63.7. DOOB MEYER DECOMPOSITION 2165
and so
E(∫
(0,t]ξ (s)dA(s)
)= E
(∫(0,t]
ξ− (s)dA(s))
which shows A is natural by Lemma 63.7.7. This proves the theorem.There is another interesting variation of the above theorem. It involves the following
definition.
Definition 63.7.13 A submartingale, {X (t)} is said to be D if
{XT : T < ∞ is a stopping time}
is equi integrable.
In this case, you can consider partitions of the entire positive real line and the mar-tingales,
{M(tnk
)}∞
k=0 and{
A(tnk
)}∞
k=0 as before. This time you don’t stop at mn. By thesubmartingale convergence theorem, you can argue there exists An
∞ = limk→∞ A(tnk
). Then
repeat the above argument using An∞ in place of An (a) . This time you get {A(t)} equi
integrable. Thus the following corollary is obtained.
Corollary 63.7.14 Let {X (t)} be a right continuous submartingale of class D. Then thereexists a right continuous martingale, {M (t)} and a right continuous increasing submartin-gale, {A(t)} such that for each t,
X (t) = M (t)+A(t) .
If {A(t)} is chosen to be natural and A(0) = 0, then with this condition, {M (t)} and{A(t)} are unique. Furthermore {M (t)} and {A(t)} are equi integrable on [0,∞).
In the above theorem, {X (t)} was a submartingale and so it has a right continuousversion. What if {X (t)} is actually continuous? Can one conclude that A(t) and M (t) arealso continuous? The answer is yes.
Theorem 63.7.15 Let {X (t)} be a right continuous submartingale of class DL. Then thereexists a right continuous martingale, {M (t)} and a right continuous increasing submartin-gale, {A(t)} such that for each t,
X (t) = M (t)+A(t) .
If {A(t)} is chosen to be natural and A(0) = 0, then with this condition, {M (t)} and{A(t)} are unique. Also, if {X (t)} is continuous, (t → X (t,ω) is continuous for a.e. ω)then the same is true of {A(t)} and {M (t)} .
Proof: The first part is done above. Let {X (t)} be continuous. As before, let{
tnk
}mnk=0
be a sequence of partitions of [0,a] such that these are equally spaced points, limn→∞ tnk+1−
tnk = 0, and
{tnk
}mnk=0 ⊆
{tn+1k
}mn+1k=0 where here a > 0 is an arbitrary positive number and let
λ > 0 be an arbitrary positive number. Define
ξn (t)≡ E
(min
(λ ,A
(tn
j))|Ft)
for tnj−1 < t ≤ tn
j