67.9. SOME REPRESENTATION THEOREMS 2305
W−11 (Rn) = Ω and the empty set. Also it is closed with respect to intersections because, in
the situation where each si is larger than every ti,(Wt1 , · · · ,Wtm1
)−1(
m1
∏i=1
Ui
)∩(
Ws1 , · · · ,Wsm2
)−1(
m2
∏i=1
Vi
)=
(Wt1 , · · · ,Wtm1
,Ws1 , · · · ,Wsm2
)−1(
m1
∏i=1
Ui×m2
∏k=1Rn
)
∩
((Wt1 , · · · ,Wtm1
,Ws1 , · · · ,Wsm2
)−1(
m1
∏i=1Rn×
m2
∏k=1
Vi
))
=(
Wt1 , · · · ,Wtm1,Ws1 , · · · ,Wsm2
)−1(
m1
∏i=1
Ui×m2
∏k=1
Vk
)In general, you would just make the obvious modification where you insert a copy of Rn
in the appropriate position after rearranging so that the indices are increasing. It was justshown that K ⊆ G where
G ≡{
U ∈ Gt :∫
Ω
gXU dP = 0}.
Now it is clear that G is closed with respect to countable disjoint unions and complements.The case of complements goes as follows. Ω ∈K and so if U ∈ G ,∫
Ω
gXUC dP+∫
Ω
gXU dP =∫
Ω
gdP
The last on the left and the integral on the right are both 0 so it follows that∫
ΩgXUC dP = 0
also. It follows from Dynkin’s lemma that G ⊇ σ (K ). Now σ (K ) is σ (W(u) : u≤ t)≡Gt . Hence, G = G t and so g is in L2 (Ω,Gt) and for every U ∈ Gt ,∫
Ω
gXU dP = 0
which requires g = 0. Thus functions of the above form are indeed dense in L2 (Ω,Gt).Note that this involves g being Gt measurable, not Ft measurable. It is not clear to
me whether it suffices to assume only that g is Ft measurable. If true, this above has notproved it. The problem is the argument at the end using Dynkin’s lemma to conclude thatg = 0.
Why such a funny lemma? It is because of the following computation which dependson Itô’s formula. Let
X =∫ t
0hT dW− 1
2
∫ t
0h ·hdτ
and g(x) = ex and consider g(X) = Y. Recall the Ito formula. Formally,
dY = g′ (X)dX +12
g′′ (X)(dX)2