Table of Contents

Example Problems Regarding Lebesgue Measurable Sets 1
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)We noted that to show that a set of Lebesgue measurable it suffices to show that:
(2)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:
 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 2 page.
Example 1
Let $E \subseteq \mathbb{R}$ be such that $m^*(E) > 0$. Prove that there exists a subset $E' \subseteq E$ that is bounded such that $m^*(E') > 0$.
Suppose not. In other words, suppose that for every bounded subset $E' \subseteq E$ we have that $m(E') = 0$.
For each $k \in \mathbb{Z}$ let:
(5)Then $(I_k)_{k \in \mathbb{Z}}$ is a countable collection of closed and bounded intervals. For each $k \in \mathbb{Z}$ let:
(6)Then $(E_k)_{k \in \mathbb{Z}}$ is a countable collection of sets whose union is all of $E$. Furthermore, for each $k \in \mathbb{Z}$ we have that $E_k$ is bounded and $E_k \subseteq E$. Therefore by hypothesis, for each $k \in \mathbb{Z}$:
(7)Now from the countable subadditivity of the Lebesgue outer measure we have that:
(8)This contradicts $m^*(E) > 0$. So there must exists a bounded subset $E' \subseteq E$ such that $m^*(E') > 0$.
Example 2
Let $E \subseteq \mathbb{R}$ be such that $m(E) = 0$. Prove that for all $a \in \mathbb{R}$, we have that $m(E + a) = 0$.
Since $m(E) = 0$ we have that $E$ is a Lebesgue measurable set. We know that the translate of every Lebesgue measurable set is Lebesgue measurable, so $E + a$ is Lebesgue measurable, and furthermore:
(9)Example 3
Prove that every open set in $\mathbb{R}$ is Lebesgue measurable. Prove that every closed in $\mathbb{R}$ is Lebesgue measurable.
Let $O \subseteq \mathbb{R}$ be an open set. We know that every open set in $\mathbb{R}$ is a countable disjoint union of open intervals, that is, $\displaystyle{O = \bigcup_{n=1}^{\infty} I_n}$ where each $I_n$ is an open interval. We have already seen that every open interval is Lebesgue measurable. So $O$ is a countable union of Lebesgue measurable sets, so $O$ is Lebesgue measurable.
Let $C \subseteq \mathbb{R}$ be a closed set. Then $C^c$ is an open set. So $C^c$ is a Lebesgue measurable set. But the complement of a Lebesgue measurable set is Lebesgue measurable. So $C = (C^c)^c$ is Lebesgue measurable.