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Since u{V qei, : Q E Q} = R, X(R) > 0 + and X is R-invariant, we must have the inequality X(V) > 0. On the other hand, for any nonzero q E Q, we can write Taking q arbitrarily small, we see that the set V does not have the Steinhaus property (see Exercise 8 of Chapter 1). This circumstance immediately yields a contradiction with X(V) > 0. The contradiction obtained finishes the proof. Remark 1. It can easily be observed that V turns out to be a Vitali type set for the group I? = Qei,. This implies at once that the set V is not Lebesgue measurable (see Theorem 1 from Chapter 1).

We now assert that the sets A and B are nonmeasurable in the Lebesgue sense. Indeed, suppose otherwise. Then at least one of these sets is Lebesgue measurable and, in view of the relation -A = B , we claim that both these sets must be Lebesgue measurable. Since we derive that X(A) = X(B) > 0. On the other hand, the metrical transitivity of the Lebesgue measure (see Exercise 7 from Chapter 1) implies which leads to a contradiction. We thus conclude that each of the sets A and B is nonmeasurable in the Lebesgue sense.

Proof. Let {ei : i E I ) be a Hamel basis of R. Fix an index io from I and consider the vector subspace of R (over Q, of course) generated by the 36 CHAPTER 3 partial family {ei : i E I\ {io}}. We denote this subspace by V. Actually, V is a vector hyperplane in R regarded as a vector space over Q. Our purpose is to show that V cannot be Lebesgue measurable. Suppose to the contrary that V E dom(X) where X denotes, as usual, the standard Lebesgue measure on R. Since u{V qei, : Q E Q} = R, X(R) > 0 + and X is R-invariant, we must have the inequality X(V) > 0.