Valuation on convex bodies

Remarks

If $K \cup L$ is convex, then $K \cap L$ is nonempty, since a convex set is connected and cannot be the union of two disjoint nonempty closed sets. So $K \cap L \in \mathcal{K}^n$ and the condition makes sense. A valuation is continuous if it is continuous for the Hausdorff distance, and invariant under rigid motions if $\varphi(gK) = \varphi(K)$ for every rigid motion $g$ of $\R^n$.

Example: $\varphi = \operatorname{vol}$, Lebesgue measure on $\R^n$. For measurable sets, $\operatorname{vol}(K \cup L) + \operatorname{vol}(K \cap L) = \operatorname{vol}(K) + \operatorname{vol}(L)$ always holds, so volume is a valuation. The constant function $\chi(K) = 1$ is a valuation as well; it is the Euler characteristic.

Non-example: $\varphi(K) = \operatorname{vol}(K)^2$ on $\mathcal{K}^1$. For $K = [0, 2]$ and $L = [1, 3]$ one gets $\varphi(K \cup L) + \varphi(K \cap L) = 9 + 1 = 10$, but $\varphi(K) + \varphi(L) = 4 + 4 = 8$.

Many authors include the empty set in $\mathcal{K}^n$ and require $\varphi(\emptyset) = 0$. Then the Euler characteristic takes the value $0$ at $\emptyset$.