2096 CHAPTER 62. STOCHASTIC PROCESSES

Thus Y (t)(ω) = X (t)(ω) a.e. for t ∈Q+. For each ω /∈ N, there exists δ > 0 such that ifr ∈Q, t < r < t +2δ , then

|Y (t,ω)−X (r,ω)|< ε/2.

Now suppose s ∈ (t, t +δ ) . Then there exists δ 1 < δ such that if s < r < s+δ 1 then

|Y (s,ω)−X (r,ω)|< ε/2.

pick r ∈Q∩ (s, t +δ ) . Then both of the above two inequalities hold and so it follows

|Y (t,ω)−Y (s,ω)| < |Y (t,ω)−X (r,ω)|+ |X (r,ω)−Y (s,ω)|< ε/2+ ε/2 = ε.

Therefore, t→ Y (t,ω) is right continuous as claimed.From the definition of Y (t,ω) , it follows ω → Y (t,ω) is measurable in Ft+ because

it is the limit of a sequence, X (rn,ω)XNC where rn→ t + . Now each X (rn, ·) is Frn mea-surable and so Y (t, ·) is Frn measurable also for each rn. Thus Y (t, ·) is Ft+ measurable.Since the filtration is normal, Ft = Ft+ and it follows Y (t, ·) is Ft measurable. Why isY (t, ·) ∈ L1 (Ω)?

From Lemma 62.8.3, the collection {X (rn)} is uniformly integrable. Therefore, fromthe Vitali convergence theorem, Theorem 11.5.3 on Page 257,

limn→∞

∫Ω

|Y (s)−X (rn)|dP = 0 (62.8.30)

and Y (s) ∈ L1 (Ω).It remains to verify {Y (s)} is a submartingale. For s < t, is it true that

E (Y (t) |Fs)≥ Y (s)?

Fix A ∈Fs. From the above construction, there exists w ∈Q and w≥ t such that∫A

Y (t)dP≥∫

AX (w)dP− ε.

Then also, there exists r ∈Q∩(s, t) such that∫A

X (r)dP≥∫

AY (s)dP− ε.

Now ∫A

E (Y (t) |Fs)dP =∫

AY (t)dP≥

∫A

X (w)dP− ε

=∫

AE (X (w) |Fr)dP− ε

≥∫

AX (r)dP− ε ≥

∫A

Y (s)dP−2ε.

Since ε was arbitrary, this shows∫A

E (Y (t) |Fs)dP≥∫

AY (s)dP