# Epic math battle of history: Grothendieck vs Nikodym

Damian Głodkowski  Institute of Mathematics of the Polish Academy of Sciences,
ul. Śniadeckich 8, 00-656 Warszawa, Poland  Faculty of Mathematics,
Informatics, and Mechanics, University of Warsaw, ul. Banacha 2, 02-097
Warszawa, Poland<EMAIL_ADDRESS>and  Agnieszka Widz  Institute of
Mathematics, Łódź University of Technology, Aleje Politechniki 8, 93-590 Łódź,
Poland<EMAIL_ADDRESS>
###### Abstract.

We define a $\sigma$-centered notion of forcing that forces the existence of a
Boolean algebra with the Grothendieck property and without the Nikodym
property. In particular, the existence of such an algebra is consistent with
the negation of the continuum hypothesis. The algebra we construct consists of
Borel subsets of the Cantor set and has cardinality $\omega_{1}$. We also show
how to apply our method to streamline Talagrand’s construction of such an
algebra under the continuum hypothesis.

###### Key words and phrases:

Grothendieck property, Nikodym property, Boolean algebras, balanced sets,
consistency result, forcing, sequences of measures, convergence of measures

###### 2020 Mathematics Subject Classification:

03E35, 03E75, 06E15, 28A33, 46E15, 46E27

The research of the first named author was partially supported by the NCN
(National Science Centre, Poland) research grant no. 2021/41/N/ST1/03682.

The research of the second named author was supported by the NCN (National
Science Centre, Poland), under the Weave-UNISONO call in the Weave programme
2021/03/Y/ST1/00124.

For the purpose of Open Access, the authors have applied a CC-BY public
copyright licence to any Author Accepted Manuscript (AAM) version arising from
this submission.

##  1\. Introduction

In 1953, Grothendieck [18, Section 4] proved that the space $l_{\infty}$ of
bounded sequences has the following property:

All weak*-convergent sequences in the dual space $l_{\infty}^{*}$ are also
weakly convergent.

The above theorem motivated the following definition.

######  Definition 1.1.

A Banach space $X$ has the Grothendieck property if all weak*- convergent
sequences in the dual space $X^{*}$ are also weakly convergent.

Research on the Grothendieck property has long history and is still ongoing
[4, 6, 10, 15, 16, 17, 19, 25, 28, 45]. If $X$ is of the form $C(K)$ for a
compact space $K$, then $X$ has the Grothendieck property if and only if each
weak*-convergent sequence of Radon measures on $K$ is also weakly convergent.
Recall that $l_{\infty}$ is isometric to the Banach space $C(\beta\mathbb{N})$
of continuous functions on the Čech-Stone compactification of the natural
numbers. Moreover, $\beta\mathbb{N}$ is the Stone space of the Boolean algebra
$\mathcal{P}(\mathbb{N})$.

Schachermayer, inspired by the Grothendieck’s result, introduced the notion of
the Grothendieck property for Boolean algebras [36, Definition 2.3].

######  Definition 1.2.

A Boolean algebra $\mathbb{A}$ has the Grothendieck property, if the Banach
space $C(\operatorname{St}(\mathbb{A}))$ of continuous functions on the Stone
space of $\mathbb{A}$ has the Grothendieck property.

Analogously, motivated by Nikodym’s article [32], Schachermayer defined the
Nikodym property for Boolean algebras [36, Definition 2.4].

######  Definition 1.3.

A Boolean algebra $\mathbb{A}$ has the Nikodym property, if every sequence
$(\mu_{n})$ of bounded finitely additive signed measures on $\mathbb{A}$,
which is pointwise convergent to zero (i.e. for all $A\in\mathbb{A}$ we have
$\lim_{n\to\infty}\mu_{n}(A)=0$) is bounded in the norm (i.e.
$\sup_{n\in\mathbb{N}}||\mu_{n}||$ is bounded, cf. Section 2).

The Nikodym property is similar to the Grothendieck property in many ways. For
example, if a Boolean algebra $\mathbb{A}$ has the Grothendieck or Nikodym
property, then its Stone space does not contain non-trivial convergent
sequences. In [18] and [2] the authors proved that the complete Boolean
algebras have both the Grothendieck and Nikodym properties. The completeness
assumption can be relaxed to some combinatorial property (SCP) introduced by
Haydon [20, Definition 1A, Proposition 1B], which is even weaker than
$\sigma$-completeness [13]. Other connections between the Grothendieck
property and the Nikodym property may be found in [31, 37]. Both of the
properties were also considered in a recent paper by Żuchowski in the context
of filters on $\mathbb{N}$ [48].

However, the Grothendieck and Nikodym properties are not equivalent. There are
Boolean algebras with the Nikodym property but without the Grothendieck
property, e.g. the Boolean algebra of Jordan measurable subsets of the unit
interval [30],[36, Propositions 3.2, 3.3]. The question if there is a Boolean
algebra with the Grothendieck property, but without the Nikodym property
turned out to be much more difficult. This is the central question of our
article.

######  Question 1.4.

Does there exist a Boolean algebra with the Grothendieck property that does
not have the Nikodym property?

So far there was only one known example of such a Boolean algebra. It was
constructed by Talagrand in [46]. However, his construction uses the continuum
hypothesis (CH) and so the question of the existence of such a Boolean algebra
in ZFC remains open. Since Talagrand’s construction there was no much progress
in this subject, so it was natural to ask the following question.

######  Question 1.5.

Is it consistent with $\neg$CH that there is a Boolean algebra with the
Grothendieck property but without the Nikodym property?

In this article we answer this problem in the affirmative. Moreover, the
algebra we construct has cardinality $\omega_{1}$.

###### Theorem 6.15.

It is consistent with $\neg$CH that there is a Boolean algebra of size
$\omega_{1}$ with the Grothendieck property but without the Nikodym property.

Very recently, there has been released a preprint by Sobota and Zdomskyy [43]
with a proof that Martin’s axiom (MA) implies the existence of such an algebra
of cardinality $\mathfrak{c}$.111We were not aware of the existence of these
results before the preprint appeared. Our results were obtained independently.

The proof of Theorem 6.15 strongly relies on the ideas behind Talagrand’s
construction. His Boolean algebra consists of Borel sets satisfying certain
symmetry property (we call such sets _balanced sets_). This ensures that the
constructed Boolean algebra will not have the Nikodym property.

We define a $\sigma$-centered forcing notion that extends a given countable
balanced Boolean algebra to a bigger one that is still balanced. Moreover,
some sequences of measures (picked by a generic filter) which were
weak*-convergent in the initial algebra are no longer weak*-convergent in the
extension. Then we show that in the model obtained from the finite support
iteration of length $\omega_{1}$ of such forcing notions, there exists a
balanced Boolean algebra with the Grothendieck property. The idea behind this
forcing comes from the work of Koszmider [27] and of Fajardo [12]. In the
former paper Koszmider introduced a notion of forcing that adds a Boolean
algebra of cardinality $\omega_{1}$, whose Stone space does not contain non-
trivial convergent sequences. Fajardo adapted this method to obtain a Banach
space $C(K)$ of small density and with few operators. In particular, this
space has the Grothendieck property. In our article we show how to combine
this approach with the theory of balanced algebras to obtain a Boolean algebra
without the Nikodym property.

Most of the results concerning fundamental properties of balanced sets (cf.
Section 4) that we use in our paper are essentially due to Talagrand. However,
our construction requires some significant changes. Since the construction is
rather complicated and technical, we decided to include detailed proofs at
each step. We also show how to construct a balanced Boolean algebra with the
Grothendieck property under CH using our modification of Talagrand’s (see
Theorem 5.8).

Another interesting related issue is the question about the possible sizes of
Boolean algebras with the Grothendieck and Nikodym properties. There always
exists such an algebra of size $\mathfrak{c}$ (e.g.
$\mathcal{P}(\mathbb{N})$). It is well-known that if
$\mathfrak{p}=\mathfrak{c}$, then $\mathfrak{c}$ is the only possible size of
such an algebra (it follows from [21, Corollary 3F]). In particular, it
happens under MA. Brech showed the consistency of the existence of a Boolean
algebra with the Grothendieck property of cardinality smaller than
$\mathfrak{c}$ [7]. In [33, Chapter 52, Question 10] Koszmider asked whether
it is consistent that there is no Boolean algebra with the Grothendieck
property of size $\mathfrak{p}$. It turned out that the answer is positive [3,
Proposition 6.18]. Recently Sobota and Zdomskyy published several articles on
cardinal characteristics related to Boolean algebras with the Grothendieck or
Nikodym property [39, 40, 41, 42, 44]. The Boolean algebra we construct is the
first example of a Boolean algebra with the Grothendieck property and without
the Nikodym property of size $\omega_{1}<\mathfrak{c}$ (in particular, our
model satisfies $\mathfrak{p}=\omega_{1}<\mathfrak{c}$, cf. Corollary 6.14)
and the first construction of such a Boolean algebra of size less than
$\mathfrak{c}$. In particular, the Stone space of this algebra is another
example of a Efimov space. In fact, if we want only to obtain a Boolean
algebra with the Grothendieck property, then our forcing can be simplified in
a natural way (by dropping some restrictions on the conditions).

It is also worth mentioning that the Grothendieck and Nikodym properties are
also discussed in the non-commutative setting in the category of C*-algebras.
The definition of the Grothendieck property for C*-algebras is the same as for
general Banach spaces. We say that a C*-algebra $A$ has the Nikodym property,
if every sequence of bounded linear functionals on $A$ that is convergent to
$0$ on projections is bounded in the norm. If $\mathbb{A}$ is a Boolean
algebra, then it has the Nikodym property if and only if
$C(\operatorname{St}(\mathbb{A}))$ has the Nikodym property, when considered
as a C*-algebra. The Nikodym property is especially interesting in the case
when given C*-algebra has many projections, e.g. when its real rank is zero.
It is well-known that von Neumann algebras have both the Grothendieck and
Nikodym properties (see [34, Corollary 7] and [9, Theorem 1]). The problem of
the existence of C*-algebras with the Grothendieck property and without the
Nikodym property is still open in ZFC even in the non-commutative case.

######  Question 1.6.

Is there a C*-algebra of real rank zero, which has the Grothendieck property,
but does not have the Nikodym property?

The structure of the paper is the following. The next section describes
notation and terminology. In Section 3 we introduce the property
$(\mathcal{G})$ of Boolean algebras and the property of being balanced. Then
we show that they imply the Grothendieck property and the negation of the
Nikodym property respectively. Section 4 is devoted to properties of finite
balanced families (this includes the behavior of balanced families under basic
operations and approximating balanced families with families of clopen subsets
of the Cantor set) and tools for extending countable balanced Boolean
algebras. In Section 5 we show a method of extending countable balanced
Boolean algebras to bigger ones in a way that destroys weak*-convergence of
given sequences of measures. Then we show how to apply this method to
construct a Boolean algebra with the Grothendieck property and without Nikodym
property assuming the continuum hypothesis. In Section 6 we describe a
$\sigma$-centered notion of forcing that forces the existence of a Boolean
algebra with the Grothendieck property and without the Nikodym property. In
the last section we include final remarks and state some open questions.

##  2\. Notation

### Basic symbols.

For the purpose of this paper we will denote by $\mathbb{N}$ the set of
positive integers. The smallest uncountable ordinal is denoted by
$\omega_{1}$. The cardinality of the set of all real numbers is denoted by
$\mathfrak{c}$.

For a sequence $s$, its $m$-th term will be denoted by $s_{m}$. By the Cantor
set we mean the set $C=\\{-1,1\\}^{\mathbb{N}}$ of the sequences with values
in $\\{-1,1\\}$ with the usual product topology. The set $\\{-1,1\\}^{n}$
consists of all sequences of length $n$ with values in $\\{-1,1\\}$.

For $s\in\\{-1,1\\}^{n}$ we put

$\langle s\rangle=\\{x\in C:x\upharpoonright n=s\\},$

where $x\upharpoonright n$ is the sequence of first $n$ elements of $x$. The
family of all Borel subsets of $C$ will be denoted by $\operatorname{Bor}(C)$.
For a set $Z\subseteq C,\chi_{Z}$ stands for the characteristic function of
$Z$.

### Boolean algebras.

For the basic terminology concerning Boolean algebras see [8, 14, 26]. We will
focus on Boolean algebras consisting of Borel subsets of the Cantor set
endowed with the standard operations $\cup,\cap,\backslash$. The symmetric
difference of sets $A$ and $B$ will be denoted by $A\triangle B$. For
$n\in\mathbb{N}$, $\mathbb{A}_{n}$ is the finite subalgebra of
$\operatorname{Bor}(C)$ generated by $\\{\langle
s\rangle:s\in\\{-1,1\\}^{n}\\}$. The family of all clopen subsets of $C$ will
be denoted by $\operatorname{Clop}(C)$. Note that
$\operatorname{Clop}(C)=\bigcup_{n\in\mathbb{N}}\mathbb{A}_{n}$. For a Boolean
algebra $\mathbb{A}$ we denote by $\operatorname{at}(\mathbb{A})$ the set of
its atoms. In particular, $\operatorname{at}(\mathbb{A}_{n})=\\{\langle
s\rangle:s\in\\{-1,1\\}^{n}\\}$. The Stone space of $\mathbb{A}$ will be
denoted by $\operatorname{St}(\mathbb{A})$. For $A\in\mathbb{A}$ we denote by
$[A]$ the corresponding clopen subset of $\operatorname{St}(\mathbb{A})$. A
family $\\{H_{n}\\}_{n\in\mathbb{N}}\subseteq\mathbb{A}$ is called an
antichain, if $H_{n}\cap H_{m}=\varnothing$ for $n\neq m$. For a Boolean
algebra $\mathbb{B}$ and a subset $B\subseteq C$ we put

$\mathcal{F}(\mathbb{B},B)=\\{A\cap B,A\backslash B:A\in\mathbb{B}\\}.$

### Measures.

We will consider measures on Boolean algebras and on their Stone spaces. For
the general theory of such measures see [38, Chapter V]. The symbol $\lambda$
will denote the normalized Haar measure on $C$ (considered as a group with
coordinate-wise multiplication). In particular, $\lambda(\langle
s\rangle)=1/2^{n}$ for $s\in\\{-1,1\\}^{n}$. Throughout the article, we will
discuss canonical measures (witnesses to the lack of the Nikodym property)
$\varphi_{n}$ for $n\in\mathbb{N}$, given by the formula

$\varphi_{n}(A)=\int_{A}\delta_{n}d\lambda$

for $A\in\operatorname{Bor}(C)$, where $\delta_{n}\colon
C\rightarrow\\{-1,1\\},\delta_{n}(x)=x_{n}$.

We will denote by $\mathcal{L}_{2}(C)$ the real Hilbert space of square-
integrable (with respect to $\lambda$) functions on $C$ with the inner product

$\langle f,g\rangle=\int_{C}fgd\lambda.$

In what follows a measure on a Boolean algebra $\mathbb{A}$ is always a
finitely additive signed bounded measure on $\mathbb{A}$. We will denote them
by the lowercase Greek letters $\mu,\nu,\vartheta$, and we will call them
concisely “measures on $\mathbb{A}$”. If $\mu$ is a measure on a Boolean
algebra $\mathbb{A}$ and $\mathbb{B}\subseteq\mathbb{A}$ is a subalgebra, then
$\mu\upharpoonright\mathbb{B}$ denotes the restriction of $\mu$ to
$\mathbb{B}$. For a measure $\mu$ on $\mathbb{A}$ we define its variation
$|\mu|$ as a measure on $\mathbb{A}$ given by

$|\mu|(X)=\sup\\{|\mu(A)|+|\mu(B)|:A,B\in\mathbb{A},A,B\subseteq X,A\cap
B=\varnothing\\},$

and its norm (total variation) as

$\|\mu\|=|\mu|(1),$

where $1$ is the biggest element of $\mathbb{A}$. Note that for every
$n\in\mathbb{N}$ we have $|\varphi_{n}|=\lambda$ and $\|\varphi_{n}\|=1$. If
$\mu$ is non-negative and $\|\mu\|=1$, then $\mu$ is called a probability
measure.

For a compact space $K$ we denote by $M(K)$ the Banach space of Radon measures
on $K$, endowed with the total variation norm. Every measure on a Boolean
algebra $\mathbb{A}$ extends uniquely to a Radon measure on the space
$\operatorname{St}(\mathbb{A})$ (cf. [38, Section 18.7]). If $\mu$ is a
measure on a Boolean algebra $\mathbb{A}$, then $\widetilde{\mu}$ denotes the
corresponding Radon measure on $\operatorname{St}(\mathbb{A})$. In particular,
$\|\widetilde{\mu}\|=\|\mu\|$ and $|\widetilde{\mu}|=|\mu|$. For a Radon
measure $\widetilde{\mu}$ a Borel set
$F\subseteq\operatorname{St}(\mathbb{A})$ is a Borel support of
$\widetilde{\mu}$, if $\widetilde{\mu}(X)=0$ for every Borel
$X\subseteq\operatorname{St}(\mathbb{A})\backslash F$. We say that a sequence
$(\widetilde{\mu}_{n})_{n\in\mathbb{N}}$ of Radon measures has pairwise
disjoint Borel supports, if there are pairwise disjoint Borel sets
$(F_{n})_{n\in\mathbb{N}}\subseteq\operatorname{St}(\mathbb{A})$ such that
$F_{n}$ is a Borel support of $\widetilde{\mu}_{n}$ for every
$n\in\mathbb{N}$. Note that, unlike the support of measure, a Borel support is
not unique.

We will consider 3 types of convergence of sequences of measures on a compact
space $K$. We say that a sequence
$(\widetilde{\mu}_{n})_{n\in\mathbb{N}}\subseteq M(K)$ converges weakly, if it
is convergent in the weak topology of the Banach space $M(K)$. We say that
$(\widetilde{\mu}_{n})_{n\in\mathbb{N}}\subseteq M(K)$ is weak*-convergent, if
it converges in the weak* topology, where $M(K)$ is treated as the dual space
to the Banach space of continuous functions $C(K)$. A sequence
$(\mu_{n})_{n\in\mathbb{N}}$ of measures on a Boolean algebra $\mathbb{A}$ is
said to be pointwise convergent, if there is a measure $\mu$ on $\mathbb{A}$
such that $\mu_{n}(A)\xrightarrow{n\to\infty}\mu(A)$ for every
$A\in\mathbb{A}$. It is a well-known fact, that a sequence of Radon measures
$(\widetilde{\mu}_{n})_{n\in\mathbb{N}}$ on $\operatorname{St}(\mathbb{A})$ is
weak*-convergent if and only if the sequence $(\mu_{n})_{n\in\mathbb{N}}$ is
bounded in the norm and pointwise convergent on $\mathbb{A}$.

### Forcing.

Most of the notation concerning forcing should be standard. For the
unexplained terminology see [5, 22, 29]. The universe of sets will be denoted
by $V$. For a forcing notion $\mathbb{P}$ we denote by $V^{\mathbb{P}}$ a
generic extension of $V$ obtained by forcing with $\mathbb{P}$. The evaluation
of a constant $c$ in the class $V$ is denoted by $c^{V}$. For the purpose of
the section involving forcing we will identify Borel subsets of $C$ with their
codes with respect to some absolute coding (cf. [22, Section 25]).

##  3\. Grothendieck and Nikodym properties

In this section we will reduce the problem of the existence of a Boolean
algebra with the Grothendieck property and without the Nikodym property by
introducing the property $(\mathcal{G})$ and the notion of a balanced Boolean
algebra.

We start with the notion of semibalanced sets that describes these subsets
$A\subseteq C$ for which the occurrences of $1$’s and $-1$’s at $r$-th
coordinate of elements of $A$ are almost equally distributed for large enough
$r$.

######  Definition 3.1.

Let $A\in\operatorname{Bor}(C);m\in\mathbb{N};\varepsilon>0$. We say that $A$
is $(m,\varepsilon)$-semibalanced if

(3.1.1) $\displaystyle\forall{r>m}\ |\varphi_{r}(A)|<\frac{\varepsilon}{r}.$

We say that $A$ is semibalanced, if for every $\varepsilon>0$ there is
$m\in\mathbb{N}$ such that $A$ is $(m,\varepsilon)$-semibalanced.

######  Definition 3.2.

Let $A\in\operatorname{Bor}(C);m,t\in\mathbb{N};t\geqslant m;\varepsilon>0$.
We say that $A$ is $(m,t,\varepsilon)$-balanced if for every
$s\in\\{-1,1\\}^{m}$

(3.2.1) $\displaystyle\frac{\lambda(A\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}<\frac{\varepsilon}{m}\ \text{or}\ \frac{\lambda(\langle
s\rangle\backslash A)}{\lambda(\langle s\rangle)}<\frac{\varepsilon}{m}$

and

(3.2.2) $\displaystyle\forall{r\in(m,t]}\ \frac{|\varphi_{r}(A\cap\langle
s\rangle)|}{\lambda(\langle s\rangle)}<\frac{\varepsilon}{r}.$

We say that A is $(m,\varepsilon)$-balanced if for every $s\in\\{-1,1\\}^{m}$
the condition (3.2.1) is satisfied and

(3.2.3) $\displaystyle\forall{r>m}\ \frac{|\varphi_{r}(A\cap\langle
s\rangle)|}{\lambda(\langle s\rangle)}<\frac{\varepsilon}{r}.$

######  Definition 3.3.

We say that a finite subfamily $\mathcal{A}\subseteq\operatorname{Bor}(C)$ is
$(m,\varepsilon)$-balanced if every $A\in\mathcal{A}$ is
$(m,\varepsilon)$-balanced.

We say that a family $\mathbb{B}\subseteq\operatorname{Bor}(C)$ is balanced,
if for every finite subfamily $\mathcal{A}\subseteq\mathbb{B}$ and every
$\varepsilon>0$ there is $m\in\mathbb{N}$ such that $\mathcal{A}$ is
$(m,\varepsilon)$-balanced. A set $A\in\operatorname{Bor}(C)$ is balanced, if
$\\{A\\}$ is balanced.

######  Remark 3.4.

If $\mathcal{A}$ is a finite balanced family and $\varepsilon>0$, then for
every $n\in\mathbb{N}$ there is $m>n$ such that $\mathcal{A}$ is
$(m,\varepsilon)$-balanced.

Note that if $\mathbb{B}$ is balanced, then every member of $\mathbb{B}$ is
balanced, but the reverse implication does not hold in general. It may happen
that $A$ and $B$ are balanced, while the sets $\\{m\in\mathbb{N}:A$ is
$(m,\varepsilon)$-balanced$\\}$ and $\\{m\in\mathbb{N}:B$ is
$(m,\varepsilon)$-balanced$\\}$ are disjoint for some $\varepsilon>0$.

######  Lemma 3.5.

Let $A\in\operatorname{Bor}(C),\varepsilon>0$ and $m,t\in\mathbb{N}$, where
$t\geqslant m$. Then $A$ is $(m,\varepsilon)$-balanced if and only if $A$ is
$(m,t,\varepsilon)$-balanced and for every $s\in\\{-1,1\\}^{m}$ the set
$A\cap\langle s\rangle$ is $(t,2^{-m}\varepsilon)$-semibalanced. In
particular, if $A$ is balanced, then it is semibalanced.

###### Proof.

If $A$ is $(m,\varepsilon)$-balanced, then it is clearly
$(m,t,\varepsilon)$-balanced and by (3.2.3) for any $s\in\\{-1,1\\}^{m}$ and
$r>t$ we have

$|\varphi_{r}(A\cap\langle s\rangle)|<\lambda(\langle
s\rangle)\frac{\varepsilon}{r}=\frac{\varepsilon}{2^{m}r},$

which shows that $A\cap\langle s\rangle$ is
$(t,2^{-m}\varepsilon)$-semibalanced.

The above inequality also shows that if $A$ is
$(t,2^{-m}\varepsilon)$-semibalanced, then (3.2.3) is satisfied for
$s\in\\{-1,1\\}^{m}$ and $r>t$. In particular, if $A$ is
$(m,t,\varepsilon)$-balanced and $(t,2^{-m}\varepsilon)$-semibalanced, then it
is $(m,\varepsilon)$-balanced.

To see that any balanced set is semibalanced fix $\varepsilon>0$ and
$m\in\mathbb{N}$ such that $A$ is $(m,\varepsilon)$-balanced. From the first
part of the lemma applied to $t=m$ we get that for $r>m$

$|\varphi_{r}(A)|\leqslant\sum_{s\in\\{-1,1\\}^{m}}|\varphi_{r}(A\cap\langle
s\rangle)|<\sum_{s\in\\{-1,1\\}^{m}}\frac{\varepsilon}{2^{m}r}=\frac{\varepsilon}{r},$

so $A$ is semibalanced. ∎

We will present a few examples to illustrate the above definitions.

######  Example 3.6.

Every clopen subset of $C$ is $(m,\varepsilon)$-balanced for every
$\varepsilon>0$ and sufficiently large $m\in\mathbb{N}$.

######  Example 3.7.

We will construct an open balanced set $U$ that is not clopen. More precisely,
$U$ will be $\left(2^{n},\frac{2^{n+2}}{2^{2^{n}}}\right)$-balanced for every
$n\in\mathbb{N}$.

For every $n\in\mathbb{N}$ consider the set $Z_{n}$ of all sequences of the
form $(a_{1},a_{2},\dots,a_{2^{n}})$ of length $2^{n}$ with values in
$\\{-1,1\\}$ with the following properties:

  1. (1)

if $n=1$, then $a_{1}=-a_{2}$,

  2. (2)

if $n\neq 1$, then $a_{1}=a_{2}$,

  3. (3)

$\forall{l<n}\ a_{2^{l-1}+1}=a_{2^{l-1}+2}=\dots=a_{2^{l}}$,

  4. (4)

$a_{2^{n-1}+1}=a_{2^{n-1}+2}=\dots=a_{2^{n}-1}=-a_{2^{n}}$,

and put $Z=\bigcup_{n\in\mathbb{N}}Z_{n}$. On the figure below, the red (dark)
sets are of the form $\langle s\rangle$, where $s\in Z_{n}$ and $s_{2^{n}}=1$
for some $n\in\mathbb{N}$, while blue (light) sets are of the form $\langle
s\rangle$, where $s\in Z_{n},s_{2^{n}}=-1$.

Let $U=\bigcup_{s\in Z}\langle s\rangle$. Consider three cases.
1\. If $s\in Z$ then

$m\frac{\lambda(\langle s\rangle\backslash U)}{\lambda(\langle
s\rangle)}=m\frac{\lambda(\varnothing)}{\lambda(\langle s\rangle)}=0.$

2\. If $s\in\\{-1,1\\}^{2^{n}}$ for $n\in\mathbb{N}\setminus\\{1\\}$ satisfy
conditions (2), (3) and

  1. (4’)

$\ a_{2^{n-1}+1}=a_{2^{n-1}+2}=\dots=a_{2^{n}-1}=a_{2^{n}}$,

then there are only 4 nonempty sets of the form $U\cap\langle s^{i}\rangle$,
for $i\in\\{1,2,3,4\\}$ where $s^{i}\in\\{-1,1\\}^{2^{n+1}}$ and $\langle
s^{i}\rangle\subseteq\langle s\rangle$. Hence

$m\frac{\lambda(U\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}<m\frac{\sum_{i=1}^{4}\lambda(\langle s^{i}\rangle)}{\lambda(\langle
s\rangle)}\leqslant 2^{n}\frac{4\cdot
2^{-2^{n+1}}}{2^{-2^{n}}}=\frac{2^{n+2}}{2^{2^{n}}}.$

3\. In other cases

$m\frac{\lambda(U\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}=m\frac{\lambda(\varnothing)}{\lambda(\langle s\rangle)}=0.$

For every $r\in\mathbb{N}$ the distribution of $1$’s and $-1$’s in the
elements of $U$ at the $r$-th coordinate is symmetric, as can be easily seen
in the figure - the blue (light) sets are symmetric to the red (dark) ones
(i.e. $\langle s\rangle$ is a blue set if and only if $\langle-s\rangle$ is
red). Thus, for every $r\in\mathbb{N}$ we have

$\frac{|\varphi_{r}(U\cap\langle s\rangle)|}{\lambda(\langle s\rangle)}=0.$

######  Proposition 3.8.

If a Boolean algebra $\mathbb{B}\subseteq\operatorname{Bor}(C)$ consists of
semibalanced sets, then $\mathbb{B}$ does not have the Nikodym property. In
particular, if $\mathbb{B}$ is balanced, then it does not have the Nikodym
property.

###### Proof.

Consider a sequence of measures $(\mu_{n})_{n\in\mathbb{N}}$ on $\mathbb{B}$
given by

$\mu_{n}(A)=n\varphi_{n}(A).$

This sequence is pointwise convergent to 0, i.e.
$\lim_{n\to\infty}|\mu_{n}(A)|=0$ for all $A\in\mathbb{B}$. Indeed, for
$A\in\mathbb{B}$ and for any $\varepsilon>0$ there exist $m\in\mathbb{N}$ such
that for all $n>m$ by (3.1.1) we have

$\displaystyle|\mu_{n}(A)|=n|\varphi_{n}(A)|\leqslant
n\frac{\varepsilon}{n}=\varepsilon.$

However, $(\mu_{n})_{n\in\mathbb{N}}$ is not bounded in the norm, because

$\sup_{n\in\mathbb{N}}||\mu_{n}||=\sup_{n\in\mathbb{N}}n\|\lambda\|=\infty.$

In particular, $\mathbb{B}$ does not have the Nikodym property.

If $\mathbb{B}$ is balanced, then by Lemma 3.5 it consists of semibalanced
sets, so it does not have the Nikodym property by the first part of the lemma.
∎

######  Example 3.9.

The following Boolean algebra considered by Plebanek222Personal communication
is interesting in the context of our considerations:

$\mathbb{B}_{P}=\\{B\in\operatorname{Bor}(C):\lim_{n\to\infty}n\psi_{n}(B)=0\\},$

where $\psi_{n}(B)=\min\\{\lambda(B\triangle C):C\in\mathbb{A}_{n}\\}$.

Each element of $\mathbb{B}_{P}$ is semibalanced, so by Proposition 3.8
$\mathbb{B}_{P}$ does not have the Nikodym property. Indeed, one needs first
to observe that if $B\in\mathbb{B}_{P}$ and $n<m$, then

$\varphi_{m}(B)\leqslant\psi_{n}(B).$

To show that every $B\in\mathbb{B}_{P}$ is semibalanced take any
$B\in\mathbb{B}_{P}$ and $\varepsilon>0$. Since $n\psi_{n}(B)$ converges to
$0$, we can find $m\in\mathbb{N}$ such that for every $r\geqslant m$ we have
$r\psi_{r}(B)<\frac{\varepsilon}{2}$. Then for every $r>m$

$r|\varphi_{r}(B)|\leqslant r\psi_{r-1}(B)\leqslant
2(r-1)\psi_{r-1}(B)<\varepsilon.$

However, $\mathbb{B}_{P}$ is not balanced. To see this, take

$B=\bigcup_{k=2}^{\infty}\langle s_{k}\rangle,$

where $s_{k}\in\\{-1,1\\}^{k}$ is of the form $s_{k}=(-1,\dots,-1,1,1)$ for
$k\geqslant 2$. Then

$n\psi_{n}(B)=n\sum_{k>n}\lambda(\langle s_{k}\rangle)=\frac{n}{2^{n}}$

converges to $0$, so $B\in\mathbb{B}_{P}$. But $B$ is not balanced. Indeed,
for $k\in\mathbb{N}$ consider $s^{\prime}_{k}\in\\{-1,1\\}^{k}$ of the form
$s^{\prime}_{k}=(-1,\dots,-1,1)$. Then

$\frac{\lambda(B\cap\langle s^{\prime}_{k}\rangle)}{\lambda(\langle
s^{\prime}_{k}\rangle)}=\frac{\lambda(\langle s^{\prime}_{k}\rangle\setminus
B)}{\lambda(\langle s^{\prime}_{k}\rangle)}=\frac{1}{2}.$

There are no non-trivial convergent sequences in
$\operatorname{St}(\mathbb{B}_{P})$. However, this Boolean algebra does not
have the Grothendieck property. The sequence of measures on $\mathbb{B}_{P}$
given by

$\vartheta_{n}(A)=2^{n}\varphi_{2^{n}}(A\cap\langle s^{\prime}_{n}\rangle),$

(where $s^{\prime}_{n}$ is as above) is weak*-convergent, but it is not weakly
convergent333The idea behind this sequence is due to Avilés..

Moreover, according to Borodulin-Nadzieja444Personal communication. no
semibalanced Boolean algebra containing $\mathbb{B}_{P}$ has the Grothendieck
property.

In order to take care of the Grothendieck property it is enough to restrict
the choice of sequences of measures to those with pairwise disjoint Borel
supports and norms equal to 1.

######  Definition 3.10.

Let $\mathbb{A}$ be a Boolean algebra. We say that a sequence
$(\nu_{n})_{n\in\mathbb{N}}$ of measures on $\mathbb{A}$ is normal, if:

  * •

$\forall n\in\mathbb{N}\ \|\nu_{n}\|=1$,

  * •

$(\widetilde{\nu}_{n})_{n\in\mathbb{N}}$ has pairwise disjoint Borel supports.

The following definition will be important throughout the article.

######  Definition 3.11.

We say that a Boolean algebra $\mathbb{B}$ satisfies $(\mathcal{G})$ if for
every normal sequence $(\nu_{n})_{n\in\mathbb{N}}$ of measures on $\mathbb{B}$
there are $G\in\mathbb{B}$, an antichain
$\\{H_{0}^{n},H_{1}^{n}:n\in\mathbb{N}\\}\subseteq\mathbb{B}$ and strictly
increasing sequences $(a_{n})_{n\in\mathbb{N}},(b_{n})_{n\in\mathbb{N}}$ of
natural numbers such that for all $n\in\mathbb{N}$

  1. (a)

$G\cap H^{n}_{1}=\varnothing$,

  2. (b)

$|\nu_{a_{n}}|(H^{n}_{0})\geqslant 0.9$ and $|\nu_{b_{n}}|(H^{n}_{1})\geqslant
0.9$,

  3. (c)

$|\nu_{a_{n}}(G\cap H^{n}_{0})|\geqslant 0.3$.

Note that Schachermayer introduced the property (G) in [36] as a name for the
Grothendieck property. Our property ($\mathcal{G}$) is different. It implies
the Grothendieck property, but the reverse implication does not hold.

To show that the property $(\mathcal{G})$ implies the Grothendieck property we
will use the following lemma known as The Kadec-Pełczyński-Rosenthal
Subsequence Splitting Lemma:

######  Lemma 3.12.

[1, Lemma 5.2.7] Let $K$ be a compact space. For every bounded sequence
$(\widetilde{\nu}_{n})_{n\in\mathbb{N}}\subseteq M(K)$ there exists a non-
negative real $r$ and a subsequence
$(\widetilde{\nu}_{n_{k}})_{k\in\mathbb{N}}$, each element of which may be
decomposed into a sum of two measures
$\widetilde{\nu}_{n_{k}}=\widetilde{\mu}_{k}+\widetilde{\theta}_{k}$, where
$\widetilde{\mu}_{k},\widetilde{\theta}_{k}\in M(K)$, satisfying the following
conditions:

  1. (1)

the measures $\widetilde{\mu}_{k}$ are supported by pairwise disjoint Borel
sets,

  2. (2)

$(\widetilde{\theta}_{k})_{k\in\mathbb{N}}$ is weakly convergent,

  3. (3)

$||\widetilde{\mu}_{k}||=r$, for every $k\in\mathbb{N}.$

######  Proposition 3.13.

If a Boolean algebra $\mathbb{B}$ satisfies $(\mathcal{G})$, then $\mathbb{B}$
has the Grothendieck property.

###### Proof.

Suppose $\mathbb{B}$ satisfies the property $(\mathcal{G})$, but does not have
the Grothendieck property. That is, there is a sequence $\widetilde{\mu}_{n}$
of measures on $\operatorname{St}(\mathbb{B})$ weak*-convergent to a measure
$\widetilde{\mu}$, which is not weakly convergent. Without loss of generality,
by passing to a subsequence, we can assume that no subsequence of
$(\widetilde{\mu}_{n})_{n\in\mathbb{N}}$ is weakly convergent. Indeed, suppose
that each subsequence of $(\widetilde{\mu}_{n})_{n\in\mathbb{N}}$ contains a
weakly convergent subsequence. Since the sequence is weak*-convergent to
$\widetilde{\mu}$, such a subsequence must be also weakly convergent to
$\widetilde{\mu}$. Then each subsequence of
$(\widetilde{\mu}_{n})_{n\in\mathbb{N}}$ has a subsequence weakly convergent
to $\widetilde{\mu}$, so the whole sequence is weakly convergent to
$\widetilde{\mu}$, which gives a contradiction.

Since $(\widetilde{\mu}_{n})_{n\in\mathbb{N}}$ is weak*-convergent, it is
bounded in the norm (cf. [11, Theorem 3.88]). By Lemma 3.12 we can find a real
$r$ and a subsequence $(\widetilde{\mu}_{n_{k}})_{k\in\mathbb{N}}$ each
element of which may be decomposed into the sum of two measures
$\widetilde{\mu}_{n_{k}}=\widetilde{\nu}_{k}+\widetilde{\theta}_{k}$
satisfying

  1. (1)

the measures $\widetilde{\nu}_{k}$ are supported by pairwise disjoint Borel
sets,

  2. (2)

$(\widetilde{\theta}_{k})_{k\in\mathbb{N}}$ is weakly convergent,

  3. (3)

$||\widetilde{\nu}_{k}||=r$, for every $k\in\mathbb{N}.$

Note that since $(\widetilde{\theta}_{n})_{n\in\mathbb{N}}$ is weakly
convergent, $(\widetilde{\nu}_{n})_{n\in\mathbb{N}}$ is not. In particular,
$r\neq 0$. Thus, without loss of generality, by the normalization, we can
assume that $r=1$. Then the sequence $(\nu_{n})_{n\in\mathbb{N}}$ is normal.
The Boolean algebra $\mathbb{B}$ satisfies the property $(\mathcal{G})$, so
there are $G\in\mathbb{B}$, an antichain
$\\{H_{0}^{n},H_{1}^{n}:n\in\mathbb{N}\\}\subseteq\mathbb{B}$ and strictly
increasing sequences $(a_{n})_{n\in\mathbb{N}},(b_{n})_{n\in\mathbb{N}}$ of
natural numbers such that for all $n\in\mathbb{N}$

  1. (a)

$G\cap H^{n}_{1}=\varnothing$,

  2. (b)

$|\nu_{a_{n}}|(H^{n}_{0})\geqslant 0.9$ and $|\nu_{b_{n}}|(H^{n}_{1})\geqslant
0.9$,

  3. (c)

$|\nu_{a_{n}}(G\cap H^{n}_{0})|\geqslant 0.3$.

Hence for $n\in\mathbb{N}$ we have

  1. (1)

$|\nu_{a_{n}}(G)|\geqslant|\nu_{a_{n}}(G\cap
H_{0}^{n})|-|\nu_{a_{n}}|(G\backslash H_{0}^{n})\geqslant 0.3-0.1=0.2$,

  2. (2)

$|\nu_{b_{n}}|(G)\leqslant|\nu_{b_{n}}|(C\backslash H_{1}^{n})\leqslant 0.1$.

So there is no $\nu$ such that $\nu_{n}(G)\to\nu(G)$. Thus,
$(\widetilde{\nu}_{k})_{k\in\mathbb{N}}$ is not weak*-convergent, which is a
contradiction, since
$(\widetilde{\nu}_{k})_{k\in\mathbb{N}}=(\widetilde{\mu}_{n_{k}}-\widetilde{\theta}_{k})_{k\in\mathbb{N}}$
is a difference of two weak*-convergent sequences. ∎

Now we will introduce the property $(\mathcal{G}^{*})$ similar to the property
$(\mathcal{G})$, which focuses on only one sequence of measures. Then we will
show that having the property $(\mathcal{G}^{*})$ for enough many sequences of
measures we can conclude that the property $(\mathcal{G})$ holds.

######  Definition 3.14.

Let $\mathbb{B}^{*}\subseteq\mathbb{B}$ be Boolean algebras and let
$\nu=(\nu_{n})_{n\in\mathbb{N}}$ be a sequence of measures on
$\mathbb{B}^{*}$. We say that $(\mathbb{B}^{*},\mathbb{B},\nu)$ satisfies
$(\mathcal{G}^{*})$, if there are: an antichain
$\\{H_{0}^{n},H_{1}^{n}:n\in\mathbb{N}\\}\subseteq\mathbb{B}^{*}$, a set
$G\in\mathbb{B}$ and strictly increasing sequences
$(a_{n})_{n\in\mathbb{N}},(b_{n})_{n\in\mathbb{N}}$ of natural numbers such
that for all $n\in\mathbb{N}$

  1. (a)

$G\cap H_{0}^{n}\in\mathbb{B}^{*}$,

  2. (b)

$G\cap H_{1}^{n}=\varnothing$,

  3. (c)

$|\nu_{a_{n}}|(H_{0}^{n}),|\nu_{b_{n}}|(H_{1}^{n})\geqslant 0.9$,

  4. (d)

$|\nu_{a_{n}}(G\cap H_{0}^{n})|\geqslant 0.3$.

The next proposition shows the relationship between the properties
$(\mathcal{G})$ and $(\mathcal{G}^{*})$.

######  Proposition 3.15.

Suppose that $\mathbb{B}$ is a Boolean algebra such that for every normal
sequence of measures $\nu=(\nu_{n})_{n\in\mathbb{N}}$ on $\mathbb{B}$ there is
a subalgebra $\mathbb{B}^{*}\subseteq\mathbb{B}$ such that the sequence
$(\nu_{n}\upharpoonright{\mathbb{B}^{*}})_{n\in\mathbb{N}}$ is normal and
$(\mathbb{B}^{*},\mathbb{B},\nu\upharpoonright{\mathbb{B}^{*}}$) satisfies
$(\mathcal{G}^{*})$. Then $\mathbb{B}$ satisfies $(\mathcal{G})$. In
particular, $\mathbb{B}$ has the Grothendieck property.

###### Proof.

Fix any normal sequence $(\nu_{n})_{n\in\mathbb{N}}$ of measures on
$\mathbb{B}$. Pick $\mathbb{B}^{*}\subseteq\mathbb{B}$ such that there exist
an antichain
$\\{H_{0}^{n},H_{1}^{n}:n\in\mathbb{N}\\}\subseteq\mathbb{B}^{*}$,
$G\in\mathbb{B}$ and sequences
$(a_{n})_{n\in\mathbb{N}},(b_{n})_{n\in\mathbb{N}}$ such that for the sequence
$(\nu_{n}\upharpoonright{\mathbb{B}^{*}})$ the conditions (a)-(d) from
Definition 3.14 are satisfied. In particular, $G\cap H_{1}^{n}=\varnothing$,
i.e. (a) of Definition 3.11 holds. To see that (b) of Definition 3.11 is
satisfied, observe that

$\displaystyle|\nu_{a_{n}}|(H^{n}_{0})$ $\displaystyle=$
$\displaystyle\sup\\{|\nu_{a_{n}}(A)|+|\nu_{a_{n}}(B)|:A,B\in\mathbb{B},A,B\subseteq
H^{n}_{0},A\cap B=\varnothing\\}\geqslant$ $\displaystyle\geqslant$
$\displaystyle\sup\\{|\nu_{a_{n}}(A)|+|\nu_{a_{n}}(B)|:A,B\in\mathbb{B}^{*},A,B\subseteq
H^{n}_{0},A\cap B=\varnothing\\}=$ $\displaystyle=$
$\displaystyle|\nu_{a_{n}}\upharpoonright{\mathbb{B}^{*}}|(H^{n}_{0})\geqslant
0.9$

and similarly
$|\nu_{b_{n}}|(H^{n}_{1})\geqslant|\nu_{b_{n}}\upharpoonright{\mathbb{B}^{*}}|(H^{n}_{1})\geqslant
0.9$.

For Definition 3.11 (c) note that

$|\nu_{a_{n}}(G\cap
H_{0}^{n})|=|\nu_{a_{n}}\upharpoonright{\mathbb{B}^{*}}(G\cap
H_{0}^{n})|\geqslant 0.3.$

Use Proposition 3.13 to conclude that $\mathbb{B}$ has the Grothendieck
property. ∎

For the reader’s convenience we provide a brief sketch of our constructions.
We describe consecutive steps of reasoning, starting from general motivations.
The parts devoted only to the case of construction under the continuum
hypothesis are tagged (CH), while the parts devoted to the forcing
construction are tagged (F).

### Construction roadmap

  1. (1)

General idea: we construct an increasing sequence
$(\mathbb{B}_{\alpha})_{\alpha<\omega_{1}}$ of balanced countable subalgebras
of $\operatorname{Bor}(C)$.

     * (CH)

For every $\alpha<\omega_{1}$ the triple
$(\mathbb{B}^{*}_{\alpha},\mathbb{B}_{\alpha+1},(\nu_{n}^{\alpha})_{n\in\mathbb{N}})$
satisfies $(\mathcal{G}^{*})$, where $(\nu_{n}^{\alpha})_{n\in\mathbb{N}}$ is
a sequence of measures (on some subalgebra $\mathbb{B}^{*}_{\alpha}$ of
$\mathbb{B}_{\alpha}$) which is given in advance (by a proper bookkeeping),
cf. Theorem 5.8.

     * (F)

We define a finite support iteration
$(\mathbb{P}_{\alpha})_{\alpha\leqslant\omega_{1}}$ of $\sigma$-centered
forcings (Definition 6.9). For every $\alpha<\omega_{1}$ the algebra
$\mathbb{B}_{\alpha}$ belongs to the $\alpha$-th intermediate model obtained
from this iteration. The triples
$(\mathbb{B}_{\alpha},\mathbb{B}_{\alpha+1},(\nu_{n})_{n\in\mathbb{N}})$
satisfy $(\mathcal{G}^{*})$ for uncountably many sequences of measures (on
$\mathbb{B}_{\alpha}$), whose choice depends on a generic filter in
$\mathbb{P}_{\alpha}$ (for the connection between the choice of sequences of
measures and a generic filter see Lemma 6.6).

We finish by taking the Boolean algebra

$\mathbb{B}=\bigcup_{\alpha<\omega_{1}}\mathbb{B}_{\alpha},$

which is balanced and satisfies $(\mathcal{G})$.

  2. (2)

We start with $\mathbb{B}_{0}=\operatorname{Clop}(C)$. At limit steps we take
unions. The only non-trivial step is the construction of
$\mathbb{B}_{\alpha+1}$ from $\mathbb{B}_{\alpha}$. In this case we extend
$\mathbb{B}_{\alpha}$ by a new set $G\in\operatorname{Bor}(C)$ which is a
union of countably many pairwise disjoint elements of $\mathbb{B}_{\alpha}$:

$G=\bigcup_{n\in\mathbb{N}}G_{n}$

that satisfy the hypothesis of Lemma 4.7 (this ensures that
$\mathbb{B}_{\alpha+1}$ is balanced). Moreover, we require that

     * (CH)

$G$ (together with some antichain
$\\{H_{0}^{n},H_{1}^{n}\\}_{n\in\mathbb{N}}\subseteq\mathbb{B}_{\alpha})$ is a
witness for the property $(\mathcal{G}^{*})$ for the triple
$(\mathbb{B}^{*}_{\alpha},\mathbb{B}_{\alpha+1},(\nu_{n}^{\alpha})_{n\in\mathbb{N}})$,

     * (F)

there is an antichain
$\\{H_{n}\\}_{n\in\mathbb{N}}\subseteq\mathbb{B}_{\alpha}$ such that for every
sequence $(\nu_{n})_{n\in\mathbb{N}}$ satisfying the hypothesis of Proposition
6.7 the set $G$ together with some subset of $\\{H_{n}\\}_{n\in\mathbb{N}}$
witnesses the property $(\mathcal{G}^{*})$ for
$(\mathbb{B}_{\alpha},\mathbb{B}_{\alpha+1},(\nu_{n})_{n\in\mathbb{N}})$.

  3. (3)

From now on we will assume that $\alpha$ is fixed and we will focus on the
construction of $G_{n}$’s, $H_{n}$’s and $H_{i}^{n}$’s for $i=0,1$.

     * (CH)

We define
$(G_{n})_{n\in\mathbb{N}},(H_{0}^{n})_{n\in\mathbb{N}},(H_{1}^{n})_{n\in\mathbb{N}}$
by induction on $n\in\mathbb{N}$ (cf. Lemma 5.6). In order to obtain the
property $(\mathcal{G}^{*})$ we need to ensure that for every $n\in\mathbb{N}$
the set $G_{n}\cap H_{0}^{n}$ is “big” in the sense of some measure from the
sequence $(\nu_{n}^{\alpha})_{n\in\mathbb{N}}$ while $G_{n}\cap
H_{1}^{k}=\varnothing$ for $k,n\in\mathbb{N}$.

     * (F)

The sets $G_{n}$’s and $H_{n}$’s appear in the forcing conditions chosen by a
generic filter. Lemma 6.6 will imply that for an appropriate sequence
$(\nu_{n})_{n\in\mathbb{N}}$ of measures, for infinitely many $n\in\mathbb{N}$
the set $G_{n}\cap H_{n}$ is “big” in the sense of some measure from this
sequence, while $G_{n}\cap H_{k}=\varnothing$ for every $n\in\mathbb{N}$ and
infinitely many $k\in\mathbb{N}$.

  4. (4)

Given finite sequences $(G_{k})_{k\leqslant n},(H_{0}^{k})_{k\leqslant
n},(H_{1}^{k})_{k\leqslant n}$ (or $(H_{k})_{k\leqslant n}$ in the case of
forcing), we extend them using Lemma 5.4 and Lemma 5.5 (applied to
$\widehat{G}=\bigcup_{k\leqslant n}G_{k}$ and $\widehat{H}=\bigcup_{k\leqslant
n}(H_{0}^{k}\cup H_{1}^{k})$ or $\widehat{H}=\bigcup_{k\leqslant n}H_{k}$).

     * (CH)

The set $G_{n+1}$ consists of 2 parts: $G_{n+1}=L\cup M$, where $L=G_{n+1}\cap
H_{0}^{n+1}$ is the part witnessing the property $(\mathcal{G}^{*})$ and $M$
is a very small set disjoint from $H_{i}^{k}$’s (so it has no influence on
$(\mathcal{G}^{*})$).

     * (F)

It follows that there are $l_{1},l_{2}>n$ such that $G_{l_{1}}=L\cup M$, where
$L=G_{l_{1}}\cap H_{l_{1}}$ witnesses the property $(\mathcal{G}^{*})$ and
$M\cap H_{k}=\varnothing$ for $k\in\mathbb{N}$, while $G_{l_{2}}\cap
H_{l_{2}}=\varnothing$ (this will ensure that $G\cap H_{l_{2}}=\varnothing$).

  5. (5)

At the same time, to make sure that the hypothesis of Lemma 4.7 is satisfied,
we need to pick $G_{n}$’s in such a way that the families
$\mathcal{F}(\mathbb{B}_{n},\bigcup_{i\leqslant k}G_{i})$ (where
$\mathbb{B}_{n}$’s are some finite subalgebras of $\mathbb{B}_{\alpha}$) will
have appropriate degrees of balance for $k,n\in\mathbb{N}$. However, we do not
have enough control over the choice of $L$, which can affect the balance.
Thus, we need to fix it with the help of $M$.

  6. (6)

The choice of $M$ depends on $L$ in the way described in Proposition 4.8. The
main idea behind this choice is to reduce the problem to finite combinatorics.
We work with some finite subalgebra $\mathbb{H}$ of $\mathbb{B}$ containing
the sets that have appeared in the construction so far (including $L$) that is
sufficiently well balanced (cf. Lemma 4.12).

  7. (7)

Lemma 4.5 shows that there is $n\in\mathbb{N}$ and a Boolean homomorphism
$h\colon\mathbb{H}\to\mathbb{A}_{n}$ (recall that $\mathbb{A}_{n}$ is a finite
Boolean algebra consisting of clopen subsets of $C$) such that every
$A\in\mathbb{H}$ is well-approximated by $h(A)$. The choice of $M$ and most of
the crucial calculations take place in $\mathbb{A}_{n}$ (see Lemma 4.11).
These include the use of techniques such as probability inequalities involving
weighted Rademacher sums (Lemma 4.9) and analysis in finite-dimensional
subspaces of the Hilbert space $\mathcal{L}_{2}(C)$.

##  4\. Properties of balanced families

This section is devoted to the combinatorics of balanced sets and families. In
a series of lemmas we will describe basic properties of balanced sets and show
how to modify a given set to a balanced one.

We start with a few simple observations.

######  Lemma 4.1.

If $A,B\in\operatorname{Bor}(C)$ are disjoint and $(m,\varepsilon)$-balanced,
then $A\cup B$ is $(m,2\varepsilon)$-balanced.

###### Proof.

First we check (3.2.1). Fix any $s\in\\{-1,1\\}^{m}$. If

$\frac{\lambda(\langle s\rangle\backslash A)}{\lambda(\langle
s\rangle)}<\frac{\varepsilon}{m}\ \text{or}\ \frac{\lambda(\langle
s\rangle\backslash B)}{\lambda(\langle s\rangle)}<\frac{\varepsilon}{m}$

then

$\frac{\lambda(\langle s\rangle\backslash(A\cup B))}{\lambda(\langle
s\rangle)}\leqslant\min\left\\{\frac{\lambda(\langle s\rangle\backslash
A)}{\lambda(\langle s\rangle)},\frac{\lambda(\langle s\rangle\backslash
B)}{\lambda(\langle s\rangle)}\right\\}<\frac{\varepsilon}{m}.$

Otherwise, since $A$ and $B$ satisfy (3.2.1) we have

$\frac{\lambda(A\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}<\frac{\varepsilon}{m}\ \text{and}\ \frac{\lambda(B\cap\langle
s\rangle)}{\lambda(\langle s\rangle)}<\frac{\varepsilon}{m}.$

Summing up the inequalities we get

$\frac{\lambda((A\cup B)\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}<\frac{2\varepsilon}{m},$

so $A\cup B$ satisfies (3.2.1) for $2\varepsilon$.

Now we check (3.2.3). Fix $r>m$. Since $A\cap B=\varnothing$ we have

$|\varphi_{r}((A\cup B)\cap\langle
s\rangle)|\leqslant|\varphi_{r}(A\cap\langle
s\rangle)|+|\varphi_{r}(B\cap\langle s\rangle)|.$

Thus, by summing up the inequalities (3.2.3) for $A$ and $B$ we get

$\frac{|\varphi_{r}((A\cup B)\cap\langle s\rangle)|}{\lambda(\langle
s\rangle)}\leqslant\frac{|\varphi_{r}(A\cap\langle s\rangle)|}{\lambda(\langle
s\rangle)}+\frac{|\varphi_{r}(B\cap\langle s\rangle)|}{\lambda(\langle
s\rangle)}<\frac{2\varepsilon}{r},$

so $A\cup B$ satisfies (3.2.3) for $2\varepsilon$. ∎

######  Lemma 4.2.

Let $A\in\operatorname{Bor}(C),t\in\mathbb{N}$ and $\varepsilon>0$. If $A$ is
$(t,\varepsilon)$-semibalanced then $C\backslash A$ is also
$(t,\varepsilon)$-semibalanced.

###### Proof.

Let $r>t$. We have

$\varphi_{r}(A)+\varphi_{r}(C\backslash
A)=\int_{A}\delta_{r}d\lambda+\int_{C\backslash
A}\delta_{r}d\lambda=\int_{C}\delta_{r}d\lambda=0.$

By (3.1.1) for $A$ we get

$|\varphi(C\backslash A)|=|\varphi(A)|<\frac{\varepsilon}{r},$

which implies (3.1.1) for $C\backslash A$ and so $C\backslash A$ is
$(t,\varepsilon)$-semibalanced. ∎

######  Lemma 4.3.

Suppose that $\mathbb{F}$ is a finite Boolean subalgebra of
$\operatorname{Bor}(C)$ and $t\in\mathbb{N}$ is such that $\mathbb{F}$ is
$(t,\varepsilon)$-balanced, where

$\varepsilon=\frac{1}{100}\inf\\{\lambda(A):A\in\mathbb{F},\lambda(A)>0\\}.$

Then for every $A\in\mathbb{F}$, if $\lambda(A)>0$, then there is
$s_{A}\in\\{-1,1\\}^{t}$ such that

$\frac{\lambda(A\cap\langle s_{A}\rangle)}{\lambda(\langle
s_{A}\rangle)}\geqslant 0.99.$

###### Proof.

Let $A\in\mathbb{F}$ be such that $\lambda(A)>0$. Then
$\lambda(A)>\varepsilon$, so there must be $s_{A}\in\\{-1,1\\}^{t}$ such that
$\lambda(A\cap\langle s_{A}\rangle)>\varepsilon\lambda(\langle s_{A}\rangle)$.
Hence

$\frac{\lambda(\langle s_{A}\rangle\backslash A)}{\lambda(\langle
s_{A}\rangle)}\leqslant 1-\varepsilon.$

Since $A$ satisfies (3.2.1) we have

$\frac{\lambda(\langle s_{A}\rangle\backslash A)}{\lambda(\langle
s_{A}\rangle)}<\frac{\varepsilon}{t}\leqslant\varepsilon.$

But $\varepsilon\leqslant 0.01$, so

$\frac{\lambda(A\cap\langle s_{A}\rangle)}{\lambda(\langle
s_{A}\rangle)}=\frac{\lambda(\langle s_{A}\rangle)-\lambda(\langle
s_{A}\rangle\backslash A)}{\lambda(\langle
s_{A}\rangle)}>1-\varepsilon\geqslant 0.99.$

∎

######  Lemma 4.4.

Let $\mathbb{H}_{0}$ be an $(n,\varepsilon)$-balanced Boolean algebra. Then
the Boolean algebra $\mathbb{H}$ generated by
$\mathbb{H}_{0}\cup\mathbb{A}_{n}$ is $(n,\varepsilon)$-balanced.

###### Proof.

Take $A\in\mathbb{H}$. Then $A$ is of the form:

$A=\bigcup_{\langle s\rangle\in\operatorname{at}(\mathbb{A}_{n})}(\langle
s\rangle\cap A_{s})=\bigcup_{s\in\\{-1,1\\}^{n}}(\langle s\rangle\cap A_{s}),$

where $A_{s}\in\mathbb{H}_{0}$ for $s\in\\{-1,1\\}^{n}$.

Let $s_{0}\in\\{-1,1\\}^{n}$. Then

$A\cap\langle s_{0}\rangle=\bigcup_{s\in\\{-1,1\\}^{n}}(\langle s\rangle\cap
A_{s})\cap\langle s_{0}\rangle=A_{s_{0}}\cap\langle s_{0}\rangle$

and

$\langle s_{0}\rangle\backslash A=\langle
s_{0}\rangle\backslash\bigcup_{s\in\\{-1,1\\}^{n}}(\langle s\rangle\cap
A_{s})=\langle s_{0}\rangle\backslash A_{s_{0}}.$

By (3.2.1) we have

$\frac{\lambda(A_{s_{0}}\cap\langle s_{0}\rangle)}{\lambda(\langle
s_{0}\rangle)}<\frac{\varepsilon}{n}\ \text{or}\ \frac{\lambda(\langle
s_{0}\rangle\backslash A_{s_{0}})}{\lambda(\langle
s_{0}\rangle)}<\frac{\varepsilon}{n}.$

If $\frac{\lambda(A_{s_{0}}\cap\langle s_{0}\rangle)}{\lambda(\langle
s_{0}\rangle)}<\frac{\varepsilon}{n}$, then

$\frac{\lambda(A\cap\langle s_{0}\rangle)}{\lambda(\langle
s_{0}\rangle)}=\frac{\lambda(A_{s_{0}}\cap\langle
s_{0}\rangle)}{\lambda(\langle s_{0}\rangle)}<\frac{\varepsilon}{n}.$

If $\frac{\lambda(\langle s_{0}\rangle\backslash A_{s_{0}})}{\lambda(\langle
s_{0}\rangle)}<\frac{\varepsilon}{n}$, then

$\frac{\lambda(\langle s_{0}\rangle\backslash A)}{\lambda(\langle
s_{0}\rangle)}=\frac{\lambda(\langle s_{0}\rangle\backslash
A_{s_{0}})}{\lambda(\langle s_{0}\rangle)}<\frac{\varepsilon}{n},$

so $A$ also satisfies (3.2.1).

Let $r>n$ and let $s_{0}\in\\{-1,1\\}^{n}$. Then by (3.2.3)

$\frac{|\varphi_{r}(A\cap\langle s_{0}\rangle)|}{\lambda(\langle
s_{0}\rangle)}=\frac{|\varphi_{r}(A_{s_{0}}\cap\langle
s_{0}\rangle)|}{\lambda(\langle s_{0}\rangle)}<\frac{\varepsilon}{r}.$

∎

The next lemma shows that while dealing with finite balanced families we can
approximate them with finite families of clopen subsets of $C$. This will
allow us to reduce many problems to the combinatorics of finite Boolean
algebras $\mathbb{A}_{n}$ for $n\in\mathbb{N}$.

######  Lemma 4.5.

Suppose that $\mathbb{H}\subseteq\operatorname{Bor}(C)$ is a finite subalgebra
that is $(n,\varepsilon)$-balanced for some $n\in\mathbb{N}$ and
$\varepsilon<1/3$. Then the function
$h_{n}\colon\mathbb{H}\rightarrow\mathbb{A}_{n}$ given by

$h_{n}(A)=\bigcup\left\\{\langle
s\rangle:{s\in\\{-1,1\\}^{n}},\frac{\lambda(\langle s\rangle\backslash
A)}{\lambda(\langle s\rangle)}<\varepsilon\right\\}$

is a homomorphism of Boolean algebras and for every $A\in\mathbb{H}$ we have

$\lambda\left(A\triangle h_{n}(A)\right)<\varepsilon/n.$

###### Proof.

For the first part of the lemma we need to show that $h_{n}(C)=C,h_{n}(A\cup
B)=h_{n}(A)\cup h_{n}(B)$ and $h_{n}(C\backslash A)=C\backslash h_{n}(A)$ for
every $A,B\in\mathbb{H}$.

The first equality holds since for every $s\in\\{-1,1\\}^{n}$ we have

$\frac{\lambda(\langle s\rangle\backslash C)}{\lambda(\langle
s\rangle)}=\frac{\lambda(\varnothing)}{\lambda(\langle s\rangle)}=0.$

The second equality follows from the fact that

$\frac{\lambda(\langle s\rangle\backslash(A\cup B))}{\lambda(\langle
s\rangle)}<\varepsilon\ \text{iff}\ \frac{\lambda(\langle s\rangle\backslash
A)}{\lambda(\langle s\rangle)}<\varepsilon\ \text{or}\ \frac{\lambda(\langle
s\rangle\backslash B)}{\lambda(\langle s\rangle)}<\varepsilon.$

Indeed, if

$\frac{\lambda(\langle s\rangle\backslash(A\cup B))}{\lambda(\langle
s\rangle)}<\varepsilon,$

then

$\frac{\lambda((A\cup B)\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}\geqslant\frac{2}{3}$

and so

$\max\left\\{\frac{\lambda(A\cap\langle s\rangle)}{\lambda(\langle
s\rangle)},\frac{\lambda(B\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}\right\\}\geqslant\frac{1}{3}>\varepsilon.$

Since $A$ and $B$ are $(n,\varepsilon)$-balanced we have

$\max\left\\{\frac{\lambda(A\cap\langle s\rangle)}{\lambda(\langle
s\rangle)},\frac{\lambda(B\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}\right\\}>1-\varepsilon$

or equivalently

$\min\left\\{\frac{\lambda(\langle s\rangle\backslash A)}{\lambda(\langle
s\rangle)},\frac{\lambda(\langle s\rangle\backslash B)}{\lambda(\langle
s\rangle)}\right\\}<\varepsilon.$

Conversely, if

$\frac{\lambda(\langle s\rangle\backslash A)}{\lambda(\langle
s\rangle)}<\varepsilon\ \text{or}\ \frac{\lambda(\langle s\rangle\backslash
B)}{\lambda(\langle s\rangle)}<\varepsilon,$

then

$\frac{\lambda(\langle s\rangle\backslash(A\cup B)}{\lambda(\langle
s\rangle)}\leqslant\min\left\\{\frac{\lambda(\langle s\rangle\backslash
A)}{\lambda(\langle s\rangle)},\frac{\lambda(\langle s\rangle\backslash
B)}{\lambda(\langle s\rangle)}\right\\}<\varepsilon.$

The equality $h_{n}(C\backslash A)=C\backslash h_{n}(A)$ holds since for
$s\in\\{-1,1\\}^{n}$

$\displaystyle\langle s\rangle\subseteq C\backslash h_{n}(A)\ \text{iff}\
\frac{\lambda(\langle s\rangle\backslash A)}{\lambda(\langle
s\rangle)}\geqslant\varepsilon\ \text{iff}\ \frac{\lambda(A\cap\langle
s\rangle)}{\lambda(\langle s\rangle)}\leqslant 1-\varepsilon\ \text{iff}$
$\displaystyle\frac{\lambda(A\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}<\varepsilon\ \text{iff}\ \frac{\lambda(\langle
s\rangle\backslash(C\backslash A))}{\lambda(\langle s\rangle)}<\varepsilon\
\text{iff}\ \langle s\rangle\subseteq h_{n}(C\backslash A).$

For the second part of the lemma we notice that for every $s\in\\{-1,1\\}^{n}$
if $\frac{\lambda(\langle s\rangle\backslash A)}{\lambda(\langle
s\rangle)}<\frac{\varepsilon}{n}$, then

$\displaystyle\lambda(A\triangle h_{n}(A)\cap\langle s\rangle)=\lambda(\langle
s\rangle\backslash A)<\lambda(\langle s\rangle)\frac{\varepsilon}{n}$

and if $\frac{\lambda(\langle s\rangle\backslash A)}{\lambda(\langle
s\rangle)}\geqslant\frac{\varepsilon}{n}$, then $\frac{\lambda(A\cap\langle
s\rangle)}{\lambda(\langle s\rangle)}<\frac{\varepsilon}{n}$, and so

$\displaystyle\lambda(A\triangle h_{n}(A)\cap\langle
s\rangle)=\lambda(A\cap\langle s\rangle)<\lambda(\langle
s\rangle)\frac{\varepsilon}{n}.$

Hence

$\displaystyle\lambda(A\triangle
h_{n}(A))=\sum_{s\in\\{-1,1\\}^{n}}\lambda(A\triangle h_{n}(A)\cap\langle
s\rangle)\leqslant\sum_{s\in\\{-1,1\\}^{n}}\lambda(\langle
s\rangle)\frac{\varepsilon}{n}=\frac{\varepsilon}{n}.$

∎

The next lemma says that small perturbations of $(m,t,\varepsilon)$-balanced
sets are still $(m,t,\varepsilon)$-balanced.

######  Lemma 4.6.

Let $m,t\in\mathbb{N},t>m,\varepsilon>0$. Then there is $\varrho>0$ such that
for every $A,B\in\operatorname{Bor}(C)$, if $A$ is
$(m,t,\varepsilon)$-balanced and $\lambda(B)<\varrho$, then $A\cup B$ and
$A\backslash B$ are $(m,t,\varepsilon)$-balanced.

###### Proof.

Let $\varepsilon_{1}<\varepsilon$ be such that $A$ is
$(m,t,\varepsilon_{1})$-balanced and let

$\varrho=\frac{\varepsilon-\varepsilon_{1}}{2^{m}t}.$

For every $s\in\\{-1,1\\}^{m}$ we have

$\displaystyle\frac{\lambda((A\cup B)\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}\leqslant\frac{\lambda(A\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}+\frac{\varrho}{\lambda(\langle
s\rangle)}<\frac{\varepsilon_{1}}{m}+\frac{\varepsilon-\varepsilon_{1}}{t}\leqslant\frac{\varepsilon}{m}$

or

$\displaystyle\frac{\lambda(\langle s\rangle\backslash(A\cup
B))}{\lambda(\langle s\rangle)}\leqslant\frac{\lambda(\langle
s\rangle\backslash A)}{\lambda(\langle s\rangle)}<\frac{\varepsilon}{m}$

and for every $s\in\\{-1,1\\}^{m}$, $m<r\leqslant t$

$\displaystyle\frac{|\varphi_{r}((A\cup B)\cap\langle
s\rangle)|}{\lambda(\langle s\rangle)}\leqslant\frac{|\varphi_{r}(A\cap\langle
s\rangle)|}{\lambda(\langle s\rangle)}+\frac{|\varphi_{r}((B\backslash
A)\cap\langle s\rangle)|}{\lambda(\langle
s\rangle)}\leqslant\frac{\varepsilon_{1}}{r}+\frac{\varepsilon-\varepsilon_{1}}{t}<\frac{\varepsilon}{r}.$

Hence $A\cup B$ is $(m,t,\varepsilon)$-balanced. Calculations showing that
$A\backslash B$ is $(m,t,\varepsilon)$-balanced are similar. ∎

In the following lemma, we provide conditions for enlarging balanced Boolean
algebras to bigger ones.

######  Lemma 4.7.

Suppose that $\mathbb{B}\subseteq\operatorname{Bor}(C)$ is a balanced Boolean
algebra and that

$\mathbb{B}=\bigcup_{n\in\mathbb{N}}\mathbb{B}_{n}$

is a representation of $\mathbb{B}$ as an increasing union of finite
subalgebras. Let $(m_{n})_{n\in\mathbb{N}}$ be a strictly increasing sequence
of natural numbers and $\\{G_{n}\\}_{n\in\mathbb{N}}\subseteq\mathbb{B}$ be an
antichain such that

$\forall k\in\mathbb{N}\ \forall n\leqslant k\
\mathcal{F}\Big{(}\mathbb{B}_{n},\bigcup_{i\leqslant k}G_{i}\Big{)}\
\text{is}\ (m_{n},2^{-n})\text{-balanced},$

where $\mathcal{F}\left(\mathbb{B}_{n},\bigcup_{i\leqslant
k}G_{i}\right)=\left\\{A\cap\bigcup_{i\leqslant
k}G_{i},A\backslash\bigcup_{i\leqslant k}G_{i}:A\in\mathbb{B}_{n}\right\\}$.

Put $G=\bigcup_{n\in\mathbb{N}}G_{n}$. Then the Boolean algebra
$\mathbb{B}^{\prime}$ generated by $\mathbb{B}\cup\\{G\\}$ is balanced.

###### Proof.

Let $\varepsilon>0$ and let $\mathcal{A}^{\prime}$ be a finite subfamily of
$\mathbb{B}^{\prime}$. Every element $A^{\prime}\in\mathcal{A}^{\prime}$ is of
the form $(A_{1}\cap G)\cup(A_{2}\backslash G)$ where
$A_{1},A_{2}\in\mathbb{B}$, so there is a finite family
$\mathcal{A}\subseteq\mathbb{B}$ such that
$\mathcal{A}^{\prime}\subseteq\\{(A_{1}\cap G)\cup(A_{2}\backslash
G):A_{1},A_{2}\in\mathcal{A}\\}$. Let $n\in\mathbb{N}$ be such that
$\mathcal{A}\subseteq\mathbb{B}_{n}$ and $1/2^{n-1}<\varepsilon$. Fix
$A_{1},A_{2}\in\mathcal{A}$. Then for every $k>n$

$A_{1}\cap\bigcup_{i\leqslant k}G_{i}\ \text{and}\
A_{2}\backslash\bigcup_{i\leqslant k}G_{i}\ \text{are}\
(m_{n},1/2^{n})-\text{balanced}.$

Hence for $s\in\\{-1,1\\}^{m_{n}}$ we have

$\displaystyle\frac{\lambda(A_{1}\cap G\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}$ $\displaystyle\leqslant$
$\displaystyle\frac{\lambda\left(A_{1}\cap\bigcup_{i\leqslant
k}G_{i}\cap\langle
s\rangle\right)+\lambda\left(\bigcup_{i>n}G_{i}\right)}{\lambda(\langle
s\rangle)}\leqslant$ $\displaystyle\leqslant$
$\displaystyle\frac{1}{m_{n}2^{n}}+\frac{\lambda\left(\bigcup_{i>k}G_{i}\right)}{\lambda(\langle
s\rangle)}\xrightarrow{k\rightarrow\infty}\frac{1}{m_{n}2^{n}}$

or

$\displaystyle\frac{\lambda(\langle s\rangle\backslash(A_{1}\cap
G))}{\lambda(\langle s\rangle)}$ $\displaystyle\leqslant$
$\displaystyle\frac{\lambda\left(\langle
s\rangle\backslash\left(A_{1}\cap\bigcup_{i\leqslant
k}G_{i}\right)\right)}{\lambda(\langle
s\rangle)}\leqslant\frac{1}{m_{n}2^{n}}$

and for $m>m_{n}$

$\displaystyle\frac{|\varphi_{m}(A_{1}\cap G\cap\langle
s\rangle)|}{\lambda(\langle s\rangle)}\\!\\!\\!\\!\\!$
$\displaystyle\leqslant\\!\\!\\!\\!\\!$
$\displaystyle\frac{\left|\varphi_{m}\left(A_{1}\cap\bigcup_{i\leqslant
k}G_{i}\cap\langle s\rangle\right)\right|}{\lambda(\langle
s\rangle)}+\frac{\left|\varphi_{m}\Big{(}A_{1}\cap\bigcup_{i>k}G_{i}\cap\langle
s\rangle\Big{)}\right|}{\lambda(\langle s\rangle)}\leqslant$
$\displaystyle\leqslant\\!\\!\\!\\!\\!$
$\displaystyle\frac{1}{m2^{n}}+\frac{\lambda\left(\bigcup_{i>k}G_{i}\right)}{\lambda(\langle
s\rangle)}\xrightarrow{k\rightarrow\infty}\frac{1}{m2^{n}},$

so $A_{1}\cap G$ is $(m_{n},1/2^{n})$-balanced. By a similar argument
$A_{2}\backslash G$ is $(m_{n},1/2^{n})$-balanced. By Lemma 4.1 the set
$(A_{1}\cap G)\cup(A_{2}\backslash G)$ is $(m_{n},1/2^{n-1})$-balanced, and so
$(m_{n},\varepsilon)$-balanced, which completes the proof. ∎

The next theorem is key for the construction in Lemma 5.5. It says how much we
can modify a given finite balanced family without losing its balance.

######  Proposition 4.8.

Let $k\in\mathbb{N}$, $\eta>0$. Let $(m_{n})_{n\leqslant k}$ be an increasing
sequence of natural numbers. Let
$\mathbb{B}^{*}\subseteq\mathbb{B}\subseteq\operatorname{Bor}(C)$ be balanced
Boolean algebras and assume that
$\operatorname{Clop}(C)\subseteq\mathbb{B}^{*}$. Let
$(\mathbb{B}_{n})_{n\leqslant k}\subseteq\mathbb{B}$ be finite subalgebras.
Suppose that $G,P\in\mathbb{B}^{*}$ and the following are satisfied:

  1. (A)

$G\subseteq P$,

  2. (B)

$\forall n\leqslant k\ \mathcal{F}(\mathbb{B}_{n},G)$ is
$(m_{n},2^{-n})$-balanced.

Then there is $\theta>0$ such that for every $L,Q\in\mathbb{B}^{*}$ satisfying

  1. (a)

$\max\\{\lambda(L),\lambda(Q)\\}<\theta$,

  2. (b)

$L\cap P=\varnothing$,

there is $M\in\mathbb{B}^{*}$ such that

  1. (1)

$M\cap(P\cup Q)=\varnothing$,

  2. (2)

$\lambda(M)<\eta$,

  3. (3)

$\forall n\leqslant k\ \mathcal{F}(\mathbb{B}_{n},G\cup L\cup M)$ is
$(m_{n},2^{-n})$-balanced.

$G$$M$$L$$Q$$P$

Before we prove the proposition, we will need a few lemmas.

The following lemma is a version of [23, Theorem 3, page 31].

######  Lemma 4.9.

Let $n\in\mathbb{N}$. Let $\lambda_{n}$ be the standard product probability
measure on the space $\\{-1,1\\}^{n}$ ($\lambda(\\{x\\})=\lambda(\\{y\\})$ for
every $x,y\in\\{-1,1\\}^{n}$). Then for all $(d_{m})_{m\leqslant
n}\in\mathbb{R}^{n}$ and any $\xi\in(0,1)$

$\lambda_{n}\left(\left\\{y\in\\{-1,1\\}^{n}:\left|\sum_{m=1}^{n}y_{m}d_{m}\right|^{2}\geqslant\xi\sum_{m=1}^{n}|d_{m}|^{2}\right\\}\right)\geqslant\frac{1}{3}(1-\xi)^{2}.$

The above lemma will allow us to pick $y$ from some big enough subset of
$\\{-1,1\\}^{n}$ satisfying the appropriate inequality.

######  Lemma 4.10.

The sequence $(\delta_{n})_{n\in\mathbb{N}}$ is orthonormal in the Hilbert
space $\mathcal{L}_{2}(C)$.

###### Proof.

We need to show that for any $n,m\in\mathbb{N}$

$\displaystyle\langle\delta_{n},\delta_{m}\rangle=\left\\{\begin{array}[]{ll}1&\mbox{if
}n=m,\\\ 0&\mbox{if }n\neq m.\end{array}\right.$

If $n=m$, then $\langle\delta_{n},\delta_{m}\rangle=\|\delta_{n}\|=1$.

Suppose that $n\neq m$. For $i,j\in\\{-1,1\\}$ let $C^{i}_{j}=\\{s\in
C:s_{n}=i,s_{m}=j\\}$ and note that if $x\in C^{i}_{j}$, then

$\displaystyle\delta_{n}(x)\delta_{m}(x)=ij.$

Since $\lambda(C_{j}^{i})=1/4$ for $i,j\in\\{-1,1\\}$, we get

$\displaystyle\langle\delta_{n},\delta_{m}\rangle=\int_{C}\delta_{n}\delta_{m}d\lambda=\sum_{i,j\in\\{-1,1\\}}\int_{C^{i}_{j}}\delta_{n}\delta_{m}d\lambda=\sum_{i,j\in\\{-1,1\\}}ij\lambda(C^{i}_{j})=0.$

∎

######  Lemma 4.11.

Let $t\in\mathbb{N}$. Let $\eta\in(0,1/2^{t+10})$. Let $n_{0}$ be large enough
so that for all $n>n_{0}$ there exists $k\in\mathbb{N}$ that

$\frac{\eta}{2}<\frac{k}{2^{n}}<\eta\ \text{and}\
\frac{n^{3}}{2^{n-1}}\leqslant\eta.$

Let $n>n_{0}$ and $k$ satisfy the above inequality. Then for all

  * •

$Q\in\mathbb{A}_{n}$ such that $\lambda(Q)<\eta$,

  * •

$F\in\mathbb{A}_{n}$ for which there exists $s\in\\{-1,1\\}^{t}$ such that
$\frac{\lambda(F\cap\langle s\rangle)}{\lambda(\langle s\rangle)}\geqslant
0.95,$

  * •

$Z\in\mathbb{A}_{n}$ such that $\lambda(Z)<\frac{\eta^{2}}{64}$ and
$Z\subseteq F$,

there exists $M\subseteq F\setminus(Q\cup Z)$ such that

  1. (a)

$M\in\mathbb{A}_{n}$,

  2. (b)

$\lambda(M)=\frac{k}{2^{n}}$,

  3. (c)

$M\cup Z$ is $(t,\eta)$-semibalanced.

${\color[rgb]{0.96,0.65,0.14}\definecolor[named]{pgfstrokecolor}{rgb}{0.96,0.65,0.14}\langle
s\rangle}$$M$$W$$Z$${\color[rgb]{0.55,0.34,0.16}\definecolor[named]{pgfstrokecolor}{rgb}{0.55,0.34,0.16}Q}$

In the figure above, the lines indicating the position of the triangles
forming Z and M at each level go to the left as many times as to the right,
which means that for every $r\in\mathbb{N}$ we have $\varphi_{r}(M\cup Z)=0$
(and so $M\cup Z$ is $(t,\eta)$-semibalanced).

###### Proof.

Let $F^{\prime}=F\setminus(Q\cup Z)$ and fix $s\in\\{-1,1\\}^{t}$ such that

$\frac{\lambda(F\cap\langle s\rangle)}{\lambda(\langle s\rangle)}\geqslant
0.95.$

In particular,

$\frac{\lambda(F^{\prime}\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}\geqslant 0.9.$

Put $\mathcal{M}=\left\\{M^{\prime}\in\mathbb{A}_{n}:M^{\prime}\subseteq
F^{\prime},\lambda(M^{\prime})=\frac{k}{2^{n}}\right\\}$. Since for
$M^{\prime}\in\mathcal{M}$ we have $M^{\prime}\cap Z=\varnothing$. It follows
that $\varphi_{m}(M^{\prime}\cup Z)=\varphi_{m}(M^{\prime})+\varphi_{m}(Z)$.

Define

$S(M^{\prime})=\sum_{m=t+1}^{n}(\varphi_{m}(M^{\prime}\cup Z))^{2}.$

Choose a set $M\in\mathcal{M}$ such that

$S(M)=\min\\{S(M^{\prime}):M^{\prime}\in\mathcal{M}\\}.$

We will show, that $M\cup Z$ is $(t,\eta)$-semibalanced. Namely, we will show,
that $S(M)<\frac{\eta^{2}}{n^{2}}$. This implies that for all $n\geqslant m>t$
we have

$|\varphi_{m}(M\cup
Z)|\leqslant\sqrt{S(M)}<\frac{\eta}{n}\leqslant\frac{\eta}{m},$

while for $m>n$ we have $|\varphi_{m}(M\cup Z)|=0$ (because
$Z,M\in\mathbb{A}_{n}$). Thus, $M\cup Z$ is $(t,\eta)$-semibalanced.
We need to show that

(1) $S(M)<\frac{k}{2^{n}}\frac{n}{2^{n-1}}.$

Indeed, if the above inequality holds, then

$S(M)<\frac{k}{2^{n}}\frac{n}{2^{n-1}}<\eta\frac{n}{2^{n-1}}=\frac{\eta}{n^{2}}\frac{n^{3}}{2^{n-1}}\leqslant\frac{\eta^{2}}{n^{2}}.$

The inequality (1) may be written as:

$\displaystyle S(M)$ $\displaystyle=$
$\displaystyle\sum_{m=t+1}^{n}\varphi_{m}(Z)\varphi_{m}(M\cup
Z)+\sum_{m=t+1}^{n}\varphi_{m}(M)\varphi_{m}(M\cup Z)<$ $\displaystyle<$
$\displaystyle\frac{k}{2^{n}}\frac{\sqrt{S(M)}}{4}+\frac{k}{2^{n}}\left(\frac{n}{2^{n-1}}-\frac{\sqrt{S(M)}}{4}\right).$

To prove it we will split it into two inequalities:

(2) $\sum_{m=t+1}^{n}\varphi_{m}(M)\varphi_{m}(M\cup
Z)\leqslant\frac{k}{2^{n}}\left(\frac{n}{2^{n-1}}-\frac{\sqrt{S(M)}}{4}\right)$

and

(3) $\sum_{m=t+1}^{n}\varphi_{m}(Z)\varphi_{m}(M\cup
Z)<\frac{k}{2^{n}}\frac{\sqrt{S(M)}}{4}.$

We will prove the inequalities (2) and (3) with the help of four claims. The
first three are necessary to show the inequality (2) while the last one will
prove inequality (3).

In Claim 1 we make use of the minimality of $S(M)$ analyzing the situation
when we change $M$ by one atom of $\mathbb{A}_{m}$.

######  Claim 1.

For any $x,y\in\\{-1,1\\}^{n}$ such that $\langle x\rangle\subseteq M$ and
$\langle y\rangle\subseteq F^{\prime}\setminus M$

(4) $\sum_{m=t+1}^{n}x_{m}\varphi_{m}(M\cup
Z)\leqslant\frac{n}{2^{n-1}}+\sum_{m=t+1}^{n}y_{m}\varphi_{m}(M\cup Z),$

where $x_{m}$ and $y_{m}$ are the $m$-th terms of the sequences $x$ and $y$
respectively.

###### Proof of the claim.

Let $M^{\prime}=(M\setminus\langle x\rangle)\cup\langle y\rangle$. We have

$\displaystyle\varphi_{m}(M^{\prime})$ $\displaystyle=$
$\displaystyle\int\limits_{M^{\prime}}\delta_{m}d\lambda\ \ \ \
=\int\limits_{M}\delta_{m}d\lambda-\int\limits_{\langle
x\rangle}\delta_{m}d\lambda+\int\limits_{\langle y\rangle}\delta_{m}d\lambda$
$\displaystyle=$ $\displaystyle\varphi_{m}(M)+\frac{1}{2^{n}}(y_{m}-x_{m}).$

Since $M$ minimizes $S$, we have $S(M^{\prime})-S(M)\geqslant 0$. Then

$\displaystyle S(M^{\prime})-S(M)$ $\displaystyle=$
$\displaystyle\sum_{m=t+1}^{n}\left((\varphi_{m}(Z)+\varphi_{m}(M^{\prime}))^{2}-(\varphi_{m}(Z)+\varphi_{m}(M))^{2}\right)$
$\displaystyle=$
$\displaystyle\sum_{m=t+1}^{n}\left(\varphi_{m}(M^{\prime})^{2}-\varphi_{m}(M)^{2}+2\varphi_{m}(Z)(\varphi_{m}(M^{\prime})-\varphi_{m}(M))\right)$
$\displaystyle=$
$\displaystyle\sum_{m=t+1}^{n}(\varphi_{m}(M^{\prime})-\varphi_{m}(M))(\varphi_{m}(M^{\prime})+\varphi_{m}(M)+2\varphi_{m}(Z))$
$\displaystyle=$
$\displaystyle\sum_{m=t+1}^{n}\frac{1}{2^{n}}(y_{m}-x_{m})\left(\frac{1}{2^{n}}(y_{m}-x_{m})+2\varphi_{m}(M)+2\varphi_{m}(Z)\right).$

Multiplying the above by $2^{n-1}$ and using the fact that
$(y_{m}-x_{m})^{2}\in\\{0,4\\}$ we get

$\displaystyle 0$ $\displaystyle\leqslant$
$\displaystyle\sum_{m=t+1}^{n}(y_{m}-x_{m})\left(\frac{1}{2^{n+1}}(y_{m}-x_{m})+\varphi_{m}(M)+\varphi_{m}(Z)\right)$
$\displaystyle\leqslant$
$\displaystyle\sum_{m=t+1}^{n}\frac{4}{2^{n+1}}+(y_{m}-x_{m})(\varphi_{m}(M)+\varphi_{m}(Z))$
$\displaystyle\leqslant$
$\displaystyle\frac{n}{2^{n-1}}+\sum_{m=t+1}^{n}(y_{m}-x_{m})\varphi_{m}(M\cup
Z).$

∎

Let $T\colon C\to C$ be a function that swaps the sign of the coordinates from
$t+1$ to $n$ given by the formula:

$T(y)(i)=\begin{cases}-y(i)&\text{if}\ i\in(t,n],\\\ y(i)&\text{if}\
i\notin(t,n].\end{cases}$

Note that for $A\in\operatorname{Bor}(C)$ we have $\varphi_{m}(A\cup T[A])=0$,
where $m\in\\{t+1,\dots,n\\}$.

######  Claim 2.

$\frac{\lambda\big{(}((F^{\prime}\cap\langle s\rangle)\cap
T[F^{\prime}\cap\langle s\rangle])\setminus(M\cup
T[M])\big{)}}{\lambda(\langle s\rangle)}\geqslant 0.75.$

###### Proof of the claim.

Since

$\lambda(M)=\frac{k}{2^{n}}<\eta<\frac{0.05}{2^{t+1}},$

we have

$\lambda(M\cup T[M])\leqslant\frac{0.05}{2^{t}}.$

Since

$\frac{\lambda(F^{\prime}\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}\geqslant 0.9,$

we have

$\frac{\lambda((F^{\prime}\cap\langle s\rangle)\cap T[F^{\prime}\cap\langle
s\rangle])}{\lambda(\langle s\rangle)}\geqslant 0.8.$

Therefore

$\displaystyle\frac{\lambda(((F^{\prime}\cap\langle s\rangle)\cap
T[F^{\prime}\cap\langle s\rangle])\setminus(M\cup T[M]))}{\lambda(\langle
s\rangle)}$ $\displaystyle\geqslant$ $\displaystyle 0.8-0.05=0.75.$

∎

It is clear that if $y\in\langle s\rangle$ then $T(y)\in\langle s\rangle$. In
the proof of the next claim we will use an obvious observation that if
$\frac{\lambda(A\cap\langle s\rangle)}{\lambda(\langle s\rangle)}\geqslant
0.75$ and $\frac{\lambda(B\cap\langle s\rangle)}{\lambda(\langle
s\rangle)}>0.25$ then there exists $y\in A\cap B\cap\langle s\rangle$.

######  Claim 3.

There exists $y^{M}\in\\{-1,1\\}^{n}$ such that $\langle
y^{M}\rangle\subseteq(F^{\prime}\setminus M)\cap T[F^{\prime}\setminus M]$ and

$\left|\sum_{m=t+1}^{n}y_{m}^{M}\varphi_{m}(M\cup
Z)\right|\geqslant\frac{\sqrt{S(M)}}{4}.$

###### Proof of the claim.

Recall that $S(M)=\sum_{m=t+1}^{n}\varphi_{m}(M\cup Z)^{2}$. By Lemma 4.9 for
$\xi=1/16$ and

$d_{m}=\begin{cases}\varphi_{m}(M\cup Z)&\text{if}\ m\in(t,n],\\\ 0&\text{if}\
m\in[0,t],\end{cases}$

we have

$\lambda_{n}\left(\left\\{y\in\\{-1,1\\}^{n}:\left|\sum_{m=t+1}^{n}y_{m}\varphi_{m}(M\cup
Z)\right|^{2}\geqslant\frac{1}{16}S(M)\right\\}\right)\geqslant\frac{1}{3}\left(\frac{15}{16}\right)^{2}.$

Hence

$\lambda_{n}\left(\left\\{y\in\\{-1,1\\}^{n}:\left|\sum_{m=t+1}^{n}y_{m}\varphi_{m}(M\cup
Z)\right|\geqslant\frac{\sqrt{S(M)}}{4}\right\\}\right)>0.25.$

Now Claim 2 implies the existence of the desired $y^{M}$. ∎

Note that $(F^{\prime}\setminus M)\cap T[F^{\prime}\setminus
M]=T[(F^{\prime}\setminus M)\cap T[F^{\prime}\setminus M]]$. So since $\langle
y^{M}\rangle\subseteq(F^{\prime}\setminus M)\cap T[F^{\prime}\setminus M]$ we
also have $T[\langle y^{M}\rangle]\subseteq(F^{\prime}\setminus M)\cap
T[F^{\prime}\setminus M]$. Moreover
$\sum_{m=t+1}^{n}y^{M}_{m}\varphi_{m}(M\cup
Z)=-\sum_{m=t+1}^{n}T(y^{M}_{m})\varphi_{m}(M\cup Z)$. Thus, by replacing
$y^{M}$ with $T(y^{M})$ if needed Claim 3 implies that

(5) $\sum_{m=t+1}^{n}y^{M}_{m}\varphi_{m}(M\cup Z)<\ -\frac{\sqrt{S(M)}}{4}.$

From inequalities (4) and (5) for $x$ as in Claim 1 we get

(6) $\sum_{m=t+1}^{n}x_{m}\varphi_{m}(M\cup
Z)\leqslant\frac{n}{2^{n-1}}+\sum_{m=t+1}^{n}y^{M}_{m}\varphi_{m}(M\cup
Z)\leqslant\frac{n}{2^{n-1}}-\frac{\sqrt{S(M)}}{4}.$

Let $\\{x^{(i)}:i\in\\{1,\dots k\\}\\}$ be an enumeration of all
$x\in\\{-1,1\\}^{n}$ such that $\langle x\rangle\subseteq M$.

$M=\bigcup_{\begin{subarray}{c}x\in\\{-1,1\\}^{n}\\\ \langle x\rangle\subseteq
M\end{subarray}}\langle x\rangle=\bigcup_{i=1}^{k}\langle x^{(i)}\rangle$

Since for all $x\in\\{-1,1\\}^{n}$ and $m\leqslant n$ we have
$x_{m}=2^{n}\int_{\langle x\rangle}\delta_{m}d\lambda$ we conclude that

(7)
$\sum_{i=1}^{k}x_{m}^{(i)}=2^{n}\int_{M}\delta_{m}d\lambda=2^{n}\varphi_{m}(M).$

By (7) and (6) we obtain

(8) $\begin{split}\sum_{m=t+1}^{n}2^{n}\varphi_{m}(M)\varphi_{m}(M\cup
Z)&=\sum_{m=t+1}^{n}\sum_{i=1}^{k}x_{m}^{(i)}\varphi_{m}(M\cup Z)=\\\
=\sum_{i=1}^{k}\sum_{m=t+1}^{n}x_{m}^{(i)}\varphi_{m}(M\cup Z)&\leqslant
k\left(\frac{n}{2^{n-1}}-\frac{\sqrt{S(M)}}{4}\right).\end{split}$

Multiplying both sides of the inequality (8) by $2^{-n}$ we get

$\sum_{m=t+1}^{n}\varphi_{m}(M)\varphi_{m}(M\cup
Z)\leqslant\frac{k}{2^{n}}\left(\frac{n}{2^{n-1}}-\frac{\sqrt{S(M)}}{4}\right),$

that is the inequality (2).

Now we will prove (3).

######  Claim 4.

$\sum_{m=t+1}^{n}\varphi_{m}(Z)\varphi_{m}(M\cup
Z)<\frac{k}{2^{n}}\frac{\sqrt{S(M)}}{4}.$

###### Proof of the claim.

By the Cauchy-Schwarz inequality we have

$\displaystyle\sum_{m=t+1}^{n}\varphi_{m}(Z)\varphi_{m}(M\cup Z)$
$\displaystyle\leqslant$
$\displaystyle\sqrt{\sum_{m=t+1}^{n}(\varphi_{m}(Z))^{2}}\sqrt{\sum_{m=t+1}^{n}\varphi_{m}(M\cup
Z)^{2}}=$ $\displaystyle=$
$\displaystyle\sqrt{\sum_{m=t+1}^{n}(\varphi_{m}(Z))^{2}}\sqrt{S(M)}.$

By Lemma 4.10 the sequence $(\delta_{n})_{n\in\mathbb{N}}$ is orthonormal in
the Hilbert space $\mathcal{L}_{2}(C)$, so by the Bessel inequality we get

$\displaystyle\sum_{m=t+1}^{n}(\varphi_{m}(Z))^{2}$ $\displaystyle=$
$\displaystyle\sum_{m=t+1}^{n}\left(\int_{Z}\delta_{m}d\lambda\right)^{2}=\sum_{m=t+1}^{n}\left(\int_{C}\chi_{Z}\delta_{m}d\lambda\right)^{2}=$
$\displaystyle=$
$\displaystyle\sum_{m=t+1}^{n}(\langle\chi_{Z},\delta_{m}\rangle)^{2}\leqslant\|\chi_{Z}\|_{2}^{2}=\lambda({Z}).$

Since $\lambda(Z)<\eta^{2}/64$, we have $\sqrt{\lambda(Z)}<\eta/8<k/2^{n+2}$
and so

$\sum_{m=t+1}^{n}\varphi_{m}(Z)\varphi_{m}(M\cup
Z)\leqslant\sqrt{\lambda(Z)}\sqrt{S(M)}<\frac{k}{2^{n}}\frac{\sqrt{S(M)}}{4}.$

∎

Adding inequalities (2) and (3) side by side we get the following estimate

$\displaystyle\sum_{m=t+1}^{n}\varphi_{m}(M\cup Z)^{2}$ $\displaystyle<$
$\displaystyle\frac{k}{2^{n}}\frac{\sqrt{S(M)}}{4}+\frac{k}{2^{n}}\left(\frac{n}{2^{n-1}}-\frac{\sqrt{S(M)}}{4}\right)=\frac{k}{2^{n}}\frac{n}{2^{n-1}},$

which shows (1) and finishes the proof. ∎

######  Lemma 4.12.

Let $\mathbb{B}^{*}\subseteq\mathbb{B}$ be balanced Boolean subalgebras of
$\operatorname{Bor}(C)$ containing $\operatorname{Clop}(C)$ and let
$\mathbb{F}$ be a finite subalgebra of $\mathbb{B}$. Let
$P\in\mathbb{F}\cap\mathbb{B}^{*}$. Let $t\in\mathbb{N},\delta>0$. Suppose
that for every $A\in\mathbb{F}$ such that $\lambda(A)>0$ there is
$s_{A}\in\\{-1,1\\}^{t}$ such that

$\begin{gathered}\frac{\lambda(A\cap\langle s_{A}\rangle)}{\lambda(\langle
s_{A}\rangle)}\geqslant 0.99.\end{gathered}$

Then there is $\theta>0$ such that for any $L,Q\in\mathbb{B}^{*}$, if
$\max\\{\lambda(L),\lambda(Q)\\}<\theta$ and $L\cap P=\varnothing$, then there
is $M\in\mathbb{B}^{*}$ such that

  1. (1)

$M\cap(P\cup Q)=\varnothing$,

  2. (2)

$\lambda(M)<\delta$,

  3. (3)

$\forall F\in\mathbb{F}\ (M\cup L)\cap F$ is $(t,\delta)$-semibalanced,

  4. (4)

$\forall F\in\mathbb{F}\ F\backslash(M\cup L)$ is $(t,\delta)$-semibalanced.

###### Proof.

Let $\eta<\min\\{\delta/(4|\mathbb{F}|),1/2^{t+10}\\},\theta<\eta^{2}/64$ and
let $n_{0}$ be large enough so that for every $n>n_{0}$

  1. (5)

there is $k\in\mathbb{N}$ such that $\frac{\eta}{2}<\frac{k}{2^{n}}<\eta$,

  2. (6)

$\frac{n^{3}}{2^{n-1}}\leqslant\eta$,

  3. (7)

$\frac{\eta}{n}<\frac{\eta^{2}}{64}-\theta.$

Fix $Q$ and $L$ satisfying the hypothesis of the lemma. Denote by
$\mathbb{H}_{0}$ the Boolean algebra generated by $\mathbb{F}\cup\\{Q,L\\}$.
Since $\mathbb{B}$ is balanced, there is $n>n_{0}$ such that $\mathbb{H}_{0}$
is $(n,\eta)$-balanced. Let $\mathbb{H}$ be the Boolean algebra generated by
$\mathbb{H}_{0}\cup\mathbb{A}_{n}$. By Lemma 4.4

$\mathbb{H}\ \text{is}\ (n,\eta)\text{-balanced}.$

Let $h_{n}$ be defined as in Lemma 4.5. By the same lemma, for
$F\in\mathbb{F}$ we have

$\frac{\lambda(h_{n}(F)\cap\langle s_{F}\rangle)}{\lambda(\langle
s_{F}\rangle)}\geqslant\frac{\lambda(F\cap\langle
s_{F}\rangle)}{\lambda(\langle s_{F}\rangle)}-\frac{\lambda(F\triangle
h_{n}(F))}{\lambda(\langle s_{F}\rangle)}\geqslant
0.99-\frac{2^{t}\eta}{n}\geqslant 0.95.$

By Lemma 4.5 and (7)

$\lambda(h_{n}(L)\cap
h_{n}(F))\leqslant\lambda(h_{n}(L))\leqslant\lambda(L)+\frac{\eta}{n}<\theta+\frac{\eta^{2}}{64}-\theta=\frac{\eta^{2}}{64}.$

By Lemma 4.11 (applied to $Z=h_{n}(L)\cap h_{n}(E))$ for every
$E\in\operatorname{at}(\mathbb{F})$ there is $M_{E}\in\mathbb{A}_{n}$ such
that

  1. (a)

$M_{E}\subseteq h_{n}(E)\backslash h_{n}(Q)$,

  2. (b)

$\lambda(M_{E})<\eta<\delta/|\mathbb{F}|$,

  3. (c)

$M_{E}\cup(h_{n}(L)\cap h_{n}(E))$ is
$(t,\delta/(4|\mathbb{F}|))$-semibalanced.

Put

$M_{0}=\bigcup_{E\in\operatorname{at}(\mathbb{F})}M_{E}\backslash
Q,M=M_{0}\backslash P.$

Then $M\cap(P\cup Q)=\varnothing$ and $\lambda(M)<\delta$. To show (3) and (4)
fix $F\in\mathbb{F}$ and $r>t$. Consider 2 cases.
Case 1. $r\leqslant n$.
For every $E\in\operatorname{at}(\mathbb{F})$ we have
$M_{E}\in\mathbb{A}_{n}$, so $h_{n}(M_{E})=M_{E}$. From (a) we get

$\displaystyle h_{n}(M_{0}\cap h_{n}(F))$ $\displaystyle=$ $\displaystyle
h_{n}(M_{0}\cap
F)=h_{n}\left(\bigcup_{E\in\operatorname{at}(\mathbb{F})}(M_{E}\backslash
Q)\cap F\right)=$ $\displaystyle=$
$\displaystyle\bigcup_{\begin{subarray}{c}E\in\operatorname{at}(\mathbb{F})\\\
E\subseteq F\end{subarray}}(h_{n}(M_{E})\cap h_{n}(E))\backslash h_{n}(Q)=$
$\displaystyle=$
$\displaystyle\bigcup_{\begin{subarray}{c}E\in\operatorname{at}(\mathbb{F})\\\
E\subseteq F\end{subarray}}((M_{E}\cap h_{n}(E))\backslash h_{n}(Q)=$
$\displaystyle=$
$\displaystyle\bigcup_{\begin{subarray}{c}E\in\operatorname{at}(\mathbb{F})\\\
E\subseteq F\end{subarray}}(M_{E}\cap h_{n}(E))\backslash
h_{n}(Q)=\bigcup_{\begin{subarray}{c}E\in\operatorname{at}(\mathbb{F})\\\
E\subseteq F\end{subarray}}M_{E}.$

By Lemma 4.5 for any $A\in\mathbb{H}$ we have

$|\varphi_{r}(A)|\leqslant|\varphi_{r}(h_{n}(A))|+\lambda(A\triangle
h_{n}(A))<|\varphi_{r}(h_{n}(A))|+\frac{\eta}{n}.$

Putting $A=(M_{0}\cup L)\cap F$ and using (c) we get

$\displaystyle|\varphi_{r}((M_{0}\cup L)\cap F)|$ $\displaystyle\leqslant$
$\displaystyle|\varphi_{r}(h_{n}((M_{0}\cup L)\cap F))|+\frac{\eta}{n}<$
$\displaystyle<$ $\displaystyle|\varphi_{r}((h_{n}(M_{0})\cup h_{n}(L))\cap
h_{n}(F))|+\frac{\delta}{4n}=$ $\displaystyle=$
$\displaystyle\Bigg{|}\varphi_{r}\Bigg{(}\bigcup_{\begin{subarray}{c}E\in\operatorname{at}(\mathbb{F})\\\
E\subseteq F\end{subarray}}M_{E}\cup(h_{n}(L)\cap
h_{n}(E))\Bigg{)}\Bigg{|}+\frac{\delta}{4n}\leqslant$ $\displaystyle\leqslant$
$\displaystyle\sum_{\begin{subarray}{c}E\in\operatorname{at}(\mathbb{F})\\\
E\subseteq F\end{subarray}}|\varphi_{r}(M_{E}\cup(h_{n}(L)\cap
h_{n}(E)))|+\frac{\delta}{4n}\leqslant$ $\displaystyle\leqslant$
$\displaystyle|\mathbb{F}|\frac{\delta}{4r|\mathbb{F}|}+\frac{\delta}{4n}\leqslant\frac{\delta}{2r}.$

Case 2. $r>n$.
In this case, since $\mathbb{H}$ is $(n,\delta/2)$-balanced, we have

$|\varphi_{r}((M_{0}\cup L)\cap
F)|\leqslant\sum_{s\in\\{-1,1\\}^{n}}|\varphi_{r}((M_{0}\cup L)\cap
F\cap\langle s\rangle)|<2^{n}\lambda(\langle
s\rangle)\frac{\delta}{2r}=\frac{\delta}{2r}.$

Hence

(8) $\displaystyle(M_{0}\cup L)\cap F\ \text{is}\
\left(t,\frac{\delta}{2}\right)\text{-semibalanced for}\ F\in\mathbb{F}.$

Since $L\cap P=\varnothing$ and $M=M_{0}\backslash P$, we get

$(M\cup L)\cap F=(M_{0}\cup L)\cap(F\backslash P).$

Since $P\in\mathbb{F}$ we get that if $F\in\mathbb{F}$, then $F\backslash
P\in\mathbb{F}$, so by (8) applied to $F\backslash P$ we get that

(9) $\displaystyle(M\cup L)\cap F\ \text{is}\
\left(t,\frac{\delta}{2}\right)\text{-semibalanced for}\ F\in\mathbb{F},$

which implies (3).

For (4) note that $C\backslash(F\backslash(M\cup L))=(C\backslash
F)\cup((M\cup L)\cap F)$. Since $C\backslash F$ and $(M\cup L)\cap F$ are
disjoint we have

$|\varphi_{r}(C\backslash(F\backslash(M\cup
L)))|\leqslant|\varphi_{r}(C\backslash F)|+|\varphi_{r}((M\cup L)\cap F)|.$

Since $\mathbb{H}$ is $(t,\delta/2)$-balanced and $C\backslash
F\in\mathbb{F}\subseteq\mathbb{H}$ we have $|\varphi_{r}(C\backslash
F)|<\delta/(2r)$ and from (9) we get that $|\varphi_{r}((M\cup L)\cap
F)|<\delta/(2r)$. Hence

$|\varphi_{r}(C\backslash(F\backslash(M\cup L)))|<\frac{\delta}{r},$

so $C\backslash(F\backslash(M\cup L))$ is $(t,\delta)$-semibalanced. By Lemma
4.2 the set $F\backslash(M\cup L)$ is $(t,\delta)$-semibalanced, which shows
(4) and finishes the proof. ∎

###### Proof of Proposition 4.8.

Denote by $\mathbb{F}$ the subalgebra of $\mathbb{B}$ generated by

$\\{G,P\\}\cup\bigcup_{n\leqslant k}\mathbb{B}_{n}\cup\mathbb{A}_{m_{k}}.$

Put

(4)
$\displaystyle\varepsilon=\min\left\\{\frac{1}{100}\inf\\{\lambda(A):A\in\mathbb{F},\lambda(A)>0\\},2^{-m_{k}-k-1}\right\\}.$

Since $\mathbb{B}$ is balanced, there is $t\in\mathbb{N},t>m_{k}$ such that

(5) $\displaystyle\mathbb{F}\ \text{is}\ (t,\varepsilon)\text{-balanced}.$

By Lemma 4.3 for every $A\in\mathbb{F}$ there is $s_{A}\in\\{-1,1\\}^{t}$ such
that

$\displaystyle\frac{\lambda(A\cap\langle s_{A}\rangle)}{\lambda(\langle
s_{A}\rangle)}\geqslant 0.99.$

By Lemma 4.6 and the assumption (B) of the proposition there is $\varrho>0$
such that

(6) $\begin{gathered}\forall n\leqslant k\ \forall
A\in\mathcal{F}(\mathbb{B}_{n},G)\ \forall B\in\operatorname{Bor}(C)\\\
\lambda(B)<\varrho\implies A\cup B,A\backslash B\ \text{are}\
(m_{n},t,2^{-n})\text{-balanced}.\end{gathered}$

By Lemma 4.12 (applied to
$\delta=\min\left\\{\eta,\varrho/2,2^{-m_{k}-k-1}\right\\}$), there is
$0<\theta<\varrho/2$ such that for any $L,Q\in\mathbb{B}^{*}$, if
$\max\\{\lambda(L),\lambda(Q)\\}<\theta$ and $L\cap P=\varnothing$, then there
is $M\in\mathbb{B}^{*}$ satisfying

  * (1’)

$M\cap(P\cup Q)=\varnothing$,

  * (2’)

$\lambda(M)<\min\left\\{\eta,\varrho/2\right\\}$,

  * (3’)

$\forall F\in\mathbb{F}\ (M\cup L)\cap F$ is
$(t,2^{-m_{k}-k-1})$-semibalanced,

  * (4’)

$\forall F\in\mathbb{F}\ F\backslash(M\cup L)$ is
$(t,2^{-m_{k}-k-1})$-semibalanced.

In particular, the conditions (1) and (2) of the proposition are satisfied.

In order to show that such $M$ satisfies (3), fix $L,Q\in\mathbb{B}^{*}$ such
that $L\cap P=\varnothing$ and $\max\\{\lambda(L),\lambda(Q)\\}<\theta$. Then
we have

$\lambda(L\cup M)\leqslant\lambda(L)+\lambda(M)<\varrho/2+\varrho/2=\varrho,$

so by (6) for every $n\leqslant k$ the family

(7) $\displaystyle\mathcal{F}(\mathbb{B}_{n},G\cup L\cup M)\ \text{is}\
(m_{n},t,2^{-n})\text{-balanced}.$

Since $\mathbb{A}_{m_{n}},\mathbb{B}_{n}\subseteq\mathbb{F}$ for $n\leqslant
k$ and $G\in\mathbb{F}$, for every $A\in\mathbb{B}_{n}$ and
$s\in\\{-1,1\\}^{m_{n}}$ we have $\langle s\rangle\cap A\cap G,A\backslash
G\in\mathbb{F}$. Hence by (4), (5) and Lemma 3.5

(8) $\displaystyle\langle s\rangle\cap A\cap G,\langle s\rangle\cap
A\backslash G\ \text{are}\ (t,2^{-m_{n}-n-1})\text{-semibalanced.}$

Since $\langle s\rangle\cap A\in\mathbb{F}$, by (3’) and (4’)

(9) $\displaystyle\langle s\rangle\cap A\cap(L\cup M),\langle s\rangle\cap
A\backslash(L\cup M)\ \text{are}\ (t,2^{-m_{n}-n-1})\text{-semibalanced.}$

Since $L\cap P=M\cap P=\varnothing$ and $G\subseteq P$, the sets $L\cup M$ and
$G$ are disjoint, so by (8) and (9) for
$A\in\mathbb{B}_{n},s\in\\{-1,1\\}^{n}$

$\displaystyle\langle s\rangle\cap A\cap(G\cup L\cup M),\langle s\rangle\cap
A\backslash(G\cup L\cup M)\ \text{are}\ (t,2^{-m_{n}-n})\text{-semibalanced}.$

By the above, (7) and Lemma 3.5

$\mathcal{F}(\mathbb{B}_{n},G\cup L\cup M)\ \text{is}\
(m_{n},2^{-n})\text{-balanced},$

which shows (3) and completes the proof. ∎

##  5\. Extensions of countable balanced Boolean algebras

In this section, we will show how to enlarge a given countable balanced
Boolean algebra $\mathbb{B}$ to a balanced Boolean algebra $\mathbb{B}^{*}$,
so that $(\mathbb{B},\mathbb{B}^{*},\nu)$ satisfies the property
$(\mathcal{G}^{*})$, where $\nu$ is a normal sequence of measures on
$\mathbb{B}$. We will also show how to deal with finitely many measures
simultaneously, which will be important in the forcing construction.

######  Lemma 5.1.

_(Folklore)_ Suppose $K$ is a compact Hausdorff space with an open basis
$\mathcal{B}$ that is closed under finite unions. Let $\widetilde{\nu}\in
M(K)$ and let $\mathcal{M}\subseteq M(K)$ be a finite set of measures such
that $\widetilde{\nu}\bot\widetilde{\mu}$ for every
$\widetilde{\mu}\in\mathcal{M}$. Then for every $\varepsilon>0$ there is
$X\in\mathcal{B}$ such that $|\widetilde{\nu}|(X)<\varepsilon$ and for every
$\widetilde{\mu}\in\mathcal{M}$ we have $|\widetilde{\mu}|(K\backslash
X)<\varepsilon$.

###### Proof.

First, we will show that the lemma holds when
$\mathcal{M}=\\{\widetilde{\mu}\\}$. Since
$\widetilde{\nu}\bot\widetilde{\mu}$ we have
$|\widetilde{\nu}|\bot|\widetilde{\mu}|$, so there exists a Borel support $A$
of $|\widetilde{\mu}|$, such that $K\setminus A$ is a Borel support of
$|\widetilde{\nu}|$. By the regularity there is closed $B\subseteq A$ such
that $|\widetilde{\mu}|(B)>|\widetilde{\mu}|(K)-\varepsilon$. We can find a
closed set $D\subseteq K\setminus B$ such that
$|\widetilde{\nu}|(D)>|\widetilde{\nu}|(K)-\varepsilon$. Note that
$|\widetilde{\mu}|(D)\leqslant\varepsilon$. Since $U=K\setminus D$ is an open
superset of $B$, there are finitely many sets $B_{1},\dots
B_{k}\in\mathcal{B},B_{i}\subseteq U$ for $i\leqslant k$, which cover $B$. Let
$X=\bigcup_{i\leqslant k}B_{i}$. Note that
$|\widetilde{\mu}|(X)\geqslant|\widetilde{\mu}|(K)-\varepsilon$, and
$X\subseteq K\setminus D$, so $|\widetilde{\nu}|(X)<\varepsilon$ and
$|\widetilde{\mu}|(K\backslash X)<\varepsilon$.

When $\mathcal{M}=\\{\widetilde{\mu}_{1},\dots\widetilde{\mu}_{n}\\}$, by the
first part of the proof for each pair $(\widetilde{\nu},\widetilde{\mu}_{i})$
where $i\leqslant n$ we can find $X_{i}\in\mathcal{B}$ such that
$|\widetilde{\nu}|(X_{i})<\frac{\varepsilon}{n}$ and
$|\widetilde{\mu}_{i}|(K\backslash X_{i})<\frac{\varepsilon}{n}$. Then for
$X=\bigcup_{i\in{1,\dots n}}X_{i}$ we have $|\widetilde{\nu}|(X)<\varepsilon$
and $|\widetilde{\mu}_{i}|(K\backslash X)<\frac{\varepsilon}{n}<\varepsilon$.
∎

######  Lemma 5.2.

_(Pełczyński)_[24, Lemma 5.3] Let $(\widetilde{\nu}_{n})_{n\in\mathbb{N}}$ be
a bounded sequence of Radon measures on a compact space $K$. Suppose there are
pairwise disjoint Borel sets $(E_{n})_{n\in\mathbb{N}}$ and $c>0$ such that
$\widetilde{\nu}_{n}(E_{n})\geqslant c$ for every $n\in\mathbb{N}$. Then for
every $\delta>0$ there is a subsequence
$(\widetilde{\nu}_{n_{k}})_{n\in\mathbb{N}}$ and a sequence of pairwise
disjoint open sets $(U_{k})_{k\in\mathbb{N}}$ such that
$\widetilde{\nu}_{n_{k}}(U_{k})\geqslant c-\delta$ for every $k\in\mathbb{N}$.

We will use the following application of the above lemma, in which $K$ is the
Stone space of a Boolean algebra.

######  Corollary 5.3.

Let $(\nu_{n})_{n\in\mathbb{N}}$ be a sequence of measures on $\mathbb{B}$ and
$(E_{n})_{n\in\mathbb{N}}$ be a sequence of disjoint Borel sets in
$\operatorname{St}(\mathbb{B})$. Let $P\in\mathbb{B}$ be such that
$E_{n}\cap[P]=\varnothing$ for every $n\in\mathbb{N}$. Let $c,\delta>0$. If
$|\widetilde{\nu}_{n}|(E_{n})\geqslant c$ for every $n\in\mathbb{N}$, then
there exist a subsequence $(\nu_{n_{k}})_{k\in\mathbb{N}}$ and a sequence
$(V_{k})_{k\in\mathbb{N}}$ of pairwise disjoint elements of $\mathbb{B}$ such
that $V_{k}\cap P=\varnothing$ and $|\nu_{n_{k}}|(V_{k})\geqslant c-\delta$
for all $k\in\mathbb{N}$.

The next lemma will let us build an antichain needed to satisfy the property
$(\mathcal{G})$.

######  Lemma 5.4.

Let $\nu=(\nu_{n})_{n\in\mathbb{N}}$ be a normal sequence of measures on a
Boolean algebra $\mathbb{B}\subseteq\operatorname{Bor}(C)$. Let $\mathcal{M}$
be a finite set of positive measures on $\mathbb{B}$ and assume that
$(|{\nu}_{n}|)_{n\in\mathbb{N}}$ has a subsequence pointwise convergent to a
measure $\mathcal{\nu_{\infty}}\in\mathcal{M}$. Let $d\in\mathbb{N}$ and
$\varepsilon>0$. Let $P\in\mathbb{B}$ be such that

$\mathcal{\nu_{\infty}}(P)<0.1.$

Then there are $H_{0},H_{1}\in\mathbb{B}$ and $a,b>d$ such that

  1. (1)

$H_{0},H_{1},P$ are pairwise disjoint,

  2. (2)

$\forall\mu\in\mathcal{M}\ \mu(H_{0}\cup H_{1})<\varepsilon$,

  3. (3)

$\lambda(H_{0}\cup H_{1})<\varepsilon$,

  4. (4)

$|\nu_{a}|(H_{0}),|\nu_{b}|(H_{1})\geqslant 0.9$.

###### Proof.

We may assume that $\lambda\upharpoonright\mathbb{B}\in\mathcal{M}$, so it is
enough to show (1), (2) and (4).

Since ${\nu}_{\infty}$ is the pointwise limit of a sequence of probability
measures, we have $\mathcal{\nu_{\infty}}(C)=1$, and so
$\mathcal{\nu_{\infty}}(C\backslash P)>0.9$. Hence there is $\delta>0$ such
that for infinitely many $n\in\mathbb{N}$ we have

$|\nu_{n}|(C\backslash P)>0.9+\delta.$

Since $(\widetilde{\nu}_{n})_{n\in\mathbb{N}}$ has pairwise disjoint Borel
supports, there are pairwise disjoint sets
$(E_{n})_{n\in\mathbb{N}}\subseteq\operatorname{Bor}(C)$ such that

$|\widetilde{\nu}_{n}|(E_{n}\backslash[P])>0.9+\delta$

for infinitely many $n\in\mathbb{N}$. By Corollary 5.3 there is an antichain
$(V_{l})_{l\in\mathbb{N}}\subseteq\mathbb{B}$ and a subsequence
$(\nu_{n_{l}})_{l\in\mathbb{N}}$ such that $V_{l}\cap P=\varnothing$ and
$|\nu_{n_{l}}|(V_{l})\geqslant 0.9$ for every $l\in\mathbb{N}$. Since
$(V_{l})_{l\in\mathbb{N}}$ is an antichain, we have for every
$\mu\in\mathcal{M}$

$\lim_{l\to\infty}\mu(V_{l})=0.$

In particular, if $l_{1},l_{2}$ are big enough, then we have
$\mu(V_{l_{1}}\cup V_{l_{2}})<\varepsilon$ for every $\mu\in\mathcal{M}$, so
it is enough to put $a=n_{l_{1}},b=n_{l_{2}},H_{0}=V_{l_{1}},H_{1}=V_{l_{2}}$
where $l_{1}\neq l_{2}$ are so big that
$\min\left\\{n_{l_{1}},n_{l_{2}}\right\\}>\max\\{d,n\\}$.

∎

In the next two lemmas we describe how to pick a sequence of sets, whose union
will be a witness for the property $(\mathcal{G}^{*})$ for a given sequence of
measures.

For the purpose of the construction under CH, in the following lemma it is
enough to take $\mathcal{M}=\\{\mathcal{\nu_{\infty}}\\}$. The case when
$\mathcal{M}$ consists of more than one measure will be used in Section 6.

######  Lemma 5.5.

Suppose we are given:

  1. (A)

natural numbers $k,d\in\mathbb{N},$

  2. (B)

subalgebras $\mathbb{B}^{*},\mathbb{B}\subseteq\operatorname{Bor}(C)$ and
finite subalgebras $\mathbb{B}_{n}\subseteq\mathbb{B}$ for $n\leqslant k+1$
such that

     * •

$\mathbb{B}$ is balanced,

     * •

$\operatorname{Clop}(C)\subseteq\mathbb{B}^{*}\subseteq\mathbb{B}$,

  3. (C)

a normal sequence of measures $(\nu_{n})_{n\in\mathbb{N}}$ on $\mathbb{B}^{*}$
and a finite set $\mathcal{M}$ of probability measures on $\mathbb{B}^{*}$
such that $(|{\nu}_{n}|)_{n\in\mathbb{N}}$ has a subsequence pointwise
convergent to a measure $\mathcal{\nu_{\infty}}\in\mathcal{M}$,

  4. (D)

a strictly increasing sequence of natural numbers $(m_{n})_{n\leqslant k}$ and
sets $\widehat{G},\widehat{H}\in\mathbb{B}^{*}$ such that

     * •

$\forall n\leqslant k\ \mathcal{F}(\mathbb{B}_{n},\widehat{G})\ \text{is}\
(m_{n},2^{-n})\text{-balanced},$

     * •

$\forall\mu\in\mathcal{M}\ \mu(\widehat{G}\cup\widehat{H})<0.1.$

Then there are
$a,b>d;m_{k+1}\in\mathbb{N};G^{\prime},H_{0},H_{1}\in\mathbb{B}^{*}$ such
that:

  1. (1)

$m_{k+1}>m_{k},$

  2. (2)

$\forall n\leqslant k+1\ \mathcal{F}(\mathbb{B}_{n},\widehat{G}\cup
G^{\prime})\ \text{is}\ (m_{n},2^{-n})\text{-balanced},$

  3. (3)

$\forall\mu\in\mathcal{M}\ \mu(\widehat{G}\cup G^{\prime}\cup\widehat{H}\cup
H_{0}\cup H_{1})<0.1,$

  4. (4)

$\widehat{H},H_{0},H_{1}\ \text{are pairwise disjoint},$

  5. (5)

$G^{\prime}\cap(\widehat{G}\cup\widehat{H}\cup H_{1})=\varnothing,$

  6. (6)

$\widehat{G}\cap(H_{0}\cup H_{1})=\varnothing,$

  7. (7)

$|\nu_{a}|(H_{0}),|\nu_{b}|(H_{1})\geqslant 0.9,$

  8. (8)

$|\nu_{a}(G^{\prime}\cap H_{0})|\geqslant 0.3.$

###### Proof.

Let $\mathbb{E}$ be the subalgebra of $\mathbb{B}$ generated by
$\mathbb{B}_{k+1}\cup\\{\widehat{G},\widehat{H}\\}$. Since $\mathbb{B}$ is
balanced, by Remark 3.4 there is $m_{k+1}\in\mathbb{N},m_{k+1}>m_{k}$ such
that

$\begin{gathered}\mathbb{E}\ \text{is}\
\left(m_{k+1},\frac{1}{2^{k+1}}\right)\text{-balanced}.\end{gathered}$

In particular,

(9) $\begin{gathered}\mathcal{F}(\mathbb{B}_{k+1},\widehat{G})\ \text{is}\
\left(m_{k+1},\frac{1}{2^{k+1}}\right)\text{-balanced}.\end{gathered}$

Put

(10)
$\begin{gathered}\xi=0.1-\max_{\mu\in\mathcal{M}}\mu(\widehat{G}\cup\widehat{H}).\end{gathered}$

For every $\mu\in\mathcal{M}$ consider its Lebesgue decomposition (cf. [35,
Theorem 6.10])

(11) $\begin{gathered}\mu=\mu_{1}+\mu_{2},\ \text{where}\ \mu_{1}\ll\lambda\
\text{and}\ \mu_{2}\bot\lambda.\end{gathered}$

In particular (cf. [35, Theorem 6.11]) there is $\eta>0$ such that

(12) $\begin{gathered}\forall\mu\in\mathcal{M}\ \forall A\in\mathbb{B}^{*}\
\lambda(A)<\eta\implies\mu_{1}(A)<\xi/4.\end{gathered}$

By (9), the first part of (D) and Proposition 4.8 (applied to
$P=\widehat{G}\cup\widehat{H}$, $G=\widehat{G}$) there is $\theta>0$ such that
whenever $L,Q\in\mathbb{B}^{*}$ and $\max\\{\lambda(L),\lambda(Q)\\}<\theta$,
there is $M_{(L,Q)}\in\mathbb{B}^{*}$ such that:

  1. (13)

$M_{(L,Q)}\cap(\widehat{G}\cup\widehat{H}\cup Q)=\varnothing$,

  2. (14)

$\lambda\left(M_{(L,Q)}\right)<\eta$,

  3. (15)

$\forall n\leqslant k+1\ \mathcal{F}\left(\mathbb{B}_{n},\widehat{G}\cup L\cup
M_{(L,Q)}\right)$ is $(m_{n},2^{-n})$-balanced.

By (11) and Lemma 5.1 there is $X\in\mathbb{B}^{*}$ such that for every
$\mu\in\mathcal{M}$

(16) $\begin{gathered}\lambda(X)<\theta,\mu_{2}(C\backslash
X)<\xi/4.\end{gathered}$

By Lemma 5.4 (applied to
$\varepsilon=\min\left\\{\theta-\lambda(X),\xi/4\right\\}$,
$P=\widehat{G}\cup\widehat{H}$) there are $a,b>d$ and
$H_{0},H_{1}\in\mathbb{B}^{*}$ such that

  * •

$\lambda(H_{0}\cup H_{1}\cup X)<\theta$,

  * •

$\forall\mu\in\mathcal{M}\ \mu(H_{0}\cup H_{1})<\xi/4$,

  * •

$H_{0}\cap H_{1}=(H_{0}\cup
H_{1})\cap(\widehat{G}\cup\widehat{H})=\varnothing$,

  * •

$|\nu_{a}|(H_{0}),|\nu_{b}|(H_{1})\geqslant 0.9$.

Let $L\in\mathbb{B}^{*}$ be such that

(17) $\begin{gathered}L\subseteq H_{0}\ \text{and}\ |\nu_{a}(L)|\geqslant
0.3.\end{gathered}$

Let $M=M_{(L,H_{0}\cup H_{1}\cup X)}$. Then, in particular,

(18) $\displaystyle M\cap(X\cup\widehat{H}\cup\widehat{G}\cup H_{0}\cup
H_{1})=\varnothing.$

We put $G^{\prime}=L\cup M$.

We need to verify that these definitions satisfy conditions $(1)$-$(8)$.

$(1)$ follows directly from the choice of $m_{k+1}$. $(2)$ follows from (15).

For $(3)$ fix any $\mu\in\mathcal{M}$. By (12) and (14) we have

$\mu_{1}(M)<\xi/4.$

By (16) and (18) we have

$\mu_{2}(M)<\xi/4.$

By (17)

$\mu(L)<\xi/4.$

Hence

$\mu(G^{\prime})=\mu(M)+\mu(L)=\mu_{1}(M)+\mu_{2}(M)+\mu(L)<3\xi/4.$

Finally, from (10) we get

$\mu(\widehat{G}\cup G^{\prime}\cup\widehat{H}\cup H_{0}\cup
H_{1})=\mu(\widehat{G}\cup\widehat{H})+\mu(G^{\prime})+\mu(H_{0}\cup
H_{1})<0.1-\xi+3\xi/4+\xi/4=0.1.$

Conditions $(4)$-$(7)$ follow directly from the choice of
$a,b,H_{0},H_{1},G^{\prime}$.

(8) follows from (17) and the fact that $G^{\prime}\cap H_{0}=L$. ∎

######  Lemma 5.6.

Let $\mathbb{B}^{*}\subseteq\mathbb{B}\subseteq\operatorname{Bor}(C)$ be
balanced countable Boolean algebras and suppose that
$(\nu_{n})_{n\in\mathbb{N}}$ is a normal sequence of measures on
$\mathbb{B}^{*}$.

Then there exists a balanced countable Boolean algebra
$\mathbb{B}^{\prime}\subseteq\operatorname{Bor}(C)$ such that
$\mathbb{B}\subseteq\mathbb{B}^{\prime}$ and
$(\mathbb{B}^{*},\mathbb{B}^{\prime},(\nu_{n})_{n\in\mathbb{N}})$ satisfies
$(\mathcal{G}^{*})$.

###### Proof.

Since $\mathbb{B}^{*}$ is countable, the dual ball in
$C(\operatorname{St}(\mathbb{B}^{*}))$ is metrizable and by the Banach-Alaoglu
theorem it is compact in the weak* topology. Hence there is a subsequence
$(|\widetilde{\nu}_{n_{k}}|)_{n\in\mathbb{N}}$ of
$(|\widetilde{\nu}_{n}|)_{n\in\mathbb{N}}$ that converges to a measure
$\widetilde{\nu}_{\infty}$ in the weak* topology. In particular,
$(|{\nu}_{n_{k}}|)_{n\in\mathbb{N}}$ is pointwise convergent to
$\nu_{\infty}$.

Let us represent $\mathbb{B}$ as an increasing union of finite subalgebras

$\mathbb{B}=\bigcup_{n\in\mathbb{N}}\mathbb{B}_{n}.$

Using Lemma 5.5 we construct by induction on $k\in\mathbb{N}$ sequences
$(m_{k})_{k\in\mathbb{N}}$, $(a_{k})_{k\in\mathbb{N}}$,
$(b_{k})_{k\in\mathbb{N}}\subseteq\mathbb{N}$ and
$(G_{k})_{k\in\mathbb{N}},(H_{0}^{k})_{k\in\mathbb{N}}$,
$(H_{1}^{k})_{k\in\mathbb{N}}\subseteq\mathbb{B}^{*}$ such that

  1. (1)

$(m_{k})_{k\in\mathbb{N}},(a_{k})_{k\in\mathbb{N}},(b_{k})_{k\in\mathbb{N}}$
are strictly increasing,

  2. (2)

$\forall k\in\mathbb{N}\ \forall n\leqslant k\
\mathcal{F}\left(\mathbb{B}_{n},\bigcup_{i\leqslant k}G_{i}\right)$ is
$(m_{n},2^{-n})$-balanced,

  3. (3)

$\forall k\in\mathbb{N}\ \mathcal{\nu_{\infty}}\left(\bigcup_{n\leqslant
k}(G_{n}\cup H_{0}^{n}\cup H_{1}^{n})\right)<0.1$,

  4. (4)

$\\{H_{0}^{k},H_{1}^{k}\\}_{k\in\mathbb{N}}$ are pairwise disjoint,

  5. (5)

$\\{G_{k}\\}_{k\in\mathbb{N}}$ are pairwise disjoint,

  6. (6)

$G_{k}\cap H_{i}^{n}\neq\varnothing$ if and only if $i=0$ and $n=k$,

  7. (7)

$\forall k\in\mathbb{N}\
|\nu_{a_{k}}|(H_{0}^{k}),|\nu_{b_{k}}|(H_{1}^{k})\geqslant 0.9$,

  8. (8)

$\forall k\in\mathbb{N}\ |\nu_{a_{k}}(G_{k}\cap H_{0}^{k})|\geqslant 0.3$.

Let $k\in\mathbb{N}\cup\\{0\\}$ and suppose we have constructed
$(m_{n})_{n\leqslant k},(a_{n})_{n\leqslant k},(b_{n})_{n\leqslant k}$,
$(G_{n})_{n\leqslant k}$, $(H_{0}^{n})_{n\leqslant k}$,
$(H_{1}^{n})_{n\leqslant k}$ (if $k=0$, then we assume that all of these
sequences are empty). We apply Lemma 5.5 to
$\mathcal{M}=\\{\mathcal{\nu_{\infty}}\\},\widehat{G}=\bigcup_{i\leqslant
k}G_{i},\widehat{H}=\bigcup_{i\leqslant k}(H_{0}^{i}\cup H_{1}^{i})$ and
$d=\max\\{a_{k},b_{k}\\}$ (or $\widehat{G}=\widehat{H}=\varnothing$ and $d=1$,
if $k=0$) to obtain $m_{k+1}$, $a_{k+1}=a$, $b_{k+1}=b$, $G_{k+1}=G^{\prime}$,
$H^{k+1}_{0}=H_{0}$, $H^{k+1}_{1}=H_{1}$ satisfying $(1)$-$(8)$.

Let

$G=\bigcup_{n\in\mathbb{N}}G_{n}$

and let $\mathbb{B}^{\prime}$ be the Boolean algebra generated by
$\mathbb{B}\cup\\{G\\}$. By (2) and Lemma 4.7 $\mathbb{B}^{\prime}$ is
balanced. To see that
$(\mathbb{B}^{*},\mathbb{B}^{\prime},(\nu_{n})_{n\in\mathbb{N}})$ satisfies
$(\mathcal{G}^{*})$ we notice that by (6) for every $n\in\mathbb{N}$ we have
$G\cap H_{i}^{n}=G_{n}\cap H_{i}^{n}$ for $i=0,1$, and so

  * •

$|\nu_{a_{n}}(G\cap H_{0}^{n})|=|\nu_{a_{n}}(G_{n}\cap H_{0}^{n})|\geqslant
0.3$,

  * •

$G\cap H_{1}^{n}=\varnothing$,

which together with (7) and (8) shows that the conditions (a)-(d) of
Definition 3.14 are satisfied and it completes the proof. ∎

Properties of measures on a Boolean algebra $\mathbb{A}$ such as norm or
disjointness of Borel supports depend only on countably many elements of
$\mathbb{A}$. In particular, the following lemma holds.

######  Lemma 5.7.

Let $\mathbb{B}=\bigcup_{\alpha<\omega_{1}}\mathbb{B}_{\alpha}$, where
$(\mathbb{B}_{\alpha})_{\alpha<\omega_{1}}$ is an increasing sequence of
countable Boolean algebras. Let $(\nu_{n})_{n\in\mathbb{N}}$ be a normal
sequence of measures on $\mathbb{B}$. Then there exists $\alpha<\omega_{1}$
such that for every $\beta>\alpha$ the sequence
$(\nu_{n}\upharpoonright\mathbb{B}_{\beta})_{n\in\mathbb{N}}$ is normal.

The next theorem shows how to construct a Boolean algebra with the
Grothendieck property and without the Nikodym property under CH.

######  Theorem 5.8.

_(Talagrand,[46])_ Assume CH. There exists a Boolean algebra with the
Grothendieck property, but without the Nikodym property.

###### Proof.

By Proposition 3.8 and Proposition 3.13 it is enough to construct a balanced
Boolean algebra $\mathbb{B}\subseteq\operatorname{Bor}(C)$ satisfying
$(\mathcal{G})$. We will define $\mathbb{B}$ as a union of a sequence of
countable subalgebras $(\mathbb{B}_{\alpha})_{\alpha<\omega_{1}}$ of
$\operatorname{Bor}(C)$, which is constructed by induction.

First, using CH we fix an enumeration

$(\nu^{\alpha},\mathbb{B}^{*}_{\alpha})_{\alpha<\omega_{1}}$

of all pairs such that each $\mathbb{B}^{*}_{\alpha}$ is a countable
subalgebra of $\operatorname{Bor}(C)$ and
$\nu^{\alpha}=(\nu^{\alpha}_{n})_{n\in\mathbb{N}}$ is a normal sequence of
measures on $\mathbb{B}_{\alpha}^{*}$. We also require each such pair to
appear cofinaly often in the sequence
$(\nu^{\alpha},\mathbb{B}_{\alpha}^{*})_{\alpha<\omega_{1}}$.

Successor stage: Suppose we have constructed $\mathbb{B}_{\alpha}$. If
$\mathbb{B}^{*}_{\alpha}$ is not a subalgebra of $\mathbb{B}_{\alpha}$, then
we put $\mathbb{B}_{\alpha+1}=\mathbb{B}_{\alpha}$. If
$\mathbb{B}^{*}_{\alpha}\subseteq\mathbb{B}_{\alpha}$, then by Lemma 5.6 there
is a balanced Boolean algebra
$\mathbb{B}^{\prime}\supseteq\mathbb{B}_{\alpha}$ such that
$(\mathbb{B}^{*}_{\alpha},\mathbb{B}^{\prime},\nu^{\alpha})$ satisfies
$(\mathcal{G}^{*})$. We put $\mathbb{B}_{\alpha+1}=\mathbb{B}^{\prime}$.

If $\gamma$ is a limit ordinal then we put
$\mathbb{B}_{\gamma}=\bigcup_{\alpha<\gamma}\mathbb{B}_{\alpha}$.

We will prove that $\mathbb{B}=\bigcup_{\alpha<\omega_{1}}\mathbb{B}_{\alpha}$
satisfies $(\mathcal{G})$. By Proposition 3.15 it is enough to show that for
every normal sequence $(\nu_{n})_{n\in\mathbb{N}}$ of measures on $\mathbb{B}$
there is $\beta<\omega_{1}$ such that the sequence
$({\nu_{n}\upharpoonright{\mathbb{B}_{\beta}}})_{n\in\mathbb{N}}$ is normal
and
$(\mathbb{B}_{\beta},\mathbb{B},(\nu_{n}\upharpoonright{\mathbb{B}_{\beta}})_{n\in\mathbb{N}})$
satisfies $(\mathcal{G}^{*})$. By Lemma 5.7 there is $\alpha<\omega_{1}$ such
that for every $\beta\geqslant\alpha$ the sequence
$({\nu_{n}\upharpoonright{\mathbb{B}_{\beta}}})_{n\in\mathbb{N}}$ is normal.
To finish the proof, pick $\beta>\alpha$ such that
$\mathbb{B}^{*}_{\beta}=\mathbb{B}_{\alpha}\subseteq\mathbb{B}_{\beta}$ and
$\nu^{\beta}=\nu\upharpoonright{\mathbb{B}_{\alpha}}$ and notice that by the
construction
$(\mathbb{B}^{*}_{\beta},\mathbb{B}_{\beta+1},(\nu_{n}\upharpoonright{\mathbb{B}^{*}_{\beta}})_{n\in\mathbb{N}})$
satisfies $(\mathcal{G}^{*})$, which implies that
$(\mathbb{B}_{\beta},\mathbb{B},(\nu_{n}\upharpoonright{\mathbb{B}_{\beta}})_{n\in\mathbb{N}})$
satisfies $(\mathcal{G}^{*})$. ∎

##  6\. Forcing

For the purpose of this section we identify Borel subsets of $C$ with their
codes (with respect to some absolute coding, cf. [22, Section 25]) i.e.
whenever we say about the same Borel sets in different models of ZFC we mean
Borel sets coded by the same code.

If $\mathfrak{c}>\omega_{1}$, then the method from the previous section does
not work, since it requires extending a given Boolean algebra $\mathfrak{c}$
many times, while this method does not allow us to enlarge uncountable Boolean
algebras keeping them balanced. Instead, we define a notion of forcing that
adds to a given balanced algebra $\mathbb{B}$ a witness for
$(\mathcal{G^{*}})$ for many sequences of measures (chosen by a generic
filter) on $\mathbb{B}$ simultaneously. However, it is not possible to pick
one extension that is suitable for every sequence, so we will iterate
$\omega_{1}$ such forcings and the final Boolean algebra will have cardinality
$\omega_{1}<\mathfrak{c}$.

######  Definition 6.1.

Let $\mathbb{B}\subseteq\operatorname{Bor}(C)$ be a balanced countable Boolean
algebra containing $\operatorname{Clop}(C)$ and fix a representation
$\mathbb{B}=\bigcup_{n\in\mathbb{N}}\mathbb{B}_{n}$, where
$(\mathbb{B}_{n})_{n\in\mathbb{N}}$ is an increasing sequence of finite
subalgebras of $\mathbb{B}$. We define a forcing notion
$\mathbb{P}_{\mathbb{B}}$ consisting of conditions of the form

$p=(k^{p},(m^{p}_{n})_{n\leqslant k^{p}},(G^{p}_{n})_{n\leqslant
k^{p}},(H^{p}_{n})_{n\leqslant k^{p}},\mathcal{M}^{p}),$

where

  1. (1)

$k^{p}\in\mathbb{N}$,

  2. (2)

$(m^{p}_{n})_{n\leqslant k^{p}}$ is a strictly increasing sequence of natural
numbers,

  3. (3)

$\mathcal{M}^{p}$ is a finite set of probability measures on $\mathbb{B}$ such
that $\lambda\upharpoonright\mathbb{B}\in\mathcal{M}^{p}$,

  4. (4)

$(G^{p}_{n})_{n\leqslant k^{p}}$ and $(H^{p}_{n})_{n\leqslant k^{p}}$ are
sequences of elements of $\mathbb{B}$ such that

    1. (a)

$G^{p}_{n}\cap G^{p}_{l}=H^{p}_{n}\cap H^{p}_{l}=G^{p}_{n}\cap
H^{p}_{l}=\varnothing$ for $n\neq l$,

    2. (b)

$\mu\left(\bigcup_{n\leqslant k^{p}}(G^{p}_{n}\cup H^{p}_{n})\right)<0.1$ for
all $\mu\in\mathcal{M}^{p}$,

    3. (c)

$\mathcal{F}\left(\mathbb{B}_{n},\bigcup_{i\leqslant k^{p}}G^{p}_{i}\right)$
is $(m_{n},2^{-n})-$balanced for $n\leqslant k^{p}$.

We put $q\leqslant p$, if

  * •

$k^{q}\geqslant k^{p}$,

  * •

$m^{q}_{n}=m^{p}_{n}$ for $n\leqslant k^{p}$,

  * •

$G^{q}_{n}=G^{p}_{n}$ for $n\leqslant k^{p}$,

  * •

$H^{q}_{n}=H^{p}_{n}$ for $n\leqslant k^{p}$,

  * •

$\mathcal{M}^{q}\supseteq\mathcal{M}^{p}$.

As expected, our forcing is ccc.

######  Lemma 6.2.

$\mathbb{P}_{\mathbb{B}}$ is $\sigma$-centered. In particular,
$\mathbb{P}_{\mathbb{B}}$ satisfies ccc.

###### Proof.

For $p\in\mathbb{P}_{\mathbb{B}}$ define

$f(p)=(k^{p},(m^{p}_{n})_{n\leqslant k^{p}},(G^{p}_{n})_{n\leqslant
k^{p}},(H^{p}_{n})_{n\leqslant k^{p}}).$

If $f(p)=f(q)$, then $r\leqslant p,q$ and $f(r)=f(p)=f(q),$ where

$r=(k^{p},(m^{p}_{n})_{n\leqslant k^{p}},(G^{p}_{n})_{n\leqslant
k^{p}},(H^{p}_{n})_{n\leqslant
k^{p}},\mathcal{M}^{p}\cup\mathcal{M}^{q})\in\mathbb{P}_{\mathbb{B}}.$

In particular, for every $x\in f[\mathbb{P}_{\mathbb{B}}]$ the set $f^{-1}(x)$
is directed. Since $\mathbb{B}$ is countable, $f[\mathbb{P}_{\mathbb{B}}]$ is
also countable, so $\mathbb{P}_{\mathbb{B}}$ is a union of countably many
directed sets. Hence $\mathbb{P}_{\mathbb{B}}$ is $\sigma$-centered. ∎

The next few lemmas concern the basic properties of $\mathbb{P}_{\mathbb{B}}$.

######  Lemma 6.3.

Let $p\in\mathbb{P}_{\mathbb{B}}$ and $k>k^{p}$. Then there is
$q\in\mathbb{P}_{\mathbb{B}},q\leqslant p$ such that $k^{q}=k$.

###### Proof.

It is enough to show that there is such $q$ for $k=k^{p}+1$ and apply an
inductive argument. For this put $\widehat{G}=\bigcup_{n\leqslant
k^{p}}G^{p}_{n},\widehat{H}=\bigcup_{n\leqslant k^{p}}H^{p}_{n}$. Since
$\lambda\upharpoonright\mathbb{B}$, by the condition (4c) of Definition 6.1 we
have $\lambda(C\backslash(\widehat{G}\cup\widehat{H}))>0$. Hence there is a
normal sequence $(\nu_{n})_{n\in\mathbb{N}}$ of measures on $\mathbb{B}$,
whose supports are included in $C\backslash(\widehat{G}\cup\widehat{H})$. We
may also assume that $(|\nu_{n}|)_{n\in\mathbb{N}}$ is pointwise convergent to
a probability measure $\nu_{\infty}$. In particular, we have

$\nu_{\infty}(\widehat{G}\cup\widehat{H})=\lim_{n\to\infty}|\nu_{n}|(\widehat{G}\cup\widehat{H})=0.$

Let $\mathcal{M}=\mathcal{M}^{p}\cup\\{\nu_{\infty}\\}$. By Lemma 5.5 there
are $m>m_{k^{p}}$ and $G^{\prime},H_{0}\in\mathbb{B}$ such that

$q=(k,((m^{p}_{n})_{n\leqslant k^{p}},m),((G^{p}_{n})_{n\leqslant
k^{p}},G^{\prime}),((H^{p}_{n})_{n\leqslant
k^{p}},H_{0}),\mathcal{M}^{p})\in\mathbb{P}_{\mathbb{B}}.$

We have $q\leqslant p$. ∎

######  Definition 6.4.

Let us introduce the following notation for a balanced Boolean algebra
$\mathbb{B}\subseteq\operatorname{Bor}(C)$ containing
$\operatorname{Clop}(C)$:

  * •

$\dot{\mathbb{G}}$ denotes the canonical name for a generic filter in
$\mathbb{P}_{\mathbb{B}}$,

  * •

$\dot{G}$ is a $\mathbb{P}_{\mathbb{B}}$-name such that

$\mathbb{P}_{\mathbb{B}}\Vdash\dot{G}=\bigcup_{p\in\dot{\mathbb{G}},n\in\mathbb{N}}\check{G}^{p}_{n},$

  * •

$\dot{\mathbb{B}}^{\prime}$ denotes a $\mathbb{P}_{\mathbb{B}}$-name such that

$\mathbb{P}_{\mathbb{B}}\Vdash\dot{\mathbb{B}}^{\prime}\ \text{is the
subalgebra of}\ \operatorname{Bor}(C)\ \text{generated by}\
\check{\mathbb{B}}\cup\\{\dot{G}\\},$

  * •

$\dot{H},\dot{H}_{n}$ for $n\in\mathbb{N}$ denote
$\mathbb{P}_{\mathbb{B}}$-names such that

$\mathbb{P}_{\mathbb{B}}\Vdash\dot{H}=(\dot{H}_{n})_{n\in\mathbb{N}}\
\text{and}\ \forall n\in\mathbb{N}\ \exists p\in\dot{\mathbb{G}}\
\dot{H}_{n}=\check{H}^{p}_{n}.$

######  Lemma 6.5.

Let $n\in\mathbb{N}$ and $p\in\mathbb{P}_{\mathbb{B}}$ be such that
$k^{p}\geqslant n$. Then

$\mathbb{P}_{\mathbb{B}}\Vdash\dot{G}\cap\dot{H}_{n}\in\check{\mathbb{B}}$

and

$p\Vdash\dot{G}\cap\dot{H}_{n}=\check{G}^{p}_{n}\cap\check{H}^{p}_{n}.$

###### Proof.

By the definitions of $\dot{G},\dot{H}_{n}$ and by (4a) from Definition 6.1 we
have

$\displaystyle\mathbb{P}_{\mathbb{B}}\Vdash\dot{G}\cap\dot{H}_{n}=\bigcup_{\begin{subarray}{c}q\in\dot{\mathbb{G}},l\in\mathbb{N}\\\
k^{q}\geqslant
n,l\end{subarray}}\check{G}^{q}_{l}\cap\check{H}^{q}_{n}=\bigcup_{\begin{subarray}{c}q\in\dot{\mathbb{G}}\\\
k^{q}\geqslant n\end{subarray}}\check{G}^{q}_{n}\cap\check{H}^{q}_{n}.$

If $q$ and $r$ are compatible and $k^{q},k^{r}\geqslant n$, then
$G^{q}_{n}=G^{r}_{n}$ and $H^{q}_{n}=H^{r}_{n}$, so the last union above is in
fact a union of one-element family consisting of a set from $\mathbb{B}$. In
particular,

$\displaystyle\mathbb{P}_{\mathbb{B}}\Vdash\dot{G}\cap\dot{H}_{n}\in\check{\mathbb{B}}$

and

$\mathbb{P}_{\mathbb{B}}\Vdash\check{p}\in\dot{\mathbb{G}}\implies\dot{G}\cap\dot{H}_{n}=\check{G}^{p}_{n}\cap\check{H}^{p}_{n}$

or equivalently

$p\Vdash\dot{G}\cap\dot{H}_{n}=\check{G}^{p}_{n}\cap\check{H}^{p}_{n}.$

∎

In the following lemma we will make use of Lemma 5.5 in the case when
$\mathcal{M}$ consists of many probability measures. This is where the
differences between our approach and that of work [46] are crucial.

######  Lemma 6.6.

Let $\mathbb{B}\subseteq\operatorname{Bor}(C)$ be a balanced Boolean algebra
containing $\operatorname{Clop}(C)$. Let $p\in\mathbb{P}_{\mathbb{B}}$ and
$\nu=(\nu_{n})_{n\in\mathbb{N}}$ be a normal sequence of measures on
$\mathbb{B}$ such that the sequence $(|\nu_{n}|)_{n\in\mathbb{N}}$ is
pointwise convergent on $\mathbb{B}$ to a measure
$\mathcal{\nu_{\infty}}\in\mathcal{M}^{p}$. Then

  1. (1)

$p\Vdash\forall k\in\mathbb{N}\ \exists n,l>k\
|\check{\nu}_{n}|(\dot{H}_{l})\geqslant 0.9\ \text{and}\
|\check{\nu}_{n}(\dot{G}\cap\dot{H}_{l})|\geqslant 0.3$,

  2. (2)

$p\Vdash\forall k\in\mathbb{N}\ \exists n,l>k\
|\check{\nu}_{n}|(\dot{H}_{l})\geqslant 0.9\ \text{and}\
\dot{G}\cap\dot{H}_{l}=\varnothing$.

###### Proof.

In the light of Lemma 6.5 it is enough to show that the sets

$\mathbb{D}_{k}=\\{q\in\mathbb{P}_{\mathbb{B}}:k^{q}>k,\exists n>k\
|\nu_{n}|(H^{q}_{k^{q}})\geqslant 0.9\ \text{and}\ |\nu_{n}(G^{q}_{k^{q}}\cap
H^{q}_{k^{q}})|\geqslant 0.3\\}$

and

$\mathbb{E}_{k}=\\{q\in\mathbb{P}_{\mathbb{B}}:k^{q}>k,\exists n>k\
|\nu_{n}|(H^{q}_{k^{q}})\geqslant 0.9\ \text{and}\ G^{q}_{k^{q}}\cap
H^{q}_{k^{q}}=\varnothing\\}$

are dense below $p$ for every $k\in\mathbb{N}$.

First, we will show that $\mathbb{D}_{k}$ is dense below $p$. Pick any
$r\leqslant p$. By Lemma 6.3 we may assume that $k^{r}\geqslant k$.

We apply Lemma 5.5 to
$\mathcal{M}=\mathcal{M}^{r},\widehat{G}=\bigcup_{i\leqslant
k^{r}}G_{i}^{r},\widehat{H}=\bigcup_{i\leqslant k^{r}}H_{i}^{r}$ and $d=k$ to
obtain $m=m_{k^{r}+1}$, $a>k$, $G^{\prime}$, $H_{0}$ so that the conditions
$(1)$-$(8)$ from Lemma 5.5 are satisfied. In particular:

  1. (a)

$m>m_{k^{r}}$,

  2. (b)

$G^{\prime}\cap G_{i}^{r}=H_{0}\cap H_{i}^{r}=G^{\prime}\cap
H_{i}^{r}=\varnothing$ for $i\leqslant k^{r}$,

  3. (c)

$\mu\left(\bigcup_{i\leqslant k^{r}}(G_{i}^{r}\cup H_{i}^{r})\cup
G^{\prime}\cup H_{0}\right)<0.1$ for all $\mu\in\mathcal{M}^{r}$,

  4. (d)

$\mathcal{F}\left(\mathbb{B}_{n},\bigcup_{i\leqslant k^{r}}G^{r}_{i}\cup
G^{\prime}\right)$ is $(m_{n},2^{-n})-$balanced for $n\leqslant k^{r}+1$,

  5. (e)

$|\nu_{a}|(H_{0})\geqslant 0.9$,

  6. (f)

$|\nu_{a}(G^{\prime}\cap H_{0})|\geqslant 0.3$.

It follows from (a)-(d) that

$q=(k^{r}+1,((m^{r}_{i})_{i\leqslant k^{r}},m),((G^{r}_{i})_{i\leqslant
k^{r}},G^{\prime}),((H^{r}_{i})_{i\leqslant
k^{r}},H_{0}),\mathcal{M}^{r})\in\mathbb{P}_{\mathbb{B}}$

and from (e), (f) that $q\in\mathbb{D}_{k}$.

We show the density of $\mathbb{E}_{k}$ in a similar way: the difference is
that instead of $H_{0}$ we pick $H_{1}$ such that

  * •

$|\nu_{a}|(H_{1})\geqslant 0.9$,

  * •

$G^{\prime}\cap H_{1}=\varnothing$.

∎

Directly from Lemma 6.5, Lemma 6.6 and Definition 3.14 we obtain

######  Proposition 6.7.

Let $\mathbb{B}\subseteq\operatorname{Bor}(C)$ be a balanced Boolean algebra
that contains $\operatorname{Clop}(C)$. Let $p\in\mathbb{P}_{\mathbb{B}}$ and
$\nu=(\nu_{n})_{n\in\mathbb{N}}$ be a normal sequence of measures on
$\mathbb{B}$ such that the sequence $(|\nu_{n}|)_{n\in\mathbb{N}}$ is
pointwise convergent to a measure $\nu_{\infty}$. Suppose that
$\mathcal{\nu_{\infty}}\in\mathcal{M}^{p}$. Then

$p\Vdash(\check{\mathbb{B}},\dot{\mathbb{B}}^{\prime},\check{\nu})\
\text{satisfies}\ (\mathcal{G}^{*}).$

######  Proposition 6.8.

Suppose that $\mathbb{B}\subseteq\operatorname{Bor}(C)$ is a balanced
countable Boolean algebra containing $\operatorname{Clop}(C)$. Then

$\mathbb{P}_{\mathbb{B}}\Vdash\dot{\mathbb{B}}^{\prime}\ \text{is balanced}.$

###### Proof.

Since the property of being balanced is absolute between transitive models of
ZFC

$V^{\mathbb{P}_{\mathbb{B}}}\models\mathbb{B}\ \text{is balanced}.$

In $V^{\mathbb{P}_{\mathbb{B}}}$ for every $n\in\mathbb{N}$ we define
$G_{n}=G^{p}_{n}$ for some $p\in\mathbb{G}$ such that $k^{p}\geqslant n$. Then
$\mathbb{B}^{\prime}$ is the Boolean algebra generated by
$\mathbb{B}\cup\\{G\\}$, where $G=\bigcup_{n\in\mathbb{N}}G_{n}$. By
Definition 6.1(4a,4c) the hypothesis of Lemma 4.7 is satisfied and so
$\mathbb{B}^{\prime}$ is balanced. ∎

######  Definition 6.9.

We define an iteration $(\mathbb{P}_{\alpha})_{\alpha\leqslant\omega_{1}}$
with finite supports and $\mathbb{P}_{\alpha}$-names
$\dot{\mathbb{B}}_{\alpha}$ for every $\alpha\leqslant\omega_{1}$ by induction
in the following way:

  * •

$\mathbb{P}_{0}$ is the trivial forcing and
$\mathbb{B}_{0}=\operatorname{Clop}(C)$,

  * •

having constructed $\mathbb{P}_{\alpha}$ and $\dot{\mathbb{B}}_{\alpha}$ we
define

$\mathbb{P}_{\alpha+1}=\mathbb{P}_{\alpha}*\mathbb{P}_{\dot{\mathbb{B}}_{\alpha}}$

and we pick a $\mathbb{P}_{\alpha+1}$-name $\dot{\mathbb{B}}_{\alpha+1}$ such
that

$\mathbb{P}_{\alpha+1}\Vdash\dot{\mathbb{B}}_{\alpha+1}=\dot{\mathbb{B}}^{\prime}_{\alpha},$

  * •

if $\gamma$ is a limit ordinal, then we define $\mathbb{P}_{\gamma}$ as the
iteration of $(\mathbb{P}_{\alpha})_{\alpha<\gamma}$ with finite supports and
we pick $\dot{\mathbb{B}}_{\gamma}$ so that

$\mathbb{P}_{\gamma}\Vdash\dot{\mathbb{B}}_{\gamma}=\bigcup_{\alpha<\gamma}\dot{\mathbb{B}}_{\alpha}$

We will identify each $\mathbb{P}_{\alpha}$ with the subset of
$\mathbb{P}_{\omega_{1}}$ consisting of those $p\in\mathbb{P}_{\omega_{1}}$,
for which $p(\beta)=\mathbbm{1}_{\beta}$ for all $\beta\geqslant\alpha$, where
$\mathbbm{1}_{\beta}$ denotes the maximal element of
$\mathbb{P}_{\dot{\mathbb{B}}_{\beta}}$.

######  Lemma 6.10.

$\mathbb{P}_{\omega_{1}}$ is $\sigma$-centered. In particular,
$\mathbb{P}_{\omega_{1}}$ satisfies ccc.

###### Proof.

This follows from Lemma 6.2 and the fact that a finite support iteration of
length $\omega_{1}$ of $\sigma$-centered forcings is $\sigma$-centered [47,
proof of Lemma 2]. ∎

Since $\mathbb{P}_{\omega_{1}}$ satisfies ccc, the standard closure argument
shows the following (cf. [12, Lemma 5.3])

######  Lemma 6.11.

Let $\dot{\nu}=(\dot{\nu}_{n})_{n\in\mathbb{N}}$ be a sequence such that

$\mathbb{P}_{\omega_{1}}\Vdash(\dot{\nu}_{n})_{n\in\mathbb{N}}\ \text{is a
sequence of measures on}\ \dot{\mathbb{B}}_{\omega_{1}}.$

Let

$C_{\dot{\nu}}=\\{\alpha<\omega_{1}:\mathbb{P}_{\omega_{1}}\Vdash\dot{\nu}\upharpoonright\dot{\mathbb{B}}_{\alpha}\in
V^{\mathbb{P}_{\alpha}}\\}.$

Then $C_{\dot{\nu}}$ is a closed and unbounded subset of $\omega_{1}$.

######  Proposition 6.12.

$\mathbb{P}_{\omega_{1}}\Vdash\dot{\mathbb{B}}_{\omega_{1}}$ is balanced and
satisfies $(\mathcal{G})$.

###### Proof.

The fact that $\mathbb{P}_{\omega_{1}}\Vdash\dot{\mathbb{B}}_{\omega_{1}}$ is
balanced follows directly from Proposition 6.8 and the fact that increasing
unions of balanced Boolean algebras are balanced.

To prove the second part of the proposition, by Proposition 3.15 it is enough
to show that for every sequence $(\dot{\nu}_{n})_{n\in\mathbb{N}}$ such that

$\mathbb{P}_{\omega_{1}}\Vdash(\dot{\nu}_{n})_{n\in\mathbb{N}}\ \text{is a
normal sequence of measures on}\ \dot{\mathbb{B}}_{\omega_{1}}$

we have

$\mathbb{P}_{\omega_{1}}\Vdash\exists_{\alpha<\omega_{1}}\
(\dot{\mathbb{B}}_{\alpha},\dot{\mathbb{B}}_{\omega_{1}},\dot{\nu}\upharpoonright{\dot{\mathbb{B}}_{\alpha}})\
\text{satisfies}\ (\mathcal{G}^{*}).$

Pick any $p\in\mathbb{P}_{\omega_{1}}$. By Lemma 5.7 there is
$\alpha_{1}<\omega_{1}$ and $p_{1}\leqslant p$ such that for every
$\beta\geqslant\alpha_{1}$

$p_{1}\Vdash(\dot{\nu}_{n}\upharpoonright{\dot{\mathbb{B}}_{\beta}})_{n\in\mathbb{N}}\
\text{is normal}.$

By Lemma 6.11 there are: $\beta\in C_{\dot{\nu}}$ such that
$\alpha_{1}<\beta$, $p_{1}\in V^{\mathbb{P}_{\beta}}$ and
$\mathbb{P}_{\beta}$-names $\dot{\nu}^{\beta},\dot{\nu}^{\beta}_{n}$ for
$n\in\mathbb{N}$ such that

$\mathbb{P}_{\beta}\Vdash\dot{\nu}^{\beta}=(\dot{\nu}^{\beta}_{n})_{n\in\mathbb{N}}=(\dot{\nu}_{n}\upharpoonright{\dot{\mathbb{B}}_{\beta}})_{n\in\mathbb{N}}.$

Without loss of generality by passing to a subsequence we may assume that
there is $\dot{\nu}_{\infty}$ such that

$\mathbb{P}_{\beta}\Vdash(|\dot{{\nu}}^{\beta}_{n}|)_{n\in\mathbb{N}}\
\text{is pointwise convergent to a measure}\ \dot{{\nu}}_{\infty}.$

Since $p_{1}(\beta)=\mathbbm{1}_{\beta}$, there is $p_{2}\leqslant p_{1}$ such
that

$p_{2}\upharpoonright\beta\Vdash\dot{\nu}_{\infty}\in\dot{\mathcal{M}}^{\check{p}_{2}({\beta})}.$

By Proposition 6.7 we have

$p_{2}\Vdash(\dot{\mathbb{B}}_{\beta},\dot{\mathbb{B}}_{\beta+1},\dot{\nu}^{\beta})\
\text{satisfies}\ (\mathcal{G}^{*})$

and hence

$p_{2}\Vdash\exists_{\alpha<\omega_{1}}\
(\dot{\mathbb{B}}_{\alpha},\dot{\mathbb{B}}_{\omega_{1}},\dot{\nu}\upharpoonright{\dot{\mathbb{B}}_{\alpha}})\
\text{satisfies}\ (\mathcal{G}^{*})$

which completes the proof. ∎

In particular, by Proposition 3.8 and Proposition 3.13 we obtain

######  Corollary 6.13.

$\mathbb{P}_{\omega_{1}}\Vdash$ there is a Boolean algebra of size
$\omega_{1}$ with the Grothendieck property but without the Nikodym property.

The existence of a Boolean algebra with the Grothendieck property of small
cardinality has influence on certain cardinal characteristics of the
continuum. Below $\mathfrak{p}$ denotes the pseudointersection number,
$\mathfrak{s}$ is the splitting number and $\operatorname{cov}(\mathcal{M})$
is the covering number of the ideal of meager sets in $\mathbb{R}$.

######  Corollary 6.14.

$\mathbb{P}_{\omega_{1}}\Vdash\mathfrak{p}=\mathfrak{s}=\operatorname{cov}(\mathcal{M})=\omega_{1}$.

###### Proof.

Apply [40, Corollary 4.3]. ∎

######  Theorem 6.15.

It is consistent with $\neg$CH that there is a Boolean algebra of size
$\omega_{1}$ with the Grothendieck property but without the Nikodym property.

###### Proof.

Start with a model $V$ of ZFC satisfying $\neg$CH. Since
$\mathbb{P}_{\omega_{1}}$ is $\sigma$-centered, it preserves cardinals and the
value of the continuum, so we have

$V^{\mathbb{P}_{\omega_{1}}}\models\mathfrak{c}=\mathfrak{c}^{V}>\omega_{1}^{V}=\omega_{1}.$

By Corollary 6.13 in $V^{\mathbb{P}_{\omega_{1}}}$ there is a Boolean algebra
with the Grothendieck property, but without the Nikodym property. ∎

##  7\. Final remarks

Let us start with a comment concerning differences between the original
Talagrand’s contruction and our approach. To obtain the Grothendieck property,
Talagrand uses CH to enumerate (in a sequence of length $\omega_{1}$) all
normalized sequences $(\nu_{n})_{n\in\mathbb{N}}$ of measures on countable
subalgebras of Borel subsets of the Cantor set, for which there exists an
antichain $(H_{n})_{n\in\mathbb{N}}$ such that $|\nu_{n}|(H_{n})\geqslant
0.95$. Then for each such sequence he constructs another antichain
$(G_{n})_{n\in\mathbb{N}}$ satisfying the hypothesis of Lemma 4.7, such that
for $G=\bigcup_{n\in\mathbb{N}}G_{n}$ we have

  * •

$|\nu_{n}(G\cap H_{n})|\geqslant 0.4$ for infinitely many $n\in\mathbb{N}$,

  * •

$|\nu_{n}(G\cap H_{n})|<0.1$ for infinitely many $n\in\mathbb{N}$.

It follows that extending a given Boolean algebra with $G$ keeps it balanced,
and in the extension the sequence $(\nu_{n})_{n\in\mathbb{N}}$ satisfies a
property similar to $(\mathcal{G^{*}})$ from Definition 3.14. Thus, the final
algebra has the Grothendieck property and does not have the Nikodym property.
The same approach was used in [43].

However, this technique applies only when we work with one sequence of
measures at a time. In our method, we construct a suitable antichain
$(H_{n})_{n\in\mathbb{N}}$ along with $(G_{n})_{n\in\mathbb{N}}$, which allows
us to pick both the antichains in a generic way, making them working for
uncountably many sequences of measures simultaneously.

The method of construction we have described relies strongly on the fact that
the Boolean algebras we extend are countable.

######  Question 7.1.

Let $\mathbb{B}\subseteq\operatorname{Bor}(C)$ be a balanced Boolean algebra
of cardinality $<\mathfrak{c}$ and $\nu$ be a normal sequence of measures on
$\mathbb{B}$.

Does there exist a balanced Boolean algebra
$\mathbb{B}\subseteq\mathbb{B}^{\prime}\subseteq\operatorname{Bor}(C)$ such
that $(\mathbb{B},\mathbb{B}^{\prime},\nu)$ satisfies $(\mathcal{G}^{*})$?

Positive answer for the above question would allow us to construct (by
induction of length $\mathfrak{c}$) a balanced Boolean algebra of size
$\mathfrak{c}$. Thus, it would imply the positive answer for the following
question:

######  Question 7.2.

Does there exist (in ZFC) a balanced Boolean algebra with the Grothendieck
property?

One may look for candidates for Boolean algebras with the Grothendieck and
without Nikodym property among maximal balanced Boolean algebras.

######  Question 7.3.

Let $\mathbb{B}\subseteq\operatorname{Bor}(C)$ be a maximal balanced Boolean
algebra. Does $\mathbb{B}$ have the Grothendieck property?

## Acknowledgments

We would like to thank our PhD supervisors Piotr Koszmider and Piotr
Borodulin-Nadzieja for introducing to the topic, careful reading and valuable
suggestions. We would like to express our gratitude towards The Fields
Institute for Research in Mathematical Sciences in Toronto for the hospitality
and excellent conditions for research during Thematic Program on Set Theoretic
Methods in Algebra, Dynamics and Geometry (January - June, 2023), during which
most of the results contained in this work were obtained. We would also like
to thank Omar Selim for sharing his notes, Julia Millhouse and Forte Shinko
for the language assistance with French.

## References

  * [1] Fernando Albiac and Nigel J. Kalton, _Topics in Banach space theory_ , second ed., Graduate Texts in Mathematics, vol. 233, Springer, [Cham], 2016, With a foreword by Gilles Godefory. MR 3526021
  * [2] Tsuyoshi Andô, _Convergent sequences of finitely additive measures_ , Pacific J. Math. 11 (1961), 395–404. MR 137806
  * [3] Antonio Avilés, Félix Cabello Sánchez, Jesús M. F. Castillo, Manuel González, and Yolanda Moreno, _Separably injective Banach spaces_ , Lecture Notes in Mathematics, vol. 2132, Springer, [Cham], 2016. MR 3469461
  * [4] Diómedes Bárcenas and Luis Gerardo Mármol, _On $C(K)$ Grothendieck spaces_, Rend. Circ. Mat. Palermo (2) 54 (2005), no. 2, 209–216. MR 2168368
  * [5] Tomek Bartoszyński and Haim Judah, _Set theory: On the structure of the real line (1st ed.)_ , A K Peters/CRC Press., 1995.
  * [6] J. Bourgain, _$H^{\infty}$ is a Grothendieck space_, Studia Math. 75 (1983), no. 2, 193–216. MR 722264
  * [7] Christina Brech, _On the density of Banach spaces $C(K)$ with the Grothendieck property_, Proc. Amer. Math. Soc. 134 (2006), no. 12, 3653–3663. MR 2240680
  * [8] H. G. Dales, F. K. Dashiell, Jr., A. T.-M. Lau, and D. Strauss, _Banach spaces of continuous functions as dual spaces_ , CMS Books in Mathematics/Ouvrages de Mathématiques de la SMC, Springer, Cham, 2016. MR 3588279
  * [9] R. B. Darst, _On a theorem of Nikodym with applications to weak convergence and von Neumann algebras_ , Bull. Amer. Math. Soc. 74 (1968), 283–284. MR 223902
  * [10] J. Diestel, _Grothendieck spaces and vector measures_ , Vector and operator valued measures and applications (Proc. Sympos., Alta, Utah, 1972), Academic Press, New York-London, 1973, pp. 97–108. MR 338774
  * [11] Marián Fabian, Petr Habala, Petr Hájek, Vicente Montesinos, and Václav Zizler, _Banach space theory_ , CMS Books in Mathematics/Ouvrages de Mathématiques de la SMC, Springer, New York, 2011, The basis for linear and nonlinear analysis. MR 2766381
  * [12] Rogério Augusto dos Santos Fajardo, _An indecomposable Banach space of continuous functions which has small density_ , Fund. Math. 202 (2009), no. 1, 43–63. MR 2457484
  * [13] Francisco José Freniche, _The Vitali-Hahn-Saks theorem for Boolean algebras with the subsequential interpolation property_ , Proc. Amer. Math. Soc. 92 (1984), no. 3, 362–366. MR 759653
  * [14] Steven Givant and Paul Halmos, _Introduction to Boolean algebras_ , Undergraduate Texts in Mathematics, Springer, New York, 2009. MR 2466574
  * [15] M. Gonzalez and V. M. Onieva, _Lifting results for sequences in Banach spaces_ , Math. Proc. Cambridge Philos. Soc. 105 (1989), no. 1, 117–121. MR 966145
  * [16] Manuel González and Tomasz Kania, _Grothendieck spaces: the landscape and perspectives_ , Jpn. J. Math. 16 (2021), no. 2, 247–313. MR 4338245
  * [17] Manuel González, Fernando León-Saavedra, and María del Pilar Romero de la Rosa, _On $\ell_{\infty}$-Grothendieck subspaces_, J. Math. Anal. Appl. 497 (2021), no. 1, Paper No. 124857, 5. MR 4190602
  * [18] A. Grothendieck, _Sur les applications linéaires faiblement compactes d’espaces du type $C(K)$_, Canad. J. Math. 5 (1953), 129–173. MR 58866
  * [19] R. G. Haydon, _Boolean rings that are Baire spaces_ , Serdica Math. J. 27 (2001), no. 2, 91–106. MR 1879874
  * [20] Richard Haydon, _A nonreflexive Grothendieck space that does not contain $l_{\infty}$_, Israel J. Math. 40 (1981), no. 1, 65–73. MR 636907
  * [21] Richard Haydon, Mireille Levy, and Edward Odell, _On sequences without weak ∗ convergent convex block subsequences_, Proc. Amer. Math. Soc. 100 (1987), no. 1, 94–98. MR 883407
  * [22] Thomas Jech, _Set theory_ , Springer Monographs in Mathematics, Springer-Verlag, Berlin, 2003, The third millennium edition, revised and expanded. MR 1940513
  * [23] Jean-Pierre Kahane, _Some random series of functions_ , second ed., Cambridge Studies in Advanced Mathematics, vol. 5, Cambridge University Press, Cambridge, 1985. MR 833073
  * [24] Ondřej F. K. Kalenda and Jiří Spurný, _Quantification of the reciprocal Dunford-Pettis property_ , Studia Math. 210 (2012), no. 3, 261–278. MR 2983464
  * [25] Jerzy Kąkol, Damian Sobota, and Lyubomyr Zdomskyy, _Grothendieck $C(K)$-spaces and the Josefson–Nissenzweig theorem_, Fund. Math. 263 (2023), no. 2, 105–131. MR 4669141
  * [26] Sabine Koppelberg, _Handbook of Boolean algebras. Vol. 1_ , North-Holland Publishing Co., Amsterdam, 1989, Edited by J. Donald Monk and Robert Bonnet. MR 991565
  * [27] Piotr Koszmider, _Forcing minimal extensions of Boolean algebras_ , Trans. Amer. Math. Soc. 351 (1999), no. 8, 3073–3117. MR 1467471
  * [28] by same author, _Banach spaces of continuous functions with few operators_ , Math. Ann. 330 (2004), no. 1, 151–183. MR 2091683
  * [29] Kenneth Kunen, _Set theory_ , Studies in Logic and the Foundations of Mathematics, vol. 102, North-Holland Publishing Co., Amsterdam-New York, 1980, An introduction to independence proofs. MR 597342
  * [30] S. López-Alfonso, _On Schachermayer and Valdivia results in algebras of Jordan measurable sets_ , Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat. RACSAM 110 (2016), no. 2, 799–808. MR 3534524
  * [31] Aníbal Moltó, _On the Vitali-Hahn-Saks theorem_ , Proc. Roy. Soc. Edinburgh Sect. A 90 (1981), no. 1-2, 163–173. MR 636070
  * [32] Otton Nikodym, _Sur les familles bornées de fonctions parfaitement additives d’ensemble abstrait_ , Monatsh. Math. Phys. 40 (1933), no. 1, 418–426. MR 1550216
  * [33] Elliott Pearl (ed.), _Open problems in topology. II_ , Elsevier B. V., Amsterdam, 2007. MR 2367385
  * [34] H. Pfitzner, _Weak compactness in the dual of a $C^{\ast}$-algebra is determined commutatively_, Math. Ann. 298 (1994), no. 2, 349–371. MR 1256621
  * [35] Walter Rudin, _Real and complex analysis_ , third ed., McGraw-Hill Book Co., New York, 1987. MR 924157
  * [36] Walter Schachermayer, _On some classical measure-theoretic theorems for non-sigma-complete Boolean algebras_ , Dissertationes Math. (Rozprawy Mat.) 214 (1982), 33. MR 673286
  * [37] G. L. Seever, _Measures on $F$-spaces_, Trans. Amer. Math. Soc. 133 (1968), 267–280. MR 226386
  * [38] Zbigniew Semadeni, _Banach spaces of continuous functions. Vol. I_ , Monografie Matematyczne, Tom 55, PWN—Polish Scientific Publishers, Warsaw, 1971. MR 0296671
  * [39] Damian Sobota, _The Nikodym property and cardinal characteristics of the continuum_ , Ann. Pure Appl. Logic 170 (2019), no. 1, 1–35. MR 3870039
  * [40] by same author, _Convergence of measures on boolean algebras and cardinal characteristics of the continuum_ , Kôkyǔroku 2164 (2020), 50–63.