Additive representation of separable preferences over infinite products

Abstract

Let $\mathcal{X }$ be a set of outcomes, and let $\mathcal{I }$ be an infinite indexing set. This paper shows that any separable, permutation-invariant preference order $(\succcurlyeq )$ on $\mathcal{X }^\mathcal{I }$ admits an additive representation. That is: there exists a linearly ordered abelian group $\mathcal{R }$ and a ‘utility function’ $u:\mathcal{X }{{\longrightarrow }}\mathcal{R }$ such that, for any $\mathbf{x},\mathbf{y}\in \mathcal{X }^\mathcal{I }$ which differ in only finitely many coordinates, we have $\mathbf{x}\succcurlyeq \mathbf{y}$ if and only if $\sum _{i\in \mathcal{I }} \left[u(x_i)-u(y_i)\right]\ge 0$. Importantly, and unlike almost all previous work on additive representations, this result does not require any Archimedean or continuity condition. If $(\succcurlyeq )$ also satisfies a weak continuity condition, then the paper shows that, for any $\mathbf{x},\mathbf{y}\in \mathcal{X }^\mathcal{I }$, we have $\mathbf{x}\succcurlyeq \mathbf{y}$ if and only if ${}^*\!\sum _{i\in \mathcal{I }} u(x_i)\ge {}^*\!\sum _{i\in \mathcal{I }}u(y_i)$. Here, ${}^*\!\sum _{i\in \mathcal{I }} u(x_i)$ represents a nonstandard sum, taking values in a linearly ordered abelian group ${}^*\!\mathcal{R }$, which is an ultrapower extension of $\mathcal{R }$. The paper also discusses several applications of these results, including infinite-horizon intertemporal choice, choice under uncertainty, variable-population social choice and games with infinite strategy spaces.

Appendix: Proofs

Proof of Theorem 1

‘${\Longleftarrow }$’ It is easy to check that the additive preorder defined by (2) is $\Pi _{\scriptscriptstyle {\mathrm{fin}}}$-invariant and separable.

‘$\Longrightarrow$’ Fix some $o\in \mathcal{X }$, and let $(\mathcal{A },+)$ be the free abelian group generated by $\mathcal{X }\setminus \{o\}$. That is, $\mathcal{A }$ consists of all formal $\mathbb{Z }$-linear combinations of the form ‘$J_1 x_1 + J_2 x_2 + \cdots + J_N x_N$’ where $N\in \mathbb{N }, J_1,\ldots ,J_N\in \mathbb{Z }\setminus \{0\}$ and $x_1,\ldots ,x_N\in \mathcal{X }\setminus \{o\}$ are distinct.

Let $\mathcal{B }^{\prime }\subset \mathcal{A }$ be the set of all elements where $J_1,\ldots ,J_N>0$. For any such $b\in \mathcal{B }'$, let us define $\mathbf{w}^b\in \mathcal{X }^\mathcal{I }$ as follows. Let $\mathcal{J }_1,\mathcal{J }_2,\ldots ,\mathcal{J }_N$ be disjoint subsets of $\mathcal{I }$, with $|\mathcal{J }_n|=J_n$ for all $n\in {\left[ 1\ldots N \right]}$. For all $n\in {\left[ 1\ldots N \right]}$ and all $j\in \mathcal{J }_n$, let $w^b_j := x_n$. Meanwhile, for all $i\in \mathcal{I }\setminus \mathcal{J }_1\sqcup \cdots \sqcup \mathcal{J }_N$, define $w^b_i:=o$. (Heuristic: if we regard the elements of $\mathcal{X }$ as ‘goods’ and ‘bads’, then $\mathbf{w}^b$ represents a ‘bundle’ containing $J_n$ units of $x_n$ for each $n\in {\left[ 1\ldots N \right]}$.) Since $(\succcurlyeq )$ is $\Pi _{\scriptscriptstyle {\mathrm{fin}}}$-invariant, it does not matter how we choose the sets $\mathcal{J }_1,\ldots ,\mathcal{J }_N$; if $\mathbf{w}^b$ and ${\widetilde{\mathbf{w}}}^b$ are two different elements of $\mathcal{X }^\mathcal{I }$ built using the above recipe, then we automatically have $\mathbf{w}^b\approx {\widetilde{\mathbf{w}}}^b$.

Finally, if $0$ denotes the identity element of $\mathcal{A }$, then let $\mathcal{B }:=\{0\}\sqcup \mathcal{B }'$, and define $\mathbf{w}^0$ by setting $w^0_i:=o$ for all $i\in \mathcal{I }$. For any $b,b^{\prime }\in \mathcal{B }$, if $b= J_1 x_1 + \cdots + J_N x_N$ and $b'= J'_1 x'_1 + \cdots + J'_M x'_M$, then say $b$ and $b'$ are disjoint if the sets $\{x_1,\ldots ,x_N\}$ and $\{x'_1,\ldots ,x'_M\}$ are disjoint.

Claim 1 Let $a\in \mathcal{A }$.

• (a) There exist unique disjoint $a_+,a_-\in \mathcal{B }$ such that $a=a_+-a_-$.

• (b) Let $b,c\in \mathcal{B }$ be any other elements such that $a=b-c$. Then $(\mathbf{w}^b\succcurlyeq \mathbf{w}^c) \iff (\mathbf{w}^{a_+} \succcurlyeq \mathbf{w}^{a_-})$.

Proof

• (a) If $a=0$, then let $0_+:=0$ and $0_-:=0$. Then $0_+,0_-\in \mathcal{B }$ are disjoint, and $0=0_+-0_-$. Now suppose that $a\ne 0$. Let $\displaystyle a= \sum _{w\in \mathcal{W }} A_w w$, where $\mathcal{W }\subseteq \mathcal{X }$ is a finite subset and $A_w\in \mathbb{Z }\setminus \{0\}$ for all $w\in \mathcal{W }$. Then $\mathcal{W }:=\mathcal{W }_-\sqcup \mathcal{W }_+$, where $\mathcal{W }_-:=\{w\in \mathcal{W }$; $A_w<0\}$ and $\mathcal{W }_+:=\{w\in \mathcal{W }$; $A_w>0\}$. Let $\displaystyle a_+:= \sum _{w\in \mathcal{W }_+} A_w w$ and $\displaystyle a_-:= \sum _{w\in \mathcal{W }_-} (-A_w) w$. (If $\mathcal{W }_+=\emptyset$, then $a_+:=0$. If $\mathcal{W }_-=\emptyset$, then $a_-:=0$.) Then $a_+,a_-\in \mathcal{B }$ are disjoint and $a=a_+-a_-$.

• (b) Suppose that $\displaystyle b:= \sum _{y\in \mathcal{Y }} B_y y$ and $\displaystyle c:= \sum _{z\in \mathcal{Z }} C_z z$, for some finite subsets $\mathcal{Y },\mathcal{Z }\subseteq \mathcal{X }$ and coefficients $B_y\in \mathbb{N }$ for all $y\in \mathcal{Y }$ and $C_z\in \mathbb{N }$ for all $z\in \mathcal{Z }$. If $b-c=a$, then we must have $\mathcal{W }\subseteq \mathcal{Y }\cup \mathcal{Z }$, $\mathcal{Y }\setminus \mathcal{Z }\subseteq \mathcal{W }_+$ and $\mathcal{Z }\setminus \mathcal{Y }\subseteq \mathcal{W }_-$. Furthermore:

  • $B_y = A_y$ for all $y\in \mathcal{Y }\setminus \mathcal{Z }$.

  • $C_z = -A_z$ for all $z\in \mathcal{Z }\setminus \mathcal{Y }$.

  • $B_w - C_w = A_w$ for all $w\in \mathcal{Y }\cap \mathcal{Z }\cap \mathcal{W }$.

  • $B_x = C_x$ for all $x\in (\mathcal{Y }\cap \mathcal{Z })\setminus \mathcal{W }$.

  Let $J:=\max \left(\{B_y; y\in \mathcal{Y }\}\cup \{C_z; z\in \mathcal{Z }\}\right)$. For all $x\in \mathcal{Y }\cup \mathcal{Z }$, let $\mathcal{J }_x\subset \mathcal{I }$ be a subset of cardinality $J$, and suppose that all these sets are disjoint. Since $(\succcurlyeq )$ is $\Pi _{\scriptscriptstyle {\mathrm{fin}}}$-invariant, we can permute the coordinates of $\mathbf{w}^b$ and/or $\mathbf{w}^c$ each in any desired finitary way without changing their $(\succcurlyeq )$-ordering. Likewise, we can finitarily permute $\mathbf{w}^{a_+}$ and/or $\mathbf{w}^{a_-}$ without changing their $(\succcurlyeq )$-ordering. Thus, without loss of generality, we can suppose:

  • $\mathbf{w}^{a_+}$ assigns the value $w$ to exactly $A_w$ coordinates in $\mathcal{J }_w$, for each $w\in \mathcal{W }_+$.

  • $\mathbf{w}^{a_-}$ assigns the value $w$ to exactly $-A_w$ coordinates in $\mathcal{J }_w$, for each $w\in \mathcal{W }_-$.

  • $\mathbf{w}^{b}$ assigns the value $y$ to exactly $B_y$ coordinates in $\mathcal{J }_y$, for each $y\in \mathcal{Y }$.

  • $\mathbf{w}^{c}$ assigns the value $z$ to exactly $C_z$ coordinates in $\mathcal{J }_z$, for each $z\in \mathcal{Z }$.

  Let us describe these as the ‘active’ coordinates. In all four cases, let us assign all other (‘inactive’) coordinates the value $o$. We can further assume that:

  • 1. For all $y\in \mathcal{Y }\setminus \mathcal{Z }$, $\mathbf{w}^{a_+}$ and $\mathbf{w}^b$ involve the same $B_y=A_y$ active coordinates.

  • 2. For all $z\in \mathcal{Z }\setminus \mathcal{Y }, \mathbf{w}^{a_-}$ and $\mathbf{w}^c$ involve the same $C_z=-A_z$ active coordinates.

  • 3. For all $x\in (\mathcal{Y }\cap \mathcal{Z })\setminus \mathcal{W }$, $\mathbf{w}^b$ and $\mathbf{w}^c$ involve the same set $\mathcal{I }_x\subseteq \mathcal{J }_x$ of $B_x=C_x$ active coordinates.

  • 4. For all $w\in \mathcal{Y }\cap \mathcal{Z }\cap \mathcal{W }_+$, the set of active coordinates in $\mathbf{w}^b$ is the disjoint union of the active coordinates of $\mathbf{w}^{a_+}$ and $\mathbf{w}^c$. In this case, let $\mathcal{I }_w\subseteq \mathcal{J }_w$ be the active coordinates of $\mathbf{w}^c$.

  • 5. For all $w\in \mathcal{Y }\cap \mathcal{Z }\cap \mathcal{W }_-$, the set of active coordinates in $\mathbf{w}^c$ is the disjoint union of the active coordinates of $\mathbf{w}^{a_-}$ and $\mathbf{w}^b$. In this case, let $\mathcal{I }_w\subseteq \mathcal{J }_w$ be the active coordinates of $\mathbf{w}^b$.

  Now suppose that $\mathbf{w}^b\succcurlyeq \mathbf{w}^c$. Let us define

  $$\begin{aligned} \mathcal{J }:= \bigsqcup _{x\in (\mathcal{Y }\cap \mathcal{Z })\setminus \mathcal{W }} \! \mathcal{I }_x \ \sqcup \ \bigsqcup _{w\in \mathcal{Y }\cap \mathcal{Z }\cap \mathcal{W }_+} \! \mathcal{I }_w \ \sqcup \ \bigsqcup _{w\in \mathcal{Y }\cap \mathcal{Z }\cap \mathcal{W }_-} \!\mathcal{I }_w. \end{aligned}$$

  Then $\mathbf{w}^b_{\mathcal{J }} = \mathbf{w}^c_{\mathcal{J }}$, by assumptions 3, 4 and 5. Meanwhile $\mathbf{w}^{a_+}_{\mathcal{J }} = \mathbf{o}_\mathcal{J }= \mathbf{w}^{a_-}_{\mathcal{J }}$, by assumptions 4 and 5. On the other hand, if $\mathcal{K }:=\mathcal{I }\setminus \mathcal{J }$, then we have $\mathbf{w}^b_\mathcal{K }= \mathbf{w}^{a_+}_\mathcal{K }$ (by assumptions 1, 4 and 5) and $\mathbf{w}^c_\mathcal{K }= \mathbf{w}^{a_-}_\mathcal{K }$ (by assumptions 2, 4 and 5). Thus, the separability of $(\succcurlyeq )$ implies that $(\mathbf{w}^b\succcurlyeq \mathbf{w}^c) \iff (\mathbf{w}^{a_+} \succcurlyeq \mathbf{w}^{a_-})$, as desired.$\diamondsuit \text { Claim } 1$

Let $\mathcal{C }_+:=\{a\in \mathcal{A }$; $\mathbf{w}^{a_+} \succ \mathbf{w}^{a_-}\}$, $\mathcal{C }_-:=\{a\in \mathcal{A }$; $\mathbf{w}^{a_+} \prec \mathbf{w}^{a_-}\}$, and $\mathcal{C }_0:=\{a\in \mathcal{A }$; $\mathbf{w}^{a_+} \approx \mathbf{w}^{a_-}\}$. Then $\mathcal{A }=\mathcal{C }_-\sqcup \mathcal{C }_0\sqcup \mathcal{C }_+$. Let $\mathcal{C }_{0+}:=\mathcal{C }_0\sqcup \mathcal{C }_+$ and $\mathcal{C }_{0-}:=\mathcal{C }_0\sqcup \mathcal{C }_-$.

Claim 2 Let $b,c\in \mathcal{A }$.

• (a) If $b,c\in \mathcal{C }_{0+}$, then $b+c\in \mathcal{C }_{0+}$. If also $b\in \mathcal{C }_+$ or $c\in \mathcal{C }_+$ then $b+c\in \mathcal{C }_+$.

• (b) If $b,c\in \mathcal{C }_{0-}$, then $b+c\in \mathcal{C }_{0-}$. If also $b\in \mathcal{C }_-$ or $c\in \mathcal{C }_-$ then $b+c\in \mathcal{C }_-$.

• (c) $b\in \mathcal{C }_{0+}$ if and only if $-b\in \mathcal{C }_{0-}$.

• (d) $\mathcal{C }_0$ is a subgroup of $\mathcal{A }$.

Proof

• (a) Claim 1(a) says that $b=b_+-b_-$ and $c=c_+-c_-$ for some disjoint $b_+,b_-,c^+,c^-\in \mathcal{B }$. Clearly, $b+c = (b_+ + c_+) - (b_- + c_-)$, and $(b_+ + c_+)$ and $(b_- + c_-)$ are also elements of $\mathcal{B }$. Define $\mathcal{J }_b^+:=\mathcal{I }(\mathbf{w}^{b_+},\mathbf{o})$, $\mathcal{J }_b^-:=\mathcal{I }(\mathbf{w}^{b_-},\mathbf{o})$, $\mathcal{J }_c^+:=\mathcal{I }(\mathbf{w}^{c_+},\mathbf{o})$ and $\mathcal{J }_c^-:=\mathcal{I }(\mathbf{w}^{c_-},\mathbf{o})$. By $\Pi _{\scriptscriptstyle {\mathrm{fin}}}$-invariance, we can assume without loss of generality that:

• $\mathcal{J }_b^+, \mathcal{J }_b^-, \mathcal{J }_c^+$, and $\mathcal{J }_c^-$ are disjoint.

• $\mathcal{I }(\mathbf{w}^{b_+ + c_+},\mathbf{o})=\mathcal{J }_b^+\sqcup \mathcal{J }_c^+$. Furthermore, $\mathbf{w}^{b_+ + c_+}_{\mathcal{J }_b^+}=\mathbf{w}^{b_+}_{\mathcal{J }_b^+}$ and $\mathbf{w}^{b_+ + c_+}_{\mathcal{J }_c^+}=\mathbf{w}^{c_+}_{\mathcal{J }_c^+}$.

• $\mathcal{I }(\mathbf{w}^{b_- + c_-},\mathbf{o})=\mathcal{J }_b^-\sqcup \mathcal{J }_c^-$. Furthermore, $\mathbf{w}^{b_- + c_-}_{\mathcal{J }_b^-}=\mathbf{w}^{b_-}_{\mathcal{J }_b^-}$ and $\mathbf{w}^{b_- + c_-}_{\mathcal{J }_c^-}=\mathbf{w}^{c_-}_{\mathcal{J }_c^-}$. Let $\mathcal{J }:=\mathcal{I }\setminus (\mathcal{J }_b^+\sqcup \mathcal{J }_b^-\sqcup \mathcal{J }_c^+\sqcup \mathcal{J }_c^-)$. Then we can indicate the three assumptions above with the following equations:

  $$\begin{aligned} \begin{array}{rcccccc} \mathbf{w}^{b_+} &{}=&{} (\mathbf{w}^{b_+}_{\mathcal{J }_b^+} &{} \mathbf{o}_{\mathcal{J }_b^-} &{} \mathbf{o}_{\mathcal{J }_c^+} &{} \mathbf{o}_{\mathcal{J }_c^-} &{}\mathbf{o}_\mathcal{J });\\ \mathbf{w}^{b_-} &{}=&{} (\mathbf{o}_{\mathcal{J }_b^+} &{} \mathbf{w}^{b_-}_{\mathcal{J }_b^-} &{} \mathbf{o}_{\mathcal{J }_c^+} &{} \mathbf{o}_{\mathcal{J }_c^-} &{}\mathbf{o}_\mathcal{J });\\ \mathbf{w}^{c_+} &{}=&{} (\mathbf{o}_{\mathcal{J }_b^+} &{} \mathbf{o}_{\mathcal{J }_b^-} &{} \mathbf{w}^{c_+}_{\mathcal{J }_c^+} &{} \mathbf{o}_{\mathcal{J }_c^-} &{}\mathbf{o}_\mathcal{J });\\ \mathbf{w}^{c_-} &{}=&{} (\mathbf{o}_{\mathcal{J }_b^+} &{} \mathbf{o}_{\mathcal{J }_b^-} &{} \mathbf{o}_{\mathcal{J }_c^+} &{} \mathbf{w}^{c_-}_{\mathcal{J }_c^-} &{}\mathbf{o}_\mathcal{J });\\ \mathbf{w}^{b_+ + c_+} &{}=&{} (\mathbf{w}^{b_+}_{\mathcal{J }_b^+} &{}\mathbf{o}_{\mathcal{J }_b^-} &{} \mathbf{w}^{c_+}_{\mathcal{J }_c^+} &{} \mathbf{o}_{\mathcal{J }_c^-} &{} \mathbf{o}_\mathcal{J });\\ \text { and }\mathbf{w}^{b_- + c_-} &{}=&{} (\mathbf{o}_{\mathcal{J }_b^+} &{} \mathbf{w}^{b_-}_{\mathcal{J }_b^-}&{} \mathbf{o}_{\mathcal{J }_c^+} &{} \mathbf{w}^{c_-}_{\mathcal{J }_c^-} &{}\mathbf{o}_\mathcal{J }). \end{array} \end{aligned}$$

  (13)

  Now, $b\in \mathcal{C }_{0+}$, so $\mathbf{w}^{b_+} \succcurlyeq \mathbf{w}^{b_-}$. Applying separability in the $\mathcal{J }_c^+$ coordinates to the first two equations of (13), we get

  $$\begin{aligned} (\mathbf{w}^{b_+}_{\mathcal{J }_b^+}, \mathbf{o}_{\mathcal{J }_b^-}, \mathbf{w}^{c_+}_{\mathcal{J }_c^+}, \mathbf{o}_{\mathcal{J }_c^-},\mathbf{o}_\mathcal{J }) \succcurlyeq (\mathbf{o}_{\mathcal{J }_b^+}, \mathbf{w}^{b_-}_{\mathcal{J }_b^-}, \mathbf{w}^{c_+}_{\mathcal{J }_c^+}, \mathbf{o}_{\mathcal{J }_c^-},\mathbf{o}_\mathcal{J }). \end{aligned}$$

  (14)

  Also, $c\in \mathcal{C }_{0+}$, so $\mathbf{w}^{c_+} \succcurlyeq \mathbf{w}^{c_-}$. Applying separability in the $\mathcal{J }_b^-$ coordinates to the third and fourth equations of (13), we get

  $$\begin{aligned} (\mathbf{o}_{\mathcal{J }_b^+},\mathbf{w}^{b_-}_{\mathcal{J }_b^-}, \mathbf{w}^{c_+}_{\mathcal{J }_c^+}, \mathbf{o}_{\mathcal{J }_c^+-},\mathbf{o}_\mathcal{J }) \succcurlyeq (\mathbf{o}_{\mathcal{J }_b^+}, \mathbf{w}^{b_-}_{\mathcal{J }_b^-}, \mathbf{o}_{\mathcal{J }_c^+}, \mathbf{w}^{c_-}_{\mathcal{J }_c^+-},\mathbf{o}_\mathcal{J }). \end{aligned}$$

  (15)

  Combining Eqs. (14) and (15) via transitivity, we get

  $$\begin{aligned} (\mathbf{w}^{b_+}_{\mathcal{J }_b^+}, \mathbf{o}_{\mathcal{J }_b^-}, \mathbf{w}^{c_+}_{\mathcal{J }_c^+}, \mathbf{o}_{\mathcal{J }_c^-},\mathbf{o}_\mathcal{J }) \succcurlyeq (\mathbf{o}_{\mathcal{J }_b^+}, \mathbf{w}^{b_-}_{\mathcal{J }_b^-}, \mathbf{o}_{\mathcal{J }_c^+}, \mathbf{w}^{c_-}_{\mathcal{J }_c^+-},\mathbf{o}_\mathcal{J }). \end{aligned}$$

  (16)

  Now, matching the two sides of Eq. (16) with the last two equations in (13), we get $\mathbf{w}^{b_+ + c_+} \succcurlyeq \mathbf{w}^{b_- + c_-}$. But $b+c = (b_+ + c_+) - (b_- + c_-)$. Thus, Claim 1(b) implies that $b+c\in \mathcal{C }_{0+}$. If $b\in \mathcal{C }_{+}$, then $\mathbf{w}^{b_+} \succ \mathbf{w}^{b_-}$, which makes Eq. (14) strict, which makes Eq. (16) strict, which means that $\mathbf{w}^{b_+ + c_+} \succ \mathbf{w}^{b_- + c_-}$, and hence $b+c\in \mathcal{C }_{+}$. Likewise, if $c\in \mathcal{C }_{+}$, then Eqs. (15) and (16) become strict, so that $b+c\in \mathcal{C }_{+}$.

  • (b) Similar to (a).

  • (c) Let $b\in \mathcal{C }_{0+}$, and write $b=b_+-b_-$. If $c=-b$, then $c = b_--b_+$, and these elements are disjoint, so the uniqueness part of Claim 1(a) implies that $c_+=b_-$ and $c_-=b_+$. We have $\mathbf{w}^{b_+}\succcurlyeq \mathbf{w}^{b_-}$ (because $b\in \mathcal{C }_{0+}$); thus, $\mathbf{w}^{c_-}\succcurlyeq \mathbf{w}^{c_+}$, so $c\in \mathcal{C }_{0-}$. By identical argument, if $-b\in \mathcal{C }_{0-}$, then $b\in \mathcal{C }_{0+}$.

  • (d) First note that (a) and (b) together imply that $\mathcal{C }_0$ is closed under addition, and (c) implies that $\mathcal{C }_0$ is closed under inverses. Finally, we have $0\in \mathcal{C }_0$, because $0_+=0_-=0$, so that $\mathbf{w}^{0_-}=\mathbf{w}^{0_+}=\mathbf{w}^0=\mathbf{o}$. $\diamondsuit \text { Claim } 2$

Define $\mathcal{R }:=\mathcal{A }/\mathcal{C }_0$; then $\mathcal{R }$ is an abelian group. Let $\phi :\mathcal{A }{{\longrightarrow }}\mathcal{R }$ be the quotient map. Then define $\mathcal{R }_+:= \phi (\mathcal{C }_+)$ and $\mathcal{R }_-:= \phi (\mathcal{C }_-)$.

Claim 3 (a) For all nonzero $r\in \mathcal{R }$, either $r\in \mathcal{R }_+$ or $-r\in \mathcal{R }_+$, but not both.

(b) For all $r,s\in \mathcal{R }_+$, we have $(r+s)\in \mathcal{R }_+$.

Proof

Mapping Claim 2(a) through $\phi$ immediately yields (b).

• To check (a), note that $\mathcal{A }= \mathcal{C }_-\sqcup \mathcal{C }_0 \sqcup \mathcal{C }_+$; thus, $\mathcal{R }= \phi (\mathcal{C }_-)\cup \phi (\mathcal{C }_0) \cup \phi (\mathcal{C }_+) = \mathcal{R }_- \cup \{0\} \cup \mathcal{R }_+$. Thus, any nonzero element of $\mathcal{R }$ is either in $\mathcal{R }_+$ or $\mathcal{R }_-$.

• It remains only to show that $\mathcal{R }_+$ and $\mathcal{R }_-$ are disjoint. By contradiction, suppose that $r\in \mathcal{R }_+\cap \mathcal{R }_-$. Find $b\in \mathcal{C }_+$ and $c\in \mathcal{C }_-$ such that $\phi (b)=\phi (c)=r$. Claim 2(c) implies that $-c\in \mathcal{C }_+$. Thus, Claim 2(a) yields $b-c\in \mathcal{C }_+$. But $\phi (b-c)=\phi (b)-\phi (c)=r-r=0$, so $b-c\in \mathcal{C }_0$. Contradiction. $\diamondsuit \text { Claim } 3$

Now define a binary relation $(>)$ on $\mathcal{R }$ by setting $r> s$ if and only if $r-s\in \mathcal{R }_+$. Claim 3(a) implies that $(>)$ is complete and antisymmetric. Claim 3(b) implies that $(>)$ is transitive. Thus $(>)$ is a total order relation.

Finally, define $u:\mathcal{X }{{\longrightarrow }}\mathcal{R }$ by $u(x) := \phi (1\cdot x)$ (where $1\cdot x$ denotes an element of $\mathcal{A }$). It remains to show that $(\succcurlyeq )$ satisfies statement (2). Let $\mathbf{x},\mathbf{y}\in \mathcal{X }^\mathcal{I }$, with $\hbox {d}(\mathbf{x},\mathbf{y})<{\infty }$. Let $\mathcal{K }:=\mathcal{I }(\mathbf{x},\mathbf{y})$ and $\mathcal{J }:=\mathcal{I }\setminus \mathcal{K }$, and define $\mathbf{x}',\mathbf{y}^{\prime }\in \mathcal{X }^\mathcal{I }$ by setting $\mathbf{x}'_\mathcal{K }:=\mathbf{x}_\mathcal{K }$ and $\mathbf{y}'_\mathcal{K }:=\mathbf{y}_\mathcal{K }$, while $x'_j=y'_j=o$ for all $j\in \mathcal{J }$.

Let $x_1,\ldots ,x_N$ be the distinct elements of $\mathcal{X }\setminus \{o\}$ which occur in $\mathbf{x}'$, and for each $n\in {\left[ 1\ldots N \right]}$, let $J_n\in \mathbb{N }$ be the number of times we see $x_n$. If $a:=J_1 x_1+\cdots J_N x_N\in \mathcal{B }$, then there exists some $\pi \in \Pi _{\scriptscriptstyle {\mathrm{fin}}}$ such that $\pi (\mathbf{x}')=\mathbf{w}^a$; thus $\mathbf{x}^{\prime }\approx \mathbf{w}^a$ by $\Pi _{\scriptscriptstyle {\mathrm{fin}}}$-invariance.

Likewise, let $y_1,\ldots ,y_M$ be the distinct elements of $\mathcal{X }\setminus \{o\}$ which occur in $\mathbf{y}'$, and for each $m\in {\left[ 1\ldots M \right]}$, let $K_m\in \mathbb{N }$ be the number of times we see $y_m$. If $b:=K_1 y_1+\cdots K_M y_M\in \mathcal{B }$, then there exists some $\tau \in \Pi _{\scriptscriptstyle {\mathrm{fin}}}$ such that $\tau (\mathbf{y}')=\mathbf{w}^b$; thus $\mathbf{y}^{\prime }\approx \mathbf{w}^b$ by $\Pi _{\scriptscriptstyle {\mathrm{fin}}}$-invariance. Now we have:

$$\begin{aligned} \left( \mathbf{x}\, {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle }}}\, \mathbf{y} \right)&\mathop {\Leftarrow \!\!\Rightarrow }\limits _{(*)}&\left( \mathbf{x}^{\prime }\, {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle }}}\,\mathbf{y}' \right) \mathop {\Leftarrow \!\!\Rightarrow }\limits _{(\dagger )} \left( \mathbf{w}^a\, {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle }}}\,\mathbf{w}^b \right) \mathop {\Leftarrow \!\!\Rightarrow }\limits _{(\diamond )} \left( a-b\in \mathcal{C }_{0+} \right) \nonumber \\&\mathop {\Leftarrow \!\!\Rightarrow }\limits _{(@)}&\left( \phi (a-b)\in \mathcal{R }_+\sqcup \{0\} \right) \mathop {\Leftarrow \!\!\Rightarrow }\limits _{(\ddagger )} \ \left( \phi (a-b)\ge 0 \right). \end{aligned}$$

(17)

$(*)$ is by separability, because $\mathbf{x},\mathbf{y},\mathbf{x}^{\prime },\mathbf{y}^{\prime }$ satisfy the separability conditions (3) by construction. Next, $(\dagger )$ is because $\mathbf{x}^{\prime }\approx \mathbf{w}^a$ and $\mathbf{y}^{\prime }\approx \mathbf{w}^b$. Finally, $(\diamond )$ is by Claim 1(b) and the definition of $\mathcal{C }_{0+}$, (@) is by definition of $\mathcal{R }_+$, and $(\ddagger )$ is by definition of $(>)$. Now,

$$\begin{aligned} \phi (a-b)&= \phi (J_1 x_1+\cdots J_N x_N) - \phi (K_1 y_1+\cdots K_M y_M)\nonumber \\&\mathop {=}\limits _{(*)} \sum _{n=1}^N J_n\,u(x_n) -\sum _{m=1}^M K_m\,u(y_m)\nonumber \\&= \sum _{i\in \mathcal{I }(\mathbf{x},\mathbf{y})} u(x_i) - \sum _{i\in \mathcal{I }(\mathbf{x},\mathbf{y})} u(y_i) = \sum _{i\in \mathcal{I }} \left( u(x_i)-u(y_i)\right). \end{aligned}$$

(18)

Here $(*)$ is because $\phi (1\cdot x_n)=u(x_n)$ and $\phi (1\cdot y_m)=u(y_m)$. Combining statements (17) and (18) yields (2). Thus, $(\succcurlyeq )$ is the additive preorder on $\mathcal{X }^\mathcal{I }$ defined by $u$.

Universal property. Let $(\mathcal{R }',+,>)$ be another linearly ordered abelian group, and let $u':\mathcal{X }{{\longrightarrow }}\mathcal{R }'$ be some function such that $(\succcurlyeq )$ is also the additive preorder defined by $u'$. Let $r':=u'(o)$, and define $u'':\mathcal{X }{{\longrightarrow }}\mathcal{R }'$ by $u''(x):=u'(x)-r'$. Thus $u''(o)=0$, and $(\succcurlyeq )$ is also the additive preorder defined by $u''$.

Define $\gamma :\mathcal{A }{{\longrightarrow }}\mathcal{R }'$ by setting $\gamma (\sum _{w\in \mathcal{W }} A_w\, w) = \sum _{w\in \mathcal{W }} A_w\,u''(w)$ for any finite $\mathcal{W }\subseteq \mathcal{X }\setminus \{o\}$ and coefficients $\{A_w\}_{w\in \mathcal{W }}\subseteq \mathbb{Z }$. Equivalently, $\gamma (a):=\sum _\mathcal{I }\, \left( u''(\mathbf{w}^{a+})- u''(\mathbf{w}^{a-}) \right)$ for all $a\in \mathcal{A }$. This is automatically a group homomorphism, because $\mathcal{A }$ is the free abelian group generated by $\mathcal{X }\setminus \{o\}$.

Claim 4 $\mathcal{C }_0\subseteq \ker (\gamma )$.

Proof

If $a\in \mathcal{C }_0$, then $\mathbf{w}^{a+}\approx \mathbf{w}^{a-}$. Thus,

$$\begin{aligned} \gamma (a) := \sum _\mathcal{I }\, \left( u''(\mathbf{w}^{a+})- u''(\mathbf{w}^{a-}) \right) \mathop {=}\limits _{(*)} ~ 0. \end{aligned}$$

Thus, $a\in \ker (\gamma )$, as desired. Here $(*)$ is by Eq. (2), because $\mathbf{w}^{a+}\approx \mathbf{w}^{a-}$, and $(\succcurlyeq )$ is the additive preorder defined by $u''$. $\diamondsuit \text { Claim } 4$

Claim 4 means that $\gamma$ factors through $\phi$ to yield a homomorphism $\psi :\mathcal{R }{{\longrightarrow }}\mathcal{R }'$ such that $\psi \circ \phi =\gamma$. (See Fig. 2.)

Claim 5 $u'' = \psi \circ u$.

Fig. 2

A commuting diagram illustrating the proof of the universal property in Theorem 1. Here $i$ represents an inclusion map (regard $\mathcal{X }$ as a subset of $\mathcal{A }$ in the obvious way) and $0$ represents a zero homomorphism

Proof

Let $x\in \mathcal{X }$. Let $a = 1\, x$ (an element of $\mathcal{A }$). Then $u''(x) = \gamma (a) = \psi \circ \phi (a) = \psi (u(x))$. $\diamondsuit \text { Claim } 5$

It remains only to show that $\psi$ is order-preserving. Let $r\in \mathcal{R }$; then $r=\phi (a)$ for some $a\in \mathcal{A }$.

$$\begin{aligned} \left( r\!\ge \! 0 \right) \!\iff \!\left( a\!\in \!\mathcal{C }_{0+} \right)&\iff \left( \mathbf{w}^{a+}\succcurlyeq \mathbf{w}^{a-} \right) \mathop {\Leftarrow \!\!\Rightarrow }\limits _{(*)} \left( \displaystyle \sum _\mathcal{I }\, \left( u''(\mathbf{w}^{a+})\!-\! u''(\mathbf{w}^{a-})\right)\!\ge \! 0 \right)\\&\mathop {\Leftarrow \!\!\Rightarrow }\limits _{(\dagger )} \left( \gamma (a)\ge 0 \right) \mathop {\Leftarrow \!\!\Rightarrow }\limits _{(\diamond )} \left( \psi (r)\ge 0 \right), \end{aligned}$$

as desired. Here, $(*)$ is by Eq. (2), because $(\succcurlyeq )$ is the additive preorder defined by $u''$. Next, $(\dagger )$ is by definition of $\gamma$ and $(\diamond )$ is because $\psi \circ \phi =\gamma$ and $\phi (a)=r$. $\square$

Remark

The proof of the ‘existence’ part of Theorem 1 is somewhat analogous to the proof strategy used by Wakker (1986). Let $\mathcal{Y }:=\mathcal{X }\setminus \{o\}$. Wakker begins with a separable, anonymous preference order $(\succcurlyeq )$ defined on all sequences in $\mathcal{X }^\mathbb{N }$ which have only finitely many nonzero entries. He introduces a set $[\mathcal{Y }]$, which is essentially the free abelian monoid generated by $\mathcal{Y }$ (denoted by $\mathcal{B }'$ in my proof). The order $(\succcurlyeq )$ induces a complete preorder on $[\mathcal{Y }]$, which is compatible with the addition operation. Wakker also assumes $(\succcurlyeq )$ is Archimedean; he then invokes Theorem 3.2.1 of Krantz et al. (1971) to obtain a real-valued, additive utility representation for $(\succcurlyeq )$, equivalent to my Proposition 5(a).

Proof of Proposition 5(a)

Let $(\mathcal{R },+,>)$ be a linearly ordered abelian group, let $u:\mathcal{X }{{\longrightarrow }}\mathcal{R }$, and let $\big ({ {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}}\big )$ be the additive preorder defined by (2). Assume without loss of generality that the image set $u(\mathcal{X })$ generates $\mathcal{R }$ (otherwise replace $\mathcal{R }$ with the subgroup generated by $u(\mathcal{X })$.)

Claim 1 $\big ({ {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}}\big )$ is Archimedean if and only if $\mathcal{R }$ is Archimedean.

Proof

Let $\mathbf{x},\mathbf{y}\in \mathcal{X }^\mathcal{I }$, with $\hbox {d}(\mathbf{x},\mathbf{o})<{\infty }$ and $\hbox {d}(\mathbf{y},\mathbf{o})<{\infty }$. Suppose that

$$\begin{aligned} r = \sum _{i\in \mathcal{I }} \left( u(x_i) - u(o)\right)\,\text { and }\, s = \sum _{i\in \mathcal{I }} \left( u(y_i) - u(o)\right). \end{aligned}$$

(19)

For any $N\in \mathbb{N }$, define $\mathbf{x}^N\in \mathcal{X }^\mathcal{I }$ as in Sect. 3. Then

$$\begin{aligned} \sum _{i\in \mathcal{I }} \left( u(x_i^N) - u(y_i)\right) \!=\! N\cdot \sum _{i\in \mathcal{I }} \left( u(x_i) \!-\! u(o)\right) \!-\! \sum _{i\in \mathcal{I }} \left( u(y_i) \!-\! u(o)\right) \!=\! N\cdot r \!-\! s. \end{aligned}$$

Thus, $\mathbf{x}^N\succ \mathbf{y}$ if and only if $N\cdot r -s > 0$.

‘${\Longleftarrow }$’ If $\mathbf{x}\succ \mathbf{o}$, then $r>0$. Since $\mathcal{R }$ is Archimedean, there exists $N\in \mathbb{N }$ such that $N\cdot r > s$, and thus, $\mathbf{x}^N\succ \mathbf{y}$.

‘$\Longrightarrow$’ For any $r,s\in \mathcal{R }$, we can construct some $\mathbf{x},\mathbf{y}\in \mathcal{X }^\mathcal{I }$ satisfying (19) (because $u(\mathcal{X })$ generates $\mathcal{R }$ by hypothesis). Since $(\succcurlyeq )$ is Archimedean, there exists $N\in \mathbb{N }$ such that $\mathbf{x}^N\succ \mathbf{y}$ and thus, $N\cdot r > s$. $\diamondsuit \text { Claim } 1$

Now combining Theorem 1, Hölder’s Theorem, and Claim 1 yields the result. $\square$

Part (b) of Proposition 5 follows immediately from part (a) and Theorem 2, which will be proved below. The proofs of Theorem 2 and the results from Sect. 4 all depend on the results from Sect. 5, so I will prove those results first.

Proofs from Sect. 5, and the proof of Theorem 2

I will now prove the results of Sect. 5, culminating in the proof of Theorem 2. First, let us define some convenient notation. For any $\mathbf{x}\in \mathcal{X }^\mathcal{I }$ and any function $u:\mathcal{X }{{\longrightarrow }}\mathcal{R }$, let us define ${}^* \sum _{_{\mathcal{I }}}\,u(\mathbf{x}):= \displaystyle {}^*\!\sum _{i\in \mathcal{I }} u(x_i)$. Also, for any $\mathcal{F }\in {\varvec{\mathcal{F }}}$, define $\sum _{\mathcal{F }}\,u(\mathbf{x}):=\displaystyle \sum _{f\in \mathcal{F }} u(x_f)$.

Proof of Lemma 13

Let $0^{\varvec{\mathcal{F }}}$ be the constant $0$ element of $\mathcal{R }^{\varvec{\mathcal{F }}}$, and define $\mathcal{Z }:=\{r\in \mathcal{R }^{\varvec{\mathcal{F }}}$; $r {\stackrel{\displaystyle \approx }{{\scriptscriptstyle \mathfrak{UF }}}} 0^{\varvec{\mathcal{F }}}\}$;

Claim 1 $\mathcal{Z }$ is a subgroup in $\mathcal{R }^{\varvec{\mathcal{F }}}$.

Proof

For any $r\in \mathcal{R }^{\varvec{\mathcal{F }}}$, let ${\varvec{\mathcal{O }}}(r):=\{\mathcal{F }\in {\varvec{\mathcal{F }}}$; $r(\mathcal{F })=0\}$. Then $r\in \mathcal{Z }$ if and only if ${\varvec{\mathcal{O }}}(r)\in \mathfrak{UF }$. But, for any $r,s\in \mathcal{R }^{\varvec{\mathcal{F }}}$, we have ${\varvec{\mathcal{O }}}(r- s) \supseteq {\varvec{\mathcal{O }}}(r)\cap {\varvec{\mathcal{O }}}(s)$; thus, if $r\in \mathcal{Z }$ and $s\in \mathcal{Z }$, then axioms (F1) and (F2) imply $r- s\in \mathcal{Z }$. $\diamondsuit \text { Claim } 1$

For any $r,s\in \mathcal{R }^{\varvec{\mathcal{F }}}$, it is easy to check that $r \, {\stackrel{\displaystyle \approx }{{\scriptscriptstyle \mathfrak{UF }}}}\, s$ if and only if $(r-s)\in \mathcal{Z }$. Thus, ${}^*\!\mathcal{R }$ is just the quotient group $\mathcal{R }^{\varvec{\mathcal{F }}}/\mathcal{Z }$. The relation $(>)$ defines a linear order on ${}^*\!\mathcal{R }$. $\square$

Proof of Lemma 14

Definition (4) says that the function $\mathcal{X }^\mathcal{I }\ni \mathbf{x}\mapsto {}^* \sum _{_{\mathcal{I }}}\,u(\mathbf{x})\in {}^*\!\mathcal{R }$ is a utility function for the preorder $\big ({}^{^*}\!{ {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}}\big )$. The value of ${}^* \sum _{_{\mathcal{I }}}\,u(\mathbf{x})$ is well-defined for all $\mathbf{x}\in \mathcal{X }^\mathcal{I }$, and ${}^*\!\mathcal{R }$ is totally ordered, so $\big ({}^{^*}\!{ {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}}\big )$ is a complete preorder on $\mathcal{X }^\mathcal{I }$.

Separable. Let $\mathcal{J }\subset \mathcal{I }$ and let $\mathcal{K }:=\mathcal{I }\setminus \mathcal{J }$. Suppose that $\mathbf{x},\mathbf{y},\mathbf{x}',\mathbf{y}^{\prime }\in \mathcal{X }^\mathcal{I }$ satisfy the separability conditions (3). I must show that $(\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\mathbf{y})\Leftrightarrow (\mathbf{x}' \,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\mathbf{y}')$.

Define ${\varvec{\mathcal{F }}}(\mathbf{x},\mathbf{y}):=\{\mathcal{F }\in {\varvec{\mathcal{F }}}$; $\sum _{\mathcal{F }}\,u(\mathbf{x})\ge \sum _{\mathcal{F }}\,u(\mathbf{y})\}$. Then $({}^* \sum _{_{\mathcal{I }}}\,u(\mathbf{x}) \ge {}^* \sum _{_{\mathcal{I }}}\,u(\mathbf{y}))\Leftrightarrow ({\varvec{\mathcal{F }}}(\mathbf{x},\mathbf{y})\in \mathfrak{UF })$ and $({}^* \sum _{_{\mathcal{I }}}\,u(\mathbf{x}') \ge {}^* \sum _{_{\mathcal{I }}}\,u(\mathbf{y}'))\Leftrightarrow ({\varvec{\mathcal{F }}}(\mathbf{x}',\mathbf{y}')\in \mathfrak{UF })$. Thus, it suffices to show that $({\varvec{\mathcal{F }}}(\mathbf{x},\mathbf{y})\in \mathfrak{UF })\Leftrightarrow ({\varvec{\mathcal{F }}}(\mathbf{x}',\mathbf{y}')\in \mathfrak{UF })$. In fact, I will show that ${\varvec{\mathcal{F }}}(\mathbf{x}',\mathbf{y}')= {\varvec{\mathcal{F }}}(\mathbf{x},\mathbf{y})$.

For any $\mathcal{F }\in {\varvec{\mathcal{F }}}$, the Eqs. (3) imply:

$$\begin{aligned} \begin{array}{rclcrcl} \mathrm{(a)} \quad \sum \nolimits _{\mathcal{F }\cap \mathcal{J }}\,u(\mathbf{x})&{}=&{}\sum \nolimits _{\mathcal{F }\cap \mathcal{J }}\,u(\mathbf{y}), &{}\quad &{} \mathrm{(b)} \quad \sum \nolimits _{\mathcal{F }\cap \mathcal{K }}\,u(\mathbf{x})&{}=&{}\sum \nolimits _{\mathcal{F }\cap \mathcal{K }}\,u(\mathbf{x}'),\\ \mathrm{(c)} \quad \sum \nolimits _{\mathcal{F }\cap \mathcal{J }}\,u(\mathbf{x}')&{}=&{}\sum \nolimits _{\mathcal{F }\cap \mathcal{J }}\,u(\mathbf{y}'), &{}\text { and }&{} \mathrm{(d)} \quad \sum \nolimits _{\mathcal{F }\cap \mathcal{K }}\,u(\mathbf{y})&{}=&{}\sum \nolimits _{\mathcal{F }\cap \mathcal{K }}\,u(\mathbf{y}'). \end{array} \end{aligned}$$

Furthermore, $\mathcal{F }=(\mathcal{F }\cap \mathcal{J })\sqcup (\mathcal{F }\cap \mathcal{K })$ (because $\mathcal{I }:=\mathcal{J }\sqcup \mathcal{K }$); thus

$$\begin{aligned} \begin{array}{rclrcl} \mathrm{(e)}\quad \sum \nolimits _{\mathcal{F }}\,u(\mathbf{x})&{}=&{}\sum \nolimits _{\mathcal{F }\cap \mathcal{J }}\,u(\mathbf{x})+\sum \nolimits _{\mathcal{F }\cap \mathcal{K }}\,u(\mathbf{x}), &{} \sum \nolimits _{\mathcal{F }}\,u(\mathbf{y})&{}=&{}\sum \nolimits _{\mathcal{F }\cap \mathcal{J }}\,u(\mathbf{y})\\ +\sum \nolimits _{\mathcal{F }\cap \mathcal{K }}\,u(\mathbf{y});\\ \mathrm{(f)}\quad \sum \nolimits _{\mathcal{F }}\,u(\mathbf{x}')&{}=&{}\sum \nolimits _{\mathcal{F }\cap \mathcal{J }}\,u(\mathbf{x}')+\sum \nolimits _{\mathcal{F }\cap \mathcal{K }}\,u(\mathbf{x}'), &{} \sum \nolimits _{\mathcal{F }}\,u(\mathbf{y}')&{}=&{}\sum \nolimits _{\mathcal{F }\cap \mathcal{J }}\,u(\mathbf{y}')\\ +\sum \nolimits _{\mathcal{F }\cap \mathcal{K }}\,u(\mathbf{y}'). \end{array} \end{aligned}$$

Thus, for all $\mathcal{F }\in {\varvec{\mathcal{F }}}$, we have:

$$\begin{aligned} \left( \mathcal{F }\in {\varvec{\mathcal{F }}}(\mathbf{x},\mathbf{y}) \right)&\iff \left( \sum \nolimits _{\mathcal{F }}\,u(\mathbf{x})\ge \sum \nolimits _{\mathcal{F }}\,u(\mathbf{y}) \right) \mathop {\Leftarrow \!\!\Rightarrow }\limits _{(*)} \left( \sum \nolimits _{\mathcal{F }\cap \mathcal{K }}\,u(\mathbf{x})\ge \sum \nolimits _{\mathcal{F }\cap \mathcal{K }}\,u(\mathbf{y}) \right)\\&\mathop {\Leftarrow \!\!\Rightarrow }\limits _{(\dagger )} \left( \sum \nolimits _{\mathcal{F }\cap \mathcal{K }}\,u(\mathbf{x}')\!\ge \! \sum \nolimits _{\mathcal{F }\cap \mathcal{K }}\,u(\mathbf{y}') \right) \mathop {\Leftarrow \!\!\Rightarrow }\limits _{(\diamond )} \left( \sum \nolimits _{\mathcal{F }}\,u(\mathbf{x}')\!\ge \! \sum \nolimits _{\mathcal{F }}\,u(\mathbf{y}') \right)\\&\iff \left( \mathcal{F }\in {\varvec{\mathcal{F }}}(\mathbf{x}',\mathbf{y}') \right). \end{aligned}$$

Thus, ${\varvec{\mathcal{F }}}(\mathbf{x}',\mathbf{y}')= {\varvec{\mathcal{F }}}(\mathbf{x},\mathbf{y})$. Here, $(*)$ is by Eqs. (a) and (e); $(\dagger )$ is by Eqs. (b) and (d); and $(\diamond )$ is by Eqs. (c) and (f). $\square$

Proof of Lemma 15

• (a) If $\epsilon \in \Pi$ is the identity permutation, then clearly ${\varvec{\mathcal{F }}}(\epsilon )={\varvec{\mathcal{F }}}\in \mathfrak{UF }$. Also, for any $\pi \in \Pi$, it is clear that ${\varvec{\mathcal{F }}}(\pi ^{-1})={\varvec{\mathcal{F }}}(\pi )$, so $\pi \in \Pi _\mathfrak{UF }$ if and only if $\pi ^{-1}\in \Pi _\mathfrak{UF }$. Finally, let $\pi _1,\pi _2\in \Pi _\mathfrak{UF }$. Then ${\varvec{\mathcal{F }}}(\pi _1\circ \pi _2) \supseteq {\varvec{\mathcal{F }}}(\pi _1)\cap {\varvec{\mathcal{F }}}(\pi _2)$. But ${\varvec{\mathcal{F }}}(\pi _1)\cap {\varvec{\mathcal{F }}}(\pi _2)\in \mathfrak{UF }$ by Axiom (F1); thus, ${\varvec{\mathcal{F }}}(\pi _1\circ \pi _2)\in \mathfrak{UF }$ by (F2); thus, $\pi _1\circ \pi _2\in \Pi _\mathfrak{UF }$.

• (b) Let $\mathbf{x}\in \mathcal{X }^\mathcal{I }$ and $\pi \in \Pi _\mathfrak{UF }$. Then for every $\mathcal{F }\in {\varvec{\mathcal{F }}}$, we have $\sum _{\mathcal{F }}\,u(\mathbf{x}) = \sum _{\pi (\mathcal{F })}\,u[\pi (\mathbf{x})]$. But if $\mathcal{F }\in {\varvec{\mathcal{F }}}(\pi )$, then $\pi (\mathcal{F })=\mathcal{F }$, so we get $\sum _{\mathcal{F }}\,u(\mathbf{x})=\sum _{\mathcal{F }}\,u[\pi (\mathbf{x})]$. If ${\varvec{\mathcal{F }}}(\pi )\in \mathfrak{UF }$, then this implies that ${}^* \sum _{_{\mathcal{I }}}\,u(\mathbf{x})={}^* \sum _{_{\mathcal{I }}}\,u[\pi (\mathbf{x})]$; thus, $\mathbf{x}\,{}^{^*}\!\!\! {\stackrel{\displaystyle \approx }{{\scriptscriptstyle u}}}\pi (\mathbf{x})$. $\square$

Proof of Lemma 17

• (a) Recall that $\mathfrak{F }_\Gamma :=\{{\varvec{\mathcal{E }}}\subseteq {\varvec{\mathcal{F }}}$; ${\varvec{\mathcal{F }}}_\Delta (\mathcal{J })\subseteq {\varvec{\mathcal{E }}}$ for some finite $\mathcal{J }\subset \mathcal{I }$ and finite $\Delta \subseteq \Gamma \}$.

Claim 1 $\mathfrak{F }$ is a free filter.

Proof

We must check axioms (F0)–(F2).

• (F0) For any finite subset $\Delta \subseteq \Gamma$, let ${\varvec{\mathcal{O }}}_\Delta$ be the orbit partition generated by ${\left\langle \Delta \right\rangle }$. Then ${\varvec{\mathcal{O }}}_\Delta$ has an infinite number of elements, because $\mathcal{I }$ is infinite, whereas each element of ${\varvec{\mathcal{O }}}_\Delta$ is a finite subset (because $\Gamma$ has locally finite orbits). For any finite subset $\mathcal{J }\subseteq \mathcal{I }$ let ${\varvec{\mathcal{O }}}_\Delta (\mathcal{J }):=\{\mathcal{O }$; $\mathcal{O }\in {\varvec{\mathcal{O }}}_\Delta$ and $\mathcal{J }\cap \mathcal{O }\ne \emptyset \}$; then ${\varvec{\mathcal{O }}}_\Delta (\mathcal{J })$ is finite, and ${\varvec{\mathcal{F }}}_\Delta (\mathcal{J }):=\{\bigsqcup _{\mathcal{P }\in {\varvec{\mathcal{P }}}} \mathcal{P }$; ${\varvec{\mathcal{P }}}\subseteq {\varvec{\mathcal{O }}}_\Delta$ any finite subset such that ${\varvec{\mathcal{O }}}_\Delta (\mathcal{J })\subseteq {\varvec{\mathcal{P }}}\}$. Thus ${\varvec{\mathcal{F }}}_\Delta (\mathcal{J })$ is infinite, because ${\varvec{\mathcal{O }}}_\Delta \setminus {\varvec{\mathcal{O }}}_\Delta (\mathcal{J })$ is infinite. Thus, if ${\varvec{\mathcal{E }}}\subseteq {\varvec{\mathcal{F }}}$ is finite, then we cannot have ${\varvec{\mathcal{F }}}_\Delta (\mathcal{J })\subseteq {\varvec{\mathcal{E }}}$ for any finite subsets $\mathcal{J }\subseteq \mathcal{I }$ and $\Delta \subseteq \Gamma$; thus, ${\varvec{\mathcal{E }}}\not \in \mathfrak{F }$.

• (F1) Let ${\varvec{\mathcal{D }}},{\varvec{\mathcal{E }}}\in \mathfrak{F }$. Then there exist finite subsets $\mathcal{J },\mathcal{K }\subset \mathcal{I }$ and $\Delta ,\mathsf{ E }\subseteq \Gamma$ such that ${\varvec{\mathcal{F }}}_\Delta (\mathcal{J })\subseteq {\varvec{\mathcal{D }}}$ and ${\varvec{\mathcal{F }}}_\mathsf{ E }(\mathcal{K })\subseteq {\varvec{\mathcal{E }}}$. Thus, $\mathcal{J }\cup \mathcal{K }$ and $\Delta \cup \mathsf{ E }$ are also finite, and we have ${\varvec{\mathcal{F }}}_{\Delta \cup \mathsf{ E }}(\mathcal{J }\cup \mathcal{K })={\varvec{\mathcal{F }}}_\Delta (\mathcal{J })\cap {\varvec{\mathcal{F }}}_\mathsf{ E }(\mathcal{K }) \subseteq {\varvec{\mathcal{D }}}\cap {\varvec{\mathcal{E }}}$. Thus, ${\varvec{\mathcal{D }}}\cap {\varvec{\mathcal{E }}}\in \mathfrak{F }$ also.

• (F2) Let ${\varvec{\mathcal{D }}}\in \mathfrak{F }$; then ${\varvec{\mathcal{F }}}_\Delta (\mathcal{J })\subseteq {\varvec{\mathcal{D }}}$ for some finite subsets $\mathcal{J }\subset \mathcal{I }$ and $\Delta \subseteq \Gamma$. Thus, for any ${\varvec{\mathcal{E }}}\subseteq {\varvec{\mathcal{F }}}$, if ${\varvec{\mathcal{D }}}\subseteq {\varvec{\mathcal{E }}}$, then ${\varvec{\mathcal{F }}}_\Delta (\mathcal{J })\subseteq {\varvec{\mathcal{E }}}$ also, so ${\varvec{\mathcal{E }}}\in \mathfrak{F }$. $\diamondsuit \text { Claim } 1$

Now apply the Ultrafilter Lemma to obtain some ultrafilter $\mathfrak{UF }$ which contains $\mathfrak{F }_\Gamma$.

• (b) Let $\pi \in \Pi _{\scriptscriptstyle {\mathrm{fin}}}$; then $\mathcal{I }(\pi )$ is finite. Thus, for any finite subset $\Delta \subseteq \Gamma$, we have ${\varvec{\mathcal{F }}}_\Delta [\mathcal{I }(\pi )]\in \mathfrak{F }_\Gamma \subseteq \mathfrak{UF }$. But ${\varvec{\mathcal{F }}}_\Delta [\mathcal{I }(\pi )]\subseteq {\varvec{\mathcal{F }}}(\pi )$; thus, axiom (F2) implies ${\varvec{\mathcal{F }}}(\pi )\in \mathfrak{UF }$, so $\pi \in \Pi _\mathfrak{UF }$.

Now let $\gamma \in \Gamma$, and fix a finite $\mathcal{J }\subseteq \mathcal{I }$. If $\mathcal{F }\in {\varvec{\mathcal{F }}}_{\{\gamma \}}(\mathcal{J })$, then $\gamma (\mathcal{F })=\mathcal{F }$, so $\mathcal{F }\in {\varvec{\mathcal{F }}}(\gamma )$. Thus, ${\varvec{\mathcal{F }}}_{\{\gamma \}}(\mathcal{J }) \subseteq {\varvec{\mathcal{F }}}(\gamma )$, so ${\varvec{\mathcal{F }}}(\gamma )\in \mathfrak{F }_\Gamma$, and thus ${\varvec{\mathcal{F }}}(\gamma )\in \mathfrak{UF }$. Thus, $\gamma \in \Pi _\mathfrak{UF }$. $\square$

Proof of Lemma 18

If $\mathbf{x},\mathbf{y}\in \mathcal{X }^\mathcal{I }$ and $\hbox {d}(\mathbf{x},\mathbf{y})<{\infty }$, then the set $\mathcal{I }(\mathbf{x},\mathbf{y})$ is finite. Recall that ${\varvec{\mathcal{F }}}[\mathcal{I }(\mathbf{x},\mathbf{y})]:=\{\mathcal{J }\in {\varvec{\mathcal{F }}}$; $\mathcal{I }(\mathbf{x},\mathbf{y})\subseteq \mathcal{J }\}$. For any $\mathcal{J }\in {\varvec{\mathcal{F }}}[\mathcal{I }(\mathbf{x},\mathbf{y})]$, we have:

$$\begin{aligned} \sum _{\mathcal{J }}\,u(\mathbf{x}) - \sum _{\mathcal{J }}\,u(\mathbf{y}) = \sum _{j\in \mathcal{J }} u(x_j)- \sum _{j\in \mathcal{J }} u(y_j) = \sum _{i\in \mathcal{I }(\mathbf{x},\mathbf{y})} u(x_i) - \sum _{i\in \mathcal{I }(\mathbf{x},\mathbf{y})} u(y_i), \end{aligned}$$

because $x_j=y_j$ for all $j\in \mathcal{J }\setminus \mathcal{I }(\mathbf{x},\mathbf{y})$.

If $\mathbf{x}\, {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\,\mathbf{y}$, then $\displaystyle \sum _{i\in \mathcal{I }(\mathbf{x},\mathbf{y})} u(x_i)\ge \sum _{i\in \mathcal{I }(\mathbf{x},\mathbf{y})} u(y_i)$; thus, $\sum _{\mathcal{J }}\,u(\mathbf{x})\ge \sum _{\mathcal{J }}\,u(\mathbf{y})$ for all $\mathcal{J }\in {\varvec{\mathcal{F }}}_\Delta [\mathcal{I }(\mathbf{x},\mathbf{y})]$. But ${\varvec{\mathcal{F }}}_\Delta [\mathcal{I }(\mathbf{x},\mathbf{y})]\in \mathfrak{UF }$ because $\mathfrak{UF }$ is a suitable ultrafilter. Thus, ${}^* \sum _{_{\mathcal{I }}}\,u(\mathbf{x})\ge {}^* \sum _{_{\mathcal{I }}}\,u(\mathbf{y})$. Thus $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\,\mathbf{y}$.

Likewise, if $\mathbf{x}\, {\stackrel{\displaystyle \succ }{{\scriptscriptstyle u}}}\,\mathbf{y}$, then $\displaystyle \sum _{i\in \mathcal{I }(\mathbf{x},\mathbf{y})} u(x_i) > \sum _{i\in \mathcal{I }(\mathbf{x},\mathbf{y})} u(y_i)$; thus, $\sum _{\mathcal{J }}\,u(\mathbf{x})> \sum _{\mathcal{J }}\,u(\mathbf{y})$ for all $\mathcal{J }\in {\varvec{\mathcal{F }}}[\mathcal{I }(\mathbf{x},\mathbf{y})]\in \mathfrak{UF }$; thus, ${}^* \sum _{_{\mathcal{I }}}\,u(\mathbf{x}) > {}^* \sum _{_{\mathcal{I }}}\,u(\mathbf{y})$. Thus $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succ }{{\scriptscriptstyle u}}}\,\mathbf{y}$. $\square$

Proof of Lemma 19

Claim 1 Let ${\varvec{\mathcal{G }}}\in \mathfrak{UF }$.

• (a) $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u,{\varvec{\mathcal{G }}}}}}\, \mathbf{y}$ if and only if $\sum _{\mathcal{J }}\,u(\mathbf{x})\ge \sum _{\mathcal{J }}\,u(\mathbf{y})$ for all $\mathcal{J }\in {\varvec{\mathcal{G }}}$.

• (b) $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succ }{{\scriptscriptstyle u,{\varvec{\mathcal{G }}}}}}\, \mathbf{y}$ if and only if $\sum _{\mathcal{J }}\,u(\mathbf{x})> \sum _{\mathcal{J }}\,u(\mathbf{y})$ for all $\mathcal{J }\in {\varvec{\mathcal{G }}}$.

Proof

Fix $\mathcal{J }\in {\varvec{\mathcal{G }}}$ and $\mathbf{z}\in \mathcal{X }^\mathcal{I }$, and let $\mathbf{x}':=\mathbf{x}_{\mathcal{J }}\mathbf{z}_{{\mathcal{I }\setminus \mathcal{J }}}$ and $\mathbf{y}':=\mathbf{y}_{\mathcal{J }}\mathbf{z}_{{\mathcal{I }\setminus \mathcal{J }}}$. Then $\mathcal{I }(\mathbf{x}',\mathbf{y}')\subseteq \mathcal{J }$, so $\hbox {d}(\mathbf{x}',\mathbf{y}')<{\infty }$ (because $\mathcal{J }$ is finite). Thus, the $\big ({ {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}}\big )$-order of $\mathbf{x}'$ and $\mathbf{y}'$ is well-defined, and Lemma 18 says that

$$\begin{aligned}&\qquad \big ( \mathbf{x}^{\prime }\, {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\,\mathbf{y}' \big )\ \iff \ \big ( \mathbf{x}^{\prime }\,\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\,\mathbf{y}' \big ) \,\,\text { and }\left( \mathbf{x}^{\prime }\, {\stackrel{\displaystyle \succ }{{\scriptscriptstyle u}}}\,\mathbf{y}' \right) \ \iff \ \left( \mathbf{x}^{\prime }\,\,{}^{^*}\! {\stackrel{\displaystyle \succ }{{\scriptscriptstyle u}}}\,\mathbf{y}' \right).\end{aligned}$$

(20)

$$\begin{aligned}&\text {Furthermore,}\quad \sum _{i\in \mathcal{I }}\left( u(x'_i)-u(y'_i)\right) = \sum _{\mathcal{J }} u(\mathbf{x})- \sum _{\mathcal{J }} u(\mathbf{y}). \end{aligned}$$

(21)

• (a) ‘$\Longrightarrow$’ Suppose that $\mathbf{x}\,{}^{^*}\!\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u,{\varvec{\mathcal{G }}}}}}\, \mathbf{y}$. Then for any $\mathcal{J }\in \mathcal{G }$ and $\mathbf{z}\in \mathcal{X }^\mathcal{I }$, if $\mathbf{x}'$ and $\mathbf{y}'$ are defined as above, then $\mathbf{x}^{\prime }\,\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\,\mathbf{y}'$. Thus, statement (20) says that $\mathbf{x}^{\prime }\, {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\,\mathbf{y}'$, so $\sum _{i\in \mathcal{I }}\left( u(x'_i)-u(y'_i)\right)\ge 0$. But then statement (21) implies that $\sum _{\mathcal{J }}\,u(\mathbf{x})\ge \sum _{\mathcal{J }}\,u(\mathbf{y})$. ‘${\Longleftarrow }$’ Fix $\mathbf{z}\in \mathcal{X }^\mathcal{I }$ and $\mathcal{J }\in {\varvec{\mathcal{G }}}$. If $\mathbf{x}'$ and $\mathbf{y}'$ are defined as above, then statement (21) implies that $\displaystyle \sum _{i\in \mathcal{I }}\left( u(x'_i)-u(y'_i)\right)\ge 0$, so $\mathbf{x}^{\prime }\, {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\,\mathbf{y}'$, so statement (20) says $\mathbf{x}^{\prime }\,\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\,\mathbf{y}'$. This holds for all $\mathcal{J }\in {\varvec{\mathcal{G }}}$ and $\mathbf{z}\in \mathcal{X }^\mathcal{I }$; thus, $\mathbf{x}\,{}^{^*}\!\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u,{\varvec{\mathcal{G }}}}}}\, \mathbf{y}$.

• (b) The proof is similar to (a); change ‘$\ge$’ to ‘$>$’ and ‘$\succcurlyeq$’ to ‘$\succ$’ everywhere. $\diamondsuit$ Claim 1

• (a) ‘$\Longrightarrow$’ Suppose that $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\, \mathbf{y}$. Thus, if ${\varvec{\mathcal{G }}}:=\{\mathcal{F }\in {\varvec{\mathcal{F }}}$; $\sum _{\mathcal{F }}\,u(\mathbf{x})\ge \sum _{\mathcal{F }}\,u(\mathbf{y})\}$, then ${\varvec{\mathcal{G }}}\in \mathfrak{UF }$. By definition, $\sum _{\mathcal{J }}\,u(\mathbf{x})\ge \sum _{\mathcal{J }}\,u(\mathbf{y})$ for all $\mathcal{J }\in {\varvec{\mathcal{G }}}$. Thus, Claim 1(a) says that $\mathbf{x}\,{}^{^*}\!\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u,{\varvec{\mathcal{G }}}}}}\, \mathbf{y}$. ‘${\Longleftarrow }$’ If $\mathbf{x}\,{}^{^*}\!\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u,{\varvec{\mathcal{G }}}}}}\, \mathbf{y}$; then Claim 1(a) says that $\sum _{\mathcal{J }}\,u(\mathbf{x})\ge \sum _{\mathcal{J }}\,u(\mathbf{y})$ for all $\mathcal{J }\in {\varvec{\mathcal{G }}}$. But ${\varvec{\mathcal{G }}}\in \mathfrak{UF }$, so this means that ${}^*\!\sum _{\mathcal{I }}\,u(\mathbf{x})\ge {}^*\!\sum _{\mathcal{I }}\,u(\mathbf{y})$, which means $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\,\mathbf{y}$.

• (b) The proof is the same as (a), but using Claim 1(b) instead of Claim 1(a). $\square$

Proof of Lemma 20

‘${\Longleftarrow }$’ follows from Lemmas 18 and 19.

‘$\Longrightarrow$’ Let $\mathbf{x},\mathbf{y}\in \mathcal{X }^\mathcal{I }$, and suppose that $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\, \mathbf{y}$. Then Lemma 19(a) yields some ${\varvec{\mathcal{G }}}\in \mathfrak{UF }$ such that, for all $\mathbf{z}\in \mathcal{X }^\mathcal{I }$ and $\mathcal{J }\in {\varvec{\mathcal{G }}}$, we have $\mathbf{x}_{\mathcal{J }}\mathbf{z}_{{\mathcal{I }\setminus \mathcal{J }}}\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\mathbf{y}_{\mathcal{J }}\mathbf{z}_{{\mathcal{I }\setminus \mathcal{J }}}$, and thus, $\mathbf{x}_{\mathcal{J }}\mathbf{z}_{{\mathcal{I }\setminus \mathcal{J }}} {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}} \mathbf{y}_{\mathcal{J }}\mathbf{z}_{{\mathcal{I }\setminus \mathcal{J }}}$, by Lemma 18. But then $\mathbf{x}_{\mathcal{J }}\mathbf{z}_{{\mathcal{I }\setminus \mathcal{J }}} {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle \mathrm fin}}}\mathbf{y}_{\mathcal{J }}\mathbf{z}_{{\mathcal{I }\setminus \mathcal{J }}}$, because $\big ( {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle \mathrm fin}}}\big )=\big ({ {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}}\big )$ by hypothesis.

This holds for all $\mathbf{z}\in \mathcal{X }^\mathcal{I }$ and $\mathcal{J }\in {\varvec{\mathcal{G }}}$; thus, (C1) forces $\mathbf{x}\succcurlyeq \, \mathbf{y}$, because $(\succcurlyeq )$ is $\mathfrak{UF }$-continuous. Thus, we have $(\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\mathbf{y})\Longrightarrow (\mathbf{x}\succcurlyeq \mathbf{y})$.

Likewise, if $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succ }{{\scriptscriptstyle u}}}\, \mathbf{y}$, then Lemma 19(b) and (C2) imply that $\mathbf{x}\succ \, \mathbf{y}$.

Lemma 14 says $\big ({}^{^*}\!{ {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}}\big )$ is a complete preorder on $\mathcal{X }^\mathcal{I }$. It follows that $(\succcurlyeq )=\big ({}^{^*}\!{ {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}}\big )$. $\square$

Proof of Theorem 2

‘${\Longleftarrow }$’ follows from Lemmas 14, 15(b), 17(b) and 19.

‘$\Longrightarrow$’ Let $\big ( {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle \mathrm fin}}}\big )$ be the finitary part of $(\succcurlyeq )$. Then $\big ( {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle \mathrm fin}}}\big )$ is $\Pi _{\scriptscriptstyle {\mathrm{fin}}}$-invariant and separable, so Theorem 1 yields a linearly ordered abelian group $(\mathcal{R },+,>)$ and $u:\mathcal{X }{{\longrightarrow }}\mathcal{R }$ such that $\big ( {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle \mathrm fin}}}\big )=\big ({ {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}}\big )$. Then Lemma 20 implies that $(\succcurlyeq )=\big ({}^{^*}\!{ {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}}\big )$.

The existence of an $\mathcal{R }$ and $u$ with the universal property follows from the construction in Theorem 1. $\square$

Proofs from Sects. 4 and 7.3

Proposition 6 is actually a special case of a more general result, which elucidates the behaviour of the hyperadditive preorder $\big ({}^{^*}\!{ {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}}\big )$ when $\mathfrak{UF }$ is constructed as in Lemma 15.

Lemma 22

Let $\Pi$ be a permutation group with locally finite orbits on $\mathcal{I }$, and let $\mathfrak{UF }$ be as in Lemma 15. Let $\mathcal{R }$ be a linearly ordered abelian group, let $u:\mathcal{X }{{\longrightarrow }}\mathcal{R }$, and define ${}^*\!\mathcal{R }$ and $\big ({}^{^*}\!{ {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}}\big )$ using $\mathfrak{UF }$. Let $\mathbf{x},\mathbf{y}\in \mathcal{X }^\mathcal{I }$. Suppose that there is some finite $\mathcal{J }\subset \mathcal{I }$ and some finite $\Delta \subset \Pi$ such that, for all finite $\mathcal{K }\subseteq \mathcal{I }$ we have:

$$\begin{aligned} \left( \mathcal{J }\subseteq \mathcal{K } \,and\, \delta (\mathcal{K })=\mathcal{K }\, for \,all \,\delta \in \Delta \right) \Longrightarrow \left( \displaystyle \sum _{k\in \mathcal{K }} u(x_k) \ge \sum _{k\in \mathcal{K }} u(y_k) \right).\qquad \quad \end{aligned}$$

(22)

Then $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\, \mathbf{y}$. If the inequalities on the right side of statement (22) are strict, then $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succ }{{\scriptscriptstyle u}}}\, \mathbf{y}$

Proof

By hypothesis, $\sum _{k\in \mathcal{K }} u(x_k) \ge \sum _{k\in \mathcal{K }} u(y_k)$ for all $\mathcal{K }\in {\varvec{\mathcal{F }}}_\Delta (\mathcal{J })$. But ${\varvec{\mathcal{F }}}_\Delta (\mathcal{J })\in \mathfrak{F }_\Gamma$ by the definition of $\mathfrak{F }_\Gamma$,  and  $\mathfrak{F }_\Gamma \subseteq \mathfrak{UF }$ by Lemma 15. Thus, $\displaystyle {}^*\!\sum _{i\in \mathcal{I }} u(x_i) \ge {}^*\!\sum _{i\in \mathcal{I }} u(y_i)$. Thus, $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\, \mathbf{y}$. $\square$

Proof of Proposition 6

Let $\Pi :=\Pi _{\scriptscriptstyle {\mathrm{fs}}}$ be the group of fixed-step permutations from Example 16(c), and then define $\mathfrak{UF }$ as in Lemma 15. Let $\pi :\mathbb{N }{{\longrightarrow }}\mathbb{N }$ be a permutation which cyclically permutes the elements of the interval $[m S+1\, \ldots \, (m+1)S]$, for every $m\in \mathbb{N }$. Then $\pi$ is a fixed-step permutation with $T_\pi =S$. Note that the only $\pi$-invariant subsets of $\mathbb{N }$ are intervals of the form $[n S\!+\!1 \,\ldots \, m S]$ for some $n\le m\in \mathbb{N }$. Let $\mathcal{J }:=[1\ldots \,M_0 S]$. Then for all finite $\mathcal{K }\subseteq \mathbb{N }$ we have:

$$\begin{aligned} \left( \mathcal{J }\subseteq \mathcal{K } \text { and } \pi (\mathcal{K })=\mathcal{K } \right)&\Longrightarrow&\left( \mathcal{K }=[1\ldots \, M S], \,for \,some \,M\ge M_0 \right)\\&\mathop {\Longrightarrow }\limits _{(*)} \left( \displaystyle \sum _{k\in \mathcal{K }} u(x_k) > \sum _{k\in \mathcal{K }} u(y_k) \right), \end{aligned}$$

where $(*)$ is by hypothesis. Thus, setting $\Delta :=\{\pi \}$ in Lemma 22, we deduce that $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succ }{{\scriptscriptstyle u}}}\, \mathbf{y}$. $\square$

Proof of Corollary 7

If $\mathcal{R }\subseteq \mathbb{R }$, then $\mathcal{R }$ is Archimedean. Thus, if condition (b) holds, then there is some $M_0\in \mathbb{N }$ such that

$$\begin{aligned} \sum _{t=1}^{t=M\,P} u(x_t) \ > \ \sum _{t=1}^{t=M\,P} u(y_t), \quad \text{ for } \text{ all } M\ge M_0\text{. } \end{aligned}$$

(23)

On the other hand, if condition (a) holds, then, set $M_0:=\lceil P/T\rceil$. Then statement (23) is true. Now set $S:=P$ in Proposition 6 to get $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succ }{{\scriptscriptstyle u}}}\, \mathbf{y}$. $\square$

Proof of Proposition 8

Recall that $\mathcal{R }:={}^*\!\mathbb{R }$, and the expected utilities of mixed strategy pairs take values in ${}^*\!\mathcal{R }$.⁴² If $\rho ^*$ is a uniformly distributed mixed strategy for Row, then there is some $r\in \mathcal{R }$ such that $EU_{\scriptscriptstyle \mathrm{col}}(\rho ^*,c)=r$ for all $c\in \mathbb{Z }$. Now define $R:={}^*\!\sum _{i\in \mathcal{I }} r$ (an element of ${}^*\!\mathcal{R }$). Then for any mixed strategy $\kappa :\mathcal{I }{{\longrightarrow }}\mathbb{Z }$ for Column, we have

$$\begin{aligned} EU_{\scriptscriptstyle \mathrm{col}}(\rho ^*,\kappa ) = {}^*\!\sum _{i\in \mathcal{I }} EU_{\scriptscriptstyle \mathrm{col}}\left(\rho , \kappa (i)\right) = {}^*\!\sum _{i\in \mathcal{I }} r = R. \end{aligned}$$

In particular, $EU_{\scriptscriptstyle \mathrm{col}}(\rho ^*,\kappa ^*)=R$. Thus, $EU_{\scriptscriptstyle \mathrm{col}}(\rho ^*,\kappa ^*)\ge EU_{\scriptscriptstyle \mathrm{col}}(\rho ^*,\kappa )$ for all mixed strategies $\kappa$, so $\kappa ^*$ is a best response to $\rho ^*$.

By an exactly symmetric argument, if $\kappa ^*$ is a uniformly distributed mixed strategy for Column, then there is some (negative) $R^{\prime }\in {}^*\!\mathcal{R }$ such that $EU_{\scriptscriptstyle \mathrm{col}}(\rho ,\kappa ^*)\!=\!R'$ for any mixed strategy $\rho$. In particular, $EU_{\scriptscriptstyle \mathrm{col}}(\rho ^*,\kappa ^*)\!=\!R'$, so $\rho ^*$ is a best response to $\kappa ^*$.

Proof of Example 9(b)

This construction is well-known, but I include it for completeness. Recall that ${\varvec{\mathcal{P }}}(\mathcal{I })$ is the set of all subsets of $\mathcal{I }$. A free ultrafilter on $\mathcal{I }$ is a subset $\varvec{\mathcal{U }\!\mathcal{F }}\subset {\varvec{\mathcal{P }}}(\mathcal{I })$ such that:

• (a) No finite subset of $\mathcal{I }$ is in $\varvec{\mathcal{U }\!\mathcal{F }}$.

• (b) If $\mathcal{J },\mathcal{K }\in \varvec{\mathcal{U }\!\mathcal{F }}$, then $\mathcal{J }\cap \mathcal{K }\in \varvec{\mathcal{U }\!\mathcal{F }}$.

• (c) For any $\mathcal{U }\in \varvec{\mathcal{U }\!\mathcal{F }}$ and $\mathcal{J }\subseteq \mathcal{I }$, if $\mathcal{U }\subset \mathcal{J }$, then $\mathcal{J }\in \varvec{\mathcal{U }\!\mathcal{F }}$ also.

• (d) For any $\mathcal{J }\subseteq \mathcal{I }$, exactly one of $\mathcal{J }$ or $\mathcal{I }\setminus \mathcal{J }$ is in $\varvec{\mathcal{U }\!\mathcal{F }}$.

As in Sect. 5, the Ultrafilter Lemma guarantees the existence of a free ultrafilter on $\mathcal{I }$. Note that properties (a) and (d) together imply that every cofinite subset of $\mathcal{I }$ is in $\varvec{\mathcal{U }\!\mathcal{F }}$. Now define $\nu :{\varvec{\mathcal{P }}}(\mathcal{I }){{\longrightarrow }}\{0,1\}$ by setting $\nu [\mathcal{J }]:=1$ for all $\mathcal{J }\in \varvec{\mathcal{U }\!\mathcal{F }}$, while $\nu [\mathcal{J }]:=0$ for any $\mathcal{J }\in {\varvec{\mathcal{P }}}(\mathcal{I })\setminus \varvec{\mathcal{U }\!\mathcal{F }}$ [in which case $\mathcal{I }\setminus \mathcal{J }\in \varvec{\mathcal{U }\!\mathcal{F }}$, by property (d)]. Using the four defining properties of $\varvec{\mathcal{U }\!\mathcal{F }}$, it is easy to verify that $\nu$ is a probability measure.

It remains to show that $\nu$ is uniform. Let $\pi \in \Pi _{\scriptscriptstyle {\mathrm{fin}}}$, and let $\mathcal{J }\subseteq \mathcal{I }$. Let $\mathcal{J }_0:=\{j\in \mathcal{J }$; $\pi (j)\ne j\}$, and let $\mathcal{J }_1:=\pi [\mathcal{J }_0]$. Then $\mathcal{J }_1$ is disjoint from $\mathcal{J }\setminus \mathcal{J }_0$ (because $\pi$ is a bijection). Also $\mathcal{J }_0$ and $\mathcal{J }_1$ are finite sets, because $\pi \in \Pi _{\scriptscriptstyle {\mathrm{fin}}}$; thus, $\nu [\mathcal{J }_0]=\nu [\mathcal{J }_1]=0$. But $\pi (\mathcal{J }) = (\mathcal{J }\setminus \mathcal{J }_0)\sqcup \mathcal{J }_1$. Thus, $\nu [\pi (\mathcal{J })] = \nu [\mathcal{J }]-\nu [\mathcal{J }_0]+\nu [\mathcal{J }_1]=\nu [\mathcal{J }]-0+0=\nu [\mathcal{J }]$, as desired. $\square$

Proof of Proposition 10

If $\mathcal{I }$ is finite, then all rankings and all probability measures on $\mathcal{I }$ are $\mathfrak{UF }$-continuous, so these conditions are vacuous. Meanwhile, $\Pi _{\scriptscriptstyle {\mathrm{fin}}}$-invariance just means invariance under all permutations of $\mathcal{I }$. It is well-known that the only permutation-invariant probability measure on $\mathcal{I }$ is the (real-valued) probability measure with $\mu [\mathcal{J }]:=|\mathcal{J }|/|\mathcal{I }|$ for all $\mathcal{J }\subseteq \mathcal{I }$, and the only $\Pi _{\scriptscriptstyle {\mathrm{fin}}}$-invariant likelihood ranking is the one defined by $\mu$. Thus, for the rest of the proof, I will suppose that $\mathcal{I }$ is infinite.

Existence of $(\succcurlyeq ):$ Let ${\varvec{\mathcal{F }}}$ be the set of all finite subsets of $\mathcal{I }$. Fix a suitable ultrafilter $\mathfrak{UF }$ on ${\varvec{\mathcal{F }}}$. Then define the preorder $(\succcurlyeq )$ on ${\varvec{\mathcal{P }}}(\mathcal{I })$ as follows: for all $\mathcal{J },\mathcal{K }\subseteq \mathcal{I }$,

$$\begin{aligned} \left( \mathcal{J }\succcurlyeq \mathcal{K } \right) \iff \left( {\left\{ \mathcal{F }\in {\varvec{\mathcal{F }}} \; ; \; |\mathcal{J }\cap \mathcal{F }| \ge |\mathcal{K }\cap \mathcal{F }| \right\} }\in \mathfrak{UF } \right). \end{aligned}$$

In particular, if $\mathcal{J },\mathcal{K }\subset \mathcal{I }$ are finite, then we have $\mathcal{J }\succcurlyeq \mathcal{K }$ if and only if $|\mathcal{J }|\ge |\mathcal{K }|$. It is easy to verify that this is a likelihood ranking. It is $\mathfrak{UF }$-continuous by construction. Finally, it is uniform because $\mathfrak{UF }$ is a suitable ultrafilter (hence $\Pi _{\scriptscriptstyle {\mathrm{fin}}}$-invariant).

Existence of $\mu$, and representation of $(\succcurlyeq )$ by $\mu$: Let $\mathcal{X }:=\{0,1\}$. Then there is an obvious correspondence between ${\varvec{\mathcal{P }}}(\mathcal{I })$ and $\mathcal{X }^\mathcal{I }$, obtained by identifying each $\mathcal{J }\subseteq \mathcal{I }$ with its indicator function $\varvec{1}_\mathcal{J }\in \mathcal{X }^\mathcal{I }$. The likelihood ranking $(\succcurlyeq )$ can thus be regarded as a complete preorder on $\mathcal{X }^\mathcal{I }$. The second condition in the definition of ‘likelihood ranking’ implies this preorder is separable. If $(\succcurlyeq )$ is uniform, then it is $\Pi _{\scriptscriptstyle {\mathrm{fin}}}$-invariant. Thus, Theorem 1 yields a linearly ordered abelian group $\mathcal{R }$ and a utility function $u:\mathcal{X }{{\longrightarrow }}\mathcal{R }$ such that the finitary part of $(\succcurlyeq )$ is the additive preorder defined by $u$ via formula (2).

However, an inspection of the proof of Theorem 1 reveals that $\mathcal{R }$ is a quotient of the free abelian group on $|\mathcal{X }|-1$ generators. Since $|\mathcal{X }|=2$, this is the free abelian group on one generator—namely $\mathbb{Z }$. Thus, $\mathcal{R }$ is a quotient of $\mathbb{Z }$. But $\mathcal{R }$ is linearly ordered, so it is torsion-free. Thus, $\mathcal{R }$ is in fact isomorphic to $\mathbb{Z }$ (with the natural order). It will be convenient to embed $\mathbb{Z }$ into $\mathbb{R }$ in the natural way. Thus, without loss of generality, we can suppose that $\mathcal{R }=\mathbb{R }$. Thus, ${}^*\!\mathcal{R }={}^*\!\mathbb{R }$.

The first condition in the definition of ‘likelihood ranking’ implies that $u(0)< u(1)$. Without loss of generality, suppose that $u(0)=0$ and $u(1)=1$. Thus, for any $\mathbf{x}\in \mathcal{X }^\mathcal{I }$, we have

$$\begin{aligned} {}^*\!\sum _{i\in \mathcal{I }} u(x_i) = {}^*\!\sum _{i\in \mathcal{I }} x_i. \end{aligned}$$

(24)

Next, the $\mathfrak{UF }$-continuity of $(\succcurlyeq )$ as a likelihood ranking is obviously equivalent to the $\mathfrak{UF }$-continuity of $(\succcurlyeq )$ as a preorder on $\mathcal{X }^\mathcal{I }$. Thus, Lemma 19 implies that $(\succcurlyeq )$ is the hyperadditive preorder defined by $u$ via Eq. (4). Combining this with Eq. (24), and the identification of each subset of $\mathcal{I }$ with its indicator function, we get:

$$\begin{aligned} \left( \mathcal{J }\succcurlyeq \mathcal{K } \right) \iff \left( \displaystyle {}^*\!\sum _{i\in \mathcal{I }}\varvec{1}_\mathcal{J }(i) \ge {}^*\!\sum _{i\in \mathcal{I }}\varvec{1}_\mathcal{K }(i) \right), \quad \text{ for } \text{ all } \mathcal{J },\mathcal{K }\subseteq \mathcal{I }\text{. } \end{aligned}$$

(25)

Define $I:={}^*\!\sum _{i\in \mathcal{I }} 1$; this is a positive (infinite) element of ${}^*\!\mathbb{R }$. Define $\epsilon :=1/I$. Then $\epsilon$ is a positive infinitesimal hyperreal. For any $\mathcal{J }\subseteq \mathcal{I }$, define

$$\begin{aligned} \mu [\mathcal{J }] := \epsilon \cdot {}^*\!\sum _{i\in \mathcal{I }} \varvec{1}_\mathcal{J }(i). \end{aligned}$$

In particular, $\mu [\mathcal{I }] = \epsilon \cdot I = 1$. We must check that $\mu$ is an ${}^*\!\mathbb{R }$-valued probability measure on $\mathcal{I }$. Properties (a) and (c) are satisfied by construction. To see (b), let $\mathcal{J },\mathcal{K }\subset \mathcal{I }$ be disjoint. Then $\varvec{1}_{\mathcal{J }\sqcup \mathcal{K }}=\varvec{1}_\mathcal{J }+ \varvec{1}_\mathcal{K }$. Thus,

$$\begin{aligned} {}^*\!\sum _{i\in \mathcal{I }} \varvec{1}_{\mathcal{J }\sqcup \mathcal{K }}(i)&= {}^*\!\sum _{i\in \mathcal{I }} \left( \varvec{1}_\mathcal{J }(i) +\varvec{1}_\mathcal{K }(i) \right) \mathop {=}\limits _{(*)} {}^*\!\sum _{i\in \mathcal{I }} \varvec{1}_\mathcal{J }(i) + {}^*\!\sum _{i\in \mathcal{I }} \varvec{1}_\mathcal{K }(i) , \end{aligned}$$

where $(*)$ follows from the definition of the summation operator ${}^*\!\sum$. Thus,

$$\begin{aligned} \mu [\mathcal{J }\sqcup \mathcal{K }]&= \epsilon \cdot {}^*\!\sum _{i\in \mathcal{I }} \varvec{1}_{\mathcal{J }\sqcup \mathcal{K }}(i) = \epsilon \cdot {}^*\!\sum _{i\in \mathcal{I }} \varvec{1}_\mathcal{J }(i) + \epsilon \cdot {}^*\!\sum _{i\in \mathcal{I }} \varvec{1}_\mathcal{K }(i) = \mu [\mathcal{J }]+\mu [\mathcal{K }], \end{aligned}$$

as desired. Thus, $\mu$ is an ${}^*\!\mathbb{R }$-valued probability measure on $\mathcal{I }$. To see that $\mu$ has full support, note that $\mu \{i\}=\epsilon$ for all $i\in \mathcal{I }$. Finally,

$$\begin{aligned} \left( \mu [\mathcal{J }]\ge \mu [\mathcal{K }] \right) \iff \left( \displaystyle {}^*\!\sum _{i\in \mathcal{I }} \varvec{1}_\mathcal{J }(i)\ge {}^*\!\sum _{i\in \mathcal{I }} \varvec{1}_\mathcal{J }(i) \right) \mathop {\Leftarrow \!\!\Rightarrow }\limits _{(*)} \left( \mathcal{J }\succcurlyeq \mathcal{K } \right), \end{aligned}$$

where $(*)$ is by statement (25). In other words, $(\succcurlyeq )$ is the likelihood ranking defined by $\mu$. Thus, $\mu$ is uniform because $(\succcurlyeq )$ is uniform. Likewise, $\mu$ is $\mathfrak{UF }$-continuous, because $(\succcurlyeq )$ is $\mathfrak{UF }$-continuous.

Uniqueness of $\mu$: Now let $\mu$ be any uniform, $\mathfrak{UF }$-continuous ${}^*\!\mathbb{R }$-valued probability measure on $\mathcal{I }$. Then for all $\mathcal{J },\mathcal{K }\subseteq \mathcal{I }$, we have

$$\begin{aligned} \left( \mu [\mathcal{J }]\ge \mu [\mathcal{K }] \right)&\mathop {\Leftarrow \!\!\Rightarrow }\limits _{(*)}&\left( {\left\{ \mathcal{F }\in {\varvec{\mathcal{F }}} \; ; \; \mu [\mathcal{J }\cap \mathcal{F }] \ge \mu [\mathcal{K }\cap \mathcal{F }] \right\} }\in \mathfrak{UF } \right) \\&\mathop {\Leftarrow \!\!\Rightarrow }\limits _{(\dagger )} \left( {\left\{ \mathcal{F }\in {\varvec{\mathcal{F }}} \; ; \; |\mathcal{J }\cap \mathcal{F }| \ge |\mathcal{K }\cap \mathcal{F }| \right\} }\in \mathfrak{UF } \right). \end{aligned}$$

Thus, $\mu$ is uniquely determined by $\mathfrak{UF }$. Here, $(*)$ is by the $\mathfrak{UF }$-continuity of $\mu$, combined with ultrafilter axiom (UF). Meanwhile, $(\dagger )$ is because the uniformity of $\mu$ implies that, for any finite subsets $\mathcal{G },\mathcal{H }\subset \mathcal{I }$, we have $\mu [\mathcal{G }]\ge \mu [\mathcal{H }]$ if and only if $|\mathcal{G }|\ge |\mathcal{H }|$.

Uniqueness of $(\succcurlyeq )$: Finally, let $(\succcurlyeq )$ be any $\mathfrak{UF }$-continuous, uniform likelihood ranking. Then we have shown that $(\succcurlyeq )$ can be represented by an $\mathfrak{UF }$-continuous, uniform ${}^*\!\mathbb{R }$-valued probability measure $\mu$. But we have also shown that there is only one such measure. Thus, there is only one such likelihood ranking. $\square$

Recall that ${}^*\!\log _2:{}^*\!\mathbb{R }_+{{\longrightarrow }}{}^*\!\mathbb{R }$ is the unique hyperreal extension of the base-2 logarithm function. The next result is needed in the proof of Proposition 11.

Lemma 23

Let $\ell :=\ln (2)$. For any positive $x<z\in {}^*\!\mathbb{R }$, we have

$$\begin{aligned} {}^*\!\log _2(z)-{}^*\!\log _2(x) < \frac{z-x}{\ell \,x }. \end{aligned}$$

Proof

First suppose that $x$ and $z$ are positive real numbers, and consider the ordinary real logarithm. The Mean Value Theorem yields some $y\in {\left( x,z \right)}$ such that

$$\begin{aligned} \frac{\log _2(z)-\log _2(x)}{z-x} = \log _2'(y) = \frac{1}{\ell \,y }. \end{aligned}$$

Thus, $\displaystyle \log _2(z)-\log _2(x)=\frac{z-x}{\ell \,y } < \frac{z-x}{\ell \,x }$, because $y>x$.

This proves the claim for any for any real $z>x>0$. Now the general statement follow by applying the Transfer Principle. $\square$

Proof of Proposition 11

Let ${\varvec{\mathcal{F }}}$ be the set of all finite subsets of $\mathcal{I }$, and let $\mathfrak{UF }$ be a suitable ultrafilter on ${\varvec{\mathcal{F }}}$ (as defined in Sect. 5). For any $\mathcal{J }\in {\varvec{\mathcal{F }}}$, we define

$$\begin{aligned} H_\mathcal{J }(\nu ):= -\sum _{j\in \mathcal{J }} \nu \{j\}\,{}^*\!\log _2[\nu \{j\}] \quad \text { and }\quad H_\mathcal{J }(\mu ):= -\sum _{j\in \mathcal{J }} \mu \{j\}\,{}^*\!\log _2[\mu \{j\}]. \end{aligned}$$

Thus, $H(\nu )$ (an element of $^{**}\mathbb{R }$) is in fact the $\mathfrak{UF }$-equivalence class of the function $[{\varvec{\mathcal{F }}}\ni \mathcal{J }\mapsto H_\mathcal{J }(\nu )\in {}^*\!\mathbb{R }]$ (and likewise for $H(\mu )$). Thus, if ${\varvec{\mathcal{G }}}:=\{\mathcal{J }\in {\varvec{\mathcal{F }}}$; $H_\mathcal{J }(\nu ) < H_\mathcal{J }(\mu )\}$, then $H(\nu )< H(\mu )$ if and only if ${\varvec{\mathcal{G }}}\in \mathfrak{UF }$. So this is what we must show.

Recall that there is some positive infinitesimal $\epsilon \in {}^*\!\mathbb{R }$ such that $\mu \{i\}=\epsilon$ for all $i\in \mathcal{I }$.

Claim 1 For any finite subset $\mathcal{J }\subset \mathcal{I }$, we have $H_\mathcal{J }(\mu )=-\mu [\mathcal{J }]\cdot {}^*\!\log _2(\epsilon )$.

Proof

$\displaystyle H_\mathcal{J }(\mu ) = -\sum _{j\in \mathcal{J }} \mu \{j\}\,{}^*\!\log _2[\mu \{j\}] = -\sum _{j\in \mathcal{J }} \epsilon \,{}^*\!\log _2(\epsilon ) = -|\mathcal{J }|\epsilon \,{}^*\!\log _2(\epsilon ) = -\mu [\mathcal{J }]\,{}^*\!\log _2(\epsilon )$. $\diamondsuit \text { Claim } 1$

For any finite subset $\mathcal{J }\subset \mathcal{I }$, let $\nu _\mathcal{J }$ and $\mu _\mathcal{J }$ be the conditional probability measures induced on $\mathcal{J }$ by $\nu$ and $\mu$. That is, let us define $\nu _\mathcal{J }[\mathcal{K }]:=\nu [\mathcal{K }]/\nu [\mathcal{J }]$ and $\mu _\mathcal{J }[\mathcal{K }]:=\mu [\mathcal{K }]/\mu [\mathcal{J }]$ for all $\mathcal{K }\subseteq \mathcal{J }$. It follows that $\mu _\mathcal{J }$ is the uniform measure on $\mathcal{J }$, because $\mu$ is the uniform measure on $\mathcal{I }$. Let us define $H(\nu _\mathcal{J }):=-\sum _{j\in \mathcal{J }} \nu _\mathcal{J }\{j\}\,{}^*\!\log _2(\nu _\mathcal{J }\{j\})$. (If $\nu$ was an ordinary real-valued probability measure, then we could remove the ‘$*$’ and this would just be the defining formula (10) for the entropy of a probability measure on a finite set.) Let us likewise define $H(\mu _\mathcal{J })$.

Claim 2 For any finite subset $\mathcal{J }\subset \mathcal{I }$, we have

• (a) $\displaystyle H(\nu _\mathcal{J }) = \frac{1}{\nu [\mathcal{J }]} H_\mathcal{J }(\nu ) + {}^*\!\log _2(\nu [\mathcal{J }])$, and

• (b) $\displaystyle H(\mu _\mathcal{J }) = \frac{1}{\mu [\mathcal{J }]} H_\mathcal{J }(\mu ) + {}^*\!\log _2(\mu [\mathcal{J }])$.

Proof

$$\begin{aligned} H(\nu _\mathcal{J })&= -\sum _{j\in \mathcal{J }}\nu _\mathcal{J }\{j\}\,{}^*\!\log _2(\nu _\mathcal{J }\{j\}) = -\sum _{j\in \mathcal{J }}\frac{\nu \{j\}}{\nu [\mathcal{J }]}\,{}^*\!\log _2\left(\frac{\nu \{j\}}{\nu [\mathcal{J }]}\right)\\&= -\frac{1}{\nu [\mathcal{J }]}\sum _{j\in \mathcal{J }} \nu \{j\}\,\left({}^*\!\log _2\left(\nu \{j\}\right)-{}^*\!\log _2\left(\nu [\mathcal{J }]\right)\right) \\&= \frac{1}{\nu [\mathcal{J }]} H_\mathcal{J }(\nu ) +\frac{1}{\nu [\mathcal{J }]}\sum _{j\in \mathcal{J }} \nu \{j\}\,{}^*\!\log _2\left(\nu [\mathcal{J }]\right) \\&= \frac{1}{\nu [\mathcal{J }]} H_\mathcal{J }(\nu ) +\frac{\nu [\mathcal{J }]}{\nu [\mathcal{J }]}\,{}^*\!\log _2\left(\nu [\mathcal{J }]\right) = \frac{1}{\nu [\mathcal{J }]} H_\mathcal{J }(\nu ) + {}^*\!\log _2(\nu [\mathcal{J }]), \end{aligned}$$

which proves (a). A similar computation proves (b). $\diamondsuit \text { Claim } 2$

Claim 3 Let ${\varvec{\mathcal{E }}}:=\{\mathcal{F }\in {\varvec{\mathcal{F }}}; \nu [\mathcal{F }]\le \mu [\mathcal{F }]\}$. Then ${\varvec{\mathcal{E }}}\in \mathfrak{UF }$.

Proof

Since $\nu$ is normal and $\mu$ is a ${}^*\!\sum$-additive probability measure, we have

$$\begin{aligned} {}^*\!\sum _{i\in \mathcal{I }} \nu \{i\} \le 1 = \mu [\mathcal{I }] = {}^*\!\sum _{i\in \mathcal{I }} \mu \{i\}. \end{aligned}$$

Thus, if ${\varvec{\mathcal{E }}}':=\left\rbrace \mathcal{J }\in {\varvec{\mathcal{F }}}; \ \sum _{j\in \mathcal{J }} \nu \{j\} \le \sum _{j\in \mathcal{J }} \mu \{j\}\right\lbrace$, then we must have ${\varvec{\mathcal{E }}}^{\prime }\in \mathfrak{UF }$, by the definition of the summation operator ${}^*\!\sum$. But for any finite subset $\mathcal{J }\subset \mathcal{I }$, we have $\nu [\mathcal{J }]=\sum _{j\in \mathcal{J }} \nu \{j\}$ and $\mu [\mathcal{J }]=\sum _{j\in \mathcal{J }} \mu \{j\}$, by finite additivity. Thus, in fact ${\varvec{\mathcal{E }}}'={\varvec{\mathcal{E }}}$. $\diamondsuit \text { Claim } 3$

Now, if $\nu \ne \mu$, then there exists some finite subset $\mathcal{K }\subset \mathcal{I }$ such that the measure $\nu _\mathcal{K }$ is not uniform on $\mathcal{K }$. Thus, for any finite subset $\mathcal{E }\subset \mathcal{I }$, if $\mathcal{K }\subseteq \mathcal{E }$, then the measure $\nu _\mathcal{E }$ is not uniform either.

Claim 4 Let $\varvec{\mathcal{J }}:=\{\mathcal{E }\in {\varvec{\mathcal{E }}}; \mathcal{K }\subseteq \mathcal{E }\}$. Then $\varvec{\mathcal{J }}\in \mathfrak{UF }$.

Proof

Let ${\varvec{\mathcal{F }}}(\mathcal{K }):=\{\mathcal{F }\in {\varvec{\mathcal{F }}}$; $\mathcal{K }\subseteq \mathcal{F }\}$. Then ${\varvec{\mathcal{F }}}(\mathcal{K })\in \mathfrak{UF }$ because $\mathfrak{UF }$ is a suitable ultrafilter. Meanwhile, ${\varvec{\mathcal{E }}}\in \mathfrak{UF }$ by Claim 3. But $\varvec{\mathcal{J }}:={\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}(\mathcal{K })$; thus, $\varvec{\mathcal{J }}\in \mathfrak{UF }$ by ultrafilter axiom (F1). $\diamondsuit \text { Claim } 4$

Claim 5 If $\mathcal{J }\in \varvec{\mathcal{J }}$, then $H(\nu _\mathcal{J })< H(\mu _\mathcal{J })$.

Proof

$\mathcal{J }$ is finite, and $\mu _\mathcal{J }$ is the uniform measure on $\mathcal{J }$; thus, it is the (unique) measure of maximal entropy on $\mathcal{J }$ (Cover and Thomas 2006). By the definition of $\varvec{\mathcal{J }}$, the measure $\nu _\mathcal{J }$ is non-uniform; thus, $H(\nu _\mathcal{J })< H(\mu _\mathcal{J })$. $\diamondsuit \text { Claim } 5$

Claim 6 If $\mathcal{J }\in \varvec{\mathcal{J }}$, then $H_\mathcal{J }(\nu ) < H_\mathcal{J }(\mu )$.

Proof

First note that

$$\begin{aligned} H_\mathcal{J }(\nu )&\mathop {=}\limits _{(*)} \nu [\mathcal{J }] \,\left( H(\nu _\mathcal{J }) - {}^*\!\log _2(\nu [\mathcal{J }])\right) \ \ \mathop {<}\limits _{(\dagger )} \ \ \nu [\mathcal{J }] \,\left( H(\mu _\mathcal{J }) - {}^*\!\log _2(\nu [\mathcal{J }])\right) \\&\mathop {=}\limits _{(\diamond )} \nu [\mathcal{J }] \,\left(\frac{1}{\mu [\mathcal{J }]} H_\mathcal{J }(\mu ) + {}^*\!\log _2(\mu [\mathcal{J }]) - {}^*\!\log _2(\nu [\mathcal{J }])\right) \\&= \frac{\nu [\mathcal{J }]}{\mu [\mathcal{J }]} H_\mathcal{J }(\mu ) +\nu [\mathcal{J }] \,\left( {}^*\!\log _2(\mu [\mathcal{J }]) - {}^*\!\log _2(\nu [\mathcal{J }])\right), \end{aligned}$$

where $(*)$ is by Claim 2(a), $(\dagger )$ is by Claim 5 and $(\diamond )$ is by Claim 2(b). It follows that

$$\begin{aligned} H_\mathcal{J }(\nu ) \!-\! H_\mathcal{J }(\mu ) \ < \ \left(\frac{\nu [\mathcal{J }]}{\mu [\mathcal{J }]} \!-\!1\right) \, H_\mathcal{J }(\mu ) \!+\!\nu [\mathcal{J }] \,\left( {}^*\!\log _2(\mu [\mathcal{J }]) - {}^*\!\log _2(\nu [\mathcal{J }])\right).\nonumber \\ \end{aligned}$$

(26)

Now,

$$\begin{aligned} \left(\frac{\nu [\mathcal{J }]}{\mu [\mathcal{J }]} -1\right) \, H_\mathcal{J }(\mu )&= \frac{\nu [\mathcal{J }]-\mu [\mathcal{J }]}{\mu [\mathcal{J }]} \, H_\mathcal{J }(\mu ) \nonumber \\&= \frac{H_\mathcal{J }(\mu )}{\mu [\mathcal{J }]}\left(\nu [\mathcal{J }]-\mu [\mathcal{J }]\right)\nonumber \\&\mathop {=}\limits _{(*)} -{}^*\!\log _2(\epsilon )\,\left(\nu [\mathcal{J }]-\mu [\mathcal{J }]\right)\nonumber \\&= {}^*\!\log _2(\epsilon )\,\left(\mu [\mathcal{J }]-\nu [\mathcal{J }]\right), \end{aligned}$$

(27)

where $(*)$ is by Claim 1. Meanwhile,

$$\begin{aligned}&\nu [\mathcal{J }] \,\left( {}^*\!\log _2(\mu [\mathcal{J }]) - {}^*\!\log _2(\nu [\mathcal{J }])\right) \nonumber \\&\mathop {<}\limits _{(*)} \nu [\mathcal{J }] \, \frac{\mu [\mathcal{J }]-\nu [\mathcal{J }]}{\ell \,\nu [\mathcal{J }]} =\frac{\mu [\mathcal{J }]-\nu [\mathcal{J }]}{\ell }, \end{aligned}$$

(28)

where $(*)$ is by Lemma 23, because $\nu [\mathcal{J }]<\mu [\mathcal{J }]$ because $\mathcal{J }\in \varvec{\mathcal{J }}\subseteq {\varvec{\mathcal{E }}}$. Substituting Eq. (27) and inequality (28) into inequality (26), we get

$$\begin{aligned} H_\mathcal{J }(\nu ) - H_\mathcal{J }(\mu )&< {}^*\!\log _2(\epsilon )\,\left(\mu [\mathcal{J }]-\nu [\mathcal{J }]\right) +\frac{\mu [\mathcal{J }]-\nu [\mathcal{J }]}{\ell } \\&= \left(\frac{1}{\ell }+{}^*\!\log _2(\epsilon )\right)\,\left(\mu [\mathcal{J }]-\nu [\mathcal{J }]\right) \, \mathop {<}\limits _{(*)} \, 0, \end{aligned}$$

and hence, $H_\mathcal{J }(\nu ) < H_\mathcal{J }(\mu )$, as desired. Here $(*)$ is because $\mu [\mathcal{J }]>\nu [\mathcal{J }]$, whereas ${}^*\!\log _2(\epsilon )<-\frac{1}{\ell }$, because $\epsilon <1/2^{1/\ell }$ (indeed, $\epsilon$ is infinitesimal). $\diamondsuit \text { Claim } 6$

Now Claims 4 and 6 implies that $H(\nu )<H(\mu )$, as desired. $\square$

The next result shows that the ‘normality’ hypothesis of Proposition 11 is not redundant. Let $\mu$ be the uniform measure on $\mathcal{I }$, from Proposition 10.

Proposition 24

If $\mathcal{I }$ is infinite, then there exists an ${}^*\!\mathbb{R }$-valued probability measure $\lambda$ on $\mathcal{I }$ which is not normal, such that $H(\lambda )>H(\mu )$.

Proof

Let $r\in {}^*\!\mathbb{R }$ be a positive value such that $\mu [\mathcal{J }]>r$ for all infinite $\mathcal{J }\subset \mathcal{I }$. (For example, we could set $r:=\mu [\mathcal{F }]$ for some nonempty but finite subset $\mathcal{F }\subset \mathcal{I }$.) Let $\nu$ be a probability measure such that $\nu [\mathcal{J }]=0$ for all finite $\mathcal{J }\subset \mathcal{I }$ [e.g. the measure from Example 9(b)]. Define the function $\lambda :{\varvec{\mathcal{P }}}(\mathcal{I }){{\longrightarrow }}{}^*\!\mathbb{R }$ by

$$\begin{aligned} \lambda (\mathcal{J }) := \frac{2\mu [\mathcal{J }] - r\,\nu [\mathcal{J }]}{2-r} , \quad \text{ for } \text{ all } \mathcal{J }\subseteq \mathcal{I }\text{. } \end{aligned}$$

Then $\lambda$ is finitely additive, because $\mu$ and $\nu$ are finitely additive. Also,

$$\begin{aligned} \lambda [\emptyset ]&= \frac{2\mu [\emptyset ]-r\,\nu [\emptyset ]}{2-r} = \frac{0}{2-r} = 0,\\ \text { and }\lambda [\mathcal{I }]&= \frac{2\mu [\mathcal{I }]-r\,\nu [\mathcal{I }]}{2-r} = \frac{2-r}{2-r} = 1. \end{aligned}$$

It remains to show that $\lambda [\mathcal{J }]\ge 0$ for all nonempty $\mathcal{J }\subseteq \mathcal{I }$. If $\mathcal{J }$ is finite, then $\lambda [\mathcal{J }]=2\mu [\mathcal{J }]/(2-r)>0$, because $\mu [\mathcal{J }]>0$ because $\mu$ has full support. If $\mathcal{J }$ is infinite, then

$$\begin{aligned} \lambda [\mathcal{J }] = \frac{2\mu [\mathcal{J }] - r\,\nu [\mathcal{J }]}{2-r} \ge \frac{2\mu [\mathcal{J }] - r }{2-r} > \frac{2r- r }{2-r} > 0, \end{aligned}$$

as desired. Thus, $\lambda$ is a probability measure. To see that $\lambda$ is not normal, note that $\lambda \{i\}=2\mu \{i\}/(2-r)$, for all $i\in \mathcal{I }$. Thus,

$$\begin{aligned} {}^*\!\sum _{i\in \mathcal{I }}\lambda \{i\} ={}^*\!\sum _{i\in \mathcal{I }} \frac{2\mu \{i\}}{2-r} =\frac{2}{2-r} {}^*\!\sum _{i\in \mathcal{I }} \mu \{i\} \ \ \mathop {=}\limits _{(*)} \ \ \frac{2}{2-r} \, \mu [\mathcal{I }] =\frac{2}{2-r} \, > \ , 1. \end{aligned}$$

Here, $(*)$ is because $\mu$ is ${}^*\!\sigma$-additive.

Finally, see that $H(\lambda )>H(\mu )$, note that

$$\begin{aligned} H(\lambda )&\mathop {=}\limits _{(*)} -{}^*\!\sum _{i\in \mathcal{I }}\lambda \{i\}\cdot {}^*\!\log _2[\lambda \{i\}] = -{}^*\!\sum _{i\in \mathcal{I }}\frac{2\,\mu \{i\} }{2-r}\cdot {}^*\!\log _2\left(\frac{2\,\mu \{i\} }{2-r}\right)\\&= -\frac{2}{2-r}{}^*\!\sum _{i\in \mathcal{I }}\mu \{i\}\cdot \left( {}^*\!\log _2[\mu \{i\}] + {}^*\!\log _2(2)\!-\!{}^*\!\log _2(2-r)\right) \\&= -\frac{2}{2-r}{}^*\!\sum _{i\in \mathcal{I }}\mu \{i\}\cdot {}^*\!\log _2[\mu \{i\}] -\frac{2}{2-r}\left( {}^*\!\log _2(2)-{}^*\!\log _2(2\!-\!r)\right){}^*\!\sum _{i\in \mathcal{I }}\mu \{i\}\\&\mathop {=}\limits _{(\diamond )} \frac{2}{2-r} H(\mu ) - \frac{2}{2-r}\left( {}^*\!\log _2(2)-{}^*\!\log _2(2-r)\right)\,\mu [\mathcal{I }]\\&\mathop {>}\limits _{(\dagger )} \left(1+\frac{r}{2-r}\right) H(\mu ) - \frac{2r}{(2-r)^2\,\ell } \qquad \text{(with } \ell :=\ln (2)\text{). } \end{aligned}$$

Here, both $(*)$ and $(\diamond )$ invoke defining formula (11), and $(\diamond )$ also uses the fact that $\mu$ is ${}^*\!\sum$-additive. Finally, $(\dagger )$ is because $\mu [\mathcal{I }]=1$ and Lemma 23 says

$$\begin{aligned} {}^*\!\log _2(2)-{}^*\!\log _2(2-r) < \frac{r}{(2-r)\,\ell }, \end{aligned}$$

Thus,

$$\begin{aligned} H(\lambda )-H(\mu )&> \frac{r}{2-r} H(\mu ) - \frac{2r}{(2-r)^2\,\ell } = \frac{r}{2-r}\left(H(\mu )-\frac{2}{(2-r)\,\ell }\right) > 0, \end{aligned}$$

because $H(\mu )>\frac{2}{(2-r)\,\ell }$ (indeed, $H(\mu )$ is infinite). Thus, $H(\lambda )>H(\mu )$, as claimed.

$\square$

Proof of Proposition 12

Any element of $r\in \mathbb{R }$ has a unique binary expansion $r=\sum _{z\in \mathbb{Z }} r_z 2^z$ for some sequence $\{r_z\}_{z\in \mathbb{Z }}$ taking values in $\{0,1\}$, such that: (i) there exists some $N\in \mathbb{N }$ such that $r_z=0$ for all $z>N$; and (ii) for any $M\in \mathbb{N }$, there exists some $z<-M$ with $r_z=0$.⁴³ For all $n\in \mathbb{Z }$, define $\pi _n:\mathbb{R }{{\longrightarrow }}\mathbb{R }$ by $\pi _n(r):=(1-r_n) 2^n + \sum _{z\in \mathbb{Z }\setminus \{n\}} r_z 2^z$ (i.e. $\pi _n$ toggles the $n$th binary digit of $r$). Let $\Gamma$ be the group generated by $\{\pi _n\}_{n\in \mathbb{Z }}$; then $\Gamma$ has locally finite orbits (because any finite subset of $\Gamma$ can only act upon a finite set of digits).

Let $\mathcal{I }:=\mathbb{R }$, and define $\mathfrak{UF }$ be as in Lemma 17; then $\mathfrak{UF }$ is a suitable ultrafilter, and $\Gamma \subseteq \Pi _\mathfrak{UF }$. Let $\mu$ be the full uniform measure on $\mathbb{R }$ defined by $\mathfrak{UF }$ in Proposition 10. Then Lemma 15(b) implies that

$$\begin{aligned} \mu [\pi _n(\mathcal{J })]=\mu [\mathcal{J }], \, \text{ for } \text{ any } \mathcal{J }\subseteq \mathbb{R } \text{ and } n\in \mathbb{Z }\text{. } \end{aligned}$$

(29)

Now let $M:=\mu {\left[ 0,1 \right)}$ and let $\lambda$ be the Lebesgue measure on $\mathbb{R }$. Statement (29) and the finite additivity of $\mu$ imply that

$$\begin{aligned} \mu {\left[ \frac{n}{2^k},\frac{m}{2^k} \right)} = \frac{(m-n)}{2^k}M = M\cdot \lambda {\left[ \frac{n}{2^k},\frac{m}{2^k} \right)}, \end{aligned}$$

(30)

for any $n,m,k\in \mathbb{Z }$ with $n<m$. An interval of this kind is called a dyadic interval. A  dyadic subset is a finite disjoint union of dyadic intervals. In particular, if $a<b\in \mathbb{R }$ have finite binary expansions, then the interval $[a,b)$ is a dyadic subset. (For example, if $a=\frac{1}{8}$ and $b=\frac{3}{4}$, then $[a,b)=[\frac{1}{8},\frac{1}{4}) \cup [\frac{1}{4},\frac{1}{2}) \cup [\frac{1}{2},\frac{3}{4})$.) If $\mathcal{D }\subset \mathbb{R }$ is any dyadic subset, then Eq. (30) and the finite additivity of $\mu$ and $\lambda$ imply that

$$\begin{aligned} \mu [\mathcal{D }] = M\cdot \lambda [\mathcal{D }]. \end{aligned}$$

(31)

Now let $a<b\in \mathbb{R }$ be arbitrary, and let $\mathcal{J }$ be either the interval $(a,b)$, or $[a,b)$, or $(a,b]$, or $[a,b]$.

Claim 1 For any (real) $\epsilon >0$, there exist dyadic subsets $\mathcal{D }_1,\mathcal{D }_2\subset \mathbb{R }$ with $\mathcal{D }_1\subseteq \mathcal{J }\subseteq \mathcal{D }_2$ and $\lambda [\mathcal{J }]-\epsilon \le \lambda [\mathcal{D }_1]\le \lambda [\mathcal{J }] \le \lambda [\mathcal{D }_2]\le \lambda [\mathcal{J }]+\epsilon$ .

Proof

Suppose that $a$ and $b$ have dyadic expansions $a=\sum _{z\in \mathbb{Z }} a_z 2^z$ and $b=\sum _{z\in \mathbb{Z }} b_z 2^z$, where $a_z,b_z\in \{0,1\}$ for all $z\in \mathbb{Z }$. Find $n\in \mathbb{N }$ such that $1/2^n<\epsilon /2$. Define

$$\begin{aligned} \begin{array}{rclcrcl} {\underline{a}}&{}:=&{}\displaystyle \sum _{z=-n}^{\infty }a_z 2^z,&{}&{} {\overline{a}}&{}:=&{} \displaystyle {\underline{a}}+ \frac{1}{2^n},\\ {\underline{b}}&{}:=&{}\displaystyle \sum _{z=-n}^{\infty }a_z 2^z,&{}\text { and }&{} {\overline{b}}&{}:=&{}\displaystyle {\underline{b}}+ \frac{1}{2^n}. \end{array} \end{aligned}$$

Then ${\underline{a}}\le a< {\overline{a}}$ and ${\underline{b}}\le b< {\overline{b}}$. Thus, if $\mathcal{D }_1:=[{\overline{a}},{\underline{b}})$ and $\mathcal{D }_2:=[{\underline{a}},{\overline{b}})$, then $\mathcal{D }_1\subseteq \mathcal{J }\subseteq \mathcal{D }_2$, and thus $\lambda [\mathcal{D }_1]\le \lambda [\mathcal{J }]\le \lambda [\mathcal{D }_2]$. By construction, ${\underline{a}}, {\overline{a}}, {\underline{b}}$ and ${\overline{b}}$ all have finite binary expansions; thus $\mathcal{D }_1$ and $\mathcal{D }_2$ are dyadic subsets. Finally, note that

$$\begin{aligned} \lambda [\mathcal{D }_2]-\lambda [\mathcal{D }_1] \!=\! ({\overline{b}}-{\underline{a}}) - ({\underline{b}}-{\overline{a}}) \!=\! ({\overline{b}}-{\underline{b}}) \!+\! ({\overline{a}}-{\underline{a}}) \!=\! \frac{1}{2^n}\!+\!\frac{1}{2^n} \!=\!\frac{2}{2^n} < \epsilon . \end{aligned}$$

The claim follows. $\diamondsuit \text { Claim } 1$

Now, if $\mathcal{D }_1,\mathcal{D }_2$ are in Claim 1, then we also have

$$\begin{aligned} \lambda [\mathcal{D }_1] ~\mathop {=}\limits _{(*)}~ \frac{\mu [\mathcal{D }_1]}{M} \underset{\dagger }{(\le )} \frac{\mu [\mathcal{J }]}{M} \underset{\dagger }{(\le )} \frac{\mu [\mathcal{D }_2]}{M} \mathop {=}\limits _{(*)} \lambda [\mathcal{D }_2], \end{aligned}$$

where both $(*)$ are invocations of statement (31), while the $(\dagger )$ inequalities are because $\mathcal{D }_1\subseteq \mathcal{J }\subseteq \mathcal{D }_2$. Since $\lambda [\mathcal{J }]-\epsilon \le \lambda [\mathcal{D }_1]$ and $\lambda [\mathcal{D }_2]\le \lambda [\mathcal{J }]+\epsilon$, we conclude that

$$\begin{aligned} \lambda [\mathcal{J }]-\epsilon \le \frac{\mu [\mathcal{J }]}{M} \le \lambda [\mathcal{J }]+\epsilon . \end{aligned}$$

Since this holds for any $\epsilon >0$, we conclude that $\mu [\mathcal{J }]/M-\lambda [\mathcal{J }]$ is either zero or infinitesimal.

This proves the theorem statement for a single interval. Any simple set is a finite union of intervals; thus, the general statement follows because both $\mu$ and $\lambda$ are finitely additive, and a finite sum of infinitesimals is still infinitesimal. $\square$

Proof of Proposition 21

Fix $\mathbf{x},\mathbf{y}\in \mathcal{X }^\mathcal{I }$. Let ${\varvec{\mathcal{F }}}_\mathbf{x}:=\{\mathcal{F }\in {\varvec{\mathcal{F }}}$; $\sum _\mathcal{F }\, u(\mathbf{x}) \ge \sum _\mathcal{F }\, u(\mathbf{y})\}$, while ${\varvec{\mathcal{F }}}_\mathbf{y}:={\varvec{\mathcal{F }}}_\mathbf{x}^\complement = \{\mathcal{F }\in {\varvec{\mathcal{F }}}$; $\sum _\mathcal{F }\, u(\mathbf{x}) < \sum _\mathcal{F }\, u(\mathbf{y})\}$.

‘$\Longrightarrow$’ If $\mathbf{x}\, {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}\!^{^{\Gamma }}\,\,\mathbf{y}$, then there exists some $\mathcal{J }\in {\varvec{\mathcal{F }}}$ and finite $\Delta \subset \Gamma$ such that $\displaystyle \sum _{\mathcal{F }} u(\mathbf{x}) \ \ge \ \sum _{\mathcal{F }} u(\mathbf{y})$ for all $\mathcal{F }\in {\varvec{\mathcal{F }}}_\Delta (\mathcal{J })$. Thus, ${\varvec{\mathcal{F }}}_\Delta (\mathcal{J })\subseteq {\varvec{\mathcal{F }}}_\mathbf{x}$. But ${\varvec{\mathcal{F }}}_\Delta (\mathcal{J })\in \mathfrak{UF }$ for any $\Gamma$-admissible free ultrafilter $\mathfrak{UF }$; thus, ${\varvec{\mathcal{F }}}_\mathbf{x}\in \mathfrak{UF }$ by (F2). Thus, $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}^{_{\mathfrak{UF }}}\, \mathbf{y}$.

‘${\Longleftarrow }$’ (by contrapositive) Suppose that . Then there is no $\mathcal{J }\in {\varvec{\mathcal{F }}}$ and finite $\Delta \subseteq \Gamma$ with ${\varvec{\mathcal{F }}}_\Delta (\mathcal{J })\subseteq {\varvec{\mathcal{F }}}_\mathbf{x}$. Thus, ${\varvec{\mathcal{F }}}_\Delta (\mathcal{J })\cap {\varvec{\mathcal{F }}}_\mathbf{y}\ne \emptyset$ for every $\mathcal{J }\in {\varvec{\mathcal{F }}}$ and finite $\Delta \subseteq \Gamma$. Thus,

$$\begin{aligned} {\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}_\mathbf{y}\ne \emptyset , \qquad \text{ for } \text{ every } {\varvec{\mathcal{E }}}\in \mathfrak{F }_\Gamma \text{. } \end{aligned}$$

(32)

Now define $\mathfrak{F }_\Gamma ^\mathbf{y}:=\{{\varvec{\mathcal{D }}}\subseteq {\varvec{\mathcal{F }}}; {\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}_\mathbf{y}\subseteq {\varvec{\mathcal{D }}}$ for some ${\varvec{\mathcal{E }}}\in \mathfrak{F }_\Gamma \}$.

Claim 1 (a) $\mathfrak{F }_\Gamma ^\mathbf{y}$ is a free filter, and (b) $\mathfrak{F }_\Gamma \subseteq \mathfrak{F }_\Gamma ^\mathbf{y}$.

Proof

(a) $\mathfrak{F }_\Gamma ^\mathbf{y}$ satisfies (F1) because $\mathfrak{F }_\Gamma$ satisfies (F1) and (32). It satisfies (F2) by construction. To verify (F0), it suffices to show that ${\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}_\mathbf{y}$ is infinite for any ${\varvec{\mathcal{E }}}\in \mathfrak{F }_\Gamma$.

By contradiction, suppose that ${\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}_\mathbf{y}$ is finite. Each element of ${\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}_\mathbf{y}$ is a finite subset of $\mathcal{I }$. Let $\mathcal{F }:= \bigcup \{\mathcal{E }$; $\mathcal{E }\in {\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}_\mathbf{y}\}$; then $\mathcal{F }$ is also finite, hence, a proper subset of $\mathcal{I }$. Let $\mathcal{D }\subset \mathcal{I }$ be a nonempty finite subset disjoint from $\mathcal{F }$; then no element of ${\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}_\mathbf{y}$ contains $\mathcal{D }$. Thus, for any finite $\Delta \subseteq \Gamma$, we have ${\varvec{\mathcal{F }}}_\Delta (\mathcal{D })\cap {\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}_\mathbf{y}=\emptyset$. But ${\varvec{\mathcal{F }}}_\Delta (\mathcal{D })\cap {\varvec{\mathcal{E }}}\in \mathfrak{F }_\Gamma$ by (F2), because ${\varvec{\mathcal{F }}}_\Delta (\mathcal{D })\in \mathfrak{F }_\Gamma$ by definition. This contradicts (32). By contradiction, ${\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}_\mathbf{y}$ must be infinite.

(b) Let ${\varvec{\mathcal{E }}}\in \mathfrak{F }_\Gamma$. Then ${\varvec{\mathcal{E }}}\supseteq {\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}_\mathbf{y}$; hence ${\varvec{\mathcal{E }}}\in \mathfrak{F }_\Gamma ^\mathbf{y}$ by definition. $\diamondsuit \text { Claim } 1$

Claim 1(a) and the Ultrafilter Lemma says there exists a free ultrafilter $\mathfrak{UF }^\mathbf{y}$ containing $\mathfrak{F }^\mathbf{y}_\Gamma$. Claim 1(b) implies that $\mathfrak{UF }^\mathbf{y}$ also contains $\mathfrak{F }_\Gamma$, so it is $\Gamma$-admissible.

Claim 2 $\mathbf{x}\, \,{}^{^*}\!\!\! {\stackrel{\displaystyle \prec }{{\scriptscriptstyle u}}}^{\!\mathfrak{UF }^\mathbf{y}}\, \mathbf{y}$.

Proof

Let ${\varvec{\mathcal{E }}}\in \mathfrak{F }_\Gamma$ be arbitrary. Then ${\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}_\mathbf{y}\in \mathfrak{F }_\Gamma ^\mathbf{y}$ by definition, and ${\varvec{\mathcal{E }}}\cap {\varvec{\mathcal{F }}}_\mathbf{y}\subseteq {\varvec{\mathcal{F }}}_\mathbf{y}$, so ${\varvec{\mathcal{F }}}_\mathbf{y}\in \mathfrak{F }_\Gamma ^\mathbf{y}$ by (F2), so ${\varvec{\mathcal{F }}}_\mathbf{y}\in \mathfrak{UF }^\mathbf{y}$, which means $\mathbf{x}\, \,{}^{^*}\!\!\! {\stackrel{\displaystyle \prec }{{\scriptscriptstyle u}}}^{\!\!\mathfrak{UF }^\mathbf{y}}\, \mathbf{y}$.$\diamondsuit \text { Claim } 2$

Thus, it is false that $\mathbf{x}\,{}^{^*}\! {\stackrel{\displaystyle \succcurlyeq }{{\scriptscriptstyle u}}}^{\!_{\mathfrak{UF }}}\,\mathbf{y}$ for every $\Gamma$-admissible ultrafilter $\mathfrak{UF }$ on ${\varvec{\mathcal{F }}}$. $\square$