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

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

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

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

(12)Therefore:

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

(15)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.