Let $E\subset\mathbb{R}$ be a Lebesgue measurable set. Is it true that there exists a set $A$ which is a countable union of intervals such that $\mu(E\Delta A) = 0$, where $\mu$ is the Lebesgue measure on $\mathbb{R}$ and $\Delta$ denotes the symmetric difference?
If not, can you provide a counter example?
What I do know is that $\mu$ is Borel-regular, which means there is a Borel set $B$ with $E\subset B$ and $\mu(B\setminus E) = 0$. But the condition on $A$ (which is a countable union of intervals) is much stronger than the condition on $B$ (which is Borel).
No - the crucial fact is that if $E$ is any nowhere dense set, and $A$ is an open interval, then $A\Delta E$ will have positive measure.
So take $E$ to be a nowhere dense set of positive measure (e.g., a fat Cantor set: http://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set#Other_fat_Cantor_sets). Then if $A$ is a countable union of open intervals, certainly $A\Delta E$ has positive measure.
If by "interval" you mean "open or closed interval," this still doesn't change anything - either one of the intervals making up $A$ are all just single points (in which case $\mu(E\Delta A)=\mu(E)>0)$ or at least one of them contains a nonempty open interval, in which case the above holds.
Do you maybe mean intersection of countably many intervals, instead of union?