The Vitali Sets

The Vitali Sets 1 Non-Measurable Sets • There are subsets of R that are non-measurable: They cannot be assigned a measure by any extension of λ, without giving up on Non-Negativity, Countable Additivity, or Uniformity. 2 The Axiom of Choice Proving that there are non-measurable sets requires: Axiom of Choice Every set of non-empty, non-overlapping sets has a choice set. (A choice set for set A is a set that contains exactly one member from each member of A.) 3 Defining the Vitali Sets 3.1 A sketch of the construction • Define an (uncountable) partition U of [0, 1). • Use the Axiom of Choice to pick a representative from each cell of U. • Use these representatives to define a (countable) partition C of [0, 1). • A Vitali Set is a cell of C. 3.2 Defining U a, b ∈ [0, 1) are in the same cell if and only if a − b ∈ Q. 1 3.3 Defining C • C has a cell Cq for each rational number q ∈ Q[0,1) . • C0 is the set of representatives of cells of U. • Cq is the set of numbers x ∈ [0, 1) which are at a “distance” of q from the representative of their cell in U. Here “distance” is measured by bending [0, 1) into a circle: 0 1 0 1 and traveling counter-clockwise. For instance, 1 4 is at “distance” 1 2 from 34 : 0 1 1 4 4 4.1 3 4 A Vitali Set Cannot Be Measured Assumptions Countable Additivity [  λ {A1 , A2 , A3 , . . .} = λ(A1 ) + λ(A2 ) + λ(A3 ) + . . . whenever A1 , A2 , . . . is a countable family of disjoint sets for each of which λ is defined. Non-Negativity λ(A) is either a non-negative real number or the infinite value ∞, for any set A in the domain of λ. Uniformity µ(Ac ) = µ(A), whenever µ(A) is well-defined and Ac is the result of adding c ∈ R to each member of A. 2 4.2 The Proof • Suppose otherwise: λ(Cq ) is well-defined for some q ∈ Q[0,1) . • By Uniformity, λ(Cq0 ) = λ(Cq ) for any q 0 ∈ Q[0,1) . • By Non-Negativity, λ(Cq ) is either 0, or a positive real number, or ∞. • By Countable Additivity, it can’t be any of these: – Suppose λ(Cq ) = 0. By Countable Additivity: λ([0, 1)) = λ(Cq ) + λ(Cq0 ) + . . . = 0| + 0 + {z0 + . .}. once for each integer = 0 – Suppose λ(Cq ) = r > 0. By Countable Additivity: λ([0, 1)) = λ(Cq ) + λ(Cq0 ) + . . . = |r + r +{zr + . .}. once for each integer = ∞ Moral: There is no way of assigning a measure to a Vitali set without giving up on Uniformity, Non-Negativity or Countable Additivity. 5 Revising Our Assumptions? • Giving up on Uniformity means changing the subject: the whole point of our enterprise is to find a way of extending the notion of Lebesgue Measure without giving up on uniformity. • Non-Negativity and Countable Additivity are not actually needed to prove the existence of non-measurable sets. • Some mathematical theories would be seriously weakened by giving up on the Axiom of Choice. 3