Under AC, the finite-second-moment posterior mean uniquely minimizes squared risk

The squared loss $L(\theta ,a)=(\theta -a{)}^2$ is continuous on ${\mathbb{R}}^2$, hence product-Borel measurable as required by [F3]. Let $s_2(y)=\int {\theta }^2{\mathrm{\Pi }}_y(d\theta )$. The function $(y,\theta )\mapsto {\theta }^2$ is product-measurable, so [F4] makes $s_2$ measurable and $D_2=\{s_2<\mathrm{\infty }\}$ measurable. On $D_2$, $2\mathrm{\mid }\theta \mathrm{\mid }\le 1+{\theta }^2$ gives a finite absolute first moment; [F4] therefore makes the signed mean $\mu$ measurable on $D_2$.

Fix $y\in D_2$ and use $(\mathbb{R},\mathcal{B}(\mathbb{R}),{\mathrm{\Pi }}_y)$ as a probability space with trivial sub-sigma-algebra $\mathcal{G}=\{\mathrm{\varnothing },\mathbb{R}\}$. The identity variable $\theta$ is a real random variable by [F5] and is square-integrable. The constant function $\mu (y)$ is $\mathcal{G}$-measurable and satisfies ${\int }_H\theta d{\mathrm{\Pi }}_y={\int }_H\mu (y)d{\mathrm{\Pi }}_y$ for both possible $H\in \mathcal{G}$; by [F2] it is a conditional-expectation version.

By [F1] under [A1], this conditional expectation uniquely minimizes squared distance over $L^2(\mathcal{G})$. The real-valued $\mathcal{G}$-measurable functions are exactly constants, and every finite constant is in $L^2({\mathrm{\Pi }}_y)$; hence $\mu (y)$ uniquely minimizes $\int (\theta -a{)}^2{\mathrm{\Pi }}_y(d\theta )$ over all real actions. This risk is finite because both $\theta$ and the constant action are square-integrable.

By step 1.1 the mean rule is measurable on $D_2$, and by step 2.1 it has finite risk and attains the minimum on every fiber. The definition [F3] therefore makes it a Bayes action rule on $D_2$.