# Measurable selection involving measure valued random variable

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!

• If $|\mu_{\omega}|(\mathbb{R}^d)<a$ wouldn't you be in trouble? Apr 7, 2021 at 19:23
• Yes, you are right, I would get $F(\omega)= \emptyset$, hence no selection. Let me edit the question a little bit. Apr 7, 2021 at 19:26
• I don't think the function $\omega\mapsto |\mu_\omega|$ will usually be measurable. Apr 7, 2021 at 20:06
• 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 $C_{0,1}$. Apr 7, 2021 at 22:35
• 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 $N^{-1}\{1,\ldots,k\}=\bigcup_{j=1}^k \{\omega\colon \int f_j\,d\mu_\omega\ge \frac 12|\mu_\omega|\}$. Apr 8, 2021 at 20:20

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 $$N^{-1}\{1,\ldots,k\}=\bigcup_{j=1}^k \{\omega\colon\int f_j\,d\mu_\omega\ge \frac12|\mu_\omega|\}$$.