Lebesgue measurability and a subgroup of $\mathbb{R}$ Let $G$ be a sugroup of $\mathbb{R}$ under addition and $G \neq \mathbb{R}$ and $G$ is Lebesgue measurable.Prove that $m(G)=0$.

One idea to solve this, is that if $m(G)&gt;0$ then $G+G$(or $G-G$) contains an interval.Also because of the fact that $G$ is a group we have that $G+G \subseteq G$.

My idea is that i can construct a non measurable set and derive a contradiction,using the fact that $G \triangleleft \mathbb{R}$ nad use the quotient group $\mathbb{R} /G$.But i cannot achieve countability.le

Maybe i have to use a different approach.

Can someone give me some ideas to solve this(not the whole answer)?

Thank you in advance!
If one can show that $G$ contains an interval, then there is some interval $(a,b)\subseteq G$.  Let $m$ be the midpoint of the interval, since $m\in G$, $-m\in G$, so we can subtract $m$ and remain in $G$ to get that $G$ contains $(a-m,b-m)$ which is an interval centered at the origin, i.e., of the form $(-c,c)$.  Therefore, $(-c,c)\subseteq G$.  Now, if you consider $G+G$ (which is a subset of $G$), you can conclude that $(-2c,2c)\subseteq G$.  By continuing this, one gets that $(-2^nc,2^nc)\subseteq G$ for all $n$, so $G$ is all of $\mathbb{R}$.
The idea the group of $R$ must contain an interval in general it is not true since we can take the rational number is sub group of  $R$ ( filed)  but it doesn't contain any interval.