Representations of positive projections 1 Introduction - Mathematics ...

Representations of positive projections
W. A. J. Luxemburg
B. de Pagter
Department of Mathematics, California Institute of
Technology, Pasadena, CA, USA
Department of Mathematics, Delft University of
Technology, Delft, The Netherlands
to Dorothy Maharam in admiration
Abstract
In this paper a number of Maharam-type slice integral representations,
with respect to scalar measures, is obtained for positive projections
in Dedekind complete vector lattices. In the approach to these
results the theory of f-algebras plays a crucial role.
1 Introduction
About half a century ago Dorothy Maharam published an important series
of papers dealing with the characterization of measure algebras and their
decomposition properties and applied them to prove a number of deep representation
theorems for positive linear transformations between spaces of
measurable functions as kernel operators by means of what is now called
Maharam's slice integral method. For a discussion of these decomposition
theorems for measure algebras and their subalgebras we refer the reader also
to D. H. Fremlin's contribution in [12].
In [5], the rst author presented a brief summary and discussion of some
of Maharam's groundbreaking work. By using the language of the theory of
Riesz spaces (vector lattices) it was suggested that an algebraization of her
work would be a worthwhile undertaking. This led to the paper [7] dealing
with Maharam extensions of positive linear operators in the setting of the
AMS Classi cation: 47B65, 46A40, 06F25
1

theory of f-modules (i.e., vector lattices which are modules over f-algebras).
For an alternative exposition of results on Maharam extensions we refer the
reader also to the book [4].
In the present paper we present a general theory of Maharam-type representation
theorems mentioned above for positive projections on unital falgebras
as well as on Dedekind complete vector lattices which are what we
call locally order separable. The main results of the paper are collected in
the theorems in Section 10.
Section 2 is devoted to some preliminary results and constructions in the
theory of Riesz spaces which will be used frequently throughout the paper.
In addition, we introduce the notion of local order separability, referred to
above, a somewhat weaker concept than that of order separability. A Riesz
space is called locally order separable if there exists a maximal disjoint system
consisting of order separable elements, i.e., the principal ideals generated by
these elements are order separable. In essence, such (Archimedean) Riesz
spaces are those who have an order dense and order separable ideal.
A representation theory for Dedekind ( -) complete Riesz spaces as special
quotient spaces is presented in Section 3. A measurable space (X ) is
called a representation space of a Dedekind -complete Riesz space L with
a weak order unit 0 < w 2 L if there exists an ideal bL in M (X ), the
Riesz space of all -measurable real functions on X, such that 1X 2 bL and
there exists a -order continuous Riesz homomorphism from bL onto L
with (1X) =w. Consequently, L is Riesz isomorphic to the quotient space
bL Ker ( ). The existence of such a representation depends on the classical
Sikorski-Loomis theorem for -complete Boolean algebras.
Using this concept we introduce the notion of a Maharam-type representation
space for a positive projection P in a Dedekind -complete Riesz
space L with weak order unit w (denoted from here by (L w)) onto a Riesz
subspace K L with w 2 K. Such a representation space is a product
space ( F) = (X Y ) of a measurable space (X ) and a -
nite measure space (Y ), such that ( F) is a representation space for
(L w) with representation homomorphism . Furthermore, there exists an
F-measurable function m 0 on satisfying R m (x y) d (y) = 1 and
R Y
m (x y) jf (x y)j d (y) < 1 for all f 2 bL, x 2 X and moreover the
Y
function Rf, de ned by
(Rf)(x y) =
for all (x y) 2 satis es Rf 2 bL and
Z
Y
m (x z) f (x z) d (z) (1)
P ( f) = (Pf)
2

for all f 2 bL. We stress the fact that in such a Maharam-type integral
representation (1) the measure is a scalar valued measure and not a vector
valued measure (as is the case in e.g. the integral representations discussed
in Section 6.3 of [4]). Actually one of the main e orts in the present paper
will be the construction of this measure (see Sections 7 and 8).
As will be shown, if (X Y ) is a representation space for the
restriction of the projection P to the principal ideal Lw generated by the
weak order unit w, then it is already a representation space for P . Since
Dedekind complete Riesz spaces with a strong order unit have a unital falgebra
structure, we switch our attention to the study of representations of
positive projections in unital f-algebras. For this reason Section 4 is devoted
to a discussion of the theory of f-algebras that plays a role in our representation
theory. First weshow that the averaging property of positive projections,
as stated in Lemma 4.1, in combination with the Radon-Nikodym theorem
for Maharam operators, implies the following interesting result that plays a
crucial role later in Section 8. Suppose that A is a Dedekind complete unital
f-algebra in which the unit element 1 is a strong order unit as well and that
B is an f-subalgebra with 1 2 B, which is the range of a strictly positive
order continuous projection P in A. If C is a complete f-subalgebra of A
such that B C, then there exists a unique positive projection Q in A onto
C such that P = P Q moreover, Q is order continuous and strictly positive.
By means of an example it is shown that the existence of a strong order unit
in A is essential.
In the theory of measure algebras Maharam ([10], Section 11) introduced
the following important concept. Suppose that A is a Dedekind complete
unital f-algebra and that B is a complete unital f-subalgebra of A. The
complete Boolean algebras of components of the unit element in A and B
are denoted by CA and CB respectively. Then p 2CA is called B-full (or, of
order zero over CB) if fe 2CA : e pg = fpq : q 2CBg. Furthermore, A is
called nowhere full with respect to B if the zero element is the only B-full
element of CA. If for every 0 < p 2 CA there exists a B-full component
0 < q 2 CA such that q p, then A is called everywhere full with respect
to B. From these de nitions it is easily seen that A splits in a part that is
everywhere full and a part that is nowhere full over B. Section 4 nishes with
atechnical result showing how families of representations of restrictions of a
positive projection can be glued together to obtain a global representation of
the projection, a technique that will yield the general representation theorems
in Section 10.
In [11] Maharam proved the following important and non-trivial version
of a Radon-Nikodym theorem (see Theorem 5.8 and its proof). Assume that
(X ) is a nite measure space with measure algebra ( )andassume
3

that is a -subalgebra of and let be the corresponding complete
Boolean subalgebra of . If is nowhere full with respect to and if
is a measure on such that on , then there exists F 0 2 such that
(D) = (D \ F 0) for all D 2 . In Section 5 we present a vector-valued
version of Maharam's theorem of the following form. Assume that B is an
f-subalgebra of the Dedekind complete unital f-algebra A with 1 2 B and
that P is a positive order continuous projection of A onto B. If A is nowhere
full with respect to B and if 0 < e 2 CA and g 2 B satisfy 0 g P (e),
then there exists p 2CA such that p e and P (p) =g. In Theorem 5.7 an
application of this result is presented, which is of independent interest.
If P is a strictly positive order continuous projection in the unital falgebra
A onto the unital f-subalgebra B and if A is nowhere full with
respect to B, then it follows from the results in Section 5 that there exist
for any 2 [0 1] components p 2 CA with the property that P (p) = 1.
The collection of all such components is denoted by S. In Section 6 the
structure of this set S and its complete Boolean subalgebras is discussed.
The construction of such special Boolean subalgebras is contained in Sections
7 and 8. We would like to point out that these special subalgebras are
the essential ingredient in the construction of the measure space (X )
occurring in the representation (1) of the projection P . For more details we
have to refer the reader to these sections. In particular the results in Section
8 use Maharam's decomposition of Boolean algebras in homogeneous parts.
Finally, in Section 10, the results of the previous sections are combined to
obtain the desired Maharam-type representations of order continuous strictly
positive projections in Dedekind complete Riesz spaces and f-algebras.
2 Preliminaries
In this section we will discuss some results concerning Riesz spaces (vector
lattices) which will be used in the sequel. For unexplained terminology and
the general theory of Riesz spaces and positive linear operators we refer the
reader to the books [1], [9], [13] and [16].
Suppose that L is an Archimedean Riesz space and that K is a Riesz
subspace of L. We will say thatK is ( -) regularly embedded in L if for any
net (sequence) ff g in K such that f # 0 in K, it follows that f # 0 in L.
Furthermore, K will be called a ( -) complete Riesz subspace of L if for any
net (sequence) ff g in K and f 2 L such that f " f in L it follows that
f 2 K. The latter terminology is inspired by the corresponding notions in
the theory of Boolean algebras (see Section 23 in [15]). If L is Dedekind ( -)
complete and K is a ( -) complete Riesz subspace of L, then it is easy to see
4

that K itself is Dedekind ( -) complete and K is ( -) regularly embedded in
L.
If L is an Archimedean Riesz space and 0 w 2 L then we will denote
by CL (w) =C (w) the set of all components of w, i.e.,
C (w) =fp 2 L : p ^ (w ; p) =0g .
With respect to the ordering induced by L, the set C (w) is a Boolean algebra.
Now we assume that L is Dedekind -complete and that w is a weak order
unit in L. Then C (w) is a Boolean -algebra, which is isomorphic to the
Boolean algebra P (L) of all band projections in L. The Boolean isomorphism
from P (L) onto C (w) isgiven by the mapping P 7! Pw. Consequently, the
Boolean algebra C (w) is complete if and only if L is Dedekind complete (see
[9], Theorem 30.6).
Next we will discuss a construction that will be used frequently in this
paper. Let L be a Dedekind -complete Riesz space with a xed weak order
unit 0 w 2 L. For f 2 L we will denote by p f : 2 R the left-continuous
spectral system of f with respect to w, i.e., p f is the component of w in the
band ( w ; f) + dd . The right-continuous spectral system of f with respect
to w is denoted by q f : 2 R , i.e., q f is the component of w in the band
(f ; w) + d . For the properties of these spectral systems we refer to [9],
Chapter 6.
De nition 2.1 Given a Boolean -subalgebra E of C (w) we de ne
L (E) = f 2 L : p f 2E for all 2 R .
We will show that L (E) is a -complete Riesz subspace of L. For this
purpose we need a result that goes back to H. Nakano ([14], Theorem 14.4)
and which is of interest in its own right. For the convenience of the reader
we will indicate the proof of this result. We need the following lemma.
Lemma 2.2 For fg 2 L and 2 R the following statements hold
(i). p f ^ p g
f +g
p +
f +g
(ii). p + ^ ; w ; pf p g .
Proof.
5

(i). For all a b 2 L we havea + b 2(a ^ b) and so (a + b) +
which implies that
a + dd + dd
\ b
(a + b) + dd .
Applying this to a = w ; f and b = w ; g we nd that
( w ; f) + dd \ ( w ; g) + dd
from which (i) follows.
(ii). It follows from
that
(( + ) w ; (f + g)) +
(( + ) w ; (f + g)) + dd
and this implies that
(( + ) w ; (f + g)) + dd \ ( w ; f) + d
2(a + ^ b + ),
(( + ) w ; (f + g)) + dd ,
( w ; f) + +( w ; g) +
( w ; f) + dd + ( w ; g) + dd
f +g
Since p + ^ ; w ; pf is the component ofw in the band
(( + ) w ; (f + g)) + dd \ ( w ; f) + d ,
f +g
it is now clear that p + ^ ; w ; pf p g .
( w ; g) + dd .
Proposition 2.3 (H. Nakano, [14]) If fg 2 L and 2 R, then
p f +g =sup p f ^ p g : 2 Q + .
Proof. Since the set Q of rationals is countable, the supremum on the
right hand side, which we denote by p, exists in L. From (ii) of Lemma 2.2
it is clear that p p f +g . To prove the reverse inequality, take 2 Q such
that and take k 2 N. De ne n =2 ;kn for all n 2 Z. Since pf # 0as
n
n !;1and pf " w as n !1,we have
n
w = _ ; _ ; f
f
p ; pf = w ; p n
n
n2Z
n+1
6
n2Z
^ pf
n+1

in C (w) and so it follows that
f +g
p ;2 ;k f +g
= p ;2 ;k ^ w = _
n2Z
f +g
p ;2 ;k ^ ; w ; p f
n
^ pf
n+1 .
Since ; ; 2 ;k ; n = ; n+1, Lemma 2.2 (ii) implies that
for all n 2 Z. Hence,
f +g
p ;2 ;k ^ ; w ; p f
n
f +g
p ;2 ;k
_
n2Z
p g
; n+1
p f
^ pg
n+1 ; n+1
as n+1 ; n+1 2 Q and n+1 +( ; n+1) = for all n 2 Z. Since
f +g
p ;2 ;k " pf +g as k ! 1, it follows that pf +g p. Finally, take a sequence
f ng 1
n=1 in Q such that f +g
f +g
n " . Then p p for all n and p n
n " pf +g , so
f +g
p p. We may conclude that p f +g = p, by which the proof is complete.
Proposition 2.4 Let L be a Dedekind -complete Riesz space with weak
order unit 0 w 2 L and suppose that E is a Boolean -subalgebra ofC (w).
Then L (E), as introduced in De nition 2.1, is a -complete Riesz subspace
of L and CL(E) (w) = E. In particular, L (E) is Dedekind -complete and
-regularly embedded in L.
Proof. First we observe that L (E) isalso given by
L (E) = f 2 L : q f 2E for all 2 R .
Indeed, this follows immediately from the fact that if n < , n " , then
q f
n " pf and if n > , n # , then p f
n # qf . Since p ;f = w ; q f
; for all
2 R, it is now clear that f 2 L (E) if and only if ;f 2 L (E). Moreover,
if 0 < 2 R then p f = p f
= and so f 2 L (E) implies that f 2 L (E).
This shows that f 2 L (E) for all f 2 L (E) and all 2 R. Furthermore, it
follows from Proposition 2.3 that f + g 2 L (E) for all fg 2 L (E), so L (E)
is a linear subspace of L. Since p f_g = p f ^ p g for all 2 R, we see that
fg 2 L (E) implies that f _ g, hence L (E) is a Riesz subspace of L. Now
assume that fn 2 L (E) for n =1 2::: and that f 2 L such that fn # f in
L. Then p fn " p f for all 2 R, which implies that f 2 L (E). Therefore,
L (E) is a -complete Riesz subspace of L. Finally, it follows immediately
from the de nition of L (E) that a component p 2 C (w) belongs to L (E) if
and only if p 2E. The proof is complete.
We will need some further properties of the space L (E). In particular,
we will give an alternative description of the space L (E). For this purpose
we give the following de nition.
7
p,

De nition 2.5
S(E) =
( nX
j=1
jej : j 2 Rej 2Ej =1:::nn2 N
)
: (2)
It is clear that S(E) is a linear subspace of L. Furthermore, every s 2 S(E)
can be written as
s =
nX
j=1
jej, with ei ^ ej =0(i 6= j),
nX
j=1
ej = w: (3)
From this observation it follows easily that S(E) is actually a Riesz subspace
of L, in which w is a strong order unit. Moreover, since E L (E), it follows
that S(E) L (E). For 0 u 2 L we will denote by Pu the band projection
in L onto fug dd .
Proposition 2.6 If L is a Dedekind -complete Riesz space with weak order
unit 0 w 2 L and E is a Boolean -subalgebra of C (w), then the following
statements hold.
(i). If 0 f 2 L, then f 2 L (E) if and only if there exists a sequence
fsng 1
n=1 in S(E) such that 0 sn " f.
(ii). All the band projections in L (E) are all of the form (Pe) jL(E) and all
the bands in L (E) are given by feg dd \ L (E) with e 2E.
(iii). L (E) is a complete Riesz subspace of L if and only if E is a complete
Boolean subalgebra of C (w).
In particular, if L is Dedekind complete and if E is a complete Boolean subalgebra
ofC (w), then L (E) is a complete Riesz subspace ofL and hence, L (E)
is Dedekind complete and regularly embedded in L.
Proof.
(i). This follows immediately from the Freudenthal spectral theorem ( [9],
Theorem 40.3) and from the fact that L (E) isa -complete Riesz subspace
of L.
(ii). If e 2 E and 0 f 2 L (E) then f ^ (ne) 2 L (E) for all n = 1 2:::
and f ^ (ne) " Pef in L. Since L (E) is a -complete Riesz subspace
of L, it follows that Pef 2 L (E). This shows that Pe [L(E)] L(E)
and now it is clear that (Pe) jL(E) is a band projection in L (E). Since
8

Pe (e) =e and Pe (f) =0whenever f 2 L (E) such that f?e, it follows
that (Pe) jL(E) is the band projection in L(E) onto the band generated
by e in L(E). Note that this also shows that the band generated by e
in L (E) is equal to feg dd \ L (E).
Now suppose that Q is a band projection in L (E). Then e = Qw 2
L (E) is a component of w in L (E) and so e 2 E. From the above we
know that (Pe) jL(E) is a band projection in L (E) and that (Pe) jL(E) (w) =
e = Qw. Since w is a weak order unit in L (E), it follows that Q =
(Pe) jL(E) . Since all bands in L (E) are principal projection bands, the
nal statement of (ii) is now obvious.
(iii). First assume that E is a complete Boolean subalgebra of C (w). Take
f 2 L (E) and f 2 L such that f # f in L. Then p f " p f in C (w)
for all 2 R. Since p f 2 E and E is a complete Boolean subalgebra
of C (w), it follows that p f 2Efor all 2 R. Hence f 2 L (E). This
shows that L (E) is a complete Riesz subspace of L.
Now assume that L (E) is a complete Riesz subspace of L. Let p 2E
and p 2 C (w) be such that p " p in C (w). Since L is Dedekind -
complete, this implies that p " p in L and so p 2 L (E), which implies
that p 2E. Hence, E is a complete Boolean subalgebra of C (w).
The nal statement of the proposition is an immediate consequence of (iii).
Now we will discuss a number of results related to order separability of
a Riesz space. Recall that the Archimedean Riesz space L is called order
separable if every non-empty subset D L for which sup D = g 0 exists
has an at most countable subset ffng D such that sup n fn = g 0 (see [9],
De nition 23.1). As is well known, L is order separable if and only if every
order bounded disjoint system in L is at most countable ([9], Theorem 29.3).
The following result will be useful. Recall that a positive operator T from
a Riesz space L into a Riesz space M is called strictly positive whenever
0 u 2 L and Tu =0imply that u =0.
Lemma 2.7 Let L and M be Archimedean Riesz spaces and suppose that M
is order separable. Let 0 T : L ! M be a strictly positive linear operator.
Then L is order separable.
Proof. Since the Dedekind completion of M is order separable as well
(see [9], Theorem 32.9), we may assume without loss of generality that M
is Dedekind complete. Take 0 < u 2 L and let fu : 2 Tg be a disjoint
9

system in [0u]. Let S be the collection of all nite subsets of T, directed by
inclusion. For 2 S we de ne
v = X
u :
Then 0 v " u and so 0 Tv " Tu in M. Let w = sup Tv . Since
M is order separable, there exists an at most countable collection f kg in S
such that w = sup k Tv k . De ne T 0 = S k k and suppose that there exists
2 T such that =2 T 0. De ning 0 k = k [f g for all k =1 2::: we nd
that
2
Tv k + Tu = Tv 0 k
for all k and hence w + Tu w. This implies that Tu =0,andsou =0,
which is a contradiction. Consequently, T = T 0 and this shows that the
disjoint system fu g is at most countable. We may conclude that L is order
separable.
Next we will discuss a property of Riesz spaces whichisweaker than order
separability.
De nition 2.8 Let L be an Archimedean Riesz space.
(i). An element 0

(i). L is locally order separable
(ii). there exists an order separable ideal I in L such that I d = f0g
(iii). for every 0 < v 2 L there exists 0 < u v such that u is order
separable.
Proof. First assume that L is locally order separable and let fu g be a
maximal disjoint system in L consisting of order separable elements. Let I be
the ideal generated by fu g. It is clear that I d = f0g. We claim that I is order
separable. Indeed, suppose that fv g is an order bounded disjoint systemin
I, so there exists 0

such that fpg dd is order separable. Let u be the component of v in fpg dd .
Then u is order separable and 0

this example also shows that the analogue of Lemma 2.7 fails for locally
order separable spaces (take for T the embedding of L into M = R X ).
However, it is not di cult to show that an Archimedean Riesz space is
locally order separable if and only if its Dedekind completion is locally
order separable.
We end this section with recalling some facts concerning order convergence.
A net ff g 2A in the Archimedean Riesz space L is called order
convergent to the element f 2 L if there exists a net fu g 2B in L satisfying
u # 0such that for every 2 B there exists 2 A such that jf ; f j u
for all . This will be denoted by f
(o)
;! f. A subset G L is called
(o)
order closed if ff g G and f ;! f 2 L imply that f 2 G. The order
closed subsets of L are the closed subsets of a topology o in L, which is
called the order topology. The o-closure of a subset D L is denoted by
D (o)
and is called the order closure of D. Although o is in general not a vector
space topology, it is not di cult to see that the order closure of any Riesz
subspace of L is a Riesz subspace. If K is an order closed Riesz subspace of
L, then it is clear that K is a complete Riesz subspace of L. Moreover, if L
is Dedekind complete, then a Riesz subspace K is order closed if and only if
K is a complete Riesz subspace of L. In particular, if K is an order closed
Riesz subspace of the Dedekind complete space L, thenKitself is Dedekind
complete and K is regularly embedded in L.
Now we assume that L is a Dedekind complete Riesz space and let
(o)
ff : 2 A g be a net in L. As is easily veri ed, f ;! f if and only if
there exists 0 2 A such that ff : 0g is order bounded and
_ ^ ^ _
f = f = f. (4)
0
Note that it follows in particular from this characterization of order convergence
that f
(o)
;! f if and only if there exist 0 2 A and a net fu : 0g
in L such that u # 0 and jf ; fj u for all 0. The following
characterization of order convergence will be useful. For a net ff : 2 A g
in L we will also consider the net ff ; f 0 :(
is given the product order.
0 ) 2 A A g, where A A
Lemma 2.13 A net ff g in the Dedekind complete Riesz space L is order
convergent if and only if f ; f 0
(o)
;! 0.
13
0

Proof. If f
(o)
(o)
;! f, then it is clear that f ; f 0
(o)
;! 0. Now suppose
that f ; f 0 ;! 0, i.e., there exist a net fu g in L satisfying u # 0such
2B
that for every 2 B there exists ; 0 2 A A such that jf ; f 0j u
; 0 0 whenever ( ) . Replacing and 0 00 0 by some , we
may assume without loss of generality that = 0 . Fix some 0 2 B . It is
clear that ff :
0
0g is order bounded. Moreover,
_ f ; ^
f 0 = _
jf ; f 0j u
for all 0. Since u # 0, it follows that
_
0
0
0
^ f = ^
0
_ f ,
and this implies that (4) holds with 0 = 0, so ff g is order convergent.
Next we will prove a simple extension result for positive operators which
will be used in a later section of the paper. Given a Riesz subspace K of the
Riesz space L we will denote by K0 the set of all f 2 L for which there exists
(o)
anetff g in K such that f ;! f in L. It is easy to see that K0 is a Riesz
subspace of L.
Lemma 2.14 Let K be a Riesz subspace of the Archimedean Riesz space L
and let M be aDedekind complete Riesz space. Suppose that 0 T : K ! M
is a positive linear operator with the property that Tf
(o)
;! 0 in M whenever
ff g is a net in K satisfying f
(o)
;! 0 in L. Then there exists a unique
positive linear operator 0 T 0 : K 0 ! M such that for every f 2 K 0 and any
net ff g K with f
(o)
;! f we have Tf
(o)
;! T 0f (in particular, T 0
jK = T ).
Proof. Take f 2 K 0 and let ff g be a net in K such that f
(o)
;! f in
L. Then f ; f 0
(o)
;! 0inL and so, by hypothesis, Tf ; Tf 0
(o)
;! 0inM.
Now it follows from Lemma 2.13 that fTf g is order convergent in M, i.e.,
(o)
Tf ;! g for some g 2 M. We claim that g does not depend on the choice
of the particular net ff g. Indeed, suppose that fh g is any other net in K
such that h
(o)
;! f in L. Then f ; h
on T it follows that Tf ; Th
(o)
(o)
;! 0 in L and by the assumption
;! 0 in M. This implies that Th
(o)
;! g.
De ning T 0 f = g, it is easy to verify that T 0 : K 0 ! M is a positive linear
operator which has, by de nition, the desired property. The uniqueness is
obvious.
14

3 Representations
Many of the familiar Dedekind ( -) complete Riesz spaces arise as quotient
spaces of ideals of measurable functions by a -ideal (usually, thenull functions
with respect to some measure). In this section we discuss the representation
of Dedekind -complete Riesz spaces as such quotient spaces.
First we introduce some notation. Given a measurable space (X ) (i.e.,
X is a non-empty set and is a -algebra of subsets of X), we will denote
by M(X ) the space of all real-valued -measurable functions on X and
Mb(X ) denotes the ideal in M(X ) consisting of all bounded functions.
The characteristic function of a set F X will be denoted by 1F .
De nition 3.1 Let L be aDedekind -complete Riesz space with weak order
unit 0 < w 2 L. The measurable space (X ) is called a representation
space for (L w) if there exist an ideal bL in M(X ) such that Mb(X ) bL
and a surjective -order continuous Riesz homomorphism : bL ! L with
(1X) =w.
Given such a homomorphism : bL ! L as in the above de nition, we
will sometimes call the representation (corresponding to (X )). It is
clear that the kernel Ker( ) is a -ideal in bL. Hence induces a Riesz
isomorphism from bL Ker ( ) onto L.
The proof of the rst result in this section will be divided into two lemmas.
The principal ideal generated by an element 0 w 2 L will be denoted by
Lw.
Lemma 3.2 Let L be a Dedekind -complete Riesz space with weak order
unit 0

(a). (g) = b(g) for all 0 g 2 Mb(X ).
(b). (g 1 + g 2)= (g 1)+ (g 2) for all 0 g 1g 2 2 bL.
(c). ( g) = (g) for all 0 g 2 b L and all 0 2 R.
(d). If 0 g 1g 2 2 bL are such thatg 1 ^ g 2 =0,then (g 1) ^ (g 2)=0inL.
Consequently, has a unique extension to a positive linear operator from
bL into L, which we will denote by again. From (d) it follows that is
a Riesz homomorphism. Furthermore, is -order continuous. Indeed,
suppose that 0 gn " g in bL. For every k 2 N we have gn ^ k1X "n
g ^ k1X in Mb(X ) and since b is -order continuous this implies that
b (gn ^ k1X) "n b (g ^ k1X) in L. Hence,
(g) = sup
k
b (g ^ k1X) =sup
kn
b (gn ^ k1X) = sup
n
(gn):
It remains to show that is surjective. To this end, let 0 f 2 L
be given. For every 1 n 2 N there exists 0 gn 2 Mb(X ) such that
b(gn) = f ^ nw. Since b is a Riesz homomorphism we may assume that
0 gn ". Let
F = fx 2 X : gn(x) "n 1g :
For every m 2 N we have (m1F ) ^ gn "n m1F and since b is -order
continuous it follows that
m b(1F ) ^ f ^ nw "n m b(1F ):
This implies that 0 m b(1F ) f for all m 2 N and so b(1F ) = 0.
Now de ne hn = gn1X F . Then b(hn) = b(gn) = f ^ nw for all n and
0 hn " h 2 M(X ). Since
b(h ^ k1X) =sup
n
b(hn ^ k1X) f
for all k, it is clear that h 2 bL. Using that is -order continuous and that
w is a weak order unit in L, we nd that
(h) = sup
n
which completes the proof of the lemma.
(hn) = sup f ^ nw = f,
n
16

Lemma 3.3 Given a Dedekind -complete Riesz space L with weak order
unit 0

Corollary 3.5 Let (X ) be a representation space for the Dedekind complete
Riesz space L with weak order unit 0 w 2 L. Then (X ) is a representation
space for (L u w) as well and we may actually take cL u = M (X ).
Proof. Let b L be an ideal such that Mb (X ) L M (X ) and
: bL ! L a surjective -order continuous Riesz homomorphism. Let b =
jMb(X ), then b : Mb (X ) ! Lw is a surjective -order continuous Riesz
homomorphism as well. Since L is Dedekind complete, (L u ) w = Lw and so
it follows immediately from Lemma 3.2 that (X ) is a representation space
for (L u w). Taking cL u as de ned by (5) we have cL u = M(X ). Indeed,
let 0 g 2 M(X ) be given and let fgkg 1
be a disjoint sequence in
k=1
Mb(X ) such that supk gk = g. Then f b(gk)g 1
is a disjoint system in L
k=1
and so supk b(gk) = f 2 Lu exists. Since b is a -order continuous Riesz
homomorphism it follows that
b(g ^ n1X) = b sup (gk ^ n1X) =sup
k
k
b (gk ^ n1X) f
for all n 2 N and so g 2 cL u . Note that, by -order continuity, the corresponding
Riesz homomorphism u : M(X ) ! L u coincides with on bL.
Next we make some remarks concerning the relation between -ideal in
ideals in M(X ) and Boolean -ideals in . We introduce some further
notation. Given a -ideal N we write
M(N) =ff 2 M(X ) : fx 2 X : f(x) 6= 0g2Ng
and Mb(N) = M(N) \ Mb(X ). Then M(N) and Mb(N) are -ideals in
M(X ) and Mb(X ) respectively. Now assume that E is an ideal in
M(X ) such thatMb(X ) E. It is clear that E \M(N) isa -ideal in E
whenever N is a -ideal in . We claim that all -ideals in E are of this form.
Indeed, given a -ideal N in E we de ne N = fF 2 :1F 2 Ng and we will
show that N = E \ M(N). Take f 2 N and let F = fx 2 X : f(x) 6= 0g.
Then n jfj ^1X 2 N for all n 2 N and n jfj ^ 1X " 1F , so 1F 2 N,
i.e., F 2 N and hence f 2 E \ M(N). Conversely, if f 2 E and F =
fx 2 X : f(x) 6= 0g2N, then jfj^n1F 2 N for all n 2 N and jfj^n1F "jfj,
so jfj 2N and hence f 2 N.
Assume that L is a Dedekind -complete Riesz space with weak order
unit 0 w 2 L and that (X ) is a representation space for (L w)
with representation : bL ! L, where Mb(X ) bL M(X ) is
an ideal. Then Ker ( ) is a -ideal in bL and it follows from the above
18

observations that Ker ( ) = bL \ M(N ), where the -ideal N is given
by N = fF 2 : (1F )g = 0. Consequently, L is Riesz isomorphic to
bL b L \ M(N ). Furthermore it is easy to see that the following three conditions
are equivalent:
(a). if fn 2 bL (n =1 2:::) and f 2 M (X ) such that 0 fn " f and if
(fn) h for all n and some h 2 L, then f 2 bL
(b). M (N ) bL
(c). if f 2 bL and g 2 M (X ) such that f = g N -a.e. (i.e., f and g are
equal except on a set belonging to N ), then g 2 bL.
Note that the ideal bL constructed in Lemma 3.2 has these properties.
Now we introduce some terminology concerning representations of positive
projections. We assume that L is an Archimedean Riesz space and that
P : L ! L is a positive projection (i.e., P is a positive linear operator such
that P 2 = P ). Furthermore we assume that K = Ran (P ) is a Riesz subspace
of L. First a simple observation.
Lemma 3.6 If in addition P is ( -) order continuous, then K is a ( -)
complete Riesz subspace of L. In particular, if L is Dedekind ( -) complete,
then K is Dedekind ( -) complete and ( -) regularly embedded in L.
Proof. We assume that P is order continuous and that ff g K and
f 2 L are such that f " f in L. This implies that f = Pf " Pf in L and
hence f = Pf 2 K, which shows that K is a complete Riesz subspace of L.
Given two measurable spaces (X ) and (Y ) we denote by F =
the product -algebra of and in the product space = X Y ,i.e.,F is
the -algebra generated by all rectangles F G with F 2 and G 2 . If
f 2 M (X ) then we de ne f 1Y 2 M ( F) by (f 1Y )(x y) =f (x)
for all (x y) 2 . The mapping f 7;! f 1Y is an f-algebra isomorphism
from M (X ) onto the -complete f-subalgebra M (X ) 1Y of M ( F),
where
M (X ) 1Y = ff 1Y : f 2 M (X )g :
It will be convenient in the sequel to identify M (X ) with this f-subalgebra
M (X ) 1Y . Similarly, M (Y ) will be identi ed with the -complete
f-subalgebra 1X M (Y ) of M ( F). Note that it is implicit in these
identi cations that the -algebra is identi ed with the -subalgebra 1
19

of F given by 1 = fF Y : F 2 g, and similarly is identi ed with
the corresponding -subalgebra 1 of F. Assuming in addition that is a
- nite measure on (Y ) and that 0 m 2 M ( F) satis es
Z
Y
m (x y) d (y) =1 8x 2 X, (6)
we de ne the linear operator Rb : Mb ( F) ! Mb ( F) by
Rbf (x y) =
Z
Y
m (x z) f (x z) d (z), (x y) 2 , (7)
for all f 2 Mb ( F). Then Rb is a -order continuous positive projection
onto Mb (X ).
De nition 3.7 Let L be aDedekind -complete Riesz space with weak order
unit 0 < w 2 L and suppose that P : L ! L is a positive projection onto
the Riesz subspace K L with w 2 K. Give a measurable space (X )
and a - nite measure space (Y ), we will say that the product space
( F) =(X Y ) is a representation space for P if:
(i). ( F) is a representation space for (L w) with representation homomorphism
:b L ! L
(ii). there exists 0 m 2 M ( F) satisfying R m (x y) d (y) =1 for all
Y
x 2 X
(iii). if f 2 bL, then
Z
Y
m (x y) jf (x y)j d (y) < 1
for all x 2 X and the function Rf on , de ned by
Rf (x y) =
for all (x y) 2 , satis es Rf 2 bL
(iv). P ( f) = (Rf) for all f 2 b L.
Z
Y
m (x z) f (x z) d (z)
If, in addition, is aprobability measure and m (x y) =1for all (x y) 2 ,
then ( F) will be called a proper representation space for P .
20

It is readily veri ed that the existence of a representation space for a
projection P implies that P is -order continuous and so we will restrict our
attention to such projections only.
Remark 3.8 Suppose that ( F) =(X Y ) isarepresentation space
for the positive projection P : L ! L and K =Ran(P ) as in the above de -
nition. De ne b LX = b L\M (X ). Then (X ) isarepresentation space for
K, where the representation is given by jbLX : bLX ! K. If there is already
given some representation K : bK ! K with Mb (X ) bK M (X )
and if K =
with K.
, then we will say that the representation jbLX
is compatible
The following lemma will be very useful.
Lemma 3.9 Let L be a Dedekind -complete Riesz space with weak order
unit 0

for all n. As we have observed before, since we assume that M (N ) bL,
this implies that Rf 2 bL. Hence Rf 2 bL \ M (X ) and it is clear that
bL\M (X ) E. Consequently, wehavea positive linear operator R : E !
E with Ran (R) =E \ M (X
all f 2 E.
). It is easy to see that P ( f) = (Rf) for
De ning 1 = jE, it is clear that 1 : E ! L is a -order continuous
Riesz homomorphism. We will show nowthat1 is surjective. Let 0 g 2 L
be given and take 0 fh 2 bL such that f = g and h = Pg. De ne
F = x 2 X :
Letting fn = f ^ (n1) we have
Z
Y
m (x y) f (x y) d (y) =1 .
(Rfn) =P ( fn) Pg = h,
which implies that (Rfn ; h) + =0. Therefore, the sets
Fn = f(x y) 2 :Rfn (x y) >h(x y)g
satisfy Fn 2 N for all n. Furthermore, it follows from 0 fn " f that
Rfn (x y) "
Z
Y
m (x z) f (x z) d (z)
for all (x y) 2 and so F Y 2 S 1
n=1 Fn. This shows that F Y 2
N . De ning ~ f = f1 (F Y ) c, it is clear that ~ f 2 E and ~ f = g, so we
may conclude that 1 is surjective. Therefore, 1 : E ! L is the desired
representation for P on L.
The above lemma shows that for the construction of representations of
positive projections we may restrict our attention to spaces with a strong
order unit. All such Dedekind complete Riesz spaces have the structure
of an f-algebra and complete Riesz subspaces containing the unit element
are f-subalgebras. Therefore, in the next sections we will study in detail
the properties and structure of positive projections in f-algebras onto fsubalgebras.
4 Some properties of f-subalgebras
For the basic theory of f-algebras we refer the reader to the books [1], [13]
and [16]. Let A be an Archimedean f-algebra with a unit element 1, which
is weak order unit in A. We denote the Boolean algebra of all components
22

of 1 by CA = CA (1). Note that CA is equal to the set of all idempotents in A
and that for all e 1e 2 2CA in A we have
e 1 ^ e 2 = e 1e 2
e 1 _ e 2 = e 1 + e 2 ; e 1e 2
Denoting by P (A) the set if all band projections in A, the mapping P 7!
P 1 is a Boolean isomorphism from P (A) onto CA. For every e 2 CA the
corresponding band projection Pe onto feg dd is given by Pef = ef for all
f 2 A. Assuming in addition that A is Dedekind complete, P(A) is complete
and hence CA is a complete Boolean algebra. Now suppose that B is an fsubalgebra
of A (i.e., B is a Riesz subspace as well as a subalgebra of A) with
1 2 B and let CB be the Boolean algebra of all components of 1 in B. Then
CB is a Boolean subalgebra of CA. If we assume that B is Dedekind complete
as well, then CB is a complete Boolean algebra in its own right, however in
general the in nite joins and meets in CB and CA need not coincide. However,
if we assume in addition that B is regularly embedded in A, then CB is a
complete Boolean subalgebra of CA (in the sense of [15], Section 23). This
is in particular the case if B is a complete f-subalgebra of A ( i.e., B is a
complete Riesz subspace of A, as de ned at the beginning of Section 2).
If A is a Dedekind -complete f-algebra with unit element 1 and if E is
a Boolean -algebra of CA, then A (E) is de ned as in De nition 2.1 and by
Proposition 2.4, A (E) is a -complete Riesz subspace of A. We claim that
A (E) isanf-subalgebra of A. Indeed, it is easy to see that S (E) asde ned
by (2) is a subalgebra of A and now it follows from Proposition 2.6 (i), in
combination with the order continuity ofthemultiplication in A, thatA (E)
is a subalgebra as well. In particular, if A is Dedekind complete and E is a
complete Boolean subalgebra of CA, then A (E) is a complete f-subalgebra
of A with CA(E) = E (see Proposition 2.6).
Next we make some comments about the results in Section 3. Given a
Dedekind -complete f-algebra with unit element 1, it follows from Proposition
3.4 that there exists a representation space (X ) for (A 1). Let bA be
an ideal such that Mb (X ) bA M (X ) with corresponding surjective
-order continuous Riesz homomorphism : bA ! A. De ning
Ab = ff 2 A : jfj n1 for some n 2 Ng : (8)
let b : Mb (X ) ! Ab be the restriction of to Mb (X ), which is
surjective aswell. Since b is a Riesz homomorphism satisfying b (1X) =1,
it is well known that b is actually an f-algebra homomorphism (i.e., a lattice
and algebra homomorphism see e.g. [2]). Replacing, if necessary, the ideal
bA by the corresponding ideal de ned by (5), which is easily seen to be a
23
.

subalgebra of M (X ), it follows that we may assume in this situation, that
bA is an f-subalgebra of M (X ) and that is an f-algebra homomorphism.
If (X ) is a representation space for (A 1), then we will simply say that
(X ) is a representation space for A.
Although the result of the following lemma is known (the result is even
true under some weaker assumptions see [3], Theorem 6.1), we indicate the
simple proof for the convenience of the reader.
Lemma 4.1 Let A be an Archimedean f-algebra with unit element 1 and
suppose that B is an f-subalgebra of A with 1 2 B. If P is a positive
projection in A onto B (i.e., Ran(P ) = B), then P (ba) = bP (a) for all
b 2 B and a 2 A (i.e., the projection P is so-called averaging).
Proof. Fix some 0 a 2 A. First we assume in addition that 0 a n1
for some n 2 N. De ne the linear operator Ta : B ! B by Ta(b) = P (ba)
for all b 2 B. Then 0 Ta(b) nb for all 0 b 2 B. Hence Ta 2 Z(B),
the center of B. Since B is an f-algebra with unit element every operator in
Z(B) is given by multiplication by some element ofB, so there exists b 1 2 B
such that Ta(b) = bb 1 for all b 2 B. Taking b = 1 we see that b 1 = P (a).
It follows that P (ba) = Ta(b) = bP (a). If 0 a 2 A is arbitrary, then
we can use that a ^ n1 "n a (a 2 -uniformly) and the result now follows via
approximation (see e.g. [16], Theorem 142.7).
The following proposition will play an important role in the sequel. As
usual, positive projection P : A ! A will be called strictly positive if Pa > 0
whenever 0 < a 2 A. The result of the following proposition, which is of
interest in its own right, will play a crucial role in the proof of Lemma 8.2.
Proposition 4.2 Let A be aDedekind complete f-algebra with unit element
1, which is a strong unit as well. Let P : A ! A be a strictly positive order
continuous projection onto the f-subalgebra B, with 1 2 B. Suppose that C
is a complete f-subalgebra ofA such that B C. Then there exists a unique
positive projection Q : A ! A onto C such that P = P Q. Moreover, this
projection Q is strictly positive and order continuous.
by
Proof. Given 0 a 2 A de ne the positive linear operator Ta : C ! B
Ta(c) =P (ac)
for all c 2 C. Since C is a regular Riesz subspace of A and P is order
continuous, it follows that Ta is order continuous as well. Furthermore, if
b 2 B then
Ta (bc) =P (bac) =bP (ac) =bTa (c)
24

for all c 2 C, bytheaveraging property ofP (see Lemma 4.1). Considering A
and C as f-modules over B via multiplication (see e.g. [7], Section 4 for the
de nition of an f-module), we see that Ta is B-linear and so Ta 2L B n (C B).
Since 1 is assumed to be a strong unit in A, there exists n 2 N such that
0 a n1, hence
0 Ta (c) =P (ac) nP (c) =nT1 (c)
for all 0 c 2 C, i.e., 0 Ta nT1 in L B n (C B). From the Radon-Nikodym
theorem in L B n (C B) (see [7], Proposition 3.10 and Examples 4.5 (3) ) it
follows that there exists a 2 Z(C) such that 0 a nIC and Ta = T1 a.
Since C is a unital f-algebra there exists Q(a) 2 C such that 0 Q(a) n1
and a(c) =Q(a)c for all c 2 C. Hence Ta(c) =T1 (Q(a)c) for all c 2 C, i.e.,
P (ac) =P (Q (a) c) for all c 2 C. (9)
The element Q(a) 2 C is uniquely determined by (9). Indeed, if f 2 C is
such that P (fc) = 0 for all c 2 C, then in particular P (f 2 ) = 0. Since
f 2 0 and P is strictly positive, it follows that f 2 = 0 and hence f =0.
Now it is clear that the mapping a 7;! Q(a) is additive on A + and that
Q(a) = a whenever 0 a 2 C. Consequently Q extends uniquely to a
linear positive projection Q : A ! A with Ran(Q) =C, satisfying (9) for all
a 2 A. Taking c = 1 in (9) it follows that P = P Q. Since P is strictly
positive, this implies that Q is strictly positive as well. To show that Q is
order continuous, suppose that a # 0 in A and assume that Q (a ) a for
all and some 0 a 2 A. This implies that P (a )=P (Q (a )) P (a) 0
for all . Since P (a ) # 0 it follows that P (a) =0,hence a =0. This shows
that Q (a ) # 0inA, so Q is order continuous.
It remains to provethatQ is unique. To this end suppose that Q 1 : A ! A
is a positive projection such that Q 1 (A) = C and P = P Q 1. By the
averaging property of Q 1it follows that
P (ac) =P (Q 1 (ac)) = P (Q 1 (a) c)
for all a 2 A and all c 2 C. As observed above, Q(a) 2 C is uniquely
determined by (9), hence Q 1 (a) =Q (a) for all a 2 A, i.e., Q 1 = Q.
Remark 4.3 If we drop the assumption that 1 is a strong order unit in A,
then the result of the above theorem does not hold in general, as can be seen
by the following example. We consider the positive real axis (0 1) equipped
with the probability measure d = e ;x dx de ned on the -algebra of all
25

Lebesgue measurable subsets. De ne the ideal A in L0 ( ) by
(
A = f 2 \
Lp ( ):jfj 1 (01) c1 (01) for some 0 c 2 R
1 p

and, using that Qf is 1-periodic,
Z
E
Qf (x) d (x) =
1X Z
k=0
E1+k
Qf (x) e ;x dx =
In combination with (11) this shows that
Z
E1
e
e ; 1 Qf (x) e;x dx =
for all measurable subsets E 1
1X
k=0
Z
E1
1X
k=0
=
1X Z
Z
k=0
E1
E1
f (x + k) e ;k
(0 1). This implies that
Qf (x) e ;(x+k) dx
e
e ; 1 Qf (x) e;x dx:
!
e ;x dx
f (x + k) e ;k = e
Qf (x) (12)
e ; 1
-a.e. on (0 1). Now we take in particular the function f 0 2 A de ned by
(10). Then
1X
k=0
and so it follows from (12) that
f 0 (x + k) e ;k = f 0 (x +1)e ;1 = jlog xj e ;1
Qf 0 (x) =
e ; 1
e 2
jlog xj
-a.e. on (0 1). Hence Qf 0 is not bounded. However, as observed above, all
functions in C are bounded, which is a contradiction. This shows that such
a projection Q does not exist.
Next we will discuss some further notation and properties concerning
subalgebras. As above we assume that A is a Dedekind complete f-algebra
with unit element 1 and we assume that B is a Dedekind complete regularly
embedded f-subalgebra of A with 1 2 B. Hence CB is a complete Boolean
subalgebra of CA. For p 2CA we will denote
A(p) =fpf : f 2 Ag (13)
which is the band in A generated by p. Note that A(p) isanf-algebra in its
own right with unit element p. The Boolean algebra of components of p in
A(p) is given by
CA(p) =fq 2CA : q pg :
27

Furthermore we de ne
B p = fpf : f 2 Bg : (14)
Then B p is a Dedekind complete regularly embedded f-subalgebra of A(p)
and it is easy to see that the Boolean algebra of components of p in B p is
given by
pCB = fpq : q 2CBg
which is a complete Boolean subalgebra of CA(p). If p 2 CB, then B p will
also be denoted by B(p).
For p 2CA we de ne
p =inffq 2CB : p qg , (15)
where the in mum is taken in CA. Since CB is a complete Boolean subalgebra
of CA, it follows that p 2CB. The element p is sometimes called the closure
of p with respect to CB ([10]). It is clear that p p.
Lemma 4.4 The mapping : B(p) ! B p, de ned by (f) = pf for all
f 2 B(p), is a surjective f-algebra isomorphism.
Proof. It is clear that is an f-algebra homomorphism. Now take
g 2 B p. Then g = pf for some f 2 B and hence (pf) = ppf = pf = g,
which shows that is surjective. It remains to proof that is injective. To
this end suppose that f 2 B(p) such that pf =0. From (15) it follows that
1 ; p =supfq 2CB : p ^ q =0g :
Since f = fp, i.e.,f (1 ; p) = 0, this implies that fq = 0 for all q 2CB \fpg d
and hence fs = 0 for all s 2 S(CB) \fpg d . Therefore, by Freudenthal's
spectral theorem we have fg =0for all g 2 B \fpg d . Since pf =0implies
that f 2 B \fpg d , it follows that f 2 =0,hence f =0,which shows that
is injective.
Remark 4.5 Note that the last argument in the above proof actually shows
that
B \fpg d(A) = B(p) d(B)
where d(A) and d(B) indicate that the disjoint complements are taken in A
and B respectively.
28

As we have seen in Example 2.12, a Riesz subspace of a locally order
separable Riesz space need not be locally order separable. However, using
Lemma 4.4 we can prove the following result, which will be used later.
Lemma 4.6 Let A be a locally order separable Dedekind complete f-algebra
with unit element 1 and let B be a Dedekind complete regular f-subalgebra
of A with 1 2 B. Then B is locally order separable as well.
Proof. Take 0 < q 2 CB. It follows from Lemma 2.11 that there exists
p 2 CA such that 0 < p q and A (p) is order separable. Since B p is a
Riesz subspace of A (p), it follows that B p is order separable. By Lemma
4.4, B(p) is Riesz isomorphic to B p, hence B(p) is order separable as well.
Since 0 < p q in CB, it follows from Lemma 2.11 that B is locally order
separable.
De nition 4.7 Let A be aDedekind complete f-algebra with unit element 1
and let B be a Dedekind complete regular f-subalgebra of A with 1 2 B. An
element p 2CA will be called B-full if CA(p) =pCB, i.e., if
fe 2CA : e pg = fpq : q 2CBg : (16)
We note that the B-full elements in CA correspond to the elements of
order zero over CB, as de ned in [10], Section 11. In the next lemma we will
use the notation introduced in (8).
Remark 4.8 (i). If p 2 CA and q 2CA (p), then q is B-full if and only if
q is B p-full in CA(p) = CA (p).
(ii). If p 2CA is B-full and q 2CA (p), then q is B-full.
Lemma 4.9 For an element p 2CA the following statements are equivalent:
(i). p is B-full
(ii). B p is an order ideal in A
(iii). Ab(p) =(Bb) p
(iv). the mapping f 7;! pf is an f-algebra isomorphism from Bb(p) onto
Ab(p)
(v). the mapping q 7;! pq is a Boolean isomorphism from CB(p) onto CA(p).
29

Proof. Suppose that p is B-full and take f 2 A satisfying 0 f pg
for some 0 g 2 B. Since f 2 A(p), there exists a sequence fskg 1
k=1 in
S (CA (p)) such that 0 sk " f. It follows from (16) that every sk can be
written as sk = ptk for some 0 tk 2 S (CB) and we may assume that
0 tk ". Hence 0 tk ^ g " g in B and so 0 tk ^ g " h 2 B. Now
p (tk ^ g) =sk ^ pg " f ^ pg = f
so f = ph 2 B p. This proves that (i) implies (ii). Since
Ab(p) =ff 2 A : jfj np for some n 2 Ng
and p 2 (Bb) p = (B p) b it is clear that (ii) implies (iii). Lemma 4.4 shows
that (iii) implies (iv) and it is also clear that (iv) implies (v), which obviously
implies (i).
It will be convenient to introduce the following de nition.
De nition 4.10 Let A be a Dedekind complete f-algebra with unit element
1 and let B be a Dedekind complete regular f-subalgebra of A with 1 2 B.
Then:
(a). A is called nowhere full with respect to B (or, nowhere B-full)
if 0 is the only B-full element in CA. In this case we will also say that
CA is nowhere full with respect to CB
(b). A is called everywhere full with respect to B (or, everywhere
B-full) if for every 0 < p 2 CA there exists 0 < q 2 CA (p) which is
B-full.
The following lemma is straightforward.
Lemma 4.11 The following three statements are equivalent:
(i). A is everywhere B-full
(ii). there exists a disjoint system fp g in CA consisting of B-full elements
such that W p = 1
(iii). sup fp 2CA : p is B-fullg = 1.
With these observations, the next result follows immediately.
30

Proposition 4.12 Let A beaDedekind complete f-algebra with unit element
1 and let B be a Dedekind complete regular f-subalgebra of A with 1 2 B.
Then there exist p 1p 2 2CA with p 1 + p 2 = 1 such that:
(i). A (p 1) is everywhere B p1-full
(ii). A (p 2) is nowhere B p2-full.
Proof. De ning p 1 = sup fp 2CA : p is B-fullg and p 2 = 1 ; p 1, the
result follows from Remark 4.8 and Lemma 4.11.
Next we will discuss restrictions of positive projections to bands in A. We
assume that A is a Dedekind complete f-algebra, that B is an f-subalgebra
of A with 1 2 B and that P : A ! A is a strictly positive order continuous
projection onto B. Let 0 < p 0 2 CA be xed. We will need the following
observation.
Lemma 4.13 The element P (p 0) is a weak order unit in B(p 0).
Proof. Since 0 p 0 p 0, it follows that 0 P (p 0) P (p 0) = p 0,
so 0 P (p 0) 2 B(p 0). Now suppose that 0 g 2 B(p 0) is such that
gP(p 0) = 0. Then P (gp 0) = 0 and so gp 0 = 0, as P is strictly positive.
Hence (g) = 0 and by Lemma 4.4 this implies that g = 0. Consequently,
P (p 0) isaweak order unit in B(p 0).
Denoting by B(p 0) u the universal completion of the f-algebra B(p 0), it is
well known that the f-algebra structure of B(p 0) extends uniquely to B(p 0) u
and that every weak order unit in B(p 0) u has an inverse. Hence, it follows
from the above lemma that in particular the element P (p 0) is invertible in
B(p 0) u , i.e., there exists 0 < w 0 2 B(p 0) u such that w 0P (p 0) = p 0. We will
denote this element w 0 by P (p 0) ;1 .
Lemma 4.14 We assume in addition that the unit element 1 in A is a strong
order unit. The mapping P p0 : A(p 0) ! A(p 0), de ned by
P p0(f) =p 0P (p 0) ;1 P (f)
for all f 2 A(p 0), is a strictly positive order continuous projection onto B p0.
Proof. First we show that P p0 is well de ned. Take 0 f 2 A(p 0).
Then 0 f np 0 for some n 2 N and so 0 P (f) nP (p 0) np 0.
Therefore, in B(p 0) u we have
0 P (p 0) ;1 P (f) nP (p 0) ;1 P (p 0)=np 0:
31

Since B(p 0) is Dedekind complete, B(p 0) is an ideal in B(p 0) u and so 0
P (p 0) ;1 P (f) 2 B(p 0). Hence 0 P p0(f) 2 B p0 A(p 0). This shows that
P p0 is a well de ned positive linear mapping with its range contained in B p0.
If f 2 B p0 then, by de nition, f = p 0g for some g 2 B. Hence
P p0(f) =p 0P (p 0) ;1 P (p 0g) =p 0P (p 0) ;1 P (p 0)g
= p 0p 0g = p 0g = f
which shows that P p0 is a projection onto B p0. It is clear that P p0 is order
continuous, so it remains to prove that P p0 is strictly positive. Suppose
that 0 f 2 A(p 0) such that P p0(f) = 0. Since 0 P (p 0) ;1 P (f) 2
B(p 0), it follows from Lemma 4.4 that P (p 0) ;1 P (f) = 0 and so p 0P (f) =
P (p 0)P (p 0) ;1 P (f) =0. Since 0 P (f) np 0 for some n 2 N, this implies
that P (f) = 0 and so f =0.
In the above situation we will say that P p0 is the restriction of P to
A(p 0) (or, the restriction to p 0). We endthis section with a technical result,
which shows how representations of restrictions of a positive projection P can
be glued together to obtain a representation of P (as de ned in De nition
3.7). We assumethatA is a Dedekind complete f-algebra in which the unit
element 1 is a strong order unit and that P : A ! A is a strictly positive order
continuous projection onto the f-subalgebra B with 1 2 B. Suppose that
(X ) is a representation space for B with representation homomorphism
B : Mb (X ) ! B. If 0 < p 0 2 CA, then (X ) is also a representation
space for B p0, the representation given by the homomorphism
p0
B : Mb (X ) ! B p0, (17)
de ned by p0
B (f) =p0 B (f) for all f 2 Mb (X ). In this situation we will
p0 call B the induced representation of B p0.
Lemma 4.15 Let B : Mb (X ) ! B be arepresentation of B. We assume
furthermore that:
(i). fpjg is an at most countable disjoint system in CA such that P j pj = 1
(ii). For every j there exists a representation space (X Yj j) for P pj
and A (pj) such that the corresponding representation
j : Mb (X Yj j) ! A (pj)
is compatible with the induced representation pj
B of B pj.
32

Then there exists a representation space (X Y ) for P and A such
that the corresponding representation :Mb (X Y ) ! A is compatible
with B.
Proof. We will denote ( j Fj) = (X Yj j). By assumption,
there exists a - nite measure j on j and for each j =12::: there is an
operator Rj : Mb ( j Fj) ! Mb ( j Fj), given by
Rjf (x y) =
Z
Yj
mj (x z) f (x z) d j (z)
for all (x y) 2 j and all f 2 Mb ( j Fj), with 0 mj R 2 M ( j Fj) and
Yj m (x z) d j (z) =1for all x 2 X, such that j Rj = P pj j.
We de ne Y = F j Yj (i.e., Y is the disjoint union of the Yj, j =12:::), = L j j and = L j j. Then (Y ) is a - nite measure space.
Denoting ( F) =(X Y ), we identify Mb ( j Fj) with the ideal in
Mb ( F) consisting of all functions of the form f1 j with f 2 Mb ; ( F). If
0 f 2 Mb ( F), then j f1 j is a disjoint order bounded sequence
in A, so we can de ne
(f) = X ;
f1 j .
j
It is easily veri ed that this de nes a -order continuous surjective f-algebra
homomorphism :Mb ( F) ! A, which is compatible with B, as all j
are compatible with pj ;
B . Note that f1 j = pj (f) for all f 2 Mb ; ( F).
For f 2 Mb ( F) we will from now on consider the function Rj f1 j as
an element of Mb ( F).The commutation relation j Rj = P pj j then
corresponds to
;
1 jRj f1 j
= P pj
j
; f1 j = P pj [pj (f)] (18)
for all f 2 Mb ( F).
For j =12::: wechoose 0 uj 2 Mb (X ) such that B (uj) =P (pj).
Take f 2 Mb ( F) and put g = (f). De ne hj 2 Mb (X ) by
;
hj = ujRj f1 j .
; ;
Since 0 P (pj) pj, Rj f1 j 2 B and hj = P (pj) Rj f1 j , it
follows that hj 2 B (pj). Furthermore, using (18) and the de nition of P pj
(see Lemma 4.14), we nd
pjhj = ; ; ;
1 j P (pj) Rj f1 j = P (pj) 1 jRj f1 j
= P (pj) P pj (pjg) =P (pj) pjP (pj) ;1 P (pjg)
33
= pjpjP (pjg) =pjP (pjg).

Since P (pjg) 2 B (pj), it follows from Lemma 4.4 that hj = P (pjg). We
thus have shown that
;
f1 j = P [pj (f)] (19)
ujRj
for all f 2 Mb ( F) and all j.
De ne the F-measurable function m : ! [0 1] by
m (x y) = X
uj (x) mj (x y) 1 j (x y) (20)
j
for all (x y) 2 and let kn 2 M ( F) be the nth partial sum of (20).
For n = 1 2::: we de ne the positive linear operators R (n) : Mb ( F) !
Mb ( F) by
R (n) f (x y) =
for all (x y) 2 and all f 2 Mb ( F). Since
it follows from (19) that
Z
R (n) f =
Y
kn (x z) f (x z) d (z)
nX
j=1
; R (n) f = P
ujRj
" nX
j=1
; f1 j ,
pj
!
(f)
#
(21)
for all f 2 Mb ( F). This implies in particular that
; R (n) f " P ( f) (22)
in A for all 0 f 2 Mb ( F).
Our next objective is to show that we can actually choose the functions
uj such that the function m, de ned by (20), satis es R m (x y) d (y) =1
Y
for all x 2 X. To this de ne h : ! [0 1] by h (x y) = R m (x z) d (z)
Y
for all (x y) 2 and put hn = R (n) 1 for all n. From (22) it follows that
(hn) " 1 in A. De ning h0 n = hn ^ 1 , we have (h0 n) = (hn) for all n
and 0 h0 n " h0 for some h0 2 Mb ( F). From the -order continuity of it
follows that (h 0 )=1 = (1 ). Hence, h 0 = 1 N -a.e. and hn = h 0 n
N -
a.e. Since hn (x y) " h (x y) forall(x y) 2 , we mayconclude that h = 1
N -a.e., which implies that there exists a set F 2 such that F Y 2 N
and R m (x y) d (y) = 1 for all x 2 X F . Now it is easy to see that we
Y
34

can adjust the values of the functions uj R on the set F Y in such away that
m (x y) d (y) = 1 for all x 2 X. With this assumption, we de ne the
Y
positive projection R : Mb ( F) ! Mb ( F) onto Mb (X ) by
Rf (x y) =
Z
Y
m (x z) f (x z) d (z)
for all (x y) 2 and all f 2 Mb ( F). It is clear that R (n) f " Rf for
all 0 f 2 Mb ( F), and so (22) implies that (Rf) = P ( f) for all
f 2 Mb ( F). Therefore, we may conclude that is a representation of P
on A and the proof is complete.
We end this section with an example which illustrates the use of the above
lemma and which will also be used later.
Example 4.16 (i). Let A be a Dedekind complete f-algebra in which the
unit element 1 is a strong order unit. The most trivial projection P :
A ! A satisfying the conditions above is the projection P = I onto
B = A. Let B : Mb (X ) ! B be a representation for B. Let
(Y ) be the one point measure space withY = fyg and (fyg) =1.
Then Mb (X Y ) = Mb (X ). De ne the representation A
of A by Af = Bf for all f 2 Mb (X Y ). This de nes
a representation for P and A, where the corresponding representing
projection Rb in Mb (X Y ) is the identity operator. This will
be called the trivial representation.
(ii). Now we assume that A is a Dedekind complete order separable f-algebra
in which the unit element 1 is a strong order unit and that P : A ! A
is a strictly positive order continuous projection onto the f-subalgebra
B with 1 2 B. Moreover, we suppose that A is everywhere B-full.
Let B : Mb (X ) ! B be a representation of B. By Lemma 4.11,
there exists a disjoint system fp g in CA consisting of B-full elements
such that sup p = 1. Since A is assumed to be order separable, this
system is at most countable, fpn : n =1 2:::g say. For each n, let
pn
B : Mb (X ) ! B pn be the induced representation de ned by (17).
As pn is B-full, it follows from Lemma 4.9 that A (pn) =B pn andsothe
restriction P pn : A (pn) ! B pn is the identity operator. Hence, by (i),
P pn has the trivial representation n : Mb (X Yn n) ! A (pn),
where Yn = fyng and n (fyng) = 1. De ning Y = fy 1y 2:::g and
=P (Y ) with counting measure , it follows from Lemma 4.15 (and
its proof) that (X Y ) is a representation space for P and A,
with a representation A : Mb (X Y ) ! A which is compatible
with B.
35

Given a Dedekind complete order separable f-algebra A in which the
unit element 1 is a strong order unit and a strictly positive order continuous
projection P : A ! A onto the f-subalgebra B with 1 2 B, letA = A (p 1)
A (p 2) be the decomposition of Proposition 4.12. As shown in the above
example, it is easy to construct a representation for P p1 on A (p 1). The main
e ort in the remainder of the present paper is to construct a representation
for the projection P p2 on the nowhere B-full part A (p 2).
5 Maharam's theorem for positive projections
In this section we will discuss a generalization of a result of D. Maharam
( [11], Lemma 3) to positive projections, which will play an important role
in the sequel. Actually, the main theorem in the present section can be
considered as a vector valued version of Maharam's lemma. As before, A
will denote a Dedekind complete f-algebra with unit element 1 and B will
be an f-subalgebra with 1 2 B. Furthermore we will assume that P : A ! A
is a positive order continuous projection onto B. As observed in Lemma 3.6,
this implies that B is a complete f-subalgebra of A. The proof of the main
result in the present section is divided intoanumber of lemmas.
Lemma 5.1 Assume that A is nowhere full with respect to B. If 0 6= e 2CA,
then there exists p1 2CA such that 0 6= p1 e and P (p1) P (e).
Proof. Since e 6= 0, the element e is not B-full and so there exists
p 2CA(e) such that p =2 eCB. Let q = e ; p and put u = p:q. First observe
that u 6= 0. Indeed, suppose that u =0,then pq =0,which implies that
so p 2 eCB, a contradiction.
ep =(p + q) p = pp = p
Let e 1 2CB be the component of p in the band
1
2
n P (p) ; 1
2 P (e) +o d
B. Then e1 P (p) ; 1 +
P (e) =0andso from Lemma 4.1 it follows that
2
P (e 1p) =e 1P (p)
1
P (e) :
2
If e 1 6= 0,then, by Lemma 4.4, e 1p 6= 0and so we can take p 1 = e 1p in this
case.
Now assume that e 1 =0. Then p 2
in
n P (p) ; 1
2 P (e) +o dd
and since 0 <
u p, it follows that u P (p) ; 1 +
P (e) > 0 and u P (p) ; 2 1
2P (e) ; = 0,
36

so
which implies that
u P (p) ; 1
1
P (e) = u P (p) ; P (e)
2 2
+
> 0,
uP (p) > 1
uP (e) : (23)
2
n
1
Let e2 2CB be the component of q in the band P (q) ; 2P (e) +od in B.
If e2 = 0, then a similar argument asabove shows that
uP (q) > 1
uP (e) : (24)
2
If (23) and (24) hold simultaneously we nd that uP (e) =uP (p)+uP (q) >
uP (e), which is impossible. Consequently, e2 6= 0. As above it now follows
that e2q 6= 0and P (e2q) 1
2P (e) and so in this case we can take p1 = e2q. Lemma 5.2 Assume that A is nowhere full with respect to B. If 0 6= e 2CA,
then for every " > 0 there exists p 2 CA such that 0 6= p e and P (p)
"P (e).
Proof. Repeated application of Lemma 5.1 immediately shows that for
every n 2 N there exists pn 2CA such that06= pn e and P (pn) 2 ;n P (e).
Lemma 5.3 Assume that A is nowhere full with respect to B. If 0 6= e 2CA
and g 2 B such that 0 < g P (e), then there exists p 2 CA such that
0 6= p e and P (p) g.
Proof. Since 0 0.
Note that e e 2CB, soP (e) e and hence
0 < [g ; "P (e)] +
g P (e) e:
Let q 2 CB be the component of e in the band generated by [g ; "P (e)] + .
From the above observation it follows that 0

Theorem 5.4 Let A be a Dedekind complete f-algebra with unit element
1 2 A and let P : A ! A be a positive order continuous projection onto the
f-subalgebra B with 1 2 B. Assume that A is nowhere full with respect to
B. If 0 6= e 2CA and g 2 B such that 0 g P (e), then there exists p 2CA
such that p e and P (p) =g.
Proof. De ne
K = fu 2CA(e) :P (u) gg :
It is clear that K 6= and by the order continuity ofP ,anyupward directed
system in K has a supremum in K. Hence, K has a maximal element p 2K.
Suppose that P (p)

Proof. First observe that'jK 2 K s n whenever ' 2 L s n ,asK is regularly
embedded in L. Take 0 2 Z(L). Then 0 nIL for some n 2 N
and so 0 (' 0 ) jK n (' 0) jK in K s n . Hence, by the Radon-Nikodym
theorem for normal functionals (see [8], Theorem 3.4 or [16], Lemma 145.1),
there exists 0 nIK in Z(K) such that
(' 0 ) jK =(' 0) jK : (26)
We claim that this 2 Z(K) is uniquely determined by (26). Indeed, suppose
that 1 2 2 Z(K) both satisfy (26) and put 3 = 1 ; 2. This implies that
('0) jK 3 =0andso
(' 0) jK j 3j = (' 0) jK 3 =0
in K s n . Therefore hj 3j u ' 0i =0and hence j 3j u =0for all 0 u 2 K, as
' 0 is strictly positive. This shows that 3 =0,bywhich the claim is proved.
Therefore we may de ne = P ( ). It is clear that P is additive on Z(L) +
and it follows that P extends to a positive projection P : Z(L) ! Z(L) onto
Z(K). Note that for every 2 Z(L) the element P ( ) is uniquely determined
by (25). Since ' 0 is strictly positive andK order dense in L, it is clear that
P is strictly positive. Finally we show that P is order continuous. To this
end assume that # 0inZ(L) and that 2 Z(L) suchthatP ( ) 0
for all . It follows from (25) that
0 h u ' 0i hP ( )u ' 0i = h u ' 0i#0
and hence u = 0 for all 0 u 2 K. Since K is order dense in L, we may
conclude that =0,and so P ( ) # 0inZ(L). This completes the proof of
the proposition.
We will call the projection P in the above proposition the conditional
expectation projection associated with' 0. Note that P(L) andP(K) are the
Boolean algebras of components of the unit element I in Z (L) and Z (K)
respectively. Recall from De nition 4.7 that Q 0 2P(L) is called Z(K)-full if
fQ 2P(L) :Q Q 0g = fQ 0R : R 2P(K)g : (27)
In the next lemma we present some characterizations of Z(K)-full band projections
in L.
Lemma 5.6 Let Q 0 2 P(L) be the projection onto the band B 0 in L. The
following statements are equivalent:
(i). Q 0 is Z(K)-full
39

(ii). for every 2 Z(B 0) there exists 2 Z(K) such that jB0 =
(iii). [0Q 0u] L = Q 0 [0u] K for all 0 u 2 K (where the subscripts L and
K indicate that the order intervals are taken in L and K respectively)
(iv). Q 0(K) is an ideal in L.
Proof. We will only show that (iv) implies (i), as the remaining implications
follow by standard arguments. So we assume that Q 0(K) isan ideal
in L. De ne Q 0 2P(K) by
Q 0 =inffR 2P(K) :Q 0
and de ne : Q 0(K) ! Q 0(K) by (f) =Q 0f for all f 2 Q 0(K). It is clear
that is a surjective Riesz homomorphism. We claim that is injective.
Suppose that 0 f 2 Q 0(K) and Q 0f =0. Since
and ; I ; Q 0 f =0,it follows that
Rg
I ; Q 0 =supfR 2P(K) :RQ 0 =0g
sup fRf : R 2P(K)RQ 0 =0g =0
i.e., Rf = 0 for all R 2P(K) withRQ 0 =0. Let Rf be the band projection
in L onto ffg dd . Since f 2 K, it is easy to see that Rf 2P(K), and Q 0f =0
implies that RfQ 0 =0. Hence f =0,which proves the claim.
Now let Q 2P(L) be given such that Q Q 0. Let Q 1 be the restriction
of Q to Q 0(K). Then Q 1 is a band projection in Q 0(K), as Q 0(K) isanideal
in L. By the rst part of the proof, there exists a band projection R in Q 0(K)
such that R = Q 1 , i.e., Q 0Rf = Q 1Q 0f = Qf for all f 2 Q 0(K).
Considering R as an element ofP(K), this shows that Q 0R = Q and we may
conclude that Q 0 satis es (27), i.e., Q 0 is Z(K)-full.
We will say that L is nowhere full with respect to K if 0 is the only
band projection in L which is full with respect to Z (K). By de nition,
this is equivalent to saying that Z(L) is nowhere full with respect to Z(K).
Therefore the following result is now an immediate consequence of Theorem
5.4.
Theorem 5.7 Let L be a Dedekind complete Riesz space and K an order
dense Dedekind complete regular Riesz subspace ofL, such that L is nowhere
full with respect to K. Suppose that 0 < '0 2 Ls n is strictly positive. If
2 Ks is such that 0 ('0) jK , then there exists a band projection
Q 2P(L) such that =('0 Q) jK , i.e., is the restriction of acomponent
of '0. 40

Proof. Let P : Z(L) ! Z(L) be the conditional expectation projection
onto Z(K) associated with ' 0. By the Radon-Nikodym theorem in K s n , it
follows from 0 (' 0) jK that = (' 0) jK for some 0 I in
Z(K). By Theorem 5.4 there exists Q 2P(L) in such that = P (Q). Now
it follows from (25) that
for all f 2 K, i.e., (' 0
hQf ' 0i = hP (Q)f' 0i = h f ' 0i = hf i
Q) jK = .
5.2 Maharam's theorem
Next we will discuss the result of Maharam by which Theorem 5.4 was inspired.
Let (X ) be a measure space. For the sake of simplicity we will
assume that is nite. We denote the corresponding measure algebra by
( ). Suppose that is a -subalgebra of and let denote the corresponding
complete Boolean subalgebra of . As before we shall say that
p 2 is -full if
fe 2 : e pg = feq : q 2 g ,
and is called nowhere full with respect to if 0 2 is the only -full
element.
Theorem 5.8 (D. Maharam, [11]) Assume that is nowhere full with
respect to . If is a measure on such that 0 (D) (D) for all
D 2 , then there exists F 0 2 such that (D) = (D \ F 0) for all D 2 .
Proof. Let A = L1 (X ) and B = L1 (X ). Then B is an order
closed f-subalgebra of A. Let P : A ! A be the conditional expectation
projection onto B, i.e., P = E ( j ), which is an order continuous positive
projection. Since CA =
with respect to B.
and CB = , it is clear that A is nowhere full
By the classical Radon-Nikodym theorem there exists g 2 B such that
0 g 1X and
(D) =
for all D 2 . It follows from Theorem 5.4 that there exists p 2 CA =
such that P (p) = g. Let F0 2 be such that 1F0 is a representative of p.
From the de ning property of P = E ( j ) it follows that
(D) =
Z
D
P (p)d =
Z
Z
D
D
gd
for all D 2 , by which the theorem is proved.
41
1F0d = (D \ F 0)

6 Special subalgebras
As before we assume that A is a Dedekind complete f-algebra with unit
element 1 and that B is an f-subalgebra with 1 2 B. Furthermore we
assume that P : A ! A is an order continuous strictly positive projection
onto B. In Section 7 we will construct complete Boolean subalgebras E of
CA with the property that for every e 2 E there exists 2 [0 1] such that
P (e) = 1. In the present section we will collect some properties of such
special Boolean subalgebras.
First we introduce some notation. We de ne the subset S = SP of CA by
S = fp 2CA : P (p) = 1 for some 2 [0 1]g : (28)
The following simple observation will be useful.
Lemma 6.1 The subset S is closed for monotone convergence in CA.
Proof. Since p 2Simplies that 1 ; p 2S, it su ces to show that S is
closed for upwards convergence in CA. To this end suppose that p 2Sand
p 2CA such thatp " p in CA. Then P (p )= 1 for all and 0 " 1.
Let = sup . By the order continuity ofP we have P (p ) " P (p) and so
P (p) = 1. Hence p 2S.
For any non-empty subset D of CA we denote by S (D) the complete
Boolean subalgebra of CA generated by D.
Lemma 6.2 Let S CA be de ned by (28).
(i). If F is a Boolean subalgebra of CA and F S , then S (F) S.
(ii). If F is a Boolean subalgebra of CA such that F S, and if we write
P (e) = (e)1 for all e 2F, then : F ! [0 1] is a strictly positive (in
general, nitely additive) measure on F.
(iii). If E is a complete Boolean subalgebra of CA such that E S, and if
we write P (e) = (e)1 for all e 2E, then : E ![0 1] is a completely
additive and strictly positive measure.
Proof.
(i). This follows via a standard argument for Boolean algebras from Lemma
6.1.
42

(ii). If e 1e 2 2F such that e 1 ^ e 2 =0,then
P (e 1 _ e 2)=P (e 1 + e 2)=f (e 1)+ (e 2)g 1
hence (e 1 _ e 2) = (e 1) + (e 2), which shows that is a nitely
additive measure on F. The measure is strictly positive as P is
strictly positive.
(iii). Since P is order continuous and E is a complete Boolean subalgebra of
CA it follows immediately that is completely additive.
We assume that E is a complete Boolean subalgebra of CA, such that
E S, so there exists a completely additive strictly positive measure :
E ! [0 1] such that P (e) = (e)1 for all e 2 E. Recall from Section 4 the
de nitions of the complete f-subalgebra A (E) generated by E and of the
f-subalgebra S (E) given by (2). If s 2 S (E) isgiven by s = Pn i=1 iei with
i 2 R and ei 2E, then P (s) ='P (s) 1, where
'P (s) =
nX
i=1
i (ei).
Clearly, 'P is a positive functional on S (E). If 0 A (E)then,byProposition 2.6, there exists a sequence fsng 1
n=1 in S (E) such that 0 sn " f. By the
order continuity of P we then have 'P (sn) 1 " P (f), which implies that
there exists 2 R such that P (f) = 1. Consequently, 'P extends to a
positive functional on A (E), denoted by 'P as well, such that
P (f) ='P (f) 1 (29)
holds for all f 2 A (E). Since P is order continuous and A (E) regularly
embedded in A, it follows that 'P is order continuous and strictly positive
on A (E).
We assume that E is a complete Boolean subalgebra of CA such that
CA Sand CA = S (CB [E). We will denote by K (CB E) the collection of
all f 2 A which can be written as
f =
nX
i=1
qifi (30)
where qi 2 CB, fi 2 A (E) for all i = 1:::n and n 2 N. It is clear that
K (CB E) is a subalgebra of A with CB K (CB E) and E A (E)
43

K (CB E). If f 2 K (CB E), then it is easy to see that f can be written
as is (30) where fqig n
i=1 a disjoint system in CB with Wn i=1 qi = 1 and fi 6= fj
whenever i 6= j. This will be called the standard form of f. We claim that
this standard form of an element f 2 K (CB E) is unique. First we observe
that if p q 2 CB and g h 2 A (E) such that pg = qh 6= 0, then p = q and
g = h. Indeed, using the averaging property of P and (29) we nd that
'P (g) p = pP (g) =P (pg) =P (qh) =qP (h) ='P (h) q.
Since 'P (g) p 6= 0,itfollows that p = q. Hence, p (g ; h) = 0, so p jg ; hj =
0, which implies that
'P (jg ; hj) p = pP (jg ; hj) =P (p jg ; hj) =0.
Therefore 'P (jg ; hj) =0,asp 6= 0, and since 'P is strictly positive wemay
conclude that g = h. From this observation the uniqueness of the standard
form of elements in K (CB E) follows by a standard argument. For future
reference we collect some properties of K (CB E) inthe next lemma.
Lemma 6.3 With the notation as introduced above, the following hold.
(i). K (CB E) is an f-subalgebra of A.
(ii). The order closure of K (CB E) in A is equal to A.
Proof.
(i). As observed already, K (CB E) is a subalgebra of A. Writing f 2
K (CB E) in standard form f = P n
i=1 qifi, it follows that
jfj = P n
i=1 qi jfij and so jfj 2K (CB E).
(ii). Let L = K (CB E) (o)
be the order closure of K (CB E) in A. Then L
is regularly embedded in A and L is Dedekind complete. Hence the
Boolean algebra CL of components of 1 in L is a complete Boolean
subalgebra of CA. Since CB CL and E CL, this implies that CA =
S (CB [E)=CL. Now it follows from the Freudenthal spectral theorem
that L = A, as L is a complete Riesz subspace of A.
Now suppose that A 1 and A 2 are two Dedekind complete f-algebras with
unit elements 1A1 and 1A2 respectively. For i = 1 2, let 1Ai 2 Bi Ai be
an f-subalgebra and let Pi : Ai ! Ai be a strictly positive order continuous
projection onto Bi. Furthermore we assume that Ei CAi is a complete
44

Boolean subalgebra such that CAi = S (CBi [Ei) and that i : Ei ! [0 1]
is a completely additive measure satisfying Pi (e) = i (e) 1Ai for all e 2 Ei.
In addition to this, we assume in the following proposition that the unit
elements 1Ai are actually strong order units in Ai (i =1 2).
Proposition 6.4 In the above situation, suppose that:
(i). : A 1 (E 1) ! A 2 (E 2) is an f-algebra homomorphisms with (1A1) =
1A2and 2 ( (e)) = 1 (e) for all e 2E 1
(ii). : B 1 ! B 2 is an order continuous f-algebra homomorphisms with
(1A1) =1A2.
Then there exists an f-algebra homomorphism :A 1 ! A 2 such that:
(a). jA1(E1) = and jB1 =
(b). P 1 = P 2 .
Moreover, is order continuous and unique. If and are surjective isomorphisms,
then is a surjective isomorphism as well.
Proof. First we show that an f-algebra homomorphism : A 1 ! A 2
satisfying (a) and (b) is necessarily order continuous. To this end suppose
that f # 0 in A 1 and that (f ) g 0 for all and some g 2 A 2. Then
P 2 (f ) P 2g 0forall and
P 2 (f )= P 1 (f )= P 1 (f ) # 0
since P 1 and are order continuous. Therefore P 2g =0,which implies that
g = 0, as P 2 is strictly positive. This shows that is order continuous. To
prove the uniqueness, assume that 0 : A1 ! A 2 is an f-algebra homomorphism
satisfying (a) and (b) as well. De ne
L = ff 2 A 1 : (f) = 0 (f)g .
Then L is an f-subalgebra of A 1 and the order continuity of and 0 implies
that L is order closed. Furthermore, it is clear that B 1 and A 1 (E 1) are
contained in L, hence K (CB1 E 1) L. It follows from Lemma 6.3 that
L = A 1, consequently = 0 .
Now we turn to the existence proof of the homomorphism . For i =1 2,
let 0 'Pi 2 Ai (Ei) n be the strictly positive functional satisfying Pi (f) =
45

'Pi (f) 1Ai for all f 2 Ai (Ei) (see (29)). Since 2 ( (e)) = 1 (e) for all
e 2E 1 it follows that 'P2 ( (f)) = 'P1 (f) for all f 2 A 1 (E 1). We de ne
by
0
0 : K (CB1 E 1) ! K (CB2 E 2)
nX
i=1
qifi
!
=
nX
i=1
(qi) (fi) (31)
for all fqig n
i=1 in CB1 and all ffig n
i=1 in A 1 (E 1). First we observe that 0 is
well de ned. Indeed, if f = P n
i=1 qifi has standard form f = P m
j=1 pjgj, then
it is easy to see that
nX
i=1
(qi) (fi) =
mX
j=1
(pj) (gj).
Since the standard form of an element inK (CB1 E1) is unique, it follows that
0 is well de ned by (31). Since and are f-algebra homomorphisms it
is now also clear that 0 is an f-algebra homomorphism.
Take f 2 K (CB1 E1) and write f = Pn i=1 qifi with fqig n
i=1 in CB1
ffig
and
n
i=1 in A1 (E1). Then
P 1f =
nX
i=1
P 1 (qifi) =
nX
i=1
qiP 1 (fi) =
nX
i=1
'P1 (fi) qi
and so P 1 [K (CB1 E 1)] K (CB1 E 1). Moreover, using that 'P2 ( (fi)) =
'P1 (fi), we nd that
0 (P 1f) =
=
nX
i=1
nX
i=1
'P1 (fi) (qi) =
(qi) P 2 ( (fi)) =
= P 2
nX
i=1
nX
i=1
nX
i=1
(qi) (fi)
'P2 ( (fi)) (qi)
P 2 ( (qi) (fi))
!
= P 2 ( 0f) :
This shows that 0 P 1 = P 2 0.
Let K 1 be the 1A1-uniform closure of K (CB1 E 1)inA 1. Then 0 extends
uniquely to an f-algebra homomorphism 1 : K 1 ! A 2. Since CB1 K 1 and
1 (q) = 0 (q) = (q) for all q 2CB1, it follows from Freudenthal's spectral
46

theorem that B 1 K 1 and 1 (g) = (g) for all g 2 B 1. Note that at this
point we use that 1A1 is a strong order unit in A 1, which guarantees that
every f 2 B 1 is a 1A1-uniform limit if a sequence in S (CB1). Moreover, it is
easy to see that 1 P 1 = P 2 1.
Now let K be the collection of all pairs (K K), where:
(i). K is an f-subalgebra of A 1 such that K 1
(ii). K : K ! A 2 is an f-algebra homomorphism such that
(iii). K P 1 = P 2 K.
( K) jK1 = 1
We note that the reason for introducing the uniformly closed f-subalgebra
K1 is that B1 K1 and K1 K imply that P1 (K) K and so (iii) makes
sense. From the above observations it is clear that (K1 1) 2K,soKis nonempty.
If (K K) (K0 K0) 2K, then wewillsaythat (K K) (K0 K0) whenever K K0 and ( K0) jK = K. It is easy to see that every chain in the
partially ordered set (K ) has an upper bound. Hence, by Zorn's lemma,
(K ) contains a maximal element (Km m). It is our aim to show that
Km = A1. Suppose that fg : 2 Tg is a net in Km such that g
(o)
;! 0 in A1. We
claim that m (g ) (o)
;! 0 in A 2. Without loss of generality we may assume
that fg g is order bounded in A 1. Since 1A1 is a strong order unit in A 1 and
m (1A1) =1A2, it is clear that f m (g )g is order bounded in A 2. For 2 T
de ne
K
u = _ jg j h = _ j m (g )j
and assume that h h 0 for all 2 T and some h 2 A 2. Fix and de ne
for any nite subset F of f : g the element hF = W 2F j m (g )j. Then
P 2hF = P 2 m
_
2F
jg j
!
= mP 1
_
2F
jg j
!
= P 1
Since hF "F h and W 2F jg j "F u , and since , P 1 and P 2 are order
continuous, it follows that P 2h = P 1u . Hence P 1u P 2h 0 for all .
By hypothesis we have u # 0 in A 1 and so P 1u # 0 in A 2. This implies
that P 2h = 0 and since P 2 is strictly positive we may conclude that h = 0,
by which our claim is proved.
47
_
2F
jg j
!
:

Now we are in a position to apply Lemma 2.14, from which it follows that
m has a unique extension
0
m : K 0 m ! A 2
with the property that m (f ) (o)
;! 0 (f) inA 2 whenever f 2 K 0 m and ff g
is a net in Km such that f
(o)
;! f in A 1. It is straightforward to verify that
K 0 m is an f-subalgebra of A 1 and that 0 m is an f-algebra homomorphism.
Hence the pair (K 0 m 0 m) satis es (i) and (ii) above. To show that condition
(iii) is satis ed as well, takeany f 2 K 0 m and let ff g be a net in Km such that
f
(o)
;! f in A1. By the order continuity of P1 of follows that P1f ;! P1f and so m (P1f ) (o)
;! 0 m (P1f). On the other hand, since mf
(o)
;! 0 and P2 is order continuous we nd that
mf
m (P1f )=P2 ( mf ) (o)
;! P2 (
0
mf)
consequently 0 m (P 1f) =P 2 ( 0 mf). This shows that 0 m P 1 = P 2
(o)
0
mand
so (K 0 m 0 m) 2K.
Since (K 0 m 0 m) is a maximal element ofK, we may conclude that K 0 m =
Km. Therefore, Km is order closed in A 1. By Lemma 6.3, the order closure
of K (CB1 E 1) is equal to A 1, so Km = A 1. Hence = m is the desired
f-algebra homomorphism.
Finally, suppose that and are surjective isomorphisms. Then we can
apply the above construction to ;1 and ;1 to obtain a corresponding falgebra
homomorphism 1 : A 2 ! A 1. Then it is clear that 1 ( f) =f for
all f 2 K (CB1 E 1). De ning
M = ff 2 A 1 : 1 ( f) =fg
it follows that M is an order closed f-subalgebra of A 1 with K (CB1 E 1) M.
Again using Lemma 6.3 we conclude that M = A 1, so 1 ( f) = f for all
f 2 A 1. Similarly we see that ( 1g) = g for all g 2 A 2, hence is a
surjective isomorphism. By this the proof of the proposition is complete.
7 The construction of special subalgebras
In this section we will discuss in some more detail the structure of positive
projections onto f-subalgebras. The construction presented in this section
are inspired by the results in [10], Section 14. We start by listing the hypotheses
which will be assumed throughout this section:
48

A is a Dedekind complete f-algebra with unit element 1 2 A
B is an f-subalgebra of A with 1 2 B
P : A ! A is a strictly positive order continuous projection onto B
A is nowhere full with respect to B.
As before, for any non-empty subset D CA we will denote by S(D) the
complete Boolean subalgebra of CA generated by D, i.e., S(D) is the smallest
complete Boolean subalgebra of CA containing the set D. The main objective
in the present section is to show that given any sequence fang 1
n=1 in CA, there
exists a complete Boolean subalgebra E of CA such that an 2 S(CB [E) for
all n and E S. Here S = SP is de ned by (28). It follows from Lemma 6.2
that the inclusion E Sis equivalenttosaying that there exists a completely
additive strictly positive measure : E ! [0 1] such that P (e) = (e)1 for
all e 2E. The proof will be divided in a number of lemmas.
Lemma 7.1 Let a p 2CA be given. Then there exist elements Hk (p a) 2CA
(k =0 1 2 3) suchthat
(i). H 0 (p a) ::: H 3 (p a) are mutually disjoint
(ii). H 0 (p a)+H 1 (p a) =ap
(iii). H 2 (p a)+H 3 (p a) =(1;a) p
(iv). P 3
k=0 Hk (p a) =p
1
(v). P (Hk (p a)) P (p) for k =01 2 3.
2
Proof. Since 0
that there exists H0 (p a) 2CA such that
1P
(ap) P (ap) in B, it follows from Theorem 5.4
2
H0 (p a) ap and P (H0 (p a)) = 1
P (ap):
2
De ne H 1 (p a) = ap ; H 0 (p a). Similarly there exists H 2 (p a) 2 CA such
that
H2 (p a) (1 ; a) p and P (H2 (p a)) = 1
P ((1 ; a) p):
2
De ne H 3 (p a) =(1;a) p ; H 2 (p a). It is clear that the elements Hk (p a)
satisfy all the requirements.
49

Observe that the elements Hk (p a) as constructed in the above lemma
are in general not uniquely determined by a and p. Before stating the next
lemma we introduce some notation. For n 2 N and 0 j 4 n ; 1wede ne
the intervals
Inj = j4 ;n (j +1)4 ;n :
For each n the intervals In0 ::: In4 n ;1 are a partition of I 00 = [0 1). Furthermore,
for all n 2 N and 0 j 4 n ; 1 we have
Inj =
3[
k=0
In+14j+k: (32)
We denote by An the algebra of subsets of [0 1) generated by the intervals
fInj :0 j 4n ; 1g. From (32) it is clear that An An+1 for all n. De ne
A1 = [ 1 n=0An, which is an algebra of subsets of [0 1). Let D denote the set
of all dyadic numbers in [0 1]. It is clear that A1 is equal to the algebra
generated by all intervals [ )with 2 D .
Let fang 1
be a n=1 xed sequence in CA.
Lemma 7.2 There exists a Boolean homomorphism h : A1 !CA such that:
(a). h(I 00) =1
(b). P [h (Inj)] 2 ;n 1 for all n 2 N and 0 j 4 n ; 1
(c). an 2 h(A1) for all n =1 2:::.
Proof. We de ne recursively Boolean homomorphisms hn : An ! CA
with (hn+1) jAn = hn for all n 2 N, as follows. The homomorphism h 0 :
A 0 ! CA is simply given by h(I 00) = 1 and h() = 0. Now assume that
hn : An ! CA has been de ned for some n 2 N. For 0 j 4 n ; 1 and
0 k 3we put
hn+1 (In+14j+k) =Hk (hn (Inj) an+1) : (33)
Using (i) and (iv) of Lemma 7.1 it is easy to see that this de nes aBoolean
homomorphism hn+1 : An+1 !CA such that (hn+1) jAn = hn. We claim that
P [hn (Inj)] 2 ;n 1 for all n 2 N and 0 j 4 n ; 1: (34)
Indeed, for n = 0 this is trivial. Furthermore, by (v) in Lemma 7.1 and (33)
we have
P [hn+1 (In+14j+k)]
50
1
2 P [hn (Inj)]

and (34) now follows by induction on n. Next we observe thatitfollows from
(33) and Lemma 7.1 (ii) that
=
4 n ;1 X
j=0
=
4 n ;1 X
j=0
hn+1
4
n
;1 [
j=0
In+14j [ In+14j+1
!
fhn+1 (In+14j)+hn+1 (In+14j+1)g
fH 0 (hn (Inj) an+1)+H 1 (hn (Inj) an+1)g
=
4 n ;1 X
j=0
hn (Inj) an+1 = h(I 00)an+1 = an+1
and so an+1 2 hn+1 (An+1) for all n 2 N.
De ning the Boolean homomorphism h : A1 ! CA by hjAn = hn for all
n 2 N it is clear that h has all the desired properties.
Lemma 7.3 There exists a function p : D !CA such that:
(i). p 0 =0and p 1 = 1
(ii). if 1 2 in D , then p 1 p 2 in CA
(iii). if 1 2 2 D such that 0 2 ; 1 < 4 ;n for some n 2 N, then
in CA
(iv). the sequence fang 1
n=1
in CA.
0 P (p 2) ; P (p 1) 2 ;n+1 1
belongs to the algebra generated by fp : 2 D g
Proof. Let h : A1 !CA be the Boolean homomorphism constructed in
the previous lemma. For 2 D we de ne p = h([0 )). It is clear that p
has properties (i) and (ii). Since h([ 1 2)) = p 2 ; p 1 for all 1 2 in D ,
the algebra generated by fp : 2 D g is equal to h(A1) and so (iv) follows
from (c) in the above lemma. It remains to proof (iii). If 1 2 2 D such
that 0 2 ; 1 < 4 ;n , then [ 1 2) Inj [ Inj+1 for some 0 j 4 n ; 1.
Hence,
0 p 2 ; p 1 = h([ 1 2)) h(Inj)+h(Inj+1)
51

and so by (b) in Lemma 7.2,
0 P (p 2) ; P (p 1) P [h(Inj)] + P [h(Inj+1)] 2 ;n+1 1:
Given the system fp : 2 D g as in Lemma 7.3 we denote by q( ) 2CB
the component of 1 in the band [P (p ) ; 1] + d for all 2 [0 1]. In the
following lemma we collect some of the relevant properties of the system
fq( ): 2 D 0 1g.
Lemma 7.4 (a). If q 2CB such that 0 q q( ), then qP(p ) q.
(b). If q 2CB such that 0 q.
(c). If 1 2 in D , then q( 1 ) q( 2 ) for all 0 1.
(d). If 0 1 2 1, then q( 1) q( 2) for all 2 D .
(e). q(0 ) = 1 for all 0 1, q(1 ) = 0 for all 0 < 1, and
q(1 1) = 1.
(f). q( 1) = 1 for all 2 D , and q( 0) is the component of 1 in fP (p )g d
for all 2 D .
(g). q( 0)p =0for all 2 D .
Proof. Properties (c), (d), (e) and (f) are obvious. To prove (a), take
q 2CB such that 0 q q( ). Then q ^ [P (p ) ; 1] + =0,so
[qP(p ) ; q] + = q [P (p ) ; 1] + =0
which shows that qP(p ) q.
Nowtake q 2CB such that 0
0. Moreover q [P (p ) ; 1] ; =0,so
q [P (p ) ; 1] =q [P (p ) ; 1] + > 0
and this proves (b).
Finally, for the proof of (g), observe that q( 0)P (p ) = 0, as q( 0) is
the component of 1 in fP (p )g d . Hence,
P (q( 0)p )=q( 0)P (p )=0
52

and since P is strictly positive itfollows that q( 0)p .
Now we de ne the system fe :0 1g in CA by
e = sup fq( )p : 2 D g : (35)
From (d) in Lemma 7.4 it is clear that e 1 e 2 whenever 0 1 2 1.
Since q( 0)p = 0 for all 2 D and since q(1 1)p 1 = 1, it follows that e 0 =0
and e 1 = 1. Our next objective is to show that P (e )= 1 for all 2 [0 1].
The proofisdividedin two lemmas.
Lemma 7.5 For all 0 1 we have P (e ) 1.
Proof. For = 1 the inequality is trivial, so we may assume that 0

Lemma 7.6 For all 0 1 we have P (e )= 1.
Proof. For =0and =1the inequality is trivial, so we may assume
that 0 <

Lemma 7.7 If 2 D and 0 < 1, then
[q ( ) ; q ( )] e p :
Proof. Let 2 D be xed. It follows from the de nition (35) that
[q ( ) ; q ( )] e =sup[q
(
2D
) ; q ( )] q ( ) p :
Since q ( ) q ( )if 2 D with , it follows that
[q ( ) ; q ( )] q ( )=q ( )[1 ; q ( )] q ( )=0
for all such 2 D . This implies that
[q ( ) ; q ( )] e = sup f[q ( ) ; q ( )] q ( ) p : 2 D g
and from this the lemma follows.
Given 2 D and a partition : 0 = 0 < 1 < ::: < n = 1 of the
interval [0 1] we de ne s 2CA by
s =
nX
k=1
Furthermore we denote j j =maxk ( k ; k;1).
[q ( k) ; q ( k;1)] e k;1 : (38)
Lemma 7.8 With the notation introduced above, we have
0 P (p ; s ) j j 1
for all 2 D and for every partition of [0 1].
Proof. Since
s =
n_
k=1
[q ( k) ; q ( k;1)] e k;1
it follows immediately from the Lemma 7.7 that 0 s p . As observed
in Lemma 7.4, q( n) =q( 1) = 1 and q( 0) =q( 0) is the component
of 1 in fP (p )g d , hence
P (p )=[1 ; q( 0)] P (p )=
nX
k=1
55
[q ( k) ; q ( k;1)] P (p ) :

Moreover, since q ( k);q ( k;1) q ( k) inCB, it follows from Lemma
7.4 (a) that
[q ( k) ; q ( k;1)] P (p ) k [q ( k) ; q ( k;1)]
for all k =1:::n. Therefore,
P (p )
nX
k=1
By Lemmas 4.1 and 7.6 we have
Consequently,
P (s )=
nX
k=1
nX
k=1
k [q ( k) ; q ( k;1)] :
k;1 [q ( k) ; q ( k;1)] :
0 P (p ; s )=P (p ) ; P (s )
( k ; k;1)[q ( k) ; q ( k;1)]
j j
by which the lemma is proved.
nX
k=1
[q ( k) ; q ( k;1)]
= j j [1 ; q( 0)] j j 1
Lemma 7.9 For all 2 D we have p 2 S (CB [fe :0 1g).
Proof. Let 2 D be xed. Take a sequence f ng 1
n=1
[0 1] such that j nj !0 as n !1and de ne s 2CA by
s =supfs n : n =1 2:::g :
of partitions of
From Lemma 7.7 and (38) it is clear that s n p for all n, so s p . Now
it follows from the above lemma that
0 P (p ; s) P (p ; s n) j nj 1
which implies that P (p ; s) =0. Since P is assumed to be strictly positive,
it follows that p = s. This shows that
p =supfs n : n =1 2:::g : (39)
56

Since q( ) 2 CB for all 0 1, it is clear from (38) that s n 2
S (CB [fe :0 1g) for all n. The result of the lemma now follows
from (39).
From the above lemma in combination with (iv) in Lemma 7.3 it follows
that
fang 1
n=1 S (CB [fe :0 1g) :
For later reference we summarize the above results in the following proposition.
Proposition 7.10 Let A, B and P satisfy the hypotheses stated at the beginning
of the present section. Given any sequence fang 1
n=1 in CA, there exists
a system fe :0 1g in CA such that:
(i). e 0 =0, e 1 = 1 and e 1 e 2 whenever 0 1 2 1
(ii). P (e )= 1 for all 0 1
(iii). fang 1
n=1 S (CB [fe :0 1g).
Let the system fe :0 1g be as in the Proposition 7.10 and let F
denote the Boolean subalgebra of CA generated by this system. It is easy
to see that F consists precisely of those elements in p 2 CA which can be
written as
p =
n_
k=1
with 0 1 1 2 2 ::: n n 1. Hence
P (p) =
nX
k=1
fP (e k ) ; P (e k )g =
(e k ; e k ) (40)
nX
k=1
( k ; k) 1
for all such p 2F. This shows that F S. Since F is a Boolean subalgebra
of CA it follows from Lemma 6.2 (i) that S(F) S. We denote E = S(F),
and we will write P (e) =
it is clear that
(e)1 for all e 2E. If p 2F is given by (40), then
(p) =
nX
k=1
( k ; k) (41)
Note that, by de nition, E is the complete Boolean subalgebra of CA generated
by fe :0 1g.
We denote by m the Lebesgue measure on [0 1] and let (Im) denote the
corresponding measure algebra.
57

Lemma 7.11 There exists a measure preserving Boolean isomorphism h
from I onto E.
Proof. Let R denote the subalgebra of I generated by all (elements
corresponding to) intervals [ ) with 0 1. De ne the Boolean
isomorphism h 0 : R ! F by h 0 ([ )) = e ; e . It follows immediately
from (40) and (41) that (h 0(R)) = m(R) for all R 2R, i.e., h 0 is measure
preserving. Since R is dense in I with respect to the metric induced by m
and F is dense in E with respect to the metric induced by , and since h 0 is
an isometry for with respect to these metrics, it now follows that h 0 extends
uniquely to a surjective measure preserving isomorphism h : I !E.
We collect the above results in the following proposition.
Proposition 7.12 Let A, B and P satisfy the hypotheses stated at the be-
ginning of the present section and let the sequence fang 1
n=1 in CA be given.
Then there exists a complete Boolean subalgebra E of CA such that:
(i). fang 1
n=1
S(CB [E)
(ii). there exists a strictly positive completely additive measure : E ! [0 1]
such that P (e) = (e)1 for all e 2E
(iii). the measure algebra (E ) is isomorphic with the measure algebra (Im)
of the Lebesgue measure on [0 1].
Remark 7.13 Observe that CB and E in the above proposition are independent
Boolean subalgebras (in the sense of [15], Section 28) of CA. Indeed, if
q 2CB and e 2E, then
P (qe) =qP(e) = (e)q:
Hence, if qe =0, then P (qe) =0and so q =0or e =0.
If there exists a sequence fang 1
n=1 in CA such thatCA = S (CB [fang 1
n=1 )
then we will say that CA is separable over CB (or, A is separable over B).
The following result is now an immediate consequence of Proposition 7.12.
Corollary 7.14 Let A, B and P satisfy the hypotheses stated at the beginning
of the present section and suppose that CA is separable over CB. Then
there exists a complete Boolean subalgebra E of CA such that:
(i). CA = S(CB [E)
58

(ii). there exists a strictly positive completely additive measure : E ! [0 1]
such that P (e) = (e)1 for all e 2E
(iii). the measure algebra (E ) is isomorphic with the measure algebra (Im)
of the Lebesgue measure on [0 1].
Hence, in the situation of the above corollary, CA is a Boolean product of
the independent Boolean subalgebras CB and E.
8 The structure of positive projections
In this section we assume again that A, B and P satisfy the hypotheses
stated at the beginning of Section 7. The main objective in this section is to
show that there exists a disjoint system fp g in CA with sup p = 1, such
that for each there exists a complete Boolean subalgebra E of CA (p )such
that:
(a). CA (p )=S (p CB [E ) in CA (p )
(b). there exists a strictly positive completely additive non-atomic measure
: E ! [0 1] such that P p (e) = (e)1 for all e 2E ,
where P p is the restriction of P to A (p ) as de ned in Lemma 4.14. We
follow the ideas of Maharam ([10]). First we introduce some terminology.
For any subset D of CA we denote by jDj the cardinal number corresponding
to D. The order of p 2CA over CB is de ned by
min fjDj : D C A such that CA(p) =pS (CB [D)g : (42)
Note that p 2 CA is of order 0 over CB if and only if CA(p) = pCB, i.e., if
and only if p is B-full (see De nition 4.7). If there exists a cardinal m such
that every 0 0. Under this
assumption we will show that there exists a complete non-atomic Boolean
subalgebra E of CA such that CA = S (CB [E) and E S, where S is given
by (28).
59

Remark 8.1 If 0 < p 2 CA has nite order over CB, then there exists 0 <
q p 2CA such that q is B-full. Indeed, let fpkg n
CA be such that
k=1
CA(p) =pS (CB [fpkg n
) : (43)
k=1
Replacing, if necessary, fpkg n
by a basis for the Boolean subalgebra gen-
k=1
erated by fpkg n
n
P we may assume that fpkg is a disjoint system with
k=1 k=1
n
k=1 pk = 1. Then it is easily veri ed that
S (CB [fpkg n
k=1 )=
( nX
k=1
pkqk : qk 2CBk =1:::n
Let 1 m n be such that q = ppm > 0. Take a 2CA(q) CA(p). It follows
from (43) that there exist qk 2CB (k =1::: n) such that a = p Pn k=1 pkqk.
Since a q pm, this implies that a = ppmqm = qqm 2 qCB, which shows
that CA(q) qCB. Hence q is B-full. From this observation it follows that,
if A is nowhere full with respect to B, then every 0 < p 2 CA has order at
least @0 over CB.
By the above remark, we may assume that A is homogeneous of order
m over B for some cardinal m @ 0. In case that m = @ 0, the desired
result has already been obtained in Corollary 7.14. Therefore in the proof of
Proposition 8.3 below we may assume that m > @ 0. But rst we prove the
following lemma.
Lemma 8.2 Let R 0 be a complete Boolean subalgebra ofCA such that R 0
S. Suppose that CA is nowhere fullwithrespect to S (CB [R 0) and let a 2CA
be given. Then there exists a complete Boolean subalgebra R 1 of CA such that:
(i). R 0
(ii). R 1
R 1 and a 2 S (CB [R 1)
S and R 1 is non-atomic
(iii). R1 is separable over R0, i.e.,there exists a sequence feng 1
n=1
that R1 = S (R0 [feng 1
n=1 ).
)
:
R 1 such
Proof. Considering the restriction of P to Ab, we may assume without
loss of generality that A = Ab, i.e., we may assume that the unit element 1
is a strong order unit in A. Now let
C = A (S (CB [R 0))
as de ned in Section 2. It follows from Proposition 2.6 (and the remarks made
at the beginning of Section 4) that C is complete f-subalgebra of A. Hence,
60

y Proposition 4.2 there exists a strictly positive order continuous projection
Q : A ! A onto C such that P = P Q. Now we apply Proposition 7.12
to A, C, Q and the constant sequence an = a for all n. Consequently there
exists a complete Boolean subalgebra E of CA such that a 2 S (CC [E), and
there exists a strictly positive completely additive measure : E ! [0 1] such
that Q(e) = (e)1 for all e 2 E, and (E ) is isomorphic with the measure
algebra (Im) of the Lebesgue measure on [0 1].
De ning R 1 = S (R 0 [E), it is clear the R 1 satis es (iii), as (Im) is a
separable measure algebra. Furthermore,
S (CC [E)=S (S (CB [R 0) [E)=S (CB [R 0 [E)=S (CB [R 1)
so R1 satis es (i) as well. For the proof of (ii) we denote by 0 : R0 ! [0 1]
the measure de ned by P (q) = 0(q)1 for all q 2R0. Let F be the Boolean
subalgebra of CA generated by R0 [E. Every p 2F is of the form
p =
n_
k=1
qkek (44)
with qk 2 R0, ek 2 E, k = 1:::n and n 2 N. It is easy to see that we
can take the elements qk in (44) mutually disjoint, so p = Pn k=1 qkek. Hence,
using Lemma 4.1 and that P = P Q, we nd
P (p) =P
= P
=
nX
k=1
nX
nX
k=1
k=1
Q (qkek)
qkQ (ek)
!
!
(ek) P (qk) =
= P
= P
nX
k=1
nX
nX
k=1
k=1
qkQ (ek)
(ek) qk
!
!
(ek) 0 (qk) 1:
(45)
This shows that for every p 2F we have P (p) = 1 for some 2 [0 1] and
so F S. Since F is a Boolean subalgebra of CA, it follows from Lemma
6.2 that R1 = S(F) S. Hence, for every p 2 R1 we have P (p) = 1(p)1
for some 1(p) 2 [0 1] and this de nes a strictly positive completely additive
measure 1 : R1 ! [0 1]. Finally, using that is non-atomic on E, it
follows easily from (45) that for every p 2Fthere exist p1p2 2Fsuch that
p1 + p2 = p, p1 ^ p2 = 0 and 1 (p1)= 1 (p2)= 1
2 1 (p). This implies that 1
is a non-atomic measure and hence, R1 is a non-atomic Boolean subalgebra
of CA. By this the proof of the lemma is complete.
Now we are in a position to prove one of the main results of the present
section.
61

Proposition 8.3 Assume that A, B and P satisfy the hypotheses stated at
the beginning of Section 7. Furthermore we assume that A is homogeneous
over B of order m for some cardinal m > 0. Then there exists a complete
Boolean subalgebra E CA such that:
(i). CA = S (CB [E)
(ii). there exists a strictly positive completely additive non-atomic measure
: E ! [0 1] such that P (e) = (e)1 for all e 2E.
Proof. As observed after Remark 8.1, we may assume that m > @ 0. Let
D = fd : 2 A g be a subset of CA such thatjA j = m and CA = S (CB [D).
Identifying m with the initial ordinal associated with m, we will write A =
f :1 @0, it follows in particular that jD j < m. If 0

S (CB [E ). Moreover, there exists a sequence feng 1
n=1 E such that
E = S (R [feng 1
n=1
). It is clear that E satis es (a), (b) and (c). De ne
G = D [feng 1
n=1 . Since jD j max fj j @ 0g, it is obviously that jG j
max fj j @ 0g. Furthermore,
E = S (R [feng 1
1
)=S (S (D ) [feng n=1 n=1 )
= S (D [feng 1
)=S (G )
n=1
which shows that (d) is satis ed as well. The construction of the collection
fE : 2 A g is complete.
De ning
F = [
E ,
it follows from (a) that F is a Boolean subalgebra of CA and by (c) we have
F S. Lemma 6.2 implies that E = S (F) S. Hence, for every e 2Ethere
exists (e) 2 [0 1] such thatP (e) = (e)1, andby Lemma 6.2, : E ! [0 1]
is a completely additive strictly positive measure. Since is non-atomic on
F, a similar argument as used in the proof of Lemma 8.2 shows that is
non-atomic on E as well. Finally, it follows from (b) that D S (CB [F),
so
CA = S (CB [D) S (CB [F)=S (CB [E),
which shows that CA = S (CB [E) and the proof is complete.
Now we will drop the assumption that A is homogeneous over B. In the
proof of the next lemma the following simple observation will be used.
Lemma 8.4 Let 0 < p 0 2 CA be xed. If p 2 CA(p 0), then the order of p
over CB is equal to the order of p over p 0CB in CA(p 0). In particular, given
any cardinal m, the following two statements are equivalent:
(i). every 0

for any non-empty subset H CA. Given p 2CA(p 0), let m be the order of p
over CB and let m 0 denote the order of p over p 0CB in CA(p 0). Take D CA
such that jDj = m and CA(p) =pS (CB [D). Using (46) it follows that
CA(p) =pp 0S (CB [D)=pS 0 (p 0CB [ p 0D)
which shows that m 0 jp 0Dj m. Now take D 0 CA(p 0) such that
jD 0j = m 0 and CA(p) =pS 0 (p 0CB [D 0). Using (46) once more, we nd that
CA(p) =pS 0 (p 0CB [D 0)=pp 0S (CB [D 0)=pS (CB [D 0)
and so m jD 0j = m 0. We may conclude that m = m 0 and this su ces to
prove the lemma.
Lemma 8.5 Assume that A, B and P satisfy the hypotheses stated at the
beginning of Section 7. Then there exists a disjoint system fp g in CA such
that:
(i). each A(p ) is homogeneous of order m > 0 over B p (equivalently,
CA(p ) is homogeneous of order m > 0 over p CB)
(ii). P p = 1.
Proof. By Zorn's Lemma there exists a maximal disjoint system fp g
with property (i). We will show that P p = 1. Put p 0 = 1 ; P p and
suppose that p 0 > 0. De ne
m 0 =minfm : 90 0 and actually
m 0 @ 0, by Remark 8.1. Take 0

Proposition 8.6 Assume that A, B and P satisfy the hypotheses stated at
the beginning of Section 7. Then there exists a disjoint system fp g in CA with
sup p = 1, such that for each there exists a complete Boolean subalgebra
E of CA (p ) such that:
(a). CA (p )=S (p CB [E ) in CA (p )
(b). there exists a strictly positive completely additive non-atomic measure
: E ! [0 1] such that P p (e) = (e)p for all e 2E .
Proof. Let the disjoint system fp g be as in Lemma 8.5. For each we
can now apply Proposition 8.3 to A(p ), B p and P p , which immediately
yields the result of the proposition.
9 Representations of positive projections:
special case
In this section we will obtain a special case of the representation of positive
projections, which will be one of the essential building blocks for the general
case, treated in the next section. Before we start, recall some of the notation
introduced in Section 3. Given two measurable spaces (X ) and (Y ) we
denote =X Y and F = . As before, we denote by M ( F) the
f-algebra of all real-valued F-measurable functions on and Mb ( F) is
the f-subalgebra of all bounded functions in M ( F). Recall furthermore
that M (X ) is identi ed with an f-subalgebra of M ( F).
Now we assume in addition that is a probability measure on (Y ) and
we de ne the linear operator Rb : Mb ( F) ! Mb ( F) by
Rbf (x y) =
Z
Y
f (x z) d (z) (47)
for all (x y) 2 and all f 2 Mb ( F), which is a positive -order continuous
projection onto Mb (X ). Furthermore suppose that B is a Dedekind
complete f-algebra in which the unit element 1 is a strong order unit and
that (X ) is a representation space for B with corresponding representation
homomorphism B : Mb (X ) ! B. De ne
S : Mb ( F) Rb
B
;! Mb (X ) ;! B,
i.e., S = B Rb, where Rb is the projection given by (47). Clearly, S is a
linear -order continuous positive operator onto B. Denote by NS the null
65

ideal of S, i.e.,
NS = ff 2 Mb ( F) :S jfj =0g ,
which is a -ideal as well as an algebra ideal in Mb ( F). Let E be the
corresponding quotient space,
E = Mb ( F) NS
and let Q : Mb ( F) ! E be the quotient f-algebra homomorphism. Then
E is Dedekind -complete and Q is -order continuous. The unit element in
E will be denoted by 1E, which is also a strong order unit in E. Furthermore
we de ne the f-subalgebra F of E by
F = Q [Mb (X )] .
Since NS Ker (S) we can de ne the linear mapping
S : E ! B (48)
by S (Qf) = Sf for all f 2 Mb ( F). In the next lemma we collect some
properties of the operator S.
Lemma 9.1 The operator S in (48) has the following properties.
(i). S is a strictly positive, -order continuous and surjective linear operator.
(ii). The restriction SjF : F ! B is an f-algebra isomorphism onto B.
Proof. The proof of (i) is easy and therefore omitted. For the proof of
(ii), we rst show that SjF is injective. Suppose that h 2 F is such that
Sh =0. Take g 2 Mb (X ) such that Qg = h. Then
S jgj = BRb jgj = B jgj = j Bgj = j BRbgj = jSgj = Sh =0,
so g 2 NS. This implies that h = Qg =0. Next weshowthatSjF is surjective
with a positive inverse. If 0 b 2 B, then there exists 0 g 2 Mb (X )
such that Bg = b. Hence, 0 Qg 2 F and
S (Qg) =Sg = b (Rbg) = bg = b.
Now it is clear that SjF is an f-algebra isomorphism from F onto B.
66

Since B is Dedekind complete, it is clear that F is Dedekind complete
as well. We claim that F is regularly embedded in E. Indeed, suppose that
ff g is a net in F such that f # 0inF and let g 2 E be such thatf g 0
for all . Since SjF is a Riesz isomorphism from F onto B, we have Sf # 0
in B and Sf Sg 0 for all , so Sg = 0. Hence, g = 0, as S is strictly
positive, which shows that f # 0 in E.
Next we introduce a second f-subalgebra G of E, de ned by
G = Q [Mb (Y )] ,
where, as before, Mb (Y ) is identi ed with the f-subalgebra 1X Mb (Y )
of Mb ( F). For any f 2 Mb (Y ) we have Rbf = ;R fd 1 and so
Y
Sf =
Z
Y
fd 1 2 B: (49)
In particular, f 2 NS if and only if R jfj d = 0, i.e., if and only if f 2
Y
Mb (N ), where N denotes the -ideal in of all -null sets. Consequently,
G is precisely the quotient space Mb (Y ) Mb (N ), as usual denoted by
L1 ( ). The integral with respect to on G = L1 ( ) will be denoted by ' ,
i.e., ' (Qf) = R Y fd for all f 2 Mb (Y ), and the corresponding measure
on CG will be denoted by , i.e., (Q1V )= (V ) for all V 2 . It follows in
particular that G is Dedekind complete and order separable. Moreover, it is
now easy to see that G is regularly embedded in E.
We denote, as before, by CE, CF and CG the Boolean algebras of components
of 1E in E, F and G respectively. Since F and G are Dedekind complete
and regularly embedded in E, it follows that CF and CG are both complete
Boolean subalgebras of CE. Observing that CE = fQ (1W ):W 2Fg,
CF = fQ (1U) :U 2 g and CG = fQ (1V ):V 2 g the next lemma follows
immediately.
Lemma 9.2 The Boolean -algebra generated by CF and CG is equal to CE
and hence CE = S (CF [CG).
Now we de ne the positive linear operator PE : E ! E by
so PE = ; SjF
;1
S.
PE : E S
;1
(SjF)
;! B ;! F ,! E
Lemma 9.3 The operator PE is a strictly positive -order continuous projection
in E onto the f-subalgebra F . Moreover, PE (Qf) =Q (Rbf) for all
f 2 Mb ( F).
67

Proof. The rst statement of the lemma follows immediately from
Lemma 9.1 and from the fact that F is regularly embedded in E. To prove
the second statement, let f 2 Mb ( F) be given. Then S (Qf) = Sf =
B (Rbf) and S (QRbf) = S (Rbf) = B (Rbf). Since Q (Rbf) 2 F , this
;1
implies that PE (Qf) = ; SjF S (Qf) =Q (Rbf).
Next we observe that, if g 2 G and we write g = Qf with f 2 Mb (Y
then it follows from the above lemma that
),
PEg = Q (Rbf) =Q
Z
Y
fd 1 = ' (g) 1E,
where 0 ' 2 G n is the integral on G = L1 ( ). In particular, PE (p) =
(p) for all p 2CG.
Nowwe assume that A is a Dedekind complete f-algebra in whichtheunit
element 1 is a strong order unit and that P : A ! A is a strictly positive
order continuous projection onto the f-subalgebra B A with 1 2 B.
Furthermore, let E CA be a complete Boolean subalgebra such that CA =
S (CB [E) and suppose that : E ! [0 1] a completely additive measure
such that P (e) = (e) 1 for all e 2 E. Let (Y ) be a representation
space for A (E) with representation A(E) : Mb (Y ) ! A (E). De ning
(V )= ; A(E)1V for all V 2 , it is clear that is a probability measure
on (Y ) and that N = N A(E) . Consequently, Ker ; A(E) = Mb (N ) and
so A(E) induces a surjective f-algebra isomorphism : L1 ( ) ! A (E)
satisfying ( (p)) = (p) for all p 2 CL1( ), where we denote by the
measure induced by on CL1( ).
Suppose that (X ) is a representation space for B with corresponding
homomorphism B : Mb (X ) ! B. Now we can apply the construction
discussed in the rst part of the present section to (Y ), (X ) and B.
We will use the same notation as introduced before. Moreover, we assume
in addition that A is order separable. This implies that B is order separable
as well. By Lemma 9.1, the operator S : E ! B is strictly positive and so
it follows from Lemma 2.7 that E is order separable. Since E is Dedekind
-complete, it follows that E is actually Dedekind complete and that the
positive projection PE : E ! E is order continuous.
Collecting the above, we are now in the following situation:
A is a Dedekind complete order separable f-algebra in which the unit
element 1 is a strong order unit, P : A ! A is an order continuous
strictly positive projection onto the f-subalgebra B A with 1 2 B
E is a complete Boolean subalgebra of CA such that CA = S (CB [E)
and there is a completely additive measure : E ! [0 1] such that
P (e) = (e) 1 for all e 2E
68

E is a Dedekind complete f-algebra in which the unit element 1E is a
strong order unit, PE : E ! E is an order continuous strictly positive
projection onto the f-subalgebra F E with 1E 2 F
CG is a complete Boolean subalgebra of CE such that CE = S (CF [CG)
and there is a completely additive measure : CG ! [0 1] such that
PE (p) = (p) 1E for all p 2CG, where G = L1 ( )
: G ! A (E) is a surjective f-algebra isomorphism with (1E) = 1
such that ( (g)) = (g) for all p 2 CG and = SjF : F ! B is a
surjective f-algebra isomorphism with (1E) =1.
Consequently, we are in a position to apply Proposition 6.4. Hence, there
exists a unique surjective f-algebra isomorphism : E ! A such that
jG = , jF = SjF and PE = P . Recalling that E is the quotient
space Mb ( F) NS with corresponding -order continuous quotient
homomorphism Q : Mb ( F) ! E, we de ne
A : Mb ( F) ! A
by A = Q. Then A is a -order continuous f-algebra homomorphism
from Mb ( F) onto A with A (1 )=1. Therefore, ( F) is a representation
space for A with representation homomorphism A. In the next lemma
we collect some properties of this representation A. Recall that we identify
Mb (X ) and Mb (Y ) with f-subalgebras of Mb ( F).
Lemma 9.4 In the above situation the following hold:
(i). the representation A is compatible with B, i.e., ( A) jMb(X ) = B
(ii). A [Mb (Y )] =A (E)
(iii). P A = A Rb.
Proof. If f 2 Mb (X ), then Qf 2 F and so A (f) = (Qf) =
S (Qf) = Sf = B (Rbf) = Rbf, which proves (i). Since jG = ,
Q [Mb (Y )] = G and (G) = A (E) it is clear that (ii) holds. For the
proof of (iii), recall from Lemma 9.3 that PE Q = Q Rb, which implies
that
P A = P Q = PE Q = Q R = A Rb,
and we are done.
A combination of the above results with Proposition 8.3 immediately
yields the following theorem.
69

Theorem 9.5 Let A be a Dedekind complete order separable f-algebra in
which the unit element 1 is a strong order unit and let P : A ! A be a
strictly positive order continuous projection onto the f-subalgebra B A
with 1 2 B. Suppose that A is homogeneous over B and nowhere B-full. Let
(X ) be a representation space for B with representation homomorphism
B : Mb (X ) ! B. Then there exists a non-atomic probability space
(Y ) such that (X Y ) is a pure representation space for P and
A with representation A : Mb (X Y ) ! A, which is compatible with
B.
10 Representations of positive projections:
general case
We start this section with an extension of Theorem 9.5.
Proposition 10.1 Let A be a Dedekind complete order separable f-algebra
in which the unit element 1 is a strong order unit and let P : A ! A be
a strictly positive order continuous projection onto the f-subalgebra B A
with 1 2 B. We assume furthermore that A is nowhere B-full. Let B :
Mb (X ) ! B be a representation of B.
Then there exists a non-atomic - nite measure space (Y ) such that
(X Y ) is a representation space for P and A such that the corresponding
representation A : Mb (X Y ) ! A is compatible with B.
Proof. Applying Proposition 8.6, let fp g be a disjoint system in CA
with the stated properties. Since A is order separable, fp g is at most countable
and we will denote this system by fpn : n =12:::g with corresponding
Boolean algebras fEng and measures f ng. Now we apply Theorem 9.5
to the restrictions P pn : A (pn) ! A (pn) and the induced representations
pn
B : Mb (X ) ! B pn. Consequently, for each n there exists a non-atomic
- nite measure space (Yn n n) such that(X Yn n) is a representation
space for P pn and A (pn), such that the corresponding representation
n : Mb (X Yn n) ! A (pn) is compatible with
pn
B
. Finally, an
application of Lemma 4.15 nishes the proof.
Acombination of Proposition 4.12, Lemma 4.15, Example 4.16 and Proposition
10.1 yields the following theorem.
Theorem 10.2 Let A be a Dedekind complete order separable f-algebra in
which the unit element 1 is a strong order unit and let P : A ! A be a
strictly positive order continuous projection onto the f-subalgebra B A
with 1 2 B. Let B : Mb (X ) ! B be a representation of B.
70

Then there exists a - nite measure space (Y ) such that (X Y )
is a representation space for P and A, where the corresponding representation
A : Mb (X Y ) ! A is compatible with B.
As already observed at the end of Section3, the next result is now an
immediate consequence of Lemma 3.9.
Theorem 10.3 Let L be a Dedekind complete order separable Riesz space
with weak order unit 0 w 2 L and let P : L ! L be a strictly positive order
continuous projection onto the Riesz subspace K L with w 2 K. Suppose
that (X ) is a representation space forK with corresponding representation
K : bK ! K.
Then there exists a - nite measure space (Y ) such that (X Y )
is a representation space forP and L, where the corresponding representation
L : bL ! A is compatible with K.
Proof. Equip Lw with the f-algebra structure for which w is the unit
element. Then Kw is an f-subalgebra of Lw and so we may apply the above
theorem to the restriction Pw = PjLw : Lw ! Lw. We only need to refer to
Lemma 3.9 to nish the proof.
Next we will consider the extension of the above result to the situation
in which L is locally order separable. Moreover, we will drop the assumption
that L has aweak order unit.
Theorem 10.4 Let L be a Dedekind complete locally order separable Riesz
space and suppose that P : L ! L is a strictly positive order continuous
projection onto the Riesz subspace K L with K d = f0g. Then there exists
a disjoint system fQ g of band projections in L such that W Q = I and
(i). PQ = Q P for all
(ii). de ning L = Ran (Q ) and P = PjL : L ! L , there exists a
representation space (X Y ) for P and L .
Proof. For 0 u 2 L we will denote by L (u) the band generated by u
and if 0 u 2 K, then K (u) will denote the band generated by u in K. As
observed in Lemma 3.6, K is a complete Riesz subspace of L and this implies
that K (u) =L (u) \ K for all 0 u 2 K. We claim that K is locally order
separable. Indeed, take 0

Let fw g be a maximal disjoint system in K consisting of order separable
elements, so K (w ) is order separable for all . Since K is a complete Riesz
subspace of L and K d = f0g it follows that fw g be a maximal disjoint system
in L as well. Denoting by Q the band projection in L onto L (w ), we have
sup Q = I. Next we will show that Q P = PQ for all . To this end take
0 f 2 L (w ). Then f = sup n f ^ nw , so Pf = sup n P (f ^ nw ) and
this shows that Pf 2 L (w ). Hence, P leaves L (w ) invariant. Using that
sup Q = I and that P is order continuous, it follows easily that P leaves
L (w ) d invariant as well. Consequently, Q P = PQ . Let P : L (w ) !
L (w ) be the restriction of P . Then P is a strictly positive order continuous
projection in L (w ) onto K (w ). By Lemma 2.7, L (w )isorder separable.
Application of Theorem 10.3 to P nishes the proof.
References
[1] C. D. Aliprantis and O. Burkinshaw, Positive Operators, Academic
Press, Orlando, 1985.
[2] C. B. Huijsmans and B. de Pagter, Subalgebras and Riesz subspaces of
an f-algebra, Proc. London Math. Soc., 48 (1984), 161-174.
[3] C. B. Huijsmans and B. de Pagter, Averaging Operators and Positive
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