Let $f:\mathbb{R}^d\rightarrow\mathbb{C-{0}}$ be Lebesgue measurable with $f(x+y)=f(x)f(y)$ for all $x,y \in \mathbb{R}^d$. Let $U\subset\mathbb{C}$ be a neighborhood of $f(0)=1$.
I want to find some Lebesgue measurable subset $A$ of $\mathbb{R}^d$ such that $m(A)>0$ and $A-A$ is contained in $f^{-1}(U)$.
I tried letting $\epsilon>0$ such that $(-\epsilon, \epsilon)\subset U$, then letting $V=(\frac{-\epsilon}{2},\frac{\epsilon}{2})$. Then $V-V\subset U$. Then, I let $A=f^{-1}(V)$, but I am not sure this will work since I am having trouble showing that $m(A)>0$ and $A-A \subset f^{-1}(U)$.
The mapping $h \colon (\mathbb{C} - \{0\})^2 \to \mathbb{C} - \{0\}, (x,y) \mapsto x/y$ is continuous, so there is an open neighbourhood $W$ of $1$ such that $W \times W \subseteq h^{-1}(U)$. Let $A = f^{-1}(W)$.
Now let $T = f(\mathbb{R}^d) \subseteq \mathbb{C} - \{0\} $. It is possible to find a countable family of points $y_i \in T$ such that $T \subseteq \bigcup y_i W$. Then if $f(x_i) = y_i$, we have $\mathbb{R}^d = \bigcup (x_i + A)$. This proves that $m(A) > 0$.