Representations of positive projections 1 Introduction - Mathematics ...

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)]