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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.

  • Definition of a Semialgebra.

  • 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*}

Giordano Ribeiro
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  • What's the definition of 'measure' in the context of semialgebras? – Berci Oct 29 '21 at 20:03
  • I edited it in. The only difference is that the measure is defined on a semialgebra instead of the usual $\sigma$-algebra. – Giordano Ribeiro Oct 29 '21 at 20:20
  • Ok. It might boil down to the difference of definitions of semialgebra. The question you linked proposes two different definitions. I think we can prove the claim using 3'. – Berci Oct 29 '21 at 21:08
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    Use the same proof that proves measures are countably subadditive. You form the disjoint sets $B_n = A_n \setminus (A_1 \cup \dots \cup A_{n - 1})$, then use countable additivity and $\mu(B_n) \leq \mu(A_n)$. It may help to first extend $\mu$ to a "measure" on the algebra generated by $C$, which is the collection of finite disjoint unions of elements of $C$. – Mason Oct 29 '21 at 21:27
  • @Mason I think I understood. The idea would be to start with the measure on a semialgebra $\mathcal{C}$, extend it to a measure on the algebra generated by $\mathcal{C}$, i.e; $\mathcal{A}(\mathcal{C})$, and since this extension satisfies countable subadditivty (I have already proven this fact), we can simply restrict the extended function to $\mathcal{C}$. Since the restriction is the same as the original function, than it must be countably subadditive. – Giordano Ribeiro Oct 29 '21 at 21:41
  • @Mason Really appreciate the insight! Thank you a lot! – Giordano Ribeiro Oct 29 '21 at 21:41
  • @GiordanoRibeiro It seems you figured it out. I put more details in an answer. – Mason Oct 29 '21 at 21:52

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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.

Mason
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