$\quad A\subset\mathbb{R}$ is measurable and $m(A)>0$. then $m\left(\mathbb{R}-\bigcup_{q\in\mathbb{Q}}q\cdot A\right)=0$. $\quad A\subset\mathbb{R}$ 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
 A: $$\mathbb{R}-\{0\}-\bigcup_{q\in\mathbb{Q}} qA\subset\bigcup_{n\in{\mathbb{Z}-\{0\}}} \left\{[1/n,n]-\bigcup_{q\in\mathbb{Q}}qA\right\}$$
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.
