Proving Countable Additivity of Rectangle Content

From Finite Additivity to Countable Additivity

Finite unions of measurable rectangles can be assigned a natural size by multiplying the measures of the two sides and adding over disjoint pieces. Finite additivity alone, however, does not establish that this size behaves like a measure when a set is written as a countable disjoint union. The missing step is control of the part left over after finitely many pieces have been removed.

Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be measure spaces with finite measures: \(\mu(X)<\infty\) and \(\nu(Y)<\infty\). No completeness assumption is needed. A measurable rectangle is a set \(E\times F\), where \(E\in\mathcal{M}\) and \(F\in\mathcal{N}\). Let \(\mathcal{R}\) be the ring generated by these rectangles. A ring is a collection closed under finite unions and set differences; here its members are finite unions of measurable rectangles and include \(X\times Y\).

For a rectangle, set \(c(E\times F)=\mu(E)\nu(F)\). For a finite disjoint union of rectangles, define \(c\) by summing these products. The rectangle semiring property ensures that ring members admit such representations. The definition does not depend on the chosen representation: partition the finitely many first-coordinate sets into their membership atoms, and do the same for the second-coordinate sets. The resulting cells \(P\times Q\) are disjoint, and their weights \(\mu(P)\nu(Q)\) determine the sum for any representation. In particular, \(c\) is finitely additive on disjoint ring members. Also \(0\le c(A)\le\mu(X)\nu(Y)<\infty\) for every \(A\in\mathcal{R}\).

Section Measures for Ring Members

For \(A\subseteq X\times Y\) and \(x\in X\), its section is \(A_x=\{y\in Y:(x,y)\in A\}\). If \(A\in\mathcal{R}\), each section is in \(\mathcal{N}\), and the function \(h_A:X\to[0,\nu(Y)]\) defined by \(h_A(x)=\nu(A_x)\) is measurable. More precisely, it is a nonnegative simple function: membership of \(x\) in the finitely many first-coordinate sets determines which finite union of second-coordinate sets forms \(A_x\).

This section function links the content to an ordinary integral on the first factor. For completeness, choose finite measurable partitions \((P_i)\) of \(X\) and \((Q_j)\) of \(Y\) that refine all coordinate sets in a finite rectangle representation of \(A\). Then \(A\) is a union of some of the disjoint cells \(P_i\times Q_j\). If \(J_i\) is the set of indices \(j\) for which that cell is included, \(h_A\) has the constant value \(\sum_{j\in J_i}\nu(Q_j)\) on \(P_i\). The simple-function integral therefore gives

This identity uses only finite additivity and the integral of a simple function. It is not an application of Tonelli's theorem, which concerns measurable functions on a product space and is not needed here.

Continuity at the Empty Set

We use continuity from below for each factor measure: if \(E_n\) is an increasing sequence of measurable sets with union \(E\), then \(\mu(E_n)\to\mu(E)\), and likewise for \(\nu\). This follows directly from countable additivity: write the union as the disjoint union of \(E_1\) and the successive differences \(E_n\setminus E_{n-1}\), and take partial sums. We now use continuity from below on the sections, together with Monotone Convergence on \(X\).

Proof. Set \(B_n=(X\times Y)\setminus A_n\). Since \(X\times Y\in\mathcal{R}\) and \(\mathcal{R}\) is a ring, each \(B_n\) belongs to \(\mathcal{R}\). The sets \(B_n\) increase, and their union is \(X\times Y\). For each fixed \(x\in X\), the sections \((B_n)_x\) increase and have union \(Y\): their union is \(Y\) because \(\bigcup_n B_n=X\times Y\). Continuity from below of \(\nu\) therefore gives \(h_{B_n}(x)=\nu((B_n)_x)\uparrow\nu(Y)\) for every \(x\).

The functions \(h_{B_n}\) are measurable and nonnegative, as established for ring members. The Monotone Convergence Theorem, applied on \((X,\mathcal{M},\mu)\), now yields \(\lim_n\int_X h_{B_n}\,d\mu=\int_X\nu(Y)\,d\mu=\mu(X)\nu(Y)\). Its hypotheses hold: the functions increase pointwise to the constant \(\nu(Y)\), and they are measurable. The section identity gives \(c(B_n)=\int_Xh_{B_n}\,d\mu\). Since \(A_n\) and \(B_n\) partition \(X\times Y\), finite additivity gives \(c(A_n)+c(B_n)=c(X\times Y)=\mu(X)\nu(Y)\). Subtracting and taking the limit proves \(c(A_n)\to0\). All these contents are finite, so the subtraction is valid. \(\square\)

The use of complements is where finiteness matters. Continuity from below for \(\nu\) gives convergence of the section measures to \(\nu(Y)\); finite total content then lets us subtract those increasing complement contents from \(c(X\times Y)\).

The Premeasure Theorem

Proof. Let \((A_n)\) be pairwise disjoint members of \(\mathcal{R}\), and suppose \(A=\bigcup_{n=1}^{\infty}A_n\) belongs to \(\mathcal{R}\). For each \(N\ge1\), define the remainder \(R_N=A\setminus\bigcup_{n=1}^{N}A_n\). It belongs to \(\mathcal{R}\), and finite additivity gives \(c(A)=\sum_{n=1}^{N}c(A_n)+c(R_N)\). The remainders decrease: \(R_{N+1}\subseteq R_N\). Their intersection is empty, because every point of \(A\) belongs to some \(A_k\) and is therefore absent from \(R_N\) for every \(N\ge k\). By the proposition, \(c(R_N)\to0\). Thus the partial sums \(\sum_{n=1}^{N}c(A_n)=c(A)-c(R_N)\) tend to \(c(A)\). By the definition of a nonnegative series, this says \(c(A)=\sum_{n=1}^{\infty}c(A_n)\). Also \(c(\varnothing)=0\) from its rectangle definition. Hence \(c\) is a premeasure. \(\square\)

A Common Pitfall and the Role of the Result

Rectangle content is additive on disjoint pieces, not on arbitrary overlapping rectangles. For example, in the unit square the rectangles \(R_1=[0,1]\times[0,1]\) and \(R_2=[1/2,1]\times[0,1]\) overlap, with \(R_2\subseteq R_1\). The incorrect calculation \(c(R_1\cup R_2)=c(R_1)+c(R_2)=1+1/2=3/2\) counts the second rectangle twice. The correction is \(R_1\cup R_2=R_1\), so its content is \(1\). Finite additivity applies only after the sets have been made disjoint.

A second shortcut to avoid is asserting continuity at the empty set for arbitrary measures. The finite total content \(c(X\times Y)=\mu(X)\nu(Y)\) was essential in the complement argument. For instance, with counting measure on \(X=\mathbb{N}\), the measurable sets \(E_n=\{n,n+1,\ldots\}\) decrease to the empty set but all have infinite measure. This does not contradict the proposition, whose factor measures are finite; it shows why finiteness cannot be omitted from that particular continuity argument.

Countable additivity on the ring is the precise property needed before extending rectangle content to a measure on a generated sigma-algebra. That extension is a separate construction: the result here establishes the premeasure hypothesis, not the extension itself.