# The Nikodym property and filters on $\omega$

Tomasz Żuchowski  Mathematical Institute, University of Wrocław, Wrocław,
Poland<EMAIL_ADDRESS>
###### Abstract.

For a free filter $F$ on $\omega$, let $N_{F}=\omega\cup\\{p_{F}\\}$, where
$p_{F}\not\in\omega$, be equipped with the following topology: every element
of $\omega$ is isolated whereas all open neighborhoods of $p_{F}$ are of the
form $A\cup\\{p_{F}\\}$ for $A\in F$. The aim of this paper is to study spaces
of the form $N_{F}$ in the context of the Nikodym property of Boolean
algebras. By $\mathcal{AN}$ we denote the class of all those ideals
$\mathcal{I}$ on $\omega$ such that for the dual filter $\mathcal{I}^{*}$ the
space $N_{\mathcal{I}^{*}}$ carries a sequence $\langle\mu_{n}\colon
n\in\omega\rangle$ of finitely supported signed measures such that
$\|\mu_{n}\|\rightarrow\infty$ and $\mu_{n}(A)\rightarrow 0$ for every clopen
subset $A\subseteq N_{\mathcal{I}^{*}}$. We prove that
$\mathcal{I}\in\mathcal{AN}$ if and only if there exists a density submeasure
$\varphi$ on $\omega$ such that $\varphi(\omega)=\infty$ and $\mathcal{I}$ is
contained in the exhaustive ideal $\mbox{Exh}(\varphi)$. Consequently, we get
that if $\mathcal{I}\subseteq\mbox{Exh}(\varphi)$ for some density submeasure
$\varphi$ on $\omega$ such that $\varphi(\omega)=\infty$ and
$N_{\mathcal{I}^{*}}$ is homeomorphic to a subspace of the Stone space
$St(\mathcal{A})$ of a given Boolean algebra $\mathcal{A}$, then $\mathcal{A}$
does not have the Nikodym property.

We observe that each $\mathcal{I}\in\mathcal{AN}$ is Katětov below the
asymptotic density zero ideal $\mathcal{Z}$, and prove that the class
$\mathcal{AN}$ has a subset of size $\mathfrak{d}$ which is dominating with
respect to the Katětov order $\leq_{K}$, but $\mathcal{AN}$ has no
$\leq_{K}$-maximal element. We show that for a density ideal $\mathcal{I}$ it
holds $\mathcal{I}\not\in\mathcal{AN}$ if and only if $\mathcal{I}$ is totally
bounded if and only if the Boolean algebra $\mathcal{P}(\omega)/\mathcal{I}$
contains a countable splitting family.

Our results shed some new light on differences between the Nikodym property
and the Grothendieck property of Boolean algebras.

###### Key words and phrases:

Nikodym property, convergence of measures, Boolean algebras, filters on
countable sets, Katětov order, density ideals.

###### 2020 Mathematics Subject Classification:

Primary: 03E75, 28A33, 03E05. Secondary: 54A20, 28A60, 06E15.

The author was supported by the Austrian Science Fund (FWF), research grant
no. I 5918-N

##  1\. Introduction

Let $\mathcal{A}$ be a Boolean algebra. We say that a sequence
$\langle\mu_{n}\colon n\in\omega\rangle$ of finitely additive signed measures
on a Boolean algebra $\mathcal{A}$ is pointwise bounded if
$\sup_{n\in\omega}|\mu_{n}(A)|<\infty$ for every $A\in\mathcal{A}$, and it is
uniformly bounded if $\sup_{n\in\omega}\|\mu_{n}\|<\infty$ (see Section $2$
for more definitions). The Nikodym Uniform Boundedness Theorem implies that if
$\mathcal{A}$ is $\sigma$-complete, then every pointwise bounded sequence of
measures on $\mathcal{A}$ is uniformly bounded. This theorem, due to its
numerous applications, is one of the most important results in the theory of
vector measures, see Diestel and Uhl [4, Section I.3], and hence motivates the
following definition.

######  Definition 1.1.

A Boolean algebra $\mathcal{A}$ has the Nikodym property if every pointwise
bounded sequence of finitely additive signed measures on $\mathcal{A}$ is
uniformly bounded.

Equivalently, $\mathcal{A}$ has the Nikodym property if, for every sequence
$\langle\mu_{n}\colon n\in\omega\rangle$ of finitely additive signed measures
on $\mathcal{A}$ such that $\mu_{n}(A)\to 0$ for all $A\in\mathcal{A}$, we
have $\sup_{n\in\omega}\|\mu_{n}\|<\infty$ (see [25, Proposition 2.4]).

The Nikodym theorem implies that $\sigma$-complete algebras have the Nikodym
property. This result has been generalized by many authors who introduced
various weaker properties of Boolean algebras that still imply the Nikodym
property, see e.g. Haydon [15], Schachermayer [24], Freniche [9].

On the other hand, countable infinite Boolean algebras do not have the Nikodym
property. Indeed, the Stone space $St(\mathcal{A})$ of a countable infinite
Boolean algebra $\mathcal{A}$ is homeomorphic to an infinite closed subset of
the Cantor space $2^{\omega}$, hence it contains a non-trivial sequence
$\langle x_{n}\colon n\in\omega\rangle$ convergent to some $x\in 2^{\omega}$.
Then, the sequence $\big{\langle}n(\delta_{x_{n}}-\delta_{x})\colon
n\in\omega\big{\rangle}$ of Borel measures on $St(\mathcal{A})$ induces a
pointwise bounded but not uniformly bounded sequence of finitely additive
measures on $\mathcal{A}$. This example is a motivation for considering the
following type of spaces.

######  Definition 1.2.

Let $F$ be a free filter on $\omega$. Endow the set
$N_{F}=\omega\cup\\{p_{F}\\}$, where $p_{F}$ is a fixed point not belonging to
$\omega$, with the topology $\tau_{F}$ defined in the following way:

  * •

every point of $\omega$ is isolated in $N_{F}$ , i.e. $\\{n\\}$ is open for
every $n\in\omega$,

  * •

a set $U\subseteq N_{F}$ is an open neighborhood of $p_{F}$ if and only if
$U=A\cup\\{p_{F}\\}$ for some $A\in F$.

For every free filter $F$ the space $N_{F}$ is a countable non-discrete zero-
dimensional space (and so a Tychonoff space). The space $N_{Fr}$, where by
$Fr$ we denote the Fréchet filter on $\omega$ (see Section $2$), is
homeomorphic to a convergent sequence of points in a given Tychonoff space
together with its limit point. Thus, for a Boolean algebra $\mathcal{A}$, the
Stone space $St(\mathcal{A})$ contains a non-trivial convergent sequence if
and only if it contains a homeomorphic copy of $N_{Fr}$. Consequently, if
$N_{Fr}$ embeds into $St(\mathcal{A})$, then $\mathcal{A}$ does not have the
Nikodym property. In this paper we study for which other free filters we also
have such an implication.

We need to introduce some notions. We say that a Borel measure $\mu$ on a
Tychonoff space $X$ is finitely supported if it can be written in the form
$\mu=\sum_{i=1}^{n}\alpha_{i}\delta_{x_{i}}$ for some distinct points
$x_{1},\ldots,x_{n}\in X$ and some real numbers
$\alpha_{1},\ldots,\alpha_{n}\in\mathbb{R}$, $n\in\omega$; in this case we
write $\|\mu\|=\sum_{i=1}^{n}|\alpha_{i}|$ (see again Section $2$).

######  Definition 1.3.

Let $X$ be a zero-dimensional (Tychonoff) space. We call a sequence
$\langle\mu_{n}\colon n\in\omega\rangle$ of finitely supported measures on $X$
anti-Nikodym, or shortly an AN-sequence, if $\|\mu_{n}\|\rightarrow\infty$ and
$\mu_{n}(A)\rightarrow 0$ for every $A\in\operatorname{Clopen}(X)$. We say
that $X$ has the finitely supported Nikodym property if there is no anti-
Nikodym sequence of measures on $X$.

We also say that a free filter $F$ on $\omega$ has the Nikodym property if the
space $N_{F}$ has a finitely supported Nikodym property. We denote by
$\mathcal{A}{\mathcal{N}}$ the class of all ideals on $\omega$ whose dual
filters do not have the Nikodym property.

We prove that if $\mathcal{A}$ is a Boolean algebra, $F$ is a free filter on
$\omega$ which does not have the Nikodym property, and $N_{F}$
homeomorphically embeds into $St(\mathcal{A})$, then $\mathcal{A}$ does not
have the Nikodym property (Corollary 8.3). This result generalizes the
aforementioned fact about embedding $N_{Fr}$ into the Stone spaces of Boolean
algebras.

For any non-trivial convergent sequence $\langle x_{n}\colon
n\in\omega\rangle$ in a given space $N_{F}$, the sequence
$\langle\mu_{n}\colon n\in\omega\rangle$ of finitely supported non-negative
measures defined for each $n\in\omega$ by the equality
$\mu_{n}=n\cdot\delta_{x_{n}}$ satisfies the conditions
$\lim_{n\to\infty}\mu_{n}(\omega)=\infty$ and
$\lim_{n\to\infty}\mu_{n}(\omega\mathbin{\backslash}A)=0$ for every $A\in F$.
We show that the Nikodym property for filters is in fact characterized by
similar conditions.

theoremtheorempositivemeasures A filter $F$ on $\omega$ has the Nikodym
property if and only if there is no (disjointly supported) sequence
$\langle\mu_{n}\colon n\in\omega\rangle$ of finitely supported non-negative
measures on $N_{F}$ such that:

  1. (1)

$\operatorname{supp}(\mu_{n})\subseteq\omega$ for every $n\in\omega$,

  2. (2)

$\lim_{n\to\infty}\mu_{n}(\omega)=\infty$,

  3. (3)

$\lim_{n\to\infty}\,\mu_{n}(\omega\mathbin{\backslash}A)=0$ for every $A\in
F$.

Theorem 1.3 can be translated in the following way, showing a connection
between filters without the Nikodym property and exhaustive ideals associated
to lower semicontinuous (lsc) submeasures on $\omega$.

theoremtheoremANexh Let $F$ be a filter on $\omega$. Then, the following are
equivalent:

  1. (1)

$F$ does not have the Nikodym property;

  2. (2)

there is a density submeasure $\varphi$ on $\omega$ such that
$\varphi(\omega)=\infty$ and $F\subseteq\operatorname{Exh}(\varphi)^{*}$;

  3. (3)

there is a non-pathological lsc submeasure $\varphi$ on $\omega$ such that
$\varphi(\omega)=\infty$ and $F\subseteq\operatorname{Exh}(\varphi)^{*}$.

The Katětov order $\leq_{K}$ is an important tool for comparing the structural
complexity of ideals on $\omega$, see e.g. [18]. As a corollary to Theorem
1.3, we get that the class of the ideals dual to filters without the Nikodym
property is bounded in this order by the ideal ${\mathcal{Z}}$ of subsets of
$\omega$ having asymptotic density zero (see Section $4$).

corollarycorollaryzet If $F$ is a filter on $\omega$ without the Nikodym
property, then $F^{*}\leq_{K}{\mathcal{Z}}$.

Extending [31, Theorem 3.16], we present several conditions for a density
ideal equivalent to the statement that its dual filter has the Nikodym
property (see Sections 4 and 5 for relevant definitions of classes and
properties of ideals).

theoremtheoremnikodymequiv Let ${\mathcal{I}}$ be the density ideal on
$\omega$ generated by a sequence of measures $\langle\mu_{n}\colon
n\in\omega\rangle$. The following are equivalent:

  1. (1)

${\mathcal{I}}$ is isomorphic to an Erdős–Ulam ideal.

  2. (2)

${\mathcal{I}}$ does not satisfy the Bolzano–Weierstrass property.

  3. (3)

The measures $\mu_{n}$ fulfill the following conditions:

    1. a)

$\sup_{n\in\omega}\|\mu_{n}\|<\infty$,

    2. b)

$\lim_{n\to\infty}\operatorname{at}^{+}(\mu_{n})=0$.

  4. (4)

The Boolean algebra ${\mathcal{P}}(\omega)/{\mathcal{I}}$ contains a countable
splitting family.

  5. (5)

${\mathcal{I}}$ is totally bounded.

  6. (6)

${\mathcal{I}}\equiv_{K}{\mathcal{Z}}$.

  7. (7)

${\mathcal{I}}^{*}$ has the Nikodym property.

We make an analysis of the cofinal structure of the family
$\mathcal{A}{\mathcal{N}}$ ordered by $\leq_{K}$ and we prove that there is a
family of size equal to the dominating number $\mathfrak{d}$ which is
$\leq_{K}$-dominating in $\mathcal{A}{\mathcal{N}}$, but there are no
$\leq_{K}$-maximal elements in $\mathcal{A}{\mathcal{N}}$.

theoremcorollarynikodymdominating There exists a family
${\mathcal{F}}\subseteq\mathcal{A}{\mathcal{N}}$ of size $\mathfrak{d}$
consisting of density ideals and having the property that for any
${\mathcal{I}}\in\mathcal{A}{\mathcal{N}}$ there is
${\mathcal{J}}\in{\mathcal{F}}$ such that
${\mathcal{I}}\leq_{K}{\mathcal{J}}$.

theoremtheoremnomaximal For every ${\mathcal{I}}\in\mathcal{A}{\mathcal{N}}$
there is ${\mathcal{J}}\in\mathcal{A}{\mathcal{N}}$ such that
${\mathcal{I}}<_{K}{\mathcal{J}}$.

Using Theorem 1.3 and results from [14] and [21], we show that the classes of
those filters $F$ on $\omega$ for which the spaces $N_{F}$ (do not) have the
Nikodym property are large. First, there are families
${\mathcal{H}}_{1},{\mathcal{H}}_{2}\subseteq\mathcal{A}{\mathcal{N}}$ of size
continuum $\mathfrak{c}$, consisting respectively of non-isomorphic summable
ideals and of non-isomorphic density ideals (Corollaries 7.3 and 7.6). Next,
there are families ${\mathcal{F}}_{1}$ and ${\mathcal{F}}_{2}$ of size
$2^{\mathfrak{c}}$, consisting of non-isomorphic filters on $\omega$ whose
dual ideals are tall and which respectively have and do not have the Nikodym
property (Corollary 7.4). Those results yield the following corollary.

corollarycorollarymanyAN There is a family ${\mathcal{F}}$ of
$2^{\mathfrak{c}}$ many pairwise non-homeomorphic countable non-discrete
spaces without non-trivial convergent sequences, and such that for every
Boolean algebra $\mathcal{A}$ and $X\in{\mathcal{F}}$, if $X$ homeomorphically
embeds into $St(\mathcal{A})$, then $\mathcal{A}$ does not have the Nikodym
property.

Our results also apply to the Grothendieck property of Boolean algebras, which
is closely related to the Nikodym property. This property has its origin in
Banach space theory, see [12].

######  Definition 1.4.

A Boolean algebra $\mathcal{A}$ has the Grothendieck property if, for every
sequence $\langle\mu_{n}\colon n\in\omega\rangle$ of finitely additive signed
measures on $\mathcal{A}$ such that $\sup_{n\in\omega}\|\mu_{n}\|<\infty$ and
$\mu_{n}(A)\to 0$ for all $A\in\mathcal{A}$, we have $\widehat{\mu}_{n}(B)\to
0$ for every Borel subset $B\subseteq St(\mathcal{A})$, where
$\widehat{\mu}_{n}$ denotes for each $n\in\omega$ the Radon extension of
$\mu_{n}$ onto the space $St(\mathcal{A})$.

The connection between the Grothendieck property and the Nikodym property of
Boolean algebras has been a subject of an extensive research (see e.g. [9],
[15] or [24]), but it is still not fully understood. It is not easy to find an
algebra having exactly one of these properties. Schachermayer [24] showed that
the algebra $\mathscr{J}$ of Jordan-measurable subsets of the unit interval
has the Nikodym property but does not have the Grothendieck property (see also
[13], [32], [26]). On the other hand, Talagrand [30] constructed under the
assumption of the Continuum Hypothesis a Boolean algebra with the Grothendieck
property and without the Nikodym property, but it is still an open problem if
one can find such an example in ZFC (cf. [27], [11]).

Marciszewski and Sobota studied in [22] a property of spaces of the form
$N_{F}$ that is connected with the Grothendieck property in a similar way as
the finitely supported Nikodym property is connected with the Nikodym property
of Boolean algebras. We need again to introduce some piece of notation. For a
topological space $X$ we denote by $C(X)$ the set of all continuous real-
valued functions on $X$ and by $C_{b}(X)$ the set of all bounded continuous
real-valued functions on $X$. If $\mu$ is a Borel measure on $X$ and $f\in
C(X)$, then we write $\mu(f)=\int_{X}fd\mu$.

######  Definition 1.5.

Let $X$ be a Tychonoff space. A sequence $\langle\mu_{n}\colon
n\in\omega\rangle$ of finitely supported measures on $X$ is a bounded
Josefson–Nissenzweig sequence, or shortly a BJN-sequence, if $\|\mu_{n}\|=1$
for every $n\in\omega$ and $\mu_{n}(f)\rightarrow 0$ for every $f\in
C_{b}(X)$. We say that $X$ has the bounded Josefson–Nissenzweig property, or
shortly the BJNP, if there is a BJN-sequence of measures on $X$.

We also say that a free filter $F$ on $\omega$ has the bounded
Josefson–Nissenzweig property, or shortly the BJNP, if the space $N_{F}$ has
the BJNP. We denote by ${\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}$
the class of all ideals on $\omega$ whose dual filters have the BJNP.

The bounded Josefson–Nissenzweig property was introduced in [19], as a tool
for investigating the linear structure of the spaces $C_{b}(X\times Y)$
endowed with the pointwise topology. It is proved in [22, Corollary D] that if
$\mathcal{A}$ is a Boolean algebra, $F$ is a free filter on $\omega$ that has
the BJNP, and $N_{F}$ homeomorphically embeds into $St(\mathcal{A})$, then
$\mathcal{A}$ does not have the Grothendieck property.

By Proposition 6.1, $\mathcal{A}{\mathcal{N}}$ is a subclass of
${\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}$. As we mentioned
earlier, by Theorem 1.3 no element of $\mathcal{A}{\mathcal{N}}$ is maximal
with respect to the order $\leq_{K}$. In contrast, the class
${\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}$ has $\leq_{K}$-maximal
elements, e.g. the asymptotic density zero ideal ${\mathcal{Z}}$ is such an
element (see [22, Theorem C]). Moreover, Theorem 1.3 shows that for a density
ideal ${\mathcal{I}}$ it is equivalent that
${\mathcal{I}}\notin\mathcal{A}{\mathcal{N}}$ and that ${\mathcal{I}}$ is a
$\leq_{K}$-maximal element in
${\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}$. For more details on
differences between the classes $\mathcal{A}{\mathcal{N}}$ and
${\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}$ see Section $6$.

The structure of the paper is as follows. In Section 2 we describe basic
notation and terminology used in this paper. In Section 3 we analyze basic
properties of anti-Nikodym sequences of measures on spaces $N_{F}$. Section 4
is devoted to proving characterizations of the Nikodym property of filters,
and studying connections between this property and submeasures on $\omega$. In
Section 5 we present several conditions equivalent to
${\mathcal{I}}\not\in\mathcal{A}{\mathcal{N}}$ for a density ideal
${\mathcal{I}}$, and we investigate the Katětov order on the class
$\mathcal{A}{\mathcal{N}}$, in particular its cofinal structure. In Section 6
we compare the Nikodym property with the bounded Josefson–Nissenzweig property
of filters, and also classes of ideals on $\omega$ related to these
properties. In Section 7 we analyze some families of ideals that are contained
in the class $\mathcal{A}{\mathcal{N}}$, which give us some rich substructures
in the order $(\mathcal{A}{\mathcal{N}},\leq_{K})$. Section 8 is devoted to
applications of the Nikodym property of filters to Boolean algebras. The last
section provides some remarks and open questions.

### Acknowledgements

The author would like to thank Damian Sobota for many inspiring discussions
that contributed to the results of this paper, as well as for helpful comments
during the preparation of the paper.

##  2\. Preliminaries

By $\omega$ we denote the first infinite (countable) cardinal number. We also
denote $\omega_{+}=\omega\mathbin{\backslash}\\{0\\}=\\{1,2,3,\ldots\\}$. For
$A,B\subseteq\omega$ we write $A\subseteq^{*}B$ if $A\mathbin{\backslash}B$ is
finite. For every $k,n\in\omega$ such that $k<n$ we set
$[k,n]=\\{k,k+1,\ldots,n\\}$. By $\mathfrak{c}$ we denote the cardinality of
the real line $\mathbb{R}$, and by $\mathfrak{c}^{+}$—the cardinal successor
of $\mathfrak{c}$.

If $A$ is a non-empty set, then ${\mathcal{P}}(A)$ denotes its power set. A
family $F\subseteq{\mathcal{P}}(A)$ is a filter on $A$ if $\emptyset\notin F$,
$A\in F$ and $F$ is closed under finite intersections and taking supersets. By
$Fr$ we denote the Fréchet filter on $\omega$, i.e.
$Fr=\\{B\subseteq\omega\colon\omega\mathbin{\backslash}B$ is finite$\\}$. A
filter $F$ on $\omega$ is free if $Fr\subseteq F$. Throughout this article,
all filters on $\omega$ are assumed to be free. For a family
$S\subseteq{\mathcal{P}}(A)$ we put $S^{*}=\\{X\colon
A\mathbin{\backslash}X\in S\\}$—$S^{*}$ is said to be dual to $S$.

A family ${\mathcal{I}}\subseteq{\mathcal{P}}(\omega)$ is an ideal on $\omega$
if ${\mathcal{I}}^{*}$ is a free filter on $\omega$. Note that this implies
$Fin\subseteq{\mathcal{I}}$, where $Fin$ is the ideal of all finite subsets of
$\omega$. An ideal ${\mathcal{I}}$ on $\omega$ is a P-ideal if for every
sequence $\langle A_{n}\in{\mathcal{I}}\colon n\in\omega\rangle$ there is
$A\in{\mathcal{I}}$ such that $A_{n}\subseteq^{*}A$ for every $n\in\omega$. We
say that an ideal ${\mathcal{I}}$ on $\omega$ is tall if for every infinite
$A\subseteq\omega$ there exists an infinite $B\subseteq A$ such that
$B\in{\mathcal{I}}$.

We denote the set of all functions from $\omega$ to $\omega$ by
$\omega^{\omega}$. We define a preorder $\leq^{*}$ on $\omega^{\omega}$ by
setting $f\leq^{*}g$ if $f(n)\leq g(n)$ for all but finitely many
$n\in\omega$. There are two cardinal invariants related to this preorder:

  * •

$\mathfrak{d}$ is the smallest size of a family
${\mathcal{F}}\subseteq\omega^{\omega}$ such that for every
$g\in\omega^{\omega}$ there is $f\in{\mathcal{F}}$ with $g\leq^{*}f$ (such a
family is called dominating in $\omega^{\omega}$);

  * •

$\mathfrak{b}$ is the smallest size of a family
${\mathcal{F}}\subseteq\omega^{\omega}$ such that for every
$g\in\omega^{\omega}$ there is $f\in{\mathcal{F}}$ with $f\nleq^{*}g$ (such a
family is called unbounded in $\omega^{\omega}$).

The invariants $\mathfrak{d}$ and $\mathfrak{b}$ are examples of the cardinal
characteristics of the continuum, i.e. cardinal invariants which are between
$\aleph_{1}$ and $\mathfrak{c}$. Such an invariants are studied extensively in
the field of infinitary combinatorics, see [3].

Let $\mathcal{A}$ be a Boolean algebra. $St(\mathcal{A})$ denotes the Stone
space of $\mathcal{A}$, i.e. the space of all ultrafilters on $\mathcal{A}$
endowed with the standard topology. If $A\in\mathcal{A}$, then
$[A]_{\mathcal{A}}$ denotes the clopen set in $St(\mathcal{A})$ corresponding
via the Stone duality to $A$.

A function $\mu\colon\mathcal{A}\to\mathbb{R}$ is called a measure on a
Boolean algebra $\mathcal{A}$ if it is a finitely additive function of bounded
total variation, that is,

$\|\mu\|=\sup\big{\\{}|\mu(A)|+|\mu(B)|\colon A,B\in\mathcal{A},A\land
B=0_{\mathcal{A}}\big{\\}}<\infty.$

Let $X$ be a Tychonoff space. By $\operatorname{Clopen}(X)$ we denote the
family of clopen subsets of $X$, i.e. subsets which are both open and closed.
When we say that $\mu$ is a measure on $X$, then we mean that $\mu$ is a
$\sigma$-additive regular signed measure defined on the Borel $\sigma$-algebra
$Bor(X)$ of $X$ which has bounded total variation, that is,

$\|\mu\|=\sup\big{\\{}|\mu(A)|+|\mu(B)|\colon A,B\in Bor(X),A\cap
B=\emptyset\big{\\}}<\infty.$

By $|\mu|(\cdot)$ we denote the variation of $\mu$; note that
$\|\mu\|=|\mu|(X)$. $\operatorname{supp}(\mu)$ denotes the support of $\mu$.
If $A\in Bor(X)$, then $\mu\restriction A$ is defined by the formula
$(\mu\restriction A)(B)=\mu(A\cap B)$ for every $B\in Bor(X)$. $\mu$ is a non-
negative measure if $\mu(A)\geq 0$ for every $A\in Bor(X)$, and it is a
probability measure if it is non-negative and $\mu(X)=1$.

We define the Radon extension $\widehat{\mu}$ of a measure $\mu$ on a Boolean
algebra $\mathcal{A}$ as the unique measure on $St(\mathcal{A})$ such that
$\widehat{\mu}\big{(}[A]_{\mathcal{A}}\big{)}=\mu(A)$ for every
$A\in\mathcal{A}$.

If $X$ is a Tychonoff space and $x\in X$, then $\delta_{x}$ denotes the one-
point measure on $X$ concentrated at $x$. We say that a measure $\mu$ on $X$
is finitely supported if it is of the form
$\mu=\sum_{i=1}^{n}\alpha_{i}\cdot\delta_{x_{i}}$ for some
$\alpha_{1},\ldots,\alpha_{n}\in\mathbb{R}\mathbin{\backslash}\\{0\\}$ and
distinct $x_{1},\ldots,x_{n}\in X$. It follows that
$\operatorname{supp}(\mu)=\\{x_{1},\ldots,x_{n}\\}$,
$\|\mu\|=\sum_{i=1}^{n}|\alpha_{i}|$, and that for every $A\in Bor(X)$ we have
$\mu\restriction A=\sum_{i\colon x_{i}\in A}\alpha_{i}\cdot\delta_{x_{i}}$ and
$|\mu|(A)=\|\mu\restriction A\|=\sum_{i\colon x_{i}\in A}|\alpha_{i}|$.

For a non-negative measure $\mu$ supported on a finite set $X$ we define the
following numbers (cf. [5, Section 1.13]):

$\operatorname{at}^{+}(\mu)=\max\big{\\{}\mu(\\{x\\})\colon x\in X\big{\\}},$
$\operatorname{at}^{-}(\mu)=\min\big{\\{}\mu(\\{x\\})\colon x\in X\big{\\}}.$

If $\mu$ is a measure on $X$ and $f\in C(X)$, then we set
$\mu(f)=\int_{X}fd\mu$ if the integral exists. Note that if $\mu$ is of the
form $\sum_{i=1}^{n}\alpha_{i}\cdot\delta_{x_{i}}$, then
$\mu(f)=\sum_{i=1}^{n}\alpha_{i}\cdot f(x_{i})$.

##  3\. Anti-Nikodym sequences of measures on spaces $N_{F}$

In this section we study the main properties of anti-Nikodym sequences of
measures on spaces $N_{F}$.

We start with a remark about the possible generalization of the finitely
supported Nikodym property. One may ask whether we could exchange the
condition that $\mu_{n}(A)\rightarrow 0$ for every
$A\in\operatorname{Clopen}(X)$ for the condition that $\mu_{n}(f)\rightarrow
0$ for every $f\in C_{b}(X)$, and in this way generalize the definition of the
finitely supported Nikodym property to non-zero-dimensional spaces. However,
this property is not satisfied for any Tychonoff space $(X,\tau)$ which has a
weaker compact Hausdorff topology $\tau^{\prime}\subseteq\tau$, as it would
imply that there is a sequence of measures on $(X,\tau^{\prime})$ which is
weak* convergent to $0$ and unbounded in norm in the dual space
$C(X,\tau^{\prime})^{*}$ of bounded linear functionals on the Banach space
$C(X,\tau^{\prime})$—that would contradict the Banach–Steinhaus theorem for
$C(X,\tau^{\prime})$. In particular, for any filter $F$ the space $N_{F}$ does
not satisfy this stronger property, as $F$ contains the Fréchet filter $Fr$
and so the topology $\tau_{Fr}$ is weaker than $\tau_{F}$, and $N_{Fr}$ is
compact being homeomorphic to a convergent sequence of real numbers together
with its limit point.

######  Remark 3.1.

If $F\subseteq G$ are filters on $\omega$, then an AN-sequence
$\langle\mu_{n}\colon n\in\omega\rangle$ of measures on $N_{G}$ is also AN-
sequence on $N_{F}$, because the topology of $N_{G}$ is finer than the
topology of $N_{F}$, so
$\operatorname{Clopen}(N_{F})\subseteq\operatorname{Clopen}(N_{G})$. The same
is true for BJN-sequences of measures.

The following result shows that we can always modify an anti-Nikodym sequence
of measures on a space $N_{F}$ so that the measures have disjoint supports. It
is known that the analogous result holds for BJN-sequences of measures on any
Tychonoff space (see [22, Lemma 5.1.(3)]).

######  Proposition 3.2.

If a filter $F$ on $\omega$ does not have the Nikodym property, then there is
an AN-sequence $\langle\mu_{n}\colon n\in\omega\rangle$ of measures on $N_{F}$
which is disjointly supported, that is,
$\operatorname{supp}(\mu_{k})\cap\operatorname{supp}(\mu_{l})=\emptyset$ for
every $k\neq l\in\omega$.

###### Proof.

Let $\langle\mu_{n}\colon n\in\omega\rangle$ be an AN-sequence of measures on
$N_{F}$.

Step 1. We will find an AN-sequence $\langle\theta_{k}:\ k\in\omega\rangle$ on
$N_{F}$ such that
$\operatorname{supp}(\theta_{k})\cap\operatorname{supp}(\theta_{l})\cap\omega=\emptyset$
for every $k\neq l\in\omega$, i.e. with supports disjoint on $\omega$.

We first define inductively a strictly increasing sequence $\langle n_{k}:\
k\in\omega\rangle$ of natural numbers and an increasing sequence $\langle
A_{k}:\ k\in\omega\rangle$ of subsets of $N_{F}$ satisfying
$A_{k}=\big{[}0,\max\big{(}\bigcup_{i=0}^{k-1}\operatorname{supp}(\mu_{n_{i}})\cap\omega\big{)}\big{]}$
and $|\mu_{n_{k}}|(A_{k})<1/k$ for every $k\in\omega$. We start with $n_{0}=0$
and $A_{0}=\emptyset$. Assume now that for some $k\in\omega_{+}$ we have
defined natural numbers $n_{0}<\ldots<n_{k-1}$ and subsets
$A_{0}\subseteq\ldots\subseteq A_{k-1}$ of $N_{F}$ as required. The set
$A_{k}=\big{[}0,\max\big{(}\bigcup_{i=0}^{k-1}\operatorname{supp}(\mu_{n_{i}})\cap\omega\big{)}\big{]}$
is a finite subset of $\omega$. As $\langle\mu_{n}\colon n\in\omega\rangle$ is
an AN-sequence and singletons are clopen, there exists $n_{k}>n_{k-1}$ such
that $|\mu_{n_{k}}|(A_{k})<1/k$.

We now define a measure

$\theta_{k}=\mu_{n_{k}}\restriction(N_{F}\mathbin{\backslash}A_{k})$

for every $k\in\omega$. The sequence $\langle\theta_{k}:\ k\in\omega\rangle$
of measures on $N_{F}$ will meet the requirements, as:

  * •

for every pair of natural numbers $k<l$ we have
$\operatorname{supp}(\theta_{k})\cap\omega\subseteq A_{k+1}$ and
$\operatorname{supp}(\theta_{l})\cap A_{k+1}=\emptyset$, by the definition of
the sequence $\langle A_{k}:\ k\in\omega\rangle$,

  * •

for all $k\in\omega_{+}$ we have

$\|\theta_{k}\|\geq\|\mu_{n_{k}}\|-\|\mu_{n_{k}}\restriction
A_{k}\|>\|\mu_{n_{k}}\|-1/k\textrm{, and so }\|\theta_{k}\|\rightarrow\infty,$

  * •

for all $A\in\operatorname{Clopen}(N_{F})$ we have $\theta_{k}(A)\rightarrow
0$, as $\mu_{n_{k}}(A)\rightarrow 0$ and

$\big{|}\theta_{k}(A)-\mu_{n_{k}}(A)\big{|}\leq\|\theta_{k}-\mu_{n_{k}}\|=\|\mu_{n_{k}}\restriction
A_{k}\|<1/k\rightarrow 0.$

Step 2. In Step $1$ we constructed an AN-sequence $\langle\theta_{n}\colon
n\in\omega\rangle$ on $N_{F}$ with supports disjoint on $\omega$, and we will
now “remove” the point $p_{F}$ from the supports. We have two cases.

  1. a)

$\liminf_{n\to\infty}\big{|}\theta_{n}(\\{p_{F}\\})\big{|}<\infty$:

We can find a subsequence $\langle\theta_{n_{k}}:\ k\in\omega\rangle$ such
that $\theta_{n_{k}}\big{(}\\{p_{F}\\}\big{)}\to\alpha$ for some
$\alpha\in\mathbb{R}$, hence
$\big{\langle}\theta_{n_{k}}\big{(}\\{p_{F}\\}\big{)}\colon
k\in\omega\big{\rangle}$ is a Cauchy sequence. We define a sequence
$\langle\nu_{k}:\ k\in\omega\rangle$ of measures by setting

$\nu_{k}=(\theta_{n_{2k}}-\theta_{n_{2k+1}})\restriction\omega$

for every $k\in\omega$. Then, $\langle\nu_{k}:\ k\in\omega\rangle$ is a
disjointly supported AN-sequence on $N_{F}$, as:

     * •

for every $k\in\omega$ we have

$\operatorname{supp}(\nu_{k})=\big{(}\operatorname{supp}(\theta_{n_{2k}})\cup\operatorname{supp}(\theta_{n_{2k+1}})\big{)}\cap\omega,$

thus $\operatorname{supp}(\nu_{k})\cap\operatorname{supp}(\nu_{l})=\emptyset$
for $k\neq l$ as $\theta_{n}$’s have disjoint supports on $\omega$,

     * •

we have
$\|\nu_{k}\|=\|\theta_{n_{2k}}\restriction\omega\|+\|\theta_{n_{2k+1}}\restriction\omega\|\rightarrow\infty$,
because $\|\theta_{n_{l}}\restriction\omega\|\rightarrow\infty$ (which follows
from $\|\theta_{n_{l}}\|\rightarrow\infty$ and
$\|\theta_{n_{l}}\restriction\\{p_{F}\\}\|\rightarrow|\alpha|$),

     * •

for all $A\in\operatorname{Clopen}(N_{F})$ we have:

$\nu_{k}(A)=\big{(}\theta_{n_{2k}}-\theta_{n_{2k+1}}\big{)}(A\cap\omega)=\big{(}\theta_{n_{2k}}-\theta_{n_{2k+1}}\big{)}(A)-\big{(}\theta_{n_{2k}}-\theta_{n_{2k+1}}\big{)}(A\cap\\{p_{F}\\})\rightarrow
0,$

as $\langle\theta_{n_{k}}:\ k\in\omega\rangle$ is an AN-sequence and
$\theta_{n_{2k}}(\\{p_{F}\\})-\theta_{n_{2k+1}}(\\{p_{F}\\})\rightarrow 0$
holds because $\big{\langle}\theta_{n_{k}}\big{(}\\{p_{F}\\}\big{)}\colon
k\in\omega\big{\rangle}$ is Cauchy.

  2. b)

$\lim_{n\to\infty}\big{|}\theta_{n}(\\{p_{F}\\})\big{|}=\infty$:

Without losing of generality we may assume that the sequence
$\big{\langle}\big{|}\theta_{n}(\\{p_{F}\\})\big{|}\colon
n\in\omega\big{\rangle}$ is strictly increasing. If we denote

$\alpha_{n}=\frac{\theta_{2n}(\\{p_{F}\\})}{\theta_{2n+1}(\\{p_{F}\\})},$

then we get that $|\alpha_{n}|<1$ for all $n\in\omega$. We define a sequence
$\langle\nu_{n}\colon n\in\omega\rangle$ of measures by setting

$\nu_{n}=(\theta_{2n}-\alpha_{n}\cdot\theta_{2n+1})\restriction\omega$

for every $n\in\omega$. Then, $\langle\nu_{n}\colon n\in\omega\rangle$ is a
disjointly supported AN-sequence on $N_{F}$, as:

     * •

for every $n\in\omega$ we have

$\operatorname{supp}(\nu_{n})=\big{(}\operatorname{supp}(\theta_{2n})\cup\operatorname{supp}(\theta_{2n+1})\big{)}\cap\omega,$

thus $\operatorname{supp}(\nu_{k})\cap\operatorname{supp}(\nu_{l})=\emptyset$
for $k\neq l$ as $\theta_{n}$’s have disjoint supports on $\omega$,

     * •

we have
$\|\nu_{n}\|=\|\theta_{2n}\restriction\omega\|+|\alpha_{n}|\cdot\|\theta_{2n+1}\restriction\omega\|\rightarrow\infty$,
because for every $k\in\omega$ it holds

$\|\theta_{k}\restriction\omega\|\geq|\theta_{k}(\omega)|=|\theta_{k}(N_{F})-\theta_{k}(\\{p_{F}\\})|\rightarrow\infty,$

which follows from $\theta_{k}(N_{F})\rightarrow 0$ and the assumption
$\big{|}\theta_{k}(\\{p_{F}\\})\big{|}\rightarrow\infty$,

     * •

for all $A\in\operatorname{Clopen}(N_{F})$ we have:

$\nu_{n}(A)=\big{(}\theta_{2n}-\alpha_{n}\cdot\theta_{2n+1}\big{)}(A\cap\omega)=\big{(}\theta_{2n}-\alpha_{n}\cdot\theta_{2n+1}\big{)}(A)-\big{(}\theta_{2n}-\alpha_{n}\cdot\theta_{2n+1}\big{)}(A\cap\\{p_{F}\\})\rightarrow
0,$

as $\theta_{2n}(\\{p_{F}\\})-\alpha_{n}\cdot\theta_{2n+1}(\\{p_{F}\\})=0$,
$\langle\theta_{n}\colon n\in\omega\rangle$ is an AN-sequence and
$\langle\alpha_{n}\colon n\in\omega\rangle$ is bounded.

∎

We will now characterize when a sequence of finitely supported measures on a
space $N_{F}$ is an AN-sequence.

######  Proposition 3.3.

Let $F$ be a filter on $\omega$ and $\langle\mu_{n}\colon n\in\omega\rangle$ a
sequence of finitely supported measures on $N_{F}$. Then $\langle\mu_{n}\colon
n\in\omega\rangle$ is an AN-sequence on $N_{F}$ if and only if the following
three conditions simultaneously hold:

  1. (1)

$\lim_{n\to\infty}\|\mu_{n}\|=\infty$,

  2. (2)

$\lim_{n\to\infty}\mu_{n}(N_{F})=0$,

  3. (3)

$\lim_{n\to\infty}\big{\|}\mu_{n}\restriction(\omega\mathbin{\backslash}A)\big{\|}=0$
for every $A\in F$.

###### Proof.

Let $\langle\mu_{n}\colon n\in\omega\rangle$ be an AN-sequence on $N_{F}$.
Then, $(1)$ and $(2)$ hold by the definition. By Proposition 3.2, without
losing of generality we may assume that the sequence $\langle\mu_{n}\colon
n\in\omega\rangle$ is disjointly supported. Assume, for the sake of
contradiction, that for some $A\in F$ there is a subsequence
$\langle\mu_{n_{k}}:\ k\in\omega\rangle$ such that
$\lim_{k\to\infty}\big{\|}\mu_{n_{k}}\restriction(\omega\mathbin{\backslash}A)\big{\|}=C$
for some $C>0$. We may additionally assume that
$\big{\|}\mu_{n_{k}}\restriction(\omega\mathbin{\backslash}A)\big{\|}\geq C/2$
for every $k\in\omega$. For each $k\in\omega$ there exists
$F_{k}\subseteq\operatorname{supp}\big{(}\mu_{n_{k}}\restriction(\omega\mathbin{\backslash}A)\big{)}$
satisfying $|\mu_{n_{k}}(F_{k})|\geq C/4$. Since $\omega\mathbin{\backslash}A$
is a discrete clopen subspace of $N_{F}$, the set
$F=\bigcup_{k\in\omega}F_{k}$ is clopen in $N_{F}$. But
$\lim_{k\to\infty}|\mu_{n_{k}}(F)|\geq C/4$, contradicting the fact that
$\langle\mu_{n_{k}}:\ k\in\omega\rangle$ is an AN-sequence on $N_{F}$.

Assume now that $\langle\mu_{n}\colon n\in\omega\rangle$ is a sequence of
finitely supported measures on $N_{F}$ satisfying conditions $(1)$–$(3)$. If
$C\in\operatorname{Clopen}(N_{F})$ is such that $p_{F}\notin C$, then it must
be of the form $\omega\mathbin{\backslash}A$ for some $A\in F$, and so
$\mu_{n}(C)\rightarrow 0$ by $(3)$. On the other hand, if
$C\in\operatorname{Clopen}(N_{F})$ is such that $p_{F}\in C$, then we already
know that $\mu_{n}(N_{F}\mathbin{\backslash}C)\rightarrow 0$, thus
$\mu_{n}(C)=\mu_{n}(N_{F})-\mu_{n}(N_{F}\mathbin{\backslash}C)\rightarrow 0$,
by $(2)$. ∎

##  4\. The Nikodym property and submeasures on $\omega$

In this section we prove our main results about filters $F$ without the
Nikodym property and their corresponding dual ideals. We will first
characterize them using a sequences of non-negative measures on $\omega$, and
next using particular ideals on $\omega$ associated with submeasures. Theorems
1.3 and 1.3, and their proofs, were strongly inspired by similar
characterizations of filters with the BJNP due to Marciszewski and Sobota [22,
Theorem B and Theorem 6.2].

###### Proof.

Assume that the filter $F$ does not have the Nikodym property. By Proposition
3.2, there is an AN-sequence $\langle\theta_{n}\colon n\in\omega\rangle$ of
measures on $N_{F}$ with pairwise disjoint supports contained entirely in
$\omega$. Let us define a measure $\mu_{n}=|\theta_{n}|$ for each
$n\in\omega$. It follows that each $\mu_{n}$ is non-negative and finitely
supported, $\mu_{n}(\omega)=\|\theta_{n}\|\rightarrow\infty$, and, by
Proposition $\ref{prop_characterization}.(3)$, that

$\lim_{n\to\infty}\,\mu_{n}(\omega\mathbin{\backslash}A)=\lim_{n\to\infty}\,\big{\|}\theta_{n}\restriction(\omega\mathbin{\backslash}A)\big{\|}=0$

for every $A\in F$. Thus, conditions $(1)$–$(3)$ are satisfied.

Assume now that there is a sequence $\langle\mu_{n}\colon n\in\omega\rangle$
of finitely supported non-negative measures on $N_{F}$ satisfying conditions
$(1)$–$(3)$. For each $n\in\omega$ let us define a measure:

$\nu_{n}=\mu_{n}(\omega)\cdot\delta_{p_{F}}-\mu_{n}.$

By Proposition 3.3, it is immediate that $\langle\nu_{n}\colon
n\in\omega\rangle$ is an AN-sequence on $N_{F}$. ∎

Note that conditions $(2)$ and $(3)$ in the above theorem imply
$\mu_{n}(A)\to\infty$ for $A\in F$.

Let us recall some standard definitions concerning submeasures on $\omega$. A
function $\varphi\colon{\mathcal{P}}(\omega)\to[0,\infty]$ is a submeasure if
$\varphi(\emptyset)=0$, $\varphi(\\{n\\})<\infty$ for every $n\in\omega$,
$\varphi(X)\leq\varphi(Y)$ whenever $X\subseteq Y$, and $\varphi(X\cup
Y)\leq\varphi(X)+\varphi(Y)$ for every $X,Y$. A submeasure $\varphi$ on
$\omega$ is lower semicontinuous (lsc) if
$\varphi(A)=\lim_{n\to\infty}\varphi(A\cap[0,n])$ for every
$A\subseteq\omega$. We consider the following ideal associated with an lsc
submeasure $\varphi$ on $\omega$:

$\operatorname{Exh}(\varphi)=\big{\\{}A\subseteq\omega\colon\lim_{n\to\infty}\varphi(A\mathbin{\backslash}[0,n])=0\big{\\}},$

called the exhaustive ideal of $\varphi$. It is not difficult to show that
$\operatorname{Exh}(\varphi)$ is an $F_{\sigma\delta}$ P-ideal for every lsc
submeasure $\varphi$. There following result of Solecki characterizes analytic
P-ideals on $\omega$ in terms of submeasures.

######  Theorem 4.1 (Solecki [28]).

Let ${\mathcal{I}}$ be an ideal on $\omega$ such that
$Fin\subseteq{\mathcal{I}}$. Then ${\mathcal{I}}$ is an analytic P-ideal if
and only if there is a lsc submeasure $\varphi$ such that
${\mathcal{I}}=\operatorname{Exh}(\varphi)$.

For submeasures $\varphi$ and $\psi$ on $\omega$ we write $\psi\leq\varphi$ if
$\psi(A)\leq\varphi(A)$ for every $A\subseteq\omega$. Following Farah [5, page
21], we call a submeasure $\varphi$ non-pathological if for every
$A\subseteq\omega$ we have:

$\varphi(A)=\sup\\{\mu(A)\colon\mu\textrm{ is a non-negative measure on
}\omega\textrm{ such that }\mu\leq\varphi\\}.$

Trivially, every non-negative measure on $\omega$ is non-pathological.

The family of density submeasures is an important subfamily of non-
pathological lsc submeasures. Recall that a submeasure $\varphi$ on $\omega$
is a density submeasure if there exists a sequence $\langle\mu_{n}\colon
n\in\omega\rangle$ of finitely supported non-negative measures on $\omega$
with pairwise disjoint supports such that

$\varphi=\sup_{n\in\omega}\mu_{n}.$

An ideal ${\mathcal{I}}$ on $\omega$ is a density ideal if there is a density
submeasure $\varphi$ such that ${\mathcal{I}}=\operatorname{Exh}(\varphi)$. We
say in this situation that ${\mathcal{I}}$ is generated by the sequence
$\langle\mu_{n}\colon n\in\omega\rangle$. Note that
$X\in\operatorname{Exh}(\varphi)$ if and only if
$\lim_{n\to\infty}\mu_{n}(X)=0$.

A prototypical ideal for the class of density ideals is the (asymptotic)
density zero ideal ${\mathcal{Z}}=\operatorname{Exh}(\varphi_{d})$, where the
asymptotic density submeasure $\varphi_{d}$ is defined for every
$A\subseteq\omega$ by:

$\varphi_{d}(A)=\sup_{n\in\omega}\frac{\big{|}A\cap[2^{n},2^{n+1})\big{|}}{2^{n}}.$

We are now ready to provide a characterization of the Nikodym property of
filters in terms of exhaustive ideals.

###### Proof.

$(1)\Rightarrow(2)$ Assume that $F$ does not have the Nikodym property. Let
$\langle\mu_{n}\colon n\in\omega\rangle$ be a sequence of finitely supported
non-negative measures on $N_{F}$ as in Theorem 1.3. By the remark after the
theorem, we may additionally assume that $\mu_{n}$’s have pairwise disjoint
supports such that
$\max\big{(}\operatorname{supp}(\mu_{n})\big{)}<\min\big{(}\operatorname{supp}(\mu_{n+1})\big{)}$
for every $n\in\omega$. For each $A\subseteq\omega$ define:

$\varphi(A)=\sup_{n\in\omega}\mu_{n}(A).$

Then $\varphi$ is a density submeasure. We have
$\varphi(\omega)=\sup_{n\in\omega}\mu_{n}(\omega)=\infty$ by condition $(2)$
from Theorem 1.3. It is fairly easy to see that
$F\subseteq\operatorname{Exh}(\varphi)^{*}$ (cf. [22, Theorem 6.2]).

$(2)\Rightarrow(3)$ Obvious.

$(3)\Rightarrow(1)$ Assume that there is a non-pathological lsc submeasure
$\varphi$ such that $\varphi(\omega)=\infty$ and
$F\subseteq\operatorname{Exh}(\varphi)^{*}$. By Remark 3.1, it is enough to
prove that $\operatorname{Exh}(\varphi)^{*}$ does not have the Nikodym
property.

As $\varphi$ is finite on finite sets and $\varphi(\omega)=\infty$, by the
subadditivity of $\varphi$ for every $n\in\omega$ we get:

($\ast$) $\varphi\big{(}\omega\mathbin{\backslash}[0,n]\big{)}=\infty.$

Let $n_{0}=0$. Since $\varphi$ is lower semi-continuous, there exists
$n_{1}>n_{0}$ such that $\varphi([n_{0},n_{1}])>2$. By ($\ast$ ‣ 4) we have
$\varphi(\omega\mathbin{\backslash}[0,n_{1}])=\infty$, so again there is
$n_{2}>n_{1}$ such that $\varphi([n_{1},n_{2}])>3$. We continue in this way
until we get a strictly increasing sequence $\langle n_{k}:\
k\in\omega\rangle$ of natural numbers satisfying for every $k\in\omega$ the
inequality

$\varphi\big{(}[n_{k},n_{k+1}]\big{)}>k+1.$

The submeasure $\varphi$ is non-pathological, so for each $k\in\omega$ there
exists a non-negative measure $\mu_{k}$ on $\omega$ such that
$\mu_{k}\leq\varphi$, $\operatorname{supp}(\mu_{k})\subseteq[n_{k},n_{k+1}]$,
and

$\mu_{k}\big{(}[n_{k},n_{k+1}]\big{)}>k.$

Measures $\mu_{k}$ are finitely supported, and
$\lim_{k\to\infty}\mu_{k}(\omega)\geq\lim_{k\to\infty}\mu_{k}\big{(}[n_{k},n_{k+1}]\big{)}=\infty$.

We claim that the sequence $\langle\mu_{k}:\ k\in\omega\rangle$ satisfies the
equality $\lim_{k\to\infty}\mu_{k}(A)=0$ for every
$A\in\operatorname{Exh}(\varphi)$, and thus, by Theorem 1.3, the filter
$\operatorname{Exh}(\varphi)^{*}$ does not have the Nikodym property. Let
$A\in\operatorname{Exh}(\varphi)$ and $\varepsilon>0$. Since
$\lim_{n\to\infty}\varphi\big{(}A\mathbin{\backslash}[0,n]\big{)}=0$, there is
$M\in\omega$ such that
$\varphi\big{(}A\mathbin{\backslash}[0,M]\big{)}<\varepsilon$. Let
$k\in\omega$ be such that $n_{k}>M$. It holds:

$\mu_{k}(A)=\mu_{k}\big{(}A\cap[n_{k},n_{k+1}]\big{)}\leq\varphi\big{(}A\cap[n_{k},n_{k+1}]\big{)}\leq\varphi\big{(}A\mathbin{\backslash}[0,M]\big{)}<\varepsilon,$

hence $\lim_{k\to\infty}\mu_{k}(A)=0$. ∎

Recall that by classical theorem of Sierpiński (see [2, Theorem 4.1.1]), every
ideal of the form $\operatorname{Exh}(\varphi)$ for some lsc submeasure
$\varphi$ is meager and of measure zero (as it is a Borel subset of
$2^{\omega}$), and every free ultrafilter on $\omega$ is non-meager.
Therefore, we get the following corollaries.

######  Corollary 4.2.

If a filter $F$ on $\omega$ does not have the Nikodym property, then $F$ is
meager and of measure zero. In particular, if $F$ is a free ultrafilter on
$\omega$, then $F$ has the Nikodym property.

It appears that the notion of a totally bounded ideal is closely related to
the Nikodym property of filters.

######  Definition 4.3 (Hernández-Hernández and Hrušák [16]).

An analytic P-ideal ${\mathcal{I}}$ on $\omega$ is said to be totally bounded
if whenever $\varphi$ is a lower semicontinuous submeasure on $\omega$ for
which ${\mathcal{I}}=\operatorname{Exh}(\varphi)$, then
$\varphi(\omega)<\infty$.

A typical example of a totally bounded ideal is ${\mathcal{Z}}$ ([16,
Proposition 3.18]), whereas, e.g., no summable ideal is totally bounded (see
Section $7.1$). We will study totally bounded ideals further in the context of
density ideals in the next section.

######  Corollary 4.4.

Let ${\mathcal{I}}=\operatorname{Exh}(\varphi)$ for an lsc submeasure
$\varphi$ on $\omega$. If ${\mathcal{I}}$ is totally bounded, then
${\mathcal{I}}^{*}$ has the Nikodym property.

###### Proof.

If ${\mathcal{I}}^{*}$ does not have the Nikodym property, then by Theorem 1.3
there is an lsc submeasure $\psi$ on $\omega$ such that $\psi(\omega)=\infty$
and $\operatorname{Exh}(\varphi)\subseteq\operatorname{Exh}(\psi)$. We define
an lsc submeasure on $\omega$ by $\varphi^{\prime}=\max(\psi,\varphi)$. Of
course, we have $\varphi^{\prime}(\omega)=\infty$ and
$\operatorname{Exh}(\varphi^{\prime})\subseteq\operatorname{Exh}(\varphi)$.
Now, if $A\in\operatorname{Exh}(\varphi)$, then also
$A\in\operatorname{Exh}(\psi)$, so

$\lim_{n\to\infty}\varphi^{\prime}(A\mathbin{\backslash}[0,n])=\max\Big{(}\lim_{n\to\infty}\varphi(A\mathbin{\backslash}[0,n]),\lim_{n\to\infty}\psi(A\mathbin{\backslash}[0,n])\Big{)}=0,$

and hence
$\operatorname{Exh}(\varphi)\subseteq\operatorname{Exh}(\varphi^{\prime})$.
Thus we have
$\operatorname{Exh}(\varphi)=\operatorname{Exh}(\varphi^{\prime})$, and so
${\mathcal{I}}$ is not totally bounded.

∎

######  Remark 4.5.

Let us note that the converse to Corollary 4.4 does not hold in general. The
counterexamples were given in [8, Theorem 4.12] (with the ideal defined in
[7]) and in [22, Example 6.15]. Both of them are of the form
${\mathcal{I}}=\operatorname{Exh}(\varphi)=\operatorname{Fin}(\varphi)$ for
some lsc submeasure $\varphi$ on $\omega$ such that ${\mathcal{I}}^{*}$ does
not have the BJNP. By Proposition 6.1, either filter ${\mathcal{I}}^{*}$ has
the Nikodym property, and $\varphi(\omega)=\infty$ because
$\omega\notin\operatorname{Fin}(\varphi)$, thus ${\mathcal{I}}$ is not totally
bounded.

##  5\. The Nikodym property and the Katětov order

We will focus in this section on the structure of the Katětov order (and its
variant) on the ideals connected with the Nikodym property.

Let ${\mathcal{I}}$ and ${\mathcal{J}}$ be ideals on $\omega$. We say that
${\mathcal{I}}$ is Katětov below  ${\mathcal{J}}$, which we denote by
${\mathcal{I}}\leq_{K}{\mathcal{J}}$, if there is a function
$f\colon\omega\to\omega$ such that $f^{-1}[A]\in{\mathcal{J}}$ for all
$A\in{\mathcal{I}}$. We call such a function $f$ a Katětov reduction.
Similarly, we say that ${\mathcal{I}}$ is Katětov–Blass below ${\mathcal{J}}$,
which we denote by ${\mathcal{I}}\leq_{KB}{\mathcal{J}}$, if there is a
finite-to-one function $f\colon\omega\to\omega$ such that
$f^{-1}[A]\in{\mathcal{J}}$ for all $A\in{\mathcal{I}}$, and we call such a
function $f$ a Katětov–Blass reduction. We write
${\mathcal{I}}<_{K}{\mathcal{J}}$ if ${\mathcal{I}}\leq_{K}{\mathcal{J}}$ and
${\mathcal{J}}\nleq_{K}{\mathcal{I}}$. We say that ${\mathcal{I}}$ and
${\mathcal{J}}$ are Katětov equivalent, which we denote by
${\mathcal{I}}\equiv_{K}{\mathcal{J}}$, if
${\mathcal{I}}\leq_{K}{\mathcal{J}}$ and ${\mathcal{J}}\leq_{K}{\mathcal{I}}$.
Trivially, ${\mathcal{I}}\subseteq{\mathcal{J}}$ implies
${\mathcal{I}}\leq_{KB}{\mathcal{J}}$, with the identity function being a
Katětov–Blass reduction. We say that ${\mathcal{I}}$ is isomorphic to
${\mathcal{J}}$, which we denote by ${\mathcal{I}}\cong{\mathcal{J}}$, if
there is a bijection $f\colon\omega\to\omega$ such that $A\in{\mathcal{I}}$ if
and only if $f[A]\in{\mathcal{J}}$ for all $A\subseteq\omega$. It is immediate
that ${\mathcal{I}}\cong{\mathcal{J}}$ implies
${\mathcal{I}}\equiv_{K}{\mathcal{J}}$. We define the Katětov order for
filters on $\omega$ in the natural way via dual ideals.

The following lemma is folklore.

######  Lemma 5.1.

Let ${\mathcal{I}}$ and ${\mathcal{J}}$ be ideals on $\omega$. If
${\mathcal{I}}\leq_{K}{\mathcal{J}}$ and ${\mathcal{J}}$ is a P-ideal, then
${\mathcal{I}}\leq_{KB}{\mathcal{J}}$.

###### Proof.

111This argument was presented to us by Jacek Tryba in personal communication.

Let $f\colon\omega\to\omega$ be a Katětov reduction witnessing
${\mathcal{I}}\leq_{K}{\mathcal{J}}$. For every $n\in\omega$ the set
$A_{n}=f^{-1}(\\{n\\})$ belongs to the ideal ${\mathcal{J}}$. By the
assumption about ${\mathcal{J}}$, there exists $A\in{\mathcal{J}}$ such that
$A_{n}\subseteq^{*}A$ for every $n\in\omega$. We define
$g\colon\omega\to\omega$ by:

$g(n)=\begin{cases}n,&\text{if }n\in A,\\\ f(n),&\text{if }n\notin
A.\end{cases}$

The function $f$ is finite-to-one on $\omega\mathbin{\backslash}A$, as
$(\omega\mathbin{\backslash}A)\cap A_{n}$ is finite for every $n\in\omega$,
thus $g$ is finite-to-one on $\omega$. The function $g$ is a Katětov–Blass
reduction, because for any $X\in{\mathcal{I}}$ we have $g^{-1}[X]\subseteq
f^{-1}[X]\cup A\in{\mathcal{J}}$, hence also $g^{-1}[X]\in{\mathcal{J}}$. ∎

First, we show how the Katětov order transfers the negation of the Nikodym
property between filters (this strengthens Remark 3.1).

######  Proposition 5.2.

If $F\leq_{K}G$ and $G$ does not have the Nikodym property, then also $F$ does
not have the Nikodym property.

###### Proof.

Let $f\colon\omega\to\omega$ be a witness for $F\leq_{K}G$ and let
$\langle\mu_{n}\colon n\in\omega\rangle$ be a sequence of finitely supported
non-negative measures on $N_{G}$ as in Theorem 1.3. For each $n\in\omega$
define a measure:

$\mu^{\prime}_{n}=\sum_{x\in\operatorname{supp}(\mu_{n})}\mu_{n}(\\{x\\})\cdot\delta_{f(x)}.$

Then, we have $\|\mu_{n}^{\prime}\|\to\infty$,
$\operatorname{supp}(\mu_{n}^{\prime})\subseteq\omega$, and
$\mu_{n}^{\prime}\geq 0$. We also claim that
$\lim_{n\to\infty}\mu_{n}^{\prime}(\omega\mathbin{\backslash}A)=0$ holds for
any $A\in F$, and thus, by Theorem 1.3, the filter $F$ does not have the
Nikodym property. Indeed, if there was $A\in F$ such that
$\limsup_{n\to\infty}\mu_{n}^{\prime}(\omega\mathbin{\backslash}A)>0$, then
$f^{-1}[A]\in G$ (as $f$ is a Katětov reduction) and
$\lim\sup_{n\to\infty}\mu_{n}\big{(}\omega\mathbin{\backslash}f^{-1}[A]\big{)}>0$,
which would contradict condition $(3)$ from Theorem 1.3. ∎

Proposition 5.2 yields the following variant of Theorem 1.3.

######  Theorem 5.3.

Let $F$ be a filter on $\omega$. Then, the following are equivalent:

  1. (1)

$F$ does not have the Nikodym property;

  2. (2)

there is a density submeasure $\varphi$ on $\omega$ such that
$\varphi(\omega)=\infty$ and $F\leq_{K}\operatorname{Exh}(\varphi)^{*}$;

  3. (3)

there is a non-pathological lsc submeasure $\varphi$ on $\omega$ such that
$\varphi(\omega)=\infty$ and $F\leq_{K}\operatorname{Exh}(\varphi)^{*}$.

###### Proof.

$(1)\Rightarrow(2)$ Let $F$ be a filter on $\omega$ without the Nikodym
property. By Theorem 1.3, there is a density submeasure $\varphi$ on $\omega$
such that $\varphi(\omega)=\infty$ and
$F\subseteq\operatorname{Exh}(\varphi)^{*}$, and so
$F\leq_{K}\operatorname{Exh}(\varphi)^{*}$.

$(2)\Rightarrow(3)$ Obvious.

$(3)\Rightarrow(1)$ Let $\varphi$ be a non-pathological lsc submeasure on
$\omega$ such that $\varphi(\omega)=\infty$ and
$F\leq_{K}\operatorname{Exh}(\varphi)^{*}$. By Theorem 1.3, the filter
$\operatorname{Exh}(\varphi)^{*}$ does not have the Nikodym property, thus by
Proposition 5.2 the filter $F$ does not have the Nikodym property. ∎

###### Proof.

If $F$ does not have the Nikodym property, then by Theorem 5.3 there exists a
density ideal ${\mathcal{I}}$ such that $F^{*}\leq_{K}{\mathcal{I}}$. By [16,
Proposition 3.6] we have ${\mathcal{I}}\leq_{K}{\mathcal{Z}}$, thus
$F^{*}\leq_{K}{\mathcal{Z}}$. ∎

We will now prove that, in the case of density ideals, several known
properties are equivalent with the Nikodym property. For the definition of
Erdős–Ulam ideal see [5, Section 1.13], for the Bolzano–Weierstrass property
see [6, Section 2], and for a splitting family in a Boolean algebra see [16,
page 589]. The following theorem extends [31, Theorem 3.16]222We omit in $(3)$
the condition $\limsup_{n\in\omega}(\omega)>0$, as the definition of an ideal
which we use in this paper already implies this condition..

###### Proof.

The equivalence of conditions $(1)$, $(2)$, and $(3)$ was already shown in
[31, Theorem 3.16].

$(2)\Leftrightarrow(4)$ By [6, Theorem 5.1].

$(4)\Rightarrow(5)$ By [16, Lemma 3.17].

$(5)\Rightarrow(7)$ By Corollary 4.4.

$(1)\Rightarrow(6)$ If there exists an Erdős–Ulam ideal ${\mathcal{J}}$ such
that ${\mathcal{I}}\cong{\mathcal{J}}$, then in particular
${\mathcal{I}}\equiv_{K}{\mathcal{J}}$. By [5, Lemma 1.13.10] we have
${\mathcal{J}}\equiv_{K}{\mathcal{Z}}$, thus also
${\mathcal{I}}\equiv_{K}{\mathcal{Z}}$.

$(6)\Rightarrow(7)$ Assume, for the sake of contradiction, that
${\mathcal{I}}\equiv_{K}{\mathcal{Z}}$ and ${\mathcal{I}}^{*}$ does not have
the Nikodym property. By Proposition 5.2 we get that ${\mathcal{Z}}$ does not
have the Nikodym property, but ${\mathcal{Z}}$ is totally bounded by [16,
Proposition 3.18], which contradicts Corollary 4.4.

$(7)\Rightarrow(3)$ Assume that condition $a)$ or condition $b)$ from $(3)$
fails. If $\sup_{n\in\omega}\|\mu_{n}\|=\infty$, then for the density
submeasure $\varphi=\sup_{n\in\omega}\mu_{n}$ we have
${\mathcal{I}}=\operatorname{Exh}(\varphi)$ and $\varphi(\omega)=\infty$,
hence by Theorem 1.3 the filter ${\mathcal{I}}^{*}$ does not have the Nikodym
property. If $\lim\sup_{n\to\infty}\operatorname{at}^{+}(\mu_{n})>0$, then
$\lim\sup_{n\to\infty}\varphi(\\{n\\})>0$, and so by [16, Lemma 1.4] the ideal
${\mathcal{I}}$ is not tall. It is well-known that this implies that
${\mathcal{I}}\equiv_{K}Fin$, and $Fin^{*}=Fr$ does not have the Nikodym
property by the discussion in Introduction about non-trivial convergent
sequences. Thus, by Proposition 5.2, the filter ${\mathcal{I}}^{*}$ does not
have the Nikodym property. In both cases we got a contradiction, hence $(3)$
is satisfied. ∎

######  Corollary 5.4.

Every Erdős–Ulam ideal is totally bounded and has the Nikodym property.

By Corollary 1.3 and Theorem 1.3 we get the following corollary.

######  Corollary 5.5.

If ${\mathcal{I}}$ is a density ideal, then
${\mathcal{I}}\in\mathcal{A}{\mathcal{N}}$ if and only if
$I<_{K}{\mathcal{Z}}$.

Let us note that the equivalence from the previous corollary does not hold
even for ideals of the form ${\mathcal{I}}=\operatorname{Exh}(\varphi)$ for a
non-pathological submeasure $\varphi$. A counterexample is given in Section
$6$.

###  5.1. Cofinal structure of $(\mathcal{A}{\mathcal{N}},\leq_{K})$

Now, we will define an operator $\Phi$ from $\omega^{\omega}$ to
$\mathcal{A}{\mathcal{N}}$ such that its image is $\leq_{K}$-cofinal in
$\mathcal{A}{\mathcal{N}}$ and which preserves dominating families. This will
enable us to obtain some properties of the order
$(\mathcal{A}{\mathcal{N}},\leq_{K})$.

Let $f\in\omega^{\omega}$ be such that $f(n)>0$ for every $n\in\omega_{+}$. We
define a sequence $\langle\mu_{n}^{f}\colon n\in\omega_{+}\rangle$ of measures
satisfying:

  * •

$\langle I_{n}\colon n\in\omega\rangle$ is an interval partition of $\omega$,
$\min(I_{1})=0$ and for every $n\in\omega_{+}$ we have
$\max(I_{n})+1=\min(I_{n+1})$,

  * •

the measure $\mu_{n}^{f}$ is supported on the interval $I_{n}$,

  * •

$\|\mu_{n}^{f}\|=n$,

  * •

$\operatorname{at}^{+}(\mu_{n}^{f})=\operatorname{at}^{-}(\mu_{n}^{f})=1/f(n)$.

Finally, let

$\Phi(f)=\operatorname{Exh}\Big{(}\sup_{n\in\omega_{+}}\mu_{n}^{f}\Big{)}.$

Then, $\Phi(f)$ is a density ideal.

The following theorem presents the most important properties of the operator
$\Phi$.

######  Theorem 5.6.

  1. (1)

For every ${\mathcal{I}}\in\mathcal{A}{\mathcal{N}}$ there exists
$f\in\omega^{\omega}$ such that ${\mathcal{I}}\leq_{K}\Phi(f)$.

  2. (2)

For every $g,h\in\omega^{\omega}$, if $2n^{2}\cdot g(n)\leq h(n)$ for all but
finitely many $n\in\omega$, then $\Phi(g)\leq_{K}\Phi(h)$.

For the proof of Theorem 5.6 we will need the following simple lemma.

######  Lemma 5.7.

Let $\lambda$ be a non-negative measure on a finite non-empty set $A$ and let
$\varepsilon>0$. If $\mu$ is a non-negative measure on a finite set $B$ such
that $\mu(B)=\lambda(A)$ and
$\operatorname{at}^{+}(\mu)\leq\varepsilon/(2|A|)$, then there exists a
function $f\colon B\to A$ such that for every $C\subseteq A$ we have:

$\big{|}\lambda(C)-\mu\big{(}f^{-1}[C]\big{)}\big{|}\leq\varepsilon.$

###### Proof.

As $\operatorname{at}^{+}(\mu)\leq\varepsilon/(2|A|)$, we can construct a
family $\langle X_{a}\colon a\in A\rangle$ of pairwise disjoint subsets of $B$
satisfying

$\lambda(\\{a\\})-\varepsilon/(2|A|)<\mu(X_{a})\leq\lambda(\\{a\\})$

for every $a\in A$. Let us define:

$Y=B\mathbin{\backslash}\bigcup\\{X_{a}\colon a\in A\\},$

and note that above conditions together with $\mu(B)=\lambda(A)$ imply that
$\mu(Y)<\varepsilon/2$.

Fix any $a_{0}\in A$ and define the function $f\colon B\to A$ in the following
way:

$f(b)=\begin{cases}a,&\text{if }b\in X_{a},\\\ a_{0},&\text{if }b\in
Y.\end{cases}$

It is not difficult to check that $f$ has the desired property. ∎

The next lemma provides a technical condition sufficient for the existence of
a Katětov reduction between density ideals.

######  Lemma 5.8.

If $\langle\mu_{n}\colon n\in\omega_{+}\rangle$ and $\langle\lambda_{n}\colon
n\in\omega_{+}\rangle$ are disjointly supported sequences of measures on
$\omega$ with finite supports such that $\|\mu_{n}\|=\|\lambda_{n}\|=n$ for
every $n\in\omega_{+}$, and
$\operatorname{at}^{+}(\mu_{n})\leq\operatorname{at}^{-}(\lambda_{n})/2n^{2}$
is satisfied for all but finitely many $n\in\omega_{+}$, then we have:

$\operatorname{Exh}\Big{(}\sup_{n\in\omega_{+}}\lambda_{n}\Big{)}\leq_{K}\operatorname{Exh}\Big{(}\sup_{n\in\omega_{+}}\mu_{n}\Big{)}.$

###### Proof.

For each $n\in\omega_{+}$ let $A_{n}=\operatorname{supp}(\lambda_{n})$. We
have $|A_{n}|\cdot\operatorname{at}^{-}(\lambda_{n})\leq\|\lambda_{n}\|=n$,
thus $\operatorname{at}^{-}(\lambda_{n})\leq n/|A_{n}|$. Let $N\in\omega_{+}$
be such that
$\operatorname{at}^{+}(\mu_{n})\leq\operatorname{at}^{-}(\lambda_{n})/2n^{2}$
for every $n\geq N$. Then, for every $n\geq N$ we get:

$\operatorname{at}^{+}(\mu_{n})\leq\operatorname{at}^{-}(\lambda_{n})/2n^{2}\leq\frac{n}{2n^{2}\cdot|A_{n}|}=\frac{1}{n}\cdot\frac{1}{2|A_{n}|}.$

Therefore, for $n\geq N$ the assumptions of Lemma 5.7 are satisfied with
$\lambda=\lambda_{n}$ on $A=A_{n}=\operatorname{supp}(\lambda_{n})$ and
$\mu=\mu_{n}$ on $B=\operatorname{supp}(\mu_{n})$, and $\varepsilon=1/n$.
Thus, for every $n\geq N$ there exists a function
$f_{n}\colon\operatorname{supp}(\mu_{n})\to A_{n}$ such that for every
$C\subseteq A_{n}$ we have:

($\ast$) $\big{|}\lambda_{n}(C)-\mu_{n}\big{(}f_{n}^{-1}[C]\big{)}\big{|}\leq
1/n.$

Now we define the function $f\colon\omega\to\omega$ by the formula

$f(l)=\begin{cases}f_{n}(l),&\text{ if }l\in\operatorname{supp}(\mu_{n})\text{
for some }n\geq N,\\\ 0,&\text{ otherwise,}\end{cases}$

for every $l\in\omega$. We will show that $f$ is a witness for
$\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\lambda_{n}\big{)}\leq_{K}\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\mu_{n}\big{)}$,
i.e.
$f^{-1}[X]\in\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\mu_{n}\big{)}$ for
any $X\in\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\lambda_{n}\big{)}$.
Let us fix
$X\in\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\lambda_{n}\big{)}$ and
$p\geq 1$. There is $m\geq\max(N,p)$ such that for every $n\geq m$ we have
$\lambda_{n}(X)<1/p$.

Then, by ($\ast$ ‣ 5.1) and the definition of $f$, for every $n\geq m$ we get:

$\mu_{n}\big{(}f^{-1}[X]\big{)}\leq 2/p,$

which shows that $\lim_{n\to\infty}\mu_{n}\big{(}f^{-1}[X]\big{)}=0$ and hence
that
$f^{-1}[X]\in\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\mu_{n}\big{)}$. ∎

We are in the position to prove Theorem 5.6.

###### Proof of Theorem 5.6.

$(1)$. Let ${\mathcal{I}}\in\mathcal{A}{\mathcal{N}}$. By Theorem 1.3, there
is a density ideal
$\operatorname{Exh}\big{(}\sup_{n\in\omega}\lambda_{n}\big{)}$ for some
sequence $\langle\lambda_{n}\colon n\in\omega\rangle$ of finitely supported
measures with disjoint supports such that $\|\lambda_{n}\|\to\infty$ and
${\mathcal{I}}\subseteq\operatorname{Exh}\big{(}\sup_{n\in\omega}\lambda_{n}\big{)}$.
Without losing of generality we may assume that $\|\lambda_{n}\|\geq n+1$ for
all $n\in\omega$, as by taking a subsequence of the sequence
$\langle\lambda_{n}\colon n\in\omega\rangle$ we can only enlarge the ideal
$\operatorname{Exh}\big{(}\sup_{n\in\omega}\lambda_{n}\big{)}$.

We define another sequence $\langle\nu_{n}\colon n\in\omega_{+}\rangle$ of
measures by setting

$\nu_{n}(A)=\lambda_{n-1}(A)\cdot n/\|\lambda_{n-1}\|$

for each $A\subseteq\omega$ and $n\in\omega_{+}$. It follows that
$\|\nu_{n}\|=n$ for every $n\in\omega_{+}$, and
${\mathcal{I}}\subseteq\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\nu_{n}\big{)}$.
Indeed, if $A\in{\mathcal{I}}$ then $\lambda_{n}(A)\to 0$, and so

$\nu_{n}(A)=\lambda_{n-1}(A)\cdot n/\|\lambda_{n-1}\|\to 0,$

since $n/\|\lambda_{n-1}\|\leq 1$ for every $n\in\omega$.

Let us define $f\in\omega^{\omega}$ by setting $f(n)=2n^{2}\cdot\lceil
1/\operatorname{at}^{-}(\nu_{n})\rceil$ for each $n\in\omega_{+}$ and
$f(0)=0$. Then, for the sequence of measures $\langle\mu_{n}^{f}\colon
n\in\omega_{+}\rangle$ generating the density ideal $\Phi(f)$ we have
$\operatorname{at}^{+}(\mu_{n}^{f})\leq\operatorname{at}^{-}(\nu_{n})/2n^{2}$
for every $n\in\omega_{+}$, and so by Lemma 5.8 we get
$\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\nu_{n}\big{)}\leq_{K}\Phi(f)$,
thus also ${\mathcal{I}}\leq_{K}\Phi(f)$.

$(2)$. Let $g,h\in\omega^{\omega}$ be such that $2n^{2}\cdot g(n)\leq h(n)$
for all but finitely many $n\in\omega$. It follows immediately from Lemma 5.8
that $\Phi(g)\leq_{K}\Phi(h)$. ∎

Theorem 5.6 has several important consequences. The next corollary shows that
$\Phi$ preserves dominating families.

######  Corollary 5.9.

If ${\mathcal{F}}$ is a dominating family in $\omega^{\omega}$, then
$\Phi[{\mathcal{F}}]$ is $\leq_{K}$-dominating in $\mathcal{A}{\mathcal{N}}$,
i.e. for any ${\mathcal{I}}\in\mathcal{A}{\mathcal{N}}$ there is
$f\in{\mathcal{F}}$ such that ${\mathcal{I}}\leq_{K}\Phi(f)$.

###### Proof.

Let us fix a dominating family ${\mathcal{F}}$ in $\omega^{\omega}$ and let
${\mathcal{I}}\in\mathcal{A}{\mathcal{N}}$. By Theorem $\ref{thm_katetov}.(1)$
there is $f\in\omega^{\omega}$ such that ${\mathcal{I}}\leq_{K}\Phi(f)$. As
${\mathcal{F}}$ is dominating, there exists $g\in{\mathcal{F}}$ such that
$g(n)\geq 2n^{2}\cdot f(n)$ for all but finitely many $n\in\omega$. By Theorem
$\ref{thm_katetov}.(2)$ we get $\Phi(f)\leq_{K}\Phi(g)$, and so
${\mathcal{I}}\leq_{K}\Phi(g)$. ∎

The following theorem follows immediately from Corollary 5.9, as by definition
there is a dominating family ${\mathcal{F}}\subseteq\omega^{\omega}$ of size
$\mathfrak{d}$.

######  Corollary 5.10.

Every subfamily of $\mathcal{A}{\mathcal{N}}$ of size less than $\mathfrak{b}$
is bounded in the order $(\mathcal{A}{\mathcal{N}},\leq_{K})$.

###### Proof.

Let $\\{{\mathcal{I}}_{\alpha}\colon\alpha<\kappa\\}$ be a subfamily of
$\mathcal{A}{\mathcal{N}}$ for some $\kappa<\mathfrak{b}$. By Theorem
$\ref{thm_katetov}.(1)$, for every $\alpha<\kappa$ there is
$f_{\alpha}\in\omega^{\omega}$ such that
${\mathcal{I}}_{\alpha}\leq_{K}\Phi(f_{\alpha})$. As $\kappa<\mathfrak{b}$,
there exists $g\in\omega^{\omega}$ such that for every $\alpha<\kappa$ we have
$g(n)\geq 2n^{2}\cdot f_{\alpha}(n)$ for all but finitely many $n\in\omega$,
so by Theorem $\ref{thm_katetov}.(2)$ we get
${\mathcal{I}}_{\alpha}\leq_{K}\Phi(g)$ for all $\alpha<\kappa$. ∎

We also have the following result, which is true in fact for all families of
tall ideals which are downwards-closed with respect to $\leq_{K}$.

######  Corollary 5.11.

The order $(\mathcal{A}{\mathcal{N}},\leq_{K})$ contains decreasing chains of
cardinality $\mathfrak{c}^{+}$ and antichains of size $\mathfrak{c}$.

###### Proof.

By [17, Theorem 4.2] and Proposition 5.2. ∎

In Section $7$ we will show that such a $\leq_{K}$-antichain of size
$\mathfrak{c}$ in $\mathcal{A}{\mathcal{N}}$ can consist of summable ideals,
or of density ideals. Finally, we prove that the order
$(\mathcal{A}{\mathcal{N}},\leq_{K})$ does not have any maximal elements.

###### Proof.

Let ${\mathcal{I}}\in\mathcal{A}{\mathcal{N}}$. By Theorem
$\ref{thm_katetov}.(1)$ there is $f\in\omega^{\omega}$ satisfying
${\mathcal{I}}\leq_{K}\Phi(f)$. Moreover, by Theorem $\ref{thm_katetov}.(2)$
we may assume that $f$ is sufficiently big so that both $f(n)\geq n^{4}$ and
$f(n)\geq\sum_{i<n}f(i)$ hold for every $n\in\omega$. Define
$g\in\omega^{\omega}$ by setting $g(n)=n\cdot f(f(n))$ for each $n\in\omega$.
We will show that $\Phi(f)<_{K}\Phi(g)$, and so ${\mathcal{I}}<_{K}\Phi(g)$.

First, we have $g(n)\geq n\cdot\big{(}f(n)\big{)}^{4}$ for every $n\in\omega$,
by the assumption that $f(m)\geq m^{4}$ for $m=f(n)$. Next, for all $n\geq 2$
we have

$n\cdot\big{(}f(n)\big{)}^{4}\geq 2n^{2}\cdot f(n)\textrm{, as
}\big{(}f(n)\big{)}^{3}\geq 2n.$

Therefore, by Theorem $\ref{thm_katetov}.(2)$, we get
$\Phi(f)\leq_{K}\Phi(g)$.

Let $\langle\lambda_{n}\colon n\in\omega_{+}\rangle$ and $\langle\mu_{n}\colon
n\in\omega_{+}\rangle$ be the sequences of measures generating the ideals
$\Phi(g)$ and $\Phi(f)$, respectively, as described in the definition of the
operator $\Phi$. Let $A_{n}=\operatorname{supp}(\lambda_{n})$ and
$B_{n}=\operatorname{supp}(\mu_{n})$ for each $n\in\omega_{+}$, so that
$|A_{n}|=n^{2}\cdot f(f(n))$ and $|B_{n}|=n\cdot f(n)$. We have
$\lambda_{n}(x)=1/\big{(}n\cdot f(f(n))\big{)}$ for every $x\in A_{n}$ and
$\mu_{n}(x)=1/f(n)$ for every $x\in B_{n}$.

To show that $\Phi(g)\nleq_{K}\Phi(f)$, assume for the sake of contradiction
that there is a Katětov reduction $\varphi\colon\omega\to\omega$ witnessing
$\Phi(g)\leq_{K}\Phi(f)$. Since every density ideal is a P-ideal, by Lemma 5.1
we may additionally assume that $\varphi$ is finite-to-one. It is enough to
find such a subset $X\subseteq\omega$ that

$X\notin\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\mu_{n}\big{)}=\Phi(f)\textrm{
and
}\varphi[X]\in\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\lambda_{n}\big{)}=\Phi(g)$

(as $X\subseteq\varphi^{-1}\big{[}\varphi[X]\big{]}$, this will imply
$\varphi^{-1}\big{[}\varphi[X]\big{]}\notin\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\mu_{n}\big{)}$,
contradicting that $\varphi$ is a Katětov reduction). We proceed in two cases.

Case 1. There is a strictly increasing sequence $\langle n_{k}:\
k\in\omega\rangle$ in $\omega_{+}$ such that for each $k\in\omega$ there
exists $F_{k}\subseteq B_{n_{k}}$ satisfying $|F_{k}|=f(n_{k})$ and
$\varphi[F_{k}]\subseteq\bigcup\big{\\{}A_{i}\colon f(i)\geq n_{k}\big{\\}}$.

Let $X=\bigcup_{k\in\omega}F_{k}$. Since for every $k\in\omega$ we have
$\mu_{n_{k}}(X)=1$, it follows that
$X\notin\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\mu_{n}\big{)}$. Next,
for every $m\in\omega_{+}$ we have

$\varphi[X]\cap A_{m}=\bigcup_{k\in\omega}\varphi[F_{k}]\cap
A_{m}\subseteq\bigcup\big{\\{}\varphi[F_{k}]\colon n_{k}\leq f(m)\big{\\}},$

and so

$\big{|}\varphi[X]\cap
A_{m}\big{|}\leq\sum\Big{\\{}\big{|}\varphi[F_{k}]\big{|}\colon n_{k}\leq
f(m)\Big{\\}}\leq\sum_{i\leq f(m)}f(i)=f(f(m))+\sum_{i<f(m)}f(i)\leq 2\cdot
f(f(m)),$

where the last inequality follows from $f(f(m))\geq\sum_{i<f(m)}f(i)$.
Therefore, for every $m\in\omega_{+}$ we have:

$\lambda_{m}\big{(}\varphi[X]\big{)}\leq\frac{2\cdot f(f(m))}{m\cdot
f(f(m))}=\frac{2}{m},$

which shows that $\lim_{m\to\infty}\lambda_{m}\big{(}\varphi[X]\big{)}=0$, and
hence that
$\varphi[X]\in\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\lambda_{n}\big{)}$.

Case 2. If we are not in Case $1$, then there exists $N\in\omega_{+}$ such
that for each $n\geq N$ there is no subset $F_{n}\subseteq B_{n}$ satisfying
$|F_{n}|=f(n)$ (i.e. $\mu_{n}(F_{n})=1$) and
$\varphi[F_{n}]\subseteq\bigcup\big{\\{}A_{i}\colon f(i)\geq n\big{\\}}$.
Then, for every $n\geq N$ there must exist $F_{n}\subseteq B_{n}$ satisfying
$\mu_{n}(F_{n})=n-1$ and $\varphi[F_{n}]\subseteq\bigcup\big{\\{}A_{i}\colon
f(i)<n\big{\\}}$. Without losing generality we may assume that $N\geq 3$.

First, let us take any $n\geq N$. As $f(i)\geq i^{4}$ for every $i\in\omega$,
we have

$\big{|}\\{i\in\omega_{+}\colon f(i)<n\\}\big{|}<\sqrt[4]{n}\leq\sqrt{n-1}.$

Thus, from $F_{n}\subseteq\bigcup\big{\\{}\varphi^{-1}[A_{i}]\colon
f(i)<n\big{\\}}$ it follows that there exists such a number
$i(n)\in\omega_{+}$ that $f(i(n))<n$ and

$\mu_{n}\big{(}F_{n}\cap\varphi^{-1}[A_{i(n)}]\big{)}\geq(n-1)\big{/}\sqrt{n-1}=\sqrt{n-1}.$

By $f(i(n))<n$ and $f(i(n))\geq i(n)^{4}$, it follows that $n\geq i(n)^{4}+1$
and so $\sqrt{n-1}\geq i(n)^{2}$. For each $n\geq N$ let us define

$E_{n}=F_{n}\cap\varphi^{-1}[A_{i(n)}].$

Then, $E_{n}\subseteq B_{n}$, $\mu_{n}(E_{n})=\sqrt{n-1}$, and
$\varphi[E_{n}]\subseteq A_{i(n)}$.

For each $n\in\omega_{+}$ let $A_{n}=\bigcup_{j=1}^{n^{2}}C_{j}^{n}$ be a
division of $A_{n}$ into $n^{2}$ disjoint parts of size $f(f(n))$, i.e.
$\lambda_{n}(C_{j}^{n})=1/n$ for every $j=1,\ldots,n^{2}$ and $C_{j}^{n}\cap
C_{i}^{n}=\emptyset$ for $i\neq j$. As we have

$E_{n}\subseteq\bigcup_{j=1}^{n^{2}}\varphi^{-1}\Big{[}C_{j}^{i(n)}\Big{]},$

for every $n\geq N$ there exists $j(n)\in\\{1,\ldots,n^{2}\\}$ such that

$\mu_{n}\Big{(}E_{n}\cap\varphi^{-1}\Big{[}C_{j(n)}^{i(n)}\Big{]}\Big{)}\geq\sqrt{n-1}\big{/}i(n)^{2}\geq
1,$

where the last inequality follows from $n\geq i(n)^{4}+1$. For each $n\geq N$
let us denote

$D_{n}=E_{n}\cap\varphi^{-1}\Big{[}C_{j(n)}^{i(n)}\Big{]}.$

We have $D_{n}\subseteq B_{n}$, $\mu_{n}(D_{n})\geq 1$, and
$\varphi[D_{n}]\subseteq C_{j(n)}^{i(n)}\subseteq A_{i(n)}$.

As $\varphi$ is finite-to-one, there is an increasing sequence $\langle
n_{k}:\ k\in\omega\rangle$ of natural numbers such that $i(n_{k})<i(n_{j})$
for every $k<j$. Let

$X=\bigcup_{k\in\omega}D_{n_{k}}.$

Since for every $k\in\omega$ we have $\mu_{n_{k}}(X)\geq 1$, it follows that
$X\notin\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\mu_{n}\big{)}$.
Finally, we have

$\varphi[X]\subseteq\bigcup_{k\in\omega}C_{j(n_{k})}^{i(n_{k})},$

where the union is disjoint, and so $\lambda_{i(n_{k})}(\varphi[X])\leq
1/i(n_{k})$ for every $k\in\omega$ (and $\lambda_{m}(\varphi[X])=0$ for other
$m\in\omega$). Thus, we get
$\lim_{n\to\infty}\lambda_{n}\big{(}\varphi[X]\big{)}=0$, due to the fact that
$i(n_{k})\to\infty$. Therefore, we have
$\varphi[X]\in\operatorname{Exh}\big{(}\sup_{n\in\omega_{+}}\lambda_{n}\big{)}$.

∎

######  Corollary 5.12.

For every family
$\\{{\mathcal{I}}_{\alpha}\colon\alpha<\kappa\\}\subseteq\mathcal{A}{\mathcal{N}}$,
where $\kappa<\mathfrak{b}$, there is
${\mathcal{J}}\in\mathcal{A}{\mathcal{N}}$ such that
${\mathcal{I}}_{\alpha}<_{K}{\mathcal{J}}$ for all $\alpha<\kappa$.

###### Proof.

If
$\\{{\mathcal{I}}_{\alpha}\colon\alpha<\kappa\\}\subseteq\mathcal{A}{\mathcal{N}}$
and $\kappa<\mathfrak{b}$, then, by Theorem 1.3, for each $\alpha<\kappa$
there is ${\mathcal{J}}_{\alpha}\in\mathcal{A}{\mathcal{N}}$ such that
${\mathcal{I}}_{\alpha}<_{K}{\mathcal{J}}_{\alpha}$. By Corollary 5.10, there
is ${\mathcal{J}}\in\mathcal{A}{\mathcal{N}}$ satisfying
${\mathcal{J}}_{\alpha}\leq_{K}{\mathcal{J}}$ for every $\alpha<\kappa$, thus
we have ${\mathcal{I}}_{\alpha}<_{K}{\mathcal{J}}$ for all $\alpha<\kappa$. ∎

##  6\. $\mathcal{A}{\mathcal{N}}$ vs
${\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}$

We will now summarize differences between the classes of ideals
$\mathcal{A}{\mathcal{N}}$ and
${\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}$. As we said in
Introduction, these are connected with differences between the Nikodym
property and the Grothendieck property of Boolean algebras. We start with the
following result, stating that $\mathcal{A}{\mathcal{N}}$ is a subclass of
${\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}$. Note that it actually
follows from Theorems 1.3 and 6.2, but for the convenience of the reader and
the self-containment of the paper we provide here a direct (and more analytic)
argument.

######  Proposition 6.1.

If a filter $F$ does not have the Nikodym property, then it has the BJNP.
Consequently,
$\mathcal{A}{\mathcal{N}}\subseteq{\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}$.

###### Proof.

Let $\langle\mu_{n}\colon n\in\omega\rangle$ be an anti-Nikodym sequence of
measures on $N_{F}$. As $\|\mu_{n}\|\rightarrow\infty$, by taking a suitable
subsequence we may assume that $\|\mu_{n}\|\geq 1$ for each $n\in\omega$. We
define a sequence $\langle\lambda_{n}\colon n\in\omega\rangle$ of measures by
setting $\lambda_{n}=\mu_{n}/\|\mu_{n}\|$ for each $n\in\omega$. We have
$\|\lambda_{n}\|=1$ for all $n\in\omega$ and
$|\lambda_{n}(A)|=|\mu_{n}(A)|/\|\mu_{n}\|\leq|\mu_{n}(A)|/1\rightarrow 0$ for
every $A\in\operatorname{Clopen}(N_{F})$.

By [22, Lemma 3.8.(2)], the Čech–Stone compactification $\beta N_{F}$ is the
Stone space of some Boolean algebra, thus in particular it is zero-
dimensional. For every $n\in\omega$ we define a measure
$\overline{\lambda_{n}}$ on $\beta N_{F}$ by setting
$\overline{\lambda_{n}}(A)=\lambda_{n}(A\cap N_{F})$ for each $A\in Bor(\beta
N_{F})$. By the definition we have $\big{\|}\overline{\lambda_{n}}\big{\|}=1$
for each $n\in\omega$ and $\overline{\lambda_{n}}(A)\rightarrow 0$ for every
$A\in\operatorname{Clopen}(\beta N_{F})$. We will show that, moreover, for
every $f\in C(\beta N_{F})$ we have$\int_{\beta
N_{F}}fd\overline{\lambda_{n}}\to 0$, i.e.
$\langle\overline{\lambda_{n}}\colon n\in\omega\rangle$ is a BJN-sequence of
measures on $\beta N_{F}$.

Let us take arbitrary $f\in C(\beta N_{F})$ and $\varepsilon>0$. By the zero-
dimensionality of $\beta N_{F}$ and the Stone–Weierstrass theorem, the linear
span of the set $\big{\\{}\chi_{A}\colon A\in\operatorname{Clopen}(\beta
N_{F})\big{\\}}$ is norm dense in $C(\beta N_{F})$ with the supremum norm.
Therefore, there exist $m\in\omega$, real numbers
$\alpha_{1},\ldots,\alpha_{m}$, and clopen subsets $A_{1},\ldots,A_{m}$ of
$\beta N_{F}$ such that
$\|f-\sum_{i=1}^{m}\alpha_{i}\chi_{A_{i}}\|_{\infty}<\varepsilon/2$. By the
convergence of the sequence $\langle\overline{\lambda_{n}}\colon
n\in\omega\rangle$ on clopen sets, there is $N\in\omega$ such that for every
$n\geq N$ we have:

$\sum_{i=1}^{m}|\alpha_{i}|\cdot|\overline{\lambda_{n}}(A_{i})|<\varepsilon/2.$

Then, for every $n\geq N$ it holds:

$\Big{|}\int_{\beta N_{F}}fd\overline{\lambda_{n}}\Big{|}\leq\int_{\beta
N_{F}}\Big{|}f-\sum_{i=1}^{m}\alpha_{i}\chi_{A_{i}}\Big{|}d\overline{\lambda_{n}}+\int_{\beta
N_{F}}\Big{|}\sum_{i=1}^{m}\alpha_{i}\chi_{A_{i}}\Big{|}d\overline{\lambda_{n}}\leq$
$\leq\Big{\|}f-\sum_{i=1}^{m}\alpha_{i}\chi_{A_{i}}\Big{\|}_{\infty}\cdot\|\overline{\lambda_{n}}\|+\sum_{i=1}^{m}|\alpha_{i}|\cdot|\overline{\lambda_{n}}(A_{i})|<\varepsilon/2+\varepsilon/2=\varepsilon,$

which proves that $\langle\overline{\lambda_{n}}\colon n\in\omega\rangle$ is a
BJN-sequence of measures on $\beta N_{F}$.

To show that $\langle\lambda_{n}\colon n\in\omega\rangle$ is a BJN-sequence of
measures on $N_{F}$, let $f\in C_{b}(N_{F})$. By the universal property of the
Čech–Stone compactification, there is a continuous extension $\overline{f}\in
C(\beta N_{F})$ of $f$. We proved that $\int_{\beta
N_{F}}\overline{f}d\overline{\lambda_{n}}\to 0$, and by the definition of
$\overline{\lambda_{n}}$’s we have $\int_{\beta
N_{F}}\overline{f}d\overline{\lambda_{n}}=\int_{N_{F}}fd\lambda_{n}$ for every
$n\in\omega$, thus $\int_{N_{F}}fd\lambda_{n}\to 0$. ∎

Let us recall that our characterization of the Nikodym property via exhaustive
ideals, theorem 1.3, was inspired by the following characterization of the
BJNP.

######  Theorem 6.2 (Marciszewski and Sobota, [22]).

Let $F$ be a free filter on $\omega$. Then, the following are equivalent:

  1. (1)

$N_{F}$ has the BJNP,

  2. (2)

there is a density submeasure $\varphi$ such that
$F\subseteq\operatorname{Exh}(\varphi)^{*}$,

  3. (3)

there is a non-pathological lsc submeasure $\varphi$ such that
$F\subseteq\operatorname{Exh}(\varphi)^{*}$,

  4. (4)

$F\leq_{K}{\mathcal{Z}}^{*}.$

The only difference between conditions $(2)$ and $(3)$ of this theorem and the
corresponding conditions of Theorem 1.3 is the requirement in Theorem 1.3 that
the submeasure $\varphi$ is not bounded, i.e. $\varphi(\omega)=\infty$.
However, this apparently small difference changes significantly the properties
of families of ideals, for example the cofinal structure of
${\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}$ is very simple, as
$({\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}},\leq_{K})$ has a
maximal element by Theorem 6.2, namely ${\mathcal{Z}}$.

By Theorem 6.2 and Corollary 4.4 we get:

######  Corollary 6.3.

If $\varphi$ is a non-pathological lsc submeasure on $\omega$ and the ideal
${\mathcal{I}}=\operatorname{Exh}(\varphi)$ is totally bounded, then
${\mathcal{I}}\in{\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}\mathbin{\backslash}\mathcal{A}{\mathcal{N}}$.

In particular, all the density ideals satisfying any of the conditions from
Theorem 1.3 are elements of
${\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}\mathbin{\backslash}\mathcal{A}{\mathcal{N}}$.

######  Corollary 6.4.

${\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}\mathbin{\backslash}\mathcal{A}{\mathcal{N}}$
contains $\mathfrak{c}$ many Borel pairwise non-isomorphic ideals on $\omega$.

###### Proof.

By [20, Propositions 5 and 6], there is a family of pairwise non-isomorphic
Erdős–Ulam ideals of size $\mathfrak{c}$. Every Erdős–Ulam as a density ideal
is Borel. The result follows then from Corollary 5.4. ∎

Another example of an element of
${\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}\mathbin{\backslash}\mathcal{A}{\mathcal{N}}$
is the ideal $\operatorname{tr}({\mathcal{N}})$ (see e.g. [16, page 580]),
which is of the form $\operatorname{Exh}(\varphi)$ for some non-pathological
submeasure $\varphi$, and is totally bounded by [16, Proposition 3.18]. The
statement of Corollary 5.5 does not hold for this ideal, as
$\operatorname{tr}({\mathcal{N}})<_{K}{\mathcal{Z}}$. It follows from the
following facts:

  * •

by [16, Proposition 3.6] we have
$\operatorname{tr}({\mathcal{N}})\leq_{K}{\mathcal{Z}}$;

  * •

there cannot hold $\operatorname{tr}({\mathcal{N}})\equiv_{K}{\mathcal{Z}}$,
as it would imply that
$\operatorname{cov}^{*}\big{(}\operatorname{tr}({\mathcal{N}})\big{)}=\operatorname{cov}^{*}({\mathcal{Z}})$
by [16, Proposition 3.1] (see [16, Section 1] for the definition of
$\operatorname{cov}^{*}$), but by [16, Theorem 3.15] this equality does not
hold in the random model, and the Katětov order between Borel ideals is
absolute (see e.g. [18, Section 1.2]).

##  7\. Examples

In this section we present some examples of classes of ideals on $\omega$
which are connected to the class $\mathcal{A}{\mathcal{N}}$.

###  7.1. Summable ideals and large families of non-isomorphic filters

Given $f\colon\omega\to[0,\infty)$ such that $\sum_{n\in\omega}f(n)=\infty$,
the summable ideal corresponding to $f$ is the ideal

${\mathcal{I}}_{f}=\Big{\\{}A\subseteq\omega\colon\sum_{n\in
A}f(n)<\infty\Big{\\}}.$

The classical example is the ideal ${\mathcal{I}}_{1/n}$ corresponding to the
function $f(n)=1/n$. It is easy to see that
${\mathcal{I}}_{f}=\operatorname{Exh}(\mu_{f})$, where $\mu_{f}$ is a non-
negative measure on $\omega$ defined for every $A\subseteq\omega$ by

$\mu_{f}(A)=\sum_{n\in A}f(n).$

Since by the definition we have $\mu_{f}(\omega)=\infty$, and $\mu_{f}$ is
obviously a non-pathological lsc submeasure, by Theorem 1.3 we get the
following corollary

######  Corollary 7.1.

The dual filter ${\mathcal{I}}_{f}^{*}$ of any summable ideal
${\mathcal{I}}_{f}$ does not have the Nikodym property.

Let us denote by $\Sigma$ the family of all summable ideals. It appears that
$(\Sigma,\leq_{K})$—and so, by Corollary 7.1,
$(\mathcal{A}{\mathcal{N}},\leq_{K})$—has a rich structure. Recall that
${\mathcal{P}}(\omega)/Fin$ is the quotient algebra of subsets of $\omega$
modulo finite sets, endowed with the ordering induced by $\subseteq^{*}$. It
is well-known that $({\mathcal{P}}(\omega)/Fin,\subseteq^{*})$ contains
increasing chains of size $\mathfrak{b}$ and antichains of size
$\mathfrak{c}$. The following result connects
$({\mathcal{P}}(\omega)/Fin,\subseteq^{*})$ and $(\Sigma,\leq_{K})$.

######  Theorem 7.2 (Guzmán-González and Meza-Alcántara [14]).

There is an order embedding of
$({\mathcal{P}}(\omega)/Fin,\subseteq^{*})$ into $(\Sigma,\leq_{K})$.

######  Corollary 7.3.

There is an order embedding of $({\mathcal{P}}(\omega)/Fin,\subseteq^{*})$
into $(\mathcal{A}{\mathcal{N}},\leq_{K})$. In particular,
$(\mathcal{A}{\mathcal{N}},\leq_{K})$ contains increasing chains of size
$\mathfrak{b}$ and antichains of size $\mathfrak{c}$, consisting of summable
ideals.

Recall that in Theorem 1.3 and Corollary 5.11 we have shown that
$(\mathcal{A}{\mathcal{N}},\leq_{K})$ contains dominating families of size
$\mathfrak{d}$ and decreasing chains of size $\mathfrak{c}^{+}$.

We now present a sketch of the construction of large families of non-
isomorphic filters with and without the Nikodym property (the ideas follow
closely [22, Section 6.1]).

######  Corollary 7.4.

There exist families ${\mathcal{F}}_{1}$ and ${\mathcal{F}}_{2}$ of size
$2^{\mathfrak{c}}$, each consisting of pairwise non-isomorphic free filters on
$\omega$ whose dual ideals are tall, such that:

  1. (1)

every $F\in{\mathcal{F}}_{1}$ does not have the Nikodym property,

  2. (2)

every $F\in{\mathcal{F}}_{2}$ has the Nikodym property.

###### Proof.

Let ${\mathcal{G}}_{2}$ be the family of all ultrafilters on $\omega$ and let
$F={\mathcal{I}}_{1/n}^{*}$. We put:

${\mathcal{G}}_{1}=\\{G\oplus F\colon G\in{\mathcal{G}}_{2}\\},$

where $G\oplus F$ denotes the disjoint sum of $F$ and $G$. By Corollary 4.2
all members of ${\mathcal{G}}_{2}$ do have the Nikodym property. Since $F$
does not have the Nikodym property by Corollary 7.1, and $G\oplus F\leq_{K}F$
for every $G\in{\mathcal{G}}_{2}$ (see [18, Section 1.2]), by Proposition 5.2
all members of ${\mathcal{G}}_{1}$ do not have the Nikodym property. Moreover,
$\big{(}G\oplus F\big{)}^{*}=G^{*}\oplus F^{*}$ is tall for each
$G\in{\mathcal{G}}_{2}$, as $G^{*}$ and $F^{*}$ are tall. As
$|{\mathcal{G}}_{2}|=|{\mathcal{G}}_{1}|=2^{\mathfrak{c}}$, and each family of
pairwise isomorphic filters has cardinality $\leq\mathfrak{c}$, for each
$i=1,2$ we can select a subfamily
${\mathcal{F}}_{i}\subseteq{\mathcal{G}}_{i}$ consisting of $2^{\mathfrak{c}}$
many pairwise non-isomorphic filters. ∎

###  7.2. Simple density ideals

The class of simple density ideals was introduced in [1] (cf. also [21]). For
every function $g\colon\omega\to[0,\infty)$ satisfying $g(n)\to\infty$ and
$n/g(n)\nrightarrow 0$ we define the simple density ideal corresponding to $g$
as:

${\mathcal{Z}}_{g}=\Big{\\{}A\subseteq\omega\colon\limsup_{n\to\infty}\frac{|A\cap
n|}{g(n)}=0\Big{\\}}.$

It is proved in [1, Theorem 3.2] that every ${\mathcal{Z}}_{g}$ is a density
ideal. [21, Proposition 10] presents a sufficient condition for a density
ideal generated by a sequence of measures $\langle\mu_{n}\colon
n\in\omega\rangle$ to be a simple density ideal. Since the statement of this
result is rather long and technical, we omit it here. The ideal $\Phi(f)$
satisfies conditions $(i),(ii)$ and $(v)$ from [21, Proposition 10] for any
$f\in\omega^{\omega}$. Conditions $(iv)$ and $(vi)$ are also satisfied for
$\Phi(f)$ when $f$ is non-decreasing and $f(n)\to\infty$, as using the
notation from that proposition we have $a_{n}=1/f(n)$. By the definition of
$\Phi(f)$ we have also
$\min\big{(}\operatorname{supp}(\mu_{n}^{f})\big{)}=\sum_{i<n}f(i)$ for every
$n\in\omega$, where $\langle\mu_{n}^{f}\colon n\in\omega\rangle$ is the
sequence of measures generating $\Phi(f)$. Thus, for condition $(iii)$ to be
satisfied, we need that

$\frac{1}{f(n)}\cdot\sum_{i<n}f(i)\to 0.$

Therefore, by [21, Proposition 10] we have the following corollary.

######  Corollary 7.5.

Let $f\in\omega^{\omega}$ be non-decreasing and satisfy the condition
$f(n)\big{/}\sum_{i<n}f(i)\to\infty$. Then, the ideal $\Phi(f)$ is equal to
the simple density ideal ${\mathcal{Z}}_{g}$, where
$g\colon\omega\to[0,\infty)$ is defined by $g(k)=f(n)$ for $k\in\omega$ when
$k\in\operatorname{supp}(\mu_{n})$.

It is proved in [21, Theorem 3] that there is a family
$\\{{\mathcal{Z}}_{g_{\alpha}}\colon\alpha<\mathfrak{c}\\}$ such that
${\mathcal{Z}}_{g_{\alpha}}$ is not $\leq_{K}$-comparable to
${\mathcal{Z}}_{g_{\beta}}$ for any $\alpha\neq\beta$. By Theorem 1.3, for
every density ideal ${\mathcal{I}}$ such that
${\mathcal{I}}\notin\mathcal{A}{\mathcal{N}}$ we have
${\mathcal{I}}\equiv_{K}{\mathcal{Z}}$, and so the $\leq_{K}$-antichain must
lie inside the $\mathcal{A}{\mathcal{N}}$ family:

######  Corollary 7.6.

$(\mathcal{A}{\mathcal{N}},\leq_{K})$ contains antichains of size
$\mathfrak{c}$ consisting of density ideals.

##  8\. The Nikodym property of Boolean algebras

This section is devoted to an analysis of the connection between embedding of
a space $N_{F}$ without the finitely supported Nikodym property into the Stone
space of a given Boolean algebra $\mathcal{A}$ and the lack of the Nikodym
property of $\mathcal{A}$.

We first need to introduce some notation (cf. [22, Section 7]). Let $X$ be a
compact Hausdorff space, $Y$ an infinite countable subset of $X$, and
$x\in\overline{Y}\mathbin{\backslash}Y$. Let $Z_{X}(x,Y)$ be the space
$Y\cup\\{x\\}$ endowed with the subspace topology. By $\mathfrak{N}_{X}(x)$ we
denote the collection of all $U\subseteq X$ such that
$x\in\operatorname{int}U$. We then put:

$\mathfrak{J}_{X}(x,Y)=\big{\\{}U\cap Y\colon
U\in\mathfrak{N}_{X}(x)\big{\\}}.$

Note that since $x\in\overline{Y}\mathbin{\backslash}Y$,
$\mathfrak{J}_{X}(x,Y)$ is a filter on the infinite countable set $Y$
containing all its co-finite subsets. If $f\colon\omega\to Y$ is a bijection,
then $F=\big{\\{}f^{-1}[V]\colon V\in\mathfrak{J}_{X}(x,Y)\big{\\}}$ is a free
filter on $\omega$—we will say in this case that $F$ is f-associated to
$\mathfrak{J}_{X}(x,Y)$. We also have a bijective continuous function
$\varphi_{f}\colon N_{F}\to Z_{X}(x,Y)$ associated with $f$, which is defined
by $\varphi_{f}\restriction\omega=f$ and $\varphi_{f}\big{(}p_{F}\big{)}=x$.

######  Proposition 8.1.

Suppose $\mathcal{A}$ is a Boolean algebra. Let $Y$ be a countable subset of
$St(\mathcal{A})$ and $x\in\overline{Y}\mathbin{\backslash}Y$. Let
$f\colon\omega\to Y$ be a bijection and $F$ the filter on $\omega$
$f$-associated to $\mathfrak{J}_{St(\mathcal{A})}(x,Y)$.

Then, if $F$ does not have the Nikodym property, then the algebra
$\mathcal{A}$ does not have the Nikodym property.

###### Proof.

Let $\langle\nu_{n}\colon n\in\omega\rangle$ be an anti-Nikodym sequence of
measures on $N_{F}$. For each $n\in\omega$ we define a measure $\mu_{n}$ on
$St(\mathcal{A})$ by setting

$\mu_{n}(B)=\nu_{n}\big{(}\varphi_{f}^{-1}[B]\big{)}$

for every $B\in Bor\big{(}St(\mathcal{A})\big{)}$. Then, we get that
$\|\mu_{n}\|=\|\nu_{n}\|\to\infty$ and that for each $n\in\omega$ the support
$\operatorname{supp}(\mu_{n})$ is finite and contained in
$Z_{St(\mathcal{A})}(x,Y)$.

Since $\varphi_{f}$ is continuous, for any
$A\in\operatorname{Clopen}\big{(}St(\mathcal{A})\big{)}$ we have
$\varphi_{f}^{-1}[A]\in\operatorname{Clopen}(N_{F})$, hence by the assumption
that $\langle\nu_{n}\colon n\in\omega\rangle$ is anti-Nikodym we get:

$\mu_{n}(A)=\nu_{n}\big{(}\varphi_{f}^{-1}[A]\big{)}\to 0\textrm{ as
}n\to\infty.$

For each $n\in\omega$ we define a measure $\lambda_{n}$ on the algebra
$\mathcal{A}$ by the formula
$\lambda_{n}(A)=\mu_{n}\big{(}[A]_{\mathcal{A}}\big{)}$ for every
$A\in\mathcal{A}$. By the properties of the sequence $\langle\mu_{n}\colon
n\in\omega\rangle$ we have $\|\lambda_{n}\|\to\infty$ and $\lambda_{n}(A)\to
0$ for every $A\in\mathcal{A}$. Therefore, $\langle\lambda_{n}\colon
n\in\omega\rangle$ is a pointwise bounded sequence of measures on
$\mathcal{A}$ which is not uniformly bounded, and hence $\mathcal{A}$ does not
have the Nikodym property. ∎

######  Theorem 8.2.

Suppose $\mathcal{A}$ is a Boolean algebra and $G$ is a filter on $\omega$.
Let $\varphi\colon N_{G}\to St(\mathcal{A})$ be a continuous function such
that $\varphi^{-1}\big{(}\varphi(p_{G})\big{)}=\\{p_{G}\\}$ (e.g. $\varphi$ is
an injection).

Then, if $G$ does not have the Nikodym property, then $\mathcal{A}$ does not
have the Nikodym property.

###### Proof.

Let us define $Y=\varphi[\omega]$ and $x=\varphi(p_{G})$. It follows that $Y$
is an infinite countable subset of $St(\mathcal{A})$, as if $Y$ was finite, in
particular closed in $St(\mathcal{A})$, then it would imply that
$\varphi^{-1}[Y]=\omega$ is closed in $N_{F}$. We also have
$x\in\overline{Y}\mathbin{\backslash}Y$. Let $f\colon\omega\to Y$ be a
bijection and let $F$ be the filter on $\omega$ $f$-associated to
$\mathfrak{J}_{X}(x,Y)$. We define $g\colon\omega\to\omega$ by the formula
$g(n)=f^{-1}\big{(}\varphi(n))$ for every $n\in\omega$, i.e.
$g=f^{-1}\circ(\varphi\restriction\omega)$.

We claim that the function $g$ is a witness for $F\leq_{K}G$. Let $A\in F$.
Since $F$ is $f$-associated to $\mathfrak{J}_{St(\mathcal{A})}(x,Y)$, we have
$f[A]\in\mathfrak{J}_{St(\mathcal{A})}(x,Y)$, and thus $f[A]\cup\\{x\\}$
contains an open neighbourhood of $x$ in $Z_{St(\mathcal{A})}(x,Y)$. Since
$\varphi$ is continuous and $\varphi^{-1}(\\{x\\})=\\{p_{G}\\}$, the set
$\varphi^{-1}\big{[}f[A]\cup\\{x\\}\big{]}=\varphi^{-1}\big{[}f[A]\big{]}\cup\\{p_{G}\\}$
contains an open neighbourhood of $p_{G}$ in $N_{G}$, and so
$g^{-1}[A]=\varphi^{-1}\big{[}f[A]\big{]}\in G$.

Therefore, $F\leq_{K}G$. By Proposition, 5.2 the filter $F$ does not have the
Nikodym property, thus by Proposition 8.1 the algebra $\mathcal{A}$ does not
have the Nikodym property. ∎

######  Corollary 8.3.

Let $\mathcal{A}$ be a Boolean algebra and let $G$ be a filter on $\omega$
that does not have the Nikodym property. If $N_{G}$ homeomorphically embeds
into $St(\mathcal{A})$, then $\mathcal{A}$ does not have the Nikodym property.

The following corollary follows from Theorem 1.3 and Corollary 7.1.

######  Corollary 8.4.

Let $\mathcal{A}$ be a Boolean algebra and let ${\mathcal{I}}$ be a summable
ideal on $\omega$, or a density ideal which is not totally bounded. If
$N_{{\mathcal{I}}^{*}}$ homeomorphically embeds into $St(\mathcal{A})$, then
$\mathcal{A}$ does not have the Nikodym property.

######  Remark 8.5.

Let us note that embedding a suitable space $N_{F}$ into the Stone space is
not the only reason for a Boolean algebra to lack the Nikodym property. There
exists an algebra ${\mathcal{S}}$ which does not have the Nikodym property and
such that $St({\mathcal{S}})$ does not contain any homeomorphic copy of a
space $N_{F}$ for which
$F\in{\mathcal{B}}{\mathcal{J}}{\mathcal{N}}{\mathcal{P}}$ (see [24, Example
4.10] and [22, Example 7.8]).

The family of filters ${\mathcal{F}}_{1}$ constructed in Corollary 7.4 gives
us a rich family of pairwise non-homeomorphic spaces $N_{F}$ which forbid the
Nikodym property of Boolean algebras. By [22, Propositions 3.10.(2) and 4.3],
for each $F\in{\mathcal{F}}_{1}$ the space $N_{F}$ contains no non-trivial
convergent sequence. Therefore, by Corollary 8.3, we get the following result.

##  9\. Final remarks and questions

###  9.1. Non-totally bounded non-pathological ideals

By Corollary 4.4, if ${\mathcal{I}}=\operatorname{Exh}(\varphi)$ is a totally
bounded ideal for some lsc submeasure $\varphi$, then ${\mathcal{I}}^{*}$ has
the Nikodym property. By Remark 4.5, the converse does not hold, but the
counterexamples are constructed only with pathological submeasures. Theorem
1.3 shows that the opposite implication is true for all density submeasures.
We do not know if this implication holds for all non-pathological submeasures.

######  Question 9.1.

Does there exist a non-pathological lsc submeasure $\varphi$ such that the
ideal $\operatorname{Exh}(\varphi)$ is not totally bounded and
$\operatorname{Exh}(\varphi)^{*}$ has the Nikodym property?

###  9.2. Tukey reductions

The analysis of the cofinal structure of the order
$(\mathcal{A}{\mathcal{N}},\leq_{K})$ shows that it is, in some way, similarly
complicated as the cofinal structure of $(\omega^{\omega},\leq^{*})$. The
standard way of comparing cofinal structures of two orders is through Tukey
reductions, see e.g. [10] and [29].

Let $(P,\leq_{P})$ and $(Q,\leq_{Q})$ be partial orders. A function $f\colon
P\to Q$ is said to be a Tukey reduction if for every $q_{0}\in Q$ there exists
$p_{0}\in P$ such that, for every $p\in P$, the following implication holds:

$f(p)\leq_{Q}q_{0}\>\Longrightarrow\>p\leq_{P}p_{0},$

i.e. if the preimages by $f$ of bounded subsets of $Q$ are bounded subsets of
$P$. We write $P\preceq_{T}Q$ if there exists a Tukey reduction from $P$ to
$Q$. We say that $P$ and $Q$ are Tukey equivalent, which we denote by
$P\equiv_{T}Q$, if $P\preceq_{T}Q$ and $Q\preceq_{T}P$.

An example of using Tukey reductions for classes of ideals ordered by
$\leq_{K}$ is the following result about the class of all $F_{\sigma}$ ideals
on $\omega$.

######  Theorem 9.2 (Minami and Sakai [23]).

$(F_{\sigma}\textit{-ideals},\leq_{K})\equiv_{T}(F_{\sigma}\textit{-ideals},\leq_{KB})\equiv_{T}(\omega^{\omega},\leq^{*}).$

Using Theorem 5.6 it is not difficult to prove the following reduction.

######  Proposition 9.3.

$(\mathcal{A}{\mathcal{N}},\leq_{K})\preceq_{T}(\omega^{\omega},\leq^{*}).$

###### Proof.

We define a Tukey reduction
$\Psi\colon\mathcal{A}{\mathcal{N}}\to\omega^{\omega}$ in the following way.
Let ${\mathcal{I}}\in\mathcal{A}{\mathcal{N}}$. By Theorem
$\ref{thm_katetov}.(1)$ there exists $f\in\omega^{\omega}$ such that
${\mathcal{I}}\leq_{K}\Phi(f)$. Let us fix any such $f\in\omega^{\omega}$, and
put $\Psi({\mathcal{I}})=g$, where $g(n)=2n^{2}\cdot f(n)$ for every
$n\in\omega$.

To prove that $\Psi$ is a Tukey reduction, let us take any
$h\in\omega^{\omega}$. We need to show that the following implication holds
for every ${\mathcal{I}}\in\mathcal{A}{\mathcal{N}}$:

$\Psi({\mathcal{I}})\leq^{*}h\>\Longrightarrow\>{\mathcal{I}}\leq_{K}\Phi(h).$

By the definition of $\Psi$, if $\Psi({\mathcal{I}})\leq^{*}h$ then there
exists $f\in\omega^{\omega}$ such that ${\mathcal{I}}\leq_{K}\Phi(f)$ and
$2n^{2}\cdot f(n)\leq h(n)$ holds for all but finitely many $n\in\omega$. By
Theorem $\ref{thm_katetov}.(2)$ we get $\Phi(f)\leq_{K}\Phi(h)$, and so
${\mathcal{I}}\leq_{K}\Phi(h)$. ∎

Note that Corollaries 1.3 and 5.10 both follow directly from Proposition 9.3
by [10, Theorem 1J]. We do not know if reverse Tukey reduction exist.

######  Question 9.4.

Does it hold
$(\omega^{\omega},\leq^{*})\preceq_{T}(\mathcal{A}{\mathcal{N}},\leq_{K})$?

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