The Mathematics of Arbitrage

14.5 Duality Results and Maximal Elements 311

Lemma 14.5.18.Ifw≥ 1 ,ifforsomeQ 0 ∈Meσwe haveEQ 0 [w]<∞,if
g
wis bounded and if
β<inf{α| there isHw-admissible andg≤α+(H·S)∞},

then there is a probability measureQ∈Mes,wsuch thatEQ[g]>β.

Proof.The hypothesis onβmeans that:
(
g−β
w

+L∞+

)

∩C∞w={ 0 }.

Because the setC∞wis weak-star-closed, we can apply Yan’s separation theorem
[Y 80] (see also Theorem 5.2.2 above) and we obtain a strictly positive measure
μ,equivalenttoPsuch that

(1)Eμ

[

g−β

w

]

> 0

(2) for allh∈Cw∞we haveEμ[h]≤0.

If we normaliseμso that the measureQdefined asdQ= w^1 dμbecomes
a probability measure, then we find that

(1)Q∼PandEQ[w]<∞,
(2)EQ[g]>β,
(3) for allh∈Cw∞we have thatEQ[hw]≤0.

The latter inequality together with the Beppo-Levi theorem then implies that
for eachw-admissible integrandHwe have thatEQ[(H·S)∞]≤0.

Lemma 14.5.19.Ifw ≥ 1 ,ifsomeQ 0 ∈Meσwe haveEQ 0 [w]<∞,if
g≥−wthen

sup

Q∈Mes,w

EQ[g]≥inf{α|there isHw-admissible andg≤α+(H·S)∞}.

Moreover if the quantity on the right hand side is finite, then the infimum is
a minimum.

Proof.For eachn≥1, we have thatg∧wnis bounded and hence we can apply
the previous lemma. This tells us that, for eachn∈N,

αn=sup

{

EQ[g∧n]|Q∈Mes,w

}

≥inf{α|there isHw-admissible andg∧n≤α+(H·S)∞}.

Because there is nothing to prove when limnαn=∞we may suppose that
supnαn= limnαn=α<∞. So, for eachn,wetakeaw-admissible integrand
Hnsuch thatg∧n≤αn+^1 n+(Hn·S)∞. Let us now fixQ 0 ∈Meσsuch that
EQ 0 [w]<∞. From Theorem 14.5.13, cited above, we deduce the existence of
Kn∈conv{Hn,Hn+1,...}as well asH^0 , such that