2
$\begingroup$

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.

$\endgroup$

0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.