# 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).

• See, the paper: arxiv.org/abs/1605.02261 Nov 14, 2019 at 16:43
• Thank you very much for the information Professor Taras. Nov 16, 2019 at 4:57

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 $$\mathcal{T}$$.

• Thank you very much Slup for your response, I'll start studying it right away. Nov 16, 2019 at 4:58
• @GabrielMedina I hope that this answer is correct, otherwise I am very sorry for wasting your time. Please let me know if I can help you with any clarifications. Thank you for your response.
– Slup
Nov 17, 2019 at 8:55
• 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 Nov 17, 2019 at 19:44
• Hi @GabrielMedina I added more details to the proof of Lemma 1 and I hope now it is more clear. In the argument I used only the fact that $m^*(G)>0$ and eventually the argument shows that $m^*(G\cap A)>0$ for every Lebesgue set $A$ of positive measure, which is a very interesting property.
– Slup
Nov 19, 2019 at 7:00
• Thanks a lot @Slup. Nov 19, 2019 at 13:24

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 $$(\mathbb{R}, \mathcal{T})$$.

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.

• (+1). It is nice that Theorem 2 is a stronger version of the lemma 2 from my answer. Can you please provide a link for the Andrew Simoson's paper? I have hard time looking for it.
– Slup
Nov 26, 2019 at 7:57
• Ok. I found it.
– Slup
Nov 26, 2019 at 10:10
• Here's an inline link to Article. Nov 26, 2019 at 12:41
• Thanks. Btw Corollary 2.1 holds for every group of positive outer measure. Indeed, if the group is measurable and of positive outer measure, then it is $\mathbb{R}$ by Corollary 1.1 and the statement holds. Also in the Proposition 3 you don't need that your group is meager.
– Slup
Nov 26, 2019 at 12:50
• You're right, thanks for the observation. Nov 26, 2019 at 12:54