Starting a Brownian motion at every point of $\mathbb Z^2$; expected number of points in a given set $A$ is $m(A)$?

Question

Suppose we start a Brownian motion at every point of $\mathbb Z^2$. Intuitively, as these points move, they should "spread out evenly", so the expected number of points in a given set $A$ should be $m(A)$? (where $m$ is Lebesgue measure on $\mathbb R^2$, and $A$ is some nice enough set). However, I'm not quite sure how one would go about proving this (dealing with infinitely many Brownian motions seems quite hard).

Remark: this question is motivated by an attempt to formalize Christian Blatter's proof of Pick's theorem https://people.math.ethz.ch/~blatter/Pick.pdf rigorously using Brownian motions. Using his symmetry arguments I think it should be easy to see that the total expected flux across any line segment (i.e. expected number of points moving from one side of line to other, minus expected number of points moving in the opposite direction) is $0$; and so for any time $t>0$ the expected number of points in a polygon $A$ should be constant. Then at sufficiently small time scales when the Brownian motions are still in an $\epsilon$-ball of their original starting point, so if a polygon $A$ has a vertex at point $z$ with interior angle $\theta$ (in radians), rotational symmetry/invariance of Brownian motion starting at $z$ tells us the probability the Brownian motion $B^z$ at $z$ is in $A$ (at that small time scale) is exactly $\frac{\theta}{2\pi}$. This gives us that $I-\frac B2 + 1$ part of Pick's formula, and my question should hopefully give the $m(A)$ part of Pick's formula.

Answer 1

The expected number of points in $A$ converges to $m(A)$ as time goes to infinity, but for most sts $A$ it is not equal to $m(A)$ at a finite time. To see the first statement, one can use the fact that the distribution of Brownian motion modulo 1 (in each coordinate) converges to the uniform measure on the unit square. For the second statement, consider $A$ to be a small disk centered at a lattice point.

Answer 2

We consider a family of independent Brownian motions $(B^\lambda)_{\lambda \in \mathbb{Z}^2}$, the Brownian motion $B^\lambda$ starting at $\lambda$. Given a Borel set $A$ with finite Lebesgue measure, the number of those Brownian motions which are in $A$ at time $t$ is $$N_t(A) = \sum_\lambda \mathbb{1}_{[B^\lambda_t \in A]}.$$ Using twice Fubini theorem for positive functions, we see that the expected number is $$\mathbb{E}[N_t(A)] = \sum_\lambda \mathbb{P}[B^\lambda_t \in A] = \int_A \sum_\lambda g_t(x-\lambda) \mathrm{d}x,$$ where $g_t$ is the density of $\mathcal{N}(0,tI_2)$. By Poisson summation formula, one checks that for every $y \in \mathbb{R}^2$, $$\sum_{\lambda \in \mathbb{Z}^2} g_t(x-\lambda) = \sum_{k \in \mathbb{Z}^2}e^{-2\pi^2k^2t}e^{i2\pi\langle k,x \rangle}.$$ One checks that as $t \to +\infty$, those sums converges uniformly (with regard to $x$) to $1$. Therefore $\mathbb{E}[N_t(A)] \to \lambda(A)$ as $t \to +\infty$.