Download Genericity: A MeasurehTheoretic Analysis

Genericity: A Measure–Theoretic Analysis
Massimo Marinacci
Draft of December 1994y
1
Introduction
In many results of Mathematical Economics the notion of smallness plays a crucial role.
Indeed, it often turns out that desirable properties do not hold in all the environment
of interest, but only in some of its parts. For example, this is usually the case in
the di¤erentiable approach to General Equilibrium (see e.g. [4]). The purpose of the
analysis then becomes to prove that the desired property fails only in a negligible part
of the environment. This is the so called generic point of view (see e.g. [14, pp. 316319]). In a similar approach it is crucial to make precise what is meant by terms like
negligible and generic. Two notions of smallness are available. On one hand, nowhere
dense subsets and meager (often called …rst category) subsets have been regarded as
small in a topological sense. To help intuition, observe that a nowhere dense subset
of the real line is a subset full of holes. On the other hand, from a measure-theoretic
viewpoint the notion of small is represented by nullsets. Consequently, quoting [15,
p.4], ”it is natural to ask whether these notions of smallness are related”. This is the
purpose of the study of residual measures, i.e. Borel measures that give measure zero
to all meager sets. With respect to a residual measure, meager subsets are small not
only in a topological sense, but also in measure–theoretic one. In this paper we will
investigate under which conditions these measures exist. In particular, it will turn
Department of Economics, Northwestern University. I wish to thank Keith Burns and Itzhak
Gilboa for helpful comments. Financial support from Ente Einaudi is gratefully acknowledged.
y
First draft: February 1993.
1
out that in many interesting spaces these measures do not exist. This means that in
such spaces for any given …nite Borel measure there always exists a meager subset
A such that (A) > 0; i.e. a topologically negligible set which is not negligible from a
measure-theoretic standpoint.
A similar situation already occurs in Euclidean spaces. In fact, in these spaces
genericity analysis integrates topological and measure-theoretic considerations. For
example, by considering as negligible the sets whose closure has Lebesgue measure zero.
However, in Euclidean spaces the topological and measure-theoretic notions employed
have usually a strong intuitive basis, and there is a natural way to integrate them. This
is not the case in abstract spaces, where the analysis is usually based on the topological
notions of smallness. Our results show that in many spaces of interest it is impossible
to have a measure-theoretic support for the topological approach. In other words, there
is no Borel measure for which all the topologically negligible sets get measure zero.
Stronger results have been obtained for nonatomic measures by Marczewski and
Sikorski [12] and [13]. In particular, they have proved that given any nonatomic …nite
Borel measure de…ned on a metric spaces with a basis whose cardinal is nonmeasurable
(see below; separable metric spaces are an example) it is always possible to decompose
the space in two disjoint subsets, one meager, the other one of measure zero. The
results presented here are weaker, but hold for any …nite Borel measure, not necessarily
nonatomic.
As an aside, it is worth noting that the results of Marczewski and Sikorski are important for the analysis of Nash re…nements in normal form games with in…nite strategy
spaces. Here the problem is to …nd the appropriate in…nite counterpart of completely
mixed strategies, on which re…nements for …nite games are based. A natural candidate
are the Borel measures with full support, i.e. measures assigning strictly positive mass
to every nonempty open subset. In Simon and Stinchcombe [19] is developed an approach to in…nite games re…nements based on full support Borel measures. However,
the results of Marczewski and Sikorski show that for every full support nonatomic measure on a metric space with a basis whose cardinal is nonmeasurable, there exists a
partition fA; Ac g such that A is meager and (Ac ) = 0. In other words, even though
puts strictly positive measure to all open sets, there exists a topologically large set like
Ac which gets zero measure under . As nonatomic measures are an important class
of mixed strategies, this shows that full support measures are a much weaker notion
2
than completely mixed strategies in …nite spaces.
Besides genericity analysis, residual measures are interesting in the theory of Boolean
algebras. Indeed, let A be a Boolean -algebra and P a probability measure (not necessarily strictly positive) de…ned on it. By the Loomis-Sikorski representation theorem, to the pair (A; P ) can be associated a measure space characterized by a residual
measure. Using techniques of this type it is possible to derive some results about probability measures de…ned on Boolean algebras from propositions established for residual
measures.
The paper is organized as follows. Section 2 contains some preliminary notions.
Section 3 gives existence results concerning residual measures. In section 4 the results
of section 3 are used in order to obtain some non-existence results for probability
measures de…ned on Boolean algebras. Finally, in section 5 the constructions of the
previous sections are used to get a result about the ranges of probability measures on
Boolean algebras.
We follow the notation of [10]. This implies, for instance, that lower and upper
cases will usually denote, respectively, sets and Boolean algebras.
2
2.1
Preliminary notions
Measures
Let X be a topological space and let cl(a) and int(a) denote, respectively, the closure
and the interior of a subset a X. A subset a is nowhere dense if int (cl(a)) = ?, i.e.
if the closure of a has empty interior. Nowhere dense subsets have no interior points.
A subset a is meager (or of the …rst category) if it can be represented as a countable
union of nowhere dense subsets. They are the subsets of X which can be approximated
by nowhere dense subsets.
We now de…ne some classes of Borel probability measures.
De…nition 1 A Borel probability measure P de…ned on a topological space X is regular
if for all Borel subsets a we have
P (a) = sup fP (c) : c
aand cclosedg = inf fP (v) : a
3
vand vopeng :
De…nition 2 A Borel probability measure P de…ned on a topological space X is additive if whenever fg g is a net of open subsets such that g
g for
, then
!
[
P
g
= sup P (g ):
Let M denote the
measures.
-ideal of all meager Borel subsets. Next we de…ne residual
De…nition 3 A Borel probability measure P de…ned on a topological space X is residual if P (a) = 0 for all a 2 M .
Given a probability measure P on a …eld A, a subset a 2 A is called a P -atom if:
(i) P (a) > 0;
(ii) if b
a and b 2 A, then either P (b) = 0 or P (a n b) = 0:
De…nition 4 A Borel probability measure P de…ned on a topological space X is nonatomic
if there are no P -atoms.
In the sequel we make use also of the notion of nonmeasurable cardinal. Indeed,
suppose that on the power set of a space X there exists a di¤use probability measure
, i.e. a measure giving measure zero to all singletons. In such a case, the cardinal
number of X is called measurable. Instead, a cardinal @ is called nonmeasurable if the
previous property does not hold for all spaces of cardinality @.
Now we turn to Boolean algebras. For basic notions and results we refer to [10]. We
P
begin with a de…nition. A bit of notation: 1 is the unit element of A, while 1
n=1 an is
1
the supremum of fan gn=1 in A.
De…nition 5 A probability measure P on a Boolean algebra A is a real-valued function
such that:
(i) P (a)
0 for all a 2 A;
P
P1
1
(ii) P ( 1
n=1 an ) =
n=1 P (an ) whenever fan gn=1
P1
elements for which n=1 an exists;
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A is a set of pairwise disjoint
(iii) P (1) = 1:
It is worth noting that P has not been de…ned as a strictly positive measure (i.e.
it is not required P (a) = 0 i¤ a is the zero element).
It is important to distinguish the previous de…nition from the usual de…nition of
probability measures on …elds of subsets. Let A be a …eld of subsets. A real–valued
set function P on A is countably additive if:
S
P1
1
(iii’) P ( 1
n=1 an ) =
n=1 P (an ) whenever fan gn=1
S1
elements such that n=1 an 2 A.
A is a set of pairwise disjoint
Condition (iii’) is quite di¤erent from condition (iii) in de…nition 5, as it was stressed
S1
P
1
in [8]. In fact, 1
n=1 an is
n=1 an , the supremum of fan gn=1 in A, can exists even when
P1
S1
S1
not in A. Instead, if n=1 an 2 A, then n=1 an = n=1 an . Therefore, if a set function
satis…es (iii’), then it satis…es (iii), while a set function which satis…es (iii’) can fail to
satisfy (iii). We illustrate this failure with what would be otherwise a counterexample
to the results of next section.
Example. Let A be the smallest …eld of subsets of [0; 1) generated by the intervals
[a; b), with 0 a < b 1: The …eld A is both separable and atomless (these notions
are introduced below). It is known that each element of A can be expressed as a …nite
and disjoint union of half-open intervals of the form [a; b). Consequently, each element
of A contains an open subset. Now, let m be the restriction of the Lebesgue measure
on A . We want to show that m does not satisfy condition (iii) of de…nition 5. In fact,
let us construct a subset of [0; 1) in a way similar to that of the Cantor set. However,
this time at stage n we delete 2n 1 intervals of the form [a; b) and of length (3 n ), with
T
0 < < 1: We denote by c ;n what remains after stage n and we set c = 1
n=1 c ;n .
The subsets c ;n belong to A for all n 1: Moreover, we have c
c 0 where c was
de…ned in section 2. Thus cl(c ) c , and so c is nowhere dense. This implies that
c does not contain any open subset. Therefore, c not belong to A because we have
already seen that each element of A has a non-empty interior. Furthermore, for the
T
same reason all subsets of c do not belong to A . To sum up, 1
n=1 c ;n = c 6 2A and
1
^n=1 c ;n = ?. In section 2 we said that m(c ) = 1
. Looking at the construction
of c , it is easy to see that the continuity of m implies m(c ) = m(c ). Therefore,
using again the continuity of m, we have limn!1 m(c ;n ) = m(c ) = 1
. But
5
1
m(^1
n=1 c ;n ) = m(?) = 0. Hence, limn!1 m(c ;n ) > m(^n=1 c
that this violates condition (iii) of de…nition 5.
2.2
;n ),
and it easy to verify
Duality
This subsection contains some preliminary material on Stone spaces that will be used
in section 4. The reader not interested in section 4 can skip this section without loss
of continuity.
Following [10], U lt(A) denotes the Stone space of A and Clop(U lt(A)) the dual
algebra of U lt(A). In the next section we will make use of a generalization of the
Loomis-Sikorski representation theorem, due to [18]. To report this generalization we
need some further notions. A subset of U lt(A) is -closed provided it is the intersection
of countably many subsets in Clop(U lt(A)). A subset of U lt(A) is -nowhere dense
if it is a subset of a nowhere dense -closed set. Clearly, a -nowhere dense subset
is nowhere dense. Finally, a subset of U lt(A) is of the -category if it is the union
of countably many subsets -nowhere dense in U lt(A). Observe that a -category
subset is meager in U lt(A). Let M0 be the -ideal of all subsets of the -category in
(Clop (U lt(A))), the -…eld generated by Clop(U lt(A)). Set
F = f(u [ v1 )
v2
with u 2 Clop(U lt(A)) and v1 ; v2 2 M0 g
F is a …eld contained in (Clop(U lt(A))). Let s : A ! Clop(U lt(A)) be the Stone
isomorphism. Let h be the -homomorphism from F onto the quotient algebra FjM0
de…ned by
h(a) = fb 2 F : b a 2 M0 g
where a 2 F. Let be a homomorphism from A onto FjM0 de…ned by (a) = h(s(a)),
where a 2 A. is called the canonical homomorphism. This is the version of Sikorski’s
result we are interested in:
Proposition 1 (Sikorski) The canonical homomorphism is a -isomorphism from A
onto FjM0 .
A Boolean algebra satis…es the countable chain condition, c.c.c. for short, if each
pairwise disjoint family in A is at most countable. For this class of Boolean algebras,
the next result is proved in [18]:
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Proposition 2 (Sikorski) A Boolean algebra A satis…es the c.c.c. if and only if every
nowhere dense subset in U lt(A) is -nowhere dense.
Therefore, if a Boolean algebra A satis…es the c.c.c., then a subset of U lt(A) is
meager i¤ it is -nowhere dense. Let M 0 be the -ideal of all meager subsets in
(Clop(U lt(A))). By proposition 2, FjM0 = FjM 0 in a Boolean algebra which satis…es
the c.c.c.
Separable Boolean algebras and atomless Boolean algebras are the two other notions
we are interested in. A Boolean algebra A is separable if there exists a countable set
D of nonzero elements of A which is dense in A, i.e. such that for every a 2 A, with
a 6= 0, there is an element a0 2 D with a0 a. A Boolean algebra is atomless if it has
no atoms. An atom is a nonzero element a 2 A such that, for every element a0 2 A
with a0 a; we have either a0 = a or a0 = 0, where 0 is the zero element of A.
In view of the results of section 3 we are particularly interested in the Boolean
algebras in separable Stone spaces. To deal with them we introduce a cardinal function.
Let s be a countable subset of the Stone space U lt(A). Let d( ) be the cardinal function
on A de…ned by
d(A) = min fjsj : s is a dense subset of U lt(A)g :
Clearly, U lt(A) is separable i¤ d(A) = @0 .
Remark. Unlike [10], we denote by Bai(X) the Baire -…eld of X and not the
-…eld of subsets with the Baire property. Moreover, we call Baire subsets the elements
of Bai(X).
3
Existence of residual measures
We begin by proving a quite useful existence result.
Proposition 3 Let X be a separable T1 space. Then there exists a residual regular
measure i¤ X has an isolated point.
Proof. If x is an isolated point, the Dirac probability measure x is a residual regular
measure. For the converse, suppose that X is perfect (i.e. X has no isolated points).
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Since X is a T1 space, all singletons fxg, for x 2 X, are closed subsets. Since X is
perfect, no singleton can be at the same time both open and closed. So int(fxg) =
? and the singletons are nowhere dense subsets. Let s be a countable dense subset of
X and let P be a regular residual measure. We have P (s) = 0 since P (fxg) = 0; the
P
singletons being nowhere dense subsets. Then P (s) = x2s P (fxg) = 0: Since s is a
dense subset, X n s does not contain any nonempty open subset. This implies that all
closed subsets c contained in X n s are such that int(c) = ?, i.e. they are nowhere
dense subsets. Then P (c) = 0: Since P is regular, we have:
P (X n s) = sup fP (c) : c
X n s and c closedg :
It follows P (X n s) = 0 and so P (X) = P (X n s) + P (s) = 0 for any regular residual
measure P:
Remark. After the …rst draft of this paper was completed, I have found a similar
result in [6, p.113]. However my proof is di¤erent and more direct than that given in
[6].
Remark. The Borel measure on [0; 1] is not residual. In fact, there exists a nowhere
dense subset with positive Borel measure. This set is constructed like the Cantor set,
except that now, at each stage, are deleted intervals of length 3n , with 0 < < 1: The
set has Borel measure 1
, and it is nowhere dense. For later reference, we denote
this set by c .
In [5, proposition 5] it is stated that in a perfect metric space cannot exist a nonzero
residual measure which is compact regular. In [2, proposition 4a] it is proved that there
exists a nonzero -additive residual measure on a metric space X i¤ X has an isolated
point. We can give a more general result.
Proposition 4 Let X be a metric space. Then there exists a nonzero residual measure
i¤ X has an isolated point.
Proof. In [5, p.249] it is proved that a perfect metric space contains a dense subset s
which is meager. Let P be a residual measure. Then P (s) = 0: Since s is dense, we
have int(X n s) = ?. Moreover, since X is a metric space, P is regular. Therefore,
in order to complete the proof it su¢ ces to follow the same reasoning employed in the
…nal part of the proof of proposition 3.
8
Proposition 4 is a partial generalization of the well known fact that an Euclidean
space can be decomposed in two sets, one of Lebesgue measure zero, the other meager.
In proposition 4 we considered metric spaces. Similar results hold for other topological spaces, as the next proposition shows. Here T3 means regular and T1 .
Proposition 5 Let X be a T3 separable space. Then there exists a residual measure
i¤ X has an isolated point.
Proof. A separable space has the countable chain property. Therefore it follows
from [2, proposition 2a] and from [1, proposition 6] that every residual measure is additive. On the other hand, by [7, theorem 5.4] every -additive probability measure
on a regular space is a regular probability measure. A simple application of proposition
3 completes the proof.
Now we turn to nonatomic residual measures. The next proposition contains some
non–existence results for this class of measures. In particular, points (i) and (iv) will
be employed in section 4. We denote by Is(X) the set of all isolated points of the space
X.
Proposition 6 (i) Let X be a T3 separable space. Then there is no nonzero nonatomic
residual measure.
(ii) Let X be a separable metric space. Then there is no nonzero nonatomic residual
measure.
(iii) Let X be a T3 space such that Is(X) forms a dense subset. If Is(X) is
nonmeasurable, then there is no nonatomic residual measure.
(iv) Let X be a compact Hausdor¤ space such that Is(X) forms a dense subset.
Then there is no nonatomic and regular residual measure.
Proof. (i) Let P be a nonatomic residual measure. Since P is nonatomic, P (fxg) = 0
for all x 2 X. In particular, if s is the countable dense subset of X which exists
P
by hypothesis, we have P (s) = x2S P (fxg) = 0: Observe that, if there are isolated
points in X, they are all contained in s. In fact int(X n s) = ?. We have seen in the
proof of proposition 5 that all residual measures on a separable T3 space are regular.
Therefore a reasoning similar to the one employed in the proof of proposition shows
that P (int(X n s)) = 0:
9
(ii) The proof is similar to the one of point (i).
(iii) Let P be a nonatomic residual measure. Clearly the cardinal number of each
pairwise disjoint family of open subsets is at most jIs(X)j. By [2, proposition 2a] every
residual measure on X is -additive. In particular, P is -additive. Let I
Is(X)
with jIj = @0 . From point (i), we know that P (I) = 0: Now suppose that I Is(X)
with jIj = @1 . Let us denote the ordinal numbers by Greek letters. It is known that
the set I can be put into a one-to-one correspondence with the ordinal numbers less
than @1 so that can be written I = fx : < @1 g (see [16, proposition 3.27]). Set
h =
[
fx :
< @1 g :
The collection fh : < @1 g is a monotone increasing net with respect to set inclusion.
Every isolated point is a clopen subset. So h is an open subset for every ordinal
S
number . By de…nition, we have I = fh : < @1 g. Therefore it holds P (I) =
supfP (h ) :
< @1 ) because P is -additive. From what we have already proved
for jIj = @0 , we can say that P (h ) = 0 for each < @1 . Therefore P (I) = 0: This
implies that every subset of Is(X) of cardinality @1 has measure zero. Using trans…nite
induction, it is easy to extend this conclusion to all cardinals @ jIs(X)j. In particular
P (Is(X)) = 0. Since Is(X) is a dense subset, we have int(X n Is(X)) = ?. Since all
-additive measures on a T3 space are regular, a reasoning similar to the one employed
in the proof of proposition 3 shows that P (int(Xns)) = 0:
(iv) A compact regular Borel probability measure is -additive. Therefore the proof
is similar to the one of point (iii).
4
The existence of probability measures on Boolean
algebras
The existence results proved in the previous section have interesting applications for
the study of probability measures de…ned on Boolean algebras. We begin with a consequence of proposition 5.
Proposition 7 Let A be an atomless Boolean algebra with d(A) = @0 . Then there is
no nonzero probability measure on A.
10
Proof. Let Bai(U lt(A)) be the Baire -…eld of U lt(A). It is known that Bai(U lt(A))
coincides with the -…eld generated by Clop(U lt(A)), i.e. (Clop(U lt(A))). Then M 0
is the collection of all meager Baire subsets. Let P be a probability measure on A. Set
1
P =P
h : F ! [0; 1] and Ps = P s 1 : Clop(U lt(A)) ! [0; 1]. By propositions
1 and 2, Ps and P are probability measure, respectively, on Clop(U lt(A)) and on F.
Let P # be the restriction of P to Clop(Ult(A)). If a 2 A, we have:
P # (s(a)) = P (s(a)) = P (
= P(
1
1
(h(s(a)))
( (a))) = P (a) = Ps (s(a)):
Therefore P # Ps , and P is the probability measure which extends Ps on F.
Clearly M 0
F and it is easy to check that P (v) = 0 for all v 2 M 0 . From
the inclusions Clop(U lt(A))
F
Bai(U lt(A)), it follows (F ) = Bai(U lt(A)).
By Caratheodory extension theorem there is a unique probability measure P 0 which
extends Ps from Clop(U lt(A)) to Bai(U lt(A)). Clearly P 0 coincides with P on F.
The Boolean algebra A satis…es the countable chain condition (c.c.c.). Suppose not.
Let fai gi2I
A be a set of pairwise disjoint and nonzero elements with jIj > @0 .
Then fs(ai )gi2I is a set of pairwise disjoint elements in U lt(A). Moreover s(ai ) 6= ?
for all i 2 I since ai is nonzero. But U lt(A) is separable and so it satis…es the c.c.c.
for topological spaces. Since the s(ai ) are clopen subsets, this implies that at most
a countable number of the s(ai ) are non-empty. This contradiction shows that A
satis…es the c.c.c.. Therefore, by proposition 2, every nowhere dense subset in U lt(A)
is -nowhere dense, i.e. it is contained in a closed nowhere dense subset c such that
T
U lt(A) be a
c= 1
i=1 ui for some ui 2 Clop(A). Clearly c 2 Bai(U lt(A)). Let a
S1
meager subset. Then a = i=1 ai , with ai nowhere dense subsets. By proposition 2,
a=
1
[
ai
i=1
1
[
ci
i=1
S
where the ci are closed and Baire nowhere dense subsets. Since 1
i=1 ci 2 Bai(U lt(A)),
it follows that every meager subset is contained in a Baire meager subset. Let Bor(U lt(A))
be the Borel -…eld of U lt(A). Since U lt(A) is compact and Hausdor¤, there exists a
unique regular extension P 00 from Bai(U lt(A)) to Bor(U lt(A)) (see [3, p.183]). Let M
be the -ideal of all meager Borel subsets of U lt(A). If a 2 M , we know that there is
a subset a0 2 M 0 such that a a0 . Since P 0 (a0 ) = 0; it follows that P 00 (a) = 0, i.e. P 00
11
is a residual measure. Furthermore, P 00 is just the unique extension of P 0 . In fact, by
now it is clear that every extension of P 0 is residual. And we have seen in the proof of
proposition 5 that each residual measure on a -…eld like Bor(U lt(A)) is regular. But,
there is a unique regular extension.
To sum up, we have proved that the existence of a nonzero probability measure
P on A implies the existence of a unique nonzero residual and regular measure P 00 on
Bor(U lt(A)). However, by proposition 5 we have P 00 0: Therefore, by contraposition,
P 0:
In [9, theorem 3.2] it was proved that on a separable atomless Boolean algebra
there is no nonzero probability measure. This result is now an easy consequence of our
proposition 7, as shown by the next corollary.
Corollary 1 Let A be a separable atomless Boolean algebra. Then there is no nonzero
probability measure on A.
Proof. In view of the proof of proposition 7, it su¢ ces to show that U lt(A) is separable.
By hypothesis there exists a countable dense subset D of A. Let s : A!U lt(A) be
the Stone isomorphism. For every f 2Clop(U lt(A)) there exists a subset v 2 s(D)
such that ? 6= v
f . Since Clop(U lt(A)) is a base for U lt(A), it follows that s(D)
is a pseudobase for U lt(A). Let xv 2 v for every v 2 s(D). It is easy to check that
fxv gv2s(D) is a countable dense subset of U lt(A).
While [9, theorem 3.2] is a consequence of proposition 7, the following result shows
that the converse is not true.
Proposition 8 There exists an atomless Boolean algebra A with d(A) = @0 , but not
separable.
Proof. Let us consider the generalized Cantor space 2I with I = 2@0 . According
to [17, theorem 14.3] its dual …eld Clop(2I ) is a free Boolean algebra with 2@0 free
generators. So Clop(2I ) is atomless (cf. [10, proposition 9.11]) and has a separable
Stone space ([10, corollary 9.7a] and [17, §14E]). However Clop(2I ) is not a separable
Boolean algebra (cf. [9, p.479]).
Remark. As shown by corollary 1 and proposition 8, proposition 7 extends the
result of Horn and Tarski. Moreover, the proof they gave was algebraic. Instead, our
proof of proposition 7 is mainly of topological nature.
12
The next proposition shows a rather interesting fact. It says that we can keep
the result of proposition 7 by replacing the atomlessness assumption on A with a
nonatomicity assumption on the probability measure.
Proposition 9 Let A be a Boolean algebra with d(A) = @0 . Then there is no nonzero
nonatomic probability measure on A.
Proof. Suppose that P is a nonzero probability measure on A. From the proof of
proposition 7 we know that this implies the existence of a nonzero, regular and residual
measure P 00 on Bor(U lt(A)). Now suppose that P is nonatomic. Set Ps = P s 1 :
Clop(U lt(A)) ! [0; 1]. We already know that Ps coincides with the restriction of
P 00 on Clop(U lt(A)). It is a simple matter to verify that Ps is nonatomic. Let g
S
be a open subset of U lt(A). We have g = i2I fi with fi 2 Clop(U lt(A) because
S
Clop(U lt(A) is a base. Let g = ff :
and
gg. Since P 00 is -additive,
we have P 00 (g) = supfP 00 (g ) :
gg. If P 00 (g) > 0; then a reasoning similar to
the one employed in the proof of proposition 6(iv) allows us to say that there is some
fi 2 Clop(U lt(A) such that Ps (fi ) > 0: Since Ps is nonatomic, fi is not a Ps -atom.
A fortiori, fi is not a P 00 -atom. Therefore g is not a P 00 -atom. Let c be a closed
subset of U lt(A). We have P 00 (c) = P 00 (int(c)) because the boundary b(c) is a nowhere
dense Borel subset. Suppose P 00 (c) > 0: Then P 00 (int(c)) > 0 and c is not a P 00 atom. Let a2Bor(U lt(A)). Since P 00 is regular we have P 00 (A)= supfP 00 (c) : c a and
c 2 U lt(A)g. If P 00 (a) > 0; then P 00 (c) > 0 for some c. Therefore a is not a P 00 -atom
and this implies that P 00 is nonatomic. Hence, by proposition 6(i) we have P 00 0 and
so, by contraposition, P 0:
Proposition 9 shows a class of Boolean algebras which do not admit nonatomic
probability measures. A special case of interest is now considered.
Corollary 2 Let A be an atomic Boolean algebra which satis…es the c.c.c.. Then there
is no nonzero nonatomic probability measure on A.
Proof. It is known that if a Boolean algebra is atomic, then the subset of the isolated
point is dense in the Stone space. Since A satis…es the c.c.c., then there are at most a
countable number of isolated points. Therefore, Is(U lt(A)) is a countable dense subset
of U lt(A), and so d(A) = @0 .
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5
The range of a nonatomic probability measure
Let P be a nonatomic probability measure de…ned on a Boolean algebra which satis…es
the c.c.c.. Using some constructions of the previous sections, we prove that the range
R(P ) of P is a dense subset of the unit interval.
To do this we need the following result, that can be proved with the techniques
used in [11].
Proposition 10 Let X be a topological space on which the open Baire subsets G(X)
form a base, and let 0 be a Baire measure which has a inner regular and weakly additive Borel extension 00 . Then R( G(X) ) is dense in R( 00 ).
The announced result is now an easy consequence of the last result.
Proposition 11 Let P be a nonatomic probability measure de…ned on a Boolean algebra A which satis…es the c.c.c.. Then R(P ) is a dense subset of [0; 1].
Proof. From the proof of proposition 9 we know that P implies the existence of a
regular, nonatomic and residual probability measure P 00 on Bor(U lt(A)). Since X is
compact, P 00 is also compact regular and so -additive. By proposition 10; R(P ) is
dense in R(P 00 ). But, being P 00 nonatomic, it is known that R(P 00 ) = [0; 1].
Remarks. (i) In looking at this result it is important to keep in mind the distinction between measures on Boolean algebras and measures on …elds outlined in section
2. In fact, the example given there shows that this result is not true for probability
measures de…ned on …elds. (ii) Let Bor(<) be the Borel -…eld of the real line. Let
N be the -ideal of the subsets with Lebesgue measure zero. The quotient algebra
Bor(<)=N does not have a separable Stone space (cf. [17, p.95]). But Bor(<)=N satis…es the c.c.c.. In fact, on Bor(<)=N there exists a nonzero strictly positive -…nite
measure induced by the Lebesgue measure. This example shows that proposition 9
does not con‡ict with proposition 11.
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