1968 CHAPTER 60. CONDITIONAL, MARTINGALES

Now note that U [a,b]M is at least as large as the number of upcrossings of {X (k)} for k ≤M.

This is because every time an upcrossing occurs, it will follow that X (τ2k+1)−X (τ2k)> 0and so a one will occur in the above sum which defines U [a,b]

M . However, this might be largerthan the number of upcrossings. The above discussion has proved the following upcrossinglemma.

Lemma 60.6.1 Let {X (k)} be a nonnegative submartingale. Let

U [a,b]M ≡ lim

ε→0

n

∑k=0

X (τ2k+1)−X (τ2k)

ε +X (τ2k+1)−X (τ2k), 2n+1 = M

ThenE(

U [a,b]M

)≤ 1

b−aE (X (M))

Suppose that there exists a constant C ≥ E (X (M)) for all M. That is, {X (k)} is boundedin L1 (Ω). Then letting

U [a,b] ≡ limM→∞

U [a,b]M ,

it follows that

E(

U [a,b])≤C

1b−a

The second half follows from the first part and the monotone convergence theorem.Now with this estimate, it is easy to prove the submartingale convergence theorem.

Theorem 60.6.2 Let {X (k)} be a submartingale which is bounded in L1 (Ω) ,

∥X (k)∥L1(Ω) ≤C

Then there is a set of measure zero N such that for ω /∈ N, limk→∞ X (k)(ω) exists. IfX (ω) = limk→∞ X (k)(ω) , then X ∈ L1 (Ω) .

Proof: Let a < b and consider the submartingale (X (k)−a)+ . Let U [0,b−a] be the ran-dom variable of the above lemma which is associated with this submartingale. Thus

E(

U [0,b−a])≤ C

b−a

It follows that U [0,b−a] is finite for a.e. ω . As noted above, U [0,b−a] is an upper bound tothe number of upcrossings of (X (k)−a)+ and each of these corresponds to an upcrossingof [a,b] by X (k). Thus for all ω /∈ Na,b where P

(Na,b

)= 0, it follows that

U [0,b−a] < ∞.

If limk→∞ X (k)(ω) fails to exist, then there exists a < b both rational such that

lim supk→∞

X (k)> b > a > lim infk→∞

X (k)

Thus ω ∈ Na,b because there are infinitely many upcrossings of [a,b]. Let

N = ∪{

Na,b : a,b ∈Q}

Then for ω /∈ N, the limit just discussed must exist. Letting X (ω) = limk→∞ X (k)(ω) forω /∈ N and letting X (ω) = 0 on N, it follows from Fatou’s lemma that X is in L1 (Ω) .