It is desirable to extend everything to stochastically square integrable functions. This will
involve localization using a suitable stopping time. First it is necessary to understand
localization for elementary functions. As above, we are in the situation described by the
following diagram.
U
↓ Q1 ∕2
U1 ⊇ JQ1 ∕2U ←J Q1 ∕2U
1−1
Φ ↘ ↓ Φ
n H
The elementary functions
{Φn }
have values in ℒ
(U1,H )
0 meaning they are restrictions
of functions in ℒ
(U1,H )
to JQ1∕2U and converge to Φ ∘ J−1 in
( ( ) )
L2 [a,T ]× Ω;ℒ2 JQ1 ∕2U,H
where Φ ∈ L2
([a,T ]×Ω; ℒ (Q1∕2U,H ))
2
is given. Let
n∑−1
Φ(t) ≡ Φ (tk)X (tk,tk+1](t)
k=0
be an elementary function. In particular, let Φ
(tk)
be ℱtk measurable as a map into
ℒ
(U1,H)
, and has finitely many values. As just mentioned, the topic of interest is the
elementary functions Φn in the above diagram. Thus Φ will be one of these elementary
functions.
Let τ be a stopping time having values from the set of mesh points
{tk}
for the
elementary function. Then from the definition of the integral for elementary
functions,
∫ t∧τ n∑−1
ΦdW ≡ Φ (tk)(W (t∧ τ ∧ tk+1) − W (t∧τ ∧ tk))
a k=0
If ω is such that τ
(ω )
= tj, then to get something nonzero, you must have tj> tk so
k ≤ j − 1. Thus the above on the right reduces to
j∑−1
Φ (tk)(W (t∧ tk+1)− W (t∧ tk))
k=0
It clearly is 0 if j = 0. Define ∑k=0−1≡ 0. Thus the integral equals
The last step occurs because of the following reasoning. The kth term of the sum in the
middle expression above equals Φ
(tk)
if and only if t > tk and τ ≥ t. If the two
conditions do not hold, then the kth term equals 0. As to the third line, if τ > tk
and t ∈ (tk,tk+1], then τ ≥ tk+1≥ t which is the same as the situation in the
second line. The term equals Φ
(tk)
. Note that X
[τ>tk]
(ω)
is ℱtk measurable,
because
[τ > tk]
is the complement of
[τ ≤ tk]
. Therefore, this is an elementary
function. Thus, from 64.8.15 -64.8.16, X
[a,τ]
(t)
Φ
(t)
is an elementary function
and
∫ t∧τ ∫ tn∑−1 ∫ t
ΦdW = X[τ>tk]Φ (tk)X(tk,tk+1](t)dW = X[a,τ](t)Φ (t)dW
a a k=0 a
From Proposition 64.1.5, if you have Φ,Ψ two of these elementary functions
( || || )
||||∫ t ∫ t ||||2
E || a X[a,τ](t)Φ (t)dW − a X [a,τ](t)Ψ (t)dW || =
H
∫ ∫
t 2
a ΩX [a,τ](t) ||(Φ (s)− Ψ (s))∘J ||ℒ2(Q1∕2U,H) dPds
∫ t∫
≤ ||(Φ (s) − Ψ (s))∘J||2ℒ2(Q1∕2U,H)dP ds (64.8.17)
a Ω