For each $r \in \mathbb N $ write $\mathbb Z/ 10^r = \{a/10^r: a \in \mathbb Z\}$ and $P(r)$ for the lattice $(\mathbb Z/10^r)^N \subset \mathbb R^N$.
Suppose the plane $P \subset \mathbb R^N$ is well-behaved with respect to the lattices $P(r)$. That means each $P \cap P(r)$ has more than one element (it follows it is a nontrivial sublattice of $P$).
Now suppose the polytope $B \subset P$ has nonempty interior when considered as a subset of $P$. By choosing an orthonormal basis for $P$ we can define the Lebesgue-measure on $B$ normalised to have $\mu(B)=1$. Note this is independent of the choice of basis since the transition matrices are measure-preserving.
But there is another way to measure subsets of $B$ using the lattices $P(r)$. For each $E \subset B$ we define
$\nu_r(E) = |P(r) \cap E|/ |P(r) \cap B|$ and $\nu(E) = \lim \nu_r(E)$
Is there any reason to believe these two measures coincide? In general I'm sure they don't since there is no reason for $B$ to intersect any $P(r)$ nontrivially. For example let $P$ be a line of irrational slope.
If it helps we can also assume $B$ is given by linear constraints with positive integer coefficients.