Set Functions Examples 1

Recall from the Set Functions page that if $\mathcal A \subseteq \mathcal P(\mathbb{R})$ then a set function on $\mathcal A$ is a function $m : \mathcal A \to [0, \infty) \cup \{ \infty \}$.

• We said that a set function defined on $\mathcal P(\mathbb{R})$ is translation invariant if for all $y \in \mathbb{R}$ we have that:

(1)

\begin{align} \quad m(E) = m(E + y) \end{align}

• We said that a set function on $\mathcal P(\mathbb{R})$ is countably additive if for every countable collection of sets $\{ E_k \}_{k=1}^{\infty}$ we have that:

(2)

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

• And countably subadditive if:

(3)

\begin{align} \quad m \left ( \bigcup_{k=1}^{\infty} E_k \right ) \leq \sum_{k=1}^{\infty} m(E_k) \end{align}

The definitions of translation invariance and countable additivity can also be defined when $\mathcal A$ is a $\sigma$-algebra.

Example 1

Prove that if $\mathcal A$ is a $\sigma$-algebra, $m : \mathcal A \to [0, \infty) \cup \{ \infty \}$ is a set function with the countable additivity property, and $A, B \in \mathcal A$ are such that $A \subseteq B$, then $m(A) \leq m(B)$.

Let $A \subseteq B$. Then the set $B \setminus A = B \cap A^c \in \mathcal A$ since $\mathcal A$ is a $\sigma$-algebra. Consider the collection $\{ A, B \setminus A \}$. This is a countable collection of sets from $\mathcal A$. Since $m$ has the countable additivity property we have that:

(4)

\begin{align} \quad m(A) \leq m(A) + m(B \setminus A) = m(B) \end{align}

Example 2

Prove that if $\mathcal A$ is a $\sigma$-algebra, $m : \mathcal A \to [0, \infty) \cup \{ \infty \}$ is a set function with the countable additivity property, and if there exists a set $A \in \mathcal A$ with $m(A) < \infty$ then $m(\emptyset) = 0$.

$\{ \emptyset, A \}$ is a countable collection of sets from $\mathcal A$. So by the countable additivity property of $m$ we have that:

(5)

\begin{align} \quad m(A) = m(A \cup \emptyset) = m(A) + m(\emptyset) \end{align}

Therefore $m(\emptyset) = 0$.