2350 CHAPTER 68. A DIFFERENT KIND OF STOCHASTIC INTEGRATION

such that the step functions given by

Φrk (t) ≡

mk

∑j=1

Φ

(tk

j

)X[tk

j−1,tkj )(t)

Φlk (t) ≡

mk

∑j=1

Φ

(tk

j−1

)X[tk

j−1,tkj )(t)

both converge to Φ in K as k→ ∞ and

limk→∞

max{∣∣∣tk

j − tkj+1

∣∣∣ : j ∈ {0, · · · ,mk}}= 0.

Also, each Φ

(tk

j

),Φ(

tkj−1

)is in Lp (Ω;E). One can also assume that Φ(0) = 0. The mesh

points{

tkj

}mk

j=0can be chosen to miss a given set of measure zero. In addition to this, we

can assume that ∣∣∣tkj − tk

j−1

∣∣∣= 2−nk

except for the case where j = 1 or j =mnk when this is so, you could have∣∣∣tk

j − tkj−1

∣∣∣< 2−nk .

Theorem 68.4.8 Let F ∈ L2 (Ω× [0,T ]) and is progressively measurable. Then it has aSkorokhod integral which coincides with the Ito integral.

Proof: From Lemma 68.4.7, there is a sequence of left step functions denoted here as{F l

k

}∞

k=1 which converges to F in L2 (Ω× [0,T ]) where F lk

(tk

j

)= F

(tk

j

). We can take a

subsequence if necessary and assume∥∥∥F lk −F

∥∥∥L2([0,T ]×Ω)

< 2−k

Here the{

tkj

}are mesh points corresponding to the kth partition described above. Thus

each F lk

(tk

j

)is in L2 (Ω). By Lemma 68.4.3 there exists a random variable Gl

k

(tk

j

)which

is a polynomial function of some W (h) for h ∈ L2(

0, tkj

)which can approximate F l

k

(tk

j

)as closely as desired in L2 (Ω). Then choosing these sufficiently close, it can be assumedthat the step functions

Glk ≡

mk−1

∑j=0

Glk

(tk

j

)X(

tkj ,t

kj+1

)also converge in L2 (Ω× [0,T ]) to F . Of course, each of these last step functions are inD(δ ).

The idea is to show that δ(Gl

k

)is Cauchy in L2 (Ω) as k→∞ and then use the fact that,

since δ is an adjoint, it must be a closed operator. This will show that F ∈ L2 (Ω× [0,T ]) ,considered as a subspace of L2

(Ω;L2 (0,∞,R)

), is in D(δ ) and δ (F) is equal to the above

2350 CHAPTER 68. A DIFFERENT KIND OF STOCHASTIC INTEGRATIONsuch that the step functions given byMrP(t) = yo(s) Bp wy)=JoiMr} (1) yo (1) Bikjal ;tk) (t)j-ljboth converge to ® in K as k + ~ andfim max { |r 1 7 € {0,--- sm} =0.k-00Also, each ® G PD G ) is in LP (Q;E). One can also assume that ® (0) = 0. The meshj-l. koints {ut}Pp J j=0can assume thatMr_ _ can be chosen to miss a given set of measure zero. In addition to this, wek _ k _ Nkks th =2except for the case where j = 1 or j =my, when this is so, you could have Gi - tty | <2",Theorem 68.4.8 Let F € L? (Q x [0,T]) and is progressively measurable. Then it has aSkorokhod integral which coincides with the Ito integral.Proof: From Lemma 68.4.7, there is a sequence of left step functions denoted here as{Fi }(_, which converges to F in L? (Q x [0,7]) where Fi (1) =F (') . We can take asubsequence if necessary and assumelA _F 2-kL?((0,T]xQ)Here the {ut} are mesh points corresponding to the k’” partition described above. Thuseach Fi (') is in L? (Q). By Lemma 68.4.3 there exists a random variable GI. (') whichis a polynomial function of some W (h) for h € L? (0.18) which can approximate F (4)as closely as desired in L? (Q). Then choosing these sufficiently close, it can be assumedthat the step functionsMga 1Gl= y Gi, (14) XaJ KO)also converge in L? (Q x [0,T]) to F. Of course, each of these last step functions are inD(6).The idea is to show that 5 (G{,) is Cauchy in L? (Q) as k —> ce and then use the fact that,since 6 is an adjoint, it must be a closed operator. This will show that F € L? (Q x [0,T]),considered as a subspace of L” (Q;L? (0,00,R)) , is in D(8) and 6 (F) is equal to the above