An axiomatization of Choquet expected utility with cominimum independence

Abstract

This paper proposes a class of independence axioms for simple acts. By introducing the ${\mathcal {E}}$-cominimum independence axiom that is stronger than the comonotonic independence axiom but weaker than the independence axiom, we provide a new axiomatization theorem of simple acts within the framework of Choquet expected utility. Furthermore, in order to provide the axiomatization of simple acts, we generalize Kajii et al. (J Math Econ 43:218–230, 2007) into an infinite state space. Our axiomatization theorem relates Choquet expected utility to multi-prior expected utility through the core of a capacity that is explicitly derived within our framework. Our result in this paper also derives Gilboa (Econometrica 57:1153–1169, 1989), Eichberger and Kelsey (Theory Decis 46:107–140, 1999), and Rohde (Soc Choice Welf 34:537–547, 2010) as a corollary.

Appendix

1.1 Appendix 1

As defined in Sect. 2, $\varOmega$ is a non-empty finite or infinite set, $\varSigma$ is a non-empty algebra of subsets of $\varOmega ,\,{\mathbb {R}}^{\varOmega }= \{ x|\, x: \varOmega \rightarrow {\mathbb {R}} \}$ denotes the set of all real-valued functions on $\varOmega$, and $\mathcal {F}$ is the collection of all non-empty subsets of $\varOmega$.

A set function $v:2^{\varOmega } \rightarrow {\mathbb {R}}$ with $v(\emptyset )=0$ is $k$-monotone for $k \ge 2$ if $v \left( \cup _{i=1} ^k A_i \right) \ge \sum _{ \left\{ I: \emptyset \ne I \subset \left\{ 1, \ldots , k \right\} \right\} } (-1)^{|I|+1} v \left( \cap _{i \in I} A_i \right)$ for all $A_1,\ldots , A_k \in 2^{\varOmega }$, it is a capacity if $v(E) \le v(F)$ for all $E \subseteq F$ and $v(\varOmega )=1$, and it is totally monotone, if it is non-negative and $k$-monotone for all $k \ge 2$. A totally monotone game $v$ with $v(\varOmega )=1$ is called a belief function.

Let $\varOmega$ be finite. For $T \in \mathcal {F}$, let $u_T \in {\mathbb {R}}^{\mathcal {F}}$ be the unanimity game on $T$ that is defined by the following: $u_T(S)=1$ if $T \subseteq S$ and $u_T(S) =0$ otherwise. The following lemma has been proved by Shapley (1953).

Lemma 3

(Shapley 1953) Let $\varOmega$ be finite. Collection $\{u_{T}\}_{T\in \mathcal {F}}$ is a linear base for ${\mathbb {R}}^{\mathcal {F}}$. The unique collection of coefficients $\left\{ \beta _T \right\} _{T \in \mathcal {F}}$ satisfying the form

$$\begin{aligned} v= \sum _{T \in \mathcal {F}} \beta _T u_T, \end{aligned}$$

(5)

or equivalently $v(E)= \sum _{T \subseteq E} \beta _T$ for all $E \in \mathcal {F}$, is provided by $\beta _{T}=\sum _{E\subseteq T}(-1)^{|T|-|E|} v(E)$.

Collection $\{ \beta _T \}_{T \in \mathcal {F}}$ is called the Möbius inversion of $v$. A totally monotone game $v$ can be characterized by coefficients $\beta _T$ for all $T \in \mathcal {F}$. The following lemma is shown by Shafer (1976).

Lemma 4

(Shafer 1976) For any $v \in {\mathbb {R}}^{\mathcal {F}},\,v$ is totally monotone if and only if in unique representation (5), $\beta _T$ is non-negative for all $T \in \mathcal {F}$.

Gilboa and Schmeidler (1994) prove the following lemma with respect to the additivity of Choquet integrals through Möbius inversions.

Lemma 5

Let $x \in {\mathbb {R}}^{\varOmega }$ and let $v= \sum _{T \in \mathcal {F}} \beta _T u_T \in {\mathbb {R}}^{\mathcal {F}}$. Then,

$$\begin{aligned} \int x \mathrm{{d}}v= \sum _{T \in \mathcal {F}} \beta _T \int x \mathrm{{d}}u_T = \sum _{T \in \mathcal {F}} \beta _T \min _T x. \end{aligned}$$

The following lemma is a special case of Proposition 3 in Chateauneuf and Jaffray (1989).¹⁷

Lemma 6

Let $v=\sum _{T \in \mathcal {F}} \beta _T u_T$. Then, for all $X,\, Y \in \mathcal {F}$,

$$\begin{aligned} v(X \cup Y) + v(X \cap Y) -v(X) - v(Y) = \sum _{T \in \mathcal {F}_{X,Y}} \beta _T, \end{aligned}$$

where $\mathcal {F}_{X,Y} \equiv \{T\,|\, T \cap (X \cap Y^c) \ne \emptyset ,\,T \cap (X^c \cap Y) \ne \emptyset$, and $T \subseteq X \cup Y \}$.

Kajii et al. (2007) characterize the Möbius inversion of a game $v$ by $v$’s ${\mathcal {E}}$-cominimum additivity.

Theorem 4

(Kajii et al. 2007) Let $\varOmega$ be finite. Let $v= \sum _{T \in \mathcal {F}} \beta _T u_T \in {\mathbb {R}}^{\mathcal {F}}$ be a game. $v$ is ${\mathcal {E}}$-cominimum additive if and only if $\beta _T =0$ for any $T \notin \Upsilon ({\mathcal {E}})$. If ${\mathcal {E}}$ is complete, that is, ${\mathcal {E}} = \Upsilon ({\mathcal {E}}),\,v$ is ${\mathcal {E}}$-cominimum additive if and only if $\beta _T =0$ for any $T \notin {\mathcal {E}}$.

1.2 Appendix 2: Proofs

1.2.1 Proof of Lemma 1

Proof

Suppose that for all pairwise ${\mathcal {E}}$-cominimum functions $a, b, c \in \varPhi _{\varSigma }$ and for all $\alpha \in (0,1),\,I(a) \ge I(b)$ implies $I(\alpha a+(1-\alpha )c) \ge I(\alpha b+(1-\alpha )c)$. First, let us prove the following claim: if $x, y \in \varPhi _{\varSigma }$ are ${\mathcal {E}}$-cominimum and $\alpha \in (0,1)$, then $I(\alpha x+(1-\alpha )y)=\alpha I(x)+(1-\alpha )I(y)$. Indeed, for any $\varepsilon >0,\,(I(x)+\varepsilon )1_{\varOmega }$ satisfies $I((I(x)+\varepsilon )1_{\varOmega })>I(x)$, and $(I(y)+\varepsilon )1_{\varOmega }$ satisfies $I((I(y)+\varepsilon )1_{\varOmega })>I(y)$ by the assumption that $I(\lambda 1_{\varOmega })=\lambda$. Hence, $\alpha I(x)+(1-\alpha )I(y)+\varepsilon = I(\alpha (I(x)+\varepsilon )1_{\varOmega }+(1-\alpha )(I(y)+\varepsilon )1_{\varOmega }) > I(\alpha x+(1-\alpha )(I(y)+\varepsilon )1_{\varOmega })> I(\alpha x+(1-\alpha )y)$. The first inequality holds, since $(I(x)+\varepsilon )1_{\varOmega },\,x$, and $(I(y)+\varepsilon )1_{\varOmega }$ are pairwise ${\mathcal {E}}$-cominimum and the second inequality holds, since $(I(y)+\varepsilon )1_{\varOmega },\,y$, and $x$ are pairwise ${\mathcal {E}}$-cominimum. Since $\varepsilon$ is any positive number, we obtain that $\alpha I(x)+(1-\alpha )I(y) \ge I(\alpha x+(1-\alpha )y)$. Furthermore, the converse inequality can be shown by using a similar argument as that for $\varepsilon <0$. Therefore, it is proved that $I(\alpha x+(1-\alpha )y)=\alpha I(x)+(1-\alpha )I(y)$. Then, our claim is proved. Next, let us use this claim twice. First, let $\alpha =1/2,\,x=2a$, and $y=0$. Then, $I(a)=(1/2)I(2a)$ for all $a \in \varPhi _{\varSigma }$. Similarly, let $\alpha =1/2,\,y=2b$, and $x=0$. Then, $I(b)=(1/2)I(2b)$ for all $b \in \varPhi _{\varSigma }$. Second, let $\alpha =1/2,\,x=2a$, and $y=2b$. Then, for all $a,\, b \in \varPhi _{\varSigma },\,I(a+b)=(1/2)I(2a)+(1/2)I(2b)=I(a)+I(b)$. Now we show Lemma 1. $\square$

1.2.2 Proof of Lemma 2

Proof

Since the proof for (ii) $\Rightarrow$ (i) has been already provided in Sect. 3, we prove the converse.

(i) $\Rightarrow$ (ii). Let $\varOmega$ be an arbitrary infinite set and ${\mathcal {E}}=\{E_1, E_2, \dots , E_n\} \subseteq \mathcal {F}$ with $|E_i|\ge 2$ for $i=1,\ldots ,n$ be a collection of events. Denote by $\varPi$ the set of all finite partitions of $\varOmega$ and by $\varPi _{{\mathcal {E}}}$ the set of all finite partitions of $\varOmega$, which separate each $E_i$ for $i=1,\ldots ,n$ to at least two non-empty subsets. That is, if $\mathcal {P}=\{P_1, P_2, \dots , P_k\} \in \varPi _{{\mathcal {E}}}$ is a partition of $\varOmega$, then for every $E_i$ for $i=1,\ldots ,n$, there are at least two non-empty sets in $P_1\cap E_i, P_2 \cap E_i, \dots , P_k \cap E_i$. Fix a partition $\mathcal {P}=\{P_1, P_2, \dots , P_k\} \in \varPi _{{\mathcal {E}}}$. Let $\sigma (\mathcal {P}, {\mathcal {E}}) \subseteq \varSigma$ be the algebra generated by $\mathcal {P}$ and ${\mathcal {E}}$: the smallest algebra containing $P_1, P_2, \dots , P_k, E_1, E_2,\dots , E_n$. Denote by $\varPhi _{\sigma (\mathcal {P}, {\mathcal {E}}) }$ the set of all $\sigma (\mathcal {P}, {\mathcal {E}})$-measurable functions in $\varPhi _{\varSigma }$. Note that $\bigcup _{\mathcal {P} \in \varPi _{{\mathcal {E}}}} \varPhi _{\sigma (\mathcal {P}, {\mathcal {E}})}=\bigcup _{\mathcal {P} \in \varPi } \varPhi _{\sigma (\mathcal {P}, {\mathcal {E}})} =\varPhi _{\varSigma }$. Let $\varOmega _{\mathcal {P}, {\mathcal {E}}}$ be the collection of minimal elements of $\sigma (\mathcal {P}, {\mathcal {E}})$, which constitutes a well defined partition of $\varOmega$. Note that $\varOmega _{\mathcal {P}, {\mathcal {E}}}$ is a collection of the subsets of $\varOmega$, not a set of states. We explain this set $\varOmega _{\mathcal {P}, {\mathcal {E}}}$ in detail. Let $E^i_1,\, E^i_2,\, \dots , E^i_{m_i}$ be non-empty sets in $\sigma (\mathcal {P}, {\mathcal {E}})$ that are subsets of $E_i$, where $m_i \ge 2$. That is, $\{E^i_1, E^i_2, \dots , E^i_{m_i}\}$ constitutes a partition of $E_i$ for $i=1,\ldots ,n$. Moreover, denote the non-empty sets in $\varOmega _{\mathcal {P}, {\mathcal {E}}}$, which are in $(\cup _{1 \le i \le n}E_i)^c$, by $Q_1, Q_2, \dots , Q_m$. Thus, the collection $\{E^1_1, E^1_2, \dots , E^1_{m_1}, \dots ,E^i_1, E^i_2, \dots , E^i_{m_i}, \dots , E^n_1, E^n_2, \dots , E^n_{m_n}, Q_1, Q_2, \dots , Q_m\}$ constitutes $\varOmega _{\mathcal {P}, {\mathcal {E}}}$. Choose one element $e^i_j$ from each $E^i_j$ for $i=1,\ldots ,n$ and $j=1,\ldots ,m_i$ arbitrarily, and choose $q_i$ from each $Q_i$ for $i=1,\ldots ,m$ arbitrarily. Denote finite set $\{e^i_j | 1 \le i \le n, 1 \le j \le m_i\} \cup \{q_i | 1 \le i \le m\}$ by $\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}$, finite set $\{e^i_j | 1 \le j \le m_i\}$ by $E^*_i$ for each $i=1,\ldots ,n$, and finite set $\{q_i | 1 \le i \le m\}$ by $Q^*$. Thus, $\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}=E^*_1 \cup \dots \cup E^*_n \cup Q^*$. Note that by construction, $\varOmega ^* _{\mathcal {P}, {\mathcal {E}}}$ is a subset of $\varOmega$. Denote by ${\mathcal {E}}^*$ collection $\{E^*_1, E^*_2, \dots , E^*_n\}$ and by $\mathcal {F}_1^*$ the collection of singleton subsets in $2^{\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}}$. Moreover, denote by $\varPhi ^*_{\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}}$ the set of real-valued functions on finite set $\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}$.

Now, we define capacity $v^*_{\mathcal {P}, {\mathcal {E}}}$ on power set $2^{\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}}$ as follows: for every $X \in 2^{\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}},\,v^*_{\mathcal {P}, {\mathcal {E}}}(X)=v(( \cup _{e^i_j \in X} E^i_j) \cup (\cup _{q_i \in X}Q_i) )$. For example, $v^*_{\mathcal {P}, {\mathcal {E}}}(\emptyset )=0, v^*_{\mathcal {P}, {\mathcal {E}}}(\{e^1_2, q_3\})=v(E^1_2 \cup Q_3), v^*_{\mathcal {P}, {\mathcal {E}}}(E^*_i) =v^*_{\mathcal {P}, {\mathcal {E}}}(\{e^i_1,\dots , e^i_{m_i}\})=v(\cup _{1 \le j \le m_i}E^i_j) =v(E_i)$.

Next, we define a function $a^*$ on $\varOmega ^* _{\mathcal {P}, {\mathcal {E}}}$ that corresponds to a function $a \in \varPhi _{\sigma (\mathcal {P}, {\mathcal {E}})}$. For every $\sigma (\mathcal {P}, {\mathcal {E}})$-measurable function $a \in \varPhi _{\sigma (\mathcal {P}, {\mathcal {E}})}$, define the function $a^* \in \varPhi ^*_{\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}}$ as follows: for any $\omega \in \varOmega ^*_{\mathcal {P}, {\mathcal {E}}},\,a^*(\omega )=a(\omega )$ naturally; that is, $a^*(e^i_j)=a(e^i_j)$ and $a^*(q_j)=a(q_j)$. The value $a^*(\omega )$ is the same on each $E^i_j$ and on each $Q_j$ regardless of the choice of the representation element, since $a$ is constant on each $E^i_j$ and on each $Q_j$ by the assumption of $a$ being $\sigma (\mathcal {P}, {\mathcal {E}})$-measurable.

Define by $I^*(a^*)$ the Choquet integral of $a^*$ with respect to capacity $v^*_{\mathcal {P}, {\mathcal {E}}}$; that is, $I^*(a^*) \equiv \int _{\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}}a^* dv^*_{\mathcal {P}, {\mathcal {E}}}$. We show that $I^*(a^*) = I(a)$ for all $a \in \varPhi _{\sigma (\mathcal {P}, {\mathcal {E}})}$. Suppose $a^*(\omega _1) \ge a^*(\omega _2) \ge \dots \ge a^*(\omega _t)$. Since $a$ is $\sigma (\mathcal {P},{\mathcal {E}})$-measurable, it follows that

$$\begin{aligned} I^*(a^*)&= (a^*(\omega _1)-a^*(\omega _2))v^*_{\mathcal {P}, {\mathcal {E}}}(\{\omega _1\}) +(a^*(\omega _2)-a^*(\omega _3))v^*_{\mathcal {P}, {\mathcal {E}}}(\{\omega _1,\omega _2\}) \\&+\,(a^*(\omega _3)-a^*(\omega _4))v^*_{\mathcal {P}, {\mathcal {E}}}(\{\omega _1,\omega _2,\omega _3\}) +\cdots = (a(\omega _1)-a(\omega _2))\\&\times \, v(\cup _{\omega _1 \in X_1 \in \Omega _{\mathcal {P}, {\mathcal {E}}}} X_1) +(a(\omega _2)-a(\omega _3))v(\cup _{\omega _j \in X_j \in \Omega _{\mathcal {P}, {\mathcal {E}}}, (j=1,2)} X_j) \\&+\,(a(\omega _3)-a(\omega _4))v(\cup _{\omega _j \in X_j \in \Omega _{\mathcal {P}, {\mathcal {E}}, (j=1,2,3)}} X_j ) +\cdots =I(a). \end{aligned}$$

Here, we claim that for every $a,b \in \varPhi _{\sigma (\mathcal {P}, {\mathcal {E}})}$, the condition that $a$ and $b$ are ${\mathcal {E}}$-cominimum on $\varOmega$ is equivalent to the condition that $a^*$ and $b^*$ are ${\mathcal {E}}^*$-cominimum on $\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}$. Indeed, suppose that $a^*$ and $b^*$ are ${\mathcal {E}}^*$-cominimum. Then, there exists an $\omega _i \in \text {argmin}_{\omega \in E^*_i}a^*(\omega ) \cap \text {argmin}_{\omega \in E^*_i} b^*(\omega )$ for every $i=1,\ldots ,n$. Such an $\omega _i$ satisfies $\omega _i \in \text {argmin}_{\omega \in E_i}a(\omega ) \cap \text {argmin}_{\omega \in E_i} b(\omega )$, since $a \in \varPhi _{\sigma (\mathcal {P}, {\mathcal {E}})}$ is a $\sigma (\mathcal {P}, {\mathcal {E}})$-measurable function. Thus, $a$ and $b$ are ${\mathcal {E}}$-cominimum. The converse also holds similarly. Hence, our claim is shown.

Now, let $a,b \in \varPhi _{\sigma (\mathcal {P}, {\mathcal {E}})}$ and suppose that $a^*$ and $b^*$ are ${\mathcal {E}}^*$-cominimum. Then, $a$ and $b$ are ${\mathcal {E}}$-cominimum, and thus, $I(a+b)=I(a)+I(b)$, since $I$ is ${\mathcal {E}}$-cominimum additive. By $I^*(a^*) = I(a)$ for all $a \in \varPhi _{\sigma (\mathcal {P}, {\mathcal {E}})}$, it holds that $I^*(a^*+b^*)=I^*(a^*)+I^*(b^*)$ since $(a+b)^*=a^*+b^*$. Thus, it holds that $I^*$ is ${\mathcal {E}}^*$-cominimum additive. Note that if $\mathcal {F}_1\cup {\mathcal {E}}$ is complete, then $\mathcal {F}_1^*\cup {\mathcal {E}}^*$ is clearly complete. Since $\varOmega _{\mathcal {P}, {\mathcal {E}}}^*$ is a finite set, we can apply Theorem 4. Thus, there exist coefficients $\{\beta _{\{\omega \}}^{\mathcal {P}, {\mathcal {E}}}\}_{\omega \in \varOmega ^*_{\mathcal {P}, {\mathcal {E}}}}, \{\beta _{E^*_i}^{\mathcal {P}, {\mathcal {E}}}\}_{1 \le i \le n}$ such that

$$\begin{aligned} v^*_{\mathcal {P}, {\mathcal {E}}} =\sum _{\omega \in \varOmega ^*_{\mathcal {P}, {\mathcal {E}}} } \beta _{\{\omega \}}^{\mathcal {P}, {\mathcal {E}}}u_{\{\omega \}} + \sum _{1 \le i \le n} \beta _{E^*_i}^{\mathcal {P}, {\mathcal {E}}}u_{E^*_i}. \end{aligned}$$

(6)

Note that for all $\omega \in \varOmega ^*_{\mathcal {P}, {\mathcal {E}}},\,\beta _{\{\omega \}}^{\mathcal {P}, {\mathcal {E}}}=v^*_{\mathcal {P}, {\mathcal {E}}} (\{\omega \}) \ge 0$ since $v^*_{\mathcal {P}, {\mathcal {E}}}$ is a capacity. Take any $a \in \varPhi _{\sigma (\mathcal {P},{\mathcal {E}})}$. Then, for $a^* \in \varPhi ^* _{\varOmega ^* _{\mathcal {P},{\mathcal {E}}}}$, by Lemma 5, it holds that

$$\begin{aligned} I^*(a^*) = \int \limits _{\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}}a^* \mathrm{{d}}v^*_{\mathcal {P}, {\mathcal {E}}} =\sum _{\omega \in \varOmega ^*_{\mathcal {P}, {\mathcal {E}}}} \beta _{\{\omega \}}^{\mathcal {P}, {\mathcal {E}}}a^*(\omega ) + \sum _{1 \le i \le n} \beta _{E^*_i}^{\mathcal {P}, {\mathcal {E}}} \min _{\omega \in E^*_i}a^*(\omega ). \end{aligned}$$

(7)

Note that $\min _{\omega \in E^*_i}a^*(\omega )=\min _{\omega \in E_i}a(\omega )$ for all $a \in \varPhi _{\sigma (\mathcal {P}, {\mathcal {E}})}$. Thus, we can use notation $\{\beta _{E_i}^{\mathcal {P}, {\mathcal {E}}}\}_{1 \le i \le n}$ instead of $\{\beta _{E^*_i}^{\mathcal {P}, {\mathcal {E}}}\}_{1 \le i \le n}$. Moreover, it holds that $a^*(\omega )=a( \omega )$ for all $\omega \in \varOmega ^*_{\mathcal {P}, {\mathcal {E}}}$. Thus, by $I^*(a^*)=I(a)$, (6) and (7) can be rewritten as follows:

$$\begin{aligned} v^*_{\mathcal {P}, {\mathcal {E}}}&= \sum _{\omega \in \varOmega ^*_{\mathcal {P}, {\mathcal {E}}}} \beta _{\{\omega \}}^{\mathcal {P}, {\mathcal {E}}}u_{\{\omega \}} + \sum _{1 \le i \le n} \beta _{E_i}^{\mathcal {P}, {\mathcal {E}}}u_{E^*_i}; \end{aligned}$$

(8)

$$\begin{aligned} I(a)&= \int \limits _{\varOmega } a \mathrm{{d}}v =\sum _{\omega \in \varOmega ^*_{\mathcal {P}, {\mathcal {E}}}} \beta _{\{\omega \}}^{\mathcal {P}, {\mathcal {E}}}a(\omega ) + \sum _{1 \le i \le n} \beta _{E_i}^{\mathcal {P}, {\mathcal {E}}} \min _{\omega \in E_i}a(\omega ). \end{aligned}$$

(9)

Now, define a finitely additive measure $\mu _{\mathcal {P}, {\mathcal {E}}}$ on $\sigma (\mathcal {P}, {\mathcal {E}})$ as follows: $\mu _{\mathcal {P}, {\mathcal {E}}}(E^i_j)= \beta _{\{e^i_j\}}^{\mathcal {P}, {\mathcal {E}}}$ for every $E^i_j \in \varOmega _{\mathcal {P}, {\mathcal {E}}}$ for $i=1,\ldots ,n$ and $j=1,\ldots ,m_i$; and $\mu _{\mathcal {P}, {\mathcal {E}}}(Q_j)= \beta _{\{q_j\}}^{\mathcal {P}, {\mathcal {E}}}$ for every $Q_j \in \varOmega _{\mathcal {P}, {\mathcal {E}}}$ for $j=1,\ldots ,m$. Then, (9) can be rewritten as follows:

$$\begin{aligned} I(a)=\int \limits _{\varOmega } a(\omega ) \mathrm{{d}} \mu _{\mathcal {P}, {\mathcal {E}}}(\omega ) +\sum _{1 \le i \le n} \beta _{E_i}^{\mathcal {P}, {\mathcal {E}}} \min _{\omega \in E_i}a(\omega ). \end{aligned}$$

(10)

Next, we shall show that finitely additive measure $\mu _{\mathcal {P}, {\mathcal {E}}}$ and coefficients $\{\beta _{E_i}^{\mathcal {P}, {\mathcal {E}}}\}_{1 \le i \le n}$ in expression (10) do not depend on the choice of partition $\mathcal {P} \in \varPi _{{\mathcal {E}}}$. In the proof, the uniqueness of the Möbius inversion plays a crucial role. Let us take another partition $\mathcal {P}^{\prime } \in \varPi _{{\mathcal {E}}}$ such that $\sigma (\mathcal {P}, {\mathcal {E}}) \subseteq \sigma (\mathcal {P}^{\prime }, {\mathcal {E}})$, and repeat the above procedure. Let us define another finite set $\varOmega ^*_{\mathcal {P}^{\prime }, {\mathcal {E}}}$. We can choose the elements in $\varOmega ^*_{\mathcal {P}^{\prime }, {\mathcal {E}}}$ such that $\varOmega ^*_{\mathcal {P}, {\mathcal {E}}} \subseteq \varOmega ^*_{\mathcal {P}^{\prime }, {\mathcal {E}}}$. For such a set $\varOmega ^*_{\mathcal {P}^{\prime }, {\mathcal {E}}}$, let us denote ${E^{\prime }}^*_i=\{\omega \in \varOmega ^*_{\mathcal {P}^{\prime }, {\mathcal {E}}}| \omega \in E_i\}$, which is the set of representation elements in $E_i$. Then, it holds that $E^*_i \subseteq {E^{\prime }}^*_i$ for all $i=1,\ldots ,n$. By defining capacity $v^{*}_{\mathcal {P}^{\prime }, {\mathcal {E}}}$ on finite power set $2^{\varOmega ^*_{\mathcal {P}^{\prime }, {\mathcal {E}}}}$, we can show that there exist coefficients $\{\beta _{\{\omega \}}^{\mathcal {P}^{\prime }, {\mathcal {E}}}\}_{\omega \in \varOmega ^*_{\mathcal {P}^{\prime }, {\mathcal {E}}}}, \{\beta _{E_i}^{\mathcal {P}^{\prime }, {\mathcal {E}}}\}_{1 \le i \le n}$ such that

$$\begin{aligned} v^*_{\mathcal {P}^{\prime }, {\mathcal {E}}} =\sum _{\omega \in \varOmega ^*_{\mathcal {P}^{\prime }, {\mathcal {E}}}} \beta _{\{\omega \}}^{\mathcal {P}^{\prime }, {\mathcal {E}}}u_{\{\omega \}} + \sum _{1 \le i \le n} \beta _{E_i}^{\mathcal {P}^{\prime }, {\mathcal {E}}}u_{{E^{\prime }}^*_i}, \end{aligned}$$

(11)

and that there exists a finitely additive measure $\mu _{\mathcal {P}^{\prime }, {\mathcal {E}}}$ on $\sigma (\mathcal {P}^{\prime }, {\mathcal {E}})$ such that for all $a \in \varPhi _{\sigma (\mathcal {P}^{\prime }, {\mathcal {E}}) }$,

$$\begin{aligned} I(a)=\int \limits _{\varOmega } a(\omega ) \mathrm{{d}} \mu _{\mathcal {P}^{\prime }, {\mathcal {E}}}(\omega ) +\sum _{1 \le i \le n} \beta _{E_i}^{\mathcal {P}^{\prime }, {\mathcal {E}}} \min _{\omega \in E_i}a(\omega ). \end{aligned}$$

(12)

To complete our proof, we have to show that $\beta _{E_i}^{\mathcal {P}^{\prime }, {\mathcal {E}}}= \beta _{E_i}^{\mathcal {P}, {\mathcal {E}}}$ for all $i=1,\ldots ,n$ and that $\mu _{\mathcal {P}^{\prime }, {\mathcal {E}}}=\mu _{\mathcal {P}, {\mathcal {E}}}$ on $\sigma (\mathcal {P}, {\mathcal {E}})$. To do so, we need to define another capacity $V^*$ on $2^{\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}}$ as follows. First, for every element $\omega \in \varOmega ^*_{\mathcal {P}, {\mathcal {E}}}$, define $\Gamma (\omega )=\{ X \in \varOmega _{\mathcal {P}^{\prime }, {\mathcal {E}}}| X \subseteq Y$ such that $\omega \in Y \in \varOmega _{\mathcal {P}, {\mathcal {E}}} \}$. In other words, $\Gamma (\omega )$ is the partition of $Y$ with respect to algebra $\sigma (\mathcal {P}^{\prime }, {\mathcal {E}})$ where $Y$ is the cell in partition $\varOmega _{\mathcal {P}, {\mathcal {E}}}$ such that $\omega \in Y$. Second, for every element $\omega \in \varOmega ^*_{\mathcal {P}, {\mathcal {E}}}$, define $\gamma _{\{\omega \}}=\sum _{X \in \Gamma (\omega )} \mu _{\mathcal {P}^{\prime }, {\mathcal {E}}}(X)$. Third, define a capacity $V^*$ on $2^{\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}}$ by

$$\begin{aligned} V^* =\sum _{\omega \in \varOmega ^*_{\mathcal {P}, {\mathcal {E}}}} \gamma _{\{\omega \}}u_{\{\omega \}} + \sum _{1 \le i \le n} \beta _{E_i}^{\mathcal {P}^{\prime }, {\mathcal {E}}}u_{E^*_i}. \end{aligned}$$

(13)

Note that coefficient $\beta _{E_i}^{\mathcal {P}^{\prime }, {\mathcal {E}}}$ has a prime symbol on $\mathcal {P}$ and that unanimity game $u_{E^*_i}$ has no prime symbol on $E$. Let us show that $V^*$ in (13) coincides with $v^*_{\mathcal {P}, {\mathcal {E}}}$ in (8). Indeed, pick any $T \in 2^{\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}}$ and define $\tilde{T} \in \sigma (\mathcal {P}, {\mathcal {E}})$ by $\tilde{T}=\bigcup _{T \cap X \ne \emptyset , X \in \varOmega _{\mathcal {P}, {\mathcal {E}}} }X$. This set $\tilde{T}$ is the minimal set in $\sigma (\mathcal {P}, {\mathcal {E}})$ that contains all the elements of $T$. Then,

$$\begin{aligned} V^* (T)&= \sum _{\omega \in T} \gamma _{\{\omega \}} + \sum _{E^*_i \subseteq T} \beta _{E_i}^{\mathcal {P}^{\prime }, {\mathcal {E}} } = \sum _{\omega \in T} \sum _{X \in \Gamma (\omega )} \mu _{\mathcal {P}^{\prime }, {\mathcal {E}}}(X) + \sum _{E_i \subseteq \tilde{T}} \beta _{E_i}^{\mathcal {P}^{\prime }, {\mathcal {E}} } \\&= \sum _{X \subseteq \tilde{T}, X \in \varOmega _{\mathcal {P}^{\prime }, {\mathcal {E}}}} \mu _{\mathcal {P}^{\prime }, {\mathcal {E}}}(X) +\sum _{E_i \subseteq \tilde{T}} \beta _{E_i}^{\mathcal {P}^{\prime }, {\mathcal {E}} } = \mu _{\mathcal {P}^{\prime }, {\mathcal {E}}}(\tilde{T})+\sum _{E_i \subseteq \tilde{T}} \beta _{E_i}^{\mathcal {P}^{\prime }, {\mathcal {E}} } \\&= I(1_{\tilde{T}})=I^*(1^*_{T})=v^*_{\mathcal {P}, {\mathcal {E}}} (T). \end{aligned}$$

Hence, $V^*$ coincides with $v^* _{\mathcal {P}, {\mathcal {E}}}$ for all $T \in 2^{\varOmega ^* _{\mathcal {P}, {\mathcal {E}}}}$. Therefore, by the uniqueness of the Möbius inversion, it must hold that $\beta _{E_i}^{\mathcal {P}^{\prime }, {\mathcal {E}}}= \beta _{E_i}^{\mathcal {P}, {\mathcal {E}}}$ for all $i=1,\ldots ,n$ and that $\mu _{\mathcal {P}^{\prime }, {\mathcal {E}}}=\mu _{\mathcal {P}, {\mathcal {E}}}$ on $\sigma (\mathcal {P}, {\mathcal {E}})$. Then, when defining $\beta _{E_i}^{\mathcal {P}, {\mathcal {E}}}=\varepsilon _i$ and $\mu _{\mathcal {P}, {\mathcal {E}}}=\mu$, we obtain our expression (1). $\square$

1.2.3 Proof of (only if part) of Theorem 3

Proof

Since A1, A3, A4, A5, and A6 hold, by Theorem 2, there exist a unique finitely additive measure $\mu$ on $(\varOmega , \varSigma )$, an affine function $u$, and a set of coefficients $\varepsilon _1,\varepsilon _2,\dots ,\varepsilon _n$, such that $f \succeq g \Leftrightarrow J(f) \ge J(g)$ where $J(f)= \int _{\varOmega } u(f(\omega ))\mathrm{{d}}\mu (\omega )+ \sum _{i=1}^n \varepsilon _i \min _{\omega \in E_i} u(f(\omega ))$. Therefore, it suffices to show that all coefficients $\varepsilon _1,\ldots , \varepsilon _n$ are non-negative. Let $U$ be the function from $L_0$ to $B_0(K)$ and $I$ be the operator on $B_0(K)$, both of which are defined in Property (3) of the proof of Theorem 2. Then, for every $a = U(f),\,I(a)$ is Choquet integral $\int a \mathrm{{d}}v$ with respect to capacity $v(T)=I(1_T)$. Moreover, by A7, it holds that $v(X \cup Y)+v(X \cap Y) \ge v(X)+v(Y)$ for all $X, Y \in 2^{\varOmega }$.

Now, pick any $E_i \in {\mathcal {E}}$. By the assumption of ${\mathcal {E}}$ being simple-complete, we can find a two-element set $\{p,q\}\subseteq E_i$ such that there is no $E_j \in {\mathcal {E}}$ satisfying $\{p,q\} \subseteq E_j \subsetneq E_i$. Moreover, take a partition $\mathcal {P} \in \varPi _{{\mathcal {E}}}$ in the proof of Lemma 2 such that $p, q \in \varOmega ^*_{\mathcal {P}, {\mathcal {E}}}$. Recall that for every $X \in 2^{\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}},\,v^*_{\mathcal {P}, {\mathcal {E}}}(X)=v(\bigcup _{e^i_j \in X} E^i_j \cup \bigcup _{q_i \in X}Q_i )$. Thus, $v(X \cup Y)+v(X \cap Y) \ge v(X)+v(Y)$ for all $X, Y \in 2^{\varOmega }$ implies that $v^*_{\mathcal {P}, {\mathcal {E}}}(X \cup Y)+v^*_{\mathcal {P}, {\mathcal {E}}}(X \cap Y) \ge v^*_{\mathcal {P}, {\mathcal {E}}}(X)+v^*_{\mathcal {P}, {\mathcal {E}}}(Y)$ for all $X,Y \in 2^{\varOmega ^*_{\mathcal {P}, {\mathcal {E}}}}$.

Here, it holds that there is no $E^*_j \in {\mathcal {E}}^*$ satisfying $\{p,q\} \subseteq E^*_j \subsetneq E^*_i$. Let $T_1=E^*_i \backslash \{p\}$ and $T_2=E^*_i \backslash \{q\}$. Thus, $T_1 \cup T_2=E_i^*$. Note that for every $S \subseteq T_1 \cup T_2,\,S \not \subseteq T_1$ and $S \not \subseteq T_2$ are equivalent to $\{p, q\} \subseteq S$. It follows that

$$\begin{aligned} 0&\le v^*_{\mathcal {P}, {\mathcal {E}}}(T_1 \cup T_2)-v^*_{\mathcal {P}, {\mathcal {E}}}(T_1) -v^*_{\mathcal {P}, {\mathcal {E}}}(T_2)+v^*_{\mathcal {P}, {\mathcal {E}}}(T_1 \cap T_2) \\&= \sum _{E_j^* \subseteq T_1 \cup T_2, E_j^* \not \subseteq T_1, E_j^* \not \subseteq T_2} \varepsilon _j =\sum _{\{p,\,q\} \subseteq E_j^* \subseteq E_i^*} \varepsilon _j = \varepsilon _i, \end{aligned}$$

where the inequality holds by the convexity of $v^*_{\mathcal {P}, {\mathcal {E}}}$ and the first equality holds by Lemma 6. $\square$