From Real Densities to a Complex Density
A complex measure can have cancellation that is invisible if we look only at the size of its value on the whole space. The right object for recording its size on every measurable set is its total variation. For measures absolutely continuous with respect to a positive measure, the Radon–Nikodym theorem identifies both the complex measure and its variation by densities.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space with \(\mu\) positive and sigma-finite; it need not be finite or complete. A complex measure \(\nu\) is a countably additive, complex-valued set function on \(\mathcal{M}\). We assume it has finite total variation, meaning \(|\nu|(X)<\infty\). For \(E\in\mathcal{M}\), its variation is defined by
Here the supremum is over finite measurable partitions of \(E\), including the trivial partition when \(E\) is empty. We write \(\nu\ll\mu\) when \(\nu(E)=0\) for every \(E\in\mathcal{M}\) with \(\mu(E)=0\). We use the standard Radon–Nikodym theorem: a finite signed measure absolutely continuous with respect to a sigma-finite positive measure has an integrable real-valued density, unique \(\mu\)-a.e. The theorem applies to the real and imaginary parts of \(\nu\).
The integrals of complex-valued functions here are defined by integrating their real and imaginary parts. Equalities of measures in the theorem hold for every measurable set; uniqueness of the density is only up to a \(\mu\)-null set.
Proof. Define the real signed measures \(\nu_R(E)=\operatorname{Re}\nu(E)\) and \(\nu_I(E)=\operatorname{Im}\nu(E)\). Countable additivity follows from that of \(\nu\). For any finite measurable partition \((E_j)\), \(\sum_j|\nu_R(E_j)|\le\sum_j|\nu(E_j)|\), and likewise for \(\nu_I\). Thus both signed measures have finite variation. If \(\mu(E)=0\), then \(\nu(E)=0\), so \(\nu_R(E)=\nu_I(E)=0\). The Radon–Nikodym theorem therefore gives real-valued \(u,v\in L^1(\mu)\) such that \(\nu_R(E)=\int_Eu\,d\mu\) and \(\nu_I(E)=\int_Ev\,d\mu\) for every \(E\). Set \(h=u+iv\). Then \(\nu(E)=\int_Eh\,d\mu\). Also, \(|h|\le |u|+|v|\), so \(h\in L^1(\mu)\).
It remains to prove the variation formula. Fix \(E\in\mathcal{M}\). For any finite measurable partition \((E_j)\) of \(E\), the triangle inequality for complex integrals gives $$ \sum_j|\nu(E_j)|=\sum_j\left|\int_{E_j}h\,d\mu\right| \le \sum_j\int_{E_j}|h|\,d\mu =\int_E|h|\,d\mu. $$ Taking the supremum over partitions proves \(|\nu|(E)\le\int_E|h|\,d\mu\).
For the reverse inequality, fix \(\varepsilon\in(0,1)\). On \(\{h\ne0\}\), the phase \(h/|h|\) takes values on the unit circle. Partition that circle into finitely many measurable arcs, each small enough that every phase in an arc differs in angle from a chosen unit number \(q_j\) by an amount whose cosine is at least \(1-\varepsilon\). The inverse images of these arcs, intersected with \(E\cap\{h\ne0\}\), form finitely many measurable sets \(E_j\); include \(E\cap\{h=0\}\) as one more cell if it is nonempty. On each nonzero cell, $$ \left|\int_{E_j}h\,d\mu\right| \ge \operatorname{Re}\left(\overline{q_j}\int_{E_j}h\,d\mu\right) =\int_{E_j}\operatorname{Re}(\overline{q_j}h)\,d\mu \ge (1-\varepsilon)\int_{E_j}|h|\,d\mu. $$ The zero cell contributes zero to both sides. Summing over this partition gives \(|\nu|(E)\ge(1-\varepsilon)\int_E|h|\,d\mu\). Letting \(\varepsilon\) decrease to zero proves the reverse inequality, including when the integral is zero. Hence \(|\nu|(E)=\int_E|h|\,d\mu\) for every \(E\).
If two integrable complex-valued functions give the same measure, their real parts give the same Radon–Nikodym derivative of \(\nu_R\), and their imaginary parts give the same derivative of \(\nu_I\). Uniqueness in the real Radon–Nikodym theorem makes both parts equal \(\mu\)-a.e.; therefore the complex densities are equal \(\mu\)-a.e.
Conversely, every complex-valued \(h\in L^1(\mu)\) defines a finite complex measure by \(\nu(E)=\int_Eh\,d\mu\). To check countable additivity, let \((E_n)\) be disjoint and put \(A_N=\bigcup_{n=1}^N E_n\), \(A=\bigcup_{n=1}^\infty E_n\). Then \(h\chi_{A_N}\to h\chi_A\) pointwise and \(|h\chi_{A_N}|\le|h|\), with \(|h|\) integrable. The Dominated Convergence Theorem, applied to the real and imaginary parts, gives \(\nu(A_N)\to\nu(A)\). The variation formula just proved gives \(|\nu|(X)=\int_X|h|\,d\mu<\infty\), and \(\nu\ll\mu\) because an integral over a \(\mu\)-null set is zero. This proves the theorem and its converse. \(\square\)
Working with the Density
Worked Example: A Complex Measure on a Finite Space
Let \(X=\{a,b,c\}\), let \(\mathcal{M}\) be all subsets of \(X\), and give the points masses \(\mu(\{a\})=2\), \(\mu(\{b\})=1\), and \(\mu(\{c\})=3\). Define \(h(a)=1+i\), \(h(b)=-2i\), and \(h(c)=0\). Compute the measure of \(X\) and its total variation.
The density formula gives \(\nu(\{a\})=2(1+i)=2+2i\), \(\nu(\{b\})=1(-2i)=-2i\), and \(\nu(\{c\})=3(0)=0\). Hence \(\nu(X)=(2+2i)+(-2i)+0=2\), so \(|\nu(X)|=2\). The variation formula instead gives $$ |\nu|(X)=\int_X|h|\,d\mu =2|1+i|+1|-2i|+3|0| =2\sqrt{2}+2. $$ As a direct partition check, the singleton partition contributes \(|2+2i|+|-2i|+|0|=2\sqrt{2}+2\). The partition \(\{a,c\},\{b\}\) contributes \(|2+2i|+|-2i|=2\sqrt{2}+2\) as well. Thus the variation is larger than \(|\nu(X)|\), as cancellation between the atoms \(a\) and \(b\) reduces the value on the whole space. Check: on \(\{a,c\}\), the density integral is \(2|1+i|+3|0|=2\sqrt{2}\), and \(\nu(\{a,c\})=2+2i\), whose modulus is \(2\sqrt{2}\).
Worked Example: An Oscillating Density on the Unit Interval
Take \(X=[0,1]\), \(\mathcal{M}\) the Lebesgue sigma-algebra, \(\mu=m\), and \(h(x)=e^{2\pi ix}\). This is integrable because \(|h(x)|=1\) on an interval of measure \(1\). Compute \(\nu([0,1])\) and \(|\nu|([0,1])\).
Using the antiderivative \(e^{2\pi ix}/(2\pi i)\), with \(e^{2\pi i}=1\), gives $$ \nu([0,1])=\int_0^1 e^{2\pi ix}\,dx =\frac{e^{2\pi i}-1}{2\pi i}=0. $$ But \(|\nu|([0,1])=\int_0^1|e^{2\pi ix}|\,dx=\int_0^1 1\,dx=1\). To see the cancellation in an explicit partition, use \([0,\tfrac12)\) and \([\tfrac12,1]\). Since \(e^{\pi i}=-1\), the integrals on these cells are respectively \(\frac{-1-1}{2\pi i}=\frac{i}{\pi}\) and \(\frac{1-(-1)}{2\pi i}=-\frac{i}{\pi}\). Their absolute values sum to \(2/\pi\), while the total variation is \(1\), the supremum over all finite partitions. Check: the two cell integrals sum to \(i/\pi-i/\pi=0\), agreeing with \(\nu([0,1])\). The density formula gives variation \(1\), not the value \(0\) obtained by taking the modulus after integrating over the entire interval.
Worked Example: A Piecewise Constant Density on the Real Line
Let \(\mu=m\) on \(\mathbb{R}\), and set \(h=2\chi_{[0,1)}-i\chi_{[1,2]}\). This density is integrable: its modulus is \(2\) on a set of measure \(1\), \(1\) on a disjoint set of measure \(1\), and zero elsewhere. Find the measure and variation of \([0,2]\).
Integrating on the two pieces gives \(\nu([0,1))=2\cdot1=2\) and \(\nu([1,2])=(-i)\cdot1=-i\). Thus \(\nu([0,2])=2-i\), with \(|\nu([0,2])|=\sqrt{2^2+(-1)^2}=\sqrt{5}\). In contrast, the variation formula gives $$ |\nu|([0,2])=\int_{[0,2]}|h|\,dm=2\cdot1+1\cdot1=3. $$ The partition \([0,1),[1,2]\) directly yields \(|2|+|-i|=2+1=3\). Check: the single-set partition gives only \(|2-i|=\sqrt{5}\le3\), consistent with the definition of variation as a supremum over partitions.
What the Hypotheses Prevent
The absolute continuity assumption is essential. For Lebesgue measure \(m\) on \(\mathbb{R}\), let \(\nu=\delta_0\), the Dirac measure at zero. Since \(m(\{0\})=0\) but \(\delta_0(\{0\})=1\), we have \(\nu\not\ll m\). No density \(h\) with \(\delta_0(E)=\int_Eh\,dm\) for every measurable \(E\) can exist: for \(E=\{0\}\), every such integral is zero. This example also has finite variation, so finite variation alone does not give a density.
The sigma-finiteness of the positive reference measure is another hypothesis of the Radon–Nikodym theorem being used here. No assumption that \(\mu(X)<\infty\) is needed: the proof requires an integrable density, not a finite reference measure. Nor must \(\mu\) be complete. The density is measurable with respect to \(\mathcal{M}\), and its uniqueness is stated \(\mu\)-a.e., not pointwise.
Common mistake. It is incorrect to compute variation by taking the modulus of the measure of the whole set. In the oscillating example, that gives \(|\nu([0,1])|=|0|=0\), whereas \(|\nu|([0,1])=\int_0^1|h|\,dm=1\). The correction is to take the modulus of the density before integrating; equivalently, variation takes a supremum over measurable partitions, allowing cancellation within a cell to be reduced by refining the partition.
Check Your Understanding
In each question, distinguish the value of the complex measure on a set from the variation of that measure on the same set.
- Why do the real and imaginary parts of a finite complex measure absolutely continuous with respect to a sigma-finite positive measure satisfy the hypotheses of the Radon–Nikodym theorem?
- If \(h=u+iv\) with real-valued \(u,v\in L^1(\mu)\), explain why \(h\in L^1(\mu)\).
- For the density \(h(x)=e^{2\pi ix}\) on \([0,1]\), compute \(\nu([0,1])\) and \(|\nu|([0,1])\), and explain why they differ.
- For \(h=2\chi_{[0,1)}-i\chi_{[1,2]}\) on \(\mathbb{R}\), compute \(\nu([0,2])\) and \(|\nu|([0,2])\).
- Give a null set that shows why the Dirac measure at zero on \(\mathbb{R}\) is not absolutely continuous with respect to Lebesgue measure.