Inner and Outer Approximation of Lebesgue Measurable Sets

Let $E\subseteq\mathbb{R}$ . Then each of the following four assertions is equivalent to the measurability of $E$ .

(Outer Approximation by Open Sets and $G_\delta$ Sets)

(i) For each $\epsilon>0$ , there is an open set $G$ containing $E$ for which $m^*(G\setminus E)<\epsilon$ .

(ii) There is a $G_\delta$ set $G$ containing $E$ for which $m^*(G\setminus E)=0$ .

(Inner Approximation by Closed Sets and $F_\sigma$ Sets)

(iii) For each $\epsilon>0$ , there is a closed set $F$ contained in $E$ for which $m^*(E\setminus F)<\epsilon$ .

(iv) There is an $F_\sigma$ set $F$ contained in $E$ for which $m^*(E\setminus F)=0$ .

Proof:
( $E$ measurable implies (i)):

Assume $E$ is measurable. Let $\epsilon>0$ . First we consider the case where $m^*(E)<\infty$ . By the definition of outer measure, there is a countable collection of open intervals $\{I_k\}_{k=1}^\infty$ which covers $E$ and satisfies $\displaystyle \sum_{k=1}^\infty l(I_k)<m^*(E)+\epsilon.$

Define $G=\bigcup_{k=1}^\infty I_k$ . Then $G$ is an open set containing $E$ . By definition of the outer measure of $G$ , $\displaystyle m^*(G)\leq\sum_{k=1}^\infty l(I_k)<m^*(E)+\epsilon.$

Since $E$ is measureable and has finite outer measure, by the excision property, $\displaystyle m^*(G\setminus E)=m^*(G)-m^*(E)<\epsilon.$

Now consider the case that $m^*(E)=\infty$ . Since $\mathbb{R}$ is $\sigma$ -finite, $E$ may be expressed as the disjoint union of a countable collection $\{E_k\}_{k=1}^\infty$ of measurable sets, each of which has finite outer measure.

By the finite measure case, for each $k\in\mathbb{N}$ , there is an open set $G_k$ containing $E_k$ for which $m^*(G_k\setminus E_k)<\epsilon/2^k$ . The set $G=\bigcup_{k=1}^\infty G_k$ is open, it contains $E$ and $\displaystyle G\setminus E=(\bigcup_{k=1}^\infty G_k)\setminus E\subseteq\bigcup_{k=1}^\infty (G_k\setminus E_k).$

Therefore
$\begin{aligned} m^*(G\setminus E)&\leq\sum_{k=1}^\infty m^*(G_k\setminus E_k)\\ &<\sum_{k=1}^\infty\epsilon/2^k\\ &=\epsilon. \end{aligned}$
Thus property (i) holds for $E$ .

((i) implies (ii)):

Assume property (i) holds for $E$ . For each $k\in\mathbb{N}$ , choose an open set $O_k$ that contains $E$ such that $m^*(O_k\setminus E)<1/k$ . Define $G=\bigcap_{k=1}^\infty O_k$ . Then $G$ is a $G_\delta$ set that contains $E$ . Note that for each $k$ , $\displaystyle G\setminus E\subseteq O_k\setminus E.$

By monotonicity of outer measure, $\displaystyle m^*(G\setminus E)\leq m^*(O_k\setminus E)<1/k.$

Thus $m^*(G\setminus E)=0$ and hence (ii) holds.

((ii) $\implies E$ is measurable):

Now assume property (ii) holds for $E$ . Since a set of measure zero is measurable, $G\setminus E$ is measurable. $G$ is a $G_\delta$ set and thus measurable. Since measurable sets form a $\sigma$ -algebra, $E=G\cap(G\setminus E)^c$ is measurable.

((i) $\implies$ (iii)):

Assume condition (i) holds. Note that $E^c$ is measurable iff $E$ is measurable. Thus there exists an open set $G\supseteq E^c$ such that $m^*(G\setminus E^c)<\epsilon$ .