The Regularity Properties of the Lebesgue Measure

# The Regularity Properties of the Lebesgue Measure

Theorem 1 (The Regularity Properties of the Lebesgue Measure): Let $E \in \mathcal P(\mathbb{R})$. Then 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 $m^*(E \setminus F) = 0$. |

**Proof:**Let $E \in \mathcal P(\mathbb{R})$.

- $a) \implies b)$: Let $E$ be Lebesgue measurable. Then either $m^*(E) < \infty$ or $m^*(E) = \infty$.

- First, suppose that $m^*(E) < \infty$ and let $\epsilon > 0$ be given. Then there exists an open set $O$ with $E \subseteq O$ such that:

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

- Since $O$ is open, $O$ is Lebesgue measurable. The inequality above also implies that $m^*(O) < \infty$. So $m^*(O \setminus E) < \epsilon$.

- Now suppose that $m^*(E) = \infty$. Then $E$ can be expressed as:

\begin{align} \quad E = \bigcup_{n \in \mathbb{Z}} (E \cap [n, n+1)) \end{align}

- In other words, $E$ can be expressed as a countable union of mutually disjoint measurable sets with $m^*(E_n) < \infty$ for all $n \in \mathbb{Z}$. By the previous case, for each $n \in \mathbb{Z}$, since the set $E_n$ is measurable and $m^*(E_n) < \infty$ for $\displaystyle{\epsilon_n = \frac{\epsilon}{4 \cdot e^{\mid n \mid}} > 0}$ we have that there exists an open set $O_n$ with $E_n \subseteq O_n$ such that:

\begin{align} \quad m^*(O_n \setminus E_n) < \epsilon_n = \frac{\epsilon}{4 \cdot 2^{\mid n \mid}} \end{align}

- Let $\displaystyle{O = \bigcup_{n \in \mathbb{Z}} O_n}$. Then $O$ is open and $E \subseteq O$. By countable subadditivity of the Lebesgue outer measure we have that:

\begin{align} \quad m^*(O \setminus E) = m^* \left ( O \setminus \bigcup_{n \in \mathbb{Z}} \right ) = m^* \left ( \bigcup_{n \in \mathbb{Z}} (O \setminus E_n) \right ) \leq \sum_{n \in \mathbb{Z}} m^*(O_n \setminus E_n) = \sum_{n \in \mathbb{Z}} \frac{\epsilon}{4 \cdot 2^{\mid n \mid}} < \epsilon \end{align}

- Thus if $E$ is measurable then there exists an open set $O$ with $E \subseteq O$ and $m^*(O \setminus E) < \epsilon$. $\blacksquare$

- $b) \implies c)$: Suppose that for all $\epsilon > 0$ there exists an open set $O$ with $E \subseteq O$ such that $m^*(O \setminus E) < \epsilon$. Then for each $n \in \mathbb{N}$ let $\displaystyle{\epsilon_n = \frac{1}{n} > 0}$ and let $O_n$ be the open set with $E \subseteq O_n$ and $m^*(O_n) < \epsilon_n$.

- Let $\displaystyle{G = \bigcap_{n=1}^{\infty} O_n}$. Then $G$ is a $G_{\delta}$ set and $E \subseteq G$. Moreover, $G \setminus E \subseteq O_n \setminus E$ for each $n \in \mathbb{N}$. Therefore, for each $n \in \mathbb{N}$ we have that:

\begin{align} \quad m^*(G \setminus E) \leq m^*(O_n \setminus E) < \epsilon_n = \frac{1}{n} \end{align}

- Since this holds for all $n \in \mathbb{N}$ we must have that $m^*(G \setminus E) = 0$. $\blacksquare$

- $c) \implies a)$ Suppose that there exists a $G_{\delta}$-set $G$ with $E \subseteq G$ and such that $m^*(G \setminus E) = 0$. Then $G \setminus E$ is a Lebesgue measurable set. Furthermore, $G$ is measurable. Notice that $E = G \setminus (G \setminus E) = G \cap (G \setminus E)^c$. This shows that $E$ is Lebesgue measurable. $\blacksquare$