We have m(D ,xA) m((R \ R )xA) = m((R ,xA)' (R xA)) =
zz z z z z
z z
m(R .xA) m(R xA) = m (A) mZ(A). Likewise, m(R zxA) =
S zzz
m (A) m 't(AA) (A) + mSt(A). Then for x E E, z0 E Z,
we have <(m -m )(A)x,z > = =
E(1A) E(1 **) = E(1 <(B ,-Bz)x, >). Tne same
s't' s't st' st
computation gives <(m m m + m )(A)x,z0>
0 0
E(1A<(AR (BO))x,z0>). Now, since p[B st] = Bs,, etc., we have
zz
0 0 0 0 a 0
p[B z B 3 B B and p[A (B )] = A (B ). In fact (we
z z z z B B ,
zz zz
give the proof for the first; the second is the same), for A E F,
x E E, z0 E Z, we have
0 0 0 0
<(Bz, Bz)x,z >1A = ****1A A) = p(1A -
0 0 0 0
) = p(lA) p(1) = 1p(A) -
0 0 0 0 0 0 0
1(A) = <(B B )xz> 1 ), so p[B B ] = B, Bz'
o
and the same for AR (B ). Then, by Proposition 1.5.6, both
zz
0 0 0 0 0
IB, Bz and AR (B )1. are P-measurable. Also, IBz Bzi
zz
0 0 0 o
SB ,zI + IB | I V + Vz 2V ,; hence |B B |z is P-integrable;
similarly, |AR (B )I | 4Vz,; hence JAR (B ) is P-integrable as
zz zz
0 0
well. Also, by properties of lifting, <(B B )x,z> and
0
<(AB (B ))x,z0> are measurable for x E E, z0 E Z (see property 2
zz
**