Let $X_1, \ldots, X_{d+1}$ be $d + 1$ i.i.d. random points in $\mathbb{R}^d$ sampled from a continuous probability $\mu$ of density $f$.
Let $x_0 \in \mathbb{R}^d$. Is it true that almost surely with respect to $\mu$, $$\mathbb{P}(x_0 \in \mathrm{Conv}(X_1, \ldots, X_{d+1})) > 0$$ where $\mathrm{Conv}$ denotes the convex hull of the points?
This assertion is part of an answer to a mathoverflow question that was given to me in comment. I suspect that am not familiar enough with the regularity properties of approximate continuity in order to conclude.
My current reasoning is as follows: if we suppose that $f$ is almost surely positive in a local neighbourhood around $x_0$, then we can find neighbourhoods of the $d+1$ vertices of a simplex having $x_0$ in its interior in the local neighbourhood of $x_0$, and the $X_i$ will be sampled in these neighbourhoods with non-zero probability. However, I am not sure if with probability 1, we can find neighbourhoods of $f$ that are almost surely positive.
@fedja's comment in the link essentially provides the answer. Let me try to add a couple of words to clarify the idea:
Choose $(d+1)$ balls $\mathsf{B}_1, \dots, \mathsf{B}_{d+1}$ in $\mathbb{R}^d$ such that $0 \in \operatorname{Conv}(x_1, \dots, x_{d+1})$ for any choices of $x_i \in \mathsf{B}_i$ for $i = 1, \dots, d+1$. For instance, fix a regular $d$-simplex centered at $0$ and replace each vertex with a small ball.
Now let $x_0$ be a Lebesgue point of $f$ such that $f(x_0) > 0$. Then it is straightforward to prove that
$$ \lim_{r \to 0^+} \frac{1}{r^d|\mathsf{B}_i|} \int_{x_0 + r \mathsf{B}_i} f(x) \, \mathrm{d}x = f(x_0) > 0 $$
for each $i = 1, \dots, d+1$. So if $r > 0$ is sufficiently small, then
$$ \mathbb{P}(x_0 \in \operatorname{Conv}(X_1,\dots,X_{d+1})) \geq \mathbb{P}\left( \cap_{i=1}^{d+1} \{ X_i \in x_0 + r\mathsf{B}_i\} \right) > 0. $$