Algebraic difference of a proper additive subgroup of $\mathbb R^n$ and itself I'm supposed to show that If $G$ is a proper additive subgroup of $\mathbb R^n$ and $G$ is a measurable set, then $\lambda(G)=0$. I have to show this using that for $\lambda(G) \gt 0$, the zero vector has to be an interior point of $G - G$, where $G - G$ is the algebraic difference of $G$ and itself.
So I just want to make sure I have this part of the proof correct, or if I'm oversimplifying the definition of interior.
$G$ is a proper additive subgroup, so $x,y \in G$ implies $x+y \in G$ and $-x \in G$. So the zero vector is in $G$ as well.
$G - G = \{g+ -g: g \in G\}$
$G - G = \{0\}$
So Since $G - G$ is only the zero vector, it has no interior.
Am I off? $\lambda$ is the Lebesgue measure, by the way.
 A: Suppose that $\lambda(G)>0$. Then $\vec{0}$ is an interior point of $G-G$. for the proof, see here.
First of all, since $\vec{0}$ is an interior point of $G-G$, one can find an open ball of radius $\epsilon$ around $\vec{0}$ which is contained in $G-G$. In other words, every vector $\vec{v}$ such that $\|\vec{v}\|<\epsilon$ is in $G-G$.
Note that the ball $B_{\epsilon}(0)$ contains a scaled basis of $\mathbb{R}^n$. Namely, it contains $\{ \frac{\epsilon}{2}  \vec{e_n}\}$.
Therefore, $G-G$ contains $n$ linearly independent vectors, $\{g_i\}$. 
Now, note that any point in $\mathbb{R}^n$ can be written in the form:
$\vec{v}=(\sum_{i=1}^n n_i\vec{g_i})+\vec{v_r}$
where $n_i \in \mathbb{Z}$ and $\| \vec{v_r}\| < \epsilon$
Since $\vec{v_r}$ is in the ball, it is an element of $G-G$. Therefore, $\vec{v} \in G-G$. This proves that $G-G=\mathbb{R}^n$. 
But since $G$ is a group, $G-G \subseteq G$ and this proves that $G=\mathbb{R}^n$. Therefore, it is not a proper subgroup.
A: As $G$ is an additive subgroup $0\in G$ and if $x\in G$ then $nx\in G$ for all $n\in \mathbb{Z}$. If $x\in int(G)$ then $\exists \epsilon>0$ such that $B(x,\epsilon)\subset G$. But then $B(0,\epsilon)\subset G$. You can show that this implies $G=\mathbb{R}^n$. So if $ G$ is a proper subgroup then $int(G)=\phi$. Hence $\lambda(G)=0$.
