Proof I adapt the proof of Proposition 1 of Predd et al. (Reference Predd, Seiringer, Lieb, Osherson, Poor and Kulkarni2009). Let ${\mathcal{F}}=\{p_1,p_2,\ldots \}$ . Let ${\mathcal{X}}$ be the collection of all nonempty sets of the form $\bigcap _{i=1}^\infty p_i^*$ , where $p_i^*$ is either $p_i$ or $p_i^c$ . Then ${\mathcal{X}}$ partitions $W$ . Also, ${\mathcal{X}}$ is in bijection with ${\mathcal{V}}_{{\mathcal{F}}}$ , the set of omniscient credence functions.

Indeed, let $f$ map $v_w$ to $\bigcap _{i=1}^\infty p_i^*$ , where $p_i^*=p_i$ if $v_w(p_i)=1$ and $p_i^*=p_i^c$ otherwise. Then for each $w$ , $w\in f(v_w)$ , and so $f(v_w)\in {\mathcal{X}}$ . Note that $f$ is onto. Indeed, let $w\in \bigcap _{i\,=1}^\infty p_i^*$ , where $\bigcap _{i=1}^\infty p_i^*\in {\mathcal{X}}$ . Then $f(v_w)=\bigcap _{i=1}^\infty p_i^*$ . Also, $f$ is injective. Indeed, assume $f(v_w)=f(v_{w'})$ . Then

$$f(v_w)=\bigcap _{i\,=1}^\infty p^1_i=\bigcap _{i\,=1}^\infty p^2_i= f(v_{w'})$$

for $p_i^j=p_i$ or $p_i^j=p_i^c$ for all $i\in {\mathbb{N}}$ and $j\in \{1,2\}$ . If $p^1_i\ne p^2_i$ for some $i$ , then without loss of generality, we may assume $p^1_i=p_i$ and $p^2_i=p_i^c$ . So $w\in p_i^1$ but $w\,\notin\, p_i^2$ , and thus $w\,\notin \bigcap _{i\,=1}^\infty p_i^2$ . But $w\in \bigcap _{i\,=1}^\infty p_i^1$ by definition of $f$ and so $\bigcap _{i\,=1}^\infty p_i^1\ne \bigcap _{i\,=1}^\infty p_i^2$ , which is a contradiction. It follows that $p^1_i=p^2_i$ for all $i$ , but then by definition of $f$ , this implies $v_w(p_i)=1$ if and only if $v_{w'}(p_i)=1$ for all $i$ , and so $v_w=v_{w'}$ .

It is easy to see that because ${\mathcal{F}}$ is countably discriminating, ${\mathcal{V}}_{{\mathcal{F}}}$ is countable. It follows that ${\mathcal{X}}$ is countable. Enumerate the elements of ${\mathcal{V}}_{{\mathcal{F}}}$ and ${\mathcal{X}}$ by $v_{w_1}, v_{w_2},\ldots$ and $e_1,e_2,\ldots$ , respectively, such that $f^{-1}(e_j)=v_{w_j}$ . We have that $p_i$ is the disjoint union of $e_j$ such that $e_j\subseteq p_i$ or, equivalently, the $e_j$ where $f^{-1}(e_j)(p_i)=1$ . Note that i) for any countably additive probability function $\mu$ on a $\sigma$ -algebra containing ${\mathcal{F}}$ (and thus containing ${\mathcal{X}}$ ) and any $p_i\in {\mathcal{F}}$ :

$$\mu (p_i)=\sum _{j\,=1}^\infty \mu (e_j) f^{-1}(e_j)(p_i). $$

Now we can prove the equivalence. Assume $c$ is countably coherent. By the definition of countable coherence, $c$ extends to a countably additive probability function $\mu$ on a $\sigma$ -algebra containing ${\mathcal{F}}$ . Then by i),

$$c(p_i)=\mu (p_i)=\sum _{j\,=1}^\infty \mu (e_j) f^{-1}(e_j)(p_i)$$

for all $p_i\in {\mathcal{F}}$ . But because $\mu (e_j)$ are nonnegative and sum to $1$ (because the $e_j$ values partition $W$ , and $\mu$ is a countably additive probability function), we have that $c$ has the form stated.

Now assume $c(p_i)=\sum _{j\,=1}^\infty \lambda _jv_{w_j}(p_i)$ for all $i$ , where $\sum _{j\,=1}^\infty \lambda _j=1$ . Let $\sigma ({\mathcal{F}})$ be the smallest $\sigma$ -algebra on $W$ containing ${\mathcal{F}}$ . Then it is easy to check that the function on $\sigma ({\mathcal{F}})$ defined by $\bar{v}_{w_j}(p)=1$ if and only if $w_j\in p$ extends $v_{w_j}$ and is a countably additive probability function on $\sigma ({\mathcal{F}})$ . Then $\sum _{j\,=1}^\infty \lambda _j \bar{v}_{w_j}$ is a countably additive probability function on $\sigma ({\mathcal{F}})$ since a countable sum of countably additive probability functions with coefficients that sum to $1$ is a countably additive probability function. Because

$$c(p_i)= \sum _{j\,=1}^\infty \lambda _j\upsilon_{w_j}(p_i)=\sum _{j\,=1}^\infty \lambda _i \bar {\upsilon}_{w_j}(p_i)$$

for all $i$ , it follows that $c$ extends to a countably additive probability function on a $\sigma$ -algebra containing ${\mathcal{F}}$ . □

Proof Let $\{\Psi (p_n)^{f(n)}\}_{n\,=1}^\infty$ be a sequence of elements of ${\mathcal{F}}^*$ or their complements, as in Definition 4.9. Case 1: For each $N$ , there is some $w_N\in W$ such that $w_N\in \bigcap _{n\,=1}^N\Psi (p_n)^{f(n)}$ . Then, since i) $\Psi (p)\cap W=p$ and ii) $\Psi (p)^c\cap W =p^c$ for any $p\in {\mathcal{F}}$ , it follows that $w_N\in \bigcap _{n\,=1}^N p^{f(n)}_n$ for each $N$ . If there is some $w'\in W$ with $w'\in \bigcap _{n\,=1}^\infty p^{f(n)}_n$ , then, by i) and ii), it follows that $w'\in \bigcap _{n\,=1}^\infty \Psi (p_n)^{f(n)}$ . Otherwise, by construction, we defined some $x_s$ to be such that $x_s \in \bigcap _{n\,=1}^\infty \Psi (p_n)^{f(n)}$ . In either case, we are done. Case 2: There is some $N$ such that $\bigcap _{n\,=1}^N\Psi (p_n)^{f(n)}\subseteq W^*\setminus W$ . I claim this implies that $\bigcap _{n\,=1}^N\Psi (p_n)^{f(n)}=\varnothing$ . Indeed, if there were some $w\in W^*\setminus W$ such that $w\in \bigcap _{n\,=1}^N\Psi (p_n)^{f(n)}$ , then that is because $\{p_n^{f(n)}\}_{n\,=1}^N$ is an initial sequence of some sequence $\{\bar{p}_n^{\bar{f}(n)}\}_{n\,=1}^\infty$ such that $\bigcap _{n\,=1}^l\bar{p}^{\bar{f}(n)}_n\ne \varnothing$ for each $l$ and thus, in particular, $\bigcap _{n\,=1}^Np^{f(n)}_n\ne \varnothing$ . Thus, there is some $w\in W$ such that $w\in \bigcap _{n\,=1}^N\Psi (p_n)^{f(n)}$ by i) and ii), which is a contradiction. Thus, we have established that $(W^*,{\mathcal{F}}^*)$ is compact. □

Proof Because $(W^*,{\mathcal{F}}^*)$ is compact, we only need to show that $c^*$ is coherent by Theorem 4.10. Thus, it suffices to show that $c^*$ can be extended to a finitely additive probability function on ${\mathcal{A}}({\mathcal{F}}^*)$ . Since $c$ is coherent, there is a finitely additive probability function $\bar{c}$ such that:

1. 1. $\bar{c}(p)=c(p)$ for $p\in {\mathcal{F}}$ ;

2. 2. $\bar{c}(p\cup q)=\bar{c}(p)+\bar{c}(q)$ for $p,q\in {\mathcal{F}}$ with $p\cap q=\varnothing$ ; and

3. 3. $\bar{c}(W)=1$ .

First, define $\Psi (p^c):=\Psi (p)^c$ for each $p\in {\mathcal{F}}$ . Then each element in ${\mathcal{A}}({\mathcal{F}}^*)$ can be represented by $\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^M\Psi (q_{ij})$ , where $q_{ij}$ or its complement is in ${\mathcal{F}}$ . We define

$$\bar {c^*}\left(\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^M\Psi (q_{ij})\right):=\bar {c}\left(\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^Mq_{ij}\right).$$

Using that $p=\Psi (p)\cap W$ and $p^c=\Psi (p)^c\cap W$ , we can show that $\bar{c^*}$ is a well-defined finitely additive probability function on ${\mathcal{A}}({\mathcal{F}}^*)$ extending $c^*$ . First, we can show that $\bar{c^*}$ is well defined. Assume that

$$\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^M\Psi (q_{ij})=\bigcup _{i\,=1}^{N'}\bigcap _{j\,=1}^{M'}\Psi (r_{ij}).$$

This clearly implies that

$$\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^M\Psi (q_{ij})\cap W=\bigcup _{i\,=1}^{N'}\bigcap _{j\,=1}^{M'}\Psi (r_{ij})\cap W,$$

which, noting that $p=\Psi (p)\cap W$ and $p^c=\Psi (p)^c\cap W$ , establishes that

$$\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^Mq_{ij}=\bigcup _{i\,=1}^{N'}\bigcap _{j\,=1}^{M'}r_{ij},$$

and so

$$\bar {c^*}\left(\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^M\Psi (q_{ij})\right)=\bar {c}\left(\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^Mq_{ij}\right)=\bar {c}\left(\bigcup _{i\,=1}^{N'}\bigcap _{j\,=1}^{M'}r_{ij}\right)=\bar {c^*}\left(\bigcup _{i\,=1}^{N'}\bigcap _{j\,=1}^{M'}\Psi (r_{ij})\right).$$

Thus, $\bar{c^*}$ is well defined. Clearly, $\bar{c^*}$ extends $c^*$ . Now, since $W\subseteq W^*$ , if

$$\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^M\Psi (q_{ij})\cap \bigcup _{i\,=1}^{N'}\bigcap _{j\,=1}^{M'}\Psi (r_{ij})=\varnothing,$$

then

$$\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^M\Psi (q_{ij})\cap W\cap \bigcup _{i\,=1}^{N'}\bigcap _{j\,=1}^{M'}\Psi (r_{ij})\cap W=\varnothing $$

and so

$$\bar {c}\left(\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^Mq_{ij}\cup \bigcup _{i\,=1}^{N'}\bigcap _{j\,=1}^{M'}r_{ij}\right)= \bar {c}\left(\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^Mq_{ij}\right)+ \bar {c}\left(\bigcup _{i\,=1}^{N'}\bigcap _{j\,=1}^{M'}r_{ij}\right).$$

Then, noting the definition of $\bar{c^*}$ in terms of $\bar{c}$ , we establish finite additivity. Lastly, if

$$W^*=\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^M\Psi (q_{ij}),$$

then

$$W=\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^M\Psi (q_{ij})\cap W,$$

and so

$$\bar {c^*}\left(\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^M\Psi (q_{ij})\right)=\bar {c}\left(\bigcup _{i\,=1}^{N}\bigcap _{j\,=1}^Mq_{ij}\right)=\bar {c}(W)=1.$$

This establishes that $c^*$ is coherent on $(W^*,{\mathcal{F}}^*)$ , and so because $(W^*,{\mathcal{F}}^*)$ is compact, $c^*$ is countably coherent. Further, $w\in p$ if and only if $w\in \Psi (p)$ for each $w\in W$ , so $v_w$ defined on ${\mathcal{F}}$ is the same as $v_w$ defined on ${\mathcal{F}}^*$ for each $w\in W$ . Since $c(p)=c^*(\Psi (p))$ for all $p \in {\mathcal{F}}$ , this establishes that ${\mathscr{I}}(c,w)={\mathscr{I}}(c^*,w)$ for each $w\in W$ . □

Proof Since $c$ is countably coherent, let $\bar{c}$ be a countably additive probability function on $\sigma ({\mathcal{F}})$ extending $c$ such that ${\mathbb{E}}_{\bar{c}}{\mathscr{I}}(c,\cdot )\lt \infty$ . Note that ${\mathfrak{d}}$ is a strictly proper inaccuracy measure for singleton opinion sets (see Remark 2.7 and Pettigrew [Reference Pettigrew2016, Theorem 4.3.5] for a precise definition), which implies by definition that $p{\mathfrak{d}}(1,x)+(1-p){\mathfrak{d}}(0,x)$ is uniquely minimized at $x=p$ .Footnote ⁴² It follows that

$${\mathbb {E}}_{\bar {c}}{\mathfrak {d}}(v_w,c_i)=c_i{\mathfrak {d}}(1,c_i)+(1-c_i){\mathfrak {d}}(0,c_i)\lt c_i{\mathfrak {d}}(1,x)+(1-c_i){\mathfrak {d}}(0,x)={\mathbb {E}}_{\bar {c}}{\mathfrak {d}}(v_w,x)$$

for any $x\ne c_i$ .

Assume toward a contradiction that there is a credence function $d$ with $d\ne c$ and ${\mathscr{I}}(d,w)\leq {\mathscr{I}}(c,w)$ for each $w$ with strict inequality for some $w$ . Then ${\mathbb{E}}_{\bar{c}}{\mathscr{I}}(d,\cdot )\leq {\mathbb{E}}_{\bar{c}}{\mathscr{I}}(c,\cdot )\lt \infty$ , so both ${\mathscr{I}}(d,\cdot )$ and ${\mathscr{I}}(c,\cdot )$ are integrable with respect to the measure space $(W,\sigma ({\mathcal{F}}),\bar{c})$ . Then let $i$ be any index such that $d_i\ne c_i$ . There must be at least one because $c\ne d$ . Then

$${\mathbb {E}}_{\bar {c}}{\mathfrak {d}}(v_w,c_i)\lt {\mathbb {E}}_{\bar {c}}{\mathfrak {d}}(v_w,d_i).$$

If $i$ is such that $d_i=c_i$ , then clearly, ${\mathbb{E}}_{\bar{c}}{\mathfrak{d}}(v_w,c_i)= {\mathbb{E}}_{\bar{c}}{\mathfrak{d}}(v_w,d_i)$ . So because ${\mathbb{E}}_{\bar{c}}{\mathscr{I}}(c,\cdot )\lt \infty$ and ${\mathbb{E}}_{\bar{c}}{\mathscr{I}}(d,\cdot )\lt \infty$ , we have

$${\mathbb {E}}_{\bar {c}}{\mathscr {I}}(c,\cdot )=\sum _{i\,=1}^\infty a_i{\mathbb {E}}_{\bar {c}}{\mathfrak {d}}(v_w,c_i)\lt \sum _{i\,=1}^\infty a_i{\mathbb {E}}_{\bar {c}}{\mathfrak {d}}(v_w,d_i)={\mathbb {E}}_{\bar {c}}{\mathscr {I}}(d,\cdot ),$$

which implies that ${\mathbb{E}}_{\bar{c}}({\mathscr{I}}(c,\cdot )-{\mathscr{I}}(d,\cdot ))\lt 0$ . Thus, there is some nonempty set $E\in \sigma ({\mathcal{F}})$ with $\bar{c}(E)\gt 0$ on which ${\mathscr{I}}(c,\cdot )-{\mathscr{I}}(d,\cdot )\lt 0$ (since the Lebesgue integral is positive). But this contradicts our assumption that $d$ weakly dominates $c$ , and so we are done. □

Proof Assume $d$ weakly dominates $c$ . Note that i) $c$ is somewhere finitely inaccurate if and only if ${\mathscr{B}}(c,w)\lt \infty$ for all $w\in W$ if and only if $\sum _{i\,=1}^\infty c_i^2\lt \infty$ . It follows by weak dominance that ${\mathscr{B}}(d,w)\lt \infty$ for all $w\in W$ , and therefore $\sum _{i\,=1}^\infty d_i^2\lt \infty$ . Let ${\mathscr{B}}(c,w)={\mathfrak{D}}(v_w,c)$ for ${\mathfrak{D}}$ a generalized quasi-additive Bregman divergence.

Because $(W,{\mathcal{F}})$ is point-finite, it is also countably discriminating as there are only countably many finite subsets of ${\mathcal{F}}$ . So by Proposition 4.7, $c=\sum _{j\,=1}^\infty \lambda _j v_{w_j}$ for $\lambda _j\in [0,1]$ with $\sum _{j\,=1}^\infty \lambda _j=1$ and ${\mathcal{V}}_{{\mathcal{F}}}=\{v_{w_j}\}_{j\,=1}^\infty$ . First, note that ${\mathfrak{D}}(\sum _{j\,=1}^\infty \lambda _jv_{w_j},c)=0$ and $(\sum _{j\,=1}^\infty \lambda _jv_{w_j}(p_i)-c(p_i))^2=0$ for all $i$ , so

$${\mathfrak {D}}\left(\sum _{j\,=1}^\infty \lambda _jv_{w_j},c\right)-{\mathfrak {D}}\left(\sum _{j\,=1}^\infty \lambda _jv_{w_j},d\right)=\sum _{i\,=1}^\infty a_i\left[\left(\sum _{j\,=1}^\infty \lambda _jv_{w_j}(p_i)-c_i\right)^2-\ \left(\sum _{j\,=1}^\infty \lambda _jv_{w_j}(p_i)-d_i\right)^2\right].$$

Using that

$${{{\left( {\sum\limits_{j{\kern 1pt} = 1}^\infty {{\lambda _j}} {v_{{w_j}}}({p_i}) - {c_i}} \right)}^2} - {{\left( {\sum\limits_{j{\kern 1pt} = 1}^\infty {{\lambda _j}} {v_{{w_j}}}({p_i}) - {d_i}} \right)}^2} = {\mkern 1mu} c_i^2 - d_i^2 + 2({d_i} - {c_i})\sum\limits_{j{\kern 1pt} = 1}^\infty {{\lambda _j}} {v_{{w_j}}}({p_i})} \\ { = \sum\limits_{j{\kern 1pt} = 1}^\infty {{\lambda _j}} c_i^2 - \sum\limits_{j{\kern 1pt} = 1}^\infty {{\lambda _j}} d_i^2 + \sum\limits_{j{\kern 1pt} = 1}^\infty {{\lambda _j}} 2({d_i} - {c_i}){v_{{w_j}}}({p_i})} \\ { = \sum\limits_{j{\kern 1pt} = 1}^\infty {{\lambda _j}} [c_i^2 - d_i^2 + 2({d_i} - {c_i}){v_{{w_j}}}({p_i})]} \\ { = \sum\limits_{j{\kern 1pt} = 1}^\infty {{\lambda _j}} [{{({v_{{w_j}}}({p_i}) - {c_i})}^2} - {{({v_{{w_j}}}({p_i}) - {d_i})}^2}]} \\$$

for each $i$ because $\sum _{j\,=1}^\infty \lambda _j=1$ , we have that

(8) $$\eqalignno{ {\mathfrak{D}}\left(\sum _{j\,=1}^\infty \lambda _jv_{w_j},c\right)-{\mathfrak{D}}\left(\sum _{j\,=1}^\infty \lambda _jv_{w_j},d\right) =\sum _{i\,=1}^\infty a_i\sum _{j\,=1}^\infty \lambda _j [(v_{w_j}(p_i)-c_i)^2-(v_{w_j}(p_i)-d_i)^2]\cr =\sum _{i\,=1}^\infty a_i\left(\sum _{j:w_j\notin p_i} \lambda _j\right) (c_i^2-d_i^2) +a_i\left(\sum _{j:w_j\in p_i}\lambda _j\right)((1-c_i)^2-(1-d_i)^2)\nonumber \cr =\sum _{i\,=1}^\infty a_i( c_i^2-d_i^2)+2a_i\left(\sum _{j:w_j\in p_i}\lambda _j\right)(d_i-c_i)\nonumber \cr =\sum _{i\,=1}^\infty a_i( -c_i^2-d_i^2)+2a_i\left(\sum _{j:w_j\in p_i}\lambda _j\right)d_i}$$

because $c_i=\sum _{j:w_j\in p_i}\lambda _j$ . We have $\sum _{i\,=1}^\infty c_i^2+d_i^2\lt \infty$ by item (i). Thus,

(9) $$ 0\leq \sum _{i\,=1}^\infty a_i2\left(\sum _{j:w_j\in p_i}\lambda _j\right)d_i\lt \infty $$

because

$$ 0\geq {\mathfrak{D}}\left(\sum _{j\,=1}^\infty \lambda _jv_{w_j},c\right)-{\mathfrak{D}}\left(\sum _{j\,=1}^\infty \lambda _jv_{w_j},d\right). $$

Having established (9), we can claim that we can use the dominated convergence theorem (see, e.g., Theorem 1.4.49 in Tao Reference Tao2011) to switch the limits in equation (8). Indeed,

$$ \sum _{i\,=1}^\infty a_i \sum _{j\,=1}^N \lambda _j[(v_{w_j}(p_i)-c_i)^2-(v_{w_j}(p_i)-d_i)^2] =\sum _{i\,=1}^\infty a_i\left(\sum _{1\leq j\leq N}\lambda _j\right)(c_i^2-d_i^2)+2a_i\left(\sum _{\substack {j:w_j\in p_i\\1\leq j\leq N}}\lambda _j\right)(d_i-c_i).$$

Letting

$$g_N(i)=a_i\left(\sum _{1\,\leq \,j\,\leq\, N}\lambda _j\right)(c_i^2-d_i^2)+2a_i\left(\sum _{\substack {j:w_j\in \,p_i\\1\,\leq \,j\,\leq \,N}}\lambda _j\right)(d_i-c_i)$$

and noting that $-\left(\sum _{\substack{j:w_j\in \,p_i\\1\,\leq \,j\,\leq \,N}}\lambda _j\right)c_i\geq - c_i^2$ because $c_i=\sum _{j:w_j\in \,p_i}\lambda _j$ , we see that

$$|g_N(i)|\leq a_i( 2c_i^2+d_i^2+2\left(\sum _{j:w_j\in \,p_i}\lambda _j\right)d_i).$$

Each of $c_i^2,d_i^2,$ and $(\sum _{j:w_j\in p_i}\lambda _j)d_i$ is summable in $i$ and $\sup _i a_i\lt \infty$ . So, the dominated convergence theorem applies, and we can switch limits.

Thus, we have

where we used that ${\mathfrak{D}}(v_{w_j},c)={\mathscr{B}}(c,w_j)\lt \infty$ and ${\mathfrak{D}}(v_{w_j},d)={\mathscr{B}}(d,w_j)\lt \infty$ for each $j$ by i) to break up the summation in the second line. Thus, we can conclude that $c=d$ because ${\mathfrak{D}}(c,d)=0$ if and only if $c=d$ . □

Proof If $c$ is coherent, then $c^*$ is countably coherent on a point-finite opinion set. Further, $c^*$ is somewhere finitely inaccurate relative to ${\mathscr{B}}$ , as $c$ is somewhere finitely inaccurate by assumption. Thus, by Proposition 4.16, $c^*$ is not weakly dominated relative to ${\mathscr{B}}$ . By W-stability, $c$ is not weakly dominated relative to ${\mathscr{B}}$ . Clearly, if $c$ is not weakly dominated, then $c$ is not strongly dominated. Finally, we show that if $c$ is incoherent, then $c$ is strongly dominated. First, if $c$ is not somewhere finitely inaccurate, then any omniscient credence function strongly dominates $c$ because ${\mathscr{I}}(v_w,w')\lt \infty$ for every $w,w'\in W$ by point-finiteness. If $c$ is somewhere finitely inaccurate, then ${\mathscr{I}}(c,w)\lt \infty$ for all $w\in W$ by point-finiteness. Thus, Theorem 3.4 establishes that $c$ is strongly dominated relative to ${\mathscr{B}}$ . □

Proof Let ${\mathcal{F}}=\{p_1, p_2,\ldots \}$ be a partition. Assume that a coherent credence function $c$ on ${\mathcal{F}}$ is weakly dominated by some credence function $d$ . We can assume $d$ is coherent by Theorem 3.4, and so $\sum _{m=1}^\infty d_m\leq 1$ . We can show that $c^*$ is weakly dominated by $d^*$ , thereby establishing that a partition is W-stable.

First, ${\mathscr{I}}(c^*,w)={\mathscr{I}}(c,w)$ and ${\mathscr{I}}(d^*,w)={\mathscr{I}}(d,w)$ for all $w\in W$ by Lemma 4.12. Thus, by the assumption of weak dominance,

$${\mathscr {I}}(c^*,w)\geq {\mathscr {I}}(d^*,w)\ {\text { for\ all\ }} w\in W$$

with a strict inequality for some $w\in W$ . We therefore only need to check what happens for $w\in W^*\setminus W$ . The compactification of a partition consists of adding one point $w^*$ that is in the complement of all $p^*\in {\mathcal{F}}^*$ ; so $W^*= W\cup w^*$ . If ${\mathscr{I}}(c^*,w^*)=\infty$ , then clearly, $d^*$ weakly dominates $c^*$ . So, we assume ${\mathscr{I}}(c^*,w^*)\lt \infty$ . Because ${\mathscr{I}}(d^*,w^*)$ and ${\mathscr{I}}(d^*,w)$ differ by a single term for any $w\in W$ and ${\mathscr{I}}(d^*,w)\lt \infty$ for some $w\in W$ by the assumption of weak dominance, we have ${\mathscr{I}}(d^*,w^*)\lt \infty$ . Now, consider

$${\mathscr {I}}(c^*,w^*)-{\mathscr {I}}(d^*,w^*)=\sum _{m\,=1}^\infty a_m( \varphi (d_m)-\varphi '(d_m)d_m)-\sum _{m\,=1}^\infty a_m(\varphi (c_m)-\varphi '(c_m)c_m),$$

which we claim is greater than or equal to $0$ . Indeed, assume toward a contradiction that

$$\sum _{m=1}^\infty a_m(\varphi (d_m)-\varphi '(d_m)d_m )\lt \sum _{m=1}^\infty a_m( \varphi (c_m)-\varphi '(c_m)c_m).$$

Then, because $d_n\to 0$ as $\sum _{n\,=1}^\infty d_n\leq 1$ , $c_n\to 0$ as $\sum _{n\,=1}^\infty c_n\leq 1$ , and $\varphi '(0)=\lim _{x\to 0}\varphi '(x)=0$ (recall Remark 3.2), we have that $\varphi '(d_n)-\varphi '(c_n)\to 0$ ; and so we can find a $K$ such that

$$|\varphi '(d_n)-\varphi '(c_n)|\lt |\sum _{m=1}^\infty a_m(\varphi (d_m)-\varphi '(d_m)d_m)-\sum _{m=1}^\infty a_m(\varphi (c_m)-\varphi '(c_m)c_m)|$$

for $n\geq K$ . Thus, for any $n\geq K$ and any $w_n\in p_n$ ,

$$ {\mathscr{I}}(c,{w_n})-{\mathscr{I}}(d,w_n)=\sum _{m=1}^\infty a_m(\varphi (d_m)-\varphi '(d_m)d_m)\\ -\sum _{m=1}^\infty a_m(\varphi (c_m)-\varphi '(c_m)c_m)+\varphi '(d_n)-\varphi '(c_n)\lt 0, $$

contradicting that $d$ weakly dominates $c$ . So indeed, $d^*$ weakly dominates $c^*$ . □

Proof The result follows from Corollary 4.22, Lemma 4.25, and the fact that ${\mathscr{I}}(c^*,\cdot )$ is bounded on $W^*$ for each coherent credence function $c$ . To see the latter, note that because $c$ is coherent, it follows that $\sum _{i\,=1}^\infty c_i=\sum _{i\,=1}^\infty c^*_i\leq 1$ . For $w\in W$ such that $w\in p_i$ , recalling that $\varphi (0)=0$ (Remark 3.2),

$$ {\mathscr{I}}(c^*,w)=a_i{\mathfrak{d}}(1,c^*_i)+\sum _{j\ne i}a_j{\mathfrak{d}}(0,c^*_j)=a_i{\mathfrak{d}}(1,c^*_i)+\sum _{j\ne i}a_j(c^*_j\varphi '(c^*_j)-\varphi (c^*_j))\\ \leq C+D\sum _{j}c^*_j\leq C+D $$

for some constants $C,D$ independent of $c^*$ . Similarly, as seen in the proof of Lemma 4.25, $W^*\setminus W=\{w^*\}$ , where

$${\mathscr {I}}(c^*,w^*)=\sum _{j\,=1}^\infty a_j {\mathfrak {d}}(0,c_j^*)=\sum _{j\,=1}^\infty a_j(c_j^*\varphi '(c_j^*)-\varphi (c_j^*))\leq C$$

for some constant $C$ independent of $c^*$ or $w$ . It follows that i) all coherent credence functions have finite expected inaccuracy, and ii) ${\mathscr{I}}(c,w)\lt \infty$ for $c\in {\mathcal{C}}$ and $w\in W$ . Thus, Lemma 4.25 and Corollary 4.22 establish the result. □

Proof Let ${\mathscr{I}}_{n}(c',w):=\sum _{i\,=1}^na_i{\mathfrak{d}}(v_w(p_i),c'(p_i))$ for each $n\in {\mathbb{N}}$ , $w\in W$ , and credence function $c'$ on ${\mathcal{F}}$ . Consider a credence function $d\ne c$ . Define

$$T^n=\{(v_w(p_1),\ldots,v_w(p_n)): {\mathscr {I}}_k(c,w)\lt {\mathscr {I}}_k(d,w)\ {\text { for\ some\ } {\it{k}}\geq {\it{n}},}\, w\in W\}$$

and $T=\{e\}\cup{\bigcup _{n\,=1}^\infty } T^n$ , where $e$ is the empty sequence. For each $s,t\in T$ , we set $s\lt t$ if and only if $s$ is an initial sequence of $t$ , and we set the height of $t\in T$ to be the length of the tuple. Then $T$ is a binary tree.

We claim $T$ is infinite. Fix $n\in {\mathbb{N}}$ . Then there is a $t\in T$ with height $n$ if and only if $T^n\ne \varnothing$ if and only if ${\mathscr{I}}_k(c,w)\lt {\mathscr{I}}_k(d,w)$ for some $k\geq n$ and $w\in W$ . Let $k$ be the maximum of $n$ and the smallest $i$ such that $c(p_i)\ne d(p_i)$ . Then, because $c$ restricted to any subset of ${\mathcal{F}}$ is coherent, by Theorem 2.9, ${\mathscr{I}}_k(c,w')\lt {\mathscr{I}}_k(d,w')$ for some $w'\in W$ , and so $(v_{w'}(p_1),\ldots,v_{w'}(p_n))\in T^n$ .

By Konig’s lemma (see, e.g., Hrbacek and Jech Reference Hrbacek and Jech1999, sec. 12.3), there exists an infinite branch

$${\mathcal {B}}=\bigcup _{n\,=1}^\infty \{ (v_{w_n}(p_1),\ldots,v_{w_n}(p_n))\}$$

through $T$ , where

$$(v_{w_n}(p_1),\ldots,v_{w_n}(p_n))\lt (v_{w_m}(p_1),\ldots,v_{w_m}(p_m))$$

whenever $n\lt m$ . For each $i$ , let $p_i^*=p_i$ if $v_{w_i}(p_i)=1$ and $p_i^*=p_i^c$ if $v_{w_i}(p_i)=0$ . Then $w_n\in \bigcap _{i\,=1}^np_i^*$ because $v_{w_i}(p_i)=1$ if and only if $v_{w_n}(p_i)=1$ for $i\lt n$ as $(v_{w_i}(p_1),\ldots, v_{w_i}(p_i))\lt (v_{w_n}(p_1),\ldots,v_{w_n}(p_n))$ . Thus, $\bigcap _{i\,=1}^np_i^*\ne \varnothing$ for each $n$ , and so by compactness, there is some $w\in \bigcap _{i\,=1}^\infty p_i^*$ . Then

$$(v_w(p_1),\ldots,v_w(p_n))=(v_{w_n}(p_1),\ldots,v_{w_n}(p_n))\in T^n$$

for each $n \in {\mathbb{N}}$ . By the definition of $T^n$ , for each $n\in {\mathbb{N}}$ , we have

$${\mathscr {I}}_{k_n}(c,w)\lt {\mathscr {I}}_{k_n}(d,w)$$

for some $k_n\geq n$ . Sending $n$ to infinity, ${\mathscr{I}}(c,w)\leq {\mathscr{I}}(d,w)$ , and thus $d$ does not strongly dominate $c$ . □

Proof Let ${\mathscr{I}}(c,w)=B_{\varphi,\mu }(v_w,c)$ . We write $B$ for $B_{\varphi,\mu }$ . Let $S$ be the set of nonnegative ${\mathcal{A}}$ -measurable functions on ${\mathcal{F}}$ . Let $E\subseteq S$ be the set of $\mu$ -coherent $\mu$ -credence functions over ${\mathcal{F}}$ . Then $E$ is convex. Let $c$ be a $\mu$ -incoherent $\mu$ -credence function. Because $\mu$ is finite and ${\mathfrak{d}}$ is bounded,

$$B(E,c)\lt \infty .$$

Thus, we can apply Theorem A.5 to get a $\pi _c\in S$ such that

(10) $$ B(s,c)\geq B(E,c)+B(s,\pi _c)\text{\ for \ every\ }s\in E. $$

In particular, (10) holds when $s$ is the omniscient credence function at world $w$ for each $w$ , so we obtain

(11) $$ {\mathscr{I}}(c,w)\geq B(E,c)+{\mathscr{I}}(\pi _c,w) $$

for all $w$ , where all numbers in (11) are finite. We show that $\pi _c$ is in fact a $\mu$ -coherent $\mu$ -credence function. It suffices to show that $\pi _c$ is $\mu$ -a.e. equal to a coherent credence function on ${\mathcal{F}}$ (because $\pi _c\in S$ , it is ${\mathcal{A}}$ -measurable). To do so, we prove the following claim: $E$ is closed under loose convergence in $\mu$ -measure.