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Background. Given a topological space $(X, \tau)$, a valuation on $X$ (with respect to $\tau$) is a map $v \colon \tau \to \mathbb{R}_{\geq 0} \cup \{ \infty \}$ which satisfies the following properties: (Strictness Property) The empty set $\emptyset$ satisfies $v(\emptyset) = 0$, (Monotonicity Property) If $U, V \in \tau$ such that $U \subseteq V$, then $v(U) \leq v(V)$, and (Modularity Property) If $U, V \in \tau$, then $v(U) + v(V) = v(U \cup V) + v(U \cap V)$.

A valuation $v$ is Scott continuous if it satisfies $v ( \bigcup_{i \in I} U_{i} ) = \sup_{i \in I}$ $v \left( U_{i} \right)$ for any directed family $\{ U_i \}_{i \in I} \in \tau$, directed in the sense that there is a $k \in I$ for each $i, j \in I$ such that $U_{i}, U_{j} \subseteq U_{k}$. A valuation $v$ is simple if for any $U \in \tau$ there is an integer $n = n(U)$ such that $v(U)$ is the linear combination $\sum_{i = 1}^{n} a_{i} \delta_{x_i}(U)$ where $a_{i} \in \mathbb{R}_{\geq 0}$ and $\delta_{x}(U)$ is the Dirac measure of $U$ with respect to $X$ defined to be $1$ if $x \in U$ and $0$ otherwise. In this case, $\{ x_1, \dots, x_{n} \} \subseteq X$ is said to be the support of the valuation $v$.

Proposition. A simple valuation is Scott continuous and can be extended to a measure $\mu(X)$ on the power set of $X$ with a unique measure $\mu^{\prime}(X)$ on the Borel $\sigma$-algebra $\mathcal{B}(X, \tau)$.

Question. Why is the uniqueness of the measure important? Does it give us integration on the Borel algebra? If not, what does it give us?

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The definition of simple seems a bit off, where did $x_i$ come from? –  Asaf Karagila May 7 '11 at 7:26
    
Then you should specify what you mean by support of the valuation, and that this is what this suppose to be in the sum. –  Asaf Karagila May 7 '11 at 10:59
    
I'm sorry, I still don't get what the support is. Is it a finite subset of atoms? (i.e. their existence implies nonzero value). Also, if you define $\delta_\bullet(U)=\chi_U$, what does the variable $x_i$ has to do with this function? –  Asaf Karagila May 7 '11 at 12:03
    
Edited and Clarified. –  user02138 May 7 '11 at 12:28
    
Much better, thanks :-) could you please give some reference about this proposition? It seems interesting and I would like to read some more about it. –  Asaf Karagila May 7 '11 at 12:33

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