I studied the article and introduced the requirements (a little measure of Lebesgue in $\mathbb{R}$ and the density topology in $\mathbb{R}$), if anyone is interested, you can send me an email to [email protected] and I will gladly send you my summary.

You're right, thanks for the observation.

Here's an inline link to Article.

Thanks a lot. I was trying, with this new characterization, the fact that additive subgroups of the real line with positive outer measure are dense in the real line with the density topology, but I still cannot prove it. Please do you think you could be more explicit in the final part of the proof of Lemma 2 and 3? Thanks

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 $A\subseteq \mathbb{R}\setminus D$, then $m(A)=0$, contradiction.

I prove the fact as follows, first 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 $\text{int}_{\mathcal{T}}(\overline{C}^{\mathcal{T}})=\text{int}_{\mathcal{T}}(C)\subseteq \text{int}_{\mathcal{T}}(\mathbb{R}\setminus D)=\emptyset$, so $C$ is $\mathcal{T}$-nowhere dense, by the Fact 2, $m(C)=0$, so $m_{*}(\mathbb{R}\setminus D)=0$.

Remember that the Lebesgue inner measure of $E\subseteq \mathbb{R}$ is defined by $$m_{*}(E)=\sup\{m(C) : C\subseteq E, C \hspace{0.1cm}\text{closed} \} $$

In the article "Topological spaces in which Blumberg's theorem holds" of White, H. E., Jr, is commented the following fact : A subset $D$ of $\mathbb{R}$ is $\mathcal{T}$-dense in $\mathbb{R}$ if and only if $m_{*}(\mathbb{R}\setminus D)=0$, where $m_{*}$ denote the Lebesgue inner measure on $\mathbb{R}$ and $\mathcal{T}$ is the density topology in $\mathbb{R}$.

Thanks a lot @Slup.

Hello @Slup, I think they are very good ideas, for the proof of Lemma 1, please do you think you could be more explicit in the final part of the proof ? Another observation is that it seems to me that you do not need the hypothesis that $G$ is of first category in the usual topology. Thanks