1
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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 $A$? $\endgroup$ Commented May 18, 2022 at 14:56
  • $\begingroup$ Yes, but that was mainly an abstract motivation. $\endgroup$
    – Cryme
    Commented May 18, 2022 at 16:08

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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
    Commented May 18, 2022 at 15:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.