Independen e-pre alibers of measure algebras
D.H.Fremlin & G.Plebanek
University of Essex, Col hester, England;Wro law University
Notation Wefollow Fremlin 06?. In parti ular,
is the usualmeasure onf0;1g
,N
its null ideal and (B
;
) its measure algebra.
1De nition(a)LetAbeaBooleanalgebraand, ardinals. Wesaythat(;)isan
independen e-pre aliberpairof Aif whenever ha
i
<
is afamilyof distin telements
of A thenthere isa 2[℄
su h thatfa
: 2 g isBoolean-independent. If (;) isan
independen e-pre aliberpairof(A;) wewillsaythatisanindependen e-pre aliber
of (A;).
(b)Let(A;) beameasurealgebraand, ardinals. Wesaythat(;)isameasure-
independen e-pre aliber pairof (A;) if whenever ha
i
<
is a familyin A su h that
inf
<<
(a
4a
) > 0 then there is a 2 [℄
su h that fa
: 2 g is Boolean-
independent. If (;) isa measure-independen e-pre aliber pair of (A;) we willsaythat
is a measure-independen e-pre aliberof (A;).
Remark Of ourseanyindependen e-pre aliber (pair)ofa measurealgebraisameasure-
independen e-pre aliber (pair). If (;) is a measure-independen e-pre aliber pair of a
totally nite measure algebra (A;), and
!
< for every < , then (;) is an
independen e-pre aliber pair of A. PPP If ha
i
<
is any family of distin t elements of A,
then fa
: <g must have density for the measure metri , and therefore must have a
metri allyisolated subset of size , whi h will have a Boolean-independent subset of size
. QQQ (See Proposition 11 below.)
2 Proposition! is a measure-independen e-pre aliber of every probability algebra.
proofLet(A;) beaprobabilityalgebraandha
n i
n2N
asequen einAsu hthat(a
m 4a
n )
Æ > 0 whenever m 6= n. Let u 2 L 2
be a luster point of h a
n i
n2N for T
s (L
2
;L 2
), and
set =[[0<u<1℄℄. Then 6=0. PPP??? Otherwise, u= a for some a. Now there must be
in nitely manyn su h that
(a4a
n )
R
(u a
n
)( a (1na))<
Æ
2 ,
whi h is impossible. XXXQQQ
We an therefore hoose indu tively a stri tly in reasing sequen e hn
k i
k2N
su h that
0 < (a
n
k
\b) < b whenever 0 6= b and b belongs to the algebra generated by
f g[fa
n
i
: i<kg; in whi h ase fa
n
k
:k2Ng will be Boolean-independent.
3 Lemma Let be an un ountable ardinal and (A;) a probability algebra. Let D
be a losed subalgebra of A su h thatA is relatively atomless over D, anda2AnD; set
Æ =(a;D). Then there are a sequen e he
n i
n2N
in Aand a 2A su h that
TypesetbyA
M S-T
E X
e
n
= 1
2
for every n
thee
n
aresto hasti ally independent of ea h other and D
a, belong to the losed subalgebra Bgenerated by D[fe
n
:n2Ng
taking : B ! B to be the measure-preserving automorphism su h that
d =d for every d2D, e
0
=1n e
0 , e
n
=e
n
for n>0, then =
\e
0
1na, ne
0
a
Æ.
proofDe neu2L 1
(D)bysayingthat R
d
u=(a \d)foreveryd 2D. Setd
0
=[[u>
1
2
℄℄.
Lete 0
0
abesu hthat(e 0
0
\d) = 1
2
(d\d
0
)foreveryd2D(331B);lete 0 0
0
1nabesu h
that(e 0
0
\d) = 1
2
(dnd
0
)for everyd 2D;sete
0
=(d
0 ne
0
0 )[e
00
0
,sothat(e
0
\d) = 1
2
d
for every d 2D. Now hoose e
n
, for n 1, su h that (e
n
\d)= 1
2
d for every d in the
algebrageneratedbyD[fe
i
:i <ng, andabelongstothe losed subalgebraB generated
byD[fe
n
: n2Ng. [To seethat thisis possible, startfromanysequen e hb
n i
n2N
in Aof
elementsof measure 1
2
sto hasti allyindependentof ea hotherandofD,andletBbethe
losed subalgebraof A generatedbyD[fa;e
0 g[fb
n
:n2Ng; useFremlin 02, 333Cto
seethat(B; B) anbeidenti edwiththeprobabilityalgebrafreeprodu tof(D
1
; D
1 )
and(B
!
;
!
),where D
1
isthe subalgebraof AgeneratedbyD[fe
0
g.℄ Be ause (B; B)
isequallyisomorphi totheprobabilityalgebrafree produ tof(D; D)and(B
!
;
! ),we
have a unique measure-preserving automorphism : B !B su h that d =d for every
d2D, e
0
=1n e
0
and e
n
=e
n
for n>0.
Set 0
=d
0 na,
0 0
=and
0
. Thenat leastone of 0
, 00
hasmeasureat least 1
2
Æ; allthis
; set =
[
, so that 2B and = .
If
= 0
then
d
0
\e
0 ,
d
0 ne
0
a, \e
0
= 0
is disjoint from a, ne
0
=
0
aand =2 0
Æ. If
= 00
then
\e
0
=0,
\e
0
=
e
0 nd
0
=e 00
0
1na,
ne
0
= 00
a and =2 00
Æ.. So we have what we need.
4 Corollary If is an in nite ardinal, (A;) a Maharam homogeneous probability
algebra, and ha
i
<
a family in A su h that inf
<<
(a
4a
) =Æ >0, then there are
a family he
n i
<;n2N
, a fun tion :! and a family h
i
<
in A su h that
e
n
= 1
2
for all , n,
he
n i
<;n2N
is sto hasti allyindependent,
a
() ,
belong to the losed subalgebra B of A generated by fe
n
: <,
n2Ng for every ,
if
: B ! B is the measure-preserving automorphism de ned by setting
(e
0
)=1ne
0
and
(e
n )=e
n
for all (;n)6=(;0), then
=
,
1
2
Æ for every ,
\e
0
1na
() , ne
0
a
()
for every .
proof If D A is a losed subalgebra with Maharam type less than , then there must
be a < su h that (a
;D) 1
2
Æ. We an therefore hoose D
, e
n ,
,
and ()
indu tively,asfollows. D
0
=f0;1g. GiventhatD
isthe losedsubalgebraofAgenerated
by fe
n
: < , n 2 Ng, let () < be su h that (a
()
;D
)
1
Æ; using Lemma 3,
hoosee
n
,of measure 1
2
,sto hasti allyindependent of ea hotherandof D
,and
,su h
that
a
() and
belong to the losed subalgebra D
+1
generatedbyD
[fe
n :
n2Ng,
1
2 Æ,
\e
0
1na
() ,
ne
0
a
() ,
taking
:D
+1
!D
+1
to be the measure-preserving automorphism su h
that d = d for every d 2 D
,
e
0
= 1n
e
0 ,
e
n
= e
n
for n> 0, then
=
.
LetB=D
be the losedsubalgebra of Ageneratedby thesto hasti allyindependent
family he
n i
<;n2N
. This works.
5 PropositionIf , are in nite ardinals, !
2
and (;) is a measure-pre aliber
pairof atotally nitemeasurealgebra(A;) thenitisameasure-independen e-pre aliber
pair of (A;).
proof (a) To begin with, suppose that (A;) is a Maharam homogeneous probability
algebra, that (;) is a measure-pre aliber pair of (A;) and that ha
i
<
is a family in
A su h that (a
4a
) Æ > 0 whenever < . Take he
n i
<;n2N
, B, : ! and
h
i
<
asinCorollary4. Forea h <letJ
betheminimalsubsetofN su hthat
belongs to the losed subalgebra generated by fe
j
:j 2Jg, and set L
=f:(;n)2Jg.
AsinFremlin06?, 534Je,(;0)2= J
. ByHajnal'sFreeSet Theoremthereis a
0 2[℄
su h that 2= L
for all distin t , 2
0
. If I, J C are disjoint nite sets and
=inf
2I[J
, then is sto hasti ally independent of the e
0
, for 2I[J, and
(inf
2J a
() n sup
2I a
()
)(inf
2I (e
0 na
() )\ inf
2J a
() ne
0 ))
(inf
2I (
\e
0
)\ inf
2J
ne
0 ))
=( \ inf
2I e
0 n sup
2J e
0 )=2
#(I[J)
:
Now,be ause(;) isameasure-pre aliberpairof(A;), thereisa 2[
0
℄
su h that
f
:2 g is entered, in whi h ase ha
() i
2
is Boolean-independent.
(b) For the general ase, we have only to note that, for a totally nite measure al-
gebra (A;), (;) is an (independen e-)measure-pre aliber pair of (A;) i it is an
(independen e-)measure-pre aliber pair of every Maharam homogeneous prin ipal ideal
of (A;) ( f. Fremlin 06?, 524Ha).
6 Proposition If ! and (;) is a measure-independen e-pre aliber pair of
every probabilityalgebra thenitis ameasure-pre aliber pair of every probabilityalgebra.
proofLet(A;) beaprobabilityalgebraandha
i
<
afamilyinAsu hthatinf
<
a
=
Æ >0. Let(C;) betheprobabilityalgebrafreeprodu tof(A;) and(B
;
),andhe
i
<
a sto hasti ally independent family in B
of elements of measure 1
2
. Set
=a
e
for
ea h ; then (
4
) = (a
[a
) Æ whenever < . There is therefore a 2 [℄
su h thath i is Boolean-independent, in whi h ase fa : 2 g must be entered.
7 Lemma Let n 1 be an integer, a regular un ountable ardinal, X (f0;1g n
)
a losed set, and f
: f0;1g n
! f0;1g a fun tion. Set f(x) = hf
(x())i
<!
1
for x 2 X.
If f[X℄ = f0;1g
, there are i < n and C 2 [℄
su h that
iC
[X℄ = f0;1g C
, where
iC
(x)()=x()(i) for x2X, 2C, i<n.
proof Indu e on n. If n = 1 then every f
has to be surje tive, therefore bije tive, and
f is a bije tion, so we an take C = , i = 0. For the indu tive step to n+1, identify
(f0;1g n+1
)
with (f0;1g n
)
f0;1g
, so that we have X (f0;1g n
)
f0;1g
and
f
: f0;1g n
f0;1g ! f0;1g su h that f[X℄ = f0;1g
, where f(x;y)()= f
(x();y())
for < , x 2 (f0;1g n
)
, y 2 f0;1g
. Let X 0
X be a minimal losed set su h that
f[X 0
℄=f0;1g
.
SetY =fy:(x;y)2X 0
g, so that Y is a losed subset of f0;1g
. Set
A=f : <; 8y 2Y 8I 2[℄
<!
9y 0
; y 00
2Y;
y 0
I =y 00
I =yI; y 0
()=0; y 00
()=1g:
If I 2 [A℄
<!
and u 2 f0;1g I
there is a y 2 Y su h that yI = u (indu e on maxI), so
fyA:y 2Yg=f0;1g A
and if #(A)= we antake C =A,i =n and stop. Otherwise,
for 2 nA take y
2 Y, I
2 [℄
<!
and v() 2 f0;1g su h that y() = v() whenever
y 2 Y and yI
= y
I
. Let J 2 [℄
<!
, u 2 f0;1g J
be su h that B = f : 2 nA,
I
=J, y
I
=ug has ardinal . For 2B, s2f0;1g n
setg
(s)=f
(s;v()).
Now onsider f(x;y): (x;y)2 X 0
, yJ = ug. This is a non-empty open subset of X 0
and fX 0
is irredu ible, so there is an open ylinder set V f0;1g
su h that yJ = u
whenever (x;y) 2 X 0
and f(x;y) 2 V. Let K 2 [℄
<!
be su h that V is determined
by oordinates in K. Set D = BnK, Z = fxD : (x;y) 2 X 0
g, g(z) = hg
(z())i
2D
for z 2 Z. If w 2 f0;1g D
, there is a w 0
2 V su h that w 0
D = w; there is an element
(x;y)2X 0
su h that f(x;y)=w 0
; now yJ =u so yB=vB and
g
(x())=f(x();y())=w 0
()=w()
for every 2D, that is, g(xD)=w. Thus g[Z℄=f0;1g D
, while Z (f0;1g n
)
.
By the indu tive hypothesis, there are C 2 [D℄
and an i < n su h that
iC [Z℄ =
f0;1g C
; now, re-interpreting the formula
iC
appropriately, we have
iC
[X℄ = f0;1g C
and the indu tion ontinues.
8 PropositionLetM beaset, aregularun ountable ardinalandX a losedsubset
of f0;1g M
. If there is a ontinuous surje tion h :X ! f0;1g
, then there is a C 2 [M℄
su h thatx 7!xC :X !f0;1g C
is surje tive.
proof Shrinking X if ne essary, we may suppose that h is irredu ible. For ea h < ,
fx : x 2 X, h(x)() = 1g is an open-and- losed set in X, so there is a nite set I
su h that h(x)() = h(x 0
)() whenever x, x 0
2 X and xI
= x 0
I
. Let A 2 [℄
be
su h that hI
i
2A
is a onstant-size -system with root I say. Let u be any member of
fxI :x2Xg;thenthereisa ylinder setV f0;1g
su h thatxI =uwheneverx2X
and h(x)2 V; shrinking A slightlyif ne essary, we an suppose that V is determinedby
oordinatesin nA. For 2A,set J
=I
nI and de nef
:f0;1g J
!f0;1gby saying
that f (v) =h(x)() whenever x 2 X, xI = u and xJ =v. Now observe that for any
w 2 f0;1g A
we have a w 0
2 V extending w, so that there is an x 2 X with h(x) = w 0
,
xI =u and f
(xJ
)=w() for every 2A.
We an therefore apply Lemma 7 to fx
S
2A J
: x 2 Xg, identifying S
2A J
with
nAwherenisthe ommonvalueof #(J
)for 2A,toseethatthereisanun ountable
C S
2A J
su h thatfxC :x2Xg=f0;1g
.
9CorollaryIf!
1
hasHaydon'spropertythen!
1
isameasure-independen e-pre aliber
of every probabilityalgebra.
proofLet (A;) be aprobabilityalgebraand ha
i
<!
1
a familyin A su h thatinf
<<!
1
(a
;a
)>0. Thenwehave a Radonprobabilitymeasure on f0;1g
!
1
de ned by saying
that
fx:xI =ug=(inf
2I;u()=1 a
n sup
2I;u()=0 a
)
whenever I 2[!
1
℄
<!
and u 2 f0;1g I
. Let Z be the support of . Sin e has Maharam
type !
1
, there is a ontinuous surje tion from Z onto [0;1℄
!
1
; so we an nd a losed
subset Z 0
of Z and a ontinuous surje tion h : Z 0
! f0;1g
!
1
. By Proposition 8, there
is a C 2 [!
1
℄
!
1
su h that fzC : z 2 Z 0
g = f0;1g C
. But this means that ha
i
2C is
Boolean-independent.
10 Proposition Suppose that there is a family hW
i
<!1 in N
!1
su h that every
losedsubset off0;1g
!
1
n S
<!
1 W
is s attered. Then!
1
isnotameasure-independen e-
pre aliber of every probability algebra.
proof As in Fremlin 06?, 534N (following Plebanek 97), there is a zero-dimensional
ompa tHausdor spa eX su hthat!
1
2Mah
R
(X)butthereisno ontinuoussurje tion
from X onto f0;1g
!
1
. Let be a Maharam homogeneous Radon probabilitymeasure on
X withMaharam type !
1
and (A;) itsmeasure algebra. Be ause X is zero-dimensional,
thereisafamilyhK
i
<!1
ofopen-and- losedsetsinX su hthat(K
4K
)
1
3
whenever
< ; now hK
i
<!
1
has no Boolean-independent subfamily of size !
1
, so hK
i
<!
1 has
no Boolean-independent subfamily of size !
1 .
11 Proposition Suppose that and are ardinals su h that max(2
;
!
) < f
2
. Then is an independen e-pre aliber of every totally nite measure algebra.
proof D
zamonja & Plebanek p04, Theorem 6.3.
Remark Note that if we have only
!
< f2
then is a measure-pre aliber of every measure algebra (D
zamonja & Plebanek p04,
4.7; Fremlin p06?, 524V), therefore a measure-independen e-pre aliber of every proba-
bility algebra (D
zamonja & Plebanek p04, 6.6). If
!
< for every <, then any
family of size in any metri spa e has a dis rete subfamily of size , so will be an
independen e-pre aliber of every probability algebra. But the result here also overs the
ase =2 2
=! .
12ProblemFromFremlin 06?,534L,andPropositions2,5and6above,weseethat
for !
2
the followingare equiveridi al:
is a measure-independen e-pre aliber of every probability algebra;
is a measure-pre aliber of every probability algebra;
has Haydon's property.
By Proposition 10 and the remarks in the notes to x534 of Fremlin 06?, it is possible
that
!
1
is a measure-pre aliber of every probabilityalgebra,
!
1
is not a measure-independen e-pre aliber of every probabilityalgebra.
ByCorollary9,weseethatif!
1
hasHaydon'spropertythenitisameasure-independen e-
pre aliber of every probabilityalgebra. Butwedo notknow whetherthe onverse istrue.
Referen es
Dzamonja M. & Plebanek G. [p04℄ `Pre alibre pairs of measure algebras', 2.4.04
Fremlin D.H. [02℄ Measure Theory, Vol. 3: Measure Algebras. Torres Fremlin, 2002.
FremlinD.H.[03℄MeasureTheory,Vol. 4: Topologi alMeasureTheory. TorresFremlin,
2003.
Fremlin D.H. [06?℄ Measure Theory, Vol. 5: Set-theoreti measure theory, in prepara-
tion. Partialdraftsavailableviahttp://www.essex.a .u k/m ath s/st aff /fr eml in/m t.h tm;
paragraph numbers indi ated here are not to be trusted.
Plebanek G. [97℄ `Non-separable Radon measures and small ompa t spa es', Funda-
menta Math. 153(1997) 25-40.
