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I came to the following question, but I don't have quite a good idea how to approach.

Can a set $A\subset \mathbb{R}^n , n\ge 2$ with nonzero measure be in a general linear position?

I believe that, since this is quite a simple question, this would already have an answer, but I could not find it.

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    $\begingroup$ Can you define precisely the concept of general linear position? $\endgroup$
    – Liviu Nicolaescu
    Commented Apr 26, 2016 at 0:00
  • $\begingroup$ I do not follow the definition. How do $m$ and $n$ relate? Say, $m=1$, how is it that: "any $2$ points...is not in a $1$-dimensional hyperplane...: ? What if $m>n$? You could google two-point set, and find related results, e.g. math.unt.edu/~mauldin/papers/no100.pdf or im0.p.lodz.pl/~sglab/szymon/BGRZ.pdf $\endgroup$
    – Mirko
    Commented Apr 26, 2016 at 1:37
  • $\begingroup$ @Mirko Sorry, I mean that for all $1\le m<n$, any $m+2$ points in $A$ is not in a $m$-domensional hyperplane of $\mathbb R ^n $. $\endgroup$
    – Gheehyun Nahm
    Commented Apr 26, 2016 at 2:08

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No, at least if $A$ is assumed to be measurable. Let $\pi \colon \mathbb{R}^n \to \mathbb{R}$ be the projection to the first coordinate. For every $t \in \mathbb{R}$, if $A$ is in general linear position, then $\pi^{-1}(t)$ is finite, and therefore has measure zero in $\mathbb{R}^{n-1}$. So Fubini's Theorem implies that $A$ has measure zero.

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