I have read that:
If $G$ be a proper measurable subgroup of the group $(\mathbb{R}, +)$ then $\mu(G)=0.$
[Here $\mu$ is the lebsgue measure function]
Now my question is that:
Is there any proper subgroup of $(\mathbb{R}, +)$ which is Non measurable?
I cannot find any example in the support of this question.
Please Help...
Thankyou...!!
Let us consider the vector space $\mathbb{R}$ over $\mathbb{Q}$. Let $\mathcal{B}$ be an uncountable basis of it and $\alpha\in \mathcal{B}$. Then $\mathcal{B}\setminus \{\alpha\}$ spans a proper subspace and hence a proper subgroup $H$ of $\mathbb{R}$. We claim that $H$ is non-measurable. If not, then by previous theorem, $m(H)=0$. Then every set of the form $H+q\alpha=\{h+q\alpha: h \in H\}$, where $q\in \mathbb{Q}$, is measurable and has measure zero, as they are just a shift of $H$. But, $\mathbb{R}$ being a countable union of such shift sets with measure zero, i.e., $$\mathbb{R}=\bigcup_{q\in \mathbb{Q}}(H+q\alpha),$$ is also of measure zero, a contradiction.
