Define for every $\omega\in\Omega$: $$f_\omega=\operatorname{argmin}_{f\in S} L(\omega,f)$$
where
- $\Omega$ is a measurable space
- $S\subseteq \big(C(\mathbb{R}^d,\mathbb{R}^n),\|{\cdot}\|_\infty\big)$ compact
- $L\colon\, \Omega\times C(\mathbb{R}^d,\mathbb{R}^n)\to [0,\infty)$
- $\omega \mapsto L(\omega,f)$ is measurable for every $f\in C(\mathbb{R}^d,\mathbb{R}^n)$
- $C(\mathbb{R}^d,\mathbb{R}^n) \ni f \mapsto L(\omega,f)$ is continuous for every $\omega\in\Omega$
Can we choose the minimizer $f_\omega$ for every $\omega$ in a way such that the mapping
$(\Omega,\mathbb{R}^d) \ni(\omega,x)\mapsto f_{\omega}(x)$
is measurable?
Idea: I think that Theorem 18.19 in Aliprantis & Border 2006 could be applicable? (Measurability of supremum over measurable set)
