Kenneth Kuttler

63.7. DOOB MEYER DECOMPOSITION 2161

Thus T n is one of those stopping times bounded by a. Since {X (t)} is DL, this shows{X (T n)} is equi integrable. Now from the definition of T n, it follows

An (T n)≤ λ

Recall T n (ω) = tnj−1 where tn

j is the first time where An(

tnj ,ω)≥ λ or T n (ω) = a if this

never happens. Thus T n is such that it is before An gets larger than λ . Thus,∫[An(a)≥2λ ]

An (a)dP≤∫[An(a)≥2λ ]

(An (a)−λ )dP

≤∫[An(a)≥2λ ]

(An (a)−An (T n))dP

≤∫

Ω

(An (a)−An (T n))dP

=∫

Ω

(X (a)−Mn (a)− (X (T n))−Mn (T n))dP

=∫

Ω

(X (a)−X (T n))dP

Because by the discrete optional sampling theorem,∫Ω

(Mn (a)−Mn (T n))dP = 0.

Remember{

Mn(tnk

)}mnk=0 was a martingale.∫

Ω

(X (a)−X (T n))dP =∫[An(a)≥λ ]

(X (a)−X (T n))dP

+∫[An(a)<λ ]

(X (a)−X (T n))dP.

The second of the integrals on the right is such that for ω in this set, T n (ω) = a and so thesecond integral equals 0. Hence from the above,∫

[An(a)≥2λ ]

An (a)dP≤∫[An(a)≥λ ]

(X (a)−X (T n))dP

and since {X (t)} is DL, this shows {An (a)}∞

n=1 is equi integrable.By Corollary 20.9.6 on Page 640 there exists a subsequence {Ank (a)}∞

k=1 which con-verges weakly in L1 (Ω) to A(a) . By Lemma 63.7.8 it also follows that E (Ank (a) |Ft)converges weakly to E (A(a) |Ft) in L1 (Ω) . Now define

M (t)≡ E (X (a)−A(a) |Ft) .

Thus it is obvious from properties of conditional expectation that {M (t)} is a martingaleadapted to Ft and without loss of generality, it is a right continuous version. Let

A(t)≡ X (t)−M (t) .

63.7. DOOB MEYER DECOMPOSITION 2161Thus 7” is one of those stopping times bounded by a. Since {X (t)} is DL, this shows{X (T")} is equi integrable. Now from the definition of T”, it followsAN(T") <ARecall 7” (@) = r/_, where ¢? is the first time where A” G o) >A or T"(@) =a if thisnever happens. Thus 7” is such that it is before A” gets larger than A. Thus,1~A" (a)dP < | A” (a) —A) dPTnvysan 2 a)aP < in ayaa] a)—4)< A” —A"(T"))dP< [yay AAO)< [a (@)—a"(r))aP= | (@) —M" (a) (x (7")) —M" (")) dP= [ (x(@)-x(r"))aPQBecause by the discrete optional sampling theorem,/ (M" (a) —M"(T")) dP =0.Q™n,_o Was a martingale.Remember {M" (t/’)[x@-x(r ar - Donon (X (a) —X (T"))dP+ Donon (X (a) —X (T")) dP.The second of the integrals on the right is such that for @ in this set, T” (@) = a and so thesecond integral equals 0. Hence from the above,1sA"(a)dP < | X (a) —X(T"))dP[ vgn EPS Jy O-XT)and since {X (t)} is DL, this shows {A” (a)}°_, is equi integrable.By Corollary 20.9.6 on Page 640 there exists a subsequence {A” (a) };-_, which con-verges weakly in L!(Q) to A(a). By Lemma 63.7.8 it also follows that E (A"« (a) |_F;)converges weakly to E (A (a) |.¥;) in L' (Q) . Now defineM (t) =E (X (a) —A(a)|F%).Thus it is obvious from properties of conditional expectation that {M (t)} is a martingaleadapted to .¥; and without loss of generality, it is a right continuous version. LetA(t) =X (t)-—M(t).