63.7. DOOB MEYER DECOMPOSITION 2163
Since ξ is bounded, you can take the limit outside. This follows from the dominatedconvergence theorem and the fact, shown above that A is increasing and nonnegative. Hereis why.
0≤∣∣∣ξ (tnk
j−1
)∣∣∣A(tnkj
)≤ A(a)C
where C is a constant larger than the values of ξ . Thus the above equals
limk→∞
E
(mnk
∑j=1
ξ
(tnk
j−1
)(A(
tnkj
)−A
(tnk
j−1
)))
= limk→∞
E
(mnk
∑j=1
ξ
(tnk
j−1
)(X(
tnkj
)−M
(tnk
j
)−(
X(
tnkj−1
)−M
(tnk
j−1
))))
= limk→∞
E
(mnk
∑j=1
ξ
(tnk
j−1
)(X(
tnkj
)−X
(tnk
j−1
)))(63.7.38)
because
E(
ξ
(tnk
j−1
)M(
tnkj
))= E
(E(
ξ
(tnk
j−1
)M(
tnkj
)|Ft
nkj−1
))= E
(ξ
(tnk
j−1
)E(
M(
tnkj
)|Ft
nkj−1
))= E
(ξ
(tnk
j−1
)M(
tnkj−1
))since M is a martingale. Now by a similar trick, this time using that
{Mnk
(tnk
j
)}mnk
j=0is a
martingale, 63.7.38 equals
limk→∞
E
(mnk
∑j=1
ξ
(tnk
j−1
)(Ank(
tnkj
)−Ank
(tnk
j−1
)))(63.7.39)
and now recall that Ank
(tnk
j
)is Ft
nkj−1
measurable. This will now be used to change the
subscript of tnkj−1 in ξ
(tnk
j−1
)to a j. 63.7.39 equals
= limk→∞
mnk
∑j=1
E(
E(
ξ
(tnk
j
)|Ft
nkj−1
)(Ank(
tnkj
)−Ank
(tnk
j−1
)))
= limk→∞
mnk
∑j=1
E(
E(
ξ
(tnk
j
)(Ank(
tnkj
)−Ank
(tnk
j−1
))|Ft
nkj−1
))
= limk→∞
mnk
∑j=1
E(
ξ
(tnk
j
)(Ank(
tnkj
)−Ank
(tnk
j−1
)))= lim
k→∞E
(mnk
∑j=1
ξ
(tnk
j
)(Ank(
tnkj
)−Ank
(tnk
j−1
)))