## 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)

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)

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 $m^*(G\setminus E^c)<\epsilon$.

Define $F=\mathbb{R}\setminus G$ which is closed. Note that $F\subseteq E$, and $m^*(E\setminus F)=m^*(G\setminus E^c)<\epsilon$.

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

Similar.

((ii) $\iff$(iv)):

Similar idea. Note that a set is $G_\delta$ iff its complement is $F_\sigma$. 