Existence of a strictly proper scoring rule on a $\sigma$-algebra that is not countably generated

Question

Given a $\sigma$-algebra $\scr F$ on $\Omega$, say that an accuracy scoring rule for $\scr F$ is a function $s$ from the set of all (countably additive) probabilities on $\scr F$ to the $\scr F$-measurable functions on $\Omega$ with values in $[-\infty,M]$ (for some fixed real $M$). A scoring rule $s$ is proper iff $\int_\Omega s(p) \, dp \ge \int_\Omega s(q) \, dp$ for all pairs of probabilities $p$ and $q$, and strictly proper if additionally equality only holds when $p=q$.

If $\scr F$ is countably generated, it has a strictly proper scoring rule (see my answer here). The same is true for any measure space that has only atomic probability measures, like the one here.

Question:

1. Are there any strictly proper scoring rules for a $\scr F$ that is not countably generated and that has a nonatomic probability measure?

2. If yes, is this true for all $\scr F$?

Note: The answer to (2) is negative if there are cases where there are more than ${\mathfrak c}^{|\Omega|}$ probability measures on $\scr F$. By a result of Paris and Koonen, it is relatively consistent with the existence of a measurable cardinal that there be a measurable cardinal $\kappa$ that has $2^{2^\kappa}$ normal measures, so it is relatively consistent with the existence of a measurable cardinal that the answer to (2) is negative.

Answer 1

There is a somewhat boring positive answer to the first question. It is shown in [K. P. S. Bhaskara Rao, and B. V. Rao. Borel spaces. Warszawa: Instytut Matematyczny Polskiej Akademi Nauk, 1981.] on page 15 that the $\sigma$-algebra on $[0,1]$ generated by analytic sets is not countably generated. However, since analytic sets are universally measurable, every Borel probability measure has a unique extension to this $\sigma$-algebra. It follows that every proper scoring rule for the probability measures on $[0,1]$ with the Borel sets works also for this larger space that is not countably generated.

Answer 2

The answer to 2 is negative. Let $\kappa$ be infinite and such that $2^\kappa > \kappa^\omega$ and let $X=2^\kappa$ with $\scr F$ being the coin-flip product $\sigma$-algebra (which is generated by sets that depend on a finite number of coordinates).

There are more probability measures on $\scr F$ than $\scr F$-measurable functions $X\to\mathbb R$, so no score can be a one-to-one function.

There are at least $2^\kappa$ probability measures (one per point, as $\scr F$ separates points).

And any measurable function depends on countably many coordinates (since the preimage of every interval with rational endpoints is is in $\scr F$ and hence depends on countably many coordinates, and the function is determined by these preimages). Since there are only continuum many real-valued measurable functions on $2^\omega$, there are at most $\kappa^\omega \times 2^\omega = \kappa^\omega$ measurable functions on $X$.

Note that if there is a strictly proper scoring rule on $2^\kappa$ for $\kappa>\lambda$, there is a strictly proper scoring rule on $2^\lambda$ (a measure on $2^\lambda$ can be extended to a measure on $2^\kappa$ by requiring the coin-flips in $\kappa\backslash\lambda$ to be all zero). Since $\kappa = \frak c$ satisfies the inequality $2^\kappa > \kappa^\omega$, it follows there is no strictly proper scoring rule on $2^\kappa$ for $\kappa \ge \frak c$.