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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 $m^*(G)>0$. These imply the equality in the statement.

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 $I$. Finally every subset $A\in \mathcal{L}$ contained in $I$ can be decomposed as a union of a Borel subset and a set of measure zero.

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.

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 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 $\mathbb{R}$. Discrete subgroup of $\mathbb{R}$ is countable and hence $m^*(G)=0$. This is contradiction with $m^*(G)>0$.

for every $g\in G$ and $g>0$. By Lemma 2 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

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 $m^*(G)>0$.

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

Proof. 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). We deduce that every point of $G$ is isolated and hence $G$ is discrete subset of $\mathbb{R}$.

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 $\mathbb{R}$. Discrete subgroup of $\mathbb{R}$ is countable and hence $m^*(G)=0$. This is contradiction with $m^*(G)>0$.