Finite Grid Refinements and Additivity

Why a Common Grid Helps

A rectangle may be split at finitely many coordinate values in many different ways. To compare two such subdivisions, or to verify the volume of a finite decomposition, it is useful to put all the relevant cuts into one grid. The key calculation is finite: the volumes of the grid cells inside any one half-open rectangle sum to the product of its side lengths.

We work in \(\mathbb{R}^n\) with its usual coordinates. This tutorial uses the coordinate volume of half-open rectangles, not a claim about countable additivity or the construction of Lebesgue measure. A half-open rectangle is a product \(R=\prod_{i=1}^n[a_i,b_i)\), where \(a_i\le b_i\) are finite real numbers. Its coordinate volume is \(v(R)=\prod_{i=1}^n(b_i-a_i)\). If any \(a_i=b_i\), then \(R\) is empty and \(v(R)=0\). All decompositions below are exact set equalities, and disjoint means pairwise disjoint as sets.

Definition: For a rectangle \(R=\prod_{i=1}^n[a_i,b_i)\), a finite grid refinement chooses, for each coordinate \(i\), finitely many cuts \(a_i=t_{i,0}<t_{i,1}<\cdots<t_{i,r_i}=b_i\). Its cells are the rectangles \(\prod_{i=1}^n[t_{i,k_i-1},t_{i,k_i})\), with \(1\le k_i\le r_i\). These cells are half-open, pairwise disjoint, and their union is \(R\).

For a nonempty rectangle, each list of cuts is finite and strictly increasing. The half-open convention assigns a cut point to the interval on its right, so neighboring cells do not overlap and no points on a cut are omitted. A finite grid is different from an arbitrary cover: it partitions the rectangle exactly into finitely many cells.

The Volume Identity for a Refinement

Theorem: Let \(R=\prod_{i=1}^n[a_i,b_i)\) be a half-open rectangle in \(\mathbb{R}^n\), and refine it by finite coordinate cuts as defined above. Then the sum of the coordinate volumes of all its cells equals \(v(R)\). The statement also holds when \(R\) is empty, with no cells and an empty sum of zero.

Proof. First suppose \(a_i<b_i\) for every coordinate \(i\). The volume of the cell indexed by \((k_1,\ldots,k_n)\) is \(\prod_{i=1}^n(t_{i,k_i}-t_{i,k_i-1})\). Summing over all choices of indices and applying finite distributivity gives

\sum_{k_1=1}^{r_1}\cdots\sum_{k_n=1}^{r_n} \prod_{i=1}^n(t_{i,k_i}-t_{i,k_i-1}) = \prod_{i=1}^n\left(\sum_{k_i=1}^{r_i}(t_{i,k_i}-t_{i,k_i-1})\right).

For each \(i\), the sum in parentheses telescopes: every interior cut appears once with a positive sign and once with a negative sign, leaving \(t_{i,r_i}-t_{i,0}=b_i-a_i\). Thus the displayed product equals \(\prod_{i=1}^n(b_i-a_i)=v(R)\). If \(R\) is empty because some \(a_i=b_i\), its coordinate volume is zero and it has no nonempty grid cells, so the empty sum is zero as required. \(\square\)

This calculation uses only finite sums and the distributive law. It does not require a measure space or an interchange of an infinite sum and an integral.

Worked Example: Refining a Rectangle in the Plane

Take \(R=[-1,3)\times[0,2)\), with cuts at \(0\) and \(2\) in the first coordinate and at \(1\) in the second. The first-coordinate interval lengths are \(0-(-1)=1\), \(2-0=2\), and \(3-2=1\). The second-coordinate interval lengths are \(1-0=1\) and \(2-1=1\).

The six cell areas, listed by row from bottom to top, are \(1\cdot1=1\), \(2\cdot1=2\), \(1\cdot1=1\), followed by \(1\cdot1=1\), \(2\cdot1=2\), \(1\cdot1=1\). Their sum is \(1+2+1+1+2+1=8\). The original coordinate volume is \((3-(-1))(2-0)=4\cdot2=8\), so the refinement preserves volume. Check: each row has area \(1+2+1=4\), and the two rows give \(4+4=8\), matching the original product.

Putting Several Rectangles on One Grid

For finitely many rectangles, collect all their endpoint coordinates in each coordinate direction and sort the distinct values. The resulting coordinate intervals form a common grid over the bounding rectangle. Each original rectangle is a union of cells of this grid: its endpoints are grid cuts, so a cell cannot cross from inside it to outside it. If a rectangle is empty, it contributes no cells and has volume zero.

When rectangles overlap, a common grid still preserves the sum of their volumes, provided a cell is counted once for each rectangle containing it. This is a statement about the original sum, which counts multiplicity. It is not generally the volume of the union, where each point is counted only once.

Worked Example: A Common Grid for Two Overlapping Rectangles

In \(\mathbb{R}^2\), let \(A=[0,2)\times[0,1)\) and \(B=[1,3)\times[0,2)\). Their coordinate volumes are \(v(A)=(2-0)(1-0)=2\) and \(v(B)=(3-1)(2-0)=4\). Use the common cuts \(0,1,2,3\) horizontally and \(0,1,2\) vertically.

Rectangle \(A\) consists of the bottom cells over \([0,1)\) and \([1,2)\), each of area \((1-0)(1-0)=1\). Their total is \(1+1=2\). Rectangle \(B\) consists of four cells: two horizontal intervals, \([1,2)\) and \([2,3)\), in each of the two vertical intervals, \([0,1)\) and \([1,2)\). Each cell has area \((2-1)(1-0)=1\) or \((3-2)(1-0)=1\), so their total is \(1+1+1+1=4\).

The sum over the two rectangles is therefore \(2+4=6\). The cell \([1,2)\times[0,1)\) belongs to both rectangles and is counted twice in this sum. If one instead counts every cell in the union just once, the cells are three at the bottom and two at the top, giving \(3\cdot1+2\cdot1=5\). Check: \(6\) is the sum of the two rectangle volumes, while \(5\) counts the grid cells in their union once; the difference \(1\) is exactly the area of the shared cell.

Finite Additivity for Rectangle Decompositions

Theorem: Let \(R\) and \(Q_1,\ldots,Q_N\) be half-open rectangles in \(\mathbb{R}^n\), where \(N\) is finite. If \(R=\bigcup_{j=1}^N Q_j\) and the \(Q_j\) are pairwise disjoint, then \(v(R)=\sum_{j=1}^N v(Q_j)\). Empty rectangles are allowed.

Proof. If \(R\) is empty, every \(Q_j\) is a subset of the empty set and is therefore empty; both sides equal zero. Now suppose \(R\) is nonempty. Because the union is \(R\), each \(Q_j\) is contained in \(R\). In each coordinate, collect the endpoints of \(R\) and of all the \(Q_j\), discard values outside the corresponding interval of \(R\), and sort the remaining distinct values. These are finite coordinate cuts for a common grid of \(R\).

By the refinement theorem, \(v(R)\) is the sum of the volumes of all grid cells in \(R\). Each \(Q_j\) is a union of cells from this grid: the endpoints of \(Q_j\) are among the cuts, so any grid cell meeting \(Q_j\) is contained in it. Conversely, every cell of \(R\) is contained in some \(Q_j\), because the \(Q_j\) cover \(R\). It cannot be contained in two different \(Q_j\), since that would make the nonempty cell a subset of their intersection, contradicting pairwise disjointness. Thus the cells of \(R\) are divided into disjoint groups, one group for each \(Q_j\), including an empty group when \(Q_j\) is empty.

Apply the refinement identity to each \(Q_j\): its volume is the sum of the volumes of its cells. Summing these finitely many equalities counts each cell of \(R\) exactly once, by the grouping just proved. The resulting sum is the sum of all cell volumes in \(R\), which equals \(v(R)\). Hence \(v(R)=\sum_{j=1}^N v(Q_j)\). \(\square\)

Worked Example: Checking a Three-Rectangle Partition

Consider \(R=[0,3)\times[0,2)\), partitioned into \(Q_1=[0,1)\times[0,2)\), \(Q_2=[1,3)\times[0,1)\), and \(Q_3=[1,3)\times[1,2)\). These rectangles are pairwise disjoint: \(Q_2\) and \(Q_3\) meet only at the cut \(y=1\), which belongs to \(Q_3\) but not \(Q_2\); the other pairs have disjoint first-coordinate intervals.

The common grid has horizontal intervals \([0,1)\), \([1,3)\) and vertical intervals \([0,1)\), \([1,2)\). Its four cell areas are \((1-0)(1-0)=1\), \((1-0)(2-1)=1\), \((3-1)(1-0)=2\), and \((3-1)(2-1)=2\). The first two cells make up \(Q_1\), and the last two are \(Q_2\) and \(Q_3\). Thus \(v(Q_1)=2\), \(v(Q_2)=2\), and \(v(Q_3)=2\); their sum is \(2+2+2=6\). The original volume is \(v(R)=(3-0)(2-0)=3\cdot2=6\). Check: the four grid-cell areas sum to \(1+1+2+2=6\), independently confirming the partition calculation.

A Common Mistake and the Finite Scope

Pitfall: treating a union as if it were a disjoint decomposition. For the overlapping rectangles \(A\) and \(B\) in the earlier example, the calculation \(v(A)+v(B)=2+4=6\) is correct. The mistaken conclusion that the union has volume \(6\) counts the shared cell \([1,2)\times[0,1)\) twice. On the common grid, that cell has area \((2-1)(1-0)=1\); counting it just once gives \(6-1=5\), agreeing with the five distinct unit cells in the union. Finite additivity applies to pairwise disjoint rectangles, not to overlapping ones.

The common-grid argument is useful because it reduces a geometric decomposition to a finite bookkeeping problem: establish which cells belong to which rectangles, then use the product identity for cell volumes. Its conclusion is deliberately finite. It neither establishes countable additivity nor proves a statement about arbitrary covers; those require additional definitions and arguments.

Check Your Understanding

In each question, distinguish the sum over rectangles from counting the cells of a union only once.

Refine \([0,4)\times[0,3)\) using cuts \(1,3\) in the first coordinate and \(1,2\) in the second. List the cell areas and verify their sum.
Explain why finite distributivity gives the product of side lengths when the cell volumes are summed over a grid in \(\mathbb{R}^n\).
For \(A=[0,2)\times[0,1)\) and \(B=[1,3)\times[0,2)\), identify the shared grid cell and its area. Why does it count twice in \(v(A)+v(B)\)?
State the hypotheses needed to conclude \(v(R)=\sum_{j=1}^N v(Q_j)\) from a finite rectangle decomposition.
What does the half-open convention ensure at a coordinate cut when two neighboring grid intervals are formed?