Is the sum of two measurable set measurable? I think it is not...

$\begingroup$ See also the previous MO question concerning sums of Borel sets mathoverflow.net/questions/48571/… $\endgroup$ – Andrey Rekalo May 27 '11 at 14:51
Evidently, there are measure zero sets with a non measurable sum. The article begins as follows:
Krzysztof Ciesielski, Hajrudin Fejzi´c, Chris Freiling,
Measure zero sets with nonmeasurable 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 nonmeasurable. Ask a typical mathematician why this should be so and you are likely to get the following response:
The Cantor middlethird 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 nonmeasurable set.
The intuition being that an interval has much more content than is needed for a nonmeasurable 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 nonmeasurable (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 (nonmeasurable) 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 middlethird 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 nonmeasurable.
The intuition relies on the fact that nonmeasurable sets can have far less content than an entire interval. Therefore, the claim should also hold when nonmeasurable 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 nonempty perfect subset of I. Constructing a set that is Bernstein in an interval is one of the standard ways of establishing nonmeasurability. 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.

$\begingroup$ "The Cantor middlethird 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 nonmeasurable set."  If I am not mistaken, is this claiming that the closed interval [o,2] is not Lebesgue measurable? $\endgroup$ – Regenbogen Mar 26 '10 at 23:04

$\begingroup$ @Regenbogen: I think the measure zero set is going to be a proper subset of the Cantor middlethird set :) $\endgroup$ – Kevin Buzzard Mar 26 '10 at 23:07

$\begingroup$ Regenbogen, I believe that the idea is to pass to subsets to get the counterexample. $\endgroup$ – Joel David Hamkins Mar 26 '10 at 23:08

3

5$\begingroup$ Here is a construction of two Borel sets whose sum is not Borel: P. Erdõs, A. H. Stone: On the sum of two Borel sets, Proc. Amer. Math. Soc. 25 (1970), 304306 available here: renyi.hu/~p_erdos/197015.pdf $\endgroup$ – Péter Komjáth Jun 13 '10 at 12:48
I think the sum of 2 Borel sets is analytic, hence measurable.
Note that the problem is trivial if you talk about subsets of the plane $\mathbb R\times \mathbb R$. Let $A\subseteq \mathbb R$ be nonmeasurable, then $A\times \{0\}$ and $\{0\}\times \mathbb R$ both have Lebesgue measure 0 in the plane, but their sum $A\times \mathbb R$ is not measurable.

$\begingroup$ I stole this argument from gowers Borel plus Borel = Borel?. (Who probably stole it from Sierpinski, or Euclid.) $\endgroup$ – Goldstern May 27 '11 at 16:06

$\begingroup$ why is $A\times \mathbb R$ non measurable? $\endgroup$ – Mathronaut Mar 1 '15 at 20:32