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 $I$?

Yet another solution:
A rational open interval is an interval of the form $(q_1, q_2)$ for $q_1<q_2$ rationals. There are countably many rational open intervals, and every nonempty open interval contains a rational open interval.
Now for each rational open interval $I$, let $S_I$ be a subset of $I$ which is uncountable but has measure zero - basically, a (thin) Cantor set in $I$. The set $$X=\bigcup_{I\mbox{ a rational open interval}} S_I$$ is then a countable union of measure-zero sets, hence has measure zero and is measurable. 


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*Since $X$ has measure zero, its complement has uncountable intersection with every nontrivial open set.

*On the other hand, by construction $X\cap U$ is uncountable for every nonempty open set $U$: any such $U$ contains a rational open interval $I$, and then $X\cap U\supseteq X\cap I\supseteq S_I$ which is uncountable.

The only weird bit about this is the choice of $S_I$, but this can be done nicely: if $I=(q_0, q_1)$, let $f: x\mapsto (q_1-q_0)x+q_0$. $f$ is a bijection from $(0, 1)$ to $I$. Let $C$ be the usual Cantor set, and let $S_I=f(C\cap (0, 1))$.
