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 $I$?
 A: 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.
A: 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. 


*

*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 $S_I=f(C\cap (0, 1))$.
A: 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 $\mu(X_m^n\cap Y)\ge 2^{-2m-n}-2^{-2(m+1)-(n-1)}>0$.
