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.


  • 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.


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).


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 $\mathcal{T}$.

  • 1
    $\begingroup$ Thank you very much Slup for your response, I'll start studying it right away. $\endgroup$ – Gabriel Medina Nov 16 '19 at 4:58
  • $\begingroup$ @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. $\endgroup$ – Slup Nov 17 '19 at 8:55
  • $\begingroup$ 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 $\endgroup$ – Gabriel Medina Nov 17 '19 at 19:44
  • $\begingroup$ 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. $\endgroup$ – Slup Nov 19 '19 at 7:00
  • $\begingroup$ Thanks a lot @Slup. $\endgroup$ – Gabriel Medina Nov 19 '19 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 $(\mathbb{R}, \mathcal{T})$.

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.

  • 1
    $\begingroup$ (+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. $\endgroup$ – Slup Nov 26 '19 at 7:57
  • $\begingroup$ Ok. I found it. $\endgroup$ – Slup Nov 26 '19 at 10:10
  • $\begingroup$ Here's an inline link to Article. $\endgroup$ – Gabriel Medina Nov 26 '19 at 12:41
  • $\begingroup$ 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. $\endgroup$ – Slup Nov 26 '19 at 12:50
  • $\begingroup$ You're right, thanks for the observation. $\endgroup$ – Gabriel Medina Nov 26 '19 at 12:54

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