In Nelson's "Internal Set Theory: A New Approach to Nonstandard Analysis", he defines a predicate holding nearly everywhere if for all standard $\epsilon > 0$, there exists an event $N$ on which the predicate holds whose complement has probability less than $\epsilon$.
Is it the case that we can upgrade this to a single event $N$ whose complement has infinitesimal probability? If not, are there any concrete counterexamples?
In general, we cannot find a single such event $N$. Basically every infinite procedure that you'd expect to stop in finite time gives rise to a counterexample by truncating it to a nonstandard number of steps. Let's see how this works.
Take your favorite nonstandard natural $\omega \in \mathbb{N}$. Consider the following procedure:
So our finite sample space $\Omega$ consists of all the natural numbers up to and including $\omega$.
Now, consider the predicate $\mathrm{st}(c)$. Intuitively, we'd expect this predicate to hold almost surely. And indeed, it satisfies the definition of nearly everywhere / almost surely given above. If you take a standard $\varepsilon > 0$, then you can find some standard $n$ so that with probability larger than $\varepsilon$, you stop before flipping $n$ times. Then, the predicate $\mathrm{st}(c)$ holds for all $c$ in the event $N = \{x < \omega \mid x < n\}$, and the complement of $N$ has its probability bounded above by $\varepsilon$.
Nonetheless, you cannot construct any single event $N$ under the procedure above which would satisfy the following two conditions:
Why? Well, notice that $N \subseteq \Omega$ is a finite set of natural numbers, so it has a largest element. If it satisfies Condition 1, then all of its elements are standard. In particular, the largest element of $N$ is standard. But the probability that $c$ exceeds the largest element of $N$ is then standard and non-zero. So the complement of $N$ cannot have infinitesimal probability.