6.9 The Closedness ofCin the CaseT≥1 under(NA)Condition 105the functions
H 1 τn
|H 1 τn|^1 Aare still inE^1 sinceE^1 is closed and stable in the sense
of lemma 6.2.1. Thereforeψ 1 ∈E 1. IndeedH 1 τn=
∑∞
m=1^1 {τn=m}H
m
1 where
H 1 m∈E 1 and theτnareF 0 -measurable. Since|Hτ 1 n|→∞onA,wehave
that a.s. ((
H 1 τn
|H 1 τn|
,∆S 1
)
+
f 2 τn
|H 1 τn|)
(^1) A−→ 0.
It follows that f
τn
2
|Hτ 1 n|^1 A→−(ψ^1 ,∆S^1 )^1 Aand hence by the closedness ofK^2
we have−(ψ 1 ,∆S 1 ) (^1) A∈K 2 .Thisimpliesthat(ψ 1 ,∆S 1 )=0sinceψ 1 ∈E 1.
But sinceψ 1 is in canonical form we must haveψ 1 = 0 a contradiction to
|ψ 1 |=1onAso thatP[A]=0.
So we get anF 0 -measurably parameterised sequence (H 1 τn)∞n=1converging
a.s. toH 1 on Ω. This implies thatf 2 τn →f−(H 1 ,∆S 1 ) and hencef 2 =
f−(H 1 ,∆S 1 )∈K 2 by the closedness ofK 2. Finallyf =(H 1 ,∆S 1 )+f 2
whereH 1 ∈E 1 andf 2 ∈K 2 i.e.f∈K.
6.9 Proof of the Closedness ofCin the CaseT≥
under the(NA)Condition
We will use the same notation as in the previous section. This means that for
the (Ft)Tt=0-adaptedRd-valued process (St)Tt=0we introduceH 1 ,K 2 ,andI^1
as in (6.14) and (6.15).
We say that a predictableRd-valued process (Ht)Tt=1is in canonical form,
if for eacht,Htis in canonical form for ∆St=(St−St− 1 ). The spacesHt
are defined in the same way asH 1 i.e.
Ht={
Ht|HtisRd-valued,Ft− 1 -measurable andPtHt=Ht}
.
HerePtis the projection inL^0 (Ω,Ft− 1 ,P;Rd) associated with the predictable
range of ∆St=St−St− 1.
Proposition 6.9.1.With the above notation and under the assumption that
Ssatisfies the (NA) condition we have
(i) I:H 1 ×H 2 ×...×HT→L^0 (Ω,FT,P),I((Ht)Tt=1)=
∑T
t=1(Ht,∆St)=
(H·S)Tis injective.
(ii) If(Hn)∞n=1is a sequence inH 1 ×H 2 ×...×HTso thatI(Hn)−=(Hn·S)−T
is bounded a.s., then(Hn)n∞=1=(H 1 n,...,HnT)∞n=1is bounded a.s.
(iii)If(fn)∞n=1 is a sequence inK which is bounded a.s., then there is a
FT-measurably parameterised subsequenceσnso thatfσn→f a.s. and
f∈K.
Proof.(i): The first statement will follow by induction onT from the fact
thatI^1 (H 1 )∈K 2 givesH 1 =0ifH 1 is in canonical form. This statement is