Measures on Algebras of Sets

# Measures on Algebras of Sets

 Definition: Let $X$ be a set and let $\mathcal A$ be an algebra of sets (not necessarily a $\sigma$-algebra) on $X$. A Measure on $\mathcal A$ is a set function $\mu : \mathcal A \to [0, \infty]$ with the following properties: 1) $\mu (\emptyset) = 0$. 2) If $(A_n)_{n=1}^{\infty}$ is any countable collection of mutually disjoint sets in $\mathcal A$ AND $\displaystyle{\bigcup_{n=1}^{\infty} A_n \in \mathcal A}$ then $\displaystyle{\mu \left ( \bigcup_{n=1}^{\infty} A_n \right) = \sum_{n=1}^{\infty} \mu (A_n)}$.

On the General Measurable Spaces and Measure Spaces page we defined a measure on a measurable space $(X, \mathcal A)$ where $\mathcal A$ is a $\sigma$-algebra similarly. Observe the small, though important differences in the definitinos. For a measurable space $(X, \mathcal A)$, since $\mathcal A$ is a $\sigma$-algebra, if $(A_n)_{n=1}^{\infty}$ is a countable collection of mutually disjoint sets in $\mathcal A$ then we are guaranteed to have $\displaystyle{\bigcup_{n=1}^{\infty} A_n \in \mathcal A}$. Now compare this with the definition above. If $X$ is any set and $\mathcal A$ is simply an algebra on $X$, if $(A_n)_{n=1}^{\infty}$ is any countable collection of mutually disjoint sets in $\mathcal A$, we are not guaranteed that the countable union of these sets is in our algebra (we are only guaranteed that finite unions are contained in our algebra).

 Definition: Let $X$ be a set and let $\mathcal A$ be an algebra of sets (not necessarily a $\sigma$-algebra) on $X$. Let $\mu$ be a measure on $\mathcal A$. The Outer Measure Induced by $\mu$ is the set function $\mu^* : \mathcal P(X) \to [0, \infty]$ defined for all subsets $E$ of $X$ by $\displaystyle{\mu^*(E) = \inf \left \{ \sum_{n=1}^{\infty} \mu (A_n) : E \subseteq \bigcup_{n=1}^{\infty} A_n \: \mathrm{and} \: A_n \in \mathcal A \right \}}$.

We will proceed to verify that $\mu^*$ is indeed an outer measure on $(X, \mathcal P(X))$. We will need to prove a few results first.

 Lemma 1: Let $X$ be a set and let $\mathcal A$ be an algebra of sets (not necessarily a $\sigma$-algebra) on $X$. If $\mu$ is a measure on $\mathcal A$, $A \in \mathcal A$, and $(A_n)_{n=1}^{\infty}$ is a countable collection of sets in $\mathcal A$ with $A \subseteq \bigcup_{n=1}^{\infty} A_n$ then $\displaystyle{\mu(A) \leq \sum_{n=1}^{\infty} \mu(A_n)}$.
• Proof: Let $A \in \mathcal A$ and let $(A_n)_{n=1}^{\infty}$ be a countable collection of sets in $\mathcal A$.
• We construct a new countable collection of sets $(A_n')_{n=1}^{\infty}$ by:
(1)
\begin{align} \quad A_1' &= A_1 \\ \quad A_n' &= A_n \setminus (A_1 \cup A_2 \cup ... \cup A_{n-1}) \quad \mathrm{if} n \geq 2 \end{align}
• Then $(A_n')_{n=1}^{\infty}$ is a countable collection of mutually disjoint sets. Furthermore, each $A_n' \in \mathcal A$ (since $\mathcal A$ is an algebra and each $A_n'$ is a finite union/intersection of sets from $\mathcal A$). Additionally we have that:
(2)
\begin{align} \quad A \subseteq \bigcup_{n=1}^{\infty} A_n = \bigcup_{n=1}^{\infty} A_n' \end{align}
• Therefore:
(3)
\begin{align} \quad A = A \cap \bigcup_{n=1}^{\infty} A_n' = \bigcup (A \cap A_n') \end{align}
• The collection of sets $(A \cap A_n')_{n=1}^{\infty}$ are also mutually disjoint, each contained in $\mathcal A$, so:
(4)
\begin{align} \quad \mu (A) = \sum_{n=1}^{\infty} \mu (A \cap A_n') \leq \mu (A_n') \leq \sum_{n=1}^{\infty} \mu (A_n) \end{align}
 Lemma 2: Let $X$ be a set and let $\mathcal A$ be an algebra of sets (not necessarily a $\sigma$-algebra) on $X$. If $\mu$ is a measure on $\mathcal A$ and $\mu^*$ is the outer measure induced by $\mu$, then for all sets $A \in \mathcal A$ we have that $\mu^*(A) = \mu(A)$.
• Proof: Let $A \in \mathcal A$. Then the set $A$ itself is a cover of $A$. So:
(5)
\begin{align} \quad \mu^*(A) = \inf \left \{ \sum_{n=1}^{\infty} \mu (A_n) : A \subseteq \bigcup_{n=1}^{\infty} A_n \: \mathrm{and} \: A_n \in \mathcal A \right \} \leq \mu(A) \quad (*) \end{align}
• Now let $(A_n)_{n=1}^{\infty}$ be a collection of sets in $\mathcal A$ such that $\displaystyle{A \subseteq \bigcup_{n=1}^{\infty} A_n}$. By Lemma 1 we have that:
(6)
\begin{align} \quad \mu (A) \leq \sum_{n=1}^{\infty} \mu(A_n) \end{align}
• By taking the infimum of both sides we get that
(7)
\begin{align} \qaud \mu(A) \leq \inf \left \{ \sum_{n=1}^{\infty} \mu (A_n) : A \subseteq \bigcup_{n=1}^{\infty} A_n \: \mathrm{and} \: A_n \in \mathcal A \right \} = \mu^*(A) \quad (**) \end{align}
• So from $(*)$ and $(**)$ we conclude that $\mu^*(A) = \mu(A)$. $\blacksquare$
 Theorem 3: Let $X$ be a set and let $\mathcal A$ be an algebra of sets (not necessarily a $\sigma$-algebra) on $X$. If $\mu$ is a measure on $\mathcal A$ then the outer measure induced by $\mu^*$ is indeed an outer measure on $(X, \mathcal P(X))$.
• Proof: Since $\emptyset \in \mathcal A$, by Lemma 2 we have that:
(8)
\begin{align} \quad \mu^*(\emptyset) = \mu(\emptyset) = 0 \end{align}
• Now let $A$ and $B$ be subsets of $X$ with $A \subseteq B$. Then every cover of $B$ by elements in $\mathcal A$ is a cover of $A$ by elements in $\mathcal A$. So
(9)
\begin{align} \quad \{ \sum_{n=1}^{\infty} \mu (A_n) : A \subseteq \bigcup_{n=1}^{\infty} A_n \: \mathrm{and} \: A_n \in \mathcal A \} \supseteq \{ \sum_{n=1}^{\infty} \mu (B_n) : B \subseteq \bigcup_{n=1}^{\infty} B_n \: \mathrm{and} \: B_n \in \mathcal A \} \end{align}
• Taking the infimum of both sides yields:
(10)
\begin{align} \quad \mu^*(A) \leq \mu^*(B) \end{align}
• Now let $(A_n)_{n=1}^{\infty}$ be any countable collection of subsets of $X$. There are two cases to consider.
• Case 1: If $\displaystyle{\sum_{n=1}^{\infty} \mu(A_n) = \infty}$ then trivially we have that:
(11)
\begin{align} \quad \mu^* \left ( \bigcup_{n=1}^{\infty} A_n \right ) \leq \sum_{n=1}^{\infty} \mu(A_n) \end{align}
• Case 2: If $\displaystyle{\sum_{n=1}^{\infty} \mu(A_n) < \infty}$ then $\mu^*(A_n) < \infty$ for each $n \in \mathbb{N}$. For each $n \in \mathbb{N}$ there exists a sequence $(A_{n, m})_{m=1}^{\infty}$ of sets in $\mathcal A$ with $\displaystyle{A_n \subseteq \bigcup_{m=1}^{\infty} A_{n, m}}$ such that:
(12)
\begin{align} \quad \sum_{m=1}^{\infty} \mu (A_{n, m}) < \mu^*(A_n) + \frac{\epsilon}{2^n} \end{align}
• Therefore:
(13)
\begin{align} \quad \mu^* \left ( \bigcup_{n=1}^{\infty} A_n \right ) \leq \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \mu (A_{n, m}) < \sum_{n=1}^{\infty} \left ( \mu^*(A_n) + \frac{\epsilon}{2^n} \right ) = \sum_{n=1}^{\infty} \mu^*(A_n) + \epsilon \end{align}
• Since $\epsilon$ is arbitrary we get that:
(14)
\begin{align} \quad \mu^* \left ( \bigcup_{n=1}^{\infty} A_n \right ) \leq \sum_{n=1}^{\infty} \mu^*(A_n) \end{align}
• So $\mu^*$ is an outer measure on $(X, \mathcal P(X))$. $\blacksquare$