Is the sum of 2 Lebesgue measurable sets measurable? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T09:03:18Z http://mathoverflow.net/feeds/question/19471 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/19471/is-the-sum-of-2-lebesgue-measurable-sets-measurable Is the sum of 2 Lebesgue measurable sets measurable? Nicolò 2010-03-26T22:45:54Z 2011-05-27T14:42:51Z <p>Is the sum of two measurable set measurable? I think it is not...</p> http://mathoverflow.net/questions/19471/is-the-sum-of-2-lebesgue-measurable-sets-measurable/19472#19472 Answer by Joel David Hamkins for Is the sum of 2 Lebesgue measurable sets measurable? Joel David Hamkins 2010-03-26T22:52:32Z 2010-03-27T15:30:45Z <p>Evidently, there are <a href="http://www.math.wvu.edu/~kcies/prepF/89A+A/89A+A.pdf" rel="nofollow">measure zero sets with a non measurable sum</a>. The article begins as follows:</p> <blockquote> <blockquote> <p>Krzysztof Ciesielski, Hajrudin Fejzi´c, Chris Freiling, </p> <p><b>Measure zero sets with non-measurable sum</b></p> <blockquote> <blockquote> <p>Abstract</p> <p>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.</p> </blockquote> </blockquote> <p>1 Introduction.</p> <p>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:</p> <blockquote> <blockquote> <p>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.</p> </blockquote> </blockquote> <p>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.</p> <p>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.</p> <ol> <li><p>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.)</p></li> <li><p>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.</p></li> <li><p>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].</p></li> <li><p>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.</p></li> </ol> <p>We will settle these four conjectures in the next four sections.</p> </blockquote> </blockquote> <p>The paper goes on to show that conjectures 1, 2 and 3 are true, but 4 is false.</p> http://mathoverflow.net/questions/19471/is-the-sum-of-2-lebesgue-measurable-sets-measurable/27903#27903 Answer by Péter Komjáth for Is the sum of 2 Lebesgue measurable sets measurable? Péter Komjáth 2010-06-12T05:35:32Z 2010-06-12T05:35:32Z <p>I think the sum of 2 Borel sets is analytic, hence measurable. </p> http://mathoverflow.net/questions/19471/is-the-sum-of-2-lebesgue-measurable-sets-measurable/66196#66196 Answer by Goldstern for Is the sum of 2 Lebesgue measurable sets measurable? Goldstern 2011-05-27T14:42:51Z 2011-05-27T14:42:51Z <p>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 non-measurable, then <code>$A\times \{0\}$</code> and <code>$\{0\}\times \mathbb R$</code> both have Lebesgue measure 0 in the plane, but their sum $A\times \mathbb R$ is not measurable. </p>