92 6 The Dalang-Morton-Willinger Theorem
As regards (ii) define∆(ω) = lim sup
n→∞d(fn(ω),f 0 (ω))so thatC={∆> 0 }. Modify the construction of the proof of Proposition
6.3.3 in a straightforward way by intersecting each of the subsetsIk(ω)ofN
with
{
n∣
∣
∣d(fn(ω),f 0 (ω))≥∆( 2 ω)}
.
6.4 The Closedness of the ConeC
The mapI:H→(H,∆S) introduced in Sect. 6.2 above is a linear map from
L^0 (Ω,F 0 ,P;Rd)toL^0 (Ω,F 1 ,P;Rd) which is continuous for the convergence in
measure. We know from Lemma 6.2.6 thatIbecomes injective when restricted
toHXwhereX=∆S=S 1 −S 0. The relevant feature shown in the subsequent
proposition is that its inverse, defined as a function fromI(L^0 (Ω,F 0 ,P;Rd))
to the subspaceHXofL^0 (Ω,F 0 ,P;Rd) defined in (6.4) above, is continuous
too. If, in addition, the processSsatisfies(NA), we may even formulate a
stronger statement.
Proposition 6.4.1.LetS=(S 0 ,S 1 )be adapted to(Ω,(Ft)^1 t=0,P)and let
(Hn)∞n=1be a sequence inL^0 (Ω,F 0 ,P;Rd)in canonical form, i.e.,Hn∈H∆S.
Then we have that
(i) (Hn)∞n=1is a.s. bounded iff((Hn,∆S))∞n=1is.
(i’) (Hn)∞n=1converges a.s. iff((Hn,∆S))∞n=1does.
If we suppose in addition thatS satisfies (NA) as defined in (6.1) we also
have
(ii) (Hn)∞n=1is a.s. bounded iff((Hn,∆S)−)∞n=1is.
(ii’) (Hn)∞n=1converges a.s. to zero iff((Hn,∆S)−)∞n=1does so.
Proof.The “only if” direction is trivial in all the above assertions and holds
true without the assumption(NA)and/or thatHnis in canonical form. Let
us now show the “if” direction.
(i) and (ii): Suppose that (Hn)∞n=1fails to be a.s. bounded and let us show
that ((Hn,∆S))∞n=1(resp. ((Hn,∆S)−)∞n=1) fails so too. We apply Proposition
6.3.4 (i) toK=Rd∪{∞}andx 0 =∞: there is a measurably parameterised
subsequence (Lk)∞k=1=(Hτk)∞k=1such that (Lk(ω)) diverges to∞on a setB
of positive measure. Note that eachLkis an integrand in canonical form.
LetL̂k=Lk (^1) B∩{|Lk|≥ 1 }/|Lk|in order to find a sequence of integrands
in canonical form such that|L̂k(ω)|=1,forω∈Bandksufficiently large.
By passing once more to a measurably parameterised subsequence we may
suppose that (L̂k)∞k=1converges to an integrand̂L, which is in canonical form
and satisfies|L̂|=1onB.