Is the sum of 2 Lebesgue measurable sets measurable? Is the sum of two measurable set measurable? I think it is not...
 A: 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 $A\times \{0\}$ and $\{0\}\times \mathbb R$ both have Lebesgue measure 0 in the plane, but their sum $A\times \mathbb R$ is not measurable. 
A: I think the sum of 2 Borel sets is analytic, hence measurable. 
A: 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 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.
