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).



Question about additive subgroups of the real line and the density topology
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](https://groupprops.subwiki.org/wiki/Every_nontrivial_discrete_subgroup_of_reals_is_infinite_cyclic) 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](https://en.wikipedia.org/wiki/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 $\mathcal{T}$.


