Euclidean Geometry: 2^(-N) × 2^(-N) Lattice-Based Definition of Area

A figure $F$ is a set of points in the plane. Let's (carefully) define an area function $A$ based on "grid squares". (Essentially, starting from the "most obvious areas possible", we can build up the area of any figure.)

Recommendation: If you ever get stuck, try working out the corresponding definitions/proofs for the one-dimensional number line.

1. For simplicity we just work with bounded $F\subset\mathbb{R}^2$.

2. For integers $n\ge0$, let $S_n$ denote the set of $2^{-n}\times 2^{-n}$ squares formed by the lattice grid $(2^{-n}\cdot\mathbb{Z})^2$. Let's say a figure $F$ completely contains $L_n(F)$ grid squares and intersects $U_n(F)$ squares. Since $F$ is bounded, $0\le L_n(F) \le U_n(F)$ are both finite.

(Intuitively, the eventual area $A(F)$ should lie between $L_n(F)(2^{-n})^2$ and $U_n(F)(2^{-n})^2$, up to scaling.)

1. Because $S_N$ perfectly tiles $S_n$ whenever $0\le n\le N$, we have $L_N(F) \ge (2^{N-n})^2 L_n(F)$ and $U_N(F) \le (2^{N-n})^2 U_n(F)$. Combined with the previous point, we have $$0 \le L_n(F)(2^{-n})^2 \le L_N(F)(2^{-N})^2 \le U_N(F)(2^{-N})^2 \le U_n(F)(2^{-n})^2 \le U_0(F)(2^{-0})^2<\infty.$$ Thus $\sup L_m(F)(2^{-m})^2 = \lim_{m\to\infty} L_m(F)(2^{-m})^2$ exists (here we use the LUB property); call this number $L(F)$. Similarly define $U(F)$ (using GLB property). Then for all $n\ge0$, we have $$0 \le L_n(F)(2^{-n})^2 \le L(F) \le U(F) \le U_n(F)(2^{-n})^2 \le U_0(F).$$

2. Let $M$ be the set of bounded figures $F\subset\mathbb{R}^2$ with $L(F) = U(F)$ (for nice figures, this is basically saying the boundary is negligible in some sense). For such $F\in M$ define $A(F) = L(F)$.

With this definition in hand, we should now verify some intuitive properties about $M$ and $A$, based on Wikipedia's entry. We'll do so in the following entries.