31.3. DOOB OPTIONAL SAMPLING CONTINUOUS CASE 837

1. Fτ ⊆Fτk and if Ft is normal, then if A ∈Fτk for all k, it follows that A ∈Fτ .

2. A ∈Fτk if and only if A∩[τk = n2−k

]∈Fn2−k for all n.

3. If τ ≤ σ , then τk ≤ σ k.

4. More generally, if τ ≤ σ are two stopping times, then Fτ ⊆Fσ .

5. If X (t) is a right continuous process adapted to the normal filtration Ft and τ is astopping time, then ω→ X (τ (ω)) is Fτ measurable. Here X has values in a Banachspace.

Proof: 1.) Let A ∈Fτ . Then for t ∈ (n2−k,(n+1)2−k], if t < (n+1)2−k,

A∩ [τk ≤ t] = A∩[τk ≤ n2−k

]= A∩

[τ ≤ n2−k

]∈Fn2−k ⊆Ft

so A∈Fτk . If t = (n+1)2−k then A∩ [τk ≤ t]∈F(n+1)2−k =Ft . Thus A∈Fτk and Fτ ⊆Fτk . Now for the other part, Consider A∩ [τ ≤ t] . If t ∈ (n2−k,(n+1)2−k], then [τ ≤ t] =[τk ≤ (n+1)2−k

]and so A∩ [τ ≤ t] =A∩

[τk ≤ (n+1)2−k

]∈F(n+1)2−k ⊆Ft+2−k . Since

the filtration is normal, and for all k, A∩ [τ ≤ t] ∈Ft+2−k , it follows that A∩ [τ ≤ t] ∈Ftand so A ∈Fτ as claimed.

2.) Note [τk = n2−k

]=

[τk ≤ n2−k

]\[τk ≤ (n−1)2−k

]=

[τ ≤ 2−kn

]\[τ ≤ (n−1)2−k

]If A ∈Fτk then by definition,

A∩[τk = n2−k

]= A∩

([τk ≤ n2−k

]\[τk ≤ (n−1)2−k

])= A∩

([τ ≤ 2−kn

]\[τ ≤ (n−1)2−k

])∈Fn2−k

Conversely, ifA∩[τk = n2−k

]∈Fn2−k

for all n, then consider A∩ [τk ≤ t] for t ∈ (n2−k,(n+1)2−k]. Is A∩ [τk ≤ t] ∈ Ft? Ift < (n+1)2−k, then A∩ [τk ≤ t] = ∑

nj=1 A∩

[τk = j2−k

]∈Fn2−k ⊆Ft . If t = (n+1)2−k,

A∩ [τk ≤ t] = ∑n+1j=1 A∩

[τk = j2−k

]∈Ft and so A ∈F(n+1)2−k ⊆Fτk . This proves 2.).

3.) The values of both τk and σ k are n2−k for some nonnegative integer n. τk equalsn2−k on τ−1

((n−1)2−k,n2−k]

). Thus on this set, σ cannot be smaller than or equal to

(n−1)2−k. Hence σ k is at least n2−k.4.) Let τ ≤ σ and let A ∈ Fτ . A∩ [σ ≤ t] = A∩ [τ ≤ t]∩ [σ ≤ t] because τ ≤ σ .

However, A∩ [τ ≤ t]∈Ft and [σ ≤ t]∈Ft so the right side is in Ft which means A∈Fσ .5.)Let U be an open set.

X (τk)−1 (U)∩ [τk < t] = ∪ j2−k≤{t}k X

(j2−k

)−1(U) ∈F{t}k .

Now say t ∈((n2−k,(n+1)2−k]

). If t < (n+1)2−k, then X (τk)

−1 (U) ∩ [τk ≤ t] wasjust given. It is in Fn2−k = F{t}k ⊆ Ft . Otherwise t = (n+1)2−k and in this case,