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$\quad A\subset\mathbb{R}$ 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

ibnAbu
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  • I think the ideas in the answer here will help. – David Mitra May 14 '19 at 08:10
  • @DavidMitrada the link you posted deals with translation of elements of $ A $ while this question deals with dilation and contractions of elements of $ A $. – ibnAbu May 14 '19 at 16:41
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    So you don't understand my proof here? https://math.stackexchange.com/questions/3214177/almost-everywhere-ae-homogeneous-function-of-degree-0-equals-to-a-constant-f/3218756#3218756 – Selene May 14 '19 at 16:59
  • @XIAODAQU yes . Please improve your proof so that I can understand. – ibnAbu May 14 '19 at 17:05
  • @ibnAbu Dilating and contracting numbers amounts to translating their logarithms. – Andreas Blass May 15 '19 at 00:29
  • @AndreasBlass what about negative real numbers ? – ibnAbu May 15 '19 at 02:54
  • If a set has positive measure then its intersection with at least one of $\mathbb R_{>0}$ and $\mathbb R_{<0}$ also has positive measure. Use that intersection and use either $\ln x$ or $\ln(-x)$ on the intersection. – Andreas Blass May 15 '19 at 02:59
  • @AndreasBlass This question is not a duplicate of [link][1] because taking logarithm $ log_{10} (qA) =log_{10}A+log_{10}q$ , $ log_{10} q $ is not always a rational number – ibnAbu May 20 '19 at 01:18
  • @ibnAbu But logarithms of rational numbers form a dense set, which is all that's needed. – Andreas Blass May 20 '19 at 10:24

1 Answers1

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$$\mathbb{R}-\{0\}-\bigcup_{q\in\mathbb{Q}} qA\subset\bigcup_{n\in{\mathbb{Z}-\{0\}}} \left\{[1/n,n]-\bigcup_{q\in\mathbb{Q}}qA\right\}$$ 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 $n<qb<3n/2$. 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 $n<q_1b<3n/2$ and $q_na<q_{n+1}b<q_n(a+b)/2$. Note that $q_{n+1}<q_n\cdot\frac{a+b}{2b}<q_1\cdot(\frac{a+b}{2b})^n\to 0$, 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. $p_1<p_2<\cdots<p_N$ 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.

Selene
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