# $\quad A\subset\mathbb{R}$ is measurable and $m(A)>0$. then $m\left(\mathbb{R}-\bigcup_{q\in\mathbb{Q}}q\cdot A\right)=0$. [duplicate]

$$\quad A\subset\mathbb{R}$$ is measurable and $$m(A)>0$$. then $$m\left(\mathbb{R}-\bigcup_{q\in\mathbb{Q}}q\cdot A\right)=0$$.

How to prove it ?

$$q \cdot A$$ is a dilation or contraction of each element in $$A$$ for example if $$A=(1 ,3], 2A= 2(1 ,3] = (2,6].$$ The proof is trivial if $$A$$ is an interval or countable union of intervals but not so for arbitrary sets.

This question is not a duplicate of link because taking logarithm $$log_{10} (qA) =log_{10}A+log_{10}q$$ , $$log_{10} q$$ is not always a rational number

• I think the ideas in the answer here will help. Commented May 14, 2019 at 8:10
• @DavidMitrada the link you posted deals with translation of elements of $A$ while this question deals with dilation and contractions of elements of $A$. Commented May 14, 2019 at 16:41
• So you don't understand my proof here? math.stackexchange.com/questions/3214177/… Commented May 14, 2019 at 16:59
• @XIAODAQU yes . Please improve your proof so that I can understand. Commented May 14, 2019 at 17:05
• @ibnAbu Dilating and contracting numbers amounts to translating their logarithms. Commented May 15, 2019 at 0:29

$$\mathbb{R}-\{0\}-\bigcup_{q\in\mathbb{Q}} qA\subset\bigcup_{n\in{\mathbb{Z}-\{0\}}} \left\{[1/n,n]-\bigcup_{q\in\mathbb{Q}}qA\right\}$$ It's sufficient to show $$m\left([1/n,n]-\bigcup_{q\in\mathbb{Q}}qA\right)=0$$, where $$n\in\mathbb{Z}-\{0\}$$.

Lemma $$\forall\alpha<1, \exists$$ an interval $$I$$ such that $$m(I\cap A)\geq\alpha\cdot m(I)$$.

$$Proof$$: By definition of outer measure, $$\exists\{I_n\}_n$$ s.t. $$A\subset\bigcup_n I_n$$ and $$m(A)\leq\sum m(I_n)\leq\alpha^{-1}m(A)\leq\alpha^{-1}\sum_n m(A\cap I_n).$$ So there must be an interval $$I_{n_0}$$ which satisfies the condition. $$\square$$

Following the lemma, we take an interval $$I:=(a,b)$$ with respect to $$\alpha$$, WLOG suppose $$0\leq|a|\leq b$$. We discuss respectively when $$a\geq 0$$ and $$a<0$$.

1) $$a\leq 0$$. Take a $$q\in\mathbb{Q}_+$$ such that $$n. Note that $$q|a|\leq qb<3n/2$$. Now $$m(q\cdot(a,b))=q(b-a)<3n$$ and $$[1/n,n]\subset q\cdot(a,b)$$.

2) $$a>0$$. Select $$\{q_n\}_n\subset\mathbb{Q}_+$$ by induction, letting $$n and $$q_na. Note that $$q_{n+1}, so $$q_n a\to 0$$. Thus, $$[1/n,n]\subset\bigcup_n q_n\cdot(a,b)$$. Since $$[1/n,n]$$ is compact, $$\exists$$ a finite collection $$\{p_k\}_{k=1}^N\subset\{q_n\}_n$$ s.t. $$p_1 and $$[1/n,n]\subset\bigcup_{k=1}^N p_k\cdot(a,b)$$. Also we have

$$\begin{eqnarray*} \sum_{k=1}^N m(p_k\cdot(a,b))&=&(b-a)\sum_{k=1}^N p_k\leq(b-a)p_1\sum_{n=0}^{+\infty} \left(\frac{a+b}{2b}\right)^n\\ &\leq&(b-a)q_1\cdot\frac{2b}{(b-a)}< 3n \end{eqnarray*}$$

To sum up these two cases, $$\exists\{q_k\}_{k=1}^N\subset\mathbb{Q}$$ such that the corresponding $$\{I_k\}_{k=1}^N:=\{q_kI\}_{k=1}^N$$ satisfies

$$[1/n,n]\subset\bigcup_{k=1}^N I_k,\quad \sum_{k=1}^n m(I_k)<3n.$$

Then we have $$\begin{eqnarray*} [1/n,n]-\bigcup_{q\in\mathbb{Q}}q_kA&\subset&[1/n,n]-\bigcup_{k=1}^Nq_kA\\ &\subset&\left\{[1/n,n]-\bigcup_{k=1}^NI_k\right\}\bigcup\left\{\bigcup_{k=1}^N[(q_kA)^c\cap I_k]\right\}\\ &=&\left\{\bigcup_{k=1}^N[(q_kA)^c\cap I_k]\right\}=\left\{\bigcup_{k=1}^Nq_k\cdot[A^c\cap I]\right\}. \end{eqnarray*}$$

Thus, $$m\left\{[1/n,n]-\bigcup_{q\in\mathbb{Q}}\{qA\}\right\}\leq m\left\{\bigcup_{k=1}^Nq_k\cdot[A^c\cap I]\right\}\leq (1-\alpha)\sum_{k=1}^N m(I_k)< (1-\alpha)3n.$$

Let $$\alpha\to 1$$, so we prove the proposition.