Remember the Steinhaus's Lemma,

Steinhaus's Lemma If $A\subseteq \mathbb{R}$, is a set of positive Lebesgue measure, then the set $A-A=\{x-y: x,y \in A\}$ contains a ball around $0$.

Corollary 1.1 If $S$ is an additive subgroup of $\mathbb{R}$, and $S$ contains a set of positive measure, then $S=\mathbb{R}$. If $T$ is a multiplicative subgroup of $]0, \infty[$ containing a set of positive measure then $T=]0, \infty[$.

Also, in the article "On two halves being two wholes" of Andrew Simoson, it is defined

Definition A subset $A$ of $\mathbb{R}$ is an Archimedean set if the set of all real numbers $r$ such that $A + r = A$ is dense in $\mathbb{R}$.

and it is shown that

Theorem 2 Let $A$ be an Archimedean set with positive outer measure. Then for any interval $I$, $$m^{*}(A \cap I) = m^{*}(I)$$

Proposition 3 Let $G$ be an additive subgroup of $\mathbb{R}$ such that $m^{*}(G)>0$. Then $G$ is dense in $(\mathbb{R}, \mathcal{T})$.

Proof. We start with the following

Lemma 3.1 Let $G$ be an additive subgroup of $\mathbb{R}$ such that $m^{*}(G)>0$, then for any interval $I$, $m^{*}(G\cap I)=m^{*}(I)$.

Proof of Lemma 3.1 Note that if $m^{*}(G)>0$, then $G$ is dense, so $G$ is an Archimedean set.

Finally, by the Lemma 3 of @Slup, we conclude the result.

Remember the Steinhaus's Lemma,

Steinhaus's Lemma If $A\subseteq \mathbb{R}$, is a set of positive Lebesgue measure, then the set $A-A=\{x-y: x,y \in A\}$ contains a ball around $0$.

Corollary 1.1 If $S$ is an additive subgroup of $\mathbb{R}$, and $S$ contains a set of positive measure, then $S=\mathbb{R}$. If $T$ is a multiplicative subgroup of $]0, \infty[$ containing a set of positive measure then $T=]0, \infty[$.

Also, in the article "On two halves being two wholes" of Andrew Simoson, it is defined

Definition A subset $A$ of $\mathbb{R}$ is an Archimedean set if the set of all real numbers $r$ such that $A + r = A$ is dense in $\mathbb{R}$.

and it is shown that

Theorem 2 Let $A$ be an Archimedean set with positive outer measure. Then for any interval $I$, $$m^{*}(A \cap I) = m^{*}(I)$$

Proposition 2.1 Any additive subgroup of the additive group of real numbers is either cyclic (i.e., equal to $c\mathbb{Z}$ for some positive number $c$) or dense.

Corollary Let $G$ be an additive subgroup of the real line such that $m^{*}(G)>0$, then $G$ is dense.

Proof. Suppose that $G$ is not dense then, by Proposition 2.1, $G$ is cyclic, then $G$ is countable, and therefore $m^{*}(G)=0$, contradiction.

Proposition 3 Let $G$ be an additive subgroup of $\mathbb{R}$ such that $m^{*}(G)>0$. Then $G$ is dense in $(\mathbb{R}, \mathcal{T})$.

Proof. We start with the following

Lemma 3.1 Let $G$ be an additive subgroup of $\mathbb{R}$ such that $m^{*}(G)>0$, then for any interval $I$, $m^{*}(G\cap I)=m^{*}(I)$.

Proof of Lemma 3.1 Note that if $m^{*}(G)>0$, then $G$ is dense, so $G$ is an Archimedean set.

Finally, by the Lemma 3 of @Slup, we conclude the result.

Remember the Steinhaus's Lemma,

Steinhaus's Lemma If $A\subseteq \mathbb{R}$, is a set of positive Lebesgue measure, then the set $A-A=\{x-y: x,y \in A\}$ contains a ball around $0$.

Corollary 1.1 If $S$ is an additive subgroup of $\mathbb{R}$, and $S$ contains a set of positive measure, then $S=\mathbb{R}$. If $T$ is a multiplicative subgroup of $]0, \infty[$ containing a set of positive measure then $T=]0, \infty[$.

Also, in the article "On two halves being two wholes" of Andrew Simoson, it is defined

Definition A subset $A$ of $\mathbb{R}$ is an Archimedean set if the set of all real numbers $r$ such that $A + r = A$ is dense in $\mathbb{R}$.

and it is shown that

Theorem 2 Let $A$ be an Archimedean set with positive outer measure. Then for any interval $I$, $$m^{*}(A \cap I) = m^{*}(I)$$

Corollary 2.1 Let $G$ be a non-measurable additive subgroup of $\mathbb{R}$, then for any interval $I$, $m^{*}(G\cap I)=m^{*}(I)$.

Proposition 3 Let $G$ be an additive subgroup of $\mathbb{R}$ such that $m^{*}(G)>0$ and $G$ is meager in $(\mathbb{R}, \mathcal{E})$. Then $G$ is dense in $(\mathbb{R}, \mathcal{T})$.

Proof. We start with the following

Lemma 3.1 For any interval $I\subseteq \mathbb{R}$ we have that $$m^{*}(G\cap I)=m^{*}(I)$$

Proof of Lemma 3.1 By @Slup's observation (Lemma 1) $G$ is dense in $(\mathbb{R}, \mathcal{E})$, as $G$ is meager, then $G$ is a proper additive subgroup of the real line. We claim that $G$ is not measurable, because otherwise, by Steinhaus's Lemma, there is a $\delta>0$ such that $]-\delta, \delta[ \subseteq G$, so $G$ contains a set of positive measure, then, by Corollary 1.1, $G=\mathbb{R}$, contradiction. Therefore $G$ is not a measurable set, then, by Corollary 2.1, we have that for any interval $I$, $m^{*}(G\cap I)=m^{*}(I)$.

Finally, by the Lemma 3 of @Slup, we conclude the result.

Remember the Steinhaus's Lemma,

Steinhaus's Lemma If $A\subseteq \mathbb{R}$, is a set of positive Lebesgue measure, then the set $A-A=\{x-y: x,y \in A\}$ contains a ball around $0$.

Corollary 1.1 If $S$ is an additive subgroup of $\mathbb{R}$, and $S$ contains a set of positive measure, then $S=\mathbb{R}$. If $T$ is a multiplicative subgroup of $]0, \infty[$ containing a set of positive measure then $T=]0, \infty[$.

Also, in the article "On two halves being two wholes" of Andrew Simoson, it is defined

Definition A subset $A$ of $\mathbb{R}$ is an Archimedean set if the set of all real numbers $r$ such that $A + r = A$ is dense in $\mathbb{R}$.

and it is shown that

Theorem 2 Let $A$ be an Archimedean set with positive outer measure. Then for any interval $I$, $$m^{*}(A \cap I) = m^{*}(I)$$

Proposition 3 Let $G$ be an additive subgroup of $\mathbb{R}$ such that $m^{*}(G)>0$. Then $G$ is dense in $(\mathbb{R}, \mathcal{T})$.

Proof. We start with the following

Lemma 3.1 Let $G$ be an additive subgroup of $\mathbb{R}$ such that $m^{*}(G)>0$, then for any interval $I$, $m^{*}(G\cap I)=m^{*}(I)$.

Proof of Lemma 3.1 Note that if $m^{*}(G)>0$, then $G$ is dense, so $G$ is an Archimedean set.

Finally, by the Lemma 3 of @Slup, we conclude the result.