_Z \ra_{Z \in \calE}$ subordinated to $\la q^{-1}(Z) \ra_{Z
\in \calE}$ is compatible. By Proposition <ref>, this family
has a gluing $r \colon Y \to \hat{X}$ and, by Lemma <ref>,
$r$ is $[(\calB, \calM), (\hat{\calA}, \hat{\calN})]$-measurable and
supremum preserving.  Call $p \defeq \coprod_{Z \in \calE}
\iota_Z$. For each $Z \in \calE$, we have

\[
q^{-1}(Z) \cap \{p \circ r \neq q\} \subset q^{-1}(Z) \cap \{r \neq
\hat{\iota}_Z \circ (q \hel q^{-1}(Z))\} \in \calM.
\]

The family $\{q^{-1}(Z) : Z \in \calE\}$ is $\calM$-generating and
$(Y, \calB, \calM)$ is locally determined, so we conclude that $\{p
\circ r \neq q\} \in \calM$, that is, $\bp \circ \br = \bq$.

As for uniqueness, let $\br$ be any morphism such that $\bp \circ
\br = \bq$, and call $r \in \br$ one of its representatives. Fix $Z
\in \calE$. Observe that $\hat{\iota}_Z (p(z)) = z$ for all $z \in
Z$. For $\calM$-almost every $x \in q^{-1}(Z)$, we have $p(r(x)) =
q(x)$ which implies that $r(x) \in Z \subset \hat{X}$. For such an
$x$, we find that $r(x) = \hat{\iota}_Z (p(r(x))) =
\hat{\iota}_Z(q(x)) = q_Z(x)$. Hence, $r$ is a gluing of the
compatible family $\la q_Z \ra_{Z \in \calE}$ and we invoke the
uniqueness part of Proposition <ref> to conclude.

Consider the following example, taken
from <cit.>. Let $X$ be a set of cardinality greater
or equal than $\aleph_2$. For each $x,y \in X$, we define $H_y = X
\times \{y\}$ and $V_x = \{x\} \times X$. Sets of this form are
respectively called horizontal and vertical lines. We define a
$\sigma$-algebra $\calA$ of $X^2$ by declaring that $A \in \calA$
iff for all $x, y \in X$, the trace $A \cap H_y$ (resp. $A \cap
V_x$) is either countable or cocountable in $H_y$
(resp. $V_x$). Also, we define the $\sigma$-ideal $\calN$ of
$\calA$ as follows: $N \in \calN$ if and only if the intersection of $N$ with any
line is countable. Clearly, $(X^2, \calA, \calN)$ is saturated.

We assert that it is not localizable. Suppose if possible that the
family of horizontal lines $\{H_y : y \in X\}$ has an
$\calN$-essential supremum $S$. Then for all $y \in X$, the
intersection $S \cap H_y$ is cocountable in $H_y$, that is, $N_y
\defeq X \cap \{x : (x, y) \not\in S\}$ is countable. Let $Z$ be a
subset of $X$ of cardinality $\aleph_1$. Then $\rmcard \bigcup_{y
\in X} N_y \leq \aleph_1$, hence the existence of $x \in X
\setminus \bigcup_{y \in X} N_y$.  This means that $V_x \subset S$.
However, $S \setminus V_x$ is easily checked to be an
$\calN$-essential upper bound of $\{H_y : y \in X\}$, as $H_y \cap
V_x$ is negligible for all $y$. Since $V_x = S \setminus (S
\setminus V_x) \not\in \calN$, we get a contradiction.

The family of all lines $\calE \defeq \{H_y : y \in X\} \cup \{V_x :
x \in X\}$ satisfies the three hypotheses of
Theorem <ref>. Applying the theorem, we see that the cccc
version of $(X^2, \calA, \calN)$ can be described as the coproduct
of all lines. Doing so, we see that each point $(x, y)$ in the base
MSN $(X^2, \calA, \calN)$ is duplicated in the cccc version: the
“fibers” $p^{-1}(\{(x,y)\})$ contains two elements, which
represent the horizontal and vertical directions emanating from the
point $(x, y)$.

If a given MSN has no obvious choice of a family satisfying the
conditions of Theorem <ref>, we can justify the existence of a
cccc version in a non constructive way.

Let $(X, \calA, \calN)$ be a saturated MSN and $\calC \subset A$ an
$\calN$-generating collection such that $(Z, \calA_Z, \calN_Z)$ is
ccc for all $Z \in \calC$. Then we can find a collection $\calE
\subset \calA \setminus \calN$ that satisfies conditions (1), (2)
and (3) of Theorem <ref> and such that each of its members is
a subset of some member of $\calC$. Moreover, we can suppose
$\rmcard \calE \leq \max\{\aleph_0, \rmcard \calC\}$.

Let $\calE$ be associated with $\calC$ in Lemma <ref>. It clearly satisfies conditions (1), (2), and (3)
of <ref>, since a subMSN of a ccc MSN is ccc as well. If $\calC$ is infinite, then for all $Z \in
\calC$, call $\calE_Z \defeq \calE \cap \{Z' : Z' \subset Z\}$. Then,
each $\calE_Z$ is at most countable, since it is an almost
disjointed family in the ccc MSN $(Z, \calA_Z, \calN_Z)$. As $\calE
= \bigcup_{Z \in \calC} \calE_Z$, we conclude that $\rmcard \calE
\leq \rmcard \calC$.

Every saturated locally ccc MSN admits a cccc version.

Apply Lemma <ref> to the family $\calC \defeq \calA \cap \{Z
: (Z, \calA_Z, \calN_Z) \text{ is ccc}\}$ and then
Theorem <ref>.

It is worth noticing that all the arguments contained in
Theorem <ref> and Corollary <ref> remain valid
provided we replace “ccc” by “strictly localizable”, “locally
ccc” by “locally strictly localizable”, and “cccc” by “strictly
localizable”. Summing up, a saturated locally strictly localizable
MSN $(X, \calA, \calN)$ has a strictly localizable version, which is
constructed as the coproduct of subMSNs whose underlying sets
belongs to a family $\calE$ that satisfies hypotheses (2), (3) of
Theorem <ref> and

(1') $(Z, \calA_Z, \calN_Z)$ is strictly localizable for
every $Z \in \calC$.

Since (1') implies (1) we can apply Theorem <ref> again to
conclude that the cccc and strictly localizable versions of $(X,
\calA, \calN)$ are the same.

As for the existence of lld versions, we have a partial result,
which applies for most locally ccc MSNs that one is likely to
encounter in analysis.

Let $(X, \calA, \calN)$ be a saturated MSN with a collection $\calC
\subset \calA$ such that

(1) $(Z, \calA_Z, \calN_Z)$ is ccc for all $Z \in \calC$;

(2) $\calC$ is $\calN$-generating;

(3) $\rmcard \calC \leq \mathfrak{c}$.

The following hold.

(A) If $(X, \calA, \calN)$ has an lld version, then it
coincides with the cccc version.

(B) If moreover $(Z, \calA_Z)$ is countably separated for all
$Z \in \calC$, then the lld version exists.

(A) Recall $(X,\calA,\calN)$ has a cccc version, according to Corollary
<ref>. Suppose $(X, \calA, \calN)$ has an lld version $[(\hat{X},
\hat{\calA}, \hat{\calN}), \bp]$. In view of Proposition <ref>(E), conclusion (A) will be established if we prove that $(\hat{X},
\hat{\calA}, \hat{\calN})$ is cccc. To this end, we need to find a
suitable decomposition in $\hat{X}$. Apply Lemma <ref> to
get an almost disjointed $\calN$-generating family $\calE$ such that
$\rmcard \calE \leq \mathfrak{c}$ and $(Z, \calA_Z, \calN_Z)$ is ccc
for all $Z \in \calE$. Choose an injection $c : \calE \to \R : Z \mapsto c_Z$ and $p \in \bp$. Let $f_Z
\colon p^{-1}(Z) \to \R$ be the constant map equal to $c_Z$; it is readily $(\hat{\calA}_{p^{-1}(Z)},\calB(\R))$-measurable. The family $\la f_Z
\ra_{Z \in \calE}$ is obviously compatible, since $\calE$ is almost disjointed. As $(\hat{X}, \hat{\calA},
\hat{\calN})$ is localizable, this family has an $(\hat{\calA},
\calB(\R))$-measurable gluing $f \colon \hat{X} \to \R$.

We now show that $\la f^{-1}\{c_Z\} \ra_{Z \in \calE}$ is a
partition of $\hat{X}$ into ccc measurable pieces.  Since $c$ is injective, the family $\la f^{-1}\{c_Z\} \ra_{Z \in \calE}$ is, indeed, a partition of $\hat{X}$ and, since $f$ is $(\hat{\calA},\calB(\R))$-measurable,
$f^{-1}\{c_Z\} \in \hat{\calA}$ for all $Z \in \calE$. As $f$ is a gluing
of $\la f_Z \ra_{Z \in \calE}$, we
have $p^{-1}(Z) \setminus
f^{-1}\{c_Z\} = p^{-1}(Z) \cap \{f \neq f_Z\}  \in \hat{\calN}$ for all $Z \in \calE$. Moreover, since $c$ is injective, for
all $Z' \in \calE$ distinct from $Z$, one has $p^{-1}(Z') \cap
(f^{-1}\{c_Z\} \setminus p^{-1}(Z)) \subset p^{-1}(Z') \cap \{f \neq
f_{Z'}\} \in \hat{\calN}$. Also, $p^{-1}(Z) \cap (f^{-1}\{c_Z\}
\setminus p^{-1}(Z)) = \emptyset$ is clearly negligible. Recalling that $(\hat{X},\hat{\calA},\hat{\calN})$ is saturated and that $p^{-1}(\calE)$ is $\hat{\calN}$-generating (because $\calE$ is $\calN$-generating and $p$ is supremum preserving), one infers from the
Distributivity Lemma <ref> that $f^{-1}\{c_Z\} \setminus
p^{-1}(Z) \in \hat{\calN}$. Thus $p^{-1}(Z) \ominus f^{-1}\{c_Z\} \in
\hat{\calN}$. As $\bp$ is a local isomorphism, according to Propositions <ref>(B) and <ref>(A), $(p^{-1}(Z),
\hat{\calA}_{p^{-1}(Z)}, \hat{\calN}_{p^{-1}(Z)})$ is ccc. By what
precedes, so is $(f^{-1}\{c_Z\}, \hat{\calA}_{f^{-1}\{c_Z\}},
\hat{\calN}_{f^{-1}\{c_Z\}})$. Therefore, the MSN

\[
(Y,\calB,\calM) \defeq \coprod_{Z \in \calE}
(f^{-1}\{c_Z\}, \hat{\calA}_{f^{-1}\{c_Z\}}, \hat{\calN}_{f^{-1}\{c_Z\}})
\]

is cccc, by definition. It remains to establish that $(\hat{X},\hat{\calA},\hat{\calN})$ and $(Y,\calB,\calM)$ are isomorphic in $\sfMSNsp$. This is a consequence of Proposition <ref> applied to the measurable partition $\calF = \{ f^{-1}\{c_Z\} : Z \in \calE \}$. Recalling that $p^{-1}(\calE)$ is $\hat{\calN}$-generating, it ensues from the preceding paragraph that so is $\calF$. Since $(\hat{X},\hat{\calA},\hat{\calN})$ is locally determined (whence, has locally determined negligible sets, recall <ref>), $\calF$ satisfies hypotheses (1) and (2)
of Proposition <ref>.

(B) Apply Lemma <ref> to $\calC$ and let
$\calE$ be the family thus obtained. By Theorem <ref>, the MSN $(\hat{X}, \hat{\calA},
\hat{\calN}) \defeq \coprod_{Z \in \calE} (Z, \calA_Z, \calN_Z)$ and