On compact spaces carrying Radon measures of uncountable Maharam type

154 (1997)

On compact spaces carrying

Radon measures of uncountable Maharam type

by

D. H. F r e m l i n (Colchester)

Abstract. If Martin’s Axiom is true and the continuum hypothesis is false, and X is a compact Radon measure space with a non-separable L

space, then there is a continuous surjection from X onto [0, 1]

.

1. Introduction. For any probability space (X, Σ, µ), its measure al- gebra is the quotient Boolean algebra Σ/N , where N is the σ-ideal of sets of measure 0. For more than fifty years we have had a complete description of these Boolean algebras. There is the two-element algebra {0, 1}; for each infinite cardinal κ there is the measure algebra B

of the usual measure on {0, 1}

; and there are countable products of these of the form PJ × Q

i∈I

B

, where I and J are countable sets and hκ

i

is a family of distinct infinite cardinals. And that is all. (See [12] and [5], §3.) Part of the interest of this classification lies in the fact that it completely describes the function spaces L

(µ), for 1 ≤ p ≤ ∞, up to Banach lattice isomorphism.

Now suppose that (X, T) is a topological space. In this case a measure on X may or may not be related to the topology in various ways. By far the most important is the idea of Radon measure, in which all open sets (and therefore all Borel sets) are measurable and µE = sup{µK : K ⊆ E is compact} for every measurable set E. It is customary, in this context, to suppose that X is Hausdorff, so that compact sets are closed. There are complications if µX = ∞; in this paper I will consider probability measures exclusively. Now, given a Hausdorff space (X, T), we can ask: which Boolean algebras can appear as the measure algebras of Radon probability measures on X? Write K(X) for the set of infinite cardinals κ for which there is a Radon probability measure µ on X such that the measure algebra of µ is isomorphic to B

. It is known that if ω ≤ λ ≤ κ ∈ K(X) then λ ∈ K(X),

1991 Mathematics Subject Classification: 28C15, 54A25.

[295]

so that K(X) must be an initial segment of the class of infinite cardinals.

It may or may not contain its supremum, but once we know K(X) we can determine all the possible measure algebras of Radon probability measures on X, as follows. If X = ∅ there are none. If X is finite and not empty, we get algebras of the form PJ where 1 ≤ #(J) ≤ #(X). If X is infinite, we get algebras of the form PJ × Q

i∈I

B

, where I and J are countable sets and hκ

i

is a family of distinct members of K(X); and these lists are complete.

So we turn to the determination of K(X). Because Radon measures are defined by their behaviour on compact sets, K(X) = S

{K(K) : K ⊆ X is compact}, so we begin by investigating compact spaces X. Now if X and Y are compact Hausdorff spaces and f : X → Y is a continuous surjection, then K(Y ) ⊆ K(X) ([7], Prop. 2.1); if Z is a closed subset of X, then K(Z) ⊆ K(X); further, sup K(X) ≤ w(X), the topological weight of X; and κ ∈ K([0, 1]

) for any infinite cardinal κ. What this means is that K([0, 1]

) must be just {λ : ω ≤ λ ≤ κ}, and if there is a continuous function from X onto [0, 1]

then κ ∈ K(X).

The question now arises: is there a converse to this result? If we know that κ ∈ K(X), when, if ever, can we deduce that there is a continu- ous surjection from X onto [0, 1]

? I will say that an infinite cardinal κ has Haydon’s property if whenever X is a compact Hausdorff space and κ ∈ K(X), then there is a continuous surjection from X onto [0, 1]

. Of course, ω has Haydon’s property, since for any compact Hausdorff space X with a non-empty perfect subset there is a continuous surjection from X onto [0, 1]

. The first investigation of the question was by R. G. Haydon, who showed that if κ is regular and λ

< κ for every λ < κ (for instance, if κ = c

), then κ has Haydon’s property ([7], Theorem 2.4). Recently, G. Plebanek extended Haydon’s result by showing that it is true for any cardinal κ such that cf(κ) ≥ ω

and κ is a precaliber of B

([14], Theo- rem 4.1).

These results leave open the basic case κ = ω

. In [8], Haydon gave an example to show that if the continuum hypothesis is true then ω

does not have Haydon’s property. This example has been refined and adapted in various ways ([10], [2]). In the present context, the best results are due to K. Kunen and J. van Mill, who showed that if [0, 1]

can be covered by ω

negligible sets then ω

does not have Haydon’s property ([11]), and to Plebanek, who showed that if κ is a cardinal of uncountable cofinality which is not a precaliber of B

then κ does not have Haydon’s property ([14], Theorem 4.2). M. R. Burke has pointed out that if we add ω

random reals to a model of ZFC+CH then ω

becomes a precaliber of B

, but ω

does not have Haydon’s property because the conditions of Theorem 6.2 of [14]

are satisfied.

All these examples have served to concentrate attention on Martin’s Ax- iom: what happens if Martin’s Axiom is true and the continuum hypothesis is false? In this paper I show that under these circumstances ω

does have Haydon’s property; and in fact the same is true of any infinite cardinal κ such that MA(κ) is true (Theorem 9 below).

2. Notation. I follow [3] in writing m for the least cardinal such that MA(m) is false, so that Martin’s Axiom becomes “m = c”. If I is a set, κ a cardinal then [I]

is the set of subsets of I of cardinal κ, and [I]

= S

n∈N

[I]

is the set of finite subsets of I. If A is a Boolean algebra, a cardinal κ is a precaliber of A if for every family ha

i

of non-zero elements of A, there is a set A ∈ [κ]

such that inf

a

6= 0 for every non-empty finite I ⊆ A. Note that if κ < m has uncountable cofinality then κ is a precaliber of every ccc Boolean algebra ([3], 41Ca), and that ω

is a precaliber of B

iff [0, 1]

is not the union of ω

negligible sets (use [6], A2U).

In a Boolean algebra A, I will use the symbol ∩ to represent “inter- section” (the “product” when A is regarded as a ring); 1 \ a will be the

“complement” of a, and ⊆ will denote the usual partial order of “inclusion”.

3. I start by recalling a well-known fact about uncountable families of sets in probability spaces.

Lemma. Let (X, Σ, ν) be a probability space and hF

i

a family of measurable sets of non-zero measure. Then there is an uncountable set A ⊆ ω

such that inf

ν(E

∩ E

) > 0 (cf. [1], Theorem 6.15; see also [4]).

4. Notation. The core of this proof is an investigation of certain prop- erties of the algebras B

.

(a) Much of the argument will be based on the following straightforward idea. Let I be a set. If E ⊆ {0, 1}

, J ⊆ I then I will say that E is determined by coordinates in J if x ∈ E whenever x ∈ {0, 1}

and there is a y ∈ E such that x¹J = y¹J; equivalently, if there is a set F ⊆ {0, 1}

such that E = π

[F ], where π

(x) = x¹J for x ∈ {0, 1}

; equivalently, if E = π

[π

[E]].

Note that

(i) the family of sets determined by coordinates in J is closed under complements and arbitrary intersections and unions;

(ii) if E is determined by coordinates in J, and J ⊆ K ⊆ I, then E is determined by coordinates in K;

(iii) if E is determined by coordinates in J, and also determined by coordinates in K, then it is determined by coordinates in J ∩ K.

(For if x ∈ {0, 1}

, y ∈ E and x¹J ∩ K = y¹J ∩ K, define z ∈ {0, 1}

by

setting z(i) = x(i) if i ∈ J, and y(i) if i ∈ I \ J; then z¹K = y¹K, so z ∈ E,

and x¹J = z¹J, so x ∈ E.)

(b) If E ⊆ {0, 1}

and J ⊆ I, set

S

(E) = {x : x ∈ {0, 1}

, x + z ∈ E whenever z ∈ {0, 1}

,

z(i) = 0 for every i ∈ I \ J}, writing + for the usual group operation on {0, 1}

derived from identifying it with Z

. Observe that

(i) S

(E) is the largest subset of E determined by coordinates in I \ J;

(ii) S

S

(E) = S

(E) ⊆ S

(E);

(iii) if K ⊆ I is such that E is determined by coordinates in K (i.e., E = S

(E)), then S

(E) is determined by coordinates in K \ J;

(iv) if J is finite and E is measurable (for the usual measure on {0, 1}

) then S

(E) is measurable;

(v) if E is closed then S

(E) is closed;

(vi) if E is a zero set (that is, in this context, E is a closed set which is determined by coordinates in some countable set), and J is countable, then S

(E) is a zero set.

5. Lemma. Let I be a set and µ the usual measure on X = {0, 1}

. Let E ⊆ X be a measurable set and hI

i

a disjoint sequence of subsets of I all of size at most n. Then lim

µS

(E) = µE.

P r o o f. Let ε > 0. There is a set F ⊆ X, determined by a finite set K of coordinates, such that µ(E 4 F ) ≤ ε. Take k

such that I

∩ K = ∅ for k ≥ k

. Then for any k ≥ k

,

E \ S

(E) ⊆ (E \ F ) ∪ [

{x : ∃z, z(i) = 0 ∀i ∈ I \ I

, x + z ∈ F \ E}

has measure at most 2

ε. As ε is arbitrary, we have the result.

6. Without further ado, I proceed to the main result.

Theorem. Suppose that κ < m, and let µ be the usual measure on X = {0, 1}

. Let h(E

, E

)i

be a family of pairs of measurable subsets of X such that

{α : µE

+ µE

≥ 1 − ε, x(α) = 0 for every x ∈ E

,

x(α) = 1 for every x ∈ E

} has cardinal κ for every ε > 0. Then there is a set D ⊆ κ, of cardinal κ, such that

µ



X ∩ \

α∈I

E

∩ \

β∈J

E



> 0 for any disjoint finite sets I, J ⊆ D.

P r o o f. Part A. I first give the argument for the case in which κ has

uncountable cofinality, as this is easier, and then turn to the modifications

required if cf(κ) = ω.

(a) Write C for

{α : µE

+ µE

> 1/2, x(α) = 0 for every x ∈ E

,

x(α) = 1 for every x ∈ E

}, so that #(C) = κ. For each α ∈ C there is a zero set Z

, determined by coordinates in κ \ {α}, such that µZ

> 0 and

{x : x ∈ Z

, x(α) = 0} ⊆ E

, {x : x ∈ Z

, x(α) = 1} ⊆ E

. (Take zero sets F ⊆ E

, F

⊆ E

such that µF + µF

> 1/2. Set Z

= S

(F ∪ F

); this works.)

(b) (i) Let P be the set of all pairs (I, F ) where I ∈ [C]

, F ⊆ T

β∈I

Z

is a zero set determined by coordinates in κ \ I and µF > 0. Order P by saying that (I, F ) ≤ (I

, F

) if I ⊆ I

and F

⊆ F ; then P is a partially ordered set. Note that q

= ({α}, Z

) belongs to P for every α ∈ C.

(ii) P is upwards-ccc. To see this, let h(I

, F

)i

be a family in P . Then we can find an uncountable set A ⊆ ω

such that

(α) hI

i

is a constant-size ∆-system with root I say;

(β) whenever ξ < η in A, α ∈ I

and β ∈ I

\ I then α < β;

(γ) whenever ξ < η in A, then F

is determined by coordinates in κ \ (I

\ I);

(δ) there is a δ > 0 such that µ(F

∩ F

) ≥ δ for every ξ, η ∈ A (using Lemma 3).

Let hξ

i

be a strictly increasing sequence in A and η a member of A greater than any ξ

. Because #(I

\ I) is the same for all k, Lemma 5 tells us that there is a k such that

µS

(F

) > µF

− δ.

Set F

= S

(F

). Then

µ(F

\ F

) < δ, µ(F

∩ F

) > 0.

Now we know that F

is determined by coordinates in κ \ I

and also by coordinates in κ \ (I

\ I), and is therefore determined by coordinates in κ \ J, where J = I

∪ I

. Similarly, F

is determined by coordinates in (κ \ I

) \ (I

\ I) = κ \ J, so F = F

∩ F

is also determined by coordinates in κ \ J, while µF > 0. Finally,

F ⊆ F

∩ F

⊆ \

β∈J

Z

.

This means that (J, F ) ∈ P , and evidently it is a common upper bound for (I

, F

) and (I

, F

).

Thus h(I

, F

)i

is not an up-antichain. As h(I

, F

)i

is arbitrary,

P is upwards-ccc.

(c) Because m > κ, there is a sequence hR

i

of upwards-directed subsets of P such that q

∈ S

n∈N

R

for every α ∈ C (see [3], 41Ca). Because

#(C) = κ and cf(κ) ≥ ω

, there is an n such that D = {α : q

∈ R

} has cardinal κ. If I, J ∈ [D]

are disjoint there is a (K, F ) ∈ R

which is an upper bound for {q

: α ∈ I ∪ J}, so that I ∪ J ⊆ K; now F is determined by coordinates in κ \ (I ∪ J), and

F ∩ E

= F ∩ Z

∩ E

= {x : x ∈ F, x(α) = 0} for every α ∈ I, F ∩ E

= F ∩ Z

∩ E

= {x : x ∈ F, x(β) = 1} for every β ∈ J, so

µ



X ∩ \

α∈I

E

∩ \

β∈J

E



≥ µ{x : x ∈ F, x(α) = 0 ∀α ∈ I, x(β) = 1 ∀β ∈ J}

= 2

µF > 0.

Thus D has the declared property.

Part B. I now turn to the adaptations required if cf(κ) ≤ ω. If κ ≤ ω the result is easy and of no importance, so I leave it as an exercise for any reader who wishes to check her understanding of the hypotheses. For the case ω = cf(κ) < κ, start by expressing κ as the union of a strictly increasing sequence hκ

i

of regular uncountable cardinals. Set

C

=



α : µE

+ µE

> m + 1 m + 2 ,

x(α) = 0 for every x ∈ E

, x(α) = 1 for every x ∈ E

 , so that C

has cardinal κ for each m ∈ N, by hypothesis. Let hC

i

be a partition of C = C

such that C

⊆ C

and #(C

) = κ

for all m, j ∈ N. Write C

= S

m∈N

C

for each j, so that hC

i

is disjoint and

#(C

∩ C

) = κ

for every j, m ∈ N.

(a) For j, m ∈ N and α ∈ C

there is a zero set Z

, determined by coordinates in κ \ {α}, such that µZ

> m/(m + 2) and {x : x ∈ Z

, x(α) = 0} ⊆ E

, {x : x ∈ Z

, x(α) = 1} ⊆ E

. (In (A-a) above, take F , F

such that µF + µF

> (m + 1)/(m + 2) and continue as before.)

(b) Define the partially ordered set P as in (A-b-i); as in (A-b-ii), P is ccc. Once again, set q

= ({α}, Z

) for α ∈ C.

(c) Now for the new idea.

(i) For any m, j ∈ N there is a finite set I

⊆ κ such that whenever (I, F ) ∈ P and I ∩ I

= ∅, #(I) ≤ m, µF ≥ 2/(m + 2) then

A

(I, F ) = {α : α ∈ C

, (I, F ) and q

are compatible in P }

has cardinal κ

. For suppose, if possible, otherwise. Then we can find a sequence h(I

, F

)i

in P such that #(I

) ≤ m, µF

≥ 2/(m + 2),

#(A

(I

, F

)) < κ

for every k and hI

i

is disjoint. For each k, let J

⊆ κ be a countable set such that F

is determined by coordinates in J

. Because κ

has uncountable cofinality, there is an α ∈ C

\ S

k∈N

(A

(I

, F

) ∪ J

).

But now Lemma 5 tells us that

k→∞

lim µS

(Z

) = µZ

> m m + 2 ,

so there is some k such that µS

(Z

) > m/(m + 2). Set F

= F

∩ S

(Z

).

Then

µF

≥ µF

+ µS

(Z

) − 1 > 0,

and F

is determined by coordinates in κ \ (I

∪ {α}), so (I

∪ {α}, F

) witnesses that (I

, F

) and q

are compatible, which is supposed to be im- possible.

(ii) Set I

= S

m,j∈N

I

, so that I

is countable. Set P

= {(I, F ) : (I, F ) ∈ P, I ∩ I

= ∅}. Then two members of P

are compatible in P

iff they are compatible in P (if (I

, F

), (I

, F

) belong to P

and have a common upper bound (I, F ) ∈ P , then (I

∪ I

, F ) ∈ P

is still an upper bound), so P

is also ccc.

(iii) Enumerate each C

as hγ

i

. For β < κ

, set

Q

= {(I, F ) : (I, F ) ∈ P

, ∃ξ, β ≤ ξ < κ

, γ

∈ I}.

Then Q

is cofinal with P

. For if (I, F ) ∈ P

, there is some m ∈ N such that #(I) ≤ m and µF ≥ 2/(m + 2). Because I ∩ I

= ∅, A

(I, F ) ⊆ C

has cardinal κ

and there must be some ξ ≥ β such that γ

∈ A

(I, F )\I

. But now (I, F ), q

have a common upper bound in P and therefore a common upper bound in P

, which is a member of Q

greater than or equal to (I, F ).

(iv) Because κ < m, there is an upwards-directed set R ⊆ P

meeting every Q

. Set D = S

{I : (I, F ) ∈ R}. Because R meets every Q

, we have #(D ∩ C

) = κ

for every j and #(D) = κ. And D has the property required by the theorem, just as in (A-c) above.

7. Corollary. Suppose that ω ≤ κ < m. Let (A, µ) be a probability alge- bra. Suppose that ha

i

is a stochastically independent family of elements of measure 1/2, and that for each α < κ we are given elements e

⊆ a

, e

⊆ 1 \ a

such that {α : µe

+ µe

≥ 1 − ε} has cardinal κ for every ε > 0.

Then there is a set D ⊆ κ, of cardinal κ, such that inf

e

∩ inf

e

6= 0 for all disjoint finite I, J ⊆ D.

P r o o f. Let A

be the closed subalgebra of A generated by {a

: α < κ},

so that A

is isomorphic (as measure algebra) to B

. Let C ⊆ κ be a set

such that C and κ \ C both have cardinal κ; let h : κ → C be a bijection.

Then we have a measure-preserving Boolean homomorphism π

: A

→ B

defined by saying that πa

= b

, where b

∈ B

is the equivalence class of G

= {x : x(β) = 0}. We can identify B

with the probability algebra free product π

[A] b ⊕ C, where C is the closed subalgebra generated by {b

: β ∈ κ \ C} (cf. [5], 2.25b). Let A

be the closed subalgebra of A generated by {a

: α < κ} ∪ {e

: α < κ} ∪ {e

: α < κ}. Then π

has an extension to a measure-preserving Boolean homomorphism π

: A

→ B

(see [5], 3.11a).

Let µ be the usual measure on {0, 1}

. For β ∈ C, take E

⊆ G

, E

⊆ {0, 1}

\ G

such that E

= π

e

, (E

)

= π

e

, where α = h

(β).

For β ∈ κ \ C, set E

= E

= ∅. Then

#({β : µE

+ µE

≥ 1 − ε}) = #({α : µe

+ µe

≥ 1 − ε}) = κ for every ε > 0.

Applying Theorem 6, there is a set D

∈ [κ]

such that µ( T

α∈I

E

∩ T

β∈J

E

) > 0 for all disjoint finite I, J ⊆ D, that is, µ(inf

e

∩inf

e

)

> 0 for all disjoint finite I, J ⊆ h

[D

]. Of course, D

⊆ C. So we take D = h

[D

].

8. Corollary. Suppose that ω ≤ κ < m. Let (X, Σ, µ) be a probability space. Suppose that hG

i

is a stochastically independent family of ele- ments of measure 1/2, and that for each α < κ we have measurable sets E

⊆ G

, E

⊆ X \ G

such that {α : µE

+ µE

≥ 1 − ε} has cardinal κ for every ε > 0. Then there is a set D ⊆ κ, of cardinal κ, such that µ(X ∩ T

α∈I

E

∩ T

β∈J

E

) > 0 for all disjoint finite I, J ⊆ D.

P r o o f. Apply Corollary 7 with (A, µ) the measure algebra of µ, a

= G

, e

= E

and e

= (E

)

for each α.

9. Theorem. Suppose that ω ≤ κ < m. Let (X, T) be a compact Haus- dorff space. Then the following are equivalent:

(i) there is a Radon probability measure µ on X with measure algebra isomorphic to B

;

(ii) there is a continuous surjection from X onto [0, 1]

.

P r o o f. The implication (ii)⇒(i) is discussed in §1 above (and does not depend on the assumption m > κ); this proof will therefore address (i)⇒(ii).

Let hG

i

be a stochastically independent family of sets of measure 1/2.

Let hC

i

be a partition of κ into sets of size κ. For each α, we can find compact sets E

⊆ G

, E

⊆ X \ G

such that µE

+ µE

> m/(m + 1) if α ∈ C

. By Corollary 8, there is a set D ∈ [κ]

such that µ(X ∩ T

α∈I

E

∩ T

β∈J

E

) > 0 for all disjoint finite I, J ⊆ D. In particular, T

α∈I

E

∩ T

β∈J

E

is non-empty for all disjoint finite I, J ⊆ D; that is, h(E

, E

)i

is independent in the sense of [7]. By Lemma 2.1 of [7], [0, 1]

is a continuous image of X.

10. R e m a r k s. (a) If you look back at the demonstration that the partially ordered set P is ccc, in (A-b-ii) of the proof of Theorem 6, you will see that what I show there is that, given a family hp

i

in P , there is an uncountable set A ⊆ ω

such that whenever hξ

i

is a strictly increasing sequence in A, and η ∈ A is greater than any ξ

, then there is some k such that p

, p

are compatible in P . But this means that there must be an uncountable A

⊆ A such that p

, p

are compatible for all ξ, η ∈ A

, by the Erd˝os–Dushnik–Miller theorem ([3], A2K). Accordingly, P satisfies Knaster’s condition ([3], 11A). In all the results of this paper, therefore, we may replace “m > κ” by “m

> κ”, where m

≥ m is the cardinal associated with “Martin’s axiom for partially ordered sets with Knaster’s condition”

([3], 11D). (In part (A-c) of the proof of Theorem 6, we must now use 31B of [3] in place of 41C.)

(b) I note that when cf(κ) ≥ ω

then the hypotheses of §§6–8 are un- necessarily elaborate. Part A of the proof of Theorem 6 demands only that µE

+ µE

> 1/2 for κ indices α. So in Corollary 7, for instance, we have:

Suppose that κ < m

is a cardinal of uncountable cofinality. Let (A, µ) be a probability algebra. Suppose that ha

i

is a stochastically independent family of elements of measure 1/2 in A, and that for each α < κ we are given elements e

⊆ a

, e

⊆ 1 \ a

such that µe

+ µe

> 1/2 for every α.

Then there is a set D ⊆ κ, of cardinal κ, such that inf

e

∩ inf

e

6= 0 for all disjoint finite I, J ⊆ D.

Acknowledgements. I should like to mention here two of the essential elements leading to the discoveries described above: first, a preview of [14], sent to me by its author; second, the courtesy and stamina of M. R. Burke, M. Dˇzamonja and M. Foreman in the face of the original exposition. As a corollary, I must thank the Mathematics Department of the University of Wisconsin for making it possible for us to gather there.

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Math. 153 (1997), 25–40.

Department of Mathematics University of Essex

Colchester, England E-mail: fremdh@essex.ac.uk

Received 24 February 1997;

in revised form 9 May 1997