838 CHAPTER 31. OPTIONAL SAMPLING THEOREMS

X (τk)−1 (U)∩ [τk ≤ t] is in F(n+1)2−k = Ft . Thus ω → X (τk)(ω) is Fτk measurable.

It follows that for a fixed k̂, X (τk) is Fτ k̂measurable for each k > k̂ because τk is de-

creasing in k so this follows from Part 4. Now let k→ ∞ and use right continuity of X toconclude that X (τ) is Fτ k̂

measurable. Thus X (τ)−1 (U) ∈Fτ k̂for each k̂ and so, by Part

1, it follows that X (τ)−1 (U) ∈Fτ . Therefore, X (τ) is Fτ measurable. ■Next is an important proposition which gives a typical example of a stopping time.

Since the process has t in an interval, one must be much more careful about the nature ofthe set which is hit.

Proposition 31.3.9 Let B be an open subset of topological space E and let X (t) be aright continuous Ft adapted stochastic process such that Ft is normal. Then define

τ (ω)≡ inf{t > 0 : X (t)(ω) ∈ B} .

This is called the first hitting time. Then τ is a stopping time. If X (t) is continuous andadapted to Ft , a normal filtration, then if H is a nonempty closed set such that H =∩∞

n=1Bnfor Bn open, Bn ⊇ Bn+1,

τ (ω)≡ inf{t > 0 : X (t)(ω) ∈ H}

is also a stopping time.

Proof: Consider the first claim. ω ∈ [τ = a] implies that for each n ∈ N, there existst ∈[a,a+ 1

n

]such that X (t) ∈ B. Also for t < a, you would need X (t) /∈ B. By right

continuity, this is the same as saying that X (d) /∈ B for all rational d < a. (If t < a, youcould let dn ↓ t where X (dn) ∈ BC, a closed set. Then it follows that X (t) is also in theclosed set BC.) Thus, aside from a set of measure zero, for each m ∈ N,

[τ = a] =(∩∞

n=m∪t∈[a,a+ 1n ][X (t) ∈ B]

)∩(∩t∈[0,a)

[X (t) ∈ BC])

Since X (t) is right continuous, this is the same as(∩∞

n=m∪d∈Q∩[a,a+ 1n ][X (d) ∈ B]

)∩(∩d∈Q∩[0,a)

[X (d) ∈ BC]) ∈Fa+ 1

m

Thus, since the filtration is normal,

[τ = a] ∈ ∩∞m=1Fa+ 1

m= Fa+ = Fa

I want to consider [τ ≤ a]. What of [τ < a]? This is equivalent to saying that X (t) ∈ B forsome t < a. Since X is right continuous, this is the same as saying that X (t) ∈ B for somet ∈Q, t < a. Thus

[τ < a] = ∪d∈Q,d<a [X (d) ∈ B] ∈Fa

It follows that [τ ≤ a] = [τ < a]∪ [τ = a] ∈Fa. Thus τ is indeed a stopping time.Now consider the claim involving the additional assumption that X (t) is continuous and

it is desired to hit a closed set H = ∩∞n=1Bn where Bn is open, Bn ⊇ Bn+1. (Note that if the

topological space is a metric space, this is always possible so this is not a big restriction.)Then let τn be the first hitting time of Bn by X (t). Then it can be shown that

[τ ≤ a] = ∩n [τn ≤ a] ∈Fa