Caratheodory Measurability of Rectangles

From Box Covers to Measurable Rectangles

Outer measure assigns a size to every subset of \(\mathbb{R}^n\), including sets that have not yet been shown measurable. The Caratheodory criterion asks whether a set \(E\) splits every test set \(A\) without losing or gaining outer measure: the portions of \(A\) inside and outside \(E\) must have outer measures adding to exactly the outer measure of \(A\). For rectangles, this splitting can be proved directly by refining each covering box into a finite grid. Once rectangles pass the criterion, the sigma-algebra property of Caratheodory measurability will give all Borel sets.

Throughout, \(n\ge1\), and \(m^*\) is the box-cover outer measure on \(\mathbb{R}^n\). A box means a bounded half-open rectangle \(Q=\prod_{i=1}^n[a_i,b_i)\), where \(a_i<b_i\); its volume is \(|Q|=\prod_{i=1}^n(b_i-a_i)\). For every \(A\subseteq\mathbb{R}^n\), define

The empty set has outer measure zero. The definition also gives monotonicity: if \(A\subseteq B\), then \(m^*(A)\le m^*(B)\), since each box cover of \(B\) covers \(A\). We use countable subadditivity, \(m^*(\bigcup_k A_k)\le\sum_km^*(A_k)\). To see why it follows from the definition, the inequality is automatic if the sum on the right is infinite. Otherwise, for any \(\varepsilon>0\), cover each \(A_k\) by boxes whose total volume is less than \(m^*(A_k)+\varepsilon 2^{-k}\); combining all these covers and then letting \(\varepsilon\) decrease to zero proves the inequality. The finite-grid refinement fact established in the earlier material on boxes will also be used: cutting a box at finitely many coordinate values partitions it into finitely many half-open cells, and the sum of their volumes is the volume of the original box. This follows by distributing the product of the side lengths over the sums of the subdivided side lengths.

Why a Half-Open Rectangle Splits Every Set

Fix a bounded half-open rectangle \(R=\prod_{i=1}^n[c_i,d_i)\). Consider any box \(Q=\prod_{i=1}^n[a_i,b_i)\). In coordinate \(i\), cut the interval \([a_i,b_i)\) at each of \(c_i,d_i\) that lies strictly between \(a_i\) and \(b_i\). The resulting finite grid partitions \(Q\). Every cell of this grid is either contained in \(R\) or disjoint from \(R\): in each coordinate, a cell lies wholly on one side of each cut. The cells contained in \(R\) cover \(Q\cap R\), and the remaining cells cover \(Q\setminus R\). By finite-grid volume conservation, the volumes of all cells, inside and outside together, sum to \(|Q|\). This also covers the cases \(Q\cap R=\varnothing\) and \(Q\subseteq R\), when one of the two groups of cells is empty.

Proof. Fix an arbitrary set \(A\subseteq\mathbb{R}^n\). Subadditivity, applied to \(A=(A\cap R)\cup(A\setminus R)\), gives \(m^*(A)\le m^*(A\cap R)+m^*(A\setminus R)\). It remains to prove the reverse inequality.

First suppose \(m^*(A)<\infty\), and fix \(\varepsilon>0\). By the definition of the infimum, there is a countable box cover \((Q_j)_{j\ge1}\) of \(A\) such that \(\sum_j|Q_j|<m^*(A)+\varepsilon\). Partition each \(Q_j\) by the coordinate cuts from \(R\), as described above. All cells lying inside \(R\), over all \(j\), form a countable box cover of \(A\cap R\); all cells outside \(R\) form a countable box cover of \(A\setminus R\). Therefore, by the definition of \(m^*\) and the volume conservation for each finite partition,

Because this holds for every \(\varepsilon>0\), the left side is at most \(m^*(A)\). If instead \(m^*(A)=+\infty\), then \(m^*(A)\le m^*(A\cap R)+m^*(A\setminus R)\) already forces the right side to be \(+\infty\), so the desired equality holds in this case too. Combining the two inequalities proves the criterion for every \(A\subseteq\mathbb{R}^n\). \(\square\)

Caratheodory-Measurable Sets Form a Sigma-Algebra

To pass from rectangles to Borel sets, we need the stability of the splitting criterion under complements and countable unions. Here is the argument for this outer measure, including the finite-union step.

Proof. The empty set satisfies the criterion, and \(E\in\mathcal{C}\) if and only if \(E^c\in\mathcal{C}\), because the two terms in the splitting equality are simply exchanged. We next verify closure under finite unions. Let \(E,F\in\mathcal{C}\), and fix \(A\subseteq\mathbb{R}^n\). Splitting first by \(E\) and then splitting \(A\setminus E\) by \(F\) gives

Set \(B_1=A\cap E\) and \(B_2=A\cap(F\setminus E)\). These sets have union \(A\cap(E\cup F)\). Subadditivity gives \(m^*(A\cap(E\cup F))\le m^*(B_1)+m^*(B_2)\). Combining this inequality with the displayed decomposition yields \(m^*(A)\ge m^*(A\cap(E\cup F))+m^*(A\setminus(E\cup F))\). The reverse inequality follows from subadditivity applied to the two parts of \(A\). Thus \(E\cup F\in\mathcal{C}\).

Now let \(E_1,E_2,\ldots\) be pairwise disjoint members of \(\mathcal{C}\), and write \(E=\bigcup_{k=1}^{\infty}E_k\). Repeatedly applying the splitting criterion to the first \(N\) sets gives, for every \(A\),

Since \(A\setminus E\subseteq A\setminus\bigcup_{k=1}^{N}E_k\), monotonicity shows that the last term is at least \(m^*(A\setminus E)\). Hence \(m^*(A)\ge\sum_{k=1}^{N}m^*(A\cap E_k)+m^*(A\setminus E)\) for every \(N\). Taking the increasing limit of the finite sums gives \(m^*(A)\ge\sum_{k=1}^{\infty}m^*(A\cap E_k)+m^*(A\setminus E)\), with extended values allowed. Countable subadditivity gives \(m^*(A\cap E)\le\sum_km^*(A\cap E_k)\), so \(m^*(A)\ge m^*(A\cap E)+m^*(A\setminus E)\). The reverse inequality again follows from subadditivity; therefore \(E\in\mathcal{C}\).

For a general sequence \(F_k\in\mathcal{C}\), define \(E_1=F_1\) and \(E_k=F_k\setminus\bigcup_{j<k}F_j\) for \(k\ge2\). Closure under complements and finite unions shows that each \(E_k\) is measurable; these sets are disjoint and have union \(\bigcup_kF_k\). The disjoint-union result proves closure under countable unions. Thus \(\mathcal{C}\) is a sigma-algebra.

Finally, if \(E_k\in\mathcal{C}\) are pairwise disjoint, apply the finite splitting identity above with \(A=\bigcup_kE_k\). For each \(N\), it gives \(m^*(A)\ge\sum_{k=1}^{N}m^*(E_k)\); letting \(N\) increase gives \(m^*(A)\ge\sum_km^*(E_k)\). Countable subadditivity gives the reverse inequality. Therefore \(m^*\) is countably additive on \(\mathcal{C}\), which is precisely the measure property. \(\square\)

Why Borel Sets Are Measurable

Every bounded half-open rectangle is in \(\mathcal{C}\), and these rectangles generate the Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}^n}\). To verify the generation statement, each half-open rectangle is Borel. Conversely, if \(U\subseteq\mathbb{R}^n\) is open, then every \(x\in U\) lies in a half-open rectangle with rational endpoints that contains \(x\) and is contained in \(U\). Such rectangles exist by choosing a sufficiently small neighborhood of \(x\) inside \(U\), then choosing rational endpoints around each coordinate of \(x\). There are only countably many rational-endpoint rectangles. Thus \(U\) is a countable union of such rectangles. The sigma-algebra they generate contains every open set and is contained in \(\mathcal{B}_{\mathbb{R}^n}\), so it equals \(\mathcal{B}_{\mathbb{R}^n}\).

Proof. The Caratheodory-measurable sets form a sigma-algebra containing every bounded half-open rectangle. They therefore contain the sigma-algebra generated by those rectangles, which is \(\mathcal{B}_{\mathbb{R}^n}\). Countable additivity on this collection follows from the proposition. \(\square\)

This result establishes Borel measurability; it does not, by itself, identify every set measurable for the completion of Borel measure. The distinction matters: the Caratheodory criterion is the step that first produces a sigma-algebra on which the outer measure becomes a measure, while questions about completion and non-Borel measurable sets require additional arguments.

A Common Pitfall: Subsets Do Not Add Automatically

In the finite-union proof, it is not enough to observe that \(B_1\) and \(B_2\) are subsets of \(A\cap(E\cup F)\) and then invoke monotonicity to conclude that their measures add to at least the measure of that union. Monotonicity compares one set with another; it gives no lower bound for the sum of two measures merely because both sets lie inside a larger set. For instance, if \(B_1=B_2=[0,1)\) and \(C=[0,2)\) in \(\mathbb{R}\), then \(B_1,B_2\subseteq C\), but \(m^*(B_1)+m^*(B_2)=2\) happens to match \(m^*(C)=2\) only because of these particular lengths. If instead both are \([0,\tfrac12)\), their sum of measures is \(1\), while \(m^*(C)=2\). The valid step in the proof uses the covering identity \(B_1\cup B_2=A\cap(E\cup F)\) and countable subadditivity: \(m^*(A\cap(E\cup F))\le m^*(B_1)+m^*(B_2)\). That is the direction and principle needed.