88 6 The Dalang-Morton-Willinger Theorem
φ 1 =|ff| (^1) A 1 is still inE 1 and it has maximal support among all elements of
E 1.
We now look at the subspaceE 2 ofE 1
E 2 ={f∈E 1 |(f, φ 1 ) = 0 a.s.}.
ClearlyE 2 is closed and satisfies the same stability property asE 1 .Wecan
therefore findφ 2 ∈E 2 , such that|φ 2 |= 1 or 0 and such thatφ 2 has maxi-
mal support among all elements ofE 2. Continuing this way we find random
variablesφ 1 ,φ 2 ,...,φdso that (φi,φj)=0,i =j,|φi|=1or0andφi
has maximal support among all elements ofEi. We claim that the procedure
necessarily stops after at mostdsteps. So we have to show that, forg∈E
and (g, φi) = 0 for alli=1,...,d,wehaveg= 0. Obviously we have by
the maximality ofφ 1 that{g=0}⊂{φ 1 =0}. Also one verifies inductively
g∈Eiand by the maximality ofφiwe have{g=0}⊂{φi=0}. This implies
that almost surely we have thatg(ω)= 0 implies thatφ 1 (ω),...,φd(ω)are
all different from zero. But thenφ 1 (ω),...,φd(ω) form an orthogonal basis of
Rdandg(ω) cannot be orthogonal toφ 1 (ω),...,φd(ω).
We now claim that
E=
{ d
∑k=1akφk∣
∣
∣
∣
∣
a 1 ,...,adareF-measurable}
. (6.2)
Indeed, denoting byF the right hand side of (6.2) we obviously haveF⊂E
by the sublemma. Conversely iff∈Ewe comparefwith
∑d
k=1(f, φk)φk.
Sinceg=f−
∑d
k=1(f, φk)φksatisfiesg∈Eand (g, φi) = 0 for alli≤d,we
must have thatg= 0. We now defineP 0 x=
∑d
i=1(x, φi)φi.Thisisclearly
a projection map. Obviouslyf∈EimpliesP 0 f=fand conversely we have
thatP 0 f=fimpliesf=
∑
(f, φi)φiwhich means thatf∈E.
It remains to check the last statement. So letP be a projection-valued
mapping so thatPf=ffor allf∈E.ThenPφi=φifor alliand hence
PP 0 =P 0 =P 0 P.
Remark 6.2.3.The reader might wonder why we gave so many details on these
rather obvious facts from linear algebra. The reason is that we had to check
the measurability. One way of doing this is to give explicit constructions.
Let us now return to the problem described in the beginning of this section.
IfX:Ω→RdisF 1 -measurable let us look at the space
EX={H|H:Ω→Rd,F 0 -measurable and (H, X)Rd=0a.s.}. (6.3)ClearlyEXsatisfies the assumption of lemma 6.2.1 and henceEXcan be
described by a projection-valued mappingP′. The projection-valued mapping
P=Id−P′defines the “orthogonal complement” ofEX. Let us define: