Two Notions of Volume
Jordan content and Lebesgue measure both assign volume to subsets of \(\mathbb{R}^n\), but they are defined using different kinds of approximation. Jordan content uses finite collections of boxes; Lebesgue outer measure allows countable box covers. For a bounded set whose boundary is sufficiently small, these different definitions yield the same number. The boundary condition is essential: a bounded set can have positive Jordan outer content and zero Lebesgue measure.
Throughout, \(m\) is Lebesgue measure on \((\mathbb{R}^n,\mathcal{L})\). This measure space is complete and sigma-finite, but not finite. Every bounded Lebesgue-measurable set has finite measure. A box is a product of bounded intervals, and its volume is the product of their side lengths; the choice of open, closed, or half-open endpoints does not change its volume. For a finite union of boxes, its Jordan content is the volume of the union, counting overlaps only once. The earlier result on finite unions of boxes identifies this content with their Lebesgue measure.
Degenerate boxes, with at least one side length zero, have volume zero. They do not affect either definition. The Jordan measurability criterion established earlier says that a bounded set is Jordan measurable exactly when its boundary has Jordan content zero. We will use that criterion to verify measurability, then compare the two volume definitions.
The Comparison Theorem
Proof. Since \(A\) is Jordan measurable, the boundary criterion gives \(J^*(\partial A)=0\). For every \(\varepsilon>0\), there is a finite union of boxes covering \(\partial A\) whose total content is less than \(\varepsilon\). The same boxes are an allowable countable box cover in the definition of Lebesgue outer measure. Therefore \(m^*(\partial A)<\varepsilon\) for every \(\varepsilon>0\), so \(m^*(\partial A)=0\). Thus the boundary is Lebesgue measurable and null.
The interior \(\operatorname{int}(A)\) is open, hence Borel and Lebesgue measurable. Also \(A\setminus\operatorname{int}(A)\subseteq\partial A\), so \(A\) is the union of a measurable set and a subset of a null set. Completeness of Lebesgue measure implies \(A\in\mathcal{L}\).
Now take any finite union \(K\) of boxes contained in \(A\). It is Lebesgue measurable, and monotonicity of measure gives \(J(K)=m(K)\le m(A)\). Taking the supremum over such \(K\) yields \(J_*(A)\le m(A)\). Conversely, for every finite union \(F\) of boxes covering \(A\), monotonicity gives \(m(A)\le m(F)=J(F)\). Taking the infimum over such \(F\) yields \(m(A)\le J^*(A)\). Since \(J_*(A)=J^*(A)=J(A)\), these inequalities imply \(m(A)=J(A)\). \(\square\)
The proof separates two issues. The boundary criterion supplies Lebesgue measurability; finite box approximations then sandwich the Lebesgue measure between the Jordan inner and outer contents. No assertion that finite covers suffice for arbitrary bounded sets is needed.
Worked Comparisons
Worked Example: A Rectangle with Any Endpoint Convention
Let \(R\subset\mathbb{R}^2\) be any rectangle with horizontal side lengths \(3\) and \(4\), allowing any choices of open, closed, or half-open edges. Its volume is \(3\cdot4=12\). Its boundary is a finite union of line segments, which has Jordan content zero: each segment can be covered by finitely many rectangles of arbitrarily small total area. Thus \(R\) is Jordan measurable, and the theorem gives \(m(R)=J(R)\).
For example, a cover by \(R\) itself shows \(J^*(R)\le12\). Any finite box cover of \(R\) has total content at least \(m(R)=12\), by subadditivity of Lebesgue measure and the fact that each covering box has Lebesgue measure equal to its volume. Hence \(J^*(R)=12\). The rectangle itself is a finite union of boxes contained in \(R\), up to its endpoint convention, and its content is \(12\); consequently \(J_*(R)=12\) as well. Check: both contents and the Lebesgue measure are \(3\cdot4=12\).
Worked Example: The Disk of Radius Two
Let \(D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\le4\}\). Its boundary is a circle. The standard Jordan area calculation for a disk gives area \(\pi r^2\); it can be seen by squeezing the disk between regular inscribed and circumscribed \(n\)-gons. For a disk of radius \(r\), their areas are, respectively, \(\frac{n r^2}{2}\sin(2\pi/n)\) and \(n r^2\tan(\pi/n)\). As \(n\to\infty\), the limits are both \(\pi r^2\), using \(\sin t/t\to1\) and \(\tan t/t\to1\). Polygon area here is its Jordan content, computed by decomposing the polygon into triangles.
For \(r=2\), the inscribed and circumscribed polygon areas are \(2n\sin(2\pi/n)\) and \(4n\tan(\pi/n)\), and both tend to \(4\pi\). For instance, at \(n=4\) these values are \(2\cdot4\sin(\pi/2)=8\) and \(4\cdot4\tan(\pi/4)=16\). The disk is Jordan measurable: its circle boundary has Jordan content zero, as is verified in the next example. Therefore \(J(D)=m(D)=4\pi\). Check: substituting the radius into the disk area formula gives \(\pi\cdot2^2=4\pi\), agreeing with the polygon limits.
Worked Example: Covering a Circle Boundary by Small Squares
Let \(C=\{(x,y):x^2+y^2=1\}\). Fix an integer \(N\ge2\), set \(\delta=1/N\), and take the finitely many squares of a grid of side \(\delta\) that meet \(C\). They cover \(C\). Any point of one of these squares is at distance at most \(\sqrt{2}\delta\) from a point of \(C\), because the square's diagonal has length \(\sqrt{2}\delta\). Hence their union lies in the annulus with radii \(1-\sqrt{2}\delta\) and \(1+\sqrt{2}\delta\); the inner radius is positive because \(N\ge2\).
The grid squares have disjoint interiors, so the sum of their areas equals the area of their union; shared edges have area zero. The annulus has area \(\pi(1+\sqrt{2}\delta)^2-\pi(1-\sqrt{2}\delta)^2=4\sqrt{2}\pi\delta\). Thus the total area of the covering squares is at most \(4\sqrt{2}\pi/N\), which tends to zero as \(N\to\infty\). It follows that \(J^*(C)=0\), and therefore \(C\) has Jordan content zero. The same covers show \(m^*(C)=0\), so \(m(C)=0\) as well. Check: the annulus-area difference is \(\pi[(1+u)^2-(1-u)^2]=4\pi u\) with \(u=\sqrt{2}/N\), giving exactly \(4\sqrt{2}\pi/N\).
Why Boundedness and Jordan Measurability Matter
The comparison theorem is not a statement that Jordan outer content equals Lebesgue measure for every bounded set. Jordan outer content restricts covers to finitely many boxes, so it can be larger than Lebesgue outer measure. A dense subset provides a direct example, and also illustrates why endpoint conventions in finite covers must be handled carefully.
Worked Example: The Rationals in the Unit Interval
Let \(A=\mathbb{Q}\cap[0,1]\subset\mathbb{R}\). It is countable, so it has Lebesgue measure zero: enumerate its points as \(q_1,q_2,\ldots\), and for any \(\varepsilon>0\) cover \(q_j\) by an interval of length \(\varepsilon/2^j\). The total length is \(\sum_{j=1}^{\infty}\varepsilon/2^j=\varepsilon\), so \(m^*(A)=0\).
In contrast, any finite union of eligible intervals covering \(A\) has total length at least \(1\). Indeed, take the closures of the finitely many intervals. Their union is a finite union of closed intervals containing the closure of \(A\), which is \([0,1]\) because \(A\) is dense there. Taking closures does not change the lengths, and the sum of the lengths of intervals covering \([0,1]\) is at least \(1\). The interval \([0,1]\) itself covers \(A\) with length \(1\). Consequently \(J^*(A)=1\), regardless of whether the original covering intervals were open, closed, or half-open.
No interval of positive length lies in \(A\), since every such interval contains irrational numbers. Thus any finite union of boxes contained in \(A\) has content zero, so \(J_*(A)=0\). In particular, \(A\) is not Jordan measurable, and the comparison theorem does not apply. Check: the two values are \(m(A)=0\) and \(J^*(A)=1\), while \(J_*(A)=0\); the unequal Jordan inner and outer contents record the failure of Jordan measurability.
A common mistake is to prove a lower bound for \(J^*(A)\) only for closed-interval covers. That misses eligible covers using other endpoint choices. The correction is to take closures of all intervals in an arbitrary finite cover: their lengths remain unchanged, and their closed union still contains the closure of the covered dense set. This supplies the required lower bound for every allowed cover.
For bounded Jordan-measurable sets, the boundary-zero criterion ensures that finite box approximations are sufficient to recover the Lebesgue measure. For arbitrary bounded sets, that conclusion cannot be assumed. The distinction is useful whenever a classical geometric volume is compared with Lebesgue measure: first verify Jordan measurability, then the two quantities agree.
Check Your Understanding
Use the definitions of Jordan inner and outer content, and identify where the comparison theorem applies.
- For a rectangle in \(\mathbb{R}^3\) with side lengths \(2\), \(3\), and \(5\), compute its Jordan content and Lebesgue measure.
- Why does the proof of the comparison theorem establish that a bounded Jordan-measurable set is Lebesgue measurable?
- For the disk of radius \(r>0\), write the inner and outer regular-polygon area formulas and state their common limit.
- Explain why taking closures in a finite interval cover does not change the total lengths, and why this matters for \(\mathbb{Q}\cap[0,1]\).
- Which condition fails for \(\mathbb{Q}\cap[0,1]\): boundedness, Lebesgue measurability, or Jordan measurability?