Interaction in Choquet inegral model

Proposition 2. Let {P, I} be an ordinal preference information on B g such that I = ∅.
Then, {P, I} is representable by a Choquet integral if and only if the binary relation
(P ∪ M) contains no strict cycle.
Proof. Necessity. Suppose that the ordinal preference information {P, I} on B g is
representable by a Choquet integral. So there exists a capacity µ such that {P, I} is
representable by C µ .
If P ∪ M contains a strict cycle, then there exists x 0 , x 1 , . . . , x r on B g such that
x 0 (P ∪ M)x 1 (P ∪ M) . . . (P ∪ M)x r (P ∪ M)x 0 and there exists two elements x i , x i+1 ∈
{x 0 , x 1 , . . . , x r } such that x i P x i+1 . Since {P, I} is representable by C µ , therefore C µ (u(x 0 )) ≥
. . . ≥ C µ (u(x i )) > C µ (u(x i+1 )) ≥ . . . ≥ C µ (u(x 0 )), then C µ (u(x 0 )) > C µ (u(x 0 )), contradiction.
Sufficiency. Assume that (P ∪M) contains no strict cycle, then there exists {B 0 , B 1 , . . . , B m }
a partition of B g , build by using a suitable topological sorting on (P ∪ M) (Gondran and
Minoux, 1995).
We construct a partition {B 0 , B 1 , . . . , B m } as follows:
B 0 = {x ∈ B g : ∀y ∈ B g , not [x(P ∪ M)y]},
B 1 = {x ∈ B g \B 0 : ∀y ∈B g \B 0 , not [x(P ∪ M)y]},
B i = {x ∈ B g \(B 0 ∪ . . . ∪ B i−1 ) : ∀y ∈ B g \(B 0 ∪ . . . ∪ B i−1 ), not [x(P ∪ M)y]}, for all
i = 1, 2, . . . , m.
Let us define the mapping φ : B g −→ P(N), f : P(N) −→ R, µ : 2 N
follows:
φ(a S ) = S { for all S ⊆ N,
0 if l = 0,
f(φ(x)) =
(2n) l if l ∈ {1, 2, . . . , m} for all l ∈ {0, 1, . . . , m}, ∀x ∈ B l,
µ(S) = f S
, where f α S = f(φ(a S )) for all S ⊆ N and α = f N = (2n) m
Let a S , a T ∈ B g such that a S P a T . We show that C µ (u(a S )) > C µ (u(a T )).
−→ [0, 1] as
Since a S , a T ∈ B g and {B 0 , B 1 , . . . , B m } is a partition of B g , then there exists r, q ∈
{0, 1, . . . , m} such that a S ∈ B r , a T ∈ B q . As a S P a T , then r > q, thus C µ (u(a S )) =
µ(S) = f S
α = (2n)r
α .
• If q = 0, then a T ∈ B 0 and a S ∈ B r with r ≥ 1.
As C µ (u(a T )) = C µ (u(a 0 )) = µ(∅) = 0, then C µ (u(a S )) > C µ (u(a T )).
• If q ≥ 1, C µ (u(a T )) = µ(T ) = f T
α = (2n)q , since 1 ≤ q ≤ m − 1.
α
But r > q therefore C µ (u(a S )) = (2n)r > (2n)q = C µ (u(a T )), then C µ (u(a S )) >
α α
C µ (u(a T )).
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