Are there any "nonrestrictive", verifiable conditions for the zeros of a nontrivial function to have measure zero? I know from this thread that this holds true for analytic functions, which is nice since analytic functions are closed under sums, products, and compositions.
I'm wondering if this property can be established for a more general class of functions, e.g., continuous differentiability plus additional nonrestrictive, verifiable assumptions under which this holds. One sufficient condition is that the function is continuously differentiable and its set of stationary/critical points has measure zero; however, this is not easily verifiable in the sense above (but there could be nonrestrictive conditions which imply this, which are of interest to me).
I apologize if this question comes across as too broad. The motivation behind this question is to establish sufficient conditions on a function $f : \mathbb{R}^n \times \mathbb{R}^d \to \mathbb{R}^m$ such that the probability function $p(x) := \mathbb{P}(f(x,\xi) \leq 0)$ is continuous - a sufficient condition for this is $f$ is continuous and $\mathbb{P}(f(x,\xi) = 0) = 0$ for each $x$.
Note: by "nonrestrictive" conditions, I wish to exclude trivial conditions such as monotonicity and conditions which essentially boil down to the function being analytic.
