Convergence of measures after adding a real

We prove that if \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {A}$$\end{document}A is an infinite Boolean algebra in the ground model V and \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {P}$$\end{document}P is a notion of forcing adding any of the following reals: a Cohen real, an unsplit real, or a random real, then, in any \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {P}$$\end{document}P-generic extension V[G], \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {A}$$\end{document}A has neither the Nikodym property nor the Grothendieck property. A similar result is also proved for a dominating real and the Nikodym property.


Introduction
Let A be a Boolean algebra.We say that A has the Nikodym property 1 if every sequence µ n : n ∈ ω of measures on A which is pointwise null, i.e. µ n (A) → 0 for every A ∈ A, is also weak* null, i.e. µ n (f ) → 0 for every continuous function f ∈ C(St(A)) on the Stone space of A, and that A has the Grothendieck property if every weak* null sequence µ n : n ∈ ω of measures on A is weakly null, i.e. µ n (B) → 0 for every Borel B ⊆ St(A) (see Section 2 for all the necessary terminology).Both of the notions have strong connections to functional analysis-the Nikodym property is closely related to the Uniform Boundedness Principle for locally convex spaces (see [31]), while the Grothendieck property is usually studied in a much more general sense in the context of dual Banach spaces (see [22], [31] or [12]).Nikodym [30] and Dieudonné [11] proved that all σ-complete Boolean algebras have the Nikodym property, while Grothendieck [22] showed that they have also the Grothendieck property.Consequently, e.g., the algebra ℘(ω) of all subsets of ω has both of the properties.On the other hand, no infinite countable Boolean algebra (or, more generally, no Boolean algebra whose Stone space contains a non-trivial convergent sequence) can have the Nikodym property or the Grothendieck property.
Since the findings of Nikodym, Dieudonné, and Grothendieck, many generalizations of the σ-completeness have been found which still give at least one of the properties, see e.g.[32], [23], [10], [29], [31], [18], [1], [24], [34].Unfortunately, none of those generalizations yields a necessary condition which a given Boolean algebra must satisfy in order to have the Nikodym property or the Grothendieck property.One of the reasons behind this is that, due to the result of Koszmider and Shelah [27], each of those generalizations implies also that an infinite Boolean algebra satisfying it contains an independent family of size continuum c and thus itself must be of cardinality at least c.Brech [7] however showed that consistently there exists a Boolean algebra of cardinality ω 1 having the Grothendieck property while at the same time c ≥ ω 2 .A similar result was also obtained by the first author [33] for the Nikodym property.Those two facts imply together that the quest for an algebraic or topological characterization of the Nikodym property or the Grothendieck property is much more demanding and requires using more sophisticated assumptions than mere existence of suprema or upper bounds of antichains in Boolean algebras.
Let us state the result of Brech [7] more precisely.She proved that if κ is a cardinal number and S(κ) is the side-by-side Sacks forcing adding simultaneously κ many Sacks reals to the ground model V , then in any S(κ)-generic extension V [G] the ground model Boolean algebra ℘(ω) ∩ V has the Grothendieck property (her argument works in fact for any infinite ground model σ-complete Boolean algebra, not only for ℘(ω)).In [35] we showed that a similar theorem may be obtained for the Nikodym property and in [36] we generalized both of the results by proving that if P is a proper notion of forcing satisfying the Laver property and preserving the reals non-meager, then in any P-generic extension V [G] every ground model σ-complete Boolean algebra has both the Nikodym property and the Grothendieck property.Recall that the class of forcings satisfying the assumptions of the latter theorem contains such classical notions like the Sacks, side-by-side Sacks, Miller, or Silver(-like) forcing, as well as their countable support iterations (see [36,Introduction] for references).
In this paper we follow the path of research described in the previous paragraph and study the case of adding just one real to the given model of set theory, however this time the results are mostly negative.Our main theorem reads to wit as follows.
Theorem 1.1.Let A ∈ V be an infinite Boolean algebra.Let P ∈ V be a notion of forcing adding one of the following reals: • a Cohen real, • an unsplit real, or • a random real.
Assume that G is a P-generic filter over V .Then, in V [G], A has neither the Nikodym property nor the Grothendieck property.
We establish the above theorem in a series of partial results.First, in Theorems 3.2 and 3.4 we prove that if P adds a Cohen real or an unsplit real, then in any P-generic extension every infinite ground model Boolean algebra A obtains a non-trivial convergent sequence in its Stone space St(A), and, consequently, it can have neither the Nikodym property nor the Grothendieck property.Theorems 3.2 and 3.4 have already been known to experts in the area (cf.e.g.page 162] and their reference to Koszmider [26]), however, it seems that their proofs have never been published anywhere.Our proof of Theorem 3.2 may be seen as a forcing counterpart of the proof of Geschke's [19,Theorem 2.1] which states that under Martin's axiom every infinite compact space of weight < 2 ω contains a non-trivial convergent sequence or, more generally, that in ZFC every infinite compact space of weight strictly less than the covering number cov(M) of the meager ideal M contains such a sequence.Geschke's argument is on the other hand a topological counterpart of Koppelberg's [25, Proposition 5] asserting that under Martin's axiom every infinite Boolean algebra of cardinality < 2 ω has countable cofinality.The argument for Theorem 3.4 is based on the idea presented in Booth [5, Theorem 2] (see also [14]) where it is showed that every infinite compact space of weight strictly less than the splitting number s is sequentially compact and thus contains a non-trivial convergent sequence.
The issue of adding random reals is more special.Recall that Dow and Fremlin [16] first proved that adding any number of random reals to the ground model does not introduce nontrivial convergent sequences to the Stone spaces of σ-complete ground model Boolean algebras (or, more generally, to the Stone spaces of ground model Boolean algebras whose Stone spaces in the ground model are F-spaces).Since not containing any non-trivial convergent sequences in the Stone space is not sufficient for an infinite Boolean algebra to have the Nikodym property or the Grothendieck property, the result of Dow and Fremlin does not say anything about the preservation of either of the properties by the random forcing.We address here this issue by proving in Theorem 4.6 that if a forcing P adds a random real, then for any infinite ground model Boolean algebra A in every P-generic extension of the ground model there are sequences of finitely supported measures on the Stone space St(A) which witness that A has neither the Nikodym property nor the Grothendieck property.
We also generalize partially the aforementioned result of Dow and Fremlin.Namely, we prove in Theorem 4.8 that for any ground model σ-complete Boolean algebra A the random forcing does not add to its Stone space St(A) any weak* null sequences of normalized measures whose supports consist of at most M points, where M ∈ ω is a fixed number.This result complements Theorem 4.6, at least in the case of σ-algebras-see Section 4.2 for more details.
As examples of forcings adding a Cohen real one can name the Hechler forcing or finite support iterations of infinite length of non-trivial posets (see [20,Example 0.2]).The Mathias forcing is a typical example of a notion adding an unsplit real.Finally, random reals are added by, e.g., the amoeba forcing.
Corollary 1.2.Let A ∈ V be an infinite Boolean algebra.Let P ∈ V be one of the following notions of forcing: Cohen, finite support iteration of infinite length of non-trivial posets, Hechler, Mathias, random, or amoeba.Assume that G is a P-generic filter over V .Then, in V [G], A has neither the Nikodym property nor the Grothendieck property.
We also study the case of adding dominating reals-following the argument presented in [33,Proposition 8.8] and based on the celebrated Josefson-Nissenzweig theorem from Banach space theory we prove in Section 4.3 that adding dominating reals kills the Nikodym property of all infinite ground model Boolean algebras.
Theorem 1.3.Let A ∈ V be an infinite Boolean algebra.Let P ∈ V be a notion of forcing adding a dominating real.Assume that G is a P-generic filter over V .Then, in V [G], A does not have the Nikodym property.
The class of forcings adding a dominating real contains such notions as Hechler, Laver, or Mathias.Thus, in addition to Corollary 1.2, we get the following result.
Corollary 1.4.Let A ∈ V be an infinite Boolean algebra.Let P ∈ V be the Laver forcing.Assume that G is a P-generic filter over V .Then, in V [G], A does not have the Nikodym property.
The case of the Laver forcing is particularly interesting as Dow [15,Theorem 11] showed that adding a single Laver real does not introduce any non-trivial converging sequences in the Stone space of the ground model Boolean algebra ℘(ω) ∩ V , yet, by Corollary 1.4, ℘(ω) ∩ V loses its Nikodym property.We do not know whether adding a Laver real (or, more generally, a dominating real) kills the Grothendieck property of ground model ℘(ω) (or any other ground model Boolean algebra)-see Section 6.
V always denotes the set-theoretic universe.
By ω we denote the first infinite countable ordinal number.If A is a set, then by ℘(A), [A] ω , and [A] <ω we denote the families of all subsets of A, all infinite countable subsets of A, and all finite subsets of A, respectively.A B denotes the family of all functions from a set B to A. If f is a function, then by ran(f ) we denote its range.If (L, ≤) is a linear order and f, g ∈ L ω , then by writing f ≤ g (f ≤ * g) we mean that for all (but finitely many) n ∈ ω we have f (n) ≤ g(n).We similarly define the strict relations < and < * on L ω .id A denotes the identity function on A. If B ⊆ A, then by χ B we denote the characteristic function of B on A.
All topological spaces considered in this paper are assumed to be Tychonoff, that is, completely regular and Hausdorff.A subset of a topological space is perfect if it is closed and contains no isolated points.A sequence x n : n ∈ ω in a topological space X is non-trivial if x n = x m for every n = m ∈ ω and, if the limit exists, x m = lim n→∞ x n for every m ∈ ω.
If A is a Boolean algebra, then St(A) denotes its Stone space (i.e. the space of all ultrafilters on A) with the usual topology which makes it a totally disconnected compact Hausdorff space.Recall that A is isomorphic to the algebra of clopen subsets of St(A).For every element A ∈ A by [A] A we denote the clopen subset of St(A) corresponding to A.
If we say that µ is a measure on a Boolean algebra A, then we mean that µ is a signed finitely additive function from A to R with bounded total variation, that is, the following holds: When we say that µ is a measure on a compact Hausdorff space K, then we mean that µ is a signed σ-additive Radon measure defined on the Borel σ-algebra Bor(K) of K-it follows automatically that µ has bounded total variation, that is: Recall that if we identify a given Boolean algebra A with the subalgebra of clopen subsets of the Borel σ-field Bor(St(A)), then every measure µ on A extends uniquely to a measure µ on St(A)-we will usually omitˆand write simply µ, too.
Let K be a compact space.For a measure µ on K and a µ-measurable function f : K → R we write µ(f ) to denote K f dµ.By C(K) we denote the Banach space of all continuous real-valued functions on K endowed with the supremum norm.Recall that by the Riesz representation theorem the dual space C(K) * is isometrically isomorphic to the Banach space M (K) of all Radon measures on K endowed with the total variation norm-M (K) acts on C(K) by the formula f, µ = µ(f ).
Let µ n : n ∈ ω be a sequence of measures on a Boolean algebra A. If lim n→∞ µ n (A) = 0 for every A ∈ A, then we say that µ n : n ∈ ω is pointwise null; if lim n→∞ µ n (f ) = 0 for every f ∈ C(St(A)), then it is weak* null; and if lim n→∞ µ n (B) = 0 for every B ∈ Bor(St(A)), then it is weakly null (cf.[13,Theorem 11,page 90]).Additionally, we say that µ n : n ∈ ω is pointwise bounded if sup n∈ω µ n (A) < ∞ for every A ∈ A, and that it is uniformly bounded if sup n∈ω µ n < ∞.

Adding a convergent sequence
In this section we prove that adding a Cohen real (Theorem 3.2) or an unsplit real (Theorem 3.4) to the ground model produces a non-trivial convergent sequence in the Stone space of every infinite ground model Boolean algebra.Notice that using the methods described in [16, page 162] one can generalize those results to any infinite ground model compact space K.
As we mentioned in Introduction, both of the theorems have been already known to some experts, but it seems that their proofs have never been published anywhere.
3.1.Cohen reals.Let us first recall the definition of a Cohen real.Let P ∈ V be a notion of forcing and G a P-generic filter over V .Then, Here F n(ω, 2) is the family of all finite partial functions from ω to 2, ordered by the reverse inclusion.
We will need the following folklore lemma.
Lemma 3.1.If K is an infinite scattered compact Hausdorff space, then K contains a nontrivial convergent sequence.
Proof.Since K is scattered and infinite, there is a countable subset A of K such that every x ∈ A is isolated in K.A must be discrete and open in K. Since K is compact, the boundary ∂A is non-empty and thus must contain an isolated point x (in ∂A).The sets {x} and (∂A) \ {x} are closed subsets of K, so there are disjoint open sets V and W such that {x} ⊆ V and Now, we are in the position to prove the main theorem of this section.
Theorem 3.2.Let P ∈ V be a notion of forcing adding a Cohen real and A ∈ V an infinite Boolean algebra.Then, for every P-generic filter G over V the Stone space St(A) Proof.We have two cases: 1) In V , the Stone space St(A) of A is scattered-by Lemma 3.1 there is a non-trivial convergent sequence in St(A).Of course, this sequence will also be convergent in the Stone space of A in any P-generic extension V [G].
2) In V , the Stone space St(A) is not scattered.Hence, there is a closed subset L of St(A) and a continuous surjection f : L → 2 ω .By the Kuratowski-Zorn lemma, we may assume that f is irreducible and hence that L is perfect.The family

partially ordered by the reverse inclusion ⊇). Indeed, given any non-empty open set
Let B be the Boolean algebra of clopen subsets of L. Of course, P ⊆ B. By the Stone duality, B is a homomorphic image of A. For every U ∈ B put: Trivially, each D U ∈ V and is dense in the poset (P, ⊇).
Fix now a P-generic filter G over V and let us work in V [G].By the assumption, there is a Cohen real c ∈ 2 ω over V .The family is a P-generic filter over V , so, in particular, G meets every D U (as D U ∈ V ).Let x ∈ St(B) be the ultrafilter with the base G. Since the ground model (perfect) set L had no isolated points (in V ) and it is dense in St(B), x is not isolated in St(B).Thus, we proved that St(B) is a perfect set containing a G δ -point.In particular, St(B) contains a non-trivial convergent sequence. In The next corollary follows from the proof of Theorem 3.2.Recall that a point x in a topological space X is a G δ -point if the singleton {x} is the intersection of a countable family of open subsets of X.
Corollary 3.3.Let P ∈ V be a notion of forcing adding a Cohen real and A ∈ V an infinite Boolean algebra such that its Stone space St(A) is not scattered.Then, for every P-generic filter G over V the Stone space St(A) V [G] contains a perfect subset L and a point x ∈ L which is a G δ -point in L.

Unsplit reals.
Let P ∈ V be a notion of forcing and G a P-generic filter over V .We say that a real The proof of the following theorem follows the idea of Booth [5, Theorem 2] (see also [14]).
Theorem 3.4.Let P ∈ V be a notion of forcing adding an unsplit real and A ∈ V an infinite Boolean algebra.Then, for every P-generic filter G over V the Stone space St(A) Proof.We work first in V .Let A ⊆ St(A) be an infinite countable set.Put:

Destroying the Nikodym property or the Grothendieck property
In this section we provide two negative results.Namely, in Theorem 4.6 we prove that adding a random real causes that no ground model Boolean algebra has the Nikodym property or the Grothendieck property, and in Theorem 1.3 we show that after adding a dominating real no ground model Boolean algebra has the Nikodym property.We do not know whether adding dominating reals kills the Grothendieck property-see Questions 6.1 and 6.2.
We start the section recalling several auxiliary facts-the first lemma provides an alternative definition for the Nikodym property (in fact, the one more commonly used in the literature, however lacking the apparent similarity to the definition of the Grothendieck property).
Lemma 4.1.Let A be a Boolean algebra.The following two conditions are equivalent: (1) every pointwise null sequence of measures on A is weak* null; (2) every pointwise bounded sequence of measures on A is uniformly bounded.
Proof.Assume (1) and suppose that there exists a sequence µ n : n ∈ ω of measures on A which is pointwise bounded but not uniformly bounded.By going to the subsequence, we may assume that µ n > n for every n ∈ ω.For each n ∈ ω define the measure ν n on A as follows: On the other hand, for every A ∈ A we have: which converges to 0 as n → ∞ (because sup n∈ω µ n (A) < ∞), which contradicts (1) as weak* null sequences are always uniformly bounded (by the virtue of the Banach-Steinhaus theorem).Hence, (2) holds.Assume now (2) and let µ n : n ∈ ω be a pointwise null sequence of measures on A. It follows immediately that µ n : n ∈ ω is pointwise bounded, hence, by (2), it is uniformly bounded.Let M > 0 be such that sup n∈ω µ n < M .Fix f ∈ C(St(A)) and let ε > 0. There are finite sequences A 1 , . . ., A k ∈ A and α Since µ n : n ∈ ω is pointwise null, there is N ∈ ω such that for every n > N we have: Thus, for every n > N it holds: It follows that µ n (f ) → 0 as n → ∞, which proves that µ n : n ∈ ω is weak* null.Consequently, (1) holds.
From the proof of implication (2)⇒(1) we immediately get the following corollary.
Corollary 4.2.Let A be a Boolean algebra.If µ n : n ∈ ω is a pointwise null uniformly bounded sequence of measures on A, then µ n : n ∈ ω is weak* null.
If X is a topological space and x ∈ X, then by δ x we denote the Borel one-point measure on X concentrated at x. Recall that a measure µ on a compact space K (a Boolean algebra A) is finitely supported or has finite support if there exist finite sequences x 1 , . . ., x n of pairwise distinct points in K (in St(A)) and α 1 , . . ., α n ∈ R such that µ = n i=1 α i δ x i .The set x 1 , . . ., x n is called the support of µ and denoted by supp(µ).We will need the following simple lemma.Lemma 4.3.Let A be a Boolean algebra.If there exists a sequence µ n : n ∈ ω of finitely supported measures on A which is pointwise null but not uniformly bounded, then A has neither the Nikodym property nor the Grothendieck property.
Proof.µ n : n ∈ ω directly witnesses the lack of the Nikodym property.Consider the sequence ν n : n ∈ ω defined as ν n = µ n / µ n for every n ∈ ω.Since it is pointwise null, too, and uniformly bounded, by Corollary 4.2 it is weak* null.Set S = n∈ω supp ν n and note that the Banach space ℓ 1 (S) of all absolutely summable sequences on the set S is a closed linear subspace of the dual space C(St(A)) * containing every ν n .Since ℓ 1 (S) has the Schur property (meaning that the weak convergence of sequences implies their norm convergence), the sequence ν n : n ∈ ω cannot be weakly null, as ν n = 1 for every n ∈ ω.In particular, A does not have the Grothendieck property.
4.1.Random reals.Destroying the Nikodym and Grothendieck properties.In order to prove Theorem 4.6, we need to recall some basic facts concerning the binomial distributions.Let (Ω, Σ, Pr) be a probability space.Given p ∈ (0, 1), for every i ∈ ω let X i be a random variable taking only two values: 0 and 1, and such that the following two equalities hold: Assume additionally that the sequence X i : i ∈ ω is independent, that is, for every n > 0 and s ∈ 2 n we have where In what follows we fix p = 1/2.Put Ω = 2 ω and let Σ denote the standard Borel σ-field on Ω and λ the standard product measure on Ω.We will now work in the probability space (Ω, Σ, λ).For every i ∈ ω and x ∈ Ω set X i (x) = x(i), i.e., the function X i is simply the projection onto the i-th coordinate.Obviously, the sequence X i : i ∈ ω of random variables is as described in the paragraph before Theorem 4.4.
Lemma 4.5.For every n ∈ ω set I n = 2 n +1, 2 n +2, . . ., 2 n+1 .Suppose that for some infinite J ⊂ ω and for every n ∈ J there is a subset Y n ⊆ I n such that η = inf η n : n ∈ J > 0, where For every n ∈ J put: Then, Proof.Applying Theorem 4.4 (for m = m n and ε = ε n ), for every n ∈ J we get: which after simplification reduces to: Observe that ( * ) actually implies that n∈J λ(A n ) < ∞ (because η > 0), and hence by the Borel-Cantelli lemma we get ( †): We are now in the position to present the proof of the main theorem of this section.We will use the following definition of a random real: Given a (forcing) extension V ′ of V , a real r ∈ 2 ω is a random real over V if for every Borel subset B of 2 ω , coded in V and such that λ(B) = 0 V , the real r does not belong to the interpretation of B in V ′ .(We will abuse the notation and denote this interpretation by B, too.)Theorem 4.6.Let P ∈ V be a notion of forcing adding a random real and A ∈ V an infinite Boolean algebra.Assume that G is a P-generic filter over V .Then, in V [G], A has neither the Nikodym property nor the Grothendieck property.
Proof.In V , let x i : i ∈ ω be a sequence of ultrafilters in St(A) such that x i = x j for i = j ∈ ω.
From now on we work exclusively in V [G].Let ϕ : ω → St(A) be such that ϕ(i) = x i for every i ∈ ω.Let r ∈ 2 ω ∩ V [G] be a random real over V .Set ω + = ω \ {0}.For every n ∈ ω + consider the measure µ n on A defined as follows: . It follows that µ n is finitely supported, supp µ n = ϕ I n , and We claim that µ n : n ∈ ω is pointwise null.Let us fix A ∈ A and for every n ∈ ω + set Of course, Y n ∈ V .Put: Assume first that J is infinite.We will prove that µ n (A) → 0 as n → ∞, n ∈ J.For every n ∈ J set also η n = Y n /2 n and let A n be the clopen subset of 2 ω such as defined in Lemma 4.5.By the definition of J, we get that η = inf η n : n ∈ J ≥ 1/2 > 0, hence equation ( †) of Lemma 4.5 together with the definition of a random real imply that r ∈ A n for all but finitely many n ∈ J, which means that i∈Yn for all but finitely many n ∈ J, and thus there is n 0 ∈ ω + such that for all n ∈ J, n ≥ n 0 we have (note that η n ≤ 1): i∈Yn which in turns implies that for all n ∈ J, n ≥ n 0 , and s ∈ {0, 1} it holds: (Just note that the values on the left hand sides of the latter two inequalities are the same.)As a result, for every n ∈ J, n ≥ n 0 , we have: If J c is finite, then we are immediately done, so assume that it is infinite.Notice that since for the unit element 1 A of the Boolean algebra A and all i ∈ ω we have 1 A ∈ x i , exactly the same reasoning as above shows that lim n→∞ µ n 1 A = 0, so in particular we have: and put: Using again the same argument as above, we show that so in particular we get that Finally, we have: We have just showed that the sequence µ n : n ∈ ω + of finitely supported measures on A is pointwise null but not uniformly bounded, so, by Lemma 4.3, A has neither the Nikodym property nor the Grothendieck property.The proof is thus finished.
Note that, normalizing measures µ n from the above proof (that is, considering the measures µ n / µ n ), by Corollary 4.2 we obtain the following result.
Corollary 4.7.Let P ∈ V be a notion of forcing adding a random real and A ∈ V an infinite Boolean algebra.Assume that G is a P-generic filter over V .Then, in V [G], St(A) carries a weak* null sequence µ n : n ∈ ω of finitely supported measures with pairwise disjoint supports such that µ n = 1 and supp µ n = 2 n for every n ∈ ω. 4.2.Random reals.Generalization of Dow-Fremlin's result.Let κ be an infinite cardinal number.By µ κ denote the standard product probability measure on the space 2 κ and let B(κ) = Bor 2 κ A ∈ Bor 2 κ : µ κ (A) = 0 be its measure algebra.B(κ) is a well-known ω ω -bounding poset adding κ many random reals (see [2, Section 3.1])2 .(A forcing P is ω ωbounding if for every P-generic filter G over V and a function f Recall again that Dow and Fremlin [16] proved that forcing with B(κ) does not introduce nontrivial convergent sequences to the Stone spaces of σ-complete ground model Boolean algebras.In this subsection we will generalize their result in the following way.
Theorem 4.8.Let A ∈ V be an infinite σ-complete Boolean algebra.Assume that G is a B(κ)-generic filter over V .Then, in V [G], St(A) does not carry any weak* null sequence µ n : n ∈ ω of finitely supported measures such that µ n = 1 for every n ∈ ω and for which there exists M ∈ ω such that supp µ n ≤ M for every n ∈ ω.
The theorem really generalizes the result of Dow and Fremlin, since if a compact space K contains a non-trivial convergent sequence x n : n ∈ ω , then the sequence µ n : n ∈ ω of measures defined as µ n = 1 2 δ x 2n − δ x 2n+1 is weak* null and such that µ n = 1 and supp µ n = 2 for every n ∈ ω.Note however that the existence of a weak* null sequence ν n : n ∈ ω of measures on a totally disconnected compact space K such that supp ν n = 2 for every n ∈ ω does not imply the existence of non-trivial convergent sequences in K (cf.[31,Example 4.10]).
Let us stress that Corollary 4.7 and Theorem 4.8 are complementary: the corollary states that the existence of a random real yields the existence of a weak* null sequence µ n : n ∈ ω of finitely supported normalized measures on the Stone space of a given infinite ground model Boolean algebra such that lim n→∞ supp µ n = ∞, while the theorem asserts that, in the case of σ-complete Boolean algebras, this is optimal-we cannot get any weak* null sequence µ n : n ∈ ω of normalized measures such that sup n∈ω supp µ n < ∞.
In order to prove Theorem 4.8, we first need to recall several auxiliary results.The first one implies that in fact we only need to deal with the case of M = 2. Lemma 4.9.Let A be an infinite Boolean algebra.If there are a weak* null sequence µ n : n ∈ ω of finitely supported measures on St(A) and a number M ∈ ω such that µ n = 1 and supp µ n ≤ M for every n ∈ ω, then M ≥ 2 and there is a weak* null sequence ν n : n ∈ ω of finitely supported measures on St(A) such that ν n = 1 and supp ν n = 2.
Proof.Let µ n : n ∈ ω be a weak* null sequence of finitely supported measures on St(A) for which there exists M ∈ ω such that µ n = 1 and supp µ n ≤ M for every n ∈ ω.If there is a subsequence µ n k : k ∈ ω such that supp µ n k = 1 for every k ∈ ω, then every µ n k is simply of the form α k • δ x k for some α k ∈ {−1, 1} and x k ∈ St(A).Consequently, for the constant unit function 1 ∈ C(St(A)) we have µ n k (1) = α k = 1 for every k ∈ ω, which contradicts the fact that µ n : n ∈ ω converges weak* to 0. It follows that for almost all n ∈ ω we have supp µ n ≥ 2 and so M ≥ 2.
We now prove the second part of the lemma.Let m ∈ ω be the minimal number such that there exists a weak* null sequence µ n : n ∈ ω of finitely supported measures on St(A) such that µ n = 1 and supp µ n = m for every n ∈ ω.By the previous paragraph, m ≥ 2. We will prove that in fact m = 2.
First note that if there are a clopen set U ⊆ St(A) and an increasing sequence n k : k ∈ ω such that µ n k ↾ U = 0 and µ n k ↾ (St(A) \ U ) = 0 for every k ∈ ω, then there is an increasing sequence k l : l ∈ ω such that at least one of the sequences µ 1 l : l ∈ ω and µ 2 l : l ∈ ω , defined for every l ∈ ω as and is weak* null.Since for every l ∈ ω and i ∈ {1, 2} it holds µ i l = 1 and supp µ i l < m, we get a contradiction with the minimality of m.It follows that for every clopen U ⊆ St(A) and almost all n ∈ ω we have either supp µ n ⊆ U , or supp µ n ∩ U = ∅.
For every n ∈ ω pick two distinct points x n , y n ∈ supp µ n and define the measure ν n simply as follows To finish the proof, we only need to show that ν n : n ∈ ω is weak* null.But this is trivial, since for every clopen subset U of St(A) and almost all n ∈ ω we have either x n , y n ∈ U , or x n , y n ∈ U ; in either case it holds ν n (U ) = 0, so ν n : n ∈ ω is pointwise null.Since ν n : n ∈ ω is also uniformly bounded, by Corollary 4.2 it is weak* null (and so, by the minimality of m, we also have m = 2).Remark 4.10.For a Boolean algebra A, if there exists a weak* null sequence µ n : n ∈ ω of measures on St(A) such that µ n = 1 and supp µ n = 2 for every n ∈ ω, then one can also easily get such a sequence but with pairwise disjoint supports.Thus, from the above proof it basically follows that for every Boolean algebra A the following two conditions are equivalent: (1) there are a weak* null sequence µ n : n ∈ ω of finitely supported measures on St(A) and M ∈ ω such that µ n = 1 and supp µ n ≤ M for every n ∈ ω, (2) there are two disjoint sequences x n : n ∈ ω and y n : n ∈ ω of distinct points in St(A) such that for every clopen set U and almost all n ∈ ω we have: The following lemma was the main tool used by Dow and Fremlin to obtain their result.We will need it, too.Lemma 4.11 Lemma 2.2]).Let A ∈ V be a Boolean algebra.Assume that ẋn : n ∈ ω is a sequence of B(κ)-names for distinct ultrafilters on A. Let G be a B(κ)-generic filter over V .Then, for every condition q ∈ B(κ) there are a condition p ≤ q and a sequence In [6] Borodulin-Nadzieja and the first author proved that, for every B(κ)-names u and v for ultrafilters on a given ground model Boolean algebra A, if B(κ) u = v, then for every ε > 0 there are a condition p ∈ B(κ) and an element C ∈ A such that µ κ (p) > 1/4 − ε and p C ∈ u△ v.By exactly the same proof, mutatis mutandis, we obtain the following formally stronger result.Lemma 4.12 (Cf.[6, Section 6.2]).Let A ∈ V be a Boolean algebra.Let u and v be B(κ)-names for ultrafilters on A. For every condition q ∈ B(κ), if q u = v, then for every ε > 0 there are a condition p ≤ q and an element C ∈ A such that µ κ (p) > µ κ (q)/4 − ε and p C ∈ u△ v.
The next lemma is folklore.Lemma 4.13.Let P be an ω ω -bounding forcing notion and G a P-generic filter over Proof.In V [G] let x n : n ∈ ω be the strictly increasing enumeration of elements of X.Since P is ω ω -bounding, there is a strictly increasing function g ∈ ω ω ∩ V such that for every k ∈ ω there is n k ∈ ω for which the following inequalities are satisfied: (cf. the proof of [33, Corollary 2.5]).In V , let F ⊆ [ω] ω be an uncountable almost disjoint family.For every Y ∈ F set: .
finite, so H is an uncountable almost disjoint family.Also, by ( * ), for every Y ∈ F the intersection X ∩ Z Y is infinite.
We are in the position to prove the main result of this section.
Proof of Theorem 4.8.Assume towards the contradiction that, in V [G], there are a weak* null sequence µ n : n ∈ ω of finitely supported measures on St(A) and a number M ∈ ω such that µ n = 1 and supp µ n ≤ M for every n ∈ ω.By Remark 4.10, we may assume that, in V , there are two sequences ẋi : i ∈ ω and ẏi : i ∈ ω of B(κ)-names for ultrafilters on A and a condition q ∈ B(κ) forcing that: (a) ẋi = ẋj and ẏi = ẏj for every i = j ∈ ω, (b) ẋi = ẏj for every i, j ∈ ω, (c) for every A ∈ A and almost all i ∈ ω we have: ẋi By Lemma 4.11, there are a condition p ≤ q and a sequence A n : n ∈ ω (in V !) of pairwise disjoint elements of A such that for every n ∈ ω we have p [A n ] A ∩ ẋi : i ∈ ω = ∅.For each n ∈ ω by Ḟn denote a B(κ)-name such that p forces that Note that, by property (c), p forces that each Ḟn is at most finite.
We need to consider two cases.
(1) p forces that for almost all n ∈ ω the set Ḟn is empty.Let r ≤ p and N ∈ ω be such that for every n ≥ N we have r Ḟn = ∅.The last formula simply means that for every n ≥ N and i ∈ ω the following holds: For each n ≥ N , let αn be a B(κ)-name such that and let un and vn be B(κ)-names for ultrafilters on A such that r un = ẋ αn and vn = ẏ αn .
Of course, for n = m ≥ N we have r αn = αm .For every n ≥ N , since r un = vn , by Lemma 4.12, we get a condition r n ≤ r and an element C n ∈ A such that µ κ r n > µ κ (r)/5 and r n C n ∈ un △ vn .Since r n un , vn ∈ [A n ] A , we may actually assume that C n ≤ A n .In particular, C n ∧ C m = 0 for every n = m ≥ N .Set: then, µ κ (s) ≥ µ κ (r)/5, so s ∈ B(κ) and s ≤ r.It follows that s forces that C n ∈ un △ vn for infinitely many n ≥ N , or, in other words, that and note that for every n ≥ N we have We claim that s forces that [C] A ∩ ẋi , ẏi = 1 for infinitely many i ∈ ω, contradicting condition (c).Indeed, observe that for every n ≥ N we have: and so, since for infinitely many n ≥ N .
(2) There is a condition r ≤ p forcing that for infinitely many n ∈ ω the set Ḟn is non-empty.Let then ḟ be a B(κ)-name for a function ω → ω such that for every n ∈ ω we have: Since r forces that Ḟn is finite for every n ∈ ω and that each i ∈ ω belongs to at most two different Ḟn 's, the above definition is valid.For every n, m ∈ ω we also have r ḟ (n) ≥ n and Since B(κ) is ω ω -bounding, there are a strictly increasing function g : ω → ω such that g(0) > 0 and a condition s ≤ r such that s ḟ (n) < g(n) for every n ∈ ω.Define the function h : ω → ω for every n ∈ ω as: , where the composition is taken n + 1 times.
We have two subcases of case (2).
(2a) There is a condition t ≤ s forcing that for infinitely many n ∈ ω the set Ḟh(n) is nonempty.Let H be a B(κ)-generic filter over V containing t.For every B(κ)-name σ by σ H we denote its evaluation in V [H].
We now work in V [H].For every n < m ∈ ω we have: so, by ( * ), it holds that exists in A. Let X ∈ [ω] ω be any set such that for every n ∈ X it holds Ḟ H h(n) = ∅, so we may pick i n ∈ Ḟ H h(n) .Note that, by condition (c), there is no X ′ ∈ [X] ω such that ẋH in , ẏH in ∩ [A] A = 1 for every n ∈ X ′ .It follows that-removing a finite number of elements of X if necessary-for every n ∈ X we have ẋH in , ẏH in ⊆ [A] A .Property ( * * ) implies that for every n ∈ X we have: that is, there is a unique point in ẋH in , ẏH in which belongs to the boundary (in St(A)) of the open set n∈ω A h(n) A .
By Lemma 4.13, there is an uncountable almost disjoint family H ⊆ [ω] ω in V such that the intersection X ∩ Z is infinite for every Z ∈ H. Since every Z ∈ H is in V , the supremum exists in A. For every Z ∈ H set: For every Z 1 = Z 2 ∈ H we have: and hence B Z 1 ∩ B Z 2 = ∅.Since H is uncountable and X is countable, it follows that there is Z ∈ H such that ẋH in , ẏH in ∩ B Z = ∅ for every n ∈ X ∩ Z, and hence ẋH in , ẏH in ∩ [A Z ] A = 1 for infinitely many n ∈ X, which contradicts condition (c).(2b) s forces that for almost all n ∈ ω the set Ḟh(n) is empty.We proceed exactly in the same way as in case (1), that is, using Lemma 4.12 we obtain N ∈ ω, an antichain C n : n ≥ N in V such that C n ≤ A h(n) for every n ≥ N , and a condition t ≤ s forcing that n∈ω C n A ∩ ẋi , ẏi = 1 for infinitely many i ∈ ω, which again contradicts condition (c).4.3.Dominating reals.Let P ∈ V be a notion of forcing and G a P-generic filter over V .Recall that a real we denote the family of all those functions f ∈ ω ω which are increasing, that is, f (n) ≤ f (n + 1) for every n ∈ ω, and lim n→∞ f (n) = ∞.Let us then also say that a real h It appears that adding a dominating real is equivalent to adding an anti-dominating real.To prove it, we need to introduce the following auxiliary operator Φ : (ω ω ) ∞ → (ω ω ) ∞ .It seems that the idea standing behind Φ, and hence also behind Propositions 4.14 and 4.15, is standard (cf.e.g.Canjar [9, Sections 1.5 and 3.6]).
Let f ∈ (ω ω ) ∞ and write ran(f Note that for every n ∈ ω there is unique i ∈ ω such that n f i ≤ n < n f i+1 .Put: The next proposition lists most basic properties of Φ. Proposition 4.14.For every f, g ∈ (ω ω ) ∞ the following conditions hold: (1) Proof.Let f, g ∈ (ω ω ) ∞ .Enumerate: and set n f 0 = −1 and n g 0 = −1.We first prove (1) and (2).Fix n ∈ ω and let i, j ∈ ω be such that n f i ≤ n < n f i+1 and n f j = f (n).We have: which proves (1).To see (2), note that the monotonicity of f implies that min f −1 n f j+1 > n and thus we have: Let us now prove (3).For every i ≥ 1 set n Finally, we shall prove (4).Assume that f ≤ * g.There exists We claim that m ≤ l, so for the sake of contradiction let us assume that l < m.We then have: , which is a contradiction.Proposition 4.15.Let P ∈ V be a notion of forcing.Then, P adds a dominating real if and only if it adds an anti-dominating real.
Proof.Let G be a P-generic filter over V .We work in V [G].Assume that there is a dominating real f ∈ ω ω over V and define an auxiliary function g ∈ ω ω as follows: where n ∈ ω.Obviously, g ∈ (ω ω ) ∞ and it is also a dominating real over V , so for every Since g ∈ ω ω and g is dominating every h ∈ (ω ω ) ∞ ∩ V , we get by ( * ) and Proposition 4.14.(4) that Φ(g) ≤ * h for every h ∈ (ω ω ) ∞ ∩ V .In other words, we get that Φ(g) is an anti-dominating real over V .
The proof in the other direction is similar.
We are ready to prove the main result of this section.
Theorem 1.3.Let A ∈ V be an infinite Boolean algebra.Let P ∈ V be a notion of forcing adding a dominating real.Assume that G is a P-generic filter over V .Then, in V [G], A does not have the Nikodym property.
Proof.We first work in V .By the Josefson-Nissenzweig theorem (see [13, Chapter XII]) and the Riesz representation theorem, there is a weak* null sequence µ n : n ∈ ω of measures on the Boolean algebra A such that µ n = 1 for every n ∈ ω.For every A ∈ A define the sequences c A , d A ∈ R ω as follows: where n ∈ ω.Then, c A (n) > 0 and for every n ∈ ω, and Finally, for every A ∈ A and n ∈ ω set e A (n) = 1/d A (n).It follows that e A ∈ (ω ω ) ∞ .
Let us now go to V [G].P adds a dominating real, so by Proposition 4.15 there is an antidominating real g ∈ (ω ω ) ∞ ∩ V [G] over V .By taking the function max(g, 1) instead of g, we may assume that g(n) > 0 for every n ∈ ω.For every A ∈ A we have g ≤ * e A , so if we define the sequence c ∈ R ω by the formula c(n) = 1/g(n), where n ∈ ω, then we get that d A ≤ * c for every A ∈ A. Of course, c(n) > 0 for every n ∈ ω and lim n→∞ c(n) = 0.
For every n ∈ ω define the measure ν n on A as follows: where A ∈ A. Note that µ n = 1 yields that On the other hand, for every A ∈ A we have for sufficiently large n ∈ ω, so sup n∈ω ν n (A) < ∞ for every A ∈ A. A close relative to the numbers nik and gr is the convergence number z defined as follows: z = min w(K) : K is an infinite compact space with no non-trivial convergent sequences .
Here w(K) denotes the weight of K.The number z was studied e.g. in Brian and Dow [8].It is immediate that z ≤ nik and z ≤ gr.By the result of Dow and Fremlin [16] stating that in any random extension V [G], for every σ-complete Boolean algebra A ∈ V , its Stone space St(A) does not contain any non-trivial convergent sequences, we have that z = ω 1 < c in the random model.Thus, by Corollary 5.3, we immediately get also the following fact.
Dow [15] proved that in the Laver model there are (totally disconnected) compact spaces of weight ω 1 containing no non-trivial convergent sequences, so z = ω 1 holds in this model.On the other hand, it is well known that the bounding number b has value ω 2 in the Laver model, and it was proved by the first author in [33,Proposition 3.2] that b ≤ nik holds in ZFC.We thus get the following corollary.
We do not know the value of gr in the Laver model (cf.Question 6.1).6. Open questions 6.1.Dominating reals and the Grothendieck property.In the introductory section we admitted that, contrary to the case of the Nikodym property, we do not know whether adding dominating reals kills the Grothendieck property of ground model σ-complete Boolean algebras.Question 6.1.Let A ∈ V be an infinite σ-complete Boolean algebra.Assume that G is a generic filter for the Laver forcing over V .Does A have the Grothendieck property in V [G]? Question 6.2.Does there exist a notion of forcing P adding dominating reals and such that in any P-generic extension V [G] any ground model σ-complete Boolean algebra has the Grothendieck property?
An affirmative answer to Question 6.1 would yield a new consistent example of a Boolean algebra with the Grothendieck property but without the Nikodym property.Recall that while there are many consistent or even ZFC examples of Boolean algebras with the Nikodym property but without the Grothendieck property, see e.g.[31], [21], [37], so far only one example of an algebra with the Grothendieck property and without the Nikodym property has been found-the construction was obtained by Talagrand [38] under the assumption of the Continuum Hypothesis.6.2.Eventually different reals.Let V [G] be a P-generic extension of the ground model V for some forcing notion P. If f ∈ ω ω ∩ V [G] is a dominating real, then obviously it is an eventually different real, that is, for every g ∈ ω ω ∩ V the set n ∈ ω : f (n) = g(n) is finite.The converse does not hold, as e.g. the random forcing or the eventually different forcing both add eventually different reals but not dominating reals.Since the latter forcing adds Cohen reals, too, by Theorems 3.2 and 4.6 both notions kill the Nikodym and Grothendieck properties of infinite ground model Boolean algebras.Thus, it seems that all the standard classical notions adding eventually different reals kill at least one of the properties.It is also a folklore fact that a forcing adds an eventually different real if and only if it makes the ground model reals meager, hence, trivially by the assumption, the notions of forcing considered in [36] (cf. the third paragraph of Introduction), which are proved therein to preserve both the Nikodym property and the Grothendieck property of ground model σ-complete Boolean algebras, do not add eventually different reals.So it seems reasonable to ask whether adding an eventually different real is solely a reason that ground model Boolean algebras lose their Nikodym property (and, in the view of Question 6.2, possibly also the Grothendieck property).Question 6.3.Does there exist a notion of forcing P adding eventually different reals and such that in any P-generic extension V [G] any ground model σ-complete Boolean algebra has the Nikodym property?Question 6.4.Does there exist a notion of forcing P adding eventually different reals and such that in any P-generic extension V [G] any ground model σ-complete Boolean algebra has the Grothendieck property?6.3.Cardinal characteristics nik and gr.We are not aware of any model in which the numbers nik and gr have different values.Thus, we pose the following question.Question 6.5.Is it consistent that nik < gr or nik > gr?
Note that an affirmative answer to Question 6.1 would imply that ω 1 = gr < nik = c holds in the Laver model (cf.Corollary 5.6).
It follows that the sequence ν Corollary 5.4.Let x ∈ {nik, gr}.Neither of the inequalities x ≤ d and x ≥ d is provable in ZFC.
n : n ∈ ω is pointwise bounded but not uniformly bounded, hence, by Lemma 4.1, A does not have the Nikodym property in V [G].