Constructing Sigma-Finite Product Measures

From Finite Pieces to a Product Measure

When both factor measures are finite, the Caratheodory construction gives a measure on the product sigma-algebra from the rectangle content. For sigma-finite factors, the total masses of the spaces may be infinite, so that finite construction cannot be applied directly to the whole product. The remedy is to cover each factor by an increasing sequence of sets of finite measure, construct a finite product measure on each corresponding product piece, and verify that those measures agree wherever the pieces overlap.

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; it does not mean that the total measure is finite. Choose increasing measurable exhaustions $X_n\uparrow X$ and $Y_n\uparrow Y$ with $\mu(X_n)<\infty$ and $\nu(Y_n)<\infty$ for every $n\ge1$. Write $W_n=X_n\times Y_n$. The product sigma-algebra is $\mathcal{M}\otimes\mathcal{N}=\sigma\{A\times B:A\in\mathcal{M},\,B\in\mathcal{N}\}$.

Finite construction used below: For finite measure spaces $(S,\mathcal{S},\alpha)$ and $(T,\mathcal{T},\beta)$, the Caratheodory construction from rectangle content gives a measure on $\mathcal{S}\otimes\mathcal{T}$ whose value on $A\times B$ is $\alpha(A)\beta(B)$. Its outer measure on any subset $H\subseteq S\times T$ is the infimum of $\sum_i\alpha(A_i)\beta(B_i)$ over countable covers $H\subseteq\bigcup_i(A_i\times B_i)$ by measurable rectangles. We use this finite construction, established in the earlier material on Caratheodory construction, and prove the compatibility needed for the sigma-finite case.

Finite Measures on the Exhaustion Pieces

Give $X_n$ the trace sigma-algebra $\mathcal{M}|_{X_n}=\{A\cap X_n:A\in\mathcal{M}\}$, and define $\mu_n(A\cap X_n)=\mu(A\cap X_n)$; define $\mathcal{N}|_{Y_n}$ and $\nu_n$ in the same way. These are finite measure spaces. Their finite product construction gives a measure on $\mathcal{M}|_{X_n}\otimes\mathcal{N}|_{Y_n}$, supported on $W_n$.

We identify this trace product sigma-algebra with $(\mathcal{M}\otimes\mathcal{N})|_{W_n}$. Indeed, both are generated on $W_n$ by the sets $(A\times B)\cap W_n=(A\cap X_n)\times(B\cap Y_n)$, for $A\in\mathcal{M}$ and $B\in\mathcal{N}$. Thus a measurable subset of $W_n$ can be treated either as a set in the finite product space or as a trace of a set in the full product space.

Lemma (Consistency on nested pieces): If $m\le n$, the outer measure from the finite construction on $W_n$, restricted to subsets of $W_m$, equals the outer measure from the finite construction on $W_m$.

Proof. Denote the two outer measures by $\lambda_n^*$ and $\lambda_m^*$. Let $H\subseteq W_m$. If $H$ is covered in $W_n$ by measurable rectangles $A_i\times B_i$, intersect each rectangle with $W_m$. The resulting rectangles $(A_i\cap X_m)\times(B_i\cap Y_m)$ cover $H$, and their costs do not increase, since $\mu(A_i\cap X_m)\le\mu(A_i\cap X_n)$ and $\nu(B_i\cap Y_m)\le\nu(B_i\cap Y_n)$. Taking the infimum over covers gives $\lambda_m^*(H)\le\lambda_n^*(H)$. Conversely, every rectangle cover of $H$ in $W_m$ is also a rectangle cover in $W_n$ with the same costs, so $\lambda_n^*(H)\le\lambda_m^*(H)$. The two outer measures are equal on every $H\subseteq W_m$. $\square$

Write $\lambda_n$ for the finite product measure on $W_n$. The lemma implies that its restriction to $W_m$ agrees with $\lambda_m$ when $m\le n$. This argument proves the needed compatibility directly from the rectangle-cover construction; it does not assume a uniqueness theorem for product measures.

Constructing the Full Measure

Theorem (Sigma-finite product construction): For sigma-finite measure spaces $(X,\mathcal{M},\mu)$ and $(Y,\mathcal{N},\nu)$, there is a measure $\lambda$ on $\mathcal{M}\otimes\mathcal{N}$ obtained from the finite product measures on $W_n=X_n\times Y_n$ by $$ \lambda(E)=\lim_{n\to\infty}\lambda_n(E\cap W_n),\qquad E\in\mathcal{M}\otimes\mathcal{N}. $$ It is sigma-finite and satisfies $\lambda(A\times B)=\mu(A)\nu(B)$, with the convention that the product is $0$ if either factor is $0$, and otherwise is the usual extended nonnegative product.

Proof. For $E\in\mathcal{M}\otimes\mathcal{N}$, define $\lambda_n(E)=\lambda_n(E\cap W_n)$, using the finite measure on $W_n$ on the right. This is a finite measure on the full product sigma-algebra, since it is the measure of the intersection with a measurable set. If $m\le n$, consistency gives $$ \lambda_n(E\cap W_m)=\lambda_m(E\cap W_m)=\lambda_m(E). $$ Since $W_m\subseteq W_n$, monotonicity of the finite measure $\lambda_n$ now gives $\lambda_m(E)\le\lambda_n(E)$. Thus $\lambda(E)=\lim_n\lambda_n(E)=\sup_n\lambda_n(E)$ is defined in $[0,\infty]$.

We verify countable additivity. Let $E_1,E_2,\ldots$ be pairwise disjoint members of $\mathcal{M}\otimes\mathcal{N}$. For each $n$, countable additivity of the finite measure $\lambda_n$ gives $$ \lambda_n\left(\bigcup_{j=1}^\infty E_j\right)=\sum_{j=1}^\infty\lambda_n(E_j). $$ For each fixed $j$, the numbers $\lambda_n(E_j)$ increase to $\lambda(E_j)$. Consequently, $$ \lim_n\sum_{j=1}^\infty\lambda_n(E_j)=\sum_{j=1}^\infty\lambda(E_j). $$ To justify this equality, the left side is at most the right side because $\lambda_n(E_j)\le\lambda(E_j)$ for every $n,j$. For the reverse inequality, fix $J$. The left side is at least $\lim_n\sum_{j=1}^J\lambda_n(E_j)=\sum_{j=1}^J\lambda(E_j)$, since this is a finite sum of convergent sequences. Taking the supremum over $J$ gives the reverse inequality, including when the series is infinite. Taking limits in the displayed countable-additivity identity proves $\lambda(\bigcup_jE_j)=\sum_j\lambda(E_j)$. Also, $\lambda(\varnothing)=0$, so $\lambda$ is a measure.

The sets $W_n$ increase to $X\times Y$, and $\lambda(W_n)=\mu(X_n)\nu(Y_n)<\infty$ by the finite construction. Hence $\lambda$ is sigma-finite. Finally, for a measurable rectangle $A\times B$, $$ \lambda(A\times B)=\lim_n \mu(A\cap X_n)\nu(B\cap Y_n). $$ Each factor increases to the measure of the corresponding full set. If either limiting factor is zero, every term in that factor's sequence is zero, so the limit is zero. If both limits are positive, the products tend to their usual extended product: this follows from convergence of finite positive factors, and if either factor tends to infinity the other is eventually bounded below by a positive number. This proves the rectangle formula with the stated zero convention. $\square$

Worked Constructions

Worked Example: Counting Measure on a Countable Product

Let $X=Y=\mathbb{N}=\{1,2,\ldots\}$, with counting measure on each factor and the full power-set sigma-algebra. Take $X_n=Y_n=\{1,\ldots,n\}$. Then $W_n$ has $n^2$ points, so its finite product measure is counting measure on $W_n$. For any $E\subseteq\mathbb{N}^2$, the construction gives $\lambda(E)=\lim_n |E\cap W_n|$.

For $E=\{(1,1),(1,2),(3,1)\}$, the first piece contains one point of $E$, the second contains two, and every piece with $n\ge3$ contains all three. Thus $\lambda_1(E)=1$, $\lambda_2(E)=2$, and $\lambda_n(E)=3$ for $n\ge3$; the constructed measure is $3$. As a second check, the three distinct points each have product measure $1$, so their disjoint union has measure $1+1+1=3$. For the diagonal $D=\{(k,k):k\in\mathbb{N}\}$, exactly $n$ diagonal points lie in $W_n$, so $\lambda(D)=\lim_n n=\infty$, as required for counting measure on an infinite set.

Worked Example: A Strip in the Product of the Line and the Integers

Let $X=\mathbb{R}$ with Lebesgue measure $m$, and let $Y=\mathbb{N}$ with counting measure. Use $X_n=[-n,n]$ and $Y_n=\{1,\ldots,n\}$. Consider the measurable set $$ E=\bigcup_{k=1}^\infty [k,k+1)\times\{k\}. $$ The rectangles in this union are pairwise disjoint because their second coordinates differ. Each has measure $m([k,k+1))\cdot 1=1$, so countable additivity gives $\lambda(E)=\sum_{k=1}^\infty1=\infty$.

The exhaustion calculation agrees. For $n\ge2$, the pieces with $1\le k\le n-1$ lie fully inside $W_n=[-n,n]\times\{1,\ldots,n\}$, contributing $n-1$. The piece with $k=n$ intersects $W_n$ only in $\{n\}\times\{n\}$, which has measure $m(\{n\})\cdot1=0$; pieces with $k>n$ do not meet $W_n$. Thus $\lambda_n(E)=n-1$ for $n\ge2$, while $\lambda_1(E)=0$. The limit is infinite, matching the disjoint-rectangle calculation.

Worked Example: Checking a Finite Piece in the Lebesgue Product

Take Lebesgue measure on both copies of $\mathbb{R}$, with $X_n=Y_n=[-n,n]$. Let $$ E=([0,1)\times[0,1))\cup([1,2)\times[0,1))\cup([0,1)\times[1,2)). $$ These three rectangles are pairwise disjoint. For every $n\ge2$, they lie in $W_n$, and each has finite product measure $1\cdot1=1$. Therefore $\lambda_n(E)=1+1+1=3$ for $n\ge2$, and the constructed measure is $3$. Independently, $E$ is the square $[0,2)\times[0,2)$, apart from its upper-right unit square $[1,2)\times[1,2)$; its measure is $2\cdot2-1\cdot1=4-1=3$. The two computations agree.

Why the Limit, Not a Sum, Is Essential

The measures $\lambda_n(E)$ already include everything in the smaller pieces, because the exhaustion sets are nested. The construction therefore takes their increasing limit, not their sum. For example, with Lebesgue measure on both factors and $X_n=Y_n=[-n,n]$, take $E=[0,1)\times[0,1)$. For every $n\ge1$, $E\subseteq W_n$ and $\lambda_n(E)=1$, so the constructed measure is $\lim_n1=1$.

A wrong calculation would sum the local values and claim $\sum_{n=1}^\infty\lambda_n(E)=\sum_{n=1}^\infty1=\infty$. This counts the same set repeatedly: $E$ is contained in every exhaustion piece. Taking the limit counts it once and gives the correct value $1$. The consistency proof is what makes that increasing-limit procedure legitimate: each finite measure agrees with all the smaller finite constructions on their common region.

Check Your Understanding

For each question, identify the finite pieces involved and distinguish an increasing limit from a sum over pieces.

Why does sigma-finiteness of both factors allow choices of increasing exhaustions with finite measure at every stage?
In the consistency lemma, why does intersecting a covering rectangle with $W_m$ not increase its cost?
For counting measure on $\mathbb{N}^2$, compute the constructed measure of $\{(1,1),(2,2),(3,3),(4,4)\}$ and of the full diagonal.
Explain why $\lambda_n(E)\le\lambda_{n+1}(E)$ follows from consistency, even when $\lambda_{n+1}(E)$ is infinite.
For $E=[0,1)\times[0,1)$ in the Lebesgue product with exhaustion $[-n,n]^2$, compare the increasing-limit calculation with the incorrect sum of all local measures.