For any $n\in\mathbb{N}$, consider the measurable space $(\mathbb{R}^n,\mathscr{B}^n)$, where $\mathscr{B}^n$ is the Borel $\sigma$-algebra generated by the Euclidean topology on $\mathbb{R}^n$.
Given a measurable set $A\in\mathscr{B}^n$, one can easily show that for every $k\in[0,1]$, the set
$k\cdot A=\{k\cdot x:x\in A\}$
is also measurable. What I am wondering is if we could prove or disprove (maybe with a counterexample) that for every measurable set $A\in\mathscr{B}^n$, the uncountable union
$\displaystyle \bigcup_{k\in[0,1]}k\cdot A$
is also measurable. It seems "intuitively clear" that this should be true, but so far I have been unable to come up with any truly rigorous argument to demonstrate it.
Take a Borel set $B \subseteq [1, 2]$ such that $\{\frac{x}{y}: x, y \in B\}$ is not Borel. Then $A = B^2$ is a counterexample.