Rational preference and rationalizable choice

Abstract

We study a decision maker characterized by two binary relations. The first reflects his judgments about well-being, his mental preferences. The second describes the decision maker’s choice behavior, his behavioral preferences. We propose axioms that describe a relation between these two preferences, so between mind and behavior, thus disentangling two different perspectives on preferences: a description of tastes (and attitudes) and a way to organize behavioral data. We obtain two representations: one in which mental preferences uniquely determine choice behavior, another for which mental preferences direct behavior but room remains for biases and framing effects. Our results also provide a foundation for a decision analysis procedure called robust ordinal regression and proposed by Greco et al. (Eur J Oper Res 191:416–436, 2008).

A Proofs

1.1 Proofs of Theorems 1 and 2 plus related results

Lemma 2

Let $\succsim ^{*}$ and $\succsim ^{\circ }$ be two binary relations on F such that $\succsim ^{{ \circ }}$ satisfies Completeness and $\left( \succsim ^{*},\succsim ^{{ \circ }}\right)$ satisfies Consistency. The following conditions are equivalent:

• (i) $f\succ ^{*}g\iff f\succ ^{\circ }g$;

• (ii) ${f\not \succsim ^{*}g\implies g\succsim ^{\circ }f}$ (Possibility) and $f\succ ^{*}g\implies f\succ ^{\circ }g$ (Strict Consistency);

• (iii) $g\succsim ^{\circ }f\iff f\not \succ ^{*}g$.

In this case, $f\sim ^{\circ }g$ if and only if either f and g are $\succsim ^{*}$ incomparable or $f\sim ^{*}g$.

Completeness of $\succsim ^{\circ }$ and its Consistency with $\succsim ^{*}$ say that the former is a behavioral completion of the latter. The lemma shows that Possibility and Strict Consistency uniquely pin down $\succsim ^{\circ }$: Its asymmetric part coincides with that of $\succsim ^{*}$, and its symmetric part is the union of $\succsim ^{*}$ indifference and incomparability.

Proof of Lemma 2

Since $\succsim ^{\circ }$ is complete, then $g\succsim ^{\circ }f\iff f\not \succ ^{\circ }g$. Thus, (i) $f\succ ^{*}g\iff f\succ ^{\circ }g$ is equivalent to $f\not \succ ^{*}g\iff f\not \succ ^{\circ }g$ which is equivalent to $f\not \succ ^{*}g\iff g\succsim ^{\circ }f$ which is (iii). This shows (i) $\iff$ (iii).¹⁹

By (i) $f\succ ^{*}g\implies f\succ ^{\circ }g$, moreover $f\not \succsim ^{*}g$ implies $f\not \succ ^{*}g$ which by (iii) implies $g\succsim ^{\circ }f$. So (i) $\implies$ (ii). Conversely, by (ii) $f\succ ^{*}g\implies f\succ ^{\circ }g$. Now if $f\succ ^{\circ }g$, then ${g\not \succsim ^{\circ }f}$, and (ii) implies $f\succsim ^{*}g$; if we had $g\succsim ^{*}f$, Consistency of $\succsim ^{\circ }$ with $\succsim ^{*}$ would imply $g\succsim ^{\circ }f$ which is impossible, then ${g\not \succsim ^{*}f}$ and $f\succ ^{*}g$. This shows (ii) $\implies$ (i), concluding the proof of the first part of the statement.

Now if $f\not \sim ^{\circ }g$, because of completeness of $\succsim ^{\circ }$ either $f\succ ^{\circ }g$ or $g\succ ^{\circ }f$. Say $f\succ ^{\circ }g$, by (i), $f\succ ^{*}g$, so that f and g are neither $\succsim ^{*}$ incomparable nor $\succsim ^{*}$ indifferent. Conversely, if f and g are neither $\succsim ^{*}$ incomparable nor $\succsim ^{*}$ indifferent, comparability implies either $f\succsim ^{*}g$ or $g\succsim ^{*}f$, non-indifference means that they cannot both hold so that $f\succ ^{*}g$ or $g\succ ^{*}f$. Say $f\succ ^{*}g$, by (i), $f\succ ^{\circ }g$, so that f and g are not $\succsim ^{\circ }$ indifferent. $\square$

Proof of Lemma 1

On constant acts, $\succsim ^{*}$ is non-trivial and satisfies the axioms of Herstein and Milnor (1953). Therefore there exists a non-constant affine $u:X\rightarrow {\mathbb {R}}$ such that, given $x,y\in X$, $x\succsim ^{*}y$ if and only if $u\left( x\right) \ge u\left( y\right)$.

• (a) Take $x,y\in X$ such that $x\succ ^{*}y$. If $f(s)\succsim ^{*} g(s)$ for all $s\in S$, then

  $$\begin{aligned} u\left( f(s)\right) \ge u\left( g(s)\right) \qquad \forall s\in S\text {.} \end{aligned}$$

  Therefore, for all $s\in S$ and all $\lambda \in \left( 0,1\right)$,

  $$\begin{aligned}&\lambda u\left( f(s)\right) +\left( 1-\lambda \right) u\left( x\right)>\lambda u\left( g(s)\right) +\left( 1-\lambda \right) u\left( y\right) \\&\quad \Longrightarrow u\left( \lambda f(s)+\left( 1-\lambda \right) x\right) >u\left( \lambda g(s)+\left( 1-\lambda \right) y\right) \\&\quad \Longrightarrow \lambda f(s)+\left( 1-\lambda \right) x\left. \succ ^{*}\right. \lambda g(s)+\left( 1-\lambda \right) y. \end{aligned}$$

  By Monotonicity, this implies $\lambda f+\left( 1-\lambda \right) x\succ ^{*}\lambda g+\left( 1-\lambda \right) y$ for all $\lambda \in \left( 0,1\right)$, and Continuity delivers $f\succsim ^{*}g$.

• (b) This proof is due to Shapley and Baucells (1998), and we report it for the sake of completeness. Let $f,g,h\in F$ and $\lambda \in \left( 0,1\right)$ be such that $\lambda f+(1-\lambda )h\succsim ^{*}\lambda g+(1-\lambda )h$. Let

  $$\begin{aligned} \bar{\alpha }=\sup \left\{ \alpha \in \left[ 0,1\right] :\alpha f+(1-\alpha )h\succsim ^{*}\alpha g+(1-\alpha )h\right\} . \end{aligned}$$

  Clearly $\bar{\alpha }\ge \lambda >0$ and, by Continuity, $\bar{\alpha } f+(1-\bar{\alpha })h\succsim ^{*}\bar{\alpha }g+(1-\bar{\alpha })h$. Now set $\beta =\dfrac{1}{1+\bar{\alpha }}$ and observe that:

  • $\beta \bar{\alpha }=\dfrac{\bar{\alpha }}{1+\bar{\alpha }}=1-\dfrac{1}{1+\bar{\alpha }}=1-\beta$ and $\beta \left( 1-\bar{\alpha }\right) =\dfrac{1-\bar{\alpha }}{1+\bar{\alpha }}$,

  • Independence yields

    $$\begin{aligned}&\beta \left( \bar{\alpha }f+(1-\bar{\alpha })h\right) +\left( 1-\beta \right) f\succsim ^{*}\beta \left( \bar{\alpha }g+(1-\bar{\alpha })h\right) +\left( 1-\beta \right) f\\&\quad =\beta \bar{\alpha }g+\beta (1-\bar{\alpha })h+\left( 1-\beta \right) f=\left( 1-\beta \right) g+\beta (1-\bar{\alpha })h+\beta \bar{\alpha }f\\&\quad =\beta \bar{\alpha }f+\beta (1-\bar{\alpha })h+\left( 1-\beta \right) g=\beta \left( \bar{\alpha }f+(1-\bar{\alpha })h\right) +\left( 1-\beta \right) g \\&\quad \succsim ^{*}\beta \left( \bar{\alpha }g+(1-\bar{\alpha })h\right) +\left( 1-\beta \right) g \end{aligned}$$

    so that, by Transitivity,

    $$\begin{aligned} \dfrac{2\bar{\alpha }}{1+\bar{\alpha }}f+\dfrac{1-\bar{\alpha }}{1+\bar{\alpha } }h&=\beta \left( \bar{\alpha }f+(1-\bar{\alpha })h\right) \\&\quad +\left( 1-\beta \right) f\succsim ^{*}\beta \left( \bar{\alpha }g+(1-\bar{\alpha })h\right) +\left( 1-\beta \right) g\\&=\dfrac{2\bar{\alpha }}{1+\bar{\alpha }}g+\dfrac{1-\bar{\alpha }}{1+\bar{\alpha }}h. \end{aligned}$$

  But then, by definition of $\bar{\alpha }$, $\dfrac{2\bar{\alpha }}{1+\bar{\alpha }}\le \bar{\alpha }$, that is, $\bar{\alpha }^{2}-\bar{\alpha } \ge 0$. Since $\bar{\alpha }>0$, we have $\bar{\alpha }=1$, and hence $f\succsim ^{*}g$.

Sufficiency of the axioms for representation (13) and its uniqueness properties follow from Theorem 1 of GMMS; necessity is easy to check. Finally, (14) is proved in Proposition 4. $\square$

The next proposition shows that if $\succsim ^{*}$ is a multiple prior (incomplete) preference à la Bewley represented by u and C as in (13), then the algebraic interior of $\succ ^{*}$ admits the representation

and moreover, it coincides with the algebraic interior of $\succsim ^{*}$.

Proposition 4

If C is a non-empty closed and convex set of probabilities on $\Sigma$, $u:X\rightarrow {\mathbb {R}}$ is a non-constant affine function, and, for every $h,l\in F$,

$$\begin{aligned} h\succsim ^{*}l\iff \int u\left( h\right) {\hbox {d}}p\ge \int u\left( l\right) {\hbox {d}}p\qquad \text {for all }p\in C, \end{aligned}$$

then the following conditions are equivalent for f and g in F:

• (i) For every $x\succ ^{*}y$ in X there exists $\varepsilon$ in $\left( 0,1\right)$ such that

  $$\begin{aligned} \left( 1-\varepsilon \right) f+\varepsilon y\succsim ^{*}\left( 1-\varepsilon \right) g+\varepsilon x. \end{aligned}$$

• (ii) There exist $x\succ ^{*}y$ in X and $\varepsilon \$in $\left( 0,1\right)$ such that

  $$\begin{aligned} \left( 1-\varepsilon \right) f+\varepsilon y\succsim ^{*}\left( 1-\varepsilon \right) g+\varepsilon x. \end{aligned}$$

• (iii) $\int u\left( f\right) {\hbox {d}}p>\int u\left( g\right) {\hbox {d}}p$ for all $p\in C$.

• (iv) For every h, l in F, there exists $\varepsilon$ in $\left( 0,1\right)$ such that

  $$\begin{aligned} \left( 1-\delta \right) f+\delta h\succ ^{*}\left( 1-\delta \right) g+\delta l\qquad \text {for all }\delta \in \left[ 0,\varepsilon \right] \end{aligned}$$

  that is, $\left( f,g\right) \in {\text {int}}\,\left( \succ ^{*}\right)$, here denoted .

• (v) For every h, l in F, there exists $\varepsilon$ in $\left( 0,1\right)$ such that

  $$\begin{aligned} \left( 1-\delta \right) f+\delta h\succsim ^{*}\left( 1-\delta \right) g+\delta l\qquad \text {for all }\delta \in \left[ 0,\varepsilon \right] \end{aligned}$$

  that is, $\left( f,g\right) \in {\text {int}}\,\left( \succsim ^{*}\right)$.

Proof of Proposition 4

(i) obviously implies (ii).

(ii) implies (iii). By (ii) there are $x\succ ^{*}y$ in X and $\varepsilon$ in $\left( 0,1\right)$ such that

$$\begin{aligned} \int u\left( \left( 1-\varepsilon \right) f+\varepsilon y\right) {\hbox {d}}p\ge \int u\left( \left( 1-\varepsilon \right) g+\varepsilon x\right) {\hbox {d}}p\qquad \forall p\in C \end{aligned}$$

but then

$$\begin{aligned} \left( 1-\varepsilon \right) \int u\left( f\right) {\hbox {d}}p+\varepsilon u\left( y\right)&\ge \left( 1-\varepsilon \right) \int u\left( g\right) {\hbox {d}}p+\varepsilon u\left( x\right) \qquad \forall p\in C\\ \left( 1-\varepsilon \right) \int u\left( f\right) {\hbox {d}}p&\ge \left( 1-\varepsilon \right) \int u\left( g\right) {\hbox {d}}p+\varepsilon \left( u\left( x\right) -u\left( y\right) \right) \qquad \forall p\in C\\ \int u\left( f\right) {\hbox {d}}p&\ge \int u\left( g\right) {\hbox {d}}p+\frac{\varepsilon }{1-\varepsilon }\left( u\left( x\right) -u\left( y\right) \right) \qquad \forall p\in C \end{aligned}$$

and so $\int u\left( f\right) {\hbox {d}}p>\int u\left( g\right) {\hbox {d}}p$ for all $p\in C$.

(iii) implies (iv). If $\int u\left( f\right) {\hbox {d}}p>\int u\left( g\right) {\hbox {d}}p$ for all $p\in C$, then

$$\begin{aligned} \int \left[ u\left( f\right) -u\left( g\right) \right] {\hbox {d}}p>0\qquad \forall p\in C. \end{aligned}$$

But, C is weak*-compact and $p\mapsto \int \left[ u\left( f\right) -u\left( g\right) \right] {\hbox {d}}p$ is weak*-continuous. Hence, we have

$$\begin{aligned} \int \left[ u\left( f\right) -u\left( g\right) \right] {\hbox {d}}p\ge \eta \qquad \forall p\in C, \end{aligned}$$

where $\eta =\min _{p\in C}\int \left[ u\left( f\right) -u\left( g\right) \right] {\hbox {d}}p>0$. For every h, l in F let $x,y\in X$ be such that $u\left( x\right) \ge u\left( l\left( s\right) \right)$ and $u\left( y\right) \le u\left( h\left( s\right) \right)$ for all $s\in S$. Without loss of generality, assume that $u\left( x\right) \ge u\left( y\right)$.²⁰ Choose $\varepsilon \in \left( 0,1\right)$ such that

$$\begin{aligned} \frac{\varepsilon }{1-\varepsilon }\left( u\left( x\right) -u\left( y\right) \right) <\eta \end{aligned}$$

and consider any $\delta \in \left[ 0,\varepsilon \right]$, then

$$\begin{aligned} \int u\left( f\right) {\hbox {d}}p-\int u\left( g\right) {\hbox {d}}p&\ge \eta>\frac{\varepsilon }{1-\varepsilon }\left( u\left( x\right) -u\left( y\right) \right) \\&\ge \frac{\delta }{1-\delta }\left( u\left( x\right) -u\left( y\right) \right) \qquad \forall p\in C\\ \int u\left( f\right) {\hbox {d}}p&>\int u\left( g\right) {\hbox {d}}p+\frac{\delta }{1-\delta }\left( u\left( x\right) -u\left( y\right) \right) \qquad \forall p\in C\\ \left( 1-\delta \right) \int u\left( f\right) {\hbox {d}}p+\delta u\left( y\right)&>\left( 1-\delta \right) \int u\left( g\right) {\hbox {d}}p+\delta u\left( x\right) \qquad \forall p\in C\\ \left( 1-\delta \right) \int u\left( f\right) {\hbox {d}}p+\delta \int u\left( h\right) {\hbox {d}}p&\ge \left( 1-\delta \right) \int u\left( f\right) {\hbox {d}}p+\delta u\left( y\right) \\&>\left( 1-\delta \right) \int u\left( g\right) {\hbox {d}}p+\delta u\left( x\right) \\&\ge \left( 1-\delta \right) \int u\left( g\right) {\hbox {d}}p+\delta \int u\left( l\right) {\hbox {d}}p\qquad \forall p\in C\\ \int u\left( \left( 1-\delta \right) f+\delta h\right) {\hbox {d}}p&>\int u\left( \left( 1-\delta \right) g+\delta l\right) {\hbox {d}}p\qquad \forall p\in C. \end{aligned}$$

(iv) implies (v) and (v) implies (i) are trivial observations. $\square$

Proofs of Theorems 1 and 2

(i) implies (ii) and (iii) and (18). By Lemma 1, there exist a non-empty closed and convex set C of probabilities on $\Sigma$ and a non-constant affine $u:X\rightarrow {\mathbb {R}}$ such that, for every $f,g\in F$,

(26)

Next we show that $\left( \succsim ^{*},\succsim ^{\circ }\right)$ satisfy Consistency. Assume $f\succsim ^{*}g$, and choose $x\succ ^{*}y$ in X, then for every $\varepsilon \in \left( 0,1\right)$

and Strong Consistency implies

$$\begin{aligned} \left( 1-\varepsilon \right) f+\varepsilon x\succ ^{\circ }\left( 1-\varepsilon \right) g+\varepsilon y\qquad \forall \varepsilon \in \left( 0,1\right) \end{aligned}$$

whence $\left( 1-\varepsilon \right) f+\varepsilon x\succsim ^{\circ }\left( 1-\varepsilon \right) g+\varepsilon y$ for all $\varepsilon \in \left( 0,1\right)$ and $f\succsim ^{\circ }g$ follows by Continuity of $\succsim ^{\circ }$.

Now, Possibility and Consistency imply that $\succsim ^{\circ }$ satisfies Completeness. In turn, Continuity and Completeness of $\succsim ^{\circ }$ imply that given any f, g, h, l in F,

$$\begin{aligned} \Big \{\lambda \in [0,1]:(1-\lambda )f+\lambda h\succ ^{\circ }(1-\lambda )g+\lambda l\Big \} \end{aligned}$$

is open in $\left[ 0,1\right]$. If $f\succ ^{\circ }g$, then 0 belongs to the set for every h, l in F, and so there is $\varepsilon >0$ such that

$$\begin{aligned} (1-\lambda )f+\lambda h\succ ^{\circ }(1-\lambda )g+\lambda l\qquad \forall \lambda \in \left[ 0,\varepsilon \right] \\ (1-\lambda )g+\lambda l\not \succsim ^{\circ }(1-\lambda )f+\lambda h\qquad \forall \lambda \in \left[ 0,\varepsilon \right] \end{aligned}$$

by Possibility

$$\begin{aligned} (1-\lambda )f+\lambda h\succsim ^{*}(1-\lambda )g+\lambda l\qquad \forall \lambda \in \left[ 0,\varepsilon \right] \end{aligned}$$

by Proposition 4,

$$\begin{aligned} \int u\left( f\right) {\hbox {d}}p>\int u\left( g\right) {\hbox {d}}p\qquad \forall p\in C, \end{aligned}$$

that is, . Summing up, $f\succ ^{\circ }g$ implies and the converse is true by Strong Consistency. This shows that (iii) holds because, as already observed, $\succsim ^{\circ }$ is complete. By (26)

which is (18) and Completeness of $\succsim ^{\circ }$ again yields (17). That is (ii) holds.

(ii) implies (iii). The properties of $\succsim ^{*}$ follow from Lemma 1, Completeness of $\succsim ^{\circ }$ from (17), in turn (17) and Completeness of $\succsim ^{\circ }$ yield

$$\begin{aligned} f\succ ^{\circ }g\iff \int u\left( f\right) {\hbox {d}}p>\int u\left( g\right) {\hbox {d}}p {\quad }\forall p\in C. \end{aligned}$$

Then (16) and Proposition 4 deliver .

(iii) implies (i) and (iv). We only have to prove that $\succsim ^{{ \circ }}$ satisfies Continuity and $\left( \succsim ^{*} ,\succsim ^{{ \circ }}\right)$ satisfies Possibility. By Lemma 1, there exist a non-empty closed and convex set C of probabilities on $\Sigma$ and a non-constant affine $u:X\rightarrow {\mathbb {R}}$ such that, for every $f,g\in F$,

$$\begin{aligned} f\succsim ^{*}g\iff \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p\qquad \forall p\in C, \end{aligned}$$

(27)

(28)

Coincidence of $\succ ^{\circ }$ with implies

$$\begin{aligned} f\succ ^{\circ }g\iff \int u\left( f\right) {\hbox {d}}p>\int u\left( g\right) {\hbox {d}}p\qquad \forall p\in C \end{aligned}$$

and Completeness of $\succsim ^{\circ }$ delivers

$$\begin{aligned} f\succsim ^{\circ }g\iff \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p\qquad \text {for some }p\in C \end{aligned}$$

(29)

and Possibility follows from (27) and (29). Moreover, (29) and (28) show that (iii) implies (iv). Given any f, g, h, l in F, set

$$\begin{aligned} \Lambda =\{\lambda \in [0,1]:\lambda f+(1-\lambda )g\succsim ^{\circ }\lambda h+(1-\lambda )l\} \end{aligned}$$

and assume $\lambda _{n}$ is a sequence in $\Lambda$ that converges to $\lambda$. Set $\varphi _{n}=u\left( \lambda _{n}f+(1-\lambda _{n})g\right)$, $\psi _{n}=u\left( \lambda _{n}h+(1-\lambda _{n})l\right)$, $\varphi =u\left( \lambda f+(1-\lambda )g\right)$, $\psi =u\left( \lambda h+(1-\lambda )l\right)$ and observe that, in supnorm, $\varphi _{n}\rightarrow \varphi$ and $\psi _{n}\rightarrow \psi$.²¹ Since $\lambda _{n}$ is a sequence in $\Lambda$, then

$$\begin{aligned} \lambda _{n}f+(1-\lambda _{n})g\succsim ^{\circ }\lambda _{n}h+(1-\lambda _{n})l\qquad \forall n\in {\mathbb {N}}, \end{aligned}$$

(30)

and by (29), for every $n\in {\mathbb {N}}$, there exists $p_{n}$ in C such that

$$\begin{aligned} \int \varphi _{n}{\hbox {d}}p_{n}\ge \int \psi _{n}{\hbox {d}}p_{n}\text {.} \end{aligned}$$

But C is a weak$^{\text {*}}$ compact subset of the space $ba\left( \Sigma \right)$ of all bounded and finitely additive set functions on $\Sigma$, then there exists a subnet $p_{n_{\beta }}$ of $p_{n}$ that weak* converges to some p in C. Since clearly $\varphi _{n_{\beta }} \rightarrow \varphi$ and $\psi _{n_{\beta }}\rightarrow \psi$ in supnorm, and $p_{n_{\beta }}$ is norm bounded in $ba\left( \Sigma \right)$, then

$$\begin{aligned} \begin{array} [c]{ccc} \int \varphi _{n_{\beta }}{\hbox {d}}p_{n_{\beta }} &{} \ge &{} \int \psi _{n_{\beta } }{\hbox {d}}p_{n_{\beta }}\\ \downarrow &{} &{} \downarrow \\ \int \varphi {\hbox {d}}p &{} \ge &{} \int \psi {\hbox {d}}p, \end{array} \end{aligned}$$

that is, $\lambda f+(1-\lambda )g\succsim ^{\circ }\lambda h+(1-\lambda )l$ and so $\lambda \in \Lambda$.²² This proves that $\Lambda$ is closed and $\succsim ^{\circ }$ is continuous.

(iv) implies (ii). By (13) of Lemma 1, $\succsim ^{*}$ admits representation (16), while (14) of Lemma 1 and coincidence of $\succsim ^{\circ }$ with delivers (17).

Uniqueness of C and cardinal uniqueness of u follow from Lemma 1. $\square$

Proof of Proposition 1

Assume there exist $l\nsucc ^{*}h$ in F, such that for every $\gamma \$in $\left( 0,1\right]$

$$\begin{aligned} \left( 1-\gamma \right) g+\gamma l\nsucc ^{*}\left( 1-\gamma \right) f+\gamma h \end{aligned}$$

then there is a sequence $\varepsilon _{n}\rightarrow 0$ such that

$$\begin{aligned} \left( 1-\varepsilon _{n}\right) g+\varepsilon _{n}l\nsucc ^{*}\left( 1-\varepsilon _{n}\right) f+\varepsilon _{n}h\qquad \forall n\in {\mathbb {N}}. \end{aligned}$$

If

$$\begin{aligned} \left( 1-\varepsilon _{n}\right) g+\varepsilon _{n}l\sim ^{*}\left( 1-\varepsilon _{n}\right) f+\varepsilon _{n}h \end{aligned}$$

for infinitely many n’s, then there exists a subsequence $\varepsilon _{n_{k}}$ of $\varepsilon _{n}$ such that

$$\begin{aligned} \left( 1-\varepsilon _{n_{k}}\right) g+\varepsilon _{n_{k}}l\sim ^{*}\left( 1-\varepsilon _{n_{k}}\right) f+\varepsilon _{n_{k}}h\qquad \forall k\in {\mathbb {N}} \end{aligned}$$

by Consistency

$$\begin{aligned} \left( 1-\varepsilon _{n_{k}}\right) f+\varepsilon _{n_{k}}h\sim ^{\circ }\left( 1-\varepsilon _{n_{k}}\right) g+\varepsilon _{n_{k}}l\qquad \forall k\in {\mathbb {N}} \end{aligned}$$

and Continuity of $\succsim ^{\circ }$ delivers $f\succsim ^{\circ }g$. Otherwise, eventually

$$\begin{aligned} \left( 1-\varepsilon _{n}\right) g+\varepsilon _{n}l\not \sim ^{*}\left( 1-\varepsilon _{n}\right) f+\varepsilon _{n}h\text { and }\left( 1-\varepsilon _{n}\right) g+\varepsilon _{n}l\nsucc ^{*}\left( 1-\varepsilon _{n}\right) f+\varepsilon _{n}h \end{aligned}$$

then eventually

$$\begin{aligned} {\left( 1-\varepsilon _{n}\right) g+\varepsilon _{n}l\not \succsim ^{*}\left( 1-\varepsilon _{n}\right) f+\varepsilon _{n}h} \end{aligned}$$

by Possibility

$$\begin{aligned} \left( 1-\varepsilon _{n}\right) f+\varepsilon _{n}h\succsim ^{\circ }\left( 1-\varepsilon _{n}\right) g+\varepsilon _{n}l \end{aligned}$$

and Continuity of $\succsim ^{\circ }$ delivers $f\succsim ^{\circ }g$.

Conversely, assume $f\succsim ^{\circ }g$, then there exists $q\in C$ such that

$$\begin{aligned} \int u\left( f\right) {\hbox {d}}q\ge \int u\left( g\right) {\hbox {d}}q \end{aligned}$$

take $x\succ ^{*}y$ (so that $y\nsucc ^{*}x$), then for every $\varepsilon$ in $\left( 0,1\right]$

$$\begin{aligned} \left( 1-\varepsilon \right) \int u\left( f\right) {\hbox {d}}q+\varepsilon u\left( x\right)&>\left( 1-\varepsilon \right) \int u\left( g\right) {\hbox {d}}q+\varepsilon u\left( y\right) \\ \int u\left( \left( 1-\varepsilon \right) f+\varepsilon x\right) {\hbox {d}}q&>\int u\left( \left( 1-\varepsilon \right) g+\varepsilon y\right) {\hbox {d}}q\\ \left( 1-\varepsilon \right) g+\varepsilon y&\nsucc ^{*}\left( 1-\varepsilon \right) f+\varepsilon x \end{aligned}$$

so that the proof is concluded by setting $l=y$ and $h=x$. $\square$

Proof of Proposition 2

Let $f,g\in F$. First observe that if $f\succsim ^{*}g$, then

$$\begin{aligned} \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p\qquad \forall p\in C. \end{aligned}$$

(31)

But then $h\succsim ^{\circ }f$ implies that there exists $q\in C$ such that

$$\begin{aligned} \int u\left( h\right) {\hbox {d}}q\ge \int u\left( f\right) {\hbox {d}}q\ge \int u\left( g\right) {\hbox {d}}q \end{aligned}$$

where the last inequality follows from (31), that is, $h\succsim ^{\circ }g$. Analogously, $g\succsim ^{\circ }l$ implies that there exists $q\in C$ such that

$$\begin{aligned} \int u\left( f\right) {\hbox {d}}q\ge \int u\left( g\right) {\hbox {d}}q\ge \int u\left( l\right) {\hbox {d}}q, \end{aligned}$$

where the first inequality follows from (31), that is, $f\succsim ^{\circ }l$. This shows that $f\succsim ^{*}g$ implies $f\succsim ^{\circ \circ }g$.

As to the converse, notice that, under the assumptions of Theorem 1, $\succsim ^{\circ }$ is represented by u on X. Consider first the case in which the interval $u\left( X\right)$ does not admit a maximum point, and assume—per contra—that there exist $f\succsim ^{\circ \circ }g$ such that ${f\not \succsim ^{*}g}$. Therefore, there exists $q\in C$ such that

$$\begin{aligned} \int u\left( g\right) {\hbox {d}}q>\int u\left( f\right) {\hbox {d}}q. \end{aligned}$$

For all $\varepsilon \in \left( 0,1\right)$, set

$$\begin{aligned} f^{\varepsilon }=\left( 1-\varepsilon \right) f+\varepsilon x, \end{aligned}$$

where $x\succ ^{\circ }f\left( s\right)$ for every $s\in S$ (such an x exists because $u\left( X\right)$ does not admit maximum). Notice that $u\left( f^{\varepsilon }\left( s\right) \right) =\left( 1-\varepsilon \right) u\left( f\left( s\right) \right) +\varepsilon u\left( x\right)$ for all $s\in S$, therefore

• (a) $\int u\left( f^{\varepsilon }\right) {\hbox {d}}q\longrightarrow \int u\left( f\right) {\hbox {d}}q$ as $\varepsilon \rightarrow 0$;

• (b) $u\left( f^{\varepsilon }\left( s\right) \right) =u\left( f\left( s\right) \right) +\varepsilon \left( u\left( x\right) -u\left( f\left( s\right) \right) \right) >u\left( f\left( s\right) \right)$ because $u\left( x\right) -u\left( f\left( s\right) \right) >0$ for all $s\in S$.

Therefore, we can choose $\varepsilon \in \left( 0,1\right)$ small enough so that

$$\begin{aligned} \int u\left( g\right) {\hbox {d}}q>\int u\left( f^{\varepsilon }\right) {\hbox {d}}q>\int u\left( f\right) {\hbox {d}}q \end{aligned}$$

and $g\succsim ^{\circ }f^{\varepsilon }$. But $f\succsim ^{\circ \circ }g$ and $g\succsim ^{\circ }f^{\varepsilon }$ imply $f\succsim ^{\circ }f^{\varepsilon }$ which is absurd because

$$\begin{aligned} \int u\left( f^{\varepsilon }\right) {\hbox {d}}p>\int u\left( f\right) {\hbox {d}}p\qquad \forall p\in C. \end{aligned}$$

Next consider the case in which the interval $u\left( X\right)$ does not admit a minimum point, and assume—per contra—that there exist $f\succsim ^{\circ \circ }g$ such that $f\not \succsim ^{*}g$. Therefore there exists $q\in C$ such that

$$\begin{aligned} \int u\left( g\right) {\hbox {d}}q>\int u\left( f\right) {\hbox {d}}q. \end{aligned}$$

For all $\varepsilon \in \left( 0,1\right)$, set

$$\begin{aligned} g^{\varepsilon }=\left( 1-\varepsilon \right) g+\varepsilon x \end{aligned}$$

where $g\left( s\right) \succ x$ for every $s\in S$ (such an x exists because $u\left( X\right)$ does not admit minimum). Notice that $u\left( g^{\varepsilon }\left( s\right) \right) =\left( 1-\varepsilon \right) u\left( g\left( s\right) \right) +\varepsilon u\left( x\right)$ for all $s\in S$, therefore

• (a) $\int u\left( g^{\varepsilon }\right) {\hbox {d}}q\longrightarrow \int u\left( g\right) {\hbox {d}}q$ as $\varepsilon \rightarrow 0$;

• (b) $u\left( g^{\varepsilon }\left( s\right) \right) =u\left( g\left( s\right) \right) -\varepsilon \left( u\left( g\left( s\right) \right) -u\left( x\right) \right) <u\left( g\left( s\right) \right)$ because $u\left( g\left( s\right) \right) -u\left( x\right) >0$ for all $s\in S$.

Therefore, we can choose $\varepsilon$ small enough so that

$$\begin{aligned} \int u\left( g\right) {\hbox {d}}q>\int u\left( g^{\varepsilon }\right) {\hbox {d}}q>\int u\left( f\right) {\hbox {d}}q \end{aligned}$$

and $g^{\varepsilon }\succsim ^{\circ }f$. But $f\succsim ^{\circ \circ }g$ and $g^{\varepsilon }\succsim ^{\circ }f$ imply $g^{\varepsilon }\succsim ^{\circ }g$ which is absurd because

$$\begin{aligned} \int u\left( g^{\varepsilon }\right) {\hbox {d}}p<\int u\left( g\right) {\hbox {d}}p\qquad \forall p\in C. \end{aligned}$$

Summing up, $f\succsim ^{\circ \circ }g$ implies $f\succsim ^{*}g$. $\square$

The next proposition promptly delivers Proposition 3 as a corollary.

Proposition 5

If C is a non-empty closed and convex set of probabilities on $\Sigma$, $u:X\rightarrow {\mathbb {R}}$ is a non-constant affine function, and, for every $h,l\in F$,

$$\begin{aligned} h\succsim ^{*}l\iff \int u\left( h\right) \mathrm{d}p\ge \int u\left( l\right) \mathrm{d}p\qquad \text {for all }p\in C \end{aligned}$$

then the following conditions are equivalent for f and g in F:

• (i) For every in F and every $\gamma$ in $\left( 0,1\right]$

  

• (ii) There exist in F such that for every $\gamma \$in $\left( 0,1\right]$

  

  that is .

• (iii) $f\succsim ^{*}g$.

In particular, under the assumptions of Theorem 1, coincides with $\succ ^{\circ }$, and the equivalence between (iii) and (ii) above means that $f\succsim ^{*}g$ if and only if there exist $h\succ ^{\circ }l$ such that

$$\begin{aligned} \left( 1-\gamma \right) f+\gamma h\succ ^{\circ }\left( 1-\gamma \right) g+\gamma l\qquad \text {for all }\gamma \in \left( 0,1\right] . \end{aligned}$$

Proof of Proposition 5

(i) obviously implies (ii) because is non-trivial.

(ii) implies (iii). Since $h,l\in F$ are such that for every $\gamma \in \left( 0,1\right)$, then

$$\begin{aligned} \int u\left( \left( 1-\gamma \right) f+\gamma h\right) {\hbox {d}}p&>\int u\left( \left( 1-\gamma \right) g+\gamma l\right) {\hbox {d}}p\qquad \forall p\in C\\ \left( 1-\gamma \right) \int u\left( f\right) {\hbox {d}}p+\gamma \int u\left( h\right) {\hbox {d}}p&>\left( 1-\gamma \right) \int u\left( g\right) {\hbox {d}}p+\gamma \int u\left( l\right) {\hbox {d}}p\qquad \forall p\in C\\ \left( 1-\gamma \right) \int u\left( f\right) {\hbox {d}}p&>\left( 1-\gamma \right) \int u\left( g\right) {\hbox {d}}p\\&\quad +\gamma \int \left[ u\left( l\right) -u\left( h\right) \right] {\hbox {d}}p\qquad \forall p\in C\\ \int u\left( f\right) {\hbox {d}}p&>\int u\left( g\right) {\hbox {d}}p+\frac{\gamma }{1-\gamma }\int \left[ u\left( l\right) -u\left( h\right) \right] {\hbox {d}}p\qquad \forall p\in C \end{aligned}$$

and so, by passing to the limits as $\gamma \rightarrow 0$, $\int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p$ for all $p\in C$.

(iii) implies (i). If $\int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p$ for all $p\in C$, then, for every in F and every $\gamma$ in $\left( 0,1\right]$

$$\begin{aligned}&\left( 1-\gamma \right) \int u\left( f\right) {\hbox {d}}p+\gamma \int u\left( h\right) {\hbox {d}}p>\left( 1-\gamma \right) \int u\left( g\right) {\hbox {d}}p+\gamma \int u\left( l\right) {\hbox {d}}p\qquad \forall p\in C\\&\quad \int u\left( \left( 1-\gamma \right) f+\gamma h\right) {\hbox {d}}p>\int u\left( \left( 1-\gamma \right) g+\gamma l\right) {\hbox {d}}p\qquad \forall p\in C \end{aligned}$$

that is, . $\square$

1.2 Proof of Theorem 3

1.2.1 Utility profiles

Here we denote by $B_{0}\left( S,\Sigma \right)$ the vector space of all simple and measurable functions $\varphi :S\rightarrow {\mathbb {R}}$, and given an element $k\in {\mathbb {R}}$, we denote by k both the real number and the constant function in $B_{0}\left( S,\Sigma \right)$ that takes value k. Given two functions $\varphi ,\psi \in B_{0}\left( S,\Sigma \right)$, we define

We also define . Consider two binary relations $\succcurlyeq ^{\circ }$ and $\succcurlyeq ^{*}$ on $B_{0}\left( S,\Sigma \right)$. Assume that $\succcurlyeq ^{*}$ is such that

$$\begin{aligned} \varphi \succcurlyeq ^{*}\psi \iff \int \varphi {\hbox {d}}p\ge \int \psi {\hbox {d}}p\qquad \forall p\in C, \end{aligned}$$

(32)

where $C\ne \varnothing$ is a convex and closed subset of $\Delta$. Define also

(33)

Assume that $\succcurlyeq ^{*}$ and $\succcurlyeq ^{\circ }$ satisfy the following properties:

• 0. $\succcurlyeq ^{\circ }$ is complete;

• 1. If $\varphi _{2}\succeq ^{\circ }\psi _{2}$ and $\lambda \in \left( 0,1\right)$,²³ then

  $$\begin{aligned} \varphi _{1}\succ ^{\circ }\psi _{1}\implies \lambda \varphi _{1}+\left( 1-\lambda \right) \varphi _{2}\succ ^{\circ }\lambda \psi _{1}+\left( 1-\lambda \right) \psi _{2} \end{aligned}$$

  and the converse is true when $\lambda =0$ and $\frac{1}{2}\psi _{1}+\frac{1}{2}\varphi _{2}=\frac{1}{2}\varphi _{1}+\frac{1}{2}\psi _{2}$ (i.e., $\varphi _{1}-\psi _{1}=\varphi _{2}-\psi _{2}$);

• 2. If and $\psi \succ ^{\circ }\zeta$, then $\varphi \succ ^{\circ }\zeta$;

• 3. If , then $\varphi \succ ^{\circ }\psi$;

• 4. If $\varphi \succ ^{\circ }\psi$, then for each $\varepsilon >0$ there exists $\hat{\lambda }\in \left( 0,1\right)$ such that $\varphi \succ ^{\circ }\lambda \psi +\left( 1-\lambda \right) \varepsilon$ for all $\lambda \in \left( \hat{\lambda },1\right)$;

• 5. There exists $\tilde{\varphi }$ such that

  $$\begin{aligned} \varphi \not \succcurlyeq ^{*}0\implies \tilde{\varphi }\succcurlyeq ^{\circ }\varphi . \end{aligned}$$

Define

$$\begin{aligned} A=\Big \{ \phi \in B_{0}\left( S,\Sigma \right) :\phi =\varphi -\psi \text { with }\varphi \succ ^{\circ }\psi \Big \} \end{aligned}$$

and

It is immediate to see that $K^{++}\supseteq B_{0}^{++}\left( S,\Sigma \right)$.

Lemma 3

The set A has the following properties:

• 1. $B_{0}^{++}\left( S,\Sigma \right) \subseteq K^{++}\subseteq A$, in particular, $A\ne \varnothing$;

• 2. A is convex;

• 3. $A+K^{++}\subseteq A$;

• 4. $A\cap -B_{0}^{++}\left( S,\Sigma \right) =\varnothing$.

Proof

We already observed that $B_{0}^{++}\left( S,\Sigma \right) \subseteq K^{++}$. Moreover, if $\phi \in K^{++}$, then , and by Property 3, $\phi \succ ^{\circ } 0$, thus $\phi =\phi -0\in A$. This proves Point 1.

Consider $\phi _{1},\phi _{2}\in A$ and $\lambda \in \left( 0,1\right)$. It follows that there exist $\varphi _{i},\psi _{i}\in B_{0}\left( S,\Sigma \right)$ such that $\left. \varphi _{i}\succ ^{\circ }\psi _{i}\right.$ and $\phi _{i}=\varphi _{i}-\psi _{i}$ for $i=1,2$. By Property 1, we have that $\lambda \varphi _{1}+\left( 1-\lambda \right) \varphi _{2}\succ ^{\circ } \lambda \psi _{1}+\left( 1-\lambda \right) \psi _{2}$, then

$$\begin{aligned} \lambda \phi _{1}+\left( 1-\lambda \right) \phi _{2}&=\lambda \left( \varphi _{1}-\psi _{1}\right) +\left( 1-\lambda \right) \left( \varphi _{2}-\psi _{2}\right) \\&=\lambda \varphi _{1}+\left( 1-\lambda \right) \varphi _{2}-\left( \lambda \psi _{1}+\left( 1-\lambda \right) \psi _{2}\right) \in A. \end{aligned}$$

This proves Point 2.

Next, consider $\eta \in K^{++}\$and $\phi \in A$. By definition of A, $\phi =\varphi -\psi$ with $\varphi \succ ^{\circ }\psi$. But then and $\varphi \succ ^{\circ }\psi$. By Property 2, it follows that $\varphi +\eta \succ ^{\circ }\psi$, then $\varphi +\eta -\psi \in A$ and $\phi +\eta \in A$. This proves Point 3.

By contradiction, notice that if $\phi \in A\cap -B_{0}^{++}\left( S,\Sigma \right)$, then there would exist $\varphi ,\psi \in B_{0}\left( S,\Sigma \right)$ such that $\varphi \succ ^{\circ }\psi$, and . But then, (because ), and, by Property 3, $\psi \succ ^{\circ }\varphi$, a contradiction with $\varphi \succ ^{\circ } \psi$. This proves Point 4. $\square$

Remark 1

Notice that if and $\phi _{2}\in A$, then $\phi _{1}\in A$. For, if we define $\eta =\phi _{1}-\phi _{2}$, then and $\eta \in K^{++}$, therefore $\phi _{1}=\phi _{2} +\eta \in A$.

Remark 2

$\varphi \succ ^{\circ }\psi \iff \varphi -\psi \in A$. In fact, by definition of A, if $\varphi \succ ^{\circ }\psi$, then $\varphi -\psi \in A$. Conversely, if $\varphi -\psi \in A$, there exist $\bar{\varphi },\bar{\psi }\in B_{0}\left( S,\Sigma \right)$ such that $\varphi -\psi =\bar{\varphi } -\bar{\psi }$ and $\bar{\varphi }\succ ^{\circ }\bar{\psi }$. Then the second part of Property 1 implies $\varphi \succ ^{\circ }\psi$.

Set

$$\begin{aligned} I\left( \phi \right) =\sup \left\{ k\in {\mathbb {R}} :\phi -k\in A\right\} \qquad \forall \phi \in B_{0}\left( S,\Sigma \right) . \end{aligned}$$

(34)

Lemma 4

If I is defined as in (34), then I is a normalized concave niveloid. Moreover,

• 1. $\varphi \succ ^{\circ }\psi$ if and only if $I\left( \varphi -\psi \right) >0$.

• 2. $\phi \in A$ if and only if $I\left( \phi \right) >0$.

• 3. $\varphi \succcurlyeq ^{*}\psi$ implies $I\left( \varphi \right) \ge I\left( \psi \right)$.

Proof

Consider $\phi \in B_{0}\left( S,\Sigma \right)$.

Since $A\supseteq B_{0}^{++}\left( S,\Sigma \right)$, $A\cap -B_{0} ^{++}\left( S,\Sigma \right) =\varnothing$, and $A+B_{0}^{++}\left( S,\Sigma \right) \subseteq A$, it follows that $A_{\phi }=\left\{ k\in {\mathbb {R}} :\phi -k\in A\right\}$ is a non-empty and bounded above half line such that

$$\begin{aligned} \left( -\infty ,\min _{s\in S}\phi \left( s\right) \right) \subseteq A_{\phi }\subseteq \left( -\infty ,\max _{s\in S}\phi \left( s\right) \right] , \end{aligned}$$

(35)

and thus, $I\left( \phi \right) =\sup A_{\phi }\in {\mathbb {R}}$, and I is well defined. Moreover, (35) implies that $I\left( \bar{k}\right) =\bar{k}$ for all $\bar{k}\in {\mathbb {R}}$, that is, I is normalized.

For every $\bar{k}\in {\mathbb {R}}$, $A_{\phi +\bar{k}}=A_{\phi }+\bar{k}$, then

$$\begin{aligned} I\left( \phi +\bar{k}\right) =\sup A_{\phi +\bar{k}}=\sup \left( A_{\phi } +\bar{k}\right) =\sup A_{\phi }+\bar{k}=I\left( \phi \right) +\bar{k}. \end{aligned}$$

Since $\bar{k}$ and $\phi$ were arbitrarily chosen, we can conclude that $I\left( \phi +\bar{k}\right) =I\left( \phi \right) +\bar{k}\$for all $\phi \in B_{0}\left( S,\Sigma \right)$ and for all $\bar{k}\in {\mathbb {R}}$. That is, I is translation invariant.

If , then $\phi _{2}-k\in A$ implies also belongs to A. This means

$$\begin{aligned} \Big \{ k\in {\mathbb {R}} :\phi _{2}-k\in A\Big \} \subseteq \Big \{ k\in {\mathbb {R}} :\phi _{1}-k\in A\Big \} , \end{aligned}$$

whence $I\left( \phi _{1}\right) \ge I\left( \phi _{2}\right)$. If $\phi _{1}\ge \phi _{2}$, then for all $n\in {\mathbb {N}}$ and so

$$\begin{aligned} I\left( \phi _{1}\right) \ge I\left( \phi _{2}-\frac{1}{n}\right) =I\left( \phi _{2}\right) -\frac{1}{n} \end{aligned}$$

for all $n\in {\mathbb {N}}$, thus $I\left( \phi _{1}\right) \ge I\left( \phi _{2}\right)$. That is, I is monotone.

Consider $\phi _{1},\phi _{2}\in B_{0}\left( S,\Sigma \right)$ and arbitrarily choose $\lambda \in \left( 0,1\right)$. If $k_{1},k_{2}\in {\mathbb {R}}$ are such that $\phi _{i}-k_{i}\in A$ for $i=1,2$ (i.e., $k_{i}\in A_{\phi _{i}}$ for $i=1,2$). Since A is convex, it follows that

$$\begin{aligned} A\ni \lambda \left( \phi _{1}-k_{1}\right) +\left( 1-\lambda \right) \left( \phi _{2}-k_{2}\right) =\left( \lambda \phi _{1}+\left( 1-\lambda \right) \phi _{2}\right) -\left( \lambda k_{1}+\left( 1-\lambda \right) k_{2}\right) . \end{aligned}$$

It follows that

$$\begin{aligned} I\left( \lambda \phi _{1}+\left( 1-\lambda \right) \phi _{2}\right) \ge \lambda k_{1}+\left( 1-\lambda \right) k_{2} \end{aligned}$$

for all $k_{1}\in A_{\phi _{1}}$, and $k_{2}\in A_{\phi _{2}}$, yielding that

$$\begin{aligned} I\left( \lambda \phi _{1}+\left( 1-\lambda \right) \phi _{2}\right) \ge \lambda I\left( \phi _{1}\right) +\left( 1-\lambda \right) I\left( \phi _{2}\right) , \end{aligned}$$

proving that I is concave.

1. If $\varphi \succ ^{\circ }\psi$, then $\varphi -\psi \in A$. Let $\varepsilon$ be such that $\varepsilon >\max \left\{ \max _{s\in S}\psi \left( s\right) ,0\right\}$. By Property 4, there exists $\hat{\lambda }\in \left( 0,1\right)$ such that $\varphi \succ ^{\circ }\lambda \psi +\left( 1-\lambda \right) \varepsilon$ for all $\lambda \in \left( \hat{\lambda },1\right)$. In particular, we have that $\varphi -\left( \lambda \psi +\left( 1-\lambda \right) \varepsilon \right) \in A$ for all $\lambda \in \left( \hat{\lambda },1\right)$, that is,

$$\begin{aligned} \left( \varphi -\psi \right) +\left( 1-\lambda \right) \left( \psi -\varepsilon \right) \in A. \end{aligned}$$

Fix such a $\lambda$ and notice that is such that

$$\begin{aligned} \left( \varphi -\psi \right) +\eta \in A. \end{aligned}$$

Now setting $d=\frac{\max _{s\in S}\eta \left( s\right) }{2}$, we have

$$\begin{aligned} 0>d=\frac{\max _{s\in S}\eta \left( s\right) }{2}>\max _{s\in S}\eta \left( s\right) \end{aligned}$$

therefore and

delivers $\left( \varphi -\psi \right) +d\in A$ or $\left( \varphi -\psi \right) -\left( -d\right) \in A$. By definition of I, we have that $I\left( \varphi -\psi \right) \ge -d>0$.

Viceversa, by definition of I, if $I\left( \varphi -\psi \right) >0$, then $\left( \varphi -\psi \right) -k\in A$ for some $k>0$. It follows that $\varphi -\psi \in A$, because . By Remark 2, $\varphi \succ ^{\circ }\psi$.

2. By Remark 2, $\phi \in A$ if and only if $\phi \succ ^{\circ }0$, which, by Point 1, is equivalent to $I\left( \phi \right) >0$.

3. Recall that $A+K^{++}\subseteq A$, assume , then and $\varphi =\psi +\eta$. Now

$$\begin{aligned} \psi -k\in A\implies \psi -k+\eta \in A\implies \varphi -k\in A \end{aligned}$$

then $A_{\psi }\subseteq A_{\varphi }$ and $I\left( \psi \right) \le I\left( \varphi \right)$. If $\varphi \succcurlyeq ^{*}\psi$, then for all $n\in {\mathbb {N}}$ and so

$$\begin{aligned} I\left( \varphi \right) \ge I\left( \psi -\frac{1}{n}\right) =I\left( \psi \right) -\frac{1}{n} \end{aligned}$$

for all $n\in {\mathbb {N}}$, thus $I\left( \varphi \right) \ge I\left( \psi \right)$. $\square$

Define $\bar{I}:B_{0}\left( S,\Sigma \right) \rightarrow {\mathbb {R}}$ as

$$\begin{aligned} \bar{I}\left( \phi \right) =-I\left( -\phi \right) \qquad \forall \phi \in B_{0}\left( S,\Sigma \right) . \end{aligned}$$

(36)

Observe that $-\bar{I}\left( \phi \right) =I\left( -\phi \right)$ for all $\phi \in B_{0}\left( S,\Sigma \right)$.

Proposition 6

If $\bar{I}$ is defined as in (36), then

$$\begin{aligned} \varphi \succcurlyeq ^{\circ }\psi \iff I\left( \psi -\varphi \right) \le 0\iff \bar{I}\left( \varphi -\psi \right) \ge 0. \end{aligned}$$

Proof

Since $\succcurlyeq ^{\circ }$ is complete (Property 0), $\varphi \succcurlyeq ^{\circ }\psi \iff \psi \not \succ ^{\circ }\varphi$, thus

$$\begin{aligned} \left. \varphi \succcurlyeq ^{\circ }\psi \right.&\iff I\left( \psi -\varphi \right) \not > 0\iff I\left( \psi -\varphi \right) \le 0\iff I\left( -\left( \varphi -\psi \right) \right) \le 0\\&\iff -\bar{I}\left( \varphi -\psi \right) \le 0 \end{aligned}$$

as wanted. $\square$

Remark 3

Maccheroni, Marinacci, and Rustichini (2006, henceforth MMR) show that if $I:B_{0}\left( S,\Sigma \right) \rightarrow {\mathbb {R}}$ is a normalized concave niveloid, there exists a unique, grounded, convex, and lower semicontinuous function $c:\Delta \rightarrow \left[ 0,\infty \right]$ such that

$$\begin{aligned} I\left( \phi \right) =\min _{p\in \Delta }\left\{ \int \phi {\hbox {d}}p+c\left( p\right) \right\} \qquad \forall \phi \in B_{0}\left( S,\Sigma \right) . \end{aligned}$$

Specifically, for each $p\in \Delta$,

$$\begin{aligned} c\left( p\right) =\sup \Big \{ I\left( \psi \right) -\left\langle \psi ,p\right\rangle :\psi \in B_{0}\left( S,\Sigma \right) \Big \} . \end{aligned}$$

(37)

Cerreia-Vioglio et al. (2011b) show that, for each $p\in \Delta$,

$$\begin{aligned} c\left( p\right) =\sup \left\{ I\left( \varphi \right) :\left\langle \varphi ,p\right\rangle =0\right\} =\sup \left\{ I\left( \phi \right) :\left\langle \phi ,p\right\rangle \le 0\right\} =\sup \left\{ I\left( \eta \right) :\left\langle \eta ,p\right\rangle <0\right\} . \end{aligned}$$

Cerreia-Vioglio et al. (2011a) show that if D is a convex and closed subset of $\Delta$ such that

$$\begin{aligned} \int \phi _{1}{\hbox {d}}p\ge \int \phi _{2}{\hbox {d}}p\quad \forall p\in D\qquad \implies \qquad I\left( \phi _{1}\right) \ge I\left( \phi _{2}\right) , \end{aligned}$$

(38)

then ${\text {cl}}\,\left( {\text {dom}}\,c\right) \subseteq D$. Finally, if I is defined as in (34), by (37), we have that

$$\begin{aligned} c\left( p\right)&=-\inf \Big \{ \left\langle \psi -I\left( \psi \right) ,p\right\rangle :\psi \in B_{0}\left( S,\Sigma \right) \Big \} \\&=-\inf \Big \{ \left\langle \psi ,p\right\rangle :\psi \in B_{0}\left( S,\Sigma \right) \text { and }I\left( \psi \right) =0\Big \} \\&=-\inf \Big \{ \left\langle \phi ,p\right\rangle :\phi \in B_{0}\left( S,\Sigma \right) \text { and }I\left( \phi \right) >0\Big \} \\ (\text {by Lemma } 4.2)&=-\inf \Big \{ \left\langle \phi ,p\right\rangle :\phi \in B_{0}\left( S,\Sigma \right) \text { and }\phi \in A\Big \} \\&=-\inf \left\{ \left\langle \varphi ,p\right\rangle -\left\langle \psi ,p\right\rangle :\varphi ,\psi \in B_{0}\left( S,\Sigma \right) \text { and }\varphi \succ ^{\circ }\psi \right\} \\&=\sup \left\{ \left\langle \psi ,p\right\rangle -\left\langle \varphi ,p\right\rangle :\varphi ,\psi \in B_{0}\left( S,\Sigma \right) \text { and }\psi \prec ^{\circ }\varphi \right\} \end{aligned}$$

for all $p\in \Delta$.

Proposition 7

Let I and $\bar{I}$ be defined as in (34) and (36). The following statements are true:

• 1. There exists a unique $c:\Delta \rightarrow \left[ 0,\infty \right]$ grounded, convex, and lower semicontinuous such that

  $$\begin{aligned} \bar{I}\left( \phi \right) =\max _{p\in \Delta }\left\{ \int \phi \mathrm{d}p-c\left( p\right) \right\} \qquad \forall \phi \in B_{0}\left( S,\Sigma \right) . \end{aligned}$$

  (39)

  Specifically, it holds

  $$\begin{aligned} c\left( p\right) =\sup \Big \{ \left\langle \psi ,p\right\rangle -\left\langle \varphi ,p\right\rangle :\varphi ,\psi \in B_{0}\left( S,\Sigma \right) \text { and }\psi \prec ^{\circ }\varphi \Big \} \end{aligned}$$

  (40)

  for all $p\in \Delta$.

• 2. ${\text {cl}}\,\left( {\text {dom}}\,c\right) \subseteq C$.

• 3. If $\varphi \succ ^{\circ }\psi$ and $\psi \succcurlyeq ^{*}\zeta$, then $\varphi \succ ^{\circ }\zeta$.

• 4. ${\text {cl}}\,\left( {\text {dom}}\,c\right) =C.$

• 5. c is bounded on ${\text {cl}}\,\left( {\text {dom}}\,c\right)$. In particular, ${\text {cl} }\,\left( {\text {dom}}\,c\right) ={\text {dom}}\,c=C$.

Proof

1. By MMR and since I is a normalized concave niveloid, there exists a unique grounded, convex, and lower semicontinuous function $c:\Delta \rightarrow \left[ 0,\infty \right]$ such that

$$\begin{aligned} I\left( \phi \right) =\min _{p\in \Delta }\left\{ \int \phi {\hbox {d}}p+c\left( p\right) \right\} \qquad \forall \phi \in B_{0}\left( S,\Sigma \right) . \end{aligned}$$

(41)

By definition of $\bar{I}$, (39) follows, while (40) descends from Remark 3.

2. We already observed that $\phi _{1}\succcurlyeq ^{*}\phi _{2}$ implies $I\left( \phi _{1}\right) \ge I\left( \phi _{2}\right)$, by Remark 3, we can conclude that $C\supseteq {\text {cl} }\,\left( {\text {dom}}\,c\right)$.

3. Consider $\varphi \succ ^{\circ }\psi$ and $\psi \succcurlyeq ^{*}\zeta$. We have that $I\left( \varphi -\psi \right) >0$ and

$$\begin{aligned} \varphi -\zeta =\left( \varphi -\psi \right) +\left( \psi -\zeta \right) \succcurlyeq ^{*}\varphi -\psi \end{aligned}$$

thus $I\left( \varphi -\zeta \right) \ge I\left( \varphi -\psi \right) >0$. By Lemma 4, $\varphi \succ ^{\circ }\zeta$.

4. By contradiction, assume that $C\supset {\text {cl}}\,\left( {\text {dom}}\,c\right)$. Thus, there exists $\bar{p}\in C\backslash {\text {cl}}\,\left( {\text {dom}}\,c\right)$. Since ${\text {cl}}\,\left( {\text {dom}}\,c\right)$ is convex and closed, there exists $\psi \in B_{0}\left( S,\Sigma \right)$, $\alpha \in {\mathbb {R}}$, and $\varepsilon >0$ such that

$$\begin{aligned} \int \psi d\bar{p}\le \alpha -\varepsilon <\alpha +\varepsilon \le \min _{p\in {\text {cl}}\,\left( {\text {dom}}\,c\right) } \int \psi {\hbox {d}}p. \end{aligned}$$

Setting $\varphi =\psi -\alpha$, we have

$$\begin{aligned} \int \varphi d\bar{p}\le -\varepsilon <\varepsilon \le \min _{p\in {\text {cl}}\,\left( {\text {dom}}\,c\right) }\int \varphi {\hbox {d}}p. \end{aligned}$$

(42)

If we define $\varphi _{n}=n\varphi$ for all $n\in {\mathbb {N}}$, then $\varphi _{n}$ satisfies (42) with $\varepsilon$ replaced by $n\varepsilon$. By (42), it follows that, for all $n\in {\mathbb {N}}$, $\varphi _{n}\not \succcurlyeq ^{*}0$ and

$$\begin{aligned} I\left( \varphi _{n}\right) =\inf _{p\in {\text {dom}}\,c}\left\{ \int \varphi _{n}{\hbox {d}}p+c\left( p\right) \right\} \ge \inf _{p\in {\text {dom}}\,c}\int \varphi _{n}{\hbox {d}}p\ge n\varepsilon>\frac{n\varepsilon }{2}>0. \end{aligned}$$

This implies that $I\left( \varphi _{n}-\frac{n\varepsilon }{2}\right) >0$, that is, $\varphi _{n}\succ ^{\circ }\frac{n\varepsilon }{2}$ for all $n\in {\mathbb {N}}$. So far we have found a sequence $\varphi _{n}$ in $B_{0}\left( S,\Sigma \right)$ such that, for all $n\in {\mathbb {N}}$, $\varphi _{n}\not \succcurlyeq ^{*}0$ and $\varphi _{n}\succ ^{\circ }\frac{n\varepsilon }{2}$. At the same time, if we choose $\bar{n}$ large enough, we have that $\frac{\bar{n}\varepsilon }{2}\ge \tilde{\varphi }$ and, in particular, $\frac{\bar{n}\varepsilon }{2}\succcurlyeq ^{*}\tilde{\varphi }$. By point 3, we have that $\varphi _{\bar{n}}\succ ^{\circ }\tilde{\varphi }$, a contradiction with Property 5 which implies $\tilde{\varphi }\succcurlyeq ^{\circ }\varphi _{\bar{n}}$ because $\varphi _{\bar{n}}\not \succcurlyeq ^{*}0$.

5. We next show that there exists $k\ge 0$ such that $c\left( p\right) \le k$ for all $p\in {\text {cl}}\,\left( {\text {dom}}\, c\right)$. By contradiction, assume that for each $n\in {\mathbb {N}}$ there exists $p_{n}\in {\text {cl}}\,\left( {\text {dom}}\, c\right)$ such that $c\left( p_{n}\right) >n$. By Remark 3, $c\left( p_{n}\right) =\sup \left\{ I\left( \phi \right) :\left\langle \phi ,p_{n}\right\rangle <0\right\}$. It follows that for each $n\in {\mathbb {N}}$ there exists $\varphi _{n}$ such that $\left\langle \varphi _{n} ,p_{n}\right\rangle <0$ and $I\left( \varphi _{n}\right) >n$. This implies that $I\left( \varphi _{n}-n\right) >0$ for all $n\in {\mathbb {N}}$. Since $\left\langle \varphi _{n},p_{n}\right\rangle <0$, $\varphi _{n}\not \succcurlyeq ^{*}0$, but $I\left( \varphi _{n}-n\right) >0$ implies that $\varphi _{n}\succ ^{\circ }n$. By Property 5, we can conclude that $\tilde{\varphi }\succcurlyeq ^{\circ }\varphi _{n}$ for all $n\in {\mathbb {N}}$. At the same time, if we choose $\bar{n}$ large enough, $\bar{n}\ge \tilde{\varphi }$, that is, $\bar{n}\succcurlyeq ^{*}\tilde{\varphi }$. By point 3 and since $\varphi _{\bar{n}}\succ ^{\circ }\bar{n}$, we have that $\varphi _{\bar{n}}\succ ^{\circ }\tilde{\varphi }$, a contradiction. $\square$

Theorem 5

Let $C\ne \varnothing$ be a convex and closed subset of $\Delta$, $\succcurlyeq ^{*}$ be the binary relation on $B_{0}\left( S,\Sigma \right)$ defined by (32), and $\succcurlyeq ^{\circ }$ be another binary relation on $B_{0}\left( S,\Sigma \right)$ that satisfies Properties 0–5, then there exists a unique function $\gamma :C\rightarrow \left[ 0,\infty \right]$ which is grounded, lower semicontinuous, convex, and bounded, such that

$$\begin{aligned} \varphi \succcurlyeq ^{\circ }\psi \iff \max _{p\in C}\left\{ \int \varphi \mathrm{d}p-\int \psi {\hbox {d}}p-\gamma \left( p\right) \right\} \ge 0. \end{aligned}$$

Specifically, it holds

$$\begin{aligned} \gamma \left( p\right) =\sup \Big \{ \left\langle \psi ,p\right\rangle -\left\langle \varphi ,p\right\rangle :\varphi ,\psi \in B_{0}\left( S,\Sigma \right) \text { and }\psi \prec ^{\circ }\varphi \Big \} \end{aligned}$$

for all $p\in C$.

Proof

Consider the normalized concave niveloid I of (34) and its conjugate functional $\bar{I}$ defined in (36). By Proposition 6,

$$\begin{aligned} \varphi \succcurlyeq ^{\circ }\psi \iff \bar{I}\left( \varphi -\psi \right) \ge 0. \end{aligned}$$

By Proposition 7, there exists $c:\Delta \rightarrow \left[ 0,\infty \right]$ grounded, convex, and lower semicontinuous such that

$$\begin{aligned} \bar{I}\left( \phi \right) =\max _{p\in \Delta }\left\{ \int \phi {\hbox {d}}p-c\left( p\right) \right\} \qquad \forall \phi \in B_{0}\left( S,\Sigma \right) . \end{aligned}$$

Proposition 7 also provides the explicit form of c. Moreover, $C={\text {cl}}\,\left( {\text {dom}}\,c\right) ={\text {dom}}\,c$ and c is bounded on C, so that,

$$\begin{aligned} \bar{I}\left( \phi \right) =\max _{p\in C}\left\{ \int \phi {\hbox {d}}p-c\left( p\right) \right\} \qquad \forall \phi \in B_{0}\left( S,\Sigma \right) . \end{aligned}$$

The function $\gamma$ in the statement is simply the restriction of c to its domain C.

Assume $\delta :C\rightarrow \left[ 0,\infty \right]$ is another grounded, lower semicontinuous, convex, and bounded function such that

$$\begin{aligned} \varphi \succcurlyeq ^{\circ }\psi \iff \max _{p\in C}\left\{ \int \varphi {\hbox {d}}p-\int \psi {\hbox {d}}p-\delta \left( p\right) \right\} \ge 0. \end{aligned}$$

Since $\succcurlyeq ^{\circ }$ is complete

$$\begin{aligned} \phi \succ ^{\circ }0\iff \min _{p\in C}\left\{ \int \phi {\hbox {d}}p+\delta \left( p\right) \right\} >0. \end{aligned}$$

(43)

Setting

$$\begin{aligned} d\left( p\right) =\left\{ \begin{array} [c]{lll} \delta \left( p\right) &{} &{} p\in C\\ \infty &{} &{} p\in \Delta {\setminus } C \end{array} \right. \end{aligned}$$

it is easy to check that $d:\Delta \rightarrow \left[ 0,\infty \right]$ is a grounded, convex, and lower semicontinuous function and so

$$\begin{aligned} J\left( \phi \right) =\min _{p\in \Delta }\left\{ \int \phi {\hbox {d}}p+d\left( p\right) \right\} \qquad \forall \phi \in B_{0}\left( S,\Sigma \right) \end{aligned}$$

defines a normalized concave niveloid $J:B_{0}\left( S,\Sigma \right) \rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} J\left( \phi \right)>0\iff \phi \succ ^{\circ }0\iff I\left( \phi \right) >0. \end{aligned}$$

But then, for all $\varphi \in B_{0}\left( S,\Sigma \right)$

$$\begin{aligned} I\left( \varphi \right)&=\sup \left\{ t\in {\mathbb {R}}:I\left( \varphi \right)>t\right\} =\sup \left\{ t\in {\mathbb {R}}:I\left( \varphi \right) -t>0\right\} \\&=\sup \left\{ t\in {\mathbb {R}}:I\left( \varphi -t\right)>0\right\} =\sup \left\{ t\in {\mathbb {R}}:J\left( \varphi -t\right) >0\right\} =J\left( \varphi \right) . \end{aligned}$$

Because of the uniqueness of the representation of concave niveloids obtained by MMR, it follows $c=d$. $\square$

1.2.2 Main body of the proof

(i) implies (ii). By Lemma 1, there exist a (unique) non-empty closed and convex set C of probabilities on $\Sigma$ and a (cardinally unique) non-constant affine $u:X\rightarrow {\mathbb {R}}$ such that, for every $f,g\in F$,

$$\begin{aligned} \left. f\succsim ^{*}g\right.&\iff \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p\qquad \forall p\in C, \end{aligned}$$

(44)

(45)

So that u represents $\succsim ^{*}$ on X. Unboundedness and Lemma 59 of Cerreia-Vioglio et al. (2011a) guarantee that $u\left( X\right) ={\mathbb {R}}$.

For any $\left( \varphi ,\psi \right) \in B_{0}\left( S,\Sigma \right)$, it is convenient to set

$$\begin{aligned} F\left( \varphi ,\psi \right) =\left\{ \left( f,g\right) \in F^{2}:u\left( f\right) =\varphi \text { and }u\left( g\right) =\psi \right\} \end{aligned}$$

and to observe that $u\left( X\right) ={\mathbb {R}}$ implies that $F\left( \varphi ,\psi \right)$ is non-empty. We also write $R^{*}$ (resp. $R^{\circ }$) to denote $\succsim ^{*}$ (resp. $\succsim ^{\circ }$) when regarded as a subset of $F^{2}$.

Lemma 5

The following conditions are equivalent for $\varphi ,\psi \in B_{0}\left( S,\Sigma \right)$:

• (a) There are $f,g\in F$ such that $u\left( f\right) =\varphi$, $u\left( g\right) =\psi$ and $f\succsim ^{*}g$ (i.e., $F\left( \varphi ,\psi \right) \cap R^{*}\ne \varnothing$).

• (b) $f^{\prime }\succsim ^{*}g^{\prime }$ for all $f^{\prime } ,g^{\prime }\in F$ such that $u\left( f^{\prime }\right) =\varphi$, $u\left( g^{\prime }\right) =\psi$ (i.e., $F\left( \varphi ,\psi \right) \subseteq R^{*}$).

• (c) $\int \varphi {\hbox {d}}p\ge \int \psi {\hbox {d}}p$ for all $p\in C$.

In this case, we write $\varphi \succcurlyeq ^{*}\psi$; this is consistent with (32), and we notice that, by points (a) and (b) above

$$\begin{aligned} f\succsim ^{*}g\iff u\left( f\right) \succcurlyeq ^{*}u\left( g\right) . \end{aligned}$$

Proof

Notice that if $f,f^{\prime },g,g^{\prime }\in F$, $u\left( f\right) =u\left( f^{\prime }\right)$, $u\left( g\right) =u\left( g^{\prime }\right)$, then

$$\begin{aligned} \left. f\succsim ^{*}g\right.&\iff \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p\qquad \forall p\in C\\&\iff \int u\left( f^{\prime }\right) {\hbox {d}}p\ge \int u\left( g^{\prime }\right) {\hbox {d}}p\qquad \forall p\in C\iff \left. f^{\prime }\succsim ^{*}g^{\prime }\right. \text {.} \end{aligned}$$

• (a) implies (b). If there are $f,g\in F$ such that $u\left( f\right) =\varphi$, $u\left( g\right) =\psi$ and $f\succsim ^{*}g$, then for any $f^{\prime },g^{\prime }\in F$ such that $u\left( f^{\prime }\right) =\varphi$, $u\left( g^{\prime }\right) =\psi$ we have $u\left( f\right) =\varphi =u\left( f^{\prime }\right)$, $u\left( g\right) =\psi =u\left( g^{\prime }\right)$, and as we observed $f^{\prime }\succsim ^{*}g^{\prime }$.

• (b) implies (c). Take $f^{\prime },g^{\prime }\in F$ such that $u\left( f^{\prime }\right) =\varphi$, $u\left( g^{\prime }\right) =\psi$, they exist because $F\left( \varphi ,\psi \right) \ne \varnothing$, by (b) we have $f^{\prime }\succsim ^{*}g^{\prime }$, by (44) we have that (c) holds.

• (c) implies (a). Take $f,g\in F$ such that $u\left( f\right) =\varphi$, $u\left( g\right) =\psi$, they exist because $F\left( \varphi ,\psi \right) \ne \varnothing$, by (c) we have $\int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p\$for all $p\in C$, by (44) we have $f\succsim ^{*}g$ and (a) holds. $\square$

An almost identical argument yields:

Lemma 6

The following conditions are equivalent for $\varphi ,\psi \in B_{0}\left( S,\Sigma \right)$:

• (a) There are $f,g\in F$ such that $u\left( f\right) =\varphi$, $u\left( g\right) =\psi$ and .

• (b) for all $f^{\prime },g^{\prime }\in F$ such that $u\left( f^{\prime }\right) =\varphi$, $u\left( g^{\prime }\right) =\psi$.

• (c) $\int \varphi {\hbox {d}}p>\int \psi {\hbox {d}}p$ for all $p\in C$.

In this case we write , this is consistent with (33), and notice that, by points (a) and (b) above

Lemma 7

The following conditions are equivalent for $\varphi ,\psi \in B_{0}\left( S,\Sigma \right)$:

• (a) There are $f,g\in F$ such that $u\left( f\right) =\varphi$, $u\left( g\right) =\psi$ and $f\succsim ^{\circ }g$ (i.e., $F\left( \varphi ,\psi \right) \cap R^{\circ }\ne \varnothing$).

• (b) $f^{\prime }\succsim ^{\circ }g^{\prime }$ for all $f^{\prime },g^{\prime }\in F$ such that $u\left( f^{\prime }\right) =\varphi$, $u\left( g^{\prime }\right) =\psi$ (i.e., $F\left( \varphi ,\psi \right) \subseteq R^{\circ }$).

In this case write $\varphi \succcurlyeq ^{\circ }\psi$ and notice that by points (a) and (b) above

$$\begin{aligned} f\succsim ^{\circ }g\iff u\left( f\right) \succcurlyeq ^{\circ }u\left( g\right) . \end{aligned}$$

Proof

(a) implies (b). If there are $f,g\in F$ such that $u\left( f\right) =\varphi$, $u\left( g\right) =\psi$ and $f\succsim ^{\circ }g$, then for any $f^{\prime },g^{\prime }\in F$ such that $u\left( f^{\prime }\right) =\varphi$, $u\left( g^{\prime }\right) =\psi$ we have $u\left( f\right) =\varphi =u\left( f^{\prime }\right)$, $u\left( g\right) =\psi =u\left( g^{\prime }\right)$, and by (44) $f\sim ^{*}f^{\prime }$ and $g\sim ^{*}g^{\prime }$. Substitution Consistency yields $f^{\prime }\succsim ^{\circ }g^{\prime }$.

(b) implies (a). Take $f^{\prime },g^{\prime }\in F$ such that $u\left( f^{\prime }\right) =\varphi$, $u\left( g^{\prime }\right) =\psi$, they exist because $F\left( \varphi ,\psi \right) \ne \varnothing$, by (b) we have $f^{\prime }\succsim ^{\circ }g^{\prime }$. $\square$

In particular, for any, $\varphi ,\psi \in B_{0}\left( S,\Sigma \right)$, taking $f,g\in F$ such that $u\left( f\right) =\varphi$ and $u\left( g\right) =\psi$, either $f\succsim ^{\circ }g$ and so $\varphi \succcurlyeq ^{\circ }\psi$, or $g\succsim ^{\circ }f$ and so $\psi \succcurlyeq ^{\circ } \varphi$. Thus $\succcurlyeq ^{\circ }$ is complete. This fact and the previous lemma readily imply the following result.

Lemma 8

The following conditions are equivalent for $\varphi ,\psi \in B_{0}\left( S,\Sigma \right)$:

• (a) There are $f,g\in F$ such that $u\left( f\right) =\varphi$, $u\left( g\right) =\psi$ and $f\succ ^{\circ }g$ (i.e., $F\left( \psi ,\varphi \right) \cap \left( R^{\circ }\right) ^{c}\ne \varnothing$).

• (b) $f^{\prime }\succ ^{\circ }g^{\prime }$ for all $f^{\prime },g^{\prime }\in F$ such that $u\left( f^{\prime }\right) =\varphi$, $u\left( g^{\prime }\right) =\psi$ (i.e., $F\left( \psi ,\varphi \right) \subseteq \left( R^{\circ }\right) ^{c}$).

• (c) $\varphi \succ ^{\circ }\psi$ (i.e., $\psi \not \succcurlyeq ^{\circ }\varphi$).

Notice that, by points (a), (b), and (c) above

$$\begin{aligned} f\succ ^{\circ }g\iff u\left( f\right) \succ ^{\circ }u\left( g\right) . \end{aligned}$$

Lemma 9

The pair $\left( \succcurlyeq ^{*},\succcurlyeq ^{\circ }\right)$ satisfies properties 0–5 (of page 26).

Proof

By Lemma 5, $\succcurlyeq ^{*}$ can be represented as in (32).

Property 0. We already observed that $\succcurlyeq ^{\circ }$ is complete.

Property 1. Let $\varphi _{1}=u\left( f_{1}\right)$, $\psi _{1}=u\left( g_{1}\right)$, $\varphi _{2}=u\left( f_{2}\right)$, $\psi _{2}=u\left( g_{2}\right)$, and $\lambda \in \left( 0,1\right)$. Observe that $\lambda \varphi _{1}+\left( 1-\lambda \right) \varphi _{2}=u\left( \lambda f_{1}+\left( 1-\lambda \right) f_{2}\right)$ and $\lambda \psi _{1}+\left( 1-\lambda \right) \psi _{2}=u\left( \lambda g_{1}+\left( 1-\lambda \right) g_{2}\right)$. If $\varphi _{2}\succ ^{\circ }\psi _{2}$ and $\varphi _{1}\succ ^{\circ }\psi _{1}$, then $f_{1}\succ ^{\circ }g_{1}$ and $f_{2}\succ ^{\circ }g_{2}$. By Strict Independence,

$$\begin{aligned} \lambda f_{1}+\left( 1-\lambda \right) f_{2}\succ ^{\circ }\lambda g_{1}+\left( 1-\lambda \right) g_{2} \end{aligned}$$

but as observed

$$\begin{aligned}&\left. \lambda f_{1}+\left( 1-\lambda \right) f_{2}\succ ^{\circ }\lambda g_{1}+\left( 1-\lambda \right) g_{2}\right. \\&\quad \iff u\left( \lambda f_{1}+\left( 1-\lambda \right) f_{2}\right) \succ ^{\circ }u\left( \lambda g_{1}+\left( 1-\lambda \right) g_{2}\right) \\&\quad \iff \lambda \varphi _{1}+\left( 1-\lambda \right) \varphi _{2}\succ ^{\circ }\lambda \psi _{1}+\left( 1-\lambda \right) \psi _{2}. \end{aligned}$$

This shows that: If $\varphi _{2}\succ ^{\circ }\psi _{2}$ and $\lambda \in \left( 0,1\right)$, then

$$\begin{aligned} \varphi _{1}\succ ^{\circ }\psi _{1}\implies \lambda \varphi _{1}+\left( 1-\lambda \right) \varphi _{2}\succ ^{\circ }\lambda \psi _{1}+\left( 1-\lambda \right) \psi _{2}. \end{aligned}$$

On the other hand, if $\varphi _{2}=\psi _{2}$ and $\lambda \in \left( 0,1\right)$, we can choose $f_{2}=g_{2}$; then, by Strict Independence again,

$$\begin{aligned} \left. \varphi _{1}\succ ^{\circ }\psi _{1}\right.&\iff f_{1}\succ ^{\circ }g_{1}\implies \lambda f_{1}+\left( 1-\lambda \right) f_{2}\succ ^{\circ }\lambda g_{1}+\left( 1-\lambda \right) g_{2}\\&\iff u\left( \lambda f_{1}+\left( 1-\lambda \right) f_{2}\right) \succ ^{\circ }u\left( \lambda g_{1}+\left( 1-\lambda \right) g_{2}\right) \\&\iff \lambda \varphi _{1}+\left( 1-\lambda \right) \varphi _{2}\succ ^{\circ }\lambda \psi _{1}+\left( 1-\lambda \right) \psi _{2}, \end{aligned}$$

and this proves the first part of the property. As to the second, if $\lambda =0$ and $\varphi _{2}\succ ^{\circ }\psi _{2}$, then $\frac{1}{2}\psi _{1}+\frac{1}{2}\varphi _{2}=\frac{1}{2}\varphi _{1}+\frac{1}{2}\psi _{2}$ allows us to choose $f_{1}$, $f_{2}$, $g_{1}$, and $g_{2}$ so that $\frac{1}{2}g_{1}+\frac{1}{2}f_{2}=\frac{1}{2}f_{1}+\frac{1}{2}g_{2}$. Since $f_{2}\succ ^{\circ }g_{2}$, another application of Strict Independence delivers $f_{1}\succ ^{\circ }g_{1}$ that is $\varphi _{1}\succ ^{\circ }\psi _{1}$, as wanted.

Property 2. Let $\varphi =u\left( f\right)$, $\psi =u\left( g\right)$, $\zeta =u\left( h\right)$. If and $\psi \succ ^{\circ }\zeta$, then and $u\left( g\right) \succ ^{\circ }u\left( h\right)$, that is, and $g\succ ^{\circ }h$, by Strong Transitive Consistency, $f\succ ^{\circ }h$ and $u\left( f\right) \succ ^{\circ }u\left( h\right)$, that is, $\varphi \succ ^{\circ }\zeta$.

Property 3. Let $\varphi =u\left( f\right)$ and $\psi =u\left( g\right)$. If , then and, by Strong Transitive Consistency, $f\succ ^{\circ }g$, thus $u\left( f\right) \succ ^{\circ }u\left( g\right)$, that is, $\varphi \succ ^{\circ }\psi$.

Property 4. Let $\varphi =u\left( f\right)$, $\psi =u\left( g\right)$, $\varepsilon =u\left( x\right)$. If $\varphi \succ ^{\circ }\psi$, then $f\succ ^{\circ }g$, thus $g\not \succsim ^{\circ }f$ and

$$\begin{aligned} 1&\notin \{\lambda \in [0,1]:\lambda g+(1-\lambda )x\succsim ^{\circ }\lambda f+(1-\lambda )f\}\\ 1&\in \{\lambda \in [0,1]:\lambda g+(1-\lambda )x\succsim ^{\circ }\lambda f+(1-\lambda )f\}^{c}\\ 1&\in \{\lambda \in [0,1]:\lambda f+(1-\lambda )f\succ ^{\circ }\lambda g+(1-\lambda )x\}, \end{aligned}$$

and, by Continuity of $\succsim ^{\circ }$, the latter set is open in [0, 1] (because it is the complement of a closed set). Therefore there exists $\hat{\lambda }\in \left( 0,1\right) \,$ such that

$$\begin{aligned} \left( \hat{\lambda },1\right] \subseteq \{\lambda \in [0,1]:\lambda f+(1-\lambda )f\succ ^{\circ }\lambda g+(1-\lambda )x\} \end{aligned}$$

that is

$$\begin{aligned} f&\succ ^{\circ }\lambda g+(1-\lambda )x\qquad \forall \lambda \in \left( \hat{\lambda },1\right] \\ u\left( f\right)&\succ ^{\circ }u\left( \lambda g+(1-\lambda )x\right) \qquad \forall \lambda \in \left( \hat{\lambda },1\right] \\ u\left( f\right)&\succ ^{\circ }\lambda u\left( g\right) +(1-\lambda )u\left( x\right) \qquad \forall \lambda \in \left( \hat{\lambda },1\right] \\ \varphi&\succ ^{\circ }\lambda \psi +(1-\lambda )\varepsilon \qquad \forall \lambda \in \left( \hat{\lambda },1\right] \end{aligned}$$

as wanted.

Property 5. Let $0=u\left( x\right)$, by Weak Possibility, there exists $\tilde{g}$ in F such that $f\not \succsim ^{*}x$ implies $\tilde{g}\succsim ^{\circ }f$. Set $\tilde{\varphi }=u\left( g\right)$, and take any $\varphi =u\left( f\right)$, then

$$\begin{aligned}&\varphi \not \succcurlyeq ^{*}0\iff \lnot \left[ u\left( f\right) \succcurlyeq ^{*}u\left( x\right) \right] \iff \lnot \left[ f\succsim ^{*}x\right] \\&\quad \implies \tilde{g}\succsim ^{\circ }f\iff u\left( g\right) \succcurlyeq ^{\circ }u\left( f\right) \iff \tilde{\varphi }\succcurlyeq ^{\circ }\varphi \end{aligned}$$

as wanted. $\square$

By Theorem 5, since $C\ne \varnothing$ is a convex and closed subset of $\Delta$, $\succcurlyeq ^{*}$ is the binary relation on $B_{0}\left( S,\Sigma \right)$ defined by (32), and $\succcurlyeq ^{\circ }$ is another binary relation on $B_{0}\left( S,\Sigma \right)$ that satisfies Properties 0–5, there exists a unique function $\gamma :C\rightarrow \left[ 0,\infty \right]$ which is grounded, lower semicontinuous, convex, and bounded such that

$$\begin{aligned} \varphi \succcurlyeq ^{\circ }\psi \iff \max _{p\in C}\left\{ \int \varphi {\hbox {d}}p-\int \psi {\hbox {d}}p-\gamma \left( p\right) \right\} \ge 0 \end{aligned}$$

then

$$\begin{aligned} \left. f\succsim ^{\circ }g\right.&\iff u\left( f\right) \succcurlyeq ^{\circ }u\left( g\right) \iff \max _{p\in C}\left\{ \int u\left( f\right) {\hbox {d}}p-\int u\left( g\right) {\hbox {d}}p-\gamma \left( p\right) \right\} \ge 0\\&\iff \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p+\gamma \left( p\right) \qquad \text {for some }p\in C. \end{aligned}$$

This concludes the proof of (i) implies (ii).

Moreover, if $\delta :C\rightarrow \left[ 0,\infty \right]$ is a grounded, convex, lower semicontinuous, and bounded function such that, for every $f,g\in F$,

$$\begin{aligned} f\succsim ^{\circ }g\iff \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p+\delta \left( p\right) \qquad \text {for some }p\in C \end{aligned}$$

then, for every $\varphi =u\left( f\right)$ and $\psi =u\left( g\right)$ in $B_{0}\left( S,\Sigma \right)$,

$$\begin{aligned} \left. \varphi \succcurlyeq ^{\circ }\psi \right.&\iff u\left( f\right) \succcurlyeq ^{\circ }u\left( g\right) \iff f\succsim ^{\circ }g\iff \int u\left( f\right) {\hbox {d}}p\\&\quad \ge \int u\left( g\right) {\hbox {d}}p+\delta \left( p\right) \qquad \text {for some }p\in C\\&\iff \max _{p\in C}\left\{ \int u\left( f\right) {\hbox {d}}p-\int u\left( g\right) {\hbox {d}}p-\delta \left( p\right) \right\} \ge 0\\&\iff \max _{p\in C}\left\{ \int \varphi {\hbox {d}}p-\int \psi {\hbox {d}}p-\delta \left( p\right) \right\} \ge 0 \end{aligned}$$

and $\delta =\gamma$, by Theorem 5. This shows that $\gamma$ is unique given u. Finally, again by Theorem 5, it holds

$$\begin{aligned} \gamma \left( p\right)&=\sup \left\{ \left\langle \psi ,p\right\rangle -\left\langle \varphi ,p\right\rangle :\varphi ,\psi \in B_{0}\left( S,\Sigma \right) \text { and }\psi \prec ^{\circ }\varphi \right\} \\&=\sup \left\{ \int u\left( g\right) {\hbox {d}}p-\int u\left( f\right) {\hbox {d}}p:g\prec ^{\circ }f\text { in }F\right\} \end{aligned}$$

for all $p\in C$, thus (22) holds.

Remark 4

If $\alpha >0$ and $\beta \in {\mathbb {R}}$, then (ii) and simple algebra yield

$$\begin{aligned} \left. f\succsim ^{*}g\right.&\iff \int \left[ \alpha u\left( f\right) +\beta \right] {\hbox {d}}p\ge \int \left[ \alpha u\left( g\right) +\beta \right] {\hbox {d}}p\qquad \forall p\in C\\ \left. f\succsim ^{\circ }g\right.&\iff \int \left[ \alpha u\left( f\right) +\beta \right] {\hbox {d}}p\ge \int \left[ \alpha u\left( g\right) +\beta \right] {\hbox {d}}p+\alpha c\left( p\right) \qquad \text {for some }p\in C\text {.} \end{aligned}$$

This means that when u is replaced by $v=\alpha u+\beta$, c must be replaced with $\alpha c$, because $\alpha c$ delivers the desired representation, and given v, there can be only one grounded, convex, lower semicontinuous, and bounded cost function with this property.

In particular, if c is not identically 0 for some u, it can never be identically zero.

(ii) implies (i). Assume that there exist a non-empty closed and convex set C of probabilities on $\Sigma$, a grounded, convex, lower semicontinuous, and bounded $c:C\rightarrow \left[ 0,\infty \right)$, and an onto affine function $u:X\rightarrow {\mathbb {R}}$, such that, for every $f,g\in F$,

$$\begin{aligned} f\succsim ^{*}g\iff \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p\qquad \forall p\in C \end{aligned}$$

and

$$\begin{aligned} f\succsim ^{\circ }g\iff \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p+c\left( p\right) \qquad \text {for some }p\in C. \end{aligned}$$

By Lemma 1, $\succsim ^{*}$ satisfies the BC, C-Completeness, Transitivity, Independence. Since u represents $\succsim ^{*}$ on X and $u\left( X\right) ={\mathbb {R}}$, Lemma 59 of Cerreia-Vioglio et al. (2011a) guarantee that Unboundedness is satisfied too.

Since c is grounded, there exists $\bar{p}\in C$ such that $c\left( \bar{p}\right) =0$. It follows that for any $f,g\in F$ either $\int u\left( f\right) d\bar{p}\ge \int u\left( g\right) d\bar{p}+c\left( \bar{p}\right)$ or $\int u\left( g\right) d\bar{p}\ge \int u\left( f\right) d\bar{p}+c\left( \bar{p}\right)$, that is, $\succsim ^{\circ }$ is complete.

Define $I:B_{0}\left( S,\Sigma \right) \rightarrow {\mathbb {R}}$ by

$$\begin{aligned} I\left( \phi \right) =\min _{p\in C}\left\{ \int \phi {\hbox {d}}p+c\left( p\right) \right\} \qquad \forall \phi \in B_{0}\left( S,\Sigma \right) . \end{aligned}$$

It is immediate to see that I is a normalized concave niveloid and

$$\begin{aligned} \bar{I}\left( \phi \right) =-I\left( -\phi \right) =\max _{p\in C}\left\{ \int \phi {\hbox {d}}p-c\left( p\right) \right\} \qquad \forall \phi \in B_{0}\left( S,\Sigma \right) \end{aligned}$$

is a normalized convex niveloid.

Moreover,

$$\begin{aligned} \left. f\succsim ^{\circ }g\right.&\iff \int u\left( f\right) {\hbox {d}}p-\int u\left( g\right) {\hbox {d}}p-c\left( p\right) \ge 0\qquad \text {for some }p\in C\\&\iff \max _{p\in C}\left\{ \int u\left( f\right) {\hbox {d}}p-\int u\left( g\right) {\hbox {d}}p-c\left( p\right) \right\} \ge 0\\&\iff \bar{I}\left( u\left( f\right) -u\left( g\right) \right) \ge 0\\&\iff -I\left( u\left( g\right) -u\left( f\right) \right) \ge 0\\&\iff I\left( u\left( g\right) -u\left( f\right) \right) \le 0 \end{aligned}$$

while

$$\begin{aligned} \left. g\succ ^{\circ }f\right. \iff \lnot \left( f\succsim ^{\circ }g\right) \iff I\left( u\left( g\right) -u\left( f\right) \right) >0. \end{aligned}$$

Continuity. Consider $f,g,h,l\in F$ and a sequence $\lambda _{n}$ in $\left[ 0,1\right]$ such that $\lambda _{n}\rightarrow \lambda$. If $\lambda _{n}f+\left( 1-\lambda _{n}\right) g\succsim ^{\circ }\lambda _{n}h+\left( 1-\lambda _{n}\right) l$ for all $n\in {\mathbb {N}}$, then

$$\begin{aligned} 0&\le \bar{I}\left( u\left( \lambda _{n}f+\left( 1-\lambda _{n}\right) g\right) -u\left( \lambda _{n}h+\left( 1-\lambda _{n}\right) l\right) \right) \\&=\bar{I}\left( \lambda _{n}u\left( f\right) +\left( 1-\lambda _{n}\right) u\left( g\right) -\left( \lambda _{n}u\left( h\right) +\left( 1-\lambda _{n}\right) u\left( l\right) \right) \right) . \end{aligned}$$

Since I is continuous, the inequality also holds for $\lambda$, proving Continuity.

Strict Independence. Assume that $g\succeq ^{\circ }l$ and $\lambda \in \left( 0,1\right)$. For all $f,h\in F$, we have that

$$\begin{aligned}&I\left( u\left( \lambda f+\left( 1-\lambda \right) g\right) -u\left( \lambda h+\left( 1-\lambda \right) l\right) \right) \\&\quad =I\left( \lambda u\left( f\right) +\left( 1-\lambda \right) u\left( g\right) -\left( \lambda u\left( h\right) +\left( 1-\lambda \right) u\left( l\right) \right) \right) \\&\quad =I\left( \lambda \left( u\left( f\right) -u\left( h\right) \right) +\left( 1-\lambda \right) \left( u\left( g\right) -u\left( l\right) \right) \right) \\&\quad \ge \lambda I\left( u\left( f\right) -u\left( h\right) \right) +\left( 1-\lambda \right) I\left( u\left( g\right) -u\left( l\right) \right) . \end{aligned}$$

If $f\succ ^{\circ }h$, then $I\left( u\left( f\right) -u\left( h\right) \right) >0$ and $g\succeq ^{\circ }l$ implies $I\left( u\left( g\right) -u\left( l\right) \right) \ge 0$ whence

$$\begin{aligned} I\left( u\left( \lambda f+\left( 1-\lambda \right) g\right) -u\left( \lambda h+\left( 1-\lambda \right) l\right) \right) >0, \end{aligned}$$

proving that $\lambda f+\left( 1-\lambda \right) g\succ ^{\circ }\lambda h+\left( 1-\lambda \right) l$. Conversely, if $\lambda =0$ and $\frac{1}{2}h+\frac{1}{2}g=\frac{1}{2}f+\frac{1}{2}l$, then $u\left( g\right) -u\left( l\right) =u\left( f\right) -u\left( h\right)$ and

$$\begin{aligned} g\succ ^{\circ }l\iff I\left( u\left( g\right) -u\left( l\right) \right)>0\iff I\left( u\left( f\right) -u\left( h\right) \right) >0\iff f\succ ^{\circ }h. \end{aligned}$$

Strong Transitive Consistency. If and $g\succeq ^{\circ }h$, then

$$\begin{aligned} \int u\left( f\right) {\hbox {d}}p>\int u\left( g\right) {\hbox {d}}p\text { and }\int u\left( h\right) {\hbox {d}}p\le \int u\left( g\right) {\hbox {d}}p+c\left( p\right) \qquad \forall p\in C \end{aligned}$$

(where the equality part in the second relation accounts for the case $g=h$ because $c\ge 0$), which implies

$$\begin{aligned} \int u\left( h\right) {\hbox {d}}p\le \int u\left( g\right) {\hbox {d}}p+c\left( p\right) <\int u\left( f\right) {\hbox {d}}p+c\left( p\right) \qquad \forall p\in C \end{aligned}$$

that is, $f\succ ^{\circ }h$.

Substitution Consistency. If $f\sim ^{*}h$, $g\sim ^{*}l$, and $f\succsim ^{\circ }g$ imply

$$\begin{aligned} \int u\left( f\right) {\hbox {d}}p&=\int u\left( h\right) {\hbox {d}}p\qquad \forall p\in C\\ \int u\left( g\right) {\hbox {d}}p&=\int u\left( l\right) {\hbox {d}}p\qquad \forall p\in C\\ \int u\left( f\right) {\hbox {d}}q&\ge \int u\left( g\right) {\hbox {d}}q+c\left( q\right) \qquad \text {for some }q\in C \end{aligned}$$

whence $\int u\left( h\right) {\hbox {d}}q\ge \int u\left( l\right) {\hbox {d}}q+c\left( q\right) \$for some $q\in C$ and $h\succsim ^{\circ }l$.

Weak Possibility. Assume that $f\not \succsim ^{*}g$, it follows that there exists $\bar{p}\in C$ such that

$$\begin{aligned} \int u\left( f\right) d\bar{p}<\int u\left( g\right) d\bar{p} \end{aligned}$$

but then, setting $k=\sup _{p\in C}c\left( p\right)$,

$$\begin{aligned} \int u\left( g\right) d\bar{p}+k\ge \int u\left( f\right) d\bar{p}+c\left( \bar{p}\right) \end{aligned}$$

and so it is sufficient to consider $\tilde{g}$ such $u\left( \tilde{g}\right) =u\left( g\right) +k$ to have $\tilde{g}\succsim ^{\circ }f$ for all $f\not \succsim ^{*}g.$

It only remains to prove that $\succsim ^{*}$ is the transitive core of $\succsim ^{\circ }$.

First assume that $f\succsim ^{*}g$, then

$$\begin{aligned} \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p\qquad \forall p\in C. \end{aligned}$$

(46)

But then $h\succsim ^{\circ }f$ implies that there exists $q\in C$ such that

$$\begin{aligned} \int u\left( h\right) {\hbox {d}}q\ge \int u\left( f\right) {\hbox {d}}q+c\left( q\right) \ge \int u\left( g\right) {\hbox {d}}q+c\left( q\right) \end{aligned}$$

where the last inequality follows from (46), that is, $h\succsim ^{\circ }g$. Analogously, $g\succsim ^{\circ }l$ implies that there exists $q\in C$ such that

$$\begin{aligned} \int u\left( f\right) {\hbox {d}}q\ge \int u\left( g\right) {\hbox {d}}q\ge \int u\left( l\right) {\hbox {d}}q+c\left( q\right) \end{aligned}$$

where the first inequality follows from (46), that is, $f\succsim ^{\circ }l$. This shows that $f\succsim ^{*}g$ implies $f\succsim ^{\circ \circ }g$.

As to the converse, assume that $f\succsim ^{\circ \circ }g$, then given $h\in F$,

$$\begin{aligned} g\succsim ^{\circ }h\implies f\succsim ^{\circ }h \end{aligned}$$

(47)

that is

$$\begin{aligned} I\left( u\left( h\right) -u\left( g\right) \right) \le 0\implies I\left( u\left( h\right) -u\left( f\right) \right) \le 0\text {.} \end{aligned}$$

Now given $\eta \in B_{0}\left( S,\Sigma \right)$ and $k\in {\mathbb {R}}$, the above relation delivers

$$\begin{aligned} \left. I\left( \eta -u\left( g\right) \right) \le k\right.&\implies I\left( \eta -k-u\left( g\right) \right) \le 0\implies I\left( u\left( h_{\eta ,k}\right) -u\left( g\right) \right) \le 0\\&\implies I\left( u\left( h_{\eta ,k}\right) -u\left( f\right) \right) \le 0\implies I\left( \eta -k-u\left( f\right) \right) \le 0\\&\implies I\left( \eta -u\left( f\right) \right) \le k \end{aligned}$$

where $h_{\eta ,k}$ is an element of F such that $u\left( h_{\eta ,k}\right) =\eta -k$. Therefore, given $\eta \in B_{0}\left( S,\Sigma \right)$ and $k\in {\mathbb {R}}$, if $I\left( \eta -u\left( g\right) \right) \le k$, then also $I\left( \eta -u\left( f\right) \right) \le k$. In particular, taking any $\eta \in B_{0}\left( S,\Sigma \right)$, since $I\left( \eta -u\left( g\right) \right) \le I\left( \eta -u\left( g\right) \right)$, then also $I\left( \eta -u\left( f\right) \right) \le I\left( \eta -u\left( g\right) \right)$. We have shown that

$$\begin{aligned} f\succsim ^{\circ \circ }g\implies I\left( \eta -u\left( g\right) \right) -I\left( \eta -u\left( f\right) \right) \ge 0\qquad \forall \eta \in B_{0}\left( S,\Sigma \right) . \end{aligned}$$

(48)

Recall that, for every $\psi \in B_{0}\left( S,\Sigma \right)$,

$$\begin{aligned} \partial I\left( \psi \right) =\left\{ p\in \Delta :I\left( \varphi \right) -I\left( \psi \right) \le \int \left( \varphi -\psi \right) {\hbox {d}}p\quad \forall \varphi \in B_{0}\left( S,\Sigma \right) \right\} . \end{aligned}$$

Then, for every $\varphi \in B_{0}\left( S,\Sigma \right)$, we have

$$\begin{aligned} \int \left( \varphi -\psi \right) {\hbox {d}}p\ge I\left( \varphi \right) -I\left( \psi \right) \qquad \forall p\in \partial I\left( \psi \right) . \end{aligned}$$

Then, for every $\eta \in B_{0}\left( S,\Sigma \right)$, setting $\psi _{\eta }=\eta -u\left( f\right)$, and $\varphi _{\eta }=\eta -u\left( g\right)$, equation (48) implies

$$\begin{aligned} \int \left( u\left( f\right) -u\left( g\right) \right) {\hbox {d}}p= & {} \int \left( \varphi _{\eta }-\psi _{\eta }\right) {\hbox {d}}p\ge I\left( \varphi _{\eta }\right) -I\left( \psi _{\eta }\right) \\= & {} I\left( \eta -u\left( g\right) \right) -I\left( \eta -u\left( f\right) \right) \ge 0 \end{aligned}$$

for all $p\in \partial I\left( \eta -u\left( f\right) \right)$. We have shown that

$$\begin{aligned} \left. f\succsim ^{\circ \circ }g\right.&\implies \int \left( u\left( f\right) -u\left( g\right) \right) {\hbox {d}}p\ge 0\qquad \forall p\in \bigcup _{\eta \in B_{0}\left( S,\Sigma \right) }\partial I\left( \eta -u\left( f\right) \right) \\&\implies \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p\qquad \forall p\in \bigcup _{\zeta \in B_{0}\left( S,\Sigma \right) }\partial I\left( \zeta \right) \\&\implies \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p\qquad \forall p\in {\text {cl}}\,\left( {\text {co}}\, \bigcup _{\zeta \in B_{0}\left( S,\Sigma \right) }\partial I\left( \zeta \right) \right) , \end{aligned}$$

but the results of MMR and Cerreia-Vioglio et al. (2015, henceforth CMMR) guarantee that ${\text {cl} }\,\left( {\text {co}}\,\bigcup _{\zeta \in B_{0}\left( S,\Sigma \right) }\partial I\left( \zeta \right) \right) =C$,²⁴ so that $f\succsim ^{*}g$. $\square$

1.3 Proof of Theorem 4

(i) implies (ii). By Lemma 1, there exist a non-empty closed and convex set C of probabilities on $\Sigma$ and a non-constant affine $u:X\rightarrow {\mathbb {R}}$ such that, for every $f,g\in F$,

$$\begin{aligned} f\succsim ^{*}g\iff \int u\left( f\right) {\hbox {d}}p\ge \int u\left( g\right) {\hbox {d}}p\qquad \forall p\in C. \end{aligned}$$

Moreover, given $x,y\in X$, by Reflexivity of $\succsim ^{\circ }$ and Transitive Consistency,

$$\begin{aligned} x\succsim ^{*}y\implies x\succsim ^{*}y\succsim ^{\circ }y\implies x\succsim ^{\circ }y. \end{aligned}$$

That is, on constant acts, $\succsim ^{*}$ is a subrelation of $\succsim ^{\circ }$, and both relations are non-trivial and satisfy the axioms of Herstein and Milnor (1953). By Corollary B.3 of Ghirardato et al. (2004), these relations coincide and are both represented by u. For this reason, we often omit the superscripts $^{*}$ and $^{\circ }$ when comparing constant acts.

Next we show that, for every $f,g\in F$,

$$\begin{aligned} g\succsim ^{\circ }f\iff \int u\left( g\right) {\hbox {d}}p\ge \int u\left( f\right) {\hbox {d}}p\qquad \text {for some }p\in C. \end{aligned}$$

Assume first that there exists $q\in C$ such that

$$\begin{aligned} \int u\left( g\right) {\hbox {d}}q\ge \int u\left( f\right) {\hbox {d}}q. \end{aligned}$$

By Proposition 4, it is not true that, for every $x\succ y$ in X, there exists $\varepsilon$ in $\left( 0,1\right)$ such that

$$\begin{aligned} \left( 1-\varepsilon \right) f+\varepsilon y\succsim ^{*}\left( 1-\varepsilon \right) g+\varepsilon x. \end{aligned}$$

Then, there are $x\succ y$ in X such that, for every $\varepsilon$ in $\left( 0,1\right)$,

$$\begin{aligned} \left( 1-\varepsilon \right) f+\varepsilon y\not \succsim ^{*}\left( 1-\varepsilon \right) g+\varepsilon x \end{aligned}$$

and by Possibility

$$\begin{aligned} \left( 1-\varepsilon \right) g+\varepsilon x\succsim ^{\circ }\left( 1-\varepsilon \right) f+\varepsilon y\qquad \forall \varepsilon \in \left( 0,1\right) \end{aligned}$$

and Continuity of $\succsim ^{\circ }$ delivers $g\succsim ^{\circ }f$. This shows that, if $\int u\left( g\right) {\hbox {d}}p\ge \int u\left( f\right) {\hbox {d}}p\$for some $p\in C$, then $g\succsim ^{\circ }f$. Conversely, assume—per contra—that there exist $f,g\in F$ such that $g\succsim ^{\circ }f$ and

$$\begin{aligned} \int u\left( g\right) {\hbox {d}}p<\int u\left( f\right) {\hbox {d}}p\qquad \forall p\in C. \end{aligned}$$

Then, by Proposition 4 again, there exist $x\succ y$ in X and $\varepsilon$ in $\left( 0,1\right)$ such that

$$\begin{aligned} \left( 1-\varepsilon \right) f+\varepsilon y\succsim ^{*}\left( 1-\varepsilon \right) g+\varepsilon x. \end{aligned}$$

By C-Independence of $\succsim ^{\circ }$, and since $g\succsim ^{\circ }f$, it follows

$$\begin{aligned} \left( 1-\varepsilon \right) f+\varepsilon y\succsim ^{*}\left( 1-\varepsilon \right) g+\varepsilon x\succsim ^{\circ }\left( 1-\varepsilon \right) f+\varepsilon x \end{aligned}$$

so that by Transitive Consistency

$$\begin{aligned} \left( 1-\varepsilon \right) f+\varepsilon y\succsim ^{\circ }\left( 1-\varepsilon \right) f+\varepsilon x \end{aligned}$$

(49)

which by Monotonicity of $\succsim ^{\circ }$ leads to a contradiction. In fact, $x\succ y$ implies

$$\begin{aligned} u\left( \left( 1-\varepsilon \right) f\left( s\right) +\varepsilon x\right) >u\left( \left( 1-\varepsilon \right) f\left( s\right) +\varepsilon y\right) \end{aligned}$$

for all $s\in S$, that is, $\left[ \left( 1-\varepsilon \right) f+\varepsilon x\right] \left( s\right) \succ \left[ \left( 1-\varepsilon \right) f+\varepsilon y\right] \left( s\right)$ for all $s\in S$, and $\left( 1-\varepsilon \right) f+\varepsilon x\succ ^{\circ }\left( 1-\varepsilon \right) f+\varepsilon y$, contradicting (49).

(ii) implies (i). By Theorem 1, $\succsim ^{*}$ satisfies the BC, C-Completeness, Transitivity, and Independence, $\succsim ^{{ \circ }}$ satisfies Continuity, and $\left( \succsim ^{*},\succsim ^{{ \circ } }\right)$ satisfies Possibility. It remains to show that $\left( \succsim ^{*},\succsim ^{{ \circ }}\right)$ satisfies Transitive Consistency, and $\succsim ^{\circ }$ satisfies Completeness, C-Transitivity, and C-Independence, Reflexivity, Monotonicity, and Non-Triviality. The verification is routinely obtained by using the representations (23) and (24) and observing that:

• given any $x,y\in X$,

  $$\begin{aligned} x\succsim ^{*}y\iff x\succsim ^{\circ }y\iff u\left( x\right) \ge u\left( y\right) \text {;} \end{aligned}$$

• given any two simple measurable functions $\varphi ,\psi :S\rightarrow {\mathbb {R}}$,

  $$\begin{aligned} \varphi \left( s\right)>\psi \left( s\right) \qquad \forall s\in S\implies \int \varphi {\hbox {d}}p>\int \psi {\hbox {d}}p\qquad \forall p\in C\text {.} \end{aligned}$$

Finally, replace C-Transitivity and Possibility with Transitivity and C-Possibility in (i) of our statement, it is then easy to check that the conditions in point (i) of Theorem 3 of GMMS are satisfied.²⁵ Representations (23) and (25) follow. The converse follows by (23) and (25). $\square$