Non-measurable set $A$ such that every measurable subset of $A$ or $A^c$ has measure zero I want to show the following:

There is $A \subset \mathbb{R}$ such that both are satisfied:
  
  
*
  
*if $E \subset A$ is a Lebesgue measurable set then $\lambda(E) = 0$
  
*if $E \subset A^c$ is a Lebesgue measurable set then $\lambda(E) = 0$
  
  
  where $\lambda$ is the Lebesgue measure on the real line.

Do you know how to show that such a set exists? Also a reference would be appreciated. Thank you :)
ps: if you find a better title feel free to modify it!
 A: Yes. A standard construction goes by building the set by transfinite recursion to ensure that neither $A$ nor its complement contains a perfect set. Note that if this is the case, then any measurable subset of $A$ has measure $0$, because (by regularity of Lebesgue measure) if $E$ is measurable, $\lambda(E)$ is the supremum of $\lambda(K)$ for $K$ compact contained in $E$. But any compact set that has no perfect subset is in fact countable, so it has measure $0$. The same happens with $A^c$, but then neither $A$ nor $A^c$ can be measurable: If $A$ is measurable, then it has measure $0$, so $A^c$ has full measure and therefore contains a perfect set.
The transfinite recursion goes as follows: There are $\mathfrak c=|\mathbb R|$ many perfect subsets of $\mathbb R$. Identifying cardinals with ordinals, list them in a well-ordering of type $\mathfrak c$. We build $A$ and its complement by recursion. At stage $\alpha$ we have a set $A_{<\alpha}$ and a disjoint set $B_{<\alpha}$. (When $\alpha=0$, both are empty.) Consider the $\alpha$-th perfect set. Pick two elements of the perfect set not in $A_{<\alpha}\cup B_{<\alpha}$. Let $A_{\alpha}=A_{<\alpha+1}$ be the union of $A_{<\alpha}$ and one of these points, and let $B_{\alpha}$ be the union of $B_{<\alpha}$ and the other point. That the points exist follows from the fact that there are $\mathfrak c$ reals in any perfect set, but at stage $\alpha$ at most $2|\alpha|<\mathfrak c$ of them have been used. (When $\lambda$ is limit, $A_{<\lambda}$ is just the union of the previous $A_\alpha$, same for $B_{<\lambda}$.) At the end, one of the sets so built is $A$. This works, because by construction, given any perfect set $P$, at least one of its points is in $A$ (so $P$ is not contained in $A^c$), and at least one of its points is not (so $P$ is not contained in $A$).
The details of the construction can be varied a bit, but any argument must use choice in a nontrivial way, since it is consistent with the usual axioms of set theory (without the axiom of choice) that all sets of reals are measurable. That said, I haven't seen a presentation of the result I'm describing that avoids the use of a well-ordering of $\mathbb R$. The sets so obtained are called Bernstein sets, and were first exhibited by Felix Berenstein (of Bernstein-Schroeder fame) in $1908$.
