2082 CHAPTER 62. STOCHASTIC PROCESSES

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

Now what of [τ < a]? This is equivalent to saying that X (t) ∈ B for some t < a. Since X isright continuous, this is the same as saying that X (t) ∈ B for some t ∈Q, t < a. Thus

[τ < a] = ∪d∈Q,d<a [X (d) ∈ B] ∈Fa

It follows that [τ ≤ a] = [τ < a]∪ [τ = a] ∈Fa.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

To show this, first note that ω ∈ [τ ≤ a] if and only if there exists t ≤ a such that X (t)(ω)∈H. This follows from continuity and the fact that H is closed. Thus τn ≤ a for all nbecause for some t ≤ a, X (t) ∈ H ⊆ Bn for all n. Next suppose ω ∈ [τn ≤ a] for all n.Then for δ n ↓ 0, there exists tn ∈ [0,a+δ n] such that X (tn)(ω) ∈ Bn. It follows there isa subsequence, still denoted by tn such that tn→ t ∈ [0,a]. By continuity of X , it must bethe case that X (t)(ω) ∈H and so ω ∈ [τ ≤ a] . This shows the above formula. Now by thefirst part, each [τn ≤ a] ∈Fa and so [τ ≤ a] ∈Fa also.

Another useful result for real valued stochastic process is the following.

Proposition 62.7.6 Let X (t) be a real valued stochastic process which is Ft adapted fora normal filtration Ft , with the property that off a set of measure zero in Ω, t → X (t) islower semicontinuous. Then

τ ≡ inf{t : X (t)> α}

is a stopping time.

Proof: As above, for each m > 0,

[τ = a] =(∩∞

n=m∪t∈[a,a+ 1n ][X (t)> α]

)∩(∩t∈[0,a) [X (t)≤ α]

)Now

∩t∈[0,a) [X (t)≤ α]⊆ ∩t∈[0,a),t∈Q [X (t)≤ α]

If ω is in the right side, then for arbitrary t < a, let tn ↓ t where tn ∈ Q and t < a. ThenX (t)≤ liminfn→∞ X (tn)≤ α and so ω is in the left side also. Thus

∩t∈[0,a) [X (t)≤ α] = ∩t∈[0,a),t∈Q [X (t)≤ α]