Lebesgue measurability and a subgroup of $\mathbb{R}$

Question

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!

Answer 1

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 $\mathbb{R}$.

Answer 2

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.

Answer 3

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.

Answer 4

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.