A subset of R that is not Lebesgue-measurable
Yongheng Zhang
When designing a measure m for R, it is desirable to make it satisfy all the four properties below.
(0) m is defined on P(R).
(1) For any interval I, mI is the length of the interval.
P
(2) (countable additivity) If hEn i is a sequence of disjoint measurable sets, then m(∪En ) = mEn .
(3) (translational invariance) If E is measurable and y ∈ R, then E + y is also measurable and
m(E + y) = mE.
However, it is impossbile to get such a measure. For example, the outer measure satisfies (0), (1),
and (3) along with a weaker version of (2) but not (2); the counting measure satisfies (0), (2) and (3)
but not (1). However, the most desirable properties are (1), (2) and (3). Lebesgure measure is a good
example and hence (0) does not hold for it. To see why, look at a subset of [0, 1] constructed below
(See Stein & Shakarchi, Real Analysis).
Let the relation ∼ on [0, 1] be defined by x ∼ y if x − y ∈ Q. It can be readily checked that ∼
is an equivalence relation and thus it partitions [0, 1]. Assuming the Axiom of Choice, we pick one
point from each equivalence class and put them into a set and call it N . We will show that N is not
Lebesgue-measurable.
Let {r1 , r2 , . . .} be an enumeration of the rational numbers in [−1, 1] and consider the collection
{N +ri }i∈Z+ . First of all, notice that the former collection is disjoint. [To see it, suppose there are i 6= j
in Z+ such that there is x ∈ (N +ri )∩(N +rj ). Thus, there are a and b in N such that a+ri = x = b+rj .
So a − b = rj − ri ∈ Q. This is ridiculous because by S
construction aPand b are in different
P∞ equivalence
classes associated with ∼ and thus a − b ∈
/ Q]. So m( (N + ri )) = m(N + ri ) = i=1 m(N ) where
the last equality follows from translational invariance.
S
Since N ⊂ [0, 1] and {r1 , r2 , . . .} ⊂ [−1, 1], for each i, N + ri ⊂ [−1, 2]. Therefore, (N + ri ) ⊂
[−1, 2]. On the other hand, for any x ∈ [0, 1], x is in some equivalence classes associated with ∼
and hence there is S
some rk ∈ Q ∩ [−1, 1] and some y ∈ N such that x = y + rk . Thus, x ∈ N + ri .
Therefore, [0, 1] ⊂ (N + ri ) ⊂ [−1, 2].
Consequently, 1 ≤ ∞ · mN ≤ 3. This makes mN neither 0 nor positive, which is impossible. Hence,
N is not measurable.
Royden, Real Analysis has a similar but more complicated construction compensated by a nicer
set. It begins with [0, 1) instead of [0, 1] but it get N in the same way. However, we defining +
to be addition modulo by 1 (m is also translational invariance with respect to the new +), it can
be shown that {N + ri } is a partition of [0, 1]Pwhere {r1 , r2 , . . .} is an enumeration of the rationals
∞
in [0, 1] (instead of [−1, 1]). So 1 = m([0, 1]) = i=1 mN which also makes mN neither 0 nor positive.
There are other examples in the literature, but most of them has the similar construction as the
one above. Nonetheless, since N is made by picking one element from each equivalence class and there
are tons of ways to do that, the number of nonmeasurable sets is enormous.
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