Lebesgue Measurable Sets

We have already established that if $E \in \mathcal P(\mathbb{R})$ then the Lebesgue outer measure of $E$ is defined to be:

(1)

\begin{align} \quad m^*(E) = \inf \left \{ \sum_{n=1}^{\infty} l(I_n) : E \subseteq \bigcup_{n=1}^{\infty} \: \mathrm{and} \: \{ I_n = (a_n, b_n) \}_{n=1}^{\infty} \right \} \end{align}

We will now distinguish a certain collection of subsets of $\mathbb{R}$ called the Lebesgue measurable sets.

Definition: A set $E \in \mathcal P(\mathbb{R})$ is Lebesgue Measurable if for all $A \in \mathcal P(\mathbb{R})$ we have that $m^*(A) = m^*(A \cap E) + m^*(A \cap E^c)$.

Notice that for all $E \in \mathcal P(\mathbb{R})$ we have that $\mathbb{R} = E \cup E^c$. By countable subadditivity of the Lebesgue outer measure we have that:

(2)

\begin{align} \quad m^*(A) = m^*(A \cap (E \cup E^c)) = m^*((A \cap E) \cup (A \cap E^c)) \leq m^*(A \cap E) + m^*(A \cap E^c) \end{align}

So to prove that $E$ is Lebesgue measurable we only need to prove that the inequality $m^*(A) \geq m^*(A \cap E) + m^*(A \cap E^c)$ holds.

Proposition 1: A set $E \in \mathcal P(\mathbb{R})$ is Lebesgue measurable if and only if $E^c$ is Lebesgue measurable.

• Proof: Notice that:

(3)

\begin{align} \quad m^*(A) = m^*(A \cap E) + m^*(A \cap E^c) = m^*(A \cap E^c) + m^*(A \cap E) \end{align}

• From this we see that $E$ is Lebesgue measurable if and only if $E^c$ is Lebesgue measurable. $\blacksquare$

Proposition 2: The sets $\emptyset$ and $\mathbb{R}$ are Lebesgue measurable.

• Proof: For all $A \in \mathcal P(\mathbb{R})$ we have that:

(4)

\begin{align} \quad m^*(A \cap \emptyset) + m^*(A \cap \emptyset^c) &= m^*(\emptyset) + m^*(A \cap \mathbb{R}) = 0 + m^*(A) = m^*(A) \end{align}

• Therefore $\emptyset$ is Lebesgue measurable. By the previous proposition, $\emptyset^c = \mathbb{R}$ is also Lebesgue measurable. $\blacksquare$

Proposition 3: Let $E \in \mathcal P(\mathbb{R})$. If $m^*(E) = 0$ then $E$ is Lebesgue measurable.

• Proof: Let $m^*(E) = 0$ and let $A \in \mathcal P(\mathbb{R})$. Then $A \cap E \subseteq E$. Therefore:

(5)

\begin{align} \quad m^*(A \cap E) \leq m^*(E) = 0 \end{align}

• So $m^*(A \cap E) = 0$. Furthermore, $A \cap E^c \subseteq A$. So $m^*(A \cap E^c) \leq m^*(A)$. Therefore:

(6)

\begin{align} \quad m^*(A) \geq m^*(A \cap E^c) + 0 = m^*(A \cap E^c) + m^*(A \cap E) \end{align}

• So $E$ is Lebesgue measurable. $\blacksquare$