In this problem you will construct a subset V of [0, 1) 
which is not Lebesgue measurable. 
(This example was discovered by Vitali in 1905.)

(a) For x, y in [0, 1) we define x(+)y to be x + y if x + y in [0, 1), 
and to be x + y ? 1 otherwise, so that always x (+) y in [0, 1). 
Make a similar definition for x (-) y, and show that your definition makes sense.

(b) Next write x ~ y if and only if x (-) y is rational. 
Show that this is an equivalence relation, i.e., 
that it is symmetric, reflexive, and transitive

(c) The equivalence relation in (b) partitions [0,1) into equivalence classes. 
Use the axiom of choice to choose one element 
from each equivalence class to form the set V. 

(d) For each rational number r in [0,1), let Vr = { r (+) x : x in V }.
 Let Q be the set of rational numbers in [0,1). Show that 

[0,1)=UNION 'r in Q' Vr (***), --V sub r--

and that the sets Vr are pairwise disjoint.

(e) Let A denote Lebesgue measure. Assume that the set V is measurable. 
Show that then each set Vr is also measurable, and 
that lamda(V) = lamda(Vr) for all r in Q. 
(Lebesgue measure has the property that it is translation invariant, i.e., 
for any measurable set E C E, and for any x in R, 
we have lamda(E) = lamda(E + x), where E + x = {x + y: y e E}).

(f) Show that all this leads to a contradiction, as follows. 
By (***), [0,1) is a countable union of disjoint measurable sets. 
Hence lamda([0,1)) = SUM 'r in Q' lamda(Vr). 
All the sets Vr have the same measure as V. 
If lamda(V) = 0, it follows that 1 = lamda([0,1)) = 0, and that's impossible. 
Hence lamda(V) > 0, and so lamda([0,1)) = oo, and that's also impossible. 
Hence V cannot be measurable.