I was reading "Measure Theory and Probability Theory" by Krishna B. Athreya and, in chapter 1, the subject of measures on semialgebras is brought up. In particular he introduces the following definition:
Given a measure $\mu$ on a semialgebra $\mathcal{C}$, the outer measure induced by $\mu$ is the set function $\mu^* : \mathcal{P}(\Omega) \longrightarrow [0, \infty]$, defined as:
\begin{equation*}
\mu^*(A) = \text{inf} \, \Bigg\{ \sum_{n = 1}^\infty \mu(A_n) \; : \; \{A_n \}_{n \geq 1} \subseteq \mathcal{C}, \; A \subseteq \bigcup_{n \geq 1} A_n \Bigg\}.
\end{equation*}
He also makes a little side comment that it is not difficult to show that $\mu^* = \mu$ on $\mathcal{C}$. This comment actually sent me in a spiral of madness that I was not expecting. My question is as follows:
Given a measure $\mu$ on a semialgebra $\mathcal{C}$, can we show that it is countably subadditive? If so, how?
I've given this problem a lot of thought but couldn't reach an answer.
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Definition of a Semialgebra.
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Definition of a Countable Subadditivity on a Semialgebra: Let $\mu$ be a measure on a semialgebra $\mathcal{C}$. If $\{ A_n \}_{n = 1}^\infty \subseteq \mathcal{C}$ such that $\bigcup_{n = 1}^\infty A_n \in \mathcal{C}$, then:
\begin{equation*}
\mu \left(\bigcup_{n = 1}^\infty A_n \right) \leq \sum_{n = 1}^\infty \mu(A_n)
\end{equation*}
- Defintion of a Measure on a Semialgebra: A set function $\mu : \mathcal{C} \rightarrow [0, \infty]$, where $\mathcal{C}$ is a semialgebra, is called a measure if it satisfies:
$(i) \; \mu(\emptyset) = 0$
$(ii)$ For any sequence of sets $\{A_n \}_{n = 1}^\infty \subseteq \mathcal{C}$ with $\bigcup_{n = 1}^\infty A_n \in \mathcal{C}$ and $A_i \cap A_j = \emptyset$ for $i \neq j$,
\begin{equation*}
\mu \left(\bigcup_{n = 1}^\infty A_n \right) = \sum_{n = 1}^\infty \mu(A_n)
\end{equation*}
Best Answer
I will assume that $C$ is a semi-algebra of subsets of $X$ means that $$\emptyset \in C,$$ $$A, B \in C \implies A \cap B \in C,$$ $$A \in C \implies A^c \text{ is a finite disjoint union of members of $C$}.$$
Now let $\mu$ be a measure on $C$. It is easy to show that the collection of finite disjoint unions of elements of $C$ is an algebra $\mathcal{A}$ (hence the algebra generated by $C$). Now extend $\mu$ to $E \in \mathcal{A}$ by $\mu(E) = \sum_{i = 1}^{N}\mu(A_i)$, where $A_i$ are disjoint sets in $C$ whose union is $E$. A common refinement argument shows that if $B_1, \dots, B_M$ are disjoint sets in $C$ whose union is $E$, then $\sum_{i = 1}^{N}\mu(A_i) = \sum_{j = 1}^{M}\mu(B_j)$, so $\mu(E)$ is well defined.
Now the standard argument that a measure on a sigma-algebra is countably subadditive goes through to show that $\mu$ is countably subadditive.