In Diestel & Uhl's "Vector measures" one finds the following definition 1 on page 1.
Definition 1 (Countably additive vector measure).
Let $X$ be a Banach space an $\mathcal F$ a $\sigma$-algebra on a set $\Omega$. A function $F \colon \mathcal F \to X$ is a countably additive vector measure if $F\left(\bigcup_{n = 1}^{\infty} E_n\right) = \sum_{n = 1}^{\infty} F(E_n)$ in the norm topology of $X$ for all pairwise disjoint sequences $(E_n)_{n \in \mathbb N} \subset \mathcal F$ with $\bigcup_{n = 1}^{\infty} E_n \in \mathcal F$.
In Yann Brenier & Dimitry Vorotnikov's "On optimal transport of matrix-valued measures", they define $\mathbb P^+$ to be "the set of $\mathcal P^+$-valued Radon measures $P$ on $\mathbb R^d$ with finite $\text{tr}$$(\text{d}P(\mathbb R^d))$", where $\mathcal P^+$ is the subspace of real $d \times d$ symmetric positive semidefinite matrices".
The notation $dP$ is explained in the answer to this question.
They don't define what "Radon" means in the context of vector valued measures, but I assume that it means some kind regularity (with respect to the Loewner order on the symmetric matrices?).
My Question
If $P \in \mathbb P^+$, doesn't that mean that $P$ maps from $\Sigma$ into $\mathcal P^+$, where $\Sigma$ is some $\sigma$-algebra on $\mathbb R^d$?
Because $\text{tr}(A) < \infty$ for all $A \in \mathcal P^+$ and $P(E) \in \mathcal P^+$ for every $E \in \Sigma$ and wouldn't we have in particular $\text{tr}(P(\mathbb R^d)) < \infty$?
In particular if $d = 1$, would the one-dimensional Lebesgue measure not belong to $\mathbb P^+$?
Question 2.
Is there some way to find out from the paper what they mean by Radon?
I only found out that in subsection 2.2. of the paper Matrix Measures and Finite Rank Perturbations of Self-adjoint Operators by Constanze Liaw and Sergei Treil, Radon measures (on $\mathbb R$) are countably additive set function defined on bounded Borel subsets of $\mathbb R$, which are bounded on bounded Borel subsets.
I have a similar issue with a similar definition in Léonard Monsaingeon & Dmitry Vorotnikov's "The Schrödinger problem on the non-commutative
Fisher-Rao space": they define a TV norm of a $H$-valued Radon measure on p. 14 (where $H$ are the Hermitian $d \times d$ matrices) and then $\mathbb H$ as the set of $H$-valued Radon measures with finite TV norm. But a norm on a set should always be finite, so why isn't this condition vacuous as well?
Best Answer
I contacted on of the authors and they said that Radon in this case means that every entry is a Radon measure in the usual sense. In this case the finite trace condition is not vacuous, for example for $d = 1$ the usual Lebesgue measure does not fulfill the finite trace condition.