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 $E$.
Best Answer
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 $m^i(E)\le \sup_{S\subseteqq E} \,m^o(S)=m^o(E).$