The Leb. Outer Measure of Open Sets and G𝛿-Sets Containing a Set

# The Lebesgue Outer Measure of Open Sets and G𝛿-Sets Containing a Set

Recall from the Countable Subadditivity of the Lebesgue Outer Measure page that if $(A_n)_{n=1}^{\infty}$ was a sequence of subsets of $\mathbb{R}$ then:

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

We will now show that given any $\epsilon > 0$ and any subset $A$ of $\mathbb{R}$ we can find an open set $O$ containing $A$ whose Lebesgue outer measure is less than or equal to the Lebesgue outer measure of $A$ plus $\epsilon$. We will then show that given any subset $A$ of $\mathbb{R}$ we can find a $G_{\delta}$-set $G$ containing $A$ whose Lebesgue outer measure is equal to the Lebesgue outer measure of $A$.

 Theorem 1: Let $A \in \mathcal P (\mathbb{R})$. Then for any $\epsilon > 0$ there exists an open set $O$ such that $A \subseteq O$ and $m^*(O) \leq m^*(A) + \epsilon$.
• Proof: There are two cases to consider.
• Case 1: If $m^*(A) = \infty$ then for all $\epsilon > 0$ and any open set $O$ with $A \subseteq O$ we have that the inequality $m^*(O)$ is either finite or $\infty$ and trivially $m^*(O) \leq m^*(A) + \epsilon$.
• Case 2: If $m^*(A) < \infty$ then since $\displaystyle{m^*(A) = \inf \left \{ \sum_{n=1}^{\infty} l(I_n) : A \subseteq \bigcup_{n=1}^{\infty} I_n \: \mathrm{and} \: \{ I_n = (a_n, b_n) \}_{n=1}^{\infty} \right \}}$ is an infimum, by definition, for all $\epsilon > 0$ there exists an open interval cover $(I_n)_{n=1}^{\infty}$ such that:
(2)
\begin{align} \quad \sum_{n=1}^{\infty} l(I_n) < m^*(A) + \epsilon \end{align}
(3)
\begin{align} \quad m^*(O) = m^* \left ( \bigcup_{n=1}^{\infty} I_n \right ) \leq \sum_{n=1}^{\infty} m^*(I_n) = \sum_{n=1}^{\infty} l(I_n) < m^*(A) + \epsilon \end{align}
• Hence, for every $\epsilon > 0$ there exists an open set $O$ with $A \subseteq O$ such that $m^*(O) \leq m^*(A) + \epsilon$. $\blacksquare$
 Theorem 2: Let $A \in \mathcal P(\mathbb{R})$. Then there exists a $G_{\delta}$-set $G$ such that $A \subseteq G$ and $m^*(A) = M^*(G)$.

Recall that a set $G \in \mathcal P(\mathbb{R})$ is a $G_{\delta}$-set if $G$ is a countable intersection of open sets.

• Proof: There are two cases to consider.
• Case 1: If $m^*(A) = \infty$ then let $G = \mathbb{R}$. Then $G$ is a $G_{\delta}$-set with $A \subseteq G$ and $m^*(A) = \infty = m^*(\mathbb{R}) = m^*(G)$.
• Case 2: If $m^*(A) < \infty$ then by the previous theorem, for each $\epsilon > 0$ there exists an open set $O$ such that $A \subseteq O$ and $m^*(A) \leq m^*(O) + \epsilon$. For each $n \in \mathbb{N}$ let $\displaystyle{\epsilon_n = \frac{1}{n} > 0}$ and let $O_n$ be an open set such that $A \subseteq O_n$ and $m^*(A) \leq m^*(O_n) + \epsilon_n$.
• Let $\displaystyle{G = \bigcap_{n=1}^{\infty} O_n}$. Then $G$ is a $G_{\delta}$-set as $G$ is a countable intersection of open sets. Furthermore, since $A \subseteq O_n$ for each $n \in \mathbb{N}$ we have that $A \subseteq G$. Therefore:
(4)
\begin{align} \quad m^*(A) \leq m^*(G) = m^* \left ( \bigcap_{n=1}^{\infty} O_n \right ) \leq m^*(O_n) < m^*(A) + \frac{1}{n} \end{align}
• Hence for each $n \in \mathbb{N}$ we have that:
(5)
\begin{align} \quad m^*(A) \leq m^*(G) \leq m^*(A) + \frac{1}{n} \end{align}
• This implies that $m^*(A) = m^*(G)$. $\blacksquare$