Consider the tiling of $\mathbb{R}^n$ by unit cubes centered at integer lattice points, i.e.
$$ \mathbb{R}^n = \bigcup_{a \in \mathbb{Z}^n} Q\left(a, \frac{1}{2}\right). $$
Color each unit cube black or white randomly with equal probability. If we now place a block $R$ of arbitrary size on $\mathbb{R}^n$ and let it expand, will it look gray in the limit, i.e. will the proportion of black squares to white squares be almost surely 1?
Formally, suppose $(X_a)_{a \in \mathbb{Z}^n}$ is an array of i.i.d. Bernoulli random variables indexed by $\mathbb{Z}^n$. Let $R \subset \mathbb{R}^n$ be an arbitrary cube. Denote by $\lambda R$ the set $\{\lambda x \in \mathbb{R}^n : x \in R \}$ and define the average
$$ S_\lambda = \frac{1}{|I_\lambda|} \sum_{a \in I_\lambda} X_a $$
where $I_\lambda = \mathbb{Z}^n \cap \lambda R$. Does $S_\lambda$ converge almost surely to $1/2$ as $\lambda \to \infty$? More generally, I would like to let $(X_a)_{a \in \mathbb{Z}^n}$ be an arbitrary integrable i.i.d. sequence of random variables. Do the averages converge to $\mathbb{E}[X_a]$ almost surely?
The Strong Law of Large Numbers cannot be applied directly, since the index sets $I_\lambda$ are not monotone increasing with respect to set inclusion. Is there a way around this? Thanks!