62.8. RIGHT CONTINUITY OF SUBMARTINGALES 2095

It follows that for λ sufficiently large the first term in 62.8.29 is smaller than ε/2 becausek is fixed. Now this shows there is a choice of λ such that for all n > k,∫

[|X(rn)|≥λ ]|X (rn)|dP < ε

There are only finitely many rn for n ≤ k and by choosing λ sufficiently large the aboveformula can be made to hold for these also, thus showing {X (rn)} is equi integrable.

Now this implies the sequence of random variables is uniformly integrable as well. Letε > 0 be given and choose λ large enough that for all n,∫

[|X(rn)|≥λ ]|X (rn)|dP < ε/2

Then let A be a measurable set.∫A|X (rn)|dP =

∫A∩[|X(rn)|≥λ ]

|X (rn)|dP+∫

A∩[|X(rn)|<λ ]|X (rn)|dP

< ε/2+∫

A∩[|X(rn)|<λ ]|X (rn)|dP≤ ε

2+λP(A)

and now you see that if P(A) is sufficiently small then for all n,∫A|X (rn)|dP < ε

which shows the set of functions is uniformly integrable as claimed. This proves the lemma.You can often consider a submartingale to be right continuous. This is the importance

of the following theorem.

Theorem 62.8.4 Let {X (t)} be a submartingale adapted to a normal filtration Ft . Thereexists a right continuous submartingale having left limits, {Y (t)} such that Y (t) = X (t)a.e. for every t ∈Q+. Furthermore {X (t)} has a right continuous left limits version if andonly if

t→ E (X (t))

is right continuous. More generally, Y (t) = X (t) a.e. at every point where the abovefunction is right continuous.

Proof: From Theorem 62.8.1, there exists a set of measure zero, N such that for ω /∈N,left and right limits of the following form exist.

limr→t+,r∈Q

X (r,ω) , limr→t−,r∈Q

X (r,ω) .

Then define for each t and ω /∈ N,

Y (t,ω)≡ limr→t+,r∈Q

X (r,ω) .

and for ω ∈ N,Y (t,ω)≡ 0