62.9. SOME MAXIMAL INEQUALITIES 2097

for any A ∈Fs and so this verifies since A is arbitrary that

E (Y (t) |Fs)≥ Y (s)

so Y is a submartingale.By Theorem 62.8.1 there exists a set of measure 0, N such that the left limits of Y (r,ω)

exist through rational numbers if ω /∈ N.Is {Y (t)} a version of {X (t)}? This is where the assumption t→E (X (t)) is continuous

is used. We know E (X (rn) |Fs)≥ X (s) and also for A ∈Fs∫A

X (rn)dP =∫

AE (X (rn) |Fs)dP≥

∫A

X (s)dP.

Hence taking a limit yields ∫A

Y (s)dP≥∫

AX (s)dP

and since A is arbitrary, Y (s)≥ X (s) . Now since t→ E (X (t)) is continuous,∫Ω

|Y (s)−X (s)|dP = E (Y (s))−E (X (s))

= limn→∞

(E (X (rn))−E (X (rn))) = 0.

It only remains to verify the only way X (t) has a right continuous version is fort → E (X (t)) to be continuous. Suppose then that {X (t)} has a right continuous ver-sion, {Y (t)} . Letting rn ↓ s, Lemma 62.8.3 implies {Y (rn)} is uniformly integrable. AlsoY (s)(ω) = limn→∞ Y (rn)(ω) a.e. and so by the Vitali convergence theorem,

limn→∞

∫Ω

|X (rn)−X (s)|dP = limn→∞

∫Ω

|Y (rn)−Y (s)|dP = 0.

This proves the theorem.Note that the condition t→ E (X (t)) being continuous holds for any martingale. There-

fore, every martingale has a right continuous version. The condition that t → E (X (t)) isright continuous is not a very stringent assumption. For {X (t)} a submartingale, this is anincreasing function. Therefore, the only points where the condition might not hold com-prise a countable set.

62.9 Some Maximal InequalitiesAs in the case of discrete martingales and submartingales, there are maximal inequalitiesavailable.

Lemma 62.9.1 Let X be right continuous and adapted such that the given filtration iscomplete in the sense that F0 contains all sets A of F such that P(A) = 0. Then there existsa set of measure zero N and a F ×B (R) measurable function Y such that if ω /∈ N, thenY (t)(ω) = X (t)(ω). Also, if f is F measurable and nonnegative then (λ ,ω)→X[ f>λ ]

is F ×B (R) measurable.