29.4. OPTIONAL SAMPLING AND STOPPING TIMES 797

Proof: [τ ≤ n] = ∪nk=1 [τ = k] = ∪n

k=1 [Xk ∈ O]∩(∪ j<k

[X j ∈ OC

]). Now

∈Fk[Xk ∈ O]∩

(∪ j<k

∈F j[X j ∈ OC]) ∈Fn

and so this is indeed a stopping time, being the union of finitely many sets of Fn. As justnoted, this means that Xn∧τ is Fn measurable.

For the claim about Fτ , it is obvious that Ω, /0 are in Fτ . Suppose A ∈Fτ . Then for

k arbitrary, AC ∩ [τ ≤ k]∪∈Fk

A∩ [τ ≤ k] =∈Fk

[τ ≤ k] and so AC ∩ [τ ≤ k] ∈Fk. It is even moreobvious that Fτ is closed with respect to countable unions. ■

Of course τ has values i, in a countable well ordered set of numbers, i≤ i+1. We havethe following about the relation with stopping times and conditional expectations.

Lemma 29.4.7 Let X be in L1 (Ω). Then

1. Fτ ∩ [τ = i] = Fi∩ [τ = i] and E (X |Fτ) = E (X |Fi) a.e. on the set [τ = i] . Also ifA ∈Fτ or Fi, then A∩ [τ = i] ∈Fi∩Fτ .

2. E (X |Fτ) = E (X |Fi) a.e. on the set [τ ≤ i] .

Proof: 1.) A typical set in Fτ ∩ [τ = i] is B ≡ A∩ [τ = i] where A ∈ Fτ . Thus A∩[τ = i] = B ∈Fi so A∩ [τ = i] = A∩ [τ = i]∩ [τ = i] = B∩ [τ = i] ∈Fi∩ [τ = i] .

A typical set in Fi∩ [τ = i] is A∩ [τ = i] where A∈Fi. Then A∩ [τ = i]∩ [τ = j]∈F jfor all j. If j ̸= i, you get /0 and if j = i, you get A∩ [τ = i] ∈Fi = F j so A∩ [τ = i] = B ∈Fτ and so A∩ [τ = i]∩ [τ = i] = B∩ [τ = i] ∈Fτ ∩ [τ = i].

For A ∈ Fτ , A∩ [τ = i] ∈ Fi by definition. However, it is also the case, from whatwas just shown that A∩ [τ = i]∈Fτ because A∩ [τ = i]∩ [τ = j]∈F j for every j. Also, ifA∈Fi, then A∩ [τ = i]∈Fi by definition of a stopping time and A∩ [τ = i]∩ [τ = j]∈F jfor every j. Thus if A is either in Fτ or Fi, then [τ = i]∩A ∈Fi∩Fτ .

Now let A ∈Fτ . Then∫A∩[τ=i]

E (X |Fi)dP =∫

A∩[τ=i]XdP≡

∫A∩[τ=i]

E (X |Fτ)dP

2.) If A ∈Fτ ,then A∩ [τ = j] ∈F j by definition of Fτ and so by definition of conditionalexpectation,

∫A∩[τ≤i]

E (X |Fi)dP =

E(E(X |Fi)|F j)=E(X |F j)i

∑j=1

∫A∩[τ= j]

E (X |Fi)dP =i

∑j=1

∫A∩[τ= j]

E (X |F j)dP

=i

∑j=1

∫A∩[τ= j]

E (X |Fτ)dP =∫

A∩[τ≤i]E (X |Fτ)dP

Since A ∈Fτ is arbitrary, it follows that E (X |Fi) = E (X |Fτ) a.e. on the set [τ ≤ i]. ■