62.10. CONTINUOUS SUBMARTINGALE CONVERGENCE THEOREM 2105

also. Therefore, there exists a set of measure 0 Nab such that if ω /∈Nab, then U[a,b] (ω)<∞.That is, there are only finitely many upcrossings. Now let

N = ∪{Nab : a,b ∈Q} .

It follows that for ω /∈ N, it cannot happen that

lim supt→∞

X (t)(ω)− lim inft→∞

X (t)(ω)> 0

because if this expression is positive, there would be arbitrarily large values of t whereX (t)(ω) > b and arbitrarily large values of t where X (t)(ω) < a where a,b are rationalnumbers chosen such that

lim supt→∞

X (t)(ω)> b > a > lim inft→∞

X (t)(ω)

Thus there would be infinitely many upcrossings which is not allowed for ω /∈N. Therefore,the limit limt→∞ X (t)(ω) exists for a.e. ω . Let X∞ (ω) equal this limit for ω /∈ N and letX∞ (ω) = 0 for ω ∈ N. Then X∞ is measurable and by Fatou’s lemma,∫

Ω

|X∞ (ω)|dP≤ lim infn→∞

∫Ω

|X (n)(ω)|dP <C.

Now here is an interesting result due to Doob.

Theorem 62.10.4 Let {M (t)} be a continuous real martingale adapted to the normal fil-tration Ft . Then the following are equivalent.

1. The random variables M (t) are equiintegrable.

2. There exists M (∞) ∈ L1 (Ω) such that limt→∞ ∥M (∞)−M (t)∥L1(Ω) = 0.

In this case, M (t) = E (M (∞) |Ft) and convergence also takes place pointwise.

Proof: Suppose the equiintegrable condition. Then there exists λ large enough that forall t, ∫

[|M(t)|≥λ ]|M (t)|dt < 1.

It follows that for all t,∫Ω

|M (t)|dP =∫[|M(t)|≥λ ]

|M (t)|dP+∫[|M(t)|<λ ]

|M (t)|dP

≤ 1+λ .

Since the martingale is bounded in L1, by Theorem 62.10.3 there exists M (∞) ∈ L1 (Ω)such that limt→∞ M (t)(ω) =M (∞)(ω) pointwise a.e. By the assumption {M (t)} are equi-integrable, it follows these functions are uniformly integrable. Letting δ > 0 be such thatif P(E)< δ , then ∫

E|M (t)|dP <

ε

5,