The Lebesgue Outer Measure of Intervals

# The Lebesgue Outer Measure of Intervals

Recall from The Lebesgue Outer Measure page that if $E \in \mathcal P(\mathbb{R})$ then the Lebesgue outer measure of $E$ is defined as:

(1)\begin{align} \quad 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 \} \end{align}

We will now prove a nice theorem which tells us that the Lebesgue outer measure of an interval $I$ will equal the length of $I$.

Theorem 1: Let $I \in \mathcal P(\mathbb{R})$ be an interval. Then $\displaystyle{m^*(I) = l(I)}$. |

**Proof:**We break the proof of the above theorem into several cases. We first cover bounded open intervals and then move on to closed intervals and unbounded intervals.

**Case 1 (Bounded Closed Intervals):**Let $I = [a, b]$ where $a, b \in \mathbb{R}$. Let $\{ I_n \}_{n=1}^{\infty}$ be an open interval cover of $I$, that is:

\begin{align} \quad I = [a, b] \subseteq \bigcup_{n=1}^{\infty} I_n \end{align}

- We first establish that $\displaystyle{m^*([a, b]) \leq l([a, b])}$. Let $\epsilon > 0$. Then $\displaystyle{\left \{ \left ( a - \frac{\epsilon}{2}, b + \frac{\epsilon}{2} \right ) \right \}}$ is an open interval cover of $[a, b]$ and so:

\begin{align} \quad m^*([a, b]) \leq l \left ( \left ( a - \frac{\epsilon}{2}, b + \frac{\epsilon}{2} \right ) \right ) = b - a + \epsilon = l([a, b]) + \epsilon \end{align}

- So for all $\epsilon > 0$ we have that $m^*([a, b]) \leq l([a, b]) + \epsilon$ which implies that:

\begin{align} \quad m^*([a, b]) \leq l([a, b]) \quad (*) \end{align}

- We now establish that $\displaystyle{m^*([a, b]) \geq l([a, b])}$. Let $\{ I_n \}_{n=1}^{\infty}$ be an open interval cover of $I$, that is, $\displaystyle{I = [a, b] \subseteq \bigcup_{n=1}^{\infty} I_n}$. Since $I$ is a closed and bounded subset of $\mathbb{R}$ we have that $I$ is compact. So the open interval cover $\{ I_n \}_{n=1}^{\infty}$ has a finite sub cover, say $\{ I_1, I_2, ..., I_n \}$. We may assume without loss of generality that this subcover consists of bounded intervals, $I_k = (a_k, b_k)$ for all $k \in \{ 1, 2, ..., n \}$ for if one of $I_k$ is unbounded we are done.

- Since $\displaystyle{I = [a, b] \subseteq \bigcup_{k=1}^{n} I_k}$ there exists a $k \in \{ 1, 2, ... n \}$ such that $a \in (a_k, b_k)$. Without loss of generality, assume that $k = 1$, i.e., $a \in (a_1, b_1)$. Then clearly $a_1 < a$. Furthermore, either $b_1 \geq b$ or $b_1 < b$.

- If $b_1 > b$ then $[a, b] \subseteq (a_1, b_1)$ and so

\begin{align} \quad \quad \sum_{k=1}^{n} l(I_k) = l(I_1) + \sum_{k=2}^{n} l(I_k) = (b_1 - a_1) + \sum_{k=2}^{n} l(I_k) > b - a = l([a, b]) \end{align}

- If $b_1 = b$ then the same inequality above holds.

- If $b_1 < b$ then since $b_1 \in [a, b]$ and $\displaystyle{[a,b] \subseteq \bigcup_{k=1}^{n} I_k}$ there exists a $k \in \{ 2, 3, ..., n \}$ such that $b_1 \in (a_k, b_k)$. Without loss of generality we may assume that $k = 2$, that is, $b_1 \in (a_2, b_2)$. Then clearly $a_2 < b_1$. Furthermore either $b_2 \geq b$ or $b_2 < b$.

- If $b_2 > b$ then $[a, b] \subseteq (a_1, b_1) \cup (a_2, b_2)$ and so:

\begin{align} \quad \quad \sum_{k=1}^{n} I_k = l(I_1) +l(I_2) + \sum_{k=3}^{n} l(I_k) = (b_1 - a_1) + (b_2 - a_2) - (b_1 - a_2) + \sum_{k=3}^{n} l(I_k) = b_2 - a_1 + \sum_{k=3}^{n} l(I_k) > b - a = l([a, b]) \end{align}

- If $b_2 = b$ then the same inequality above holds.

- If $b_2 < b$ then since $b_2 \in [a, b]$ and $\displaystyle{[a, b] \subseteq \bigcup_{k=1}^{n} I_k}$ there exists a $k \in \{ 3, 4, ..., n \}$ such that $b_2 \in (a_k, b_k)$. Without loss of generality we may assume that $k = 3$, that is, $b_2 \in (a_3, b_3)$. We continue the process above which will eventually terminate at some $t \in \{ 1, 2, ..., n \}$ with $[a, b] \subseteq \bigcup_{k=1}^{t} (a_k, b_k)$ with $a_1 < a$ and $b < b_t$ with:

\begin{align} \quad \quad \sum_{k=1}^{n} l(I_k) = \sum_{k=1}^{t} l(I_k) + \sum_{k=t}^{n} l(I_k) = (b_t - a_1) + \sum_{k=t}^{n} l(I_k) > b - a = l([a, b]) \end{align}

- So for any arbitrary open interval cover $\{ I_n \}_{n=1}^{\infty}$ of $[a, b]$ we have that:

\begin{align} \quad \quad \sum_{n=1}^{\infty} l(I_n) \geq \sum_{k=1}^{n} l(I_k) > b - a = l([a, b]) \end{align}

- If we take the infimum of this set of numbers we have that

\begin{align} \quad m^*([a, b]) \geq l([a, b]) \quad (**) \end{align}

- From $(*)$ and $(**)$ we conclude that $m^*([a, b]) = l([a, b])$.

**Case 2 (Bounded Open and Half-Open Intervals):**Let $I$ be a half-open or open interval that is bounded with endpoints $a$ and $b$, $a < b$. Then the closure of $I$ is $\overline{I} = [a, b]$. Note that $I \subset \overline{I}$ and so $m^*(I) \leq m^*(\overline{I}) = b - a = l(I)$ by the previous case.

- Furthermore, we note that for all $\epsilon > 0$ that:

\begin{align} \quad \left [ a + \frac{\epsilon}{2}, b - \frac{\epsilon}{2} \right ] \subset I \end{align}

- So for all $\epsilon > 0$ we have:

\begin{align} \quad m^*\left ( \left [ a + \frac{\epsilon}{2}, b - \frac{\epsilon}{2} \right ] \right ) = b - a - \epsilon \leq m^*(I) \end{align}

- Therefore $l(I) \leq m^*(I)$ and we can conclude that $m^*(I) = l(I)$ for every bounded open and half-open interval $I$.

**Case 3 (Unbounded Intervals):**Let $I$ be unbounded. Then $l(I) = \infty$.

- Since $I$ is unbounded, for every $M \in \mathbb{R}$, $M > 0$ there exists a bounded interval $I_M$ with $l(I_M) = M$ and $I_M \subset I$. So for all $M > 0$:

\begin{align} \quad M = l(I_M) \leq l(I) \end{align}

- Therefore $M \leq m^*(I)$ for all $M > 0$ which implies that $m^*(I) = \infty$, i.e., $m^*(I) = l(I)$. $\blacksquare$