842 CHAPTER 31. OPTIONAL SAMPLING THEOREMS

Proof: To see this is so, let

Y (t)(ω) = ∥X (t ∨ τ)(ω)−X (τ (ω))∥

Then Y (t) is Ft∨τ measurable. It is desired to show that Y is Ft adapted. Hence if U isopen in R, then

Y (t)−1 (U) =(

Y (t)−1 (U)∩ [τ ≤ t])∪(

Y (t)−1 (U)∩ [τ > t])

The second set in the above union on the right equals either /0 or [τ > t] depending onwhether 0 ∈ U. If τ > t, then Y (t) = 0 and so the second set equals [τ > t] if 0 ∈ U. If0 /∈U, then the second set equals /0. Thus the second set above is in Ft . It is necessary toshow the first set is also in Ft . The first set equals

Y (t)−1 (U)∩ [τ ≤ t] = Y (t)−1 (U)∩ [τ ∨ t ≤ t]

because [τ ∨ t ≤ t] = [τ ≤ t]. However, Y (t)−1 (U) ∈Ft∨τ and so the set on the right in theabove is in Ft . For A to be in Ft∨τ means A∩ [τ ∨ t ≤ s] ∈Fs for each s. In particular, thisis true for s = t. Therefore, Y (t) is adapted. Then σ is just the first hitting time for Y (t) toequal the closed set a. Therefore, σ is a stopping time by Proposition 31.3.9. ■

The following corollary involves the same argument. Just replace

∥X (t ∨ τ)(ω)−X (τ (ω))∥

with g(X (t ∨ τ)(ω)−X (τ (ω))) .

Corollary 31.3.14 Let τ be a stopping time and let X be continuous and adapted to thefiltration Ft . Also let g be a continuous real valued function. Then for a > 0, define σ as

σ (ω)≡ inf{t > τ (ω) : g(X (t)(ω)−X (τ (ω))) = a}

Then σ is also a stopping time.

Next I want a version of the Doob optional sampling theorem which applies to martin-gales defined on [0,L],L≤∞. First recall the fundamental property of conditional expecta-ton that ∥E ( f |G )∥ ≤ E (∥ f∥G ).

Here is a lemma for an optional sampling theorem for the continuous case.

Lemma 31.3.15 Let X (t) have the property that it is a right continuous nonnegativesub-martingale, t ≥ 0 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{∣∣tn

k − tnk+1

∣∣ : k = 1,2, · · · ,mn}

. Then let

τn (ω)≡mn

∑k=0

tnk+1Xτ−1((tn

k ,tnk+1])

(ω) , tnmn+1 = T