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)
which contradicts (4).
Thus cannot
be measurable.