31.3. DOOB OPTIONAL SAMPLING CONTINUOUS CASE 843

It follows that τn is a stopping time and also the functions |X (τn)| are uniformly integrable.Furthermore, |X (τ)| is integrable. Also X (0) ,X (τn) ,X (T ) is a sub-martingale for thefiltration F0,Fτn ,FT . If instead X (t) is a martingale having values in a separable Banachspace, X (0) ,X (τn) ,X (T ) is a martingale for the filtration F0,Fτn ,FT . In this case,the conclusions about integrability and uniform integrability apply to the sub-martingale∥X (t)∥ .

Proof: First of all, say t ∈ (tnk , t

nk+1]. If t < tn

k+1, then

[τn ≤ t] = [τ ≤ tnk ] ∈Ftn

k⊆Ft

and if t = tnk+1, then [

τn ≤ tnk+1]=[τ ≤ tn

k+1]∈Ftn

k+1= Ft

and so τn is a stopping time. Thus from Proposition 31.3.11, X (τn) is in L1 (Ω) the mea-surability being resolved from Proposition 31.3.8.

Now

X (τn) = X

(mn

∑k=0

tnk+1Xτ−1((tn

k ,tnk+1])

(ω)

)=

mn

∑k=0

X(tnk+1)X

τ−1((tnk ,t

nk+1])

(ω)

Now 0,τn,T is an increasing list of stopping times. Is it the case that it is a sub-martingale for F0,Fτn ,FT ?

E (X (τn) |F0) =mn

∑k=0

E(

X(tnk+1)X

τ−1((tnk ,t

nk+1])|F0

)

=mn

∑k=0

Xτ−1((tn

k ,tnk+1])

E(X(tnk+1)|F0

)≥

mn

∑k=0

Xτ−1((tn

k ,tnk+1])

X (0) = X (0)

Now also X (T ) = ∑mnk=0 X (T )X

τ−1((tnk ,t

nk+1])

(ω) so

E (X (T ) |Fτn) =mn

∑k=0

E(X

τ−1((tnk ,t

nk+1])

X (T ) |Fτn

). (31.1)

What is the value of τn on[τ ∈ (tn

k , tnk+1]

]? It is tn

k+1, and so from Lemma 31.1.4

E(X

τ−1((tnk ,t

nk+1])

X (T ) |Fτn

)= E

(X

τ−1((tnk ,t

nk+1])

X (T ) |Ftnk+1

)= X

τ−1((tnk ,t

nk+1])

E(

X (T ) |Ftnk+1

)because X

τ−1((tnk ,t

nk+1])

is Ftnk+1

measurable. Now since X is a sub-martingale,

E(

X (T ) |Ftnk+1

)≥ X

(tnk+1)

and so 31.1 implies

E (X (T ) |Fτn) =mn

∑k=0

E(X

τ−1((tnk ,t

nk+1])

X (T ) |Fτn

)