Basic Properties of Measure Spaces

Recall from the General Measurable Spaces and Measure Spaces page that if $X$ is a set and $\mathcal A$ is a $\sigma$-algebra on $X$ then the pair $(X, \mathcal A)$ is called a measurable space. We said that the sets $E \in \mathcal A$ are called measurable sets.

We said that a function $\mu : \mathcal A \to [0, \infty]$ is called a measure on $\mathcal A$ if $\mu (\emptyset) = 0$ and if for every countable collection of disjoint measurable sets $(E_n)$ we have that $\displaystyle{\mu \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \sum_{n=1}^{\infty} \mu (E_n)}$, and we defined the triple $(X, \mathcal A, \mu)$ to be a measure space.

We will now give some basic properties of measure spaces.

Theorem 1 (The Finite Additivity Property): Let $(X, \mathcal A, \mu)$ be a measure space. Then if $\{ E_1, E_2, ..., E_n \}$ is a finite collection of mutually disjoint measurable sets then $\displaystyle{\mu \left ( \bigcup_{k=1}^{n} E_k \right ) = \sum_{k=1}^{n} \mu(E_k)}$.

• Proof: Define a countably infinite collection of mutually disjoint measure sets $(E_k)_{k=1}^{\infty}$ for $k \geq n+1$ by $E_k = \emptyset$. Then by the countable additivity property of measures we have that:

(1)

\begin{align} \quad \mu \left ( \bigcup_{k=1}^{n} E_k \right ) = \mu \left ( \bigcup_{k=1}^{\infty} E_k \right ) = \sum_{k=1}^{\infty} \mu (E_k) = \sum_{k=1}^{n} \mu (E_k) \quad \blacksquare \end{align}

Theorem 2 (The Monotonicity Property): Let $(X, \mathcal A, \mu)$ be a measure space. Then if $A$ and $B$ are measurable sets with $A \subseteq B$ then $\mu (A) \leq \mu (B)$.

• Proof: Let $A$ and $B$ be measurable sets with $A \subseteq B$. Then we have that:

(2)

\begin{align} \quad B = (B \setminus A) \cup A \end{align}

• By Theorem 2 we have that:

(3)

\begin{align} \quad \mu (B) = \mu (B \setminus A) + \mu (A) \geq \mu (A) \end{align}

• Hence $\mu (A) \leq \mu (B)$. $\blacksquare$

Theorem 3 (The Excision Property): Let $(X, \mathcal A, \mu)$ be a measure space. Then if $A$ and $B$ are measurable sets with $A \subseteq B$ and $\mu (A) < \infty$ then $\mu (B \setminus A) = \mu (B) - \mu (A)$.

• Proof: If $\mu (A) < \infty$ as above, we can rearrange the equality from the previous theorem to get:

(4)

\begin{align} \quad \mu (B \setminus A) = \mu(B) - \mu(A) \quad \blacksquare \end{align}

Theorem 4 (The Continuity of Measure): Let $(X, \mathcal A, \mu)$ be a measure space. a) If $(A_n)_{n=1}^{\infty}$ is a sequence of ascending measurable sets ($A_n \subseteq A_{n+1}$ for all $n \in \mathbb{N}$) then $\displaystyle{\mu \left ( \bigcup_{n=1}^{\infty} A_n \right ) = \lim_{n \to \infty} \mu (A_n)}$. b) If $(B_n)_{n=1}^{\infty}$ is a sequence of descending measurable sets ($B_n \supseteq B_{n+1}$ for all $n \in \mathbb{N}$) and $\mu (B_1) < \infty$ then $\displaystyle{\mu \left ( \bigcap_{n=1}^{\infty} B_n \right ) = \lim_{n \to \infty} \mu (B_n)}$.