A maximal-mass action is Bayes under finite-class zero–one loss

Since the parameter and action spaces are finite with all subsets measurable, the zero–one loss is product-measurable as required by [F1]. For $j=1,\dots ,k$, put $p_j(y)={\mathrm{\Pi }}_y(\{j\})$. By [F2], every $p_j$ is measurable. The zero–one posterior risk of action $j$ is ${\rho }_y(j)={\sum }_{\mathrm{\ell }\mathrm{\ne }j}{p}_{\mathrm{\ell }}(y)=1-p_j(y)\mathrm{.}$ Thus the risk is minimized exactly by maximizing $p_j(y)$.

Since $k\ge 1$ is finite, the nonempty finite list $p_1(y),\dots ,p_k(y)$ has a maximum. Define $j^{\ast }(y)$ to be the least index attaining it. For each $j$, $\{y:j^{\ast }(y)=j\}={\bigcap }_{\mathrm{\ell }>j}\{y:p_j(y)\ge p_{\mathrm{\ell }}(y)\}\cap {\bigcap }_{\mathrm{\ell }<j}\{y:p_j(y)>p_{\mathrm{\ell }}(y)\}\mathrm{.}$ Both intersections are finite, and each set is measurable because the $p_{\mathrm{\ell }}$ are measurable; hence $j^{\ast }$ is measurable. Its risk is finite and it attains the pointwise minimum from step 1.1, so [F1] makes it a Bayes action rule. Choosing the least maximizing index uses only the displayed finite order, not AC.