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.



Measurable sets of $\mathbb R^n$ forming unique absolutely continuous convex combinations?
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.


