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



Is the sum of 2 Lebesgue measurable sets measurable?
Evidently, there are [measure zero sets with a non measurable sum](http://www.math.wvu.edu/~kcies/prepF/89A+A/89A+A.pdf). The article begins as follows:



> 
> 
> > 
> > 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.
> > 
> > 
> > 1. 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.)
> > 2. 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.
> > 3. 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].
> > 4. 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.
> > 
> > 
> > 
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> 


The paper goes on to show that conjectures 1, 2 and 3 are true, but 4 is false.


