Is there any necessary condition that a non-measurable subset of $[0,1]$ with outer measure 1 should satisfy?

I was going through the Vitali set with outer measure 1 construction

The family F of open subsets of [0,1] with measure < 1 has cardinality c. So it can be well-ordered in such a way that every member of F has fewer than c predecessors. Since the complement of each member of F has cardinality c, there is a function f: F -> [0,1] such that for each A in F, f(A) is not a member of A and is not in {f(B)+q: B a predecessor of A, q in Q}. Thus { f(A): A in F } contains at most one representative of each coset. Complete V by putting in representatives of all cosets not already represented. Then the outer measure of this V is 1.

It is not clear to me why it has outer measure 1

  • 1
    $\begingroup$ "Clearly, it need not be dense" is false. If $A \subset [0,1]$ is not dense, then $\overline{A} \subsetneq [0,1]$ which already implies $\overline{A} \cap (0,1) \subsetneq (0,1)$. But then $(0,1) \setminus A$ is a nonempty open set and thus has positive measure and (up to nullsets), we have $A \subset (0,1) \setminus [(0,1) \setminus \overline{A}]$ and the right-hand side has measure strictly smaller than $1$. Hence, $A$ can not have full measure in $[0,1]$. $\endgroup$ – PhoemueX Apr 7 '15 at 17:56
  • $\begingroup$ @PhoemueX: correct, thanks. $\endgroup$ – Anonymous Apr 8 '15 at 13:18

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