Integrals Under Counting and Dirac Measures

Two Measures That Make Integration Concrete

Under counting measure, each point has mass one, so integrating a nonnegative function amounts to summing its values. Under a Dirac measure, all mass is concentrated at one point, so the integral depends only on the value there. These descriptions are useful only when the underlying measure is kept in view: the same function can have very different integrals for counting measure, a Dirac measure, and Lebesgue measure.

For counting measure, use a set \(S\), its power set \(\mathcal{P}(S)\), and the measure \(\#\) defined by \(\#(E)=|E|\) when \(E\) is finite and \(\#(E)=\infty\) when \(E\) is infinite. Thus \((S,\mathcal{P}(S),\#)\) is complete: every subset of every measurable set is measurable. The measure is finite exactly when \(S\) is finite. It is sigma-finite exactly when \(S\) is countable: a countable set is a countable union of singletons, each of measure one; conversely, a countable union of finite-measure sets is a countable union of finite sets and hence countable.

Counting Measure: Integration Is Summation

Proof. First suppose \(f\) is nonnegative. A nonnegative simple function \(\varphi\) takes finitely many finite values. Its integral under counting measure is the sum of its values over \(S\): on each of its finitely many level sets, the integral is the level value times the number of points in that set, with an infinite level set contributing infinity when the level value is positive. Since \(\varphi\le f\), this gives \(\int\varphi\,d\#\le\sum_{s\in S}f(s)\). Taking the supremum over such \(\varphi\) proves \(\int f\,d\#\le\sum_{s\in S}f(s)\).

For the reverse inequality, fix a finite \(F\subseteq S\). For each \(s\in F\), choose a finite \(c_s\) with \(0\le c_s\le f(s)\). The simple function \(\varphi=\sum_{s\in F}c_s\chi_{\{s\}}\) satisfies \(\varphi\le f\), so \(\int f\,d\#\ge\int\varphi\,d\#=\sum_{s\in F}c_s\). Taking the supremum first over all such finite coefficients and then over all finite \(F\) gives \(\int f\,d\#\ge\sum_{s\in S}f(s)\). In particular, if \(f(s)=\infty\) at a point, the finite coefficient there can be made arbitrarily large; no infinite coefficient is used in a simple function. This proves the nonnegative formula.

Now let \(f\) be real-valued. Its positive and negative parts \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\) are nonnegative measurable functions, with \(|f|=f^++f^-\). By the nonnegative formula, \(\int|f|\,d\#=\sum_s|f(s)|\). The definition of integrability for a real-valued function is \(\int|f|\,d\#<\infty\), so this proves the criterion. When it holds, both \(\sum_s f^+(s)\) and \(\sum_s f^-(s)\) are finite, and the definition of the signed integral gives \(\int f\,d\#=\int f^+\,d\#-\int f^-\,d\#=\sum_s f(s)\). The series is absolutely convergent because its absolute-value sum is finite. \(\square\)

Dirac Measure: Integration Is Point Evaluation

On a measurable space \((X,\mathcal{M})\), fix \(x\in X\) and define the Dirac measure \(\delta_x\) by \(\delta_x(E)=1\) if \(x\in E\), and \(\delta_x(E)=0\) otherwise, for \(E\in\mathcal{M}\). This is a measure with \(\delta_x(X)=1\), so \((X,\mathcal{M},\delta_x)\) is finite. No completeness assumption is made; for example, a Dirac measure on the Borel sets need not be complete.

Proof. For a nonnegative simple function \(\varphi\), the definition of its integral gives \(\int\varphi\,d\delta_x=\varphi(x)\): among its disjoint level sets, exactly one contains \(x\), and all other level sets have Dirac measure zero. For nonnegative measurable \(f\), every simple \(\varphi\le f\) therefore satisfies \(\int\varphi\,d\delta_x=\varphi(x)\le f(x)\), giving \(\int f\,d\delta_x\le f(x)\). For every finite \(c\) with \(0\le c\le f(x)\), the simple function \(c\chi_X\) is bounded above by \(f\) at \(x\), but not necessarily elsewhere; instead use \(c\chi_{\{x\}}\) when \(\{x\}\in\mathcal{M}\). More generally, to avoid requiring \(\{x\}\) measurable, use the defining simple approximations to \(f\): the supremum of \(\varphi(x)\) over simple \(\varphi\le f\) equals \(f(x)\), as follows by taking a simple lower approximation to the value \(f(x)\) on the measurable level set \(\{y:f(y)>c\}\), for each finite \(c<f(x)\). The integral definition then gives \(\int f\,d\delta_x\ge c\) for every such \(c\), hence \(\int f\,d\delta_x\ge f(x)\), including \(f(x)=\infty\). Thus equality holds. For real \(f\), \(\int|f|\,d\delta_x=|f(x)|\) by the nonnegative formula. The integrability criterion and signed integral formula follow from the definition of integrability and the positive- and negative-part definition. \(\square\)

Measures with Several Atoms

Let \(x_1,\ldots,x_r\) be distinct points of a measurable space \((X,\mathcal{M})\), and let \(a_1,\ldots,a_r\) be positive finite numbers. Define \(\mu=\sum_{j=1}^{r}a_j\delta_{x_j}\). This is a finite measure: for disjoint measurable sets \(E_k\), each point \(x_j\) belongs to at most one \(E_k\), so \(\mu(\bigcup_k E_k)=\sum_j a_j\delta_{x_j}(\bigcup_kE_k)=\sum_k\sum_j a_j\delta_{x_j}(E_k)=\sum_k\mu(E_k)\); the exchange here is only of a finite sum over \(j\) with a nonnegative series over \(k\). Its total mass is \(\mu(X)=\sum_{j=1}^r a_j\).

Proof. Write \(A=\sum_{j=1}^r a_j\delta_{x_j}\). For a nonnegative simple \(\varphi\), finite additivity of its simple-function integral gives \(\int\varphi\,d\mu=\sum_{j=1}^r a_j\int\varphi\,d\delta_{x_j}=\sum_{j=1}^r a_j\varphi(x_j)\). For general nonnegative \(f\), the integral is the supremum over simple \(0\le\varphi\le f\). This supremum equals \(\sum_j a_j f(x_j)\): the upper bound follows from \(\varphi(x_j)\le f(x_j)\); for the lower bound, the finitely many point values can be approximated simultaneously from below by simple approximations to \(f\), or equivalently by partitioning the range of \(f\) into finitely many measurable level sets and refining the approximation. Taking increasingly fine simple lower approximations gives the stated sum. Finally, \(\int|f|\,d\mu=\sum_j a_j|f(x_j)|\). Since there are finitely many atoms and all \(a_j>0\), this is finite exactly when every \(f(x_j)\) is finite; applying the positive- and negative-part definition yields the signed formula. \(\square\)

Common Pitfalls and Why the Measure Matters

Counting measure on an uncountable set is not sigma-finite. A common incorrect argument says that any infinite set is a countable union of singletons. That only holds for countable sets. If \(S\) is uncountable and \(\#\) were sigma-finite, then \(S=\bigcup_n E_n\) with each \(E_n\) finite, since finite counting measure means finite cardinality. Such a union is countable, a contradiction. Thus counting measure is sigma-finite exactly on countable sets, and is not sigma-finite on an uncountable set.

Do not confuse counting measure with Lebesgue measure. On \(\mathbb{R}\), the rationals \(\mathbb{Q}\) are countably infinite. For counting measure on \((\mathbb{Q},\mathcal{P}(\mathbb{Q}),\#)\), \(\int\chi_{\mathbb{Q}}\,d\#=\sum_{q\in\mathbb{Q}}1=\infty\). For Lebesgue measure \(m\) on \((\mathbb{R},\mathcal{L},m)\), \(\int\chi_{\mathbb{Q}}\,dm=m(\mathbb{Q})=0\), since a countable set has Lebesgue measure zero. The function is the same; the measure is not.

For Dirac measure, check the atom, not the whole function. Under \(\delta_x\), changing a function away from \(x\) does not change its integral. But a non-finite value at \(x\) prevents a real-valued function from being integrable. Under counting measure, by contrast, every point contributes, and integrability requires absolute summability over all points.