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.
Best Answer
There is no such set. Given an $A\subseteq\mathbb R^n$, we can pick arbitrarily many disjoint positive measure subsets $A_j$, $j=1,\ldots ,N$, and consider measures of the form $$ d\mu = \left( \sum_j c_j \chi_{A_j}\right)\, dx . $$ The conditions we're trying to satisfy lead to a homogeneous linear system on the $c_j$. We have $N$ variables and $n+1$ equations. A homogeneous linear system with more variables than equations always has a non-trivial solution.