It is a bit surprising to see that this post is still alive... but here we go. The question is:
What happens for $S=V$?
In a nutshell, and as was already explained in the comments, what happens is that nothing guarantees that the sets $[N_V=\infty]$ or $[N_V=0]$ are measurable, hence neither $P(N_V=\infty)$ nor $P(N_V=0)$ is defined. Thus, asking if these probabilities are $0$ or $\lt1$ or whatever has no sense.
Let us recall why (the function) $N_S$ is measurable when (the subset) $S$ is measurable. One writes $ N_S=\sum_{i=1}^{+\infty}\mathbf 1_{A^S_i},\qquad A^S_i=[X_i\in S], $ and, by the definition of the measurability of $X_i$ and $S$, each set $A^S_i$ is measurable hence each function $\mathbf 1_{A^S_i}$ is measurable and, by measurability of pointwise limits, the function $N_S$ is measurable.
When $V$ is not measurable, the reasoning above breaks down at the moment when one needs each $A^V_i$ to be measurable. For example, $ [N_V=0]=\bigcap_{i=0}^{+\infty}(\Omega\setminus A_i^V)=\bigcap_{i=0}^{+\infty}[X_i\notin V], $ and none of the subsets $[X_i\notin V]$ is measurable, a priori. If you find a way to prove that these subsets are in fact measurable, or only that their whole intersection is measurable (something which could happen without every $[X_i\notin V]$ being measurable), please go ahead. Otherwise...