Krzysztof Ciesielski,
Hajrudin Fejzi´c, Chris Freiling,
Measure zero sets with non-measurable sum
Abstract
For any C ⊆ R there is a subset A ⊆ C such that A + A has inner
measure zero and outer measure the same as C + C. Also, there is a
subset A of the Cantor middle third set such that A+A is Bernstein in
[0, 2]. On the other hand there is a perfect set C such that C + C is an
interval I and there is no subset A ⊆ C with A + A Bernstein in I.
1 Introduction.
It is not at all surprising that there should be measure zero sets, A, whose sum
A+A = {x+y : x ∈ A, y ∈ A} is non-measurable. Ask a typical mathematician
why this should be so and you are likely to get the following response:
The Cantor middle-third set, when added to itself gives an entire
interval, [0, 2]. So certainly there exists a measure zero set that
when added to itself gives a non-measurable set.
The intuition being that an interval has much more content than is needed for
a non-measurable set.
Indeed such sets do exist (in ZFC). Sierpi´nski (1920) seems to be the first
to address this issue. Actually, he shows the existence of measure zero sets
X, Y such that X+Y is non-measurable (see [7]). The paper by Rubel (see [6])
in 1963 contains the first proof that we could find for the case X = Y (see also
[5]). Ciesielski [3] extends these results to much greater generality, showing
that A can be a measure zero Hamel basis, or it can be a (non-measurable)
Bernstein set and that A+A can also be Bernstein. He also establishes similar
results for multiple sums, A + A + A etc.
This paper is mainly about the statement above and the intuition behind
it. Below we list four conjectures, each of which seems justified by extending
this line of reasoning.
Not only does such a set exist, but it can be taken to be a subset of the
Cantor middle-third set, C. (This does not seem to immediately follow
from any of the above proofs. Thomson [9, p. 136] claims this to be
true, but without proof.)
The intuition really has nothing to do with the precise structure of the
Cantor set, which might lead one to conjecture the following. Suppose
C is any set with the property that C + C contains a set of positive
measure. Then there must exist a subset A ⊆ C such that A + A is
non-measurable.
The intuition relies on the fact that non-measurable sets can have far
less content than an entire interval. Therefore, the claim should also
hold when non-measurable is replaced by other similar qualities. Recall
that if I is a set then a set S is called Bernstein in I if and only if
both S and its complement intersect every non-empty perfect subset
of I. Constructing a set that is Bernstein in an interval is one of the
standard ways of establishing non-measurability. Certainly, any set that
is Bernstein in an interval has far less content than the interval itself.
Therefore, we might conjecture that there is a subset A ⊆ C
with A+A
Bernstein in [0,2].
Combining the reasoning behind the Conjectures 2 and 3, let C be any
set with the property that C + C contains an interval, I. We might
conjecture that there must exist a subset A ⊆ C such that A + A is
Bernstein in I.
We will settle these four conjectures in the next four sections.
The paper goes on to show that conjectures 1, 2 and 3 are true, but 4 is false.
A set $E$ with positive Lebesgue measure can be decomposed as a union $E = A \cup B$ where each of $A$ and $B$ have zero inner measure, and therefore each of $A$ and $B$ are nonmeasurable with $m^*(A) = m^*(B) = m(E)$.
An example for this construction is a Bernstein set.
It seems that your example of bijection that sends one Cantor set with positive measure to another Cantor set with zero measure can be made $C^\infty$.
Best Answer
Evidently, there are measure zero sets with a non measurable sum. The article begins as follows:
The paper goes on to show that conjectures 1, 2 and 3 are true, but 4 is false.