Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and let $\mathcal{M}(\mathbb{R}^d)$ be the space of finite signed measures on $\mathbb{R}^d$ endowed with the narrow topology (i.e. the initial topology w.r.t. $C\_b(\mathbb{R}^d)$, the set of real valued, continuous and bounded functions on $\mathbb{R}^d$) and the corresponding Borel $\sigma$-algebra. Let $\mu: \Omega \to \mathcal{M}(\mathbb{R}^d)$ be measurable and let us define the multifunction
$$F : \Omega \rightrightarrows C\_{0,1}(\mathbb{R}^d):=\{ \varphi \in C\_0(\mathbb{R}^d) \mid |\varphi|\_{\infty} \le 1 \}$$
(where $C\_0(\mathbb{R}^d)$ is the Banach space of continuous functions vanishing at infinity with the supremum norm) as
$$F(\omega) := \left \{ \varphi \in C\_{0,1}(\mathbb{R}^d) \mid \int\_{\mathbb{R}^d} \varphi \text{ d} \mu\_{\omega} \ge \frac{|\mu\_{\omega}|}{2} \right \}, $$
where $|\mu\_{\omega}|$ is the total variation norm of $\mu\_{\omega}$.


Can we find a measurable selection $f: \Omega \to C\_{0,1}(\mathbb{R}^d)$ of $F$, meaning that $f$ is measurable and $f(\omega) \in F(\omega)$ for every $\omega \in \Omega$?


I tried with the [Kuratowski–Ryll-Nardzewski measurable selection theorem](https://en.wikipedia.org/wiki/Kuratowski_and_Ryll-Nardzewski_measurable_selection_theorem) but I am not able to prove that $\{ \omega \in \Omega \mid F(\omega) \cap U \}$ is measurable for every $U \subset C\_{0,1}(\mathbb{R}^d)$ open.


Any hint would be really appreciated!



Measurable selection involving measure valued random variable
I think this is straightforward by what I call the "lexicographic" method: fix a countable dense sequence $(f\_n)$ in $\_{0,1}$. Then as you point out, $\omega\mapsto|\mu\_\omega|=\sup\int f\_n\,d\mu\_\omega$. Let $N(\omega)=\min\{n:\int f\_n\,d\mu\_\omega\ge\frac12|\mu\_\omega|\}$. This is measurable because $N^{-1}\{1,\ldots,k\}=\bigcup\_{j=1}^k \{\omega\colon\int f\_j\,d\mu\_\omega\ge \frac12|\mu\_\omega|\}$.


