62.8. RIGHT CONTINUITY OF SUBMARTINGALES 2093

are increasing in n and each has measure less than

1M

(1

b−aE((X (t)−a)+

))and so

P(∪∞

n=1

[Un[a,b] [0, t]≥M

])≤ 1

M

(1

b−aE((X (t)−a)+

)).

which shows that

P(N[a,b]

)≤ 1

M

(1

b−aE((X (t)−a)+

))for every M and therefore, P

(N[a,b]

)= 0.

Therefore, corresponding to a < b, there exists a set of measure 0, N[a,b] such that forω /∈ N[a,b] 62.8.28 is not true for any s ∈ [0, t). Let N ≡ ∪a,b∈QN[a,b], a set of measure 0with the property that if ω /∈ N, then 62.8.28 fails to hold for any pair of rational numbers,a < b for any s ∈ [0, t). Thus for ω /∈ N,

limr→s+,r∈Q

X (r,ω)

exists for all s ∈ [0, t). Similar reasoning applies to show the existence of the limit

limr→s−,r∈Q

X (r,ω) .

for all s ∈ (0, t] whenever ω is outside of a set of measure zero. Of course, this exceptionalset depends on t. However, if this exceptional set is denoted as Nt , one could considerN ≡ ∪∞

n=1Nn. It is obvious there is no change if Q is replaced with any countable densesubset. This proves the theorem.

Of course the above theorem does not say the left and right limits are equal, just thatthey exist in some way for ω not in some set of measure zero. Also it has not been shownthat limr→s+,r∈QX (r,ω) = X (r,ω) for a.e. ω .

Corollary 62.8.2 In the situation of Theorem 62.8.1, let s > 0 and let D1 and D2 be twocountable dense subsets of R. Then

limr→s−,r∈D1

X (r,ω) = limr→s−,r∈D2

X (r,ω) a.e. ω

limr→s+,r∈D1

X (r,ω) = limr→s+,r∈D2

X (r,ω) a.e. ω

Proof: Let{

rin}

be an increasing sequence from Di converging to s and let N be theexceptional set corresponding to the countable dense set D1 ∪D2. Then for ω /∈ N, andi = 1,2,

limr→s−,r∈D1∪D2

X (r,ω) = limn→∞

X(ri

n,ω)= lim

r→s−,r∈DiX (r,ω)

The other claim is similar. This proves the corollary.Now here is an impressive lemma about submartingales and uniform integrability.