Product Measures

From Two Measure Spaces to One Product Space

Tonelli’s Theorem and Fubini’s Theorem concern integration on a product space. To use them, we need a measure on that space: a measure that assigns the expected size to a rectangle $A\times B$ from the measures of its two sides. This tutorial constructs that measure and explains the roles of the product sigma-algebra and sigma-finiteness.

The central construction uses sections. For a set $E\subseteq X\times Y$ and a point $x\in X$, its section at $x$ is the set of points in $Y$ paired with $x$ that belong to $E$. For a rectangle, every section is either all of $B$ or empty. For more complicated sets, sections still allow us to build a measure by integrating their measures.

Definition: Let $(X,\mathcal{F})$ and $(Y,\mathcal{G})$ be measurable spaces. The product sigma-algebra, denoted $\mathcal{F}\otimes\mathcal{G}$, is the sigma-algebra on $X\times Y$ generated by the measurable rectangles $A\times B$, where $A\in\mathcal{F}$ and $B\in\mathcal{G}$. For $E\subseteq X\times Y$, $x\in X$, and $y\in Y$, define the sections $E_x=\{y'\in Y:(x,y')\in E\}$ and $E^y=\{x'\in X:(x',y)\in E\}$.

A measurable rectangle is a generator of the product sigma-algebra, but a product-measurable set need not itself be a rectangle. Sections are useful because the section of a product-measurable set is measurable in the corresponding factor space. We will also need a stronger fact when one factor has finite measure: the measure of the sections varies measurably with the point in the other factor.

Constructing a Product Measure with a Finite Factor

Start with measure spaces $(X,\mathcal{F},\mu)$ and $(Y,\mathcal{G},\nu)$, where $\nu(Y)<\infty$. Consider the class of sets $E\subseteq X\times Y$ for which every $E_x$ belongs to $\mathcal{G}$ and the function $x\mapsto\nu(E_x)$ is $\mathcal{F}$-measurable. This class contains every measurable rectangle: for $E=A\times B$, the section $E_x$ is $B$ when $x\in A$ and empty otherwise, so its measure function is $\nu(B)\mathbf{1}_A(x)$.

The class is closed under complements because $\nu$ is finite: $\nu((E^c)_x)=\nu(Y)-\nu(E_x)$. It is closed under countable disjoint unions because sections of disjoint sets are disjoint, and the measure of their union is the sum of their measures. Thus the class is a lambda-system containing the measurable rectangles, which form a pi-system. The pi-lambda theorem implies that it contains $\mathcal{F}\otimes\mathcal{G}$. In particular, product-measurable sets have measurable sections and measurable section-measure functions.

Theorem (Existence of the Product Measure): Let $(X,\mathcal{F},\mu)$ and $(Y,\mathcal{G},\nu)$ be measure spaces, and suppose both are sigma-finite. There is a measure $\mu\otimes\nu$ on $\mathcal{F}\otimes\mathcal{G}$ such that $$ (\mu\otimes\nu)(A\times B)=\mu(A)\nu(B) $$ for all $A\in\mathcal{F}$ and $B\in\mathcal{G}$, with the convention that $0\cdot\infty=0$. The measure $\mu\otimes\nu$ is sigma-finite.

Proof. First suppose $\nu(Y)<\infty$; $\mu$ need not be finite. For $E\in\mathcal{F}\otimes\mathcal{G}$, the section-measure function $g_E(x)=\nu(E_x)$ is measurable by the argument above. Define $$ (\mu\otimes\nu)(E)=\int_X g_E\,d\mu. $$ This is a measure. Indeed, the empty set has measure zero. If $E_1,E_2,\ldots$ are pairwise disjoint product-measurable sets, then, for every $x$, their sections are pairwise disjoint and $$ \nu\left(\left(\bigcup_{n=1}^{\infty}E_n\right)_x\right) =\sum_{n=1}^{\infty}\nu((E_n)_x). $$ The Monotone Convergence Theorem therefore gives $$ (\mu\otimes\nu)\left(\bigcup_{n=1}^{\infty}E_n\right) =\sum_{n=1}^{\infty}(\mu\otimes\nu)(E_n). $$ For a rectangle, its section-measure function is $\nu(B)\mathbf{1}_A$, so the Integral of an Indicator theorem gives $$ (\mu\otimes\nu)(A\times B) =\int_X\nu(B)\mathbf{1}_A\,d\mu =\mu(A)\nu(B). $$ Here the formula also covers zero measures: if $\nu(B)=0$, the integrand is identically zero, even if $\mu(A)=\infty$.

Now suppose both spaces are sigma-finite. Choose measurable partitions $X=\bigcup_{i=1}^{\infty}X_i$ and $Y=\bigcup_{j=1}^{\infty}Y_j$ such that $\mu(X_i)<\infty$ and $\nu(Y_j)<\infty$. The sets $X_i\times Y_j$ partition $X\times Y$. On each such block, use the finite-factor construction with the restricted measures on $X_i$ and $Y_j$; both restricted measures are finite. For $E\in\mathcal{F}\otimes\mathcal{G}$, define its measure to be the sum of the measures of $E\cap(X_i\times Y_j)$ on these blocks. A countable sum of measures on disjoint measurable blocks is a measure: countable additivity follows by applying countable additivity on each block and regrouping nonnegative sums. The rectangle formula follows by summing over $i,j$: $$ (\mu\otimes\nu)(A\times B) =\sum_{i,j}\mu(A\cap X_i)\nu(B\cap Y_j) =\mu(A)\nu(B). $$ The last equality follows by countable additivity in each factor and the distributive law for nonnegative sums. Finally, each block has measure $\mu(X_i)\nu(Y_j)<\infty$, and the blocks cover $X\times Y$. Thus the product measure is sigma-finite.

Worked Examples with Product Measures

Worked Example: Measuring a Rectangle

Take $X=Y=[0,1]$ with Lebesgue measure $\lambda$, and let $A=[0,\tfrac13]$ and $B=[\tfrac14,1]$. Their lengths are $\lambda(A)=\tfrac13$ and $\lambda(B)=\tfrac34$. The rectangle formula gives $$ (\lambda\otimes\lambda)(A\times B) =\lambda(A)\lambda(B) =\frac13\cdot\frac34 =\frac14. $$ The result is the area of the rectangle, including or excluding its boundary edges makes no difference to this measure.

Worked Example: Measuring a Strip

Let $X=[-2,2]$ and $Y=[0,3]$, both with Lebesgue measure, and consider the strip $E=[-1,1]\times[1,3]$. The first side has measure $2$, and the second has measure $2$. Hence $$ (\lambda\otimes\lambda)(E)=2\cdot2=4. $$ The same answer can be read from the sections: for $x\in[-1,1]$, $E_x=[1,3]$ has measure $2$; for other $x\in X$, the section is empty. Integrating this section-measure function gives $\int_{-2}^{2}2\mathbf{1}_{[-1,1]}(x)\,dx=4$.

Worked Example: The Diagonal Has Product Measure Zero

In $[0,1]^2$, the diagonal $D=\{(x,y):x=y\}$ is closed, and therefore Borel and product-measurable. For each $x\in[0,1]$, its vertical section is $D_x=\{x\}$, which has Lebesgue measure zero. The section construction yields $$ (\lambda\otimes\lambda)(D) =\int_{[0,1]}\lambda(D_x)\,d\lambda(x) =\int_{[0,1]}0\,d\lambda(x) =0. $$ The diagonal contains infinitely many points, but it still has zero two-dimensional product measure. A set’s number of points alone does not determine its measure.

Why the Product Measure Is Unique

The rectangle formula determines the measure on every measurable rectangle. The uniqueness theorem below shows that, under sigma-finiteness, it determines the measure on every set in the product sigma-algebra as well. The sigma-finiteness hypothesis lets us reduce the argument to finite measures on blocks; without a finiteness condition, agreement on a generating class need not determine two measures.

Theorem (Uniqueness of the Product Measure): Suppose $\mu$ and $\nu$ are sigma-finite. If two measures on $\mathcal{F}\otimes\mathcal{G}$ agree with $\mu(A)\nu(B)$ on every measurable rectangle $A\times B$, then they agree on all of $\mathcal{F}\otimes\mathcal{G}$.

Proof. Choose the finite-measure partitions $X_i$ and $Y_j$ used in the existence proof. Fix $i,j$, and restrict each of the two measures to the block $X_i\times Y_j$. These restricted measures are finite, since the measure of the whole block is $\mu(X_i)\nu(Y_j)<\infty$. They agree on every rectangle of the form $(A\cap X_i)\times(B\cap Y_j)$, because this is itself a measurable rectangle and both measures obey the rectangle formula.

Within the block, these rectangles form a pi-system and generate the trace of $\mathcal{F}\otimes\mathcal{G}$. The uniqueness theorem for finite measures on a generating pi-system follows from the pi-lambda theorem: the sets on which the two finite measures agree form a lambda-system, since they agree on the whole block, are closed under relative complements, and are closed under countable disjoint unions. Consequently, the restricted measures agree on every product-measurable subset of the block. Since the blocks form a countable disjoint partition of $X\times Y$, countable additivity shows that the original measures agree on every product-measurable set. This proves uniqueness.

Product Measurability and a Useful Distinction

The product sigma-algebra is generated by measurable rectangles, and the product measure assigns each rectangle the product of its side measures. Sigma-finiteness ensures that this measure can be assembled from finite-measure blocks and that it is unique. These are the foundations for the product-space integration results in Fubini’s Theorem and Tonelli’s Theorem.

One distinction is worth keeping in view: the product sigma-algebra $\mathcal{F}\otimes\mathcal{G}$ is not automatically the completion of the product measure. Completion adds all subsets of product-measurable null sets. Thus a subset of a null set need not belong to the product sigma-algebra, even though it belongs to the completed measure space. When measurability is required in an application, it matters which of these spaces is being used.

A second common pitfall is to use the rectangle formula without checking the measure spaces’ hypotheses. The existence and uniqueness result here assumes both factor measures are sigma-finite. In applications of Tonelli’s Theorem and Fubini’s Theorem, that same hypothesis ensures the product measure used for integration is available and uniquely determined by its values on rectangles.

Key takeaway: For sigma-finite measure spaces, there is a unique sigma-finite measure on the product sigma-algebra whose value on each measurable rectangle is the product of the two factor measures. Sections provide a direct construction when one factor is finite; finite-measure blocks extend it to the sigma-finite case.

Check Your Understanding

Use the definitions and results in this tutorial to answer the following questions.

How is the product sigma-algebra $\mathcal{F}\otimes\mathcal{G}$ generated?
Why does finiteness of $\nu(Y)$ help show that the section-measure function $x\mapsto\nu(E_x)$ is measurable under complements?
How does the Monotone Convergence Theorem establish countable additivity in the finite-factor construction?
Why do finite-measure blocks allow the uniqueness proof to use the pi-lambda theorem for finite measures?
What is the product measure of the diagonal in $[0,1]^2$, and what property of its sections gives that value?

What You'll Learn