# On the definition of vector measures

I'm working through a model of transportation taking place in a good domain $\Omega$ of $\mathbb{R}^d$, you can assume that it's open, connected, bounded and with as-smooth-as-you-wish boundary. At some point, a vector meausure $\boldsymbol{\operatorname{y}}$ in $\Omega$ (I assume that it's on the Borel sigma algebra of $\Omega$ with the topology inherited from $\mathbb{R}^d$) is defined by means of an operator $$\langle\, \boldsymbol{\operatorname{y}} , \cdot \; \rangle : C(\overline{\Omega},\mathbb{R}^d)\longrightarrow \mathbb{R}\;.$$ I don't know if this is a standard way of defining a vector measure in the sense of this definition. But in that case, what vector measure are they defining? If they are not using that definition of vector measure, which definition are they using?

I haven't seen this exact notation before, but this is what I would assume it means. Assume $\overline{\Omega}$ is compact. Given an $\mathbb{R}^d$-valued vector measure $\mathbf{y}$ on $\overline{\Omega}$, write $\mathbf{y}(A)=(y_1(A),y_2(A),\dots,y_d(A))$ for any Borel set $A$. Then each $y_i$ is a signed measure on $\overline{\Omega}$. Given a continuous function $f=(f_1,f_2,\dots,f_d):\overline{\Omega}\to\mathbb{R}^d$, we can then define $$\langle\mathbf{y},f\rangle=\sum_{i=1}^d\int_{\overline{\Omega}} f_i dy_i .$$
This gives a bounded linear map $\langle\mathbf{y},\cdot\rangle:C(\overline{\Omega},\mathbb{R}^d)\to\mathbb{R}$. Conversely, a vector version of the Riesz-Markov theorem says that any bounded linear map $C(\overline{\Omega},\mathbb{R}^d)\to\mathbb{R}$ comes from a unique vector measure on $\overline{\Omega}$ in this way. So if your vector measure $\mathbf{y}$ is presumably being defined as (the restriction to $\Omega$ of) the unique vector measure on $\overline{\Omega}$ that gives rise to the specified bounded linear map $C(\overline{\Omega},\mathbb{R}^d)\to\mathbb{R}$.
• Great. So you can recover the vector measure coordinate by coordinate evaluating the operator at an indicator function in the $i$-th place. Thanks! – Pipicito Aug 17 '16 at 1:15