Lemma 5.17

Let H be a sub-restriction of G and (A’₉) be satisfied for H . Suppose also that (A ₁)–(A ₄), (A ₆) and (A ₇) hold true. Assume that f and h are two H -allocations such that Vt(h(t,⋅))>Vt(f(t,⋅)) µ-a.e. on some coalition S. Then (f,h,S,H) satisfies the property “ (P) ” if either of the following two conditions is satisfied:

1. a ^S  > 0.

2. a ^S  = 0, there is an ɛ > 0 such that e(t,ω)+B(0,ε)⊆Xt(ω) for all ( t,ω)∈T×Ω , and, moreover, there are some i0∈ΛS and δ > 0 such that

   2μ(Si0)∫D(h−e)dμ∈Hi0

   for any measurable subset D of S with µ(D) < δ.

Proof

Let {δm:m≥1}⊆(0,1) be a decreasing sequence converging to 0. For all i ∈ Λ ^S and r≥1 , define the set

Consider the function ψ:S×Ω→R , defined by ψ(t,ω):=diste(t,ω),Rℓ∖ Xt(ω) . Since ψ( · , ω) is T -measurable for all ω ∈ Ω, the set

is T -measurable. Moreover, {Sir:r≥1} is increasing and Si∼⋃{Sir:r≥1} for all i ∈ Λ ^S .^[12] If Λ ^S is finite, choose an integer r′≥1 with μ(Sir′)>2μ(Si)3 for all i ∈ Λ ^S . Define κ to be 1r′ if (i) is satisfied; and ɛ, otherwise. Let

Choose some ζ<κ3 such that d−B(0,ζ)⊆R++ℓ . In case (ii) holds, pick a finite subset Θ of Λ ^S such that i ₀ ∈ Θ, ∑i∈ΛS∖Θμ(Si)<δ and

Define Ψ: = Λ ^S if (i) is satisfied; and Θ, if (ii) is satisfied. Let E:=⋃{Sir′:i∈Ψ} if (i) is satisfied; and S, otherwise. For all m≥1 , define the function hm:S×Ω→Rℓ by letting  

Put,

By (A ₄) and (A ₆), the mapping ξk:S→R , defined by

is T -measurable and so is R ^m . It is obvious that {Rm:m≥1} is increasing and S∼⋃{Rm:m≥1} . It follows from the absolute continuity of Lebesgue integral that there is some γ > 0 such that

for any measurable subset D of S with µ(D) < γ, where b:=min{μ(Si)|i∈Ψ} . For each i ∈ Ψ, define Gi:=Sir′ if (i) is satisfied; and S _i , if (ii) is satisfied. Choose some m' such that μ(S∖Rm′)<min{γ,c4} , where c:=min{μ(Gi)|i∈Ψ} . Hence,

and

for all i ∈ Ψ. Thus, one obtains

for all i ∈ Ψ. Put, λ: = δ _m' and

Let z∈⋂{ηHi:i∈K}∩B0,ηΩ and define

where |Ψ| denotes the number of elements in Ψ. Assuming z = ηz' for some z′∈⋂{Hi:i∈K} , one obtains

for some 0<αi≤1 and for all i ∈ Ψ. Since Hi is convex and 0∈Hi , one obtains zˆi∈Hi for all i ∈ Ψ. On the other hand, z(ω)∈B(0,η) implies zˆi(ω)∈B(0,ζ) for all i ∈ Ψ. By Lemma 5 in Shitovitz (1973), one has

for all i ∈ Ψ. Since μ(Si∖Rm′)<μ(Gi∩Rm′) , the above together with the fact that Hi is convex and 0∈Hi further yield

for all i ∈ Ψ. Define

for all i∈Ψ∖{i0} ; and

if (ii) is satisfied. Obviously, ciz∈3Hi∩B0,κΩ for all i ∈ Ψ. Thus, e(t,ω)−ciz∈Xt(ω) for all ( t,ω)∈Gi×Ω and i ∈ Ψ. For all i ∈ Ψ, consider an assignment giz:Si×Ω→R+ defined by

It is easy to verify that giz(t,ω)∈Xt(ω) for all ( t,ω)∈Si×Ω . Since −ciz∈3Hi , one must have giz(t,⋅)−e(t,⋅)∈4Hi for all t∈Gi∩Rm′ . Likewise, it can be shown that giz(t,⋅)−e(t,⋅)∈4Hi for all t ∈ S _i . Moreover, giz(t,ω)≫hm′(t,ω) for all t∈Gi∩Rm′ . It follows that

for all t∈Gi∩Rm′ . Hence, Vt(giz(t,⋅))>Vt(f(t,⋅)) for all t ∈ S _i . It can be checked that

for all i∈Ψ∖{i0} ; and

if (ii) is satisfied. Define the assignment yz:T×Ω→Rℓ by letting

It can be simply checked that y ^z satisfies the required condition.                   □

Proposition 5.18

Suppose that H is a sub-restriction of G that satisfies (A’₉). Assume further that f and h are two H -allocations such that Vt(h(t,⋅))>Vt(f(t,⋅)) µ-a.e. on some coalition S∈T0 . Then (f,h,S,H) satisfies the property (Q) .

Proof

Choose an i0∈K such that μ(S∩Ii0)>0 . By Lemma 5.17, (f,h,S∩Ii0,H) satisfies the property (P) . Then there exist some 0 < λ < 1 and some 0 < η < 1 such that for all

there is an assignment y ^z such that yz(t,⋅)−e(t,⋅)∈4Ht and Vt(yz(t,⋅))>Vt(f(t,⋅)) µ-a.e. on S∩Ii0 and

Let η0=η2 and take any z∈⋂{η0Hi:i≥1}∩B(0,η0)Ω . Then zˆ=2z∈⋂{ηHi:i≥1}∩B(0,η)Ω . So, there is an assignment yzˆ such that yzˆ(t,⋅)−e(t,⋅)∈4Ht and Vt(yzˆ(t,⋅))>Vt(f(t,⋅)) µ-a.e. on S∩Ii0 and

By the Lyapunov convexity theorem, there is a coalition R1⊆S∩Ii0 such that

Define the correspondence Γf:R1⇉(Rℓ)Ω by letting

Clearly, ∫R1yzˆdμ∈∫R1Γfdμ and ∫R1hdμ∈∫R1Γfdμ . Since ∫R1Γfdμ is convex, one obtains

So, there is an assignment φzˆ such that ∫R1φzˆdμ∈∫R1Γfdμ and

This implies that

Again, by the Lyapunov convexity theorem, there are coalitions R2⊆(S∩Ii0)∖R1 and R3⊆S∖Ii0 such that

and

Let R:=R1∪R2∪R3 and define an assignment ξz:T×Ω→Rℓ such that

Thus, ξz(t,⋅)−e(t,⋅)∈4Ht and Vt(ξz(t,⋅))>Vt(f(t,⋅)) µ-a.e. on R, and

                                                 □

Proof of Theorem 3.4.

Choose an ɛ ∈ (0, µ(S)). Let α ∈ (0, 1) be such that ɛ = α µ(S). It follows from the Lyapunov convexity theorem that there exists a coalition E such that µ(E) = α µ(S) and ∫E(h−e)dμ=α∫S(h−e)dμ=0 . Thus, there is a coalition E with µ(E) = ɛ and ∫E(h−e)dμ=0 . If µ(S) = µ(T) then nothing is remaining to prove. So, assume that µ(S) < ɛ < µ(T). By Proposition 5.18, there exist an η > 0 and a sub-coalition R of S such that for all z∈⋂{ηHi:1≤i≤n}∩B0,ηΩ , there is an assignment y ^z such that yz(t,⋅)−e(t,⋅)∈4Ht and Vt(yz(t,⋅))>Vt(f(t,⋅)) µ-a.e. on R, and

Let

By the Lyapunov convexity theorem, there is a coalition B⊆T∖R such that μ(B)=(1−β)μ(T∖R) and

Let

and define g:T×D→R by

Since g( · , z) is T -measurable for all z ∈ D and g(t,  · ) is continuous with respect to the subspace topology of the usual topology on D for all t ∈ T, g is T⊗B(D) -measurable. Define the correspondence Λf:T⇉D by letting

By our assumption, Λf(t)≠∅ for all t ∈ T. Moreover, GrΛf is T⊗B(D) -measurable. Consider the correspondence Φf:T⇉D defined by

As X _t (ω) is closed, Φ _f (t) can be equivalently expressed as

Since 0 ∈ D, Φf(t)≠∅ for all t ∈ T. As before, one can show that GrΦf is T⊗B(D) -measurable. Let Ψf:T⇉D be a correspondence defined by Ψf(t):=Λf(t)∩Φf(t) for all t ∈ T. By our assumptions, Ψf(t)≠∅ for all t ∈ T. Moreover, GrΨf is T⊗B(D) -measurable. By the Aumann-Saint-Beuve measurable selection theorem, there is a T -measurable selection ξ of Ψ _f such that g(t, ξ(t)) > 0 for all t ∈ T. Define

By Lemma 5 in Shitovitz (1973), one has

So,

and

Thus, by eq. (1), there is an assignment yc0 such that yc0(t,⋅)−e(t,⋅)∈4Ht and Vt(yc0(t,⋅))>Vt(f(t,⋅)) µ-a.e. on R, and

As in the proof of Proposition 5.18, one can show that there is an assignment φ such that φ(t,⋅)−e(t,⋅)∈4Ht and Vt(φ(t,⋅))>Vt(f(t,⋅)) µ-a.e. on R, and

Let E: = R ∪ B. Consider an assignment ψ:T×Ω→Rℓ defined by^[13]

Obviously, ψ is an assignment with ψ(t,⋅)−e(t,⋅)∈4Ht µ-a.e. on T and µ(E) = ɛ. Furthermore, Vt(ψ(t,⋅))>Vt(f(t,⋅)) µ-a.e. on E. It can be easily verified that ∫E(ψ−e)dμ=0 . This completes the proof.                                                 □

Proof of Theorem 3.9.

Suppose that the hypothesis of the theorem is satisfied. The rest of the proof is decomposed into two cases:

Case 1. μ(S1∪S2)=μ(T) . It is not difficult to show that there is an assignement φ such that φ(t,⋅)−e(t,⋅)∈Ht and Vt(φ(t,⋅))>Vt(f(t,⋅)) µ-a.e. on S ₁, and

This implies that

This is a contradiction to that fact that f is in the ex ante core.

Case 2. μ(S1∪S2)<μ(T) . By the Lyapunov convexity theorem, there is a coalition B⊆T∖S2 such that μ(B)=12μ(T∖S2) and

By Proposition 5.18, there exist an η > 0 and a sub-coalition R of S such that for all z∈⋂{ηHi:1≤i≤n}∩B0,ηΩ , there is an assignment y ^z such that yz(t,⋅)−e(t,⋅)∈4Ht and Vt(yz(t,⋅))>Vt(f(t,⋅)) µ-a.e. on R, and

Applying an argument similar to that in the proof of Theorem 3.4, one can show that there exists an element

such that c=∫Bξdμ , where

1. ξ(t)∈⋂ηHi:1≤i≤n∩B0,ηΩ ;

2. f(t, · ) + ξ(t) ∈ X _t ; and

3. Vt(f(t,⋅)+ξ(t))>Vt(f(t,⋅))

for all t ∈ B. Thus, by eq. (2), there exists an assignment y ^c such that yc(t,⋅)−e(t,⋅)∈4Ht and Vt(yc(t,⋅))>Vt(f(t,⋅)) µ-a.e. on R, and

Thus,

which further implies

Define the assignment ψ:T×Ω→Rℓ by letting

Note that ψ(t,⋅)−e(t,⋅)∈4Ht µ-a.e. on T. If µ(R ∩ B) = 0 then f is ex ante blocked by R ∪ B via ψ, which is a contradiction. So, assume that µ(R ∩ B) > 0 and rewrite the above equality as

Thus, there exist R1,R2∈T such that R1⊆R∖B and R2⊆B∖R such that

Thus, the coalition E:=R1∪(R∩B)∪R2 ex ante blocks f via ψ, which is a contradiction.

Proof of Theorem 3.12.

Suppose that the hypothesis of the theorem is true. The rest of the proof is completed by considering the following two cases:

Case 1. μ(S1∪S2)=μ(T) . This case can be done analogous to Case 1 in the proof of Theorem 3.9.

Case 2. μ(S1∪S2)<μ(T) . By Lemma 5.17, there exist a λ ∈ (0, 1) and an η > 0 such that for all z∈⋂{ηGi:1≤i≤n}∩B0,ηΩ , there is an assignment y ^z such that yz(t,⋅)−e(t,⋅)∈4Ht and Vt(yz(t,⋅))>Vt(f(t,⋅)) µ-a.e. on S ₁, and

By the Lyapunov convexity theorem, there are coalitions B1⊆S2 and B2⊆T∖(S1∪S2) such that

and

Let B:=B1∪B2 . Applying an argument similar to that in the proof of Theorem 3.4, one can show that there exists an element

such that c=∫Bξdμ , where

1. ξ(t)∈⋂η2Hi:1≤i≤n∩B0,η2Ω ;

2. f(t, · ) + ξ(t) ∈ X _t ; and

3. Vt(f(t,⋅)+ξ(t))>Vt(f(t,⋅))

for all t ∈ B. Define

Thus, by eq. (3), there exists an assignment yc0 such that yc0(t,⋅)−e(t,⋅)∈4Ht and Vt(yc0(t,⋅))>Vt(f(t,⋅)) µ-a.e. on S ₁, and

Thus, one has

Hence, one can find an assignment φ such that φ(t,⋅)−e(t,⋅)∈4Ht and Vt(φ(t,⋅))>Vt(f(t,⋅)) µ-a.e. on S ₁, and

As a result, one obtains

which is equivalent to

Thus, the coalition R: = S ₁ ∪ B ex ante blocks f via ψ, defined by

Since ψ(t,⋅)−e(t,⋅)∈4Ht µ-a.e. on T, one arrives at a contradiction.   □

Proof of Theorem 4.4.

Suppose that g is a feasible assignment in E(f) not belonging to the ex post core of E(f) . Then there are some state ω ₀, coalition S and assignment h in E(f;ω0) such that ut(ω0,h(t))>ut(ω0,g(t,ω0)) µ-a.e. on S, and

Since Corollary 3.5 is valid when Ω = {ω ₀} and Gt=Rℓ for all t ∈ T, we can conclude the following: for any 0 < ɛ < µ(T), there must exist some coalition R and assignment y in E(f;ω0) such that ut(ω0,y(t))>ut(ω0,g(t,ω0)) µ-a.e. on R, and

This completes the proof.                                    □

Proof of Theorem 4.11.

Suppose that g∈C(E;f) . Without loss of generality, suppose that g is not ex post C(T0,T1)(E;f) -fair. Thus, there exist a state ω ₀, two disjoint elements S1∈T0 , S2∈T1 and an assignment h in E(f;ω0) such that ut(ω0,h(t))>ut(ω0,g(t,ω0)) µ-a.e. on S ₁, and

Since Corollary 3.5 is valid when Ω = {ω ₀} and Gt=Rℓ for all t ∈ T, we can conclude that g is not in the core of E(f;ω0) , which is is a contradiction. This completes the proof.                                                          □