Unions of Countable Collections of Lebesgue Measurable Sets

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

Recall from The Union of a Finite Collection of Lebesgue Measurable Sets is Lebesgue Measurable that the union of a finite collection of Lebesgue measurable sets is also Lebesgue measurable.

We will now establish a stronger result in that the union of a countable collection of Lebesgue measurable sets is also Lebesgue measurable.

 Theorem 1: The union of a countable collection of Lebesgue measurable sets is Lebesgue measurable.
• Proof: Let $(E_n)_{n=1}^{\infty}$ be any sequence of Lebesgue measurable sets and let $\displaystyle{E = \bigcup_{n=1}^{\infty} E_n}$. We want to show that $E$ is also a Lebesgue measurable set.
• We have already established that the collection of all Lebesgue measurable sets is a $\sigma$-algebra. From one of the properties on the Properties of Algebras of Sets 2 we have that there exists a sequence $(E_n')_{n=1}^{\infty}$ of Lebesgue measurable sets that are mutually disjoint and such that $\displaystyle{\bigcup_{n=1}^{\infty} E_n = \bigcup_{n=1}^{\infty} E_n'}$. So without loss of generality we can assume that $(E_n)_{n=1}^{\infty}$ itself is a sequence of mutually disjoint Lebesgue measurable sets.
• Let $A \in \mathcal P (\mathbb{R})$. Since the union of a finite collection of Lebesgue measurable sets is Lebesgue measurable we have that for each $n \in \mathbb{N}$:
(1)
\begin{align} \quad m^* (A) = \underbrace{m^* \left ( A \cap \left [ \bigcup_{k=1}^{n} E_k \right ] \right )}_{(**)} + \underbrace{m^* \left ( A \cap \left [ \bigcup_{k=1}^{n} E_k \right ]^c \right )}_{(*)} \end{align}
• For the term $(*)$, note that $\displaystyle{\bigcup_{k=1}^{n} E_k \subset \bigcup_{k=1}^{\infty} E_k = E }$. Therefore $\displaystyle{\left ( \bigcup_{k=1}^{n} E_k \right )^c \supset \left ( \bigcup_{k=1}^{\infty} E_k \right )^c = E^c}$. So $\displaystyle{A \cap \left ( \bigcup_{k=1}^{n} E_k \right )^c \supset A \cap E^c}$ and:
(2)
\begin{align} \quad m^*\left (A \cap \left [ \bigcup_{k=1}^{n} E_k \right ]^c \right ) \geq m^* (A \cap E^c) \end{align}
(3)
\begin{align} \quad m^* \left ( A \cap \left [ \bigcup_{k=1}^{n} E_k \right ] \right ) = \sum_{k=1}^{n} m^* (A \cap E_k) \end{align}
• Hence for each $n \in \mathbb{N}$ we have that:
(4)
\begin{align} \quad m^*(A) \geq \sum_{k=1}^{n} m^* (A \cap E_k) + m^* (A \cap E^c) \end{align}
• We take the limit as $n \to \infty$ of both sides above and use the countable subadditivity of the Lebesgue outer measure to get:
(5)
\begin{align} \quad m^*(A) \geq \sum_{k=1}^{\infty} m^* (A \cap E_k) + m^* (A \cap E^c) & \geq m^* \left ( \bigcup_{n=1}^{\infty} (A \cap E_n) \right ) + m^* (A \cap E^c) \\ & \geq m^* \left ( A \cap \left [ \bigcup_{n=1}^{\infty} E_n \right ] \right ) + m^* (A \cap E^c) \\ & \geq m^* (A \cap E) + m^*(A \cap E^c) \\ \end{align}
• Therefore $E$ is Lebesgue measurable. $\blacksquare$
 Corollary 2: The collection $\mathcal M$ of Lebesgue measurable sets is a $\sigma$-algebra.
• Proof: From above, $\mathcal M$ is closed under countable unions and we have already proven that if $E \in \mathcal M$ then $E^c \in \mathcal M$. So $\mathcal M$ is a $\sigma$-algebra. $\blacksquare$