Basically, this question arises when I am trying to see Bayes Action is measurable. Bayes action is defined as follows:
Let $A$ be an action space and $L(\theta, a)\geq 0$ be a loss function. For any $x\in \mathcal{X}$, a Bayes action is any $\delta(x)\in A$ such that \begin{align*} E[L(\theta, \delta(x))|X=x] = \min_{a\in A}E[L(\theta, a)|X=x] \end{align*}
So Bayes action is defined pointwisely for each $x$. It is not clear to me that $\delta(X)$ is $X$ measurable. This problem can be rewrited as the following.
Let $f(x,y): \Omega\times \mathbb{R}^n\longrightarrow \mathbb{R}$ be a function such that it is measurable with respect to $x$ and continuous with respect to $y$. Consider the set function \begin{align*} \psi: x\longmapsto \{y: f(x,y) \text{ attains the minimum }\} \end{align*} (assumed this minimum are always attained). Then this set function admits measurable selection. i.e. there exists measurable $\phi$ from $\Omega$ to $\mathbb{R}^n$ such that $\phi(x)$ is in the argmin set.
I have read some previous posts, such as measurability of argmin. A comment says applying Kuratowski--Ryll-Nardzewski selection theorem. However, I am having trouble to verify the weakly measurable assumption... which is \begin{align*} \{x\in \Omega, \psi(x)\cap U\neq \varnothing\}\in \mathcal{F}, \qquad \forall \text{open } U\subseteq \mathbb{R}^n \end{align*} where $\mathcal{F}$ is the $\sigma$-algebra of $\Omega$. (Honestly, I am not even sure this is indeed true or not) Does anyone have any ideas?
Thanks in advance!
