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).