Polar Decomposition of a Complex Measure

What Polar Decomposition Says

A complex measure can assign values with different phases to different sets, so its total mass may be small even when its total variation is large. Polar decomposition separates these two features: the total variation records the magnitude of the measure, while a measurable complex-valued function records its phase. This is the measure-theoretic analogue of writing a nonzero complex number as its modulus times a unit complex number.

Throughout, $(X,\mathcal{M})$ is a measurable space and $\nu:\mathcal{M}\to\mathbb{C}$ is a countably additive complex measure of finite variation. No completeness assumption on $\mathcal{M}$ is needed. Its total variation $|\nu|$ is the positive measure defined by taking, for each $E\in\mathcal{M}$, the supremum of $\sum_{j=1}^k|\nu(E_j)|$ over all finite measurable partitions $E=\bigcup_{j=1}^k E_j$. We use the standard total-variation result that $|\nu|$ is a measure; here it is finite. Write $\lambda=|\nu|$. A complex-valued function $h$ is a polar density if $\nu(E)=\int_E h\,d\lambda$ for every $E\in\mathcal{M}$, and $|h|=1$ $\lambda$-almost everywhere.

Theorem (Polar Decomposition): For every complex measure $\nu$ of finite variation on $(X,\mathcal{M})$, there is a complex-valued $\mathcal{M}$-measurable function $h$, integrable with respect to $\lambda=|\nu|$, such that for every $E\in\mathcal{M}$, $$ \nu(E)=\int_E h\,d\lambda, $$ and $|h|=1$ $\lambda$-almost everywhere. The density is unique up to equality $\lambda$-almost everywhere. If $\lambda=0$, the almost-everywhere assertion is vacuous.

The Variation of a Measure Given by a Density

The key step in the theorem is to identify the variation of a measure that already has a density. Let $(X,\mathcal{M},\lambda)$ be any measure space and let $g:X\to\mathbb{C}$ be measurable with $\int_X|g|\,d\lambda<\infty$. Define the complex measure $\nu_g(E)=\int_Eg\,d\lambda$. We will show that its total variation is exactly the measure $E\mapsto\int_E|g|\,d\lambda$. This identity explains why the polar density must have modulus one.

Proposition: For every $E\in\mathcal{M}$, $$ |\nu_g|(E)=\int_E|g|\,d\lambda. $$

Proof. Fix $E\in\mathcal{M}$. For any finite measurable partition $E=\bigcup_{j=1}^kE_j$, the triangle inequality for the complex integral gives $$ \sum_{j=1}^k|\nu_g(E_j)| =\sum_{j=1}^k\left|\int_{E_j}g\,d\lambda\right| \leq\sum_{j=1}^k\int_{E_j}|g|\,d\lambda =\int_E|g|\,d\lambda. $$ Taking the supremum over partitions proves $|\nu_g|(E)\leq\int_E|g|\,d\lambda$.

For the reverse inequality, fix an integer $N\geq3$ and partition $E$ into measurable sets $E_0,\ldots,E_{N-1}$ according to which of the $N$ angular sectors contains the nonzero value $g(x)$. Choose each sector to have central angle $\theta_j=2\pi j/N$ and angular width $2\pi/N$; assign points where $g=0$ to $E_0$. For $x\in E_j$ with $g(x)\ne0$, the angle between $g(x)$ and $e^{i\theta_j}$ is at most $\pi/N$, so $$ \operatorname{Re}\!\left(e^{-i\theta_j}g(x)\right)\geq\cos(\pi/N)|g(x)|. $$ The inequality also holds when $g(x)=0$. Integrating on $E_j$, and using that the modulus of a complex number is at least its real part, gives $$ |\nu_g(E_j)|=\left|\int_{E_j}g\,d\lambda\right| \geq\operatorname{Re}\!\left(e^{-i\theta_j}\int_{E_j}g\,d\lambda\right) \geq\cos(\pi/N)\int_{E_j}|g|\,d\lambda. $$ Summing over this partition yields $|\nu_g|(E)\geq\cos(\pi/N)\int_E|g|\,d\lambda$. The integral is finite because $g$ is integrable. Letting $N\to\infty$, so that $\cos(\pi/N)\to1$, proves $|\nu_g|(E)\geq\int_E|g|\,d\lambda$. The two inequalities give the claimed identity for every measurable $E$. $\square$

Proof of Polar Decomposition

We apply the Radon–Nikodym theorem to the real and imaginary parts of $\nu$, using the finite positive measure $\lambda=|\nu|$. Write $\nu_R(E)=\operatorname{Re}\nu(E)$ and $\nu_I(E)=\operatorname{Im}\nu(E)$; both are finite signed measures. For every finite measurable partition $E=\bigcup_jE_j$, $$ \sum_j|\nu_R(E_j)|\leq\sum_j|\nu(E_j)|\leq\lambda(E), \qquad \sum_j|\nu_I(E_j)|\leq\sum_j|\nu(E_j)|\leq\lambda(E). $$ Taking suprema over partitions shows that the total variations of $\nu_R$ and $\nu_I$ are bounded above by $\lambda$. In particular, both signed measures are absolutely continuous with respect to $\lambda$. The Radon–Nikodym theorem for finite signed measures therefore gives real-valued integrable functions $u$ and $v$ such that $\nu_R(E)=\int_Eu\,d\lambda$ and $\nu_I(E)=\int_Ev\,d\lambda$ for every $E$. Set $h=u+iv$. Then $h$ is complex-valued, measurable and integrable, and $\nu(E)=\int_Eh\,d\lambda$ for every measurable $E$.

Apply the proposition to this $h$. It gives $|\nu|(E)=\int_E|h|\,d\lambda$ for every $E$. Since $\lambda=|\nu|$, we have $$ \int_E(|h|-1)\,d\lambda=0 $$ for every $E\in\mathcal{M}$. The real-valued function $q=|h|-1$ is integrable: $\int|h|\,d\lambda<\infty$ and $\lambda(X)<\infty$. It need not be nonnegative on all of $X$, so we examine its positive and negative regions separately. For each integer $n\geq1$, on $A_n=\{|h|-1\geq1/n\}$ we have $q\geq1/n$. Thus $$ 0=\int_{A_n}q\,d\lambda\geq\frac{1}{n}\lambda(A_n), $$ and $\lambda(A_n)=0$. Since $\{|h|>1\}=\bigcup_{n\geq1}A_n$, this set is null. On $B_n=\{1-|h|\geq1/n\}$, we have $-q=1-|h|\geq1/n$; also $\int_{B_n}(-q)\,d\lambda=0$. The same inequality gives $\lambda(B_n)=0$, so $\{|h|<1\}=\bigcup_{n\geq1}B_n$ is null. Therefore $|h|=1$ $\lambda$-almost everywhere. If $\lambda=0$, these sets are still $\lambda$-null and the representation holds with any measurable choice of $h$, for example $h=1$.

For uniqueness, suppose $h_1$ and $h_2$ are two integrable densities representing $\nu$. The Radon–Nikodym uniqueness theorem, applied to the real and imaginary parts with respect to $\lambda$, gives $h_1=h_2$ $\lambda$-almost everywhere. This proves the theorem, including the zero-measure case. $\square$

Worked Atomic Examples

Worked Example: Two Atoms with Different Phases

Let $X=\{a,b\}$, $\mathcal{M}$ be its power set, and define $\nu(\{a\})=3i/5$ and $\nu(\{b\})=-4/5$. On a finite space, the total variation of an atomic measure is the sum of the moduli of its weights on the atoms. Hence $\lambda(\{a\})=|3i/5|=3/5$, $\lambda(\{b\})=|-4/5|=4/5$, and $\lambda(X)=3/5+4/5=7/5$.

Define $h(a)=i$ and $h(b)=-1$. Then $|h(a)|=|i|=1$ and $|h(b)|=|-1|=1$. For every $E\subseteq X$, the integral is the sum over the atoms in $E$, so $$ \int_Eh\,d\lambda =\frac35i\,\chi_E(a)-\frac45\chi_E(b) =\nu(E). $$ In particular, $\int_Xh\,d\lambda=3i/5-4/5$, which is the total complex mass. Check: its modulus is $\sqrt{(3/5)^2+(4/5)^2}=1$, while $\lambda(X)=7/5$. The modulus of the total mass is not the total variation.

Worked Example: A Geometric Sequence of Atomic Masses

On $X=\mathbb{N}=\{1,2,\ldots\}$ with $\mathcal{M}$ the power set, assign the atom $n$ the mass $\nu(\{n\})=(i/2)^n$. The series of absolute weights is $\sum_{n\geq1}2^{-n}=1$, so this defines a complex measure of finite variation. Its total variation is the finite measure $\lambda$ with $\lambda(\{n\})=2^{-n}$; consequently $\lambda(X)=1$. The polar density at each atom is $$ h(n)=\frac{(i/2)^n}{2^{-n}}=i^n, $$ and $|h(n)|=1$ for every $n$. For every $E\subseteq\mathbb{N}$, the series is absolutely convergent and $$ \int_Eh\,d\lambda=\sum_{n\in E}i^n2^{-n}=\sum_{n\in E}(i/2)^n=\nu(E). $$ The total complex mass is computed from the geometric series with ratio $i/2$: $$ \nu(X)=\sum_{n=1}^{\infty}(i/2)^n =\frac{i/2}{1-i/2} =\frac{(i/2)(1+i/2)}{(1-i/2)(1+i/2)} =\frac{-1/4+i/2}{5/4} =\frac{-1+2i}{5}. $$ Check: the modulus of this total mass is $\sqrt{1+4}/5=1/\sqrt{5}$, while the total variation is $1$. The density $i^n$ reproduces each atom exactly, and its modulus is one at every point.

Worked Example: Two Dirac Masses

On $X=\mathbb{R}$ with its Borel sigma-algebra, let $\nu=(1+i)\delta_0-2i\delta_1$, where $\delta_x(E)=1$ if $x\in E$ and $0$ otherwise. The atoms are distinct, so their variation weights are the moduli of their respective masses: $\lambda(\{0\})=|1+i|=\sqrt{2}$ and $\lambda(\{1\})=|-2i|=2$; $\lambda$ is zero away from these atoms. Thus $\lambda(\mathbb{R})=\sqrt{2}+2$. Set $h(0)=(1+i)/\sqrt{2}$, $h(1)=-i$, and $h(x)=1$ elsewhere. This is Borel measurable and has modulus one $\lambda$-almost everywhere. For any Borel set $E$, $$ \int_Eh\,d\lambda =\frac{1+i}{\sqrt{2}}\sqrt{2}\,\chi_E(0)+(-i)2\,\chi_E(1) =(1+i)\chi_E(0)-2i\chi_E(1) =\nu(E). $$ Check: the total mass is $\nu(\mathbb{R})=1+i-2i=1-i$, whose modulus is $\sqrt{2}$; the total variation is $\sqrt{2}+2$. The integral of the phase density against $\lambda$ gives the complex mass, not the sum of the atom magnitudes.

A Common Pitfall

A frequent error is to treat the modulus of the total mass as though it were the total variation. For the countable example, replacing the variation $\sum_{n\geq1}|(i/2)^n|$ by $\left|\sum_{n\geq1}(i/2)^n\right|$ would give $1/\sqrt{5}$ instead of $1$. These quantities answer different questions: total variation adds the magnitudes across a partition, while the total mass adds the complex values first, allowing cancellation.

Another possible error is to assign the density $h(n)=(i/2)^n$ relative to $\lambda$ in that example. The density must be divided by the reference measure’s atom weight: $\nu(\{n\})/\lambda(\{n\})=(i/2)^n/2^{-n}=i^n$. Substitution then gives $h(n)\lambda(\{n\})=i^n2^{-n}=(i/2)^n$, as required.

Why the Decomposition Matters

The polar density isolates the part of a complex measure responsible for cancellation. The positive measure $|\nu|$ supplies the size, and the unit-modulus function $h$ supplies the phase. The condition $|h|=1$ is relative to $|\nu|$, not necessarily everywhere on $X$: where $|\nu|$ vanishes, changing $h$ does not change any integral against $|\nu|$. This distinction is especially important for measures supported on a small set, such as the Dirac examples above.

Check Your Understanding

In each question, distinguish identities of measures from statements that hold only almost everywhere.

For a measure supported on two distinct atoms with weights $3$ and $-4i$, compute its total variation and give a unit-modulus polar density on those atoms.
Explain why $|\nu(X)|$ need not equal $|\nu|(X)$, using the two-atom example in this tutorial.
For the geometric atomic measure, verify the polar representation at the atom $n=3$, including both $\nu(\{3\})$ and $h(3)\lambda(\{3\})$.
In the proof, why must the sets where $|h|>1$ and $|h|<1$ be treated separately rather than declaring $q=|h|-1$ nonnegative?
What does the assertion $|h|=1$ $\lambda$-almost everywhere mean when $\lambda=0$?