63.7. DOOB MEYER DECOMPOSITION 2167
Thus[T n (ε)≤ s] = ∩∞
N=1∪r∈D∩[s,s+ 1N ) [ξ
n (r,ω)−λ ∧A(r,ω)> ε]
and each ∪r∈D∩[s,s+ 1N ) [ξ
n (r,ω)−λ ∧A(r,ω)> ε] ∈Fs+1/N and so
[T n (ε)≤ s] ∈ ∩r∈D,r≥sFr = Fs+ = Fs
due to the assumption that the filtration is normal. What if s≥ a? Then from the definition,[T n (ε)≤ a] = Ω ∈Fa. Thus this really is a stopping time.
Now let B j ≡[tn
j−1 < T n (ε)≤ tnj
]. Note that T n (ε)∧ tn
j is also a stopping time.
∫Ω
ξnT n(ε)dP =
mn
∑j=1
∫B j
ξnT n(ε)dP =
mn
∑j=1
∫B j
ξnT n(ε)∧tn
jdP
=mn
∑j=1
∫B j
E(
ξnT n(ε)∧tn
j|FT n(ε)∧tn
j
)dP.
This is because B j ∈FT n(ε)∧tnj. Thus from the definition, the above equals
=mn
∑j=1
∫B j
E(
E(
λ ∧A(tn
j)|FT n(ε)∧tn
j
)|FT n(ε)∧tn
j
)dP
=mn
∑j=1
∫B j
E(
λ ∧A(tn
j)|FT n(ε)∧tn
j
)dP
=mn
∑j=1
∫B j
λ ∧A(tn
j)
dP =∫
Ω
λ ∧A(⌈T n (ε)⌉)dP (63.7.40)
where on (tnj−1, t
nj ], ⌈T n (ε)⌉ ≡ tn
j . Now ⌈T n (ε)⌉ is also a bounded stopping time. Here iswhy. Suppose s ∈ (tn
j−1, tnj ]. Then
[⌈T n (ε)⌉ ≤ s] =[T n (ε)≤ tn
j−1]∈Ftn
j−1⊆Fs.
Now letQn ≡ sup
t∈[0,a]|ξ n (t)−λ ∧A(t)| .
Then first note that
[Qn > ε] =
[sup
t∈[0,a)|ξ n (t)−λ ∧A(t)|> ε
]
because Qn (a) = 0 follows from the definition of ξn (t) as
E(λ ∧A
(tn
j)|Ft)
for tnj−1 < t ≤ tn
j
and soξ
n (a) = E (λ ∧A(a) |Fa) = λ ∧A(a) .