# Lebesgue outer measure equals Lebesgue Inner Measure

Definition. (Lebesgue Measurable)

A set $$E$$ is said to be Lebesgue measurable if for each $$\epsilon >0$$ there exists an open set $$G$$ and a closed set $$F$$ such that $$F\subset E\subset G: m^*(G\setminus F)<\epsilon$$.

How would I go about showing the equivalence between the above definition and the definition of Lebesgue measurable set which states that a set $$E$$ is Lebesgue measurable if the Lebesgue outer measure equals the Lebesgue inner measure?

where Lebesgue inner measure for set $$E\subset\mathbb{R}$$ of a bounded interval $$[a,b]$$ is defined as $$m_*(E):=b-a-m^*([a,b]\setminus E)$$

Not too sure where to start attacking this problem. The hint that is offered in the book says to note that a open superset of $$[a,b]\setminus E$$ supplies a closed set of $$E$$.

• If $E$ is bounded then ($E$ is measurable)$\iff (m^i(E)=m^o(E)).$ But suppose $D \subset [0,1]$ and $D$ is not measurable, and $E=D\cup [1,\infty).$ Then $m^i(E)=m^o(E)=\infty$ but $E$ is not measurable. An unbounded set is measurable iff its intersection with every bounded open set (or with every bounded measurable set) is measurable. Commented Jul 1, 2016 at 3:26

## 2 Answers

Partial answer. Let $$m^i$$ denote inner measure and let $$m^o$$ denote outer measure. For each $$n\in\Bbb N$$ suppose $$F_n$$ is open and $$G_n$$ is closed with $$G_n\subseteqq E\subseteqq F_n$$ and $$m^o(F_n-G_n)<1/n.$$

Then $$m^o(G_n)+m^o(F_n-G_n)\ge m^o(G_n\cup (F_n-G_n))=m^o(F_n).$$

So $$m^o(G_n)\ge m^o(F_n)-m^o(F_n-G_n)\ge m^o(F_n)-1/n\ge m^o(E)-1/n.$$ Therefore $$m^i(E)\ge \sup_n m^o(G_n)\ge \sup_n \, m^o(E)-1/n=m^o(E).$$ And for any $$E$$ we have $$m^i(E)\le \sup_{S\subseteqq E} \,m^o(S)=m^o(E).$$

• In case any of the measures are $\infty$ we define $\infty \pm r=\infty +\infty=\infty$ for any $r\in\Bbb R$. Commented Jun 27, 2022 at 18:33

Well, for open sets, this is clear, since there is nothing to prove. If we are talking about closed sets on the other hand, then we can consider the bounded ones so by the Heinie Borel thm, we have compactness as is requested by the definition of the Outer measure by Lebesgue. Then, all that is left is to show what happens when the set is nether open nor closed.