62.7. DOOB OPTIONAL SAMPLING CONTINUOUS CASE 2085

and this shows E(τ (t) |Fτ(s)

)≥ τ (s) .

Now here is an important example. First note that for τ a stopping time, so is t ∨ τ .Here is why.

[t ∨ τ ≤ s] = [t ≤ s]∩ [τ ≤ s] ∈Fs.

Example 62.7.10 Let τ be a stopping time and let X be continuous and adapted to thefiltration Ft . Then for a > 0, define σ as

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

Then σ is also a stopping time.

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 . 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 62.7.5.

62.7.2 The Optional Sampling Theorem Continuous Case

Next I want a version of the Doob optional sampling theorem which applies to martingalesdefined on [0,L],L≤∞. First recall Theorem 61.1.2 part of which is stated as the followinglemma.

Lemma 62.7.11 Let f ∈ L1 (Ω;E,F ) where E is a separable Banach space. Then if G isa σ algebra G ⊆F ,

||E ( f |G )|| ≤ E (∥ f∥|G ) .

Here is a lemma which is the main idea for the proofs of the optional sampling theoremfor the continuous case.