# There is a P-measure in the random model

Piotr Borodulin–Nadzieja  Mathematical Institute, University of Wrocław, pl.
Grunwaldzki 2, 50-384 Wrocław, Poland<EMAIL_ADDRESS>and  Damian
Sobota  Kurt Gödel Research Center for Mathematical Logic, Department of
Mathematics, University of Vienna, Kolingasse 14-16, 1090 Vienna, Austria
<EMAIL_ADDRESS>
###### Abstract.

We say that a finitely additive probability measure $\mu$ on $\omega$ is _a
P-measure_ if it vanishes on points and for each decreasing sequence $(E_{n})$
of infinite subsets of $\omega$ there is $E\subseteq\omega$ such that
$E\subseteq^{*}E_{n}$ for each $n\in\omega$ and
$\mu(E)=\lim_{n\to\infty}\mu(E_{n})$. Thus, P-measures generalize in a natural
way P-points and it is known that, similarly as in the case of P-points, their
existence is independent of $\mathsf{ZFC}$. In this paper we show that there
is a P-measure in the model obtained by adding any number of random reals to a
model of $\mathsf{CH}$. As a corollary, we obtain that in the classical random
model $\omega^{*}$ contains a nowhere dense ccc closed P-set.

###### Key words and phrases:

P-points, P-measures, Additive Property, random forcing, random model

###### 2010 Mathematics Subject Classification:

Primary: 03E05, 03E35. Secondary: 03E75, 28E15.

The first author was supported by the National Science Center project no.
2018/29/B/ST1/00223. The second author was supported by the Austrian Science
Fund (FWF) grant no. 4570-N

##  1\. Introduction

We say that a (finitely additive) probability measure $\mu$ on $\omega$ is _a
P-measure_ if it vanishes on points and for each decreasing sequence $(E_{n})$
of infinite subsets of $\omega$ there is $E\subseteq\omega$ such that
$E\subseteq^{*}E_{n}$ for each $n\in\omega$ and
$\mu(E)=\lim_{n\to\infty}\mu(E_{n})$ (see Section 2 for the unexplained
terminology).

The importance of this notion lies in the fact that it generalizes the notion
of $P$-points in $\omega^{*}$ in a natural way. Indeed, consider an
ultrafilter $\mathcal{U}$ on $\omega$ and let
$\mu\colon\wp(\omega)\to\\{0,1\\}$ be the one-point measure concentrated in
$\mathcal{U}$, i.e., for every $A\in\wp(\omega)$, $\mu(A)=1$ if and only if
$A\in\mathcal{U}$. Then, $\mu$ is a P-measure if and only if $\mathcal{U}$ is
a P-point. Moreover, if $\mathcal{U}$ is a P-point, then the measure $\nu$ on
$\omega$, defined for every $A\in\wp(\omega)$ by the formula

$\nu(A)=\lim_{n\to\mathcal{U}}\frac{|A\cap n|}{n}$

is a non-atomic P-measure (extending the asymptotic density). So, every
P-point _is_ a P-measure and, moreover, it induces a P-measure which
additionally is non-atomic.

The converse issue however, that is, whether the existence of P-measures
implies the existence of P-points, is so far unresolved. It was considered by
Mekler in [Mek84] (where the P-measures were introduced for the first time,
under the name of _measures with the additive property_ , or _with (AP)_ in
short), by Blass _et al._ in [BFPRN01] (where P-measures are called _measures
with AP(*)_) and by Grebík in [Gre19] (where P-measures are also called
_measures with AP(*)_). Unfortunately, all the constructions of P-measures
presented in those articles involve more or less directly P-points.

Recall that basically only two models of $\mathsf{ZFC}$ without P-points are
known: the classical construction due to Shelah and its variants (see [She98],
cf. also [FSZ93] and [FGZ05]), and the Silver model (see [CG19]). Mekler
[Mek84] showed than in the model used by Shelah there are no P-measures. It is
unknown if there is a P-measure in the Silver model.

Also, it is still open if there is a P-point in the classical random model,
i.e. the model obtained by adding $\omega_{2}$ random reals to a model
satisfying the Continuum Hypothesis ($\mathsf{CH}$). Although in [Coh79] the
author claimed to answer this problem in positive, it has recently appeared
that his proof contains an essential gap (see [CG19]). Recall that Kunen
proved that if we first add $\omega_{1}$ Cohen reals to the ground model and
then $\omega_{2}$ random reals, then we get a model with a P-point (see
[Bre04]). Recently, Dow [Dow20] also showed that if we add $\omega_{2}$ random
reals to a model of $\mathsf{CH}+\square_{\omega_{1}}$, then the resulting
model contains a P-point.

The main purpose of this paper is to show that in the classical random model
there is a P-measure (Corollary 4.16). Our measure is not defined using any
P-point from the extension (we still do not know if there exists one!) and it
seems to be the first example of this kind. The construction exploits some
ideas of Solovay from [Sol71] (and so we call the resulting measure _a Solovay
measure_). It proceeds more or less as follows. We start with an ultrafilter
$\mathcal{U}$ in the ground model $V$ and we generate, in a certain way, a
name $\dot{\mu}_{\mathcal{U}}$ for a finitely additive measure on $\omega$. In
the random extension $V[G]$, the resulting measure
$(\dot{\mu}_{\mathcal{U}})_{G}$ in a sense extends the ground model one-point
measure $\delta_{\mathcal{U}}$ concentrated in $\mathcal{U}$, and we show that
whenever $\mathcal{U}$ is a ground model P-point, then
$(\dot{\mu}_{\mathcal{U}})_{G}$ is a P-measure in $V[G]$ which do not extend
the asymptotic density (Theorem 4.15 and Proposition 4.14). So, perversely, we
do use a P-point in the construction of our P-measure, but this P-point lives
in another model.

We show that all Solovay measures are non-atomic (Theorem 5.3). Since the
supports of non-atomic measures cannot contain P-points (cf. Proposition 3.7),
the same is true for Solovay measures. But, as the support of a P-measure is a
nowhere dense ccc closed P-subset of $\omega^{*}$, we get that in the random
model (or in any random extension of a model of $\mathsf{CH}$) there is a
nowhere dense ccc closed P-set in $\omega^{*}$ (Corollary 4.17). This is an
interesting result as there are models without such subsets of $\omega^{*}$,
see e.g. [FSZ93].

It turns out however that supports of Solovay measures in random extensions
can be used to obtain characterizations of various classes of ultrafilters on
$\omega$ in the ground model. Notice that every ultrafilter $\mathcal{U}$ on
$\omega$ from the ground model $V$ generates a closed subset $K_{\mathcal{U}}$
of $\beta\omega$ in the extension $V[G]$ (consisting of all ultrafilters
extending the filter $\mathcal{U}$). We show that an ultrafilter $\mathcal{U}$
is semi-selective (in $V$) if and only if the measure
$(\dot{\mu}_{\mathcal{U}})_{G}$ is strictly positive on $K_{\mathcal{U}}$, and
so its support is the whole $K_{\mathcal{U}}$ (Theorem 5.6). This result sheds
some new light on the well-known theorem, due to Kunen [Kun76], saying that no
P-point in the random model can extend a ground model semi-selective
ultrafilter. Indeed, as we already said, no P-point can be an element of the
support of a non-atomic measure, so, if $\mathcal{U}$ is semi-selective, then
no element of $K_{\mathcal{U}}$ can be a P-point (see Corollary 5.7).

In the context of aforementioned Kunen’s theorem it is natural to ask whether
a P-point $\mathcal{V}$ in the random extension (assuming it exists) can at
all extend a ground model ultrafilter $\mathcal{U}$. If this is the case, then
it is easy to observe that $\mathcal{U}$ itself must be a P-point. However, by
Kunen’s result $\mathcal{U}$ cannot be semi-selective and so, as explained in
the beginning of Section 5.2, there exists a decreasing sequence $(E_{n})$ of
infinite subsets of $\omega$ (in the extension) such that $\mathcal{U}$ and
$\\{E_{n}\colon n\in\omega\\}$ together generate a filter which cannot be
further extended to a bigger filter by _any_ pseudointersection of the
sequence $(E_{n})$. On the other hand, if $\mathcal{U}$ is a Canjar
ultrafilter, then for every decreasing sequence $(E_{n})$ of infinite subsets
of $\omega$, if $\mathcal{F}=\mathcal{U}\cup\\{E_{n}\colon n\in\omega\\}$ is a
base of a filter, then there is a pseudointersection $E$ of $(E_{n})$ such
$\mathcal{F}\cup\\{E\\}$ is still a base of a filter (Proposition 5.10).
Hence, if there are ultrafilters in the ground model which can be extended to
a P-point in the random extension, then Canjar ultrafilters should be
considered as promising candidates.

##  2\. Preliminaries

$\wp(\omega)$ denotes the power set of $\omega$, that is, the family of all
subsets of $\omega$. If $A,B\in\wp(\omega)$, then we write $A\subseteq^{*}B$
if the set $A\setminus B$ is finite, and $A=^{*}B$ if $A\subseteq^{*}B$ and
$B\subseteq^{*}A$. _An antichain_ in $\wp(\omega)$ is a sequence $(A_{n})$ of
pairwise disjoint subsets of $\omega$. If $(A_{n})$ is a collection of subsets
of $\omega$, then if $A_{n}\subseteq^{*}A$ (resp. $A\subseteq^{*}A_{n}$) for
each $n\in\omega$, then $A$ is called _a pseudounion_ (resp. _a
pseudointersection_) of $(A_{n})$. Note that if
$\bigcup_{n\in\omega}A_{n}\subseteq A$, then $A$ is a pseudounion of $(A_{n})$
and hence $\omega$ is trivially a pseudounion of every family.

$Fin$ denotes the ideal of all finite subsets of $\omega$. Also, for every
$A\in\wp(\omega)$ by $[A]^{<\omega}$ we denote the set of all finite subsets
of $A$ and by $[A]^{\omega}$ the set of all infinite subsets of $A$.

If $\mathbb{A}$ is a Boolean algebra, then by $St(\mathbb{A})$ we denote its
Stone space. Recall that $\mathbb{A}$ is isomorphic to the algebra
$Clopen(St(\mathbb{A}))$ of all clopen subsets of $St(\mathbb{A})$; for every
$A\in\mathbb{A}$ by $[A]_{\mathbb{A}}$ we will denote the corresponding clopen
subset of $St(\mathbb{A})$. Recall also that the Stone spaces
$St(\wp(\omega))$ and $St(\wp(\omega)/Fin)$ are homeomorphic to $\beta\omega$,
the Čech–Stone compactification of $\omega$, and
$\omega^{*}=\beta\omega\setminus\omega$, respectively. For every
$A\in\wp(\omega)$ we will simply write $[A]$ for $[A]_{\wp(\omega)}$.

Let $X$ be a topological space. We say that $X$ is _an almost P-space_ if
every non-empty $\mathbb{G}_{\delta}$ subspace of $X$ has non-empty interior.
A subset $Y$ of $X$ is _a P-set_ if the intersection of countably many open
sets containing $Y$ contains $Y$ in its interior. A point $x\in X$ is _a
P-point_ in $X$ if the singleton $\\{x\\}$ is a P-set in $X$. If
$X=\omega^{*}$, this definition of course coincides with the standard one of a
P-point.

All _measures_ considered on Boolean algebras are assumed to be non-negative,
finitely additive, and finite (i.e. if $\mu$ is a measure on a Boolean algebra
$\mathbb{A}$, then $\mu(1_{\mathbb{A}})<\infty$). A measure $\mu$ on a Boolean
algebra $\mathbb{A}$ is _probability_ if $\mu(1_{\mathbb{A}})=1$. The space of
all probability measures on $\mathbb{A}$ will be denoted by $P(\mathbb{A})$.
All _measures_ on compact spaces are meant to be non-negative,
$\sigma$-additive, and _Radon_ , that is, Borel, regular, and finite (i.e. if
$\mu$ is a measure on a compact space $K$, then $\mu(K)<\infty$). Similarly as
above, a measure $\mu$ on a compact space $K$ is _probability_ if $\mu(K)=1$.
The space of all probability measures on $K$ will be denoted by $P(K)$. _The
support_ of a measure $\mu$ on a compact space $K$ is denoted by
$\operatorname{supp}(\mu)$.

If $\mu$ is a measure on the Boolean algebra $\wp(\omega)$, then we simply say
that _$\mu$ is a measure on $\omega$_. If $\mu$ is a measure on $\omega$ and
$\mu(\\{n\\})=0$ for every $n\in\omega$, then we also say that _$\mu$ vanishes
on points_.

Recall that each measure $\mu$ on a Boolean algebra $\mathbb{A}$ extends to a
unique measure $\check{\mu}$ on the Stone space $St(\mathbb{A})$ such that
$\mu(A)=\check{\mu}([A]_{\mathbb{A}})$ for every $A\in\mathbb{A}$. Also, for
every measure $\nu$ on a totally disconnected compact space $K$, there is a
measure $\mu$ on the Boolean algebra of clopen subsets of $K$ such that
$\nu=\check{\mu}$. In particular, every measure on $\beta\omega$ is induced by
a measure on $\omega$ and every measure on $\omega^{*}$ is induced by a
measure on $\wp(\omega)/Fin$, or, equivalently, by a measure on $\omega$ which
vanishes on points.

##  3\. Basic facts on P-measures

In this section we will collect several general facts on P-measures (some of
them are standard and folklore), which will be useful in the sequel. We start
with the following simple lemma.

######  Lemma 3.1.

  1. (1)

If $\mu$ is a measure on $\omega$ vanishing on points, then for every
$A\in[\omega]^{\omega}$ there is $B\in[A]^{\omega}$ such that $\mu(B)=0$.

  2. (2)

The support of a measure on $\omega^{*}$ vanishing on points is nowhere dense
in $\omega^{*}$.

  3. (3)

There is no strictly positive measure on $\omega^{*}$, that is, a measure
$\mu$ on $\omega^{*}$ such that $\mu(U)>0$ for every clopen subset $U$ of
$\omega^{*}$.

###### Proof.

(1) Let $\mu$ be a measure on $\omega$ which vanishes on points. Fix
$A\in[\omega]^{\omega}$ and let $\mathcal{A}\subseteq[A]^{\omega}$ be an
uncountable family of pairwise almost disjoint infinite subsets of $A$. If
$B,C\in\mathcal{A}$ are distinct, then $\mu(B\cup C)=\mu(B)+\mu(C)$, and so
there exists $D\in\mathcal{A}$ such that $\mu(D)=0$.

(2) and (3) follow immediately from (1). ∎

Clearly, measures on $\omega$ which vanish on points cannot be
$\sigma$-additive. However, P-measures are ’almost’ $\sigma$-additive as the
following proposition shows. We leave the proof of this standard fact to the
reader.

######  Proposition 3.2.

Let $\mu$ be a measure on $\omega$ vanishing on points. Then, the following
are equivalent:

  1. (1)

$\mu$ is a P-measure, that is, for every decreasing sequence $(A_{n})$ in
$\wp(\omega)$ there is a pseudointersection $A\in\wp(\omega)$ of $(A_{n})$
such that $\mu(A)=\lim_{n\to\infty}\mu(A_{n})$;

  2. (2)

for every increasing sequence $(A_{n})$ in $\wp(\omega)$ there is a
pseudounion $A\in\wp(\omega)$ of $(A_{n})$ such that
$\mu(A)=\lim_{n\to\infty}\mu(A_{n})$;

  3. (3)

for every antichain $(A_{n})$ in $\wp(\omega)$ there is a pseudounion
$A\in\wp(\omega)$ of $(A_{n})$ such that $\mu(A)=\sum_{n\in\omega}\mu(A_{n})$.

######  Lemma 3.3.

Assume $\mu$ is not a P-measure. Then, there is a decreasing sequence
$(A_{n})$ of subsets of $\omega$ and $\varepsilon>0$ such that
$\mu(A)<\lim_{n\to\infty}\mu(A_{n})-\varepsilon$ for every pseudointersection
$A$ of $(A_{n})$.

###### Proof.

Let $(A_{n})$ be a decreasing sequence witnessing that $\mu$ is not a
P-measure. Suppose for the contradiction that for every $k\in\omega$ there is
a pseudointersection $B_{k}$ of $(A_{n})$ such that
$\mu(B_{k})\geq\lim_{n\to\infty}\mu(A_{n})-1/k$. Then the sequence
$(B_{0}\cup\dots\cup B_{n},A_{n})$ forms an $(\omega,\omega)$-pregap. Since
there is no $(\omega,\omega)$-gap (see e.g. [Sch93]), it can be interpolated,
i.e. there is a set $B\subseteq\omega$ such that
$B_{n}\subseteq^{*}B\subseteq^{*}A_{n}$ for each $n\in\omega$. Clearly, $B$ is
a pseudointersection of $(A_{n})$ and $\mu(B)=\lim_{n\to\infty}\mu(A_{n})$, a
contradiction. ∎

Recall that a finitely additive measure $\mu$ on a Boolean algebra
$\mathbb{A}$ is _non-atomic_ if for each $\varepsilon>0$ there is a finite
partition of the unit element $1_{\mathbb{A}}$ into elements of $\mu$-measure
$\leq\varepsilon$ (note that in some sources such measures are called
’strongly non-atomic’, see e.g. [BRBR83]). Note that a measure $\mu$ on
$\mathbb{A}$ is non-atomic if and only if its Radon extension $\check{\mu}$
onto the Stone space $St(\mathbb{A})$ vanishes on points (that is,
$\check{\mu}(\\{x\\})=0$ for every $x\in St(\mathbb{A})$). A classical theorem
of Sierpiński [Sie34] asserts that every non-atomic $\sigma$-additive measure
$\mu$ on a $\sigma$-complete Boolean algebra $\mathbb{A}$ attains every value
from the interval $[0,\mu(1_{\mathbb{A}})]$ (or, more precisely, for every
$A\in\mathbb{A}$ and $\alpha\in[0,\mu(A)]$ there is $B\subseteq A$ in
$\mathbb{A}$ such that $\mu(B)=\alpha$). It appears however that the proof of
Sierpiński’s theorem presented in [BRBR83, Theorem 5.1.6] also works _mutatis
mutandis_ for P-measures on $\omega$.

######  Corollary 3.4.

Let $\mu$ be a non-atomic P-measure on $\omega$. Then, for every
$A\in\wp(\omega)$ and $\alpha\in[0,\mu(A)]$ there is $B\subseteq A$ such that
$\mu(B)=\alpha$.

The next proposition shows that atoms of P-measures give rise to P-points in
$\omega^{*}$.

######  Proposition 3.5.

Let $\mu$ be a P-measure on $\omega$. If there is $x\in\omega^{*}$ such that
$\check{\mu}(\\{x\\})>0$, then $x$ is a P-point.

###### Proof.

Assume that $x\in\omega^{*}$ is such that $\delta=\check{\mu}(\\{x\\})>0$.
There exists $A\in\wp(\omega)$ such that $x\in[A]$ and
$\check{\mu}([A]\setminus\\{x\\})<\delta$. Let $(A_{n})$ be a sequence of
elements of $x$ decreasing to $\emptyset$. Since $\mu$ is a P-measure, there
is $B\in\wp(\omega)$ such that $B\subseteq^{*}A_{n}\cap A$ for every
$n\in\omega$ and $\mu(B)=\lim_{n\to\infty}\mu(A_{n}\cap A)$. As $x$ is an
element of each clopen $[A_{n}\cap A]$, we have $\mu(B)\geq\delta$. It follows
that $x\in[B]$. Indeed, if $x\not\in[B]$, then, since $B\subseteq^{*}A$, it
holds
$\delta\leq\mu(B)=\check{\mu}([B])\leq\check{\mu}([A]\setminus\\{x\\})<\delta$,
a contradiction. So, $B\in x$ and $B\subseteq^{*}A_{n}$ for each $n\in\omega$.
It follows that $x$ is a P-point. ∎

######  Corollary 3.6.

If there is no P-point, then every P-measure is non-atomic.

The following fact, which we will need later, is folklore, but we include its
proof for the sake of self-containment of the paper.

######  Proposition 3.7.

Suppose that $\mu$ is a non-atomic measure on $\beta\omega$. If $\mathcal{U}$
is a P-point on $\omega$, then $\mathcal{U}$ does not belong to the support of
$\mu$.

###### Proof.

The non-atomicity of $\mu$ implies that there is a family
$\\{N_{\tau}\colon\tau\in 2^{<\omega}\\}\subseteq\mathcal{P}(\omega)$ such
that

  * •

$\mu([N_{\tau}])>0$ for each $\tau\in 2^{<\omega}$,

  * •

$N_{\tau}$ is a disjoint union of $N_{\tau^{\smallfrown}0}$ and
$N_{\tau^{\smallfrown}1}$ for each $\tau\in 2^{<\omega}$, and

  * •

$\lim_{n\to\infty}\mu([N_{s\restriction n}])=0$ for each $s\in 2^{\omega}$.

For each $n\in\omega$ there is $\tau_{n}\in 2^{n}$ such that
$N_{\tau_{n}}\in\mathcal{U}$. Clearly, $\\{\tau_{n}\colon n\in\omega\\}$ forms
a branch in the tree $2^{<\omega}$. If $\mathcal{U}$ is a P-point, then there
is $N\in\mathcal{U}$ such that $N\subseteq^{*}N_{\tau_{n}}$ for each
$n\in\omega$. Since $\mu$ is non-atomic, we have $\mu([N])=0$ and so the
clopen $[N]$ witnesses that $\mathcal{U}$ lies outside the support of $\mu$. ∎

##  4\. Measures on $\omega$ in random extensions

Fix a cardinal number $\kappa\geq\omega$. Let $\mathrm{Bor}(2^{\kappa})$
denote the $\sigma$-field of all Borel subsets of the space $2^{\kappa}$. By
$\lambda_{\kappa}$ denote the standard product measure on $2^{\kappa}$ and set
$\mathcal{N}_{\kappa}=\\{B\in\mathrm{Bor}(2^{\kappa})\colon\lambda_{\kappa}(B)=0\\}$.
Let $\mathbb{M}_{\kappa}$ be the measure algebra of type $\kappa$, i.e.
$\mathbb{M}_{\kappa}=\mathrm{Bor}(2^{\kappa})/{\mathcal{N}_{\kappa}}$ with the
standard Boolean operations $\wedge$, $\vee$, c, zero $0$ and unit $1$
elements, and ordering $\leq$. The measure $\lambda_{\kappa}$ induces in the
natural way the measure on $\mathbb{M}_{\kappa}$ which we will also denote by
$\lambda_{\kappa}$.

By $V$ we denote the set-theoretic universe. Speaking about _forcing with
$\mathbb{M}_{\kappa}$_, we formally mean the partial ordering
$(\mathbb{M}_{\kappa}\setminus\\{0\\},\leq)$—this abuse of notation should not
lead to any confusion. When forcing with $\mathbb{M}_{\kappa}$ (i.e. _adding
$\kappa$ random reals_), if $G$ is a generic filter over $V$, then for every
$\mathbb{M}_{\kappa}$-name $\sigma$ by $\sigma_{G}$ we denote its
interpretation through $G$ in the extended model $V[G]$. Note that
$\mathbb{M}_{\kappa}$ is forcing equivalent to the poset
$(\mathrm{Bor}(2^{\kappa})\setminus\mathcal{N}_{\kappa},\subseteq)$.

###  4.1. Names for subsets of $\omega$

In this section we will present a handy way of dealing with names for subsets
of $\omega$ in random extensions.

Consider the product Boolean algebra $\mathbb{M}_{\kappa}^{\omega}$ with the
Boolean operations defined coordinate-wise, i.e. for every
$M,N\in\mathbb{M}_{\kappa}^{\omega}$ and $k\in\omega$ we set:

  * •

$(M\wedge N)(k)=M(k)\wedge N(k)$,

  * •

$(M\vee N)(k)=M(k)\vee N(k)$,

  * •

$M^{c}(k)=M(k)^{c}$.

As above, the zero and unit elements of $\mathbb{M}_{\kappa}^{\omega}$ will be
denoted by $0$ and $1$, respectively. On $\mathbb{M}_{\kappa}^{\omega}$ we
consider two orderings: the standard one $\leq$ originating from
$\mathbb{M}_{\kappa}$, and the ’almost’ one $\leq^{*}$, defined for every
$M,N\in\mathbb{M}_{\kappa}^{\omega}$ as follows: $M\leq^{*}N$ if $M(k)\leq
N(k)$ for almost all $k\in\omega$.

For $M\in\mathbb{M}_{\kappa}^{\omega}$ define an $\mathbb{M}_{\kappa}$-name
$\dot{M}$ by the formula:

$\dot{M}=\\{\langle k,M(k)\rangle\colon k\in\omega\\}.$

Clearly, $\dot{M}$ is _an $\mathbb{M}_{\kappa}$-name for a subset of
$\omega$_, that is, $\Vdash_{\mathbb{M}_{\kappa}}\dot{M}\subseteq\omega$. The
proof of the next lemma is left to the reader.

######  Lemma 4.1.

For each $M,N\in\mathbb{M}_{\kappa}^{\omega}$ we have
$\Vdash_{\mathbb{M}_{\kappa}}\dot{M}\cap\dot{N}=(M\wedge N)\dot{\ }$, and
$\Vdash_{\mathbb{M}_{\kappa}}\dot{M}^{c}=(M^{c})\dot{\ }$.

######  Lemma 4.2.

For every $M\in\mathbb{M}_{\kappa}^{\omega}$ and every $k\in\omega$ we have
$M(k)=\llbracket k\in\dot{M}\rrbracket$.

###### Proof.

Fix $M\in\mathbb{M}_{\kappa}^{\omega}$ and $k\in\omega$. If
$M(k)\in\\{0,1\\}$, then we are trivially done, so assume the opposite. Since
$M(k)\Vdash k\in\dot{M}$, $M(k)\leq\llbracket k\in\dot{M}\rrbracket$.
Similarly, by Lemma 4.1, $M(k)^{c}\leq\llbracket k\in(M^{c})\dot{\
}\rrbracket=\llbracket k\in\dot{M}^{c}\rrbracket=\llbracket
k\not\in\dot{M}\rrbracket=\llbracket k\in\dot{M}\rrbracket^{c}$. It follows
thus that $M(k)=\llbracket k\in\dot{M}\rrbracket$. ∎

We will use Lemma 4.2 frequently without direct referring to it.

######  Corollary 4.3.

For every $M,N\in\mathbb{M}_{\kappa}^{\omega}$ it holds:

  1. (1)

if $M\leq N$, then $\Vdash_{\mathbb{M}_{\kappa}}\dot{M}\subseteq\dot{N}$;

  2. (2)

if $M\wedge N=0$, then
$\Vdash_{\mathbb{M}_{\kappa}}\dot{M}\cap\dot{N}=\emptyset$.

Actually, it appears that each subset of $\omega$ in $V^{\mathbb{M}_{\kappa}}$
can be described in the above way.

######  Lemma 4.4.

For each $\mathbb{M}_{\kappa}$-name $\dot{N}$ there is
$M\in\mathbb{M}_{\kappa}^{\omega}$ such that

$\Vdash_{\mathbb{M}_{\kappa}}(\dot{N}\subseteq\omega\ \Rightarrow\
\dot{N}=\dot{M}).$

###### Proof.

For $k\in\omega$ let $M(k)=\llbracket k\in\dot{N}\rrbracket$. ∎

So, in what follows, we will assume that all names for subsets of $\omega$ are
basically elements of $\mathbb{M}_{\kappa}^{\omega}$. In particular, when
writing that _$\dot{N}$ is an $\mathbb{M}_{\kappa}$-name for a subset of
$\omega$_, we will assume that $\dot{N}$ comes from the element
$N\in\mathbb{M}_{\kappa}^{\omega}$ constructed as in the proof of the above
lemma.

######  Lemma 4.5.

For every $M,N\in\mathbb{M}_{\kappa}^{\omega}$ we have:

  * •

$\bigwedge_{l\in\omega}\bigvee_{k>l}M(k)=0$ if and only if
$\Vdash_{\mathbb{M}_{\kappa}}$“$\dot{M}$ is finite”;

  * •

$\bigwedge_{l\in\omega}\bigvee_{k>l}(M(k)\setminus N(k))=0$ if and only if
$\Vdash_{\mathbb{M}_{\kappa}}\dot{M}\subseteq^{*}\dot{N}$;

  * •

if $M\leq^{*}N$, then
$\Vdash_{\mathbb{M}_{\kappa}}\dot{M}\subseteq^{*}\dot{N}$.

###### Proof.

Just notice that $\llbracket\forall l\ \exists k>l\
k\in\dot{M}\rrbracket=\bigwedge_{l\in\omega}\bigvee_{k>l}M(k)$. ∎

We need the following piece of notation in the sequel.

######  Definition 4.6.

For $X\subseteq\omega$ define $\check{X}\in\mathbb{M}^{\omega}_{\kappa}$ for
every $k\in\omega$ by

$\check{X}(k)=\begin{cases}1,\mbox{ if }k\in X,\\\ 0,\mbox{
otherwise.}\end{cases}$

For $q\in\mathbb{M}_{\kappa}$ define $\vec{q}\in\mathbb{M}^{\omega}_{\kappa}$
for every $k\in\omega$ by

$\vec{q}(k)=q.$

Notice that if we interpret $\check{X}$ as an $\mathbb{M}_{\kappa}$-name
$(\check{X})\dot{}$ for a subset of $\omega$, then it is just the canonical
$\mathbb{M}_{\kappa}$-name for $X$ (so the usage of $\check{\ }$ is
justified). We will thus also write $\check{X}$ for the interpretation of
$\check{X}$ as an $\mathbb{M}_{\kappa}$-name, omitting the dot $\dot{}$.

######  Lemma 4.7.

Let $M,N\in\mathbb{M}_{\kappa}^{\omega}$ and $q\in\mathbb{M}_{\kappa}$ be such
that $q\Vdash\dot{M}\subseteq\dot{N}$. Then, for $M^{\prime}=M\wedge\vec{q}$
and $N^{\prime}=N\wedge\vec{q}$ we have:

  1. (1)

$M^{\prime}\leq N^{\prime}$, and

  2. (2)

$q\Vdash\dot{M}=\dot{M}^{\prime},\dot{N}=\dot{N}^{\prime}$.

###### Proof.

We first prove (1). For the sake of contradiction, assume there is
$k\in\omega$ such that $K=M^{\prime}(k)\setminus N^{\prime}(k)\neq 0$. Since
$K\leq M^{\prime}(k)=\llbracket k\in\dot{M}^{\prime}\rrbracket$, we have
$K\Vdash k\in\dot{M}^{\prime}$, so by Corollary 4.3 it holds $K\Vdash
k\in\dot{M}$. Similarly, as $K\wedge N^{\prime}(k)=0$ and $K\leq q$, we get
that $K\wedge N(k)=0$ and hence that $K\Vdash k\not\in\dot{N}$. Summing up,
$K\leq q$ and $K\Vdash k\in\dot{M}\setminus\dot{N}$.

Let now $G$ be an $\mathbb{M}_{\kappa}$-generic filter containing both $q$ and
$K$. In $V[G]$, we have $k\in\dot{M}_{G}\setminus\dot{N}_{G}$ (since $K\in G$)
and $\dot{M}_{G}\subseteq\dot{N}_{G}$ (since $q\in G$), which is impossible.
It follows that $M^{\prime}(k)\leq N^{\prime}(k)$ for every $k\in\omega$ and
thus that $M^{\prime}\leq N^{\prime}$.

To prove (2), first note that by Corollary 4.3 we have
$q\Vdash\dot{M}^{\prime}\subseteq\dot{M}$. If
$q\not\Vdash\dot{M}\subseteq\dot{M}^{\prime}$, then there is $r\leq q$ and
$k\in\omega$ such that $r\Vdash k\in\dot{M}\setminus\dot{M}^{\prime}$. Since
$r\leq\llbracket k\in\dot{M}\rrbracket=M(k)$ and $r\leq q$, we have $r\leq
M^{\prime}(k)$, whence $r\Vdash k\in\dot{M}^{\prime}$, which is a
contradiction. The case of $q\Vdash\dot{N}=\dot{N}^{\prime}$ is identical. ∎

###  4.2. Names for reals.

It is well known that there is a correspondence between
$\mathbb{M}_{\kappa}$-names for reals and the ground model Borel functions
$f\colon 2^{\kappa}\to\mathbb{R}$ (see e.g. the classical article of Scott
[Sco67], or [Abr80]). In this section we will exploit this approach, although
we will rather encode names for reals using measures than functions.

Consider the $\mathbb{M}_{\kappa}$-name

$\dot{r}=\big{\\{}\langle\langle\alpha,i\rangle,C^{i}_{\alpha}\rangle\colon\alpha<\kappa,i\in\\{0,1\\}\big{\\}},$

where $C^{i}_{\alpha}=\\{x\in 2^{\kappa}\colon x(\alpha)=i\\}$. We say that
$\dot{r}$ is the name for _the generic element of $2^{\kappa}$_ (or _the
generic random real_ in case $\kappa=\omega$). Notice that
$\llbracket\dot{r}\in C^{i}_{\alpha}\rrbracket=C^{i}_{\alpha}$ for every
$\alpha<\kappa$ and $i\in\\{0,1\\}$. In a generic extension $V[G]$,
$\dot{r}_{G}$ is the unique element of $2^{\kappa}$ contained in every element
of the generic filter. Notice that the generic filter on $\mathbb{M}_{\kappa}$
induces the unique generic element of $2^{\kappa}$ and vice versa (see e.g.
[FGS06, page 6]).

We will need the following standard lemma. (As usual, a.e. is the abbreviation
for almost everywhere.)

######  Lemma 4.8.

Let $f\colon 2^{\kappa}\to\mathbb{R}$ be a Borel function in the ground model
$V$, $A\in\wp(\mathbb{R})\cap V$111Note that speaking about $f$ and $A$ we
mean their ground model Borel codes, not the actual sets., and
$y\in\mathbb{R}\cap V$. Let $p\in\mathbb{M}_{\kappa}$. Then

  1. (1)

$(p\Vdash f(\dot{r})\in A)\iff p\setminus\\{z\in 2^{\kappa}\colon f(z)\in
A\\}\in\mathcal{N}_{\kappa}$;

  2. (2)

$(p\Vdash f(\dot{r})=y)\iff f=y$, $\lambda_{\kappa}$-a.e. on $p$.

###### Proof.

For (1), just notice that $p\Vdash f(\dot{r})\in A$ is equivalent to
$p\Vdash\dot{r}\in f^{-1}[A]$ which is on the other hand equivalent to $p\leq
f^{-1}[A]$. (2) follows from (1) by setting $A=\\{y\\}$. ∎

######  Corollary 4.9.

Let $f_{1},f_{2}\colon 2^{\kappa}\to\mathbb{R}$ be Borel functions in the
ground model $V$. Then, $f_{1}\leq f_{2}$ $\lambda_{\kappa}$-a.e if and only
if $\Vdash_{\mathbb{M}_{\kappa}}f_{1}(\dot{r})\leq f_{2}(\dot{r})$.

###### Proof.

Let $\dot{x}$ be an $\mathbb{M}_{\kappa}$-name for an element of $\mathbb{R}$
such that $\Vdash_{\mathbb{M}_{\kappa}}\dot{x}=f_{2}(\dot{r})-f_{1}(\dot{r})$.
Let $A=(-\infty,0)$.

Assume that $f_{1}\leq f_{2}$ $\lambda_{\kappa}$-a.e. If there is
$p\in\mathbb{M}_{\kappa}$ such that $p\Vdash\dot{x}\in A$, then by Lemma
4.8.(1) applied for $f=f_{2}-f_{1}$ we have $\lambda_{\kappa}(p)=0$ (since
$\lambda_{\kappa}(f^{-1}[A])=0$), which is impossible, so $1\Vdash
f_{1}(\dot{r})\leq f_{2}(\dot{r})$.

Conversely, if $\Vdash_{\mathbb{M}_{\kappa}}f_{1}(\dot{r})\leq
f_{2}(\dot{r})$, then for every $p\in\mathbb{M}_{\kappa}$ we have
$p\Vdash\dot{x}\not\in A$, hence by Lemma 4.8.(1) we get that
$p\setminus\\{z\in 2^{\kappa}\colon(f_{2}-f_{1})(z)\in
A\\}\not\in\mathcal{N}_{\kappa}$. It follows that for the condition $p=\\{z\in
2^{\kappa}\colon(f_{2}-f_{1})(z)\in A\\}$ we have $p\in\mathcal{N}_{\kappa}$
and hence $f_{1}\leq f_{2}$ $\lambda_{\kappa}$-a.e. ∎

To each ground model $\sigma$-additive Borel measure on $2^{\kappa}$ which is
absolutely continuous with respect to $\lambda_{\kappa}$ one can linearly
assign a real in any $\mathbb{M}_{\kappa}$-generic extension.

######  Definition 4.10.

Let $\mu$ be a Borel measure on $2^{\kappa}$ such that
$\mu\ll\lambda_{\kappa}$. Then, by $\dot{r}(\mu)$ we will mean an
$\mathbb{M}_{\kappa}$-name for a real prepared in the following way. By
Radon–Nikodym theorem there is a Borel function $f\colon
2^{\kappa}\to\mathbb{R}$ such that $\mu(A)=\int_{A}f\ d\lambda_{\kappa}$ for
every $A\in Bor(2^{\kappa})$. Now, let $\dot{r}(\mu)$ be an
$\mathbb{M}_{\kappa}$-name such that
$\Vdash_{\mathbb{M}_{\kappa}}\dot{r}(\mu)=f(\dot{r})$, where $\dot{r}$ is the
generic random real.

######  Remark 4.11.

Notice that $\dot{r}(\mu)$ is uniquely determined by $\mu$ (despite the fact
that the Radon–Nikodym derivative $f$ is determined modulo
$\mathcal{N}_{\kappa}$). Also, if $\mu$, $\nu_{0}$, $\nu_{1}$ are
$\sigma$-additive Borel measures which are absolutely continuous with respect
to $\lambda_{\kappa}$, and $\alpha_{0}$, $\alpha_{1}$ are non-negative real
numbers, $\mu=\alpha_{0}\nu_{0}+\alpha_{1}\nu_{1}$, then by the linearity of
the Radon–Nikodym derivative it holds
$\Vdash_{\mathbb{M}_{\kappa}}\dot{r}(\mu)=\alpha_{0}\dot{r}(\nu_{0})+\alpha_{1}\dot{r}(\nu_{1})$.

###  4.3. Names for measures induced by ultrafilters.

All ultrafilters on $\omega$ considered in this and following sections are
assumed to be non-principal.

Fix an ultrafilter $\mathcal{U}$ on $\omega$ (in the ground model). Following
the ideas of Solovay [Sol71], we are going to define an
$\mathbb{M}_{\kappa}$-name for a finitely additive measure on $\omega$, which
is in a sense generated by $\mathcal{U}$.

Let $\dot{M}$ be an $\mathbb{M}_{\kappa}$-name for a subset of $\omega$
(induced by an element $M\in\mathbb{M}_{\kappa}^{\omega}$ as described in
Section 4.1) and define the function
$\mu_{M}\colon\mathbb{M}_{\kappa}\to[0,1]$ in the following way:

$\mu_{M}(B)=\lim_{k\to\mathcal{U}}\lambda_{\kappa}(\llbracket
k\in\dot{M}\rrbracket\wedge B).$

It is immediate that $\mu_{M}$ is a finitely additive measure on
$\mathbb{M}_{\kappa}$. Obviously, $\mu_{\check{\emptyset}}=0$ and
$\mu_{\check{\omega}}=\lambda_{\kappa}$.

######  Lemma 4.12.

For each $\mathbb{M}_{\kappa}$-name $\dot{M}$ for a subset of $\omega$, the
measure $\mu_{M}$ is $\sigma$-additive and $\mu_{M}\ll\lambda_{\kappa}$.

###### Proof.

Clearly, $\mu_{M}\leq\lambda_{\kappa}$. It implies, of course, that
$\mu_{M}\ll\lambda_{\kappa}$ but also that $\mu_{M}$ is continuous from above
at $\emptyset$. Indeed, let $(B_{n})$ be a decreasing sequence of Borel
subsets of $2^{\kappa}$ such that $\bigcap_{n\in\omega}B_{n}=\emptyset$. Then,
$\lambda_{\kappa}(B_{n})\geq\mu_{M}(B_{n})$ for every $n\in\omega$ and so, by
the continuity of $\lambda_{\kappa}$, $\lim_{n\to\infty}\mu_{M}(B_{n})=0$. It
follows that $\mu_{M}$ is $\sigma$-additive, too. ∎

Now, with $\mu_{M}$ we can associate an $\mathbb{M}_{\kappa}$-name for a real
$\dot{r}(\mu_{M})$ in the way described in Definition 4.10. Finally, let
$\dot{\mu}_{\mathcal{U}}$ be such an $\mathbb{M}_{\kappa}$-name for a function
$\wp(\omega)\to[0,1]$ such that for every $\mathbb{M}_{\kappa}$-name $\dot{M}$
for a subset of $\omega$ we have

$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(\dot{M})=\dot{r}(\mu_{M}).$

The following fact is proved in [Sol71] (see also [Kun76]).

######  Proposition 4.13.

$\Vdash_{\mathbb{M}_{\kappa}}``\dot{\mu}_{\mathcal{U}}\mbox{ is a finitely
additive probability measure on }\omega".$

###### Proof.

Let $G$ be an $\mathbb{M}_{\kappa}$-generic filter over the ground model $V$.
We need to show that in the extension $V[G]$ the interpretation
$\rho=(\dot{\mu}_{\mathcal{U}})_{G}$ of the name $\dot{\mu}_{\mathcal{U}}$
through $G$ is a finitely additive probability measure on $\omega$. It is
immediate that $\rho(\emptyset)=0$ and $\rho(\omega)=1$.

Let $A,B\in V[G]$ be two subsets of $\omega$ such that $A\cap B=\emptyset$. By
Lemma 4.4, there are in $V$ two elements $M,N\in\mathbb{M}_{\kappa}^{\omega}$
such that $\dot{M}_{G}=A$ and $\dot{N}_{G}=B$. By taking $N\setminus M$
instead of $N$ and appealing to Corollary 4.3, we may assume that $M\wedge
N=0$ , hence $\lambda_{\kappa}(M(k)\vee
N(k))=\lambda_{\kappa}(M(k))+\lambda_{\kappa}(N(k))$ for each $k\in\omega$. By
the additivity of the limit $\lim_{n\to\mathcal{U}}$, we have $\mu_{M\lor
N}=\mu_{M}+\mu_{N}$. By Remark 4.11 we are done. ∎

We call $\dot{\mu}_{\mathcal{U}}$ _the $\mathbb{M}_{\kappa}$-name for the
Solovay measure associated to the ultrafilter $\mathcal{U}$_. If $G$ is an
$\mathbb{M}_{\kappa}$-generic filter over $V$, then the evaluation
$(\dot{\mu}_{\mathcal{U}})_{G}$ in $V[G]$ is called _the Solovay measure
(associated to $\mathcal{U}$)_. Usually, we will simply write
$\dot{\mu}_{\mathcal{U}}$ instead of $(\dot{\mu}_{\mathcal{U}})_{G}$, in
particular, when we do not have in mind any specific extension $V[G]$.

The next proposition shows that, in $V^{\mathbb{M}_{\kappa}}$,
$\dot{\mu}_{\mathcal{U}}$ extends the ground model one-point measure
$\delta_{\mathcal{U}}$ on $\omega$, that is, the measure defined for every
$A\in\wp(\omega)\cap V$ as follows: $\delta_{\mathcal{U}}(A)=1$ if
$A\in\mathcal{U}$, and $\delta_{\mathcal{U}}(A)=0$ otherwise.

######  Proposition 4.14.

For every $A\in\wp(\omega)\cap V$ we have
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(A)=\delta_{\mathcal{U}}(A)$.

###### Proof.

For every $A\in\mathcal{U}$, $\mu_{\check{A}}=\lambda_{\kappa}$ and
$\mu_{(\omega\setminus A)\check{\ }}=0$, so
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(A)=1$ and
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(\omega\setminus A)=0$. ∎

Consequently, we obtain that if, in $V$, $\mathcal{A}\subseteq\wp(\omega)$,
$\varphi\colon\mathcal{A}\to[0,1]$ is a function, and there is
$A\in\mathcal{A}\cap\mathcal{U}$ such that $0<\varphi(A)<1$, then, in
$V^{\mathbb{M}_{\kappa}}$, $\dot{\mu}_{\mathcal{U}}$ does not extend
$\varphi$. In particular, $\dot{\mu}_{\mathcal{U}}$ is not an extension of any
non-atomic ground model measure on $\omega$.

###  4.4. P-measure in the random model.

For an ultrafilter $\mathcal{U}$ on $\omega$ by $\dot{\mu}_{\mathcal{U}}$ we
denote the $\mathbb{M}_{\kappa}$-name for the Solovay measure on $\omega$
associated to $\mathcal{U}$. Our main result is the following.

######  Theorem 4.15.

If $\mathcal{U}$ is a P-point, then
$\Vdash_{\mathbb{M}_{\kappa}}$“$\dot{\mu}_{\mathcal{U}}$ is a P-measure”.

Since there are P-points under the assumption of the Continuum Hypothesis, the
following result follows.

######  Corollary 4.16.

In the random model, obtained by forcing with $\mathbb{M}_{\omega_{2}}$ over a
model satisfying $\mathsf{CH}$, there is a P-measure.

By the discussion after Proposition 4.14 we get also that
$\Vdash_{\mathbb{M}_{\kappa}}$“$\dot{\mu}_{\mathcal{U}}$ does not extend the
asymptotic density”.

Recall that the support of a P-measure on $\omega$ is a nowhere dense ccc
closed P-set (see Lemma 3.1).

######  Corollary 4.17.

In the random model, there is a nowhere dense ccc closed P-set in
$\omega^{*}$.

To prove Theorem 4.15 we will need a lemma.

######  Lemma 4.18.

Suppose $\mathcal{U}$ is a P-point. Let $\nu$ be the measure on
$\mathbb{M}^{\omega}_{\kappa}$ defined for every
$M\in\mathbb{M}_{\kappa}^{\omega}$ by the formula
$\nu(M)=\lim_{i\to\mathcal{U}}\lambda_{\kappa}(M(i))$. Then, $\nu$ satisfies
the following property: whenever $(M_{n})$ is a sequence from
$\mathbb{M}^{\omega}_{\kappa}$ such that $M_{n+1}\leq M_{n}$ for each
$n\in\omega$, then there is $M\in\mathbb{M}^{\omega}_{\kappa}$ such that

  1. (1)

$M\leq^{*}M_{n}$ for each $n\in\omega$, and

  2. (2)

$\nu(M)=\lim_{n\to\infty}\nu(M_{n})$.

###### Proof.

Let $(M_{n})$ be a decreasing sequence in $\mathbb{M}^{\omega}_{\kappa}$. For
each $n\in\omega$ let $r_{n}=\nu(M_{n})$ and put

$U_{n}=\big{\\{}k\in\omega\colon\
|\lambda_{\kappa}(M_{n}(k))-r_{n}|<1/n\big{\\}}.$

Notice that $U_{n}\in\mathcal{U}$ for every $n\in\omega$.

Since $\mathcal{U}$ is a P-point, there is $U\in\mathcal{U}$ such that
$U\subseteq^{*}U_{n}$ for every $n\in\omega$. It follows that for every
$n\in\omega$ there is $l_{n}\in U$ such that
$U\setminus\\{0,\ldots,l_{n}-1\\}\in U_{n}$. In particular, $l_{n}\in U_{n}$.
Increasing sufficiently each $l_{n}$, we may assume that $l_{n+1}>l_{n}\geq n$
for every $n\in\omega$.

For each $k<l_{0}$ set $M(k)=1$. For each $k\geq l_{0}$ there is
$n_{k}\in\omega$ such that $l_{n_{k}}\leq k<l_{n_{k}+1}$, so we may set
$M(k)=M_{n_{k}}(k)$.

We will check that the sequence $M=(M(k))$ is as desired.

Since the sequence $(M_{n})$ is decreasing, for each $n\in\omega$ and $k\geq
l_{n}$ we have $M(k)\leq M_{n}(k)$, so (1) holds. We claim that (2) holds,
too, that is, that $\nu(M)=\lim_{n\to\infty}\nu(M_{n})$. Let $r=\nu(M)$. By
(1), for every $n\in\omega$ we have:

$\nu(M_{n})=\lim_{k\to\mathcal{U}}\lambda_{\kappa}(M_{n}(k))\geq\lim_{k\to\mathcal{U}}\lambda_{\kappa}(M(k))=\nu(M),$

which yields that $r_{n}\geq r$. It follows that $\lim_{n\to\infty}r_{n}\geq
r$ (note that the sequence $(r_{n})$ is decreasing). Assume that there exists
$\varepsilon>0$ such that $r_{n}>r+\varepsilon$ for every $n\in\omega$. Put

$A=\big{\\{}k\in\omega\colon\
|\lambda_{\kappa}(M(k))-r|<\varepsilon/2\big{\\}}.$

Of course, $A\in\mathcal{U}$, so $B=A\cap U\in\mathcal{U}$. Let $k\in B$ be
sufficiently large, so that there exists $n_{k}\in\omega$ such that
$l_{n_{k}}\leq k<l_{n_{k}+1}$ and $1/n_{k}<\varepsilon/2$. It follows that
$k\in U_{n_{k}}$, so

$|\lambda_{\kappa}(M_{n_{k}}(k))-r_{n_{k}}|<1/n_{k}<\varepsilon/2,$

hence

$\lambda_{\kappa}(M_{n_{k}}(k))>r_{n_{k}}-\varepsilon/2>r+\varepsilon/2.$

On the other hand, since $k\in A$, we also have that

$|\lambda_{\kappa}(M_{n_{k}}(k))-r|<\varepsilon/2,$

so

$\lambda_{\kappa}(M_{n_{k}}(k))<r+\varepsilon/2.$

But then, by ($*$) and ($**$), we get that

$\lambda_{\kappa}(M_{n_{k}}(k))>r+\varepsilon/2>\lambda_{\kappa}(M_{n_{k}}(k)),$

a contradiction. Thus, $\lim_{n\to\infty}r_{n}=r$, and the proof is finished.
∎

We are in the position to prove the theorem.

###### Proof of Theorem 4.15.

We work in the ground model $V$. Suppose for the sake of contradiction that
there is a condition $p\in\mathbb{M}_{\kappa}$ forcing that
$\dot{\mu}_{\mathcal{U}}$ is not a P-measure. Then, by Lemma 3.3, the Maximum
Principle, and Lemma 4.4, there is a condition $q\leq p$, a rational
$\varepsilon>0$, and a sequence $(M_{n})$ of elements of
$\mathbb{M}_{\kappa}^{\omega}$ such that

$q\Vdash``(\dot{M}_{n})\mbox{ is a decreasing sequence of subsets of
}\omega",$

and

$q\Vdash``\mbox{if }\dot{M}\mbox{ is a pseudointersection of
}(\dot{M}_{n})\mbox{, then }\forall n\in\omega\
\dot{\mu}_{\mathcal{U}}(\dot{M})<\dot{\mu}_{\mathcal{U}}(\dot{M}_{n})-\varepsilon".$

For each $n\in\omega$ define $N_{n}\in\mathbb{M}^{\omega}_{\kappa}$ by
$N_{n}=M_{n}\wedge\vec{q}$. Then, by Lemma 4.7.(1), the sequence $(N_{n})$ is
decreasing and we can apply Lemma 4.18 to find
$M\in\mathbb{M}^{\omega}_{\kappa}$ such that

  1. (1)

$M\leq^{*}N_{n}$ for each $n\in\omega$, and

  2. (2)

$\nu(M)=\lim_{n\to\infty}\nu(N_{n})$, where the measure $\nu$ is defined as in
Lemma 4.18.

Since, by Lemmas 4.5 and 4.7.(2), $q\Vdash``\dot{M}\mbox{ is a
pseudointersection of }(\dot{M}_{n})"$, for every $n\in\omega$ we have:

$q\Vdash\dot{\mu}_{\mathcal{U}}(\dot{M})<\dot{\mu}_{\mathcal{U}}(\dot{M}_{n})-\varepsilon.$

Let $f,f_{n}\colon 2^{\kappa}\to\mathbb{R}$ be Borel functions associated to
$\mu_{M}$ and $\mu_{M_{n}}$, $n\in\omega$, via the Radon–Nikodym theorem (see
Definition 4.10). Then, for every $n\in\omega$ we have

$f<f_{n}-\varepsilon\quad\mbox{almost everywhere on }q,$

which follows from Lemma 4.8.(1) (applied to the function $f-f_{n}$ and the
set $A=(-\infty,-\varepsilon)$).

Integrating both sides of inequality ($*$) (over $q$), we get for every
$n\in\omega$ that

$\int_{q}f\ d\lambda_{\kappa}<\int_{q}f_{n}\
d\lambda_{\kappa}-\varepsilon\lambda_{\kappa}(q).$

Then, by the definitions of $\mu_{M}$ and $\mu_{M_{n}}$, we have:

$\mu_{M}(q)<\mu_{M_{n}}(q)-\varepsilon\lambda_{\kappa}(q),$

and so

$\nu(M\wedge\vec{q})<\nu(M_{n}\wedge\vec{q})-\varepsilon\lambda_{\kappa}(q).$

Since for each $n\in\omega$ we have $N_{n}=M_{n}\wedge\vec{q}$ and
$M\leq^{*}N_{n}\leq\vec{q}$, it holds

$\nu(M)<\nu(N_{n})-\varepsilon\lambda_{\kappa}(q),$

for every $n\in\omega$, which contradicts (2) above. ∎

The converse to Theorem 4.15 also holds.

######  Theorem 4.19.

Let $\mathcal{U}$ be a ground model ultrafilter on $\omega$ such that
$\Vdash_{\mathbb{M}_{\kappa}}$“$\dot{\mu}_{\mathcal{U}}$ is a P-measure”.
Then, $\mathcal{U}$ is a P-point in $V$.

###### Proof.

Suppose for the sake of contradiction that $\mathcal{U}$ is not a P-point in
$V$ and let $(A_{n})$ be a sequence witnessing it, i.e. $(A_{n})$ is a
decreasing sequence such that:

  * (a)

$A_{0}=\omega$ and $A_{n}\in\mathcal{U}$ for each $n\in\omega$,

  * (b)

for every $A\in\wp(\omega)\cap V$, if $A\setminus A_{n}$ is finite for each
$n\in\omega$, then $A\notin\mathcal{U}$.

Since $\Vdash_{\mathbb{M}_{\kappa}}$“$\dot{\mu}_{\mathcal{U}}$ is a
P-measure”, by (a) and Proposition 4.14, there is
$M\in\mathbb{M}_{\kappa}^{\omega}$ such that
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(\dot{M})=1$ and
$\Vdash_{\mathbb{M}_{\kappa}}$“$\dot{M}\setminus A_{n}$ is finite for each
$n\in\omega$”. In particular, for every $n\in\omega$ we have that
$\Vdash_{\mathbb{M}_{\kappa}}$“$\dot{M}\cap(A_{n}\setminus A_{n+1})$ is
finite”. Then, by Lemma 4.5,

$\lambda_{\kappa}\Big{(}\bigwedge_{l\in\omega}\bigvee_{\begin{subarray}{c}k\in
A_{n}\setminus A_{n+1}\\\ k>l\end{subarray}}M(k)\Big{)}=0$

for each $n\in\omega$. So, for each $n\in\omega$, there is a finite subset
$F_{n}\subseteq A_{n}\setminus A_{n+1}$ such that

$\lambda_{\kappa}\Big{(}\bigvee_{k\in(A_{n}\setminus A_{n+1})\setminus
F_{n}}M(k)\Big{)}<1/2^{n+2}.$

Let $A=\bigcup_{n\in\omega}F_{n}$. Obviously, $A\in V$ and $A\setminus A_{n}$
is finite for every $n\in\omega$ (since $A\setminus
A_{n}\subseteq\bigcup_{i=0}^{n-1}F_{i}$), so by (b) we have
$A\not\in\mathcal{U}$ and hence $A^{c}\in\mathcal{U}$. By Proposition 4.14,
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(A^{c})=1$. Let

$p=\bigvee_{k\in A^{c}}M(k)=\bigvee_{n\in\omega}\bigvee_{k\in(A_{n}\setminus
A_{n+1})\setminus F_{n}}M(k).$

Notice that $\lambda(p)\leq 1/2$ and so, in particular, $p\neq 1$. Hence,
setting $q=1\setminus p$, we have $q\neq 0$. Then $q\Vdash\dot{M}\cap
A^{c}=\emptyset$, and so $q\Vdash\dot{M}\subseteq A$. Since
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(\dot{M})=1$, we get that
$q\Vdash\dot{\mu}_{\mathcal{U}}(A)=1$. But
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(A^{c})=1$, so
$q\Vdash\dot{\mu}_{\mathcal{U}}(A^{c})=1$, too, which is a contradiction. ∎

######  Corollary 4.20.

Let $\mathcal{U}$ be a ground model ultrafilter on $\omega$. Then,
$\mathcal{U}$ is a P-point in $V$ if and only if
$\Vdash_{\mathbb{M}_{\kappa}}$“$\dot{\mu}_{\mathcal{U}}$ is a P-measure”.

##  5\. Non-atomicity of Solovay measures

Recall that each P-point induces a P-measure which is a one-point measure. So,
if we could show that the P-measures whose existence we have shown in the
previous section can be one-point measures, then we would prove the existence
of a P-point in the random model. Unfortunately, as we will see, all Solovay
measures are non-atomic.

Suppose $\mathcal{U}$ is an ultrafilter on $\omega$, in $V$. Let $G$ be an
$\mathbb{M}_{\kappa}$-generic filter over $V$. Then, in $V[G]$, $\mathcal{U}$
generates a filter on $\omega$, and so a closed subset of $\beta\omega$.
Denote this set by $K_{\mathcal{U}}$. In other words, in $V[G]$,

$K_{\mathcal{U}}=\bigcap_{X\in\mathcal{U}}[X].$

(Recall, that here $[X]$ is the clopen subset of $\beta\omega$ associated to a
subset $X$ of $\omega$, i.e. $[X]=[X]_{\wp(\omega)}$.) Now, still in $V[G]$,
the finitely additive Solovay measure $(\dot{\mu}_{\mathcal{U}})_{G}$
associated with $\mathcal{U}$ can be uniquely extended to a regular Borel
$\sigma$-additive probability measure $\nu_{\mathcal{U}}$ on $\beta\omega$.
Let $\dot{K}_{\mathcal{U}}$ and $\dot{\nu}_{\mathcal{U}}$ denote such
$\mathbb{M}_{\kappa}$-names in $V$ that in every $\mathbb{M}_{\kappa}$-generic
extension $V[G]$ we have $(\dot{K}_{\mathcal{U}})_{G}=K_{\mathcal{U}}$ and
$(\dot{\nu}_{\mathcal{U}})_{G}=\nu_{\mathcal{U}}$. It appears that each
measure of the form $\nu_{\mathcal{U}}$ is carried by $K_{\mathcal{U}}$, i.e.,
that $\operatorname{supp}(\nu_{\mathcal{U}})\subseteq K_{\mathcal{U}}$.

######  Proposition 5.1.

For each ultrafilter $\mathcal{U}$ on $\omega$,

$\Vdash_{\mathbb{M}_{\kappa}}\dot{\nu}_{\mathcal{U}}(\dot{K}_{\mathcal{U}})=1.$

###### Proof.

By Proposition 4.14 for every $X\in\mathcal{U}$ we have
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(\check{X})=1$, so
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\nu}_{\mathcal{U}}([\check{X}])=1$. The
measure $\dot{\nu}_{\mathcal{U}}$ is a Radon measure in
$V^{\mathbb{M}_{\kappa}}$, hence it is $\tau$-additive (by [Bog07, Proposition
7.2.2.(i)]). Since $\\{[X]\colon X\in\mathcal{U}\\}$ is a decreasing net of
closed sets (indexed by the directed set $(\mathcal{U},\supseteq)$),
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\nu}_{\mathcal{U}}(\dot{K}_{\mathcal{U}})=1$
(by [Bog07, Proposition 7.2.5]). ∎

We will now show that Solovay measures are always non-atomic, regardless of an
associated ground model ultrafilter. This fact will be an immediate
consequence of the existence of a family
$\mathcal{M}=\\{M_{s}\in\mathbb{M}_{\kappa}^{\omega}\colon s\in
2^{<\omega}\\}$ having for every $s\in 2^{<\omega}$ the following properties:

  1. (1)

$M_{s}$ is a disjoint union of $M_{s^{\smallfrown}0}$ and
$M_{s^{\smallfrown}1}$, and

  2. (2)

if $\mathcal{U}$ is a ground model ultrafilter, then
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(\dot{M}_{s})=1/2^{|s|}$.

The definition of $\mathcal{M}$ is as follows. We let $M_{\epsilon}=\vec{1}$
and for each $s\in 2^{<\omega}$, $|s|>0$, and $k\in\omega$, we put:

$M_{s}(k)=\big{\\{}x\in 2^{\kappa}\colon\ x(k+i)=s(i)\text{ for all }0\leq
i<|s|\big{\\}}.$

Note that trivially $\lambda_{\kappa}(M_{s}(k))=1/2^{|s|}$.

Property (1) of $\mathcal{M}$ is obviously satisfied. We will now show that
(2) also holds.

######  Lemma 5.2.

Let $\mathcal{U}$ be a ground model ultrafilter. For every $n\in\omega$ and
$s\in 2^{n}$, the Radon–Nikodym derivative $f_{M_{s}}$ of $\mu_{M_{s}}$ is
$\lambda_{\kappa}$-a.e. equal to $1/2^{n}$, and so it holds

$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(\dot{M}_{s})=1/2^{n}.$

###### Proof.

Let $C\subseteq 2^{\kappa}$ be a clopen. Then it depends on finitely many
coordinates, i.e. there is a finite set $I\subseteq\kappa$ such that for every
$x\in C$ and $y\in 2^{\kappa}$, if $x(\alpha)=y(\alpha)$ for all $\alpha\in I$
then $y\in C$.

It follows that there is $K\in\omega$ such that for every $k\in\omega$, $k>K$,
we have $k\not\in I$. Hence, for every such $k$ it holds
$\lambda_{\kappa}(M_{s}(k)\wedge C)=1/2^{n}\lambda_{\kappa}(C)$, and so

$\lim_{k\to\infty}\lambda_{\kappa}(M_{s}(k)\wedge
C)=1/2^{n}\lambda_{\kappa}(C).$

Now, let $B\subseteq 2^{\kappa}$ be a Borel set. Fix $\varepsilon>0$. There is
a clopen $C\subseteq 2^{\kappa}$ such that $\lambda_{\kappa}(B\triangle
C)<\varepsilon$. Let $K\in\omega$ be as in the previous paragraph, so that
$\lambda_{\kappa}(M_{s}(k)\wedge C)=1/2^{n}\lambda_{\kappa}(C)$ for every
$k\in\omega$, $k>K$. Then, for every such $k$ we have:

$|\lambda_{\kappa}(M_{s}(k)\wedge B)-1/2^{n}\lambda_{\kappa}(B)|\leq$
$\leq|\lambda_{\kappa}(M_{s}(k)\wedge B)-\lambda_{\kappa}(M_{s}(k)\wedge
C)|+|\lambda_{\kappa}(M_{s}(k)\wedge C)-1/2^{n}\lambda_{\kappa}(B)|\leq$
$\leq\lambda_{\kappa}(M_{s}(k)\wedge(B\triangle
C))+|1/2^{n}\lambda_{\kappa}(C)-1/2^{n}\lambda_{\kappa}(B)|<$
$<(1+1/2^{n})\varepsilon.$

Since $\varepsilon$ was arbitrary, we get that

$\lim_{k\to\infty}\lambda_{\kappa}(M_{s}(k)\wedge
B)=1/2^{n}\lambda_{\kappa}(B).$

It means that $\mu_{M_{s}}=1/2^{n}d\lambda_{\kappa}$, so the Radon–Nikodym
derivative $f_{M_{s}}$ of $\mu_{M_{s}}$ is almost everywhere constantly equal
to $1/2^{n}$. Hence,
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(\dot{M}_{s})=1/2^{n}$. ∎

######  Corollary 5.3.

For every ultrafilter $\mathcal{U}$ on $\omega$,

$\Vdash_{\mathbb{M}_{\kappa}}``\dot{\mu}_{\mathcal{U}}\text{ is non-atomic}".$

###### Proof.

Fix $\varepsilon>0$ and let $n\in\omega$ be such that $\varepsilon>1/2^{n}$.
Consider $\\{\dot{M}_{s}\colon\ s\in 2^{n}\\}$ to obtain a desired partition
of $\omega$ in $V^{\mathbb{M}_{\kappa}}$. ∎

In Lemma 5.2 we showed that, for every $s\in 2^{<\omega}$, the function
$f_{M_{s}}$ is constantly equal to $1/2^{|s|}$ ($\lambda_{\kappa}$-a.e.). It
appears that, exploiting this fact, one can show that for every ground model
$\alpha\in[0,1]$ there exists an element
$M_{\alpha}\in\mathbb{M}_{\kappa}^{\omega}$ such that the function
$f_{M_{\alpha}}$ is constantly equal to $\alpha$ ($\lambda_{\kappa}$-a.e) and
hence, for every ground model ultrafilter $\mathcal{U}$,
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(\dot{M}_{\alpha})=\alpha$.

We construct elements $M_{\alpha}$’s as follows. Set $M_{0}=\check{\emptyset}$
and notice that for every $\alpha\in[0,1]$ there is a (finite or infinite)
antichain $I_{\alpha}\in 2^{<\omega}$ such that $\alpha=\sum_{s\in
I_{\alpha}}1/2^{|s|}$ and set for every $k\in\omega$:

$M_{\alpha}(k)=\bigvee_{s\in I_{\alpha}}M_{s}(k).$

Note that for each distinct $s,t\in I_{\alpha}$ we have $M_{s}(k)\cap
M_{t}(k)=\emptyset$. Additionally, put

$\dot{\mathcal{M}}_{\sigma}=\\{M_{\alpha}\colon\ \alpha\in[0,1]\\}.$

######  Theorem 5.4.

Let $\mathcal{U}$ be a ground model ultrafilter. For every $\alpha\in[0,1]\cap
V$, the Radon–Nikodym derivative $f_{M_{\alpha}}$ of $\mu_{M_{\alpha}}$ is
$\lambda_{\kappa}$-a.e. equal to $\alpha$, and so we have:

$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}(\dot{M}_{\alpha})=\alpha.$

In particular,
$\Vdash_{\mathbb{M}_{\kappa}}\dot{\mu}_{\mathcal{U}}[\dot{\mathcal{M}}_{\sigma}]=[0,1]\cap
V$.

###### Proof.

Fix $\alpha\in[0,1]\cap V$ and a Borel $B\subseteq 2^{\kappa}$. We will show
that

$\lim_{k\to\infty}\lambda_{\kappa}(M_{\alpha}(k)\wedge
B)=\alpha\lambda_{\kappa}(B).$

For $\alpha=0$ and $M_{0}$ we are obviously done, so assume that $\alpha>0$.
If $I_{\alpha}$ is finite, then the equality easily follows from the
additivity of $\lambda_{\kappa}$ and $\lim_{k\to\infty}$. So assume that
$I_{\alpha}$ is infinite. Enumerate $I_{\alpha}=(s_{n})$ and for each
$n\in\omega$ set $\alpha_{n}=1/2^{|s_{n}|}$. Then,
$\alpha=\sum_{n\in\omega}\alpha_{n}$. For each $n,k\in\omega$ let

$f_{k}(n)=\lambda_{\kappa}(M_{s_{n}}(k)\wedge B).$

Hence, for each $k\in\omega$ we have $f_{k}\colon\omega\to[0,1]$ and
$\sum_{n\in\omega}f_{k}(n)\leq 1$. For each $n\in\omega$, by (the proof of)
Lemma 5.2, we also have:

$\lim_{k\to\infty}f_{k}(n)=\alpha_{n}\lambda_{\kappa}(B).$

Finally, for each $k,n\in\omega$ it holds $|f_{k}(n)|\leq
1/2^{|s_{n}|}=\alpha_{n}$, so for the function $g\colon\omega\to[0,1]$, given
for every $n\in\omega$ by the formula $g(n)=\alpha_{n}$, we have
$|f_{k}(n)|\leq g(n)$ for every $k,n\in\omega$ and $\sum_{n\in\omega}g(n)\leq
1$.

Using the $\sigma$-additivity of $\lambda_{\kappa}$ and Lebesgue’s Dominated
Convergence Theorem (applied to the functions $f_{k}$’s and $g$ and the
counting measure on $\omega$), we get:

$\lim_{k\to\infty}\lambda_{\kappa}(M_{\alpha}(k)\wedge
B)=\lim_{k\to\infty}\sum_{n\in\omega}\lambda_{\kappa}(M_{s_{n}}(k)\wedge
B)=\sum_{n\in\omega}\lim_{k\to\infty}\lambda_{\kappa}(M_{s_{n}}(k)\wedge B)=$
$=\sum_{n\in\omega}\alpha_{n}\lambda_{\kappa}(B)=\lambda_{\kappa}(B)\sum_{n\in\omega}\alpha_{n}=\alpha\lambda_{\kappa}(B),$

which is what we wanted.

To finish the proof, note that as previously the Radon–Nikodym derivative
$f_{M_{\alpha}}$ is constantly equal to $\alpha$ ($\lambda_{\kappa}$-a.e.). ∎

Theorem 5.4 asserts that the chunk
$\dot{\mu}_{\mathcal{U}}\restriction\dot{\mathcal{M}}_{\sigma}$ of every
Solovay measure $\dot{\mu}_{\mathcal{U}}$ attains every ground model real
value and is described by simple $\mathbb{M}_{\kappa}$-names (the elements of
$\dot{\mathcal{M}}_{\sigma}$). It also shows that every two Solovay measures,
whatever their associated ground model ultrafilters are, have the same values
on $\dot{\mathcal{M}}_{\sigma}$. Note that by Proposition 4.14 the
interpretations of names from $\dot{\mathcal{M}}_{\sigma}$ through an
$\mathbb{M}_{\kappa}$-generic filter do not belong to the ground model (except
for $M_{0}$ and $M_{1}$).

Let us finish this section with the following remark. In Corollary 3.4,
referring to the proof of the classical Sierpiński theorem, we claimed that
every non-atomic P-measure on $\omega$ attains every value in the interval
$[0,1]$. It appears however that in the case of Solovay P-measures one can
obtain the same result without actually using Sierpiński’s theorem—it is
enough to repeat the (slightly adjusted) argument presented in the proof of
Theorem 5.4.

###  5.1. Semi-selective ultrafilters

Now, we will show that for a particular subclass of P-points their Solovay
measures are as far from being one-point measures as possible, in the sense
that their Radon extensions are strictly positive on the sets
$K_{\mathcal{U}}$’s (and so their supports are as big as possible in the light
of Proposition 5.1).

Recall the following standard definition of a semi-selective ultrafilter.

######  Definition 5.5.

An ultrafilter $\mathcal{U}$ on $\omega$ is called _semi-selective_ if
whenever $(a_{n})$ is a sequence of non-negative reals such that
$\lim_{n\to\mathcal{U}}a_{n}=0$, then there is $X\in\mathcal{U}$ such that
$\sum_{n\in X}a_{n}<\infty$.

Note that an ultrafilter $\mathcal{U}$ is semi-selective if and only if it is
a rapid P-point (see e.g. [Can90]). In particular, every selective ultrafilter
is semi-selective. The next theorem provides two characterizations of semi-
selective ultrafilters: one in terms resembling the classical Borel–Cantelli
lemma, and the other in terms of the supports of Solovay measures (cf. also
[Gre19, Theorem 3.2]).

######  Theorem 5.6.

Suppose $\mathcal{U}$ is an ultrafilter on $\omega$. Then the following are
equivalent:

  1. (1)

$\mathcal{U}$ is semi-selective;

  2. (2)

for every probability measure $\mu$ on $2^{\kappa}$ and sequence $(P_{k})$ in
$\mathrm{Bor}(2^{\kappa})/\mathcal{N}(\mu)$, if
$\lim_{k\to\mathcal{U}}\mu(P_{k})=0$, then there is $X\in\mathcal{U}$ such
that

$\bigwedge_{k\in X}\bigvee_{\begin{subarray}{c}i\in X\\\
i>k\end{subarray}}P_{i}=0;$

  3. (3)

$\Vdash_{\mathbb{M}_{\kappa}}$“$\dot{\nu}_{\mathcal{U}}$ is strictly positive
on $\dot{K}_{\mathcal{U}}$”.

($\mathcal{N}(\mu)$ in (2) denotes the ideal of $\mu$-null Borel subsets of
$2^{\kappa}$.)

###### Proof.

(1) $\implies$ (2). Let $\mu$ be a probability measure on $2^{\kappa}$ and
$(P_{k})$ a sequence in $\mathrm{Bor}(2^{\kappa})/\mathcal{N}(\mu)$ such that
$\lim_{k\to\mathcal{U}}\mu(P_{k})=0$. Since $\mathcal{U}$ is semi-selective,
there is $X\in\mathcal{U}$ such that $\sum_{k\in X}\mu(P_{k})<\infty$. The
Borel–Cantelli lemma implies that

$\bigwedge_{k\in X}\bigvee_{\begin{subarray}{c}i\in X\\\
i>k\end{subarray}}P_{i}=0.$

(2) $\implies$ (3). Recall that
$1\Vdash\operatorname{supp}(\dot{\nu}_{\mathcal{U}})\subseteq\dot{K}_{\mathcal{U}}$.
Fix $p\in\mathbb{M}_{\kappa}$ and $M\in\mathbb{M}_{\kappa}^{\omega}$ such that
$p\Vdash\dot{\nu}_{\mathcal{U}}([\dot{M}])=0$. We will show that
$p\Vdash[\dot{M}]\cap\dot{K}_{\mathcal{U}}=\emptyset$. Set $P=M\wedge\vec{p}$.

We claim that $\lim_{k\to\mathcal{U}}\lambda_{\kappa}(\llbracket
k\in\dot{P}\rrbracket)=0$. Indeed, since $P\leq M$, $\mu_{P}\leq\mu_{M}$, so
$f_{P}\leq f_{M}$ $\lambda_{\kappa}$-a.e., and hence, by Corollary 4.9,
$1\Vdash\dot{\mu}_{\mathcal{U}}(\dot{P})\leq\dot{\mu}_{\mathcal{U}}(\dot{M})$.
It follows that $p\Vdash\dot{\mu}_{\mathcal{U}}(\dot{P})=0$. Since
$P\leq\vec{p}$, by the definition of $f_{P}$ we have $f_{P}=0$
$\lambda_{\kappa}$-a.e. on $1\setminus p$, so by Lemma 4.8.(2) we have
$1\setminus p\Vdash\dot{\mu}_{\mathcal{U}}(\dot{P})=0$. In total,
$1\Vdash\dot{\mu}_{\mathcal{U}}(\dot{P})=0$. Finally, again by Lemma 4.8.(2),
we have that $f_{P}=0$ $\lambda_{\kappa}$-a.e., so
$\lim_{k\to\mathcal{U}}\lambda_{\kappa}(\llbracket
k\in\dot{P}\rrbracket)=\mu_{P}(1)=0$.

By (2) there is $X\in\mathcal{U}$ such that

$\bigwedge_{k\in X}\bigvee_{\begin{subarray}{c}i\in X\\\
i>k\end{subarray}}\llbracket i\in\dot{P}\rrbracket=0.$

Since $\llbracket i\in\dot{P}\rrbracket=\llbracket i\in\dot{M}\rrbracket\wedge
p$ for each $i\in\omega$, ($*$) means that $p\Vdash
X\cap\dot{M}=^{*}\emptyset$, and thus $p\Vdash[X]\cap[\dot{M}]=\emptyset$.
From $1\Vdash\dot{K}_{\mathcal{U}}\subseteq[X]$ (recall that
$X\in\mathcal{U}$) we get that
$p\Vdash\dot{K}_{\mathcal{U}}\cap[\dot{M}]=\emptyset$.

(3) $\implies$ (1). If $\mathcal{U}$ is not semi-selective, then there is a
sequence $(a_{n})\in[0,1]^{\omega}$ such that $\lim_{n\to\mathcal{U}}a_{n}=0$
and $\sum_{n\in X}a_{n}=\infty$ for each $X\in\mathcal{U}$. By [BJ95, Lemma
1.3.23], there is $M\in\mathbb{M}^{\omega}_{\kappa}$ such that
$\lambda_{\kappa}(M(k))=a_{k}$ for each $k\in\omega$ and the sequence $(M(k))$
is $\lambda_{\kappa}$-independent. It follows that
$\lim_{k\to\mathcal{U}}\lambda_{\kappa}(\llbracket k\in\dot{M}\rrbracket)=0$
and so $1\Vdash\dot{\mu}_{\mathcal{U}}(\dot{M})=0$. We claim that
$1\Vdash[\dot{M}]\cap K_{\mathcal{U}}\neq\emptyset$. Indeed, if
$X\in\mathcal{U}$, then $\sum_{k\in X}a_{k}=\infty$ and so, by the second
Borel–Cantelli lemma,

$\bigwedge_{k\in X}\bigvee_{\begin{subarray}{c}n\in X\\\
n>k\end{subarray}}M(n)=1.$

But this precisely means that $1\Vdash\dot{M}\cap\check{X}\neq^{*}\emptyset$.
As $\mathcal{U}$ is a directed family and $1\Vdash$“$\dot{K}_{\mathcal{U}}$ is
compact and non-empty”,
$1\Vdash[\dot{M}]\cap\dot{K}_{\mathcal{U}}\neq\emptyset$, which is a
contradiction since we have $1\Vdash\dot{\mu}_{\mathcal{U}}(\dot{M})=0$ and
$1\Vdash\operatorname{supp}(\dot{\nu}_{\mathcal{U}})=\dot{K}_{\mathcal{U}}$. ∎

As a corollary we get the following well-known theorem, proved by Kunen for
$\kappa\geq\omega_{2}$ (see [Kun76, Lemma 5.2]).

######  Corollary 5.7.

In $V^{\mathbb{M}_{\kappa}}$ no P-point extends a semi-selective ultrafilter
from the ground model.

###### Proof.

Let $G$ be an $\mathbb{M}_{\kappa}$-generic filter over $V$. Let $\mathcal{U}$
be a semi-selective ultrafilter in $V$. Suppose that, in $V[G]$, an
ultrafilter $\mathcal{W}$ extends $\mathcal{U}$. It means that $\mathcal{W}\in
K_{\mathcal{U}}$. By Corollary 5.3 and Theorem 5.6, $K_{\mathcal{U}}$ supports
a non-atomic measure. So, according to Proposition 3.7, $\mathcal{W}$ cannot
be a P-point. ∎

###  5.2. Canjar ultrafilters

The results of Section 5.1 say that if we want to look for a P-point in the
random model which is an extension of a ground model ultrafilter, then we have
to avoid semi-selective ultrafilters. In particular, whenever $\mathcal{U}$ is
a ground model semi-selective ultrafilter, then there is a decreasing sequence
$(E_{n})$ of infinite subsets of $\omega$ (in the extension) such that
$\mathcal{U}\cup\\{E_{n}\colon n\in\omega\\}$ generates a filter which cannot
be further extended to a bigger filter by _any_ pseudointersection of
$(E_{n})$. (It is enough to take $\dot{E}_{n}$’s such that
$\dot{\mu}_{\mathcal{U}}(\dot{E}_{n})<1/(n+1)$ for each $n\in\omega$ (cf.
Lemma 5.2) and then use Theorem 5.6). In this section we will show that such a
sequence $(E_{n})$ cannot exist for any Canjar ultrafilter $\mathcal{U}$.
Therefore, perhaps the ground model Canjar ultrafilters have chances to be
’extendable’ to a P-point in the random extension.

For an ultrafilter $\mathcal{U}$ and a condition $p\in\mathbb{M}_{\kappa}$,
say that $M\in\mathbb{M}^{\omega}_{\kappa}$ is $(\mathcal{U},p)$-_full_ if for
each $X\in\mathcal{U}$ we have

$\lambda_{\kappa}\big{(}\bigvee_{k\in X}M(k)\wedge
p\big{)}=\lambda_{\kappa}(p).$

Equivalently, $M$ is $(\mathcal{U},p)$-full if and only if $\bigvee_{k\in
X}M(k)\geq p$ for every $X\in\mathcal{U}$.

######  Proposition 5.8.

Assume that $\mathcal{U}$ is an ultrafilter on $\omega$ and
$p\in\mathbb{M}_{\kappa}$ is a condition. Let
$M\in\mathbb{M}_{\kappa}^{\omega}$. Then, $M$ is $(\mathcal{U},p)$-full if and
only if $p\Vdash$“$\mathcal{U}\cup\\{\dot{M}\\}$ forms a base of a filter”.

###### Proof.

Suppose first that $M$ is $(\mathcal{U},p)$-full but
$p\nVdash$“$\mathcal{U}\cup\\{\dot{M}\\}$ forms a base of a filter”. It means
that there is $q\leq p$ and $X\in\mathcal{U}$ such that $q\Vdash
X\cap\dot{M}=\emptyset$. But then $\bigvee_{k\in X}M(k)\land q=0$ and so $M$
is not $(\mathcal{U},p)$-full.

Conversely, suppose that $M$ is not $(\mathcal{U},p)$-full. Then, let
$X\in\mathcal{U}$ be such that for $q=\bigvee_{k\in X}M(k)\wedge p$ we have
$q<p$. Then, $p\setminus q\Vdash\dot{M}\cap X=\emptyset$, and so $p\setminus
q\Vdash$ “there is no filter containing $\mathcal{U}$ and $\dot{M}$”. ∎

In the following definition we treat $\mathcal{U}$ as a subset of
$2^{\omega}$. Recall that a sequence $(I_{n})$ of finite subsets of $\omega$
is _an interval partition_ if $\omega=\bigcup_{n\in\omega}I_{n}$ and $\max
I_{n}=\min I_{n+1}-1$ for every $n\in\omega$.

######  Definition 5.9.

An ultrafilter $\mathcal{U}$ is called _Canjar_ 222The original definition of
Canjar ultrafilters is different but equivalent to the following one (cf.
[BHV13]). if for every sequence $(\mathcal{C}_{n})$ of compact subsets of
$\mathcal{U}$ there is an interval partition $(I_{n})$ of $\omega$ such that
for every sequence $(X_{n})$, where $X_{n}\in\mathcal{C}_{n}$ for every
$n\in\omega$, we have

$X=\bigcup_{n\in\omega}(X_{n}\cap I_{n})\in\mathcal{U}.$

Note that every Canjar ultrafilter is a P-point: if $(X_{n})$ is a decreasing
sequence in $\mathcal{U}$, then $X$ defined as above for the sequence
$\mathcal{C}_{n}=\\{X_{n}\\}$ is a pseudointersection of $(X_{n})$. The above
definition is however stronger than that of a P-point—it says that we can
diagonalize _many_ sequences of elements of an ultrafilter at the same time.

The hope that Canjar ultrafilters may have an extension to a P-point in the
random model comes from the following result and its consequence, Theorem
5.11.

######  Proposition 5.10.

Let $\mathcal{U}$ be a Canjar ultrafilter and fix a condition
$p\in\mathbb{M}_{\kappa}$. Suppose that $(E_{n})$ is a decreasing sequence of
elements of $\mathbb{M}_{\kappa}^{\omega}$ which are $(\mathcal{U},p)$-full.
Then, there is $(\mathcal{U},p)$-full $E\in\mathbb{M}_{\kappa}^{\omega}$ such
that $E\leq^{*}E_{n}$ for each $n\in\omega$.

###### Proof.

For every $n\in\omega$ put

$\mathcal{C}_{n}=\Big{\\{}X\in\mathcal{U}\colon\
\lambda_{\kappa}\Big{(}\bigvee_{k\in\omega\setminus X}E_{n}(k)\wedge
p\Big{)}\leq\lambda_{\kappa}(p)-1/(n+1)\Big{\\}}.$

Notice that the set $\mathcal{C}_{n}$ is a compact subset of $\mathcal{U}$
such that $\omega\in\mathcal{C}_{n}$. Indeed, suppose that
$X\in\wp(\omega)\setminus\mathcal{C}_{n}$. If $X\notin\mathcal{U}$, then
$\omega\setminus X\in\mathcal{U}$ and so

$\lambda_{\kappa}\Big{(}\bigvee_{k\in\omega\setminus X}E_{n}(k)\wedge
p\Big{)}=\lambda_{\kappa}(p),$

as $E_{n}$ is $(\mathcal{U},p)$-full. If
$X\in\mathcal{U}\setminus\mathcal{C}_{n}$, then

$\lambda_{\kappa}\Big{(}\bigvee_{k\in\omega\setminus X}E_{n}(k)\wedge
p\Big{)}>\lambda_{\kappa}(p)-1/(n+1).$

In any case, there is $N\in\omega$ such that

$\lambda_{\kappa}\Big{(}\bigvee_{k\in N\setminus X}E_{n}(k)\wedge
p\Big{)}>\lambda_{\kappa}(p)-1/(n+1).$

Then, $\\{Y\subseteq\omega\colon Y\cap N=X\cap N\\}$ is an open neighbourhood
of $X$ disjoint with $\mathcal{C}_{n}$. It follows that $\mathcal{C}_{n}$ is a
closed subset of $2^{\omega}$, hence it is compact.

As $\mathcal{U}$ is Canjar, there is an interval partition $(I_{n})$ such
that, for every $(X_{n})$ with $X_{n}\in\mathcal{C}_{n}$ for each
$n\in\omega$, we have $\bigcup_{n\in\omega}(X_{n}\cap I_{n})\in\mathcal{U}$.
For every $n\in\omega$ and $k\in I_{n}$ set

$E(k)=E_{n}(k).$

Clearly, $E\leq^{*}E_{n}$ for each $n\in\omega$. We are going to show that $E$
is $(\mathcal{U},p)$-full.

Suppose that $Y\subseteq\omega$ and $n_{0}\in\omega$ are such that

$\lambda_{\kappa}\Big{(}\bigvee_{k\in Y}E(k)\wedge
p\Big{)}\leq\lambda_{\kappa}(p)-1/(n_{0}+1)<\lambda_{\kappa}(p).$

Let $X_{n}=(I_{n}\cap Y)^{c}$ for each $n\in\omega$. Since $X_{n}^{c}$ is
finite, $X_{n}\in\mathcal{U}$. Then, for every $n>n_{0}$ we have:

$\lambda_{\kappa}\Big{(}\bigvee_{k\in\omega\setminus X_{n}}E_{n}(k)\wedge
p\Big{)}=\lambda_{\kappa}\Big{(}\bigvee_{k\in I_{n}\cap Y}E_{n}(k)\wedge
p\Big{)}=$ $=\lambda_{\kappa}\Big{(}\bigvee_{k\in I_{n}\cap Y}E(k)\wedge
p\Big{)}\leq\lambda_{\kappa}\Big{(}\bigvee_{k\in Y}E(k)\wedge
p\Big{)}\leq\lambda_{\kappa}(p)-1/(n+1),$

so $X_{n}\in\mathcal{C}_{n}$. Hence, $X=\bigcup_{n>n_{0}}(X_{n}\cap
I_{n})\in\mathcal{U}$. But $X\cap Y=\emptyset$ and so $Y\notin\mathcal{U}$,
and we are done. ∎

Recall that a topological space is _an almost P-space_ if its every non-empty
$\mathbb{G}_{\delta}$ subset has non-empty interior.

######  Theorem 5.11.

Suppose $\mathcal{U}$ is a Canjar ultrafilter in $V$. Then,
$\Vdash_{\mathbb{M}_{\kappa}}$“$\dot{K}_{\mathcal{U}}$ is an almost P-space”.

###### Proof.

Suppose that $p\in\mathbb{M}_{\kappa}$ is a condition and $\dot{A}$ is an
$\mathbb{M}_{\kappa}$-name such that $p\Vdash$“$\dot{A}$ is a non-empty
$\mathbb{G}_{\delta}$ subset of $\dot{K}_{\mathcal{U}}$”. Then, by the Maximum
Principle and Lemma 4.7, there is a decreasing sequence $(E_{n})$ of elements
of $\mathbb{M}^{\omega}_{\kappa}$ such that

$p\Vdash\emptyset\neq\bigcap_{n}[\dot{E}_{n}]\cap\dot{K}_{\mathcal{U}}\subseteq\dot{A}.$

Since for each $n\in\omega$ we have
$p\Vdash[\dot{E}_{n}]\cap\dot{K}_{\mathcal{U}}\neq\emptyset$, Proposition 5.8
implies that each $E_{n}$ is $(\mathcal{U},p)$-full. By Proposition 5.10 there
is a $(\mathcal{U},p)$-full $E\in\mathbb{M}^{\omega}_{\kappa}$ such that
$E\leq^{*}E_{n}$ for every $n\in\omega$. Then, again by Proposition 5.8,

$p\Vdash[\dot{E}]\cap\dot{K}_{\mathcal{U}}\neq\emptyset\mbox{ and
}\dot{E}\subseteq^{*}\dot{E_{n}}\mbox{ for each }n\in\omega,$

and so

$p\Vdash\emptyset\neq[\dot{E}]\cap\dot{K}_{\mathcal{U}}\subseteq\bigcap_{n}[\dot{E}_{n}]\cap\dot{K}_{\mathcal{U}}\subseteq\dot{A}.$

Since $p$ and $\dot{A}$ were arbitrary, we are done. ∎

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