Finite Box Approximation and Boundaries
Jordan measurability asks whether a bounded set can be approximated in volume by finite unions of boxes. The obstruction is not whether the set itself is large or small: it is whether boxes near its boundary can be made to have arbitrarily small total volume. For a bounded set in \(\mathbb{R}^n\), this obstruction is measured exactly by the Lebesgue measure of the boundary.
Throughout, \(n\ge1\), and \(m\) denotes Lebesgue measure on \(\mathbb{R}^n\). A box means a bounded axis-parallel rectangle; its volume is the product of its side lengths. We use finite unions of boxes, whose volume is well-defined by finite grid refinement and additivity. The boundary \(\partial E\) of a set \(E\) is its closure with its interior removed. A set is bounded if it is contained in some bounded box.
The sandwich \(A\subset E\subset B\) forces \(A\subset B\), and the difference in volumes measures how much room the approximation leaves around \(E\). The theorem below identifies when this difference can be made arbitrarily small.
The Boundary-Zero Criterion
Proof. First suppose \(E\) is Jordan measurable. Fix \(\varepsilon>0\), and choose finite unions of closed boxes \(A\subset E\subset B\) such that \(\operatorname{vol}(B)-\operatorname{vol}(A)<\varepsilon\). If \(x\in\partial E\), then \(x\in\overline E\subset B\), since \(B\) is closed. Also \(x\notin\operatorname{int}A\), because \(\operatorname{int}A\subset E\) would make \(x\) an interior point of \(E\). Consequently, \[ \partial E\subset B\setminus\operatorname{int}A =(B\setminus A)\cup(A\setminus\operatorname{int}A). \] The second set on the right is \(\partial A\). The boundary of a finite union of boxes is contained in the union of the boundaries of those boxes: at a point outside all the box boundaries, membership in each box is constant in a neighborhood, and hence membership in their union is constant there. Each box boundary is a finite union of faces and has Lebesgue measure zero. Indeed, each bounded face can be covered by a slab of arbitrarily small thickness, with volume tending to zero with that thickness. Thus \(m(\partial A)=0\). Finite additivity and the inclusion \(A\subset B\) give \(m(B\setminus A)=\operatorname{vol}(B)-\operatorname{vol}(A)<\varepsilon\). Therefore \(m(\partial E)<\varepsilon\). Since this holds for every \(\varepsilon>0\), \(m(\partial E)=0\).
Conversely, suppose \(E\) is bounded and \(m(\partial E)=0\). Fix \(\varepsilon>0\). By the box-cover definition of Lebesgue outer measure, the null set \(\partial E\) has a countable cover by open boxes whose total volume is less than \(\varepsilon\). Compactness of \(\partial E\) gives a finite subcover \(O\) consisting of open boxes; its total volume is still less than \(\varepsilon\). If \(\partial E=\varnothing\), take \(O=\varnothing\). Since \(\partial E\) is compact and \(O\) is an open neighborhood of it, there is a \(\delta>0\) such that every point at distance less than \(\delta\) from \(\partial E\) belongs to \(O\). If the boundary is empty, any \(\delta>0\) suffices.
Choose a closed box \(Q\) with \(E\subset\operatorname{int}Q\), and partition \(Q\) into finitely many closed grid cells of diameter less than \(\delta\), with disjoint interiors. Call a cell mixed if it meets both \(E\) and \(E^c\). Every mixed cell meets \(\partial E\): choose a point of \(E\) and a point of \(E^c\) in that cell; the line segment joining them lies in the cell and passes from \(E\) to \(E^c\), so it meets the boundary. Every point of a mixed cell is therefore within distance less than \(\delta\) of \(\partial E\), and the whole cell lies in \(O\).
Let \(A\) be the union of the grid cells contained in \(E\), and let \(B\) be the union of the cells meeting \(E\). These are finite unions of closed boxes and satisfy \(A\subset E\subset B\). Any cell contributing to \(B\setminus A\) is mixed; cell boundaries have volume zero, so they do not affect the volume sum. The mixed cells have disjoint interiors and lie in \(O\). By finite additivity and monotonicity of Lebesgue measure, their total volume is at most \(m(O)\), which is at most the sum of the volumes of the finite boxes forming \(O\), hence is less than \(\varepsilon\). Thus \(\operatorname{vol}(B)-\operatorname{vol}(A)<\varepsilon\). This proves that \(E\) is Jordan measurable. \(\square\)
The proof also explains the criterion geometrically. A grid cell that meets both \(E\) and its complement must touch the boundary. If the boundary has measure zero, the cells that can cause uncertainty fit inside a box cover of arbitrarily small total volume. Conversely, if inside and outside approximations are close in volume, the boundary must fit inside their small difference, apart from the zero-volume boundaries of the approximating boxes.
Worked Example: A Closed Box
Let \(E=\prod_{j=1}^n[a_j,b_j]\subset\mathbb{R}^n\), where \(a_j<b_j\) for each \(j\). Its boundary is contained in the union of the \(2n\) faces obtained by fixing one coordinate at \(a_j\) or \(b_j\). Each face has measure zero: for a fixed face, cover it by a slab of thickness \(2t\) in its fixed coordinate, while keeping the other coordinates inside bounded intervals. The slab volume is \(2t\prod_{k\ne j}(b_k-a_k)\), which tends to zero as \(t\downarrow0\). A finite union of null sets is null, so \(m(\partial E)=0\). The theorem gives that \(E\) is Jordan measurable.
In fact, the inside and outside approximations can be exact: take \(A=B=E\). Then \(\operatorname{vol}(B)-\operatorname{vol}(A)=0\), which verifies the defining approximation directly. The criterion and the definition agree in this case.
Worked Example: A Euclidean Ball
Let \(E=\{x\in\mathbb{R}^n:\|x\|\le r\}\) with \(r>0\). Its boundary is the sphere of radius \(r\). Write \(v_n=m(\{x:\|x\|\le1\})\), which is finite because the unit ball is bounded. The scaling property of Lebesgue measure gives \(m(\{x:\|x\|\le s\})=v_ns^n\) for \(s>0\). For \(0<t<r\), the sphere of radius \(r\) is contained in the closed shell between radii \(r-t\) and \(r+t\). The shell’s measure is bounded by \[ v_n\big((r+t)^n-(r-t)^n\big). \] This quantity tends to zero as \(t\downarrow0\): the polynomial difference has limit \(v_n(r^n-r^n)=0\). Hence the sphere has Lebesgue measure zero, and the ball is Jordan measurable by the theorem.
For the degenerate radius \(r=0\), the ball is the singleton \(\{0\}\). It can be covered by a box of side length \(2t\) in each coordinate, of volume \((2t)^n\), which tends to zero as \(t\downarrow0\). Its boundary is itself and is null, so this case also satisfies the criterion.
Worked Example: Rational Points in a Cube
In \(\mathbb{R}\), let \(E=\mathbb{Q}\cap[0,1]\). The rational numbers are dense in \([0,1]\), so \(\overline E=[0,1]\). The set \(E\) has empty interior: every nonempty open interval contains an irrational number and therefore cannot be contained in \(E\). Thus \(\partial E=\overline E\setminus\operatorname{int}E=[0,1]\), and \(m(\partial E)=1\). The boundary-zero criterion shows that \(E\) is not Jordan measurable.
This example also exposes a tempting error. Since \(E\) is countable, one might reason that it has measure zero and conclude that it is Jordan measurable. Even if one uses the valid fact that a countable set has Lebesgue measure zero, this does not establish Jordan measurability: the criterion concerns \(m(\partial E)\), not \(m(E)\). Here the boundary has measure \(1\). For every fine partition of \([0,1]\), each grid interval meets both \(E\) and its complement, so the uncertain cells do not have small total length.
Using the Criterion Carefully
When applying the theorem, first identify the boundary as a set, then determine its Lebesgue measure. A set may be null while its boundary has positive measure, as the rational-point example shows. Conversely, a bounded set can have positive measure and still be Jordan measurable: the criterion requires a null boundary, not a null set.
One common mistake is to infer that a dense boundary is automatically null because it is “thin.” Density says that every neighborhood of a point in the closure meets the boundary; it gives no small-volume estimate. In the rational-point example the boundary is the entire interval, not merely a countable dense subset. Another mistake is to check only that the boundary is contained in a convenient bounded box. Boundedness ensures the boundary is bounded, but does not make its measure zero. The needed test is whether its Lebesgue measure is exactly zero.
The boundary-zero test is useful because it replaces the search for finite inside and outside approximations by a geometric question. For boxes and balls, faces and spheres have zero measure, so the criterion applies. For sets whose boundary fills a region, the criterion fails, even if the set itself is sparse. In each case, the boundary is the precise part of the set that controls the Jordan approximation.
Check Your Understanding
Use the definition and the boundary-zero criterion, stating which set is being measured in each answer.
- For \(E=[0,1]\cup[2,3]\subset\mathbb{R}\), identify \(\partial E\) and explain why \(E\) is Jordan measurable.
- Let \(E=\mathbb{Q}\cap[0,1]\). Why does \(m(E)=0\), if used, not imply that \(E\) is Jordan measurable?
- For a ball of positive radius in \(\mathbb{R}^n\), what shell contains its boundary, and why does the shell volume tend to zero?
- In the converse proof, why must a grid cell meeting both \(E\) and \(E^c\) meet \(\partial E\)?
- Does boundedness alone imply \(m(\partial E)=0\)? Give a bounded counterexample from the tutorial.