Converse to Steinhaus? Does there exist a Lebesgue null set $A$ such that $\{ x-y : x,y \in A \}$ contains an interval?
Under CH, yes.  Let $\langle C_\alpha \rangle_{\alpha < \omega_1}$ list the closed nowhere dense sets and $\langle r_\alpha \rangle_{\alpha < \omega_1}$ the reals.  At stage $\alpha$ choose a real $x_\alpha \notin \bigcup \{ C_\beta + nr_\alpha : \beta < \alpha, n \in \mathbb Z \} \cup \{ x_\beta + nr_\alpha : \beta < \alpha, n \in \mathbb Z \}$.  Then both $x_\alpha$ and $x_\alpha + r_\alpha$ are not previously chosen and not in the previously listed nwd sets and the translations.  Any closed nowhere dense set has only countable intersection with $A = \{ x_\alpha, x_\alpha + r_\alpha : \alpha < \omega_1 \}$.  So let $\epsilon$ be arbitrary and let $O$ be a dense open set of size $\epsilon$.  Then $A \setminus O$ is countable, and so $A$ must have measure zero.  Finally, note $\{ x-y : x,y \in A \} = \mathbb R$.  CH is overkill; clearly Martin's Axiom suffices for this.
 A: Let $A$ be the middle-thirds Cantor set
$$A=\left\{\sum_{j=1}^\infty a_j3^{-j} : a_1,a_2,\dots \in\{0,2\}\right\}$$
Claim: $\{a-b: a,b\in A\} \supseteq [-1,1]$. To prove this, let $x\in[-1,1]$ be given. We will recursively construct sequences $(a_j)$ and $(b_j)$, with $a_j,b_j \in\{0,2\}$, such that $\left|x-\sum_{j=1}^n (a_j-b_j)3^{-j}\right| \leq 3^{-n}$. It then follows that $x=\sum_{j=1}^\infty a_j3^{-j}-\sum_{j=1}^\infty b_j3^{-j} \in A$.
Suppose that $a_1,\dots,a_n$ and $b_1,\dots,b_n$ are defined, satisfying the inequality $\left|x-\sum_{j=1}^n (a_j-b_j)3^{-j}\right| \leq 3^{-n}$ (for $n=0$ this just reduces to the assumption $|x|\leq 1$). In other words, if we let $d=x-\sum_{j=1}^n (a_j-b_j)3^{-j}$, then $|d| \leq 3^{-n}$. Now we have three cases:


*

*If $|d|\leq 3^{-n-1}$, then define $a_{n+1}=0$ and $b_{n+1}=0$.

*If $d>3^{-n-1}$, then define $a_{n+1}=2$ and $b_{n+1}=0$.

*If $d<-3^{-n-1}$, then define $a_{n+1}=0$ and $b_{n+1}=2$.


In each case, it follows that $|d-(a_{n+1}-b_{n+1})3^{-n-1}|\leq 3^{-n-1}$. This completes the proof.
