Solution to selected problems.

For this, let t 0 = 0 and t n+1 = ∞ and write
[
∏ n
E
j=1
1 {Ztj ∈B j }
] n+1
∣ ∑
∣G T =
∏
1 {T ∈[tk−1 ,t k )}
k=1
j 0, ∀t > 0, {|△Z t | > ε} = ∪ n ∩ n≥m {|Z t − Z T −1/n | > ε}.
Therefore,
P (|△Z t | > ε) ≤ lim inf
n→∞ P (|Z t − Z T −1/n | > ε) = 0 (33)
Since this is true for all ε > 0, P (|△Z t | > 0) = 0, ∀t.
26. To apply results of exercise 24 and 25, we first show following almost trivial lemma.
Lemma Let T be a stopping time and t ∈ R + . If T ≡ t, then G T = G t and G T − = G t− .
Proof. G T − = {A ∩ {T > s} : A ∈ G s } = {A ∩ {t > s} : A ∈ G s } = {A : A ∈ G s , s < t} = G t− . Fix
A ∈ G t . A ∩ {t ≤ s} = A ∈ G t ⊂ G s (t ≤ s), or ∅ ∈ G s (t > s) and hence G t ⊂ G T . Fix A ∈ G T ,
A ∩ {t ≤ s} ∈ G s , ∀s > 0. Especially, A ∩ {t ≤ t} = A ∈ G t and G T ⊂ G t . Therefore G T = G t .
Fix t > 0. By exercise 25, Z t = Z t− a.s.. By exercise 24 (d), G t = G t− since t is a bounded stopping
time.
27. Let A ∈ F t . Then A ∩ {t < S} = (A ∩ {t < S}) ∩ {t < T } ∈ F T − since A ∩ {t < S} ∈ F t .
Then F S− ⊂ F T − .
Since T n ≤ T , F Tn− ⊂ F T − for all n as shown above. Therefore, ∨ n F Tn− ⊂ F T − . Let A ∈ F t .
A ∩ {t < T } = ∪ n (A ∩ {t < T n }) ∈ ∨ n F Tn − and F T − ⊂ ∨ n F Tn −. Therefore, F T − = ∨ n F Tn −.
28. Observe that the equation in theorem 38 depends only on the existence of a sequence of simple
functions approximation f1 Λ ≥ 0 and a convergence of both sides in E{ ∑ j a jN Λ j
j
} = t ∑ j a jν(Λ j ).
For this, f1 Λ ∈ L 1 is enough. (Note that we need f1 Λ ∈ L 2 to show the second equation.)
9