Exact Outer Measure of Rectangles

The Exact Formula for a Rectangle

A box cover immediately gives an upper bound for the outer measure of a rectangle: cover it by a slightly larger box and let the enlargement shrink. The lower bound is more subtle. Outer measure is an infimum over all countable covers, so it is not enough to calculate the volume of one particular cover. The lower bound comes from compactness: every open cover of a compact inner rectangle has a finite subcover, and finite grid refinements show that the total volume of that finite cover cannot be smaller than the inner rectangle's volume.

Work in \(\mathbb{R}^n\), for a fixed positive integer \(n\). A half-open box is a set \(Q=\prod_{i=1}^n[u_i,v_i)\), where \(u_i<v_i\) are finite, and its volume is \(|Q|=\prod_{i=1}^n(v_i-u_i)\). The outer measure \(m^*(A)\) of \(A\subseteq\mathbb{R}^n\) is the infimum of \(\sum_{j=1}^{\infty}|Q_j|\) over countable half-open box covers \(A\subseteq\bigcup_{j=1}^{\infty}Q_j\). This is an outer measure, not yet a measure on a sigma-algebra.

The Finite-Cover Inequality

We first prove the finite geometric fact used for the lower bound. The boxes in this fact are open: \(U_j=\prod_{i=1}^n(\alpha_{ji},\beta_{ji})\), with finite endpoints and \(\alpha_{ji}<\beta_{ji}\). Their volumes are \(|U_j|=\prod_i(\beta_{ji}-\alpha_{ji})\). The rectangle \(K=\prod_{i=1}^n[c_i,d_i]\) is closed and has positive side lengths.

Proof. In each coordinate \(i\), form a finite grid on \([c_i,d_i]\) using \(c_i,d_i\) and every endpoint \(\alpha_{ji}\) or \(\beta_{ji}\) that lies strictly between them. Sort these distinct points as \(x_{i,0}<x_{i,1}<\cdots<x_{i,r_i}\), with \(x_{i,0}=c_i\) and \(x_{i,r_i}=d_i\). The open grid cells are products \(\prod_i(x_{i,k_i-1},x_{i,k_i})\), one for each choice \(1\le k_i\le r_i\). The sum of their volumes is \(\prod_i(d_i-c_i)\): in each coordinate the subinterval lengths sum to \(d_i-c_i\), and multiplying these finite sums gives the sum of the cell-volume products.

Take the midpoint of any grid cell. It belongs to \(K\), so it lies in at least one covering box \(U_j\). In each coordinate, no endpoint of \(U_j\) lies strictly inside the corresponding grid interval. Since the midpoint lies between \(\alpha_{ji}\) and \(\beta_{ji}\), the entire open grid interval lies between those endpoints. Thus the open grid cell lies in \(U_j\). Assign that cell to one such box. For a fixed \(U_j\), the total volume of cells assigned to it is at most \(|U_j|\): the assigned coordinate intervals lie within its coordinate intervals, and the sum of the volumes of any such grid cells is bounded by the product of the box's side lengths, by finite grid refinement and additivity. Summing over \(j\), the volume of all grid cells is at most \(\sum_{j=1}^N|U_j|\). This is the claimed inequality. \(\square\)

Proving the Lower Bound for Countable Covers

The lemma applies to finite open covers. To use it for the countable half-open covers in the definition of \(m^*\), we enlarge each covering box by an arbitrarily small amount. For \(Q=\prod_i[u_i,v_i)\), let \(U(\eta)=\prod_i(u_i-\eta,v_i+\eta)\), where \(\eta>0\). Then \(Q\subset U(\eta)\), and \(|U(\eta)|=\prod_i((v_i-u_i)+2\eta)\). As \(\eta\) decreases to zero, this volume tends to \(|Q|\), including when a side length is zero.

Proof. Write \(\ell_i=b_i-a_i\). For the upper bound, for every \(\eta>0\), the half-open box \(\prod_i[a_i-\eta,b_i+\eta)\) covers \(R\), whatever the endpoint conventions of its intervals. Therefore \(m^*(R)\le\prod_i(\ell_i+2\eta)\). Letting \(\eta\) decrease to zero gives \(m^*(R)\le\prod_i\ell_i\).

For the lower bound, first suppose every \(\ell_i>0\). Choose \(\delta>0\) small enough that \(2\delta<\ell_i\) for every \(i\), and define the closed inner rectangle \(K_\delta=\prod_i[a_i+\delta,b_i-\delta]\). Each of its points lies in the interior of every corresponding interval \(I_i\), so \(K_\delta\subseteq R\). Its volume is \(\prod_i(\ell_i-2\delta)\).

Consider any countable half-open box cover \(R\subseteq\bigcup_{j=1}^{\infty}Q_j\). If \(\sum_j|Q_j|=\infty\), it is already at least the finite number \(\prod_i(\ell_i-2\delta)\). Otherwise, fix \(\varepsilon>0\). For each \(j\), enlarge \(Q_j\) to an open box \(U_j\supset Q_j\) such that \(|U_j|\le |Q_j|+\varepsilon 2^{-j}\). Such an enlargement exists because the product of the enlarged side lengths tends to \(|Q_j|\) as the common enlargement tends to zero. The open boxes \(U_j\) cover \(K_\delta\). The rectangle \(K_\delta\) is compact by the Heine–Borel theorem, so finitely many of these open boxes cover it. Apply the finite-cover lemma to that finite subcover. Since all volumes are nonnegative, this gives

Let \(\varepsilon\) decrease to zero. The resulting lower bound holds for every countable box cover of \(R\), so taking the infimum over covers gives \(m^*(R)\ge\prod_i(\ell_i-2\delta)\). Now let \(\delta\) decrease to zero; the finite product tends to \(\prod_i\ell_i\). This proves the lower bound when all side lengths are positive.

If some \(\ell_i=0\), then the desired lower bound is just \(m^*(R)\ge0\), which follows from the definition of outer measure. The upper bound already gives \(m^*(R)\le\prod_i\ell_i=0\). Thus equality holds in this case as well, including an empty rectangle. The two bounds prove the formula in all cases. \(\square\)

Worked Applications

A Common Pitfall: One Cover Gives Only an Upper Bound

For \(R=[0,1]\times[0,1]\), the box \(Q=[0,1)\times[0,1)\) is a cover of the rectangle except for its right and top boundary. Enlarging it slightly gives an actual cover of area \((1+2\eta)^2\), which tends to \(1\). This establishes an upper bound. A mistaken argument might say, “The covering box has area \(1\), so the outer measure is at least \(1\).” That reverses the role of an infimum: one cover of area \(1\) shows at most \(1\), not at least \(1\).

The correction is to prove that every countable box cover has total volume at least the volume of each compact inner rectangle. Enlarge the covering boxes by a total added volume less than \(\varepsilon\), take a finite subcover of the compact inner rectangle, and apply the finite grid inequality. Then let \(\varepsilon\) decrease to zero and let the inner rectangle expand to the original rectangle. This establishes the lower bound without assuming the rectangle has already been shown measurable. The exact outer measure formula is the geometric input used next to prove that rectangles are Carathéodory measurable.