Uniqueness for Sigma-Finite Product Measures

From Finite Uniqueness to Sigma-Finite Uniqueness

Finite uniqueness applies when the measures being compared have finite total mass. For sigma-finite factors, however, the product space may have infinite measure, so that argument cannot be applied directly to the whole space. The remedy is to compare the measures on an increasing sequence of rectangles whose measures are finite. On each such rectangle, finite uniqueness applies; continuity from below then recovers equality on the entire product space.

Let \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) be sigma-finite measure spaces. Here sigma-finite means that each space is a countable union of measurable sets of finite measure. The product sigma-algebra \(\mathcal{M}\otimes\mathcal{N}\) is the sigma-algebra generated by the measurable rectangles \(A\times B\), where \(A\in\mathcal{M}\) and \(B\in\mathcal{N}\). The product measure \(\mu\times\nu\), constructed earlier, satisfies \((\mu\times\nu)(A\times B)=\mu(A)\nu(B)\), with the usual nonnegative extended-real convention for the right-hand side.

The rectangles form a pi-system: the intersection of \(A_1\times B_1\) and \(A_2\times B_2\) is \((A_1\cap A_2)\times(B_1\cap B_2)\), and the whole space \(X\times Y\) is itself a rectangle. Their generated sigma-algebra is \(\mathcal{M}\otimes\mathcal{N}\). The point of the proof below is to arrange that the measures to which finite uniqueness is applied are finite, even when the original measures are not.

Building the Finite-Measure Rectangles

Proof. Sigma-finiteness gives a sequence \((A_k)\) of measurable sets with \(\mu(A_k)<\infty\) and \(\bigcup_{k=1}^{\infty}A_k=X\). Define \(X_n=\bigcup_{k=1}^{n}A_k\). Then \(X_n\subseteq X_{n+1}\), and finite subadditivity gives \(\mu(X_n)\le\sum_{k=1}^{n}\mu(A_k)<\infty\). Also, every point of \(X\) belongs to some \(A_k\), hence to \(X_n\) for every \(n\ge k\). Therefore \(\bigcup_nX_n=X\). If \(X\) is empty, the same conclusion holds by taking every \(X_n\) empty. \(\square\)

Apply the lemma to both factor spaces. We obtain increasing sequences \((X_n)\) and \((Y_n)\) of measurable sets with finite measures and unions \(X\) and \(Y\), respectively. Set \(C_n=X_n\times Y_n\). Then \(C_n\in\mathcal{M}\otimes\mathcal{N}\), \(C_n\subseteq C_{n+1}\), and \(\bigcup_n C_n=X\times Y\). The last equality follows because for any \((x,y)\), there are indices \(r,s\) with \(x\in X_r\) and \(y\in Y_s\), so \((x,y)\in C_{\max(r,s)}\).

Uniqueness for Sigma-Finite Product Measures

Proof. Choose increasing finite-measure exhaustions \((X_n)\) and \((Y_n)\) as above, and write \(C_n=X_n\times Y_n\). For each \(n\), define two measures on \(\mathcal{M}\otimes\mathcal{N}\) by \(\eta_n(E)=\eta(E\cap C_n)\) and \(\rho_n(E)=(\mu\times\nu)(E\cap C_n)\). They are finite because \(\eta_n(X\times Y)=\eta(C_n)=\mu(X_n)\nu(Y_n)<\infty\), and the same rectangle formula gives \(\rho_n(X\times Y)=\mu(X_n)\nu(Y_n)<\infty\). Thus the assumed rectangle identity also shows that \(\eta\) is sigma-finite: the sets \(C_n\) cover the space and have finite \(\eta\)-measure.

For any measurable rectangle \(A\times B\), its intersection with \(C_n\) is \((A\cap X_n)\times(B\cap Y_n)\). Both factor sets have finite measure, so the rectangle formula gives \(\eta_n(A\times B)=\mu(A\cap X_n)\nu(B\cap Y_n)=\rho_n(A\times B)\). The rectangles are a pi-system generating \(\mathcal{M}\otimes\mathcal{N}\) and containing \(X\times Y\). The finite-measure uniqueness theorem therefore gives \(\eta_n(E)=\rho_n(E)\) for every \(E\in\mathcal{M}\otimes\mathcal{N}\).

Now fix any \(E\in\mathcal{M}\otimes\mathcal{N}\). The sets \(E\cap C_n\) increase to \(E\). By continuity from below for measures, \(\eta(E)=\lim_n\eta(E\cap C_n)\) and \((\mu\times\nu)(E)=\lim_n(\mu\times\nu)(E\cap C_n)\). The terms in these limits agree for every \(n\), since they are \(\eta_n(E)\) and \(\rho_n(E)\). Hence \(\eta(E)=(\mu\times\nu)(E)\), with infinite values allowed. Since \(E\) was arbitrary, the measures agree on the whole product sigma-algebra. \(\square\)

For completeness, continuity from below is a direct consequence of countable additivity: if \(E_n\uparrow E\), write \(E\) as the disjoint union of \(E_1\) and the differences \(E_k\setminus E_{k-1}\) for \(k\ge2\). The measure of \(E_n\) is the partial sum of the measures of these disjoint pieces, and those partial sums increase to the measure of their union. This argument allows the limit to be \(+\infty\).

A Common Pitfall

It is not valid to apply finite-measure uniqueness to the full product space merely because both candidate measures agree on rectangles. For example, in the Lebesgue product example, \((m\times m)(\mathbb{R}^2)=\infty\): the squares \(C_n\) have measures \(4n^2\), which grow without bound. The finite uniqueness theorem therefore does not apply directly to the two measures on all of \(\mathbb{R}^2\).

The correction is to restrict both measures to the same \(C_n\). The restrictions have total mass \(4n^2<\infty\), agree on every rectangle after intersection with \(C_n\), and hence agree on the full product sigma-algebra by finite uniqueness. Only after that local comparison do we let \(n\) increase. This avoids invalid expressions such as subtracting one infinite measure from another and makes explicit where sigma-finiteness enters the uniqueness proof.