This question arose a few years back when I was an assistant teacher on a course of basic (Lebesgue) measure theory, but I didn't find an answer or anyone able to solve the problem. The setting of the problem is as follows:


We say that $A$ has full outer measure in the unit interval $[0,1]$ if 
$$m^\ast(A) = m^\ast([0,1]).$$
How many disjoint subsets of the unit interval with full outer measure can there exist? By the addivity property of the Lebesgue measure these sets have to be non-measurable.
In [this MO post](https://mathoverflow.net/questions/12935/non-lebesgue-measurable-subsets-with-large-outer-measure) Gerald Edgar shows that by using the [Bernstein set](http://en.wikipedia.org/wiki/Bernstein_set) you can compose any measurable subset $A$ of $\mathbb{R} ^n$ into two disjoint subsets, both with full outer measure in $A$. This gives a partial answer of "at least two" to my question. 



> 
> **Question:** Can the 'Bernstein set -construction' be modified (or is there some other method) to create $n$ disjoint subsets of $[0,1]$ with full outer measure in the unit interval? 
>  Can there exist infinitely many disjoint subsets of the unit interval with full outer measure? If so, then can this family be uncountable?
> 
> 
> 



Is there a maximum to the amount of disjoint non-measurable subsets of the unit interval with full outer measure?
In [this article](http://cs.pwr.edu.pl/cichon/Math/Bernstein.pdf), J. Cichon shows that the real line (and hence the unit interval) can be partitioned into continuum many Bernstein subsets. Of course, any Bernstein set has full outer measure.


