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 $I$?

This construction might be rather tedious but the nice thing is that it works also when 'uncountable' is replaced by 'of positive measure' in the question.
Fix some enumeration $(q_i)_i$ of ${\mathbb Q}$ and define $X_i^j:=B(2^{-2i-j},q_j)$ (where $B(\epsilon,x)$ is the open $\epsilon$-ball around $x$), $X_i:=\cup_j X_i^j$, $Y_0:=\emptyset$ and $$Y_{i+1}:=\begin{cases}Y_i\cup X_{i+1}&i\text{ is even}\\Y_i\setminus X_{i+1}&\text{else.}\end{cases}$$  
Then the set $Y:=\liminf_i Y_i$ satisfies our demands.  (Note that $\liminf_i Y_i=\limsup_i Y_i$ almost everywhere since all points outside of $X_i$ become stable after the $i$th construction step and the measure of $X_i$ converges to $0$.)
$Y$ is measureable as $\liminf$ of measurable sets.
Now I want to show that $\mu(Y\cup I)>0$ for any open interval $I$, the proof works analogously for $Y^\complement$ because one can construct $Y^\complement$ analogously to $Y$ by switching $\emptyset$ with ${\mathbb R}$, even with odd and $\liminf$ with $\limsup$.
There exists some odd $m$ and (arbitrary) $n$ with $X_m^n\subseteq I$ and $X_m^n\cap X_{m+1}^j=\emptyset$ for $j<n$.  So $X_m^n\cap Y\supseteq X_m^n\setminus(\cup_{j\ge n} X_{m+1}^j)$ hence $\mu(X_m^n\cap Y)\ge 2^{-2m-n}-2^{-2(m+1)-(n-1)}>0$.
