61.1. CONDITIONAL EXPECTATION IN BANACH SPACES 1987

Thus, if A ∈ G , ∫A

ZndP =m

∑k=1

xk

∫A

E(XFk |G

)dP

=m

∑k=1

xk

∫AXFk dP

=m

∑k=1

xkP(Fk ∩A) =∫

AXndP (61.1.5)

Then since E(XFk |G

)≥ 0,

||Zn|| ≤m

∑k=1||xk||E

(XFk |G

)Thus if A ∈ G ,

E (||Zn||XA) ≤ E

(m

∑k=1||xk||XAE

(XFk |G

))=

m

∑k=1||xk||

∫A

E(XFk |G

)dP

=m

∑k=1||xk||

∫AXFk dP = E (XA ||Xn||) . (61.1.6)

Note the use of ≤ in the first step in the above. Although the Fk are disjoint, all that isknown about E

(XFk |G

)is that it is nonnegative. Similarly,

E (||Zn−Zm||)≤ E (||Xn−Xm||)

and this last term converges to 0 as n,m→∞ by the properties of the Xn. Therefore, {Zn} isa Cauchy sequence in L1 (Ω;E;G ) . It follows it converges to some Z in L1 (Ω;E,G ) . Thenletting A ∈ G , and using 61.1.5,∫

AZdP =

∫XAZdP = lim

n→∞

∫XAZndP = lim

n→∞

∫A

ZndP

= limn→∞

∫A

XndP =∫

AXdP.

Then define Z ≡ E (X |G ).It remains to verify ||E (X |G )|| ≡ ||Z|| ≤ E (||X || |G ) . This follows because, from the

above,||Zn|| → ||Z|| , ||Xn|| → ||X || in L1 (Ω)

and so if A ∈ G , then from 61.1.6,

1P(A)

∫A||Zn||dP≤ 1

P(A)

∫A||Xn||dP

and so, passing to the limit,

1P(A)

∫A||Z||dP≤ 1

P(A)

∫A||X ||dP =

1P(A)

∫A

E (∥X∥|G )dP