Set-valued measure

Remarks

Many authors also require the values to be closed, compact or convex, and some work in a Banach space instead of $\R^n$. Example: let $\mu$ be a finite measure on $\mathcal{F}$, let $K \subseteq \R^n$ be nonempty, compact and convex, and let $M(A) = \mu(A) K$. Then $M(\emptyset) = \set{0}$. Given $x_i \in \mu(A_i) K$, the series converges absolutely since $K$ is bounded, and its sum lies in $\mu(A) K$ because $K$ is convex and closed. Conversely $\mu(A) k = \sum_i \mu(A_i) k$ for $k \in K$, so clause 2 holds. Non-example: let $\Omega = \set{a, b}$ with counting measure $\mu$ and $K = \set{0, 1}$, which is not convex. Then $M(\Omega) = 2K = \set{0, 2}$, while the sums from $M(\set{a})$ and $M(\set{b})$ form $\set{0, 1, 2}$, so clause 2 fails. A vector measure $\nu$ gives the set-valued measure $A \mapsto \set{\nu(A)}$.