840 CHAPTER 31. OPTIONAL SAMPLING THEOREMS

Proof: From Proposition 31.3.8 M (τ) is measurable. Since τ is bounded, this is alwaysa finite sum for τk. Then each ω is in exactly one τ−1

((n2−k,(n+1)2−k]

)for some n. Say

ω ∈ τ−1((n2−k,(n+1)2−k]

). Then for that ω,M (τk (ω)) = M

((n+1)2−k

)(ω). Thus

M (τk (ω)) is given by

M (τk (ω)) =∞

∑n=0

Xτ−1((n2−k,(n+1)2−k]) (ω)M

((n+1)2−k

)(ω)

and

∥M (τk)(ω)∥ ≤∞

∑n=0

Xτ−1(In) (ω)∥∥∥M((n+1)2−k

)∥∥∥ ,In ≡ (n2−k,(n+1)2−k]

Then, since M (t) is a martingale, E(M((n+1)2−k

)|Fn2−k

)= M

(n2−k

)and so∥∥∥M

(n2−k

)∥∥∥ =∥∥∥E(

M((n+1)2−k

)|Fn2−k

)∥∥∥≤ E

(∥∥∥M((n+1)2−k

)∥∥∥ |Fn2−k

).

Therefore, iterating this gives∥∥∥M((n+1)2−k

)∥∥∥≤ E(∥M (Tk)∥|F(n+1)2−k

)where Tk is the smallest number greater than or equal to T which is of the form m2−k for ma positive integer. Then, since Xτ−1(In) is F(n+1)2−k measurable, it is FTk measurable andso

∥M (τk)∥ ≤∞

∑n=0

Xτ−1(In) (ω)E(∥M (Tk)∥|F(n+1)2−k

)=

∞

∑n=0

E(Xτ−1(In) ∥M (Tk)∥|F(n+1)2−k

)because Xτ−1(In) is F(n+1)2−k measurable. Thus

∫Ω

∥M (τk)∥dP≤∞

∑n=0

∫Ω

Xτ−1(In) ∥M (Tk)∥dP =∫

Ω

∥M (Tk)∥dP.

Thus ∫Ω

∥M (τk)∥dP≤∫

Ω

∥M (Tk)∥dP

Now use right continuity and Fatou’s lemma.∫Ω

∥M (τ)∥dP≤ lim infk→∞

∫Ω

∥M (τk)∥dP≤ lim infk→∞

∫Ω

∥M (Tk)∥dP

Pick T̂ > Tk for all k = 1,2, .... Then

M (Tk) = E(M(T̂)|FTk

)