Unions of Finite Collections of Lebesgue Measurable Sets

# The Union of a Finite Collection of Lebesgue Measurable Sets is Lebesgue Measurable

Recall from the Lebesgue Measurable Sets page that a set $E \in \mathcal P(\mathbb{R})$ is Lebesgue measurable if for all $A \in \mathcal P(\mathbb{R})$ we have that:

(1)
\begin{align} \quad m^*(A) = m^*(A \cap E) + m^*(A \cap E^c) \end{align}

The following theorem tells us that the union of every finite collection of Lebesgue measurable sets is also Lebesgue measurable.

 Theorem 1: The union of a finite collection of Lebesgue measurable sets is Lebesgue measurable.
• Proof: Let $E_1$ and $E_2$ be Lebesgue measurable sets. Then for all $A \in \mathcal P(\mathbb{R})$ we have that:
(2)
\begin{align} \quad m^*(A) = m^*(A \cap E_1) + m^*(A \cap E_1^c) \quad \mathrm{and} \quad m^*(A) = m^*(A \cap E_2) + m^*(A \cap E_2^c) \end{align}
• Then:
(3)
\begin{align} \quad m^*(A) &= m^*(A \cap E_1) + m^*(A \cap E_1^c) \\ &= m^*(A \cap E_1) + m^*([A \cap E_1^c] \cap E_2) + m^*([A \cap E_1^c] \cap E_2^c) \\ &= m^*(A \cap E_1) + m^*(A \cap E_1^c \cap E_2) + m^*(A \cap [E_1 \cup E_2]^c) \\ & \geq m^*(A \cap E_1^c \cap E_2) + m^*(A \cap [E_1 \cup E_2]^c) \end{align}
• Note the following identity: $[A \cap E_1] \cup [[A \cap E_1^c \cap E_2] = A \cap [E_1 \cup E_2]$. Therefore $A \cap E_1^c \cap E_2 \subseteq A \cap [E_1 \cup E_2]$. So $m^*(A \cap E_1^c \cap E_2) \leq m^*(A \cap [E_1 \cup E_2])$. Thus:
(4)
\begin{align} \quad m^*(A) & \geq m^*(A \cap [E_1 \cup E_2]) + m^*(A \cap [E_1 \cup E_2]^c) \end{align}
• So $(E_1 \cup E_2)$ is Lebesgue measurable.
• From this and by induction it is easy to show that the union of an finite collection of Lebesgue measurable sets is also Lebesgue measurable. $\blacksquare$
 Corollary 2: The set $\mathcal M$ of Lebesgue measurable sets is an algebra.
• Proof: For all $E_1, E_2 \in \mathcal M$ we have by the previous theorem that $(E_1 \cup E_2) \in \mathcal M$, and we've already proven that if $E \in \mathcal M$ then $E^c \in \mathcal M$. So $\mathcal M$ is an algebra. $\blacksquare$