What a Complex Measure Records
A nonnegative measure assigns a nonnegative extended real number to each measurable set. A complex measure instead assigns a complex number, so its values can encode both real and imaginary contributions and can cancel. The definition still requires countable additivity: the value on a disjoint union must be the sum of the values on its pieces. This tutorial develops that definition and checks it directly for point masses, their finite linear combinations, and integrable complex densities.
Let \((X,\mathcal{M})\) be a measurable space. No measure on \(X\) is needed to define a complex measure. We use the convention that a countable sum of complex numbers is required to converge absolutely in the countable-additivity condition. When a reference measure \(\mu\) is introduced for a density, it will be specified separately. For complex-valued integrable functions, the integral is defined by integrating real and imaginary parts; the standard inequalities and convergence theorems for the Lebesgue integral apply to these complex integrals.
Here and below, “pairwise disjoint” means \(E_j\cap E_k=\varnothing\) whenever \(j\ne k\). Absolute convergence matters because complex terms can cancel: merely having a particular ordering of partial sums converge would not by itself provide the robust countable-additivity condition used here.
Proof. Let \(A,B\in\mathcal{M}\) be disjoint. Apply countable additivity to the sequence \(E_1=A\), \(E_2=B\), and \(E_n=\varnothing\) for \(n\ge3\). Since \(\nu(\varnothing)=0\), the resulting sum is \(\nu(A)+\nu(B)\), and therefore \(\nu(A\cup B)=\nu(A)+\nu(B)\). For the complement identity, \(E\) and \(E^c\) are disjoint and have union \(X\). Finite additivity gives \(\nu(X)=\nu(E)+\nu(E^c)\), which rearranges to the asserted formula. \(\square\)
Dirac Measures and Finite Atomic Measures
For any measurable space \((X,\mathcal{M})\) and any point \(x\in X\), define \(\delta_x:\mathcal{M}\to\mathbb{C}\) by \(\delta_x(E)=\chi_E(x)\). This definition does not require the singleton \(\{x\}\) to be measurable. If \((E_n)\) is a pairwise disjoint sequence, \(x\) belongs to at most one \(E_n\). Thus \(\sum_n\chi_{E_n}(x)\) is either zero or one, and it equals \(\chi_{\cup_n E_n}(x)\). The series is absolutely convergent, so \(\delta_x\) is a complex measure (in fact, a nonnegative measure).
Worked Example: A Dirac Measure Without a Measurable Singleton
Take \(X=\{a,b\}\) and \(\mathcal{M}=\{\varnothing,X\}\), the sigma-algebra consisting only of the empty set and the whole space. The singleton \(\{a\}\) is not in \(\mathcal{M}\). Nevertheless, for \(x=a\), the formula \(\delta_a(E)=\chi_E(a)\) is defined on every \(E\in\mathcal{M}\).
Its values are \(\delta_a(\varnothing)=0\) and \(\delta_a(X)=1\). To check countable additivity, consider any pairwise disjoint sequence in \(\mathcal{M}\). If one of its sets is \(X\), all other sets must be empty, and the sum of their \(\delta_a\)-values is \(1\), equal to the value on their union \(X\). If none is \(X\), every set is empty, and both the sum and the value on the union are zero. Thus \(\delta_a\) is a measure on this measurable space, although its defining point singleton is not measurable.
A finite linear combination of Dirac measures gives a basic finite atomic complex measure. For points \(x_1,\ldots,x_N\in X\) and coefficients \(c_1,\ldots,c_N\in\mathbb{C}\), define \(\nu(E)=\sum_{j=1}^N c_j\delta_{x_j}(E)\). Repeated points are allowed; their coefficients simply contribute separately in this formula. The coefficients need not be positive, real, or nonzero.
Proof. Since \(\chi_{\varnothing}(x_j)=0\) for every \(j\), \(\nu(\varnothing)=0\). Let \((E_n)\) be pairwise disjoint measurable sets. For each \(j\), the point \(x_j\) belongs to at most one \(E_n\), so \(\sum_n\chi_{E_n}(x_j)=\chi_{\cup_n E_n}(x_j)\). Moreover, \[ \sum_{n=1}^{\infty}|\nu(E_n)| \le \sum_{n=1}^{\infty}\sum_{j=1}^N |c_j|\chi_{E_n}(x_j) =\sum_{j=1}^N |c_j|\sum_{n=1}^{\infty}\chi_{E_n}(x_j) \le\sum_{j=1}^N|c_j|<\infty. \] The interchange of sums here is only between a finite sum and a nonnegative sum; the displayed bound also proves absolute convergence. Using the pointwise indicator identity and the finite sum over \(j\), \[ \sum_{n=1}^{\infty}\nu(E_n) =\sum_{j=1}^N c_j\sum_{n=1}^{\infty}\chi_{E_n}(x_j) =\sum_{j=1}^N c_j\chi_{\cup_n E_n}(x_j) =\nu\!\left(\bigcup_{n=1}^{\infty}E_n\right). \] Both requirements of the definition hold. \(\square\)
Worked Example: Evaluating a Finite Atomic Measure
On \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), let \(\nu=(2+i)\delta_0+(-1+2i)\delta_1+(3-i)\delta_2\). Compute \(\nu([0,1])\), \(\nu((1,2])\), and \(\nu([0,2])\). Membership at each endpoint is determined by the indicated closed or open interval.
The interval \([0,1]\) contains \(0\) and \(1\), but not \(2\). Hence \[ \nu([0,1])=(2+i)+(-1+2i)=1+3i. \] The interval \((1,2]\) excludes \(1\), includes \(2\), and excludes \(0\), so \[ \nu((1,2])=3-i. \] The interval \([0,2]\) contains all three atoms, and therefore \[ \nu([0,2])=(2+i)+(-1+2i)+(3-i)=4+2i. \] Check: The two sets \([0,1]\) and \((1,2]\) are disjoint and their union is \([0,2]\). Their values add to \((1+3i)+(3-i)=4+2i\), agreeing with the direct calculation.
Measures from Integrable Complex Densities
Let \((X,\mathcal{M},\mu)\) be a measure space, and let \(g:X\to\mathbb{C}\) be measurable with \(\int_X |g|\,d\mu<\infty\). The condition \(g\in L^1(\mu)\) means precisely that \(g\) is measurable and this integral is finite. Define \(\nu(E)=\int_E g\,d\mu\) for \(E\in\mathcal{M}\). These set values are finite, since \(\left|\int_Eg\,d\mu\right|\le\int_E|g|\,d\mu\le\int_X|g|\,d\mu\). The next result verifies countable additivity rather than presuming it.
Proof. The integral over the empty set is zero, so \(\nu(\varnothing)=0\). Let \((E_n)\) be pairwise disjoint measurable sets, and put \(U=\bigcup_{n=1}^{\infty}E_n\). For each \(N\), \[ \sum_{n=1}^{N}|\nu(E_n)| \le \sum_{n=1}^{N}\int_{E_n}|g|\,d\mu =\int_{\cup_{n=1}^{N}E_n}|g|\,d\mu \le\int_X|g|\,d\mu. \] The equality uses finite additivity of the integral over disjoint sets. The partial sums of the nonnegative series \(\sum_n|\nu(E_n)|\) are increasing and bounded above by the finite number \(\int_X|g|\,d\mu\); hence the series converges, and \(\sum_n\nu(E_n)\) converges absolutely.
To identify its sum, set \(U_N=\bigcup_{n=1}^N E_n\). The indicators \(\chi_{U_N}\) converge pointwise to \(\chi_U\), and \(|g\chi_{U_N}|\le |g|\) everywhere, with \(|g|\) integrable by assumption. The Dominated Convergence Theorem, applied to the real and imaginary parts (each bounded in absolute value by \(|g|\)), gives \[ \nu(U)=\int_X g\chi_U\,d\mu =\lim_{N\to\infty}\int_X g\chi_{U_N}\,d\mu =\lim_{N\to\infty}\sum_{n=1}^N\int_{E_n}g\,d\mu =\sum_{n=1}^{\infty}\nu(E_n). \] The last limit is the sum of the already established absolutely convergent series. Thus \(\nu\) satisfies countable additivity and is a complex measure. \(\square\)
Worked Example: A Complex Density on a Finite Interval
Let \(X=[0,2]\), let \(\mathcal{M}\) be the Lebesgue-measurable subsets of \(X\), and let \(\mu\) be Lebesgue measure restricted to \(X\). This is a finite measure space, with \(\mu(X)=2\). Set \(g(x)=1+ix\). It is measurable and \(\int_X|g|\,d\mu\le\int_0^2(1+x)\,dx=4<\infty\), since \(|1+ix|=\sqrt{1+x^2}\le1+x\) for \(x\ge0\). Thus \(\nu(E)=\int_E(1+ix)\,dx\) is a complex measure by the theorem.
For \(E=[0,1]\), integrating the real and imaginary parts gives \[ \nu([0,1])=\int_0^1 1\,dx+i\int_0^1x\,dx=1+\frac{i}{2}. \] For \(F=(1,2]\), the endpoint \(1\) has Lebesgue measure zero, so the integral is the same as over \([1,2]\): \[ \nu((1,2])=(2-1)+i\frac{2^2-1^2}{2}=1+\frac{3i}{2}. \] The whole interval has value \[ \nu([0,2])=2+i\frac{2^2}{2}=2+2i. \] Check: The disjoint sets \([0,1]\) and \((1,2]\) partition \(X\), and their values sum to \((1+\frac{i}{2})+(1+\frac{3i}{2})=2+2i\), as computed directly. Also, the bound for integrability used the actual interval: \(\int_0^2(1+x)\,dx=2+2=4\).
How to Read These Examples—and a Common Pitfall
For a density measure, the set value is an integral over that set, not necessarily a nonnegative number and not generally determined by the set's measure alone. For an atomic measure, only the atoms that belong to the set contribute. In either construction, the set value can be zero even when the set is nonempty, because contributions can cancel.
A common mistake is to apply a property of nonnegative measures—such as monotonicity—to a complex measure. For example, with \(\nu=\delta_0-\delta_1\) on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), one has \(\nu(\{0\})=1\), \(\nu(\{1\})=-1\), and \(\nu(\{0,1\})=0\). A mistaken calculation might claim \(\nu(\{0,1\})=2\) by adding the magnitudes of the two coefficients. The correction is to add the complex coefficients with their signs: \(1+(-1)=0\). Absolute values are used to test absolute convergence in the definition, not to replace the actual set values.
Another potential confusion concerns Dirac measures: measurability of \(\{x\}\) is not required for \(\delta_x(E)=\chi_E(x)\) to define a measure on \((X,\mathcal{M})\). The definition asks only for values on sets in \(\mathcal{M}\), and the disjoint-union identity follows by checking whether the point belongs to one of those sets. In contrast, when the space is \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\), singletons are Borel sets, but that fact is not needed for the construction.
Check Your Understanding
For each question, distinguish the set function's complex value from its absolute value, and check the hypotheses of any result you use.
- On an arbitrary measurable space \((X,\mathcal{M})\), prove directly that \(\delta_x(E)=\chi_E(x)\) is countably additive, and decide whether \(\{x\}\in\mathcal{M}\) is necessary.
- Let \(\nu=2\delta_a-i\delta_b\), where \(a\ne b\). Compute \(\nu(\{a\})\), \(\nu(\{b\})\), and \(\nu(\{a,b\})\), assuming these sets are measurable.
- For a complex measure \(\nu\), derive \(\nu(E^c)=\nu(X)-\nu(E)\) from finite additivity.
- On \([0,1]\) with Lebesgue measure, let \(g(x)=2+ix\). Compute \(\int_{[0,1]}g\,dm\) by integrating real and imaginary parts.
- In the density theorem, identify the integrable dominating function used to pass from the integrals over \(U_N\) to the integral over \(U\).