Sigma-Finite Product Measure Construction

Why Sigma-Finite Spaces Need a Block Construction

For finite measure spaces, the product measure can be constructed directly: measurable rectangles receive the product of their side measures, and the finite product construction supplies a measure on the product sigma-algebra. Sigma-finite spaces may have infinite total measure, so that direct finite-space construction does not apply to the whole space at once.

Sigma-finiteness provides a way to reduce the problem to finite pieces. Cover each space by countably many measurable sets of finite measure, make each cover disjoint, and pair the pieces. Each resulting product block has finite measure and carries a finite product measure. The global measure is assembled by adding these block measures.

Construction principle: Partition both spaces into countably many measurable pieces of finite measure. Construct the finite product measure on each paired block, then sum the resulting measures over all blocks.

Partitions into Finite-Measure Pieces

Let $(X,\mathcal A,\mu)$ be sigma-finite. By definition, $X$ is covered by countably many sets $C_n\in\mathcal A$ with $\mu(C_n)<\infty$. Such a cover can be made into a disjoint partition by setting $A_1=C_1$ and, for $n\ge 2$, setting $A_n=C_n\setminus\bigcup_{k=1}^{n-1}C_k$. Then the $A_n$ are pairwise disjoint, cover $X$, and satisfy $\mu(A_n)<\infty$.

Apply the same procedure to a second sigma-finite measure space $(Y,\mathcal B,\nu)$. We obtain pairwise disjoint measurable sets $B_m$ covering $Y$, with $\nu(B_m)<\infty$. Empty pieces cause no difficulty and may be retained in the indexing.

The product blocks $A_n\times B_m$ are measurable in the product sigma-algebra $\mathcal A\otimes\mathcal B$. They are pairwise disjoint and cover $X\times Y$. Each block has finite side measures, so the finite product construction applies to the restricted measure spaces on $A_n$ and $B_m$.

The Sigma-Finite Product Construction

Definition: For each pair $(n,m)$, let $\eta_{nm}$ be the finite product measure on $A_n\times B_m$ obtained from the restrictions of $\mu$ and $\nu$ to those pieces. For $E\in\mathcal A\otimes\mathcal B$, define $$ \lambda(E)=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \eta_{nm}\bigl(E\cap(A_n\times B_m)\bigr). $$ The sum is a sum of nonnegative extended real numbers, so its value is allowed to be $+\infty$.

The intersection in this definition is measurable for the product measure on the block: the trace of $\mathcal A\otimes\mathcal B$ on $A_n\times B_m$ is the product of the trace sigma-algebras on $A_n$ and $B_m$. This follows from the generating rectangles: intersecting $C\times D$ with the block gives $(C\cap A_n)\times(D\cap B_m)$.

Theorem: The function $\lambda$ defined above is a measure on $(X\times Y,\mathcal A\otimes\mathcal B)$. It is sigma-finite, and for measurable $C\in\mathcal A$ and $D\in\mathcal B$, it satisfies $$ \lambda(C\times D)=\mu(C)\nu(D), $$ where the product on the right is interpreted as $0$ if either factor is zero, and as $+\infty$ if both factors are positive and at least one is infinite.

Proof. For each $n,m$, define $$ \lambda_{nm}(E)=\eta_{nm}\bigl(E\cap(A_n\times B_m)\bigr). $$ This is a measure on $\mathcal A\otimes\mathcal B$: if $E_k$ are pairwise disjoint measurable sets, their intersections with the block are pairwise disjoint, so countable additivity of $\eta_{nm}$ gives $$ \lambda_{nm}\left(\bigcup_{k=1}^{\infty}E_k\right) =\sum_{k=1}^{\infty}\lambda_{nm}(E_k). $$ The function $\lambda=\sum_{n,m}\lambda_{nm}$ is also a measure. Indeed, for pairwise disjoint $E_k$, all terms are nonnegative, and rearranging the countable sums gives $$ \lambda\left(\bigcup_{k=1}^{\infty}E_k\right) =\sum_{n,m}\sum_{k=1}^{\infty}\lambda_{nm}(E_k) =\sum_{k=1}^{\infty}\sum_{n,m}\lambda_{nm}(E_k) =\sum_{k=1}^{\infty}\lambda(E_k). $$ Also, $\lambda(\varnothing)=0$, since every $\eta_{nm}(\varnothing)=0$. Each block has finite measure: $$ \lambda(A_n\times B_m)=\eta_{nm}(A_n\times B_m) =\mu(A_n)\nu(B_m)<\infty. $$ The blocks cover $X\times Y$, so $\lambda$ is sigma-finite. For measurable $C,D$, the finite product property on each block gives $$ \lambda(C\times D) =\sum_{n,m}\mu(C\cap A_n)\nu(D\cap B_m). $$ Put $a_n=\mu(C\cap A_n)$ and $b_m=\nu(D\cap B_m)$. These numbers are finite, and disjointness of the partitions gives $\sum_n a_n=\mu(C)$ and $\sum_m b_m=\nu(D)$. If either sum is zero, all its terms are zero, so the double sum is zero. If both sums are positive, there is a positive term in each sequence. If both sums are finite, the double sum is their product by multiplication of nonnegative series. If either sum is infinite, fixing a positive term from the other sequence shows that the double sum is infinite. Thus the double sum has exactly the stated value in every case. This proves the rectangle formula and completes the proof. $\square$

Worked Examples

Worked Example: Lebesgue Measure on the Plane

Let both spaces be $\mathbb R$ with Lebesgue measure. Use the partition $I_k=[k,k+1)$, $k\in\mathbb Z$, of each copy of $\mathbb R$. Every interval has measure $1$, so each block $I_k\times I_\ell$ has product measure $1$. Consider $C=[0,2]$ and $D=[-1,1]$. The intersections with the partition intervals have length $1$ for $k=0,1$ on the first axis and for $\ell=-1,0$ on the second axis; other intersections have length zero. Therefore $$ \lambda(C\times D)= \sum_{k\in\{0,1\}}\sum_{\ell\in\{-1,0\}}1\cdot 1=4. $$ The inclusion of endpoints does not change any of these Lebesgue measures. This calculation recovers the area of the rectangle from four finite-measure blocks.

Worked Example: Counting Measure Times Lebesgue Measure

Take $X=\mathbb N$ with counting measure and $Y=\mathbb R$ with Lebesgue measure. Partition $X$ into the singletons $\{n\}$, and partition $Y$ into $I_k=[k,k+1)$, $k\in\mathbb Z$. For $$ E=\bigcup_{n=1}^{3}\bigl(\{n\}\times[0,n)\bigr), $$ the pieces for different $n$ lie in disjoint first-coordinate blocks. On the block $\{n\}\times I_k$, the contribution is $1$ if $I_k\subseteq[0,n)$, and $0$ otherwise. For $n=1,2,3$, exactly $1,2,3$ intervals, respectively, contribute. Hence $$ \lambda(E)=1+2+3=6. $$ This is the block construction in a mixed discrete and continuous setting.

Worked Example: Counting Measure on Both Factors

Let $X=Y=\mathbb N$, each with counting measure, and partition each space into its singleton sets. Each product block $\{m\}\times\{n\}$ has measure $1$. For $$ F=\{(m,n)\in\mathbb N^2:m\le n\le m+2\}, $$ there are exactly three points in $F$ for each fixed $m$, namely $(m,m)$, $(m,m+1)$, and $(m,m+2)$. These points are all distinct as $m$ varies. Thus $F$ is a countable union of singleton blocks, with three blocks for every $m$, and $$ \lambda(F)=\sum_{m=1}^{\infty}3=+\infty. $$ The construction accommodates an infinite total measure because it sums the finite contributions from countably many blocks.

What the Construction Does—and Does Not—Require

The key step is partitioning each space before forming products. The blocks are disjoint, and every block has finite measure. Countable additivity then makes the sum of the block measures a measure on the entire product sigma-algebra. The theorem also verifies the expected rectangle formula, including cases where a rectangle has zero or infinite measure.

A common pitfall is to treat the expression $\mu(C)\nu(D)$ as an ordinary product of extended real numbers without checking the zero case. When one side has measure zero and the other has infinite measure, the numerical expression $0\cdot\infty$ is otherwise ambiguous. In the construction, every block contribution is zero in that case, so the rectangle has measure zero.

This argument establishes existence. It does not rely on iterated integration or on switching the order of integration; those are separate results. The construction also makes clear why sigma-finiteness is useful here: it supplies a countable collection of finite-measure blocks whose measures can be assembled by a countable sum.

Check Your Understanding

Use the construction and its block interpretation to answer the following questions.

How can a countable cover by finite-measure sets be changed into a disjoint partition?
Why is each product block $A_n\times B_m$ assigned a finite product measure?
Why does the countable sum of the block measures define a measure?
How does the block construction establish sigma-finiteness of the resulting product measure?
What value should the rectangle formula give when one side has measure zero and the other has infinite measure, and why?