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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 $m(G)>0$ 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!

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    If $G$ contains an interval, then it contains an interval around $0$. If $G$ contains an interval around $0$, then it is $\mathbb{R}^2$. – 2017-01-18
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    i did not understand that.Can you explain it a bit more? – 2017-01-18
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    ok thank you.Whenever you have the time...:) – 2017-01-18
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    That $G$ has positive Lebesgue measure doesn't imply it contains an interval. There are sets of positive measure that contain no non-trivial intervals. – 2017-01-18
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    i use the fact that $G$ is a group and it is closed under addition,thus $G+G \subseteq G$.I did not assume that every set with possitive measure contains an interval – 2017-01-18

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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 $\mathbb{R}$.

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    isn't enough for jestification that $G+G \subseteq G$? We know of course that if a set $A \subseteq \mathbb{R}$ has a possitive lebesgue measure then $A+A$ contains an interval. – 2017-01-18
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    I agree, it does follow, see [Positive measure of a sum](http://math.stackexchange.com/questions/596273/sum-of-positive-measure-set-contains-an-open-interval). – 2017-01-19
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If the group is numerable, the resuslt is true since it is the union of a numerable sets (the singletons) and the measure of each of these sets is zero. If the group is not numerable, a classic result shows it is dense in $R$. Let $c\in ]0,1[\cap G$, If $m(G\cap [0,c[)=c>0$ implies $m(G)=\sum_{n\in Z}m(G\cap [nc,nc+c[)=\sum_{n\in Z}m(G\cap [0,c[)=+\infty$. We deduce that $m(G\cap [0,c[)=0$ and since $\bigcup_{n\in Z}G\cap [nc,nc+c[)=G$ is a disjoint union of a numerable set of measure zero and $m(G\cap [0,c[)=m(G\cap [nc,nc+c[)$, its measure is zero.

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    So if $G$ is measurable then $m(G)=0.$ Can we show that $G$ is a measurable set?...........+1 – 2017-01-25
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Let $G$ be a Lebesgue measurable proper subgroup of $(\mathbb R,+).$ Assume for a contradiction that $m(G)\gt0.$

Choose an interval $I$ with $m(G\cap I)\gt\frac23m(I).$ Let $\varepsilon=\frac13m(I).$

Choose $t\in(0,\varepsilon)\setminus G.$

Observe that $m((G+t)\cap I)\gt\frac13m(I).$

Since $G\cap(G+t)=\emptyset$ it follows that $m(I)\ge m(G\cap I)+m((G+t)\cap I)\gt m(I),$ which is absurd.

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    This shows very clearly that if $G$ is measurable then $m(G)=0.$ Can we show that $G$ is a measurable set?..........+1 – 2017-01-25
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    Yes it occurred to me later that I knew that. – 2017-01-26
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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.

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    That should be a comment – 2017-01-25
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    yes but i assumed that the group has positive measure – 2017-01-30
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    the rationals have zero measure and a lebesgue measurable set $A$ with positive measure has this property : $A+A$ contains an interval – 2017-01-30