Define volume as $\text{vol}(A) = \int _A 1$, and A has volume if this exists.
A couple questions similar to this has been asked, however, I (and everyone else) was unsure of their definition of volume. So, using the above definition show the following:
If $E'$ is a subset of $E^n$ has volume, then the interior of $E'$ has the same volume as $E'$.
I believe the best way to prove this is to prove that the boundary has volume of $0$ through a method making an $\epsilon$ containing the boundary arbitrarily small. Could someone help me prove this?
Useful Definition: Let $S$ be a subset of the metric space $E$. A point $p\in S$ is called an interior point of $S$ if there is an open ball in $E$ of center $p$ which is contatined in $S$. The set of interior points of $S$ is an open subset of $E$, called the interior, that contains all other open subsets of $E$ that are contained in $S$.
I think you're trying to prove the following. Your definition of "volume" is Jordan measure: $$ \mbox{vol}(A) = \int_{\mathbb{R}^n} \mathbb{1}_A, $$ where we are taking the Reimann integral of the indicator function $$ \mathbb{1}_A(x) = \begin{cases} 1 & \mbox{if } x \in A \\ 0 & \mbox{if } x \notin A \end{cases} $$ of a bounded set $A \subseteq \mathbb{R}^n$. It turns out that $$ A \mbox{ "has volume" (is Jordan measurable)} \iff \mbox{vol}(\partial A) = 0 $$ is a theorem, where \begin{align*} \partial A &= \bar A \backslash A^\circ \mbox{ is the boundary of } A, \\ \bar A &= \mbox{closure of } A, \\ A^\circ &= \mbox{interior of } A. \end{align*} Hence if $A$ "has volume", $\mbox{vol}(A)$ exists and is squeezed by $\mbox{vol}(A^\circ) \leq \mbox{vol}(A) \leq \mbox{vol}(\bar A)$ since $A^\circ \subseteq A \subseteq \bar A$. If $\bar A$ "has volume", write $\bar A$ as the disjoint union $\bar A = \partial A \cup A^\circ$, and compute $$ \mbox{vol}(\bar A) = \int_{\mathbb{R}^n} \mathbb{1}_{\bar A} = \underbrace{\int_{\mathbb{R}^n} \mathbb{1}_{\partial A}}_{0} + \int_{\mathbb{R}^n} \mathbb{1}_{A^\circ} = \int_{\mathbb{R}^n} \mathbb{1}_{A^\circ} = \mbox{vol}(A^\circ). $$ The step $\int_{\mathbb{R}^n} \mathbb{1}_{\partial A} = 0$ follows from the theorem under the hypothesis that $A$ "has volume".
By the "squeezed" inequality, we've computed $\mbox{vol}(A) = \mbox{vol}(A^\circ)$. Of course, we've made some extra assumptions by using $\bar A$, but these can be relaxed.
If this is indeed what you mean, we can discuss the proof of the theorem.