The posterior forecast minimizes strictly proper scoring risk

By [F1], choose a measurable posterior probability kernel ${\mathrm{\Pi }}_x$ on the finite label set; identify it with the measurable vector $G(x)=({\mathrm{\Pi }}_x(1),\dots ,{\mathrm{\Pi }}_x(K))\in {\mathrm{\Delta }}_K$. For each $x$, the posterior expected score of reporting $q$ is $\bar{S}(G(x),q)$, which is minimized at $q=G(x)$ by [F2]. The posterior risk of this rule is measurable by [F4], so the rule meets the measurable-minimizer hypothesis in [F6].

To apply the Bayes-risk interface to the original joint law, let ${\pi }_j=P(Y=j)$. For ${\pi }_j>0$, define $K_j(A)=P(X\in A,Y=j)\mathrm{/}{\pi }_j$ on feature events $A$; for ${\pi }_j=0$, set $K_j=P_X$. Each $K_j$ is a probability measure, and the finite map $j\mapsto K_j$ is a probability kernel. The joint prior-data law built from prior $({\pi }_j)$ and kernel $K_j$ agrees with the original joint law because for every $A$ and $j$ its $(A,j)$ mass is $P(X\in A,Y=j)$, including zero-prior labels. Consequently the Bayes-risk theorem [F6] states that the posterior rule $G$ minimizes the original expected score risk over all measurable randomized forecasts, hence over deterministic forecasts.

Now suppose $S$ is strictly proper and another measurable forecast $q$ has the same risk as $G$, whose risk is finite by hypothesis. Write $R_q(x)=\bar{S}(G(x),q(x))$ and $R_G(x)=\bar{S}(G(x),G(x))$. They are nonnegative measurable functions by [F3,F4], and posterior disintegration gives their integrals as the two risks, so both integrals are finite. By [F7], the measurable set $N=\{R_q=+\mathrm{\infty }\}\cup \{R_G=+\mathrm{\infty }\}$ is null. Set ${\tilde{R}}_q=R_q$ and ${\tilde{R}}_G=R_G$ off $N$, and set both to zero on $N$; these modified functions are measurable by [F10]. By [F8], the changes preserve their integrals: each original function is the sum of its restriction to $N^c$ and its restriction to the null set $N$. The functions ${\tilde{R}}_q,{\tilde{R}}_G$ are finite and integrable. By propriety, $D={\tilde{R}}_q-{\tilde{R}}_G$ is nonnegative, and it is measurable by [F10]; [F9] gives $\mathbb{E}D=\mathbb{E}{\tilde{R}}_q-\mathbb{E}{\tilde{R}}_G=0$. Thus [F5] makes $D=0$ almost surely. Off $N$ this is equality of the finite posterior risks, so strict propriety in [F2] gives $q(X)=G(X)$ almost surely.