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 > = <mz (A)x,z > <m (A)x,z0> =

 E(1A<Bz x,z >) E(1 <B X,0 >) = 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 = <B x,z >1A <Bzx,z01 A E L(P)



 a 0 0
since each term is. Also, p(<(B Bz)x,zo> A) = p(<Bzx,zo>1A -

 0 0 0 0
<Bzx,z > ) = p(<Bzx,z >lA) p(<Bzx,z>1) = <Bx,0 >1p(A) -

 0 0 0 0 0 0 0
<BX, 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