898 CHAPTER 33. QUADRATIC VARIATION AND STOCHASTIC INTEGRATION
33.3 More on Stopping TimesNext I want to consider
∫ t0 f X[0,σ ]dM in the case of elementary functions for f a continuous
in t, adapted, and bounded. This involves the same process as earlier and it is analogous tothe traditional definition of the Stieltjes integral. First we approximate with an elementaryfunction and then pass to a limit.
Proposition 33.3.1 Let σ be a stopping time and let {tk}mnk=0 be partition points of
[0,T ] . Also define the discrete approximation of σ
σn (ω)≡mn−1
∑k=0
tkXσ−1((tk,tk+1])(ω)
Then for f an elementary function with respect to {tk}nk=0 points as described above, it
follows that f X[0,σn] is an elementary function and σn is a stopping time. Also∫ t
0f X[0,σn]dM =
∫ t∧σn
0f dM
Proof: First, why is σn a stopping time? Consider [σn ≤ t] . Say t ∈ (tk, tk+1]. In case,t = tk+1, [σn ≤ t] = [σ ≤ t] ∈Ft . Otherwise, [σn ≤ t] = [σ ≤ tk] ∈Ftk ⊆Ft . Thus σn isindeed a stopping time. Now
f X[0,σn] (t) =mn−1
∑k=0
fkX[0,σn] (t)X(tk,tk+1] (t) .
t is somewhere. Say t ∈ (tk, tk+1]. Then consider fkX[0,σn] (t) . For this t, this term isnonzero if and only if ω ∈ [t ≤ σn] if and only if ω ∈ [tk < σn] ∈Ftk . Thus this term is theindicator function of a set in Ftk for all t ∈ (tk, tk+1]. It follows f X[0,σn] is an elementaryfunction and can be written as ∑
mn−1k=0 fkX[tk<σn]X(tk,tk+1] (t) , so its integral is
mn−1
∑k=0
fkX[tk<σn] (M (t ∧ tk+1)−M (t ∧ tk))
=mn−1
∑k=0
fk (M (t ∧ tk+1∧σn)−M (t ∧ tk ∧σn))
because if σn > tk in the kth term and the term is nonzero, then σn = tk+1. If σn ≤ tk, thekth term on the left is 0 and on the right that term is
fk (M (t ∧σn)−M (t ∧σn)) ,
also zero. Now the right side in the above is just∫ t∧σn
0 f dM and the left side is definedearlier as
∫ t0 f X[0,σn]dM. ■
Lemma 33.3.2 Let f be bounded, adapted, and continuous in t and let σ be a stoppingtime with finite values in [0,T ]. Also assume [M] (T ) ∈ L1 (Ω,P). Then letting σn be asabove,
σn (ω)≡mn−1
∑k=0
tkXσ−1((tk,tk+1])(ω) ,