214 10 Counter-Example
We takeu 1 small enough so thatR[νu 1 <∞]> 1 −ε 1. From the definition
ofνu 1 it follows that〈N, N〉∞>u 1 on the set{νu 1 <∞}.Foreachkwe look
at the exponentialE(k·N) and we letfk=(E(k·N))νu 1 .Since〈N, N〉νu 1 > 0
we have thatfktends to zero a.s. asktends to∞.
Takek 1 big enough to haveR[fk 1 <^12 ]> 1 −ε 1. We now define
τ 1 =inf{
t∣
∣(E(k 1 ·N))t>2or<^1
2}
∧νu 1.ClearlyR[τ 1 <νu 1 ]> 1 −ε 1 and hence
R[
(E(k 1 ·N))τ 1 ∈{ 1
2 ,^2
}]
> 1 −ε 1.For later use we defineX 1 =(E(k 1 ·N))τ 1 and we observe thatR[X 1 =
2]>^13 −ε 1 andR[X 1 =^12 ]>^23 −ε 1.
We now repeat the construction at timeνu 1. Of course this can only be
done on the set{νu 1 <∞}={〈N, N〉∞>u 1 }.Takeu 2 >u 1 small enough
so that
R[νu 2 <∞]>R[νu 1 <∞](1−ε 2 ).
We definefk=(E(k·(N−Nνu^1 )))νu 2 and observe thatfktends to zero on the
set{νu 1 <∞}asktends to infinity. Indeed this follows from the statement
that〈N−Nνu^1 ,N−Nνu^1 〉∞>0ontheset{νu 1 <∞}.
So we takek 2 big enough to guarantee thatR[fk 2 <^12 ]>R[νu 1 <∞](1−
ε 2 ). We defineτ 2 =inf
{
t>νu 1∣
∣(E(k·(N−Nνu (^1) )))t>2or<^1
2
}
∧νu 2.
ClearlyR[τ 2 <νu 2 ]>R[νu 1 <∞](1−ε 2 ). We defineX 2 =(E(k·(N−
Nνu^1 )))τ 2 and we observe that^12 ≤X 2 ≤2,R
[
X 2 ∈{^12 , 2 }|νu 1 <∞]
>
1 −ε 2.
SinceER[X 2 |νu 1 <∞] = 1 we therefore have thatR[X 2 =2|νu 1 <
∞]>^13 −ε 2 andR[X 2 =^12 |νu 1 <∞]>^23 −ε 2.
Continuing this way we construct sequences of
(1) stopping timesνunwithR[νun<∞]>Rνun− 1 <∞,ν 0 =0
(2) real numberskn
(3) stopping timesτnwithνun− 1 ≤τn≤νun
(4)Xn=(E(kn·(N−Nνun−^1 )))τn,
so that
(1)^12 ≤Xn≤ 2
(2)Xn=1ontheset{νun− 1 =∞}
(3)R[Xn=2|νun− 1 <∞]>^13 −εn
(4)R[Xn=^12 |νun− 1 <∞]>^23 −εn.
Let nowK=∑
n≥ 1 kn^1 [[νun− 1 ,τn]]. ClearlyE(K·N) is defined and (E(K·
N))τn=
∏n
k=1Xk. We claim thatER[(E(K·N))∞]<1, showing thatE(K·N)
is not uniformly integrable.
Obviously (E(K·N))∞=
∏
k≥ 1 Xk. From the strong law of large numbers
for martingale differences we deduce that a.s.