Accuracy, probabilism and Bayesian update in infinite domains

Abstract

Scoring rules measure the accuracy or epistemic utility of a credence assignment. A significant literature uses plausible conditions on scoring rules on finite sample spaces to argue for both probabilism—the doctrine that credences ought to satisfy the axioms of probabilism—and for the optimality of Bayesian update as a response to evidence. I prove a number of formal results regarding scoring rules on infinite sample spaces that impact the extension of these arguments to infinite sample spaces. A common condition in the arguments for probabilism and Bayesian update is strict propriety: that according to each probabilistic credence, the expected accuracy of any other credence is worse. Much of the discussion needs to divide depending on whether we require finite or countable additivity of our probabilities. I show that in a number of natural infinite finitely additive cases, there simply do not exist strictly proper scoring rules, and the prospects for arguments for probabilism and Bayesian update are limited. In many natural infinite countably additive cases, on the other hand, there do exist strictly proper scoring rules that are continuous on the probabilities, and which support arguments for Bayesian update, but which do not support arguments for probabilism. There may be more hope for accuracy-based arguments if we drop the assumption that scores are extended-real-valued. I sketch a framework for scoring rules whose values are nets of extended reals, and show the existence of a strictly proper net-valued scoring rules in all infinite cases, both for f.a. and c.a. probabilities. These can be used in an argument for Bayesian update, but it is not at present known what is to be said about probabilism in this case.

Appendix: Proofs

Proof of Proposition 1

To obtain a contradiction, suppose that s is continuous at a fixed $p\in {\mathcal {P}}$. For $x \in [M,\infty )$ and $n>0$, let $B_n(x) = (x-1/n,x+1/n)$ and let $B_n(\infty )=(\max (M,n),\infty ]$. Then $B_n(x)$ is a neighborhood of x and $\bigcap _n B_n(x) = \{ x \}$. Let

$$\begin{aligned} V_{\omega ,n}=\{ s' \in [M,\infty ]^\Omega : s'(\omega ) \in B_n(s(p)(\omega )) \}. \end{aligned}$$

This is a neighborhood of s(p) in in the product topology on $[M,\infty ]^\Omega$. Note that

$$\begin{aligned} \bigcap _{(\omega ,n) \in \Omega \times {\mathbb {N}}} V_{\omega ,n} = \{ s(p) \}. \end{aligned}$$

For a finite subset F of ${\mathbb {P}}\Omega$, a credence c and an $r>0$, let

$$\begin{aligned} B(c,F,r) = \{ c' \in {\mathcal {C}} : \forall A\in F(|c(A)-c'(A)| < r) \}. \end{aligned}$$

The collection of the B(c, F, r) forms a neighborhood basis for ${\mathcal {C}}$ with the product topology.

Since s is continuous at $p\in {\mathcal {P}}$, for each $(\omega ,n) \in \Omega \times {\mathbb {N}}$, using the Axiom of Choice, select a finite $F_{\omega ,n}\subseteq {\mathbb {P}}\Omega$ and an $r_{\omega ,n}>0$ such that if $W_{\omega ,n}=B(p,F_{\omega ,n},r_{\omega ,n})$, then $s[W_{\omega ,n}] \subseteq V_{\omega ,n}$. Observe that:

$$\begin{aligned} s\left[ \bigcap _{(\omega ,n)\in \Omega \times {\mathbb {N}}} W_{\omega ,n}\right] \subseteq \bigcap _{(\omega ,n)\in \Omega \times {\mathbb {N}}} s[W_{\omega ,n}] \subseteq \bigcap _{(\omega ,n)\in \Omega \times {\mathbb {N}}} V_{\omega ,n} = \{ s(p) \}. \end{aligned}$$

Since s is ${\mathcal {P}}$-distinguishing, nothing in ${\mathcal {C}}-{\mathcal {P}}$ can have the same score as p. Hence, $W=\bigcap _{\omega ,n} W_{\omega ,n}$ must be a subset of ${\mathcal {P}}\subseteq {\mathcal {P}}_{\textrm{f}}$. But we shall see this is impossible. For let $F = \bigcup _{\omega ,n} F_{\omega ,n}$. This is a union of $|\Omega \times {\mathbb {N}}|=|\Omega |$ many finite sets and hence $|F|\le |\Omega |$ (here we used the Axiom of Choice twice). Thus by Cantor’s Theorem, $|F|<|{\mathbb {P}}\Omega |$ and since F is a subset of ${\mathbb {P}}\Omega$, it must be a proper subset of it. Now, $p\in W$, and any credence c such that $c(A)=p(A)$ for all $A\in F$ will be a member of each of the $W_{\omega ,n}$ and hence of W. Fix any $B\notin F$. Let c be any credence function such that $c(A)=p(A)$ for all $A\in F$, but such that c(B) is chosen so as to be different from $1-c(\Omega -B)$ (if $\Omega -B$ is in F, we have no choice as to the value of $c(\Omega -B)$, and if $\Omega -B \notin F$, we can let $c(\Omega -B)$ be whatever we like). Then c is not finitely additive, but is in W, a contradiction. $\square$

We now show that ${\text {Sph}}^*$ from Sect. 2 is strictly proper. Suppose first that p and q are probabilities. Restricted to the probabilities, ${\text {Sph}}^*$ is just the spherical scoring rule, which is well-known to be strictly proper [Gneiting and Raftery (2007, Sect. 4.1)], but a proof is given for completeness. We have:

$$\begin{aligned} E_p {\text {Sph}}^*(q) = -\frac{\Vert {\hat{p}}{\hat{q}} \Vert _1}{\Vert {\hat{q}} \Vert _2} \end{aligned}$$

and

$$\begin{aligned} E_p {\text {Sph}}^*(p) = -\frac{\sum _n {\hat{p}}_n^2}{\Vert {\hat{p}} \Vert _2} = -\frac{ \Vert {\hat{p}} \Vert _2^2 }{ \Vert {\hat{p}} \Vert _2 } = -\Vert {\hat{p}} \Vert _2. \end{aligned}$$

The Cauchy-Schwarz inequality says that $\Vert {\hat{p}}{\hat{q}} \Vert _1 \le \Vert {\hat{p}} \Vert _2 \Vert {\hat{q}} \Vert _2$, with equality if and only if one of ${\hat{p}}$ and ${\hat{q}}$ is a scalar multiple of the other. For probabilities p and q, the latter condition is met only if $p=q$. Thus:

$$\begin{aligned} E_p {\text {Sph}}^*(q) \ge -\frac{\Vert {\hat{p}} \Vert _2 \Vert {\hat{q}} \Vert _2}{\Vert {\hat{q}} \Vert _2} = -\Vert {\hat{p}} \Vert _2 = E_p {\text {Sph}}^*(p), \end{aligned}$$

with equality only when $p=q$.

Now suppose $c\notin {\mathcal {P}}_{\textrm{c}}{}$ and $p\in {\mathcal {P}}_{\textrm{c}}{}$. We need to show that $E_p {\text {Sph}}^*(p) < E_p {\text {Sph}}^*(c)$. Now

$$\begin{aligned} \Vert {\text {Sph}}^*(c) \Vert _2^2 = \sum _{n=0}^\infty \frac{1}{(2(n+1))^{2}} = \frac{\pi }{24}, \end{aligned}$$

because of the Euler series formula $\sum _{n=1}^\infty n^{-2} = \pi /6$. By Cauchy-Schwarz again and as $(\pi /24)^{1/2} < 1$,

$$\begin{aligned} -E_p {\text {Sph}}^*(c)&= \sum _{n=0}^\infty {{\hat{p}}(n)}{{\text {Sph}}^*(c)(n)} \\&\le \Vert {\text {Sph}}^*(c) \Vert _2 \Vert {\hat{p}} \Vert _2 < \Vert {\hat{p}} \Vert _2 = -E_p {\text {Sph}}^*(p), \end{aligned}$$

and so $E_p {\text {Sph}}^*(p) < E_p {\text {Sph}}^*(c)$.

Now for $u \in \ell ^1(\Omega )$, let

$$\begin{aligned} \Vert u \Vert _* = \sup _{A\in {\mathbb {P}}\Omega } \left| \sum _{\omega \in \Omega } u(\omega ) \right| . \end{aligned}$$

Two norms $\Vert \cdot \Vert _A$ and $\Vert \cdot \Vert _B$ on a vector space are said to be equivalent if there is a positive real constant c such that for every u in the space $\Vert u \Vert _A \le c\Vert u \Vert _B$ and $\Vert u \Vert _B \le c\Vert u \Vert _A$. Credences are members of $\ell ^\infty ({\mathbb {P}}\Omega )$, the space of bounded real-valued functions on ${\mathbb {P}}\Omega$ with norm given by the supremum of the absolute value.

Lemma 1

The function $\Vert \cdot \Vert _*$ on $\ell ^1(\Omega )$ is a vector space norm such that $\Vert u \Vert _* \le \Vert u \Vert _1 \le 2 \Vert u \Vert _*$ for all u. Moreover, for any $p\in {\mathcal {P}}_{\textrm{c}}$, we have $\Vert {\hat{p}} \Vert _* = \Vert p \Vert _{\ell ^\infty ({\mathbb {P}}\Omega )}$.

Proof

That $\Vert \cdot \Vert _*$ is a vector space norm is easy to check using the triangle inequality and the fact that $\sup _x (f(x)+g(x)) \le \sup _x f(x) + \sup _x g(x)$.

Furthermore,

$$\begin{aligned} \Vert u \Vert _* = \sup _{A\in {\mathbb {P}}\Omega } \left| \sum _{\omega \in A} u(\omega ) \right| \le \sup _{A\in {\mathbb {P}}\Omega } \sum _{\omega \in A} |u(\omega )| = \sum _{\omega \in \Omega } |u(\omega )| = \Vert u \Vert _1. \end{aligned}$$

Next, given $u \in \ell ^1(\Omega )$, let $B = \{ \omega : u(\omega )\ge 0 \}$ and $C = \Omega -B$. Then

$$\begin{aligned} \Vert u \Vert _1&= \sum _{\omega \in B} u(\omega ) - \sum _{\omega \in C} u(\omega ) = \left| \sum _{\omega \in B} u(\omega ) \right| + \left| \sum _{\omega \in C} u(\omega )\right| \\&\le 2\max _{A \in \{B,C\}} \left| \sum _{\omega \in A} u(\omega ) \right| \le 2\Vert u \Vert _*. \end{aligned}$$

Finally, if $p\in {\mathcal {P}}_{\textrm{c}}$:

$$\begin{aligned} \Vert p \Vert _{\ell ^\infty ({\mathbb {P}}\Omega )} = \sup _{A\subseteq \Omega } p(A) = \sup _{A\subseteq \Omega } \left| \sum _{\omega \in A} {\hat{p}}(\omega ) \right| = \Vert {\hat{p}} \Vert _*. \end{aligned}$$

$\square$

Our observations related to the continuity of ${\text {Sph}}^*$ on ${\mathcal {P}}_{\textrm{c}}$ in Sect. 2.1 together with Lemma 1 yield:

Lemma 2

The scoring rule ${\text {Sph}}^*$ is continuous on ${\mathcal {P}}_{\textrm{c}}$ in the $\ell ^\infty ({\mathbb {P}}\Omega )$ topology with respect to the ${\mathcal {P}}_{\textrm{c}}$-weak topology on $[M,\infty ]^\Omega$.

To prove Proposition 2 we need one more crucial result.

Lemma 3

The sets ${\mathcal {P}}_{\textrm{c}}$ and ${\mathcal {P}}_{\textrm{f}}$ are closed subsets of $\ell ^\infty ({\mathbb {P}}\Omega )$.

Proof of Lemma 3

We need to show that every member u of $\ell ^\infty ({\mathbb {P}}\Omega )-{\mathcal {P}}_{\textrm{c}}$ has a neighborhood that doesn’t intersect ${\mathcal {P}}_{\textrm{c}}$ and every member u of $\ell ^\infty ({\mathbb {P}}\Omega )-{\mathcal {P}}_{\textrm{f}}$ has a neighborhood that doesn’t intersect ${\mathcal {P}}_{\textrm{f}}$.

If $u\notin {\mathcal {C}}$, then there is an $A\subseteq \Omega$ such that $u(A)<0$ or $u(A)>1$. Then

$$\begin{aligned} \Vert u-p \Vert _{\ell ^\infty ({\mathbb {P}}\Omega )} \ge |u(A)-p(A)| \end{aligned}$$

for any $p\in {\mathcal {P}}_{\textrm{f}}$. Moreover, we have $p(A) \in [0,1]$, so if $u(A)<0$, the right hand side is at least $-u(A)$ and if $u(A)>1$, the right hand side is at least $u(A)-1$. In the former case, a ball of radius $-u(A)$ around u does not intersect ${\mathcal {P}}_{\textrm{f}}$ and in the latter, one of radius $u(A)-1$ does not intersect ${\mathcal {P}}_{\textrm{f}}$.

Next suppose that $u\in {\mathcal {C}}-{\mathcal {P}}_{\textrm{f}}{}$. Since we’ve assumed credences take values between 0 and 1, there are two possible ways for u not to be a f.a. probability: either the total probability or the finite additivity axiom fails for u.

If the total probability axiom fails, then $u(\Omega )<1$. But for any $p\in {\mathcal {P}}_{\textrm{f}}{}$ we have $p(\Omega )=1$, so $\Vert u-p \Vert _{\ell ^\infty ({\mathbb {P}}\Omega )} \ge |p(\Omega )-u(\Omega )|=1-u(\Omega )$, and so a ball of radius $1-u(\Omega )$ around u does not intersect ${\mathcal {P}}_{\textrm{f}}$.

Suppose now that finite additivity fails, so $u(A\cup B)\ne u(A)+u(B)$ for some disjoint A and B. Let $\varepsilon = |u(A)+u(B)-u(A\cup B)|>0$. Suppose that $p\in {\mathcal {P}}_{\textrm{f}}$. I will show that $\Vert u-p \Vert _{\ell ^\infty ({\mathbb {P}}\Omega )} \ge \varepsilon /3$ for any $p\in {\mathcal {P}}_{\textrm{f}}$, from which it follows that a ball of radius $\varepsilon /3$ around u does not intersect ${\mathcal {P}}_{\textrm{f}}$. To obtain a contradiction, suppose $\Vert u-p \Vert _{\ell ^\infty ({\mathbb {P}}\Omega )} < \varepsilon /3$. Then $|u(D)-p(D)| < \varepsilon /3$ for every event D. Thus,

$$\begin{aligned} \varepsilon = |u(A)+u(B)-u(A\cup B)| < 3(\varepsilon /3) + |p(A)+p(B)-p(A\cup B)| = \varepsilon , \end{aligned}$$

where the inequality follows from the triangle inequality and the last equality from the finite additivity of p. And that’s a contradiction.

So, far we have shown that any $u\notin {\mathcal {P}}_{\textrm{f}}$ has a neighborhood that doesn’t intersect ${\mathcal {P}}_{\textrm{f}}$, and hence doesn’t intersect ${\mathcal {P}}_{\textrm{c}}$. In particular, we’ve shown that ${\mathcal {P}}_{\textrm{f}}$ is closed.

Finally, suppose that $u\in {\mathcal {P}}_{\textrm{f}}{}-{\mathcal {P}}_{\textrm{c}}{}$ so that we can show that there is a neighborhood of u that doesn’t intersect ${\mathcal {P}}_{\textrm{c}}$. Finite but not countable additivity must hold for u. Thus we have disjoint $A_i$ such that

$$\begin{aligned} \varepsilon = \left| u\left( \bigcup _{i=1}^\infty A_i\right) - \sum _{i=1}^\infty u(A_i) \right| > 0. \end{aligned}$$

But we have:

$$\begin{aligned} \sum _{i=1}^\infty u(A_i)=\lim _{n\rightarrow \infty } \sum _{i=1}^n u(A_i) = \lim _{n\rightarrow \infty } u\left( \bigcup _{i=1}^n A_i\right) \end{aligned}$$

by finite additivity. Thus:

$$\begin{aligned} \varepsilon = \left| u\left( \bigcup _{i=1}^\infty A_i\right) - \lim _{n\rightarrow \infty } u\left( \bigcup _{i=1}^n A_i\right) \right| . \end{aligned}$$

To obtain a contradiction, suppose $\Vert u-p \Vert _{\ell ^\infty ({\mathbb {P}}\Omega )} < \varepsilon /2$ for some $p\in {\mathcal {P}}_{\textrm{c}}$. Then $|u(D)-p(D)|<\varepsilon /3$ for any D and hence by the triangle inequality again:

$$\begin{aligned} \varepsilon \le \left| p\left( \bigcup _{i=1}^\infty A_i\right) - \lim _{n\rightarrow \infty } p\left( \bigcup _{i=1}^n A_i\right) \right| +2\varepsilon /3 = 2\varepsilon /3, \end{aligned}$$

by the countable additivity of p, and contradicting the positivity of $\varepsilon$. Hence, again, a ball of radius $\varepsilon /3$ around u does not intersect ${\mathcal {P}}_{\textrm{c}}$, and our proof of the closedness of ${\mathcal {P}}_{\textrm{c}}$ is complete. $\square$

Proof of Proposition 2

By the Dugundji extension theorem Dugundji (1951, Theorem 4.1), together with the closedness (Lemma 3) and convexity of ${\mathcal {P}}$, and the local convexity of ${\mathcal {P}}$-weak topology, there is an $\ell ^\infty ({\mathbb {P}}\Omega )$-continuous function $\psi :{\mathcal {C}}\rightarrow {\mathcal {P}}$ whose restriction to ${\mathcal {P}}$ is the identity function.

Define $s_1(c)=s(\psi (c))$. Since this agrees with s on ${\mathcal {P}}$ and has the same range as s while s is proper, this is a proper scoring rule on ${\mathcal {C}}$.

Now, let $\delta (c) = \inf _{p \in {\mathcal {P}}} \Vert c-p \Vert _{\ell ^\infty ({\mathbb {P}}\Omega )}$ be the distance from c to ${\mathcal {P}}$. By the triangle inequality, this is a continuous function, and since ${\mathcal {P}}$ is closed (Lemma 3), $\delta (c_0)>0$ for our $c_0 \in {\mathcal {C}}-{\mathcal {P}}$. Let $\phi (c)=\max (1, \delta (c)/\delta (c_0))$ and define

$$\begin{aligned} {\bar{s}}(c)(\omega ) = \phi (c) f_0(\omega ) + (1-\phi (c)) s_1(c)(\omega ) \end{aligned}$$

for any credence c. Note that s, $s_1$ and ${\bar{s}}$ all agree on ${\mathcal {P}}$ (since $\phi (p)=0$ for $p\in {\mathcal {P}}$). Then if $c\notin {\mathcal {P}}$, we have $\phi (c)>0$ and so for any $p\in {\mathcal {P}}$:

$$\begin{aligned} E_p {\bar{s}}(c)&= \phi (c) E_p f_0 + (1-\phi (c)) E_p s_1(c) \\&> \phi (c) E_p s(p) + (1-\phi (c)) E_p s_1(c) \\&\ge \phi (c) E_p s(p) + (1-\phi (c)) E_p s_1(p) \\&=\phi (c) E_p s_1(p) + (1-\phi (c)) E_p s_1(p) \\&= E_p s_1(p) = E_p {\bar{s}}(p) \end{aligned}$$

by our assumption on $f_0$ and the propriety of $s_1$. Hence $E_p$ is ${\mathcal {P}}$-strictly proper. Moreover, it is continuous because $\delta$ is continuous. And $\phi (c_0)=1$ so ${\bar{s}}(c_0)=f_0$. $\square$

For the proof of Proposition 3, we will need the following.

Lemma 4

Assume the Axiom of Choice. Let p and q be any f.a. probabilities on ${\mathbb {P}}\Omega$ where $\kappa =|\Omega |$ is infinite. If p and q differ on any event, they differ on $2^\kappa$ events.

Proof

Suppose A is such that $p(A) \ne q(A)$. Either A or its complement $A^c$ has cardinality $\kappa$, and if p and q differ on A, they differ on $A^c$ by finite additivity. Thus, replacing A with $A^c$ if necessary, assume $|A^c|=\kappa$. Let $F = \{ B \subseteq A^c : p(B) = q(B) \}$. If $|F| < 2^\kappa$, then $|{\mathbb {P}}(A^c) - F| = 2^\kappa$ (here we use Choice and the fact that $|{\mathbb {P}}(A^c)|=2^\kappa$), and so there are $2^\kappa$ many events on which p and q differ. Thus suppose $|F| = 2^\kappa$. But then any $B\in F$ is disjoint from A and so we have

$$\begin{aligned} p(A\cup B)=p(A)+p(B)=p(A)+q(B)\ne q(A)+q(B)=q(A\cup B) \end{aligned}$$

and so p and q differ on the event $A\cup B$, and there are $2^\kappa$ such events. $\square$

Proof of Proposition 3

Fix $p\in {\mathcal {P}}_{\textrm{c}}$ such that $p(\{\omega \})>0$ for all $\omega \in \Omega$. If we have quasi-strict propriety, then $E_p(s(p))<E_p(s(c))$ for any non-probability c, and hence $\infty > E_p(s(p)) = \sum _\omega p(\{\omega \}) s(p)$. Since p is non-zero on every singleton, it follows that $s(p)(\omega )<\infty$ for every $\omega$. Since (inaccuracy) scores cannot equal $-\infty$, we must have $s(p)(\omega )$ finite for every $\omega$.

Now, a sum of uncountably many non-zero values is either undefined or infinite. Thus for any fixed $\omega$, the summand $s_A(p(A),1_A(\omega ))$ is zero except for at most countably many $A\in {\mathbb {P}}\Omega$. Let

$$\begin{aligned} N_p = \{ A \in {\mathbb {P}}\Omega : \exists \omega \in \Omega (s_A(p(A),\omega )\ne 0) \}. \end{aligned}$$

Let $\kappa = |\Omega |$. Then $N_p$ is the union of $\kappa$ sets, each of which is countable, and $\kappa$ is infinite, so $|N_p| \le \aleph _0 \times \kappa \le \kappa \times \kappa = \kappa$, where we used the Axiom of Choice in both inequalities.

Let q be any c.a. probability other than p such that $q(\{\omega \})>0$ for all $\omega$. Let $N=N_p\cup N_q$. Then for any $A\notin N$ and $\omega \in \Omega$, we have $s_A(p,1_A(\omega ))=0=s_A(q,1_A(\omega ))$ and $|N| \le \kappa$, as $\kappa$ is infinite.

By Lemma 4, p and q differ on $2^\kappa$ events. Since $|N| \le \kappa <2^\kappa$, by the pigeonhole principle choose a $C\notin N$ for which $p(C)\ne q(C)$. Let $c(A)=p(A)$ if $A\ne C$ and $c(C)=q(C)$. Then finite additivity is violated by c:

$$\begin{aligned} c(C)+c(C^c)=q(C)+p(C^c)=q(C)+1-p(C)\ne q(C)+1-q(C)=1. \end{aligned}$$

But

$$\begin{aligned} s(c)(\omega ) = \sum _{A\in {\mathbb {P}}\Omega } s_A(c(A),1_A(\omega )) = \sum _{A\in {\mathbb {P}}\Omega } s_A(p(A),1_A(\omega )) = s(p)(\omega ), \end{aligned}$$

for all $\omega$, since $c(A)=p(A)$ except when $A=C$, in which case $s_A(c(A),1_A(\omega ))=s_A(q(A),1_A(\omega ))=0=s_A(p(A),1_A(\omega ))$ since $C\notin N$. Therefore, $s(c)=s(p)$ everywhere, and hence we cannot have $E_p s(c) > E_p s(p)$ as would be necessary for either f.a. or c.a. quasi-propriety since c is not a f.a. probability. $\square$

To prove Proposition 4, we need the following improvement on Cantor’s Theorem, whose proof was essentially given to me by my colleague Daniel Herden.

Lemma 5

If X is a set, then there is no function $f:{\mathbb {P}}X\rightarrow X$ such that $f(A)\ne f(B)$ whenever $A\subset B$.

Proof

Suppose we have such a function f. Let ON be the class of ordinals. Define $F:ON\rightarrow X$ by transfinite induction: $F(0)=f(\varnothing )$ and $F(\alpha )=f(\{F(\beta ):\beta <\alpha \})$ whenever $\alpha$ is a successor or limit ordinal.

I prove by transfinite induction that F is one-to-one on $\alpha$ for any ordinal $\alpha$. This is trivially true for $\alpha =0$.

Suppose F is one-to-one on $\beta$ for all $\beta <\alpha$. If $\alpha$ is a limit ordinal, then it immediately follows that F is one-to-one on $\alpha$ as well.

If $\alpha =\beta +1$ for some ordinal $\beta$, I claim that it is one-to-one as well. For given that F is one-to-one on $\beta$, the only possible failure of injectivity would be if $F(\beta )=F(\gamma )$ for some $\gamma <\beta$. Suppose that happens. Let

$$\begin{aligned} H_\delta = \{ F(\varepsilon ) : \varepsilon < \delta \}. \end{aligned}$$

Note that $F(\beta )=f(H_\beta )$ and $F(\gamma )=f(H_\gamma )$. Furthermore, $H_\gamma \subset H_\beta$ since F is one-to-one on $\beta$. Thus, $F(\gamma )=f(H_\gamma ) \ne f(H_\beta )=F(\beta )$, a contradiction.

So, by transfinite induction, F is one-to-one on ON, and hence embeds ON in the set X, which is impossible by Burali-Forti. $\square$

Proof of Proposition 4

Let s be a strictly truth-directed scoring rule defined for all extreme credences. Fix $\omega _0 \in \Omega$. Given a subset U of ${\mathbb {P}}\Omega$, let $c_U$ be the extreme credence function that is correct at all and only the members of U. Thus, $c_U(A)$ is 1 if both $\omega _0\in A$ and $A\in U$ or both $\omega _0\notin A$ and $A\notin U$, and is 0 otherwise. Note that if $U\subset V$, then $c_V$ is strictly truer at $\omega _0$ than $c_U$, and hence $s(c_U)(\omega _0)>s(c_V)(\omega _0)$. Let $h(U)=s(c_U)(\omega _0)$. Thus, h is a function from ${\mathbb {P}}{\mathbb {P}}\Omega$ to $[M,\infty ]$ such that $h(A)\ne h(B)$ whenever $A\subset B$. But if $\Omega$ is infinite, then by the Axiom of Countable Choice, $\aleph _0 \le |\Omega |$, and $|[M,\infty ]|=2^{\aleph _0} \le |{\mathbb {P}}\Omega |$, and so there is a one-to-one function g from $[M,\infty ]$ to ${\mathbb {P}}\Omega$. Letting $f=g\circ h$ and $X={\mathbb {P}}\Omega$, we get a function whose existence contradicts Lemma 5. $\square$

For the proof of Proposition 5, we need this easy fact.

Lemma 6

Assume the Axiom of Countable Choice. If ${\mathcal {F}}$ is an infinite $\sigma$-algebra on $\Omega$, then there is a countably infinite partition of $\Omega$ by non-empty members of ${\mathcal {F}}$.

Proof

Let ${\mathcal {F}}_0 = {\mathcal {F}}$ and $\Omega _0 = \Omega$. Given an infinite $\sigma$-algebra ${\mathcal {F}}_{n}$ on $\Omega _{n}$, I claim there is a a non-empty member $\Omega _{n+1}$ of ${\mathcal {F}}_n$ such that ${\mathcal {F}}_n \cap {\mathbb {P}}(\Omega _n-\Omega _{n+1})$ is infinite. To see this, given the infinitude of ${\mathcal {F}}_n$, let B be any member of ${\mathcal {F}}_n$ such that neither B nor $\Omega _n-B$ is empty. Then every member of ${\mathcal {F}}_n$ is the union of a member of ${\mathcal {F}}_n \cap {\mathbb {P}}B$ and a member of ${\mathcal {F}}_n \cap {\mathbb {P}}(\Omega _n-B)$. Hence at least one of these two sets is infinite. If ${\mathcal {F}}_n \cap {\mathbb {P}}B$ is infinite, let $\Omega _{n+1}=\Omega _n-B$, and otherwise let $\Omega _{n+1}=B$. Let ${\mathcal {F}}_{n+1} = {\mathcal {F}}_n \cap {\mathbb {P}}(\Omega _n-\Omega _{n+1})$. Note that getting such a sequence of ${\mathcal {F}}_n$ and $\Omega _n$ requires the Axiom of Countable Choice since the choice of B was not determinate.

Let $A_n = \Omega _{n-1} - \Omega _{n}$ for $n\ge 1$. Then $A_1,A_2,...$ are non-empty pairwise disjoint members of ${\mathcal {F}}$. If their union is $\Omega$, we are done, and otherwise just add $\Omega - \bigcup _{n=1}^\infty \Omega _n$ to the sequence. $\square$

Sketch of Proof of Proposition 5

Let $A_1,A_2,\ldots$ be a countably infinite partition of $\Omega$ by members of ${\mathcal {F}}$ by Lemma 6. Let $B_1,B_2,\ldots$ be an algebra that $\sigma$-generates ${\mathcal {F}}$ (we assumed ${\mathcal {F}}$ is countably generated, and then the algebra generated by the generating set will also be countable). Let $C_{mn} = A_{m} \cap B_n$. Given an event A, let

$$\begin{aligned} b_A(p)(\omega ) = -(1_A(\omega )-p(A))^2 \end{aligned}$$

be the A-Brier score of a (c.a.) probability p. It is elementary to check that

$$\begin{aligned} E_p b_A(q) = -p(A)(1-q(A))^2 - (1-p(A))(q(A))^2 \ge E_p b_A(p) \end{aligned}$$

with equality if and only if $p(A)=q(A)$. In particular, $b_A$ is a proper score.

Next, define

$$\begin{aligned} b(p) = \sum _{m=1}^\infty \sum _{n=1}^\infty 2^{-m-n} b_{C_{mn}}(p) \end{aligned}$$

The scoring rule b is continuous on ${\mathcal {P}}_{\textrm{c}}({\mathcal {F}})$ in our topology. It is proper being the sum of proper scoring rules.

Moreover, b is strictly proper on the probabilities. For suppose that p and q are probabilities that are not identical.

I now claim that p and q differ on $B_n$ for some n. For suppose p and q are equal on all the $B_n$. Now $\{ B_1,B_2,\ldots \}$ is an algebra that $\sigma$-generates ${\mathcal {F}}$. If c.a. probabilities agree on $U_1 \subseteq U_2 \subseteq \cdots$, they agree on the union, and similarly if they agree on $U_1 \supseteq U_2 \supseteq \cdots$, they agree on the intersection. Let ${\mathcal {G}}$ be the sets on which p and q agree. This is thus a monotone class that contains $\{ B_1,B_2,\ldots \}$, and hence by the Monotone Class Theorem, ${\mathcal {F}} \subseteq {\mathcal {G}}$, and hence they agree on ${\mathcal {F}}$, a contradiction.

Since $B_n$ is in turn partitioned into $C_{1n},C_{2n},...$, it follows that p and q differ on $C_{mn}$ for some m. Thus by the propriety of $b_{C_{mn}}$ and the condition for equality in the propriety inequality for A-Brier scores, we have $E_p b(p) < E_p b(q)$ as desired.

Now, define

$$\begin{aligned} s_1(c)(\omega ) = {\left\{ \begin{array}{ll} - \frac{\sum _{n=1}^\infty 1_{A_n}(\omega ) c(A_n)}{\left( \sum _{n=1}^\infty (c(A_n))^2 \right) ^{1/2}} &{} \text {if }c \in {\mathcal {P}}_{\textrm{c}}({\mathcal {F}}) \\ -\sum _{n=1}^\infty \frac{1_{A_n}(\omega )}{2(n+1)} &{} \text {otherwise.} \end{array}\right. } \end{aligned}$$

Given a probability $p\in {\mathcal {P}}_{\textrm{c}}({\mathcal {F}})$, let $\phi (p)$ be the unique c.a. probability on ${\mathbb {P}}{\mathbb {N}}$ such that $\phi (p)(\{ n \})=p(A_{n+1})$. Note that $\phi$ is a continuous function from ${\mathcal {P}}_{\textrm{c}}({\mathcal {F}})$ to ${\mathcal {P}}_{\textrm{c}}({\mathbb {P}}{\mathbb {N}})$ with respect to the $\ell ^\infty$ topologies. Extend $\phi$ to all of ${\mathcal {C}}({\mathcal {F}})$ by letting $\phi (c)=d_0$ for all $c\in {\mathcal {C}}({\mathcal {F}})\backslash {\mathcal {P}}_{\textrm{c}}({\mathcal {F}})$ and any fixed non-probability credence $d_0$ on ${\mathbb {P}}{\mathbb {N}}$. Then it is easy to verify that:

$$\begin{aligned} E_q s_1(c) = E_{\phi (q)} {\text {Sph}}^*(\phi (c)). \end{aligned}$$

The ${\mathcal {P}}_{\textrm{c}}({\mathcal {F}})$-weak continuity and quasi-strict propriety of $s_1$ then follow from those of ${\text {Sph}}^*$ (we do not get strict propriety, because $\phi$ is not one-to-one on the c.a. probabilities, though we do get quasi-strictness because it maps probabilities to probabilities and non-probabilities to non-probabilities).

Fix some non-probability $c_0$ in ${\mathcal {C}}({\mathcal {F}})$. Let $f_0 = s_1(c_0)$. Note that $E_p s_1(p) < E_p f_0$ for all c.a. probabilities p.

Let $s_2(p) = b(p) + s_1(p)$ for a probability p. This is the sum of a strictly proper and a proper scoring rule on ${\mathcal {P}}_{\textrm{c}}({\mathcal {F}})$, so it is strictly proper there. It is continuous and uniformly bounded. Using Proposition 2, extend $s_2$ to a continuous strictly proper scoring rule on all of ${\mathcal {C}}({\mathcal {F}})$ with $s_2(c_0) = f_0$. It remains to check that we do not have even non-strict domination. To that end, choose $\omega _n \in A_n$ (this uses the Axiom of Countable Choice). Observe that

$$\begin{aligned} \sum _{n=1}^\infty f_0(\omega _n) = -\sum _{n=1}^\infty \frac{1}{2(n+1)} = -\infty \end{aligned}$$

but

$$\begin{aligned} \sum _{n=1}^\infty s_1(p)(\omega _n) = - \sum _{n=1}^\infty \frac{p(A_n)}{\left( \sum _{n=1}^\infty (p(A_n))^2 \right) ^{1/2}} > -\infty \end{aligned}$$

for any probability p by countable additivity. Moreover,

$$\begin{aligned} \sum _{n=1}^\infty b(p)(\omega _n) \ge -\sum _{n=1}^\infty 2^{-n} > -\infty , \end{aligned}$$

since $b(p)(\omega ) \in [-2^{n},0]$ for $\omega \in A_n$. It follows that $f_0$ cannot be even non-strictly dominated by $b(p)+s_1(p)$ for any probability p. $\square$

Proof of Proposition 6

Say that a function f depends only on the coordinates in $C\subseteq \kappa$ if for any $\omega$ and $\omega '$, if $\omega (x)=\omega '(x)$ whenever $x\in C$, then $f(\omega )=f(\omega ')$.

Let Q be the set of all intervals (a, b) with a and b rational numbers. If f is a real-valued measurable function on $\Omega$ and $I \in Q$, then the preimage $f^{-1}[I]$ is a member of ${\mathcal {F}}$, and hence depends only on countably many coordinates. Suppose that $f^{-1}[I]$ depends only on the coordinates in a countable set $C_I$ (use the Axiom of Choice to choose $C_I$), then let C be the union of $C_I$ as I ranges over Q. Then C is a countable set and $f^{-1}[I]$ depends only on the coordinates in C for any such interval with rational number endpoints. Since $f(x)=y$ if and only if x is a member of $f^{-1}[I]$ for all $I\in Q$ such that $y \in I$, it follows that f depends only on the coordinates in C.

For any countable subset C of $\kappa$, the number of measurable functions that depend only on the coordinates in C is the number of measurable functions on $\{0,1\}^\omega$, with respect to the product $\sigma$-algebra, and that is just $\mathfrak {c}$, since this product $\sigma$-algebra is countably generated. There are at most $\kappa ^\alpha$ subsets of $\kappa$ with cardinality $\alpha$ and so at most $\kappa ^\omega$ countable subsets of $\kappa$ (here we use the Axiom of Choice twice). Thus, again using the Axiom of Choice, there are at most $\kappa ^\omega \times \mathfrak {c} = \kappa ^\omega$ measurable functions on $\{0,1\}^{\kappa }$, and given the assumption that $2^{\kappa } > \kappa ^\omega$, and the fact that there $2^\kappa$ extreme probabilities concentrated at singletons, the proof is complete. $\square$

Finally, if ${\mathcal {Q}}$ is a convex subset of a vector space, say that a topology on ${\mathcal {Q}}$ is line segment compatible provided that for any x and y in ${\mathcal {Q}}$, the function $t \mapsto (1-t)x+ty$ from [0, 1] to ${\mathcal {Q}}$ is continuous in that topology. Any topology on ${\mathcal {P}}_{\textrm{f}}$ which is derived from embedding in a topological vector space of real-valued functions on ${\mathcal {F}}$ is line segment compatible.

Proposition 9

Let $\Omega$ be any non-empty set. Suppose ${\mathcal {P}}$ is either ${\mathcal {P}}_{\textrm{f}}$ or ${\mathcal {P}}_{\textrm{c}}$ and has a line segment compatible topology. Let s be a proper scoring rule on ${\mathcal {P}}$ such that $p\mapsto s(p)(\omega )$ is continuous for every fixed $\omega \in \Omega$. Then if $E_p s(p) = E_p s(q)$ and both $E_p s(p)$ and $E_q s(q)$ are finite, we have $s(p)=s(q)$ everywhere on $\Omega$.

Proof

Fix $\omega \in \Omega$. Let $u_\omega$ be the probability measure such that $u_\omega (\{\omega \})=1$. Let $p_t = (1-t)p + tu_\omega$ for $t\in [0,1]$. Then:

$$\begin{aligned} (1-t) E_p s(p) + t E_{u_\omega } s(q)&= (1-t) E_p s(q) + t E_{u_\omega } s(q) \\&= E_{p_t} s(q) \\&\ge E_{p_t} s(p_t) \\&= (1-t) E_p s(p_t) + t E_{u_\omega } s(p_t) \\&\ge (1-t) E_p s(p) + t E_{u_\omega } s(p_t), \end{aligned}$$

where the first equality follows from the assumed equality in the statement of the proposition and the two inequalities follow from propriety. Since $E_p s(p)$ is finite, it follows that

$$\begin{aligned} E_{u_\omega } s(q) \ge E_{u_\omega } s(p_t) \end{aligned}$$

for all $t\in (0,1]$. Since $E_{u_\omega } f = f(\omega )$ for any f, taking the limit as $t\rightarrow 0+$, it follows by continuity and line segment compatibility that

$$\begin{aligned} s(q)(\omega ) = E_{u_\omega } s(q) \ge E_{u_\omega } s(p) = s(p)(\omega ). \end{aligned}$$

We have thus shown that if p and q are such that $E_p s(p) = E_p s(q)$, then $s(q) \ge s(p)$ everywhere. It follows that $E_q s(q) \ge E_q s(p)$ and by propriety that $E_q s(q) = E_q s(p)$. Applying what we have just proved with p and q swapped, we conclude that $s(p) \ge s(q)$ everywhere, and hence that $s(p) = s(q)$ everywhere. $\square$