The Lebesgue Outer Measure

The Lebesgue Outer Measure

We are now ready to define the Lebesgue outer measure set function. This is a set function defined for all subsets of $\mathbb{R}$.

Definition: The Lebesgue Outer Measure Function is the function $m^* : \mathcal{P}(\mathbb{R}) \to [0, \infty) \cup \{ \infty \}$ defined for all sets $E \in \mathcal P(\mathbb{R})$ by $\displaystyle{m^*(E) = \inf \left \{ \sum_{n=1}^{\infty} l(I_n) : E \subseteq \bigcup_{n=1}^{\infty} I_n \: \mathrm{and} \: \{ I_n = (a_n, b_n) \}_{n=1}^{\infty} \right \}}$. The Lebesgue Outer Measure of the Set $E$ is $m^*(E)$.

In other words, if $E \subseteq \mathcal P(\mathbb{R})$ then $m^*(E)$ is defined by taking all open-interval covers of $E$, summing the lengths of all of the open intervals in those covers, and then taking the infimum of these values.

Sometimes for brevity we simply say "outer measure" to refer to the Lebesgue outer measure of a set.

We now describe some important properties of the Lebesgue outer measure function.

Theorem 1: Let $E \in \mathcal P(\mathbb{R})$ be a finite set. Then $m^*(E) = 0$.
  • Proof: Let $E = \{ x_1, x_2, ..., x_n \}$. Then for every $\epsilon > 0$ and for all $k \in \{ 1, 2, ..., n \}$ we have that $\displaystyle{x_k \in \left ( x_k - \frac{\epsilon}{2}, x_k + \frac{\epsilon}{2} \right )}$. So:
\begin{align} \quad E \subseteq \bigcup_{k=1}^{n} \left ( x_k - \frac{\epsilon}{2}, x_k + \frac{\epsilon}{2} \right ) \end{align}
  • Furthermore, the length corresponding to the open interval cover of $E$ for each given $\epsilon$ is given by:
\begin{align} \quad \sum_{k=1}^{n} l \left ( \left ( x_k - \frac{\epsilon}{2}, x_k + \frac{\epsilon}{2} \right ) \right ) = \sum_{k=1}^{n} \epsilon = n\epsilon \end{align}
  • So for all $\epsilon > 0$ we have that
\begin{align} \quad m^*(E) < n \epsilon \end{align}
  • Since $n$ is fixed, the inequality above implies that $m^*(E) = 0$. $\blacksquare$
Theorem 2 (Monotonicity of the Lebesgue Outer Measure): Let $A, B \in \mathcal P(\mathbb{R})$. If $A \subseteq B$ then $m^*(A) \leq m^*(B)$.
  • Proof: Let $A, B \in \mathcal P(\mathbb{R})$ be such that $A \subseteq B$. If $\{ I_n = (a_n, b_n) \}_{n=1}^{\infty}$ is such that $\displaystyle{B \subseteq \bigcup_{n=1}^{\infty} I_n}$ then $\displaystyle{A \subseteq \bigcup_{n=1}^{\infty} I_n}$. Therefore:
\begin{align} \quad \left \{ \sum_{n=1}^{\infty} l(I_n) : A \subseteq \bigcup_{n=1}^{\infty} I_n \: \mathrm{and} \: \{ I_n = (a_n, b_n) \}_{n=1}^{\infty} \right \} \supseteq \left \{ \sum_{n=1}^{\infty} l(I_n) : B \subseteq \bigcup_{n=1}^{\infty} I_n \: \mathrm{and} \: \{ I_n = (a_n, b_n) \}_{n=1}^{\infty} \right \} \end{align}
Unless otherwise stated, the content of this page is licensed under Creative Commons Attribution-ShareAlike 3.0 License