Integration at a Single Atom
A Dirac measure concentrates all its mass at one specified point. Consequently, integrating a simple function against it does not average the function over an interval or compare the sizes of its level sets in the usual geometric way: the integral is determined by the value at that point. For a finite weighted sum of Dirac measures, each atom contributes its weight multiplied by the function’s value there.
Let \((X,\mathcal{M})\) be a measurable space, and fix \(x\in X\). No topology, completeness, or finiteness assumption on a background measure is needed. The Dirac measure at \(x\), denoted by \(\delta_x\), assigns mass according to whether a measurable set contains \(x\). In this tutorial, simple functions are measurable functions \(\varphi:X\to[0,\infty)\) taking finitely many finite values. Their integrals are defined using a disjoint measurable representation: if \(\varphi=\sum_{i=1}^r a_i\chi_{E_i}\), where \(a_i\ge0\) and the sets \(E_i\in\mathcal{M}\) are disjoint, then \(\int\varphi\,d\mu=\sum_{i=1}^r a_i\mu(E_i)\).
To see why this is a measure, first note that \(\delta_x(\varnothing)=0\). If \(E_1,E_2,\ldots\) are pairwise disjoint measurable sets, at most one of them contains \(x\). If none contains \(x\), both \(\delta_x(\bigcup_n E_n)\) and \(\sum_n\delta_x(E_n)\) are zero. If \(x\in E_k\) for one index \(k\), the union contains \(x\), and exactly one term of the sum is one; both sides are one. Thus \(\delta_x\) is countably additive. It is not necessary that the singleton \(\{x\}\) belong to \(\mathcal{M}\) for this definition to work. When \(\{x\}\) is measurable, it has \(\delta_x\)-measure one.
The Evaluation Formula
Proof. Write \(\varphi=\sum_{i=1}^r a_i\chi_{E_i}\) with disjoint \(E_i\in\mathcal{M}\) and finite coefficients \(a_i\ge0\). By the definition of the simple-function integral and the definition of \(\delta_x\), $$ \int_X\varphi\,d\delta_x =\sum_{i=1}^r a_i\delta_x(E_i) =\sum_{i=1}^r a_i\chi_{E_i}(x). $$ Because the sets \(E_i\) are disjoint, at most one indicator in the final sum equals one. If \(x\in E_k\), the sum is \(a_k=\varphi(x)\); if \(x\) belongs to none of the listed sets, all its terms are zero, and \(\varphi(x)=0\). The equality follows in both cases. \(\square\)
The formula does not depend on which disjoint representation of \(\varphi\) is chosen: each representation gives the same function value at \(x\). Also, changing \(\varphi\) away from \(x\) cannot affect this integral, while changing its value at \(x\) can. This is a pointwise feature of an atom, not a statement that all measurable functions that agree almost everywhere for some other measure must have the same Dirac integral.
Worked Example: One Atom on a Partition Boundary
On \((\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_0)\), let \(\varphi=3\chi_{(-2,0)}+8\chi_{[0,5)}+4\chi_{[5,\infty)}\). Compute its integral.
The three sets are disjoint and cover \(\mathbb{R}\). The point \(0\) is not in \((-2,0)\), because that interval excludes its right endpoint. It is in \([0,5)\), because that interval includes its left endpoint, and it is not in \([5,\infty)\). Thus \(\varphi(0)=8\). Equivalently, the three Dirac masses of the partition sets are respectively \(0,1,0\), so the simple-function definition gives $$ \int_{\mathbb{R}}\varphi\,d\delta_0 =3\cdot0+8\cdot1+4\cdot0 =8. $$ Check: the evaluation formula gives \(\varphi(0)=8\), agreeing with the partition calculation.
Finite Weighted Combinations of Atoms
Let \(x_1,\ldots,x_N\in X\) and let \(c_1,\ldots,c_N\) be finite nonnegative real numbers. Define a set function \(\mu\) on \(\mathcal{M}\) by $$ \mu(E)=\sum_{j=1}^N c_j\delta_{x_j}(E) =\sum_{j=1}^N c_j\chi_E(x_j). $$ This is a measure: for pairwise disjoint sets, countable additivity follows by applying the countable additivity of each \(\delta_{x_j}\) and summing over the finitely many indices \(j\). Its total mass is \(\mu(X)=\sum_{j=1}^N c_j\), so it is finite and sigma-finite. It is a probability measure exactly when \(\sum_j c_j=1\). Zero weights are allowed. Points in the list need not be distinct; repeated points represent repeated contributions to the same location.
Proof. First suppose \(N\ge1\). List the distinct values taken by \(\varphi\) as \(b_1,\ldots,b_s\), including zero if it is attained, and set \(A_\ell=\{y\in X:\varphi(y)=b_\ell\}\). Each \(A_\ell\) is measurable, the sets are disjoint, and their union is \(X\). By the definition of the simple-function integral, $$ \int_X\varphi\,d\mu =\sum_{\ell=1}^s b_\ell\mu(A_\ell) =\sum_{\ell=1}^s b_\ell\sum_{j=1}^N c_j\chi_{A_\ell}(x_j). $$ All sums here are finite, so they can be regrouped: $$ \int_X\varphi\,d\mu =\sum_{j=1}^N c_j\sum_{\ell=1}^s b_\ell\chi_{A_\ell}(x_j). $$ For each \(j\), exactly one set \(A_\ell\) contains \(x_j\). Therefore the inner sum equals \(\varphi(x_j)\), proving the formula. If \(N=0\), then \(\mu(E)=0\) for every \(E\in\mathcal{M}\), so the simple-function integral is zero by its defining finite sum; the empty sum on the right is also zero. \(\square\)
The proposition also explains how to deal with repeated atoms. If \(x_j=x_k\), their two contributions are \(c_j\varphi(x_j)\) and \(c_k\varphi(x_j)\), whose sum is \((c_j+c_k)\varphi(x_j)\). Thus the two weights can be combined at that point. A zero weight contributes nothing, regardless of the function value there.
Worked Example: Three Weighted Atoms and a Simple Function
On \((\mathbb{R},\mathcal{B}_{\mathbb{R}},\mu)\), take \(\mu=3\delta_0+\tfrac12\delta_1+2\delta_3\) and \(\varphi=1\chi_{[-1,2)}+6\chi_{[2,4]}+0\chi_{\mathbb{R}\setminus[-1,4]}\). Calculate \(\int\varphi\,d\mu\).
The sets \([-1,2)\), \([2,4]\), and \(\mathbb{R}\setminus[-1,4]\) are disjoint and cover \(\mathbb{R}\). The atoms \(0\) and \(1\) lie in \([-1,2)\), so \(\varphi(0)=1\) and \(\varphi(1)=1\). The atom \(3\) lies in \([2,4]\), so \(\varphi(3)=6\). The weighted evaluation formula gives $$ \int\varphi\,d\mu =3\cdot1+\frac12\cdot1+2\cdot6 =3+\frac12+12 =\frac{31}{2}. $$ For a check using the level sets, \(\mu([-1,2))=3+\tfrac12=\tfrac72\), \(\mu([2,4])=2\), and the remaining set has measure zero. Hence $$ \int\varphi\,d\mu =1\cdot\frac72+6\cdot2+0\cdot0 =\frac72+12 =\frac{31}{2}, $$ as before.
Worked Example: Combining Repeated Atoms
Let \(\nu=2\delta_{-1}+\delta_2+3\delta_{-1}+0\delta_5\) on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), and let \(\psi=9\chi_{(-\infty,0)}+4\chi_{[0,3]}+0\chi_{\mathbb{R}\setminus(-\infty,3]}\). Find its integral.
The sets in the definition of \(\psi\) are disjoint and cover \(\mathbb{R}\). The values at the listed points are \(\psi(-1)=9\), \(\psi(2)=4\), and \(\psi(5)=0\). Applying the formula to the original list of atoms, $$ \int\psi\,d\nu =2\cdot9+1\cdot4+3\cdot9+0\cdot0 =18+4+27+0 =49. $$ Combining the repeated atoms first gives \(\nu=5\delta_{-1}+\delta_2\), with the zero-weight term omitted. This gives \(5\cdot9+1\cdot4=45+4=49\). As a level-set check, \(\nu((-\infty,0))=2+3=5\), \(\nu([0,3])=1\), and the remaining set has \(\nu\)-measure zero; therefore the integral is \(9\cdot5+4\cdot1+0=49\).
A Common Confusion: Dirac Mass Is Not Interval Length
A frequent error is to calculate a Dirac integral as though the measure were Lebesgue measure. For example, on \(\mathbb{R}\) let \(\varphi=3\chi_{[0,1]}+5\chi_{(1,3]}\) and integrate against \(\delta_2\). A wrong calculation treats the two coefficients as heights over intervals and uses their lengths, obtaining \(3\cdot1+5\cdot2=13\). Those lengths are Lebesgue measures of the intervals, not their \(\delta_2\)-measures.
The correction is to check which partition set contains the atom. Since \(2\in(1,3]\) and \(2\notin[0,1]\), the two Dirac masses of those sets are \(0\) and \(1\). Thus $$ \int\varphi\,d\delta_2=3\cdot0+5\cdot1=5=\varphi(2). $$ The same simple function has Lebesgue integral \(13\), but that is an integral with respect to \(m\), a different measure. The set-membership test must respect endpoint conventions: an atom on a boundary belongs to exactly the partition set whose interval includes that endpoint.
For a finite atomic measure, the practical procedure is therefore to identify the values of the function at the listed atoms, multiply each by its corresponding weight, and add. This works because each atom lies in exactly one level set of a simple function, so the defining sum for the integral reduces to those finitely many weighted evaluations. It does not replace the definition for other measures, where the masses of all relevant level sets must be calculated using that measure.
Check Your Understanding
For each question, specify which partition set contains each atom before calculating the integral.
- On \((\mathbb{R},\mathcal{B}_{\mathbb{R}},\delta_1)\), compute the integral of \(2\chi_{(-\infty,1]}+7\chi_{(1,\infty)}\). Which endpoint convention determines the answer?
- Let \(\mu=2\delta_0+4\delta_2\) and \(\varphi=3\chi_{[-1,1)}+5\chi_{[1,3]}\). Compute \(\int\varphi\,d\mu\) by weighted evaluation and by the level-set definition.
- For \(\nu=\delta_0+3\delta_0\), explain why replacing the two atoms by \(4\delta_0\) leaves every simple-function integral unchanged.
- Give a finite nonnegative simple function whose integral with respect to \(\delta_0\) is zero but whose value at some other point is positive. Verify both facts.
- For a finite measure \(\mu=\sum_{j=1}^N c_j\delta_{x_j}\), state the condition on the weights that makes \(\mu\) a probability measure, and explain how a zero weight affects the integral formula.