A semialgebra $\mathcal S$ is a Set of subsets such that:
- $\emptyset, X \in\mathcal S$ where $X$ is the Universe.
- If $A,B \in\mathcal S$, then $A \cap B \in\mathcal S$
- If $A \in\mathcal S$, then $A^c = A_1 \cup A_2 \cup ...... \cup A_n$ where $A_i \in\mathcal S, \forall i$ and $A_j \cap A_m = \emptyset$, $\forall m \neq j$
Then:
Let $\mathcal S$ be a semialgebra and let $\mu_s : \mathcal S \to \mathbb R$ a measure of semialgebras. Suppose that exists $S_n \in\mathcal S$ such that $X = \bigcup_{i=0}^n S_n$ and $\mu_s (S_n) < \infty $. Prove that there is one only measure of $\sigma$-algebra
$\mu :\sigma(\mathcal S)$ $\to \mathbb R$ such that $\mu (S) = \mu_s(S)$ $\forall S \in\mathcal S$.
Here $\sigma(\mathcal S)$ denotes the $\sigma$-algebra generated by semialgebra $\mathcal S$.
I have no clue about how to prove this.
There's a suggestion: Use that in the case that $\mu_s(X) < \infty$, the result is valid.
