Question about additive subgroups of the real line and the density topology I am new studying additive subgroups of the real line, I would like to know if someone could give me an idea for the next question.
Let $m$ be the Lebesgue measure in $\mathbb{R}$. A measurable set $E\subseteq\mathbb{R}$ has density $d$ at $x$ if $$\lim_{h\to 0} \frac{m(E\cap [x-h, x+h])}{2h} $$ exists and equals $d$. Denote by $\phi(E)$, $\{x\in\mathbb{R} : d(x, E)=1\}$.
The family of all measurable sets $E$ such that $E\subseteq\phi(E)$ is a topology on $\mathbb{R}$, henceforth denoted by $(X, \mathcal{T})$ or just X if confusion is unlikely. Clearly $\mathcal{T}$ is stronger that the usual topology $(\mathbb{R}, \mathcal{E})$, that is, $\mathcal{E}\subseteq\mathcal{T}$. This topology is called the density topology in $\mathbb{R}$.
Some properties of the density topology.
FACT 1

*

*The Borel subsets of $X$ are precisely the measurable sets.


*

*Every Borel subset of $X$ is a $G_{\delta}$.


*

*Every regular open set is a Euclidean $F_{\sigma \delta}$.

*$X$ satisfies the countable chain condition.

*$X$ is neither separable nor first countable, but every subspace of $X$ is Baire.

FACT 2
The following conditions on a subset $Y$ of $X$ are equivalent:

*

*$Y$ is a nullset (i.e. has measure zero)


*

*$Y$ is a nowhere dense


*

*$Y$ is a first category

*$Y$ is closed discrete.

My question is the following :
Suppose $G$ is an additive subgroup of $\mathbb{R}$ of positive Lebesgue outer measure such that $G$ is of the first category in $(\mathbb{R}, \mathcal{E})$. How can I conclude that $G$ is dense in $(\mathbb{R}, \mathcal{T})$?

Remember that the Lebesgue inner measure of $E\subseteq \mathbb{R}$ is defined as
$$m_{*}(E)=\sup\{m(C) : C\subseteq E, C\hspace{0.1cm} \text{is}\hspace{0.1cm}\mathcal{E}-\text{closed}    \}   $$
In general, we have the following characterization for dense subsets in the density topology on $\mathbb{R}$.
Theorem. A subset $D$ of $\mathbb{R}$ is $\mathcal{T}$-dense in $\mathbb{R}$ iff $m_{*}(\mathbb{R}\setminus D)=0$.
Proof. Suppose that $D$ is $\mathcal{T}$-dense in $\mathbb{R}$, then $\text{int}_{\mathcal{T}}(\mathbb{R}\setminus D)=\emptyset$. Let $C$ be a closed set of $(\mathbb{R}, \mathcal{E})$ such that $C\subseteq \mathbb{R}\setminus D$, in particular $C$ is $\mathcal{T}$-closed, then $\text{int}_{\mathcal{T}}(\overline{C}^{\mathcal{T}})=\text{int}_{\mathcal{T}}(C)\subseteq \text{int}_{\mathcal{T}}(\mathbb{R}\setminus D)=\emptyset$, by FACT 2, $m(C)=0$, then $m_{*}(\mathbb{R}\setminus D)=0$.
Now, suppose that $D$ is not $\mathcal{T}$-dense, then there is $A\in \mathcal{T}\setminus \{\emptyset \}$ such that $A\cap D=\emptyset$, so $A\subseteq \mathbb{R}\setminus D$, therefore $m(A)=0$, contradiction (because every non-empty $\mathcal{T}$-open subset of $\mathbb{R}$ has positive measure).
 A: I think that below I manage to answer the first of your questions. I will be very grateful for verification of this argument. Unless it is correct, I will delete this answer.
I denote Lebesgue outer measure by $m^*$ and $\sigma$-algebra of Lebesgue measurable sets by $\mathcal{L}$. Let $\mathcal{B}(\mathbb{R})$ be the $\sigma$-algebra of Borel sets on $\mathbb{R}$ with respect to the usual topology. Suppose that $G$ is an additive subgroup of $\mathbb{R}$ such that $m^*(G)>0$.
Lemma 1.
$G$ is dense in $\mathbb{R}$ with respect to the usual topology.
Proof.  In the proof of this lemma we consider $\mathbb{R}$ with the usual topology.
Suppose that for every $\epsilon >0$ there exists $g\in G$ such that $g\in (-\epsilon,\epsilon)\setminus \{0\}$. Then for every $x\in \mathbb{R}$ there exists $n\in \mathbb{Z}$ such that $|n\cdot g - x|<\epsilon$. Thus $G$ is dense in $\mathbb{R}$.
Now if $G$ is not dense in $\mathbb{R}$, then by what we said above, there must be $\epsilon>0$ such that $\{0\} = G\cap (-\epsilon, \epsilon)$. This means that $0$ is isolated in $G$. Since $G$ is a topological group, it is homogeneous (any two points have homeomorphic neighborhoods - just pick translation). We deduce that every point of $G$ is isolated and hence $G$ is a discrete subset of $\mathbb{R}$. A discrete subgroup of $\mathbb{R}$ is countable and hence $m^*(G)=0$. This is contradiction with $m^*(G)>0$.
Lemma 2.
There exists a constant $c>0$ depending only on $G$ such that for every bounded interval $I\subseteq \mathbb{R}$ we have
$$m^*\left(G\cap I\right) = c\cdot m^*(I)$$
Proof. Fix $a\in \mathbb{R}$ and consider a function $f_a:(a,+\infty)\rightarrow \mathbb{R}$ given by formula 
$$f_a(x) = m^*\left((a,x]\cap G\right)$$
Next for any $h>0$ pick a function $D_{a,h}:(a,+\infty)\rightarrow \mathbb{R}$ given by formula 
$$D_{a,h}(x)= f_a(x+h)-f_a(x) =m^*\left((a,x+h]\cap G\right) - m^*\left((a,x]\cap G\right) =$$
$$= m^*\left((x,x+h]\cap G\right)$$
The last equality follows from the Caratheodory's criterion. Since $G$ is an additive subgroup of $\mathbb{R}$ and Lebesgue outer measure is translation invariant, we derive that 
$$D_{a,h}(x+g) = D_{a,h}(x)$$
for every $g\in G$ and $g>0$. By Lemma 1 we know that $G_+ = \{g>0|g\in G\}$ is dense in $\mathbb{R}_+$. Now the fact that $D_{a,h}$ is continuous implies that 
$$D_{a,h}(x+t) = D_{a,h}(x)$$
for every $t>0$. Thus $D_{a,h}$ is constant for every $h>0$ and this implies that monotone function $f_a$, which has derivative almost everywhere, has constant derivative everywhere. Say $f_a'(x) = c$ for every $x\in (a,+\infty)$. Next $c$ does not depend on $a<0$, because for $a_1,a_2$ functions $f_{a_1}$ and $f_{a_2}$ have the same slope on $\left(\max\{a_1,a_2\},+\infty\right)$. Moreover, $c>0$, since $m^*(G)>0$. 
Now for given $a\in \mathbb{R}$ we have $f_a(a) = 0$ and $f_a'(x) = c$ for $x\in (a,+\infty)$. This implies that
$$m^*\left(G\cap (a,x]\right) = f_a(x) = c\cdot x - c\cdot a$$
for every $a\in \mathbb{R}$. If $I = (\xi_1,\xi_2]$ is an interval, then pick $a< \xi_1$ and then
$$m^*(G\cap I) = m^*(G\cap (\xi_1,\xi_2]) = m^*(G\cap (a,\xi_2]) - m^*(G\cap (a,\xi_1]) =$$
$$= f_a(\xi_2) - f_a(\xi_1) = \left(c\cdot \xi_2 - c\cdot a\right) - \left(c\cdot \xi_1 - c\cdot a\right) = c\cdot (\xi_2 - \xi_1) = c\cdot m^*(I) $$
Thus we proved that the statement of the lemma holds for intervals open from the left and closed from the right. Since every interval is up to (endpoints) a set of measure zero of the above form, we deduce that the result holds for all intervals. 
Lemma 3.
Let $E$ be a subset of $\mathbb{R}$ and let $c>0$. Suppose that $m^*(E\cap I) = c\cdot m^*(I)$ for every bounded interval $I\subseteq \mathbb{R}$. Then for every bounded set $A\in \mathcal{L}$ we have
$$m^*(E\cap A) = c\cdot m^*(A)$$
Sublemma.
Let $\{A_n\}_{n\in \mathbb{N}}$ be a family of pairwise disjoint members of $\mathcal{L}$ and let $E\subseteq \mathbb{R}$ be a bounded subset. Then
$$m^*\left(E\cap \bigcup_{n\in \mathbb{N}}A_n\right) = \sum_{n\in \mathbb{N}}m^*(E\cap A_n)$$
Proof of the sublemma.
It is a consequence of the fact that $\mathcal{L}$ is constructed via Caratheodory's criterion that this result holds for finite family $\{A_n\}_{n=0}^N$ of pairwise disjoint sets in $\mathcal{L}$. We use this in the proof of the general case. We have
$$m^*(E) \leq m^*\left(E\cap \bigcup_{n\in \mathbb{N}}A_n\right) + m^*\left(E\setminus \bigcup_{n\in \mathbb{N}}A_n\right)\leq \sum_{n\in \mathbb{N}}m^*(E\cap A_n) + m^*\left(E\setminus \bigcup_{n\in \mathbb{N}}A_n\right) =$$
$$\leq \lim_{N\rightarrow +\infty}\left(\sum_{n=0}^Nm^*(E\cap A_n) + m^*\left(E\setminus \bigcup_{n=0}^NA_n\right)\right) =$$
$$ = \lim_{N\rightarrow +\infty}\left(m^*\left(E\cap \bigcup_{n=0}^NA_n\right) + m^*\left(E\setminus \bigcup_{n=0}^NA_n\right)\right)=\lim_{N\rightarrow +\infty}\mu^*(E) =\mu^*(E)$$
Thus in the inequality above, we have equality everywhere. In particular, we have 
$$ m^*\left(E\cap \bigcup_{n\in \mathbb{N}}A_n\right) + m^*\left(E\setminus \bigcup_{n\in \mathbb{N}}A_n\right)= \sum_{n\in \mathbb{N}}m^*(E\cap A_n) + m^*\left(E\setminus \bigcup_{n\in \mathbb{N}}A_n\right)$$
The boundness of $E$ implies that we can cancel out $m^*\left(E\setminus \bigcup_{n\in \mathbb{N}}A_n\right) < +\infty$.
Proof of the lemma 3.
We prove that for every bounded interval $I$ and $A\in \mathcal{B}(\mathbb{R})$ such that $A\subseteq I$ we have
$$m^*(E\cap A) = c\cdot m^*(A)$$
Note that the family $\mathcal{F}$ of all such $A$ is a Dynkin system in the power set $\mathcal{P}(I)$. Indeed, if $A\in \mathcal{F}$ that is  $m^*(E\cap A)= c\cdot m^*(A)$, then
$$c\cdot m^*(I) = m^*(E\cap I) = m^*(E\cap A) + m^*(E\cap (I\setminus A)) =$$
$$= c\cdot m^*(A) + m^*(E\cap (I\setminus A)) $$
Hence $m^*(E\cap (I\setminus A)) = c\cdot m^*(I\setminus A)$ and thus $I\setminus A\in \mathcal{F}$. Moreover, from Sublemma we derive that $\mathcal{F}$ is closed under countable unions of pairwise disjoint sets. By the assumption $\mathcal{F}$ contains all subintervals of $I$ and they form $\pi$-system. Now by Dynkin's $\pi\lambda$-theorem $\mathcal{F}$ contains all subsets in $\mathcal{B}(\mathbb{R})$ which are contained in $I$. Suppose that $A\in \mathcal{L}$ is contained in $I$, then $A = B\cup Z$, where $B\subseteq I$ and $B\in \mathcal{B}(\mathbb{R})$ and $m^*(Z)=0$. Hence
$$m^*(E\cap B) \leq m^*(E\cap A) = m^*\left((E\cap B)\cup (E\cap Z)\right) \leq m^*(E\cap B) + m^*(E\cap Z) = m^*(E\cap B) + 0 = m^*(E\cap B)$$
Thus $m^*(E\cap B) = m^*(E\cap A)$ and
$$m^*(E\cap A) = m^*(E\cap B) = c\cdot m^*(B) = c\cdot m(B) = c\cdot m(A) = c\cdot m^*(A)$$
This proves the lemma.
Now by Lemma 2 and Lemma 3, we derive that there exists $c>0$ such that $m^*(G\cap A) = c\cdot m^*(A)$ for every bounded measurable subset $A$ of $\mathbb{R}$. Now pick any nonempty subset $A\in \mathcal{T}$. Intersect $A$ with some open interval $I$ such that $A\cap I\neq \emptyset$. Then $B = A\cap I\in \mathcal{T}$ is bounded. Thus
$$m^*(G\cap A) = m^*(G\cap B)= c\cdot m^*(B)>0$$
because $c>0$ and $m^*(B)>0$. Thus $G\cap A\neq \emptyset$. Therefore, $G$ is dense in $\mathcal{T}$.
A: 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.
