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Jordan Content of Finite Unions of Boxes

Learn to compute the Jordan content of a finite union of boxes by refining their coordinates into a grid and summing the volumes of occupied cells.

Advanced 11 min read

What You'll Learn

  • Define Jordan content for a bounded finite union of axis-parallel boxes using a finite coordinate partition.
  • Prove that refining the partition does not change the computed content.
  • Compute unions of boxes with overlaps by counting each occupied grid cell once.
  • Handle shared faces and lower-dimensional intersections as zero-content boundaries.
  • Apply finite inclusion-exclusion to box unions and check it against a grid calculation.
  • Identify why simply adding the volumes of overlapping boxes gives the wrong answer.

From Box Volumes to Jordan Content

The volume of one axis-parallel box is the product of its side lengths. For a finite union, however, adding the individual volumes can count the overlap more than once. A finite coordinate partition resolves this: cut along every box endpoint, inspect the resulting cells, and count each cell in the union exactly once.

Work in \(\mathbb{R}^n\), with \(n\ge1\). An axis-parallel box is a product of bounded intervals, with any choice of endpoint inclusion; its volume is the product of the interval lengths. A side of length zero gives volume zero. A bounded finite union of such boxes will be called a finite box union. The value defined below is its Jordan content, not a claim about arbitrary bounded sets.

Definition: Let \(E\subset\mathbb{R}^n\) be a bounded finite box union. For each coordinate, collect the endpoints of all the intervals defining the boxes, together with the endpoints of a bounded enclosing box. Sort these finitely many numbers. The resulting coordinate cuts form a finite grid. Each open grid cell is a product \(\prod_{i=1}^n(a_i,b_i)\). Its volume is \(\prod_{i=1}^n(b_i-a_i)\). Define \(J(E)\) to be the sum of the volumes of the positive-volume grid cells whose interiors lie in \(E\). Grid boundary points are not counted as cells.

Every open grid cell is either entirely in \(E\) or entirely outside \(E\): none of the box endpoints lies inside a grid interval, so membership in each box is constant throughout the cell. Choices about whether to include endpoints can affect membership on grid boundaries, but not this cell classification. The next result shows that the sum does not depend on how finely the grid is cut.

Why the Finite-Partition Definition Is Well-Defined

Proposition: For a bounded finite box union \(E\subset\mathbb{R}^n\), every coordinate grid containing all box endpoints gives the same value \(J(E)\). Inserting additional cuts cannot change the content.

Proof. Consider two grids that both contain the endpoints of the boxes defining \(E\). Form a common refinement by taking, in each coordinate, the union of the two finite lists of cuts. Each positive-volume cell in either original grid is divided into finitely many smaller cells by this refinement. Every smaller cell has the same membership status in \(E\) as its parent cell, because the parent lies between consecutive box endpoints. The sum of the volumes of the smaller cells equals the parent volume: in each coordinate, the lengths of the subintervals add to the original length, and the product of these sums is the sum of the product volumes of the subcells. Thus refinement preserves the sum of occupied-cell volumes. Both original grids therefore give the sum from their common refinement, so their values agree. \(\square\)

Finite unions of coordinate boundary faces have zero Jordan content. To see the relevant covering estimate, enclose the faces in \([-R,R]^n\), where \(R>0\). A face lying in the hyperplane \(x_i=c\) is covered inside this box by a slab of width \(2\delta\), whose volume is \(2\delta(2R)^{n-1}\). For \(N\) such faces, the total volume of these finitely many slabs is at most \(2N\delta(2R)^{n-1}\), which can be made arbitrarily small by choosing \(\delta>0\) sufficiently small. This explains why including or excluding finitely many box faces does not affect the content. It does not assert that every set of Lebesgue measure zero has Jordan content zero.

Theorem (Finite additivity and inclusion-exclusion): If \(E,F\subset\mathbb{R}^n\) are bounded finite box unions, then \(J(E\cup F)=J(E)+J(F)-J(E\cap F)\). More generally, for finitely many bounded finite box unions \(E_1,\ldots,E_r\), the content of their union is the alternating sum of the contents of all nonempty intersections.

Proof. Use a common grid containing the endpoints of all the boxes in question; the proposition permits this choice. For each positive-volume cell \(Q\), let \(e(Q)\) be \(1\) if its interior lies in \(E\), and \(0\) otherwise; define \(f(Q)\) in the same way for \(F\). The cell lies in \(E\cup F\) precisely when at least one of \(e(Q),f(Q)\) is \(1\), and it lies in \(E\cap F\) precisely when both are \(1\). Hence \(e(Q\text{ in }E\cup F)=e(Q)+f(Q)-e(Q)f(Q)\). Multiply this identity by the volume of \(Q\) and sum over all grid cells. By the definition of content, the resulting equality is the stated formula. For \(r\) sets, on each cell let \(k\) be the number of sets containing it. Its contribution to the alternating sum is \(\sum_{j=1}^{k}(-1)^{j+1}\binom{k}{j}=1\) if \(k\ge1\), and is \(0\) if \(k=0\), by the binomial identity \((1-1)^k=0\). Thus the alternating sum counts exactly the cells in the union, proving the general formula. \(\square\)

Worked Partitions and Overlaps

Worked Example: Two Rectangles Sharing a Boundary

In \(\mathbb{R}^2\), let \(A=[0,1]\times[0,2]\) and \(B=[1,3]\times[0,2]\). Find the Jordan content of their union, noting that the rectangles share a vertical edge.

The side lengths of \(A\) are \(1\) and \(2\), so \(J(A)=1\cdot2=2\). The side lengths of \(B\) are \(2\) and \(2\), so \(J(B)=2\cdot2=4\). Their intersection is \(\{1\}\times[0,2]\), a line segment. It has zero content: for any \(\delta>0\), it is covered by the rectangle \([1-\delta,1+\delta]\times[0,2]\), of area \(4\delta\), which can be made arbitrarily small.

The grid has \(x\)-cuts \(0,1,3\) and \(y\)-cuts \(0,2\). The occupied cell volumes are \(1\cdot2=2\) and \(2\cdot2=4\). Thus \(J(A\cup B)=2+4=6\). Check: the union is the rectangle \([0,3]\times[0,2]\), whose area is \(3\cdot2=6\). The shared edge belongs to both closed rectangles, but does not add positive area.

Worked Example: Three Overlapping Rectangles

Let \(A=[0,2]\times[0,1]\), \(B=[1,3]\times[0,2]\), and \(C=[2,4]\times[1,2]\). Compute \(J(A\cup B\cup C)\) both by inclusion-exclusion and by a grid.

The individual areas are \(J(A)=2\cdot1=2\), \(J(B)=2\cdot2=4\), and \(J(C)=2\cdot1=2\). The pairwise intersections are \(A\cap B=[1,2]\times[0,1]\), of area \(1\); \(B\cap C=[2,3]\times[1,2]\), of area \(1\); and \(A\cap C=\{(2,1)\}\), of area \(0\). The triple intersection is also \(\{(2,1)\}\), of area \(0\): this point lies on the boundary of \(A\), belongs to \(B\), and lies on the boundary of \(C\). Therefore inclusion-exclusion gives

$$ J(A\cup B\cup C)=2+4+2-1-1-0+0=6. $$

For an independent grid check, the \(x\)-cuts are \(0,1,2,3,4\), and the \(y\)-cuts are \(0,1,2\). In the strip \(0<x<1\), only the cell \(0<y<1\) is occupied, contributing \(1\cdot1=1\). In \(1<x<2\), both unit-height cells are occupied, contributing \(1\cdot1+1\cdot1=2\). In \(2<x<3\), both unit-height cells are again occupied, contributing \(1+1=2\). In \(3<x<4\), only \(1<y<2\) is occupied, contributing \(1\cdot1=1\). The total is \(1+2+2+1=6\), agreeing with inclusion-exclusion.

Worked Example: Overlapping Boxes in Three Dimensions

In \(\mathbb{R}^3\), let \(Q_1=[0,2]\times[0,1]\times[0,1]\) and \(Q_2=[1,3]\times[0,2]\times[0,1]\). Compute the content of their union by refining the grid.

The volumes are \(J(Q_1)=2\cdot1\cdot1=2\) and \(J(Q_2)=2\cdot2\cdot1=4\). Their intersection is \([1,2]\times[0,1]\times[0,1]\), with volume \(1\cdot1\cdot1=1\). The two-set inclusion-exclusion formula gives \(2+4-1=5\).

The grid has \(x\)-cuts \(0,1,2,3\), \(y\)-cuts \(0,1,2\), and \(z\)-cuts \(0,1\). For \(0<x<1\), just the cell \(0<y<1\) is occupied, with volume \(1\cdot1\cdot1=1\). For \(1<x<2\), both \(y\)-cells are occupied, with total volume \(1\cdot1\cdot1+1\cdot1\cdot1=2\). For \(2<x<3\), both \(y\)-cells are occupied, again with total volume \(2\). The sum is \(1+2+2=5\). Check: the intersection volume subtracted in inclusion-exclusion is exactly the one cell with \(1<x<2\), \(0<y<1\), and \(0<z<1\), whose volume is \(1\).

A Common Counting Error

A frequent mistake is to add the volumes of overlapping boxes without subtracting the overlap. For \(Q_1,Q_2\) in the three-dimensional example, this gives \(2+4=6\), which is incorrect because the intersection of volume \(1\) was counted once in each box. The correction is \(J(Q_1\cup Q_2)=2+4-1=5\), verified directly by the grid. By contrast, boxes sharing only a face have an intersection of zero content, so adding their contents is valid. The deciding issue is whether the overlap contains a positive-volume grid cell, not whether the sets have any points in common.

The partition method matters because it turns union volume into a finite, exact computation. Each occupied cell is counted once, regardless of how many original boxes contain it. For a manageable number of boxes, inclusion-exclusion can be faster; the grid provides a transparent check and handles shared boundaries without special corrections. These conclusions concern bounded finite box unions; criteria for when more general bounded sets have Jordan content require additional ideas.

Check Your Understanding

Use a finite coordinate grid in each problem, and distinguish positive-volume overlaps from boundary-only intersections.

  1. Find the Jordan content of \([0,2]\times[0,1]\ \cup\ [1,3]\times[0,1]\). Verify the answer by both a grid and inclusion-exclusion.
  2. Two boxes in \(\mathbb{R}^3\) share a face but have disjoint interiors. Explain why their union content is the sum of their individual volumes.
  3. For three finite box unions, explain why a point belonging to all three sets contributes with coefficient \(1\) in the alternating inclusion-exclusion sum.
  4. Give a finite grid for \([0,1]\times[0,2]\ \cup\ [0,2]\times[1,2]\), and compute the volume of each occupied cell.
  5. Why does adding the volumes of two boxes that overlap in a positive-volume box generally overcount their union?

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