The Leb. Mea. of a Count. Union of Mut. Dis. Leb. Measurable Sets

The Lebesgue Measure of a Countable Union of Mutually Disjoint Lebesgue Measurable Sets

Recall from the Lebesgue Measurable Sets page that if $\mathcal M$ denotes the set of Lebesgue measurable sets then the Lebesgue measure on $\mathbb{R}$ is the Lebesgue outer measure function restricted to $\mathcal M$, that is, $m : \mathcal M \to [0, \infty) \cup \{ \infty \}$ where $m(E) = m^*(E)$ for all $E \in \mathcal M$.

All of the results that we have proven for the Lebesgue outer measure hold when we restrict the hypotheses of the results to be Lebesgue measurable sets.

We now state a nice result concerning the Lebesgue measure of a countable union of mutually disjoint Lebesgue measurable sets. We first begin with a lemma which tells us that the Lebesgue measure of a finite union of mutually disjoint Lebesgue measurable sets is the sum of the Lebesgue measures of each disjoint set in the union.

Theorem 1: Let $(E_n)_{n=1}^{\infty}$ be a sequence of mutually disjoint Lebesgue measurable sets. Then $\displaystyle{m \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \sum_{n=1}^{\infty} m(E_n)}$.

• Proof: Let $(E_n)_{n=1}^{\infty}$ be a sequence of mutually disjoint Lebesgue measurable sets. Then $\displaystyle{\bigcup_{n=1}^{\infty} E_n}$ is Lebesgue measurable. By the Countable Subadditivity of the Lebesgue Outer Measure we have that:

(1)

\begin{align} \quad m \left ( \bigcup_{n=1}^{\infty} E_n \right ) = m^* \left ( \bigcup_{n=1}^{\infty} E_n \right ) \leq \sum_{n=1}^{\infty} m^*(E_n) = \sum_{n=1}^{\infty} m(E_n) \quad (*) \end{align}

• On the other hand we note that $\displaystyle{\bigcup_{k=1}^{n} E_k \subset \bigcup_{k=1}^{\infty} E_k}$. By applying the corollary on the A Property of Finite Mutually Disjoint Collections of Lebesgue Measurable Sets page we have that for all $n \in \mathbb{N}$:

(2)

\begin{align} \quad m \left ( \bigcup_{k=1}^{\infty} E_k \right ) = m^* \left ( \bigcup_{k=1}^{\infty} E_k \right ) \geq m^* \left ( \bigcup_{k=1}^{n} E_k \right ) = \sum_{k=1}^{n} m^*(E_k) = \sum_{k=1}^{n} m(E_k) \end{align}

• Taking the limit as $n \to \infty$ yields:

(3)

\begin{align} \quad m \left ( \bigcup_{k=1}^{\infty} E_k \right ) \geq \sum_{k=1}^{\infty} m(E_k) \quad (**) \end{align}

• Combining $(*)$ and $(**)$ show us that $\displaystyle{m \left ( \bigcup_{n=1}^{\infty} E_n \right ) = \sum_{n=1}^{\infty} m(E_n)}$. $\blacksquare$