62.8. RIGHT CONTINUITY OF SUBMARTINGALES 2091

Applying the maximal inequality for submartingales, Theorem 60.6.4,

P(max

{∣∣X j∣∣ : j = n, · · · ,1

}≥ λ

)≤ 1

λ(E (|X0|)+E (|X∞|))≤

Cλ

and taking sup for all n,

P(sup{∣∣X j

∣∣}≥ λ)≤ C

λ

It follows that for all λ large enough,∫[|Xn|≥λ ]

|Xn|dP≤ 2ε

and since ε is arbitrary, this shows {Xn} for n > k is equiintegrable. Since there are onlyfinitely many X j for j ≤ k, this shows {Xn} is equiintegrable. Hence {Xn} is uniformlyintegrable.

62.8 Right Continuity Of SubmartingalesThe following theorem is an attempt to consider the question of right continuity. It turnsout that you can always assume right continuity of a submartingale by going to a suitableversion and this theorem is a first step in this direction.

Theorem 62.8.1 Let {X (t)} be a real valued submartingale adapted to the filtration Ft .Then there exists a set of measure zero N, P(N) = 0, such that if ω /∈ N then,

limr→t+,r∈Q

X (r,ω) , limr→t−,r∈Q

X (r,ω)

both exist. There also exists a set of measure zero N such that for Q+ the nonnegativerationals and ω /∈ N,

supt∈Q+∩[0,M]

|X (t,ω)|< ∞

is bounded for each M ∈ N. Q can be replaced with any countable dense subset of R.

Proof: Let {rk}∞

k=1 be an enumeration of the nonnegative rationals. Let t > 0 be given.Then let {s1,s2, · · · ,sn} be such that these are in order and {s2, · · · ,sn−1} are the first n−2rationals less than t listed in order and s1 = 0 while sn = t. Then let Yk ≡ X (sk) . It follows{Yk} is a submartingale and so from the maximal inequality in Theorem 60.6.4,

P([

max1≤k≤n

|Yk| ≥ 2m])

≤ 12m (2E (|Yn|+ |Y1|))

= 2−m (2E (|X (t)|+ |X (0)|))

Then letting n→ ∞, it follows upon letting Ct = 2E (|X (t)|+ |X (0)|) ,

P

([sup

r∈Q+∩[0,t]|X (r)| ≥ 2m

])≤ 2−mCt .