Probability zero vs impossible I understand that probability $0$ does not mean 'impossible' - because if we look for instance at a uniform distribution over $[0, 1]$ then while each of the singleton events $\{r\}$ for $0\leq r \leq 1$ has probability $0$, if we carried out the experiment then we would get exactly one of the numbers in $[0,1]$, and so these events are not actually impossible, even though they have probability $0$.
However, if we look at a distribution defined by a density function which is zero on $[0,\frac{1}{2}]$ and non-zero on $(\frac{1}{2},1]$ (let's say with a continuous transition between them), then we know that for any $0\leq r\leq \frac{1}{2}$ the event $\{r\}$ is impossible, and for $\frac{1}{2}<r\leq 1$ the event is possible, yet still has probability $0$.
Both have probability $0$, but one of them is possible, and the other is not. Is there a definition that captures this distinction between the two cases?


However, if we look at a distribution defined by a density function which is zero on $[0,\frac{1}{2}]$ and non-zero on $(\frac{1}{2},1]$ (let's say with a continuous transition between them), then we know that for any $0\leq r\leq \frac{1}{2}$ the event $\{r\}$ is impossible, and for $\frac{1}{2}<r\leq 1$ the event is possible, yet still has probability $0$.

Why do you say that?  Consider the uniform distribution on $[0,1]$.  Let $r$ be any point of $[0,1]$.  Use this density:
$$
f(x) = \begin{cases}
0,\qquad x <0 \text{ or }x=r \text{ or } x>1\\
1,\qquad \text{otherwise.}
\end{cases}
$$
Of course this is still the same distribution: uniform on $[0,1]$.  But now your reasoning says $r$ is impossible.  So your notion of "impossible" depends not on the distribution itself, but in your choice of density function.
