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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.

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  • $\begingroup$ 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 $A$? $\endgroup$ May 18, 2022 at 14:56
  • $\begingroup$ Yes, but that was mainly an abstract motivation. $\endgroup$
    – Cryme
    May 18, 2022 at 16:08

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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.

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  • $\begingroup$ 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. $\endgroup$
    – Cryme
    May 18, 2022 at 15:59

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