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Suppose $\scr F$ is a countably generated $\sigma$-algebra on any $\Omega$ (in particular, the Borel $\sigma$-algebra on any Polish space will work) and $\scr P$ is all countably additive probabilities on $\scr F$. Suppose $A_1,A_2,...$ generate $\scr F$. Let $S_n(P,\omega) = -(1_{A_n}(\omega)-P(A_n))^2$ be a Brier accuracy score with respect to $A_n$ and let $S=\sum_n 2^{-n} S_n$. Then it's elementary that $S_n$ has the following strict propriety condition: $S_n(P,P) \ge S_n(Q,P)$ with equality iff $P(A_n)=Q(A_n)$. Hence, $S$ is proper, and $S(P,P) > S(Q,P)$ if $P$ and $Q$ differ on at least one of the $A_n$. But since the $A_n$ generate $\scr F$, if $P$ and $Q$ differ anywhere, they differ on at least one of the $A_n$, so $S$ is strictly proper. Moreover, it's regular, and in fact it's uniformly bounded, and $P\mapsto S(P,\omega)$ is continuous for each fixed $\omega$ in the $\ell^\infty(\scr F)$ metric. Finding $G$ should then be an easy exercise.