Example Problems Regarding Lebesgue Measurable Sets 2

So far we have defined a subset $E$ of $\mathbb{R}$ to be a Lebesgue measurable set if for all subsets $A$ of $\mathbb{R}$ we have that:

(1)

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

We noted that to show that a set of Lebesgue measurable it suffices to show that:

(2)

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

We have already proven the following results regarding Lebesgue measurable sets.

• 1. If $E$ is Lebesgue measurable then $E^c$ is Lebesgue measurable. Furthermore, $\emptyset$ and $\mathbb{R}$ are Lebesgue measurable.

• 2. If $m^*(E) = 0$ then $E$ is a Lebesgue measurable set.

• 3. If $A$ is any set and $\{ E_k \}_{k=1}^{n}$ is a finite collection of disjoint Lebesgue measurable sets then:

(3)

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

(4)

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

• 4. The union of a finite collection of Lebesgue measurable sets is Lebesgue measurable.

• 5. The union of a countable collection of Lebesgue measurable sets is Lebesgue measurable.

• 6. Every interval $I$ is Lebesgue measurable.

• 7. If $E$ is a Lebesgue measurable set then for all $a \in \mathbb{R}$ the translate $E + a$ is a Lebesgue measurable set.

• 8. The following statements are equivalent:
  • a) $E$ is Lebesgue measurable.
  • b) For all $\epsilon > 0$ there exists an open set $O$ with $E \subseteq O$ such that $m^*(O \setminus E) < \epsilon$.
  • c) There exists a $G_{\delta}$ set $G$ such that $E \subseteq G$ and $m^*(G \setminus E) = 0$.
  • d) For all $\epsilon > 0$ there exists a closed set $F$ with $F \subseteq E$ such that $m^*(E \setminus F) < \epsilon$.
  • e) There exists an $F_{\sigma}$ set $F$ such that $F \subseteq E$ and $m^*(E \setminus F) = 0$.

We will now look at some example problems regarding Lebesgue measurable sets. More problems can be found on the Example Problems Regarding Lebesgue Measurable Sets 1 page.

Example 1

Prove that a $E \subseteq \mathbb{R}$ is Lebesgue measurable if and only if for all $\epsilon > 0$ there exists a closed set $F$ and an open set $O$ with $F \subseteq E \subseteq O$ such that $m^*(O \setminus C) < \epsilon$.

$\Rightarrow$ Suppose that $E$ is Lebesgue measurable. Then for $\displaystyle{\epsilon^* = \frac{\epsilon}{2} > 0}$ there exists an open set $O$ with $E \subseteq O$ such that:

(5)

\begin{align} \quad m^*(O \setminus E) < \epsilon^* = \frac{\epsilon}{2} \end{align}

And there exists a closed set $F$ with $F \subseteq E$ such that:

(6)

\begin{align} \quad m^*(E \setminus F) < \epsilon^* = \frac{\epsilon}{2} \end{align}

So $F \subseteq E \subseteq O$, and by countable subadditivity of the Lebesgue outer measure we have that:

(7)

\begin{align} \quad m^*(O \setminus F) = m^*([O \setminus E] \cup [E \setminus F]) \leq m^*(O \setminus E) + m^*(E \setminus F) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon \end{align}

$\Leftarrow$ Suppose that for all $\epsilon > 0$ there exists a closed set $F$ and an open set $O$ with $F \subseteq E \subseteq O$ such that $m^*(O \setminus F) < \epsilon$.

Let $\epsilon > 0$ be given and let $O$ be an open set with the property above. Note that $O \setminus E \subseteq O \setminus F$. By the monotonicity of the Lebesgue outer measure we have that:

(8)

\begin{align} \quad m^*(O \setminus E) \leq m^*(O \setminus F) < \epsilon \end{align}

So for every $\epsilon > 0$ there exists an open set $O$ with $E \subseteq O$ such that $m^* (O \setminus E) < \epsilon$. So $E$ is Lebesgue measurable.

Example 2

Let $E \subseteq \mathbb{R}$ with $m^*(E) < \infty$. Prove that $E$ is Lebesgue measurable if and only if for every open bounded interval $(a, b)$ we have that $b - a = m^*((a, b) \cap E) + m^*((a, b) \setminus E)$.

$\Rightarrow$ Suppose that $E$ is Lebesgue measurable. Let $(a, b)$ be any open bounded interval. Then:

(9)

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

Since $m^*((a, b)) = l((a, b)) = b - a$ and since $(a, b) \cap E^c = (a, b) \setminus E$ we have that:

(10)

\begin{align} \quad b - a = m^*((a, b) \cap E) + m^*((a, b) \setminus E) \end{align}

$\Leftarrow$ Suppose that for every open bounded interval $(a, b)$ we have that $b - a = m^*((a, b) \cap E) + m^*((a, b) \setminus E))$.

Let $\epsilon > 0$ be given. Since $m^*(E) < \infty$, by the definition of the Lebesgue outer measure being an infimum, there must exist a collection of open, bounded intervals $(I_n)_{n=1}^{\infty}$ such that:

(11)

\begin{align} \quad \sum_{n=1}^{\infty} l(I_n) < m(E) + \epsilon \end{align}

Since each $I_n$ is an open, bounded interval we have that for each $n \in \mathbb{N}$ that:

(12)

\begin{align} \quad l(I_n) = m^*(I_n \cap E) + m^*(I_n \setminus E) \end{align}

Therefore:

(13)

\begin{align} \quad m(E) + \epsilon & > \sum_{n=1}^{\infty} l(I_n) \\ & > \sum_{n=1}^{\infty} [m^*(I_n \cap E) + m^*(I_n \setminus E)] \\ & > \sum_{n=1}^{\infty} m^*(I_n \cap E) + \sum_{n=1}^{\infty} m^*(I_n \setminus E) \\ & > m^* \left ( \bigcup_{n=1}^{\infty} I_n \cap E \right ) + m^* \left ( \bigcup_{n=1}^{\infty} I_n \setminus E \right ) \\ & > m^* \left ( \left ( \bigcup_{n=1}^{\infty} I_n \cap E \right ) \cup \left ( \bigcup_{n=1}^{\infty} I_n \cap E^c \right ) \right ) \\ & > m^* \left ( \bigcup_{n=1}^{\infty} I_n \right ) \end{align}

Let $\displaystyle{O = \bigcup_{n=1}^{\infty} I_n}$. Then $O$ is an open set as it is a countable union of open intervals. Furthermore, by definition, $E \subseteq O$ since $\{ I_n \}_{n=1}^{\infty}$ is an open cover of $E$. The above inequality becomes:

(14)

\begin{align} \quad m(E) + \epsilon \geq m^*(O) \end{align}

Since $E \subseteq O$ and $m^*(E) < \infty$ we have that $m^*(O \setminus E) = m^*(O) - m^*(E)$. Therefore:

(15)

\begin{align} \quad m^*(O \setminus E) < \epsilon \end{align}

So for every $\epsilon > 0$ there exists an open set $O$ with $E \subseteq O$ such that $m^*(O \setminus E) < \epsilon$, so $E$ is Lebesgue measurable.