Download Products of completion regular measures

FUNDAMENTA
MATHEMATICAE
147 (1995)
Products of completion regular measures
by
D. H. F r e m l i n (Colchester) and S. G r e k a s (Athens)
Abstract. We investigate the products of topological measure spaces, discussing conditions under which all open sets will be measurable for the simple completed product
measure, and under which the product of completion regular measures will be completion
regular. In passing, we describe a new class of spaces on which all completion regular Borel
probability measures are τ -additive, and which have other interesting properties.
1. Introduction. Suppose that (X, T, Σ, µ) and (Y, S, T, ν) are topological probability spaces, that is, probability spaces with topologies such
that every open set is measurable. We can form product measures on X × Y
in various ways. First, we have the ordinary completed product measure λ
derived by Carathéodory’s method from the outer measure λ∗ , where
nX
o
[
λ∗ C = inf
µEi νFi : Ei ∈ Σ, Fi ∈ T ∀i ∈ N, C ⊆
Ei × Fi .
i∈N
i∈N
It can happen that λ is again a topological measure in that every open set in
X × Y , for the product topology, is λ-measurable, but even for apparently
well-behaved spaces (e.g., completion regular compact Radon probability
spaces) this is not necessarily true (see [4]); the conditions under which it
occurs are not well understood.
For a wide variety of important cases, it is known that λ does at least have
an extension to a topological measure.
S Recall that a topological probability
space (X, T, Σ, µ) is τ -additive if µ( G) = supG∈G µG for every non-empty
upwards-directed family G of open sets. If (X, T, Σ, µ) is a topological probability space, and (Y, S, T, ν) is a τ -additive topological probability space,
then we have a topological measure λR on X × Y , given by setting
λR C =
R
νCx µ(dx)
for Borel sets C ⊆ X × Y , writing Cx = {y : (x, y) ∈ C} for C ⊆ X × Y ,
x ∈ X; for consistency with other constructions, we will take it that λR is
1991 Mathematics Subject Classification: Primary 28C15.
[27]
28
D. H. Fremlin and S. Grekas
to be the completion of its restriction to the Borel sets of X × Y . If now µ
is also τ -additive then we have a similar measure λ0R given by
λ0R C =
R
µC y ν(dy)
for Borel C ⊆ X × Y , writing C y = {x : (x, y) ∈ C}, and under these
circumstances λR and λ0R are both τ -additive, therefore equal (since they
agree on open rectangles). (See [18]. We remark that if µ and ν are both
Radon measures, then so is λR = λ0R .) On Borel sets in the domain of λ, λ
agrees with λR and λ0R if these are defined (using Fubini’s theorem on λ);
so if µ and ν are inner regular for the Borel sets—for instance, if they are
completions of Borel measures—then λR , if it is defined, will extend λ, and
λ will be a topological measure iff it is equal to λR .
b σ T , the
Yet another aspect of the problem concerns the algebra Σ ⊗
σ-algebra of subsets of X × Y generated by {E × F : E ∈ Σ, F ∈ T }. λ is
b σ T —that is, it is inner regular for
the completion of its restriction to Σ ⊗
b
Σ ⊗σ T —so (if ν is τ -additive and µ, ν are inner regular for the Borel sets)
bσ T.
λ = λR iff λR is inner regular for Σ ⊗
While the original impulse to study such questions arose from a simple
desire to understand the nature of product Radon measures, there are important problems in functional analysis which depend on the analysis here;
see, for instance, the distinction between “stable” and “R-stable” set which
is necessary in [20], §9.
An allied question refers to “completion regular” spaces. Let us say that
a topological measure space (X, T, Σ, µ) is completion regular if µ is inner
regular for the zero sets (that is, sets of the form f −1 [{0}] for some continuous f : X → R). Note that if B0 = B0 (X) is the Baire σ-algebra of X, that
is, the σ-algebra generated by the zero sets in X, then µ¹B0 is necessarily inner regular for the zero sets ([6], Theorem 4.2), so (X, T, Σ, µ) is completion
regular iff µ is inner regular for B0 . Many important spaces are completion
regular; in particular, if X is a locally compact topological group and µ is
(left or right) Haar measure on X, then µ is completion regular ([13], or [10],
§64, Theorem I), and if X is a product of compact metric spaces, and µ is
a product of strictly positive Radon probabilities on the factors, then µ is
completion regular ([12], Theorem 3, or [1], Theorem 3). Now the question
is, when is the product of completion regular measures again completion
b σ B0 (Y ) (and these
regular? Because B0 (X × Y ) always includes B0 (X) ⊗
are equal if X and Y are compact), the “product” in this question should
be taken to be a topological product measure; normally, of course, the product measure λR discussed above. If (X, T, Σ, µ) and (Y, S, T, ν) are compact completion regular topological probability spaces, with ν τ -additive,
then we see that λR is completion regular precisely when all open sets are
λ-measurable.
Products of completion regular measures
29
A variety of conditions have been found which are sufficient to ensure
that open sets are λ-measurable, or that λR is completion regular. Most of
those with which we are acquainted amount to special cases of the main
result of the present paper, which is based on a particular topological condition on one of the factors. Recall that a Hausdorff space is dyadic if it is
a continuous image of {0, 1}I for some set I. For the elementary properties
of dyadic spaces see [3], 3.12.12 and 4.5.9–11; the essential ones relevant to
us here are that
(i)
(ii)
(iii)
(iv)
(v)
compact metric spaces are dyadic,
finite unions of dyadic spaces are dyadic,
zero sets in dyadic spaces are dyadic,
all products of dyadic spaces are dyadic,
compact Hausdorff topological groups are dyadic
([16]; see also [21] and [2], pp. 93–94). We generalise this concept by saying that a topological space is quasi-dyadic if it is the continuous image
of a product of separable metrizable spaces. Now our first result is that if
(X, T, Σ, µ) and (Y, S, T, ν) are τ -additive topological probability spaces,
and one of them is quasi-dyadic and completion regular, then every open
set in X × Y is λ-measurable (Theorem 5). It follows that if both factors
are completion regular, so is λR . The argument of Theorem 5 depends on
both factors having τ -additive measures. As it happens, however, this assumption can be omitted in the case of the quasi-dyadic factor, because
any completion regular topological probability on a quasi-dyadic space is
τ -additive (Theorem 4).
Of course, these results have consequences for the product of many factors. If we have finitely many completion regular quasi-dyadic topological
Q
probability spaces (Xα , Tα , Σα , µα ), then every open set in Z = α Xα is
measurable for the simple product measure on Z; if we have infinitely many
such spaces, and all but countably many of the µα are strictly positive (that
is, give non-zero measure to every non-empty open set), the same will be
true on the infinite product (Corollary 6).
We should mention the following known special cases of these results.
(i) Kakutani’s theorem on the products of measures on compact metric
spaces, already discussed, is of course the fundamental version of Corollary 6.
(ii) [1], Theorem 3, allows one factor to be non-metrizable, and introduces the asymmetry to be found in the hypotheses of our Theorem 5.
(iii) [9], Theorem 3.1, covers the case in which the quasi-dyadic factor is
actually a product of separable metric spaces; it strengthens Theorem 2 of
[7] and the main theorem of [8].
(iv) Concerning our Theorem 4, the case in which X is itself a product
of separable metric spaces is given in [9], Theorem 3.1.
30
D. H. Fremlin and S. Grekas
For simplicity, we express our results in terms of probability measures. Of
course, everything we say about finite products applies at least to all totally
finite measures, and much of it can be extended to non-σ-finite measures if
care is taken over the definitions of the product measures.
2. Lemma. If hSα iα∈I is a family of separable
metric spaces, then every
Q
member E of the Baire σ-algebra of S = α∈I Sα depends on a countable
Q
subset
J of Q
I, that is, E = πJ−1 [F ] for some F ⊆ α∈J Sα , writing πJ :
Q
α∈I Sα →
α∈J Sα for the canonical map.
P r o o f. See [19], or [3], 2.7.12.
3. To help give meaning to the concept of “quasi-dyadic” space, we give
some elementary properties of these spaces.
Proposition. (a) A continuous image of a quasi-dyadic space is quasidyadic. Any product of quasi-dyadic spaces is quasi-dyadic.
(b) A space with countable network is quasi-dyadic.
(c) If X is a quasi-dyadic space and Y belongs to the Baire σ-algebra of
X, then Y is quasi-dyadic.
(d) A countable union of quasi-dyadic subspaces of a given topological
space is quasi-dyadic.
P r o o f. (a) Immediate from the definition.
(b) (See [3], 3.1.J and elsewhere, for basic facts concerning countable
networks.) Let D be a countable network for the topology of X. On X let
∼ be the equivalence relation in which x ∼ y if they belong to just the
same members of D; let Y be the space X/∼ of equivalence classes, and
θ : X → Y the canonical map. Y has a separable metrizable topology with
base {θ[D] : D ∈ D} ∪ {θ[X \ D] : D ∈ D}. Let I be any set such that
#({0, 1}I ) ≥ #(X), and for each y ∈ Y let fy : {0, 1}I → y be a surjection.
Then we have a continuous surjection f : Y × {0, 1}I → X given by saying
that f (y, z) = fy (z) for y ∈ Y , z ∈ {0, 1}I .
(c) Let hSα iα∈I be a family of separable metric spaces with product S
and f : S → X a continuous surjection. Let A be the family of subsets of
S which factor through countable sub-products, and E the set {E : E ⊆ X,
f −1 [E] ∈ A}. Then A and E are σ-algebras. By Lemma 2, every zero set
in S belongs to A; consequently, every zero set in X belongs to E and E
includes the Baire σ-algebra
of X. But if E ∈ E, there
Q is a countable J ⊆ I
Q
such that E = f [F × α∈I\J Sα ] for some F ⊆ α∈J Sα ; and F , being a
subset of a countable product of separable metric spaces, is separable and
metrizable, so E is quasi-dyadic.
Q
(d) If En ⊆ X is quasi-dyadic
for
each
n
∈
N,
then
Z
=
N
×
n∈N En
S
is quasi-dyadic, and f : Z → n∈N En is a continuous surjection, where
f (n, hxi ii∈N ) = xn .
Products of completion regular measures
31
R e m a r k. We include (b) to emphasize that the ideas here may be applied to non-Hausdorff spaces. Any continuous image of a product of spaces
with countable networks will be quasi-dyadic.
4. Theorem. Let (X, T, Σ, µ) be a quasi-dyadic completion regular topological probability space. Then µ is τ -additive.
P r o o f. Suppose, if possible, otherwise.
(a) The first step is the standard reduction to the case in which X
is covered by open sets of zero measure. In detail: suppose that X0 is a
quasi-dyadic space and µ0 is a completion regular topological probability
measure on X0 which is not τ -additive.
S Let G be an upwards-directed family
of open sets in X0 such that µ0 ( G) > supG∈G µ0 G. Set µ1 E = µ0 E −
supG∈G µ0 (E ∩ G) for every Borel set E ⊆ X0 ; then µ1 is a completion
S
regular Borel
S measure on X0 , µ1 G = 0 for every G ∈ G, and µ1 ( G) > 0.
Let X ⊆ G be a Baire set such that µ1 X > 0. Then X is quasi-dyadic
(3c). For Borel sets E ⊆ X, set µE = µ1 E/µ1 X; then µ is a completion
regular Borel probability measure on X, and {X ∩ G : G ∈ G} is a cover of
X by open negligible sets.
(b) Now let hSα iα∈I be a family of separable metricQspaces such that
there is a continuous surjection f : S → X, where S = α∈I Sα . For each
α ∈ I let Bα be a countable base for the topology of Sα . For J ⊆ I let C(J)
be the family of all subsets of S expressible in the form
{s : s(α) ∈ Bα ∀α ∈ K},
where K is a finite subset of J and Bα ∈ Bα for each α ∈ K; thus C(I) is
a base for the topology of S. Set C0 (J) = {U : U ∈ C(J), µ∗ f [U ] = 0} for
each J ⊆ I.
For each negligible set E ⊆ X, let hFn (E)in∈N be a family of zero subsets
of X \ E such that supn∈N µFn = 1. Then each f −1 [Fn (E)] is a zero subset
of S, so there is a countable set M (E) ⊆ I such that all the sets f −1 [Fn (E)]
depend on M (E) (Lemma 2). Let J be the family of countable subsets J of
I such that M (f [U ]) ⊆ J for every U ∈ C0 (J); then J is cofinal with [I]≤ω ,
that is, every countable subset of I is included in some member of J . (If we
start from any countable subset J0 of I and set
[
Jn+1 = J0 ∪ {M (f [U ]) : U ∈ C0 (Jn )}
S
for each n ∈ N, then n∈N Jn ∈ J .)
(c) For each J ∈ J , set
o
\n [
QJ =
Fn (f [U ]) : U ∈ C0 (J) .
n∈N
Then µQJ = 1 and f
−1
[QJ ] depends on J.
32
D. H. Fremlin and S. Grekas
If G ⊆ X is a negligible open set, then G ∩ QJ = ∅ whenever J ∈ J
−1
and there is a negligible Baire set Q ⊇ G such that
Q f [Q] depends on J.
−1
−1
For set H = πJ [πJ [f [G]]], where πJ : S → α∈J Sα is the canonical
map; then H is a union of members of C(J), because f −1 [G] is open. Also,
because f −1 [Q] depends on J, H ⊆ f −1 [Q], so f [H] ⊆ Q and µ∗ f [H] = 0;
thus all the members of C(J) included in H actually belong to C0 (J), and
H ∩f −1 [QJ ] = ∅. But this means that f −1 [G]∩f −1 [QJ ] = ∅ and G∩QJ = ∅,
as claimed.
In particular, if G is a negligible open set in X, then G∩QJ = ∅ whenever
J ∈ J and J ⊇ M (G).
(d) If J ∈ J , there are s, s0 ∈ f −1 [QJ ] such that s¹J = s0 ¹J and f (s),
f (s0 ) can be separated by open sets in X. To see this, start from any x ∈ QJ
and take a negligible open set G including x (recall that our hypothesis is
that X is covered by negligible open sets). For each n ∈ N let hn : X → R be
a continuous function
such that Fn (G) = h−1
n [{0}]. We know that G∩QJ 6= ∅,
S
while G ⊆
X
\
(
F
(G)
∩
Q
),
which
is a negligible Baire set; by (c),
n
J
n∈N
S
f −1 [X \ ( n∈N Fn (G) ∩ QJ )] does not depend on J, and there must be some
n such that f −1 [Fn (G) ∩ QJ ] does not depend on J. Accordingly, there must
be s, s0 ∈ S such that s¹J = s0 ¹J, s ∈ f −1 [Fn (G) ∩ QJ ] and s0 6∈ f −1 [Fn (G)
∩ QJ ]. Now s and s0 both belong to f −1 [QJ ], because f −1 [QJ ] depends
on J; while f (s) ∈ Fn (G) and f (s0 ) 6∈ Fn (G), so hn (f (s)) 6= hn (f (s0 )) and
f (s), f (s0 ) can be separated by open sets.
(e) We are now ready to embark on the central construction of the argument. We may choose inductively, for ordinals ξ < ω1 , sets Jξ ∈ J , negligible
open sets Gξ , G0ξ ⊆ X, points sξ , s0ξ ∈ S, sets Uξ , Vξ , Vξ0 ∈ C(I) such that
• Jη ⊆ Jξ , Uη , Vη , Vη0 all belong to C(Jξ ), Gη ∩ QJξ = ∅ whenever η <
ξ < ω1 (using the results of (b) and (c) to choose Jξ );
• sξ ¹Jξ = s0ξ ¹Jξ , sξ ∈ f −1 [QJξ ], f (sξ ) and f (s0ξ ) can be separated by
open sets in X (using (d) to choose sξ , s0ξ );
• Gξ , G0ξ are disjoint negligible open sets containing f (sξ ), f (s0ξ ) respectively (choosing Gξ , G0ξ );
• Uξ ∈ C(Jξ ), Vξ , Vξ0 ∈ C(I \ Jξ ), sξ ∈ Uξ ∩ Vξ ⊆ f −1 [Gξ ], s0ξ ∈ Uξ ∩ Vξ0 ⊆
f −1 [G0ξ ] (choosing Uξ , Vξ , Vξ0 , using the fact that sξ ¹Jξ = s0ξ ¹Jξ ).
On completing this construction, take for each ξ < ω1 a finite set Kξ ⊆
Jξ+1 such that Uξ , Vξ and Vξ0 all belong to C(Kξ ). By the ∆-system lemma
([14], II.1.5), there is an uncountable A ⊆ ω1 such that hKξ iξ∈A is a ∆system with root K say, that is, Kξ ∩ Kη = K for all distinct ξ, η ∈ A.
eξ ∩ U 0 where U
eξ ∈ C(K), U 0 ∈ C(Kξ \ K). Then
For ξ ∈ A, express Uξ as U
ξ
ξ
eξ , so there is an uncountable
there are only countably many possibilities for U
eξ is constant for ξ ∈ B; write U
e for the constant value.
B ⊆ A such that U
Products of completion regular measures
33
Let C ⊆ B be an uncountable set, not containing min A, such that Kξ \ K
does not meet Jη whenever ξ, η ∈ C and η < ξ. Let D ⊆ C be such that D
and C \ D are both uncountable.
Note that K ⊆ Kη ⊆ Jξ whenever η, ξ ∈ A and η < ξ, so that K ⊆ Jξ
for every ξ ∈ C. Consequently, Uξ0 , Vξ and Vξ0 all belong to C(Kξ \ K) for
every ξ ∈ C.
(f) Consider the open set
[
G=
Gξ ⊆ X.
ξ∈D
At this point the argument divides.
e ]) > 0. Then there is a Baire set Q ⊆ G
C a s e 1. Suppose µ∗ (G ∩ f [U
e ]) > 0. Let J ⊆ I be a countable set such that f −1 [Q]
such that µ∗ (Q ∩ f [U
depends on J. Let γ ∈ C \ D be so large that Kξ \ K does not meet
e ] is not empty; take
J for any ξ ∈ A with ξ ≥ γ. Then Q ∩ QJγ ∩ f [U
e ∩ f −1 [Q ∩ QJ ]. Because the Kξ \ K are disjoint from each other
s ∈ U
γ
and from J ∪ Jγ for ξ ∈ D, ξ > γ, we may modify s to form s0 such that
e
s0 ¹J ∪ Jγ = s¹J ∪ Jγ and s0 ∈ Uξ0 ∩ Vξ0 for every ξ ∈ D, ξ > γ; now s0 ∈ U
e ∩ U 0 ∩ V 0 ⊆ f −1 [G0 ] and f (s0 ) 6∈ Gξ whenever
(because K ⊆ Jγ ), so s0 ∈ U
ξ
ξ
ξ
ξ ∈ D, ξ > γ. On the other hand, if ξ ∈ D and ξ < γ, Gξ ∩ QJγ = ∅, while
s0 ∈ f −1 [QJγ ] (because f −1 [QJγ ] depends on Jγ ), so again f (s0 ) 6∈ Gξ .
Thus f (s0 ) 6∈ G. But s0 ¹J = s¹J so f (s) ∈ Q ⊆ G, which is impossible.
e ]) > 0.
This contradiction disposes of the possibility that µ∗ (G ∩ f [U
e ]) = 0. In this case there is a negligible
C a s e 2. Suppose that µ∗ (G ∩ f [U
e
Baire set Q ⊇ G ∩ f [U ]. Let J ⊆ I be a countable set such that f −1 [Q]
depends on J. Let ξ ∈ D be such that Kξ \ K does not meet J. Then
e ⊆ f −1 [G ∩ f [U
e ]] ⊆ f −1 [Q],
e ∩ Uξ0 ∩ Vξ ⊆ f −1 [Gξ ] ∩ U
U
e ⊆ f −1 [Q], because U 0 ∩ Vξ is a non-empty member of C(I \ J). But
so U
ξ
∗
e
this means that µ f [U ] = 0 and µ∗ f [Uξ ] = 0. On the other hand, we have
sξ ∈ Uξ ∩ f −1 [QJξ ], so Uξ 6∈ C0 (Jξ ) and µ∗ f [Uξ ] > 0.
Thus this route is also blocked and we must abandon the original hypothesis that there is a quasi-dyadic space with a completion regular topological
probability measure which is not τ -additive.
5. Theorem. Let (X, T, Σ, µ) and (Y, S, T, ν) be topological probability
spaces; suppose that (X, T, Σ, µ) is completion regular and quasi-dyadic, and
that (Y, S, T, ν) is τ -additive. Then every open set in X × Y is measurable
for the ordinary completed product measure λ on X × Y .
34
D. H. Fremlin and S. Grekas
P r o o f. Suppose, if possible, otherwise. By Theorem 4, we know that
µ is τ -additive. We may suppose that the domain T of ν is precisely the
algebra of Borel subsets of Y , since restricting the domain of ν will tend to
decrease the domain of λ.
(a) There must be a closed set C ⊆ X × Y which is not λ-measurable.
Because λR is τ -additive, λR ((X × Y ) \ C) ≤ λ∗ ((X × Y ) \ C); but also
λR C ≤ λ∗ C, so in fact we must have λR C = λ∗ C.
We are supposing that X is quasi-dyadic; take a family hSα iα∈I of
separable
metric spaces and a continuous surjection f : S → X, where
Q
S = α∈I Sα . Choose Bα and define C(J), for J ⊆ I, as in part (b) of the
proof of Theorem 4.
(b) If J ⊆ I is countable, there are G, U , V such that G ⊆ Y is open, U ∈
C(J), V ∈ C(I \J), V 6= ∅, C ∩(f [U ∩V ]×G) = ∅ and λ∗ (C ∩(f [U ]×G)) > 0.
To see this, write
GU = {G : G ⊆ Y is open,
∃V ∈ C(I \ J), V 6= ∅, C ∩ (f [U ∩ V ] × G)) = ∅},
[
GU =
GU ,
and choose FU ⊇ f [U ] such that FU ∈ Σ and µFU = µ∗ (f [U ]), for each
U ∈ C(J). We have GU ∈ T for every U , and C(J) is countable, so
[
bσ T;
C1 = (X × Y ) \
FU × GU ∈ Σ ⊗
U ∈C(J)
also C1 ⊆ C because C is closed and
{f [U ∩ V ] × G : G ⊆ Y is open, U ∈ C(J), V ∈ C(I \ J)}
is a network for the topology of X × Y . So λR (C \ C1 ) = λR C − λR C1 =
λ∗ (C \ C1 ) > 0, because C is not λ-measurable. Accordingly, there is a
U ∈ C(J) such that λR (C ∩ (FU × GU )) > 0. Next,
S because ν is τ -additive,
there isSa countable G ⊆ GU such that ν(GU \ G) = 0, and now λR (C ∩
(FU × G)) = λR (C ∩ (FU × GU )) > 0, so there is a G ∈ G such that
λR (C ∩ (FU × G)) > 0, that is,
R
ν(Cx ∩ G) µ(dx) > 0.
FU
But this means that µ{x : x ∈ FU , ν(Cx ∩ G) > 0} > 0, so that µ∗ {x : x ∈
f [U ], ν(Cx ∩ G) > 0} > 0, and λ∗ (C ∩ (f [U ] × G)) > 0. Finally, because
G ∈ GU , there is a V ∈ C(I \ J) such that C ∩ (f [U ∩ V ] × G) = ∅.
(c) We may therefore choose inductively families hJξ iξ<ω1 , hGξ iξ<ω1 ,
hUξ iξ<ω1 , hVξ iξ<ω1 in such a way that, for every ξ < ω1 ,
Products of completion regular measures
•
•
•
•
•
•
•
35
Jξ is a countable subset of I,
Gξ is an open subset of Y ,
Uξ ∈ C(Jξ ), Vξ ∈ C(I \ Jξ ), Vξ 6= ∅,
C ∩ (f [Uξ ∩ Vξ ] × Gξ ) = ∅,
∗
λ
S (C ∩ (f [Uξ ] × Gξ )) > 0,
η<ξ Jη ⊆ Jξ ,
Vξ ∈ C(Jξ+1 ).
For each ξ < ω1 , let Kξ be a finite subset of I such that Uξ , Vξ both belong
to C(Kξ ). By the ∆-system lemma, there is an uncountable set A ⊆ ω1 such
eξ ∩ U 0
that hKξ iξ∈A is a ∆-system with root K say. Express each Uξ as U
ξ
eξ ∈ C(K) and U 0 ∈ C(Kξ \ K); because C(K) is countable, there is
where U
ξ
e such that B = {ξ : ξ ∈ A, ξ 6= min A, U
eξ = U
e } is uncountable. Note
aU
e ]) ≥ µ∗ (f [Uξ ]) > 0 for any ξ ∈ B. Also, K ⊆ Kmin A ⊆ Jξ for
that µ∗ (f [U
each ξ ∈ B, so Vξ S
∈ C(Kξ \ Jξ ) ⊆ C(Kξ \ K) for each ξ ∈ B.
(e) Set Hζ = ξ∈B, ξ≥ζ Gξ for each ζ < ω1 . The family hνHζ iζ<ω1 is
non-increasing in R, so there must be a δ < ω1 such that νHζ = νHδ
whenever δ ≤ ζ < ω1 . Now consider
D = C ∩ (X × Hδ ).
Because D ⊇ C ∩ (f [Uξ ] × Gξ ) for any ξ ∈ B \ δ, λR D > 0. Set
F = {x : ν(Cx ∩ Hδ ) > 0} ⊆ X;
because Fubini’s theorem applies to λR , µF > 0. At this point we use the
hypothesis that µ is completion regular to see that there is a zero set F0 ⊆ F
e ]). Now W = U
e ∩ f −1 [F0 ] is non-empty because
with µF0 > µF − µ∗ (f [U
e ] is non-empty. Let J ⊆ I be a countable set such that K ⊆ J and
F0 ∩ f [U
W depends on J (see §§2–3 above). Let ζ < ω1 be such that ζ ≥ δ and
Kξ \ K = ∅ for every ξ ∈ B \ ζ; this exists because hKξ \ Kiξ∈B is disjoint.
Take any w ∈ W , and modify it to obtain w0 ∈ S such that
w0 ¹J = w¹J
and w0 ∈ Uξ0 ∩ Vξ
for every ξ ∈ B \ ζ;
this is possible because Uξ0 ∩Vξ ∈ C(Kξ \K) for each ξ ∈ B\ζ, and hKξ \Kiξ∈B
is disjoint. Set x = f (w0 ); then x ∈ F0 ⊆ F , so νDx > 0.
e ∩ U 0 ∩ Vξ ] = f [Uξ ∩ Vξ ], so (x, y) 6∈ C for
If ξ ∈ B \ ζ, then x ∈ f [U
ξ
y ∈ Gξ and Cx ∩ Gξ = ∅. Thus
Dx ⊆ Hδ \ Hζ ,
and µHζ < µHδ , contrary to the choice of δ.
This contradiction proves the theorem.
6. Corollary. Let h(Xα , Tα , Σα , µα )iα∈I be a family of quasi-dyadic
completion regular topological probability spaces, and suppose that all but
36
D. H. Fremlin and S. Grekas
countably many of the µα Q
are strictly positive. Then the simple (completed )
product measure on Z = α∈I Xα is also a completion regular topological
measure.
P r o o f. For finite I, this is a simple induction on #(I), using Theorems 4 and 5. For infinite I, use Theorem 2.9 of [9], which says that if all
the τ -additive topological product measures on finite subproducts are completion regular, and all but countably many of the factor measures µα are
strictly positive, the τ -additive topological product measure on Z will also
be completion regular, and therefore will coincide with the simple completed
product measure.
7. R e m a r k s. In Theorems 4 and 5 it is not of course necessary that
X itself should be quasi-dyadic. If (X, T, Σ, µ) is a completion regular τ additive topological probability space in which µ is inner regular for the
quasi-dyadic subsets of X, this will do just as well. This will be so, for
instance, if X is an open subset of a dyadic space and µ is a completion
regular Radon measure on X.
In the same way, it will be enough, in Corollary 6, if every µα is inner
regular for the quasi-dyadic subsets of Xα .
Theorem 4 can be generalized as follows: Let X be a quasi-dyadic space
and I a proper σ-ideal of subsets of X such that for every open G ⊆ X there
are Baire subsets E, F of X such that E ⊆ G ⊆ F and F \SE ∈ I. Then X
cannot be covered by the open sets belonging to I; indeed, {G : G ∈ I, G
is open} ∈ I.
We do not know of a ZFC example of a completion regular measure on
a completely regular Hausdorff space which is not τ -additive. (We note that
there is a probability measure on the Baire σ-algebra of Rc which is not
τ -additive, in the strong sense that Rc is covered by the negligible cozero
sets; see [17], or [6], 15.4; but this measure is not completion regular.)
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MATHEMATICS DEPARTMENT
SECTION OF MATHEMATICAL ANALYSIS
UNIVERSITY OF ESSEX
PANEPISTEMIOPOLIS
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Received 15 February 1994