$F \subset E \subset G$ and $V(G \setminus F ) < \epsilon$ then $E$ is Jordan measurable I'm trying to prove the following statement: 
$E \subseteq \mathbb R^n$ is a set such that for all $\epsilon >0$ there are bounded Jordan measurable sets $F,G$ such that $F \subset E \subset G$ and $V(G \setminus F) < \epsilon$. Then $E$ is Jordan measurable.
I falsly claimed that $\partial E \subset G\setminus F$ and then used that the volume is less than epsilon, but apparently that's not true.
Looking for a correct answer.
 A: For the bounded set $A \subset \mathbb{R}^n$ let $\mathcal{R}_*(A)$ and $\mathcal{R}^*(A)$ denote the collections of all elementary sets (finite unions of rectangles) that are contained in and are containing $A$, respectively.  Jordan inner and outer measure are defined as 
$$|A|_* = \sup_{\cup_jR_j \in \mathcal{R_*(A)}} \sum_j vol(R_j) \quad\text{(inner measure)} \\ |A|^* = \inf_{\cup_jR_j \in \mathcal{R^*(A)}} \sum_j vol(R_j) \quad\text{(outer measure)}$$
The inner and outer Jordan measure of a bounded set always exist and satisfy $|A|_* \leqslant |A|^*$. The set is said to be Jordan measurable with measure $|A| = |A|_* = |A|^*$ when the inner and outer measures are equal.
If $\cup_j R_j \in \mathcal{R}_*(F)$ then $\cup_j R_j  \subset F \subset E$, and it follows that $\mathcal{R}_*(F) \subset \mathcal{R}_*(E)$, and
$$ |F|_* = \sup_{\cup_jR_j \in \mathcal{R_*}(F)} \sum_j vol(R_j) \leqslant \sup_{\cup_jR_j \in \mathcal{R_*}(E)} \sum_j vol(R_j) = |E|_*$$
Similarly we have $|E|^* \leqslant |G|^*$ since $E \subset G.$
Since $F$ and $G$ given here are Jordan measurable we have
$$|F| = |F|_* \leqslant |E|_* \leqslant |E|^* \leqslant |G|^* = |G|.$$
and 
$$0 \leqslant |E|^* - |E|_* \leqslant |G|- |F| = |G \setminus F| < \epsilon$$
Since, this holds for every $\epsilon > 0$ it follows that $|E|_* = |E|^*$ and $E$ is Jordan measurable.
