2) ** is P-integrable for all f e L ( mZ) and z Z,
and = fdP.
so
3) If p is a lifting of L (P), we can choose (B ) uniquely
0 0
P-a.s. such that p[Bz] = Bz, i.e., for all A E F, x c E, z0 E Z,
0 0 0
we have 1A L (P), and p(1 A) = p(A). If, in
addition, there exists a>0 such that |mZ S aP, then we can choose
0 0 0 0
Bz uniquely everywhere such that p(B ) = B, i.e., E L (P)
0 0
for all x E E, z0 E Z, and p() = for all x,z0.
4) If one of the conditions in 3(a) or 3(b) is satisfied, then
B takes values in L(E,F).
z
Now, in particular, taking i = 1A, A c F in (1) we obtain
1') ImZI(A) = E(1AIBl ) for A E F.
Also, taking first f E x, x E E, and then f = x1A, A e F, we get
0
2') is Integrable for x e E, z0 E Z, and
= = f****dP = E(1 ), for
A E F, x E E, z0 E Z.
(Notice also that from (1'), if B is bounded, the condition
ImZI S aP in (3) is satisfied.)
From (1') and the inequality ImI Zl Imlz we obtain
E(1AIBj) = ImZI(A) ImlZ(A) = Iml(RzxA) E(J5R xAdVu) = E(1AVz),
z
0 0
i.e., E(1AIBz ) S E(1AV ) for all A e F; hence IB I 5 V P-a.s.
2
Let z = (s,t), z' = (s',t') be points in R z < z'. Denote by
D the set R ,\ R and by R the rectangle ((s,t),(s',t')].
zz Z a zz
**