62.7. DOOB OPTIONAL SAMPLING CONTINUOUS CASE 2087
To verify that |X (τ)| is integrable, note that by right continuity, X (τn)→ X (τ) point-wise. Apply the Vitali convergence theorem to obtain∫
Ω
|X (τ)|dP = limn→∞
∫Ω
|X (τn)|dP≤∫
Ω
X (T )dP < ∞.
In fact, you do not need to assume X is nonnegative.
Lemma 62.7.13 Let X (t) be a right continuous submartingale such that the filtration {Ft}is normal. Recall this includes
Ft = ∩s>tFs.
Also let τ be a stopping time with values in [0,T ] . Let Pn ={
tnk
}mn+1k=1 be a sequence of
partitions of [0,T ] which have the property that
Pn ⊆Pn+1, limn→∞||Pn||= 0,
where||Pn|| ≡ sup
{∣∣tnk − tn
k+1
∣∣ : k = 1,2, · · · ,mn}
Then let
τn (ω)≡mn
∑k=0
tnk+1Xτ−1((tn
k ,tnk+1])
(ω)
It follows that τn is a stopping time and also the functions |X (τn)| are uniformly integrable.Furthermore, |X (τ)| is integrable.
Proof: It was shown above that τn is a stopping time. Also, tnk → X
(tnk
)is a discrete
submartingale. Then by Theorem 62.6.6 there is a martingale tnk →M
(tnk
)and an increasing
submartingale tnk → A
(tnk
)such that A≥ 0 and is increasing
X (tnk ) = M (tn
k )+A(tnk )
You define A(tm0)≡ 0 and for n≥ 1,
A(tmn )≡
n
∑k=1
E(
X (tmk )−X
(tmk−1)|Ftm
k−1
)and repeat the arguments in that theorem. You know that A(0) ,A(τn) ,A(T ) is a submartin-gale by the optional sampling theorem given earlier, Theorem 62.6.7, and so
P(A(τn)> λ )≤ 1λ
∫[A(τn)>λ ]
A(τn)dP≤ 1λ
∫[A(τn)>λ ]
A(T )dP≤ ∥A(T )∥L1
λ
It also follows from the definition of A that
∥A(T )∥L1 =∫
Ω
X (T )−X (0)dP < ∞
Hencelim
λ→∞
∫[A(τn)>λ ]
A(τn)dP≤ limλ→∞
∫[A(τn)>λ ]
A(T )dP = 0