Sorry I never got to learn about Hausdorff measures so this may come as simple. Suppose $$ A = \{x\in [0,1)^d: f(x)=0\} $$ where $f$ is analytic and nontrivial. Then $A$ is an analytic variety of dimension $d-1$. My intuition is that
(1) If we make a uniform subdivision of $[0,1)^d$ using $N^d$ points, then $A$ will contain $\sim N^{d-1}$ points of the subdivision.
(2) I also think we have more precisely $$ \lim_{N\to\infty} \frac{ \# \{n\in \{0,\dots,N-1\}^d: f(\frac{n}{N})=0\}}{N^{d-1}} = \mathcal{H}^{d-1}(A) .$$
Are these facts true? Thanks very much.
Edit: That's not true as stated, as commented by GEdgar. I am interested more specifically in $f(x) = \cos(2\pi x_1) + \cos(2\pi x_2) + \dots+ \cos(2\pi x_d)$.