Equivalent Definition of Measurable set $E$  is measurable (lebesgue and on Real line) if for each $\epsilon >0$ ,there is a closed set  $F$ and open set $O$ for which $F\subseteq E \subseteq O $ and $ m^{*}(O-F)<\epsilon $ ?
Edit:-
Definition of Measurable set:-
A set $E$ is measurable if for each set $A$ of Real number , $m^{*}(A)=m^{*}(A\cap E)+m^{*}(A\cap E^{c})$.
 A: Yes they are equivalent. Consider $E \subseteq \mathbb{R}$.
1.(Definition due to Carathéodory) $E$ is measurable if for each set $A \subset \mathbb{R}$,  $m^{*}(A)=m^{*}(A\cap E)+m^{*}(A\cap E^{c})$.
The following are equivalent:
i. E is measurable (according to definition 1)
ii. For each $\epsilon >0$ , there is a closed set  $F$ and open set $O$ for which $F\subseteq E \subseteq O $ and $ m^{*}(O-F)<\epsilon $ 
Let us also have the definition of Lebesgue outer measure $m^{*}: \mathscr{P}(\mathbb{R}) \rightarrow \bar{\mathbb{R}}^{+}$ given by, 
$ m^{*}(E) = \inf \{ \sum l(I_{n}) : (I_n)_{n \in \mathbb{N}}$ is a collection of open intervals such that $E \subset \cup I_{n} \}$
When $m^{*}$ is defined on $\mathscr{P}(\mathbb{R})$ it is only countably sub-additive. But if we restrict $m^{*}$ to $\mathscr{M}\subset \mathscr{P}(\mathbb{R})$, where $\mathscr{M}$ is the collection of all sets that satisfies 1, then it has been proved that $m^{*}|_{\mathscr{M}}$ is countably additive and that $\mathscr{M}$ is a $\sigma$-algebra (containing the intervals).
Proof: i. $\implies$ ii. (Case $m^{*}(E) < +\infty$ )
Given an $\epsilon > 0$ there is a collection $\{I_{n} \}$ of open intervals such that $m^{*}(E) < l(\cup \{I_{n} \}) < m^{*}(E) + \epsilon $, this is so because $m^{∗}(E)$ is an infimum. Define $O = \bigcup_{n =1}^{\infty} \{I_{n} \}$. Note that $O$ is an open set, $O \in \mathscr{M}$, $E \subset O$, and $m^{*}(O) = l(O)$. Now we have,
$ m^{*}(O) < m^{*}(E) + \epsilon \implies m^{*}(O\backslash E) = m^{*}(O) - m^{*}(E) < \epsilon $
Let us find now that closed $F$.
Notice that $ E \subset \mathscr{M} \implies E^{c} \subset \mathscr{M}$. By the first part of this proof there is an open set $O \in \mathscr{M}$ such that $E^{c} \subset O$ and $ m^{*}(O\backslash E^{c}) < \epsilon$.
Take $ F = O^{c}$ closed. Then, $ F \subset E$ and,
$m^{*}(E \backslash F) = m^{*}((E^{c})^{c} \cap F^{c}) =  m^{*}((E^{c})^{c}\cap O) =  m^{*}(O \backslash E^{c}) < \epsilon$
Finally we observe that,
$(O \backslash F) = (O \backslash E)  \bigcup (E \backslash F)$ (disjoint)
hence,
$m^{*}(O \backslash F) = m^{*}(O \backslash E) + m^{*}(E \backslash F) < \epsilon + \epsilon = 2 \epsilon$.
To prove ii. $ \implies$ i. first prove that ii. implies that exists a $G \in \mathscr{G}_{\delta}$ such that $m^{*}(G \backslash E) = 0$. Then use it to prove i.
Once this has been done, we have to think on the case $m^{*}(E) = \infty$. 
I hope this could help a bit. I recommend reading Chapter 3 from Royden's book, the overall structure of the proof I tried to sketch here are suggested there.
