2274 CHAPTER 66. THE INTEGRAL∫ t
0 (Y,dM)H
Are the right sides of 66.0.1 and 66.0.2 equal?Begin with the right side of 66.0.1 and consider τ = t j. Then to get something nonzero
in the terms of the sum in 66.0.1, you would need to have t j ≥ ti+1. Otherwise, t j ≤ ti andthe difference involving M would give 0. Hence, for such ω you would need to have thesum in 66.0.1 equal to
j−1
∑i=0
(Yi,M (t ∧ ti+1)−M (t ∧ ti))H
Thus this sum in 66.0.1 equals
m
∑j=0
X[τ=t j]
j−1
∑i=0
(Yi,M (t ∧ ti+1)−M (t ∧ ti))H
Of course when j = 0 the term in the sum in 66.0.1 equals 0 so there is no harm in defining∑−1i=0 ≡ 0. Then from the sum, you have i ≤ j−1 and so when you interchange the order,
you get that∫ t∧τ
0 (Y,dM) =
m−1
∑i=0
m
∑j=i+1
X[τ=t j] (Yi,M (t ∧ ti+1)−M (t ∧ ti))H
=m−1
∑i=0
(X[τ>ti] (ω)Yi,M (t ∧ ti+1)−M (t ∧ ti)
)H
Thus the right side of 66.0.1 equals the right side of 66.0.2.
∫ t
0
(X[0,τ]Y,dM
)=
m−1
∑i=0
(X[τ>ti] (ω)Yi,M (t ∧ ti+1)−M (t ∧ ti)
)=∫ t∧τ
0(Y,dM)
This has proved the first part of the following lemma.
Lemma 66.0.4 For an elementary function Y, and a stopping time τ having values in the{ti} , the points of discontinuity of Y, it follows that X[0,τ]Y is also an elementary functionand ∫ t∧τ
0(Y,dM) =
∫ t
0
(X[0,τ]Y,dM
)=∫ t
0(Y,dMτ)
Proof: Consider the second equal sign. By definition,
∫ t∧τ
0(Y,dM) =
m−1
∑i=0
(Yi,M (t ∧ ti+1∧ τ)−M (t ∧ ti∧ τ))H
=m−1
∑i=0
(Yi,Mτ (t ∧ ti+1)−Mτ (t ∧ ti))H ≡∫ t
0(Y,dMτ)
Next is another lemma about these integrals of elementary functions. First recall thefollowing definition
M∗ ≡ sup{∥M (t)∥ : t ∈ [0,T ]}