Can every measurable set (with positive measure) be represented as a countable union of non-measurable set? Let $A$ be a measurable set with positive measure. I know that you can always find a non-measurable subset $E\subset A$ using a construction similar to Vitali sets.
But I have a stronger question: can $A$ always be written as a countable union of non-measurable sets? I tried to use the old trick of using $\mathbb{R} /\mathbb{Q}$ but what I did didn't exactly work.
Any ideas?
NOTE: I'm talking only about lebesgue measure in $\mathbb{R}$.
 A: Generalize the construction of a Bernstein set.
Enumerate $A=\{x_\xi:\xi<2^\omega\}$. Let $\mathscr{F}=\{F_\xi:\xi<2^\omega\}$ be the family of uncountable closed subsets of $A$. Recursively construct distinct $y_{\xi,n}\in A$ for $\langle\xi,n\rangle\in 2^\omega\times\omega$ as follows. 
If $\eta<2^\omega$, and we have $y_{\xi,n}$ for each $\langle\xi,n\rangle\in\eta\times\omega$, let $Y_\eta=\{y_{\xi,n}:\langle\xi,n\rangle\in\eta\times\omega\}$, and let $X_\eta(0)=\{\xi<2^\omega:x_\xi\notin Y_\eta\}$. Recursively construct $\xi_n$ for $n\in\omega$ by setting
$$\xi_n=\min\{\xi\in X_\eta(n):x_\xi\in F_\eta\}$$
and $X_\eta(n+1)=X_\eta\setminus\{\xi_n\}$. In other words, the set $\{\xi_n:n\in\omega\}$ contains precisely the indices of the first $\omega$ members of $F_\eta$ (in the enumeration of $A$) that aren’t in $Y_\eta$. Now let $y_{\eta,n}=x_{\xi_n}$ for each $n\in\omega$.
For $n\in\omega$ let $E_n=\{y_{\xi,n}:\xi<2^\omega\}$; each $E_n$ is a Bernstein set in $A$ and is therefore non-measurable, and the sets are certainly pairwise disjoint. Finally, the construction ensures that $A=\bigcup_{n\in\omega}E_n$, since each point of $A$ is in $2^\omega$ members of $\mathscr{F}$ and must therefore eventually be picked. (Alternatively, one could index $\mathscr{F}$ so that each member is listed $2^\omega$ times.)
