Many points on hyperplane with probability zero Let $m$ be a finite measure on $X \subseteq \mathbb{R}^n$, so that $m(\mathbb{R}^n) < \infty$.
Define the hyperplanes on $\mathbb{R}^n$, parametrized by $A \in \mathbb{R}^{n \times n}$ and $b \in \mathbb{R}^n$, as
$$ H(A,b) := \{ x \in \mathbb{R}^n \mid A x = b \}. $$
Take many "extractions" $y_1, y_2, ...$ from $m$. 
Say under what conditions on $m$ no more than $n$ i.i.d. extractions $y_i$'s belong to the same hyperplane almost surely, i.e.
$$ m^{n+1}\left( \left\{ (y_1, y_2, ..., y_{n+1}) \in X^{n+1} \mid \\
\exists (A,b) \in (\mathbb{R}^{n \times n} \times \mathbb{R}^n) \text{ such that } y_1, y_2, ..., y_{n+1} \in H(A,b)  \right\} \right) = 0,$$
where $m^{n+1} := m \times m \times \cdots \times m$ ($n+1$ times) is the product measure.
I was thinking about merely atomless $m$, but I then thought it may not be enough.
 A: Assuming that what you call "extractions" is an i.i.d. sample with distribution $m$, a necessary and sufficient condition is that no hyperplane $H(A,b)$ has positive measure with respect to $m$ (in particular, $m$ being atomless is not enough). 
On the one hand, if $m(H(a,b))\gt0$ for some $(A,b)$ then the $n+1$ values $y_i$ are in $H(A,b)$ with positive probability.
On the other hand, if $m(H(a,b))=0$ for every $(A,b)$, consider the event $C_k$ that the $k$ first values $y_i$, $1\leqslant i\leqslant k$, are affinely independent, that is, that  $(y_i)_{i\leqslant k}$ generates a random affine subspace $H_k$ of dimension $k-1$. 
Let $1\leqslant k\leqslant n$. Conditionally on $C_k$ and $H_k$, $C_{k+1}$ fails if and only if $y_{k+1}$ is in $H_k$. This happens with conditional probability $m(H_k)$. Since the dimension of the affine subspace $H_k$ is at most $n-1$, $H_k\subseteq H(A_k,b_k)$ for some $(A_k,b_k)$, hence $m(H_k)=0$ almost surely on $(y_i)_{i\leqslant k}$, conditionally on $C_k$. Thus, $P[\Omega\setminus C_{k+1}\mid C_k]=0$. Since $C_1=\Omega$, this shows recursively that $P[C_k]=1$ for every $1\leqslant k\leqslant n+1$, in particular $C_{n+1}$ is almost sure.
