60.8. A REVERSE SUBMARTINGALE CONVERGENCE THEOREM 1979

and so the above

≤∫[Xn≥λ ]

E (Xk|Fn)dP−∫

Ω

XkdP+ ε +∫[−Xn<λ ]

E (Xk|Fn)dP

=∫[Xn≥λ ]

XkdP−∫

Ω

XkdP+ ε +∫[−Xn<λ ]

XkdP

=∫[Xn≥λ ]

XkdP−∫

Ω

XkdP+ ε +∫[Xn>−λ ]

XkdP

=∫[Xn≥λ ]

XkdP+

(∫Ω

(−Xk)dP−∫[Xn>−λ ]

(−Xk)dP)+ ε

=∫[Xn≥λ ]

XkdP+∫[Xn≤−λ ]

(−Xk)dP+ ε =∫[|Xn|≥λ ]

|Xk|dP+ ε

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 since the single function, Xk is equiintegrable 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. This proves the lemma.

Now with this lemma and the upcrossing lemma it is easy to prove an important con-vergence theorem.

Theorem 60.8.3 Let {Xn,Fn}∞

n=0 be a backwards submartingale as described above andsuppose supn≥0 E (|Xn|)< ∞. Then {Xn} converges a.e. and in L1 (Ω) to a function, X∞.

Proof: By the upcrossing lemma applied to the submartingale {Xk}Nk=0 , the number

of upcrossings (Downcrossings is probably a better term. They are upcrossings as n getssmaller.) of the interval [a,b] satisfies the inequality

E(

UN[a,b]

)≤ 1

b−aE((X0−a)+

)Letting N → ∞, it follows the expected number of upcrossings, E

(U[a,b]

)is bounded.

Therefore, there exists a set of measure 0 Nab such that if ω /∈ Nab,U[a,b] (ω) < ∞. LetN = ∪{Nab : a,b ∈Q}. Then for ω /∈ N,

lim supn→∞

Xn (ω) = lim infn→∞

Xn (ω)