2276 CHAPTER 66. THE INTEGRAL∫ t

0 (Y,dM)H

One also sees that E(∥Y∥2 N̂ (t)

)= 0.

Now it follows from Corollary 63.3.3 that

[(Ms−Mr)] = [Ms]− [Mr] = [M]s− [M]r

Hence

[(Y,(Ms−Mr))] (t)≤ ∥Y∥2 [Ms−Mr] (t) = ∥Y∥2 ([M]s (t)− [M]r (t))

as claimed.The last claim is easy. Let τ p be a localizing sequence for which Mτ p is a martingale.

Then ∫ t∧τ p

0(Y,dM) ≡

m−1

∑i=0

(Yi,M (t ∧ ti+1∧ τ p)−M (t ∧ ti∧ τ p))H

=m−1

∑i=0

(Yi,Mτ p (t ∧ ti+1)−Mτ p (t ∧ ti))H

a finite sum of martingales.Note that this is just a definition and did not use the above localization lemma. In

particular, τ p is not restricted to having only the partition points as values.Next one needs to generalize past the elementary functions.Continue writing M in place of Mτ p in what follows. Consider an elementary function

Y ≡mn−1

∑k=0

YkX(tk,tk+1] (t)

where YkM∗ ∈ L2 (Ω). Consider

∫ t

0(Y,dM)≡

mn−1

∑k=0

(Yk,M (t ∧ tk+1)−M (t ∧ tk)) (66.0.3)

Then it is routine to verify that

E

(mn−1

∑k=0

(Yk,M (t ∧ tk+1)−M (t ∧ tk))H

)2

=mn−1

∑k=0

E((Yk,M (t ∧ tk+1)−M (t ∧ tk))

2H

)(66.0.4)

This is because the mixed terms all vanish. This follows from the following reasoning. Lett j < tk

E((Yk,M (t ∧ tk+1)−M (t ∧ tk))H

(Yj,M

(t ∧ t j+1

)−M

(t ∧ t j+1

))H

)= E

(E((Yk,∆kM (t))H (Yj,∆ jM (t))H |Ftk

))

2276 CHAPTER 66. THE INTEGRAL {\ (Y,dM),,One also sees that E (iv? 9) =0.Now it follows from Corollary 63.3.3 that[(M* — M")]| = [M*] — [M"] = [M)’ — [MJ’Hence[(¥, (MS —M"))] (t) < ||¥|)? [MSM] (0) = [IF 1? (IMB (2) - [MI (0)as claimed.The last claim is easy. Let T, be a localizing sequence for which M*? is a martingale.ThentT p m—1[ (Y.dM) = VY (¥,M(t tii At) —M(tAGiAtp)) yi=0m—1= LY YM? (t Ati) -—M® (tAti)) 4i=0a finite sum of martingales. §fNote that this is just a definition and did not use the above localization lemma. Inparticular, T, is not restricted to having only the partition points as values.Next one needs to generalize past the elementary functions.Continue writing M in place of M’? in what follows. Consider an elementary functionmn—1Y= Ve ty tess] (t)k=0where ¥.M* € L? (Q). Considerot myn—1| (¥.dM)= Yo (¥%,M(t tsi) —M(tAK)) (66.0.3)9 k=0Then it is routine to verify thatmy—1 2E Yo MoM (ttn) —M(tAt))yk=0mn—1-y E (Yee M((\tes1) —M(tt))ir) (66.0.4)k=0This is because the mixed terms all vanish. This follows from the following reasoning. Letti < tkE ((YesM (t\tey1) —M (Ate) (VieM (tAti41) —M (Ati) y)=E(E ((Y¥e,AcM (t)) xy (Vj, AM (1) |Fry))