Math 828 Fall 1998
Non-measurable sets
1 Introduction
The purpose of these notes is to show how our original construction of a nonmeasur-
able set can be refined to yield a set with even more remarkable properties. I’ll also
show how our original construction leads to results that are even more paradoxical
than we might at first have thought.
We work on the half-open interval G =[0, 1) which we regard as a commutative
group with addition mod 1. In what follows the operations of addition, subtraction,
and multiplication on elements of G will be denoted by the usual symbols, but the
results of these operations are always understood to be reduced modulo 1.
The goal here is to produce a non-measurable subset NM of G that has outer
measure 1 and inner measure zero (in the sense that every measurable subset has
measure zero). Necessarily such a set is “universally non-measurable” in the sense
that its intersection with any measurable set of positive measure is non-measurable.
To see why this is so, note first that NM
0
also has outer measure 1 and inner
measure zero. To prove the former assertion, suppose E is measurable and contains
NM
0
. Then E
0
is measurable and contained in NM ,soE
0
has measure zero. Therefore
(by measurability) E has measure one. The statement about inner measure is proved
similarly.
Now suppose E is a measurable subset of G whose intersection with NM is mea-
surable. Then E ∩ NM
0
= E\(E ∩ NM ) is measurable, so because both NM and
NM
0
have inner measure zero, both E ∩ NM and E ∩ NM
0
have measure zero, hence
so does their union E. Thus any set whose intersection with NM is measurable has
measure zero. Equivalently, the intersection of a set of positive measure with NM
must be nonmeasurable. ¤
2 Dense Subgroups of G
The rationals in G form a subgroup, and they are dense in G. The main result of
this section shows that this density can be viewed within a larger context.
2.1 Theorem. Every infinite subgroup of G is dense.
Proof. Suppose H is a subgroup of G that has infinitely many elements. Let I
be any interval in G. Our goal is to show that some element of H lies in I. Fix a
positive integer k with 1/k < length of I. Let h
1
<h
2
<...<h
k+1
be k + 1 distinct
elements of G, arranged in increasing order. Then for some index j we must have
d = h
j+1
− h
j
< 1/k.Thusnd ∈ I for some positive integer n.Nowd ∈ H, hence
nd ∈ H, and we are done. ¤
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Math 828 Fall 1998
2.2 Exercise. let T denote the unit circle with complex multiplication, taken in
the metric of R
2
. Show that every infinite subgroup of T is dense in T .
2.3 Exercise. Show that for every irrational number ξ:
(a) The sequence {e
2πinξ
: n ∈ Z} is dense in the unit circle.
(b) The sequences {sin(2πnξ):n ∈ N} and {cos(2πnξ):n ∈ N} are dense in
[−1, 1].
3 Difference sets
If A ⊂ G we define
A − A
def
= {a − b : a, b ∈ A}.
Note that for x ∈ g:
x ∈ A − A ⇐⇒ A ∩ (A + x) 6= ∅.
3.1 Theorem. If E is a measurable subset of G and E has positive measure, then
E − E contains an interval.
Proof. By the Lebesgue Density Theorem there is a point 0 6= d ∈ E such that
µ(E ∩ I(d, h))
2h
> 3/4, (1)
for all sufficiently small h>0, where I(d, h)=(d − h, d + h). Now fix a “sufficiently
small” h that is also small enough that both d − h and d +2h lie in G (without
carrying out any reductions mod 1).
Claim: The interval [0,h) lies in E − E.
To see this, fix x ∈ (0,h) (there’s no doubt that 0 lies in every difference set). We’ll
show that (E + x) ∩ E 6= ∅. To simplify notation, let I = I(d, h) and set F = E ∩ I.
Then because 0 <x<h, both F and F + x lie in the interval [d − h, d +2h]. By (1)
both F and F +x have measure > 3h/2, so if they were disjoint, their union—which is
contained in [d − h, d +2h]—would have measure > 2 × 3h/2=3h, which contradicts
the fact that the containing interval only has measure 3h.ThusF and F + x have
nontrivial intersection, hence the same is true of their supersets E and E + x. ¤
4 Construction of NM
Fix an irrational number ξ ∈ G and let
A = {nξ : n ∈ Z} and B = {2nξ : n ∈ Z}
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Math 828 Fall 1998
Then A and B are subgroups of G, both infinite because of the irrationality of ξ.
Therefore by Theorem 2.1, both A and B are dense in G.NowA = B ∪ C, a disjoint
union, where
C = {(2n +1)ξ : n ∈ Z}.
C is also dense in G since it equals B + ξ, and B is dense.
Let NM
0
be the set you get by choosing one element from each of the cosets
of G modulo A
1
. In other words, G is partitioned into equivalence classes by the
equivalence relation
g
1
≡ g
2
⇐⇒ g
1
− g
2
∈ A,
and we are choosing one element from each equivalence class. Then, as we saw in
our original construction of a non-measurable set, not only is G the disjoint union
of sets g + A as g runs through NM
0
, it is also the countable disjoint union of the
sets a + NM
0
as a runs through A. Just as for our original example, NM
0
has inner
measure zero. For if F is a subset of NM
0
that is measurable, then G contains the
countable disjoint union f sets F + a as a runs through A. Since all these sets have
the same measure, translation-invariance (modulo 1) of Lebesgue measure (on G), F
must have measure zero, else G would have infinite measure, an absurdity.
Here is another proof that uses the translation-invariance of Lebesgue measure
more subtlely—can you see where it shows up? NM
0
is constructed so that its differ-
ence with itself contains no element of A other than 0. Thus the same will be true of
any measurable subset F of NM
0
. Since A is dense, F −F cannot contain an interval,
and so by Theorem 3.1 F cannot have positive measure.
We use NM
0
to construct the non-measurable set that simultaneously has inner
measure zero and outer measure 1. We claim that
NM
def
= NM
0
+ B =
[
b∈B
(NM
0
+ b).
has the desired property. We phrase these properties like this:
4.1 Theorem. Every measurable subset of NM has measure zero. Every measur-
able superset of NM has measure 1.
Proof. To see that NM has inner measure zero, note that
NM − NM = NM
0
− NM
0
+ B
and since NM
0
− NM
0
does not contain any non-zero element of A, one checks easily
that NM − NM does not intersect C. Since C is dense, NM − NM contains no
interval, so as above, any measurable subset of NM must have measure zero, which
establishes the desired “inner measure zero” property of NM .
For the “outer measure one” property note that the complement of NM in G is just
the translate NM + ξ, which—by the translation-invariance of Lebesgue measure—
also has inner measure zero. Thus if F is a measurable superset of NM , then F
0
is a
1
We use the Axiom of Choice here
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Math 828 Fall 1998
measurable subset of NM
0
, and so has measure zero, so F itself has measure 1. Thus
NM has outer measure 1 as advertised. ¤
The following property of NM was noted in the Introduction:
4.2 Corollary. NM is “universally non-measurable” in the sense that its intersec-
tion with any measurable subset of G having positive measure is non-measurable.
4.3 Corollary. Every measurable subset of R having positive measure contains a
non-measurable subset.
Proof. Exercise. ¤
5 Paradoxical decompositions
Suppose we had just defined Lebesgue outer measure on G, and were asking if that
definition gave a reasonable extension of the notion of “length” to all subsets of G.
The set NM and its complement NM + ξ show that this approach needs refinement,
since they give a “paradoxical” decomposition of G—which has length one—into two
disjoint subsets, both of which also, according to our new definition, have “length
one.”
The set NM
0
(or equally effectively, our original nonmeasurable set) gives another
kind of paradoxical decomposition that is, in some sense, even stranger. Recall that
G is the disjoint union of sets NM
0
+ a where a runs through the dense subgroup
A generated by the irrational number ξ. Let’s enumerate the elements of a as a
0
=
0,a
1
,a
2
,..., and write M
j
= NM
0
+ a
j
(in particular M
0
= NM
0
). Consider the
pairwise disjoint sets M
0
,M
2
,M
4
,..., whose union is our universally nonmeasurable
set NM . Now observe that M
2
can be translated onto M
1
, M
4
can be translated onto
M
2
, M
6
onto M
3
, etc. In other words, the disjoint pieces {M
2n
}
∞
0
can be individually
translated so they can be reassembled into all of G!
Similarly, the disjoint pieces {M
2n+1
}
∞
0
can be individually translated and re-
assembled into all of G.ThusG can be broken into a countable collection of pairwise
disjoint which can be appropriately translated and reassembled into two copies of G!
It’s perhaps better to think of G as being the unit circle here, in which case the
translations become rotations: isometric motions of the circle. Even more amazing
results were proved in the period 1915–25 by Hausdorff, Banach, and Tarski. Here is
the most famous of these:
The Banach-Tarski Paradox. The unit ball of R
3
can be decomposed into a finite
collection of pairwise disjoint sets which can be reassembled by rigid motions into a
ball of radius 2.
In fact, Hausdorff, Banach, and Tarski showed that given any two bounded subsets
A and B of R
3
that have nonempty interior, you can cut A into finitely many pairwise
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Math 828 Fall 1998
disjoint sets which you can then rearrange by rigid motions to form B. This has to
be one of the strangest results ever proved by mathematicians!
For more on the Banach-Tarski paradox and its implications for mathematics, see
Stan Wagon’s book The Banach-Tarski Paradox, Cambridge Univ. Press 1985, and
a little (4 page) article by Frank Wikstr¨om titled the Banach-Tarski Theorem which
is available on the Web at
{\tt http://abel.math.umu.se/~frankw/articles/index.html}
(or you can get a copy from me).
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