2088 CHAPTER 62. STOCHASTIC PROCESSES

Because P(A(τn)> λ )→ 0 and a single function in L1 is uniformly integrable. Thusthese functions A(τn) are equi-integrable. Hence they are uniformly integrable. Now tn

k →∣∣M (tnk

)∣∣ is also a nonnegative submartingale. Thus

|M (0)| , |M (τn)| , |M (T )|

is a submartingale by the optional sampling theorem for discrete submartingales given ear-lier. Therefore,

P(|M (τn)|> λ )≤ 1λ

∫[|M(τn)|>λ ]

|M (τn)|dP≤ 1λ

∫[|M(τn)|>λ ]

|M (T )|dP≤ ∥M (T )∥L1

λ

Of course ∥M (T )∥L1 is finite because it is dominated by∫Ω

A(T )+ |X (T )|dP < ∞

Hence

limλ→∞

supn

∫[|M(τn)|>λ ]

|M (τn)|dP≤ limλ→∞

supn

∫[|M(τn)|>λ ]

|M (T )|dP = 0

because a single function in L1 is uniformly integrable and the above estimate shows thatP([|M (τn)|> λ ])→ 0 uniformly in n. Thus, in fact X (τn) must be uniformly integrablesince it is the sum of two which are.

Theorem 62.7.14 Let {M (t)} be a right continuous martingale having values in E a sepa-rable real Banach space with respect to the increasing sequence of σ algebras, {Ft} whichis assumed to be a normal filtration satisfying,

Ft = ∩s>tFs,

for t ∈ [0,L] , L≤∞ and let σ ,τ be two stopping times with τ bounded. Then M (τ) definedas

ω →M (τ (ω))

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

Proof: Since M (t) is a martingale, ∥M (t)∥ is a submartingale. Let

τn (ω)≡∞

∑k=0

2−n (k+1)TXτ−1((k2−nT,(k+1)T 2−n]) (ω) .

By Lemma 62.7.13, τn is a stopping time and the functions ||M (τn)|| are uniformly in-tegrable. Also ||M (τ)|| is integrable. Similarly ||M (τn∧σn)|| are uniformly integrablewhere σn is defined similarly to τn.

Consider the main claim now. Letting σ ,τ be stopping times with τ bounded, it followsthat for σn and τn as above, it follows from Theorem 62.6.5

M (σn∧ τn) = E (M (τn) |Fσn)

2088 CHAPTER 62. STOCHASTIC PROCESSESBecause P(A(t,) > 2) — 0 and a single function in L! is uniformly integrable. Thusthese functions A (T,,) are equi-integrable. Hence they are uniformly integrable. Now t >|M (1) | is also a nonnegative submartingale. Thus\M (0)|,|M (tn)| 1M (7)|is a submartingale by the optional sampling theorem for discrete submartingales given ear-lier. Therefore,| 1 |M(T) ||P(\M(t,)| >A <;/ M (T, ap<; | M(T)|dP < ——"=(IM (tu)I > 4) A Imcenioa| (=n) A Iceal>al! I AOf course ||M(T)||,1 is finite because it is dominated by[ A(t) +|x (D)|aP <JoHencelim sup [ M(t,)|dP < lim su M(T)|dP =0PSAP fiacen isa odlAP S Den sYP Siayeyyisay MObecause a single function in L! is uniformly integrable and the above estimate shows thatP(||M(tn)| >A]) + 0 uniformly in n. Thus, in fact X (t,) must be uniformly integrablesince it is the sum of two which are. JjTheorem 62.7.14 Let {M (t)} be a right continuous martingale having values in E a sepa-rable real Banach space with respect to the increasing sequence of 6 algebras, { F;} whichis assumed to be anormal filtration satisfying,F, = Asst Fs,fort €[0,L], L < and let o,t be two stopping times with t bounded. Then M (1) definedas@ — M(t(@))is integrable andM(oAt)=E(M(t)|F¥o).Proof: Since M (t) is a martingale, ||M (r)|| is a submartingale. LetcoT(@) = P2"(k+1) T 2-1 ((K2-"7 (k4-1)T2-n}) (@) -k=0By Lemma 62.7.13, T, is a stopping time and the functions ||M(t,)|| are uniformly in-tegrable. Also ||M(t)|| is integrable. Similarly ||M(t, A 0o,)|| are uniformly integrablewhere O,, is defined similarly to T,.Consider the main claim now. Letting o, tT be stopping times with T bounded, it followsthat for 0, and T, as above, it follows from Theorem 62.6.5M (0nA Tn) = E (M (tn) |-Fo, )