Set Functions
Table of Contents

Set Functions

We are about to develop an important function called the Lebesgue outer measure function. This special function is classified as a set function on $\mathbb{R}$. We define such functions below.

Definition: Let $\mathcal A \subseteq \mathcal P (\mathbb{R})$. A Set Function defined on $\mathcal A$ is a function $m : \mathcal A \to [0, \infty) \cup \{ \infty \}$. That is, every set in $\mathcal A$ is mapped to a nonnegative extended real-number.

Perhaps the simplest set function is the counting set function which we define below.

Definition: The Counting Set Function is the function $c : \mathcal P(\mathbb{R}) \to [0, \infty) \cup \{ \infty \}$ defined for all subsets $E \subseteq \mathbb{R}$ by $c(E) = |E|$.

So the counting set function maps sets containing $n$ elements to the nonnegative extended real-number $n$. For example, $c(\{ 1, 2, 4\}) = 3$, $c([0, 1]) = \infty$, etc…

We now define two important properties that a set function may or may not possess.

Definition: A set function $m : \mathcal P(\mathbb{R}) \to [0, \infty) \cup \{ \infty \}$ is said to be Translation Invariant if for all $y \in \mathbb{R}$ we have that $m(E) = m(E + y)$.

The set $E + y$ is defined as $E + y = \{ x + y : x \in E \}$.

Definition: A set function $m : \mathcal P(\mathbb{R}) \to [0, \infty) \cup \{ \infty \}$ is said to be Countably Additive if for any countable collection of sets $\{ E_k \}_{k=1}^{\infty}$ we have that $\displaystyle{m \left ( \bigcup_{k=1}^{\infty} E_k \right ) = \sum_{k=1}^{\infty} m(E_k)}$.
Definition: A set function $m : \mathcal P(\mathbb{R}) \to [0, \infty) \cup \{ \infty \}$ is said to be Countably Subadditive if for any countable collection of sets $\{ E_k \}_{k=1}^{\infty}$ we have that $\displaystyle{m \left ( \bigcup_{k=1}^{\infty} E_k \right ) \leq \sum_{k=1}^{\infty} m(E_k)}$.

The counting set function defined above has both of these properties and it's fairly straightforward to show that this is true.

For example, let $E \subseteq \mathbb{R}$. If $E$ is a finite set then clearly $E + y$ is a finite set and $|E| = |E + y|$, so $m(E) = m(E + y)$. If $E$ is an infinite set then clearly $E + y$ is an infinite set and $|E| = |E + y| = \infty$, so again, $m(E) + m(E + y)$.

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