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

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.

• What do you mean by "uniqueness of convex decomposition?" Is it the uniqueness of representations of points of the convex hull of $A$ as convex combinations of points from $A$? Commented May 18, 2022 at 14:56
• Yes, but that was mainly an abstract motivation. Commented May 18, 2022 at 16:08

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.
• Nice answer, thank you. I just figured the same thing but using polynomials, this is cleaner. This result also holds more generally for $\lambda$ non-atomic actually, since we can partition it this way. Commented May 18, 2022 at 15:59