The Mathematics of Arbitrage

14.3 One-period Processes 289

follows. For an elementD∈E⊗B(Rd)oftheformD=A×B, we define
λF(D)=

∫

AFη(B)π(dη).
For eachη∈Ewe define the set Supp(Fη) as the support of the measure
Fη, i.e. the smallest closed set of fullFη-measure. The setSis defined as
{(η, x)|x∈Supp(Fη)}.ThesetSis an element ofE⊗B(Rd). Indeed, take
a countable base (Un)n≥ 1 of the topology ofRdand write the complement as:

Sc=

⋃

n≥ 1

({η|Fη(Un)=0}×Un).

Ifx:E →Rdis a measurable function thenφ:E→R+∪{+∞}defined
asφ(η)=‖(xη,.)−‖L∞(Fη)isE-measurable. Indeed take a countable dense
selection (fn)n≥ 1 ofSand observe thatφ(η) = inf{(xη,fn(η))−|n≥ 1 }.
For each η ∈ E we denote by Adm(η) the cone inRd consisting of
elements x ∈ Rd so that (x, .)− ∈ L∞(Fη). The set Adm is then de-
fined as{(η, x)|x∈Adm(η)}. This set is certainly inE⊗B(Rd). Indeed
Adm ={(η, x)|infn≥ 1 (x, fn)>−∞}where the sequence (fn)n≥ 1 is a count-
able dense selection ofS.

Lemma 14.3.4.IfF is a measurable mapping from(E,E,π)into the proba-
bility measures onRd, then the following are equivalent:

(1)For almost everyη ∈ E, the probability measureFη satisfies the no-
arbitrage property.
(2)For every measurable selectionxηofAdm, we haveλF[(η, a)|xη(a)<0]>
0 as soon asλF[(η, a)|xη(a)>0]> 0.

Proof.The implication 1⇒2 is almost obvious since for eachη∈Ewe have
thatFη[{a|(xη,a)< 0 }]>0assoonasFη[{a|(xη,a)> 0 }]>0. Therefore
ifλF[(η, a)|xη(a)>0]>0, we have thatπ(B)>0whereBis the set

B={η∈E|Fη[{a|(xη,a)> 0 }]> 0 }.

For the elementsη∈Bwe then also have thatFη[{a|(xη,a)< 0 }]>0and
integration with respect toπthen gives the result:

λF[(η, a)|(xη,a)<0] =

∫

E

π(dη)Fη[{a|(xη,a)< 0 }]> 0.

Let us now prove the reverse implication 2⇒1.

We consider the set

A={(η, x)|Fη[a|(x, a)≥0] = 1 andFη[a|(x, a)>0]> 0 }.

The reader can check that this set is inE⊗B(Rd) and therefore the set
B=pr(A)∈E. Suppose thatπ(B)>0 and take a measurable selection
xηofA. OutsideBwe definexη= 0. ClearlyλF({(η, a)|(xη,a)> 0 })> 0
and hence we have thatλF({(η, a)|(xη,a)< 0 })>0, a contradiction since
(xη,a)≥0,λFa.s.. So we see thatB=∅a.s. or what is the same for almost
everyη∈Ethe measureFηsatisfies the no-arbitrage property.