We let $m^*(E)$ denote the outer measure of a set $E$ in $\mathbb R^n$.
Given $E$ in $\mathbb R^n$, there exists an open $G$ containing $E$ such that $m^*(G)\le m^*(E) + \epsilon$.
In general, since $G=E\cup (G-E)$ where $E\subset G$, we only have $m^*(G) \le m^*(E) + m^*(G-E)$, and we cannot conclude from $m^*(G)\le m^*(E) + \epsilon$ that $m^*(G-E) < \epsilon$.
Why $m^*(G-E) < \epsilon$ may not hold?
If this is possible then, for each $n$ we can find an open set $G_n$ such that $m^{*}(G_n\setminus E) <\frac 1 n$ . From this it follows that $m^{*}(\cap_n G_n\setminus E)=0$. Since $E \subset \cap_n G_n$ we can conclude that $E$ is Lebesgue measurable. But not every subset of $\mathbb R$ is Lebesgue measurable.