62.6. OPTIONAL SAMPLING THEOREMS 2075

= ∪k−1i=1

∈Fi︷ ︸︸ ︷A∩ [σ ∧ τ = i]∩

∈Fk︷ ︸︸ ︷[τ = k]∪∪∞

i=kA∩ [σ = i]∩ [τ = k]

= ∪k−1i=1

∈Fi︷ ︸︸ ︷A∩ [σ ∧ τ = i]∩

∈Fk︷ ︸︸ ︷[τ = k]∪

∈Fk︷ ︸︸ ︷A∩ [σ ∧ τ = k]∩

∈Fk︷ ︸︸ ︷[τ = k]

and so this is in Fk. Thus A∩ [τ ≤ k] ∈ Fk being the finite union of sets which are.Similarly A∩ [σ ≤ k] ∈Fk for all k and so A ∈Fτ ∩Fσ .

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

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 . First consider the claim

it is in Fτ .[τ = k]∩ [τ ≤ l] = /0 if l < k

which is in Fl . If l ≥ k, it reduces to [τ = k] ∈Fk ⊆Fl so it is in Fτ . [τ = k] is obviouslyin Fk.

I need to showFτ ∩ [τ = k] = Fk ∩ [τ = k]

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

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

Therefore, there exists B ∈Fk such that B = A∩ [τ = k] and so

B∩ [τ = k] = A∩ [τ = k]

which shows Fτ ∩ [τ = k]⊆Fk ∩ [τ = k]. Now let A ∈Fk so that

A∩ [τ = k] ∈Fk ∩ [τ = k]

ThenA∩ [τ = k]∩ [τ ≤ j] ∈F j

because in case j < k, the set on the left is /0 and if j ≥ k it reduces to A∩ [τ = k] and bothA and [τ = k] are in Fk ⊆F j. Thus A∩ [τ = k] = B ∈Fτ and so

A∩ [τ = k] = B∩ [τ = k] ∈Fτ ∩ [τ = k] .

Therefore, the two σ algebras of subsets of [τ = k] ,

Fτ ∩ [τ = k] ,Fk ∩ [τ = k]

are equal. Thus for A in either Fτ or Fk, A∩ [τ = k] is a set of both Fτ and Fk because ifA ∈Fk, then from the above, there exists B ∈Fτ such that

A∩ [τ = k] = B∩

∈Fτ∩Fk︷ ︸︸ ︷[τ = k] ∈Fτ