Products with Counting and Dirac Measures

What Counting and Dirac Factors Measure

Counting measure assigns to a set its number of elements, with $+\infty$ for an infinite set. A Dirac measure concentrates all its mass at one point. Their products are especially concrete: a counting factor records how many discrete coordinates occur, while a Dirac factor restricts attention to its specified point. This makes it possible to compute some product measures directly, without integrating general functions or using the later theory of sections.

For a set $X$, write $\#A$ for the cardinality of $A\subseteq X$, interpreted as a nonnegative integer or $+\infty$ when $A$ is infinite. On a countable set $X$ equipped with its power-set sigma-algebra $\mathcal P(X)$, counting measure is $\#$. It is sigma-finite: $X$ is a countable union of finite-measure singletons. A Dirac measure at $y_0$ on a measurable space $(Y,\mathcal N)$ is defined by $\delta_{y_0}(B)=1$ if $y_0\in B$, and $0$ otherwise. We assume $\{y_0\}\in\mathcal N$; then $\delta_{y_0}$ is a finite measure.

Definition: The product measure of two sigma-finite measure spaces is the measure on their product sigma-algebra whose value on each measurable rectangle $A\times B$ is the product of the factor measures, with nonnegative extended-real multiplication understood to satisfy $0\cdot\infty=0$. In the cases below the factors are sigma-finite, so the product measure exists and is unique.

The product formulas say exactly which coordinates survive. In $X\times Y$, the measure $\#\times\delta_{y_0}$ is concentrated on $X\times\{y_0\}$; the measure $\delta_{x_0}\times\#$ is concentrated on $\{x_0\}\times Y$. A singleton rectangle has measure zero unless its Dirac coordinate is the selected point. When it is selected, the counting coordinate determines whether its measure is zero or one.

Formulas on Rectangles and Finite Unions

Theorem: Let $X$ be countable with sigma-algebra $\mathcal P(X)$ and counting measure $\#$. Let $(Y,\mathcal N)$ be a measurable space, and let $y_0\in Y$ have measurable singleton. On $\mathcal P(X)\otimes\mathcal N$, $$ (\#\times\delta_{y_0})(A\times B)= \begin{cases} \#A,&y_0\in B,\\ 0,&y_0\notin B. \end{cases} $$ In particular, for a finite union $E=\bigcup_{i=1}^r(A_i\times B_i)$ of measurable rectangles, $$ (\#\times\delta_{y_0})(E)=\#\left(\bigcup_{\{i:\,y_0\in B_i\}}A_i\right). $$ The union over no indices is empty and has measure zero; the cardinality on the right may be infinite.

Proof. Define $j:X\to X\times Y$ by $j(x)=(x,y_0)$. This map is measurable: the preimage of a generating rectangle $A\times B$ is $A$ when $y_0\in B$, and is empty when $y_0\notin B$. Define a measure $\lambda$ on the product sigma-algebra by $\lambda(E)=\#(j^{-1}(E))$. The preimage operation preserves the empty set and disjoint countable unions, and counting measure is countably additive, so $\lambda$ is a measure. On a rectangle, $j^{-1}(A\times B)=A$ if $y_0\in B$, and is empty otherwise. Thus $\lambda(A\times B)=\#A\,\delta_{y_0}(B)$. Since both factors are sigma-finite, uniqueness of the sigma-finite product measure gives $\lambda=\#\times\delta_{y_0}$. For the finite union, its preimage under $j$ is exactly $\bigcup_{\{i:\,y_0\in B_i\}}A_i$: a point $x$ belongs to that preimage precisely when $(x,y_0)$ lies in at least one rectangle. Taking its counting measure proves the formula, including the empty-union case. $\square$

Interchanging the coordinates gives the corresponding formula for $\delta_{x_0}\times\#$, when $X$ is a countable discrete space and the other coordinate carries counting measure. More generally, a Dirac measure times a Dirac measure is a point mass: $(\delta_{x_0}\times\delta_{y_0})(E)$ is $1$ if $(x_0,y_0)\in E$, and $0$ otherwise. These formulas follow from the same pushforward construction, or from the rectangle values and uniqueness.

Worked Example: A Finite Union Under Counting Times Dirac

Let $X=\mathbb N=\{1,2,\ldots\}$ with counting measure, and let $Y=\{0,1,2\}$ with its power-set sigma-algebra. Use $\#\times\delta_0$ to measure $E=(\{1,2\}\times\{0,1\})\cup(\{2,3\}\times\{0,2\})$.

Both second-coordinate sets contain $0$, so both first-coordinate sets contribute. The contributing union is $\{1,2\}\cup\{2,3\}=\{1,2,3\}$. The overlap at $2$ is counted once, not twice. Therefore $$ (\#\times\delta_0)(E)=\#\{1,2,3\}=3. $$ As a direct check, $E\cap(\mathbb N\times\{0\})=\{(1,0),(2,0),(3,0)\}$, and each of these three distinct points has measure one under the product. Points in $E$ with second coordinate $1$ or $2$ have measure zero under this product.

Worked Example: Singleton Rectangles and an Infinite Value

On $\mathbb N\times\{a,b\}$, use $\#\times\delta_a$. For the singleton rectangle $\{4\}\times\{a\}$, the first factor has measure $1$ and the second factor has measure $1$, so the product measure is $1$. For $\{4\}\times\{b\}$, the second factor has measure $0$, so the product measure is $0$.

Now let $E=(\mathbb N\times\{a\})\cup(\{2,5\}\times\{b\})$. Only the first rectangle meets the support $\mathbb N\times\{a\}$. The formula for a finite union therefore gives $$ (\#\times\delta_a)(E)=\#\mathbb N=+\infty. $$ For comparison, the rectangle $\mathbb N\times\{b\}$ has measure zero: its intersection with $\mathbb N\times\{a\}$ is empty. Thus the product measure can be infinite on a set containing infinitely many supported points, yet zero on an infinite rectangle lying entirely away from the Dirac point. The formulas use strict membership: $a\in\{a\}$, whereas $a\notin\{b\}$.

Products of Counting Measures

Counting measure also combines simply with counting measure when both underlying sets are countable and discrete. Every singleton $\{(x,y)\}$ is the rectangle $\{x\}\times\{y\}$, so it has product measure $1$. More generally, the product measure of any subset of $X\times Y$ is its cardinality, with $+\infty$ for an infinite subset. To justify this, the product of the two counting measures is sigma-finite and agrees on singleton rectangles with counting measure on $X\times Y$. Uniqueness of sigma-finite product measures identifies them.

Worked Example: A Finite Union in a Counting Product

Take $X=Y=\mathbb N$, both with counting measure, and set $F=(\{1,2\}\times\{1,2\})\cup(\{2,3\}\times\{2,3\})$. The first rectangle contains four points: $(1,1),(1,2),(2,1),(2,2)$. The second contains four points: $(2,2),(2,3),(3,2),(3,3)$. Their intersection is the single point $(2,2)$. Hence the union has $4+4-1=7$ points, and $$ (\#\times\#)(F)=7. $$ Checking by listing the union gives $(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)$, again seven distinct points. The subtraction is necessary because the overlapping point belongs to both rectangles but only once to the union.

For a countably infinite set $D\subseteq\mathbb N\times\mathbb N$, the same singleton description gives $(\#\times\#)(D)=+\infty$. For example, the diagonal $D=\{(n,n):n\in\mathbb N\}$ is a countable union of distinct singleton rectangles. Its measure is infinite, not zero: each singleton has measure one, and countable additivity gives the sum $\sum_{n=1}^\infty 1=+\infty$.

Common Pitfall: Multiplying Zero by Infinity

A tempting but unsafe calculation is to write the measure of $\mathbb N\times\{b\}$ under $\#\times\delta_a$ as “$\infty\cdot0$” and leave it unresolved. The direct rectangle formula settles the matter: since $a\notin\{b\}$, the product measure is $0$. Under the standard convention for nonnegative extended-real multiplication, $0\cdot\infty=0$, but the support-based explanation is more informative: none of the points in that rectangle lies on the Dirac support.

In contrast, $(\#\times\delta_a)(\mathbb N\times\{a\})=+\infty$, because every $n\in\mathbb N$ contributes one point $(n,a)$. Confusing these cases can lead to the false conclusion that every rectangle with an infinite counting factor has infinite measure. The Dirac factor must be checked first: if it vanishes on the rectangle, the product measure is zero.

The sigma-finiteness assumptions also matter when using uniqueness. Counting measure on a countable set is sigma-finite, but counting measure on an uncountable set is not: every finite-measure set is finite, and countably many finite sets cannot cover an uncountable set. The direct support construction for counting times a Dirac measure still defines a measure in many such settings, but the sigma-finite uniqueness argument above is not available. The formulas in this tutorial rely on the stated countable discrete setting for their product-measure identification.

Check Your Understanding

For each problem, identify which points lie on the Dirac support before counting.

Under $\#\times\delta_3$ on $\mathbb N\times\mathbb N$, find the measure of $\{2,4\}\times\{3,5\}$.
Under $\#\times\delta_0$, find the measure of $(\{1,2\}\times\{0,1\})\cup(\{2,3\}\times\{2\})$.
Under $\#\times\delta_a$, find the measure of $\mathbb N\times\{b\}$ when $a\ne b$, and state why it is not infinite.
Under $\#\times\#$ on $\mathbb N\times\mathbb N$, find the measure of $\{(1,1),(1,2),(2,1)\}$.
Explain why a finite union formula counts a point in overlapping rectangles only once.

Products with Counting and Dirac Measures

What You'll Learn