# Lebesgue outer measure equals Lebesgue Inner Measure

Definition. (Lebesgue Measurable)

A set $E$ is said to be Lebesgue measurable if 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. – DanielWainfleet Jul 1 '16 at 3:26