Why Rectangle Values Can Determine a Measure
On a product measurable space, measurable rectangles are the basic sets whose measures are usually specified first. A typical rectangle is \(A\times B\), where \(A\) is measurable in the first space and \(B\) is measurable in the second. The central question is whether knowing the measure of every such rectangle determines the measure of every set in the product sigma-algebra.
The answer relies on two ingredients. Rectangles form a pi-system that generates the product sigma-algebra, and a countable collection of rectangles of finite measure can localize the question to finite measures. The pi-lambda theorem then gives equality on the full sigma-algebra within each finite region.
Rectangles and the Product Sigma-Algebra
Let \((X,\mathcal A)\) and \((Y,\mathcal B)\) be measurable spaces. The product sigma-algebra \(\mathcal A\otimes\mathcal B\) is the sigma-algebra generated by the measurable rectangles.
A pi-system is a convenient generating family because the pi-lambda theorem applies to it. A lambda-system on a space \(\Omega\) contains \(\Omega\), is closed under complements relative to \(\Omega\), and is closed under countable disjoint unions. The pi-lambda theorem says that a lambda-system containing a pi-system also contains the sigma-algebra generated by that pi-system.
The Finite-Measure Uniqueness Principle
The basic uniqueness step is for finite measures. The whole space must be included in the generating pi-system, so that agreement there also gives equality of the two finite total masses.
Proof. Since \(\Omega\in\mathcal P\), the measures have equal total mass: \(\rho(\Omega)=\eta(\Omega)<\infty\). Define \[ \mathcal D=\{E\in\mathcal F:\rho(E)=\eta(E)\}. \] This is a lambda-system. It contains \(\Omega\). If \(E\in\mathcal D\), finiteness and equality of total masses give \[ \rho(E^c)=\rho(\Omega)-\rho(E) =\eta(\Omega)-\eta(E)=\eta(E^c). \] If \(E_1,E_2,\ldots\) are pairwise disjoint members of \(\mathcal D\), countable additivity gives \[ \rho\!\left(\bigcup_{k=1}^{\infty}E_k\right) =\sum_{k=1}^{\infty}\rho(E_k) =\sum_{k=1}^{\infty}\eta(E_k) =\eta\!\left(\bigcup_{k=1}^{\infty}E_k\right). \] Thus \(\mathcal D\) is a lambda-system. It contains \(\mathcal P\) by hypothesis, so the pi-lambda theorem implies \(\sigma(\mathcal P)\subseteq\mathcal D\). Therefore the measures agree throughout \(\mathcal F\). \(\square\)
Uniqueness from a Finite Rectangle Cover
For measures on a product space, the finite-measure principle can be applied locally. A rectangle cover need not be disjoint or increasing; it only needs to be countable and to consist of sets on which both measures are finite.
Proof. Fix \(n\). Define finite measures on \(\mathcal A\otimes\mathcal B\) by \[ \mu_n(E)=\mu(E\cap R_n), \qquad \nu_n(E)=\nu(E\cap R_n). \] Their total masses are finite and equal, because \(\mu_n(X\times Y)=\mu(R_n)=\nu(R_n)=\nu_n(X\times Y)\).
Write \(R_n=A_n\times B_n\). For any measurable rectangle \(Q=A\times B\), the intersection \[ Q\cap R_n=(A\cap A_n)\times(B\cap B_n) \] is again a measurable rectangle. Since \(\mu\) and \(\nu\) agree on rectangles, \(\mu_n(Q)=\nu_n(Q)\). The finite-measure uniqueness principle, applied to the rectangle pi-system, now gives \(\mu_n(E)=\nu_n(E)\) for every \(E\in\mathcal A\otimes\mathcal B\).
To combine these local equalities without double-counting overlaps, set \(U_0=\varnothing\), \(U_n=\bigcup_{k=1}^n R_k\), and \(D_n=R_n\setminus U_{n-1}\). The \(D_n\) are pairwise disjoint, measurable, and cover \(X\times Y\); also \(D_n\subseteq R_n\). For any \(E\in\mathcal A\otimes\mathcal B\), the equality of the restricted measures on all measurable sets gives \[ \mu(E\cap D_n)=\nu(E\cap D_n). \] Countable additivity on the disjoint pieces therefore yields \[ \mu(E)=\sum_{n=1}^{\infty}\mu(E\cap D_n) =\sum_{n=1}^{\infty}\nu(E\cap D_n)=\nu(E). \] The sums may be infinite, which causes no problem for countable additivity. Hence \(\mu=\nu\). \(\square\)
Worked Example: Sigma-Finite Factors Give a Rectangle Cover
Worked Example: Sigma-Finite Factors Give a Rectangle Cover
Let \(\alpha\) and \(\beta\) be sigma-finite measures on \((X,\mathcal A)\) and \((Y,\mathcal B)\). By sigma-finiteness, choose measurable sets \(A_i\) and \(B_j\) such that \(X=\bigcup_i A_i\), \(Y=\bigcup_j B_j\), \(\alpha(A_i)<\infty\), and \(\beta(B_j)<\infty\).
The countable family \(R_{ij}=A_i\times B_j\) covers \(X\times Y\). For the product measure \(\alpha\times\beta\), its rectangle values satisfy \[ (\alpha\times\beta)(R_{ij})=\alpha(A_i)\beta(B_j)<\infty. \] Suppose another measure \(\nu\) on the product sigma-algebra agrees with \(\alpha\times\beta\) on every measurable rectangle. Then \[ \nu(R_{ij})=(\alpha\times\beta)(R_{ij})<\infty \] for every \(i,j\). Enumerating the pairs \((i,j)\) gives a countable finite-measure rectangle cover for both measures. The uniqueness theorem proves \(\nu=\alpha\times\beta\) on all of \(\mathcal A\otimes\mathcal B\).
The important point is that sigma-finiteness of the factor measures supplies rectangles of finite product measure that cover the product space.
Worked Example: A Countable Product of Discrete Spaces
Worked Example: A Countable Product of Discrete Spaces
Take \(X=Y=\mathbb N_0\), with each power-set sigma-algebra, and let \(\mu\) and \(\nu\) be measures on the product sigma-algebra that agree on every rectangle. The singleton rectangles \(R_{ij}=\{i\}\times\{j\}\) cover \(\mathbb N_0\times\mathbb N_0\).
Agreement on rectangles gives \(\mu(R_{ij})=\nu(R_{ij})\). If these common values are finite for every \(i,j\), the singleton rectangles form a countable finite-measure cover. For any set \(E\subseteq\mathbb N_0\times\mathbb N_0\), its elements are exactly the singleton cells it contains, and countable additivity gives \[ \mu(E)=\sum_{(i,j)\in E}\mu(\{(i,j)\}) =\sum_{(i,j)\in E}\nu(\{(i,j)\})=\nu(E). \] The sum is understood as a sum of nonnegative terms and may equal infinity. This direct calculation is the countable discrete version of the rectangle-cover uniqueness theorem.
Worked Example: Equality Despite Infinite Total Mass
Worked Example: Equality Despite Infinite Total Mass
Suppose \(R_n=[-n,n]\times[-n,n]\) for \(n\in\mathbb N\), and \(\mu,\nu\) are measures on the Borel sigma-algebra of \(\mathbb R^2\) that agree on all Borel rectangles and have finite, equal values on each \(R_n\). These rectangles cover \(\mathbb R^2\), although the measures of the whole space need not be finite.
The finite restriction argument shows that \(\mu(E\cap R_n)=\nu(E\cap R_n)\) for every Borel set \(E\) and every \(n\). To see why infinite total mass is harmless, disjointify the cover: let \(D_1=R_1\), and for \(n\ge2\) let \(D_n=R_n\setminus R_{n-1}\). Then the \(D_n\) are disjoint and cover the plane. For each Borel \(E\), \[ \mu(E)=\sum_{n=1}^{\infty}\mu(E\cap D_n) =\sum_{n=1}^{\infty}\nu(E\cap D_n)=\nu(E). \] Either sum may be infinite. The conclusion concerns equality of measures, not finiteness of their total mass.
What the Finite-Cover Condition Is Doing
A subtle point is that sigma-finiteness on the whole product space means there is a countable cover by measurable sets of finite measure. Those sets need not be rectangles. The pi-lambda argument above needs a cover drawn from the generating rectangles, because restricting to such a rectangle preserves agreement on the rectangle pi-system.
In the product-measure setting, sigma-finite factor measures provide exactly the needed cover: sets of finite factor measure produce rectangles of finite product measure. More generally, the theorem applies whenever a countable rectangle cover of finite measure is available. One should not replace that condition in the proof by a cover consisting of arbitrary finite-measure sets; the finite-measure uniqueness argument must still be anchored to the generating family.
This distinction explains why rectangle values are especially effective in product measure theory. The rectangles both generate the product sigma-algebra and, under the sigma-finite factor hypothesis, supply the finite regions needed to extend local equality to the entire space.
Check Your Understanding
Use the definitions and arguments above to answer the following questions.
- Why do measurable rectangles form a pi-system, and which sigma-algebra do they generate?
- Where in the finite-measure uniqueness proof is finiteness used?
- Why does restricting both measures to a rectangle preserve their agreement on rectangles?
- How do sigma-finite factor measures produce a countable cover by finite-measure rectangles?
- Why is it useful to disjointify an overlapping rectangle cover before comparing the measures of an arbitrary measurable set?