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Let $G$ be a non-measurable subgroup of $(\mathbb R,+)$ , $I$ be a bounded interval , then is it true that $m^*(G \cap I)=m^*(I)(=|I|)$ ? where $m^*$ denotes the Lebesgue outer measure

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    Note that $G$ is dense, since it cannot be countable. – 2017-01-17
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    @LeGrandDODOM : yeah I can see that $G$ is dense and uncountable ... but does that help much ? – 2017-01-18
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    Do you know if this is true for hyperplanes of $\mathbf R$ ? (viewed as a $\mathbf Q $-vector space) – 2017-01-25
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    This one seems to be related: https://www.jstor.org/stable/2325061?seq=1#page_scan_tab_contents – 2017-01-25

2 Answers 2

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The answer "yes" follows from results in the article "On Two Halves Being Two Wholes" by Andrew Simoson in The American Mathematical Monthly Vol. 91, No. 3 (Mar., 1984), pp. 190-193. Some readers might not have access, but here is the JSTOR link: http://www.jstor.org/stable/2322357 The article was cited in the article i707107 linked to in a comment.

In the article a subset $A\subseteq \mathbb R$ is called Archimedean if $\{r\in\mathbb R:A+r=A\}$ is dense in $\mathbb R$. Theorem 2 states that if $A$ is Archimedean with positive outer measure, then for every interval $I$, $m^*(A\cap I)=m^*(I)$.

A dense subgroup of $\mathbb R$ is Archimedean, and a nonmeasurable subgroup of $\mathbb R$ is dense with positive outer measure, so the theorem applies to affirm your conjecture.

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From the article provided by @Jonas Meyer: http://www.jstor.org/stable/2322357

An alternative approach to this problem is possible. This approach is already provided by @Syang Chen at MathOverflow: https://mathoverflow.net/questions/59978/additive-subgroups-of-the-reals/59980

Let $f(x) = m^*(G\cap [0,x])$. Then, for any $x, y \in G \cap (0,\infty)$, we have the additivity: $$ f(x+y)=f(x)+f(y). $$ Since $G$ is dense and $f$ is monotone, the additivity is generalized to all positive real numbers $x, y$. Moreover, $f(x) = ax$ with $a= f(1)$ is satisfied.

Thus, if $I$ is any open interval, then $$ \frac{m^*(G\cap I)}{m^*(I)} = a . $$

Now, we find $I$ by the Theorem 1 in the article. This gives $a=1$.

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    There is (at least) one other interesting answer on that MO question, apparently for every $a\in(0;1)$ there is a non measurable subgroup of $\mathbf R$ with Hausdorf dimension $a$. – 2017-01-27