Let $A,B\in\mathcal A_{\Bbb R}^*$ be given with $\overline{\lambda}(A)<\infty$ and $\overline{\lambda}(B)<\infty$. Let $\overline{\lambda}(A)>0$ I want to prove that: $$\exists\ \delta>0\;\text{s.t.}\;\;\text{if}\;\;|x|<\delta\ \Rightarrow\ \overline{\lambda}(A\cap (A+x))>0\;\;\text{and}\;\;(-\delta,\delta)\subset \{a-a':a,a'\in A\}=:A-A$$
Proof:
I've already show that $\overline{\lambda}_{A,B}(x)=\overline{\lambda}(A\cap(B+x))$ is continuous $\forall x\in\Bbb R$ here.
So, since $\overline{\lambda}_{A,A}$ is continuous and $\overline{\lambda}_{A,A}(0)=\overline{\lambda}(A)>0$
$$\Rightarrow\ \exists\ \delta>0\;\; \text{s.t.}\;\; \overline{\lambda}_{A,A}(x)>0\;\; \text{if}\;\; |x|<\delta$$
where $\overline{\lambda}_{A,A}(x)=\overline{\lambda}(A\cap (A+x))$
$$\Rightarrow\ \text{if}\; x\in(-\delta,\delta),\;\; \overline{\lambda}(A\cap (A+x))>0\\ \Rightarrow\ \text{if}\; x\in(-\delta,\delta),\;\;A\cap (A+x)\ne\emptyset\\ \Rightarrow\ \text{if}\; x\in(-\delta,\delta),\;\;\exists\ a\in A\cap (A+x)\\ \Rightarrow\ \text{if}\; x\in(-\delta,\delta),\;\; \exists\ a\in A\;\text{and}\; a=a'+x\;\text{for an}\; a'\in A \\ \Rightarrow\ \text{if}\; x\in(-\delta,\delta),\;\;\exists\ a,a'\in A\;\text{s.t.}\;\ x=a-a'$$
thereby $x\in \{a-a':a,a'\in A\}$ and hence $(-\delta,\delta)\subset A-A$.
What do you think, did I miss something?