The Lebesgue Measurability of Intervals

# The Lebesgue Measurability of Intervals

Recall from The Union of a Countable Collection of Lebesgue Measurable Sets is Lebesgue Measurable page that the set of all Lebesgue measurable sets is a $\sigma$-algebra. This means that:

• 1. For any sequence $(A_n)_{n=1}^{\infty}$ of Lebesgue measurable sets we have that $\displaystyle{\bigcup_{n=1}^{\infty} A_n}$ is Lebesgue measurable.
• 2. For all Lebesgue measurable sets $A$ we have that $A^c$ is Lebesgue measurable.

We use this fact to prove that every interval is Lebesgue measurable.

 Theorem 1: Every interval $I \subseteq \mathcal P(\mathbb{R})$ is Lebesgue measurable.
• Proof: We break this theorem up into cases.
• Case 1 (Intervals of the form $(a, \infty)$): Let $A \in \mathcal P(\mathbb{R})$.
• If $m^*(A) = \infty$ then trivially $\infty = m^*(A) \geq m^*(A \cap I) + m^*(A \cap I^c)$.
• So, assume that $m^*(A) < \infty$. Let $\epsilon > 0$ be given. Since $\displaystyle{m^*(A) = \inf \left \{ \sum_{n \in \mathbb{N}} l(I_n) : A \subseteq \bigcup_{n \in \mathbb{N}} I_n \: \mathrm{and} \: \{ I_n = (a_n, b_n) : n \in \mathbb{N} \} \right \}}$, by the properties of infimum, there exists an open interval cover $(I_n)_{n=1}^{\infty}$ such that:
(1)
\begin{align} \quad \sum_{n=1}^{\infty} l(I_n) \leq m^*(A) + \epsilon \end{align}
• For each $n \in \mathbb{N}$ define $I_n' = I \cap I_n$ and $I_n'' = I^c \cap I_n$. Notice that each $I_n'$ and $I_n''$ are either intervals or the emptyset and that:
(2)
\begin{align} \quad l(I_n) = l(I_n') + l(I_n'') = m^*(I_n') + m^*(I_n'') \end{align}
• Hence:
(3)
\begin{align} \quad \sum_{n=1}^{\infty} m^*(I_n') + \sum_{n=1}^{\infty} m^*(I_n'') \leq m^*(A) + \epsilon \end{align}
• Now notice that:
(4)
\begin{align} \quad A \cap I \subseteq \bigcup_{n=1}^{\infty} I_n' \quad \mathrm{and} \quad A \cap I^c \subseteq \bigcup_{n=1}^{\infty} I_n'' \end{align}
• This implies that:
(5)
\begin{align} \quad m^*(A \cap I) \leq m^* \left ( \bigcup_{n=1}^{\infty} I_n' \right ) \quad \mathrm{and} \quad m^*(A \cap I^c) \leq m^* \left ( \bigcup_{n=1}^{\infty} I_n'' \right ) \end{align}
• By countable subadditivity of the Lebesgue outer measure we have that:
(6)
\begin{align} \quad m^*(A \cap I) \leq \sum_{n=1}^{\infty} m^*(I_n') \quad \mathrm{and} \quad m^*(A \cap I^c) \leq \sum_{n=1}^{\infty} m^*(I_n'') \end{align}
• Therefore $m^*(A \cap I) + m^*(A \cap I^c) < m^*(A) + \epsilon$. Since $\epsilon > 0$ is arbitrary, this implies that $m^*(A) \geq m^*(A \cap I) + m^*(A \cap I^c)$ for all $A \in \mathcal P(\mathbb{R})$. So $I = (a, \infty)$ is Lebesgue measurable.
• For the remaining cases of this theorem we use the fact that the set of Lebesgue measurable sets is a $\sigma$-algebra on $\mathbb{R}$.
• Case 2 (Intervals of the form $(-\infty, a]$): Since every interval of the form $(a, \infty)$ is Lebesgue measurable, the complement, $(a, \infty)^c = (-\infty, a]$, is Lebesgue measurable.
• Case 3 (Intervals of the form $(-\infty, a)$): Since every interval of the form $(-\infty, a]$ is Lebesgue measurable, we note that:
(7)
\begin{align} \quad (-\infty, a) = \bigcup_{n=1}^{\infty} \left ( -\infty, a - \frac{1}{n} \right ] \end{align}
• A countable union of Lebesgue measurable sets is Lebesgue measurable, so $(-\infty, a)$ is Lebesgue measurable.
• Case 4 (Intervals of the form $[a, \infty)$): Since every interval of the form $(-\infty, a)$ is Lebesgue measurable, the complement, $(-\infty, a)^c = [a, \infty)$, is Lebesgue measurable.
• Case 5 (Intervals of the form $(a, b)$, $[a, b]$, $[a, b)$, and $(a, b]$): We have that:
(8)
\begin{align} \quad (a, b) &= (-\infty, b) \cap (a, \infty) \\ \quad [a, b] &= (-\infty, b] \cap [a, \infty) \\ \quad [a, b) &= (-\infty, b) \cap [a, \infty) \\ \quad (a, b] &= (-\infty, b] \cap (a, \infty) \end{align}
• The intersection of two Lebesgue measurable sets is Lebesgue measurable, so $(a, b)$, $[a, b]$, $[a, b)$, and $(a, b]$ are all Lebesgue measurable. $\blacksquare$