2285
and for all ω, including the one of interest, the above equals∫ t
0
(X[0,τ p]X[0,τq]Y,dMτ p
)=∫ t
0
(X[0,τ p]Y,dMτ p
)thus for this particular ω, you get the same for both p and q. Thus the definition is welldefined because for a given ω,
∫ t0
(X[0,τ p]Y,dMτ p
)is constant for all p large enough.
Next consider the claim about this process being a local martingale. Is∫ t∧τ p
0(Y,dM)
is a martingale? From the definition,∫ t∧τ p
0(Y,dM) = lim
q→∞
∫ t∧τ p
0
(X[0,τq]Y,dMτq
)
= limq→∞
∫ t
0
(X[0,τq]Y,d (M
τq)τ p)= lim
q→∞
∫ t∧τ p
0
(X[0,τq]Y,dMτ p
)= lim
q→∞
∫ t
0
(X[0,τ p]X[0,τq]Y,dMτ p
)=∫ t
0
(X[0,τ p]Y,dMτ p
)(66.0.6)
which is known to be a martingale since X[0,τ p]Y ∈ G . This is what it means to be a localmartingale. You localize and get a martingale.
Next consider the claim about an arbitrary stopping time. Why is X[0,σ ]X[0,τ p]Y ∈ G ?
This is part of a more general question. Suppose Ŷ ∈ G . Then why is X[0,σ ]Ŷ ∈ G . Itsuffices to show this. Let {Y n} be the sequence of elementary functions which converge toŶ as in the definition. Also let σn be the stopping time with discreet values which equalstnk+1 when σ ∈ (tn
k , tnk+1],
{tnk
}mnk=0 being the partition associated with Y n. Then, as explained
earlier, X[0,σn]Yn is an acceptable elementary function and also
{E(∫ T
0
∥∥X[0,σn]Yn−X[0,σ ]Ŷ
∥∥2 d [M]
)}1/2
≤{
E(∫ T
0
∥∥X[0,σn]Yn−X[0,σn]Ŷ
∥∥2 d [M]
)}1/2
+
{E(∫ T
0
∥∥X[σ ,σn]Ŷ∥∥2 d [M]
)}1/2
≤{
E(∫ T
0
∥∥Y n− Ŷ∥∥2 d [M]
)}1/2
+
{E(∫ T
0
∥∥X[σ ,σn]Ŷ∥∥2 d [M]
)}1/2
which converges to 0 from the definition of Ŷ ∈ G and the dominated convergence theorem.Thus X[0,σ ]Ŷ ∈ G .