Existence of measurable set $A\subseteq{\mathbb R}$ which is locally uncountable and so is its complement Is there a measurable set $A\subseteq{\mathbb R}$ such that $|A\cap I|$ and $|A^\complement\cap I|$ are both uncountable for any open interval $I$?

The set of real numbers whose decimal representation has a finite number of ones (let's agree a number can't end with infinite 9's, even if it doesn't really change anything.)
It is measurable because you can write it as a countable union of countable intersections of intervals (it's kind of tedious to write down); it is uncountable on every interval because you can truncate the decimal representation and then put only $2$'s and $3$'s as you want; the complement is uncountable because you can truncate the decimal representaiton and then put $1$'s at odd positions and $2$'s or $3$'s at even positions as you want. As a side note this set is of measure zero.
