The following theorem is an attempt to consider the question of right continuity. It turns
out that you can always assume right continuity of a submartingale by going to a suitable
version and this theorem is a first step in this direction.
Theorem 61.8.1Let
{X (t)}
be a real valued submartingale adapted to the filtrationℱ_{t}. Then there exists a set of measure zero N,P
(N )
= 0, such that if ω
∕∈
Nthen,
r→tli+m,r∈ℚX (r,ω ),r→ltim−,r∈ℚX (r,ω)
both exist. There also exists a set of measure zero N such that for ℚ^{+}the nonnegativerationals and ω
∕∈
N,
su+p |X (t,ω)| < ∞
t∈ℚ ∩[0,M ]
is bounded for each M ∈ ℕ. ℚ can be replaced with any countable dense subset ofℝ.
Proof:Let
{rk}
_{k=1}^{∞} be an enumeration of the nonnegative rationals. Let t > 0 be
given. Then let
{s1,s2,⋅⋅⋅,sn}
be such that these are in order and
{s2,⋅⋅⋅,sn− 1}
are the
first n − 2 rationals less than t listed in order and s_{1} = 0 while s_{n} = t. Then let
Y_{k}≡ X
(sk)
. It follows
{Yk}
is a submartingale and so from the maximal inequality in
Theorem 59.5.4,
([ m ]) 1
P 1m≤akx≤n |Yk| ≥ 2 ≤ 2m-(2E (|Yn |+ |Y1|))
− m
= 2 (2E (|X (t)|+ |X (0)|))
Then letting n →∞, it follows upon letting C_{t} = 2E
(|X (t)|+ |X (0)|)
,
([ ])
P sup |X (r)| ≥ 2m ≤ 2−mC .
r∈ℚ+∩ [0,t] t
By the Borel Cantelli lemma, there exists a set of measure 0, N_{t} such that for ω
∕∈
N_{t},ω
is contained in only finitely many of the sets
[ ]
m
su+p |X (r)| ≥ 2
r∈ℚ ∩[0,t]
which shows that for ω
∕∈
N_{t},sup_{r∈ℚ+∩[0,t]
}
|X (r)|
is bounded. Now let N = ∪_{j=1}^{∞}N_{j}.
This proves the second claim.
Next consider the first claim. By the upcrossing estimate, Theorem 59.5.9 or Lemma
59.2.6, and letting a < b and U_{}
[a,b]
^{n}
[c,d]
the upcrossings of Y_{k} from a to b on
[c,d]
for
d ≤ t and c ≥ 0,
( ) 1 ( )
E Un[a,b][0,t] ≤ -----E (Yn − a)+
b−1a ( )
= -----E (X (t)− a)+ .
b− a
Hence
( )
P ([U n [0,t] ≥ M ]) ≤ 1- -1--E ((X (t) − a)+) . (61.8.27)
[a,b] M b− a
(61.8.27)
Suppose for some s < t,
lim sup X (r,ω ) > b > a > lim r→isn+f,r∈ℚ X (r,ω). (61.8.28)
r→s+,r∈ℚ
(61.8.28)
If this is so, then in
(s,t)
∩ ℚ there must be infinitely many values of r ∈ ℚ such that
X
(r,ω)
≥ b as well as infinitely many values of r ∈ ℚ such that X
(r,ω)
≤ a. Note this
involves the consideration of a limit from one side. Thus, since it is a limit from one
side only, there are an arbitrarily large number of upcrossings between s and t.
Therefore, letting M be a large positive number, it follows that for all n sufficiently
large,
U n [0,t](ω ) ≥ M
[a,b]
which implies
[ n ]
ω ∈ U [a,b][0,t] ≥ M
which from 61.8.27 is a set of measure no more than
1 ( 1 ( + ))
M-- b−-aE (X (t) − a) .
This has shown that the set of ω such that for some s ∈ [0,t) 61.8.28 holds is contained in
the set
[ ]
N [a,b] ≡ ∩ ∞M=1 ∪∞n=1 U[na,b][0,t] ≥ M
Now the sets,
[ ]
Un[a,b][0,t] ≥ M
are increasing in n and each has measure less than
( )
1-- --1--E( (X (t)− a)+)
M b− a
and so
( [ ]) ( ( ))
P ∪∞n=1 U n[a,b][0,t] ≥ M ≤ 1-- --1-E (X (t)− a)+ .
M b− a
which shows that
( ) 1 ( 1 ( ))
P N [a,b] ≤ --- -----E (X (t)− a)+
M b − 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 for
ω
∕∈
N_{[a,b]
}61.8.28 is not true for any s ∈ [0,t). Let N ≡∪_{a,b∈ℚ}N_{[a,b]
}, a set of measure 0
with the property that if ω
∈∕
N, then 61.8.28 fails to hold for any pair of rational
numbers, a < b for any s ∈ [0,t). Thus for ω
∕∈
N,
lim X (r,ω)
r→s+,r∈ℚ
exists for all s ∈ [0,t). Similar reasoning applies to show the existence of the
limit
lim X (r,ω).
r→s −,r∈ℚ
for all s ∈ (0,t] whenever ω is outside of a set of measure zero. Of course, this exceptional
set depends on t. However, if this exceptional set is denoted as N_{t}, one could consider
N ≡∪_{n=1}^{∞}N_{n}. It is obvious there is no change if ℚ is replaced with any countable dense
subset. This proves the theorem.
Of course the above theorem does not say the left and right limits are equal, just that
they exist in some way for ω not in some set of measure zero. Also it has not been shown
that lim_{r→s+,r∈ℚ}X
(r,ω )
= X
(r,ω )
for a.e. ω.
Corollary 61.8.2In the situation of Theorem 61.8.1, let s > 0 and let D_{1}and D_{2}betwo countable dense subsets of ℝ. Then
r→sli−m,r∈D X (r,ω) = r→sl−im,r∈D X (r,ω) a.e. ω
1 2
r→sli+m,r∈D1X (r,ω) = r→sl+im,r∈D2X (r,ω) a.e. ω
Proof: Let
{ }
rin
be an increasing sequence from D_{i} converging to s and let N be the
exceptional set corresponding to the countable dense set D_{1}∪ D_{2}. Then for ω
∈∕
N, and
i = 1,2,
lim X (r,ω ) = lim X (ri,ω) = lim X (r,ω)
r→s−,r∈D1∪D2 n→∞ n r→s −,r∈Di
The other claim is similar. This proves the corollary.
Now here is an impressive lemma about submartingales and uniform integrability.
Lemma 61.8.3Let X
(t)
be a submartingale adapted to a filtration ℱ_{t}. Let
{rk}
⊆ [s,t) be a decreasing sequence converging to s. Then
{X (rj)}
_{j=1}^{∞}isuniformly integrable.
Proof: First I will show the sequence is equiintegrable.I need to show that for all
ε > 0 there exists λ large enough that for all n
∫
|X (rn)|dP < ε.
[|X(rn)|≥ λ]
Let ε > 0 be given. Since
{X (r)}
_{r≥0} is a submartingale, E
(X (rn))
is a decreasing
sequence bounded below by E
(X (s))
. This is because for r_{n}< r_{k},
E (X (rn)) ≤ E (E (X (rk)|ℱn)) = E (X (rk))
Pick k such that
E (X (rk))− lim E (X (rn))
| n→∞ |
= ||E (X (rk)) − lim E (X (rn))|| < ε∕2.
n→ ∞
Then for n > k,
∫ ∫ ∫
|X (rn)|dP = X (rn)dP + − X (rn)dP
[|X(rn)|≥λ] [X(rn)≥λ] [X (rn)≤ −λ]
∫ ∫ ∫
= X (r )dP + X (r )dP − X (r )dP
[X(rn)≥λ] n [X(rn)>− λ] n Ω n
∫ ∫ ∫
≤ X (rn)dP + E (X (rk)|ℱn )dP − X (rn)dP
∫ [X(rn)≥λ] ∫[X(rn)>− λ] ∫ Ω
≤ X (rk)dP + X (rk)dP − X (rk)dP + ε∕2
∫ [X(rn)≥λ] ∫[X(rn)>− λ] Ω
= [X(rn)≥λ]X (rk)dP + [X(rn)≤− λ](− X (rk))dP + ε∕2
∫
= |X (rk)|dP +ε∕2
∫ [|X (rn)|≥λ]
≤ |X (rk)|dP + ε∕2 (61.8.29)
[sup{|X(r)|≥ λ:r∈{rj}∞j=1}]
It follows that for λ sufficiently large the first term in 61.8.29 is smaller than
ε∕2 because k is fixed. Now this shows there is a choice of λ such that for all
n > k,
∫
|X (rn)|dP < ε
[|X(rn)|≥λ]
There are only finitely many r_{n} for n ≤ k and by choosing λ sufficiently large the
above formula 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. Let
ε > 0 be given and choose λ large enough that for all n,