TITLE: Billingsley's solution to 3.18 (b)
QUESTION [5 upvotes]: Problem 3.18(b) in Billingsley's Probability and Measure (3e) is

Show that if $\lambda^*(E)>0$, then $E$ contains a nonmeasurable
  subset.

[Here $\lambda^*$ is the Lebesgue outer measure.]
In the ‘Notes on  the Problems’, on page. 556 the following solution is given

If the $E \cap (H \oplus r)$ are all Borel sets, they all have
  Lebesgue measure $0$, and so $E$ is a Borel set of measure $0$.

[Here $H$ is the Vitali set, $r$ implicitly ranges over all rational numbers and $\oplus$ denotes addition modulo 1.]
It seems to me that the solution offered is wrong since the quoted sentence only implies that either $E$ is not Borel or one of $E \cap (H \oplus r)$ is not Borel. From this it does not follow that $E$ has a nonmeasurable subset.
Am I right in thinking this? 
It also seems to me that it is not possible to solve this question given the material covered so far in Billingsley. Am I right in this?
Just to clarify: I am aware that the statement to be proved is a standard theorem and I have found its proof in other books. My question is specifically regarding Billingsley solution/hint and whether the problem is solvable at this point in the text.

REPLY [2 votes]: Found an answer to my own question. In the section before the question, on page 45, Billingsley uses ‘nonmeasurable’ as equivalent to ‘not-Borel’. While I’m not sure if this is standard, given this interpretation of ‘nonmeasurable’ the question and the hint do make perfect sense. 
Not deleting the question as it might be valuable to others who may have the same confusion as I had.