If we consider a finite set $A\subset\mathbb R^n$, uniqueness of the convex decomposition of points in $A$ is equivalent to the absence of $\mu\neq0$ signed measure supported on $A$ such that $\mu(\mathbb R^n) = 0$ and, $$ \int_{\mathbb R^n}x\mathrm d\mu(x)=0. $$ My question is, what happens when $A$ is a measurable set of non-null measure and we restrict combinations to be absolutely continuous? More precisely:

Is there a Borel set $A \subset R^n$ of positive (Lebesgue) measure such that there exists

no$\mu\neq0$ signed measure verifying $|\mu|\leq\lambda_A$ (noting $\lambda$ the Lebesgue measure, and $\lambda_A$ its trace on $A$), $\mu(\mathbb R^n) = 0$ and, $$\int_{\mathbb R^n}x\mathrm d\mu(x)=0?$$

Typically, as soon as $A$ contains an open set, there exists such $\mu$. On the other hand, $A$ does not need to contain an open set to have non-null measure.