Kenneth Kuttler

24.12. CONDITIONAL EXPECTATION IN BANACH SPACES 703

andlimn→∞

∫Ω

∥X (ω)−Xn (ω)∥dP = 0. (24.56)

Now let {Xn} be the simple functions just defined and let Xn (ω) = ∑mk=1 xkXFk (ω)

where Fk ∈F , the Fk being disjoint. Then define Zn ≡ ∑mk=1 xkE

(XFk |G

). Thus, if A ∈ G ,

∫A

ZndP =m

∑k=1

∫A

E(XFk |G

)dP =

∑k=1

∫AXFk dP

∑k=1

xkP(Fk ∩A) =∫

AXndP (24.57)

Then since E(XFk |G

)≥ 0, it follows that ∥Zn∥ ≤ ∑

mk=1 ∥xk∥E

(XFk |G

). Thus if A ∈ G ,

E (∥Zn∥XA) ≤ E

∑k=1∥xk∥XAE

(XFk |G

))=

∑k=1∥xk∥

∫A

E(XFk |G

)dP

∑k=1∥xk∥

∫AXFk dP = E (XA ∥Xn∥) . (24.58)

Note the use of≤ in the first step in the above. Although the Fk are disjoint, all that is knownabout 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} is aCauchy sequence in L1 (Ω;E;G ) . It follows it converges to some Z in L1 (Ω;E,G ) . Thenletting A ∈ G , and using 24.57,∫

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 24.58,

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

Since A is arbitrary, this shows that ∥E (X |G )∥ ≡ ∥Z∥ ≤ E (∥X∥ |G ) . ■In the case where E is reflexive, one could also use Corollary 24.7.6 on Page 682 to

get the above result. You would define a vector measure on G , ν (F) ≡∫

F XdP and thenyou would use the fact that reflexive separable Banach spaces have the Radon Nikodymproperty to obtain Z ∈ L1 (Ω;E,G ) such that ν (F) =

∫F XdP =

∫F ZdP.

The function, Z whose existence and uniqueness is guaranteed by Theorem 24.12.2 iscalled E (X |G ).

24.12. CONDITIONAL EXPECTATION IN BANACH SPACES 703andlim I \IX (@) — X,(@)|| dP = 0. (24.56)Nn—yooNow let {X,} be the simple functions just defined and let X,(@) = V7", x 2A, (@)where Fy € F, the Fy being disjoint. Then define Z, = V7. x,E (2%, |G ). Thus, if A € Y,[mar = Yon |e (ani9)aP= Yom | anarA k=1 7A k=1 7Am= )YxP(ROA)= | X,dP (24.57)k=1 AThen since E (.2%,|Y) > 0, it follows that ||Z,|| < LZ, ||xe|| E (2%,|Y) . Thus if A € Y,E(||Zn|| 2a) e(E [xxl] 2aE (2m 1) = Yell B (Zl) aPk=1 k=1¥ [sil | 2iaP =E (2a). (24.58)k=1Note the use of < in the first step in the above. Although the F; are disjoint, all that is knownabout E (.2%,|%) is that it is nonnegative. Similarly, E (|Z —Zm||) < E (||Xn —Xm||) andthis last term converges to 0 as n,m — © by the properties of the X,. Therefore, {Z,,} is aCauchy sequence in L! (Q;E;@%). It follows it converges to some Z in L' (Q;E,Y). Thenletting A € Y, and using 24.57,| zar = [azar = fin lim 2aZ, dP = lim ) Sn dPAn—oo= tim | X,dP = | XaP.noo JA AThen define Z = E (X|¥).It remains to verify ||E (X|Y)|| = ||Z|| < E (||X|| |G). This follows because, from theabove, ||Zn|| — ||Z|| , |[Xn|| > ||X|| in L' (Q) and so if A € GY, then from 24.58,1 1Bay I loll? < pray | IXellaPand so, passing to the limit,1 ‘ 1 1 °Bay | Vl? < Bey [MAP = Bray [ EUIKIIM APSince A is arbitrary, this shows that ||E (X|Y)|| = ||Z|| < E (||X|| |Y).In the case where E is reflexive, one could also use Corollary 24.7.6 on Page 682 toget the above result. You would define a vector measure on Y, v(F) = f,,XdP and thenyou would use the fact that reflexive separable Banach spaces have the Radon Nikodymproperty to obtain Z € L' (Q;E,Y) such that v(F) = J, XdP = J, ZdP.The function, Z whose existence and uniqueness is guaranteed by Theorem 24.12.2 iscalled E (X|Y).