1962 CHAPTER 60. CONDITIONAL, MARTINGALES

for all k. This will be done if I can show

A∩ [σ ≤ j]∩ [τ ≤ k] ∈Fk

for each j ≤ k because

∪ j≤kA∩ [σ ≤ j]∩ [τ ≤ k] = A∩ [σ ≤ τ]∩ [τ ≤ k]

However, since A ∈Fσ , it follows A∩ [σ ≤ j] ∈F j ⊆Fk for each j≤ k and [τ ≤ k] ∈Fkand so this has shown what I wanted to show, A∩ [σ ≤ τ] ∈Fτ .

Now replace the stopping time, τ with the stopping time τ ∧σ in what was just shown.Note

[τ ∧σ ≤ n] = [τ ≤ n]∪ [σ ≤ n] ∈Fn

so τ ∧σ really is a stopping time. This yields

A∩ [σ ≤ τ ∧σ ] ∈Fτ∧σ

However the left side equals A∩ [σ ≤ τ] . Thus

A∩ [σ ≤ τ] ∈Fτ∧σ

This has shown the first part of 2.), Fσ ∩ [σ ≤ τ] ⊆ Fτ∧σ . Now 60.5.10 implies if A ∈Fσ∧τ ,

A = A∩all of Ω︷ ︸︸ ︷

[σ ∧ τ ≤ τ] ∈Fτ

and so Fσ∧τ ⊆Fτ . Similarly, Fσ∧τ ⊆Fσ which shows

Fσ∧τ ⊆Fτ ∩Fσ .

Next let A ∈Fτ ∩Fσ . Then is it in Fσ∧τ ? Is A∩ [σ ∧ τ ≤ k] ∈Fk? Of course this is sobecause

A∩ [σ ∧ τ ≤ k] = A∩ ([σ ≤ k]∪ [τ ≤ k])

= (A∩ [σ ≤ k])∪ (A∩ [τ ≤ k]) ∈Fk

since both σ ,τ are stopping times. This proves part 2.).Now consider part 3.). Note that [τ = k] is in both Fk and Fτ . This is because τ is a

stopping time so it is in Fk. Why is it in Fτ ? Is [τ = k]∩ [τ ≤ j] ∈F j for all j? If j < k,then the intersection is /0 ∈F j. If j ≥ k, then the intersection reduces to [τ = k] and this isin Fk ⊆F j so yes, [τ = k] is in both Fk and Fτ .

Let A ∈Fk. I need to show

Fτ ∩ [τ = k] = Fk ∩ [τ = k]

where G∩ [τ = k] means all sets of the form A∩ [τ = k] where A ∈ G . Let A ∈Fτ . Then

A∩ [τ = k] = (A∩ [τ ≤ k])\ (A∩ [τ ≤ k−1]) ∈Fk