63.7. DOOB MEYER DECOMPOSITION 2155

and by right continuity, it follows

limp→∞

np

∑k=1

ξ(t pk

)X(t p

k−1,tpk ](s) = ξ (s)

and so the dominated convergence theorem applies and it follows

limp→∞

np

∑k=1

ξ(t pk

)(A(t pk

)−A

(t pk−1

))=∫(0,t]

ξ (s)dA(s)

where this is a random variable. Thus

E(∫

(0,t]ξ (s)dA(s)

)=∫

(limp→∞

∫(0,t]

np

∑k=1

ξ(t pk

)X(t p

k−1,tpk ](s)dA(s)

)dP (63.7.32)

Now as mentioned above,∫(0,t]

np

∑k=1

ξ(t pk

)X(t p

k−1,tpk ](s)dA(s) =

np

∑k=1

ξ(t pk

)(A(t pk

)−A

(t pk−1

))and since A is increasing, this is bounded above by an expression of the form CA(t) , afunction in L1. Therefore, by the dominated convergence theorem, 63.7.32 reduces to

limp→∞

∫Ω

∫(0,t]

np

∑k=1

ξ(t pk

)X(t p

k−1,tpk ](s)dA(s)dP

= limp→∞

∫Ω

np

∑k=1

ξ(t pk

)(A(t pk

)−A

(t pk−1

))dP

= limp→∞

∫Ω

(np

∑k=1

ξ(t pk

)A(t pk

)−

np−1

∑k=0

ξ(t pk+1

)A(t pk

))dP

= limp→∞

np−1

∑k=1

∫Ω

(ξ(t pk

)−ξ

(t pk+1

))A(t pk

)dP+

∫Ω

ξ (t)A(t)dP. (63.7.33)

Since ξ is a martingale,∫Ω

ξ(t pk+1

)A(t pk

)dP =

∫Ω

E(

ξ(t pk+1

)A(t pk

)|Ft p

k

)dP

=∫

A(t pk

)E(

ξ(t pk+1

)|Ft p

k

)dP

=∫

A(t pk

)ξ(t pk

)dP

and so in 63.7.33 the term with the sum equals 0 and it reduces to

E (ξ (t)A(t)) .

This is sufficiently interesting to state as a lemma.

63.7. DOOB MEYER DECOMPOSITIONand by right continuity, it followsNpjim 26 (2) 2p apis) =6 (s)and so the dominated convergence theorem applies and it followshin § (8) (4 (t?) —A(tP_,)) = [5 & (s)dA(spr éwhere this is a random variable. Thuse( | Eloaals )= Li(vm fds i?) eT)Now as mentioned above,Np Np[XS @) Mem 4A) = FE (2) (A) AL)(0,t] K=1 k=l2155(63.7.32)and since A is increasing, this is bounded above by an expression of the form CA (t), afunction in L!. Therefore, by the dominated convergence theorem, 63.7.32 reduces totim [28 C8) Zig. an @)4alsya= lim pe (1?) (A(@) -A (4) aPpe JO,np Np—1- inf (Lemaw-Leee. veNp—l= jim Y [E@) (ha le)ars | EoacarSince € is a martingale,[saadL ve (‘2) Fp) aP= [awe (E (2.1) |Fip) aP[aw E (t?) )dPand so in 63.7.33 the term with the sum equals 0 and it reduces toE(§(t)A(t)).This is sufficiently interesting to state as a lemma.(63.7.33)