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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 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!

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  • $\begingroup$ If $|\mu_{\omega}|(\mathbb{R}^d)<a$ wouldn't you be in trouble? $\endgroup$ Apr 7, 2021 at 19:23
  • $\begingroup$ Yes, you are right, I would get $F(\omega)= \emptyset$, hence no selection. Let me edit the question a little bit. $\endgroup$
    – Bremen000
    Apr 7, 2021 at 19:26
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    $\begingroup$ I don't think the function $\omega\mapsto |\mu_\omega|$ will usually be measurable. $\endgroup$ Apr 7, 2021 at 20:06
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    $\begingroup$ I would forget about the KRN Thm and do instead the toy model: let $f_n$ be a sequence of measurable functions $\Omega\rightarrow \mathbb{R}$ with pointwise finite sups, find a measurable map $\phi:\Omega\rightarrow \mathbb{N}$ such that $f_{\phi(\omega)}(\omega)\ge \sup_n f_n(\omega) -1$ for all $\omega$. Then adapt to the present question by leveraging the separability of $C_{0,1}$. $\endgroup$ Apr 7, 2021 at 22:35
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    $\begingroup$ I think this is straightforward by what I call the "lexicographic" method: fix a countable dense sequence $(f_n)$ in $C_{0,1}$. Then as you point out, $\omega\mapsto|\mu_\omega|=\sup\int f_n\,d\mu_\omega$. Let $N(\omega)=\min\{n\colon \int f_n\,d\mu_\omega\ge \frac 12|\mu_\omega|\}$. This is measurable because $N^{-1}\{1,\ldots,k\}=\bigcup_{j=1}^k \{\omega\colon \int f_j\,d\mu_\omega\ge \frac 12|\mu_\omega|\}$. $\endgroup$ Apr 8, 2021 at 20:20

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I think this is straightforward by what I call the "lexicographic" method: fix a countable dense sequence $(f_n)$ in $𝐶_{0,1}$. 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 $N^{-1}\{1,\ldots,k\}=\bigcup_{j=1}^k \{\omega\colon\int f_j\,d\mu_\omega\ge \frac12|\mu_\omega|\}$.

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