60.5. OPTIONAL STOPPING TIMES AND MARTINGALES 1963

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. 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∩ [τ = k] ∈Fτ

with similar reasoning holding if A∈Fτ . In other words, if g is Fτ or Fk measurable, thenthe restriction of g to [τ = k] is measurable with respect to Fτ ∩ [τ = k] and Fk ∩ [τ = k] .Let Y be an arbitrary random variable in L1 (Ω,F ) . It follows∫

A∩[τ=k]E (Y |Fτ)dP ≡

∫A∩[τ=k]

Y dP

≡∫

A∩[τ=k]E (Y |Fk)dP

Since this holds for an arbitrary set in Fτ ∩ [τ = k] = Fk ∩ [τ = k] , it follows

E (Y |Fτ) = E (Y |Fk) a.e. on [τ = k]

This proves the third claim and the Lemma.With this lemma, here is a major theorem, the optional sampling theorem of Doob. This

one is special for martingales.

Theorem 60.5.4 Let {M (k)} be a real valued martingale with respect to the increasingsequence of σ algebras, {Fk} and let σ ,τ be two stopping times such that τ is bounded.Then M (τ) defined as

ω →M (τ (ω))

is integrable andM (σ ∧ τ) = E (M (τ) |Fσ ) .

Proof: By Proposition 62.6.3 M (τ) is Fτ measurable.Next note that since τ is bounded by some l,∫

Ω

||M (τ (ω))||dP≤l

∑i=1

∫[τ=i]||M (i)||dP < ∞.