Nonmeasurable Sets

Theorem

There exist sets which are not Lebesgue measurable.

Proof: (Vitali 1905)

Define an equivalence relation ~ on by .

This relation divides into equivalence classes . Clearly each equivalence class contains a point in the interval . Let be a set which contains exactly one point from each equivalence class. (By strong axiom of choice exist.)

For all , define . (1)

We observe that . (2)

The second inclusion is clear. To see the first we note that if , then there must be a s.t. (because must belong some equivalence class and by our definition contains one element from every equivalence class). Then for some and thus we have .

We note also that (3)

Otherwise for some we would have , which is impossible by our choice of .

Now suppose is measurable. Then every are.

Using (2) and monotonicity property of the Lebesgue measure we would have (4)

Also by translation invariance of the Lebesgue measure we have (5)

Therefore, using the fact that are pairwise disjoint and that the Lebesgue measure is -additive, we get that (6)

Thus cannot be measurable.