Simple-Function Integrals Under Counting Measure

Counting Points in a Simple-Function Integral

For counting measure, integrating a nonnegative function means adding its values over the points of the space. If the space is finite, this is an ordinary finite sum. On a countably infinite space, the sum may be finite or infinite; in particular, infinitely many points at which a simple function has the same strictly positive value force its integral to be infinite. The level sets of a simple function make this distinction precise.

Throughout, let \(X\) be a finite or countable set, with measurable space \((X,\mathcal{P}(X))\), where \(\mathcal{P}(X)\) is the collection of all subsets of \(X\). Let \(\#\) denote counting measure, defined by \(\#E\) equal to the number of points of \(E\) if \(E\) is finite and \(+\infty\) if \(E\) is infinite. Thus \((X,\mathcal{P}(X),\#)\) is a measure space. It is finite exactly when \(X\) is finite; it is sigma-finite because \(X\) is a countable union of singletons, each of measure \(1\). It is complete because its only null set is the empty set.

This is the simple-function integral, using the definition established in the earlier material on simple-function integrals. The omission of the zero level is important: it avoids an expression such as \(0\cdot(+\infty)\), which is not a defined product of extended real numbers. Since every subset of \(X\) is measurable, all the level sets in this definition are measurable.

Summing Values over a Countable Set

For a nonnegative function \(f\) on a countable set, define its sum over \(X\) to be the supremum of its sums over finite subsets: \(\displaystyle \sum_{x\in X}f(x)=\sup\left\{\sum_{x\in F}f(x):F\subseteq X,\ F\text{ finite}\right\}\). This definition applies when \(X\) is finite, countably infinite, or empty, and allows the value \(+\infty\). If \(X\) is enumerated without repetition as \(x_1,x_2,\ldots\), this supremum equals \(\lim_{N\to\infty}\sum_{k=1}^N f(x_k)\): each initial segment is finite, and every finite subset of \(X\) lies in some initial segment. Thus the value does not depend on the chosen enumeration. For finite \(X\), the sum is simply the sum over all its points.

Proof. If \(X=\varnothing\), there are no positive level sets, \(f\) is identically zero on its empty domain, and both the integral and the sum are zero. Now suppose \(X\) is nonempty. For any finite \(F\subseteq X\), the disjoint level sets give \(\displaystyle \sum_{x\in F}f(x)=\sum_{j=1}^r a_j\,\#(F\cap E_j)\). Each term is at most \(a_j\,\#E_j\), so taking the supremum over finite \(F\) gives \(\sum_{x\in X}f(x)\le\sum_{j=1}^r a_j\,\#E_j\). If some positive level set \(E_j\) is infinite, then for every positive integer \(N\) it contains a finite subset of \(N\) points. Choosing those points as \(F\) gives \(\sum_{x\in F}f(x)=Na_j\). These values are unbounded as \(N\) increases, so the sum over \(X\) is \(+\infty\), as is the level-set expression. If every positive level set is finite, their finite union is finite. Taking \(F=\bigcup_{j=1}^rE_j\) gives \(\sum_{x\in F}f(x)=\sum_{j=1}^r a_j\,\#E_j\), so the supremum is at least that expression. Combined with the upper bound, this proves equality. By the definition of the simple-function integral, the level-set expression is \(\int f\,d\#\). \(\square\)

Finite and Infinite Integrals on Countable Sets

A countable domain does not by itself make a simple-function integral infinite: the decisive issue is how many points have positive values. Because a simple function has only finitely many distinct positive values, if even one of its positive level sets is infinite, that level alone makes the integral infinite. If all positive level sets are finite, their finite union is finite, and the integral is a finite sum.

Proof. If the support is finite, it is a finite set of points, and \(f\) has a finite value at each point. Its sum over the support is therefore finite, so the theorem gives \(\int f\,d\#<+\infty\). Conversely, suppose the support is infinite. Since \(f\) is simple, its strictly positive range is a nonempty finite set \(\{a_1,\ldots,a_r\}\). Let \(a=\min\{a_1,\ldots,a_r\}\), so \(a>0\). For every positive integer \(N\), the infinite support contains a finite subset \(F\) of \(N\) points, and \(f(x)\ge a\) for every \(x\in F\). Hence \(\sum_{x\in F}f(x)\ge Na\). These lower bounds are unbounded as \(N\) increases, so the supremum defining the sum over \(X\) is \(+\infty\). The theorem then gives \(\int f\,d\#=+\infty\). If the support is empty, it is finite and \(f=0\) everywhere, so the first direction includes that case. \(\square\)

A Common Counting Pitfall

Do not count the zero level as if it contributed a positive amount. For example, let \(X=\mathbb{N}\), and define \(u(1)=2\) and \(u(n)=0\) for \(n\ge2\). Its zero level set is infinite. A tempting but invalid calculation is \(2\cdot1+0\cdot(+\infty)=+\infty\). The product \(0\cdot(+\infty)\) is undefined, and the conclusion is wrong. The definition sums only over strictly positive levels: here the value \(2\) occurs once, so \(\int u\,d\#=2\cdot1=2\). Directly, every partial sum containing the first term equals \(2\), because all later terms are zero. The support is the singleton \(\{1\}\), so the finiteness criterion gives the same answer.

A related error is to count distinct values rather than points. In the finite-set example, the value \(5\) occurs three times; its contribution is \(5\cdot3\), not merely \(5\). Under counting measure, each point has measure \(1\), and the level-set cardinality records exactly how many times each positive value appears. This is the discrete version of the level-set definition of the simple-function integral.