A Property of Finite Mutually Disjoint Collections of Leb. Meas. Sets

A Property of Finite Mutually Disjoint Collections of Lebesgue Measurable Sets

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}

We now prove a result that will be used later regarding the Lebesgue outer measure of a set $A$ intersected with the union of a finite collection $\{ E_k \}_{k=1}^{n}$ of mutually disjoint Lebesgue measurable sets.

Theorem 1: Let $A \in \mathcal P(\mathbb{R})$ and let $\{ E_k \}_{k=1}^{n}$ be a finite collection of mutually disjoint Lebesgue measurable sets. Then $\displaystyle{m^* \left ( A \cap \bigcup_{k=1}^{n} E_k \right ) = \sum_{k=1}^{n} m^*(A \cap E_k)}$.

• Proof: Let $A \in \mathcal P(\mathbb{R})$. We carry this proof out by induction.

• Let $E_1$ be a Lebesgue measurable set. Then:

(2)

\begin{align} \quad m^* \left ( A \cap \bigcup_{k=1}^{1} E_k \right ) = m^* (A \cap E_1) = \sum_{k=1}^{1} m^* (A \cap E_k) \end{align}

• So the theorem is true for $n = 1$. Suppose that for $n > 1$ we have that for any finite collection $\{ E_k \}_{k=1}^{n}$ of mutually disjoint Lebesgue measurable sets that $\displaystyle{m^* \left ( A \cap \bigcup_{k=1}^{n} E_k \right ) = \sum_{k=1}^{n} m^*(A \cap E_k)}$. We want to then show that for any finite set $\{ E_k \}_{k=1}^{n+1}$ of mutually disjoint Lebesgue measurable sets satisfies $\displaystyle{m^* \left ( A \cap \bigcup_{k=1}^{n+1} E_k \right ) = \sum_{k=1}^{n+1} m^*(A \cap E_k)}$. Let $\{ E_k \}_{k=1}^{n+1}$ be a finite collection of mutually mutually disjoint Lebesgue measurable sets.

• Let $\displaystyle{A' = A \cap \bigcup_{k=1}^{n+1} E_k}$. Since $E_{n+1}$ is Lebesgue measurable, we have that:

(3)

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

(4)

\begin{align} \quad m^* \left ( A \cap \bigcup_{k=1}^{n+1} E_k \right ) &= \underbrace{m^* \left ( \left ( A \cap \bigcup_{k=1}^{n+1} E_k \right ) \cap E_{n+1} \right )}_{(*)} + \underbrace{m^* \left ( \left ( A \cap \bigcup_{k=1}^{n+1} E_k \right ) \cap E_{n+1}^c \right )}_{(**)} \\ \end{align}

• For the first term $(*)$, since $\{ E_k \}_{k=1}^{n+1}$ is a mutually disjoint collection of Lebesgue measurable sets we see that:

(5)

\begin{align} \quad m^* \left ( \left ( A \cap \bigcup_{k=1}^{n+1} E_k \right ) \cap E_{n+1} \right ) = m^* (A \cap E_{n+1}) \end{align}

• For the second term $(**)$, since $\{ E_k \}_{k=1}^{n+1}$ is a mutually disjoint collection of Lebesgue measurable sets we see that $\displaystyle{\bigcup_{k=1}^{n+1} E_k \cap E_{n+1}^c}$ is the set of all real numbers contained in $E_1$, $E_2$, …, $E_{n+1}$ and NOT contained in $E_{n+1}$, i.e., $\displaystyle{\bigcup_{k=1}^{n+1} E_k \cap E_{n+1}^c = \bigcup_{k=1}^{n} E_k}$. So we see that:

(6)

\begin{align} \quad m^* \left ( \left ( A \cap \bigcup_{k=1}^{n+1} E_k \right ) \cap E_{n+1}^c \right ) = m^* \left ( A \cap \bigcup_{k=1}^{n} E_k \right ) \overset{I.H.} = \sum_{k=1}^{n} m^*(A \cap E_k) \end{align}

• Hence:

(7)

\begin{align} \quad m^* \left ( A \cap \bigcup_{k=1}^{n+1} E_k \right ) &= m^*(A \cap E_{n+1}) + \sum_{k=1}^{n} m^*(A \cap E_k) \\ &= \sum_{k=1}^{n+1} m^* (A \cap E_k) \end{align}

• So by the principle of mathematical induction, for any $A \in \mathcal P (\mathbb{R})$ and for any finite mutually disjoint collection $\{ E_k \}_{k=1}^{n}$ of Lebesgue measurable sets we have that $\displaystyle{m^* \left ( A \cap \bigcup_{k=1}^{n} E_k \right ) = \sum_{k=1}^{n} m^*(A \cap E_k)}$. $\blacksquare$

Corollary 2: If $\{ E_k \}_{k=1}^{\infty}$ is a finite collection of disjoint Lebesgue measurable sets then $\displaystyle{m^* \left ( \bigcup_{k=1}^{n} E_k \right ) = \sum_{k=1}^{n} m^* (E_k)}$.

• Proof: Let $A = \mathbb{R}$. Then by the previous theorem the result holds. $\blacksquare$