Adding a single Cohen real makes the set of reals from the ground model strong measure zero (see [this question](https://mathoverflow.net/questions/63497/cohen-reals-and-strong-measure-zero-sets)).


The notion of strong measure zero sets has its dual concept in the category branch -- strongly meager sets. A set $X\subseteq \mathbb{R}$ is strongly meager if for any null set $Y$ there exists $t\in \mathbb{R}$ such that $(t+X)\cap Y=\varnothing$. 
One can see duality of these notions due to Galvin-Mycielski-Solovay Theorem which states that a set $X\subseteq \mathbb{R}$ is strong measure zero if and only if for any meager set $Y$ there exists $t\in \mathbb{R}$ such that $(t+X)\cap Y=\varnothing$. 


Random real forcing is dual to Cohen forcing in the sense of measure and category. Therefore it makes sense to ask, whether:



> 
> The set of reals from generic model $ \mathbb{R}\cap V$ is strongly meager after adding a single random real?
> 
> 
> 


I have heard that the answer is affirmative, but I have not been able to find any published proof. 
Note that $\mathbb{R}\cap V$ is meager after adding a random real (see [this question](https://mathoverflow.net/questions/8269/adding-a-random-real-makes-the-set-of-ground-model-reals-meager)).



Random reals and strongly meager sets
As I have written above the affimative answer itself was known to many people including T. Bartoszyński. The following proof is due to T. Weiss (my advisor).


Proof. We follow closely the proof and notation of Lemma 3.2.42 from [1]. Let $A$ be a Borel measure zero set in $M[r]$, where $r$ is a random real over $M$. There exists $\dot{A}\subseteq 2^{\omega}\times 2^{\omega}$ measure zero set coded in $M$, such that $\dot{A}\_{r}=A$ (notation: $\dot{A}\_{r}=\{y\colon \left<r,y\right>\in \dot{A}\}$).


Then 
$$\dot{A}\subseteq\bigcap\_{m\in\omega}\bigcup\_{n\geq m} [s\_{n}]\times[t\_{n}]$$ 
where $s\_n, t\_n\in 2^{<\omega}$, $\sum\_{n=0}^{\infty}\frac{1}{2^{2|s\_{n}|}}<\infty$ and we can assume that $|t\_{n}|=|s\_{n}|$ for any $n\in\omega$.


Let $z\in 2^{\omega}\cap M$ and $f\in\omega^\omega$ be increasing. Then
$$\mu(\{x\colon\left<x,x\_f+z\right>\in [s]\times[t]\})\leq \frac{2^{f^{-1}(|s|)}}{2^{|s|+|t|}}$$
(where $x\_f\in 2^{\omega}$ such that $x\_{f}(n)=x(f(n))$). By induction on length $|s\_{n}|$ we define an increasing function $f\_{A}\in\omega^{\omega}$ such that 
$$\sum\_{n=0}^{\infty}\frac{2^{f\_{A}^{-1}(|s\_{n}|)}}{2^{2|s\_{n}|}}<\infty.$$
It is easy to see that such function exists as for any $\varepsilon>0$ we can find $N\_{\varepsilon}\in\omega$ such that $\sum\_{n\geq N\_{\varepsilon}}\frac{1}{2^{2|s\_{n}|}}<\varepsilon$.


Notice also that $\left< x,x\_{f}+z\right>\in [s]\times[t] $ if and only if $\left<x,x\_{f}\right>\in[s]\times [t+z]$.


The set
$$H\_{z}=\{x\in\left<x,x\_{f\_A}+z\right>\in\bigcap\_{m\in\omega}\bigcup\_{n\geq m} [s\_{n}]\times[t\_{n}]\}$$ has measure zero and is coded in $M$ for every $z\in 2^{\omega}\cap M$. Thus $r\notin H\_{z}$ and $\left<r,r\_{f\_{A}}+z\right>\notin \dot{A}$. This implies that $r\_{f\_{A}}\notin A+z$ for every $z\in 2^{\omega}\cap M$, so $(2^{\omega}\cap M)+A\neq 2^{\omega}$ and so $2^{\omega}\cap M$ is strongly meager. 


$\square$


References


[1] T. Bartoszyński, H. Judah, Set thoery: on the structure of the real line, A K Peters, 1995


