TITLE: Dual Borel conjecture in Laver's model
QUESTION [18 upvotes]: A set $X\subseteq 2^\omega$
of reals is of strong measure zero (smz) if $X+M\not=2^\omega$
for every meager set $M$.  (This is a theorem of Galvin-Mycielski-Solovay,
but for the question I am going to ask we may as well take it as a definition.)
A set $Y$ is strongly meager (sm) if $Y+N\not=2^\omega$ for every Lebesgue null
set $N$.
The Borel conjecture (BC) says that every smz set is countable; the dual Borel conjecture (dBC) says that every sm set is countable.
In Laver's model (obtained by a countable support iteration of Laver reals
of length $\aleph_2$)
the BC holds.   Same for the Mathias model.
In a paper that I (with Kellner+Shelah+Wohofsky) just sent to arxiv.org, we claim that it is not clear if Laver's model satisfies the dBC.
QUESTION:
 Is that correct?  Or is it perhaps known that Laver's model has uncountable sm
 sets?
Additional remark 1: Bartoszynski and Shelah (MR 2020043) proved in 2003
that in Laver's model there are no sm sets of size continuum ($\aleph_2$).
(The MR review states that the paper proves that the sm sets are exactly
$[\mathbb R] ^{\le \aleph_0}$.  This is obviously a typo in the review.)
Additional remark 2: If many random reals are added to Laver's model (either
during the iteration, or afterwards), then BC still holds, but there will be sm
sets of size continuum, so dBC fails in a strong sense.

REPLY [5 votes]: In analogy with the fact that Lusin set are strongly null, Pawlikowski has shown that Sierpinksi sets are strongly meagre, so one might ask if there are 
are Sierpinski sets in the iterated Laver Model. The answer is no; indeed
 any set of positive outer measure contains an uncountable null set after adding a Laver real
Let $\psi:\omega\to\omega$ be a function growing so quickly  that
$$\lim_{n\to\infty} \prod_{j=n}^\infty \left(1- 2^{-\psi(j)} \right) =1  $$
Then let $L:\omega\to \omega$ be the generic real added by $\mathbb L$ and define
$A$ to be the  set of those $f\in 2^\omega $ such that
$f$ is constantly $0$ on the interval $[L(n), L(n) + \psi(n)] $  for all but finitely many  $n$. 
 $A$ is Lebesgue null and it will be shown that $|A\cap X|=\aleph_1$.
Suppose not and let $T\in \mathbb L$ and $C\in [X]^{\aleph_0}$ be such that
$T$ forces that $X\cap A\subseteq C$.  For $s\in T$ extending the root of $T$
define $S_s $  to be the set of all $f\in 2^\omega$ for which there exists infinitely many $ n\in \omega$ such that $ s^\frown n\in T$ and $ f(n+j) = 0 $ for all $j\leq \psi(|s|) $. Then each $S_s$ is of measure one. Since $X$ has positive outer measure  there is some $x\in \bigcap_{s\supseteq \text{root}(T)}S_s\cap(X\setminus C)$. It follows that there is then
$\tilde{T}\subseteq T$ such that $\tilde{T}\in \mathbb L$ and
$x$ is constantly $0$ on the interval $[m,m+\psi(|s|)]$
for all $s^\frown m\supseteq \text{root}(\tilde{T})$.
In other words, $\tilde{T}$  forces that $x$ is in $A$ contradicting that $x\in X\setminus  C$.